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Games Problems Go Pro! Find a, b, and c if a + b + c = 5; a2 + b2 + c2 = 29, and a3 + b3 + c3 = 83. This is an interesting problem, and some clever manipulation may be required. EQ 1: a + b + c = 5 EQ 2: a2 + b2 + c2 = 29 EQ 3: a3 + b3 + c3 = 83 Let's start by observing that if we square the first equation, we get: EQ 4: a2 + b2 + c2 + 2ab + 2bc + 2ac = 25. Subtracting (2) from (4), and simplifying, gives: EQ 5: ab + bc + ac = -2 Now let's (3), and rewrite it with a3 + b3 factored: EQ 6: (a + b)(a2 - ab + b2) +c3 = 83. Ideally, it would be nice if we could rewrite (6) in terms of c. We note the following: (1) can be rewritten as: a + b = 5 - c (5) can be rewritten as ab = - 2 - bc - ac = -2 - c(a + b) = -2 - c(5 - c) = c2 - 5c - 2 (2) can be rewritten as a2 + b2 = 29 - c2 Substituting these into (6) gives us: EQ 7: (5 - c)(29 - c2 - c2 + 5c + 2) + c3 = 83 We simplify this equation as follows: (5 - c)(-2c2 + 5c + 31) + c3 = 83 2c3 - 15c2 - 6c + 155 + c3 = 83 3c3 - 15c2 - 6c + 72 = 0 c3 - 5c2 - 2c + 24 = 0 This is a cubic equation, so we naturally expect three answers. And this makes perfect sense, in light of the problem; the equations are symmetric with respect to a, b, and c, so once we find three values for c, we know that any combinations of those three values make ordered triples (a, b, c) that work. I don't see a quick solution by grouping, so I resort to the rational root theorem until I find a value that works; then I factor the resulting quadratic to obtain -2, 3, and 4, in any order. Thanks for the problem; that was a fun one! # Featured Games on This Site Match color, font, and letter in this strategy game Mastermind variation, with words # Blogs on This Site Reviews and book lists - books we love! The site administrator fields questions from visitors.
Algebra Tutorials! Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: # Reducing Numerical Fractions to Simplest Form There is one simplification process that is usually called cancelling common factors. You write the numerator and denominator as products of prime factors. Then, if they have a prime factor in common, simply “cancel” it out in each. This cancelling is the equivalent of dividing the numerator and denominator by that same value, so the fraction that is left after cancelling a prime factor common to both numerator and denominator will be equivalent to the original fraction, but of course, its denominator will be smaller. Then, just repeat this cancelling process for every common factor in the numerator and denominator. Example: Simplify solution: Very briefly, resolving the numerator and denominator into products of prime factors gives 220 = 2 Ã— 2 Ã— 5 Ã— 11 and 88 = 2 Ã— 2 Ã— 2 Ã— 11 So, Thus, the simplest form of 220 / 88 is 5 / 2. (Notice that this simplification method works even for “improper fractions” – fractions in which the numerator is larger than the denominator.) Example: Reduce to lowest terms. solution: 234 = 2 Ã— 3 Ã— 3 Ã—13 and 315 = 3 Ã— 3 Ã— 5 Ã— 7 So, Since 26 and 35 share no factors in common, this is the simplest form to which we can reduce the original fraction. If you recognize a common factor in the numerator and denominator of a fraction, it is permissible to cancel that factor as a way to get the fraction into a simpler form immediately. However, to guarantee that your final result is the simplest form you must perform this systematic procedure. It will be essential to use the systematic approach when we deal with the simplification of algebraic fractions.
# Help me with math rules. 18÷9(2)=? A) 1 or B) 4 Please explain. Feb 11, 2016 1 #### Explanation: you will apply the BODMAS rule . so according to the rule we will first we will solve the bracket first which is 9(2)= 18 then after solving the bracket divide according to the rule and u will get 1 by dividing 18/18.. just remember BODMAS is Bracket Or Division Multiplication Addition Subtraction Aug 3, 2016 #### Explanation: There are 2 operations going on here - multiplication and division. Either can be done first, but we have to know which is the multiply and which is the divide.... It can be written like this: $18 \div 9 \times 2 = \frac{18 \times 2}{9} = \frac{36}{9} = 4$ Or $\frac{{\cancel{18}}^{2} \times 2}{\cancel{9}} = 4$ This is NOT the same as $18 \div \left(9 \times 2\right)$ which can be shown as $\frac{18}{2 \times 9} = \frac{18}{18} = 1$ Aug 3, 2016 $1$ or $4$ #### Explanation: If you are using PEMDAS, BODMAS or BIDMAS in pure form, then what we have here is the same as: $18 \div 9 \times 2$ This is evaluated left to right (since multiplication and division have the same priority). So we perform the division first to get: $2 \times 2$ then the multiplication to get: $4$ If you are not restricted to PEMDAS/BODMAS/BIDMAS, then there are at least a couple of justifications for performing the multiplication first: • By juxtaposing $9$ and $\left(2\right)$, the writer intended to convey that $9 \left(2\right)$ should be treated as one term, like $9 a$ with $a = 2$. • The obelus $\div$ is historically shorthand for dividing the whole expression on the left by the whole expression on the right. If you understood the expression one of these ways then you would perform $9 \times 2$ first giving $18$, then evaluate $18 \div 18$ to get the value $1$ This is not "wrong". The purpose of conventions like PEMDAS is to try and disambiguate such cases, but such conventions can result in counterintuitive results. The bottom line is that the writer of the expression should either be specific about what conventions they are using or add parentheses to make it unambiguous.
Medians! Medians! Everywhere! Assignment 6 By Amber Candela _______________________________________________________________________________________________________________________________________________ The Problem Construct a triangle and its medians. Construct a second triangle with the three sides having the lengths of the three medians from your first triangle. Find some relationship between the two triangles. What is a median? A median is the segment that connects the vertex to the midpoint of the opposite side. The medians intersect to create segments that are in a ratio of 2:1. The median bisects the triangle, creating two smaller triangles that have the same area. How can you construct this? First create triangle ABC with medians AD, FC and BE. We need to create a triangle with the medians, so start the new triangle with median FC. We then need to use circles to construct the other two sides of the new triangle. Construct circle with center F and the radius the same length as median AD. Then construct circle with center C and the radius the same length as median BE. The two circles intersect at point G. Create triangle FGC which will be constructed with the median lengths from triangle ABC. Side FG = AD and side GC = BE by construction and side FC = FC by the reflexive property. The construction can be seen below. What are the relationships between triangle ABC and triangle FGC? Are the triangles congruent? No. The sides of the original triangle are not equal to the sides of the new triangle, thus by SSS the triangles are not congruent. This would mean they do not have the same perimeter. The areas of the triangles are also not equal because they do not have the same base or height. The ratio areas of the two triangles is constant. The areas are in a ratio of 4:3. Use the GSP file below to explore the area ratios. Area Ratio Explorer Proof of the ratios. While GSP is useful to show how the areas are in the ratio of 4:3, it cannot prove such statements. Let's look at a the triangles in a different way. Notice this time we have two pairs of parallel lines, GF is parallel to EB and GC is parallel to AD. This created two parallelograms, FBEG and AGCD. How do these parallelograms help?
Question Video: Using the Determinant of a Matrix to Find Missing Values \| Nagwa Question Video: Using the Determinant of a Matrix to Find Missing Values \| Nagwa # Question Video: Using the Determinant of a Matrix to Find Missing Values Mathematics • First Year of Secondary School ## Join Nagwa Classes Solve for 𝑥: \|0, 5, −5𝑥 and 𝑥, 4, 5 and 4, 1, 3\| = 280. 06:14 ### Video Transcript Solve for 𝑥: the determinant of the matrix zero, five, negative five 𝑥, 𝑥, four, five, four, one, three is equal to 280. So the first thing we need to do to solve this problem is work out the determinant of our matrix. And if we try to find the determinant of the three-by-three matrix, what we use is the first row. And we use the first row to be our coefficients or the values that we multiply our submatrices by, which are gonna be two-by-two submatrices. And I’m gonna show you how it’s gonna work. It’s worth noting also that these values are gonna be positive, negative, or positive, determined by the pattern that we have. Where the first column is positive, so it’ll be positive zero, the second one will be negative, the third one will be positive, et cetera. So what this means is where we have a negative above the column, it’s gonna change the sign of the value that we’ve got from our first row. If you’ve got a positive, then the sign is gonna stay the same. So first of all, we’re gonna have zero multiplied by the determinant of the submatrix four, five, one, three. And we find that submatrix by deleting the row on column that the zero is in. And then, it’s what we’re left with afterwards. So 𝑥 would be left with four, five, one, three. Then next, what we have is negative five multiplied by the determinant of the submatrix 𝑥, five, four, three. Again, finding that in the same way, and it’s negative five because, as we said, the second column is negative. And what we mean by this is that the coefficient or our five is multiplied by negative one. So we get negative five. Then, finally, we have minus five 𝑥 multiplied by the determinant of the submatrix 𝑥, four, four, one. And again, we’ve done this the same way. And this time, we’ve still got negative five 𝑥. And that’s because the third column is positive. So we multiply negative five 𝑥 by positive one, which gives us the same sign because it doesn’t change. And then, this is all equal to 280. So now, what we need to do is evaluate our submatrices, well, the determinant of our submatrices. And the way we do that is using this method. So if we’ve got two-by-two submatrix, we can call this one 𝑎, 𝑏, 𝑐, 𝑑. Then, what we do is we multiply diagonally. So we have 𝑎 multiplied by 𝑑 and 𝑏 multiplied by 𝑐. And then, we take away 𝑏 multiplied by 𝑐 from 𝑎 multiplied by 𝑑. Well, for the first term, we don’t have to worry about the submatrix. And that’s because we’ve got zero multiplied by the determinant of the submatrix. So anything multiplied by zero is just zero. And then, we have minus five multiplied by. Now, we’ve got three multiplied by 𝑥 or 𝑥 multiplied by three, which gives us three 𝑥 minus. Then, we’ve got five multiplied by four, which is 20. And then, we have minus five 𝑥 multiplied by. Then, we’ve got 𝑥 and that’s because we had 𝑥 multiplied by one which is just 𝑥 minus 16 because we’ve four multiplied by four, which is 16. And this is all equal to 280. So now, what we need to do is distribute across our parentheses. So first of all, we have zero minus. And then, we’ve got five multiplied by three 𝑥, which is 15𝑥 and then plus a 100. And that’s because we had negative five multiplied by negative 20. And a negative multiplied by a negative is a positive. And then, we have negative five 𝑥 squared plus 80 𝑥. That’s cause we’ve got negative five 𝑥 multiplied by 𝑥, which is negative five 𝑥 squared. And we’ve got negative five 𝑥 multiplied by negative 16. Negative multiplied by a negative is a positive. So it gives us positive 80𝑥. This is equal to 280. So now, if we simplify, we get negative five 𝑥 squared plus 65𝑥 plus 100 equals 280. So now, what I wanna do is make this so that I have positive five 𝑥 squared or positive 𝑥 squared. So I’m going to add five 𝑥 squared to both sides of the equation, subtract 65𝑥 from both sides of the equation, and subtract 100 from both sides of the equation. So I wanna make our quadratic equal to zero. And when I do that, I get zero is equal to five 𝑥 squared minus 65𝑥 plus 180. So now, what we wanna do is solve this to find 𝑥. But the first thing we do to make it easier is divide through by five because five is a factor of each of our terms. So we’ve got zero is equal to 𝑥 squared minus 13𝑥 plus 36. So now, we can solve this using factoring. And to factor, just to remind us what we need to do, we need to find two factors whose product is positive 36 and whose sum is negative 13. So we’re gonna have zero is equal to 𝑥 minus nine multiplied by 𝑥 minus four. And we get that because nine multiplied by four is 36. And we’ve got negative nine and negative four because we got positive 36. But I need to sum to negative 13. So we know they both need to be negative. And then, you’ve got negative nine and negative four gives us negative 13. So great, we factored. So therefore, we can say that the solution for 𝑥 is gonna be equal to positive nine or positive four. And the way we got this is because we want to find the value where the quadratic is equal to zero. So that means that one of our parentheses needs to be equal to zero. So we can set them equal to zero. So we get 𝑥 minus nine equals zero and 𝑥 minus four equals zero. So therefore, if we add nine to each side of the equation, we get 𝑥 equals nine. If we add four to each side of the equation, we get 𝑥 equals four. So therefore, we’ve got a final answer 𝑥 equals nine or four. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
# Euclidean division In arithmetic, Euclidean division – or division with remainder – is the process of dividing one integer (the dividend) by another (the divisor), in a way that produces an integer quotient and a natural number remainder strictly smaller than the absolute value of the divisor. A fundamental property is that the quotient and the remainder exist and are unique, under some conditions. Because of this uniqueness, Euclidean division is often considered without referring to any method of computation, and without explicitly computing the quotient and the remainder. The methods of computation are called integer division algorithms, the best known of which being long division. Euclidean division, and algorithms to compute it, are fundamental for many questions concerning integers, such as the Euclidean algorithm for finding the greatest common divisor of two integers,[1] and modular arithmetic, for which only remainders are considered.[2] The operation consisting of computing only the remainder is called the modulo operation,[3] and is used often in both mathematics and computer science. ## Division theorem Euclidean division is based on the following result, which is sometimes called Euclid's division lemma. Given two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r and 0 ≤ r < \|b\|, where \|b\| denotes the absolute value of b.[4] In the above theorem, each of the four integers has a name of its own: a is called the dividend, b is called the divisor, q is called the quotient and r is called the remainder. The computation of the quotient and the remainder from the dividend and the divisor is called division, or in case of ambiguity, Euclidean division. The theorem is frequently referred to as the division algorithm (although it is a theorem and not an algorithm), because its proof as given below lends itself to a simple division algorithm for computing q and r (see the section Proof for more). Division is not defined in the case where b = 0; see division by zero. For the remainder and the modulo operation, there are conventions other than 0 ≤ r < \|b\|, see § Other intervals for the remainder. ## History Although "Euclidean division" is named after Euclid, it seems that he did not know the existence and uniqueness theorem, and that the only computation method that he knew was the division by repeated subtraction.[citation needed] Before the discovery of Hindu–Arabic numeral system, which was introduced in Europe during the 13th century by Fibonacci, division was extremely difficult, and only the best mathematicians were able to do it. Presently, most division algorithms, including long division, are based on this notation or its variants, such as binary numerals. A notable exception is Newton–Raphson division, which is independent from any numeral system. The term "Euclidean division" was introduced during the 20th century as a shorthand for "division of Euclidean rings". It has been rapidly adopted by mathematicians for distinguishing this division from the other kinds of division of numbers.[citation needed] ## Intuitive example Suppose that a pie has 9 slices and they are to be divided evenly among 4 people. Using Euclidean division, 9 divided by 4 is 2 with remainder 1. In other words, each person receives 2 slices of pie, and there is 1 slice left over. This can be confirmed using multiplication, the inverse of division: if each of the 4 people received 2 slices, then 4 × 2 = 8 slices were given out in total. Adding the 1 slice remaining, the result is 9 slices. In summary: 9 = 4 × 2 + 1. In general, if the number of slices is denoted ${\displaystyle a}$ and the number of people is denoted ${\displaystyle b}$, then one can divide the pie evenly among the people such that each person receives ${\displaystyle q}$ slices (the quotient), with some number of slices ${\displaystyle r being the leftover (the remainder). In which case, the equation ${\displaystyle a=bq+r}$ holds. If 9 slices were divided among 3 people instead of 4, then each would receive 3 and no slice would be left over, which means that the remainder would be zero, leading to the conclusion that 3 evenly divides 9, or that 3 divides 9. Euclidean division can also be extended to negative dividend (or negative divisor) using the same formula; for example −9 = 4 × (−3) + 3, which means that −9 divided by 4 is −3 with remainder 3. ## Examples • If a = 7 and b = 3, then q = 2 and r = 1, since 7 = 3 × 2 + 1. • If a = 7 and b = −3, then q = −2 and r = 1, since 7 = −3 × (−2) + 1. • If a = −7 and b = 3, then q = −3 and r = 2, since −7 = 3 × (−3) + 2. • If a = −7 and b = −3, then q = 3 and r = 2, since −7 = −3 × 3 + 2. ## Proof The following proof of the division theorem relies on the fact that a decreasing sequence of non-negative integers stops eventually. It is separated into two parts: one for existence and another for uniqueness of ${\displaystyle q}$ and ${\displaystyle r}$. Other proofs use the well-ordering principle (i.e., the assertion that every non-empty set of non-negative integers has a smallest element) to make the reasoning simpler, but have the disadvantage of not providing directly an algorithm for solving the division (see § Effectiveness for more).[5] ### Existence Consider first the case b < 0. Setting b' = –b and q' = –q, the equation a = bq + r may be rewritten as a = b'q' + r and the inequality 0 ≤ r < \|b\| may be rewritten as 0 ≤ r < \|b\|. This reduces the existence for the case b < 0 to that of the case b > 0. Similarly, if a < 0 and b > 0, setting a' = –a, q' = –q – 1, and r' = br, the equation a = bq + r may be rewritten as a' = bq' + r, and the inequality 0 ≤ r < \|b\| may be rewritten as 0 ≤ r' < \|b\|. Thus the proof of the existence is reduced to the case a ≥ 0 and b > 0 — which will be considered in the remainder of the proof. Let q1 = 0 and r1 = a, then these are non-negative numbers such that a = bq1 + r1. If r1 < b then the division is complete, so suppose r1b. Then defining q2 = q1 + 1 and r2 = r1b, one has a = bq2 + r2 with 0 ≤ r2 < r1. As there are only r1 non-negative integers less than r1, one only needs to repeat this process at most r1 times to reach the final quotient and the remainder. That is, there exist a natural number kr1 such that a = bqk + rk and 0 ≤ rk < \|b\|. This proves the existence and also gives a simple division algorithm for computing the quotient and the remainder. However, this algorithm is not efficient, since its number of steps is of the order of a/b. ### Uniqueness The pair of integers r and q such that a = bq + r is unique, in the sense that there can be no other pair of integers that satisfy the same condition in the Euclidean division theorem. In other words, if we have another division of a by b, say a = bq' + r' with 0 ≤ r' < \|b\|, then we must have that q' = q and r' = r. 0 ≤ r < \|b\| 0 ≤ r' < \|b\| a = bq + r a = bq' + r' Subtracting the two equations yields b(qq) = rr. So b is a divisor of rr. As \|rr\| < \|b\| by the above inequalities, one gets rr = 0, and b(qq) = 0. Since b ≠ 0, we get that r = r and q = q, which proves the uniqueness part of the Euclidean division theorem. ## Effectiveness In general, an existence proof does not provide an algorithm for computing the existing quotient and remainder, but the above proof does immediately provide an algorithm (see Division algorithm#Division by repeated subtraction), even though it is not a very efficient one as it requires as many steps as the size of the quotient. This is related to the fact that it uses only additions, subtractions and comparisons of integers, without involving multiplication, nor any particular representation of the integers such as decimal notation. In terms of decimal notation, long division provides a much more efficient algorithm for solving Euclidean divisions. Its generalization to binary and hexadecimal notation provides further flexibility and possibility for computer implementation. However, for large inputs, algorithms that reduce division to multiplication, such as Newton–Raphson, are usually preferred, because they only need a time which is proportional to the time of the multiplication needed to verify the result—independently of the multiplication algorithm which is used (for more, see Division algorithm#Fast division methods). ## Variants The Euclidean division admits a number of variants, some of which are listed below. ### Other intervals for the remainder In Euclidean division with d as divisor, the remainder is supposed to belong to the interval [0, d) of length \|d\|. Any other interval of the same length may be used. More precisely, given integers ${\displaystyle m}$, ${\displaystyle a}$, ${\displaystyle d}$ with ${\displaystyle m>0}$, there exist unique integers ${\displaystyle q}$ and ${\displaystyle r}$ with ${\displaystyle d\leq r such that ${\displaystyle a=mq+r}$. In particular, if ${\displaystyle d=-\left\lfloor {\frac {m}{2}}\right\rfloor }$ then ${\displaystyle -\left\lfloor {\frac {m}{2}}\right\rfloor \leq r . This division is called the centered division, and its remainder ${\displaystyle r}$ is called the centered remainder or the least absolute remainder. This is used for approximating real numbers: Euclidean division defines truncation, and centered division defines rounding. ### Montgomery division Given integers ${\displaystyle a}$, ${\displaystyle m}$ and ${\displaystyle R,}$ with ${\displaystyle m>0}$ and ${\displaystyle \gcd(R,m)=1,}$ let ${\displaystyle R^{-1}}$ be the modular multiplicative inverse of ${\displaystyle R}$ (i.e., ${\displaystyle 0 with ${\displaystyle R^{-1}R-1}$ being a multiple of ${\displaystyle m}$), then there exist unique integers ${\displaystyle q}$ and ${\displaystyle r}$ with ${\displaystyle 0\leq r such that ${\displaystyle a=mq+R^{-1}\cdot r}$. This result generalizes Hensel's odd division (1900).[6] The value ${\displaystyle r}$ is the N-residue defined in Montgomery reduction. ## In Euclidean domains Euclidean domains (also known as Euclidean rings)[7] are defined as integral domains which support the following generalization of Euclidean division: Given an element a and a non-zero element b in a Euclidean domain R equipped with a Euclidean function d (also known as a Euclidean valuation[8] or degree function[7]), there exist q and r in R such that a = bq + r and either r = 0 or d(r) < d(b). Uniqueness of q and r is not required.[1] It occurs only in exceptional cases, typically for univariate polynomials, and for integers, if the further condition r ≥ 0 is added. Examples of Euclidean domains include fields, polynomial rings in one variable over a field, and the Gaussian integers. The Euclidean division of polynomials has been the object of specific developments. ## Notes 1. ^ a b "Division and Euclidean algorithms". www-groups.mcs.st-andrews.ac.uk. Retrieved 2019-11-15. 2. ^ "What is modular arithmetic?". Khan Academy. Retrieved 2019-11-15. 3. ^ "Fun With Modular Arithmetic – BetterExplained". betterexplained.com. Retrieved 2019-11-15. 4. ^ Burton, David M. (2010). Elementary Number Theory. McGraw-Hill. pp. 17–19. ISBN 978-0-07-338314-9. 5. ^ Durbin, John R. (1992). Modern Algebra : an Introduction (3rd ed.). New York: Wiley. p. 63. ISBN 0-471-51001-7. 6. ^ Haining Fan; Ming Gu; Jiaguang Sun; Kwok-Yan Lam (2012). "Obtaining More Karatsuba-Like Formulae over the Binary Field". IET Information Security. 6 (1): 14–19. CiteSeerX 10.1.1.215.1576. doi:10.1049/iet-ifs.2010.0114. 7. ^ a b Rotman 2006, p. 267 8. ^ Fraleigh 1993, p. 376 ## References • Fraleigh, John B. (1993), A First Course in Abstract Algebra (5th ed.), Addison-Wesley, ISBN 978-0-201-53467-2 • Rotman, Joseph J. (2006), A First Course in Abstract Algebra with Applications (3rd ed.), Prentice-Hall, ISBN 978-0-13-186267-8
Courses Courses for Kids Free study material Offline Centres More Store If $\cos \theta =0.6$, show that: $5\sin \theta -3\tan \theta =0$ Last updated date: 08th Sep 2024 Total views: 418.5k Views today: 10.18k Verified 418.5k+ views Hint: In this question, from the given values of cos function by using the trigonometric identities we can find the values of the sin and tan functions ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. Then on substituting the respective values in the given expression of the question we can calculate the left hand side value and the right hand side value. Then on comparing the values obtained, we get the result. Now, from the given question we have $\cos \theta =0.6\ \ \ \ \ ...(a)$ Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following substitute the value from the question and as well as from equation (a) $\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Now substitute the $\cos \theta$ value and find the $\sin \theta$ value in the above equation then we get $\Rightarrow {{\sin }^{2}}\theta =1-{{\left( 0.6 \right)}^{2}}$ $\Rightarrow {{\sin }^{2}}\theta =1-0.36$ On simplifying we get as follows $\Rightarrow {{\sin }^{2}}\theta =0.64$ Since the perfect square of $0.64$ is $0.8$ $\Rightarrow \sin \theta =0.8$ Therefore, the value of $\sin \theta =0.8$ Now, again from the hint, we know that we can get the tan function as follows $\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }$ And now let us substitute the $\sin \theta$ and $\cos \theta$ values in $\tan \theta$ then we get as follows $\Rightarrow \tan \theta =\dfrac{0.8}{0.6}$ On simplifying we get as follows $\Rightarrow \tan \theta =\dfrac{4}{3}$ Now, from the given expression in the question, on substituting the values, we have $5\sin \theta -3\tan \theta =0$ Let us first consider the left hand side and calculate its value $L.H.S=5\sin \theta -3\tan \theta$ And now substitute the $\sin \theta$ and $\tan \theta$ values then we get as follows $L.H.S=5\times 0.8-3\times \dfrac{4}{3}$ On simplifying the above calculation we get as follows $L.H.S=4-4=0$ Thus, the value of right hand side is equal to left hand side Therefore, it is verified that $5\sin \theta -3\tan \theta =0$ Hence proved. Note: The other way of solving the above problem is as follows: We have given $\cos \theta =0.6$ so writing the value of $\cos \theta$ in terms of fraction we get, $\cos \theta =\dfrac{6}{10}=\dfrac{3}{5}$ Now, we are going to draw a right angled triangle with angle $\theta$ as follows: The length of AB is calculated using Pythagoras theorem which is equal to: ${{\left( Hypotenuse \right)}^{2}}={{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}}$ Substituting hypotenuse as AC, base as BC and perpendicular as AB in the above equation we get, ${{\left( AC \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ $\Rightarrow {{5}^{2}}={{\left( AB \right)}^{2}}+{{3}^{2}}$ $\Rightarrow 25-9={{\left( AB \right)}^{2}}$ $\Rightarrow 16={{\left( AB \right)}^{2}}$ Taking square root on both the sides we get, $4=AB$ Now, we can find $\sin \theta \And \tan \theta$ by using trigonometric ratios. $\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$ Substituting perpendicular as AB and hypotenuse as AC in the above we get, $\sin \theta =\dfrac{AB}{AC}$ $\Rightarrow \sin \theta =\dfrac{4}{5}$ $\tan \theta =\dfrac{Perpendicular}{Base}$ $\Rightarrow \tan \theta =\dfrac{AB}{BC}$ $\Rightarrow \tan \theta =\dfrac{4}{3}$ Now, substitute the values of $\sin \theta \And \tan \theta$ in $5\sin \theta -3\tan \theta =0$ we get, $5\left( \dfrac{4}{5} \right)-3\left( \dfrac{4}{3} \right)=0$ $\Rightarrow 4-4=0$ $\Rightarrow 0=0$ As L.H.S is equal to R.H.S so we have proved the given equation.
# find the 1991st term • Jan 13th 2011, 06:37 PM chris86 find the 1991st term Let 2,3,5,6,7,10,11,... be an increasing sequence of positive integers that are neither the perfect squares nor the cubes. Find the 1991st term. • Jan 13th 2011, 06:57 PM chiph588@ Quote: Originally Posted by chris86 Let 2,3,5,6,7,10,11,... be an increasing sequence of positive integers that are neither the perfect squares nor the cubes. Find the 1991st term. Let's define $\displaystyle f(n)$ to be the number of numbers less than or equal to $\displaystyle n$ such that they are neither the perfect squares or cubes. We then have $\displaystyle \displaystyle f(n) = n - \lfloor \sqrt{n}\rfloor - \lfloor \sqrt[3]{n}\rfloor + \lfloor \sqrt[6]{n}\rfloor$. The intuition behind $\displaystyle f(n)$ is we take every number up to $\displaystyle n$ and then subtract all of the perfect squares and cubes. We must then account for any duplicates. Now solve $\displaystyle f(n) = 1991$. • Jan 13th 2011, 07:05 PM chiph588@ Quote: Originally Posted by chiph588@ Let's define $\displaystyle f(n)$ to be the number of numbers less than or equal to $\displaystyle n$ such that they are neither the perfect squares or cubes. We have $\displaystyle \displaystyle f(n) = n - \lfloor \sqrt{n}\rfloor - \lfloor \sqrt[3]{n}\rfloor + \lfloor \sqrt[6]{n}\rfloor$. Intuition behind $\displaystyle f(n)$ is we take every number up to $\displaystyle n$ and then subtract all of the perfect squares and cubes. We then must account for duplicate counting. Now solve $\displaystyle f(n) = 1991$. Don't know a great way to solve that though. Perhaps binary search in the interval $\displaystyle [0,4000]$ or something like that. Or you could just plot it and solve via inspection (that's how I did it).
Search IntMath Close 3. Values of the Trigonometric Functions by M. Bourne In the last section, Sine, Cosine, Tangent and the Reciprocal Ratios, we learned how the trigonometric ratios were defined, and how we can use x-, y-, and r-values (r is found using Pythagoras' Theorem) to evaluate the ratios. Now we'll see some examples of these ratios. Finding Exact Values of Trigonometric Ratios Find the exact values indicated. What this means is don't use your calculator to find the value (which will normally be a decimal approximation). Keep everything in terms of surds (square roots). You will need to use Pythagoras' Theorem. Example 1 Find the exact value of sin θ if the terminal side of θ passes through (7, 4). This is what the question means by "the terminal side passes through (7, 4)". We need to know r. Using Pythagoras, we have r=sqrt(7^2+4^2)=sqrt(65) So sin theta=y/r=4/sqrt(65) Example 2 Find the exact values of all 6 trigonometric ratios of θ if the terminal side of θ passes through (2, 10). "Find all 6 trigonometric ratios of θ" means "find sin θ, cos θ, tan θ, csc θ, sec θ and cot θ". This is what the question means by "the terminal side passes through (2, 10)". First we need to find r: r=sqrt(2^2+10^2)=sqrt104=2sqrt26 We now use r to find the required trigonometric ratios. sin theta=y/r=10/(2sqrt26)=5/sqrt26 cos theta=x/r=2/(2sqrt26)=1/sqrt26 tan theta=y/x=10/2=5 csc theta=r/y=(2sqrt26)/10=sqrt26/5 sec theta=r/x=(2sqrt26)/2=sqrt26 cot theta=x/y=2/10=1/5 The following two cases are very common in the study of exact trigonometric ratios. 45o - 45o Triangle sin 45^text(o)=text(opp)/text(hyp)=1/sqrt2 cos 45^text(o)=text(adj)/text(hyp)=1/sqrt2 tan 45^text(o)=text(opp)/text(adj)=1/1=1 30o - 60o Triangle sin 30^text(o)=text(opp)/text(hyp)=1/2 cos 30^text(o)=text(adj)/text(hyp)=sqrt3/2 tan 30^text(o)=text(opp)/text(adj)=1/sqrt3 Memory Aid In the 30-60 triangle, it is easy to forget where to put the 1, 2, sqrt(3) sides and the angles. You could remember it like this: Take an equilateral triangle, sides 2 units: Equilateral triangle, sides 2 units Now, cut it in half horizontally: Equilateral triangle, cut in 2 Take the top half only. The unknown side is the sqrt(3) and the 30^@ and 60^@ angles are as indicated: 30-60 triangle Example 3 - Exact Values Find exact values of the following: a. sin 60o b. cos 60o c. tan 60o d. csc 30o e. cot 45o f. sec 45o Using the 30-60 triangle: 30-60 triangle a. sin 60^"o"=sqrt3/2 b. cos 60^"o"=1/2 c. tan 60^"o"=sqrt3/1=sqrt3 d. csc 30^"o" = 1/(sin 30^"o") = 2 Questions (e) and (f) need the 45-45 triangle: 45-45 triangle e. cot 45^"o" = 1/ (tan 45^"o") = 1 f. sec 45^"o"= = 1/ (cos 45^"o") = sqrt2 Finding Trigonometric Ratios Using Calculator Suggestion: Go find the instruction book for your calculator. You are sure to need it in this section. Each calculator brand and model is a bit different - please don't expect your teacher to know how to use every model of every brand of calculator! Caution: Make sure your calculator is set correctly to degree mode (not radian mode!) for this section. [We learn about radians later. It is very easy to mess up these problems when you are mixing degrees and radians - always check that your answer is reasonable before moving on.] Example 4 Find using caclulator. Answer correct to 4 decimal places. a. sin 49o b. cos 27.53o c. tan 26o35'57" d. csc 18.34o e. sec 5o34'72" f. cot 73o a. This is just one step on the calculator. sin 49o = 0.7547 b. This question is also just one step on calculator, since it is in decimal degree form. cos 27.53o = 0.8868 c. For this next one, you need to make sure that you know how to enter an angle in DMS form (degrees - minutes - seconds). tan 26o35'57" = 0.5007 d. You do not have a "csc" button on your calculator, so you need to do this in 2 steps. Find the sin of 18.34o first, then press the "1/x" button (or "x-1" button) to find the reciprocal. csc 18.34o = 3.1781 e. Likewise with this one, you need to find cos 5o34'72" first, then press the "1/x" button. sec 5o34'72" = 1.0048 f. Similar to numbers 4) and 5), you need to find tan 73o first, then press the "1/x" button. cot 73o = 0.3057 Finding Angles Given The Trig Ratio We are now going to work the other way around. We may know the final trigonometeric ratio, but we don't know the original angle. Example 5 Find θ, given that tan θ = 0.3462 and that 0oθ < 90o. Solution: We need to use the inverse tangent function (not the reciprocal function, as we did for cot θ). Our answer will be an angle. So we use the "tan^-1" button on our calculator, and we have: θ = tan-1 0.3462 = 19.096o. Check: We can use our calculator to check our answer: tan 19.096o = 0.3462. Checks OK. NOTE 1: It is very common (and better) to use "arctan" instead of "tan^-1". You will often see "arctan" throughout this site, rather than tan-1. It helps us to remember the difference between the reciprocal ratio (cot) and the inverse function (arctan). In the above example, we would write: θ = arctan 0.3462 = 19.096o. You'll also see "arcsin", "arccos", ""arccsc"" etc. See more on this in Trigonometry Functions of Any Angle. NOTE 2: Be very careful with the difference between (eg) "sin^-1" and "csc". They are NOT the same! Example: sin-1 0.935 = 69.23o (this gives us an angle). But csc 0.935 = 1.2429 (there is no degree sign on 0.935, so it must be in radians). This is the csc of the angle 0.935 radians. It is a ratio, not an angle, and as you can see, it has a different value. We meet radians later in 7. Radians. For the record, csc 0.935 means: csc 0.935=1/(sin 0.935)=1.2429 (in radians) Exercises - Finding Angles Find θ for 0oθ < 90o, given that 1. sin θ = 0.6235 2. tan θ = 3.689 3. csc θ = 8.32 4. sec θ = 6.96 (I have restricted the domain for θ from 0^"o" to 90^"o" because we haven't seen how to solve it for angles greater than 90^"o" yet.) 1. This is straightforward - use the sin^-1 button on your calculator: θ = sin-1 0.6235 = 38.572o This is equivalent to (and better): θ = arcsin 0.6235 = 38.572o 2. This one uses the tan^-1 button: θ = tan-1 3.689 = 74.833o This is equivalent to: θ = arctan 3.689 = 74.833o Can you draw a triangle to illustrate what this means? Go on, try - it really helps to understand it. 3. We have to do some thinking for this one. There is no csc^-1 button on our calculators, so we need to proceed as follows. csc θ = 8.32, so sin θ = 1/8.32 = 0.12019. (since sin θ is the reciprocal of csc θ). Now we can use the sin^-1 button to obtain: θ = sin-1 0.12019 = arcsin 0.12019 = 6.9032o 4. Similar to Q3, we need to find the reciprocal first. sec\ θ = 6.96, giving us cos θ = 1/6.96 = 0.143678. Therefore θ = cos-1 0.143678 = arccos 0.143678 = 81.739o
Sei sulla pagina 1di 19 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis CHAPTER 6- COORDINATE GEOMETRY 6.1 DISTANCE BETWEEN TWO POINTS y y Q (x 2 , y 2 ) 2 y 2 - y 1 y P(x 1 , y 1 ) 1 x 2 - x 1 How to obtain the formula? x To find the distance or length of PQ, use the concept of Pythagoras’ Theorem. PQ 2 = (x 2 x ) 1 2 + ( y 2 y ) 1 2 Therefore: D = (x 2 x ) 1 2 + ( y 2 y ) 1 2 where D is distance. Example 1: The distance point A(6, 3t) and point B(12, -t) is 10 units. Find the possible values of t. Solution: 2 2 10 = (12 6) + ( t 3 ) t 2 2 100 = 6 + ( 4 ) t Square the both sides 2 100 = 36 + 16 t 2 16 t = 64 t 2 = 4 t = ± 2 Example 2: Point A(h, 2h) and point B(h -3 , 2h + 1) are two points which are equidistant from the origin. Find the value of h. Solution: (h 0) 2 + (2h 0) 2 = (h 3 0) 2 + (2h +10) 2 Page \| 60 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 2 2 2 2 h + (2 h ) = ( h 3) + (2 h + 1) 2 2 2 2 h + 4 h = h 6 h + 9 + 4 h + 4 h + 2 2 5 h = 5 h 2 h + 10 2 h = 10 h = 5 EXERCISE 6.1 1 1. Find the perimeter of triangle ABC with points A(2, 1), B(4, 5) and C(-2, 8). 2. The points (-3, -5) lies on the circumference of a circle with centre (2, 1). Calculate the radius of the circle. 3. Given that the distance between points A(1,3) and B(7, k) is 10 units . Find the possible value of k. 4. Given point A( 1, 7) and B(p, 2) and the distance between the points A and B is 13 units. Find the value of p. 6.2 DIVISION OF A LINE SEGMENT Line segment is a line that has distant. 6.2.1 Mid-point The formula to find mid-points is just the same as we have learned in Form Two that is: 2 , 2 ( x , y ) = ( x 1 + x 2 y 1 + y 2 ) Example: Given C(2, 5) is the mid-point of the point B(h, 3) and point D(-4, k). Find the values of h and k. Solution: Use the formula, 2 = 4 + h 5 = k + 3 2 2 h − 4 = 4 k + 3 = 10 h = 8 k = 7 EXERCISE 6.2.1 1. The coordinates of A and B are (m, 5) and (6, n) respectively. Find the values of m and n if the mid- point of the points is (4, 10). 2. The coordinates of M and N are (4, 2) and (6, 5) respectively. Find the mid-point of these two points. 3. The coordinates of X and Y are (-1, b) and (a, 7) respectively. Find the values of a and b if the mid- point of the points is (1, 2). Page \| 61 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 6.2.2 Point which divides a line segment in the ratio m: n y 2 - y y- y 1 P(x 1 , y 1 ) n X(x, y) m x- x 1 x 2 - x Q (x 2 , y 2 ) y 2 - y y- y 1 If the line PQ moves downwards, it will reach the horizontal line. From that we know that: x x 1 m = x 2 n nx nx x ( ( x x x 1 ) n = m ( x 2 x mx mx 2 mx 2 ) x nx mx n 1 mx + = = nx 1 2 + m = + nx 1 ) + = nx mx 2 1 + m + n If the line PQ moves upwards, it will reach the vertical line y y 1 m = y n ny ny ( y 2 y n = ( ) y + m y 1 ny my + n 1 y = ny 1 m my ny 1 = ny my 2 ( 2 + 1 + = = ) + y 2 y my my 2 my 2 m + n ) Hence, the formula to find the point that divides the line segment in ratio m : n is ( x , y ) =  , m + n m + n nx 1 + mx 2 ny 1 + my 2 When m = n , the point will be the midpoint of the line segment. Page \| 62 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis Example: Find the coordinates of point P that divides the straight line that joins E(-6, 10) and F(4, -5) in the ratio 2: 3 Solution: m: n = 2 : 3 ( x , y )  nx =   1 + m + mx n 2 , ny 1 m + + my n 2     P(x, y) =   (3)( − 6) + (2)(4) (3)(10 ) + (2)( − 5)   2 + 3 , 2 + 3   P(x, y) =   − 18 5 + 8 , 30 − 5 10    = (2,4) E(-6, 10) o 2 P(x, y) o 3 o F(4, -5) EXERCISE 6.2.2 1. Point R divides the line segment joining J(-1, -7) and Q(10 ,7) internally in the ration PR: RQ = 1: 3. Find the coordinate of R. 2. The points P(t, 2t), Q(2a, b) and R(4a, 3b) are on a straight line. Q divides PR internally in the ratio 1: 4. Show that b = 6a . 3. Given points A(k, 5), B(0, 3) and C(5, 4). Find the possible values of k if the length of AB is twice the length of BC. 6.3 AREA OF POLYGONS 6.3.1 Area Of Triangle C(x 3 , y 3 ) B(x 2 , y 2 ) A(x 1 , y 1 ) K L M How to obtain the formula? The area of ABC = Area of trapezium ACLK + Area of trapezium BCLM - Area of trapezium ABMK Page \| 63 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis = = = 1 3 2 1 2 3 1 [ 2 3 x [ x ( y + y 1 y y 3 3 + + )( x x 3 3 x 3 y y 1 1 x 1 x x 1 1 ) + y y 1 1 1 2 ( x x y 2 1 1 y y 3 3 + y 3 + x + x )( 2 2 y y x 2 2 2 x + + x x 2 2 3 ) y y 3 3 1 2 x x ( 3 3 y y y 1 2 2 = 1 ( 2 y + x y + x y x y x y x y ) x 1 2 2 3 3 1 2 1 3 2 1 3 = 1 [( 2 x y + x y + x y ) ( x y + x y + x y )] 1 2 2 3 3 1 2 1 3 2 1 3 1 x x x x 1 2 3 1 = Area of ∆ ABC = 2 y y y y 1 2 3 1 1 = [( x y + x y + x y ) ( x y + x 1 2 2 3 3 1 2 1 3 2 y 2 The formula of the area of triangle is 1 x x x y y 2 3 2 3 x 1 2 y 1 1 y 1 6.3.2 Area of Quadrilateral The formula of the area of quadrilateral is + y 2 x x 3 3 )( x y y 3 3 2 x ( x x 2 1 ) 2 y y 2 2 + x x 2 Simplify + x 1 y 3 )] 2 y y 1 1 x x x x 2 3 4 y y y 2 3 4 x 1 2 y 1 1 y 1 1 = ( x y + x y 1 2 2 3 2 Example 1: + x y 3 4 + x y 4 1 ) ( x y + x y 2 1 3 2 + x y 4 3 + x y 1 4 ) 1 x + x 1 1 y y 1 1 x + x 1 1 y y 2 2 ) ) Find the value of m if the point P(m, 2), Q(4, -3) and R(-2, 5) lie on a straight line. Solution: Method 1- Using Concept Area Of Triangle 1 m 4 − 2 m 2 − 3 2 5 2 = 0 The area of a straight line is zero. Page \| 64 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis ( − 3 m + 20 − 4) − (8 + 6 + 5 m ) = 0 − 3 m + 16 − (14 + 5 m = 0 − 3 m + 16 − 14 − 5 m ) = 0 − 8 m + 2 = 0 − 8 m + 2 = 0 − 8 m + 2 = 0 8 m = 2 1 m = 4 Method 2 Using the concept of gradient of straight line m m QR QR = = 3 5 4 8 6 ( 2) = − 4 3 m 2 PQ ( = m 3) QR = − 4 4 4 m m m 4 = 1 16 1 m = 4 = − 3 15 Point P, Q and R lie on the same line, so m PQ = m QR Example 2: Find the possible values of k if the area of triangle with vertices A (9, 2), B(4, 12) and C(k, 6) is 30 unit 2 Solution: 1 9 4 k 9 = 30 2 2 12 6 2 (108 + 24 + 2 k ) (8 + 70 10 k = 60 12 k + 54) = 60 Page \| 65 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis (i) 70 10k = −60 10 k = 1 k = 10 (ii) 70 10k = −60 10 k = 130 k = 13 The modulus sign will always result a positive number. If result 60, there will be two values of k. If the value of k is 1, the value in the modulus is 60. If the value of k is 13, the value in the modulus is -60 and at last becomes 60 because the modulus sign will always result a positive number. 70 10 k would Example 3: Find the area of quadrilateral KLMN given K (1, 3), L(-1, 2) and M(-4, -3) and N(6, -9). Solution: 1 x 1 x x x x 2 3 4 1 2 y y y y 1 y 2 3 4 1 1 1 1 − 4 6 1 = 2 3 2 3 − 9 3 = = = (2 + 3 + 36 + 18) − ( − 3 − 8 − 18 − 9) 59 − ( − 38) 53 + 38 1 = 97 2 = 48 .5 unit EXERCISE 6.3 2 1. Find the possible values of p if the area of triangle with the vertices D(p, -p), E(1, 0) and F(-3, 6) is 10 unit 2 . 2. Find the area of the rhombus PQRS if the coordinates of the points P, Q and R are (6, 4). (8, 7) and (-6, 3) respectively. 3. The points G(4, -2), F(1, 1) and H(-2, p) lie on a straight line. Find the value of p. 4. Find the area of the triangle PQR if the coordinates of the vertices are: (a) P(1, 3), Q (4, 2) and R (7, 0) (b) P(-2, 5), Q (7, -2) and R (-3, 1) (c) P(-1, -5),Q (1, -4) and R(-1, -3) Page \| 66 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 6.4 EQUATION OF A STRAIGHT LINE 6.4.1 x-intercept and y-Intercept of a straight line y b 0 a x 1. The line intersects with the x-axis at a and the line intersects with the y-axis at b 2. a is called x-intercept and b is called y-intercept. EXERCISE 6.4.1 1. Determine the x-intercept for the following straight lines: (a) (b) (c) x y + 2 3 3x 4 y 9 = 0 y = 6x 6 = 1 2. Determine the y-intercept for the following straight lines: (a) (b) x 1 + y = 1 5 5x 7 y 14 = 0 (c) 3y = 5x 6 6.4.2 (i) The Gradient of a Straight Line (ii) (iii) (iv) Page \| 67 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis Finding the gradient (m) of a straight line y − y 1- 2 1 m = x − x 2 1 Example: y A(0,5) 0 B(0, 4) y − y 2 1 m = x − x 2 1 5 − 0 = m AB 0 − 4 5 = − 2- y 4 (0, b) 0 (a, 0) b − 0 m = 0 − a x x Page \| 68 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis = − b a Hence, m = − x int ercept y int ercept Example: y A(0,5) 0 B(0, 4) y − int ercept m = x − int ercept 5 = − m AB 4 3- m = tanθ Example: y A(0,5) 5 θ 0 4 B(0, 4) x x Page \| 69 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis m m AB AB = tanθ = − 5 4 EXERCISE 6.4.2 Find the gradient of the following points. (a) A (2, 3) and B (4, 5) (b) M (-3, 1) and N (4, -2) (c) P (2, 3) and Q (4, 3) (d) C (2, 5) and D (2, 8) (e) F(0, 6) and G( 3, 0) 6.4.3 The Equation of a Straight Line 1-General Form The equation of general form is ax + by + c = 0 Example: Given the equation of a straight line is 2 y = 4x + 5 . Change the equation into the general form. Solution: 2y = 4x + 5 4x 2y + 5 = 0 2-Gradient Form The equation of gradient form is y = mx + c where m is the gradient and c is y-intercept Example: Given the equation of a straight line is 2 y = 4x + 5 . Determine the gradient and the y-intercept of the straight line. Solution: 2y = 4x + 5 5 y = 2x + 2 Hence, the gradient of the straight line is 2 while and the y-intercept of the straight line is 5 . 2 Page \| 70 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 3-Intercept form The equation of intercept form is x + a y b = 1 where a is x-intercept and b is y-intercept. Example: Given the equation of a straight line is 2 y = 3x + 6 . Convert the equation into the intercept form. Hence, state the x-intercept and y-intercept of the straight line. Solution: 2y = 3x + 6 = 3 x 3 x + + y 2 y 2 6 = 1 6 6 x y + 2 x 3 y + 2 3 = 1 = 1 Hence, the x-intercept of the straight line is -2 and y-intercept of the straight line is 3. Example: Find the equation of the straight line which has a gradient of -3 and passes through the mid-point of the line joining A (1, 4) and B(7, -2). Solution: Mid-point of AB =  7 + 1 , 4 + 2 ( 2) = (4,1) 2 The equation of the straight line which has a gradient of -3 and passes through (4, -1) is y 1 = − 3 x 4 1 = 12 3 x y 3x + y 13 = 0 or y = −3x +13 Use the general point (x, y) and specific point (4, -1) to find the gradient of the line and it is equal to the given gradient. EXERCISE 6.4.3 1. Write each of the following equations to intercept form. Hence, state the gradient of the straight line. (a) 2x + y − 4 = 0 (b) x − 3y = 6 (c) 4x + 3y = 2 (d) 2x + 3y = 1 2. Find the equation of the straight line which has a gradient of -2 and passes through point B (7, -2). Page \| 71 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 6.4.4 The Point of intersection of two Straight lines 1. When two lines intersect, the point of intersection lies on both lines. 2. This means the coordinates of the point satisfy both the equations of the lines. 3. Therefore, we need to solve the equation simultaneously in order to determine the point of intersection. Example 1: The straight line which has a gradient of 2 and passes through the point (4, -1) intersects with the straight line x + y = −6 at the point P. Find the coordinates of the point P. Solution: First of all, we have to find the equation of the straight line. y − ( − 1) = 2 y x − 4 + = 1 2 x − 8 2 x − y = 9 1 x + y = −6 2 1 + 2 , 3 x = 3 x = 1 Substitute x = 1into 2(1) y = − 7 y = 9 1 , Hence the coordinates of P is (1, -7) Example 2: The straight line x + y 4 = 0 and 2x + 3y 11 = 0 intersect at point A. Find the equation of the straight line which passes through the point A and point B (5, 2). Solution: x + y − 4 = 0 2 x 2x + 3y −11 = 0 + 2 y − 8 1 = 0 2 - , y − 3 = 0 y = 3 1 2 Page \| 72 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis Substitute y = 3 into 2 2 x = 1 x x = + 2(3) 2 8 = 0 1 , Hence the coordinates of A is (1, 3) m m AB AB = 2 3 5 1 = − 1 4 The equation of the straight line that passes through point A and B is y − 3 1 = − x − 1 4 We can write the equation in any form. Either the general form or 4 y − 12 = 1 − x 1 13 intercept form or gradient form. x + 4 y −13 = 0 or x + 4 y = 13 or y = − x + 4 4 EXERCISE 6.4.4 1. Find the coordinates of the point of intersection of the line 4x + 3y 11 = 0 and 2x 6 y +17 = 0 . 2. Find the points of intersection of the following pairs of straight lines. (a) x 5y 2 = 0 2x y + 5 = 0 (b) y = 2x + 4 y = x + 5 3. Find the equation of that is parallel to the line y = 2x + 5 and passing through the point of intersection of lines 2x y 9 = 0 and x + 2y = 2 . 6.5 PARALLEL AND PERPENDICULAR LINES Parallel Lines 1. When two lines are parallel, they have the same gradient. B A P Q 2. If line AB and line PQ are parallel, so m AB = m PQ . Page \| 73 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis Example: The straight line AB passes through the point (6, 3) is parallel to the straight line PQ. Given point P (0, 2) and point Q(4, 0). Find the equation of the straight line AB. Solution: First of all, we have to find the gradient of straight line PQ m PQ = − = 1 2 2 4 m m AB AB = m PQ = 1 2 The equation of the straight line that passes through point (6, 3) is y 3 x 6 = − 1 2 2 y 6 = 6 x x + 2 y 12 = 0 or x + 2 y = 12 or y = − We can write the equation in any form. Either the general form or 1 2 intercept form or gradient form. x + 6 Perpendicular Lines y B m 1 m 2 θ α A C x 1. Given that line AB and BC are perpendicular to each other. 2. We already know m = tanθ . m 1 = tanθ BC m 2 = tanα AB = = − AB BC Page \| 74 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 3. When m 1 × m 2 = m 1 × m 2 = − 1 BC AB × − AB BC Hence, if two lines are perpendicular to each other, then the product of their gradient is 1. Example: Given the straight line y = tx 9 and y = 2x + 3 is perpendicular to each other. Find the value of t. Solution: y m y m = 2x + 3 = 2 1 = tx 9 = t 2 × m m 2 2 1 × 2 = − 1 t t = − t = − 1 2 = − 1 1 Use the concept EXERCISE 6.5 1. The equation of the straight line PQ is 6x 8x + 7 = 0 . Each of the following straight line is parallel to PQ. Find the value of t in each case. (a) tx + 4 y 6 = 0 y = t (b) x + 8 2 (c) 2x ty 1 = 0 2. Find the equation of the straight line which passes through point B (2, -5) and perpendicular to the straight line y = −3x + 1 . 3. PQRS is a rhombus with P (0, 5) and the equation of QS is y = 2x + 1. Find the equation of diagonal of PR. 4. Find the value of h if the straight line y hx + 2 = 0 is perpendicular to the straight line 5y + x + 3 = 0 . 5. Given that the equation of the line PQ is 2 y = 3x +15 and point Q lies on the y-axis. Point R is (4, 1) lies on line QR. Find the equation of QR if the line PQ and QR are perpendicular to each of other. Page \| 75 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 6.6 LOCUS OF A MOVING POINT Locus represents the path followed by a moving point with the reference to one or more fixed points, satisfying certain conditions. 6.6.1 Equation of Locus Equidistant from a fixed point y P (x, y) 3 unit A (1, 1) The equation of locus is x 2 2 ( x 1) + ( y 1) = 3 Square the both sides 2 2 ( x 1) + ( y 1) = 9 2 2 x 2 x + 1 + y 2 y + 1 = 9 2 2 x + y 2 x 2 y 7 = 0 Equation of locus is actually involving the distance between two points. So we have to use the formula of the distance between two points to find the equation of locus. There is no specific formula to find the formula to find the equation of locus. In this case, P is the moving point such that its distance is always 3 unit from point A. Equidistant from two fixed points Find the equation of the locus of a moving point P such that its distance from the point A (1, 2) and point B (3, 4) are equal. B (3, 4) P(x, y) locus A (1, 2) AP = BP ( x − 1) 2 + ( y − 2) 2 = ( x − 1) 2 + ( y − 2) 2 = ( x x 2 − 2 x + 1 + y 2 − 4 y + x 2 + y 2 − 2 x − 4 y + 5 = 4 x + 4 y − 20 = 0 x + y − 5 = 0 x 3) = 2 2 4 x + x 2 3) + 2 y ( 2 ( + y 6 x 6 y 4) 2 + x 9 + 8 y 2 y + 4) 2 Square the both sides y + 16 8 25 Page \| 76 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis Constant ratio between two fixed points Find the equation of locus of a moving point R such that its distance from the point E (4, 3) and the point F (1, 5) is in the ration 3: 1. RE 3 = RF 1 3 RF = RE 2 2 2 2 Square the both sides 3 ( x 1) + ( y 5) = ( x 4) + ( y 3) 2 2 2 2 9( x 2 x + 1 + y 10 y + 25) = x 8 x + 16 + y 6 y + 9 2 2 2 2 9 x 18 x + 9 + 9 y 90 y + 225 = x 8 x + 16 + y 6 y + 9 2 2 8 x + 8 y 10 x 84 y + 209 = 0 EXERCISE 6.6 1. Given the point A (0, 3) and the point B (1, 4).Find the equation of locus of a moving point Q such that AQ= 2QB. 2. Given A (5, -2) and B (2, 1) are two fixed points. Point Q moves such that the ratio of AQ: QB = 2: 1. Show that the equation of the locus of point Q is x 2 + y 2 2 x 4 y 3 = 0 . 3. P is a moving point such that its distances from the points A(2, 5) and B(0, 3) is in the ratio of 2: 1. Find the equation of locus P. 4. N is a locus which moves in such a way that NP=NQ. Given that P and Q are coordinates (-3, 6) and (6,- 4) respectively, find the equation of locus N. 5. Show that the equation of the locus of a point that moves in such way that is distance from a fixed point (3, -1) is 6 units, is by x 2 + y 2 6 x + 2 y 26 CHAPTER REVIEW EXERCISE 1. Given the equation of straight lines AB and CD are x 6 = + 0 . y k = 1 and 2x + 3 y 4 = 0 respectively, find the value of k if AB is perpendicular to CD. 2. The coordinates of the point A and B are (-2, 3) and (7, -3) respectively. Find (a) the coordinates of C given that AB: BC = 1: 2. (b) the equation of the straight line that passes through B and is perpendicular to AB. 3. ABCD is a parallelogram with coordinates A (-2, 3), B (3,4), C (2, -1) and D (h, k). (a) Find the value of h and k. (b) Find the equations of the diagonals AC and BD. (c) State the angle between the diagonals AC and BD. (d) Find the area of the parallelogram ABCD. Page \| 77 Chapter 6- Coordinate Geometry Additional Mathematics Module Form 4 SMK Agama Arau, Perlis 4. P, Q and R are three points on a straight line. The coordinates of P and R are (-2, 3) and (3, 5) respectively. Point Q lies on the y-axis. Find (a) the ratio PQ: QR (b) the coordinates of point Q 5. H is a point which moves such that its distance from point P (1, -2) and Q (-3, 4) is always equal. Show that the equation of the locus H is given by the equation 2x 3 y + 5 = 0 . 6. Find the equation of straight line that passes through point P( 1 -2) and parallel to 7. In diagram below, PRS = 90° . 4x – 2y = 8. y P(−6,9) S Q 3 0 R Find (a) the coordinates of R (b) the ratio of PQ: QR (c) the equation of RS. x 8. Diagram below shows a triangle BCD. The point A lies on the straight line BD.\ y D(17,k) A(13,8) C(−1,7) B(5,2) x 0 Find (a) the value of k (b)
# Fraction Arithmetic Fraction Arithmetic I hope everyone had a good holiday season. I certainly enjoyed mine. Over this season, I had a chance to speak with some youth and their parents. Funny that whenever we discuss that I have a PhD in applied mathematics, the topics of the children struggling in mathematics and the possibility of tutoring their children always seem to come up. I have no problem with tutoring and I actively participate in such sessions in my spare time. However I will say that it is sometimes a difficult task to do this job over such a short time period. Needless to say, I felt bad that I couldn’t have been of more assistance. So, this being the holiday season and all, I decided to make somewhat of a new years resolution to focus this site more towards some of the things that the youth struggle with to hopefully be able to answer some of their questions. With that being said, the first area that I decided to look at was fractions. This is one of the first areas where the youth begin to dislike mathematics. I feel like regardless of how much teachers and professors speak of the importance of understanding these processes, many students simply never grasp the procedures involved, partially because they never get used to the rules associated with these matters. In this first script on fractions, I’ve focused on four types of problems corresponding to the four basic operations of arithmetic: Addition, Subtraction, Multiplication and Division. To add two fractions of the form num1 den1 + num2 den2 We use the formula num1 den1 + num2 den2 = num1 den1 + num2 den2 = num1den2 + num2den1 den1den2 Lets take a moment to consider where this formula comes from. In order to be able to add fractions we first need to obtain a common denominator for the two fractions. One way that always works to obtain a common denominator is to multiply the denominators of the two fractions. So in the formula above, the denominator on the right hand side of the equals sign is the product of the two denominators on the left hand side. Once we have a common denominator, we need to rewrite each of the two fractions in terms of this common denominator. num1 den1 + num2 den2 = num1den2 den1den2 + num2den1 den1den2 The formula for subtracting fractions is similar, with the notable difference of a subtraction in the place of addition. num1 den1 num2 den2 = num1den2 – num2den1 den1den2 To multiply two fractions (also known as taking the product of two fractions, the resulting numerator is the product of the two initial numerators, and likewise the resulting denominator is the product of the two initial denominators. num1 den1 * num2 den2 = num1num2 den1den2 Finally, remembering that division is the inverse of multiplication, we can derive the formula to divide two fractions by multiplying by the inverse of the fractions: num1 den1 ÷ num2 den2 = num1 den1 * den2 num2 = num1den2 den1num2 The next step in each of these operations is to reduce the fraction to lowest terms. One way of doing this is by considering Euclid’s GCD algorithm which is available here. The script is available to practice your work on fractions at http://www.learninglover.com/examples.php?id=31 # Learning Math through Set Theory In grade school, we’re taught that math is about numbers. When we get to college (the ones of us who are still interested in math), we’re taught that mathematics is about sets, operations on sets and properties of those sets. Understanding Set Theory is fundamental to understanding advanced mathematics. Iv wrote these scripts so that users could begin to play with the different set operations that are taught in a basic set theory course. Here, the sets are limited to positive integers and we’re only looking at a few operations, in particular the union, intersection, difference, symmetric difference, and cross product of two sets. I will explain what each of these is below. The union of the sets S1 and S2 is the set S1 [union] S2, which contains the elements that are in S1 or S2 (or in both). Note: S1 [union] S2 is the same as S2 [union] S1. The intersection of the sets S1 and S2 is the set S1 [intersect] S2, which contains the elements that are in BOTH S1 and S2. Note: S1 [intersect] S2 is the same as S2 [intersect] S1. The difference between the sets S1 and S2 is the set S1 / S2, which contains the elements that are in S1 and not in S2. . Note. S1 / S2 IS NOT the same as S2 / S1. Note. S1 / S2 is the same as S1 [intersect] [not]S2. The symmetric difference between the sets S1 and S2 is the set S1 [symm diff] S2, which contains the elements that are in S1 and not in S2, or the elements that are in S2 and not in S1. Note. S1 [symm diff] S2 is the same as S2 [symm diff] S1. Note. S1 [symm diff] S2 is the same as (S1 [intersect] [not] S2) [union] (S2 [intersect] [not] S1). The cartesian product of the two sets S1 and S2 is the set of all ordered pairs (a, b), where a [in] S1 and b [in] S2.
