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# Clenshaw递推公式 ## 切比雪夫多项式 N次切比雪夫多项式，是下面形式的多项式p(x) ${\displaystyle p(x)=\sum _{n=0}^{N}a_{n}T_{n}(x)}$ ## Clenshaw递推公式 Clenshaw递推公式可以用来计算切比雪夫多项式的值。给定 ${\displaystyle p(x)=\sum _{n=0}^{N}a_{n}T_{n}(x)}$ ${\displaystyle b_{N}\,\!}$ ${\displaystyle :=a_{N}\,}$ ${\displaystyle b_{N-1}\,\!}$ ${\displaystyle :=2xb_{N}+a_{N-1}\,}$ ${\displaystyle b_{N-n}\,\!}$ ${\displaystyle :=2xb_{N-n+1}+a_{N-n}-b_{N-n+2}\,,\;n=2,\ldots ,N-1\,}$ ${\displaystyle b_{0}\,\!}$ ${\displaystyle :=xb_{1}+a_{0}-b_{2}\,}$ ${\displaystyle p(x)=\sum _{n=0}^{N}a_{n}T_{n}(x)=b_{0}.}$ (注)上面的公式在 ${\displaystyle N=0,1}$的情况下无意义。此时我们可以用下面的公式： ${\displaystyle b_{N+2}:=0\,}$ ${\displaystyle b_{N+1}:=0\,}$ ${\displaystyle b_{j}:=2xb_{j+1}-b_{j+2}+a_{j}\,,\;j=N,\ldots ,1}$ (downward, omit if N=0) ${\displaystyle p(x):=xb_{1}-b_{2}+a_{0}\,}$ ${\displaystyle q(x):=2xb_{1}-b_{2}+a_{0}\,}$ ${\displaystyle p(x)=\sum _{n=0}^{N}a_{n}T_{n}(x)}$ ${\displaystyle q(x)=\sum _{n=0}^{N}a_{n}U_{n}(x)}$
# IDENTIFY PARTS OF AN EXPRESSION ## About "Identify parts of an expression" Identify parts of an expression : Here we are going to see how to identify the parts of an algebraic expression. In an algebraic expression, we may find the following three parts. (i) Terms (ii) Factors (iii) Coefficients What is term ? A single variable or a constant or a combination of these as a product or quotient forms a term. Examples of terms : 5, -a, 3ab, 21/7, ........... etc Terms can be added or subtracted to form an expression. What is factor ? Consider the expression 3ab – 5a. It has two terms 3ab and -5a. The term 3ab is a product of factors 3, a and b. The term -5a is a product of -5 and a. The coefficient of a variable is a factor or factors. Example : In the term 3ab; (i) the coefficient of ab is 3 (ii) the coefficient of a is 3b (iii) the coefficient of b is 3a. In the term –5a the coefficient of a is –5 What is constant ? A number which is not having any variable with it is known as constant. ## Identifying parts of an expression using flow chart Example 1 : Identify the parts of the following expression 2x + 3 Solution : In the expression 2x + 3 the term 2x is made of 2 factors and 2 and x while 3 is a single factor. Example 2 : Identify the parts of the following expression 3ab - 5a Solution : Example 3 : Identify the number of terms and coefficient of each term in the expression. x2 y2 - 5 x2 y + (3/5) xy2 - 11 Solution : In the given expression, we have four terms. Term 1 ==> x2 y2 Term 2 ==> - 5 x2 y Term 3 ==> (3/5) xy2 Term 4 ==> -11 Coefficient of 1st term = 1 Coefficient of 2nd term = -5 Coefficient of 3rd term = 3/5 Since the last term is not having any variable, it is a constant term. Example 4 : Identify the number of terms, coefficient and factors of each term in the expression. 3abc - 5ca Solution : The given expression contains two terms. Term 1 ==> 3abc Term 2 ==> -5ca Terms Coefficients Factors 1) 3abc 3 a, b and c 2) -5ca -5 c and a Example 5 : Identify the number of terms, coefficient and factors of each term in the expression. 1 + x + y2 Solution : The given expression contains three terms. Term 1 ==> 1 Term 2 ==> x Term 2 ==> y2 Terms Coefficients Factors 1) 1 - - 2) x 1 x 3) y2 1 y and y Example 6 : Identify the number of terms, coefficient and factors of each term in the expression. 3x2 y2 - 3xyz + z Solution : The given expression contains three terms. Term 1 ==> 3x2 y2 Term 2 ==> - 3xyz Term 2 ==> z Terms Coefficients Factors 1) 3x2 y2 3 x2 and y2 2) - 3xyz -3 x, y and z 3) z3 1 z, z and z Example 7 : The coefficient of x4 in -5x7 + (3/7) x4 - 3x3 + 7x2 - 1 Solution : The coefficient of x4 is 3/7. 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# Differentiate each of the following functions from the first principal : Question: Differentiate each of the following functions from the first principal : $\log \cos x$ Solution: We have to find the derivative of $\log \cos x$ with the first principle method, so, $f(x)=\log \cos x$ by using the first principle formula, we get, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \cos (x+h)-\log \cos x}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\cos (x+h)}{\cos x}\right)}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\cos (x+h)}{\cos x}-1\right)}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\cos (x+h)-\cos x}{\cos x}\right)}{h}$ [Rationalising] $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\cos (x+h)-\cos x}{\cos x}\right)}{h} \times \frac{\frac{\cos (x+h)-\cos x}{\cos x}}{\frac{\frac{\cos (x+h)-\cos x}{\cos x}}{h}}$ [By using $\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$ ] $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{\cos (x+h)-\cos x}{\cos x}}{h}$ $\left[\cos C-\cos D=-2 \sin \frac{C-D}{2} \sin \frac{C+D}{2}\right]$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{-2 \sin \frac{2 x+h}{2} \sin \frac{h}{2}}{\cos x}}{\frac{2 h}{2}}$ [By using $\left.