## Simplification A process which convert complex arithmetical expression into simple expression is called simplification. Order of Operations When we simplify expression and equation we used order of operation. ### VBODMAS Rule In this rule, we use operations in order V - Vinculum or Bar B - Brackets - Order of brackets First- Small brackets ( ) second- Middle brackets{ } Third- Square brackets[ ] O - Of D - Division M - Multiplication A - Addition S - Subtraction Order of above mentioned operations is same as the order of letters in the VBODMAS from left to right Example- Simplify 5 - [ 4 - {9 - ( 7 - 5 - 3 )}] Solution:- 5 - [ 4 - {9 - ( 7 - 5 - 3 )}] = 5 - [ 4 - { 9 - ( 7 - 5 + 3 )}] = 5 - [ 4 - { 9 - ( 7 - 2 )}] = 5 - [ 4 - { 9 - 5 }] = 5 - [ 4 - 4 ] = 5 - 0 = 5 Example- Simplify 34500 ÷ 30 ÷ 10 Solution:- 34500 ÷ 30 ÷ 10 = 34500/30 ÷ 10 = 1150 ÷ 10 = 1150/10 = 115 Example = - 5 1 + 3 2 + 2 3 7 7 7 Solution:- = 5 1 + 3 2 + 2 3 7 7 7 = (5 + 2 + 2 ) + 1 + 2 + 3 7 7 7 = 10 5 7 = 10 5 7 Example- Simplify 34500 ÷ 30 ÷ 10 Solution:- 34500 ÷ 30 ÷ 10 = 34500/30 ÷ 10 = 1150 ÷ 10 = 1150/10 = 115 Example = - 5 1 + 3 2 + 2 1 7 7 7 Solution:- = 5 1 + 3 2 + 2 1 7 7 7 = (5 + 3 + 2) + 1 + 2 + 1 7 7 7 = 10 5 7 ## Basic Formulae (a + b)² = a² + b² + 2ab Example- Find the square of 107. Solution:- ( 107 )2 = ( 100 + 7 )2 = 1002 + 72 + 2 × 100 × 7 = 10000 + 49 + 1400 = 11449 (a - b)² = a² + b² - 2ab Example- Find the square 999. Solution:- ( 999 )2 = ( 1000 - 1 )2 = 10002 + 12 - 2 × 1000 × 1 = 1000000 + 1 - 2000 = 998001 a² - b² = (a + b)(a - b) Example- Find the product of 998 and 1002. Solution:- ( 1000 - 2 ) ( 1000 + 2 ) = 1000² - 2² = 1000000 - 4 = 999996 Example- Find the value of 48² - 46² Solution:- 48² - 46² = ( 48 + 46 ) ( 48 - 46 ) = 94 × 2 = 188 (a + b)³ = a³ + b³ + 3ab (a - b) Example- Find the value of ( 103 )³ Solution:- ( 103 )³ = ( 100 + 3 )³ = 100³ + 3³ + 3 × 100 × 3 ( 100 + 3 ) = 1000000 + 27 + 900 × 103 = 1000000 + 27 + 92700 = 109272 (a - b)³ = a³ - b³ - 3ab (a - b) Example- Find the value of ( 98 )³. Solution:- ( 98 )³ = ( 100 - 2 )³ = 100³ - 2³ - 3 × 100 × 2 ( 100 - 2 ) = 1000000 - 8 - 600 × 98 = 1000000 - 8 - 58800 = 1000000 - 58808 = 941192 a³ + b³ = (a + b) - (a² - ab + b²) Example- Find the value of ( 4.8 )³ + ( 1.2 )³ / ( 4.8 )² - 4.8 × 1.2 + ( 1.2 )² Solution:- Let a = 4.8 and b = 1.2 a³ + b³ / (a² - ab + b² ) = {( a + b ) ( a2 - ab + b2 ) }/ (a2 - ab + b2) = ( a + b ) = 4.8 + 1.2 =6 7. a³ - b³ = (a - b) - (a² + ab + b²) Example- Find the value of ( 11.3 )³ -( 1.3 )³ / ( 11.3 )³ +11.3 ×1.3 +( 1.3 )³ Solution:- Let a = 11.3 and b = 1.3 = a³ - b³ / a² + ab + b² = ( a - b ) ( a² + ab + b² ) / a² + ab + b² = ( a - b ) = 11.3 - 1.3 =10 8. (a + b + c)² = a² + b² + C² +2ab + 2bc + 2ca Example- If a + b + c = 14 and a² + b² + c² = 96, then find the value of ab + bc + ca . Solution:- a + b + c = 14 Squaring on both sides or, ( a + b + c )² = (14 )² or, a² + b² + c² + 2ab + 2bc + 2ca = 196 or, 96 + 2ab + 2bc + 2ca = 196 or, 96 + 2 ( ab + bc + ca ) = 196 or, 2 ( ab + bc + ca ) = 196-96 or, 2 ( ab + bc + ca ) = 100 ∴ ab + bc + ca = 100/2 = 50 9. (a + b)² + (a - b)² = a² + b² - 2ba = 2a² + 2b² = 2 ( a² + b²) ∴ ( a + b )² + ( a - b )² = 2 ( a² + b²) Example- Find the value of (745 + 123)² + (745 - 123)² 745 × 745 + 123 × 123 Solution:- Let a = 745 and b = 123 = (a + b)² + (a - b)² a² + b² = 2 (a² + b²) a² + b² = 2 10. (a + b)² - (a - b)² = a² + b² - 2ba - (a² + b² - ab) = a² + b² + 2ab - a² - b² + 2ab =2ab+2ab = 4ab ∴ ( a + b )² - ( a - b )² = 4ab 11. a³ + b³ c³ - 3abc = (a + b + c) = (a² + b² + c² - ab - bc - ca) = 1 (a + b + c)[( a - b )² + ( b - c )² + ( c - a )² ] 2 If a + b + c = 0, then a3 + b3 + c3 - 3abc = 0 or, a3 + b3 + c3 = 3abc Example- Find the value of a3 + b3 + c3 - 3abc when a = 23, b = 24 and c = 25 Solution:- a3 + b3 + c3 - 3abc = 1/2 ( a + b + c ) [ ( a - b )2 + ( b - c )2 + ( c - a )2 ] = 1 (23 + 24 + 25)[( 23 - 24 )² + ( 24 - 25 )² + ( 25 - 23 )²] 2 = 1 × 72 [( -1 )2 + ( -1 )2 + ( 2 )2] 2 = 36 [ 1 + 1 + 4 ] = 36 × 6 = 261 Example- Solution:- Let a = 5.3, b = 3.5 and c = 1.2 = (5.3)3 + (3.5)3 + (1.2)3 - 3 × 5.3 × 3.5 × 1.2 (5.3)2 + (3.5)2 + (1.2)2 - 5.3 × 3.5 - 3.5 × 1.2 - 1.2 × 5.3 = a3 + b3 + c3 - 3abc a2 + b2 + c2 - ab - bc - ca = 5.3 + 3.5 + 1.2 = 10 Example- If a = 25, b = 50 and c = -75, then find the value of a3 + b3 + c3 - 3abc. Solution:- Here, a = 25, b = 50 and c = -75 a + b + c = 25 + 50 + ( -75 ) = 75 - 75 = 0 a3 + b3 + c3 - 3abc = ( a + b + c ) ( a2 + b2 + c2 - ab - bc - ca ) or, a3 + b3 + c3 - 3abc = 0 ( a2 + b2 + c2 - ab - bc - ca ) ∴ a3 + b3 + c3 - 3abc = 0 ### Some Important Rule Rule- If x + 1/x = a, then x2 + 1/x2 = a2 - 2 ( x + 1/x )2 = x2 + 1/x2 + 2 × x × 1/x or, ( x + 1/x )2 = x2 + 1/x2 + 2 ∴ x2 + 1/x2 = ( x + 1/x )2 - 2 Example- If x + 1/x = 3, then find the value x2 + 1/x2. Solution:- x + 1/x = 3 ∴ x2 + 1/x2 = ( x + 1/x )2 - 2 = 32 - 2 = 9 - 2 = 7 Example- If x + 1/x = 3, then find the value x4 + 1/x4. Solution:- x + 1/x = 3 ∴ x2 + 1/x2 = ( x + 1/x )2 - 2 = 32 - 2 = 9 - 2 = 7 ∴ x4 + 1/x4 = ( x2 + 1/x2 )2 - 2 = 72 - 2 = 49 - 2 = 47 Example- If x + 1/x = 3, then find the value x8 + 1/x8. Solution:- x + 1/x = 3 ∴ x2 + 1/x2 = ( x + 1/x )2 - 2 = 32 - 2 = 9 - 2 = 7 ∴ x4 + 1/x4 = ( x2 + 1/x2 )2 - 2 = 72 - 2 = 49 - 2 = 47 ∴ x8 + 1/x8 = ( x4 + 1/x4 )2 - 2 = 472 - 2 = 2209 - 2 = 2207 Rule- If x + 1/x = a, then x3 + 1/x3 = a3 - 3a ( x + 1/x )3 = x3 + 1/x3 + 3 × x × 1/x ( x + 1/x ) or, ( x + 1/x )3 = x3 + 1/x3 + 3( x + 1/x ) ∴ x3 + 1/x3 = ( x + 1/x )3 - 3( x + 1/x ) Example- If x + 1/x = 3, then find the value of x3 + 1/x3. Solution:- x + 1/x = 3 ∴ x3 + 1/x3 = ( x + 1/x )3 - 3( x + 1/x ) = 33 - 3( x + 1/x ) = 27 - 3 × 3 = 27 - 9 = 18 Example- If x + 1/x = 3, then find the value of x9 + 1/x9. Solution:- x + 1/x = 3 ∴ x3 + 1/x3 = ( x + 1/x )3 - 3( x + 1/x ) = 33 - 3( x + 1/x ) = 27 - 3 × 3 = 27 - 9 = 18 ∴ x9 + 1/x9 = ( x3 + 1/x3 )3 - 3( x + 1/x ) = 183 - 3( x3 + 1/x3 ) = 5832 - 3 × 18 = 5832 - 54 = 5778 Rule- If x + 1/x = a, then x5 + 1/x5 = [{ ( a3 - 3a ) ( a2 - 2 ) } - a ] x5 + 1/x5 = ( x3 + 1/x3 ) ( x2 + 1/x2 ) - ( x + 1/x ) Example- If x + 1/x = 3, then find the value of x5 + 1/x5. Solution:- x + 1/x = 3 ∴ x3 + 1/x3 = ( x + 1/x )3 - 3( x + 1/x ) = 33 - 3( x + 1/x ) = 27 - 3 × 3 = 27 - 9 = 18 ∴ x2 + 1/x2 = ( x + 1/x )2 - 2 = 32 - 2 = 9 - 2 = 7 ∴ x5 + 1/x5 = ( x3 + 1/x3 ) ( x2 + 1/x2 ) - ( x + 1/x ) = 18 × 7 - 3 = 126 - 3 = 123 Rule- If x + 1/x = a, then x - 1/x = √a2 - 4 Example- If x + 1/x = 5, then find the value of x - 1/x. Solution:- x + 1/x = 5 ∴ x - 1/x = √( x + 1/x )2 - 4 = √( 5 )2 - 4 = √21 Example- If x3 + 1/x3 = 10, then find the value of x3 - 1/x3. Solution:- x3 + 1/x3 = 10 ∴ x3 - 1/x3 = √( x3 + 1/x3 )2 - 4 = √( 10 )2 - 4 = √100 - 4 = √96 = 4 √6 Rule- If x - 1/x = a, then x + 1/x = √a2 + 4 Example- If x - 1/x = 7, then find the value of x + 1/x. Solution:- x - 1/x =7 ∴ x + 1/x = √( x - 1/x )2 + 4 = √( 7 )2 + 4 = √49 + 4 = √53 Example- If x = 17, then find the value of x5 - 18x4 + 18x3 - 18x2 + 18x - 19. Solution:- x5 - 18x4 + 18x3 - 18x2 + 18x - 19. = x5 - 17x4 - x4 + 17x3 + x3 - 17x2 - x2 + 17x + x - 19 = 175 - 17 ( 17 )4 - 174 + 17 ( 17 )3 + 173 - 17 ( 17 )2 - 172 + 17 ( 17 ) + 17 - 19 = 175 - 175 - 174 + 174 + 173 - 173 - 172 + 172+ 17 - 19 = 17 - 19 = -2
Home » Math Theory » Geometry » Inscribed Angles # Inscribed Angles ## Introduction Many examples of basic plane Euclidean geometry make use of the inscribed angle. Other theorems relating to the power of a point with respect to a circle are based on the inscribed angle. The inscribed angle plays an important role in the circle theorems. The inscribed angle is the foundation for several circle theorems. In this article, we will define inscribed angles and other associated concepts, measure inscribed angles, and learn inscribed angles conjectures using examples that have been solved. ## What is an inscribed angle? ### Definition An angle is said to be inscribed if its vertex is on the circumference of the circle and both of its sides are chords of the same circle. The arc created by the inscribed angle is the intercepted arc. The intercepted arc is the arc that lies in the interior of an inscribed angle and has endpoints on the angle. In the circles below, we have circles P, S and G. In circle P, the inscribed angle is ∠ XYZ. Notice that the vertex Y is on the circumference of the circle while the sides of the angle are the chords $\overline{YX}$ and $\overline{YZ}$. The intercepted arc formed is arc XZ ( XZ). The center is in the interior of the inscribed angle. In circle S, the intercepted arc is ∠ LMN. The intercepted formed is arc LN. The sides of the intercepted arc are the chords $\overline{MN}$ and $\overline{ML}$. The chord $\overline{MN}$ is also the circle’s diameter since it passes through the center of circle S. In circle G, the intercepted arc of the inscribed angle ∠ KRC is arc KC. The circle’s center is outside of the inscribed angle or exterior of the inscribed angle. ## Identifying and Measuring Inscribed Angles By examining two chords that share an endpoint, it is necessary to identify the inscribed angle of the circle. The vertex of the inscribed angle is where the two chords’ common endpoints meet. We use three letters to name the inscribed angle, with the vertex being the middle letter and the other endpoints being the other two letters. For example, we have circle M below. We have the chords $\overline{AR}$ and $\overline{AE}$. The common endpoint or the vertex is point A. Therefore, the inscribed angle of the circle is ∠ EAR or ∠ RAE. Get ½ of the intercepted arc’s measure in order to determine the measure of the inscribed angle formed by the two chords. Simply expressed, we can multiply by ½ or divide by 2 the measurement of the intercepted arc to get the measurement of the inscribed angle. Inscribed Angle Theorem The measurement of an angle equals one-half the size of its intercepted arc if the angle is inscribed in a circle. So, in the given circle M, since the measure of the intercepted arc RE is 50°, the measure of the inscribed angle ∠ RAE is 25°. m ∠ RAE= ½ ( m RE ) m ∠ RAE= ½( 50° ) m ∠ RAE= $\frac{50^o}{2}$ m ∠ RAE= 25° ### Inscribed Angle Conjecture An inscribed angle in a circle has a measure that is half that of the central angle with the same intercepted arc. m ∠ TDS=½( m ∠ TLS ) In the figure, ∠ TDS is an inscribed angle, while ∠ TLS is a central angle. Both angles intersect the same arc, which is TS. Let us say, for example, the measure of ∠ TLS is 220°, then the measure of ∠ TDS is 110°. m ∠ TDS= ½( m ∠ TLS ) m ∠ TDS= ½( 220° ) m ∠ TDS= $\frac{220^o}{2}$ m ∠ TDS= 110° ### Inscribed Angles Intercepting Arcs Conjecture Two inscribed angles in a circle that share the same intercepted arc are congruent. m ∠TDS≅m ∠ TCS ∠ TDS and ∠TCS are inscribed angles and share the same intercepted arc TS. Let us say, for instance, the measure of ∠ TDS is 50°, then the measure of ∠ TCS is also 50°. ### Inscribed Angles in a Semi-circle Conjecture Any angle that is inscribed in a semi-circle is a right angle. m ∠ DCS=90° In the figure, ∠DCS is an inscribed angle, and $\overline{DS}$ is a diameter of the circle L ## More Examples Example 1 Identify the inscribed angles and their intercepted arc in the given circle below. Solution: The circle has four inscribed angles: ∠ RAC or ∠ CAR with intercepted arc CR or RC, ∠ ACF or ∠ FCA with intercepted arc AF or FA, ∠ CFR or ∠ RFC with intercepted arc CR or RC, and ∠ ARF or ∠ FRA with intercepted arc AF or FA. Example 2 Find the measure of the inscribed angles ∠ XWY and ∠ VYW. Solution: The intercepted arc of ∠ XWY is the XY which measures 160°. To find the measure of ∠ XWY, we get ½ of 160°. Here is the math, m ∠ XWY= ½ ( m XY ) m ∠ XWY = ½( 160° ) m ∠ XWY= $\frac{160^o}{2}$ m ∠ XWY= 80° The inscribed angle ∠ VYW has an intercepted arc VW which measures 60°. To get the measure of ∠ VYW, we get half of 60°. That is, m ∠ VYW=½ ( m VW ) m ∠ VYW=½( 60° ) m ∠ VYW=$\frac{60^o}{2}$ m ∠ VYW=30° Thus, the measures of the inscribed angles ∠XWY and ∠VYW are 80° and 30°, respectively. Example 3 Find m ∠ RVC in circle F. Solution: ∠ RVC is an intercepted arc. We can get its measurement by first identifying the measurement of its intercepted arc RC. It can be observed that the chord $\overline{RC}$ is a diameter since it passes through the center of circle F. ∠ RFC forms a straight angle which measures 180°. Since ∠ RFC is a central angle, an angle whose vertex is the center of the circle, then the measure of RC is also 180°. To get the measure of the inscribed angle ∠RVC, we must divide 180° by 2. We have, m ∠ RVC=½ m RC m ∠ RVC=½( 180° ) m ∠ RVC=$\frac{180^o}{2}$ m ∠ RVC=90° Therefore, the measure of the ∠ RVC is 90°. Example 4 Complete the table below. Solution: If the inscribed angle’s measurement is what is missing, we must multiply the intercepted arc’s given measurement by ½ . On the other hand, if the intercepted arc’s measurement is missing, we must multiply the given measurement of the inscribed angle by 2. To find m BS, given that m ∠ BTS=80°, m BS= 2 (m ∠ BTS ) m BS= 2( 80° ) m BS= 160° To find m ∠DLS, given that m DS}=78°, m ∠DLS= ½ m BS m∠ DLS= ½ 78° m ∠ DLS= $\frac{78^o}{2}$ m ∠ DLS= 39° To find m LR, given that m ∠LOR=55°, m BS= 2( m ∠LOR ) m LR= 2( 55° ) m LR= 110° To find m ∠MID, given that m MD=12.5°, m ∠ MID=½ ( m MD ) m ∠ MID= ½( 12.5° ) m ∠ MID= $\frac{12.5^o}{2}$ m ∠ MID= 6.25° Therefore, here is the completed table. Example 5 Find the measure of ∠ CVG and m CG. Solution: Any two inscribed angles in a circle that share the same intercepted arcs are congruent. In this example, the inscribed angles ∠ CHG and ∠ CVG have the same intercepted arc CG. Thus, we may say that ∠ CHG is congruent with ∠ CVG or ∠ CHG≅∠ CVG. Since m ∠ CHG=75°, then ∠ CVG measures 75° too. To find the measure of CG, we must multiply the measure of ∠ CHG by 2. So, we have, m CG=2( m ∠CHG ) m CG=2( 75° ) m CG=150° Therefore, the m ∠ CVG=75° and m CG=150°. Example 6 In each case, find the value of x. a ) b ) c ) d ) Solution: a ) The inscribed angle in circle R that we need to find the value of x is ∠ LKM. Notice that the chord $\overline{LM}$ is a diameter of circle R, which divides the circle into two semi-circles. Thus, the measure of arc LM is 180°. ∠ LKM measures 90° because the measure of an inscribed angle is half the measure of its intercepted arc, then m ∠LKM= ½( m LM ) m ∠ LKM= ½( 180° ) m ∠ LKM= $\frac{180^o}{2}$ m ∠ LKM= 90° Thus, x= 90°. b ) In circle Y, ∠ GXT is an inscribed angle, while ∠ GYT is a central angle. Since both angles share the same arc, which is GT, then the measure of ∠ GYT is twice the measure of ∠ GXT. That is, m ∠ GYT=2( m ∠ GXT ) m ∠ GYT=2( 30° ) m ∠ GYT=60° Therefore, the value of x is 60°. c ) ∠ ZES is a central angle that measures 130°, and we are asked to find the measure of the inscribed angle ∠ ZQS. Since both ∠ ZES and ∠ ZQS share the same intercepted arc, then m ∠ ZQS is half the m ∠ ZES. m ∠ ZQS= ½ m( ∠ ZES) m ∠ ZQS= ½( 130° ) m ∠ ZQS= $\frac{130^o}{2}$ m ∠ ZQS= 65° d ) In circle M, both ∠ PRO and ∠ PWO are inscribed angles that share the same intercepted arc PO. Thus, m ∠ PRO≅m ∠ PWO. Since it is given that m ∠PRO=40°, then the m ∠ PWO is also 40°. Example 7 Solve for the value of x. Solution: In the given illustration, we have m LC=120° and m ∠ LRC=(4x+8)°. Since ∠LRC is an inscribed angle, its measure is ½ of the measure of its intercepted arc LC. m ∠LRC= ½( m LC ) m ∠ LRC= ½( 120° ) m ∠ LRC= $\frac{120^o}{2}$ m ∠ LRC= 60° Since we already know that the measure of the inscribed angle is 60°, we may now solve for the value of x, that is, $4x + 8 = 60$ $4x = 60-8$ $4x = 52$ $\frac{4x}{4} = \frac{52}{4}$ $x= \frac{52}{4}$ $x=13$ Therefore, the value of x is 13. To check, we may substitute 13 to the given expression, and it should equate to 60. 4(13)+8= 60 52+8= 60 60= 60 Example 8 Find the value of x given that the measure is arc AB is 100°. Solution: In circle M, ∠ ACB is an inscribed angle with intercept arc AB. Since m AB=100°, the measure of ∠ ACB is half of m AB, which is 50°. We have, m ∠ACB= ½( m AB ) m ∠ ACB= ½( 100° ) m ∠ ACB= $\frac{100^o}{2}$ m ∠ ACB= 50° Notice that AC is an arc of a semi-circle; therefore, since m AB=100°, the measure of BC is 80°. m BC=180°-m AB m BC=180°-100° m BC=80° Since ∠ BMC is a central angle with the intercepted arc BC, its measure is also 80°. Notice that angles ∠ B, ∠ C, and ∠ M form a triangle. Remember that the sum of the angles in a triangle is equal to 180°. Thus, we can find the value of x, the m ∠ B, by subtracting the sum of angles ∠ C and ∠ M from 180°. Hence, m ∠ B=180°- ( m ∠ C+m ∠ M ) m ∠ B=180°- ( 50°+ 80° ) m ∠ B=180°-130° m ∠ B=50° Therefore, the value of x is 50°. Example 9 Solve for x. Solution: In the given circle with center P, ∠ VFD is an inscribed angle that measure 105°. To get the measure of the central angle ∠ DPV, we must multiply the m ∠ VFD by 2. Remember that the measure of a central angle is twice the measure of the inscribed angle if they share the same intercepted arc. Hence, m ∠ DPV=2 ( m ∠ VFD ) m ∠ DPV=2( 105 ) m ∠ DPV=210° Let us now equate ( 3x+6 )° to 210° to find the value of x. We have, $3x+6=210$ $3x=210-6$ $3x=204$ $\frac{3x}{3}=\frac{204}{3}$ $x=68$ Therefore, the value of x is 68°. Example 10 In the figure below, points D, V, P, and F are on the circle. Find the measure of ∠ VDF if m VP=105° and m PF=48°. Solution: ∠ VDF is an inscribed angle since the points VDF are on the circumference of the circle. Adding the m VP and m PF gives the m VF which is the intercepted arc of the angle ∠ VDF. To find the measure of ∠ VDF, we must get half the sum of the measures of VP and PF. Thus, m ∠ VDF=½( m VP+m PF ) m ∠ VDF=½( 105°+48° ) m ∠ VDF=½( 153° ) m ∠ VDF=$\frac{153^o}{2}$ m∠ VDF=76.5° ## Summary An angle is said to be inscribed if its vertex is on the circumference of the circle and both of its sides are chords of the same circle. The intercepted arc is the arc that lies in the interior of an inscribed angle and has endpoints on the angle. Inscribed Angle Theorem The measurement of an angle equals one-half the size of its intercepted arc if the angle is inscribed in a circle. The term “central angle” refers to an angle whose vertex is at the center of the circle. Its measure is equal to its intercepted arc. A chord that passes the circle’s center is referred to as the diameter. Identifying Inscribed Angles By examining two chords that share an endpoint, it is necessary to identify the inscribed angle of the circle. The vertex of the inscribed angle is where the two chords’ common endpoints meet. We use three letters to name the inscribed angle, with the vertex being the middle letter and the other endpoints being the other two letters. Measuring Inscribed Angles Given its Intercepted Arc Get 1/2 of the intercepted arc’s measure in order to determine the measure of the inscribed angle formed by the two chords. Simply expressed, we can multiply by 1/2 or divide by 2 the measurement of the intercepted arc to get the measurement of the inscribed angle. Inscribed Angle Conjecture An inscribed angle in a circle has a measure that is half that of the central angle with the same intercepted arc. Inscribed Angles Intercepting Arcs Conjecture Two inscribed angles in a circle that share the same intercepted arc are congruent. Inscribed Angles in a Semi-circle Conjecture Any angle that is inscribed in a semi-circle is a right angle. ### What is meant by inscribed angle? An angle is said to be inscribed if its vertex is on the circumference of the circle and both of its sides are chords of the same circle. ### What is the difference between a central angle and an inscribed angle? An angle is said to be inscribed if its vertex is on a circle and both of its sides are chords of that circle. A central angle, on the other hand, has a vertex at the center of the circle and two radii as its sides. For example, we have the image below with circle P and its central and inscribed angles. The central angle in circle V is ∠ AVM or ∠ MVA. The vertex of ∠ AVM is the center of the circle, and the two radii $\overline{VM}$ and $\overline{VA}$ are the sides. The inscribed angle in circle V is ∠ XCM or ∠ MCX. The vertex of ∠ XCM lies on the circle, and the two chords $\overline{CX}$ and $\overline{CM}$ are the sides. ### How are inscribed and central angles related to one another? The measure of a central angle would be double the inscribed angle if a central angle and an inscribed angle both intersect the same arc. To put it simply, the measure of an inscribed angle is half of the measure of the central angle. For example, the figure below shows circle V with the measure of arc PR equals 80°. Both the central angle( ∠PVR )and the inscribed angle (∠PQR ) intersect the same arc PR, then the measure of ∠ PVR is twice the measure of ∠ PQR The measure of the central angle ∠ PVR=80° is the measure of its intercepted arc PR. For the measure or ∠ PQR, we have, m ∠ PQR=½ m PR m ∠ PQR=½( 80° ) m ∠ PQR=$\frac{80^o}{2}$ m ∠ PQR=40° Therefore, m ∠ VPR=80° while m ∠ PQR=40°. ### How can I find the measure of an inscribed angle? To find the measure of an inscribed angle, get ½ of the intercepted arc’s measure formed by the two chords. Simply expressed, we can multiply by ½ or divide by 2 the measurement of the intercepted arc to get the measurement of the inscribed angle. Let us say, for example, that the measure of the intercepted arc in the given image below is 210° or m AF=210°. To find the measure of the inscribed angle, which is ∠ ARF, we get ½ of 210°. Thus, m ∠ ARF=105°. Here is the solution: m ∠ ARF=½ m AF m ∠ ARF=½( 210° ) m ∠ ARF=$\frac{210^o}{2}$ m ∠ ARF=105° ### How do you calculate the measure of an intercepted arc of an inscribed angle? By multiplying the inscribed angle’s measure by two or doubling it, you can get the measure of an intercepted arc from the inscribed angle. Let us say, for instance, that the measure of the inscribed angle in the given image below is 35° or m ∠ GHI=35°. In finding the measure of the intercepted arc, GI, we get twice 35°. Hence, m GI=70°. Here is the solution: m GI=2( m ∠GHI ) m GI=2( 35° ) m GI=70° ### What are the inscribed angles conjectures? An angle is said to be inscribed if its vertex is on the circumference of the circle and both of its sides are chords of the same circle. The intercepted arc is the arc that lies in the interior of an inscribed angle and has endpoints on the angle. The term “central angle” refers to an angle whose vertex is at the center of the circle. Its measure is equal to its intercepted arc. The conjectures about inscribed angles are given below. Inscribed Angle Conjectures Inscribed Angle Conjecture An inscribed angle in a circle has a measure that is half that of the central angle with the same intercepted arc. Inscribed Angles Intercepting Arcs Conjecture Two inscribed angles in a circle that share the same intercepted arc are congruent. Inscribed Angles in a Semi-circle Conjecture Any angle that is inscribed in a semi-circle is a right angle. ### What is the measure of an angle inscribed in a semi-circle? A 90° is the measure of an angle inscribed in a semi-circle. m ∠ DCS=90° In the figure, ∠DCS is an inscribed angle, and DS is a diameter of circle L. ### What is the total measure of the opposite angles of a quadrilateral inscribed in a circle? The total measure of the opposite angles of a quadrilateral inscribed in a circle is 180°. It means that they are supplementary angles. Let us say, for example, in the figure below, the points Q, U, A, and D form an inscribed quadrilateral. ∠Q, ∠U, ∠A, and ∠D are all inscribed angles. ∠Q and ∠A are supplementary ∠D and ∠U are supplementary
# Analyzing Triangle Congruence with AngLegs I’ve tried to explain why AngLegs are a must have for high school geometry and should not only be considered a tool for younger students. Here is an example of how I find them indispensable in teaching triangle congruence. This lesson is adapted from MARS: Evaluating Conditions for Congruency. “Ok class, you are sitting in pairs and at each table is a bag of AngLegs. On the board I have written the question we are trying to answer for each of the scenarios I will present.” “I made a triangle that includes a blue AngLeg. Can you make a different triangle that also has a blue AngLeg?” “Make one. Hold it up.” “How do you know that these triangles are different?” “So, you are saying that keeping one side the same does not mean that the triangles must be congruent.” Next, lets look at Card 3: “I made a triangle out of a blue, a purple and a yellow AngLeg. Can you make a different triangle using the same three AngLegs?” “What if you put them in a different order? …or move the purple between the blue and the yellow? Are you sure they have to be the same? How can you tell?” Student: “The triangles still fit perfectly on top of each other”. Student: “If the three sides on one triangle are the same as three sides of another triangle, then the triangles must be the same.” Look at Card 7: “How can you tell if an angle is the same in two different triangles?” Student: “They fit perfectly on top of each other!” “Is there a way to make these triangles so that they are not congruent?” Student: “No way. These have to be the same” Student: “Wait! I made two different triangles with all three angles the same and one side the same.” Does this count? Look!” Student:”If two angles are the same, then the third angle always has to be the same because they add up to 180 degrees!” “So what is the conclusion for this one?” Student:”The triangles can still be different sizes, but their angles are all the same.” “For the remainder of this class period, individually analyze card 5 and any other card, so 2 additional cards. Take good notes and write down your conclusions for tomorrow, where you will be randomly assigned a partner to complete the triangle activity. ” From here the lesson continues as described in the SHELL center teacher guide linked above and described further in a blog post here. Removing the hurdle of constructions allows students to focus on the learning goal for this activity: determining the minimum information required to guarantee triangle congruence, and what congruence means. It also connects nicely to congruence proofs through transformations as students are physically checking of the triangles fit on top of each other.
Posted in: Basic Maths Skills # Repeating decimal to fraction and number facts Revision for the Grade 8 maths exam does not have to be stressful. Making flashcards, memorising maths key facts, or memorising the repeating decimal to fraction number can help you as you prepare for the coming maths exams. Below are key facts and maths skill that you can master in 10 minutes. The multiplication and division tricks and tips below will make a world of difference when you take the Grade 8 maths exam. ## 1 is not a prime number For a number to be a prime number it must have only two factors, 1 and itself. 1 has only 1 factor, 1. First 10 prime numbers: 2,3,5,7,11, 13,17, 19, 23,… • 9 is odd number, not prime number with factors {1,3,9} • 15 is odd number, not prime number with factors {1,3,5,15} ## List of common and repeating decimal to fraction and percentages ½=50%=0.5 ⅓=33⅓%=33.3… ¼=25%=0.25 ⅕=20%=0.2 ⅛=12.5%=0.125 ¹/10=10%=0.1 ⅔=66⅔%=66.6… ⅖=40%=0.4 ¾=75%=0.75 ⅗=60%=0.6 Take the online Fraction Quiz to test your knowledge. ## ×12 Recall Tip 12×1=1 2 12×2=2 4 12×3=3 6 12×4=4 8 12×5=6 0 12×6=7 2 12×7=8 4 12×8=9 6 12×9=10 8 12×10=12 0 12×11=13 2 12×12=14 4 • Starting numbers increase in 1,2,3…* • Multiples of ×12 end in 0,2,4,6 or 8 and reoccur after 5 numbers • Only applies to ×12 123 + 573 =(100 + 20 + 3) + (500 + 70 + 3) = 100 + 500 + 20 + 70+ 3 + 3 = 600 + 90 + 6 = 696 ## Mental tips for division by a decimal 270 ÷ 0.3 🚦make 0.3 whole number by × 10 I.e. 270 × 10 = 2700; 0.3 ×10= 3 Now, 2700 ÷ 3 = 900 Top tip: In division by decimals, multiply the dividing numbers by 10, 100,… to change decimal numbers to whole numbers #NumberSkills #CoolTips ## Dividing with decimal numbers E.g. 80÷1.5 🚦×10 make 1.5 whole 80×10=800; 1.5×10=15 –‐‐—————-‐– 🚦simplify 800÷15… ÷5 =160÷3 🚦partition 160…150+10 (150÷3)+(10÷3) =50+3⅓ =53⅓ (or 53.33…) *Tip: make the decimal a whole number; then simplify, partition and solve. ## Dividing 3-digit numbers 836 ÷ 32 🚦split 836 into easily divisible numbers \|640 \| 160 \| 32 \| 4 …sum is 836 ÷\| 32 \| 32 \| 32 \| 32 …divisor is 32 ——–‐‐——————- 20 + 5 + 1 + ⅛ = 26⅛ or 26.125 Tip: Go by 20, 10, 5 to reduce 3 digit numbers like 836. You can also apply the same technique to simplify even bigger numbers. #NumberFacts ## Mental multiplication of big numbers 53 × 37 ○ 50 \| 30 ● 3 \| 7 -‐——————-‐– ○50×30=1,500 ○50×7= 350 ●3×30= 90 ●3×7= 21 ————————- 1,500 + 350 + 90 + 21 = 1,961 #MentalMathTips 53 × 37 ———– 53 ×37 ””””””’ 371 + 159 ”””””””””” ## Divide with Common and Repeating Decimal to Fraction E.g.280 ÷ 3½ 🚦Convert fraction to decimal 3½=3.5 280÷3.5 🚦multiply 280 and 2.5 by 10 to make 3.5 whole numbers before dividing. = 2,800÷35 💡35 × 8 = 280 If 35×8=280🔁280÷35=8, so 2800÷35=80 ———————- So, 280÷3½ =280÷3.5 =2,800÷35 =80 #MentalDivision ## Basic maths skills: Tips and Study Guides We have a collection of Grade 8 topics and study guides that you might find useful when planning for the year or just brainstorming the maths course. Check it out.
# Rounding and Estimation Estimating and rounding are very important math skills. It provides a tool for people to judge the reasonableness of answers to math problems and provides a framework for making an educational and informed guess when necessary. ## Estimating Sums A sum is the answer to an addition problem. It is best to estimate before adding decimals and fractions. You can use an estimate to check the accuracy of the sum when adding decimals or fractions. 1. Find the sum of 24.86 + 15.15. 2. Round each decimal to the nearest whole number. 3. 24.86 is rounded to 25 and 15.15 is rounded to15 4. Add: 25 + 15 = 40 , so the actual sum should be close to 40. Use this to check the actual sum. 24.86+15.15 = 40.01 5. The sum 40.01 is very close to the estimate 40, so the answer is reasonable and probably correct. ## Estimating Differences A difference is the answer to a subtraction problem. It is best to estimate before subtracting decimals and fractions. You can use an estimate to check the accuracy of the difference when adding decimals or fractions. ### Find the difference of 95.16 – 16.73. 1. Round each decimal to the nearest whole number. 2. 95.16 is rounded to 95 and 16.73 is rounded to 17. 3. Subtract: 95-17 = 78 , so the actual difference should be close to 78. 4. Use this to check the actual difference. 5. 95.16 – 16.73 = 78.43 The difference 78.43 is very close to the estimate 78, so the answer is reasonable and probably correct. ## Use Compatible Numbers In mathematics, compatible numbers are the numbers that are easy to add, subtract, multiply, or divide mentally. Compatible numbers are close in value to the actual numbers that make estimating the answer and computing problems easier. 500 + 300 = 800 The numbers 500 and 300 are compatible for addition, since the sum of 800 can be easily calculated mentally. 519 + 293 = 812 The numbers 519 and 293 are not compatible for addition, since the sum (812) cannot be easily calculated mentally. To estimate 513 + 299, replace 513 and 299 with the compatible numbers 500 and 300. An estimate of 513 + 299 is found by mentally calculating 500 + 300 = 800. Example 2 (Subtraction) 19.4 − 3.8 = 15.6 The numbers 19.4 and 3.8 are not compatible for subtraction, since the difference (15.6) cannot be easily calculated mentally. To estimate 19.4 − 3.8, replace 19.4 and 3.8 with the compatible numbers 19 and 4. An estimate of 19.4 − 3.8 is found by mentally calculating 19 − 4 = 15. Example 3 (Multiplication) 19.4 × 3.8 = 73.72 The numbers 19.4 and 3.8 are not compatible for multiplication since the product (73.72) can’t be easily calculated mentally. To estimate 19.4 × 3.8, replace 19.4 and 3.8 with the compatible numbers 20 and 4. An estimate of 19.4 × 3.8 is found by mentally calculating 20 × 4 = 80. Example 4 (Division) 721 ÷ 70 = 10.3 The numbers 721 and 70 are not compatible for division, since the quotient (10.3) can’t be easily calculated mentally. To estimate 721 ÷ 70, replace 721 with 700 so that the numbers involved are compatible. An estimate of 721 ÷ 70 is found by mentally calculating 700 ÷ 70 = 10.
# Volume of Revolution Solid of revolution This image is a solid of revolution # Basic Description This image shows the solid formed after revolving the region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$, about the $x$-axis # A More Mathematical Explanation Note: understanding of this explanation requires: *Pre-calculus and elementary calculus When finding the volume of revolution of solids, in many cases the problem is not with the calculus, [...] When finding the volume of revolution of solids, in many cases the problem is not with the calculus, but with actually visualizing the solid. To find the volume of a solid like a cylinder, usually we use the formula ${\pi} {r^2} h$. Alternatively we can imagine chopping up the cylinder into thin cylindrical plates, much like slicing up bread, computing the volume of each slice, each of which is ${\Delta x }$ units thick , then summing up the volumes of all the slices. The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html In general, given a function, we can graph it then revolve the area under the curve between two specific coordinates about a fixed axis to obtain a solid called the solid of revolution. The volume of the solid can then be computed using the disc method. Note: There are other ways of computing the volumes of complicated solids other than the disc method. In the disc method, we imagine chopping up the solid into thin cylindrical plates calculating the volume of each plate, then summing up the volumes of all plates. For example, let's consider a region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$ <-------Plotting the graph of this area, If we revolve this area about the x axis ($y=0$), then we get the main image on the right hand side of the page This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm To find the volume of the solid using the disc method: Volume of one disc = ${\pi} y^2{\Delta x}$ where $y$- which is the function- is the radius of the circular cross-section and $\Delta x$ is the thickness of each disc. Using the analogy of the bread, computing the volume of one disc would correspond to computing the volume of one slice of bread. With this in mind, the area of one disc would correspond to the area of a slice of bread, while the thickness of a disc would correspond to the thickness of a slice of bread. To find the total volume of the bread, we would have to sum up the volumes of each of the slices. Volume of all discs: Volume of all discs = ${\sum}{\pi}y^2{\Delta x}$, with $X$ ranging from 0 to 1 If we make the slices infinitesmally thick, the Riemann sum becomes the same as: $\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx$ Evaluating this intergral, ${\pi}\int_0^1 x^4 dx$ =$[{{x^5\over 5} + C\|}_0^1] {\pi}$ =$[{1\over 5} + {0\over 5}] {\pi}$ =${\pi}\over 5$ volume of solid= ${\pi\over 5} units^3$ # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Have questions about the image or the explanations on this page?
# Problem Solving Division Day 1 7 teachers like this lesson Print Lesson ## Objective SWBAT solve word problems using division skills. #### Big Idea In this fun partner based lesson, students solve multistep word problems and practice discussing division and multiplication strategies. ## Number Hook 5 minutes I start today's math lesson with this number hook. I tell students that I will be able to tell them their age and address simply by having them write down some numbers and perform a few simple calculations. I tell the students these instructions: 1) Write down their street number: 2979 2) Double it: 5958 4) Multiply by 50: 298150 5) Add the person’s age (31): 2981181 6) Add the number of days in a year (365): 298546 Students tell me their result. I subtract 15 from the last two numbers and get the student's age. Then I subtract 6 from the remaining other digits to get the person’s address. For example, let's suppose someone lived at 1234 Apple Street who was 54. Their numbers would look like this from the above directions. 1. 1234 2. 2468 3. 2468 + 5 = 2473 4. 2473 * 50 = 123650 (note, my students did get nervous about multiplying four digits by two digits, but a few students were able to tell the rest of the class that that they could multiply the number by five and then multiply that number by 10) 5. 123650 + 54 = 123704 6. 123704 + 365 = 124069 Result: 69 - 15 = 54 (the person's age) and 1240 - 6 = 1334 My students love this trick. They think it's so "cool" that math can tell their age and address. They do understand that there is some sort of pattern happening to the numbers in order to get their age and address, but they don't really know what. I do not think it is necessary they know the "how" just that math is everywhere - even in magic! ## Warm Up 5 minutes For this warm up, I challenge students to write a division problem using a 3-digit dividend and one digit divisor that results in a quotient and a remainder of 1. I ask them to show all their work. This is an example of a student's work. I think the written explanation is somewhat confusing, but you can see by the division problem that this student's work seems to have a solid understanding of multiples of ten with division and how to result with a remainder of 1. This is another student's thinking. This student admitted that this division problem was a lucky guess. This student was amazed at the other strategies his/her peers discussed when talking about this problem because the student really thought that his was purely a guess and check type of problem. That told me a lot about this student's number sense right now and where this student is at in thinking about numbers. One of the most surprising aspects to this journal activity was the amount of thinking my students did. One student reported that he knew that if the divisor was 2, the biggest remainder in that problem would be a one. He went on to say that all he needed was an odd number since all even numbers are divisible by 2. I was really in awe of his thinking and so proud of the way he communicated it. (So sad I didn't have my video rolling to capture it) Another student reported that she knew her multiples of ten and that she could simply add one more any multiple of ten to get a remainder of 1. She gave this example, "I know five times six equal thirty. So, fifty times six equals three hundred which means five hundred times 6 equals three thousand. I know I just need to make three thousand one and divide it by six to get a remainder of 1!" ## Concept Development 45 minutes Partner work is a strategy that promotes participation and interaction that I try to incorporate on a regular basis. It fosters a deeper and more active learning process, and it also provides me with valuable information about what students understand regarding topics or concepts. In addition to exposing students to different approaches and ways of thinking, working with another student can promote a sense of belonging. Working together with a partner also gives students the opportunity to learn from and teach each other. For this lesson, students will use a strategy called Thinking- Aloud Partner Problem Solving. I wanted to incorporate partner work in this lesson, but also use problem solving situations in which students would have opportunities to apply the division skills they have gained from this division unit. Problem solving is an integral part of my students learning. Students acquire ways of thinking, habits of persistence and curiosity, and confidence in unfamiliar situations that will serve them well outside the mathematics classroom. In everyday life and in the workplace, being a good problem solver can lead to great advantages. The goal for today's lesson is for partners to work on the problems presented on the 3 ÷ 1 (With Remainder) problem set worksheet. I know that students will not finish this worksheet in the time allotted, but I encourage them to work wisely and efficiently so they are able to continue to work during tomorrow's lesson. As students work, I am constantly moving around the classroom to observe students thinking. As much as I can, I resist the urge to re-direct students thinking until I am certain they have talked with their learning partner about the problem. By doing this, students partners often help correct misunderstandings and errors. Students utilize Math Practice Standards 1 and 3 in this lesson. CCSS MP1 states mathematically proficient students start by explaining to themselves the meaning of a problem and looking for entry points to its solution. They analyze givens, constraints, relationships, and goals. They make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt. They consider analogous problems, and try special cases and simpler forms of the original problem in order to gain insight into its solution. They monitor and evaluate their progress and change course if necessary. Mathematically proficient students check their answers to problems using a different method, and they continually ask themselves, “Does this make sense?” They can understand the approaches of others to solving complex problems and identify correspondences between different approaches. One of the strengths in this lesson is directly related to the above text. The think aloud partner problem solving (TAPPS) instructional strategy allows and encourages students to share their thinking with peers and allows peers to help or guide checking work and the reasonableness of answers. Student partners can continually check each other's work and ask each other if their answers makes sense. Often times, one student's strategy or method differs from their partners, which creates an opportunity to see and hear other approaches and build students' abilities to think flexibly. Below, you can see two students using the TAPPS strategy and talking through a problem. In this student's work, you can see that a question this student had was, "How did we get such different answers?" The above example is exactly what I wanted my students to be able to do in this lesson, thoughtfully reflect on his or her partners thinking in order to deepen understanding of division. Students work until the end of the class period.
# NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials. This Exercise contains 4 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 2 or other Chapters, you can click the link at the end of this Note. ### NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials 1. Find the value of the polynomial 5x – 4x2 + 3 at (i) x = 0 (ii) x = –1 (iii) x = 2 Solution: Let, p(x) = 5x – 4x2 + 3 Then value of polynomial at (i) x = 0 is given by p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3 (ii) x = –1 is given by p(-1) = 5(-1) – 4(-1)2 + 3 = – 5 – 4 + 3 = -6 (iii) x = 2 is given by P(2) = 5(2) – 4(2)2 + 3 = 10 – 16 + 3 = -3 2. Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3 (iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1) Solution: To find p(x) at x = a , put x = a in the equation and evaluate the expression obtained as follows: (i) Given p(y) = y2 – y + 1 Therefore, p(0) = 02 – 0 + 1 = 0 – 0 +1 = 1 p(1) = 12 – 1 + 1 = 1 – 1 + 1 = 1 p(2) = 22 – 2 + 1 = 4 – 2 + 1 = 3 (ii) Given p(t) = 2 + t + 2t2 – t3 Therefore, p(0) = 2 + 0 + 2(0)2 – 03 = 2 + 0 + 0 – 0 = 2 p(1) = 2 + 1 + 2(1)2 – 13 = 2 + 1 + 2 – 1 = 4 p(2) = 2 + 2 + 2(2)2 – 23 = 2 + 2 + 8 – 8 = 4 (iii) Given p(x) = x3 Therefore, p(0) = 03 = 0 p(1) = 13 = 1 p(2) = 83 = 8 (iv) Given , p(x) = (x – 1) (x + 1) Therefore, p(0) = (0 – 1)(0 + 1) = -1 p(1) = (1 – 1)(1 + 1) = 0 p(2) = (2 – 1)(2 + 1) = 3 3. Verify whether the following are zeroes of the polynomial, indicated against them. (i) p(x) = 3x + 1, x = (-1/3) (ii) p(x) = 5x – π, x = (4/5) (iii) p(x) = x2 – 1, x = 1, –1 (iv) p(x) = (x + 1) (x – 2), x = – 1, 2 (v) p(x) = x2 , x = 0 (vi) p(x) = lx + m, x = -m/l (vii) p(x) = 3x2 – 1, x = −(1/√3), (2/√3) (viii) p(x) = 2x + 1, x = 1/2 Solution : Zero of a polynomial p(x) is a number a such that p(a) = 0. (i) p(x) = 3x + 1, x = (-1/3) We get, p(1/3) = 3 x (-1/3) + 1 = -1 + 1 = 0 Hence, x = (-1/3) is a zero of the given polynomial. (ii) p(x) = 5x – π, x = 4/5 We get, p(4/5) = 5 x (4/5) – π = 4 – π ≈ 0.8584 ≠ 0 Hence, x = (4/5) is not a zero of the given polynomial. (iii) p(x) = x2 – 1, x = 1, –1 We get, p(1) = 12 – 1 = 1 – 1 = 0 p(-1) = (-1)2 – 1 = 1 – 1 = 0 Hence, x = 1, –1 both are the zeroes of the given polynomial. (iv) p(x) = (x + 1) (x – 2), x = – 1, 2 We get, p(-1) = (-1 + 1) (-1 – 2) = 0 p(2) = (2 + 1) (2 – 2) = 0 Hence, x = –1 , 2 both are the zeroes of the given polynomial. (v) p(x) = x2 , x = 0 We get, p(0 ) = 02 = 0 Hence, x = 0 is a zero of the given polynomial. (vi) p(x) = lx + m, x = (-m/l) We get p(-m/l ) = l(-m/l) + m = -m + m = 0 Hence, x = -m/l is a zero of the given polynomial. (vii) p(x) = 3x2 – 1, x = −(1/√3), (2/√3) We get, p[−(1/√3)] = 3(−1/√3)2 – 1 = 3 x (1/3) – 1 = 1 – 1 = 0 p(2/√3) = 3(2/√3)2 – 1 = 4 – 1 = 3 Hence, x = −(1/√3) is a zero of the given polynomial, but x = 2/√3 is not the zero of the given polynomial. (viii) p(x) = 2x + 1, x = ½ We get, p(½ ) = 2 x (1/2) + 1 = 1 + 1= 2 Hence, x = (1/2) is not a zero of the given polynomial. 4. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. Solution: Solving the equation p(x) = 0 will give us the zero of the given polynomial. (i) p(x) = x + 5 Solving p(x) = 0 we get , x + 5 = 0 Or, x = -5 Thus we obtain x = -5 as the zero of the given polynomial. (ii) p(x) = x – 5 Solving p(x) = 0 we get , x – 5 = 0 Or, x = 5 Thus we obtain x = 5 as the zero of the given polynomial. (iii) p(x) = 2x + 5 Solving p(x) = 0 we get , 2x + 5 = 0 Or, x = (-5/2) Thus we obtain x = (-5/2) as the zero of the given polynomial. (iv) p(x) = 3x – 2 Solving p(x) = 0 we get , 3x – 2 = 0 Or, x = (2/3) Thus we obtain x = 2/3 as the zero of the given polynomial. (v) p(x) = 3x Solving p(x) = 0 we get , 3x = 0 Or, x = 0 Thus we obtain x = 0 as the zero of the given polynomial. (vi) p(x) = ax, a ≠ 0 Solving p(x) = 0 we get , ax = 0 Or, x = 0 Thus we obtain x = 0 as the zero of the given polynomial. (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. Solving p(x) = 0 we get , cx + d = 0 Or, x = (-d/c) Thus we obtain x = (-d/c) as the zero of the given polynomial. NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials, has been designed by the NCERT to test the knowledge of the student on the topic – Zeroes of a Polynomial The next Exercise for NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 – Number System can be accessed by clicking here. Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials
# How to Identify the Change, Price, or Amount Paid Hello young mathematicians! Have you ever wondered how to figure out the cost of your favorite candy, how much money you've spent, or the change you get back when you pay with a bigger note or coin? Well, today, we are embarking on a fun-filled math adventure that will make these puzzles easy as pie. Let's get started, shall we?" Let’s explain these concepts in a way an elementary student can understand. Let’s start with each term: 1. Price: The price is how much something costs. If you want to buy a toy and the tag on it says $15, that means the price of the toy is$15. That’s how much money you need to give to the cashier to get the toy. 2. Amount Paid: The amount paid is the total amount of money you give to the cashier when you buy something. Sometimes, the amount paid is exactly the same as the price. But let’s say you’re buying a toy that costs $15, and you only have a$20 bill. You give the $20 bill to the cashier. The$20 you give is the amount paid. 3. Change: Change is the money you get back from the cashier when the amount paid is more than the price. If the toy costs $15 and you give the cashier a$20 bill, the cashier will give you some money back. In this case, the cashier will give you $5 back ($20 – $15 =$5). This $5 is your change. The Absolute Best Book for 4th Grade Students ## A Step-by-step Guide to Identifying the Change, Price, or Amount Paid Here is a simple step-by-step guide that can help a 4th grader understand how to identify the price, amount paid, and change: ### Step 1: Identify the Price The price is the amount of money listed on the item you want to buy. • Look at the price tag or sticker on the item or the amount listed in a menu or catalog. • Sometimes, you might be buying more than one item, so you need to add up the prices of all the items to find the total price. A Perfect Book for Grade 4 Math Word Problems! ### Step 2: Identify the Amount Paid The amount paid is the total money you give to the cashier when you are ready to pay for your items. • This could be the exact amount of the total price if you have the exact money. • If you don’t have the exact money, you might give more money than the total price, expecting to receive change. ### Step 3: Calculate the Change Change is the money you receive back when the amount paid is more than the price. • If you paid more than the total price, the cashier should give you a change. • To calculate the change, subtract the total price from the amount paid. The Best Math Books for Elementary Students ## Related to This Article ### More math articles ### What people say about "How to Identify the Change, Price, or Amount Paid - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 51% OFF Limited time only! Save Over 51% SAVE$15 It was $29.99 now it is$14.99
# What is the probability of two or more from n events occuring? A number of independent events, say $A$, $B$, $\ldots\,$, $E$, can happen with associated probabilities $P(A)$, $P(B)$, $\ldots$ For each event that happens I have to pay £10. The likelihood I have to pay at least £10 is straightforward to work out: $$1 - \bigl(1-P(A)\bigr)\bigl(1-P(B)\bigr)\bigl(1-P(C)\bigr)\bigl(1-P(D)\bigr)\bigl(1-P(E)\bigr)$$ But what is the likelihood that I have to pay out at least £20? ie what is the likelihood of at least two events occuring? For 5 events I could list out all 32 possible outcomes (from none of the events happen through to all 5 events happen) and work out the probability of each outcome. However this approach is not feasible if I have 10s of events. We first give an answer to the question as put. Then we deal with $n$ events. In a sense, the general case turns out to be simpler than the $A,\dots,E$ case, since the structure is more evident. For simplicity let $\Pr(A)=a$, $\Pr(b)=b$, and so on. The probability that at least $2$ of $A,\dots, E$ happen is $1$ minus the probability that at most $1$ of our events happens. The probability of at most $1$ is the sum of two probabilities: (i) the probability of none and (ii) the probability of exactly $1$, (a) You calculated this already: it is $(1-a)(1-b)(1-c)(1-d)(1-e)$. (b) We get exactly $1$ if $A$ happens but the others don't, or if $B$ happens but the others don't, and so on. These events are pairwise disjoint, so the probability of exactly $1$ is $$a(1-b)(1-c)(1-d)(1-e)+(1-a)(b)(1-c)(1-d)(1-e)+\cdots$$ (the sum contains $5$ terms, of which we have shown the first two). The general case: Let the independent events be $A_1,A_2,\dots,A_n$ Let $\Pr(A_i)=p_i$. Then the probability $A_i$ doesn't happen is $1-p_i$. It is convenient to let $1-p_i=q_i$. (This is a standard abbreviation.) The probability that $0$ of the events happen is then given by the pleasant expression $q_1q_2q_3\cdots q_n$. Call this $P_0$. The probability exactly $1$ of the $A_i$ happens is the sum of $n$ terms, where the $i$-th term is $q_1q_2\cdots q_{i-1}p_iq_{i+1}\cdots q_n$. Call this sum $P_1$. Then the probability that $2$ or more of our events happen is $1-P_0-P_1$. Computational note: The calculation is not nearly as ugly as it seems. Assume that none of the $q_i$ is $0$. Then the probability of exactly $1$ can be rewritten as $$\left(q_1q_2\cdots q_n\right)\left(\frac{p_1}{q_1}+\frac{p_2}{q_2}+\cdots+\frac{p_n}{q_n} \right).$$ So the probability of at least $2$ is $$1-\left(q_1q_2\cdots q_n\right)\left(1+\frac{p_1}{q_1}+\frac{p_2}{q_2}+\cdots+\frac{p_n}{q_n} \right).$$
# How do you find the roots, real and imaginary, of y=8x^2+7x-(3x-1)^2 using the quadratic formula? Apr 10, 2016 Zeros of function are $x = \frac{13}{2} - \frac{\sqrt{165}}{2}$ and $x = \frac{13}{2} + \frac{\sqrt{165}}{2}$ #### Explanation: The roots of general form of equation $a {x}^{2} + b x + c = 0$ are zeros of the general form of equation $y = a {x}^{2} + b x + c$. Simplifying the given equation $y = 8 {x}^{2} + 7 x - {\left(3 x - 1\right)}^{2}$, gives us $y = 8 {x}^{2} + 7 x - \left(9 {x}^{2} - 6 x + 1\right) = - {x}^{2} + 13 x - 1$ we see that $a = - 1$, $b = 13$ and $c = - 1$. As discriminant ${b}^{2} - 4 a c = {\left(13\right)}^{2} - 4 \left(- 1\right) \left(- 1\right) = 169 - 4 = 165$ As discriminant is positive but not a complete square, we have real but irrational roots. Hence, using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, the zeros are the given equation are $x = \frac{- \left(13\right) \pm \sqrt{165}}{2 \cdot \left(- 1\right)}$ or Zeros of function are $x = \frac{13}{2} - \frac{\sqrt{165}}{2}$ and $x = \frac{13}{2} + \frac{\sqrt{165}}{2}$
#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 36 Maths Textbook Solution. $\frac{1}{2}log\; \; 6$ Hint: Use indefinite formula and put limits to solve this integral Given: $\int_{1}^{2}\frac{x+3}{x\left ( x+2 \right )}dx$ Solution: \begin{aligned} &\int_{1}^{2} \frac{x+3}{x(x+2)} d x=\int_{1}^{2} \frac{x}{x+2} d x+\int_{1}^{2} \frac{3}{x(x+2)} d x \\ &=\int_{1}^{2} \frac{1}{x+2} d x+3 \int_{1}^{2} \frac{1}{x(x+2)} d x \end{aligned} ..............(1) Now $\int_{1}^{2}\frac{1}{x+2}dx$ Putting $x+2=u$ $\Rightarrow dx=du$ when x=1 then u=1+2=3and when x=2 then u=2+2=4 Then \begin{aligned} \int_{1}^{2} \frac{1}{x+2} d x &=\int_{3}^{4} \frac{1}{u} d u=[\log u]_{3}^{4} \quad\left[\because \int \frac{1}{x} d x=\log x\right] \\ &=[\log 4-\log 3] \end{aligned} ...............(2) and $\int_{1}^{2}\frac{1}{x\left ( x+2 \right )}dx$ To solve this integral, first we need to find its partial fraction then integrate it and put the given limits. So \begin{aligned} &\frac{1}{x(x+2)}=\frac{A}{x}+\frac{B}{(x+2)} \\ &\Rightarrow 1=\frac{A}{x}(x)(x+2)+\frac{B}{(x+2)} x(x+2) \\ &\Rightarrow 1=(x+2) A+B x \\ &\Rightarrow 1=A x+2 A+B x \\ &\Rightarrow 1=(A+B) x+2 A \end{aligned} Equating coefficient of x from both sides, then $0=A+B\Rightarrow B=-A$ Again, Equating coefficient of constant term from both sides then $1=2A\Rightarrow A=\frac{1}{2}$ $\Rightarrow B=-\frac{1}{2}$ $\therefore A=\frac{1}{2}$&$B=-\frac{1}{2}$ Then \begin{aligned} &\frac{1}{x(x+2)}=\frac{1}{2 x}-\frac{1}{2(x+2)} \\ &\therefore \int_{1}^{2} \frac{1}{x(x+2)} d x=\int_{1}^{2}\left(\frac{1}{2 x}-\frac{1}{2(x+2)}\right) d x \\ &=\frac{1}{2} \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{(x+2)}\right) d x \\ &=\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{(x+2)} d x \end{aligned} Put $x+2=u$ $\Rightarrow dx=du$ in the 2nd integral then \begin{aligned} &\int_{1}^{2} \frac{1}{x(x+2)} d x=\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{u} d x \\ &=\frac{1}{2}[\log x]_{1}^{2}-\frac{1}{2}[\log u]_{1}^{2} \\ &{\left[\because \int \frac{1}{x} d x=\log x\right]} \\ &=\frac{1}{2}[\log x]_{1}^{2}-\frac{1}{2}[\log (x+2)]_{1}^{2} \\ &{[\because u=x+2]} \end{aligned} \begin{aligned} &=\frac{1}{2}[\log 2-\log 1]-\frac{1}{2}[\log (2+2)-\log (1+2)] \\ &=\frac{1}{2}[\log 2-0]-\frac{1}{2}[\log (4)-\log (3)] \\ &=\frac{1}{2}[\log 2]-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \end{aligned} ..............(3) Putting the value of integrals from eq(2) and (3) in (1), we get \begin{aligned} &\int_{1}^{2} \frac{x+3}{x(x+2)} d x=\log 4-\log 3+3\left[\frac{1}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3\right] \\ &=\log 4-\log 3+\frac{3}{2} \log 2-\frac{3}{2} \log 4+\frac{3}{2} \log 3 \\ &=\left(1-\frac{3}{2}\right) \log 4-\left(1-\frac{3}{2}\right) \log 3+\frac{3}{2} \log 2 \\ &=\left(\frac{2-3}{2}\right) \log 4-\left(\frac{2-3}{2}\right) \log 3+\frac{3}{2} \log 2 \end{aligned} \begin{aligned} &=-\frac{1}{2} \log 4-\left(-\frac{1}{2}\right) \log 3+\frac{3}{2} \log 2 \\ &=\frac{1}{2}\left[-\log 2^{2}+\log 3+3 \log 2\right] \\ &=\frac{1}{2}[-2 \log 2+\log 3+3 \log 2] \\ &=\frac{1}{2}[\log 2+\log 3] \end{aligned} \begin{aligned} &=\frac{1}{2} \log (2 \times 3) \\ &{[\because \log (m n)=\log m+\log n]} \\ &=\frac{1}{2} \log 6 \end{aligned}
# IGCSE Mathematics Paper-1: Specimen Questions with Answers 166 - 168 of 324 ## Question 166 Essay▾ Simplify (a) (b) , (c) ### Explanation Here, (a) Because 0 powers always gives value 1 and minus power of denominator will become plus when denominator multiply by its numerator. (b) Because division of two terms with same base will subtract denominator՚s exponent from numerator՚s exponent. (c) Because it is an exponent expression that is raised to a power, so we can multiply the exponent and power which is second exponent rule. ## Question 167 ### Write in Short The area of a square is 42.25 . Find length of one side of the square. ### Explanation Here, formula for area of a square = where, a is side of square. So, here, ## Question 168 ### Describe in Detail Essay▾ In the diagram AC is the diameter of a circle, center O. The length of AC is 12 cm. (a) Write down the size of angle ABC. (b) Angle BAC = 40°. Calculate the length of BC. (c) Calculate the area of the circle. ### Explanation Here, AC is the diameter of a circle which is of 12 cm and O is center. (a) Size of Here, edge AC of triangle is diameter of the circle which will be diagonal of triangle ABC. So, we can see in the figure, it is a right angled triangle, so The triangle formed by the diameter and the inscribed angle (triangle ABC above) is always a right triangle. (b) Here, Angle BAC = 40° and length of AC is 12 cm. So, using cosine formula of triangle: (C) Area of circle = , where r = radius of circle. So, here diameter AC = 12 cm, so radius = . So, area of circle = . Developed by:
# What is 1/164 as a decimal? ## Solution and how to convert 1 / 164 into a decimal 1 / 164 = 0.006 Converting 1/164 to 0.006 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Once we've decided the best way to represent the number, we can dive into how to convert 1/164 into 0.006 ## 1/164 is 1 divided by 164 The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 1 is being divided into 164. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! We use this as our equation: numerator(1) / denominator (164) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's how you set your equation: ### Numerator: 1 • Numerators are the top number of the fraction which represent the parts of the equation. Small values like 1 means there are less parts to divide into the denominator. 1 is an odd number so it might be harder to convert without a calculator. Ultimately, having a small value may not make your fraction easier to convert. Let's look at the fraction's denominator 164. ### Denominator: 164 • Denominators are located at the bottom of the fraction, representing the total number of parts. Larger values over fifty like 164 makes conversion to decimals tougher. And it is nice having an even denominator like 164. It simplifies some equations for us. Have no fear, large two-digit denominators are all bark no bite. So grab a pen and pencil. Let's convert 1/164 by hand. ## Converting 1/164 to 0.006 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 164 \enclose{longdiv}{ 1 }$$ To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 164 \enclose{longdiv}{ 1.0 }$$ We've hit our first challenge. 1 cannot be divided into 164! Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 164 into 10. ### Step 3: Solve for how many whole groups you can divide 164 into 10 $$\require{enclose} 00.0 \\ 164 \enclose{longdiv}{ 1.0 }$$ How many whole groups of 164 can you pull from 10? 0 Multiple this number by our furthest left number, 164, (remember, left-to-right long division) to get our first number to our conversion. ### Step 4: Subtract the remainder $$\require{enclose} 00.0 \\ 164 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$ If there is no remainder, youâ€™re done! If you still have numbers left over, continue to the next step. ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable. ### Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Here are examples of when we should use each. ### When you should convert 1/164 into a decimal Contracts - Almost all contracts leverage decimal format. If a worker is logging hours, they will log 1.0 hours, not 1 and 1/164 hours. Percentage format is also used in contracts as well. ### When to convert 0.006 to 1/164 as a fraction Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4). ### Practice Decimal Conversion with your Classroom • If 1/164 = 0.006 what would it be as a percentage? • What is 1 + 1/164 in decimal form? • What is 1 - 1/164 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.006 + 1/2?
1. ## Find the derivative! Find the derivative of by product rule and quotient rule. (Thanks in advance!) 2. Originally Posted by Affinity Find the derivative of by product rule and quotient rule. (Thanks in advance!) Define $\displaystyle f(x) = x - 3x \sqrt{x}$ and $\displaystyle g(x) = \sqrt{x}$ Then $\displaystyle f'(x) = 1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}}$ and $\displaystyle g'(x) = \frac{1}{2} \frac{1}{\sqrt{x}}$ Quotient Rule: So $\displaystyle h(x) = \frac{f(x)}{g(x)}$ $\displaystyle h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ $\displaystyle h'(x) = \frac{ \left ( 1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}} \right ) \sqrt{x} - (x - 3x \sqrt{x}) \frac{1}{2} \frac{1}{\sqrt{x}} }{x}$ $\displaystyle h'(x) = ...= \frac{\sqrt{x} - 6x}{2x}$ ================================================== ======== Define $\displaystyle f(x) = x - 3x \sqrt{x}$ and $\displaystyle g(x) = \frac{1}{\sqrt{x}}$ Then $\displaystyle f'(x) = 1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}}$ and $\displaystyle g'(x) = -\frac{1}{2} \frac{1}{\sqrt{x^3}} = -\frac{1}{2x\sqrt{x}}$ Product rule: So $\displaystyle h(x) = f(x)g(x)$ $\displaystyle h'(x) = f'(x)g(x) + f(x)g(x)$ $\displaystyle h'(x) = \left (1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}} \right ) \frac{1}{\sqrt{x}} + (x - 3x \sqrt{x}) \left ( -\frac{1}{2x\sqrt{x}} \right )$ $\displaystyle h'(x) = ... = \frac{1}{2 \sqrt{x}} - 3 = \frac{\sqrt{x} - 6}{2x}$ -Dan 3. Hello, Affinity! Differentiate: .$\displaystyle f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}}$ by the product rule and quotient rule. Product Rule We have: .$\displaystyle f(x)\;=\;x^{-\frac{1}{2}}\left(x - 3x^{\frac{3}{2}}\right)$ Then: .$\displaystyle f'(x)\;=\;x^{-\frac{1}{2}}\left(1 - \frac{9}{2}x^{\frac{1}{2}}\right) + \left(-\frac{1}{2}x^{-\frac{3}{2}}\right)\left(x - 3x^{\frac{3}{2}}\right)$ . . . . . $\displaystyle f'(x) \;= \;x^{-\frac{1}{2}} - \frac{9}{2} - \frac{1}{2}x^{-\frac{1}{2}} + \frac{3}{2}$ . . . . . $\displaystyle \boxed{f'(x) \;= \;\frac{1}{2}x^{-\frac{1}{2}} - 3}$ Quotient Rule We have: .$\displaystyle f(x) \;= \;\frac{x - 3x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$ Then: .$\displaystyle f'(x) \;= \;\frac{x^{\frac{1}{2}}\left(1 - \frac{9}{2}x^{\frac{1}{2}}\right) - \left(\frac{1}{2}x^{-\frac{1}{2}}\right)\left(x - 3x^{\frac{3}{2}}\right)}{\left(x^{\frac{1}{2}}\rig ht)^2}$ . . . . . $\displaystyle f'(x) \;= \;\frac{x^{\frac{1}{2}} - \frac{9}{2}x - \frac{1}{2}x^{\frac{1}{2}} + \frac{3}{2}x}{x}$ . . . . . $\displaystyle f'(x) \;= \;\frac{\frac{1}{2}x^{\frac{1}{2}} - 3x}{x}$ . . . . . $\displaystyle \boxed{f'(x) \;= \;\frac{x^{\frac{1}{2}} - 6x}{2x}}$
# TCS NQT Ninja and Digital aptitude questions - 25 1. How many liters of a 90% of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 liter solution of 78% concentrated acid? a. 3 b. 4 c. 6 d. 10 Answer: c Explanation: Let $n_1$ liters of 90% concentration has to be mixed with $n_2$ liters of 75% concentration to get 78% concentration solution. Using weighted average formula $a_x=\dfrac{n_1\times a_1+n_2 \times a_2}{n_1 + n_2}$ $78=\dfrac{n_1\times 90+n_2 \times 75}{n_1 + n_2}$ $\dfrac{n_1}{n_2}=\dfrac{1}{4}$ by dividing 30 in the ratio 1:4 we get 6 : 24. So we need 6 liters. 2. Find the ratio of the area of square to area of triangle. a. 1:2 b. 2:1 c. 2:3 d. 3:2 Answer: b Explanation: Have a look at the diagram below. Let the side of the square = 2 units. Now the area of the square = 2= 4. Area of the triangle = $\dfrac{1}{2} \times 2 \times 2 = 2$ Ratio = 4 : 2 = 2 : 1. 3. In this question A^B means A raised to the power B. If f(x) = ax^4 - bx^2 + x + 5 and f(-3) = 2, then f(3) = a. 1 b. - 2 c. 3 d. 8 Answer: d Explanation: f(-3) = a(-3)4 - b(-3)2 + (-3) + 5 = 81a - 9b + 2 = 2 So 81a - 9b = 0 f(3) = a(3)4 - b(3)2 + (3) + 5 = 81a - 9b + 8 Substituting the value of 81a - 9b = 0 in the above we get f(3) = 8 4. Of a set of 30 numbers, average of first 10 numbers = average of last 20 numbers. Then the sum of the last 20 numbers is? a. Cannot be determined. b. 2 x sum of last ten numbers c. 2 x sum of first ten numbers d. sum of first ten numbers Answer: c Explanation: We know that sum = average x number of observations. Let the common average = x Now sum of first 10 numbers = 10x Sum of the last 20 numbers = 20x. So sum of the last 20 numbers = 2 × sum of the first ten numbers. 5. A play school has chocolates which can supply 50 students for 30 days. For the first ten days only 20 students were present. How many more students can be accommodated into the earlier group such that the entire chocolates get consumed in 30 days. Assume each student takes the same number of chocolates. a. 45 b. 60 c. 55 d. 70 Answer: a Explanation: Let each students gets 1 chocolate. Now total chocolates = 50 x 30 = 1500 If first 10 days only 20 students were present, then total chocolates consumed = 10 × 20 = 200 Now we are left with 1500 - 200 = 1300 chololates. These were to be consumed in 20 days. So each day 1300/20 = 65 chocolates were to be distributed. So we can add 65 - 20 = 45 students. 6. In the town of Unevenville, it is a tradition to have the size of the front wheels of every cart different from that of the rear wheels. They also have special units to measure cart wheels which is called uneve. The circumference of the front wheel of a cart is 133 uneves and that of the back wheel is 190 uneves. What is the distance traveled by the cart in uneves, when the front wheel has done nine more revolutions than the rear wheel? a. 570 b. 1330 c. 3990 d. 399 Answer: c Explanation: LCM of 133 and 190 is 1330. So to cover this distance, front wheel takes 10 rounds, and back wheel takes 7 rounds. So for 3 rounds extra, 1330 uneves distance has to be travelled. To take 9 rounds extra, 1330 × 3 = 3990 uneves has to be traveled. 7. There are 20 persons sitting in a circle. In that there are 18 men and 2 sisters. How many arrangements are possible in which the two sisters are always separated by a man? a. 18!x2 b. 17! c. 17x2! d. 12 Answer: a Explanation: Let the first sister name is A. Now she can sit any where in the 20 places (Symmetrical). Now her sister B can sit to her left or right in 2 ways. Now the remaining 18 persons can be sit in 18 places in 18! ways. Total = 18! × 2 8. A number plate can be formed with two alphabets followed by two digits, with no repetition. Then how many possible combinations can we get? a. 58500 b. 67600 c. 65000 d. 64320 Answer: a Explanation: Easy. 26 × 25 × 10 × 9 = 58500 9. A alone can do 1/4th of the work in 2 days. B alone can do 2/3th of the work in 4 days. If all the three work together, they can complete it in 3 days so what part of the work will be completed by C in 2 days? a. 1/12 b. 1/8 c. 1/16 d. 1/20 Answer: a Explanation: A can do the total work in 8 days, and B can do it in 6 days. Let the total work be 24 units. Now capacities are A = 24/8 = 3, B = 24/6 = 4, A + B + C = 24/3 = 8 So Capacity of C = 1 unit. In two days C will do 2 units which is 2/24th part of the total work. So 1/12th part. 10. How many prime numbers are there which are less than 100 and greater than 3 such that they are of the following forms • 4x + 1 • 5y - 1 a. 11 b. 12 c. 7 d. None of the above Answer: d Explanation: Let the number be N. So N = 4x + 1 = 5y - 1 $\Rightarrow x = \dfrac{{5y - 2}}{4}$ y = 2 satisfies the equation. So minimum number satisfies both the equations is 9 and general format of the numbers which satisfies the equation = k. LCM (4, 5) + 9 = 20k + 9. Now by putting values 1, 2, 3 . . . . for k, we get 29, 49, 69, 89. Of which only 29, 89 are primes. 11. Babla alone can do a piece of work in 10 days. Ashu alone can do it in 15 days. The total wages for the work is Rs.5000. How much should be Babla be paid if they work together for an entire duration of work. a. 2000 b. 4000 c. 5000 d. 3000 Answer: d Explanation: Money should always be divided in the inversely proportion way. So Babla will get $\dfrac{{15}}{{15 + 10}} \times 5000 = 3000$ 12. The shopkeeper charged 12 rupees for a bunch of chocolate. but i bargained to shopkeeper and got two extra ones, and that made them cost one rupee for dozen less then first asking price. How many chocolates I received in 12 rupees ? a. 10 b. 16 c. 14 d. 18 Answer: b Explanation: Let the number of chocolates bought = n or n/12 dozens Assume this would cost x rupees. Now given that (n+2)/12 dozens cost x - 1 rupee. So $\dfrac{{12}}{{n/12}} = x$ - - - (1) and $\dfrac{{12}}{{(n + 2)/12}} = x - 1$ - - - (2) (1) - (2) = $\dfrac{{144}}{n} - \dfrac{{144}}{{n + 2}} = 1$ From the options, 16 satisfies.
# Tables of 19 Let us know about Tables of 19. Friends, as you must be aware that how much usefulness of mountains is in calculation. We memorize the hills in small classes and we use these mountains in our everyday life. Through today’s blog, we will learn how to remember the table of 19 by the following methods. ## What is the table of 19? Tables of 19 can be described using repeated additions of 19. For example, in a basket there are 19 balls of 6 different colors. In this case, the total number of balls in the basket can be counted using the multiplication table. This means there are 19 balls in each color. Thus, the total number of balls is 19 × 6 = 114. It can also be expressed as 19+19+19+19+19+19=114 So, as explained above, we can get all the multiplications of 19. ## table of 19 in english How the table of 19 is written in English is given below. ## 19 as a table word The table of 19 is given here in the form of words, so that students do not face difficulties in remembering it- ## Table of 19 in the form of maths How to write the table of 19 in mathematics is given below- ## Tips to remember table of 19 Children can easily memorize the table of 19 according to some tips- Tips 1: The first 10 multiples of 19 are: 19, 38, 57, 76, 95, 114, 133, 152, 171, 190. Here the digit in the units place is decreasing from 9 to 0, while the tens digit is increasing from 1 to 19 at intervals of 2. Tips 2: The first 10 multiples of 19, that is, the results of the 19 multiplication table from 1 to 10, can also be written as: Tips 3: • When a number is to be tabled, the number is continuously multiplied from 1 to 10. • Follow the rule of the above chart to memorize the tables. • Any table will be easily remembered in two to three attempts. Teachers believe so. • just remember the series of multiplication • Tables 1 to 20 are easy to remember, if the regular effort is continued. ## worksheet Now you have seen how we can make the table of 19. Now you need practice. Write the table of 19 over and over again, repeating it over and over again. Make a time table and keep practicing continuously. With this you will be able to memorize the table of 19 very soon. For your brain exercise, we have given below some mathematical questions related to the table of 19. Solve these and assess your ability. ## FAQs Is there any trick to memorize the table of 19? Answer – While reading the table of 19, you can notice that here the digits in the units place are decreasing from 9 to 0, while the digits in the tens place are increasing from 1 to 19 at an interval of 2. eg- 19 × 1 = 19 19 = 1 × 19 19 × 2 = 3 8 3 8 = 2 × 19 19 × 3 = 5 7 5 7 = 3 × 19 19 × 4 = 7 6 7 6 = 4 × 19 19 How much is a power? Answer – 19 Satte is 133. What can students do to memorize the table? Answer – Students can memorize the table by practicing it again and again, by reading it again and again. If there are 19 flowers in a garden bed, then how many more flowers will be in 9 such beds? Answer – 19 * 9 = 171 Fill in the blanks below: 19×3=___ 19+19=____ 57-19=____
Select Page Remainders: Part-2 Type: Difference between divisor and remainder is same On dividing by p, q and r the number gives different remainders a, b and c where p-a=q-b=r-c Let’s take an example to understand this Example What is the minimum two-digit number which when divided by 3, 4, 5 leaves 1, 2, 3 as the remainder respectively? Solution: Let P be the required two digit number .As given in the question P = 3 d + 1 =3 (d + 1) -2 P = 4 e + 2= 4( e + 1) -2 P = 5 f + 3= 5 (f+1) – 2 In each of above cases we observe difference between divisor and remainder is same. Hence we have same remainder -2 when divided by 3,4,5 Thus number N must be in the form L.C.M. (3,4,5) k – 2 =60 k – 2 Since we are looking for two digit number putting k =1 we get 58 as minimum two-digit number which when divided by 3, 4, 5 leaves 1, 2, 3 as the remainder respectively. Example What is the minimum three-digit number which when divided by 2, 5, 6 leaves 1, 4, 5 as the remainder respectively? Solution: Let P be the required three digit number .As given in the question P = 2 d + 1 =2 (d + 1) -1 P = 5 e + 4 = 5( e + 1) -1 P = 6 f + 5 = 6 (f+1) – 1 In each of above cases we observe difference between divisor and remainder is same. Hence we have same remainder -1 when divided by 2,5,6 Thus number N must be in the form L.C.M. (2,5,6) k – 1 =30 k – 1 Since we are looking for two digit number putting k =34 we get 1019 as minimum three-digit number which when divided by 2, 5, 6 leaves 1, 4, 5 as the remainder respectively. Assignment: QUESTIONS: 1. What is the remainder when 72 + 51 is divided by 8 ? a)1 b)2 c)3 d)5 Solution : Since 72 is multiple of 8 hence it leaves remainder zero and 51 leaves remainder 3 when divided by 8.Hence 72 + 51 leaves remainder 3 when divided by 8. 2. What is the remainder when 111 + 68 + 48 is divided by 11 ? a)5 b)7 c)9 d)None Solution: 111 leaves remainder 1 when divided by 11 68 leaves remainder 2 when divided by 11 48 leaves remainder 4 when divided by 11 Thus 111 + 68 + 48 leaves remainder 1+2+4 =7 when divided by 11 3. What is the remainder when 42×56 is divided by 8 ? a)0 b)1 c)2 d)3 Solution : 56 is a multiple of 8 hence remainder is zero. 4.What is the remainder when 48×65 is divided by 7 ? a)1 b)2 c)3 d)5 48 leaves remainder 6 and 65 leaves remainder 2 so 48 65 divided by 7 gives 6×2=12 but 12 is greater than 7 hence we further divide by 7 and we obtain 5 as remainder. 5. What is the remainder of 84x71x41 by 9? a)2 b)3 c)6 d)8 84 leaves remainder 3, 71 leaves remainder 8 and 41 leaves remainder 5 when divided by 9. So 84x71x42 divided by 9 gives 3 8 =120 but 120 is greater than 9 hence we further divide by 9 and we obtain 3 as remainder. FREE CHEAT SHEET Learn How to Master VA-RC This free (and highly detailed) cheat sheet will give you strategies to help you grow No thanks, I don't want it.
# Carpentry T Chart Duty Hand Tools Task Layout and Document Sample ``` Carpentry T-Chart Radius = Radius PSSA Eligible Content – M11.C.1.1.1 Duty: Hand Tools Identify and/or use the properties of a radius, Task: Layout and cut patters diameter and/or tangent of a circle (given numbers should be whole) MATH ASSOCIATED WORDS: CARPENTRY ASSOCIATED WORDS: TERM - BRACE, AUGER, BORING BIT, PERPENDICULAR, CONCENTRIC CIRCLES. Formula (type purpose of formula here): Formula (type purpose of formula here): PDE/BCTE Math Council TEACHER’S SCRIPT FOR BRIDGING THE GAP In both Carpentry and in Math Class you learn about the relationship between radius and diameter. How are the concepts similar? These concepts are very similar in both classes. How do the concepts differ? In Carpentry there are many additional terms and tools you use associated with radius and diameter. In Math Class we study the concept of tangent as well as radius and diameter. A tangent touches the circle in exactly one point and is perpendicular to the radius of the circle at that point. Common Mistakes Students Make A common mistake that students make in interpreting drawings in both mathematics and carpentry is confusing the radius of a circle to the radius of a corner in a diagram. Another common error is to include the diameter of a circle as a part of the length the rectangular portion of an irregular figure that contains a rectangle with a semi-circle on each end. PDE/BCTE Math Council Occupational (Contextual) Math Concepts 1. If a circle has a radius of 7 yards what is the diameter? 2. If a circle has a diameter of 18 feet what is the radius? 3. If the radius of a circle is 4 inches and the corner radius is 9 inches, how far should the circle measure from the corner of the pattern? Related, Generic Math Concepts For the following problems use the figure below. 1. If AB = 15, then what is AC? 2. If BO = 3, and AO = 5, what is the length of AB? 3. If BO = 12, what is the length of CO? <"http://etc.usf.edu/clipart/""Visit Clipart ETC for a great collection of clipart for students and teachers." PSSA Math Concept Look 1. If the radius of one circle is 3 feet and the radius of another circle is 10 feet, what is the difference in their diameters? 2. If concentric circles have diameter of 16 inches and 22 inches, what would be the shortest distance connecting the two circles? 3. If a tangent segment to a circle with a radius of 3 feet measures 4 feet, what is the distance from the endpoint of the tangent segment outside the circle? PDE/BCTE Math Council Occupational (Contextual) Math Concepts 1. If the radius of a circle is 7 yards, then the diameter is 14 yards. D = 2r Write formula D = 2 (7yards) Substitute known values D = 14 yards Evaluate answer including correct units 2. If the diameter radius of a circle is 18 feet, then the radius is 9 feet. D = 2r Write formula 18 feet = 2r Substitute known values 9 feet = r Divide each side of the equation by 2. 3. If the radius is 4 inches and the corner radius is 9 inches, then the distance from the circle to the corner of the pattern would be the difference of these two radii, 5 inches. Related, Generic Math Concepts 1. Since AC and AB are tangents to the same circle they are congruent so, AC = 15. 2. Since BO is a radius of circle O and AB is a tangent to circle O. AO would be a leg of a right triangle ABO. This would be a 3-4-5 right triangle. If you did not recognize it as a 3-4-5 right triangle, you could use the Pythagorean Theorem to find the length. Since AO is the hypotenuse of this right triangle its length would be 5. 3. CO = 12 since BO and CO are radii of the same circle. PSSA Math Concept Look 1. First you need to find the diameters of the circle, then you need to subtract those diameters to get the difference. D = 2r D = 2r Write formula D = 2(3 feet) D = 2(10 feet) Substitute known values D = 6 feet D = 20 feet Calculate Diameters 20 feet – 6 feet = 14 feet Subtract The difference would be 14 feet. 2. Since concentric circles have the same center, the shortest distance between the points would be the difference in the radii. D = 2r D = 2r Write formula 22 inches = 2r 16 inches = 2r Substitute known values 11 inches = r 8 inches = r Divide each side by 2. 11 inches – 8 inches = 3 inches Subtract The difference is 3 inches. 3. Since a tangent always meets a radius at a right angle this would be a 3-4-5 right triangle and CA = 5. If you did not recognize it as a 3-4-5 right triangle, you could use the Pythagorean Theorem to find the length. PDE/BCTE Math Council ``` DOCUMENT INFO Shared By: Categories: Tags: Stats: views: 46 posted: 3/13/2009 language: English pages: 4 How are you planning on using Docstoc?
# Number Theory: Problem 2 ### Model solution Let p, q, and r be integers (p non-zero). Suppose that $$p \mid 3q$$ and $$3q \mid r$$. By the definition of divides, $$p \| 3q$$ means pm = 3q for some integer m. Similarly, $$3q \| r$$ means 3qn = r for some integer n. Now substitute these two equations into the expression 3q+r: $$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$ We know m(1+n) is an integer, since m and n are integers. So let s=m(1+n). Then 3q+r = ps. Therefore, by the definition of divides, $$p \| (3q + r)$$, which is what we wanted to show. ### Also ok It's ok to write the last paragraph without introducing the extra variable s. For example: We know m(1+n) is an integer, since m and n are integers. So the above equation shows that 3q+r can be written as p times an integer. Therefore, by the definition of divides, $$p \| (3q + r)$$, which is what we wanted to show. ### Self-check At the start of the proof, did you use two different variables m and n when expanding the definition of "divides"? When you invoke the definition of divides near the end, it's critical that m(1+n) is an integer. You need to make this clear to the reader, by specifying m and n as integers when you introduce them and briefly saying why m(1+n) is an integer later on. ### Poor style 1 Proofs like the following will lose points due to lack of connector words and brief explanations. How is the poor grader supposed to know what you are trying to do at each step? Let p, q, and r $$\in \mathbb{Z}$$ $$p \mid 3q$$ and $$3q \mid r$$. pm = 3q, $$m \in \mathbb{Z}$$ 3qn = r, $$n \in \mathbb{Z}$$ $$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$ $$m(1+n) \in \mathbb{Z}$$ So $$p \| (3q + r)$$ ### Poor style 2 Here's a more subtle example of poor style. It does have connector words, but the author has forgotten everything he learned in high school English class. Paragraphs, punctuation, and whitespace are critical to making technical arguments readable. Everyone makes occasional mistakes, and you'll make a few extras if your first language isn't English. But all of you can do a better job than this. let p, q, and r be integers (p non-zero) suppose that $$p \mid 3q$$ and $$3q \mid r$$. by definiton divides, $$p \| 3q$$ means pm = 3q, m integer. $$3q \| r$$ means 3qn = r for some integer n. now substitute these two equations into the expression 3q+r $$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$ We know m(1+n) is an integer, since m and n are integers solet s=m(1+n) then 3q+r = ps therefore by the definiton of divides $$p \| (3q + r)$$ QED
# 3.2.2 Circles II, PT3 Practice 3.2.2 Circles II, PT3 Practice Question 6: Diagram below shows a circle with centre O. POR is a straight line. Find the value of x and of y. Solution: $\begin{array}{l}x=40×2\\ \text{}=80\\ \\ y=\frac{180-80}{2}\\ \text{}=50\end{array}$ [adinserter block="5"] Question 7: Diagram below shows a circle with centre O. Find the value of x. Solution: $\begin{array}{l}\angle QRO={38}^{o},\angle ORS={44}^{o}\\ \angle QRS={38}^{o}+{44}^{o}={82}^{o}\\ {x}^{o}={82}^{o}×2\\ {x}^{o}={164}^{o}\\ x=164\end{array}$ [adinserter block="5"] Question 8: Diagram below shows a circle. QTS is the diameter of the circle and PTR is a straight line. Find the value of x. Solution: $\begin{array}{l}\angle RPS={90}^{o}-{42}^{o}={48}^{o}\\ {x}^{o}={48}^{o}+{37}^{o}={85}^{o}\end{array}$ [adinserter block="5"] Question 9: Diagram below shows a circle with centre O. PQR is a straight line. Find the value of x and of y. Solution: In cyclic quadrilateral, Angle at Q = 180o – 70o = 110o x2 = exterior angle at Q = 180o – 110o = 70o x = 70 yo = reflex angle at O yo = 360o – (70o × 2) yo = 220o y = 220 Question 10: In diagram below, ABCD and DEFG are straight lines. Find the value of x and of y. Solution:
Notice: Undefined offset: 0 in /var/www/content_farm_zero/html/index.php on line 24 Two step inequality solver - Mathematic # Two step inequality solver Two step inequality solver can help students to understand the material and improve their grades. We can solving math problem. ## The Best Two step inequality solver Apps can be a great way to help students with their algebra. Let's try the best Two step inequality solver. Trigonometry is used in a wide variety of fields, including architecture, engineering, and even astronomy. While the concepts behind trigonometry can be challenging, there are a number of resources that can help students to understand and master this important subject. Trigonometry textbooks often include worked examples and practice problems, while online resources can provide interactive lessons and quizzes. In addition, many math tutors offer trigonometry help specifically designed to address the needs of individual students. With a little effort, anyone can learn the basics of trigonometry and unlock its power to solve complex problems. First, you should try to understand the problem. This means that you should try to figure out what the problem is and what is causing it. Once you understand the problem, you can start looking for a solution. Another tip is to break the problem down into smaller pieces. This math is all about solving linear functions. A linear function is any function that can be written in the form y = mx + b, where m and b are constants. The most basic linear function is the equation of a straight line, which has the form y = mx + b. The slope of the line is m, and b is the y-intercept, the point where the line crosses the y-axis. To solve a linear function, you need to find the Triple integrals are often used in physics and engineering to solve problems involving three-dimensional objects or systems. The triple integral solver can be used to calculate the volume of a three-dimensional object, the moment of inertia of a three-dimensional object, or the center of mass of a three-dimensional object. There are a lot of math apps out there for free, but not many of them are good. Math is hard and it’s important to have a math app that will make it easier for you to understand the material and get better at solving problems. Math apps should have things like an interactive graph, video tutorials and expert-level explanations in case you need help. There are no free math apps with all these features, but there are some great ones that you can try out. One of the best free math apps is Khan Academy . This app has tons of videos with step-by-step explanations and helpful tips so that even if you’re not sure what’s going on in a lesson, you can still get it. It also has a community where people can ask questions about math and help each other out when they get stuck. ## Math solver you can trust I love this app because I'm terrible at math. And normally, I don't care about the steps or explanations, I just want the answer. However, right now, I'm a parent suddenly having to teach my 5th grader math I don't understand, so the explanations and steps are necessary. And the app is offering a much-needed free subscription until May 31, to help out during this time. Thank you!! Itzel Brown This is an amazing app. Thank you so much to the developers and everyone else involved, I’ve always struggled with factoring and negative exponents and this app uses the exact same methods we were taught. But with any sum or example!!! THANK YOU! from all the students ever Yulianna Lopez
# 483 and Level 3 4 + 8 = 12, a multiple of 3, so 483 can be evenly divided by 3. (It wasn’t necessary to add 4 + 8 + 3 because we already know 3 is divisible by 3.) We can easily see that 483 is divisible by 7 if we apply the divisibility trick for 7. Separate the last number, 3, from the rest, double it, and subtract the double from the remaining numbers: 48 – (3 x 2) = 48 – 6 = 42, which is a multiple of 7. Thus 483 is divisible by 7. Since 483 is 21 x 23, we can predict that 484 will be 22 x 22. Print the puzzles or type the solution on this excel file: 12 Factors 2015-05-04 ————————————————————————————————— • 483 is a composite number. • Prime factorization: 483 = 3 x 7 x 23 • The exponents in the prime factorization are 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 = 8. Therefore 483 has exactly 8 factors. • Factors of 483: 1, 3, 7, 21, 23, 69, 161, 483 • Factor pairs: 483 = 1 x 483, 3 x 161, 7 x 69, or 21 x 23 • 483 has no square factors that allow its square root to be simplified. √483 ≈ 21.97726 —————————————————————————————————
Edit Article # How to Write a Congruent Triangles Geometry Proof Congruent triangles are triangles that are identical to each other, having three equal sides and three equal angles.[1] Writing a proof to prove that two triangles are congruent is an essential skill in geometry. Since the process depends upon the specific problem and givens, you rarely follow exactly the same process. This can be frustrating; however, there is an overall pattern to solving geometric proofs and there are specific guidelines for proving that triangles are congruent. Once you know them, you’ll be able to prove them on your own with ease. ### Part 1 Proving Congruent Triangles 1. 1 Draw a diagram. A diagram may already be provided, but if one is not, it’s essential to draw one. Try to draw it as accurately as you can. Include all of the given information in your diagram. If two sides or angles are congruent (equal), mark them as such.[2] • It may be beneficial to sketch a first diagram that is not accurate and re-draw it a second time to look better. • If your diagram has two overlapping triangles, try redrawing them as separate triangles. It will be much easier to find and mark the congruent pieces. • If your diagram does not have two triangles, you might have a different kind of proof. Double check to make sure the problem asks you to prove congruency of two triangles. 2. 2 Identify the known information. Using the givens and your knowledge of geometry, you can start to prove some things and determine if any sides and/or angles of two triangles are congruent. Think about the parts of the proof logically and determine step-by-step how to get from the givens to the final conclusion.[3] • For example: Using the following givens, prove that triangle ABC and CDE are congruent: C is the midpoint of AE, BE is congruent to DA. If C is the midpoint of AE, then AC must be congruent to CE because of the definition of a midpoint. This allows you prove that at least one of the sides of both of the triangles are congruent. • If BE is congruent to DA then BC is congruent to CD because C is also the midpoint of AD. You now have two congruent sides. • Also, because BE is congruent to DA, angle BCA is congruent to DCE because vertical angles are congruent. 3. 3 Choose the correct theorem to prove congruency. There are five theorems that can be used to prove that triangles are congruent. Once you have identified all of the information you can from the given information, you can figure out which theorem will allow you to prove the triangles are congruent.[4] • Side-side-side (SSS): both triangles have three sides that equal to each other. • Side-angle-side (SAS): two sides of the triangle and their included angle (the angle between the two sides) are equal in both triangles. • Angle-side-angle (ASA): two angles of each triangle and their included side are equal. • Angle-angle-side (AAS): two angles and a non-included side of each triangle are equal. • Hypotenuse leg (HL): the hypotenuse and one leg of each triangle are equal. This only applies to right triangles. • For example: Because you were able to prove that two sides with their included angle were congruent, you would use side-angle-side to prove that the triangles are congruent. ### Part 2 Writing a Proof 1. 1 Set up a two-column proof. The most common way to set up a geometry proof is with a two-column proof. Write the statement on one side and the reason on the other side. Every statement given must have a reason proving its truth. The reasons include it was given from the problem or geometry definitions, postulates, and theorems.[5] 2. 2 Write down the givens. The easiest step in the proof is to write down the givens. Write the statement and then under the reason column, simply write given. You can start the proof with all of the givens or add them in as they make sense within the proof.[6] • Write down what you are trying to prove as well. If you want to prove that triangle ABC is congruent to XYZ, write that at the top of your proof. This will also be the conclusion of your proof. 3. 3 Use the appropriate theorems, definitions, and postulates as reasons. When developing a proof, you need a solid foundation in geometry before you can begin. Knowing the relevant theorems, definitions, and postulates is essential. A working knowledge of these will help you to find reasons for your proof.[7] • Some good definitions and postulates to know involve lines, angles, midpoints of a line, bisectors, alternating and interior angles, etc. • You cannot prove a theorem with itself. If you're trying to prove that base angles are congruent, you won't be able to use "Base angles are congruent" as a reason anywhere in your proof. 4. 4 Order the proof logically. When constructing a proof, you want to think through it logically. Try to order all of your steps so that they naturally follow each other. Sometimes it helps to work the problem backwards: start with the conclusion and work your way back to the first step.[8] • Every step must be included even if it seems trivial. • Read through the proof when you are done to check to see if it makes sense. ## Community Q&A Unanswered Questions • How can I construct a formal proof in order to prove congruent line segments? Show more unanswered questions Ask a Question If this question (or a similar one) is answered twice in this section, please click here to let us know. ## Tips • If your givens include the word "perpendicular," do not say that an angle is 90 degrees due to definition of perpendicular lines. Instead, write a statement saying such angle is a right angle because of "definition of perpendicular lines" and then write another statement saying said angle is 90 degrees because of "definition of right angle." ## Article Info Categories: Geometry Thanks to all authors for creating a page that has been read 195,577 times. Did this article help you?
# Lesson 17 True or False? ## Warm-up: True or False: Fraction Addition (10 minutes) ### Narrative The purpose of this True or False is for students to demonstrate strategies and understandings they have for adding fractions with unlike denominators. These understandings help students deepen their understanding of the properties of operations and will be helpful later in this lesson when students will develop their own true or false activity. In the synthesis it is important to discuss things the writer had to pay attention to when they designed this activity. ### Launch • Display one equation. • “Give me a signal when you know whether the equation is true and can explain how you know.” • 1 minute: quiet think time ### Activity • Share and record answers and strategy. • Repeat with each equation. ### Student Facing Decide if each statement is true or false. Be prepared to explain your reasoning. • $$\frac{3}{4}+\frac{3}{8} = \frac{3}{8}+\frac{3}{8}$$ • $$\frac{7}{5}+\frac{2}{3} = \frac{21}{15}+\frac{8}{15}$$ • $$\frac{8}{9}+\frac{5}{12}=\frac{32}{36}+\frac{15}{36}$$ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • “What did the writer of this activity have to pay attention to when they designed this activity?” (Some equations are true and some are false. Some terms on both sides are equal. Pay attention to the unlike and like denominators.) • “Where do we see those things in how the equations change during the True or False?” (The first two equations are false, but they use an appropriate common denominator.) ## Activity 1: True or False: Design 1 (15 minutes) ### Narrative The purpose of this activity is for students to reason about subtracting fractions with unlike denominators to add one equation to a partially-completed True or False activity. If there is time, students can facilitate their True or False with another group. MLR8 Discussion Supports. Prior to solving the problems, invite students to make sense of the situations and take turns sharing their understanding with their partner. Listen for and clarify any questions about the context. Representation: Internalize Comprehension. Synthesis: Use multiple examples and non-examples to emphasize the importance of finding like units in order to subtract fractions. Supports accessibility for: Conceptual Processing, Memory ### Launch • Groups of 2 or 4 • “Now you will work with your group to complete a True or False activity. This activity has one equation missing. Decide on an equation that would complete the True or False and write it on the blank line.” ### Activity • 10 minutes: small-group work time • As students work, monitor for groups who discuss and design an equation based on some of the following: • They create equations where the fractions on one side of the equation have unlike denominators and the other side consists of fractions with like denominators. • They create an equation where the common denominator is appropriate, but the corresponding fraction used is not equivalent. • They create an equation that is similar to a previous equation. For example: $$\frac{5}{6}-\frac{4}{9}=\frac{30}{36}-\frac{16}{36}$$ • They create an equation where a similar strategy can be used to determine if it’s true. ### Student Facing Write an equation to complete the True and False task. Be prepared to share your reasoning for the last equation. • $$\frac{5}{6}-\frac{4}{9}=\frac{45}{54}-\frac{24}{54}$$ • $$\frac{5}{6}-\frac{4}{9}=\frac{15}{36}$$ • _______________________ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Choose small groups to share that had different reasons for their fourth equation. • Ask students to share their completed True or False and ask the class to share reasons for the last equation. • As each group shares, continually ask others in the class if they agree or disagree and the reasons why. ## Activity 2: Design 2 (15 minutes) ### Narrative The purpose of this activity is for students to reason about adding fractions with unlike denominators to add two equations to a partially-completed True or False activity. If there is time, students can facilitate their True or False with another group. ### Launch • Groups of 2 or 4 • “Now you will work with your group to complete a True or False activity. This activity has two equations missing. Decide on equations that would complete the True or False and write them on the blank lines.” ### Activity • 10 minutes: small-group work time • As students work, monitor for groups who discuss and design equations based on some of the following: • They create equations where the fractions on one side of the equation have unlike denominators and the other side consists of fractions with like denominators. • They create an equation where the common denominator is appropriate, but the corresponding fraction used is not equivalent. • They create an equation that is similar to a previous equation. • They create an equation where a similar strategy can be used to determine if it’s true. ### Student Facing Write two equations to complete the True or False task. Be prepared to share your reasoning for the equations. • $$\frac{8}{14}+\frac{3}{7}=\frac{4}{7}+\frac{3}{7}$$ • _______________________ • _______________________ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Choose small groups to share that had different reasons for their equations. • Ask students to share their completed True or False and ask the class to share reasons for the equations. • As each group shares, continually ask others in the class if they agree or disagree and the reasons why. ## Activity 3: Design 3 (15 minutes) ### Narrative The purpose of this activity is for students to reason about adding and subtracting fractions with unlike denominators. Students create equations to complete a True or False activity. If there is time, students can facilitate their True or False with another group. ### Launch • Groups of 2 or 4 • “Now you will work with your group to complete a True or False activity. This activity has all three equations missing. Decide on equations that would complete the True or False and write them on the blank lines.” ### Activity • 10 minutes: small-group work time • As students work, monitor for groups who discuss and design equations based on some of the following: • They create equations where the fractions on one side of the equation have unlike denominators and the other side consists of fractions with like denominators. • They create an equation where the common denominator is appropriate, but the corresponding fraction used is not equivalent. • They create an equation that is similar to a previous equation. • They create an equation where a similar strategy can be used to determine if it’s true. ### Student Facing Write three equations to complete the True or False task. Be prepared to share your reasoning for the equations. • _______________________ • _______________________ • _______________________ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Choose small groups to share that had different reasons for their equations. • Ask students to share their completed True or False and ask the class to share reasons for the equations. • As each group shares, continually ask others in the class if they agree or disagree and the reasons why. ## Lesson Synthesis ### Lesson Synthesis “What were the most important things about your equations you had to consider as you created your True or False? Why were these things important?” (I needed to find equations where you could figure out whether or not it was true with a mental strategy. So the numbers had to be related in a nice way and there needed to be some structure to help see if the expressions were equal or not, without finding the sums or differences.) ## Cool-down: Reflection (5 minutes) ### Cool-Down For access, consult one of our IM Certified Partners.
We all know that a circle is a geometrical shape which is a closed figure; a two-dimensional curved shape where every point from the center of the circle is equidistant. The circumference of a circle is also known as the perimeter of the circle. We can define it as a measurement used to measure the bounding line of the circle. The circumference can be defined as the length of the circle when opened widely. Mathematically, the formula given for the circumference of the circle is, 2 πr where ‘r’ is the radius of the circle. In this article, we may further discuss the circumference of a circle and try to solve as many questions as possible, in order to understand the topic in a detailed manner. ## Study Math Online From the Best in this Field Sometimes, the subject of mathematics gets boring, and you don’t feel like learning and understanding it. In that case, you should visit cuemath.com as it helps and makes the subject mathematics more interesting. Study math online from Cuemath and get your concepts cleared by the best tutors from around the world. Studying math online also helps the student to record and revisit the lecture as much as he wants, so that he or she might see it when needed. ## Area Of Circle And Calculations Based on It As mentioned above, the measurement of the length of the circle is known as the circumference of the circle. If we talk about the area of a circle, it is the pace or region which is enclosed by the boundaries of the circle. Mathematically, the area of a circle is given by; πr.πr where r is the radius of the circle, and the value of pi is generally given by 3.14 or 22/7. Let us solve some calculations based on the area of the circle in order to understand this topic in a better way. Example 1: Find the area of the circle if the radius given is 10 cm and the value of pi to be used as 3.14? Provided that, Radius of the Circle = 10 cm Value of π given is = 3.14 Using the formula of the area of circle; πr.πr; 3.14 * 10.10 = 314 cm square. Thus, the area of the circle = 314 cm square. Example 2: Find the area of the circle if the radius given is 12 cm and the value of pi to be used as 3.14? Provided that, Radius of the Circle = 12 cm Value of π given is = 3.14 Using the formula of the area of circle; πr.πr; 3.14 * 12.12 = 452.16 cm square. Thus, the area of the circle = 452.16 cm square. ## Calculations Based on the Perimeter/Circumference of The Circle The formula given for the circumference/perimeter of the circle is 2 πr, where ‘r’ is the radius of the circle. In the next paragraph, we will try to solve some calculations based on the circumference of the circle. Some of the examples are as follows: Example 1: Calculate the perimeter of the circle if the value given for the radius is 5 cm and the value of π to be used 3.14? Given that, Radius of The Circle = 5 cm Value of π give is = 3.14 Using the formula of the perimeter of the circle; 2 πr, 2 * 3.14 * 5 = 31.4 cm. Example 2: Calculate the perimeter of the circle if the value given for the radius is 8 cm and the value of π to be used 3.14? Given that, Radius of The Circle = 8 cm Value of π give is = 3.14 Using the formula of the perimeter of the circle; 2 πr, 2 * 3.14 * 8 = 50.24 cm. If you want to learn about the circumference of a circle in a detailed manner, in a fun way, and in an interesting manner, you may visit the Cuemath website. ## What Is Civil Service Exam in India (CSE)? Eligibility, Syllabus, Age Limit, Fees, Registration As per UPSC, Civil Service Exams are a set of competitive examinations… ## What is GRE Exam in India? Eligibility, Syllabus, Age Limit, Fees, Registration The Graduate Record Examination (GRE) is a standardized test designed to assess… ## Swami Vivekananda- The Wandering Monk The story of Swami Vivekananda is one of suffering and triumph. The… ## Essay, IELTS Cue Card, Speech, Paragraph The Economics of Everyday Life Last week the Canadian government led by…
##### Motivating Questions • How can we use a Riemann sum to estimate the area between a given curve and the horizontal axis over a particular interval? • What are the differences among left, right, middle, and random Riemann sums? • How can we write Riemann sums in an abbreviated form? In Section 4.1, we learned that if we have a moving object with velocity function $v\text{,}$ whenever $v(t)$ is positive, the area between $y = v(t)$ and the $t$-axis over a given time interval tells us the distance traveled by the object over that time period; in addition, if $v(t)$ is sometimes negative and we view the area of any region below the $t$-axis as having an associated negative sign, then the sum of these signed areas over a given interval tells us the moving object's change in position over the time interval. For instance, for the velocity function given in Figure 4.2.1, if the areas of shaded regions are $A_1\text{,}$ $A_2\text{,}$ and $A_3$ as labeled, then the total distance $D$ traveled by the moving object on $[a,b]$ is \begin{equation} D = A_1 + A_2 + A_3, \end{equation} while the total change in the object's position on $[a,b]$ is \begin{equation} s(b) - s(a) = A_1 - A_2 + A_3. \end{equation} Because the motion is in the negative direction on the interval where $v(t) \lt 0\text{,}$ we subtract $A_2$ when determining the object's total change in position. Of course, finding $D$ and $s(b)-s(a)$ for the situation given in Figure 4.2.1 presumes that we can actually find the areas represented by $A_1\text{,}$ $A_2\text{,}$ and $A_3\text{.}$ In most of our work in Section 4.1, such as in Activities 4.1.3 and Activity 4.1.4, we worked with velocity functions that were either constant or linear, so that by finding the areas of rectangles and triangles, we could find the area bounded by the velocity function and the horizontal axis exactly. But when the curve that bounds a region is not one for which we have a known formula for area, we are unable to find this area exactly. Indeed, this is one of our biggest goals in Chapter 4: to learn how to find the exact area bounded between a curve and the horizontal axis for as many different types of functions as possible. To begin, we expand on the ideas in Activity 4.1.2, where we encountered a nonlinear velocity function and approximated the area under the curve using four and eight rectangles, respectively. In the following preview activity, we focus on three different options for deciding how to find the heights of the rectangles we will use. ##### Preview Activity4.2.1 A person walking along a straight path has her velocity in miles per hour at time $t$ given by the function $v(t) = 0.25t^3-1.5t^2+3t+0.25\text{,}$ for times in the interval $0 \le t \le 2\text{.}$ The graph of this function is also given in each of the three diagrams in Figure 4.2.2. Note that in each diagram, we use four rectangles to estimate the area under $y = v(t)$ on the interval $[0,2]\text{,}$ but the method by which the four rectangles' respective heights are decided varies among the three individual graphs. 1. How are the heights of rectangles in the left-most diagram being chosen? Explain, and hence determine the value of \begin{equation} S = A_1 + A_2 + A_3 + A_4 \end{equation} by evaluating the function $y = v(t)$ at appropriately chosen values and observing the width of each rectangle. Note, for example, that \begin{equation} A_3 = v(1) \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1. \end{equation} 2. Explain how the heights of rectangles are being chosen in the middle diagram and find the value of \begin{equation} T = B_1 + B_2 + B_3 + B_4. \end{equation} 3. Likewise, determine the pattern of how heights of rectangles are chosen in the right-most diagram and determine \begin{equation} U = C_1 + C_2 + C_3 + C_4. \end{equation} 4. Of the estimates $S\text{,}$ $T\text{,}$ and $U\text{,}$ which do you think is the best approximation of $D\text{,}$ the total distance the person traveled on $[0,2]\text{?}$ Why? # Subsection4.2.1Sigma Notation It is apparent from several different problems we have considered that sums of areas of rectangles is one of the main ways to approximate the area under a curve over a given interval. Intuitively, we expect that using a larger number of thinner rectangles will provide a way to improve the estimates we are computing. As such, we anticipate dealing with sums with a large number of terms. To do so, we introduce the use of so-called sigma notation, named for the Greek letter $\Sigma\text{,}$ which is the capital letter $S$ in the Greek alphabet. For example, say we are interested in the sum \begin{equation} 1 + 2 + 3 + \cdots + 100, \end{equation} which is the sum of the first 100 natural numbers. Sigma notation provides a shorthand notation that recognizes the general pattern in the terms of the sum. It is equivalent to write \begin{equation} \sum_{k=1}^{100} k = 1 + 2 + 3 + \cdots + 100. \end{equation} We read the symbol $\sum_{k=1}^{100} k$ as “the sum from $k$ equals 1 to 100 of $k\text{.}$” The variable $k$ is usually called the index of summation, and the letter that is used for this variable is immaterial. Each sum in sigma notation involves a function of the index; for example, \begin{equation} \sum_{k=1}^{10} (k^2 + 2k) = (1^2 + 2\cdot 1) + (2^2 + 2\cdot 2) + (3^2 + 2\cdot 3) + \cdots + (10^2 + 2\cdot 10), \end{equation} and more generally, \begin{equation} \sum_{k=1}^n f(k) = f(1) + f(2) + \cdots + f(n). \end{equation} Sigma notation allows us the flexibility to easily vary the function being used to track the pattern in the sum, as well as to adjust the number of terms in the sum simply by changing the value of $n\text{.}$ We test our understanding of this new notation in the following activity. ##### Activity4.2.2 For each sum written in sigma notation, write the sum long-hand and evaluate the sum to find its value. For each sum written in expanded form, write the sum in sigma notation. 1. $\sum_{k=1}^{5} (k^2 + 2)$ 2. $\sum_{i=3}^{6} (2i-1)$ 3. $3 + 7 + 11 + 15 + \cdots + 27$ 4. $4 + 8 + 16 + 32 + \cdots + 256$ 5. $\sum_{i=1}^{6} \frac{1}{2^i}$ # Subsection4.2.2Riemann Sums When a moving body has a positive velocity function $y = v(t)$ on a given interval $[a,b]\text{,}$ we know that the area under the curve over the interval is the total distance the body travels on $[a,b]\text{.}$ While this is the fundamental motivating force behind our interest in the area bounded by a function, we are also interested more generally in being able to find the exact area bounded by $y = f(x)$ on an interval $[a,b]\text{,}$ regardless of the meaning or context of the function $f\text{.}$ For now, we continue to focus on determining an accurate estimate of this area through the use of a sum of the areas of rectangles, doing so in the setting where $f(x) \ge 0$ on $[a,b]\text{.}$ Throughout, unless otherwise indicated, we also assume that $f$ is continuous on $[a,b]\text{.}$ The first choice we make in any such approximation is the number of rectangles. If we say that the total number of rectangles is $n\text{,}$ and we desire $n$ rectangles of equal width to subdivide the interval $[a,b]\text{,}$ then each rectangle must have width $\Delta x = \frac{b-a}{n}\text{.}$ We observe further that $x_1 = x_0 + \Delta x\text{,}$ $x_2 = x_0 + 2 \Delta x\text{,}$ and thus in general $x_{i} = a + i\Delta x,$ as pictured in Figure 4.2.3. We use each subinterval $[x_i, x_{i+1}]$ as the base of a rectangle, and next must choose how to decide the height of the rectangle that will be used to approximate the area under $y = f(x)$ on the subinterval. There are three standard choices: use the left endpoint of each subinterval, the right endpoint of each subinterval, or the midpoint of each. These are precisely the options encountered in Preview Activity 4.2.1 and seen in Figure 4.2.2. We next explore how these choices can be reflected in sigma notation. If we now consider an arbitrary positive function $f$ on $[a,b]$ with the interval subdivided as shown in Figure 4.2.3, and choose to use left endpoints, then on each interval of the form $[x_{i}, x_{i+1}]\text{,}$ the area of the rectangle formed is given by \begin{equation} A_{i+1} = f(x_i) \cdot \Delta x, \end{equation} as seen in Figure 4.2.4. If we let $L_n$ denote the sum of the areas of rectangles whose heights are given by the function value at each respective left endpoint, then we see that \begin{align} L_n =\mathstrut \amp A_1 + A_2 + \cdots + A_{i+1} + \cdots + A_n\\ =\mathstrut \amp f(x_0) \cdot \Delta x + f(x_1) \cdot \Delta x + \cdots + f(x_i) \cdot \Delta x + \cdots + f(x_{n-1}) \cdot \Delta x. \end{align} In the more compact sigma notation, we have \begin{equation} L_n = \sum_{i = 0}^{n-1} f(x_i) \Delta x. \end{equation} Note particularly that since the index of summation begins at $0$ and ends at $n-1\text{,}$ there are indeed $n$ terms in this sum. We call $L_n$ the left Riemann sum for the function $f$ on the interval $[a,b]\text{.}$ There are now two fundamental issues to explore: the number of rectangles we choose to use and the selection of the pattern by which we identify the height of each rectangle. It is best to explore these choices dynamically, and the applet 1 found at http://gvsu.edu/s/a9 is a particularly useful one. There we see the image shown in Figure 4.2.5, but with the opportunity to adjust the slider bars for the left endpoint and the number of subintervals. By moving the sliders, we can see how the heights of the rectangles change as we consider left endpoints, midpoints, and right endpoints, as well as the impact that a larger number of narrower rectangles has on the approximation of the exact area bounded by the function and the horizontal axis. To see how the Riemann sums for right endpoints and midpoints are constructed, we consider Figure 4.2.6. For the sum with right endpoints, we see that the area of the rectangle on an arbitrary interval $[x_i, x_{i+1}]$ is given by $B_{i+1} = f(x_{i+1}) \cdot \Delta x,$ so that the sum of all such areas of rectangles is given by \begin{align} R_n =\mathstrut \amp B_1 + B_2 + \cdots + B_{i+1} + \cdots + B_n\\ =\mathstrut \amp f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x + \cdots + f(x_{i+1}) \cdot \Delta x + \cdots + f(x_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(x_i) \Delta x. \end{align} We call $R_n$ the right Riemann sum for the function $f$ on the interval $[a,b]\text{.}$ For the sum that uses midpoints, we introduce the notation \begin{equation} \overline{x}_{i+1} = \frac{x_{i} + x_{i+1}}{2} \end{equation} so that $\overline{x}_{i+1}$ is the midpoint of the interval $[x_i, x_{i+1}]\text{.}$ For instance, for the rectangle with area $C_1$ in Figure 4.2.6, we now have \begin{equation} C_1 = f(\overline{x}_1) \cdot \Delta x. \end{equation} Hence, the sum of all the areas of rectangles that use midpoints is \begin{align} M_n =\mathstrut \amp C_1 + C_2 + \cdots + C_{i+1} + \cdots + C_n\\ =\mathstrut \amp f(\overline{x_1}) \cdot \Delta x + f(\overline{x_2}) \cdot \Delta x + \cdots + f(\overline{x}_{i+1}) \cdot \Delta x + \cdots + f(\overline{x}_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(\overline{x}_i) \Delta x, \end{align} and we say that $M_n$ is the middle Riemann sum for $f$ on $[a,b]\text{.}$ When $f(x) \ge 0$ on $[a,b]\text{,}$ each of the Riemann sums $L_n\text{,}$ $R_n\text{,}$ and $M_n$ provides an estimate of the area under the curve $y = f(x)$ over the interval $[a,b]\text{;}$ momentarily, we will discuss the meaning of Riemann sums in the setting when $f$ is sometimes negative. We also recall that in the context of a nonnegative velocity function $y = v(t)\text{,}$ the corresponding Riemann sums are approximating the distance traveled on $[a,b]$ by the moving object with velocity function $v\text{.}$ There is a more general way to think of Riemann sums, and that is to not restrict the choice of where the function is evaluated to determine the respective rectangle heights. That is, rather than saying we'll always choose left endpoints, or always choose midpoints, we simply say that a point $x_{i+1}^$ will be selected at random in the interval $[x_i, x_{i+1}]$ (so that $x_i \le x_{i+1}^ \le x_{i+1}$), which makes the Riemann sum given by \begin{equation} f(x_1^) \cdot \Delta x + f(x_2^) \cdot \Delta x + \cdots + f(x_{i+1}^) \cdot \Delta x + \cdots + f(x_n^) \cdot \Delta x = \sum_{i=1}^{n} f(x_i^) \Delta x. \end{equation} At http://gvsu.edu/s/a9, the applet noted earlier and referenced in Figure 4.2.5, by unchecking the “relative” box at the top left, and instead checking “random,” we can easily explore the effect of using random point locations in subintervals on a given Riemann sum. In computational practice, we most often use $L_n\text{,}$ $R_n\text{,}$ or $M_n\text{,}$ while the random Riemann sum is useful in theoretical discussions. In the following activity, we investigate several different Riemann sums for a particular velocity function. ##### Activity4.2.3 Suppose that an object moving along a straight line path has its velocity in feet per second at time $t$ in seconds given by $v(t) = \frac{2}{9}(t-3)^2 + 2\text{.}$ 1. Carefully sketch the region whose exact area will tell you the value of the distance the object traveled on the time interval $2 \le t \le 5\text{.}$ 2. Estimate the distance traveled on $[2,5]$ by computing $L_4\text{,}$ $R_4\text{,}$ and $M_4\text{.}$ 3. Does averaging $L_4$ and $R_4$ result in the same value as $M_4\text{?}$ If not, what do you think the average of $L_4$ and $R_4$ measures? 4. For this question, think about an arbitrary function $f\text{,}$ rather than the particular function $v$ given above. If $f$ is positive and increasing on $[a,b]\text{,}$ will $L_n$ over-estimate or under-estimate the exact area under $f$ on $[a,b]\text{?}$ Will $R_n$ over- or under-estimate the exact area under $f$ on $[a,b]\text{?}$ Explain. # Subsection4.2.3When the function is sometimes negative For a Riemann sum such as \begin{equation} L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x, \end{equation} we can of course compute the sum even when $f$ takes on negative values. We know that when $f$ is positive on $[a,b]\text{,}$ the corresponding left Riemann sum $L_n$ estimates the area bounded by $f$ and the horizontal axis over the interval. For a function such as the one pictured in Figure 4.2.7, where in the first figure a left Riemann sum is being taken with 12 subintervals over $[a,d]\text{,}$ we observe that the function is negative on the interval $b \le x \le c\text{,}$ and so for the four left endpoints that fall in $[b,c]\text{,}$ the terms $f(x_i) \Delta x$ have negative function values. This means that those four terms in the Riemann sum produce an estimate of the opposite of the area bounded by $y = f(x)$ and the $x$-axis on $[b,c]\text{.}$ In Figure 4.2.7, we also see evidence that by increasing the number of rectangles used in a Riemann sum, it appears that the approximation of the area (or the opposite of the area) bounded by a curve appears to improve. For instance, in the middle graph, we use 24 left rectangles, and from the shaded areas, it appears that we have decreased the error from the approximation that uses 12. When we proceed to Section 4.3, we will discuss the natural idea of letting the number of rectangles in the sum increase without bound. For now, it is most important for us to observe that, in general, any Riemann sum of a continuous function $f$ on an interval $[a,b]$ approximates the difference between the area that lies above the horizontal axis on $[a,b]$ and under $f$ and the area that lies below the horizontal axis on $[a,b]$ and above $f\text{.}$ In the notation of Figure 4.2.7, we may say that \begin{equation} L_{24} \approx A_1 - A_2 + A_3, \end{equation} where $L_{24}$ is the left Riemann sum using 24 subintervals shown in the middle graph, and $A_1$ and $A_3$ are the areas of the regions where $f$ is positive on the interval of interest, while $A_2$ is the area of the region where $f$ is negative. We will also call the quantity $A_1 - A_2 + A_3$ the net signed area bounded by $f$ over the interval $[a,d]\text{,}$ where by the phrase “signed area” we indicate that we are attaching a minus sign to the areas of regions that fall below the horizontal axis. Finally, we recall from the introduction to this present section that in the context where the function $f$ represents the velocity of a moving object, the total sum of the areas bounded by the curve tells us the total distance traveled over the relevant time interval, while the total net signed area bounded by the curve computes the object's change in position on the interval. ##### Activity4.2.4 Suppose that an object moving along a straight line path has its velocity $v$ (in feet per second) at time $t$ (in seconds) given by \begin{equation} v(t) = \frac{1}{2}t^2 - 3t + \frac{7}{2}. \end{equation} 1. Compute $M_5\text{,}$ the middle Riemann sum, for $v$ on the time interval $[1,5]\text{.}$ Be sure to clearly identify the value of $\Delta t$ as well as the locations of $t_0\text{,}$ $t_1\text{,}$ $\cdots\text{,}$ $t_5\text{.}$ In addition, provide a careful sketch of the function and the corresponding rectangles that are being used in the sum. 2. Building on your work in (a), estimate the total change in position of the object on the interval $[1,5]\text{.}$ 3. Building on your work in (a) and (b), estimate the total distance traveled by the object on $[1,5]\text{.}$ 4. Use appropriate computing technology 2 to compute $M_{10}$ and $M_{20}\text{.}$ What exact value do you think the middle sum eventually approaches as $n$ increases without bound? What does that number represent in the physical context of the overall problem? # Subsection4.2.4Summary • A Riemann sum is simply a sum of products of the form $f(x_i^) \Delta x$ that estimates the area between a positive function and the horizontal axis over a given interval. If the function is sometimes negative on the interval, the Riemann sum estimates the difference between the areas that lie above the horizontal axis and those that lie below the axis. • The three most common types of Riemann sums are left, right, and middle sums, plus we can also work with a more general, random Riemann sum. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed in the sum. For a left Riemann sum, we evaluate the function at the left endpoint of each subinterval, while for right and middle sums, we use right endpoints and midpoints, respectively. • The left, right, and middle Riemann sums are denoted $L_n\text{,}$ $R_n\text{,}$ and $M_n\text{,}$ with formulas \begin{align} L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x,\\ R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x,\\ M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x, \end{align} where $x_0 = a\text{,}$ $x_i = a + i\Delta x\text{,}$ and $x_n = b\text{,}$ using $\Delta x = \frac{b-a}{n}\text{.}$ For the midpoint sum, $\overline{x}_{i} = (x_{i-1} + x_i)/2\text{.}$ ##### 4 Consider the function $f(x) = 3x + 4\text{.}$ 1. Compute $M_4$ for $y=f(x)$ on the interval $[2,5]\text{.}$ Be sure to clearly identify the value of $\Delta x\text{,}$ as well as the locations of $x_0, x_1, \ldots, x_4\text{.}$ Include a careful sketch of the function and the corresponding rectangles being used in the sum. 2. Use a familiar geometric formula to determine the exact value of the area of the region bounded by $y = f(x)$ and the $x$-axis on $[2,5]\text{.}$ 3. Explain why the values you computed in (a) and (b) turn out to be the same. Will this be true if we use a number different than $n = 4$ and compute $M_n\text{?}$ Will $L_4$ or $R_4$ have the same value as the exact area of the region found in (b)? 4. Describe the collection of functions $g$ for which it will always be the case that $M_n\text{,}$ regardless of the value of $n\text{,}$ gives the exact net signed area bounded between the function $g$ and the $x$-axis on the interval $[a,b]\text{.}$ ##### 5 Let $S$ be the sum given by \begin{equation} S = ((1.4)^2 + 1) \cdot 0.4 + ((1.8)^2 + 1) \cdot 0.4 + ((2.2)^2 + 1) \cdot 0.4 + ((2.6)^2 + 1) \cdot 0.4 +((3.0)^2 + 1) \cdot 0.4. \end{equation} 1. Assume that $S$ is a right Riemann sum. For what function $f$ and what interval $[a,b]$ is $S$ an approximation of the area under $f$ and above the $x$-axis on $[a,b]\text{?}$ Why? 2. How does your answer to (a) change if $S$ is a left Riemann sum? a middle Riemann sum? 3. Suppose that $S$ really is a right Riemann sum. What is geometric quantity does $S$ approximate? 4. Use sigma notation to write a new sum $R$ that is the right Riemann sum for the same function, but that uses twice as many subintervals as $S\text{.}$ ##### 6 A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. 1. Plot the given data on a set of axes with time on the horizontal axis and the velocity on the vertical axis. 2. Estimate the total distance traveled during the car the time brakes using a middle Riemann sum with 3 subintervals. 3. Estimate the total distance traveled on $[0,1.8]$ by computing $L_6\text{,}$ $R_6\text{,}$ and $\frac{1}{2}(L_6 + R_6)\text{.}$ 4. Assuming that $v(t)$ is always decreasing on $[0,1.8]\text{,}$ what is the maximum possible distance the car traveled before it stopped? Why? ##### 7 The rate at which pollution escapes a scrubbing process at a manufacturing plant increases over time as filters and other technologies become less effective. For this particular example, assume that the rate of pollution (in tons per week) is given by the function $r$ that is pictured in Figure 4.2.9. 1. Use the graph to estimate the value of $M_4$ on the interval $[0,4]\text{.}$ 2. What is the meaning of $M_4$ in terms of the pollution discharged by the plant? 3. Suppose that $r(t) = 0.5 e^{0.5t}\text{.}$ Use this formula for $r$ to compute $L_5$ on $[0,4]\text{.}$ 4. Determine an upper bound on the total amount of pollution that can escape the plant during the pictured four week time period that is accurate within an error of at most one ton of pollution.
Custom Search If you want to discuss more on this or other issues related to Physics, feel free to leave a message on my Facebook page. ## Tuesday, November 24, 2009 ### What is a vector and how to add vector? A vector is an object that has a magnitude, size or length and direction. It is generally represented by an arrow that starts from a initial point (tail) to an end at a terminal point (tip). Here the initial point is A and the terminal point is B. The magnitude is the length of the arrow. The longer the arrow the greater the magnitude of the vector. And the direction of the arrow is the direction of the vector. However in physics the direction of the vector is often represented by an angle θ1 with respect to the horizontal or an angle θ2 with respect to the vertical. A vector is named by using the initial and terminal point as shown in the diagram below. Hence the vector shown is . The vector can also be name by using a single letter that may be bold or using the arrow on the letter as shown in the diagram below. Hence the vector can be named either using B or . We are now going to see how vectors are subtracted, added and multiplied. In order to show you how to add vectors we are going to use vectors that horizontal. Let us have a look at two vectors A and B. Vector A has magnitude 4 and vector B has magnitude 2. Here it is important to note that as vector A has a magnitude twice that of vector B then the length of vector A must be twice that of B. A + B To add vector A and B you place the tail of vector B on the tip of vector A. The result is as shown in the diagram. The two vectors drawn as such is A + B. As you can see the length of the two vectors together is 6. Hence the A + B = 6 What would happen if the two vectors are as shown below. Then A + B is obtained by placing the tail of vector B on the tip of vector A as shown below. The vector that starts from the tail of A to the tip of B is the vector (A + B). But as you can see here the magnitude of the vector A + B must be calculated using Pythagoras theorem. If the two vectors A and B are not at right angle to each other, then the angles a and b that the two vectors make to the vertical or the horizontal must be known as shown in the diagram below. Hence to add these two vectors we just place the tail of vector B on the tip of vector A as shown below. Then the vector that starts from the tail of vector A to the tip of vector B is the vector (A + B) In this example the two vectors A and B have angle a and b relative to the horizontal respectively. Hence using the two angles a and b, the angle between the two vectors can be found and as a result the vector A + B can be found using the cosine rule. #### 1 comment: 1. This is really good you should go further like teach how to use measuring instruments like micrometer screw gauges and vernier calipers,etc good jb thoe.:-)
Right Triangle Trigonometry Worksheets with application, special right triangles trigonometry problems. In this article we have covered wide variety of Right Triangle Trigonometry Worksheets that are suitable for middle schoolers. For each section, we have given the methodology that can be used to solve the problems in the worksheet. Feel free to download and print (for personal use) and try these intuitive trigonometry problems. ### Applications of Right Triangle Trigonometry Worksheet To solve these problems, you will have to first learn the concept of the Pythagorean Theorem and the law of tangents. The Pythagorean Theorem Is about the relationship between the three sides of a right-angle triangle. So, if ABC is a right-angle triangle in which the three sides are AB, BC, and AC, AB2 + BC2 =AC2. Here, AC is the hypotenuse-the longest side of the right angle triangle. You are needed to use the above formula to determine the unknown side of the right-angled triangle when two sides are already given. Law of tangents It is the relationship between any two sides and the two angles opposite to these two sides of a right-angled triangle ABC. Remember, the law of tangent applies to only right-angle triangles. Again tangent of a given angle of a right-angled triangle is the ratio of its opposite side to its adjacent side. So, here tanC=AB/BC Example of finding the length of a side. Calculate the length of x and y of two right-angle triangles in the figure below that share two vertices and one common straight-line base. Here the length of one side AB is 12, D=600 and angle C=300. To solve the above problem, you have to do the following steps. Step 1. For triangle ABD, you already have AB=12, B=900 and D=600 So, tan 600 =opposite side/adjacent side =AB/BD =12/BD =12/x Since it is known from the trigonometry table that tan 600 =√3, you can write √3=12/x So, x=12/√3 =3×4/√3 =√3x√3 x4/√3 =4√3 Step 2 For triangle ABC, you already have AB=12, B=900 and C=300 So, tan 300 =opposite side/adjacent side =AB/BC =12/(BD +DC) =12/(x+y) Since it is known from the trigonometry table that tan 300 =1/√3, you can write 1/√3=12/(4√3 +y) So, (4√3 +y)=12x√3 y =12√3-4√3 =√3(12-4) =8√3 Now since √3=1.732, you have x=4×1.732 =6.928=7 (approximately ) and y=8×1.732 =13.85=14 (approximately) Worksheet for you to try : ### Basic Right Triangle Trigonometry Worksheets To solve these problems, you have to use the law of tangents, sines, and cosines as per the given adjacent side, opposite side and the hypotenuse. Example of finding the angle of a right angle triangle when the adjacent side and the hypotenuse are already given Calculate the angle x of the right-angle triangle in the figure below. Here, the length of one side BC =3 and the hypotenuse AC=6. To solve the above problem, you have to do the following steps. Since you have the length of the adjacent side of the angle C and the hypotenuse, you have to use the law of cosine. =BC/AC =3/6 =1/2 Now, as we know cos 600 =1/2, so , x=600 Example of finding the angle of a right-angle triangle when the opposite side and the hypotenuse are already given Calculate the angle x of the right-angle triangle in the figure below. Here the length of one side BC =14 and the hypotenuse AC=21. To solve the above problem, you have to do the following steps. Since, you have the length of the opposite side of the angle A and the hypotenuse, you have to use the law of sine So, sin x=opposite side/hypotenuse =BC/AC =14/21 =2/3 =0.666 Now as sin 41.80 =0.666, so, x=41.80 Worksheets to try: ### Practice Worksheet – Right Triangle Trigonometry Example of finding sin 45 degrees and cos 45 degrees of the right angle triangle in fraction form when all the three sides are given. Sin 450 =opposite side/hypotenuse = 4/7.2 =40/72 =10/18 =5/9 = 6/7.2 =60/72 =10/12 =5/6 ### Precalculus Right Triangle Trigonometry Worksheet Example A swimmer is 210 meter below the surface of the ocean and begins to descend at an angle of 30 degrees from the vertical. How far will be the swimmer travel before he breaks the surface of the water. You know here Sin 300 =opposite side/hypotenuse 1/2 =210/hypotenuse [As sin30 degree=1/2] So hyotenuse =210 ÷0.5 =420 So, the swimmer travels 420 meter before he breaks the ocean surface. ### Right Triangle Trigonometry – Angle of Elevation and Depression Worksheet To solve the problems in the worksheet, you need to first know the concept of the angle of elevation and the angle of depression. The angle of elevation-The angle formed when an observer looks at an object above his horizontal line of sight. For example, if you stand on a plateau and look at the peak of a nearby mountain, an angle of elevation is formed. The angle of depression-The angle formed when an observer looks down at an object below his horizontal line of sight. For example, if you stand on a plateau and look at a house in the plains, an angle of depression is formed. Example- A bird is sitting on an iceberg 100 feet above the water. If a sea lion in water is 220 feet from the base of the iceberg, find the angle of depression. Let x is the angle of elevation that the sea lion at a distance of 220 meter from the base of the iceberg makes when he looks at the bird sitting on the iceberg at a vertical height of 100 meters. From the above figure, it becomes apparent that tan x0 =adjacent side /opposite side =220/100 =11/5 tan 660=2.2 Now as tan 660=2.2,you have angle of elevation x= 660 Next, we know that the sum of all three angles of a triangle equals to 1800 So, the remaining angle will be =1800 –(660 + 900) = 1800 – 1560=240 Hence, the angle of depression = 900 – 240=660 Example A 25 feet ladder leans against a house so that the base of the ladder is 7 feet from the base of the house. What will be the angle of elevation of the ladder. =7/14 =1/2 [cos 600=1/2] So, angle of elevation of the ladder= 600 ### Special Right Triangle Trigonometry Worksheet Example-Half of an equilateral triangle is often called “30-60” right or “30-60-90” triangle. Explain why it is called with that name? Since, an equilateral triangle can be split into two right angle triangles with the remaining angles being 30 degrees and 60 degrees, half of an equilateral triangle is often called “30-60” right or “30-60-90” triangle. Let us prove this with the help of the law of sine and cosine. Let ABC is an equilateral triangle with sides x, y and z where x=y=z. Now you draw a perpendicular from vertex A on the side BC, you will have two right angle triangles-ABD and ACD with angle D=900. Next, as per the law of cosine, you know in the right-angle triangle ADC =DC/AC =Half of the side z/y =Half of the side y/y [since y=z] =y/2 ÷ y =1/2 Now from the trigonometry table, you know that cos 600=1/2 So, ∠C=600 Similarly, as per the law of sine, you know that in the right-angle triangle ADC Sin A=opposite side/hypotenuse = DC/AC =Half of the side z/y =Half of the side y/y [since y=z] =y/2 ÷ y =1/2 Now from the trigonometry table, you know that sin300=1/2 So, ∠A=300 Hence, you have in the right-angle triangle ACD, A=300, C=600, and D=900 Similarly, in the right-angle triangle ABD, A=300, B=600, and D=900
# 4.3: The Binomial Distribution $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ The vast majority of the content in this book relies on one of five distributions: the binomial distribution, the normal distribution, the t distribution, the F distribution, and the χ2 (“chi-square”) distribution. Let's start with the binomial distribution, since it’s the simplest of the five. The normal distribution will be discussed later this chapter, and the others will be mentioned in other chapters. Our discussion of the binomial distribution will start with figuring out some probabilities of events happening. Let’s imagine a simple “experiment”: in my hot little hand I’m holding 20 identical six-sided dice. On one face of each die there’s a picture of a skull; the other five faces are all blank. If I proceed to roll all 20 dice, what’s the probability that I’ll get exactly 4 skulls? Let's unpack that last sentence. First, in case you missed it, rolling of the 20 dice is your sample. N = 20 (20 scores in this distribution). Second, it's asking about a probability, so we'll be looking back at the last section and realizing that we should do some division. Finally, this is a question, some might even call it a Research Question. Okay, back to the "experiment." Assuming that the dice are fair, we know that the chance of any one die coming up skulls is 1 in 6; to say this another way, the skull probability for a single die is approximately .1667, or 16.67%. $${\dfrac1 6} = 0.166\overline {7} * 100 = 16.67%$$ The probability of rolling a skull with one die is 16.67%, but I'm doing this 20 times so that's $$0.166\overline {7} * 20 = 3.33$$ skulls So, if I rolled 20 dice, I could expect about 3 of them to come up skulls. But that's not quite the Research Question is it? Let’s have a look at how this is related to a binomial distribution. Figure $$\PageIndex{1}$$ plots the binomial probabilities for all possible values for our dice rolling experiment, from X=0 (no skulls) all the way up to X=20 (all skulls). On the horizontal axis we have all the possible events (the number of skulls coming up when all 20 dice are rolled), and on the vertical axis we can read off the probability of each of those events. If you multiple these by 100, you'd get the probability in a percentage form. Each bar depicts the probability of one specific outcome (i.e., one possible value of X). Because this is a probability distribution, each of the probabilities must be a number between 0 and 1 (or 0% to 100%), and the heights of the all of the bars together must sum to 1 as well. Looking at Figure $$\PageIndex{1}$$, the probability of rolling 4 skulls out of 20 times is about 0.20 (the actual answer is 0.2022036), or 20.22%. In other words, you’d expect to roll exactly 4 skulls about 20% of the times you repeated this experiment. This means that, if you rolled these 20 dice for 100 different repetitions, you'd get exactly 4 skulls in about 20 of your attempts. ## Sample Size Matters We'll be talking a lot about how sample size (N) affects distributions in this chapter, starting now! To give you a feel for how the binomial distribution changes when we alter the probability and N, let’s suppose that instead of rolling dice, I’m actually flipping coins. This time around, my experiment involves flipping a fair coin repeatedly, and the outcome that I’m interested in is the number of heads that I observe. In this scenario, the success probability is now 1/2 (one out of two options). Suppose I were to flip the coin N=20 times. In this example, I’ve changed the success probability (1 out of 2, instead of 1 out of 6 in the dice example), but kept the size of the experiment the same (N=20). What does this do to our binomial distribution? Well, as Figure $$\PageIndex{2}$$ shows, the main effect of this is to shift the whole distribution higher (since there's more chance of getting a heads (one out of two options) than a 4 (one out of six options). Okay, what if we flipped a coin N=100 times? Well, in that case, we get Figure $$\PageIndex{3}$$. The distribution stays roughly in the middle, but there’s a bit more variability in the possible outcomes (meaning that there are more extreme scores); with more tosses, you are more likely to get no heads but also more likely to get all heads than if you only flipped the coin 20 times. And that's it on binomial distributions! We are building understanding about distributions and sample size, so nothing too earth-shattering here. One thing to note is that both the coin flip and the dice toss were measured as discrete variables. Our next distributions will cover continuous variables. The difference is that discrete variables can be one or the other (a heads or a tails, a four or not-four), while continuous variables can have gradations shown as decimal points. Although the math works out to 3.33 skulls when you throw 30 dice, you can't actually get 3.33 skulls, you can only get 3 or 4 skulls. We're building our knowledge, so keep going! This page titled 4.3: The Binomial Distribution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Michelle Oja.
# Algebra Word Problem by Abby Hope (Henderson, NC) The product of two fractions is 1/9. The larger fraction divided by the smaller fraction is 4. What is the sum of the two fractions? Express your answer as a common fraction. ### Comments for Algebra Word Problem Average Rating Sep 29, 2010 Rating Algebra Word Problem by: Karin Hi Abbey, This is a great problem! Let's first identify the variables. You know that you are dealing with 2 numbers and one is larger than the other. Let x = the larger number Let y = the smaller number Now we can write two equations: The product of the two numbers is 1/9. Product means multiply. xy = 1/9 The larger number divided by the smaller number is 4. x/y = 4 So, we have: xy = 1/9 and x/y = 4 Since we have two equations, we actually have a system of equations. We need to solve the system. I am going to use the substitution method. Step 1: Solve one of the equations for a variable. I will solve x/y = 4 for x. Multiply both sides by y. y(x/y)= 4y x = 4y Now, we'll substitute 4y for x into the equation: xy = 1/9 4y(y) = 1/9 4y^2 = 1/9 Now divide by 4 on both sides. 4y^2/4 = 1/9 / 4 y^2 = 1/36 Now take the square root of both sides and you end up with: y = 1/6 Now substitute 1/6 for y into the equation x = 4y x = 4y x = 4(1/6) x = 4/6 x = 2/3 So, the two numbers are 1/6, & 2/3. Now let's check before we finish the problem: xy = 1/9 (1/6)(2/3) = 1/9 2/18 = 1/9 1/9 = 1/9 AND x/y = 4 2/3 / 1/6 = 4 2/3 * 6/1 = 4 12/3 = 4 4 = 4 Yes, our answers are correct. So, now the question asked for the sum of the numbers, so we must add the numbers together. 1/6 + 2/3 = 5/6 The correct answer is 5/6 Hope this helps, Karin xy = 1/9 (1/6)(2/3) = Need More Help With Your Algebra Studies?
Paul's Online Notes Home / Calculus I / Limits / Continuity Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. ### Section 2-9 : Continuity 1. The graph of $$f\left( x \right)$$ is given below. Based on this graph determine where the function is discontinuous. Solution 2. The graph of $$f\left( x \right)$$ is given below. Based on this graph determine where the function is discontinuous. Solution For problems 3 – 7 using only Properties 1 – 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the given function is continuous or discontinuous at the indicated points. 1. $$\displaystyle f\left( x \right) = \frac{{4x + 5}}{{9 - 3x}}$$ 1. $$x = - 1$$ 2. $$x = 0$$ 3. $$x = 3$$ Solution 2. $$\displaystyle g\left( z \right) = \frac{6}{{{z^2} - 3z - 10}}$$ 1. $$z = - 2$$ 2. $$z = 0$$ 3. $$z = 5$$ Solution 3. $$g\left( x \right) = \left\{ {\begin{array}{rl}{2x}&{x < 6}\\{x - 1}&{x \ge 6}\end{array}} \right.$$ 1. $$x = 4$$ 2. $$x = 6$$ Solution 4. $$h\left( t \right) = \left\{ {\begin{array}{rl}{{t^2}}&{t < - 2}\\{t + 6}&{t \ge - 2}\end{array}} \right.$$ 1. $$t = - 2$$ 2. $$t = 10$$ Solution 5. $$g\left( x \right) = \left\{ {\begin{array}{rc}{1 - 3x}&{x < - 6}\\7&{x = - 6}\\{{x^3}}&{ - 6 < x < 1}\\1&{x = 1}\\{2 - x}&{x > 1}\end{array}} \right.$$ 1. $$x = - 6$$ 2. $$x = 1$$ Solution For problems 8 – 12 determine where the given function is discontinuous. 1. $$\displaystyle f\left( x \right) = \frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}$$ Solution 2. $$\displaystyle R\left( t \right) = \frac{{8t}}{{{t^2} - 9t - 1}}$$ Solution 3. $$\displaystyle h\left( z \right) = \frac{1}{{2 - 4\cos \left( {3z} \right)}}$$ Solution 4. $$\displaystyle y\left( x \right) = \frac{x}{{7 - {{\bf{e}}^{2x + 3}}}}$$ Solution 5. $$g\left( x \right) = \tan \left( {2x} \right)$$ Solution For problems 13 – 15 use the Intermediate Value Theorem to show that the given equation has at least one solution in the indicated interval. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval. 1. $$25 - 8{x^2} - {x^3} = 0$$ on $$\left[ { - 2,4} \right]$$ Solution 2. $${w^2} - 4\ln \left( {5w + 2} \right) = 0$$ on $$\left[ {0,4} \right]$$ Solution 3. $$4t + 10{{\bf{e}}^t} - {{\bf{e}}^{2t}} = 0$$ on $$\left[ {1,3} \right]$$ Solution
Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line Find the coordinates of the foot of the perpendicular drawn from the point to the line . Also, find the length of the perpendicular from the given point to the line. Given: Equation of line is . To find: coordinates of foot of the perpendicular from to the line. And find the length of the perpendicular. Formula Used: 1. Equation of a line is Cartesian form: where is a point on the line and is the direction ratios of the line. 2. Distance between two points and is $\sqrt{{\left({\mathrm{x}}_{1}–{\mathrm{x}}_{2}\right)}^{2}+{\left({\mathrm{y}}_{1}–{\mathrm{y}}_{2}\right)}^{2}+{\left({\mathrm{z}}_{1}–{\mathrm{z}}_{2}\right)}^{2}}$ \sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}+\left(\mathrm{z}_{1}-\mathrm{z}_{2}\right)^{2}} Explanation: Let $\frac{x–6}{\mathrm{\not}3}=\frac{y–7}{3}=\frac{z–7}{–3}=\lambda$ \frac{x-6}{\not 3}=\frac{y-7}{3}=\frac{z-7}{-3}=\lambda So the foot of the perpendicular is Direction ratio of the line is Direction ratio of the perpendicular is $\begin{array}{l}⇒\left(3\lambda +6–1\right):\left(2\lambda +7–2\right):\left(–2\lambda +7–3\right)\\ ⇒\left(3\lambda +5\right):\left(2\lambda +5\right):\left(–2\lambda +4\right)\end{array}$ \begin{array}{l} \Rightarrow(3 \lambda+6-1):(2 \lambda+7-2):(-2 \lambda+7-3) \\ \Rightarrow(3 \lambda+5):(2 \lambda+5):(-2 \lambda+4) \end{array} Since this is perpendicular to the line, $\begin{array}{l}3\left(3\lambda +5\right)+2\left(2\lambda +5\right)–2\left(–2\lambda +4\right)=0\\ ⇒9\lambda +15+4\lambda +10+4\lambda –8=0\\ ⇒17\lambda =–17\\ ⇒\lambda =–1\end{array}$ \begin{array}{l} 3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4)=0 \\ \Rightarrow 9 \lambda+15+4 \lambda+10+4 \lambda-8=0 \\ \Rightarrow 17 \lambda=-17 \\ \Rightarrow \lambda=-1 \end{array} So the foot of the perpendicular is \begin{array}{l} \text { Distance }=\sqrt{(3-1)^{2}+(5-2)^{2}+(9-3)^{2}} \\ =\sqrt{4+9+36} \\ =7 \text { units } \end{array}
# SUM OF ALL 3 DIGIT NUMBERS DIVISIBLE BY 6 To get the sum of 3 digit numbers divisible by 6, first we have to find the first and last 3 digit numbers divisible by 6. ## First 3 Digit Number Exactly Divisible by 6 The first and also the smallest 3 digit number is 100. To find the first 3 digit number divisible by 6, we have to divide the very first 3 digit number 100 by 6 100/6 = 16.67 We have decimal in the result of 100/6. Clearly the first 3 digit number 100 is not exactly divisible by 6 Let us divide the second 3 digit number 101 by 6 101/6 = 16.83 We have decimal in the result of 101/6 also. So, the second 3 digit number 101 is also not exactly divisible by 6 Here, students may have some questions on the above process. They are, 1. Do we have to divide the 3 digit numbers by 6 starting from 100 until we get a 3 digit number which is exactly divisible by 6 ? 2. Will it not take a long process ? 3. Is there any shortcut instead of dividing the 3 digit numbers 100, 101, 102.... one by one ? There is only one answer for all the above three questions. That is, there is a shortcut to find the first three digit number which is exactly divisible by 6. SHORTCUT What has been done in the above shortcut ? The process which has been done in the above shortcut has been explained clearly in the following steps. Step 1 : To get the first 3 digit number divisible by 6, we have to take the very first 3 digit number 100 and divide it by 6. Step 2 : When we divide 100 by 6 using long division as given above, we get the remainder 4. Step 3 : Now, the remainder 4 has to be subtracted from the divisor 6. When we subtract the remainder 4 from the divisor 6, we get the result 2 (That is 6 - 4 = 2). Step 4 : Now, the result 2 in step 3 to be added to the dividend 100. When we add 2 to 100, we get 102 Now, the process is over. So, 102 is the first 3 digit number exactly divisible by 6 This is how we have to find the first 3 digit number exactly divisible by 6 Important Note : This method is not only applicable to find the first 3 digit number exactly divisible by 6. It can be applied to find the first 3 digit number exactly divisible by any number, say k. ## Last 3 Digit Number Exactly Divisible by 6 The last and also the largest 3 digit number is 999. To find the last 3 digit number divisible by 6, we divide the very last 3 digit number 999 by 6. 999/6 = 166.5 We have decimal in the result of 999/6. Clearly the last 3 digit number 999 is not exactly divisible by 6. Let us divide the preceding 3 digit number 998 by 6. 998/6 = 166.33 We have decimal in the result of 998/6 also. So, the preceding 3 digit number 998 also is not exactly divisible by 6 Here, students may have some questions on the above process. They are, 1. Do we have to divide the 3 digit numbers .........997, 998, 999 by 6 until we get a 3 digit number which is exactly divisible by 6 ? 2. Will it not take a long process ? 3. Is there any shortcut instead of dividing the 3 digit numbers ...........997, 998, 999 one by one ? There is only one answer for all the above three questions. That is, there is a shortcut to find the last three digit number which is exactly divisible by 6. SHORTCUT What has been done in the above shortcut ? The process which has been done in the above shortcut has been explained clearly in the following steps. Step 1 : To get the last 3 digit number divisible by 6, we have to take the very last 3 digit number 999 and divide it by 6. Step 2 : When we divide 999 by 6 using long division as given above, we get the remainder 3. Step 3 : Now, the remainder 3 has to be subtracted from the dividend 999. When we subtract the remainder 3 from the dividend 999, we get the result 996 (That is 999 - 3 = 996). Now, the process is over. So, 996 is the last 3 digit number exactly divisible by 6. This is how we have to find the last 3 digit number exactly divisible by 6. Important Note : The process of finding the first 3 digit number exactly divisible by 6 and the process of finding the last 3 digit number exactly divisible by 6 are completely different. Be careful! Both are not same. The methods explained above are not only applicable to find the first 3 digit number and last 3 digit number exactly divisible by 6. They can be applied to find the first 3 digit number and last 3 digit number exactly divisible by any number, say k. ## Sum of all 3 Digit Numbers Divisible by 6 Let us see how to find the sum of all 3 digit numbers divisible by 6 in the following steps. Step 1 : The first 3 digit number divisible by 6 is 102. After 102, to find the next 3 digit number divisible by 6, we have to add 6 to 102. So the second 3 digit number divisible by 6 is 108. In this way, to get the succeeding 3 digit numbers divisible by 6, we just have to add 6 as given below. 102, 108, 114, 120,.......................996 Clearly, the above sequence of 3 digit numbers divisible by 6 forms an arithmetic sequence. And our aim is to find the sum of the terms in the above arithmetic sequence. Step 2 : In the arithmetic sequence 102, 108, 114, 120,.......................996 we have first term = 102 common difference = 6 last term = 996 That is, a = 102 d = 6 l = 996 Step 3 : The formula to find the numbers of terms in an arithmetic sequence is given by n = [(l - a) / d] + 1 Substitute a = 102, l = 996 and d = 6. n = [(996 - 102) / 6] + 1 n = [894/6] + 1 n = 149 + 1 n = 150 So, number of 3 digit numbers divisible by 6 is 150. Step 4 : The formula to find the sum of 'n' terms in an arithmetic sequence is given by = (n/2)(a + l) Substitute a = 102, d = 6, l = 996 and n = 150. = (150/2)(102 + 996) = 75 x 1098 = 82350 So, the sum of all 3 digit numbers divisible by 6 is 82350. Note : The method explained above is not only applicable to find the sum of all 3 digit numbers divisible by 6. This same method can be applied to find sum of all 3 digit numbers divisible by any number, say k. Kindly mail your feedback to v4formath@gmail.com ## Recent Articles 1. ### Equation of Tangent Line to Inverse Function May 30, 23 11:19 AM Equation of Tangent Line to Inverse Function 2. ### Derivative of Inverse Functions May 30, 23 10:38 AM Derivative of Inverse Functions May 26, 23 12:27 PM Adaptive Learning Platforms: Personalized Mathematics Instruction with Technology
# Math in Focus Grade 2 Chapter 19 Practice 1 Answer Key Plane Shapes This handy Math in Focus Grade 2 Workbook Answer Key Chapter 19 Practice 1 Plane Shapes detailed solutions for the textbook questions. ## Math in Focus Grade 2 Chapter 19 Practice 1 Answer Key Plane Shapes Look at the shapes. Question 1. Color the circles green, the triangles yellow, the rectangles purple, the trapezoids blue, the hexagons red, and the pentagons orange. Explanation: I colored the circles green, the triangles yellow, the rectangles purple, the trapezoids blue, the hexagons red, and the pentagons orange. There are 2 circles, 2 rectangles, 2 hexagons, 3 trapezoids, 2 triangles and 1 pentagon. Question 2. Which of these are quadrilaterals? Write the numbers: _____________ 2, 4, 10 and 12 are quadrilaterals. Draw lines on each shape to show the smaller shapes. Example Question 2. Explanation: I drew 2 lines to make smaller shapes in the given rectangle. Question 3. Explanation: I drew 2 lines to make smaller shapes in the given square. Question 4. Cut out the shapes. Then glue them on top of the shape given. Here are two simple rules to follow: a) All cut-outs must be used. b) Cut-outs cannot overlap. Question 5. Draw lines on each figure to show how it is made with these shapes: triangle, square, rectangle, trapezoid, hexagon, and pentagon. Explanation: I drew lines on each figure to show how it is made with these shapes: triangle, square, rectangle, trapezoid, hexagon, and pentagon. Each figure is made with two shapes. Name the shapes. Example The figure is made with a triangle and a rectangle. Question 6. The figure is made with a ___________ and a ___________. The figure is made with a trapezium a and a rectangle. Question 7. The figure is made with a ___________ and a ___________. The figure is made with a hexagon and a circle. Question 8. The figure is made with a ___________ and a ___________. The figure is made with a pentagon and a rectangle. A part is missing from each figure. Color the shape that makes the figure complete. Example Question 9. Explanation: A Hexagon is made up of a square and a rectangle. A part is missing from each figure. Color the shape that makes the figure complete. Question 10. Explanation: The missing part in the above shape is a triangle So, i colored the missing part. Question 11. Explanation: A square is always a trapezoid So, i colored square. Question 12. Explanation: To make a pentagon we need to join a triangle to a quadrilateral So, i colored triangle. Question 13. Cut out the shapes. Then glue them on top of the figure. Here are two simple rules to follow. (a) All cut-outs must be used. (b) Cut-outs cannot overlap. Copy each figure. Example Question 14. Explanation: I drew the same shape as shown in the first grid. It is a pentagon. Copy each figure. Example Question 15. Explanation: I drew the same shape as shown in the first grid. It is a Hexagon. Draw a quadrilateral on the dot grid paper. Circle the angles. Write how many angles. Question 16. Explanation: I drew a rectangle and circled all the angles A rectangle has 4 angles. Draw a shape with three angles on the dot grid paper. Question 17.
or Find what you need to study Light # 5.7 Using the Second Derivative Test to Determine Extrema You’ve probably noticed by now that Unit 5 deals with analytical applications of differentiation; that means that a function’s derivatives can tell us something about its behaviors. We learned from 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema that the first derivative of a function tells us information on where its relative extrema (aka local maximum and minimum) will be. What does the second derivative tell us, then? Answer: Whether a critical point we found yields a local maximum or minimum (it can’t be both)! 🧠 ## ✏️ Warm-up: Finding Critical Points Let’s do a quick recap of critical points from key topic 5.2! To refresh, a critical point is a point on a function where… • the first derivative equals zero or • fails to exist Analytically, the first bullet point is easier to wrap your head around. Let’s take a look at a quick example: $f(x)=\frac{2}{3}x^3-\frac{5}{2}x^2-3x$ Computing the first derivative gives us: $f'(x)=2x^{2}-5x-3$ To find where the first derivative equals zero, we can factor and then set f’(x) = 0: $0=2x^2-5x-3$ $0=(2x+1)(x-3)$ Looks familiar, right? We proceed as if we’re finding the roots (x) in a regular equation. In this case, the two x values are our critical points! $x=-\frac{1}{2}, x=3$ Not bad, right? What do we do with these critical point values, then? We learned two things from 5.6 Determining Concavity of Functions over Their Domains about what the second derivative of a function can tell us information of: 1. 🌈 The function’s intervals of upward or downward concavity. 1. $f’’(x)$ is positive (or greater than zero, aka $f’’(x)>0$) = concave up 2. $f’’(x)$ is negative (or less than zero, aka $f’’(x)<0$) = concave down 2. 🪝 The function graph’s inflection points (where concavity changes). ## 🥈 Extending the Second Derivative Test Now, let’s connect these ideas to the critical points we mentioned earlier: by knowing the concavities before and after the critical points, we can determine where our local minima and maxima are! 🗺️ ### 🪜 Second Derivative Test Steps Here are some steps that we’ll go through: 1. Find the critical points of $f(x)$ using $f'(x)$. 2. Plug the critical points into $f''(x)$ to get a y-value. 3. Determine whether you have a local min or max! ### 📝 Second Derivative Test Walkthrough To illustrate, let’s continue with our example while we walk through the steps pertaining to the Second Derivative Test. #### 🥇 Step 1: Find f(x)’s critical points using f’(x) Done! We found our critical points: $x = -0.5$ and $x = 3$ earlier. #### 🥈 Step 2: Plug critical points into f’’(x) to get a y-value. From above, we found $f’(x)$ to be: $f'(x)=2x^2-5x-3$ Calculating the second derivative: $f''(x)=4x-5$ Then, we plug our critical points: $f''(-\frac{1}{2})=4(-\frac{1}{2})-5=-7$ $f''(3)=4(3)-5=7$ These numbers give us the y-values for our local extrema and thus the whole coordinates: $(-0.5, -7)$ and $(3, 7)$. We’re ready for our final step… #### 🥉 Step 3: Determine whether you have a local min or max! In our case, $(-0.5, -7)$ is a local maximum and $(3, 7)$ is a local minimum. This is because of the work we did in step 2! Conceptually, why does this make sense? Tap into your imagination… imagine a hill and a bowl! • 🗻 If a point represents a maximum and is situated at the peak of a hill, the point appears to lie on the concave-down surface of said hill. • 🥣 Conversely, if a point signifies a minimum and is located at the bottom of an upright bowl, it suggests that the point lies on the concave-up surface of the bowl. Therefore, when the function f is concave down at a critical point (we’ll call it “c”), it implies a maximum, whereas f being concave up at “c” indicates a minimum. Feel more confident? Try the following examples on your own before looking at the solutions to see which areas you should do more practice on! 📣 ## 💭 Second Derivative Test: Let’s Practice! For each of the following functions f(x), determine whether each of their critical points are local maxima or minima. ### Second Derivative Test: Example 1 $f(x)=4\sin(x), 0 Let’s find our first and second derivatives: $f'(x)=4\cos(x)$ $f''(x)=-4\sin(x)$ Setting $f’(x) = 0$ to find critical points x: $0=4\cos(x)$ $0=\cos(x)$ Thinking back to the Unit Circle from trigonometry, at which angles $x$ will $cos(x) = 0$? We’ll get our critical points: $x=\frac{\pi}{2}, x=\frac{3\pi}{2}$ Plugging these values into our second derivative: $f''(\frac{\pi}{2})=-4\sin(\frac{\pi}{2})=-41=-4$ $f''(\frac{3\pi}{2})=-4\sin(\frac{3\pi}{2})=-4-1=4$ Since $-4 < 0$, $(\frac{\pi}{2}, -4)$ is a local maximum of $f(x)$. On the other hand, since $4 > 0$, $(\frac{3\pi}{2}, 4)$ is a local minimum of f(x). We’re done! ⭐ ### Second Derivative Test: Example 2 $f(x)=247ln(x^2)$ Calculating f’(x) and f’’(x) gives us: $f'(x)=\frac{247}{x}$ $f''(x)=-\frac{247}{x^2}$ Setting f’(x) = 0… $0=\frac{247}{x}$ Wait a second. This doesn’t give us a solution. What do we do in this case, then? This brings up another element of the Second Derivative Test: ## 💫 Closing Great work! Now we applied the second derivative test to different functions to determine extrema. There’s one key point that we want to leave you with… When dealing with a continuous function that has just one critical point across its entire domain and this particular point serves as a local extremum within a specific interval, it also stands as the global extremum for the function within that particular interval. This is because there are no other critical points to challenge its extremum status within that specific interval. This is due to the function having only one peak or valley (extremum) within that interval. 🗻 # Key Terms to Review (8) Critical Point : A critical point is a point on a function where the derivative is either zero or undefined. It represents a potential maximum, minimum, or inflection point. First Derivative : The first derivative represents the rate at which a function is changing at any given point. It measures the slope of the tangent line to the graph of a function at that point. Local Maximum : A local maximum refers to the highest point of a function within a specific interval. It is higher than all nearby points but may not be higher than all other points on the entire function. Local Minimum : A local minimum refers to the lowest point of a function within a specific interval. It is lower than all nearby points but may not be lower than all other points on the entire function. Minima : Minima refers to the lowest points on a graph or the smallest values of a function. These points represent the minimum values of a function. Second Derivative : The second derivative of a function represents the rate at which the slope of the original function is changing. It measures how the rate of change of the first derivative changes. Second Derivative Test : The second derivative test is used to determine whether critical points correspond to local maxima, minima, or neither. It involves analyzing the concavity of a function at those critical points. Twice Differentiable : A function is said to be twice differentiable if it has both a first and second derivative defined for all points in its domain. # 5.7 Using the Second Derivative Test to Determine Extrema You’ve probably noticed by now that Unit 5 deals with analytical applications of differentiation; that means that a function’s derivatives can tell us something about its behaviors. We learned from 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema that the first derivative of a function tells us information on where its relative extrema (aka local maximum and minimum) will be. What does the second derivative tell us, then? Answer: Whether a critical point we found yields a local maximum or minimum (it can’t be both)! 🧠 ## ✏️ Warm-up: Finding Critical Points Let’s do a quick recap of critical points from key topic 5.2! To refresh, a critical point is a point on a function where… • the first derivative equals zero or • fails to exist Analytically, the first bullet point is easier to wrap your head around. Let’s take a look at a quick example: $f(x)=\frac{2}{3}x^3-\frac{5}{2}x^2-3x$ Computing the first derivative gives us: $f'(x)=2x^{2}-5x-3$ To find where the first derivative equals zero, we can factor and then set f’(x) = 0: $0=2x^2-5x-3$ $0=(2x+1)(x-3)$ Looks familiar, right? We proceed as if we’re finding the roots (x) in a regular equation. In this case, the two x values are our critical points! $x=-\frac{1}{2}, x=3$ Not bad, right? What do we do with these critical point values, then? We learned two things from 5.6 Determining Concavity of Functions over Their Domains about what the second derivative of a function can tell us information of: 1. 🌈 The function’s intervals of upward or downward concavity. 1. $f’’(x)$ is positive (or greater than zero, aka $f’’(x)>0$) = concave up 2. $f’’(x)$ is negative (or less than zero, aka $f’’(x)<0$) = concave down 2. 🪝 The function graph’s inflection points (where concavity changes). ## 🥈 Extending the Second Derivative Test Now, let’s connect these ideas to the critical points we mentioned earlier: by knowing the concavities before and after the critical points, we can determine where our local minima and maxima are! 🗺️ ### 🪜 Second Derivative Test Steps Here are some steps that we’ll go through: 1. Find the critical points of $f(x)$ using $f'(x)$. 2. Plug the critical points into $f''(x)$ to get a y-value. 3. Determine whether you have a local min or max! ### 📝 Second Derivative Test Walkthrough To illustrate, let’s continue with our example while we walk through the steps pertaining to the Second Derivative Test. #### 🥇 Step 1: Find f(x)’s critical points using f’(x) Done! We found our critical points: $x = -0.5$ and $x = 3$ earlier. #### 🥈 Step 2: Plug critical points into f’’(x) to get a y-value. From above, we found $f’(x)$ to be: $f'(x)=2x^2-5x-3$ Calculating the second derivative: $f''(x)=4x-5$ Then, we plug our critical points: $f''(-\frac{1}{2})=4(-\frac{1}{2})-5=-7$ $f''(3)=4(3)-5=7$ These numbers give us the y-values for our local extrema and thus the whole coordinates: $(-0.5, -7)$ and $(3, 7)$. We’re ready for our final step… #### 🥉 Step 3: Determine whether you have a local min or max! In our case, $(-0.5, -7)$ is a local maximum and $(3, 7)$ is a local minimum. This is because of the work we did in step 2! Conceptually, why does this make sense? Tap into your imagination… imagine a hill and a bowl! • 🗻 If a point represents a maximum and is situated at the peak of a hill, the point appears to lie on the concave-down surface of said hill. • 🥣 Conversely, if a point signifies a minimum and is located at the bottom of an upright bowl, it suggests that the point lies on the concave-up surface of the bowl. Therefore, when the function f is concave down at a critical point (we’ll call it “c”), it implies a maximum, whereas f being concave up at “c” indicates a minimum. Feel more confident? Try the following examples on your own before looking at the solutions to see which areas you should do more practice on! 📣 ## 💭 Second Derivative Test: Let’s Practice! For each of the following functions f(x), determine whether each of their critical points are local maxima or minima. ### Second Derivative Test: Example 1 $f(x)=4\sin(x), 0 Let’s find our first and second derivatives: $f'(x)=4\cos(x)$ $f''(x)=-4\sin(x)$ Setting $f’(x) = 0$ to find critical points x: $0=4\cos(x)$ $0=\cos(x)$ Thinking back to the Unit Circle from trigonometry, at which angles $x$ will $cos(x) = 0$? We’ll get our critical points: $x=\frac{\pi}{2}, x=\frac{3\pi}{2}$ Plugging these values into our second derivative: $f''(\frac{\pi}{2})=-4\sin(\frac{\pi}{2})=-41=-4$ $f''(\frac{3\pi}{2})=-4\sin(\frac{3\pi}{2})=-4-1=4$ Since $-4 < 0$, $(\frac{\pi}{2}, -4)$ is a local maximum of $f(x)$. On the other hand, since $4 > 0$, $(\frac{3\pi}{2}, 4)$ is a local minimum of f(x). We’re done! ⭐ ### Second Derivative Test: Example 2 $f(x)=247ln(x^2)$ Calculating f’(x) and f’’(x) gives us: $f'(x)=\frac{247}{x}$ $f''(x)=-\frac{247}{x^2}$ Setting f’(x) = 0… $0=\frac{247}{x}$ Wait a second. This doesn’t give us a solution. What do we do in this case, then? This brings up another element of the Second Derivative Test: ## 💫 Closing Great work! Now we applied the second derivative test to different functions to determine extrema. There’s one key point that we want to leave you with… When dealing with a continuous function that has just one critical point across its entire domain and this particular point serves as a local extremum within a specific interval, it also stands as the global extremum for the function within that particular interval. This is because there are no other critical points to challenge its extremum status within that specific interval. This is due to the function having only one peak or valley (extremum) within that interval. 🗻 # Key Terms to Review (8) Critical Point : A critical point is a point on a function where the derivative is either zero or undefined. It represents a potential maximum, minimum, or inflection point. First Derivative : The first derivative represents the rate at which a function is changing at any given point. It measures the slope of the tangent line to the graph of a function at that point. Local Maximum : A local maximum refers to the highest point of a function within a specific interval. It is higher than all nearby points but may not be higher than all other points on the entire function. Local Minimum : A local minimum refers to the lowest point of a function within a specific interval. It is lower than all nearby points but may not be lower than all other points on the entire function. Minima : Minima refers to the lowest points on a graph or the smallest values of a function. These points represent the minimum values of a function. Second Derivative : The second derivative of a function represents the rate at which the slope of the original function is changing. It measures how the rate of change of the first derivative changes. Second Derivative Test : The second derivative test is used to determine whether critical points correspond to local maxima, minima, or neither. It involves analyzing the concavity of a function at those critical points. Twice Differentiable : A function is said to be twice differentiable if it has both a first and second derivative defined for all points in its domain. ###### Stay Connected AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
How do you write 2.5xx10^7 in expanded form? 1 Answer Sep 27, 2016 $2.5 \times {10}^{7} = 25000000.0$ Explanation: In scientific notation, we write a number so that it has single digit to the left of decimal sign and is multiplied by an integer power of $10$. In other words, in scientific notation, a number is written as $a \times {10}^{n}$, where $1 \le a < 10$ and $n$ is an integer and $1 \le a < 10$. To write the number in normal or standard notation one just needs to multiply by the power ${10}^{n}$ (or divide if $n$ is negative). This means moving decimal $n$ digits to right if multiplying by ${10}^{n}$ and moving decimal $n$ digits to left if dividing by ${10}^{n}$ (i.e. multiplying by ${10}^{- n}$). In the given case, as we have the number as $2.5 \times {10}^{7}$, we need to move decimal digit to the right by seven points. For this, let us write $2.5$ as $2.50000000$ and moving decimal point three points to right means $25000000.0$ Hence in standard notation $2.5 \times {10}^{7} = 25000000.0$
# Mathematics: Algebraic Reasoning Study Guide for the TSIA2 Page 1 ## General Information Both of the TSIA2 tests, the CRC and the Diagnostic Test, contain a math section with questions covering these topics: • Quantitative Reasoning • Algebraic Reasoning • Geometric and Spatial Reasoning • Probabilistic and Statistical Reasoning The skills tested are the same on both tests and you won’t know if you’ll need to take the Diagnostic Test until after you complete the CRC. So, it’s a great idea to go ahead and prepare as if you’ll need to take both of them. On the CRC, you’ll find seven questions related to algebraic reasoning, whereas there will be 12 of them on the Diagnostic Test. You’ll find an outline of all of the algebra skills covered on both tests in this study guide, as well as some information about each. We also provide practice questions and flashcards to help you study, and encourage you to seek additional assistance for any problem areas. Be sure to also use our other three study guides for the TSIA2 Mathematics tests as you prepare. The only question type on these math tests is multiple-choice. ## Solving Equations and Inequalities Keep in mind that the term solving means finding values that, when substituted for variables, make the statement (equation or inequality) true. The procedures are similar for both equations and inequalities, with the exception of multiplying or dividing both sides by a negative number. In an inequality, these operations require you to change the direction of the inequality sign. ### Linear Equations Linear equations result in a straight line when they are graphed. Linear equations can be written in several forms, and can be converted from one form to another with algebra. One common form is $$y=mx + b$$, where b is the y-intercept (value of y where the line crosses the y-axis) and m is the slope of the line (how much the y value changes for a given change in x). Other forms reveal other characteristics of the line. #### Word Problems and Linear Equations Many real-life situations can be described with linear models. Often, the information given in a problem can be used to form an equation that describes the situation. This is the model. Usually, the equation can be solved to find the required solution. For example, suppose a car is traveling at a rate of $$50$$ miles per hour. How long does it take to travel $$120$$ miles? If $$y$$ is miles traveled, and $$x$$ is the hours traveled, then the model is: $y = 50x$ In this case we know that $$y$$ is 120 miles, so: $120 = 50x$ To solve for $$x$$, we have to divide both sides by 50, so: $x = 120 \div 50$ or $$2.4$$ hours. #### General Solving Procedure Solving an equation for a variable means to find an equivalent equation with that variable alone on one side. Do this by adding, subtracting, multiplying, or dividing both sides of the equation by the same quantity. #### Example Solve $$4x + 2y = 8$$ for $$y$$: $4x + 2y = 8$ subtract 4x from both sides $2y = 8 - 4x$ divide both sides by 2 $y = 4 - 2x$ re-order to put it in the form $$y \;=mx \;+\; b$$ $y = -2x + 4$ ### Inequalities An inequality is a statement of comparison between quantities that is not an equality. There are four possible ways to state an inequality. $$a<b$$ $$\quad a$$ is less than $$b$$. $5<11 \text{ is true.}$ $$a>b$$ $$\quad a$$ is greater than $$b$$ $11>5 \text{ is true.}$ $$a\leq b$$ $$\quad a$$ is less than or equal to $$b$$ $3 \leq 6 \text{ is true, and so is } 3 \leq 3 .$ $$a\geq b$$ $$\quad a$$ is greater than or equal to $$b$$ $8 \geq 6 \text{ is true, and so is } 8 \geq 8 .$ #### General Procedure Solving linear inequalities is very similar to solving equalities. There is one important change. As mentioned before, if you multiply or divide by a negative number, the direction of the inequality sign must be reversed. #### Example Consider the following linear inequality: $-3y - 1 \ge 2x + 2$ To solve, begin by isolating y: $-3y \ge 2x + 3$ and dividing both sides by $$-3$$, remembering to switch the direction of the inequality: $y \le -\frac{2}{3}x - 1$ which graphs as: Graphically, the linear inequality is a bold line (because the inequality is less than/greater than or equal to, so the points along the line are also included in the solution set), with all values below and along the line shaded. All of these points are shaded because, when plugged into the original inequality, they yield true statements; consequently, they make up the solution set. ### Systems of Linear Equations A set of linear equations where the number of equations matches the number of variables is a linear system of equations. A value of (x,y) that works in both equations is the solution of the system. This is the spot where the lines intersect. If the lines are parallel, there are no solutions. In the rare case where both equations have the same graph, there are infinitely many solutions. #### General Procedure Finding a solution to the system of equations can be done as shown below. Graphing the two equations is helpful for visualizing the solution, but it can be a bit time-consuming, and can involve estimating a result. Substitution and elimination are often the preferred ways to solve systems of linear equations. #### Example Consider the following system of linear equations: $$-x + y = 1$$, and $$2x + y = 4$$, which graphs as: The point, $$(1,2)$$, marked in red, indicates the solution to the system of equations. It is the point that lies along both lines and marks their intersection. Here are two methods for solving systems of equations: ##### Substitution The system $$x + y = 5$$ and $$x - 2y = 2$$ will be solved by substitution: $x + y = 5$ $x = 5 - y$ This will be substituted into the second equation for x: $x - 2y = 2$ $(5 - y) - 2y = 2$ Now, solve for y: $5 - 3y = 2$ $-3y = 2 - 5$ $-3y = -3$ $y = 1$ Now, substitute this value back into the first equation and solve for x: $x = 5 - 1$ $x = 4$ The solution of the system is $$x = 4$$, $$y = 1$$. These values work in both equations. ##### Elimination In elimination, the equations are added or subtracted to eliminate one of the variables. An equation can be multiplied as well. The goal is to get coefficients of the same variable to be the same or opposite in both equations. For the same system, multiply the first equation by 2 to get: $2x + 2y = 10$ The other equation is $$x - 2y = 2$$. Notice that the coefficient of y is opposite. If the equations are added, y will be eliminated: $2x + 2y = 10$ $x - 2y = 2$ $3x = 12$ $x = 4$ Substitute this back into either equation to get $$y = 1$$. All Study Guides for the TSIA2 are now available as downloadable PDFs
# 10.1 - Introduction to Analysis of Variance Let's use the following example to look at the logic behind what an analysis of variance is after. ## Application: Tar Content Comparisons Section We want to see whether the tar contents (in milligrams) for three different brands of cigarettes are different. Two different labs took samples, Lab Precise and Lab Sloppy. #### Lab Precise Lab Precise took six samples from each of the three brands and got the following measurements: Sample Brand A Brand B Brand C 1 10.21 11.32 11.60 2 10.25 11.20 11.90 3 10.24 11.40 11.80 4 9.80 10.50 12.30 5 9.77 10.68 12.20 6 9.73 10.90 12.20 Average $$\bar{y}_1= 10.00$$ $$\bar{y}_2= 11.00$$ $$\bar{y}_3= 12.00$$ #### Lab Sloppy Lab Sloppy also took six samples from each of the three brands and got the following measurements: Sample Brand A Brand B Brand C 1 9.03 9.56 10.45 2 10.26 13.40 9.64 3 11.60 10.68 9.59 4 11.40 11.32 13.40 5 8.01 10.68 14.50 6 9.70 10.36 14.42 Average $$\bar{y}_1= 10.00$$ $$\bar{y}_2= 11.00$$ $$\bar{y}_3= 12.00$$ ##### Lab Sloppy Dotplot The sample means from the two labs turned out to be the same and thus the differences in the sample means from the two labs are zero. From which data set can you draw more conclusive evidence that the means from the three populations are different? We need to compare the between-sample-variation to the within-sample-variation. Since the between-sample-variation from Lab Sloppy is large compared to the within-sample-variation for data from Lab Precise, we will be more inclined to conclude that the three population means are different using the data from Lab Precise. Since such analysis is based on the analysis of variances for the data set, we call this statistical method the Analysis of Variance (or ANOVA).
# Percent word problem: recyclingÂ cans CCSS Math: 6.RP.A.3c ## Video transcript In the United States, 13 out of every 20 cans are recycled. What percent of cans are recycled? So 13 out of every 20 are recycled. So 13/20, or 13 over 20, could also be viewed as 13 divided 20, or 13 divided by 20. And if we do this, we'll get a decimal, and it's fairly straightforward to convert that decimal into a percentage. So 13 divided 20. We have the smaller number in this case being divided by the larger number. So we're going to get a value less than 1. Since we're going to get a value less than 1, let's put a decimal right over here. And let's add a couple of zeroes, as many zeroes as we would need. And we could say, hey, look, 20 goes into 13 zero times. 0 times 20 is 0. And then 13 minus 0 is 13. Now you bring down a 0. 20 goes into 130. Let's see, it goes in-- 5 times 20 is 100. So 6 times 20 is 120. So it's going to six times. 6 times 20 is 120. You subtract. You get a 10. Let's bring down another 0. 20 goes into 100 five times. 5 times 20 is 100. And we are done. So this, written as a decimal, is 0.65. So as a decimal, it's 0.65. And if you want to write it as a percentage you essentially multiply this number by 100. Or another way you could say is you shift the decimal point over two spots to the right. So this is going to be equal to 65%. Now, there's another way you could have done it. You could have said, look, percent literally means per hundred. So 13 out of 20 is going to be equal to what over 100? Well, to go from 20 to a 100-- forget the denominator. To go from 20 to 100, you would multiply by 5. So let's multiply the numerator by 5 as well. And 13 times 5, let's see, that's 15 plus 50, which is 65. So this would have been a faster way to do it, especially if you recognize it's pretty easy to go from 20 to 100. You multiply it by 5. So we would do the same thing with the 13. And so you would get 65/100, which is the same thing as 65 per-- let me write this percent symbol-- 65%. And just a reminder, percent literally means per hundred, 65 per hundred, 65%.
## Algebra 1 $\frac{12 - 2x}{3x}$ To solve: $\frac{4}{x} + \frac{2}{3}$, you will first need to make sure both fractions have common denominators. To do this, we must find the LCD, or the Least Common Multiple, by writing out each as products of prime factors. Since x and 3 have no common factors, the Least Common Multiple, or LCD, then is $3\times x = 3x$ Now\times we can rewrite the fractions using the new LCD we found. $\frac{4}{x} - \frac{2}{3}$ = $\frac{4 \times 3}{x \times 3} - \frac{2 \times x}{3 \times x}$ =$\frac{12}{3x} - \frac{2x}{3x}$ Now that we have common denominators, we can subtract the numerators: $\frac{12}{3x} - \frac{2x}{3x}$ = $\frac{12 - 2x}{3x}$
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SfC Home > Physics > Force > # Proof that Work Can be the Change in Kinetic Energy by Ron Kurtus When you accelerate an object, you are doing work against inertia. That work equals the change in kinetic energy of the object. Proof that the relationship is true can be determined by starting with the standard equation for work against inertia. You then define the constant acceleration in terms of velocity and time, as well as the distance traveled. By some simple algebraic substitution, you can derive the equation for work in terms of change in kinetic energy. Questions you may have include: • What do you want to prove? • Where do you start? • How are terms manipulated to get the result? This lesson will answer those questions. Useful tool: Units Conversion ## Show work equals change in KE You want to prove that the equation for work in terms of the change in kinetic energy of an object is: W = ΔKE or W = KEf− KEi where • W is the work done against the resistance of inertia • ΔKE is the change in kinetic energy (Δ is Greek letter capital delta) • KEf is the final kinetic energy of the object (KEf = mvf2/2) • KEi is the initial kinetic energy of the object (KEi = mvi2/2) ## Start with standard work against inertia equation When you accelerate an object, you are doing work against inertia over the distance that the object is accelerated: W = mad where • W is the work in joules (J or kg-m²/s²) • m is the mass of the object in kg • a is the acceleration of the object in m/s² • d is the distance the object moves in meters (m) ## Break into components to prove equation If the acceleration is constant, it is equal to the change in velocity over time: a = (vf − vi)/t Multiply both sides of the equation by t and divide by a: t = (vf − vi)/a Also, the distance traveled is the product of the average of the velocities and time: d = t(vf + vi)/2 Substitute t = (vf − vi)/a in the equation: d = (vf − vi)(vf + vi)/2a Since (vf − vi)(vf + vi) = v2f − v2i, you get: 2ad = v2f − v2i Multiply both sides of the equation by m and divide by 2: mad = m(v2f − v2i)/2 mad = mv2f/2 − mv2i/2 W = KEf− KEi ∴ W = ΔKE ( means "therefore") ## Summary You can show that work equals the change in kinetic energy of the object by starting with the standard equation for work against inertia. You define the constant acceleration in terms of velocity and time, as well as the distance traveled. Then by simple algebraic substitution, you can prove the equation for work in terms of change in kinetic energy. Be clever ## Resources and references Ron Kurtus' Credentials ### Websites Physics Resources ### Books (Notice: The School for Champions may earn commissions from book purchases) ## Share this page Click on a button to bookmark or share this page through Twitter, Facebook, email, or other services: ## Students and researchers The Web address of this page is: www.school-for-champions.com/science/ work_kinetic_energy.htm Please include it as a link on your website or as a reference in your report, document, or thesis. Copyright © Restrictions ## Where are you now? School for Champions Physics topics ## Force topics ### Let's make the world a better place Be the best that you can be. Use your knowledge and skills to help others succeed. Don't be wasteful; protect our environment. ### Live Your Life as a Champion: Take care of your health Seek knowledge and gain skills Do excellent work Be valuable to others Have utmost character #### Be a Champion! The School for Champions helps you become the type of person who can be called a Champion.
Class 10 ### Topic Covered ♦ Section Formula ### Section Formula Let us, Suppose a telephone company wants to position a relay tower at P between A and B is such a way that the distance of the tower from B is twice its distance from A. If P lies on AB, it will divide AB in the ratio 1 : 2 (see Fig. 7.9). If we take A as the origin O, and 1 km as one unit on both the axis, the coordinates of B will be (36, 15). In order to know the position of the tower, we must know the coordinates of P. Let's find these coordinates Let the coordinates of P be (x, y). Draw perpendiculars from P and B to the x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by the AA similarity criterion, studied in Chapter 6, Delta POD and Delta BPC are similar. Therefore , (OD)/(PC) = (OP)/(PC) =1/2 , and (PD)/(BC) = (OP)/(PB) = 1/2 So, x/(36-x) = 1/2 and y/(15- y) = 1/2 . You can check that P(12, 5) meets the condition that OP : PB = 1 : 2. Now let us use the understanding that you may have developed through this example to obtain the general formula. Consider any two points A(x_1, y_1) and B(x_2, y_2) and assume that P (x, y) divides AB internally in the ratio m_1 : m_2, i.e., (PA)/(PB) = (m_1)/(m_2) (see Fig. 7.10). Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to the x-axis. Then, by the AA similarity criterion, Delta PAQ ~ Delta BPC Therefore, (PA)/(BP) = (AQ)/(PC) = (PQ)/(BC) ........(1) Now , AQ = RS = OS – OR = x – x_1 PC = ST = OT – OS = x_2 – x PQ = PS – QS = PS – AR = y – y_1 BC = BT– CT = BT – PS = y_2 – y Substituting these values in (1), we get (m_1)/(m_2) = (x- x_1)/( x_2 -x ) = (y -y_1)/( y_2 - y) Taking (m_1)/(m_2) = (x-x_1)/(x_2 - x) , we get x = (m_1 x_2 + m_2 x_1)/( m_1 +m_2) Similarly, taking (m_1)/(m_2) = (y-y_1)/( y_2 -y) , we get y = ( m_1 y_2 + m_2 y_1)/( m_1 + m_2) So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x_1, y_1) and B(x_2, y_2), internally, in the ratio m_1 : m_2 are ( (m_1 x_2 + m_2 x_1)/( m_1 + m_2) , ( m_1 y_2 + m_2 y_1)/( m_1 + m_2) ) ............(2) This is known as the section formula. This can also be derived by drawing perpendiculars from A, P and B on the y-axis and proceeding as above. If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be ( ( k x_2 + x_1)/( k+1) , ( k y_2 + y_1)/(k+1) ). Special Case : The mid-point of a line segment divides the line segment in the ratio 1 : 1. Therefore, the coordinates of the mid-point P of the join of the points A(x_1, y_1) and B(x_2, y_2) is ( (1 * x_1 + 1 * x_2)/( 1+1) , ( 1 * y_1 + 1 * y_2)/( 1+1) ) = ( (x_1 + x_2 )/2, ( y_1 + y_2)/2 ) . Let us solve a few examples based on the section formula. Q 3109580418 Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally. Class 10 Chapter 7 Example 6 Solution: Let P(x, y) be the required point. Using the section formula, we get x= ( 3(8) + 1(4) )/(3+1) =7 , y = ( 3(5) +1(-3) )/( 3+1) =3 Therefore, (7, 3) is the required point. Q 3129680511 In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8) ? Class 10 Chapter 7 Example 7 Solution: Let (– 4, 6) divide AB internally in the ratio m_1 : m_2. Using the section formula, we get (-4, 6) = ( (3 m_1 - 6 m_2)/( m_1 + m_2 ) , ( -8 m_1 +10 m_2)/( m_1 + m_2) ) ...................(1) Recall that if (x, y) = (a, b) then x = a and y = b. So, -4 = (3 m_1 -6 m_2)/( m_1 + m_2) and 6 = ( -8 m_1 +10 m_2)/( m_1+ m_2) Now, -4 = (3 m_1 -6 m_2)/( m_1 + m_2) gives us – 4m_1 – 4m_2 = 3m_1 – 6m_2 i.e., 7m_1 = 2m_2 i.e., m_1 : m_2 = 2 : 7 You should verify that the ratio satisfies the y-coordinate also. Now, (-8 m_1 +10 m_2)/(m_1 + m_2) = ( -8 (m_1)/(m_2) +10 )/( (m_1)/(m_2) +1) (Dividing throughout by m_2) = (-8 xx 2/7 +10 )/( 2/7 +1) =6 Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7. Alternatively : The ratio m_1 : m_2 can also be written as (m_1)/(m_2) :1 , or k :1 . Let (-4 ,6 ) divide AB internally in the ratio k : 1. Using the section formula, we get (-4 ,6) = ( (3k -6)/( k+1) , ( -8k +10)/(k+1) ) ...........(2) So, -4 = (3k -6 )/(k+1) i.e., – 4k – 4 = 3k – 6 i.e., 7k = 2 i.e., k : 1 = 2 : 7 You can check for the y-coordinate also. So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7. Note : You can also find this ratio by calculating the distances PA and PB and taking their ratios provided you know that A, P and B are collinear. Q 3149680513 Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4). Class 10 Chapter 8 Example 8 Solution: Let P and Q be the points of trisection of AB i.e., AP = PQ = QB (see Fig. 7.11). Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are ( (1 (-7) +2 (2) )/( 1+2) , ( 1(4) +2(-2) )/(1+2) ) , i.e., (-1, 0 ) Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are ( ( 2 (-7) + 1(2) )/( 2+1) , ( 2(4) + 1 ( -2) )/( 2+1) ) i.e, ( -4,2 ) Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2). Note : We could also have obtained Q by noting that it is the mid-point of PB. So, we could have obtained its coordinates using the mid-point formula. Q 3189680517 Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection. Class 10 Chapter 7 Example 9 Solution: Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are ( ( -k +5)/(k+1 ) , (-4k -6)/(k+1)). This point lies on the y-axis, and we know that on the y-axis the abscissa is 0. Therefore, (-k+5)/(k+1) = 0 So, k =5 That is, the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as (0, (-13)/3) . Q 3119680519 If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p. Class 10 Chapter 7 Example 10 Solution: We know that diagonals of a parallelogram bisect each other. So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD i.e., ( (6+9)/2 , (1+4)/2 ) = ( (8+p)/2 , (2+3)/2) i.e., (15/2, 5/2) =( (8+p)/2 , 5/2 ) so, 15/2 = (8+p)/2 i.e., p =7
Convergence and Divergence of Sequences # Convergence and Divergence of Sequences We will now look at two very important terms when it comes to categorizing sequences. Definition: A sequence $(a_n)$ is said to be convergent to the real number $L$ and we write $\lim_{n \to \infty} a_n = L$ if $\forall \epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. If $\lim_{n \to \infty} a_n$ does not exist (that is the limit is infinity, negative infinity, or just doesn't converge in general) then we say that the sequence $(a_n)$ is divergent. From this definition of convergence, we immediately have the following theorem of equivalence statements. Theorem 1: Let $(a_n)$ be a convergent sequence. Then the following statements are equivalent: 1. The sequence $(a_n)$ converges to the real number $L$. 2. For every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. 3. For every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ the terms satisfy $L - \epsilon < a_n < L + \epsilon$. 4. For every $\epsilon > 0$, for the $\epsilon$-neighbourhood $V_{\epsilon}(L)$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_n \in V_{\epsilon} (L)$. • Proof: $1 \Leftrightarrow 2$ Suppose that the sequence $(a_n )$ converges to the real number $L$. Then $\lim_{n \to \infty} a_n = L$ and by the definition, for every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. • $2 \Leftrightarrow 3$ Suppose that for every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. Taking the last inequality, by an earlier theorem we note that $\mid a_n - L \mid < \epsilon$ if and only if: (1) \begin{align} -\epsilon < a_n - L < \epsilon \\ \Leftrightarrow L - \epsilon < a_n < L + \epsilon \end{align} • $3 \Leftrightarrow 4$ If for every $\epsilon > 0$, for the $\epsilon$-neighbourhood $V_{\epsilon}(L)$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_n \in V_{\epsilon} (L)$, then $x_n \in V_{\epsilon} (L)$ for all $n ≥ N$. $\blacksquare$ The following theorem tells us that the sequence $\displaystyle{\left \{ \frac{1}{n} \right \}_{n=1}^{\infty}}$ converges to $0$. Theorem 1: The sequence $\displaystyle{\left \{ \frac{1}{n} \right \}_{n=1}^{\infty}}$ converges to $0$. • Proof: Let $\epsilon > 0$ be given. Observe that: (2) \begin{align} \quad \biggr \lvert \frac{1}{n} - 0 \biggr \rvert = \biggr \lvert \frac{1}{n} \biggr \rvert = \frac{1}{n} \end{align} • We want $\frac{1}{n} < \epsilon$, i.e., $\frac{1}{\epsilon} < n$. So choose $N > \frac{1}{\epsilon}$. Then if $n \geq N$ we have that $n > \frac{1}{\epsilon}$, and so $\frac{1}{n} < \epsilon$, so from above: (3) \begin{align} \quad \biggr \vert \frac{1}{n} - 0 \biggr \rvert = \frac{1}{n} < \epsilon \end{align} • Therefore $\displaystyle{\left \{ \frac{1}{n} \right \}_{n=1}^{\infty}}$ converges to $0$. $\blacksquare$
# Volume of Pyramid The volume of pyramid is the space occupied by a pyramid and also the number of cubes we can put in that pyramid. Pyramids can be of different types some of those are square pyramid, triangular pyramid, pentagonal pyramid, rectangular pyramid, and many more all the pyramids were named as per their shapes for example if the shape of the pyramid is a rectangle then we will represent it as a rectangular pyramid. The faces of the pyramid were made up of polygon so we can also represent it as a polyhedron. The volume of pyramid = 1/3* Bh ## Formula of Volume of Pyramid As we discussed above the theoretical part of the pyramid now lets us know how to solve the questions with the help of a formula. So the formula of volume of a pyramid is 1/3 Bh where B stands for the base of the pyramid and h is represented as the height of the pyramid. Example1. The height of the pyramid is 6 ft. its base is 2 find the volume of the pyramid? Solution: So to find the volume of the pyramid first we use the formula then put the given value in it. Volume of Pyramid = 1/3 Bh This means 1/363 so the volume of a pyramid is 6. This is how we can easily come up with our solution. ## Facts of Pyramid The sides of each type of pyramid look like a triangle that merges with shape from its sides. The pyramid’s volume is always one-third of the volume of the prism, the height of the pyramid and prism will be the same. Three rectangular pyramids can be placed in the form of a rectangular pyramid. Hope you can understand well about the pyramid if not then we have an amazing activity that help you to understand more about the topic. So first take a rectangular pyramid and fill the pyramid with water then take a rectangular prism and pour that water from the rectangular pyramid into the rectangular prism now you can see that the water is one-third of the size of the prism. Hope so this fun activity has cleared all your doubts. ## Math Classes Online Due to pandemics nowadays it is not possible to be regular with your school and physical classes you should find a way to digitally clear your doubts. Even in school, many students face lots of problems due to their lack of interest, concentration, etc. In classes, if the student misses any lesson due to any reason then he can’t grasp the topic again. He can request his teacher to go through it again but through the online classes, you can go through your lectures again and again whenever you face any doubt. At every step of your practice you can go through the recordings it seems like your teachers were 247 available for you in each step, in each doubt. Many students lose their interest in the subject just because their base is not strong, they miss some of the concepts but these online math classes are so astonishing that it helps in regenerating the student’s interest in their subject like Cuemath were based on the concept of learning with fun, they provide lots of fun activity so that if a student can’t understand the theoretical concept then he can grasp it the help of fun activities, these programs also provide practice papers, quizzes, concept sheets and many more so that student can understand what he is learning and it builds confidence in him that he is going in a right track.
# How do you solve for y? #### Understand the Problem The question is about the process of isolating the variable 'y' in an equation, which can often involve algebraic manipulation. The context or specific equation is not provided, but typically, it would require rearranging the equation to express 'y' in terms of other variables or constants. y = 3 - \frac{3x}{2} #### Steps to Solve 1. Identify the equation with y The first step is to write down the equation where you need to solve for $y$. The equation should contain $y$, along with other variables and constants. Let's use an example equation: $3x + 2y = 6$. 2. Isolate the term with y Move all other terms to the opposite side of the equation to isolate the $y$ term. This often involves adding or subtracting terms from both sides of the equation. For our example: $3x + 2y - 3x = 6 - 3x$ Simplified, you get: $2y = 6 - 3x$ 3. Isolate y by dividing Divide both sides of the equation by the coefficient of $y$ to solve for $y$. Here, the coefficient is 2. So: $$y = \frac{6 - 3x}{2}$$ 4. Simplify the equation if possible Simplify the equation by performing the division if it is possible. In this case, you get: $$y = 3 - \frac{3x}{2}$$ So the solution for $y$ is: $y = 3 - \frac{3x}{2}$ Solving for $y$ is a common algebraic task that involves isolating $y$ on one side of the equation. This requires understanding and applying inverse operations such as addition/subtraction and multiplication/division. Common mistakes include not properly adding or subtracting terms from both sides of the equation and forgetting to divide by the coefficient of $y$.
# How do you solve the following linear system: 8x - 7y = -3 , 6x - 5y = -1 ? Mar 7, 2016 $\left(4 , 5\right)$ #### Explanation: In order to solve systems of linear equations, we can use either of the three methods (1) elimination, (2) substitution, and (3) graphing. For this question I would use the elimination because I am more used to it and I think that it is quicker to do but note that using any of the three methods would give you the same result. [Solution] $8 x - 7 y = - 3$ $6 x - 5 y = - 1$ Multiplying the first equation by 3 and multiplying the second equation by 4 would make the coefficients of the first terms of both equations 24. This would allow us to eliminate the $x$ variable so we can solve for the $y$. $24 x - 21 y = - 9$ $24 x - 20 y = - 4$ Subtracting the two equations... $- y = - 5$ $y = 5$ Once we get the value of $y$ we would then substitute it to any of the two equations to solve for $x$. In this case, I would use the second equation since it has a smaller coefficient hence would produce smaller values. $6 x - 5 y = - 1$ $6 x - 5 \left(5\right) = - 1$ $6 x - 25 = - 1$ $6 x = - 1 + 25$ $6 x = 24$ $x = 4$ $\left(4 , 5\right)$ [Checking byt substituting $\left(4 , 5\right)$] $8 x - 7 y = - 3$ $6 x - 5 y = - 1$ First Equation $8 \left(4\right) - 7 \left(5\right) = - 3$ $32 - 35 = - 3$ $- 3 = - 3$ Second Equation $6 x - 5 y = - 1$ $6 \left(4\right) - 5 \left(5\right) = - 1$ $24 - 25 = - 1$ $- 1 = - 1$ Since both equations were satisfied by the computed values, we know that our answer is correct!
# Prime (and Other) Factors of 72 The number 72 is frequently used to demonstrate deep mathematical structure. Not only does it have factors other than 1 and itself (it is composite, or not prime), it has quite a lot of them, and its prime factors repeat themselves. It is also a small enough number to be accessible to most learners but large enough to be interesting. Some of these features even make it appear “in the real world” more often than some other numbers (for example, it is the number of hours in 3 days). The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. As the only prime numbers in this list, the prime factors of 72 are 2 and 3. The prime factorization of 72 is $2\times 2\times 2\times 3\times 3$ or, using exponents, $2^3 \times 3^2$. ## Factor pairs of 72 A factor pair of a number is two other whole numbers that multiply together to equal that same number. One way to visualize a factor pair is as the sides of a rectangle with the target area. One factor pair of a number is always $1$ times the number itself; in this case, $1\times 72$. From the times tables so many of us were asked to memorize, we can also see that $6\times 12=72$ and $8\times 9=72$, so $6\times 12$ and $8\times 9$ are also factor pairs. Finally, if we start back at $2$ and just try numbers to see what divides 72 evenly, we can find $2\times 36$, $3\times 24$ (did you remember that 3 days = 72 hours?), and $4\times 18$ are all factor pairs of 72. ## Factors of 72 If we make a list of the individual numbers from the factor pairs, then we have a list of the factors of our target number. In our case, the factors of 72 are 1, 72, 6, 12, 8, 9, 2, 36, 3, 24, 4, and 18; or, in order from smallest to largest: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. ## Prime Factors of 72 The only prime numbers in our list above are 2 and 3. Thus, 2 and 3 are the only prime factors of 72. Prime factors are useful in examining the structure of a number and its relationship to other numbers. ## Prime Factorization of 72 The prime factorization of a number shows us not only what the prime factors are, but how many copies of each must be multiplied together to get to the target number. There are a variety of tools for finding prime factorizations, but the two most common are factor trees and factor ladders. A factor tree allows us to start with any factor pair we know, whether using a prime number or not, in order to eventually get to the prime factors. A factor ladder more directly looks for prime numbers by using divisibility rules for prime numbers in order to divide them out. (Factor ladders are also excellent tools when we learn GCF and LCM as well!) Either method shows us that $72=2\times 2\times 2\times 3\times 3=2^3\times 3^2$, so the prime factorization of 72 can be written as $2\times 2\times 2\times 3\times 3$ or as $2^3\times 3^2$, depending on whether the student has learned about exponents or not. For more about factoring numbers, check out my blog post Factoring Numbers: How Do I Help My Child? You’ve Got This! Barbra Barbie has taught math, supported students, and volunteered in classrooms for over 20 years. Her daughter is currently learning math in a Common Core state.
Just remember: Always the same distance apart and never touching. Parallel lines are useful in understanding the relationships between paths of objects and sides of various shapes. Learn how to identify parallel and perpendicular lines. Two lines are parallel and do not intersect for longer than they are prolonged. Properties of Parallel Lines RS Aggarwal Class 7 Maths Solutions. Use the image shown below to answer Questions 9- 12. Created by Sal Khan. Theorem and Proof Statement for Alternate Interior Angles: The Alternate interior angle theorem states that “ if a transversal crosses the set of parallel lines, then the alternate interior angles … c) If two parallel lines are cut by a transversal, then the same-side interior angles are supplementary. Property 2 : Let us consider the general form of equation of a straight line. And this line that intersects both of these parallel lines, we call that a transversal. When corresponding angles are equal, the lines are parallel. Recall that two lines are parallel if its pair of consecutive exterior angles add up to $\boldsymbol{180^{\circ}}$. If we have two parallel lines and have a third line that crosses them as in the ficture below - the crossing line is called a transversal. 5. Properties of Parallel Lines – The basic characteristic of Parallel Lines is they never meet at any point The distance between two Parallel Lines would always be same at every point Any third line will cut Parallel Lines at same angles Any straight line cutting Parallel Lines is called Transversal Two Transversals cutting Parallel lines… Stay Home , Stay Safe and keep learning!!! Property 1 : Let m1and m2be the slopes of two lines. Parallel lines are a fixed distance apart and will never meet, no matter how long they are extended. Use the image shown below to answer Questions 4 -6. This means that $\boldsymbol{\angle 1 ^{\circ}}$ is also equal to $\boldsymbol{108 ^{\circ}}$. Fill in the blank: If the two lines are parallel, $\angle c ^{\circ}$, and $\angle f ^{\circ}$ are ___________ angles. We will begin with corresponding angles. If corresponding angles are equal, then the lines are parallel. 15 Name Class Date Parallel Lines, and Pairs of Angles Parallel Lines. Add $72$ to both sides of the equation to isolate $4x$. Missing angles (CA geometry) Our mission is to provide a … The notation for parallel lines,, is read: "line AB is parallel to line CD." So, we can say that, 2x + 20 + 4x + 40 = 180 6x + 60 = 180 6x = 180-60 = 120 X = 20. Properties Of Angles Associated with … Let’s summarize what we’ve learned so far about parallel lines: The properties below will help us determine and show that two lines are parallel. Let’s see, the value of consecutive interior angles The 3 properties that parallel lines have are the following: Properties of Angles Formed by Two Parallel Lines and a Transversal - Axiom: If a Transversal Intersects Two Parallel Lines, Then Each Pair of Interior Angles on the Same Side of the Transversal is Supplementary. If the two lines are parallel, all of the angles, corresponding, alternate interior, alternate exterior and same side interior have new properties. Since the lines are parallel and $\boldsymbol{\angle B}$ and $\boldsymbol{\angle C}$ are corresponding angles, so $\boldsymbol{\angle B = \angle C}$. Add the two expressions to simplify the left-hand side of the equation. Corresponding angles are equal. Consecutive exterior angles are consecutive angles sharing the same outer side along the line. The terms parallel and perpendicular occur both in algebra (equations of lines) and in geometry (properties of many geometric figures). Topic: Angles. Just a very quick reference sheet for parallel lines, in colour with examples provided. The angles $\angle WTS$ and $\angle YUV$ are a pair of consecutive exterior angles sharing a sum of $\boldsymbol{180^{\circ}}$. Two lines are said to be parallel when they do not meet at any point in a plane or which do not intersects each other. Substitute this value of $x$ into the expression for $\angle EFA$ to find its actual measure. Angles 1 and 5 constitutes one of the pairs. The students should download … Exam questions are included as an extension task. Parallel lines. Properties Of Parallel Lines. Properties of Parallel Lines RS Aggarwal Class 7 Maths Solutions. Roadways and tracks: the opposite tracks and roads will share the same direction but they will never meet at one point. This property holds good for more than 2 lines also. Created: Sep 10, 2015. View parcel number, acreage, and owner name and search by any of these dimensions. Learn. Find the value of angle x using the given angles. 1 8 0 ∘. 9. keyboardsmash8826. At the intersection of two parallel lines of the third line, the angles formed by them crosswise are equal to: The following property is a special case for each previous one: 4. For More Resources Angles in Parallel Lines. Example: $\angle a^{\circ} + \angle g^{\circ}=$180^{\circ}$,$\angle b ^{\circ} + \angle h^{\circ}=$180^{\circ}$. It is a basic tool in descriptive geometry.The projection is called orthographic if the rays are perpendicular (orthogonal) to the image plane, and oblique … A line that intersects two lines in a distinct point is known as Transversal line’ Parallel lines. The vertically opposite angles are equal. The alternate interior angles are equal. If a transversal obliquely intersects two parallel lines, then: All the acute angles are congruent. Transversal lines are lines that cross two or more lines. By accessing this content, you agree to not share or disseminate this content electronically or by any other means and to … Introducing Transversals & Parallel Lines First, students will need to be able to identify angle pairs, then know the properties and relationships that exist when the lines that the tranversal intersects happen to be parallel. 11. The alternate exterior angles are equal. Author: GeoGebra Materials Team. 18. Padmini Roy. The two pairs of angles shown above are examples of corresponding angles. Transcript Parallel lines never intersect, and perpendicular lines intersect at a 90 degree angle. Which of the following term/s do not describe a pair of parallel lines? Properties of Parallel Lines Practice. There is another line B which is parallel to the same line. Following are the properties: Vertically opposite angles are equal. Lessons on Vectors: Parallel Vectors, how to prove vectors are parallel and collinear, conditions for two lines to be parallel given their vector equations, Vector equations, vector math, with video lessons, examples and step-by-step solutions. Parallel lines are equidistant lines (lines having equal distance from each other) that will never meet. The options in b, c, and d are objects that share the same directions but they will never meet. 180^\circ. Properties of Parallel Lines. Perpendicular Line. If lines are parallel, corresponding angles are equal. – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow.com - id: 6a278e-ZDI0Z Are the two lines cut by the transversal line parallel? 5. Prentice Hall Foundations Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. 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Hall Foundations geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its.! Image shown to the right shows how a transversal line are parallel if they do not describe a pair parallel... Supplementary, then the two straight lines are parallel when the alternate exterior are. Sheet for parallel lines RS Aggarwal Class 7 Maths Solutions angles parallel lines are parallel, diagonally, pairs.$ \overline { AB } $and$ \overline { WX } \$ are parallel property and is determined 0.69. Running Protection Knife, Mayak Nuclear Disaster 2017, Batman Ninja Full Movie, Correlation Coefficient Formula, South Carolina Building Fire Code, Multiplayer Games For Couples Iphone, Snl Karen Potato Salad, Emperor Of Japan 2020, Fast Five Google Drive,
# What is meant by the limit of an infinite sequence? The limit of an infinite sequence tells us about the long term behaviour of it. Given a sequence of real numbers ${a}_{n}$, it's limit ${\lim}_{n \to \infty} {a}_{n} = \lim {a}_{n}$ is defined as the single value the sequence approaches (if it approaches any value) as we make the index $n$ bigger. The limit of a sequence does not always exist. If it does, the sequence is said to be convergent, otherwise it's said to be divergent. Two simple examples: • Consider the sequence $\frac{1}{n}$. It's easy to see that it's limit is $0$. In fact, given any positive value close to $0$, we can alway find a great enough value of $n$ such that $\frac{1}{n}$ is less than this given value, wich means that it's limit must be less or equal to zero. Also, every term of the sequence is greater then zero, so it's limit must be greater or equal to zero. Therefore, it is $0$. • Take the constant sequence $1$. That is, for any given value of $n$, the term ${a}_{n}$ of the sequence is equal to $1$. It's clear that no matter how big we make $n$ the value of the sequence is $1$. So it's limit is $1$. For a more rigorous definition, let ${a}_{n}$ be a sequence of real numbers (that is, $\forall n \in \mathbb{N} : {a}_{n} \in \mathbb{R}$) and $\epsilon \in \mathbb{R}$. Then the number $a$ is said to be the limit of the sequence ${a}_{n}$ if and only if: $\forall \epsilon > 0 \exists N \in \mathbb{N} : n > N \implies \| {a}_{n} - a \| < \epsilon$ This definition is equivalent to the informal definition given above, except that we don't need to impose unicity for the limit (it can be deduced).
## Elementary Statistics (12th Edition) Published by Pearson # Chapter 4 - Probability - 4-4 Multiplication Rule: Basics - Basic Skills and Concepts - Page 167: 30 #### Answer a)0.99 b)0.98 c)0.779 #### Work Step by Step a) We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, hence $probability=\frac{492+306}{492+306+8}=0.99.$ b)Two events are dependent, if the outcome of one effects the outcome of the other. Here the events are independent, because the sampling was done with replacement, also $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, hence $probability=\left (\frac{2518-252}{2518}\right )^{50}=0.00493.$ b) Events are dependent, if the outcome of one effects the outcome of the other. Here the events are dependent, because the choice of the first person effects the opportunities of the second one. $P(A \cap B)=P(A)\cdot P(B\|A)$. Hence $P(A \cap B)=\frac{798}{806}\cdot\frac{797}{805}=0.98.$ c)Events are dependent, if the outcome of one effects the outcome of the other. Here the events are independent, because the sampling was done with replacement, also $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, hence $probability=\left (\frac{798}{806}\right )^{25}=0.779.$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
# DAV Class 4 Maths Chapter 8 Brain Teasers Solutions The DAV Class 4 Maths Solutions and DAV Class 4 Maths Chapter 8 Brain Teasers Solutions of Time and Calendar offer comprehensive answers to textbook questions. ## DAV Class 4 Maths Ch 8 Brain Teasers Solutions Question 1. Tick (✓) the correct answer. (a) 3 hrs 33 min = ________ min. (i) 333 (ii) 213 (iii) 180 (iv) 210 Answer: (ii) 213 (b) Today is Monday. After 61 days it will be- (i) Wednesday (ii) Saturday (iii) Tuesday (iv) Thursday Answer: Today is Monday then after every 7 days it will be Monday so we have to divide by 7. = 8y mean after five day of Monday i.e. Saturday so option (ii) (c) Which of the following months has 30 days? (i) July (ii) December (iii) October (iv) April Answer: (iv) i.e. April (d) A train leaves Delhi at 12:05 p.m. At what time will it reach Agra, if the journey takes 3 hrs 42 minutes? (i) 1:37 p.m. (ii) 1:47 p.m. (iii) 3:47 p.m. (iv) 4:37 p.m. Answer: Time of departure = 12 hrs 05 min Time to reach Agra = 3hrs 42 min 15:47 or 3:47 p.m. so option (iii) (e) The number of days in two non-leap years is- (i) 720 (ii) 732 (iii) 722 (iv) 730 Answer: (iv) i.e. 730 days. (365 × 2 = 730) Question 2. Is the year in which India became Republic, a leap year? Answer: India became Republic in 1950 Divide it by 4 So, it was not a leap year. Question 3. If 4th of February is a Sunday, what day is 25th of February? Answer: 4th Feb — Sunday so, every 7th day is Sunday. 11th Feb — Sunday 18th Feb — Sunday 25th Feb — Sunday Question 4. Fill in the blanks. (a) A leap year has ________ days. Answer: 366 (b) There are ________ seconds in one hour. Answer: 3600 (c) The shortest month of a year is ________. Answer: February (d) 5 hours = ________ minutes. Answer: 300 minutes (e) 3 hours 35 minutes = ________ minutes. Answer: 215 minute (f) 1 century = ________ years. Answer: 100 (g) 180 minutes = ________ hours. Answer: 3 (h) 3 hours after 5 p.m. is ________. Answer: 8 p.m. (i) 2 hours before 9 a.m. is ________. Answer: 7 a.m. (j) 2 hrs 20 min after 3:30 a.m. is ________ Answer: 5: 50 a.m. (k) 4 hrs 5 min before 10:40 a.m. is ________ Answer: 6:35 a.m. Question 5. Find the sum: (a) 3 hrs 15 min and 5 hrs 20 min Answer: 8hrs 35 min (b) 6 hrs 47 min and 45 min Answer: 7 hrs 32 min Question 6. Find the difference between: (a) 6 hrs 40 min and 5 hrs 5 min Answer: 1 hr 35 min (b) 3 hrs 47 min and 55 min Answer: 2 hrs 52 min Question 7. What time was it 3600 seconds before 12 midnight? Answer: As 3600 sec = 1 hr 11: 00 p.m. Question 8. What time will it be 3600 seconds after 11 a.m. ? Answer: 3600 sec = 1 hr 12: 00 noon Question 9. A. T.V. was switched on at 7:30 p.m. and switched off at 11:45 p.m. How long did it run? Answer: T.V. was switched on at 7:30 p.m. T.V. was switched off at 11:45 p.m. Duration of T.V. run = 4hrs15min 4 hrs 15 min. Question 10. Renu studied for 1 hour 25 minutes. If she stopped studying at 8:30 p.m., when did she start studying? Answer: Duration of study = 1hr 25 min Stopped studying = 8: 30 p.m She starts studying at 7: 05 p.m Question 11. Find the names of the famous personalities whose date of birth is 29 February. Answer: Moraraji Desai, Rukmani Devi. Additional Questions Question 1. Which of the following are leap years? (a) 2009 (b) 1996 (c) 2014 (d) 2015 Answer: (b) 1996 It is a leap year (a) 2009 It is not a leap year (c) 2014 It is not a leap year (d) 2015 It is not a leap year Question 2. Fill in the blanks: (a) 5 days = ________ hours. Answer: 120 (24 × 5 = 120) (b) 3 decades = ________ years. Answer: 30 (c) 10 years = ________ month. Answer: 120 (d) 3600 second = ________ hour. Answer: 1 (e) 1 hour = ________ minutes. Answer: 60 Question 3. Find the sum: (a) 6 hrs 5 min and 5 hrs 20 min. Answer: (b) 9 hrs 15 min and 35 min. Answer: (c) 3 hrs and 2 hrs 15 min. Answer: (d) 10 hrs and 3 hrs 7 min. Answer: (e) 5 hrs and 45 min and 6 hrs 55 min. Answer: Question 4. Find the difference? (a) 3 hrs 25 min and 6 hrs 15 min. Answer: 2 hrs 50 min (b) 9 hrs 28 min and 7 hrs 35 min. Answer: 1 hr 53 min (c) 5 hrs 25 min and 55 min. Answer: 4 hrs 30 min. Question 5. What time will it be? (a) 4 hrs before 5:30 p.m.? Answer: 1: 30 p.m (b) 2 hrs after 8:20 p.m.? Answer: 10:20 p.m (c) 2 hrs before 2:30 p.m.? Answer: 12:30 p.m (d) 5 hrs after 11:30 a.m.? Answer: 4:30 p.m
# RD Sharma Chapter 4 Class 10 Maths Exercise 4.9 Solutions RD Sharma Chapter 4 Class 10 Maths Exercise 4.1 Solutions: Quadratic Equations have been compiled to help you score well in the CBSE board exam. It is a compulsory subject of CBSE class 10 syllabus as questions of this subject always get sufficient coverage in the question papers every year. Our RD Sharma Class 10 Maths Solutions Quadratic Equation tells you about quadratic equations, including factoring methods like finding a solution to a quadratic equation, and completing class methods. Here you will also understand the nature of the root and the applications of quadratic equations in daily life. RD Sharma Chapter 4 Class 10 Maths Exercise 4.9 Solutions has 4 questions and all of them are problems based on word age calculation. All these questions test your analytical skills. You have to first draw a quadratic equation and then solve to find the answer. ## Download RD Sharma Chapter 4 Class 10 Maths Exercise 4.9 Solutions RD SHARMA Solutions Class 10 Maths Chapter 8 Ex 8.9 ## Important Definition for RD Sharma Chapter 4 Class 10 Maths Exercise 4.9 Solutions Here you will learn than an equation with the form ax2 + bx + c = 0, with x as the variable and a, b, c as real numbers (where a ≠ 0) is known as a quadratic equation. A quadratic equation is obtained upon equating the quadratic polynomial of the form ax2 + bx + c, a ≠ 0 to zero. Some equations may be of higher degree and need to be simplified before deciding whether it is quadratic or not. We often come across quadratic equations in real-life situations. • Solving Quadratic Equation by Factorisation This part of the Chapter tells you that the first method of solving a quadratic equation is the factorisation method. Here values of the variable x are determined, for which a quadratic equation is satisfied. These particular values of x are called the roots of the quadratic equation. • Solving Quadratic Equation by Completing the Square This section teaches you that a quadratic equation can also be solved by the method of completing the square like- (x + a)2 – b2 = 0 • Nature of Roots This part of Cass 10 Maths Quadratic Equations teaches you that values of real roots vary for a quadratic equation of the form ax2 + bx + c = 0.vWhen b2 – 4ac > 0, then the equation possesses two distinct real roots. In case b2 – 4ac = 0, then the equation has two equal roots (known as coincident roots). Know more at the official website.
# LCD Calculator (Least Common Denominator) Find the least common denominator for a set of fractions using the LCD calculator below. Two or more fractions, separated by a comma ## Least Common Denominator: 12 ### Steps to Solve 12 is the smallest number that is evenly divisible by the denominators 2, 4, & 3 without a remainder 12 ÷ 2 = 6 12 ÷ 4 = 3 12 ÷ 3 = 4 Learn how we calculated this below ## How to Find the Least Common Denominator A denominator is the bottom number of a fraction, or the number below the fraction line. For a fraction 1/3, the denominator is 3. A common denominator is a denominator that is common to the fractions being operated on. For the denominator to be common, it must be the same in all fractions. For instance, 1/3 and 2/3 have common denominators – the denominators are the same. 1/3 and 2/5 do not have common denominators – the denominators are different. The least common denominator is the smallest common denominator. It is the smallest whole number that is evenly divisible by all uncommon denominators. The least common denominator is also referred to as the lowest common denominator or the least common multiple. The least common denominator for the fractions 1/3 and 2/5 is 15. 15 ÷ 3 = 5 15 ÷ 5 = 3 Note that there can be no remainder when dividing one of the denominators by the least common denominator. There are a few methods that you can use to find the least common denominator. The easiest is to use the calculator above. If you want to do the work yourself, follow along to learn two methods to solve it. ### Method One: Find the Least Common Denominator Using Factorization One way to find the least common denominator is to use prime factorization. Find the prime factors of each denominator. Then multiply all of the prime factors, taking the common factors only once, to find the least common denominator. You can use our prime factorization calculator to find the prime factors for each number. #### Example: Find the Least Common Denominator of 10 and 15 Step one: find the prime factors of 10. To find the prime factors of 10, you can begin by dividing 10 by the smallest prime number, which is 2. Since 10 is even, it is divisible by 2. 10 divided by 2 is equal to 5. Since 5 is a prime number, we cannot divide it any further. Therefore, the prime factors of 10 are 2 and 5. Step two: find the prime factors of 15. To find the prime factors of 15, you can begin by dividing 15 by the smallest prime number, which is 2. However, 15 is an odd number and is not divisible by 2. Then you can divide 15 by the next smallest prime number, which is 3. 15 divided by 3 is equal to 5. Since 5 is a prime number, we cannot divide it any further. Therefore, the prime factors of 15 are 3 and 5. Step three: find the prime factors that are common to both 10 and 15, then remove the ones that are common to both from the list. The prime factors of 15 and 10 are 5, 3, & 2. Step four: multiply the factors together to find the least common denominator. Note that since 5 is common between 10 and 15 it is only used once. LCD = 5 × 3 × 2 LCD = 30 ### Method Two: Find the Least Common Denominator by Finding all Multiples You can also find the least common denominator by finding all of the multiples of each denominator, then identifying the smallest one that is common to both of them. To find multiples, start by multiplying it by 2 to find the first multiple, then multiply it by 3 to find the second, and so on. Continue by multiplying by larger numbers to find all of the multiples. Then, repeat this for each denominator. For example, let’s find the least common denominator of 10 and 15 by finding all multiples. The multiples of 10 are [10,20,30,40,50,60,…] The multiples of 15 are [15,30,45,60,75,…] Observe that the multiples that are common to both 10 and 15 are 30 and 60. The smallest of these is 30, which makes it the least common denominator. Thus, the least common denominator of 10 and 15 is 30. ### Method Three: Find the LCD Using Division You can also find the least common denominator using division. Begin by writing the denominators as factors and divide out the common factors. For example, if we want to find the LCD of 1/3 and 2/5: 3 = 3 × 1 5 = 5 × 1 Multiply the remaining factors together. 3 × 5 = 15 The product 15 is the least common multiple (LCM) of 3 and 5, so it is also the least common denominator (LCD). You can then rewrite each fraction with the LCD as the denominator by multiplying both the numerator and denominator by the factor needed to obtain the LCD. 1 / 3 = 5 / 15 2 / 5 = 6 / 15 So the LCD is 15, and the equivalent fractions with the LCD as the denominator are 5/15 and 6/15. ### What is the least common denominator used for? The least common denominator (LCD) is the smallest number that is a multiple of the denominators of two or more fractions. It is used in fraction operations and is used to convert fractions to equivalent fractions with the same denominator. When you find the LCD and convert the fractions, you can add or subtract fractions easily without changing their values. ### Are LCD and LCM the same thing? LCD (least common denominator) and LCM (least common multiple) are not the same, although they are related. The LCD is the smallest common multiple of the denominators of two or more fractions, which is used to find a common denominator for the fractions, whereas the LCM is the smallest multiple that is common to two or more numbers. ### What is the difference between LCD and GCF? The GCF is the largest factor common to two or more numbers, and it is used to simplify fractions, while the LCD is used to find a common denominator in fraction operations. Although both concepts are used in math, they serve different purposes.
Q: # According to the Environmental Protection Agency, the mean miles per gallon of all large cars manufactured in 2012 driven under highway driving conditions is 25.1 mpg. A researcher claims that a new fuel additive will increase the miles per gallon of cars. After obtaining a random sample of 35 large cars from 2012, the researcher adds the fuel additive. The sample mean is found to be 26.8 mpg, with a standard deviation of 3.9. Test the researcher's claim at the .05 level of significance. Accepted Solution A: Step 1: Writing Null and Alternate Hypothesis First we need to write the Null and Alternate Hypothesis for this problem. Researcher claims that mileage will be increased on addition of Fuel additive. So the null and alternate hypothesis will be: $$H_{o}$$: μ ≤ 25.1 (Null Hypothesis) $$H_{a}$$: μ > 25.1 (Alternate Hypothesis) This is a Right Tailed Test. Since population standard deviation is not know, we will use t-test to check the researchers claim. Step 2: Finding Test Statistic Sample Mean = x = 26.8 Standard Deviation = s = 3.9 Sample Size = n = 35 Degrees of Freedom = df = n - 1 = 34 Test statistic(t) is given by: $$t= \frac{x-u}{ \frac{s}{ \sqrt{n} } } \\ \\ t= \frac{26.8-25.1}{ \frac{3.9}{ \sqrt{35} } } \\ \\ t=2.579$$ Step 3: Finding p value Using t table or calculators find the p value for t=2.579 with 34 degrees of freedom for one tailed test. P value comes out to be: p = 0.0072 Step 4: Conclusion Since the p value is lesser than the significance level of 0.05, we reject the Null Hypothesis. We have enough evidence to support the researcher's claim that the new additive increases the miles per gallon of the cars.
1.4 Equations of Lines and Linear Models 1 / 14 # 1.4 Equations of Lines and Linear Models - PowerPoint PPT Presentation ##### 1.4 Equations of Lines and Linear Models Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. 1.4 Equations of Lines and Linear Models 2. Quiz • If two distinct lines, y=m1x+b1, y=m2x+b2 , are parallel with each other, what’s the relationship between m1 and m2? 3. Point-Slope Form Given the slope m of a linear function and a point (x1,y1) on the graph of the linear function. We write the equation of the linear function as y-y1=m(x-x1) We call a linear equation in this form as point-slope form of a linear function. 4. Standard Form A linear equation written in the form Ax+By=C, where A,B, and C are real numbers(A and B not both 0), is said to be in standard form. Notice: When A≠0, B=0, the linear equation will be Ax=C, which is a vertical line and is not a linear function. 5. Point-Slope Form Exercise: Write the equation of the line through (-1,3) and (-2,-3). Does it matter which point is used? 6. Parallel and Perpendicular Lines Parallel Lines: Two distinct non-vertical lines are parallel if and only if they have the same slope. y=m2x+b2 y y=m1x+b1 m1=m2 x 7. Parallel and Perpendicular Lines Perpendicular Lines: Two lines, neither of which is vertical, are perpendicular if and only if their slopes have product -1 y y=m2x+b2 y=m1x+b1 m1 × m2=-1 x 8. Parallel and Perpendicular Lines Exercises: 1, Write the equation of the line through (-4,5) that is parallel to y=(1/2)x+4 2, Write the equation of the line through (5,-1) that is perpendicular to 3x-y=8. Graph both lines by hand and by using the GC. 3, Write the equation of the line through (2/3,-3/4) that is perpendicular to y=1. Graph both lines by hand and by using the GC. 9. Linear Applications Example 1: The cellular Connection charges \$60 for a phone and \$29 per month under its economy plan, Write an equation that can model the total cost, C, of operating a Cellular Connection phone for t months. Find the total cost for six months. 10. Linear Application Example 2: The number of land-line phones in the US has decreased from 101 million in 2001 to 172 million in 2006. What is the average rate of change for the number of land-line phones over that time? Predict how many land-line phones are in use in 2010. 11. Linear Regression Why Linear Regression? In most real-life situations data seldom fall into a precise line. Because of measurement errors or other random factors, a scatter plot of real-world data may appear to lie more or less on a line, but not exactly. Fitting lines to data is one of the most important tools available to researchers who need to analyze numerical data.
## 1. Introduction In Computer Science, a linked list is a linear data structure in which a pointer in each element determines the order. In this tutorial, we’ll show how to reverse a linked list. Each element of a linked list contains a data field to store the list data and a pointer field to point to the next element in the sequence. We can use a pointer to point to the start element of a linked list: After we reverse the linked list, the will point to the last element of the original linked list, and the pointer of each element will point to the previous element of the original linked list: ## 3. Iterative Solution Firstly, let’s solve a simpler version of the problem: reverse a linked list with two elements: Suppose the pointer points to the second element and the pointer points to the element before the element, we can switch the link order between them with two operations: = = For a linked list with more than two elements, we can traverse the linked list and use the same strategy to reverse the current element’s next pointer: \end{tikzpicture} algorithm reverseListIteratively(head): // INPUT // OUTPUT previous <- null while current is not null: nextElement <- current.next current.next <- previous previous <- current current <- nextElement return previous In this iterative algorithm, we first set the pointer as a pointer and the as the . Then, in each iteration of the loop, we reverse the linked pointer of these two elements and shift the and pointers to the next two elements. In the end, the pointer will point to the new head element of the reversed linked list. Since each element only has one reference to the next element, we need another pointer, , to store the next element before changing the pointer. The loop traverses the whole linked list once. Therefore, the running time of the iterative algorithm is , where is the total number of elements of the linked list. ## 4. Recursive Solution We can also solve the problem with a recursive solution. Let’s first consider a simpler case where we have reversed the rest of the linked list after the element: We only need to reverse two elements: and . At the beginning, the pointer of the is . We should change that to make it point to . Then, we need to change the next pointer of element to to finish the reversal: = = We can extend this solution to a recursive algorithm of reversing a linked list staring with a element. Firstly, we can reverse the linked list starting with element by recursively calling our reversal function. Then, the linked list becomes our simpler case. Therefore, we can reverse the last two elements, and , with the above two operations. We can construct a recursive algorithm based on this approach: algorithm reverseListRecursively(head): // INPUT // OUTPUT return null return node
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 9.1: Parts of Circles & Tangent Lines Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Define the parts of a circle. • Discover the properties of tangent lines. ## Review Queue 1. Find the equation of the line with m=25\begin{align}m = \frac{2}{5}\end{align} and \begin{align}y-\end{align}intercept of 4. 2. Find the equation of the line with \begin{align}m = -2\end{align} and passes through (4, -5). 3. Find the equation of the line that passes though (6, 2) and (-3, -1). 4. Find the equation of the line perpendicular to the line in #2 and passes through (-8, 11). Know What? The clock to the right is an ancient astronomical clock in Prague. It has a large background circle that tells the local time and the “ancient time” and the smaller circle rotates to show the current astrological sign. The yellow point is the center of the larger clock. How does the orange line relate to the small and large circle? How does the hand with the moon on it relate to both circles? ## Defining Terms Circle: The set of all points that are the same distance away from a specific point, called the center. The center of the circle is point \begin{align}A\end{align}. We call this circle, “circle \begin{align}A\end{align},” and it is labeled \begin{align}\bigodot A\end{align}. Radii (the plural of radius) are line segments. There are infinitely many radii in any circle and they are all equal. Radius: The distance from the center to the circle. Chord: A line segment whose endpoints are on a circle. Diameter: A chord that passes through the center of the circle. Secant: A line that intersects a circle in two points. The tangent ray \begin{align}\overrightarrow{TP}\end{align} and tangent segment \begin{align}\overline{TP}\end{align} are also called tangents. The length of a diameter is two times the length of a radius. Tangent: A line that intersects a circle in exactly one point. Point of Tangency: The point where the tangent line touches the circle. Example 1: Find the parts of \begin{align}\bigodot A\end{align} that best fit each description. b) A chord c) A tangent line d) A point of tangency e) A diameter f) A secant Solution: a) \begin{align}\overline{HA}\end{align} or \begin{align}\overline{AF}\end{align} b) \begin{align}\overline{CD}, \ \overline{HF}\end{align}, or \begin{align}\overline {DG}\end{align} c) \begin{align}\overleftrightarrow{BJ}\end{align} d) Point \begin{align}H\end{align} e) \begin{align}\overline{HF}\end{align} f) \begin{align}\overleftrightarrow{BD}\end{align} ## Coplanar Circles Example 2: Draw an example of how two circles can intersect with no, one and two points of intersection. You will make three separate drawings. Solution: Tangent Circles: When two circles intersect at one point. Concentric Circles: When two circles have the same center, but different radii. Congruent Circles: Two circles with the same radius, but different centers. If two circles have different radii, they are similar. All circles are similar. Example 3: Determine if any of the following circles are congruent. Solution: From each center, count the units to the circle. It is easiest to count vertically or horizontally. Doing this, we have: \begin{align}\text{Radius of} \ \bigodot A & = 3 \ units\\ \text{Radius of} \ \bigodot B & = 4 \ units\\ \text{Radius of} \ \bigodot C & = 3 \ units\end{align} From these measurements, we see that \begin{align}\bigodot A \cong \bigodot C\end{align}. Notice the circles are congruent. The lengths of the radii are equal. ## Internally & Externally Tangent If two circles are tangent to each other, then they are internally or externally tangent. Internally Tangent Circles: When two circles are tangent and one is inside the other. Externally Tangent Circles: When two circles are tangent and next to each other. Internally Tangent Externally Tangent If circles are not tangent, they can still share a tangent line, called a common tangent. Common Internal Tangent: A line that is tangent to two circles and passes between the circles. Common External Tangent: A line that is tangent to two circles and stays on the top or bottom of both circles. Common Internal Tangent Common External Tangent Let’s investigate a tangent line and the radius drawn to the point of tangency. Investigation 9-1: Tangent Line and Radius Property Tools Needed: compass, ruler, pencil, paper, protractor 1. Using your compass, draw a circle. Locate the center and draw a radius. Label the radius \begin{align}\overline{AB}\end{align}, with \begin{align}A\end{align} as the center. 2. Draw a tangent line, \begin{align}\overleftrightarrow{BC}\end{align}, where \begin{align}B\end{align} is the point of tangency. To draw a tangent line, take your ruler and line it up with point \begin{align}B\end{align}. \begin{align}B\end{align} must be the only point on the circle that the line passes through. 3. Find \begin{align}m\angle ABC\end{align}. Tangent to a Circle Theorem: A line is tangent to a circle if and only if the line is perpendicular to the radius drawn to the point of tangency. \begin{align}\overleftrightarrow{BC}\end{align} is tangent at point \begin{align}B\end{align} if and only if \begin{align}\overleftrightarrow{BC} \perp \overline{AB}\end{align}. This theorem uses the words “if and only if,” making it a biconditional statement, which means the converse of this theorem is also true. Example 4: In \begin{align}\bigodot A, \ \overline{CB}\end{align} is tangent at point \begin{align}B\end{align}. Find \begin{align}AC\end{align}. Reduce any radicals. Solution: \begin{align}\overline{CB}\end{align} is tangent, so \begin{align}\overline{AB} \perp \overline{CB}\end{align} and \begin{align}\triangle ABC\end{align} a right triangle. Use the Pythagorean Theorem to find \begin{align}AC\end{align}. \begin{align}5^2+8^2&=AC^2\\ 25+64&=AC^2\\ 89&=AC^2\\ AC&=\sqrt{89}\end{align} Example 5: Find \begin{align}DC\end{align}, in \begin{align}\bigodot A\end{align}. Round your answer to the nearest hundredth. Solution: \begin{align}DC = AC - AD\!\\ DC = \sqrt{89}-5 \approx 4.43\end{align} Example 6: Determine if the triangle below is a right triangle. Solution: Again, use the Pythagorean Theorem. \begin{align}4\sqrt{10}\end{align} is the longest side, so it will be \begin{align}c\end{align}. \begin{align}8^2+10^2 \ & ? \ \left ( 4\sqrt{10} \right )^2\\ 64+100 & \ne 160\end{align} \begin{align}\triangle ABC\end{align} is not a right triangle. From this, we also find that \begin{align}\overline{CB}\end{align} is not tangent to \begin{align}\bigodot A\end{align}. Example 7: Find \begin{align}AB\end{align} in \begin{align}\bigodot A\end{align} and \begin{align}\bigodot B\end{align}. Reduce the radical. Solution: \begin{align}\overline{AD} \perp \overline{DC}\end{align} and \begin{align}\overline{DC} \perp \overline{CB}\end{align}. Draw in \begin{align}\overline{BE}\end{align}, so \begin{align}EDCB\end{align} is a rectangle. Use the Pythagorean Theorem to find \begin{align}AB\end{align}. \begin{align}5^2+55^2&=AC^2\\ 25+3025&=AC^2\\ 3050&=AC^2\\ AC&=\sqrt{3050}=5\sqrt{122}\end{align} ## Tangent Segments Theorem 9-2: If two tangent segments are drawn from the same external point, then they are equal. \begin{align}\overline{BC}\end{align} and \begin{align}\overline{DC}\end{align} have \begin{align}C\end{align} as an endpoint and are tangent; \begin{align}\overline{BC} \cong \overline{DC}\end{align}. Example 8: Find the perimeter of \begin{align}\triangle ABC\end{align}. Solution: \begin{align}AE = AD, \ EB = BF,\end{align} and \begin{align}CF = CD\end{align}. Therefore, the perimeter of \begin{align}\triangle ABC=6+6+4+4+7+7=34\end{align}. \begin{align}\bigodot G\end{align} is inscribed in \begin{align}\triangle ABC\end{align}. A circle is inscribed in a polygon, if every side of the polygon is tangent to the circle. Example 9: If \begin{align}D\end{align} and \begin{align}A\end{align} are the centers and \begin{align}AE\end{align} is tangent to both circles, find \begin{align}DC\end{align}. Solution: \begin{align}\overline{AE} \perp \overline{DE}\end{align} and \begin{align}\overline{AE} \perp \overline{AC}\end{align} and \begin{align}\triangle ABC \sim \triangle DBE\end{align}. To find \begin{align}DB\end{align}, use the Pythagorean Theorem. \begin{align}10^2+24^2&=DB^2\\ 100+576&=676\\ DB&=\sqrt{676}=26\end{align} To find \begin{align}BC\end{align}, use similar triangles. \begin{align}\frac{5}{10}=\frac{BC}{26} \longrightarrow BC=13. \ DC=AB+BC=26+13=39\end{align} Example 10: Algebra Connection Find the value of \begin{align}x\end{align}. Solution: \begin{align}\overline{AB} \cong \overline{CB}\end{align} by Theorem 9-2. Set \begin{align}AB = CB\end{align} and solve for \begin{align}x\end{align}. \begin{align}4x-9&=15\\ 4x&=24\\ x&=6\end{align} Know What? Revisited The orange line is a diameter of the smaller circle. Since this line passes through the center of the larger circle (yellow point), it is part of one of its diameters. The “moon” hand is a diameter of the larger circle, but a secant of the smaller circle. ## Review Questions • Questions 1-9 are similar to Example 1. • Questions 10-12 are similar to Example 2. • Questions 13-17 are similar to Example 3. • Questions 18-20 are similar to Example 6. • Questions 21-26 are similar to Example 4, 5, 7, and 10. • Questions 27-31 are similar to Example 9. • Questions 32-37 are similar to Example 8. • Question 38 and 39 use the proof of Theorem 9-2. • Question 40 uses Theorem 9-2. Determine which term best describes each of the following parts of \begin{align}\bigodot P\end{align}. 1. \begin{align}\overline{KG}\end{align} 2. \begin{align}\overleftrightarrow{FH}\end{align} 3. \begin{align}\overline{KH}\end{align} 4. \begin{align}E\end{align} 5. \begin{align}\overleftrightarrow{BK}\end{align} 6. \begin{align}\overleftrightarrow{CF}\end{align} 7. \begin{align}A\end{align} 8. \begin{align}\overline{JG}\end{align} 9. What is the longest chord in any circle? Copy each pair of circles. Draw in all common tangents. Coordinate Geometry Use the graph below to answer the following questions. 1. Find the radius of each circle. 2. Are any circles congruent? How do you know? 3. Find all the common tangents for \begin{align}\bigodot B\end{align} and \begin{align}\bigodot C\end{align}. 4. \begin{align}\bigodot C\end{align} and \begin{align}\bigodot E\end{align} are externally tangent. What is \begin{align}CE\end{align}? 5. Find the equation of \begin{align}\overline{CE}\end{align}. Determine whether the given segment is tangent to \begin{align}\bigodot K\end{align}. Algebra Connection Find the value of the indicated length(s) in \begin{align}\bigodot C\end{align}. \begin{align}A\end{align} and \begin{align}B\end{align} are points of tangency. Simplify all radicals. \begin{align}A\end{align} and \begin{align}B\end{align} are points of tangency for \begin{align}\bigodot C\end{align} and \begin{align}\bigodot D\end{align}. 1. Is \begin{align}\triangle AEC \sim \triangle BED\end{align}? Why? 2. Find \begin{align}CE\end{align}. 3. Find \begin{align}BE\end{align}. 4. Find \begin{align}ED\end{align}. 5. Find \begin{align}BC\end{align} and \begin{align}AD\end{align}. \begin{align}\bigodot A\end{align} is inscribed in \begin{align}BDFH\end{align}. 1. Find the perimeter of \begin{align}BDFH\end{align}. 2. What type of quadrilateral is \begin{align}BDFH\end{align}? How do you know? 3. Draw a circle inscribed in a square. If the radius of the circle is 5, what is the perimeter of the square? 4. Can a circle be inscribed in a rectangle? If so, draw it. If not, explain. 5. Draw a triangle with two sides tangent to a circle, but the third side is not. 6. Can a circle be inscribed in an obtuse triangle? If so, draw it. If not, explain. 7. Fill in the blanks in the proof of Theorem 9-2. Given: \begin{align}\overline{AB}\end{align} and \begin{align}\overline{CB}\end{align} with points of tangency at \begin{align}A\end{align} and \begin{align}C\end{align}. \begin{align}\overline{AD}\end{align} and \begin{align}\overline{DC}\end{align} are radii. Prove: \begin{align}\overline{AB} \cong \overline{CB}\end{align} Statement Reason 1. 2. \begin{align}\overline{AD} \cong \overline{DC}\end{align} 3. \begin{align}\overline{DA} \perp \overline{AB}\end{align} and \begin{align}\overline{DC} \perp \overline{CB}\end{align} 4. Definition of perpendicular lines 5. Connecting two existing points 6. \begin{align}\triangle ADB\end{align} and \begin{align}\triangle DCB\end{align} are right triangles 7. \begin{align}\overline{DB} \cong \overline{DB}\end{align} 8. \begin{align}\triangle ABD \cong \triangle CBD\end{align} 9. \begin{align}\overline{AB} \cong \overline{CB}\end{align} 1. Fill in the blanks, using the proof from #38. 1. \begin{align}ABCD\end{align} is a _____________ (type of quadrilateral). 2. The line that connects the ___________ and the external point \begin{align}B\end{align} __________ \begin{align}\angle ABC\end{align}. 2. Points \begin{align}A, \ B,\end{align} and \begin{align}C\end{align} are points of tangency for the three tangent circles. Explain why \begin{align}\overline{AT} \cong \overline{BT} \cong \overline{CT}\end{align}. 1. \begin{align}y = \frac{2}{5} x + 4\end{align} 2. \begin{align}y = -2x + 3\end{align} 3. \begin{align}{\;} \ m = \frac{2 - (-1)}{6 - (-3)} = \frac{3}{9} = \frac{1}{3}\!\\ {\;} \ y = \frac{1}{3} x+b \rightarrow \text{plug in} (6, 2)\!\\ {\;} \ 2 = \frac{1}{3} (6)+b\!\\ {\;} \ 2 =2+b \rightarrow b=0\!\\ {\;} \ y = \frac{1}{3} x\end{align} 4. \begin{align}{\;} \ \ m_\perp = -3\!\\ {\;} \quad 11 = -3(-8)+b\!\\ {\;} \quad 11 = 24+b \rightarrow b=-13\!\\ {\;} \quad \ \ y = -3x-13 \end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
Courses Courses for Kids Free study material Offline Centres More Store # The difference between the two numbers is 26 and one of the numbers is three times the other. Find the sum of these numbers. Last updated date: 20th Jul 2024 Total views: 449.1k Views today: 12.49k Verified 449.1k+ views Hint: Using the equation of difference between two numbers and one number is thrice the other number, determine the two numbers. Then, find the sum of these two numbers. Let us assume the two numbers to be x and y with x greater than y. It is given that the difference between the two numbers is 26. Hence, we have: $x - y = 26.........(1)$ It is also given that one of the numbers is three times the other number. Since, x is the greatest number of the two, x is 3 times y. Hence, we have: $x = 3y............(2)$ We have two equations with two unknowns x and y. Hence, we can solve them Substitute equation (2) in equation (1) to get: $3y - y = 26$ We know that 3y – y is 2y, hence, we have: $2y = 26$ Taking 2 to the other side and dividing with 26, we get 13. $y = \dfrac{{26}}{2}$ $y = 13...........(3)$ Using equation (3) in equation (2), we get the value of x. $x = 3(13)$ We know the value of 3(13) is 39, hence, we have: $x = 39..........(4)$ We now need to find the sum of these two numbers x and y. From equation (3) and equation (4), we can add x and y to get the desired answer. $x + y = 39 + 13$ $x + y = 52$ Hence, the value of the sum of the two numbers is 52. Note: We can cross check your answer by substituting the value of the variables in the equations and see if they satisfy the expressions. This will help in understanding whether we got the correct answer or not.
# Formation of Numbers with the Given Digits \| How to Form Different Numbers with Given Digits? Numbers is the word that plays the main role in every aspect of our lives. We use numbers not only in mathematics to solve problems but, also for comparisons in various ways and measuring quantities, and so on. To solve or to compare any things or objects first we should form the numbers. Forming numbers may be done with or without repetition of given digits. While forming the numbers if you are confused to know the smallest and greatest numbers then take a look at the Formation of Greatest and Smallest Numbers here. For now, we are going to discuss the Formation of Numbers with the given digits. This article will surely help you to clear your confusion in every aspect, so without having any delay let’s move ahead and check what is it and how to form the numbers with given digits. ## What is meant by the Formation of Numbers with Given Digits? Numbers Formation is created by increasing the digits number. It is done in various directions where we form the digits in our brains. First and foremost, the kids should clearly know about all types of digits to form the numbers. The numbers are arranged as a group of digits, with or without repetition of digits. Forming numbers can also be done in ascending order or decreasing order. By using or imagining the number line we arrange the given digits in order. After the formation of numbers, we can also know the highest and lowest values. If we arrange the given numbers in ascending order, then it is the smallest number. For the greatest number, it is vice-versa, because you write the numbers in descending order. Let’s witness some of the formation of numbers with or without the repetition of digits in the below section. ### Forming Numbers with Given Digits Formation of Two-Digit Numbers with Given Digits (i) With the single digits 5 and 3, the formed numbers are 35 and 53. (ii) With the single digits 8 and 6, the formed numbers are 68 and 86. (iii) With the single digits 4 and 7, the formed numbers are 47 and 74. (iv) With the single digits 0 and 2, the formed numbers are 20, because, by starting with the digit 0 we get 02 and that is not a two digits number, and 0 has no value at the start of the number. Formation of Three-Digit Numbers with Given Digits (i) With the digits 2, 4, and 5, the formed numbers are 245, 254, 425, 452, 524, 542. (ii) With the digits 3, 6, and 9, the formed numbers are 369, 396, 639, 693, 936, 963. (iii) With the digits 7, 0, and 8, the formed numbers are 708, 780, 807, 870. Here, we get only 4 values of three digits. Some, of the numbers formed by the digits 1, 2, 3, and 4 are as follows: 1234, 1243, 1324, 1342, 1423, 1432 2134, 2143, 2314, 2341, 2413, 2431 3124, 3142, 3421, 3412, 3214, 3241 4123, 4132, 4213, 4231, 4312, 4321. In a similar way for forming the numbers, when the digits of numbers are increased, the formation of numbers is also increased. Then, it is also easy to know the smallest and the greatest values of those numbers you formed. ### Forming Numbers with Given Digits Examples Example 1: Find the smallest and greatest four digits number using the digits 6, 2, 7, 1 such that the number thus formed has 7 at its hundreds place and 2 at its one’s place. Solution: First, we form the smallest and greatest four digits number from the given single digits 6, 2, 7, and 1. The smallest digit number is 1267, and the greatest digit number is 7621. But, we have certain rules to form the smallest and greatest numbers from the question. It says, the number in hundreds place should be 7 and the number at one’s place should be 2. We have to form the smallest number and the greatest number values by applying these rules. So, arrange the remaining numbers in ascending order i.e., 1 and 6, and descending order, i.e., 6 and 1. The smallest number is 1762. The greatest number is 6712. Example 2: Write the formation of three digits numbers from the given digits 4, 6, and 9? Solution: The given digits are 4, 6, and 9 Now, form the possible three digits formation of those given numbers 469, 496, 649, 694, 946, 964. ### FAQs on Formation of Numbers with the Given Digits of a Number 1. What is number formation? In the formation of numbers with the given digits, we say that the number is arranged as a group of digits. Numbers may be formed in ascending or descending order with or without repetition of digits. 2. How numbers are formed? Numbers are formed by using digits and various mathematical symbols. The number system is the numeral representation. In numbers, there are four types of number systems namely the Decimal number system (Base-10), Binary number system (Base-2), Hexadecimal number system (Base-16), and Octal number system (Base-8). 3. How many digits we use for the formation of numbers? We use ten digits to form numbers in different ways. The ten digits we use to make numerals are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Scroll to Top
# 8.2 Trigonometric Integrals Trigonometric functions are useful for describing periodic behavior. This section describes several techniques for finding antiderivatives of certain combinations of trigonometric functions. ## Integrals of the form $\displaystyle\int\sin^{m}x\cos^{n}x\operatorname{d}\!x$ In learning the technique of Substitution, we saw the integral $\int\sin x\cos x\operatorname{d}\!x$ in Example 5.5.4. The integration was not difficult, and one could easily evaluate the indefinite integral by letting $u=\sin x$ or by letting $u=\cos x$. This integral is easy since the power of both sine and cosine is 1. We generalize this and consider integrals of the form $\int\sin^{m}x\cos^{n}x\operatorname{d}\!x$, where $m,n$ are nonnegative integers. Our strategy for evaluating these integrals is to use the identity $\cos^{2}x+\sin^{2}x=1$ to convert high powers of one trigonometric function into the other, leaving a single sine or cosine term in the integrand. We summarize the general technique in the following Key Idea. ###### Key Idea 8.2.1 Integrals Involving Powers of Sine and Cosine Consider $\displaystyle\int\sin^{m}x\cos^{n}x\operatorname{d}\!x$, where $m,n$ are nonnegative integers. 1. (a) If $m$ is odd, then $m=2k+1$ for some integer $k$. Rewrite $\sin^{m}x=\sin^{2k+1}x=\sin^{2k}x\sin x=(\sin^{2}x)^{k}\sin x=(1-\cos^{2}x)^{k% }\sin x.$ Then $\int\sin^{m}x\cos^{n}x\operatorname{d}\!x=\int(1-\cos^{2}x)^{k}\sin x\cos^{n}x% \operatorname{d}\!x=-\int(1-u^{2})^{k}u^{n}\operatorname{d}\!u,$ where $u=\cos x$ and $\operatorname{d}\!u=-\sin x\operatorname{d}\!x$. 2. (b) If $n$ is odd, then using substitutions similar to that outlined above we have $\int\sin^{m}x\cos^{n}x\operatorname{d}\!x=\int u^{m}(1-u^{2})^{k}\operatorname% {d}\!u,$ where $u=\sin x$ and $\operatorname{d}\!u=\cos x\operatorname{d}\!x$. 3. (c) If both $m$ and $n$ are even, use the half-angle identities $\cos^{2}x=\frac{1+\cos(2x)}{2}\quad\text{and}\quad\sin^{2}x=\frac{1-\cos(2x)}{2}$ to reduce the degree of the integrand. Expand the result and apply the principles of this Key Idea again. We practice applying Key Idea 8.2.1 in the next examples. ###### Example 8.2.1 Integrating powers of sine and cosine Evaluate $\displaystyle\int\sin^{5}x\cos^{8}x\operatorname{d}\!x$. SolutionThe power of the sine factor is odd, so we rewrite $\sin^{5}x$ as $\sin^{5}x=\sin^{4}x\sin x=(\sin^{2}x)^{2}\sin x=(1-\cos^{2}x)^{2}\sin x.$ Our integral is now $\displaystyle\int(1-\cos^{2}x)^{2}\cos^{8}x\sin x\operatorname{d}\!x$. Let $u=\cos x$, hence $\operatorname{d}\!u=-\sin x\operatorname{d}\!x$. Making the substitution and expanding the integrand gives $\displaystyle\int(1-\cos^{2}x)^{2}\cos^{8}x\sin x\operatorname{d}\!x$ $\displaystyle=-\int(1-u^{2})^{2}u^{8}\operatorname{d}\!u$ $\displaystyle=-\int\bigl{(}1-2u^{2}+u^{4}\bigr{)}u^{8}\operatorname{d}\!u$ $\displaystyle=-\int\bigl{(}u^{8}-2u^{10}+u^{12}\bigr{)}\operatorname{d}\!u$ $\displaystyle=-\frac{1}{9}u^{9}+\frac{2}{11}u^{11}-\frac{1}{13}u^{13}+C$ $\displaystyle=-\frac{1}{9}\cos^{9}x+\frac{2}{11}\cos^{11}x-\frac{1}{13}\cos^{1% 3}x+C.$ ###### Example 8.2.2 Integrating powers of sine and cosine Evaluate $\displaystyle\int\sin^{5}x\cos^{9}x\operatorname{d}\!x$. SolutionBecause the powers of both the sine and cosine factors are odd, we can apply the techniques of Key Idea 8.2.1 to either power. We choose to work with the power of the sine factor since that has a smaller exponent. We rewrite $\sin^{5}x$ as $\displaystyle\sin^{5}x$ $\displaystyle=\sin^{4}x\sin x$ $\displaystyle=(1-\cos^{2}x)^{2}\sin x$ $\displaystyle=(1-2\cos^{2}x+\cos^{4}x)\sin x.$ This lets us rewrite the integral as $\int\sin^{5}x\cos^{9}x\operatorname{d}\!x=\int\bigl{(}1-2\cos^{2}x+\cos^{4}x% \bigr{)}\sin x\cos^{9}x\operatorname{d}\!x.$ Substituting and integrating with $u=\cos x$ and $\operatorname{d}\!u=-\sin x\operatorname{d}\!x$, we have $\displaystyle\int\bigl{(}1-2\cos^{2}x+\cos^{4}x\bigr{)}$ $\displaystyle\sin x\cos^{9}x\operatorname{d}\!x$ $\displaystyle=-\int\bigl{(}1-2u^{2}+u^{4}\bigr{)}u^{9}\operatorname{d}\!u$ $\displaystyle=-\int u^{9}-2u^{11}+u^{13}\operatorname{d}\!u$ $\displaystyle=-\frac{1}{10}u^{10}+\frac{1}{6}u^{12}-\frac{1}{14}u^{14}+C$ $\displaystyle=-\frac{1}{10}\cos^{10}x+\frac{1}{6}\cos^{12}x-\frac{1}{14}\cos^{% 14}x+C.$ Instead, another approach would be to rewrite $\cos^{9}x$ as $\displaystyle\cos^{9}x$ $\displaystyle=\cos^{8}x\cos x$ $\displaystyle=(\cos^{2}x)^{4}\cos x$ $\displaystyle=(1-\sin^{2}x)^{4}\cos x$ $\displaystyle=(1-4\sin^{2}x+6\sin^{4}x-4\sin^{6}x+\sin^{8}x)\cos x.$ We rewrite the integral as $\int\sin^{5}x\cos^{9}x\operatorname{d}\!x=\int\bigl{(}\sin^{5}x\bigr{)}\bigl{(% }1-4\sin^{2}x+6\sin^{4}x-4\sin^{6}x+\sin^{8}x\bigr{)}\cos x\operatorname{d}\!x.$ Now substitute and integrate, using $u=\sin x$ and $\operatorname{d}\!u=\cos x\operatorname{d}\!x$. $\displaystyle\int$ $\displaystyle\bigl{(}\sin^{5}x\bigr{)}\bigl{(}1-4\sin^{2}x+6\sin^{4}x-4\sin^{6% }x+\sin^{8}x\bigr{)}\cos x\operatorname{d}\!x$ $\displaystyle=\int u^{5}(1-4u^{2}+6u^{4}-4u^{6}+u^{8})\operatorname{d}\!u$ $\displaystyle=\int\bigl{(}u^{5}-4u^{7}+6u^{9}-4u^{11}+u^{13}\bigr{)}% \operatorname{d}\!u$ $\displaystyle=\frac{1}{6}u^{6}-\frac{1}{2}u^{8}+\frac{3}{5}u^{10}-\frac{1}{3}u% ^{12}+\frac{1}{14}u^{14}+C$ $\displaystyle=\frac{1}{6}\sin^{6}x-\frac{1}{2}\sin^{8}x+\frac{3}{5}\sin^{10}x-% \frac{1}{3}\sin^{12}x+\frac{1}{14}\sin^{14}x+C.$ #### Technology Note: The work we are doing here can be a bit tedious, but the skills developed (problem solving, algebraic manipulation, etc.) are important. Nowadays problems of this sort are often solved using a computer algebra system. The powerful program Mathematica® integrates $\int\sin^{5}x\cos^{9}x\operatorname{d}\!x$ as $f(x)=\\ -\frac{45\cos(2x)}{16384}-\frac{5\cos(4x)}{8192}+\frac{19\cos(6x)}{49152}+% \frac{\cos(8x)}{4096}-\frac{\cos(10x)}{81920}-\frac{\cos(12x)}{24576}-\frac{% \cos(14x)}{114688},$ which clearly has a different form than our second answer in Example 8.2.2, which is margin: Figure 8.2.1: A plot of $f(x)$ and $g(x)$ from Example 8.2.2 and the Technology Note. Λ $g(x)=\frac{1}{6}\sin^{6}x-\frac{1}{2}\sin^{8}x+\frac{3}{5}\sin^{10}x-\frac{1}{% 3}\sin^{12}x+\frac{1}{14}\sin^{14}x.$ Figure 8.2.1 shows a graph of $f$ and $g$; they are clearly not equal, but they differ only by a constant: $g(x)=f(x)+C$ for some constant $C$. So we have two different antiderivatives of the same function, meaning both answers are correct. ###### Example 8.2.3 Integrating powers of sine and cosine Evaluate $\displaystyle\int\sin^{2}x\operatorname{d}\!x$. SolutionThe power of sine is even so we employ a half-angle identity, algebra and a u- substitution as follows: $\displaystyle\int\sin^{2}x\operatorname{d}\!x$ $\displaystyle=\int\frac{1-\cos(2x)}{2}\operatorname{d}\!x$ $\displaystyle=\frac{1}{2}\int 1-\cos(2x)\operatorname{d}\!x$ $\displaystyle=\frac{1}{2}\left(x-\frac{1}{2}\sin(2x)\right)+C$ $\displaystyle=\frac{1}{2}x-\frac{1}{4}\sin(2x)+C.$ ###### Example 8.2.4 Integrating powers of sine and cosine Evaluate $\displaystyle\int\cos^{4}x\sin^{2}x\operatorname{d}\!x$. SolutionThe powers of sine and cosine are both even, so we employ the half-angle formulas and algebra as follows. $\displaystyle\int\cos^{4}x\sin^{2}x\operatorname{d}\!x$ $\displaystyle=\int\left(\frac{1+\cos(2x)}{2}\right)^{2}\left(\frac{1-\cos(2x)}% {2}\right)\operatorname{d}\!x$ $\displaystyle=\int\frac{1+2\cos(2x)+\cos^{2}(2x)}{4}\cdot\frac{1-\cos(2x)}{2}% \operatorname{d}\!x$ $\displaystyle=\int\frac{1}{8}\bigl{(}1+\cos(2x)-\cos^{2}(2x)-\cos^{3}(2x)\bigr% {)}\operatorname{d}\!x$ The $\cos(2x)$ term is easy to integrate. The $\cos^{2}(2x)$ term is another trigonometric integral with an even power, requiring the half-angle formula again. The $\cos^{3}(2x)$ term is a cosine function with an odd power, requiring a substitution as done before. We integrate each in turn below. $\displaystyle\int\cos(2x)\operatorname{d}\!x=\frac{1}{2}\sin(2x)+C.$ $\displaystyle\int\cos^{2}(2x)\operatorname{d}\!x=\int\frac{1+\cos(4x)}{2}% \operatorname{d}\!x=\frac{1}{2}\bigl{(}x+\frac{1}{4}\sin(4x)\bigr{)}+C.$ Finally, we rewrite $\cos^{3}(2x)$ as $\cos^{3}(2x)=\cos^{2}(2x)\cos(2x)=\bigl{(}1-\sin^{2}(2x)\bigr{)}\cos(2x).$ Letting $u=\sin(2x)$, we have $\operatorname{d}\!u=2\cos(2x)\operatorname{d}\!x$, hence $\displaystyle\int\cos^{3}(2x)\operatorname{d}\!x$ $\displaystyle=\int\bigl{(}1-\sin^{2}(2x)\bigr{)}\cos(2x)\operatorname{d}\!x$ $\displaystyle=\int\frac{1}{2}(1-u^{2})\operatorname{d}\!u$ $\displaystyle=\frac{1}{2}\Bigl{(}u-\frac{1}{3}u^{3}\Bigr{)}+C$ $\displaystyle=\frac{1}{2}\Bigl{(}\sin(2x)-\frac{1}{3}\sin^{3}(2x)\Bigr{)}+C$ Putting all the pieces together, we have $\displaystyle\int$ $\displaystyle\cos^{4}x\sin^{2}x\operatorname{d}\!x$ $\displaystyle=\int\frac{1}{8}\bigl{(}1+\cos(2x)-\cos^{2}(2x)-\cos^{3}(2x)\bigr% {)}\operatorname{d}\!x$ $\displaystyle=\frac{1}{8}\Bigl{[}x+\frac{1}{2}\sin(2x)-\frac{1}{2}\bigl{(}x+% \frac{1}{4}\sin(4x)\bigr{)}-\frac{1}{2}\Bigl{(}\sin(2x)-\frac{1}{3}\sin^{3}(2x% )\Bigr{)}\Bigr{]}+C$ $\displaystyle=\frac{1}{8}\Bigl{[}\frac{1}{2}x-\frac{1}{8}\sin(4x)+\frac{1}{6}% \sin^{3}(2x)\Bigr{]}+C.$ The process above was a bit long and tedious, but being able to work a problem such as this from start to finish is important. ## Integrals of the form $\displaystyle\int\tan^{m}x\sec^{n}x\operatorname{d}\!x$ When evaluating integrals of the form $\int\sin^{m}x\cos^{n}x\operatorname{d}\!x$, the Pythagorean Theorem allowed us to convert even powers of sine into even powers of cosine, and vice versa. If, for instance, the power of sine was odd, we pulled out one $\sin x$ and converted the remaining even power of $\sin x$ into a function using powers of $\cos x$, leading to an easy substitution. The same basic strategy applies to integrals of the form $\int\tan^{m}x\sec^{n}x\operatorname{d}\!x$, albeit a bit more nuanced. The following three facts will prove useful: • $\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\tan x)=\sec^{2}x$, • $\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\sec x)=\sec x\tan x$ , and • $1+\tan^{2}x=\sec^{2}x$ (the Pythagorean Theorem). If the integrand can be manipulated to separate a $\sec^{2}x$ term with the remaining secant power even, or if a $\sec x\tan x$ term can be separated with the remaining $\tan x$ power even, the Pythagorean Theorem can be employed, leading to a simple substitution. This strategy is outlined in the following Key Idea. ###### Key Idea 8.2.2 Integrals Involving Powers of Tangent and Secant Consider $\displaystyle\int\tan^{m}x\sec^{n}x\operatorname{d}\!x$, where $m$ and $n$ are nonnegative integers. 1. (a) If $n$ is even, then $n=2k$ for some integer $k$. Rewrite $\sec^{n}x$ as $\sec^{n}x=\sec^{2k}x=\sec^{2k-2}x\sec^{2}x=(1+\tan^{2}x)^{k-1}\sec^{2}x.$ Then $\int\tan^{m}x\sec^{n}x\operatorname{d}\!x=\int\tan^{m}x(1+\tan^{2}x)^{k-1}\sec% ^{2}x\operatorname{d}\!x=\int u^{m}(1+u^{2})^{k-1}\operatorname{d}\!u,$ where $u=\tan x$ and $\operatorname{d}\!u=\sec^{2}x\operatorname{d}\!x$. 2. (b) If $m$ is odd and $n>0$, then $m=2k+1$ for some integer $k$. Rewrite $\tan^{m}x\sec^{n}x$ as $\tan^{m}x\sec^{n}x=\tan^{2k+1}x\sec^{n}x=\tan^{2k}x\sec^{n-1}x\sec x\tan x=(% \sec^{2}x-1)^{k}\sec^{n-1}x\sec x\tan x.$ Then $\int\tan^{m}x\sec^{n}x\operatorname{d}\!x=\int(\sec^{2}x-1)^{k}\sec^{n-1}x\sec x% \tan x\operatorname{d}\!x=\int(u^{2}-1)^{k}u^{n-1}\operatorname{d}\!u,$ where $u=\sec x$ and $\operatorname{d}\!u=\sec x\tan x\operatorname{d}\!x$. 3. (c) If $n$ is odd and $m$ is even, then $m=2k$ for some integer $k$. Convert $\tan^{m}x$ to $(\sec^{2}x-1)^{k}$. Expand the new integrand and use Integration By Parts, with $\operatorname{d}\!v=\sec^{2}x\operatorname{d}\!x$. 4. (d) If $m$ is even and $n=0$, rewrite $\tan^{m}x$ as $\tan^{m}x=\tan^{m-2}x\tan^{2}x=\tan^{m-2}x(\sec^{2}x-1)=\tan^{m-2}\sec^{2}x-% \tan^{m-2}x.$ So $\int\tan^{m}x\operatorname{d}\!x=\underbrace{\int\tan^{m-2}\sec^{2}x% \operatorname{d}\!x}_{\text{\small apply rule \#1}}\quad-\underbrace{\int\tan^% {m-2}x\operatorname{d}\!x}_{\text{\small apply rule \#4 again}}.$ The techniques described in items 1 and 2 of Key Idea 8.2.2 are relatively straightforward, but the techniques in items 3 and 4 can be rather tedious. A few examples will help with these methods. ###### Example 8.2.5 Integrating powers of tangent and secant Evaluate $\displaystyle\int\tan^{2}x\sec^{6}x\operatorname{d}\!x$. SolutionSince the power of secant is even, we use rule #1 from Key Idea 8.2.2 and pull out a $\sec^{2}x$ in the integrand. We convert the remaining powers of secant into powers of tangent. $\displaystyle\int\tan^{2}x\sec^{6}x\operatorname{d}\!x$ $\displaystyle=\int\tan^{2}x\sec^{4}x\sec^{2}x\operatorname{d}\!x$ $\displaystyle=\int\tan^{2}x\bigl{(}1+\tan^{2}x\bigr{)}^{2}\sec^{2}x% \operatorname{d}\!x$ Now substitute, with $u=\tan x$, with $\operatorname{d}\!u=\sec^{2}x\operatorname{d}\!x$. $\displaystyle=\int u^{2}\bigl{(}1+u^{2}\bigr{)}^{2}\operatorname{d}\!u$ We leave the integration and subsequent substitution to the reader. The final answer is $\displaystyle=\frac{1}{3}\tan^{3}x+\frac{2}{5}\tan^{5}x+\frac{1}{7}\tan^{7}x+C.$ We derived integrals for tangent and secant in Section 5.5 and will regularly use them when evaluating integrals of the form $\tan^{m}x\sec^{n}x\operatorname{d}\!x$. As a reminder: $\displaystyle\int\tan x\operatorname{d}\!x$ $\displaystyle=\ln\left\lvert\sec x\right\rvert+C$ $\displaystyle\int\sec x\operatorname{d}\!x$ $\displaystyle=\ln\left\lvert\sec x+\tan x\right\rvert+C$ ###### Example 8.2.6 Integrating powers of tangent and secant Evaluate $\displaystyle\int\sec^{3}x\operatorname{d}\!x$. SolutionWe apply rule #3 from Key Idea 8.2.2 as the power of secant is odd and the power of tangent is even (0 is an even number). We use Integration by Parts; the rule suggests letting $\operatorname{d}\!v=\sec^{2}x\operatorname{d}\!x$, meaning that $u=\sec x$. \begin{aligned} u&=\sec x&\operatorname{d}\!v&=\sec^{2}x\operatorname{d}\!x\\ \operatorname{d}\!u&=\text{?}&v&=\text{?}\end{aligned}\qquad\Rightarrow\qquad% \begin{aligned} u&=\sec x&\operatorname{d}\!v&=\sec^{2}x\operatorname{d}\!x\\ \operatorname{d}\!u&=\sec x\tan x\operatorname{d}\!x&v&=\tan x\end{aligned} Figure 8.2.2: Setting up Integration by Parts. Employing Integration by Parts, we have $\displaystyle\int\sec^{3}x\operatorname{d}\!x$ $\displaystyle=\int\underbrace{\sec x}_{u}\cdot\underbrace{\sec^{2}x% \operatorname{d}\!x}_{\operatorname{d}\!v}$ $\displaystyle=\sec x\tan x-\int\sec x\tan^{2}x\operatorname{d}\!x.$ This new integral also requires applying rule #3 of Key Idea 8.2.2: $\displaystyle=\sec x\tan x-\int\sec x\bigl{(}\sec^{2}x-1\bigr{)}\operatorname{% d}\!x$ $\displaystyle=\sec x\tan x-\int\sec^{3}x\operatorname{d}\!x+\int\sec x% \operatorname{d}\!x$ $\displaystyle=\sec x\tan x-\int\sec^{3}x\operatorname{d}\!x+\ln\left\lvert\sec x% +\tan x\right\rvert$ margin: Note: Remember that in Example 5.5.8, we found that $\int\sec x\operatorname{d}\!x=\ln\left\lvert\sec x+\tan x\right\rvert+C$ Λ In previous applications of Integration by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding $\int\sec^{3}x\operatorname{d}\!x$ to both sides, giving: $\displaystyle 2\int\sec^{3}x\operatorname{d}\!x$ $\displaystyle=\sec x\tan x+\ln\left\lvert\sec x+\tan x\right\rvert$ $\displaystyle\int\sec^{3}x\operatorname{d}\!x$ $\displaystyle=\frac{1}{2}\Bigl{(}\sec x\tan x+\ln\left\lvert\sec x+\tan x% \right\rvert\Bigr{)}+C.$ We give one more example. ###### Example 8.2.7 Integrating powers of tangent and secant Evaluate $\displaystyle\int\tan^{6}x\operatorname{d}\!x$. SolutionWe employ rule #4 of Key Idea 8.2.2. $\displaystyle\int\tan^{6}x\operatorname{d}\!x$ $\displaystyle=\int\tan^{4}x\tan^{2}x\operatorname{d}\!x$ $\displaystyle=\int\tan^{4}x\bigl{(}\sec^{2}x-1\bigr{)}\operatorname{d}\!x$ $\displaystyle=\int\tan^{4}x\sec^{2}x\operatorname{d}\!x-\int\tan^{4}x% \operatorname{d}\!x$ We integrate the first integral with substitution, $u=\tan x$ and $\operatorname{d}\!u=\sec^{2}x\operatorname{d}\!x$; and the second by employing rule #4 again. $\displaystyle=\int u^{4}\operatorname{d}\!u-\int\tan^{2}x\tan^{2}x% \operatorname{d}\!x$ $\displaystyle=\frac{1}{5}\tan^{5}x-\int\tan^{2}x\bigl{(}\sec^{2}x-1\bigr{)}% \operatorname{d}\!x$ $\displaystyle=\frac{1}{5}\tan^{5}x-\int\tan^{2}x\sec^{2}x\operatorname{d}\!x+% \int\tan^{2}x\operatorname{d}\!x$ Again, use substitution for the first integral and rule #4 for the second. $\displaystyle=\frac{1}{5}\tan^{5}x-\frac{1}{3}\tan^{3}x+\int\bigl{(}\sec^{2}x-% 1\bigr{)}\operatorname{d}\!x$ $\displaystyle=\frac{1}{5}\tan^{5}x-\frac{1}{3}\tan^{3}x+\tan x-x+C.$ ## Integrals of the form $\displaystyle\int\cot^{m}x\csc^{n}x\operatorname{d}\!x$ Not surprisingly, evaluating integrals of the form $\int\cot^{m}x\csc^{n}x\operatorname{d}\!x$ is similar to evaluating $\int\tan^{m}x\sec^{n}x\operatorname{d}\!x$. The guidelines from Key Idea 8.2.2 and the following three facts will be useful: $\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\cot x)$ $\displaystyle=-\csc^{2}x$ $\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\csc x)$ $\displaystyle=-\csc x\cot x,\qquad\text{and}$ $\displaystyle\csc^{2}x$ $\displaystyle=\cot^{2}x+1$ ###### Example 8.2.8 Integrating powers of cotangent and cosecant Evaluate $\displaystyle\int\cot^{2}x\csc^{4}x\operatorname{d}\!x$ SolutionSince the power of cosecant is even we will let $u=\cot x$ and save a $\csc^{2}x$ for the resulting $\operatorname{d}\!u=-\csc^{2}x\operatorname{d}\!x$. $\displaystyle\int\cot^{2}x\csc^{4}x\operatorname{d}\!x$ $\displaystyle=\int\cot^{2}x\csc^{2}x\csc^{2}x\operatorname{d}\!x$ $\displaystyle=\int\cot^{2}x(1+\cot^{2}x)\csc^{2}x\operatorname{d}\!x$ $\displaystyle=-\int u^{2}(1+u^{2})\operatorname{d}\!u.$ The integration and substitution required to finish this example are similar to that of previous examples in this section. The result is $-\frac{1}{3}\cot^{3}x-\frac{1}{5}\cot^{5}x+C.$ ## Integrals of the form $\displaystyle\int\sin(mx)\sin(nx)\operatorname{d}\!x,$$\displaystyle\int\cos(mx)\cos(nx)\operatorname{d}\!x$, and $\displaystyle\int\sin(mx)\cos(nx)\operatorname{d}\!x$. Functions that contain products of sines and cosines of differing periods are important in many applications including the analysis of sound waves. Integrals of the form $\int\sin(mx)\sin(nx)\operatorname{d}\!x,\quad\int\cos(mx)\cos(nx)\operatorname% {d}\!x\quad\text{and}\quad\int\sin(mx)\cos(nx)\operatorname{d}\!x$ are best approached by first applying the Product to Sum Formulas of Trigonometry found in the back cover of this text, namely $\displaystyle\sin(mx)\sin(nx)$ $\displaystyle=\frac{1}{2}\Bigl{[}\cos\bigl{(}(m-n)x\bigr{)}-\cos\bigl{(}(m+n)x% \bigr{)}\Bigr{]}$ $\displaystyle\cos(mx)\cos(nx)$ $\displaystyle=\frac{1}{2}\Bigl{[}\cos\bigl{(}(m-n)x\bigr{)}+\cos\bigl{(}(m+n)x% \bigr{)}\Bigr{]}$ $\displaystyle\sin(mx)\cos(nx)$ $\displaystyle=\frac{1}{2}\Bigl{[}\sin\bigl{(}(m-n)x\bigl{)}+\sin\bigl{(}(m+n)x% \bigr{)}\Bigr{]}$ ###### Example 8.2.9 Integrating products of $\sin(mx)$ and $\cos(nx)$ Evaluate $\displaystyle\int\sin(5x)\cos(2x)\operatorname{d}\!x$. SolutionThe application of the formula and subsequent integration are straightforward: $\displaystyle\int\sin(5x)\cos(2x)\operatorname{d}\!x$ $\displaystyle=\int\frac{1}{2}\Bigl{[}\sin(3x)+\sin(7x)\Bigr{]}\operatorname{d}\!x$ $\displaystyle=-\frac{1}{6}\cos(3x)-\frac{1}{14}\cos(7x)+C.$ ## Integrating other combinations of trigonometric functions Combinations of trigonometric functions that we have not discussed in this chapter are evaluated by applying algebra, trigonometric identities and other integration strategies to create an equivalent integrand that we can evaluate. To evaluate “crazy” combinations, those not readily manipulated into a familiar form, one should use integral tables. A table of “common crazy” combinations can be found at the end of this text. These latter examples were admittedly long, with repeated applications of the same rule. Try to not be overwhelmed by the length of the problem, but rather admire how robust this solution method is. A trigonometric function of a high power can be systematically reduced to trigonometric functions of lower powers until all antiderivatives can be computed. The next section introduces an integration technique known as Trigonometric Substitution, a clever combination of Substitution and the Pythagorean Theorem. ## Exercises 8.2 ### Terms and Concepts 1. 1. T/F: $\displaystyle\int\sin^{2}x\cos^{2}x\operatorname{d}\!x$ cannot be evaluated using the techniques described in this section since both powers of $\sin x$ and $\cos x$ are even. 2. 2. T/F: $\displaystyle\int\sin^{3}x\cos^{3}x\operatorname{d}\!x$ cannot be evaluated using the techniques described in this section since both powers of $\sin x$ and $\cos x$ are odd. 3. 3. T/F: This section addresses how to evaluate indefinite integrals such as $\displaystyle\int\sin^{5}x\tan^{3}x\operatorname{d}\!x.$ 4. 4. T/F: Sometimes computer programs evaluate integrals involving trigonometric functions differently than one would using the techniques of this section. When this is the case, the techniques of this section have failed and one should only trust the answer given by the computer. ### Problems In Exercises 5–32., evaluate the indefinite integral. 1. 5. $\displaystyle\int\sin^{3}x\cos x\operatorname{d}\!x$ 2. 6. $\displaystyle\int\cos^{2}x\operatorname{d}\!x$ 3. 7. $\displaystyle\int\cos^{4}x\operatorname{d}\!x$ 4. 8. $\displaystyle\int\sin^{3}x\cos^{2}x\operatorname{d}\!x$ 5. 9. $\displaystyle\int\sin^{3}x\cos^{3}x\operatorname{d}\!x$ 6. 10. $\displaystyle\int\sin^{6}x\cos^{5}x\operatorname{d}\!x$ 7. 11. $\displaystyle\int\cos^{2}x\tan^{3}x\operatorname{d}\!x$ 8. 12. $\displaystyle\int\sin^{2}x\cos^{2}x\operatorname{d}\!x$ 9. 13. $\displaystyle\int\sin^{3}x\sqrt{\cos x}\operatorname{d}\!x$ 10. 14. $\displaystyle\int\sin(x)\cos(2x)\operatorname{d}\!x$ 11. 15. $\displaystyle\int\sin(3x)\sin(7x)\operatorname{d}\!x$ 12. 16. $\displaystyle\int\sin(\pi x)\sin(2\pi x)\operatorname{d}\!x$ 13. 17. $\displaystyle\int\cos(x)\cos(2x)\operatorname{d}\!x$ 14. 18. $\displaystyle\int\cos\left(\frac{\pi}{2}x\right)\cos(\pi x)\operatorname{d}\!x$ 15. 19. $\displaystyle\int\tan^{2}x\operatorname{d}\!x$ 16. 20. $\displaystyle\int\tan^{2}x\sec^{4}x\operatorname{d}\!x$ 17. 21. $\displaystyle\int\tan^{3}x\sec^{4}x\operatorname{d}\!x$ 18. 22. $\displaystyle\int\tan^{3}x\sec^{2}x\operatorname{d}\!x$ 19. 23. $\displaystyle\int\tan^{3}x\sec^{3}x\operatorname{d}\!x$ 20. 24. $\displaystyle\int\tan^{5}x\sec^{5}x\operatorname{d}\!x$ 21. 25. $\displaystyle\int\tan^{4}x\operatorname{d}\!x$ 22. 26. $\displaystyle\int\sec^{5}x\operatorname{d}\!x$ 23. 27. $\displaystyle\int\tan^{2}x\sec x\operatorname{d}\!x$ 24. 28. $\displaystyle\int\tan^{2}x\sec^{3}x\operatorname{d}\!x$ 25. 29. $\displaystyle\int\csc x\operatorname{d}\!x$ 26. 30. $\displaystyle\int\cot^{3}x\csc^{3}x\operatorname{d}\!x$ 27. 31. $\displaystyle\int\cot^{3}x\operatorname{d}\!x$ 28. 32. $\displaystyle\int\cot^{6}x\csc^{4}x\operatorname{d}\!x$ In Exercises 33–40., evaluate the definite integral. 1. 33. $\displaystyle\int_{0}^{\pi}\sin x\cos^{4}x\operatorname{d}\!x$ 2. 34. $\displaystyle\int_{-\pi}^{\pi}\sin^{3}x\cos x\operatorname{d}\!x$ 3. 35. $\displaystyle\int_{-\pi/2}^{\pi/2}\sin^{2}x\cos^{7}x\operatorname{d}\!x$ 4. 36. $\displaystyle\int_{0}^{\pi/2}\sin(5x)\cos(3x)\operatorname{d}\!x$ 5. 37. $\displaystyle\int_{-\pi/2}^{\pi/2}\cos(x)\cos(2x)\operatorname{d}\!x$ 6. 38. $\displaystyle\int_{0}^{\pi/4}\tan^{4}x\sec^{2}x\operatorname{d}\!x$ 7. 39. $\displaystyle\int_{-\pi/4}^{\pi/4}\tan^{2}x\sec^{4}x\operatorname{d}\!x$ 8. 40. $\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^{2}x\operatorname{d}\!x$ 1. 41. Find the area between the curves $y=\sin^{2}x$ and $y=\cos^{2}x$ on the interval $[\pi/4,3\pi/4]$.
## Movie Lines • Lesson 9-12 1 This lesson allows students to apply their knowledge of linear equations and graphs in an authentic situation. Students plot data points corresponding to the cost of DVD rentals and interpret the results. This lesson provides students with the description of an authentic situation and data points that fit that situation. Students graph the line that contains those points; determine its y‑intercept, slope, and equation; and interpret the slope and y‑intercept in the context of the problem. Finally, students will extrapolate to determine the a functional value that is out of the domain they have graphed. To begin, present students with the following situation: While surfing the Internet, you find a site that claims to offer "the most popular and the cheapest DVDs anywhere." Unfortunately, the website isn’t clear about the how much they charge for each DVD, but it does give you the following information: Number of DVDs Ordered 1 2 3 Total Cost (includes S&H) $15$24 $33 Although you can present the problem to students verbally, it will likely be necessary to project the table of costs on the overhead projector or draw it on the chalkboard. Students will need to refer to it throughout the first part of the lesson. Ask students to plot the data shown in the table and to connect the points. Discuss what the graph of these points looks like [A line.]. When students justify that the points form a line, ask them to write an equation for the line. Then, ask them to answer the following questions: 1. Assume that total cost is a linear function of number of DVDs ordered. • What is the slope of the line that contains the data points? [9.] • What does that slope represent in the context of this problem? [You will be charged$9.00 for each additional movie purchased.] • What is the y‑intercept of the line that contains the data points? [6.] • What does that y‑intercept represent in the context of this problem? [It would cost you $6.00 for the store to send you an empty box.] 2. Your friend says that he can get a dozen DVDs from this web site for$90. Is he correct? Explain. [No. The equation for this line y=9x+6. If you buy a dozen DVDs, then y=9(12)+6, or y=114. It would cost you $114, not$90, to buy a dozen DVDs.] 3. How much would it cost to order 50 DVDs from this web site? [y=9(50)+6=456. It would cost $456.00 to buy 50 DVDs.] Students may work individually at first, but then they should share their answers with a partner. Each pair should then compare their results with another pair. A whole class review of the solutions should follow. Then, tell students that there is another web site offering movies at a cost of$10 per DVD (including S&H). Allow students time to follow the same steps in analyzing the cost for DVDs: plot data points (if necessary), draw the line, and write an equation. The graph for this situation should be drawn on the same set of axes. The equation of the second web site offering movies is: y=10x. When students graph both lines, the results will look like this: Students can then compare the results from both situations to answer the following questions: • How do the graphs differ? [The have different slopes; thus, at first, the original website is more expensive, but over time, the second website is more cost efficient. They also have different y-intercepts.] • How many movies would you have to rent for the price to be exactly the same at both sites? [6 movies (solve 9x+6=10x for x).] • What would be the price difference if you bought a dozen movies at each site? [The first website, as stated before, would cost you $114. The second website would cost you: y=10(12), or$120.] Conclude the lesson by having a whole-class discussion about how these two situations are similar and how they are different. • Computer with a spreadsheet program or graphing calculators (optional) Assessment Options 1. Give students other linear equations with m and b greater than 0 and ask them to assume the context of the problem is the same as in the DVD problems above. • For each line, ask students to state the y‑intercept and slope and interpret both in the context of the problem. From these values, have them generate an equation for the line. • Ask students to explain the meaning of a y‑intercept of 0 in the context of these problems. • Ask students to explain the meaning of a slope of 0 in the context of these problems. • Ask students to find the number of DVDs that they would need to purchase for both websites to be equal in cost. 2. Ask students to come up with two linear equations, and a real world context, where one had a positive slope, while the other had a negative slope. Have students repeat finding the answers to the four bullet points in Assessment Question #1. Extensions 1. Allow students to write authentic problems to fit given linear equations and to interpret slope and y‑intercept in the context of each problems. 2. Allow students to search for similar data on the web, in catalogs, and elsewhere. Then, have them provide the graphs, equations, and tables to represent those situations. Questions for Students 1. How did you find the slope of each equation? What does the slope mean in this problem? [From the graph of each situation, the slope can be identified by determining the ratio of the vertical change to the horizontal change (that is, rise to run). Additionally, students should realize that the rate of change for this situation is equivalent to the cost per movie. In the first situation, the total cost increases by $9 each time the number of DVD’s increases by 1, so the slope is 9/1 = 9. Similarly for the second situation, the slope is 10/1 = 10.] 2. How did you find the y‑intercept? What does the y‑intercept mean in this problem? [The y‑intercept is the point at which the line crosses the y‑axis. For the first situation, the y‑intercept is 6, and for the second situation, the y‑intercept is 0. In the context of these problems, the y‑intercept represents the surcharge applied to any order; for the first web site, there is a$6 surcharge, but the second web site has no surcharge.] Teacher Reflection • Do students need more practice in interpreting slope and y‑intercept in the context of an authentic problem? • Do students understand the power of algebra in checking "their friend’s calculation"? Did they rely on the graph? • Were clear expectations discussed so that students knew what was expected of them? If not, how can you make expectations more clear in the future? • This lesson is very open‑ended. Once given the scenario, students are expected to make progress by themselves, with very little guidance. How might this lesson be structured for low‑ability students to minimize frustration? ### Learning Objectives Students will: • Graph a line based on data points. • Find the equation of a line. • Identify y‑intercept and slope and state their significance in the context of the problem. • Extrapolate data. ### NCTM Standards and Expectations • Analyze functions of one variable by investigating rates of change, intercepts, zeros, asymptotes, and local and global behavior. • Draw reasonable conclusions about a situation being modeled. • Approximate and interpret rates of change from graphical and numerical data.
# LABELLING SIDE LENGTHS in RIGHT ANGLE TRIANGLES In this section we learn how to label side lengths in a right angle triangle. Given a right angle triangle, like the one shown here, its sides are named either: • Hypotenuse, • Opposite Like every right angle triangle, it has two acute interior angles, labelled $$a$$ and $$b$$. Looking at the triangle we see here, we have no trouble in seeing that the hypotenuse is the side of length $$5$$, the side length opposite the right angle. On the other hand, the side we call adjacent and the side we call opposite depends entirely on whether we're focusing on angle $$a$$ or $$b$$. ### Tutorial: Labelling Sides in Right Angle Triangles In the following tutorial we learn how to label the sides in a right angle triangle. ## HOW TO LABEL SIDE LENGTHS • ### HYPOTENUSE By definition the hypotenuse is the side opposite the right angle. Since there is only one right angle in a right angle triangle, the hypotenuse "never changes", regardless of whether we are focusing on interior angle $$a$$ or $$b$$ it will always be the side length opposite the right angle and is the longest side of the right angle triangle. Unlike the hypotenuse, the adjacent and opposite side lengths depend on the interior angle we're focusing on. This is illustrated in the diagrams we see here: • ### relative to angle $$a$$ We can see that the: • adjacent side is of length $$(4$$, $$A = 4$$ • opposite side is of length $$(3$$, $$O = 3$$ • ### relative to angle $$b$$ We can see that the: • adjacent side is of length $$(3$$, $$A = 3$$ • opposite side is of length $$4$$, $$O = 4$$ ## Exercise 1 Given the right angle triangle $$ABC$$ shown here, name the: 1. Hypotenuse 2. Adjacent side to $$b$$ 3. Opposite side to $$a$$ 4. Opposite side to $$b$$ 5. Adjacent side to $$a$$ Note: this exercise, as well as exercise 2, can be downloaded as a worksheet to practice with: ## Solution Without Working 1. Hypotenuse is $$AB$$. 2. Adjacent side to $$b$$ is $$BC$$. 3. Opposite side to $$a$$ is $$BC$$. 4. Opposite side to $$b$$ is $$AC$$. 5. Adjacent side to $$a$$ is $$AC$$ ## Exercise 2 Given the right angle triangle shown here, state the length of: 1. the side opposite angle $$a$$ 2. the side opposite angle $$b$$ 3. the hypotenuse 4. the side adjacent to angle $$b$$ 5. the side adjacent to angle $$a$$ Note: this exercise, as well as exercise 1 can be downloaded as a worksheet to practice with: ## Solution Without Working 1. the length of the opposite to $$a$$ is $$4$$ 2. the length of the opposite to $$b$$ is $$3$$ 3. the length of the hypotenuse $$5$$ 4. the length of the adjacent to $$b$$ is $$4$$ 5. the length of the adjacent to $$a$$ is $$3$$ ### Subscribe to Our Channel Subscribe Now and view all of our playlists & tutorials.
# Prove that $\sum\limits_{cyc}\frac{a^2-bd}{b+2c+d}\geq0$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\frac{a^2-bd}{b+2c+d}+\frac{b^2-ca}{c+2d+a}+\frac{c^2-db}{d+2a+b}+\frac{d^2-ac}{a+2b+c}\geq0$$ This inequality is a similar to the following inequality of three variables. Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^2-bc}{b+c+2a}+\frac{b^2-ca}{c+a+2b}+\frac{c^2-ab}{a+b+2c}\geq0,$$ which we can prove by the following reasoning. $$\sum\limits_{cyc}\frac{a^2-bc}{b+c+2a}=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{b+c+2a}=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a+c}{b+c+2a}-\frac{b+c}{c+a+2b}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{(b+c+2a)(c+a+2b)}\geq0,$$ but this idea does not help for the starting inequality. We can make the following. By Holder $$\sum_{cyc}\frac{a^2}{b+2c+d}=\sum_{cyc}\frac{a^3}{ab+2ac+ad}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(ab+2ac+ad)}=\frac{(a+b+c+d)^3}{8\sum\limits_{cyc}(ab+ac)}$$ and $$\sum_{cyc}\frac{bd}{b+2c+d}=2(a+b+c+d)\left(\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}\right).$$ Thus, it remains to prove that $$\frac{(a+b+c+d)^2}{16\sum\limits_{cyc}(ab+ac)}\geq\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}$$ and I don't see, what is the rest. The following quantity is clearly positive \begin{eqnarray} ((2(d^3+b^3)+8bd(b+d))(b-d)^2+(2(a^3+c^3)+8ca(c+a))(c-a)^2)+ ((a+c)(5(d^2+b^2)(b-d)^2+12bd(b-d)^2)+(b+d)(5(a^2+c^2)(a-c)^2+12ac(a-c)^2))+ 2((b^3+d^3)(a-c)^2+(a^3+c^3)(b-d)^2)+ 14(bd(a+c)(a-c)^2+ac(b+d)(b-d)^2) \end{eqnarray} After doing some algebra, this is the same as \begin{eqnarray} (a^2-bd)(c+2d+a)(d+2a+b)(a+2b+c)+(b^2-ca)(d+2a+b)(a+2b+c)(b+2c+d)+(c^2-db)(a+2b+c)(b+2c+d)(c+2d+a)+(d^2-ac)(b+2c+d)(c+2d+a)(d+2a+b) \geq 0 \end{eqnarray} Now divide by $(c+2d+a)(d+2a+b)(a+2b+c)(b+2c+d)$ and the result follows. (a^2-bd)(c+2d+a)(d+2a+b)(a+2b+c)+(b^2-ca)(d+2a+b)(a+2b+c)(b+2c+d)+(c^2-db)(a+2b+c)(b+2c+d)(c+2d+a)+(d^2-ac)(b+2c+d)(c+2d+a)(d+2a+b)-(((2(d^3+b^3)+8bd(b+d))(b-d)^2+(2(a^3+c^3)+8ca(c+a))(c-a)^2)+((a+c)(5(d^2+b^2)(b-d)^2+12bd(b-d)^2)+(b+d)(5(a^2+c^2)(a-c)^2+12ac(a-c)^2))+2((b^3+d^3)(a-c)^2+(a^3+c^3)(b-d)^2)+14(bd(a+c)(a-c)^2+ac(b+d)(b-d)^2)); The above computer algebra can be copied & pasted into reduce & serves as justification for the above claim. The above solution does not use any well known theorems or tricks. I am sure that more elegant solutions exist & We would be interested to see them. • Woah. How did you come up with the first quantity? That's work on a humongous level! May 15, 2017 at 16:54 • @Lohith-kumar I like to call this the "brute force" method for showing inequalities. If you look in the copy & paste box, the first half is what we are aiming to show (rearranged a bit) & the second half is the what I think you refer to as the "first quantity". This is formed by subtracting the "largest monomials" times squares of "special combinations", you then look carefully at what is left & subtract the next largest monomial ... keep doing this until you hit zero (If you are lucky). Largest m $a^{x} b^{y} c^{z}$ ... largest $x$; special quantity = sum of its coefficients is zero. May 19, 2017 at 20:14 Update By chance, I saw a nice proof as follows. \begin{align} \sum_{\mathrm{cyc}} \frac{a^2-bd}{b+2c+d} &\ge \sum_{\mathrm{cyc}} \frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d}\\ &= \sum_{\mathrm{cyc}} \Big(\frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d} + \frac{b+d-2c}{4}\Big)\\ &= \sum_{\mathrm{cyc}} \frac{a^2-c^2}{b+2c+d}\\ &= \Big(\frac{a^2-c^2}{b+2c+d} + \frac{c^2-a^2}{d+2a+b}\Big) + \Big(\frac{b^2-d^2}{c+2d+a} + \frac{d^2-b^2}{a+2b+c}\Big)\\ &= \frac{2(a+c)(a-c)^2}{(b+2c+d)(d+2a+b)} + \frac{2(b+d)(b-d)^2}{(c+2d+a)(a+2b+c)}\\ &\ge 0. \end{align} Previously written Donald Splutterwit gave a nice SOS (Sum of Squares) solution. Actually, the Buffalo Way works. After clearing the denominators, we need to prove that $$f(a,b,c, d)\ge 0$$ where $$f(a,b,c,d)$$ is a homogeneous polynomial of degree four. WLOG, assume that $$d = \min(a,b,c,d)$$. Let $$c = d+s, \ b=d+t, \ a = d+r; \ s, t, r\ge 0$$. We have $$f(d+r, d+t, d+s, d) = Ad^2 + Bd + C$$ where \begin{align} A &= 24 r^2-48 r s+24 s^2+24 t^2, \\ B &= 16 r^3-16 r^2 s+8 r^2 t-16 r s^2-16 r s t+8 r t^2+16 s^3+8 s^2 t+8 s t^2+16 t^3, \\ C &= 2 r^4+2 r^3 s+3 r^3 t-8 r^2 s^2-3 r^2 s t-r^2 t^2+2 r s^3-3 r s^2 t \\ &\quad +4 r s t^2+3 r t^3+2 s^4+3 s^3 t-s^2 t^2+3 s t^3+2 t^4. \end{align} It suffices to prove that $$A, B, C\ge 0$$. Clearly $$A\ge 0$$. It is easy to prove that $$B\ge 0$$ using discriminant. The proof of $$C\ge 0$$ may be a little harder. Omitted here. However I also verified it by Mathematica Resolve. Maybe someone can find a nice proof of $$B, C\ge 0$$.
Question # A coin is tossed twice. Find the probability distribution of the number of heads. Hint: In this question first find out the probability of getting a head as well as not getting a head when a coin is tossed, later on in the solution construct all possible cases for getting a head as well as not getting a head because this time coin is tossed two times, so use this concept to reach the solution of the question. Given a coin is tossed twice. As we know a coin has two sides. So, total number of outcomes $= 2$ To get a head when a coin is tossed (p)$= \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{1}{2}$ For not getting a head when a coin is tossed $\left( q \right) = 1 - \left( p \right)$ As the total probability is always 1. $\Rightarrow \left( q \right) = 1 - \left( p \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$ As the coin is tossed twice. So, construct different cases for getting a head. $1.$For not getting any head i.e. when we tossed the coin twice head does not appears$= q.q = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}$ $2.$Head appears only once, it can happen in two ways when we tossed the first time we got head and in the second time head does not appear, and when we tossed again the first time we did not got head but in second time we got a head. $= pq + qp = \dfrac{1}{2}.\dfrac{1}{2} + \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = \dfrac{1}{2}$ $3.$Always get a head i.e. when we tossed the coin twice head appears all the time $= p.p = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}$ So, the required probability distribution is shown below. $\begin{array}{*{20}{c}} {\left( H \right)}&0&1&2 \\ {\left( {{P_h}} \right)}&{\dfrac{1}{4}}&{\dfrac{1}{2}}&{\dfrac{1}{4}} \end{array}$ So, this is the required probability distribution of the number of heads (H). Note: In such types of questions the key concept we have to remember is that always recall the property of probability which is stated above, then construct different cases as above to get the required probability distribution of the number of heads as shown above which is the required answer.
# 5.05 Compare decimal numbers Lesson ## Are you ready? Remembering how to compare whole numbers is going to help us compare decimals in this lesson. Let's try this practice problem to review. ### Examples #### Example 1 Which symbol, greater than (\gt) or less than (\lt) will make the following statements true? a 91 \, ⬚ \, 97 A \gt B \lt Worked Solution Create a strategy Use a place value table and compare the numbers. Apply the idea As we can see in the table, both numbers in the tens column have the same number so we move to the units column. In the units column, we can see that 1 is smaller than 7. So the correct answer is 91 \lt 97, option B. b 682 \, ⬚ \, 782 A \gt B \lt Worked Solution Create a strategy Use a place value table and compare the numbers. Apply the idea In the hundreds column, we can see that 6 is smaller than 7. We do not need to compare the remaining columns since we already know which is greater. So the correct answer is 682 \lt 782, option B. Idea summary \gt means "greater than". \lt means "less than". ## Compare decimals with thousandths We can use place value columns to help us compare numbers with decimals. If we have 0.342, and we are comparing it to 0.458, we know that 342 is smaller than 458, so 342 thousandths is smaller than 458 thousandths. In this video we work through some examples, and see how we can compare decimals. ### Examples #### Example 2 Fill in the box the greater than (\gt) or less than (\lt) symbol that would make this number sentence true. 0.082 \, ⬚ \, 0.088 Worked Solution Create a strategy Use a place value table and compare the numbers. Apply the idea Write the decimals in a place value table and use zeros as place holders. As we can see in the table, both numbers in the units, tenths, and hundredths columns have the same number so we move to the thousandths column. In the thousandths column, we can see that 2 is smaller than 8. We can write the number sentence as:0.082 \lt 0.088 Idea summary We can use a place value table to compare decimals. ## Compare numbers beyond thousandths This video shows how to compare numbers that have many decimal places. ### Examples #### Example 3 Fill in the box the greater than (\gt) or less than (\lt) symbol that would make this number sentence true. 0.754 \, ⬚ \, 0.6094 Worked Solution Create a strategy Use a place value table and compare the numbers. Apply the idea Write the decimals in a place value table and use zeros as place holders. In the tenths column, we can see that 7 is larger than 6. We do not need to compare the remaining columns since we already know which is larger. We can write the number sentence as:0.754 \gt 0.6094 Idea summary When using a place value table to compare decimals we can start at the place value column furthest to the left and work to the right, comparing the values in each column to work out the bigger/smaller number. Or, we can add zeros at the end of a decimal so we can compare the same number of place value columns. For example, if we wanted to compare 0.45 and 0.672, we could write 0.45 as 0.450. This means we can compare thousandths to thousandths. ### Outcomes #### MA3-7NA compares, orders and calculates with fractions, decimals and percentages
RS Aggarwal Solutions: Heights and Distances- 1 # RS Aggarwal Solutions: Heights and Distances- 1 \| RS Aggarwal Solutions for Class 10 Mathematics PDF Download ## Exercise: 14 Q.1. A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower. [Take √3 = 1.732.] Let AB be the tower and C be the point on the ground 20 m away from the foot of the tower B from where the angle of elevation of the top of the tower is 60°. Now, draw a line from C to A. Join B and C. We get a triangle ABC with right angle at B. We are to find the height of the tower, that is AB. We will be using trigonometric ratios involving AB(height) and BC(base). So, ∠ACB = 60° BC = 20m. Now in ∆ABC, or, AB = 20√3 = 34.64m. The height of the tower is 34.64 m. Q.2. A kite is flying at a height of 75 m from the level ground, attached to a string inclined at 60° to the horizontal. Find the length of the string, assuming that there is no slack in it. [Take √3 = 1.732.] Let A be the position of the kite in the sky. Let C be the position on the ground from where a string is attached to the kite. Since we have assumed that there is no slack in the string, we take AC to be a straight line making 60° angle with the ground. Draw a perpendicular from A on the ground which meets at point B. The kite is flying at a height of 75 m above the ground. So, AB = 75 m. Join B and C. We thus get a triangle ABC with right angle at B. We are to find the length of the string that is AC. We will use the trigonometric ratio sine which uses the perpendicular AB and the hypotenuse AC to find AC. Now, ∠ ACB = 60° From ∆ABC, or, The length of the string is 86.60 m. Q.3. An observer 1.5 m tall is 30 m away from a chimney. The angle of elevation of the top of the chimney from his eye is 60°. Find the height of the chimney. In the figure, let BD be the height of the man, i.e. BD = 1.5m. Let AE be the chimney in the figure. Join B and C. We get a triangle which is right angled at C. Clearly, BD = CE. Also, given that ∠ ABC = 60°. We use the trigonometric ratio tan which uses AC as height and BC as a base to find the height of the chimney AE. Now, clearly CE = 1.5m. The height of the chimney is AE. From ∆ABC, or, Thus, AC = 51.96m and CE = 1.5m. The height of the chimney AE = AC + CE = 51.96 + 1.5 = 53.46 m. Hence, the answer is 53.46 m. Q.4. The angles of elevation of the top of a tower from two points at distances of 5 meters and 20 meters from the base of the tower and in the same straight line with it are complementary. Find the height of the tower. Let AB be the tower in the figure and C and D be two points on the straight line BD, at distances 5m and 20m from the foot of the tower AB. Now, join B, C, and C, D. We get two right-angled triangles both right angled at B. We use the trigonometric ratio tan by using AB as height and BC as a base(for ∆ABC) and AB as height and BD as a base(for ∆ABD) to find the height of the tower AB. By the problem we have, ∠ACB + ∠ADB = 90°. In ∆ABC, we have or, or, Also, from ∆ABD, or, Putting the value of tan ∠ABD in the equation we get or, or, a positive value is taken since the height of the tower can’t be negative. The height of the tower AB = 10m. Q.5. The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff. [Use √3 = 1.732.] Let BC be the tower and CD be the flagstaff. Join C, A and D, A and A, B. We get two right-angled triangles ABC and BAD which are right-angled at B. By the problem, it is clear that ∠ BAC = 45° and ∠ BAD = 60°. We use trigonometric ratio tan for both the triangles using BC as height and AB as a base(for ∆ABC) and BD as height and AB as a base(for ∆ABD) to find the height of the flagstaff CD. Let BC be x. In ∆ABC we have, or, or, x = 120 So, we get BC = 120m. In ∆ABD, or, or, or, So, height of the flagstaff = DC = 87.84m. Q.6. From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank (on the top of the tower) is 45°. Find (i) the height of the tower, (ii) the depth of the tank. In the figure assume, A to be the point 40m away from the foot of the tower BC. Join A,B and A,C and A,D. Let DC be the water tank and BC be the tower. We get two right-angled triangles ABC and ABD, right angled at B. Also, ∠ BAC = 30° and ∠ BAD = 45°. We use trigonometric ratios tan using AB as base and BC as height(for ∆ABC) and AB as base and CD as height(for ∆ABD). Let BC be x. In ∆ABC, or, So, BC = 23.09m. In ∆ABD, or, i. Height of the tower = BC = 23.1m ii. Depth of the tank = DC = 16.9m Q.7. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 60°. Find the height of the tower. [Use √3 = 1 .732.] Let AD be the Flagstaff, 6m high and DC be the tower. B be the point on the plane from where the angles of elevation of the top(A) and bottom of the flagstaff(D) are 60° and 30° respectively. Join C and B. Then we get two triangles ABC and BCD with right angle at C. We are to find the height of the tower DC. Now, to summarize, we are given, AD = 6m, ∠ ABC = 60°, and ∠ DBC = 30°. To find DC we will first find BC from the triangle ABC and again from triangle BCD and equate the expression for BC in both the cases. Then we will solve for DC from the equation. In ∆ABC, or, In ∆BCD, or, Equating the values of BC, or, 3DC = 6 + DC or, DC = 3m Hence the height of the tower is 3m. Q.8. A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45° Find the height of the pedestal. [√3 = 1.73.] In the figure, let AB be the statue of height 1.46 m, BC the pedestal. Let D be a point on the ground from which the angles of elevation of the top of the statue and the top of the pedestal are 60° and 45° respectively. We are to find the height of the pedestal, which is BC. Join C and D. We get two triangles ACD and BCD with right angle at C. To find BC, we use trigonometric ratios to find the expression of CD from ACD and BCD and then equate them. Now, ∠BDC = 45°, ∠ADC = 60°, AB = 1.46 m. or, Again, from ∆BCD, or, BC = CD or, or, or, or, or, Q.9. The angle of elevation of the top of an unfinished tower at a distance of 75 m from its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°? [Take √3 = 1.732.] In the above figure, let DB be the unfinished tower. Let C be the point on the ground, 75 m from the base B from which the angle of elevation of the top D is 30°. Let A be the point to which the tower must be raised such that the angle of elevation of it from C becomes 60°. Join B and C. We get two triangles ABC and BCD with right angle at B. We need to find the excess height that is an AD. We have, BC = 75 m, ∠BCD = 30° and ∠ACB = 60°. To find the AD, we will find AB and BD from the triangles ABC and BCD respectively using the trigonometric ratio tan and then subtract BD from AB. Now, from ∆BCD, or, Again, from ∆ABC, or, AB = 75 x √3 Now, we have got both AB and DB. Then we are asked to find the AD. Hence the answer is 86.6 m. Q.10. On a horizontal plane, there is a vertical tower with a flagpole on the top of the tower. At a point, 9 meters away from the foot of the tower, the angle of elevation of the top and bottom of the flagpole is 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it. [Take √3 = 1.73.] In the above figure, let BC be the tower, AB is the flagpole. Let D be the point 9 m away from C such that the angle of elevations of the top and bottom of the flagpole AB are 60° and 30° respectively. Join C and D. We get two triangles ADC and BCD with right angle at C. We have to find the height of the tower BC and the height of the flagpole AB. For this, we use trigonometric ratio tan for triangles ACD and BCD to find AC and BC respectively. Subtract BC from AC to find AB. Now we have, DC = 9 m, ∠ADC = 60°, and ∠BDC = 30°. We are to find AB and BC. From ∆ACD, or, AC = 9 x √3 From ∆BDC, or, BC = 9/√3 Now, And, BC = 9/1.73 = 5.20 m Q.11. Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of one pole is 60° and the angle of depression from the top of another pole at P is 30°. Find the height of each pole and distances of the point P from the poles. Let AD and BC be the two poles of equal height standing on the two sides of the road of width 80 m. Join DC. Then DC = 80 m. P is a point on the road from which the angle of elevation of the top of tower AD is 60°. Also, the angle of depression of the point P from the point B is 30°. Draw a line BX from B parallel to the ground. Then ∠XBP = ∠BPC = 30°. We are to find the height of the poles and the distance of the point P from both the poles. Also, ∠APD = 60°. Let DP = x. Then PC = 80-x. In ∆APD, or, Now, from ∆APD, or, So, or, 3x = 80 - x or, x = 20 m So, PD = 20 m. Hence, PC = 80-20 = 60 m. Also, BC = AD = 20√3 m. Q.12. Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 meters, find the distance between the two men. [Take √3 = 1.732.] Let the positions of the two men are B and C. Let AP represent the tower in the figure. Join B, P, and C, P.We have, ∠ABP = 30° and ∠ACP = 45°. Given that AP = 50 m.We get two right-angled triangles ∆APB and ∆APC, both right angled at P. We use trigonometric ratio tan for both the triangles using BP as a base and AP as height(for ∆APB) and PC as base and AP as height(for ∆APC). We get BP and PC. The distance between the men is BC. We have to find BC. Let BP = x. In ∆APB, or, Again, in ∆APC, or, PC = 50 So, distance between the two men is BC = BP + PC = 50√3 + 50 = 50(√3 + 1) = 136.6m. Q.13. From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. [Take √3 = 1.73.] In the figure, let AC be the tower of height 100m. The angles of depression are ∠BAE and ∠DAF. Given, AC = 100m. Since the line EF is parallel to the line BD, we have ∠BAE = ∠BAC = 30°, ∠DAF = ∠DAC = 45°, where B, D are the positions of the cars. Join B, C, and C, D. We get two right-angled triangles ∆ABC and ∆ADC, both right angled at C. We use trigonometric ratio tan for both the triangles, where BC is the base and Ac is the height (for ∆ABC) and CD is the base and AC is the height in ∆ACD. We are to find the distance between the two cars, i.e. BD. In ∆ABC, or, or, DC = 100 Therefore, the distance between the two cars is BD = BC + CD = 100√3 + 100 = 100(√3 + 1) = 273 m. Q.14. A straight highway leads to the foot of a tower. A man standing on the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower form this point. In the figure, let AB be the tower and D be the position of the car. Let C be the position of the car 6seconds later. The angles of depression are ∠DAE and ∠CAE. Join B, C, and C, D and A, E. Since AE is parallel to BD, we must have, ∠ADB = ∠DAE = 30°, ∠CAE = ∠ACB = 60°. We get two right-angled triangles ∆ABC and ∆ABD, both right angled at B. We use trigonometric ratio tan for both the triangles using BC as base and AB as height for ∆ABC and BD as base and AB as height in ∆ABD. We find the CD. Let BC = x. In ∆ABC, or, AB = xtan60° = x√3 Again, in ∆ABD, or, CD = 2x Now, the speed of the car = CD/6 = 2x/6 = x/3. Hence, time taken by the car to reach the foot of the tower is = BC/(Speed of the car) = 3 secs. Q.15. A TV tower stands vertically on a bank of the canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. In the figure, let AB be the tower and BC be the canal. C is the point on the other side of the canal directly opposite the tower. Join B and C. Let D be another point, 20m away from C. Join C and D. We get two right-angled triangles ∆ABC and ∆ABD, both right angled at B. Given that, ∠ACB = 60° and ∠ADB = 30°. DC = 20m. We use trigonometric ratio tan using AB as height and BC as a base(for ∆ABC) and AB as height and BD as a base(for ∆ABD) to find AB and BC. In ∆ABC, tan ∠ACB = tan60° = AB/BC or, AB = BC x tan60° = BC√3 In ∆ABD, or, Equating the values of AB in ∆ABC and ∆ABD, or, 3BC = BC + 20 or, 2BC = 20 or, BC = 10 So, the width of the canal = BC = 10m. We found that AB = BC√3 = 10√3m. So, height of the tower = AB = 10√3m. Q.16. The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building. In the figure, let CD be the tower and AB be the building. Join the points A, D, and B, C. We get two right-angled triangles ∆ABD and ∆BCD, which are right-angled at B and D respectively. We are given that the angle of elevation of the top of the building from the foot of the tower is 30° and that of the top of the tower from the foot of the building is 60°. So, ∠CBD = 60°, ∠ADB = 30°. We are also given that the height of the tower is 60 m. Hence CD = 60 m. We need to find the height of the building, that is AB. For this, we will first find BD from ∆BDC using the trigonometric ratio tan. Using this value of BD, we will find the value of AB from ∆ABD using the trigonometric ratio tan. In ∆BCD, or, So, BD = 60/√3m. In ∆ABD, or, So, height of the building = AB = 20m. Q.17. The horizontal distance between two towers is 60 metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90 metres, find the height of the first tower. [Use √3 = 1.732.] In the figure, let CD be the first tower and AB, the second tower. We are given that the distance between the two towers is 60 m. Join C and B. We thus get CB = 60 m. Again, we are given that the angle of depression of the top of the first tower from the top of the second tower is 30°. We draw a horizontal line parallel to BC from A to F. Then we get ∠FAD = 30°. Now, we draw a line from the top of the first tower onto the second tower parallel to BC to the point E. We get a right-angled triangle ADE with right angle at E. Then ∠ ADE = 30°.We are also given that the height of the second tower is 90 m. So, AB = 90m. We are told to find the height of the first tower, that is CD. Note that, BC = ED = 60m. To find CD, we use the trigonometric ratio tan on ∆ADE to find AE. In ∆AED, or, AE = 60 x tan30° = 60/√3m Now we just see that we can get the height of the first tower by subtracting the value of AE from AB to get BE which is equal to CD. Height of the first tower = DC = AB-AE = 90-60/√3 = 55.35m. Q.18. The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 metres, find the height of the chimney. According to pollution control norms, the minimum height of a smoke-emitting chimney should be 100 metres. State if the height of the above-mentioned chimney meets the pollution norms. What value is discussed in this question? Let AB be the chimney and DC be the tower. We are given that the angle of elevation of the top of the chimney from the foot of the tower DC is 60°. Join A and C. We get a right-angled ∆ABC with right angle at B and ∠ACB = 60°. We are again told that the angle of depression of the foot of the chimney from the top of the tower is 30°. If we join B and D, then we get a right-angled ∆BCD with right angle at C. Draw a line from D to a point on AB parallel to BC. So, ∠EDB = 30° and ∠DBC = 30°. The height of the tower is given to be 40 m. So, DC = 40m. We will first find the height of the chimney AB. If AB is less then 100 m, then it does not meet the pollution norms and if otherwise, then it does not meet the pollution norms. First, we will find the value of BC from ∆BCD using the trigonometric ratio tan. Then using the value of BC, we will find the value of AB from ∆ABC using the trigonometric ratio tan. In ∆BDC, or, In ∆ABC, or, So, the height of the chimney is AB = 120m, which is greater than 100 m. The height satisfies the pollution norms. Q.19. From the top of a 7-meter-high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. [Use √3 = 1.732.] Let AB be the building and CE be the cable tower. We are given that, the angle of elevation of the top of the cable tower from the top of the building is 60° and the angle of depression of the bottom of the tower from the top of the building is 45°. Join C, E to A. Also draw a line from A to EC at the point D parallel to BC. We get two right-angled triangles ABC and AED with right angles at B and D respectively. Also, ∠EAD = 60°. ∠DAC = ∠ACB = 45°. Also given that the height of the building is 7 m. That is AB = 7m. We are to find the height of the tower CE. We first find the value of BC from ∆ABC using the trigonometric ratio tan. In ∆ABC, or, In ∆DAE, or, DE = 7 x tan60° = 7√3m So, the height of the cable tower = CE = CD + DE = 7m + 7√3m = 19.12m. Q.20. The angle of depression from the top of a tower of a point A on the ground is 30°. On moving a distance of 20 meters from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60°. Find the height of the tower and its distance from the point A. Let DC be the tower. Given that the angle of depression of the point A on the ground from the top of the tower DC is 30°. Join C and A. Now draw a line DE parallel to CA. Also join A and D. Then, ∠ EDA = 30°. We get a right-angled triangle ACD with right angle at C and ∠DAC = 30°. If we move 20 m from A to B towards the foot of the tower C, then the angle of depression changes to 60°. Then, AB = 20 m. Join D and B. Then we get a right-angled triangle ∆DCB with right angle at C. We are to find the height of the tower that is DC and its distance from A, that is, AC. Let DC = x. In ∆DCB, or, or, or, or, x = 10√3 = 17.32 m The height of the tower = DC = x = 17.32m. We have, BC = x/√3 = 10m. So, a distance of A from the tower = AC = 20 + 10 = 30m. Q.21. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower. In the figure, let AB be the tower and C be the point on the ground from which the angle of elevation of the top of the tower AB is 60°. Join A and C and B and C. Then we get a right-angled triangle ABC with right angle at B and ∠ACB = 60°. Let D be the point 10 m vertically above C. Then, CD = 10 m. Given that the angle of elevation of the top of the tower from the point D is 30°. Join D and A. Also draw a line DE from D to AB, parallel to BC. Then we get a right-angled triangle AED with right angle at E and ∠ADE = 30°. We are to find the height of the tower, that is AB. Clearly, ED = BC. Let AE = x In ∆AED, or, tan30° = x/DE or, Again, in ∆ABC, or, or, or, 3x = x + 10" or, x = 5m Height of the tower = AB = AE + EB = 5m + 10m = 15m. Q.22. The angles of depression of the top and bottom of a tower as seen from the top of a 60√3-m-high cliff are 45° and 60° respectively. Find the height of the tower. Let AB be the cliff and CD be the tower. Join B and C. Draw a line AF parallel to BC. Now, given that the angles of depression of the top and bottom of the tower from the top of the cliff are 45° and 60° respectively. Hence, ∠FAD = 45° and ∠FAC = 60°. Draw a line DE from CD to AB parallel to BC. Also, join A, D and A, C. We get two right-angled triangles ADE and ABC with right angles at E and B respectively. Also, ∠ADE = ∠FAD = 45°, and ∠ACB = ∠FAC = 60°. We are also given that the height of the cliff AB is 60√3 m. We are to find the height of the tower, that is, CD. We first find the value of BC from the ∆ABC, using the trigonometric ratio tan. In ∆ABC, or, Clearly, ED = BC. Then, we will find the value of AE from ∆ADE using the trigonometric ratio tan. In ∆ADE, or, AE = 60m. The height of the tower = DC = 60√3m-60m = 43.92 m. Q.23. A man on the deck of a ship, 16 m above water level, observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60° and 30°. Calculate the distance of the cliff from the ship and height of the cliff. [Take √3 = 1.732.] In the above figure, let B be the position of the man and AE be the cliff. We are given that the position of the man is 16 m above water level. Draw a line BD to a point D on the water level. So, BD = 16 m. Now, we are given that the angle of elevation of the top of the tower from the position of the man is 60°. Join A and B. Also, draw a line BC on to the line AE parallel to the water level. Then we get a right angled triangle ABC with right angle at B and ∠ABC = 60°. Also, the angle of depression of the bottom of the tower from the position of the man is 30°. So, ∠CBE = 30°. Joining B and E we get a right angled triangle BDE. And, ∠BED = ∠CBE = 30°. We need to find the distance of the ship from the cliff, that is DE and the height of the cliff AE. We first find DE from the ∆BDE by using the trigonometric ratio tan. In ∆BDE, tan ∠BED = BD/DE or, tan 30° = 16/DE or, DE = 16√3m = 27.71 m. Now, BC = DE = 16√3 m. From ∆ABC, tan ∠ABC = AC/ BC or, or, Also, CE = BD = 16 m. Hence, the height of the cliff is AE = AC + CE = 48 + 16 = 64 m. Q.24. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of tower PQ. [Take √3 = 1.73.] PQ is the tower and from a point X on the ground, the angle of elevation of the top of the tower is 60°. Join A and P, and Q and X. Then we get a right-angled triangle QPX with right angle at P and ∠QXP = 60°. Y is a point vertically above X. Then YX = 40 m. Draw a line YZ from Y onto the line QP parallel to PX. We are also given that the angle of elevation of the top of the tower from the point Y is 45°. Join Y and Q. We get a right-angled triangle QYZ with right angle at Z and ∠QYZ = 45°. We are to find the height of the tower PQ. Clearly, ZY = XP, ZP = XY. Let QZ = x In ∆QZY, or, tan 45° = x/ZY or, In ∆PQX, or, or, √3x = x + 40 or, QZ = ZY = x = 54.79 m. The height of the tower PQ = PZ + ZQ = 54.79m + 40m = 94.79m. Q.25. The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, find the speed of the aeroplane. Let D be the initial position of the aeroplane, 2500 m above the ground. After 15 s, let A be the position of the aeroplane flying in the same height. Draw two lines AB and DE from A and D perpendicular to the ground. Then, AB = DE = 2500m. Let C be the point on the ground from which the angles of elevation of the two positions of the plane, viz., D and A are 45° and 30° respectively. Join A and D with C. Also join the points B,E and C on the ground. We then get two right-angled triangles ABC and DEC with right angles at B and E respectively. So, we get ∠ACB = 30° and ∠DCE = 45°. We are to find speed of the plane. The plane travels a distance AD in 15 s. Since speed = (distance/time), we need to find AD only. We first find CE from ∆DEC using the trigonometric ratio tan θ. In ∆DEC, tan ∠DCE = DC/EC or, tan 45° = DE/EC or, DE = EC tan 45° = EC or, DE = EC = 2500 m. Now, we will again use tan to find the value of BE from ∆ABC. In ∆ABC, or, or, or, Now, BE = AD = 1830 m. Speed of the aeroplane = . Q.26. The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres. [Given √3 = 1.732] In the given figure, let AB be the tower. Let D be the point on the same level as the foot of the tower from which the angle of elevation of the top of the tower is 30°. Join B,C and A,D. Then we get a right-angled triangle ABD with right angle at B and ∠ADB = 30°. Let C be the point on the same level of the ground as Band D, 150 m from D towards B, from which the angle of elevation of the top of the tower is 60°. Joining A and C, we get a right angled triangle ABC, with right angle at B and ∠ACB = 60°. We are to show that the height of the tower AB is 129.9 m. In ∆ABC, tan ∠ACB = tan 60° = AB/BC or, In ∆ABD, or, 3BC = BC + 150 So, BC = 75m. Now, AB = BC√3 = 75 × 1.732m = 129.9m. Hence proved. Q.27. As observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation. [Use √3 = 1.732.] In the above figure, let AD be the lighthouse and B and C be the two consecutive positions of the ship in view from it. The top of the lighthouse is 100 m above sea level. So, AD = 100 m. Also given that the angles of depression of the ship from the point A at the positions B and C are 30° and 60° respectively. Join D, C, B. Also join B, C to A. Then we get two right angled triangles ABD and ACD with right angles at D. Now, draw a line AE parallel to the sea level. Then, ∠EAB = ∠ABD = 30° and ∠EAC = ∠ACD = 60°. We are to find the distance travelled by the ship during the period of observation, that is, BC. We just use the trigonometric ratio tan from triangles ABD and ADC to find DB and DC respectively. From ∆ACD, or, Again, from ∆ABD, or, tan 30° = 100/BD or, BD = 100√3 Hence, distance travelled by ship during the observation is Q.28. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find the width of the river. [Take √3 = 1.732. In the above figure, let XY be the bridge and A be the point on the bridge from which two points, say B and D, on opposite sides of the river are observed. Join B and C. Given that the angles of depression of B and C from the point A are 30° and 45° respectively. Join B, D to A. So, ∠XAB = ∠ABD = 30° and also ∠YAD = ∠ADB = 45°. Again, draw a line AC from A perpendicular to the ground. Then, we get two right-angled triangles ABC and ACD with right angles at C. Now, given that the height of the bridge is AC = 2.5 m. We have to find the width of the river, that is BD. From ∆ACD, CD = 2.5m. Again from ∆ABC, or, Hence, the width of the river is BD = BC + CD = 4.33 + 2.5 = 6.83m, which is the required solution. Q.29. The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is 6 metres. In the given figure, let AB be the tower. Let P and C be the points on the ground 4 m and 9 m from the foot of the tower. Join B, P and C. So, BC = 9m, BP = 4m. Again join P and C with A. Then we get two right-angled triangles ABP and ABC with right angle at B. Also, ∠APB and ∠ACB are given to be complementary. So, ∠APB + ∠ACB = 90°. We have to show that the height of the tower is 6 m. We find the value of AB from ∆ABC using tan and also using the fact that tan(90-θ) = cotθ. In ∆ABC, or, We will now use the above-found value of AB in ∆APB. From ∆APB, or, or, tan ∠APB = 3/2 or, cot ∠APB = 2/3 Now, AB = 9 cot ∠APB = 9 x (2/3) = 3 x 2 = 6m. Hence, proved. Q.30. A ladder of length 6 metres makes an angle of 45° with the floor while leaning against one wall of a room. If the foot of the ladder is kept fixed on the floor and it is made to lean against the opposite wall of the room, it makes an angle of 60° with the floor. Find the distance between two walls of the room. In the above figure, AD and BC are the two opposite walls of the same room. The ladder is fixed at the position say P on the floor. Join D,P and C. At first let it leans against the wall BC making an angle of 45° with the floor. Join B and P. We get a right-angled triangle BPC with right angle at C and ∠BPC = 45°. Again, when the ladder leans against the second wall AD, it makes an angle of 60° with the floor. Joining A and P, we get a right-angled triangle APD with right angle at D and ∠APD = 60°. We are to find the distance between the two walls, that is DC. Also given that, AP = PB = length of the ladder = 6 m. To find DC, we separately find DP and PC from triangles APD and BPC respectively by using the trigonometric ratio cosine. From ∆APD, or, DP = 1/2 x 6 = 3m. Again, from ∆BPC, or, Hence, the distance between the two walls is, DC = DP + PC = 3 + 4.23 = 7.23m. Q.31. From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower. In the above figure, let AB be the tower and P and D be the positions of the two cars at an instant, observed from A. join A and E. The angles of depression are ∠DAE and ∠PAE. Given that, PD = 100 m. Since AE is parallel to BD, so, ∠ADB = ∠DAE = 45° and ∠APB = ∠PAE = 60°. Join P,D and A,B. We get two right-angled triangles ∆ABD and ∆ABP. We are to find AB. We use trigonometric ratio tan for both the triangles, using AB as height and BP as base for ∆ABP and AB as height and BD as base for ∆ABD. From ∆ABD, tan ∠ADB = tan 45° = AB= BD or, AB = BD From ∆APB, tan ∠APB = tan 60° = AB/BP or, BD = BP√3 or, BP + 100 = BP√3 or, BP(√3-1) = 100 or, Hence, the height of the tower is, AB = BD = BP + 100 = 136.61 + 100 = 236.61m Q.32. An electrician has to repair an electric fault on a pole of height 4 metres. He needs to reach a point 1 metre below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use, which when inclined at an angle of 60° to the horizontal would enable him to reach the required position? [Use √3 = 1.73.] Let AB be the pole and D be the 1m below A. The person needs to reach the point D. So, BD = 3m. Let DC be the ladder. Join D and C. We get a right-angled triangle ∆DBC, right angled at B. The angle of elevation of the ladder, ∠BCD = 60°. We use trigonometric ratio sin, using DC as hypotenuse and DB as height to find the length of the ladder DC. In ∆BDC, or, Hence, the length of the ladder will be 3.46m. Q.33. From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find (i) the horizontal distance between AB and CD, (ii) the height of the lamp post, (iii) the difference between the heights of the building and the lamp post. In the above figure, let AB be the building such that AB = 60 m. Join A and X. The angles of depression are ∠CAX = 30° and ∠DAX = 60°. Let CD be the lamp post. Join C,D and B,D and C,E. Since AX is parallel to DB, we must have ∠XAC = ∠ACE = 30° and ∠XAD = ∠ADB = 60°. We get two right-angled triangles ∆ABD and ∆CAE. We use trigonometric angle tan in both the triangles with AB as height and DB as base (for ∆ABD) and AE as height and CE as base(for ∆CAE). We have to find, (i)BD, (ii)CD, and (iii) AB-CD. From ∆ABD, or, DB = CE. From ∆ACE, tan ∠ACE = AE/CE or, or, Hence, And, the difference between the heights of the building and the lamp post is, AE = 20m. Thus our solutions are, (i) The horizontal distance between AB and CD = 34.64 m. (ii) The height of the lamp post = 40 m. (iii) Difference between the heights of the building and the lamp post = 20 m. The document RS Aggarwal Solutions: Heights and Distances- 1 \| RS Aggarwal Solutions for Class 10 Mathematics is a part of the Class 10 Course RS Aggarwal Solutions for Class 10 Mathematics. All you need of Class 10 at this link: Class 10 53 docs\|15 tests ## RS Aggarwal Solutions for Class 10 Mathematics 53 docs\|15 tests ### Up next Explore Courses for Class 10 exam ### Top Courses for Class 10 Signup to see your scores go up within 7 days! 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$\DeclareMathOperator{\asin}{asin}$ $\DeclareMathOperator{\acos}{acos}$ $\DeclareMathOperator{\atan}{atan}$ < Previous Section Home Next Section> # Section 2.0From Trigonometry for High School to Trigonometry for Calculus ## Trigonometry in High School Encyclopedia Brittanica: "The word trigonometry comes from the Greek words trigonon (“triangle”) and metron (“to measure”)." ### SOH-CAH-TOA You might recall this mnemonic for sine, cosine, and tangent of an acute angle in a right triangle: • SOH—Sine is Opposite over Hypotenuse • CAH—Cosine is Adjacent over Hypotenuse • TOA—Tangent is Opposite over Adjacent As a pnemonic, SOH-CAH-TOA works well to remind you of what to divide by what. But it hides the ideas triangle trigonometry is all about—similarity and proportionality. ## The Origins of Trigonometry ### Similar Triangles Ancient Babylonians and Egyptians developed a practical trigonometry for surveying and map making. They needed to compute distances between places they could not actually measure. They did this by drawing a quasi-accurate map of actual locations, then reasoning proportionally. For example, Figure 2.0.1 shows a triangle Egyptians drew with side lengths $s_1$, $s_2$, and $s_3$, which they measured. They drew the triangle to model a situation involving a much larger actual triangle for which they could measure only one side ($s'_1$). Figure 2.0.1 Using proportional reasoning to find unknown lengths. The quotient $\dfrac{s_2}{s_1}$ being 0.48 means $s_2$ is 48% as large as $s_1$. The quotient $\dfrac{s_3}{s_1}$ being 0.65 means $s_3$ is 65% as large as $s_1$. Since the two triangles are similar, $s'_2$ will be 48% as large as $s'_1$ and $s'_3$ will be 65% as large as $s'_1$ Reflection 2.0.1. Suppose the larger triangle is made by three actual villages separated by a river, and $s'_1$ is approximately 29,348 cubits ($1\text{ cubit}\approx 0.523\text{ meters}$). What, approximately, are the other two distances (in cubits)? Similarity, and thus proportionality, is the heart of high school trigonometry. Once you know quotients among sides of one triangle, you know quotients among sides of any triangle similar to it. ### Hipparchus, Ptolemy, and Astronomy Ancient Greeks were fascinated by the stars. They noticed that any star appeared in different positions in the sky when viewed at the same time of night from different locations. They used this observation (now called parallax) as motivation for determining distances from Earth to a star. We will not delve into their methods for determining distances to stars except to point out they realized that the important aspect of measuring triangles ("trigonometry") was to focus on the measure of an arc on a unit circle in which a right triangle is embedded. They used the length of arc subtended by the angle, measured in units of 1/360 the circle's circumference, as the angle's measure. The relative sizes among one triangle's sides tells the relative sizes of sides in any triangle similar to it. Hipparchus, and later Ptolemy, used this fact (and geometric theorems from Euclid and Archimedes) to approximate the relative sizes of sides in right triangles with a hypotenuse of length 60 and angles subtending arcs of $\frac{1}{2}^\circ$ and multiples of $\frac{1}{2}^\circ$. Ptolemy created a table of these values that was used for centuries by anyone needing to reason proportionally about lengths of sides in a right triangle. All they needed to know was the measure of one angle (other than the right angle) in it. They could calculate the (approximate) relative sizes among all three sides of one triangle and then use Ptolemy's table to reason proportionally about any similar triangle's sides based on knowing the length of just one. David Bressoud's Historical Reflections on Teaching Trigonometry gives a more detailed history of trigonometry. ## Exercise Set 2.0.1 1. Triangles A and B are similar. Triangles C and D are similar. Use relative sizes of sides in Triangle A to determine the approximate lengths of Triangle B's sides. Do the same for Triangles C and D. 2. Christian Carmen gives a detailed explanation of Ptolemy's calculation of the distance from Earth to Sun. The figure below is adapted from his Figure 1. It shows Earth, Moon, and Sun during a total eclipse of the sun. 1. Ptolemy determined angle MEH has a measure of approximately 0.26 degree. His table of values said a right triangle with one angle measuring 0.26 degrees has 0.0046 as the quotient of opposite and adjacent lengths. Ptolemy also determined the Sun's radius is approximately 401 times Moon's's radius. According to these approximations, how far away is Sun from Earth (in Moon radii)? 2. Earth's radius is 3.66 times as large as Moon's radius. According to these approximations, how far is Sun from Earth in Earth radii? 3. Earth's radius is 3950 miles (6356 km). According to these approximations, how far is Sun from Earth in miles? in kilometers? ## What High School Trigonometry Became Trigonometry for centuries was about triangles centered in a circle made by a chord of a circle, and angle measure historically was about the length of subtended arcs measured in units proportional to the circle's circumference (e.g., 1/360th the circumference). Starting in the early 1900's, textbooks' emphasis moved to ratios of sides in a right triangle and the ideas of circle and arc length eventually disappeared from high school trigonometry. ### But Trigonometric Identities Can Be Useful A trigonometric identity is an equation involving trigonetric functions that is always true. The most basic identity, called the Pythagorean Identity, is $(\cos A)^2 + (\sin A)^2 = 1$ for any acute angle A in a right triangle. Figure 2.0.2 shows its derivation. It also shows the central role of Pythagoras' Theorem—that in any right triangle with sides of length a and b and hypotenuse c, $a^2+b^2=c^2$. Figure 2.0.2. An important trigonometric identity based on Pythagoras' Theorem: $(\cos X)^2+(\sin X)^2=1$ for any acute angle X in a right triangle. Other trigonometric functions are tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). Their definitions, using the triangle in Figure 2.0.2 are: \begin{align} \tan A &= \frac{a}{b}=\frac{\left(\dfrac{a}{c}\right)}{\left(\dfrac{b}{c}\right)}=\frac{\sin A}{\cos A}\\[1ex] \cot A &= \frac{b}{a} = \frac {1}{\tan A}\\[1ex] \csc A &= \frac{c}{a} = \frac {1}{\sin A}\\[1ex] \sec A &= \frac{b}{a} = \frac {1}{\cos A} \end{align} Some identities follow directly from the identity $(\cos A)^2+(\sin A)^2=1$. For example: \begin{align} (\sin A)^2&=1-(\cos A)^2\\[1ex] (\cos A)^2&=1-(\sin A)^2\\[1ex] \end{align} More identities can be derived from the ones above. For example: \begin{align} (\sin A)^2&=1-(\cos A)^2\\[1ex] \frac {(\sin A)^2} {(\cos A)^2} &= \frac {1}{(\cos A)^2} - \frac{(\cos A)^2}{(\cos A)^2}\\[1ex] (\tan A)^2 &= (\sec A)^2 - 1 \end{align} This PDF contains an extensive collection of trigonometric identites. Please note that in this list, θ and x stand for angle measures, not angles themselves. This will make more sense in the next sections. ## Exercise Set 2.0.2 1. Start with the Pythagorean Identity $(\sin A)^2+(\cos A)^2=1$ to end with the identity $(\tan A)^2 + 1 = (\sec A)^2$. 2. Start with the identity $\sin(A+B)=\sin A\cos B+\sin B\cos A$ to show: 1. $\sin 2A=2\sin A\cos A$ 2. $\sin 3A=3\sin A-4(\sin A)^3$ Hint: Think of 3A as (2A + A). 3. $\sin 4A=4\sin A\cos A-8(\sin A)^3\cos A$ 3. Start with the identities $(\sin A)^2+(\cos A)^2=1$ and $\cos (A+B)=\cos A \cos B - \sin A \sin B$ to show: 1. $\cos 2A = (\cos A)^2 - (\sin A)^2$ 2. $\cos 2A = 1 - 2(\sin A)^2$ 3. $\cos 2A = 2(\cos A)^2 - 1$ ## Triangle Trigonometry is Important, but Inadequate for Calculus Calculus is about relationships quantities quantities whose values vary. High school trigonometry is about sides of triangles. Also, the idea of angle measure in high school trigonometry is hardly treated seriously. It is common to find expressions like $\sin(\angle A)$ or $\sin(A)$ in high school textbooks, where A is a vertex of a triangle. Put another way, sine, cosine, tangent, etc. are portrayed in high school trigonometry as having geometric objects as their arguments. But you cannot put angles on a number line; you can only put angle measures (i.e., numbers) on a number line. Thus, graphs of trigonetric functions (e.g., $y=\sin A$) make no sense when A is an angle or a vertex in a triangle. The remainder of this chapter will build up the idea of angle measure and will develop trigonometric functions as functions of angle measure. < Previous Section Home Next Section>>
# Square and square roots 1. Examples0/4 watched 2. Practice0/11 practiced 1. 0/4 2. 0/11 Now Playing:Square and square roots – Example 1a Examples 0/4 watched 1. Understanding the negative square roots of the following 1. $\sqrt{225}$ $-\sqrt{225}$ $\sqrt{-225}$ Practice 0/11 Square and square roots Notes To square is to raise the number to the second power. In other words, to square is to multiply the number by itself. Square root is the inverse operation of squaring. To square root is to find the two identical factors of a number. ## What is a square root? To explain square roots, let's take a step back and remember what it means to square a number. To square is to raise the number to the second power. Square roots are the opposite of that, and is actually the inverse operation of squaring. To square root is to find the two identical factors of a number. ## How to find the square root of a number For numbers that are perfect squares, you can find whole numbers as answers. However, for numbers that aren't perfect squares, you'll have to use a method that involves estimation (or you can use a table of square and square roots). ## Finding square root of perfect square numbers Let's first take a look at this question here: What is the square root of 64? If you have a calculator, you can always just punch in it and get the answer. But do you know how to find the square root of a number without a calculator? Now, if you do remember your perfect squared numbers, the root of 64 is just eight. Eight times eight gives you 64. But let's say you can't freely recall perfect numbers. How would do we do this from scratch? First, you will have to find all the prime factors of 64. So, let's go ahead and do that: Imagine that the question now becomes 2x2x2x2x2x2— 2 is multiplied 6 times here. So we've just determined that 64 is just a square root of six 2s, all multiplied together. Before we move on, we must remember that the radical sign actually means "the square root". The square root symbol should really be written with a tiny little two here: Since it's a square root, you can pick a pair of identical numbers to work with and bring them out from under the radical. In this case, we'll take out a 2 from the first pair of 2s, another 2 from the second pair, and another 2 from the last pair. It should look something like this: Now if you multiply the 2s with one another, what do you get? You'll find that you get 8, which is exactly the same as what you would have remember if you knew your perfect squares. However, this is the correct way to find the square root of a number without memorization. ## Finding square root of numbers that aren't perfect squares The basic method to find the square root of a number that is not a perfect square is as follows: 1. Estimate: Pick a number that if you square comes close to, but is less than, the square root of the number you're trying to find. 2. Divide: Divide the number that you are finding the squared root for with the number you picked in step 1 3. Average: Take the average of the number you got in step 2 and the square root 4. Repeat: Repeat steps 2 and 3 until the number is accurate enough for you Now you've learned how to find the square root for numbers that both are and are not perfect squares. Continue on with our lessons to learn how to deal with different radical numbers examples. To square: Raise the number to the second power Ex: ${5^2}$= $5\times 5 = 25$ ${8^2}$= $8\times 8 = 64$ To square root: Finding the two identical factors Ex: $\sqrt{16}$ = $\sqrt{4\times 4}$ = 4 $\sqrt{49}$ = $\sqrt{7\times 7}$ = 7 Perfect squares numbers: ${0^2}$ = 0 ${1^2}$ = 1 ${2^2}$ = 4 ${3^2}$ = 9 ${4^2}$ = 16 ${5^2}$ = 25 ${6^2}$ = 36 ${7^2}$ = 49 ${8^2}$ = 64 ${9^2}$ = 81 & so on... {100, 121, 144, 169, 196...}
## Characteristic Roots Question 1 In this page characteristic roots question1 we are going to see how to find characteristic roots of any given matrix. Definition : Let A be any square matrix of order n x n and I be a unit matrix of same order. Then \|A-λI\| is called characteristic polynomial of matrix. Then the equation \|A-λI\| = 0 is called characteristic roots of matrix. The roots of this equation is called characteristic roots of matrix. Another name of characteristic roots: characteristic roots are also known as latent roots or eigenvalues of a matrix. Question 1 : Determine the characteristic roots of the matrix 5 0 1 0 -2 0 1 0 5 Solution: Let A = 5 0 1 0 -2 0 1 0 5 The order of A is 3 x 3. So the unit matrix I = 5 0 1 0 -2 0 1 0 5 Now we have to multiply λ with unit matrix I. λI = λ 0 0 0 λ 0 0 0 λ A-λI= 5 0 1 0 -2 0 1 0 5 - λ 0 0 0 λ 0 0 0 λ = (5-λ) (0-0) (1-0) (0-0) (-2-λ) (0-0) (1-0) (0-0) (5-λ) = (5-λ) 0 1 0 (-2-λ) 0 1 0 (5-λ) A-λI= (5-λ) 0 1 0 (-2-λ) 0 1 0 (5-λ) = (5-λ)[(-2-λ) (5-λ) - 0] - 0 [(0 - 0)] + 1 [0- (-2 -λ)] = (5-λ)[ -10 + 2 λ - 5 λ + λ²] - 0 + 2 + λ = (5-λ)[ -10 - 3 λ + λ²] - 0 + 2 + λ = (5-λ)[λ² -3 λ-10] + 2 + λ = 5 λ² - 15 λ - 50 - λ³ + 3 λ² + 10 λ + 2 + λ = - λ³ + 5 λ² + 3 λ² - 15 λ + 10 λ + λ - 50 + 2 = - λ³ + 8 λ² - 4 λ - 48 = λ³ - 8 λ² + 4 λ + 48 To find roots let \|A-λI\| = 0 λ³ - 8 λ² + 4 λ + 48 = 0 For solving this equation first let us do synthetic division.characteristic roots question1 By using synthetic division we have found one value of λ that is λ = -2. Now we have to solve λ² - 10 λ + 24 to get another two values. For that let us factorize λ² - 10 λ + 24 = 0 λ² - 6 λ - 4 λ + 24 = 0 λ (λ - 6) - 4 (λ - 6) = 0 (λ - 6) (λ - 4) = 0 λ - 6 = 0 λ = 6 λ - 4 = 0 λ = 4 Therefore the characteristic roots (or) Eigen values are x = -2,4,6 Questions Solution Question 2 : Determine the characteristic roots of the matrix 1 1 3 1 5 1 3 1 1 Question 3 : Determine the characteristic roots of the matrix -2 2 -3 2 1 -6 -1 -2 0 Question 4 : Determine the characteristic roots of the matrix 4 -20 -10 -2 10 4 6 -30 -13 Question 5 : Determine the characteristic roots of the matrix characteristic roots question 1 characteristic roots question 1 11 -4 -7 7 -2 -5 10 -4 -6 characteristic roots question 1 characteristic roots question 1 Characteristic Roots Question1 to Examples 1. Click on the HTML link code below. Featured Categories Math Word Problems SAT Math Worksheet P-SAT Preparation Math Calculators Quantitative Aptitude Transformations Algebraic Identities Trig. Identities SOHCAHTOA Multiplication Tricks PEMDAS Rule Types of Angles Aptitude Test
# Equation of a Sphere A sphere is a 3 – dimensional object. It is the collection of all the points equidistant from a fixed point, the center. It can be referred to as the 3-d version of a circle. So its equation is almost similar to that of a circle with an added variable for the extra dimension. ## In Standard Form The standard equation of a sphere given the center and radius is: (x – h)2 + (y – k)2 + (z – l)2 = r2, here (x, y, z) = any point on the sphere, (h, k, l) = center, r = radius When the center is at the origin (0, 0, 0), the equation of a sphere in standard form becomes: (x – 0)2 + (y – 0)2 + (z – 0)2 = r2 x+ y2 + z2 = r2 Let us derive the equation. ### Derivation Let (h, k, l) be a fixed point in the 3 – D plane. Let (x, y, z) be another point from (h, k, l) at a distance r such that r is constant and a real positive number. Squaring both sides, (x – h)2 + (y – k)2 + (z – l)2 = r2 (applying the Pythagorean theorem) So the equation of sphere with center and radius is: (x – h)2 + (y – k)2 + (z – l)2 = r2 r = √[(x – h)2 + (y – k)2 + (z – l)2 ] ——— (applying the distance formula) If the center is at the origin (0, 0, 0) Then the equation of the sphere is: x+ y2 + z2 = r2 ∴r = √( x+ y2 + z2) Let us solve some examples to understand the above concept. Find the equation of sphere with center and radius as (1, 1, 2) and 5 units respectively. Solution: As we know, The equation of a sphere in standard form is: (x – h)2 + (y – k)2 + (z – l)2 = r2, here (h, k, l) = (1, 1, 2), r = 5 units ∴ The equation is: (x – 1)2 + (y – 1)2 + (z – 2)2 = 52 Write the equation of a sphere given the center as (2, 4, 6) and radius 3 units. Solution: As we know, The equation of a sphere in standard form is: (x – h)2 + (y – k)2 + (z – l)2 = r2, here (h, k, l) = (2, 4, 6), r = 3 units ∴ The equation is: (x – 2)² + (y – 4)² + (z – 6)² = 3² Finding the center and radius of a sphere when the EQUATION is known Find the center and radius of a sphere having the equation x² + (y – 1)² + (z – 4)² = 112. Solution: Given the equation of the sphere: x² + (y – 1)² + (z – 4)² = 11² ∴ Center (h, k, l) = (0, 1, 4), Given r2 = 11² ∴ r = 11 units ## In General Form The equation of a sphere in general (expanded) form is written as: x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, here (-u, -v, -w) = center, and u, v, w, & d = constants Radius (r) = ${\sqrt{u^{2}+v^{2}+w^{2}-d}}$, here (-u, -v, -w) = center, and u, v, w, & d = constants, then d = u2 + v2 + w2 – r2 Let us solve some examples to understand the above concept. Write the equation of a sphere in expanded form centered at (1, 2, 3) with a radius of 6 units. Solution: As we know, The equation of a sphere in expanded form is: x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, here (-u, -v, -w) = (1, 2, 3) ∴ u = -1, v = -2, w = -3 Now, d = u2 + v2 + w2 – r2, here u = -1, v = -2, w = -3, r = 6 units ∴ d = 1 + 4 + 9 – 36 = -22 So, the equation is: x² + y² + z² – 2x – 4y – 6z – 22
# 1.9: Numerical Expression Evaluation with Basic Operations Difficulty Level: At Grade Created by: CK-12 Estimated24 minsto complete % Progress Practice Numerical Expression Evaluation with Basic Operations MEMORY METER This indicates how strong in your memory this concept is Progress Estimated24 minsto complete % Estimated24 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Have you ever been to the White Mountains? Take a look at this dilemma. The first few days of Teen Adventure were spent at the base of the White Mountains in the Lafayette Place Campground. There the teens worked on survival skills, map skills and basic first aid. The students were kept all together and Kelly made some terrific new friends. Her favorite part was all of the team building and trust building exercises that they learned. She even surprised herself by being able to complete some tasks that she wasn’t sure that she could do. Kelly and the other students learned to rely on themselves and each other over those first few days. The night before their first hike, the group leaders told them that they would be divided up into hiking groups. Each week they would meet altogether once again to talk about their adventures, but during the actual week each group would be on a different trail. There would be three groups out of the 30 students. The 6 leaders would be divided into the 3 groups. For the first few days, there would be a manager along and two first aid persons. After the first few days, the manager and the two first aid persons would head back to the base office to be checked in with weekly or in case of emergency. So they would start with one number of people and after the first few days the number would change. Kelly is curious about how many people will start and how many will be together after the first few days. This is where you come in. You can write an expression to figure this out-using grouping symbols and the order of operations you should be able to calculate two different group numbers. This Concept will teach you all that you need to know. Pay attention because you will see this problem once again. ### Guidance To begin this Concept, you have to blast back in math. Think back to last year in math, do you remember using the order of operations to evaluate numerical expressions? Remember that a numerical expression is an expression that has numbers and operations in it. Let’s review. We could evaluate this expression in two different ways. In the first way, we simply perform the operations in order from left to right. Wait a minute!! That is incorrect because we did not use the order of operations. Let’s use the order of operations as we evaluate this expression a second way. Good, now we can apply this information to our example. We multiply first. We can apply this information to variable expressions. How can we do that? Remember that a variable expression has variables, numbers, operations and sometimes exponents in it. We can start by thinking about evaluating variable expressions. When we evaluate a variable expression, we are finding the value of that expression. If given a value for a variable, we substitute the numerical value of the variable into the expression. Then we can find the total value of the expression. Some variable expressions will have more than one variable in it. We can evaluate the expression if we have been given values for each of the variables. Evaluate the expression: if First, we substitute the given value in for the variable. Next, we perform multiplication and division in order from left to right. Now we add and subtract in order from left to right. Sometimes, you will see a problem with two variables in it. Evaluate the expression: if and First, we substitute the given values for and into the expression. Next, we perform multiplication and division in order from left to right. Notice that this problem has a fraction bar in it which means division. Now we add and subtract in order from left to right. Evaluate the following variable expressions by following the order of operations. if is 5 Solution: 37 if is 4 Solution: 27 #### Example C if is 6 and is 9 Solution: 46 Now back to the hiking trip and the groupings. Kelly is curious about how many people will start and how many will be together after the first few days. First, let’s work with the numbers involved. 30 students 1 manager 2 first aid persons Now we can write an expression to show how one group is created out of the whole. Next, we solve it for the number of students and leaders per group. Remember to follow the order of operations. persons For the first few days, there will also be a manager and two first aid persons. 15 people in a group for the first few days. Then the manager and two first aid persons leave. The core group will consist of 12 people. ### Vocabulary Here are the vocabulary words in this Concept. Numerical Expression an expression that uses numbers and operations. Variable Expression an expression that uses numbers, variables and operations. Order of Operations the order that you perform each operation when evaluating an expression. ### Guided Practice Here is one for you to try on your own. if is 4 First, remember to follow the order of operations. Then substitute 4 into the expression for x. Now we can complete multiplication and division in order from left to right. Next, we can complete addition and subtraction in order from left to right. ### Video Review Here are videos for review. http://www.youtube.com/watch?v=moUaatNssoQ - This is a James Sousa video on evaluating an expression using the order of operations. ### Practice Directions: Use the order of operations to evaluate each numerical expression. 1. 2. 3. 4. 5. Directions: Evaluate the following expressions when is 4. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:
LESSON 2: What does ‘equation’ mean? Introduction to Elementary AlgebraHelp on solving algebra math problemsSimple step by step method An equation looks like this And your main goal is to solve the equation so you get the value of the x (not known at this time) that makes the formula correct. You will learn later that the correct answer in this case is: If you replace x by 11 in the above equation you will get 3 times 11 less 4 = 7 plus 2 times 11 (left side of the equation) (right side of the equation) Which is the same as 29 = 29 True isn’t it? Let’s start from the very beginning. Looking at the equation, we don’t know the value of ‘x’ but we know one important thing: The LEFT side is EQUAL to the RIGHT side, and we have to believe this and move on with our lives. this side is equal to this side Now we arrive at the first confusion about equations: Both sides are equal “in value” only, not the way they look in the paper. This is similar as stating: this side is equal to this side Very easy to believe because you know that 2 plus 3 adds 5, even if 2 + 3 is written differently than the number 5. The ‘=’ sign is saying both sides are equal in value not in appearance. So, when you get an equation to solve, you start by believing that the equal sign is telling the absolute truth: both sides are equal in value.
GCSE Maths Geometry and Measure Area Area Of An Isosceles Triangle # Area Of An Isosceles Triangle Here we will learn about the area of an isosceles triangle including how to find the area of an isosceles with given lengths and how to calculate those lengths if they are not given. There are also area of a triangle worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck. ## What is an isosceles triangle? An isosceles triangle is a type of triangle with two equal sides. The base angles which are opposite to the equal sides are also equal. There are different types of isosceles triangles: Isosceles right triangle -two equal angles (45° each) -two equal side lengths A special case of an isosceles triangle is an equilateral triangle where all three sides and angles of the triangle are equal. ### How to name a triangle We can identify a triangle by putting a capital letter on each vertex (corner). We can then refer to each of the sides of the triangle by using two letters to describe where the line starts and ends. We can refer to the entire triangle by using all three letters. E.g. Name of sides: side AB, side AC, side BC Name of triangle: triangle ABC ### How do we find the area of a triangle? In order to find the area of a triangle, we need to start with the area of a rectangle. To find the area of a rectangle you must multiply adjacent sides together. The area of the rectangle below would be calculated by multiplying the base x height (b x h). We can create an isosceles triangle by drawing two sides from the midpoint of a side of the rectangle to the corners. The rectangle has been split into an isosceles triangle and two congruent (identical) right angled triangles. If we combine the area of the two right angled triangles they will form the pink isosceles triangle. The area of the isosceles triangle is exactly half the area of the rectangle. ### Area of an isosceles triangle formula $\text { Area of a triangle }=\frac{\text { base } \times \text { height }}{2}$ This can be shortened to $A=\frac{1}{2} b h$ where b is the base length and h is the height of the triangle. To find the area of an isosceles triangle when only the side lengths have been given, we will need to calculate the height of the triangle using Pythagoras’ Theorem Finding the height using Pythagoras’ Theorem Let the base of the isosceles triangle be b and the two equal sides be l. Drawing a perpendicular line from the top vertex to the base forms the height, h, and also forms a right angled triangle with hypotenuse l and short sides h and \cfrac{1}{2} \, b. h=\sqrt{{{l}^{2}}-{{\left( \cfrac{1}{2} \, b \right)}^{2}}} We can also find the area of the isosceles triangle using the area of a triangle formula Area=\cfrac{1}{2} \, ab\sin C If we know (or can find) the two equal sides, l and the angle between them, \theta , the formula would give the area as A=\cfrac{1}{2} \, {{l}^{2}}\sin \left( \theta \right) ## How to find the area of an isosceles triangle In order to find the area of a isosceles triangle: 1Identify the height and base length of your triangle if given. (You might need to calculate these values) 2Write the appropriate formula A=\cfrac{1}{2} \,b h or A=\frac{1}{2}{{l}^{2}}\sin \left( \theta \right) 3Substitute the values into the formula 4Calculate ## Related lessons on area Area of isosceles triangles is part of our series of lessons to support revision on area. You may find it helpful to start with the main area lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include: ## Area of isosceles triangles examples ### Example 1: given base length and height Find the area of the triangle below: 1. Identify the height and base length of the triangle if given. h = 6cm b = 10cm 2Write down the appropriate formula $A=\frac{1}{2} b h$ 3Substitute the values into the formula \begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(6)(10) \end{aligned} 4Calculate \begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(6)(10)\\ &=30cm^2 \end{aligned} ### Example 2: given base length and height Calculate the area of the triangle below: Identify the height and base length of the triangle if given. Write down the appropriate formula Substitute the values into the formula. Calculate ### Example 3: worded question Shown below is a triangular shaped field. Each cow needs 2000m^{2} to graze. How many cows can fit into this field? Identify the height and base length of the triangle if given. Write down the appropriate formula Substitute the values into the formula. Calculate ### Example 4: missing height Find the area of the triangle below: Identify the height and base length of the triangle if given. Write down the appropriate formula Substitute the values into the formula. Calculate ### Example 5: compound shapes Below is the floor plan for a new house. Calculate the area of the plan. Identify the height and base length of the triangle if given. Write down the appropriate formula Substitute the values into the formula. Calculate ### Example 6: Using the ½ab sin C formula Below shows an isosceles triangle with two sides of length 10 \, cm and a base angle of 75^{\circ}. Find the area. Identify the height and base length of your triangle if given. (You might need to calculate these values) Write down the appropriate formula Substitute the values into the formula. Calculate ### How to find a missing side length given the area Sometimes a question might give you the area and ask you to work out the height or missing length. In order to do this you must rearrange the formula. To find a missing length given the area: 1Rearrange the formula 2Substitute in the values you know 3Calculate Step by step guide: Rearranging equations ### Example 7: calculating base length Triangle ABC is an isosceles triangle with an area of 18cm^{2}. The height of the triangle is 6cm. Find the length of the base of the triangle. Rearrange the formula Now substitute the given values. Calculate ### Common misconceptions • Identifying the correct information to use A question may give extra information that is not needed to answer it. Carefully identify the relevant pieces of information. E.g. To calculate the area here we only need the base and height. Base= 12cm Height = 8cm $Area = \frac{12 \times 8}{2} = 48cm^{2}$ We can ignore the values of the other sides (10cm) • Units It is a common error to forget the units for area in the final answer. When calculating area, your answer must always have units squared. ### Practice area of an isosceles triangle questions 1. Find the area of the triangle below 90cm^{2} 45cm^{2} 180cm^{2} 21cm^{2} Using the lengths for base and height, the calculation we must perform is \frac{1}{2} \times 15 \times 6 2. Find the area of the triangle below giving your answer in cm^{2} 225cm^{2} 2250cm^{2} 22500cm^{2} 45000cm^{2} After converting the height to the appropriate units, the calculation becomes \frac{1}{2} \times 300 \times 150 3. Shown below is a triangular shaped chicken enclosure cage. Each chicken needs 8m^{2} to roam around. How many chickens can fit into this enclosure? 125 chickens 15 chickens 16 chickens 30 chickens The area of the enclosure is \frac{1}{2} \times 20 \times 12.5 = 125m^{2}. We then consider multiples of 8 to work out how many chickens will fit. Or use a division method: 125 \div 8 = 15.625 So 15 chickens will fit. 4. Find the area of the triangle below (give your answer to two decimal places) 18.00cm^{2} 9.00cm^{2} 8.71cm^{2} 17.42cm^{2} The height of the triangle can be found using Pythagoras’ Theorem. Height = \sqrt{6^{2}-1.5^{2}} Height = 5.809… The area of the triangle is then given by \frac{1}{2} \times 3 \times 5.809… 24.6m^{2} 18.3m^{2} 12.3m^{2} 19.4m^{2} The shape can be split into a rectangle and a triangle. The area of the rectangle is 3 \times 4 = 12m^{2} The area of the triangle is \frac{1}{2} \times 3 \times 4.2 = 6.3m^{2} The total area is 12 + 6.3 = 18.3m^{2} 6. Triangle MNP is an isosceles triangle with an area of 20cm^{2} . The base length of the triangle is 4cm . Find the height of the triangle. 80cm 5cm 2.5cm 10cm Starting with the formula and the information we know already, Area = \frac{1}{2} \times base \times height 20 = \frac{1}{2} \times 4 \times height 20 = 2 \times height So the height is 10cm ### Area of an isosceles triangle GCSE questions 1. This pattern is made from three identical isosceles triangles. Find the total area. (3 marks) 10 \div 2 = 5 (1) One triangle: \begin{aligned} A&= \frac{1}{2} \times 5 \times 12\\ A&=30cm^{2} \end{aligned} (1) Total area: 3 \times 30 = 90cm^{2} (1) 2. (a) Find the area of the following isosceles triangle: (b) The isosceles triangle below has the same area as the triangle in part (a). Work out the height of this triangle. (3 marks) \begin{aligned} A&= \frac{1}{2} \times 8 \times 12\\ A&=48cm^{2} \end{aligned} (1) Rearrange area of triangle: \begin{aligned} A &= \frac{1}{2} bh \\ 2A &= bh\\ h & = \frac{2A}{b} \end{aligned} Substitute in values: h= \frac{2 \times 48}{16} (1) \begin{array}{l} h= \frac{96}{16}\\ h=6cm \end{array} (1) 3. Calculate the area of the isosceles triangle. (4 marks) Find height of triangle using Pythagoras: h^{2}=5^{2}-3^{2} (1) \begin{aligned} h^{2}&=16\\ h&=4 \end{aligned} (1) A=\frac{1}{2} \times 6 \times 4 (1) A=12cm^{2} (1) ## Learning checklist You have now learned how to: • Apply formula to calculate and solve problems involving the area of triangles • Use Pythagoras’ Theorem to solve problems involving triangles ## Still stuck? Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors. Find out more about our GCSE maths tuition programme.
# 1TTO 12.4 Linear formulas §12.4 Linear formulas The main goal in this paragraph is: Learning how to make a FORMULA for a LINEAR GRAPH! See next slide! 1 / 26 Slide 1: Slide WiskundeMiddelbare schoolhavo, vwoLeerjaar 1 This lesson contains 26 slides, with interactive quizzes and text slides. Lesson duration is: 45 min ## Items in this lesson §12.4 Linear formulas The main goal in this paragraph is: Learning how to make a FORMULA for a LINEAR GRAPH! See next slide! #### Slide 1 -Slide timer 0:25 What is a linear graph? #### Slide 2 -Mind map A linear graph is a straight line in a coordinate system. #### Slide 3 -Slide Point of attention: Everybody can learn how to make a formula for a linear graph. However, it takes several steps - 4, in fact! - to do this. So concentrate and work hard on all the coming slides. Only then you will get to know 'the whole trick'! #### Slide 4 -Slide Take before you: + Notebook and Textbook 1B, p.182 pencil, pen, geo, eraser timer 0:25 #### Slide 5 -Slide Something new about the Coordinate system(1): #### Slide 6 -Slide Something new about the Coordinate system (2): Schermafbe #### Slide 7 -Slide In which quadrant have you been working up to now? timer 0:15 A 4 B 3 C 2 D 1 #### Slide 8 -Quiz Watch: the POSITIVE NUMBERS are used for the 1st Quadrant. And the 1st Quadrant is the one that you've used up to now. But this gonna change soon... #### Slide 9 -Slide Now more about Formulas and Graphs: #### Slide 10 -Slide Here you see the graph that belongs to the Formula! #### Slide 11 -Slide Making a FORMULA for this graph... is exactly what you're going to learn now! #### Slide 12 -Slide We study the Method together now. After this you are going to work out these steps yourself! om 19.56. #### Slide 14 -Slide Make a FORMULA for the GRAPH! How? + Open your Textbook on p.184 so you can look back at the METHOD + You're going WORK OUT the 4 STEPS that we just studied #### Slide 15 -Slide Step 1=FOTOVRAAG (next slide) Step 1: Make a TABLE with values you can (for this graph only 3 values for x and 3 for y) Take a pic and send it in. timer 0:45 #### Slide 16 -Slide The TABLE (step 1): Solution Step 1: #### Slide 18 -Slide Step 2: Do this from the Table you made! timer 0:15 A 6 B 2 C -6 D 3 #### Slide 19 -Quiz Solution Step 2: The goes through the Vertical axis at height -6. That's why. The coordinates (0, -6) from the table also give you this information. #### Slide 20 -Slide Step 3: How much does the graph rise or fall per horizontal step of 1? timer 0:25 A 2 B 3 C -6 D 6 Solution step 3: #### Slide 22 -Slide Step 4: The Formula is: .................... timer 0:30 A y = 6x - 3 B 3x - 6 = y C x = 3y - 6 D y = - 6 + 3x #### Slide 23 -Quiz Solution step 4: the Formula is: y = 3x - 6 or: 3x - 6 = y or also: -6 + 3x = y #### Slide 24 -Slide Looking back .... - did you do great, or - did you miss out on these tasks? - most important is to try hard, so that you learn a lot, - you probably really begin to understand it when doing your homework! #### Slide 25 -Slide Homework: § 12.4 Linear formulas: 27, 28 and 29.
# 2013 AMC 10B Problems/Problem 7 ## Problem Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle? $\textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2}$ ## Solution 1 unitsize(8); draw((0,0)--(1/2,sqrt(3)/2)); draw(1/2,sqrt(3)/2)--(3/2,sqrt(3)/2)); (Error compiling LaTeX. draw(1/2,sqrt(3)/2)--(3/2,sqrt(3)/2)); ^ 3bd139aed9da3832a56caf57561ed2683df0306f.asy: 5.37: syntax error error: could not load module '3bd139aed9da3832a56caf57561ed2683df0306f.asy') If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a $30-60-90$ triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is $1$, so the side opposite the thirty degree angle in the triangle is also $1$. From the properties of $30-60-90$ triangles, the area is $1\cdot\sqrt3/2$=$\boxed{\textbf{(B) } \frac{\sqrt3}{2}}$ ## Solution 2—Similar to Solution 1 As every point on the circle is evenly spaced, the length of each arc is $\frac{\pi}{3}$, because the circumference is $2\pi$. Once we draw the triangle (as is explained in solution 1), we see that one angle in the triangle subtends one such arc. Thus, the measure of that angle is thirty degrees. Similarly, another angle in the triangle subtends an arc of twice the length, and thus equals 60 degrees. The last angle is equal to 90 degrees and the triangle is a $30-60-90$ triangle. We know that as the diameter, the length of the hypotenuse is 2, and thus, the other sides are 1 and $\sqrt{3}$. We then find the area to be $\boxed{\textbf{(B) } \frac{\sqrt{3}}{2} }$.
# Sine WaveRunner 3 Act Math ## Act 1 1. Who will win the race? 2. Write down a guess. My initial thought for this lesson was to focus on what it could look like in a Pre-Cal or Calculus class. However, I’m intrigued by the possibilities for earlier subjects as well. More on that in Act 2. ## Act 2 3. What information would be useful to figure this out? 4. Write down some questions you have in your head right now. The students need to know the speeds. Proportional reasoning can be used to figure out the meters per second for each vehicle. Next, we need to know the distances that each will travel. This is where teacher options are presented. First, let’s see an image. The rower distance is 500 meters. However, work needs to be done to find the Jet Ski distance. Honestly, I didn’t know how to calculate this, but thankfully, Michael Fenton was a huge help. So, here is my vision for the task. If you’re a Pre-Cal or Calculus teacher, let your kids try to figure out the distance with the image above and/or this Desmos graph (includes the equation I used). However, if you teach an earlier subject, here are some other routes through the lesson. ### Middle School There’s a lot of potential for ratios and proportional reasoning in this task. For example, here is how I figured out the Jet Ski distance (thanks to Michael). WolframAlpha did the calculation for me. 15.2808 represents the arc length on the graph. From there, I used a proportion to determine the distance in meters for the problem. Granted, it might be a lot to ask for a middle schooler to figure this out. Therefore, this picture could be helpful (here’s the link to the graph as well if you’re interested). Once the students figure out the Jet Ski distance, they can divide the rates for each and determine the solution. ### Elementary In the Elementary grades, this lesson can be modified to work on division and division with decimals. Show the kids this image. Also, if they aren’t ready to calculate the speeds, show them this image. # Sequel: 5.What would happen if the Jet Ski was going 50 mph, and the course looked like this? Common Core Standards 7.RP.A.2 Understand ratio concepts and use ratio reasoning to solve problems. HSF.TF.B.5 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. TEKS 6.5(A) represent mathematical and real-world problems involving ratios and rates using scale factors, tables, graphs, and proportions Pre-Cal 2(G) graph functions, including exponential, logarithmic, sine, cosine, rational, polynomial, and power functions and their transformations, including af(x), f(x) + d, f(x – c), f(bx) for specific values of a, b, c, and d, in mathematical and real-world problems ## 2 thoughts on “Sine WaveRunner” 1. This is so slick. Nice work! Thanks for the shoutout, though I feel like I did about 0.00001% of the work here. 🙂 I’d love to hear how this plays out with students. Will you have an opportunity to use this in the classroom? 1. Thanks Michael! You helped me more than you know. Unfortunately, I don’t think I’ll get a chance to use this for a while. It doesn’t really intersect with the learning goals for the rest of the year. I’m curious to see what an upper level class could do with it though. This site uses Akismet to reduce spam. Learn how your comment data is processed.
## Learning goals how to graph vertex form factoring standard form to factored first and second difference ## what is vertex form and how to graph an equation, written out in vertex form, is like this: y= a(x-h)^2 +k. in this equation, a, tells us if the parabola will be stretched or compressed, h, tells us where the parabola will be left to right, and k tells us where it would be up or down. When given an equation like 3(x−1)^2+1, a, which is 3, would tell you that it is stretched by a factor of 3, while h, which is -1, tells you that it is moved 1 unit to the right, since the h sign flips, and k, which is also 1, tells you that the parabola is translated 1 unit up. ## First and second difference using the first and second difference will help you determine if a graph is linear, quadratic, or neither. to do this, you make a chart, with labels, x and y. write all numbers on x in order from 1 to any number, and for y, you can make a pattern, like 6,3,-3,-12. then multiply each number by the next. if you get a pattern, there is a first difference, indicating that the graph is linear, but if it is not, subtract the first differences the same way to find t he second difference. it is the same, it is quadratic, but if now, it is neither. ## Factoring/simple trinomials there are 2 methods of factoring that are used the most, and that is simple and complex trinomials. for simple trinomials, it's as simple as finding what multiplies to give you what ever number c is, and what adds up to give you whatever b is. for example, if given an equation like, x^2+8x+16, you have to find what adds to give you 8, but multiplies to get 16, which is 4 and 4. then you put it in brackets like this: (x+ ) (x+ ) and fill it in (x+4) (x+4). since the numbers were the same, you can simplify by doing (x+4)^2. ## Factoring/complex trinomials When solving a complex trinomial, you always want to look for a common factor first. if there is none, continue. what you do next, is multiply a and b to get a product. for example, if given an equation like 6x^2+7x-3, you multiply 6 and -3 to get a product, which is -18. then, you want to find 2 numbers that multiply to give you the product, but add up to give you b, which in this case, is 7. those numbers are 9 and -2. now, rewrite the equation, replacing b with the 2 numbers like this : 6x^2 +9x-2x-3. now there are 4 variables. split them into to, giving you [6x^2+9x] [-2x-3]. now you find the individual factors of what's in those brackets, giving you: 3x{2x+3} and -1{2x+3}. the last step is to take whatever is in the brackets as your first factored equation, and then, the 2 numbers outside the brackets as your second, like this: {2x+3} {3x-1} Factoring Complex Trinomials.mp4
# Rate, Ratio, and Proportional Reasoning A rate is a ratio. A ratio is a comparison of two numbers. A ratio can be expressed three ways: 1. Using the fraction bar as in 2/3. 2. Using a colon symbol as in 2:3. 3. Using the word “to” as in 2 to 3. When the denominator of a rate is 1, we call the rate a unit rate. We usually use the key word “per” or the division symbol ( / ) to indicate a unit rate. A ratio table is another way of finding answers to ratio problems. With a ratio table, you start with a known ratio and use it to find other numbers with the same ratio. You can keep using the numbers you find to discover more numbers until you solve the problem. A proportion is an equation which states that two ratios are equal. To solve a proportion, you simply cross multiply to find the missing value. There are three parts to every proportion: 2 fractions and an equal sign. Each of the fractions by itself is a ratio. The numerator of the first ratio and the denominator of the second ratio are called the extremes. The denominator of the first ratio and the numerator of the second ratio are called the means. We can designate equivalent amounts of a whole by using a fraction, a decimal number, or a percent. To change a fraction to a decimal, divide the numerator by the denominator. To change a decimal to a percent, we move the decimal point two places to the right. To change a percent to a decimal, move the decimal two places to the left. Remember, percent means out of (per) 100, so the denominator of the percent ratio will always be 100. ## Practice 1) Write the ratio in another form. 9:6 2) Write the ratio of ones to zeros. 1110000 3) There are 145 students in your school chorus program. 25 students sing soprano and 15 sing alto. Write a ratio of the number of altos to all the students in the chorus program. Write your answer in lowest terms. 4) Make the ratios equal. 9/25 = 36/? 5) The ratio of apples to oranges in a fruit bowl is 1 to 4. If there are 8 apples in the bowl, how many oranges are there? 6) 2 pounds of apples cost \$0.76. How much would 5.5 pounds cost? 7) If it is \$55 for 20 people, how much for one? 8) Which is the better deal: 5 cans of soda for \$2.27 or 7 cans of soda for \$2.89? 9) Which is traveling faster? Traveling 350 miles in 5 hours or traveling 215 miles in 3 hours? 10) Which is traveling faster? Traveling 170 yards in 45 seconds or traveling 416 feet in 35 seconds (source)
# Question Video: Using the Multiplication Table to Find a Pair of Numbers that Satisfies an Addition and Multiplication Condition Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. 03:34 ### Video Transcript Among all the pairs of numbers whose sum is six, find the pair with the largest product. There are two keywords in this problem that we need to know what they mean before we can work out the answer, and they are sum and product. Sometimes in maths, you might hear the word sum used to describe a calculation. I’ve just completed a page of sums. But really that’s not the correct way to use the word sum. The word sum means something very definite. The sum of two or more numbers is what we get when we add them together. It’s the total of some numbers. So, the first part of our problem could read, among all the pairs of numbers whose total is six. The product of a pair of numbers is what we get when we multiply them together. It’s the answer to a multiplication. Let’s read our problem using our new phrases. Among all the pairs of numbers whose total is six, find the pair with the largest answer when multiplied. To start with, we need to find all the pairs of numbers whose total is six. So that we don’t get mixed up, let’s call one number 𝑥 and one number 𝑦. When we’re trying to find a list of numbers like this. It’s sometimes a good idea to start either at the very bottom or the very top and to work our way methodically through them. So, let’s start with the very lowest possible value of 𝑥 and go from there. The lowest number that 𝑥 could be is zero. And if it makes a pair of numbers whose total is six when we add it to 𝑦, 𝑦 must be six. Zero plus six equals six. The next highest number is one. One plus five equals six. You’ll see a pattern here. Two plus four equals six. Three plus three equals six. Four plus two equals six. But we’ll stop there because now we’re starting to repeat ourselves. The numbers two and four, if we multiply them together have the same product as four and two. And so, there’s no point writing them down twice. The only pairs of numbers that will make different products are these four. So, our problem asks to look among these pairs whose total is six and find the pair that has the largest product, the largest answer when we multiply them together Let’s multiply each pair to see what the answers are. Zero multiplied by six equals zero. One multiplied by five equals five. Two fours are eight. And the product of three and three equals nine. So, the pair with the largest product is when both numbers are three. First, we found all the pairs of numbers whose sum was six. We found four different pairs. Then, we multiplied the two numbers in each pair together to see what the products were. And among all the pairs of numbers whose sum was six, the pair with the largest product was when 𝑥 equals three and 𝑦 equals three.
GeoGebra Classroom # Intro: GeoGebra skills, terms, vocabulary & postulates The objectives of this activity: With the GeoGebra app: 1. you will become familiar with some of the tools used in GeoGebra. 2. you will explore what you can and can't change once you have constructed a figure. 3. You will work with 2 of the three undefined terms, point & line, in order to illustrate some postulates. With the follow-up questions: 1. You will review and extend your understanding of the postulates and definitions you encountered 2. You will review the vocabulary used in this activity. 3. You will use the Applet to make some observations and write some conclusions. What you should you do in your notebook: 1. Add any new vocabulary to your vocabulary section. 2. Write down questions and correct answers for any questions that you weren't sure about and/or got incorrect. 3. Keep track of new observations and facts that helps you extend your understanding of the material covered. ## Some of the questions will talk about congruence. Watch this short video before you answer the questions. We encountered 2 postulates in this activity: 1. Through 2 points there is exactly one line 2. if two lines intersect, then they intersect in exactly one point. In your own words describe what each postulate is saying. ## 2. Definition circle is all points in the same plane that lie at an equal distance from a center point. A radius of a circle is a segment whose endpts. are the center of the circle and a point on the circle. A diameter is a chord that contains the center of the circle. Define, chord (of a circle) ## 3. Definition Define, ray. In plane geometry we have three undefined terms one is "plane" ( a 2-dimensional surface that extends out indefinitely into space), what are the other two undefined terms mentioned in this activity. ## 5. Definition Define, postulate. Define, segment. ## 7. Explain Explain why the following statement must be true: in a circle or in congruent circles, radii or diameters are congruent. (make sure to watch the video that explains congruent figures) ## 8. Fill-in a. A circle has ________ radii and diameters. b. A segment has ________ midpoint(s). c. A line has ______ midpoint(s). possible answer choices: zero, one, two, three, infinite (unlimited) Define, midpoint ## 10. Definition Define, congruent (figures) ## 11. Vocabulary List all the vocabulary used in this activity.
PRINTABLE FOR KIDS XII (12) HSC XI (11) FYJC X (10) SSC ### Question 1. Complete the following activity to solve the simultaneous equations. 5x + 3y = 9 ……. (i) 2x + 3y = 12 ……… (ii) Question 1. Complete the following activity to solve the simultaneous equations. 5x + 3y = 9 ……. (i) 2x + 3y = 12 ……… (ii) 5x + 3y = 9 ……. (i) 2x + 3y = 12 ……… (ii) Subtracting equation (ii) from (i), we get, (5x + 3y ) - (2x + 3y) = 9 - 12 5x - 2x + 3y - 3y = -3 3x = -3 x = -1 Putting the value of x in equation (i), 5(-1) + 3y = 9 -5 + 3y = 9 3y = 14 y = 14/3 Let’s add equations (I) and (II). Hence, x = -1 and y = 14/3 is the solution of the equation. MATHS I SSC NEW SYLLABUS Chapter No. 1. Linear Equations in Two Variables Practice Set 1.1 Practice Set 1.2 Practice Set 1.3 Practice Set 1.4 Practice Set 1.5 Problem Set No. 1 Practice Set 2.1 Practice Set 2.2 Practice Set 2.3 Practice Set 2.4 Practice Set 2.5 Practice Set 2.6 Problem Set No. 2 Chapter No. 3: Arithmetic Progression Practice Set 3.1 Practice Set 3.2 Practice Set 3.3 Practice Set 3.4 Problem Set No 3 Chapter 4: Financial Planning. Practice set 4.1 Practice set 4.2 Practice set 4.3 Practice set 4.4 Problem set 4 A Problem set 4 B CHAPTER NO. 5. PROBABILITY PRACTICE SET 5.1 PRACTICE SET 5.4 PROBLEM SET NO. 5 Chapter No. 6: STATISTICS PRACTICE SET 6.1 PRACTICE SET 6.2 PRACTICE SET 6.3 PRACTICE SET 6.4 PRACTICE SET 6.5 PRACTICE SET 6.6 PROBLEM SET NO. 6 BOARD PAPERS WITH COMPLETE SOLUTION Mathematics Part - i Paper Pattern and Guidelines SSC Maths I March 2019 Board Paper Solution 10th Standard 11th, March, 2019 GENERAL MATHS ALGEBRA BOARD PAPERS MATHS I PRACTICE PAPER FOR SCHOOL PRELIMINARY EXAM. [AS PER NEW SYLLABUS] Paper No. 1 Paper No. 2 Paper No. 3 ALGEBRA SSC OLD SYLLABUS ARITHMETIC PROGRESSION LINEAR EQUATIONS PROBABILITY STATISTICS ONE STATISTICS TWO EXTRA HOTS SUMS FORMULA ALGEBRAIC FORMULAE ## PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra) SUBJECTS HINDI ENTIRE PAPER SOLUTION MARATHI PAPER SOLUTION SSC MATHS I PAPER SOLUTION SSC MATHS II PAPER SOLUTION SSC SCIENCE I PAPER SOLUTION SSC SCIENCE II PAPER SOLUTION SSC ENGLISH PAPER SOLUTION SSC & HSC ENGLISH WRITING SKILL HSC ACCOUNTS NOTES HSC OCM NOTES HSC ECONOMICS NOTES HSC SECRETARIAL PRACTICE NOTES 2019 Board Paper Solution HSC ENGLISH SET A 2019 21st February, 2019 HSC ENGLISH SET B 2019 21st February, 2019 HSC ENGLISH SET C 2019 21st February, 2019 HSC ENGLISH SET D 2019 21st February, 2019 SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019 HSC XII PHYSICS 2019 25th February, 2019 CHEMISTRY XII HSC SOLUTION 27th, February, 2019 OCM PAPER SOLUTION 2019 27th, February, 2019 HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019 HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019 SSC ENGLISH STD 10 5TH MARCH, 2019. HSC XII ACCOUNTS 2019 6th March, 2019 HSC XII BIOLOGY 2019 6TH March, 2019 HSC XII ECONOMICS 9Th March 2019 SSC Maths I March 2019 Solution 10th Standard11th, March, 2019 SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019 SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019. SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019. SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019 SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019 XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019 HSC Maharashtra Board Papers 2020 (Std 12th English Medium) HSC ECONOMICS MARCH 2020 HSC OCM MARCH 2020 HSC ACCOUNTS MARCH 2020 HSC S.P. MARCH 2020 HSC ENGLISH MARCH 2020 HSC HINDI MARCH 2020 HSC MARATHI MARCH 2020 HSC MATHS MARCH 2020 SSC Maharashtra Board Papers 2020 (Std 10th English Medium) English MARCH 2020 HindI MARCH 2020 Hindi (Composite) MARCH 2020 Marathi MARCH 2020 Mathematics (Paper 1) MARCH 2020 Mathematics (Paper 2) MARCH 2020 Sanskrit MARCH 2020 Important-formula THANKS
# Parts of Circle : Learn Definition with Properties, Formula and Diagrams 0 Save Parts of a Circle is an important topic of Geometry because knowing the Parts of a Circle helps us to solve problems based on them easily and quickly. An understanding of the Parts of a Circle helps us to draw the circle and its different parts. In this article, we are going to learn about the properties of parts of a Circle, parts like radius, diameter, chord, secant, tangent, arc, sector, segment, annulus, point of tangency, with solved examples and fun facts. ## What is a Circle? A circle is a curved line that runs around a centre point. Every part of the curved line is the same distance from the centre. In other words, it is the locus of all points that have equidistant from the origin. In the two-dimensional plane, the circle is the shape enclosing the most area per unit perimeter squared. A circle may also be defined as a special kind of ellipse in which the two foci are coincident, the eccentricity is 0, and the semi-major and semi-minor axes are equal. Examples of the circle in daily life are the wheel, coin, compact disc, etc. The following figure represents the shape of a circle. ## Symmetry of Circle A circle can be folded into two halves that are exactly the same, which means that it is Symmetrical. The line of this fold is an important part of the circle, called the diameter. One half of this line (from the centre to the edge of the circle) is known as the radius. ## What are the Parts of a Circle? A circle consists of many parts. A part of a circle is called the elements of a circle which defines the circle. Following are the different parts of a circle. 1. Center 3. Diameter 4. Chord 5. Secant 6. Tangent 7. Arc 8. Sector 9. Segment 10. Point of Tangency 11. Annulus of the Circle 12. Circumference Let’s study them in detail. ## Center of the Circle The center in the middle of the circle that is equidistant from all the points on the circle. It is also known as origin. It is denoted by O. In the following figure, O denotes the origin or center of the circle. The radius of the circle is one of the essential parts of a circle. It is the length between the center of the circle to any location on its boundary. In simple words, we can say that, when we join the center of a circle to any point over the circumference applying a straight line, then that line is termed as the radius of that circle. The diameter of a circle is a vertical line, crossing through the center and connecting a point from one end of the circle to a point on the other end of the circle. The diameter is twice the measure of the radius. Mathematically,$$Radius={Diameter\over{2}}$$ ### Properties of the Radius of the Circle • Radius is always perpendicular to the tangent at the point where it touches the circle. • It is half the diameter. • The circles are said to be congruent if they have equal radii. ## Diameter of Circle The length of the line through the center that touches two points on the edge of the circle is called the Diameter of the circle. Mathematically,$$Diameter=2\times{radius}$$ ### Properties of the Diameter of the Circle • The chord of a circle which passes through the centre of the circle is called the diameter of the circle. • The diameter of a circle is the distance across a circle. • The diameter is also the longest chord of a circle. 1. If the radius of a circle is 6 cm, find out its diameter. 2. $$Diameter=2\times{radius}=2\times6=12cm$$ ## Chord of the Circle A chord of a circle is a straight line segment whose endpoints both lie on a circular arc thus dividing a circle into two segments. In short, the chord of a circle can be stated as a line segment joining two points on the circumference of the circle. The diameter is the longest chord of the circle. ### Chord Length Formula There are two fundamental formulas to determine the length of the chord of a circle: Chord Length by Perpendicular Distance from the Center is equivalent to: $$Chord\ Length=2\times\sqrt{\left(r^2−d^2\right)}$$ Length of a chord of circle formula applying trigonometry: Consider the above figure where r denotes the radius of the circle and c is the angle subtended at the center. $$Chord\ length=2r\sin\left(\frac{c}{2}\right)$$ ### Properties of Chord of the Circle • In a circle, or congruent circles, congruent chords are equidistant from the center OR in a circle, or congruent circles, chords equidistant from the center are congruent. • Chords that are equal in measure subtend equal angles at the center of the circle. This is also known as the equal chords equal angles Theorem. • In a circle, or congruent circles, congruent chords have congruent arcs OR in a circle, or congruent circles, congruent arcs have congruent chords. • In a circle, parallel chords intercept congruent arcs. • If two chords intersect in a circle, the product of the lengths of the segments of one chord equal the product of the segments of the other. • In a circle, a radius perpendicular to a chord bisects the chord i.e. in a circle, the perpendicular bisector of a chord passes through the center of the circle. It can also be stated as, in a circle, a diameter perpendicular to a chord bisects the chord and its arc Converse: In a circle, a radius that bisects a chord is perpendicular to the chord. Here is the proof of how to find the length of a chord of a circle. Proof: In the above diagram r is the radius of the circle, d is the perpendicular distance from the chord to the center of the circle. A triangle is formed where r will be the hypotenuse of the triangle. As we know a perpendicular bisector from the chord to the center of the circle bisects the chord in two halves. $$By\ Pythagoras\ theorem:$$ $$\left(\frac{1}{2}chord\right)^2+d^2=r^2$$ $$\Rightarrow\frac{1}{2}of\ Chord\ length=\sqrt{\left(r^2−d^2\right)}$$ $$\Rightarrow Chord\ length=2\times\sqrt{\left(r^2−d^2\right)}$$ ## Secant of the Circle A secant intersects the circle at exactly two points. It can be said that, a secant is a line that intersects the circle no more than two times. ### Properties of Secant of the Circle • In a circle, or congruent circles, congruent chords are equidistant from the center OR in a circle, or congruent circles, chords equidistant from the center are congruent. • It is an extended chord or a coplanar straight line. • If a secant segment and tangent segment are drawn to a circle from the same external point, the length of the tangent segment is the geometric mean between the length of the secant segment and the length of the external part of the secant segment. Thus, the product of the length of the secant segment and its external part equals the square of the length of the tangent segment. • If two secant segments are drawn to a circle from the same external point, the product of the length of one secant segment and its external part is equal to the product of the length of the other secant segment and its external part. ### Tangent of the Circle Tangent of the Circle is a coplanar straight line that has one single point in common with a circle. The tangent line that joins two infinitely close points from a point on the circle. Common tangents are lines, rays or segments that are tangent to more than one circle at the same time. ### Fun Fact about Tangents of Circle The discus throw in the Olympics is a track and field event. During discus throw an athlete attempts to throw a heavy disc that is called a discus. He or she aims to throw the discus further than his or her competitors. The athlete spins counterclockwise around one and a half times through a circle, then releases the throw. When released, the discus travels on a path that is tangent to the circular spin orbit! 4 Common Tangents 3 Common Tangents(2 completely tangent circles) 2 Common Tangents(2 overlapping circles) (2 completely separate circles) 1 Common Tangent(2 internally tangent circles) ### Properties of Tangent of the Circle • If a line is tangent to a circle, it is perpendicular to the radius drawn to the point of tangency. • Tangent segments to a circle from the same external point are congruent. ## Arc of the Circle An arc of a circle is related to a curve, which is a section/portion of its circumference. It is any connected part of a circle. Specifying two end points of an arc and a centre allows for two arcs that together make up a full circle. It is also known as a segment of the circumference of a circle. ### Arc Measure In a circle, the degree measure of an arc is equal to the measure of the central angle that intercepts the arc. ### Arc Length In a circle, the length of an arc is a portion of the circumference. The letter “s” is used to represent arc length. ### Relationship between Arc Measure and Arc Length Consider the following equality $${\text{arc measure}\over{360^o}}={\text{arc length}\over{\text{circumference}}}$$ 1. In circle O, the radius is 12 inches and the minor arc is intercepted by a central angle of 135 degrees. Find the length of the minor arc to the nearest integer. 2. $$\text{Arc Length}={\text{arc measure}\over{360^o}}\times{\text{circumference}}$$ =$${135\over360}\times{2\pi(12)}$$ =28.26inches. ### Measurement of Arc of Circle in Radians The radian measure, θ, of a central angle is defined as the ratio of the length of the arc the angle subtends, s, divided by the radius of the circle, r. Mathematically, it gives arc length, s: s = θr ### Fun Fact • One radian is the central angle that subtends an arc length of one radius (s = r). Since all circles are similar, one radian is the same value for all circles. • The length of the arc intercepted by a central angle is proportional to the radius. $${r_1\over{r_2}}={s_1\over{s_2}}$$ This proportion shows that the ratio of the arc length intercepted by a central angle to the radius of the circle will always yield the same (constant) ratio. ## Sector of the Circle The sector of a circle is the area circumscribed by two radii and the corresponding arc in a circle. There are two types of sectors; minor and major. A “pizza” slice is a sector of the pizza! ### Minor Sector The smaller part of the arc is called the minor arc. In the following figure, green arc shows the minor Sector. ### Major Sector The larger part of the arc is called the major arc. In the following figure, red arc shows the major Sector. ### Finding the Area of the Sector When finding the area of a sector, we are actually finding a fractional part of the area of the entire circle. This fraction is determined by the ratio of the central angle of the sector to the “entire central angle” of 360 degrees. Mathematically, it is given by, $$n\over360^o$$ The area of a sector can be found out by expressing the fraction as the ratio of the arc length (s) to the entire circumference. $${\text{central angle}\over{360^o}}={\text{arc length}\over{\text{circumference}}}$$ $${\text{central angle}\over{\pi{r^2}}}={\text{arc length}\over{2\pi{r}}}$$ A = area of the sector, and s = arc length, $${A\over{\pi{r^2}}}={s\over{2\pi{r}}}$$ $$A={s\over{2\pi{r}}}.{\pi{r^2}}={1\over2}sr$$ ### Area of Sector Circle Semi-circle Quarter Circle Any Sector $$A=\pi{r^2}$$Central angle = 360Fractional part:360/ 360 = 1 $$A={1\over{2}}\pi{r^2}$$Central angle = 180Fractional part:180/ 360 = 1/2 $$A={1\over{4}}\pi{r^2}$$Central angle = 90Fractional part:90/ 360 = ¼ $$A={1\over{4}}\pi{r^2}$$Central angle = n $$A={n\over{360}}\pi{r^2}$$$$A={s\over{2\pi{r}}}$$ ## Segment of the Circle The area contained by the chord and the corresponding arc of a circle is a segment. The two types of segments – minor and major segments. It should be noted that segments do not include the centre. If a sector looks like a “pizza” slice, a segment looks like a crust section of the pizza slice. The segment is the small partially curved figure left when the triangular portion of the sector is removed. ### Finding the Area of the Segment of the Circle Area of the Segment of the Circle = Area of the Sector of the Circle – Area of Triangle Area of the Sector of the Circle= $$A={s\over{2\pi r}}.{\pi r^2}={1\over2}sr$$ Area of triangle can be found by the formula Area = (½) Base x Height Therefore, Area of the Segment of the Circle = $$A={s\over{2\pi {r}}}.{\pi {r^2}}={1\over2}sr$$ – (½) Base x Height 1. Find the area of a segment of a circle with a central angle of 120 degrees and a radius of 8 cm. 2. Let’s find the Area of the Sector of the Circle = $$A={\text{central angle}\over{360^o}}\times{\text{circumference}}={120\over360}\times{2\pi{r}}={120\over360}\times{2\pi{8}}=67.02cm^2$$ Using the 30-60-90 rules (or trigonometry), find the altitude, which is 4, and the other leg, which is $$4\sqrt{3}$$ Area of Triangle = (½) Base x Height = (½) x $$4\sqrt{3}$$ x 4 = 27.71 $$cm^2$$ Area of the Segment of the Circle = 67.02 – 27.71 = 39.31 $$cm^2$$ ## Circumference of the Circle The circumference is the perimeter of a circle or ellipse. That is, the circumference of a circle would be the arc length of the circle as if it were opened up and straightened out to a line segment. A circle is described as a form where all the locations are equidistant from a point at the centre. π shows the ratio of the perimeter of a circle to the diameter. The value of pi(π also pronounced as Pi ) is roughly 3.1415926535897…. which is a non-terminating/never-ending value. $$The\ Circumference\ \ of\ a\ circle=2\pi R$$ $$where,\ R\ is\ the\ radius\ of\ the\ circle,\ \pi\ is\ equal\ to\ 3.14\ or\ \frac{22}{7}$$ ## Other Parts ### Point of Tangency on a Circle The point where a tangent line touches the circle. It is single for a tangent with specific slope. ### Annulus of the Circle The area bounded by two concentric circles is called the annulus. It resembles a ring-shaped object as shown in the image. Hope this article on Parts of a Circle was informative. Get some practice of the same on our free Testbook App. Download Now! If you are checking Parts of a Circle article, also check the related maths articles in the table below: First principles of derivatives Derivatives of logarithmic functions Addition and subtraction of polynomials Finding square root by division method Prove that square root 11 is irrational Prove that square root 2 is irrational ## Parts of a Circle FAQs Q.1 What are the 8 parts of the circle? Ans.1 A circle consists of many parts. Following are the parts of a circle. • Center • Diameter • Chord • Secant • Tangent • Arc • Sector • Segment • Point of Tangency • Annulus of the Circle Q.2 Is Centre a part of the circle? Ans.2 Yes, The centre of is a point in the middle of the circle that has equidistant from all the points on the circle. It is also known as origin. It is denoted by O. Q.3 What is a piece of a circle called? Ans.3 A piece of circle is called a sector. The sector of a circle is the area circumscribed by two radii and the corresponding arc in a circle. There are two types of sectors; minor and major. A “pizza” slice is a sector of the pizza! Q.4 What is the chord of the circle? Ans.4 A chord of a circle is a straight line segment whose endpoints both lie on a circular arc thus dividing a circle into two segments. In short, the chord of a circle can be stated as a line segment joining two points on the circumference of the circle. The diameter is the longest chord of the circle. In a circle, or congruent circles, congruent chords are equidistant from the center. Chords that are equal in measure subtend equal angles at the center of the circle. In a circle, or congruent circles, congruent chords have congruent arcs OR in a circle, or congruent circles, congruent arcs have congruent chords. In a circle, parallel chords intercept congruent arcs. Q.5 What is a tangent in circles? Ans.5 Tangent of the Circle is a coplanar straight line that has one single point in common with a circle. The tangent line that joins two infinitely close points from a point on the circle. Common tangents are lines, rays or segments that are tangent to more than one circle at the same time.
# Teaching Math and the Obvious [Total: 2 Average: 5/5] My #1 goal, when I teach a math class, is to convey a certain way of thinking about math. It’s quite different from what my students have done before, and many of them find it difficult and frustrating. But in the end, many of them start to see math the way I do–and I count that as a major success even if they eventually forget all the details, facts, and techniques we learned along the way. I begin on Day 1 by writing this on the board: 3 + 7 = 7 + 3 This is called the “Commutative Property of Addition,” but I have another word. I call it: obvious. If you have three apples and I have seven…or, if you have seven apples and I have three…the total we have together is the same either way, isn’t it? Now, I put something else on the board: 3 × 7 = 7 × 3 You can guess that this is the “Commutative Property of Multiplication.” But I call it: not obvious. Let’s think about what that’s saying. 3×7 means you have three groups, each of which has seven items. Let’s count that out: 7, 14, 21. 7×3 means seven groups of three each. Let’s count that out: 3, 6, 9, 12, 15, 18…21. Hey, it came out the same. It worked! But is it just a coincidence? Will it also work for 6×9, and for 12×137, and for every other possible multiplication we could do? What I’m asking is, can you think of a reason that will make it obvious that these two things–three groups of seven, and seven groups of three–had to come out the same? This is not a rhetorical question. I give the class some time to think about it, and I get a variety of answers. You can look at this as three groups (rows) with seven squares each: 7+7+7. Or, you can look at it as seven groups (columns) with three squares each: 3+3+3+3+3+3+3. Either way, you get the same number of squares. That picture convinces me that it will also work for 12×137: I don’t have to count it out, and I don’t have to take anyone’s word for it either. It’s…well, obvious. So I’m not using the word “obvious” the way that most people do. When I say something is “obvious” I do not mean “anyone would figure it out quickly.” Sometimes I spend hours and hours trying to make something obvious to myself. But if I succeed, I eventually get to the point where I can “Well, of course it’s that way. It couldn’t possibly be any other way.” That, to me, is what math is all about. Math is often compared to a foreign language. I’ve even heard math teachers say “math is just another language.” I think this is a very misleading analogy. Math has a language. Certain words (“root”) are used very differently from how they are conventionally used, and other words (“polynomial”) are never used outside of math. But these words express the math: they are not the math itself. If we started saying “fizbot” instead of root, the math (“you can’t take the square fizbot of a negative number”) would be the same. What math has in common with a foreign language is that you have to memorize rules of how things fit together. “Je vais, tu vas, il va…” “Negative b plus or minus the square root of b-squared minus…” Apply these rules scrupulously or you will go in the wrong direction. But here’s the key difference. If you ask your French teacher “Why does it go Je vais, tu vas, il va?” the only answer you might expect is historical (“Here’s how it evolved from the Latin roots”). You’re not looking for “why it makes sense” because it isn’t really supposed to. It’s just how they talk in France. In math, on the other hand, there is a reason why you can’t take the square root of a negative number. There is a reason why 12÷4 is smaller than 12, but 12÷(1/4) is bigger than 12. And that reason has to go deeper than saying “When we divide by a fraction, we flip-and-multiply.” It has to go to the heart of what division means, until you say “Well of course 12÷(1/4) has to be 48, and nothing but 48 would make sense. It’s obvious!” (Hint: what does this problem mean in terms of pizza?) To put it another way: if all the English speakers in the world decided tomorrow that we would use the word “thurple” to mean “lunch,” they would all be correct–by definition they would be correct, simply because they agreed on it–and the world would go on just as before. But if all the mathematicians in the world decided that 37 was the same thing as 73, they would all, unanimously, be wrong. If they convinced the engineers, bridges would fall down. No one decided that there is no “Commutative Property of Exponents”: someone figured it out. You can figure it out too. So why don’t people get that? Why do my students look terrified and cry out “Just tell us the answer!” when I suggest they figure something out? For the most part, I blame their elementary school teachers. Very few people go into elementary school teaching because they love math. They go because they love children, they love playing games and telling stories, or (in some cases) it was the only professional job they could possibly get and keep. So the teacher learns the rules of math from the book, and teaches them from the book. And if a student asks “Why do we need a common denominator when we add fractions, but not when we multiply them,” the teacher has no clue. Rather than risk embarrassment in front of the class, she glowers and says “Because that’s how we do it.” It doesn’t take long for the students to get the idea: don’t try to think, just learn the rules. Or, to put it another way, all of math is glorified long division. We divide, multiply, subtract, bring down, divide, multiply, subtract, bring down, and some day when we’re in high school we’ll do pages and pages of dividing, multiplying, subtracting, bring downing. Is it any wonder that they come out thinking math is pointless and boring? In five years, my current crop of Common Core students will not remember the laws of logarithms, and my Calculus students will not remember the quotient rule. I don’t mind that a bit. If they ever need those things, they can look them up. What I do hope they remember–and many of them have told me that they do–is that math is not at all what they used to think it was. Math is not a foreign language, or a set of rules that you learn and apply, or glorified long division. Math is the most perfect elaboration of common sense. It has no rules except the rules that we were all born with, built into our brains. And any time anyone tells you “This is the way it is” in math, you have the right–even the obligation–to ask why, and to keep asking why. You’re not done when you can say “I can do it now.” You’re done when you say “Of course. Now that I see it this way…it’s obvious.” Want to read more? Here are a couple of links from excellent educators. At the level of details, their messages are quite different from mine (and from each other’s). But in terms of the big pictures, I think we all have the same core message: it’s more important to get students actively engaged in formulating the math, and applying their own thinking, than working through given processes to find a right answer. Kenny Felder http://www.felderbooks.com Tags: 20 replies 1. MidgetDwarf says: symbolipoint They (those teachers) are over-worked. Some kids take longer to reach the stage that some things in arithmetic or algebra become obvious. What I saw as a student, was that there were several SMART kids who had trouble with what is obvious. Being overworked should not be an excuse to not execute one's job at full capacity all the time.. Let's be honest here. Are the parents responsible? Yes. Should mom and dad be more involved? Yes. It comes down to economics and parents level of education. People in low- income areas: have less money for tutoring services, lack the education themselves, or may not see the value of education. A recent google search of tutoring jobs in my state of California revealed some amazing results. In the Malibu, Brentwood, Palos Verdes, or even middle class areas, such as, Culver City, people are paying at minimum 60,000 yearly salary for a personal teacher for their kids. I have seen postings for as much as 120,000. Won't the kids whose parents can afford to higher a private instructor have the greatest chance of success? Or even middle class households were at least one parent completed the educational system, will have a the know how to get their children to college. Another factor are the textbooks used in the American public school system. Many inner city kids do not have books. Most do not get math books until 6 or 7th grade. Therefore, a child has not practiced how to read mathematical books until the age of 12 or 13 in most cases. Kids are given atrocious work sheets by their teacher. Even if the kids received textbooks, they are of inferior quality. Richard Feynman mentions, "In Surely You're Joking," how the schools adopt the textbooks. Im sure everyone is familiar with this book on physics forum so I will cut that part short. Books are extremely dumbed down. Language barrier of immigrant groups also plays a factor in students success. In the USA hispanic are the 2nd largest population. Caucasians are first place and third are African-Americans. With the influx of asian and middle-eastern immigration groups, this is becoming a greater problem. A lot of immigrant parents, both the wife and father, work 2 jobs to support their family structure. Oftentimes, children rarely see their parents thoughtout the day. Kids are usually taken care of by an older brother or sister. The most important in my opinion are the teachers. Many teachers had dreams of grandeur while attending college. Once completing their education and getting a degree, most but not all, failed to live out their dreams. Instead, the only jobs they could get was teaching. I have a friend who recently got her masters in English about 2 yrs ago. She got a degree that she could not find employment with. Her highest math course taken was statistics. We took pre-calculus together. The reason she took pre-calculus was that she wanted to go into the medical field and she lacked the science/math for the MCAT etc. She could not hang dropped the course. Today she teaches algebra to students. She is not qualified to teach children, but the state of California says she is. Instead of prepping her lesson plan/ lectures she goes bar hopping…. I know many other stories like this. 2. symbolipoint says: MidgetDwarf, you give enough information that suggests both that teachers are overworked, and that some of them are less qualified than necessary to do some parts of their jobs. Maybe both things occur. Part of the problem is how the determination of who should teach what is made. Also, administrators have ways of ignoring rules (but they don't always want to). 3. snowman_ says: MidgetDwarf, even if you work full-time, if you care about your child's education then you'll take part in it…there are more than 40-50 hours in a week. You don't need a tutoring service to help your child with their math/reading homework. A lot of people just have poor time management skills and then blame it on teachers. You get out of education what you put into it. Teachers go above and beyond a 40 hour week with many working beyond the "clock" with no compensation. In districts with very little money the teachers are even expected to buy all the school supplies for the students…when you get paid so little it's rough buying 30 composition notebooks, folders, pens and pencils etc.. It's insulting to imply that teachers don't put all their effort into their job…I've met very few teachers who are that lazy, many are incredibly passionate and put in an extreme amount of effort. In my experience it's not that the parents are too busy, many simply don't get a crap…you get a lot of kids living with their grandparents (because parents ditched out) who most of the time participate very little in the child's education. Also if you get a degree in elementary education you actually do have to take a math class which explains all the "whys" behind things…it also covers math topics at a very intuitive level, definitely not memorization or "do this because that's the way we do it". Just because you know a handful of slackers doesn't mean that elementary education in general is like that. The teacher is the last thing in the chain of issues that is a child's early education…the administration policies (which includes stricter hiring policies), parent involvement, and district funding are much bigger issues. 4. MidgetDwarf says: snowman_ MidgetDwarf, even if you work full-time, if you care about your child's education then you'll take part in it…there are more than 40-50 hours in a week. You don't need a tutoring service to help your child with their math/reading homework. A lot of people just have poor time management skills and then blame it on teachers. You get out of education what you put into it. Teachers go above and beyond a 40 hour week with many working beyond the "clock" with no compensation. In districts with very little money the teachers are even expected to buy all the school supplies for the students…when you get paid so little it's rough buying 30 composition notebooks, folders, pens and pencils etc.. It's insulting to imply that teachers don't put all their effort into their job…I've met very few teachers who are that lazy, many are incredibly passionate and put in an extreme amount of effort. In my experience it's not that the parents are too busy, many simply don't get a crap…you get a lot of kids living with their grandparents (because parents ditched out) who most of the time participate very little in the child's education. Also if you get a degree in elementary education you actually do have to take a math class which explains all the "whys" behind things…it also covers math topics at a very intuitive level, definitely not memorization or "do this because that's the way we do it". Just because you know a handful of slackers doesn't mean that elementary education in general is like that. The teacher is the last thing in the chain of issues that is a child's early education…the administration policies (which includes stricter hiring policies), parent involvement, and district funding are much bigger issues. Believe it or not, there are people whose parents still can not read in this day and age. Stop looking at from a first world perspective. It is not just one factor but many, just how symbo pointed out. I had a friend growing up who was from the middle east. His parents brought him here to escape a tyrannical government. His parents could not read or write english. My friend had to repeat grades several times because of the language barier. Or even the immigration problem that is facing the United States. Like I said many parents have to work multiple jobs to provide for their families. You do understand that the 2nd generation American has a better chance at success than the first? Remember it is not the kids fault, rather the adults who placed them on this earth or are running things. What teachers have you seen going above and beyond 40 hour a week? Are you talking about teachers in middle class and afluent areas or poor areas? Like i said, the major problem is economic. People in better areas will have better resources, because they can afford to do so. For every 1 good teacher there are 10 that should not be teaching. I do not speak in absolutes. I was playing devil's advocate to start a discussion. You have to be aware that not everyone grew up in the same circumstances you did. You may or may not have had people who took an interest in your education. If you did, then had a blessed child hood. You cannot just isolate one factor or factors. Rather, you have to look at the picture as the whole. It is a chicken and egg argument. Improving textbooks and getting them into the hands of younger students is a must. Look at what the Soviet Union did. (lets not get into the whole civil rights violations argument). Truth be told. A lot of people do not care about others if it does not effect them directly. Not that they should. It is the sad reality in which we live. 5. snowman_ says: MidgetDwarf You cannot just isolate one factor or factors. Rather, you have to look at the picture as the whole. It is a chicken and egg argument. Improving textbooks and getting them into the hands of younger students is a must. Look at what the Soviet Union did. (lets not get into the whole civil rights violations argument). Truth be told. A lot of people do not care about others if it does not effect them directly. Not that they should. It is the sad reality in which we live. I unfortunately did not have my parents help whatsoever in my primary education. I was at risk of repeating every grade from 6th to 12th, early on it was because my parents were too busy doing things that they shouldn't and then later on it was due to poor habits I had developed. If anything I speak from experience seeing what happens with lack of parent involvement. Ok, so parents can't read…then why the hell are they having kids if they can't help facilitate an education? So maybe parents aren't to blame, but why the heck are we placing the blame with the teachers? The fault is definitely not with the kids…it's with the parents who brought them into this world when they are unable to fully provide for their children along with other factors. As for the teachers working over 40 hours, it's about every school in my district (not a well off district) that has plenty of teachers doing so. I bolded what I did because my point was specifically that there are multiple things at play and the teacher is the least of our concerns. The kids are only with the teacher a few hours each day. I don't care to derail this thread, but I just wanted to point out how stupid the comment on elementary teachers was in the article. If anything it distracts from real issues. btw I appreciate the polite reply. 6. MidgetDwarf says: snowman_ I unfortunately did not have my parents help whatsoever in my primary education. I was at risk of repeating every grade from 6th to 12th, early on it was because my parents were too busy doing things that they shouldn't and then later on it was due to poor habits I had developed. If anything I speak from experience seeing what happens with lack of parent involvement. Ok, so parents can't read…then why the hell are they having kids if they can't help facilitate an education? So maybe parents aren't to blame, but why the heck are we placing the blame with the teachers? The fault is definitely not with the kids…it's with the parents who brought them into this world when they are unable to fully provide for their children along with other factors. As for the teachers working over 40 hours, it's about every school in my district (not a well off district) that has plenty of teachers doing so. I bolded what I did because my point was specifically that there are multiple things at play and the teacher is the least of our concerns. The kids are only with the teacher a few hours each day. I don't care to derail this thread, but I just wanted to point out how stupid the comment on elementary teachers was in the article. If anything it distracts from real issues. btw I appreciate the polite reply. It has been proven that American education is a joke. So the schools are a major contributor to lack of student success, but not the sole factor. Are you aware that most schools do not even textbooks that children can take home and look at? Even if they did have textbooks, the books are of such low quality. No major mathematics is done at a young age compared to other nations. Even developing nations have a stronger educational system than the US. Are you an educator or have family members that are? You are also aware, that unlike other nations, such as: Japan, Sweeden, Denmark, China, Costa Rica etc. are homogenous populations. Race is also a contributing factor in America. Are you American? Not naturalized American but actually born and raised in the US. 7. snowman_ says: MidgetDwarf It has been proven that American education is a joke. So the schools are a major contributor to lack of student success, but not the sole factor. Are you aware that most schools do not even textbooks that children can take home and look at? Even if they did have textbooks, the books are of such low quality. No major mathematics is done at a young age compared to other nations. Even developing nations have a stronger educational system than the US. Are you an educator or have family members that are? You are also aware, that unlike other nations, such as: Japan, Sweeden, Denmark, China, Costa Rica etc. are homogenous populations. Race is also a contributing factor in America. Are you American? Not naturalized American but actually born and raised in the US. I agree with what you're saying. The article however specifically called out the teachers as a main issue, not the school in general. From the article "For the most part, I blame their elementary school teachers. " That's what I was commenting on, I do realize the school system is far less than ideal…I just think it's silly to partially single out teachers as the problem. I don't think I made myself clear, I apologize. The school system is a joke, but the teachers in my experience do their best to make do with what resources they have available. Blaming the teachers and not the system itself or the parents (who imo carry much more responsibility in a child's education) is just wrong imo. Yes I was born and raised in the US. Btw I was taught algebra in elementary school in the 3rd grade, non-gifted classes took it in 4th. Standard run of mill public school in an ok district. Then again, teachers don't set curriculums so if I didn't have "advanced math" it certainly wouldn't be the teachers fault. 8. Guapa says: I think every person has it very own way to understand the world that sourrounds him/her. I studied law some years ago, not my desire but by default, due to some circumstances. Along with highschool I took English, then allong with college I took French, German and I could not finish Iitalian due to family and personal circumstances. My native tongue is Spanish. A few years ago, having some spare time (or Sabbatical), I decided to learn Math as a language, I defined that the numbers is the substance and the other symbols as plus or minus signs are the procedure ordered or requested to be performed, I also defined that de the actual problem is telling me what it wants or what is need it to be done, by means of observation, analysis and practice I could understand how to solve from simple to complex equations. so after all this learning, I got my certificate on Calculus Derivatives and soon I hope given time I should be taking Integration. I love being able of seeing math simbols and understand them in the way I do, it is a great skill or knowledge. And above all it is fun because it is challenging. 9. 256bits says: Is it any wonder that they come out thinking math is pointless and boring?Well math is boring if one is not interested in math. I do wonder though if that is the teacher's fault for not making it exciting for all students, or if it is rather an innate feature of a student him/herself. Or does it become boring because the student is not excelling and receiving positive reinforcement by encouragement and marks. Large classrooms do not allow as much time spent per child as necessary for detection of problem areas that a student can have, and they fall behind and this becomes only noticeable at quiz time or examination, when it is surely sometimes too late. Every student does not learn at the same rate, nor ready to learn the same material at the same age. Yet, the standard is to group all students by age and give to them a curriculum to infuse into their brains. One can definitely see the deviation from the "norm" in physical appearance of young people, so why does it not seem also obvious that the mental maturity and intellect also has a variance amongst the young people. The simplest explanation I know is this picture:A part I especially like. Knowing that, a student can feel "one up" on the subject. Other tiny tidbits placed in strategic locations surely should, I think, feel that the student is being let in on little secrets ( and what student doesn't like secrets, which they will remember better than the main textual explanations ) of math.
You are on page 1of 8 # 25 7. HOMOGENEOUS EQUATIONS Homogeneous Functions A function , is said to be homogeneous of degree if , , The differential equation , , , is said to be homogeneous if , and , are homogeneous of the same degree. Alternatively a differential equation in standard form is homogeneous if , , for every real number . Homogeneous equation can be transformed into a separable equation by making the substitution ## along with its corresponding derivative The resulting equation in the variables and is solved as a separable differential equation; the required solution to original equation is obtained by back substitution. Note: The word Homogeneous used for a function and for a differential equation are completely different. Example: Solve Solution: Let us consider, , , ## Therefore, the given DE is a homogeneous one. Let us make the following transformation, and Saba Fatema 26 Example: Solve Solution: Let us consider, , , ## Therefore, the given DE is a homogeneous one. Let us make the following transformation, ## After the transformation given DE becomes, ln ln ln ln Example: Solve sin Solution: Let us consider, , sin and Saba Fatema 27 , , Therefore, the given DE is a homogeneous one. Let us make the following transformation, sin sin sin csc ln tan # ln ln 2 tan 2 2 tan \$ Problems: Solve ## % 2 cos , % 3 sec , % 4 tan , % 5 cot , % 6 csc , 2 3 4 5 % , % cos , % sec , % tan , % cot , 6 % csc Example: Solve , 3 , 2 0 Solution: Let us first write the differential equation into derivative form as follows, 3 , , 2 Now consider, 3 , , , 2 3 , , , , 3 , , , , 2 2 Therefore, the given DE is a homogeneous one. and Saba Fatema 28 ## After the transformation given DE becomes, 3 , , , 2 , 3 , 1 2 , 1 2 Now separating variables and integrating, we obtain, 2 , 1 ln , 1 ln Taking exponentials on both sides, we get, /01 2 \$3 /0 45 /01 2 \$3 5 /0 , 1 6 , , 1 6 Example: Solve , , 2 Solution: Let us consider, , , , 2 , , , , , , , , , 2 ,2 2 Therefore, the given DE is a homogeneous one. Let us make the following transformation, ## After the transformation given DE becomes, , , 2 1 , and Saba Fatema 29 1 , 1 , 2 2 2 1 , Integrating we obtain, 2 1 1 , ,\| ln\|1 ln\| \| ln ln\| \| ln\|1 ,\| ln ln \| \|\|1 ,\| ln , 8 91 , :8 , 91 ,: , 91 ,: , 1 , , ;1 < % =;1 < Example: Solve > 2 , , > Solution: Let us consider, > 2 , , > , ? > 2 , , > , ? , ## Therefore, the given DE is a homogeneous one. Let us make the following transformation, and Saba Fatema 30 > 2 , , > 1 2 , > 1 2 , > 1 , > ## Now separating variables and integrating both sides, we obtain, 1 , > ln 1 , , , @ , , @ 2 1 @ ln 1 @ @, 1 , @ ln 1 , 3 @ 2# 92: 2 2@ 1 tan \$ ; < ln 3 3 2@ 1 3 tan \$ ; < ln B 3 2 2@ 1 3 ; < tan 9 ln B: 3 2 3 2@ 1 3 tan 9 ln B: 2 3 2 , 1 3 tan 9 ln B: 2 , 3 2 1 3 tan 9 ln B: , 2 3 2 , , 3 , tan 9 ln B: 2 Example: Solve ? 3 , ? 3 , Solution: Let us consider, and Saba Fatema 31 ? 3 , , ? 3 , ? 3 , , ? 3 , , , ? 3 , ? 3 , ## Therefore, the given DE is a homogeneous one. Let us make the following transformation, ? ? 3 , ? 3 , , ? 3 1 3 , ? 3 1 3 , ? 3 3 ? 1 3 , 2 ? 1 3 , ## Now separating variables and integrating both sides, we get, 1 3 , 2 ? 1 3 , ? 2 ln 1 Now the integral on the left side may be evaluated in the following way 1 3 , 6 C 1 1 1 1 1 3 , 6 1 1 C 1 1 Now putting 0 we get, 6 1 Similarly, by putting 1, and 1, we get, C 2, 2 1 3 , 1 1 1 2 2 1 1 1 1 ln 2 ln 1 2 ln 1 ln ln 1 , ln 1 , ln 1 , 1 , and Saba Fatema 32 ln 2 ln 1 1 , 1 , 6 , 1 , 1 , , , 6 , 1 # 1 # , , 6 Problems 2 2 5 , , , , , , 2 2 2 5 > > 2 , , , , , E , ? , , , 0, , ? ?
Short Tricks to Solve Problems on Area FUNDAMENTAL CONCEPTS 1. Results on Triangles: 1. Sum of the angles of a triangle is 180°. 2. The sum of any two sides of a triangle is greater than the third side. 3. Pythagoras Theorem:In a right-angled triangle, (Hypotenuse)2 = (Base)2 + (Height)2. 4. The line joining the mid-point of a side of a triangle to the positive vertex is called the median. 5. The point where the three medians of a triangle meet, is called centroid. The centroid divided each of the medians in the ratio 2 : 1. 6. In an isosceles triangle, the altitude from the vertex bisects the base. 7. The median of a triangle divides it into two triangles of the same area. 8. The area of the triangle formed by joining the mid-points of the sides of a given triangle is one-fourth of the area of the given triangle. 1. The diagonals of a parallelogram bisect each other. 2. Each diagonal of a parallelogram divides it into triangles of the same area. 3. The diagonals of a rectangle are equal and bisect each other. 4. The diagonals of a square are equal and bisect each other at right angles. 5. The diagonals of a rhombus are unequal and bisect each other at right angles. 6. A parallelogram and a rectangle on the same base and between the same parallels are equal in area. 7. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area. IMPORTANT FORMULAE 1. 1. Area of a rectangle = (Length x Breadth). 2. Perimeter of a rectangle = 2(Length + Breadth). 2. Area of a square = (side)2 = (diagonal)2. 3. Area of 4 walls of a room = 2 (Length + Breadth) x Height. 4. 1. Area of a triangle = x Base x Height.2. Area of a triangle = s(sa)(sb)(sc) where abc are the sides of the triangle and s = (a + b + c). 3. Area of an equilateral triangle = 3 x (side)2. 4 4. Radius of incircle of an equilateral triangle of side a = a . 23 5. Radius of circumcircle of an equilateral triangle of side a = a . 3 6. Radius of incircle of a triangle of area and semi-perimeter r = . s 5. 1. Area of parallelogram = (Base x Height).2. Area of a rhombus = x (Product of diagonals).3. Area of a trapezium = x (sum of parallel sides) x distance between them. 6. 1. Area of a circle = R2, where R is the radius.2. Circumference of a circle = 2R. 3. Length of an arc = 2R , where is the central angle. 360 4. Area of a sector = 1 (arc x R) = R2 . 2 360 7. 1. Circumference of a semi-circle = R. 2. Area of semi-circle = R2 . 2 Area – General Questions 1. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is: A. 15360 B. 153600 C. 30720 D. 307200 Explanation: Perimeter = Distance covered in 8 min. = 12000 x 8 m = 1600 m. 60 Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320) m2 = 153600 m2. An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is: A. 2% B. 2.02% C. 4% D. 4.04% Explanation: 100 cm is read as 102 cm. A1 = (100 x 100) cm2 and A2 (102 x 102) cm2. (A2 – A1) = [(102)2 – (100)2] = (102 + 100) x (102 – 100) = 404 cm2. Percentage error = 404 x 100 % = 4.04% 100 x 100 The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? Correct! Wrong! The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is Correct! Wrong! A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road Correct! Wrong! A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is: Correct! Wrong! A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges? Correct! Wrong! QUIZ ON PROBLEMS OF AREA EXCELLENT
# Write an exponential function ## Presentation on theme: "Write an exponential function"— Presentation transcript: Write an exponential function EXAMPLE 1 Write an exponential function Write an exponential function y = ab whose graph passes through (1, 12) and (3, 108). x SOLUTION STEP 1 Substitute the coordinates of the two given points into y = ab . x 12 = ab 1 Substitute 12 for y and 1 for x. 108 = ab 3 Substitute 108 for y and 3 for x. STEP 2 Solve for a in the first equation to obtain a = , and substitute this expression for a in the second equation. 12 b Write an exponential function EXAMPLE 1 Write an exponential function 108 = b3 12 b Substitute for a in second equation. 12 b 108 = 12b 2 Simplify. 2 9 = b Divide each side by 12. 3 = b Take the positive square root because b > 0. Determine that a = 12 b = 3 = 4. so, y = x STEP 3 EXAMPLE 2 Find an exponential model A store sells motor scooters. The table shows the number y of scooters sold during the xth year that the store has been open. Scooters • Draw a scatter plot of the data pairs (x, ln y). Is an exponential model a good fit for the original data pairs (x, y)? • Find an exponential model for the original data. Find an exponential model EXAMPLE 2 Find an exponential model SOLUTION STEP 1 Use a calculator to create a table of data pairs (x, ln y). x 1 2 3 4 5 6 7 ln y 2.48 2.77 3.22 3.58 3.91 4.29 4.56 STEP 2 Plot the new points as shown. The points lie close to a line, so an exponential model should be a good fit for the original data. Find an exponential model EXAMPLE 2 Find an exponential model STEP 3 Find an exponential model y = ab by choosing two points on the line, such as (1, 2.48) and (7, 4.56). Use these points to write an equation of the line. Then solve for y. x ln y – 2.48 = 0.35(x – 1) Equation of line ln y = 0.35x Simplify. y = e 0.35x Exponentiate each side using base e. y = e (e ) 2.13 0.35 x Use properties of exponents. y = 8.41(1.42) x Exponential model Use exponential regression EXAMPLE 3 Use exponential regression Use a graphing calculator to find an exponential model for the data in Example 2. Predict the number of scooters sold in the eighth year. Scooters SOLUTION Enter the original data into a graphing calculator and perform an exponential regression. The model is y = 8.46(1.42) . x Substituting x = 8 (for year 8) into the model gives y = 8.46(1.42) scooters sold. 8 Substitute the coordinates of the two given points into y = ab . GUIDED PRACTICE for Examples 1, 2 and 3 Write an exponential function y = ab whose graph passes through the given points. x 1. (1, 6), (3, 24) SOLUTION STEP 1 Substitute the coordinates of the two given points into y = ab . x 6 = ab 1 Substitute 6 for y and 1 for x. 24 = ab 3 Substitute 24 for y and 3 for x. Solve for a in the first equation to obtain GUIDED PRACTICE for Examples 1, 2 and 3 STEP 2 Solve for a in the first equation to obtain a = , and substitute this expression for a in the second equation. 6 b 24 = b3 6 b Substitute for a in second equation. 6 b 24 = 6b 2 Simplify. 2 4 = b Divide each side by 6. 2 = b Take the positive square root because b > 0. Determine that a = 6 b = 2 = 3. so, y = x STEP 3 Substitute the coordinates of the two given points into y = ab . GUIDED PRACTICE for Examples 1, 2 and 3 Write an exponential function y = ab whose graph passes through the given points. x 2. (2, 8), (3, 32) SOLUTION STEP 1 Substitute the coordinates of the two given points into y = ab . x 8 = ab 1 Substitute 8 for y and 2 for x. 32 = ab 3 Substitute 32 for y and 3 for x. Solve for a in the first equation to obtain GUIDED PRACTICE for Examples 1, 2 and 3 STEP 2 Solve for a in the first equation to obtain a = , and substitute this expression for a in the second equation. 8 b 2 32 = b3 8 b 2 Substitute for a in second equation. 8 b 2 32 = 8b Divide each side by 4. 4 = b Take the positive square root because b > 0. Determine that a = = 8 b 2 4 16 1 . STEP 3 1 2 So, y = 4 x Substitute the coordinates of the two given points into y = ab . GUIDED PRACTICE for Examples 1, 2 and 3 Write an exponential function y = ab whose graph passes through the given points. x 3. (3, 8), (6, 64) SOLUTION STEP 1 Substitute the coordinates of the two given points into y = ab . x 8 = ab 3 Substitute 8 for y and 3 for x. 64 = ab 6 Substitute 64 for y and 6 for x. Solve for a in the first equation to obtain GUIDED PRACTICE for Examples 1, 2 and 3 STEP 2 Solve for a in the first equation to obtain a = , and substitute this expression for a in the second equation. 8 b 3 64 = b6 8 b 3 Substitute for a in second equation. 8 b 3 64= 8b 3 Divide each side by 8. b = = 8 3 64 8 Simplify. b = 2 Take the positive square root because b > 0. GUIDED PRACTICE for Examples 1, 2 and 3 8 b Determine that a = = 2 . 1 STEP 3 So, b = 2, a = 1, y = = 2 . x GUIDED PRACTICE for Examples 1, 2 and 3 4. WHAT IF? In Examples 2 and 3, how would the exponential models change if the scooter sales were as shown in the table below? SOLUTION The initial amount would change to and the growth rate to 1.45.
# Evaluate the integral? : int x^3e^(x^2) dx Aug 4, 2017 The answer is $= \frac{1}{2} {e}^{{x}^{2}} \left({x}^{2} - 1\right) + C$ #### Explanation: We need the integration by parts $\int p ' q \mathrm{dx} = p q - \int p q ' \mathrm{dx}$ We perform the substitution Let $u = {x}^{2}$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$ Therefore, $\int {x}^{3} {e}^{{x}^{2}} \mathrm{dx} = \frac{1}{2} \int u {e}^{u} \mathrm{du}$ We apply the integration by parts $p ' \left(u\right) = {e}^{u}$, $\implies$, $p \left(u\right) = {e}^{u}$ $q \left(u\right) = u$, $\implies$, $q ' \left(u\right) = 1$ Therefore, $\frac{1}{2} \int u {e}^{u} \mathrm{du} = \frac{1}{2} \left(u {e}^{u} - \int {e}^{u} \mathrm{du}\right) = \frac{1}{2} \left(u {e}^{u} - {e}^{u}\right)$ $= \frac{1}{2} {e}^{{x}^{2}} \left({x}^{2} - 1\right) + C$ Aug 4, 2017 We seek: $\int \setminus {x}^{3} {e}^{{x}^{2}} \setminus \mathrm{dx} = \frac{1}{2} \left({x}^{2} - 1\right) {e}^{{x}^{2}} + C$ #### Explanation: We seek: $I = \int \setminus {x}^{3} {e}^{{x}^{2}} \setminus \mathrm{dx}$ Note as a helper that: $\frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right) = 2 x {e}^{{x}^{2}} \iff \int \setminus 2 x {e}^{{x}^{2}} \setminus \mathrm{dx} = {e}^{{x}^{2}}$ So we can write the integral as: $2 I = \int \setminus \left({x}^{2}\right) \left(2 x {e}^{{x}^{2}}\right) \setminus \mathrm{dx}$ We can now use the formula for Integration By Parts (IBP): $\int \setminus u \frac{\mathrm{dv}}{\mathrm{dx}} \setminus \mathrm{dx} = u v - \int \setminus v \frac{\mathrm{du}}{\mathrm{dx}} \setminus \mathrm{dx}$, or less formally $\text{ } \int \setminus u \setminus \mathrm{dv} = u v - \int \setminus v \setminus \mathrm{du}$ I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation. Let $\left\{\begin{matrix}u & = {x}^{2} & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = 2 x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = 2 x {e}^{{x}^{2}} & \implies v & = {e}^{{x}^{2}}\end{matrix}\right.$ Then plugging into the IBP formula: $\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$ gives us $\int \setminus \left({x}^{2}\right) \left(2 x {e}^{{x}^{2}}\right) \setminus \mathrm{dx} = \left({x}^{2}\right) \left({e}^{{x}^{2}}\right) - \int \setminus \left({e}^{{x}^{2}}\right) \left(2 x\right) \setminus \mathrm{dx}$ $\therefore 2 I = {x}^{2} {e}^{{x}^{2}} - \int \setminus 2 x {e}^{{x}^{2}} \setminus \mathrm{dx} + A$ $\text{ } = {x}^{2} {e}^{{x}^{2}} - {e}^{{x}^{2}} + A$ $\text{ } = \left({x}^{2} - 1\right) {e}^{{x}^{2}} + A$ Hence $I = \frac{1}{2} \left({x}^{2} - 1\right) {e}^{{x}^{2}} + C$

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