\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{2 x+h}{2}}{2 \cos x}$ $f^{\prime}(x)=-\tan x$
Question 1. # According To The Rational Root Theorem, The Following Are Potential Roots Of F(X) = 2X2 + 2X – 24 The rational root theorem is a mathematical principle that tells us the potential roots of an equation. This theorem helps us to understand and identify the rational zeros of polynomials. Recently, I came across a polynomial equation – F(X) = 2X2 + 2X – 24 – which I wanted to use the rational root theorem to find out the potential roots for. After some research, I discovered the potential roots for this particular polynomial equation and today, in this article, I’m here to share my findings with you all! So let’s get started… ## What is the Rational Root Theorem? The Rational Root Theorem is a theorem that states that if a polynomial equation has integer coefficients, then any rational roots of the equation must be expressible as a fraction whose numerator is a factor of the constant term and whose denominator is a factor of the leading coefficient. In other words, if F(x) = x + x – is a polynomial with integer coefficients, then any rational roots of the equation must be of the form m/n where m is a factor of -1 and n is a factor of 1. ## What are the potential roots of F(X)? There are a few potential roots of F(X), according to the Rational Root Theorem. These roots could be X+X-, X-X+, or -X+X. However, these are not the only possible roots; any rational number that satisfies the equation F(X)=0 is a potential root of F(X). So, if we plug in different rational numbers for X, we might be able to find more potential roots. ## How to find potential roots of F(X)? To find potential roots of F(X), we can use the Rational Root Theorem. This theorem states that if F(X) is a polynomial with integer coefficients, and if P/Q is any rational number where P is a factor of the leading coefficient and Q is a factor of the constant term, then P/Q is a potential root of F(X). In our example, F(X) = X + X –, so the leading coefficient is 1 and the constant term is –1. Therefore, any rational number P/Q where P is a factor of 1 and Q is a factor of –1 is a potential root of F(X). Some possible values for P and Q are: P = ±1, Q = ±1 P = ±2, Q = ±2 P = ±3, Q = ±3 This means that the following are potential roots of F(X): 1/1, –1/1, 2/2, –2/2, 3/3, –3/3. ## Conclusion The Rational Root Theorem can be a great tool for finding potential roots of a polynomial. By applying the theorem, we have found that the potential roots of F(X) = 2X2 + 2X – 24 are ±1, ±2, ±3, and ±4. It is important to remember that these numbers are only possible solutions; they may not in fact be actual roots. To determine whether or not these values are indeed valid solutions for this equation requires further investigation into F(x). In any case, understanding and applying the rational root theorem has been helpful in qualifying our search further down to just four possible roots! 2. The Rational Root Theorem is an important part of algebra that can help students determine the potential roots of a polynomial equation. This theorem states that if a polynomial has integer coefficients, then any rational number can be expressed as a ratio of two integers, and those two integers must be factors of the constant term in the original equation. In other words, when looking to solve for all possible roots of an equation like 2×2 + 2x – 24 = 0, the Rational Root Theorem can be used to identify which numbers may provide solutions. When applying this theorem to equations like f(x) = 2×2 + 2x – 24, it’s important to note that the constant term (in this case 24) must be factored first. 3. Have you ever heard of the Rational Root Theorem? It’s a powerful tool used to find the potential roots of a polynomial equation. ‍♀️ In this article, we’re going to dive a bit deeper into the Rational Root Theorem and explore an example of how it can be applied. To start, let’s take a look at the equation F(x) = 2×2 + 2x – 24. By applying the Rational Root Theorem, we can determine the potential roots of this equation. Essentially, the Rational Root Theorem states that all potential roots of a polynomial equation can be expressed as a fraction whose numerator is a factor of the constant term (in this case, the number 24) and whose denominator is a factor of the leading coefficient (in this case, the number 2). Using this information, we can determine that the potential roots of F(x) = 2×2 + 2x – 24 are -4, -2, 1, 3, 6, and 12. And there you have it – that’s the Rational Root Theorem in a nutshell! Of course, this isn’t the only way to find potential roots of a polynomial equation. ‍♀️ But the Rational Root Theorem is a helpful tool that can make the task a bit easier. So if you ever find yourself in need of some help finding the potential roots of a polynomial equation, don’t forget about the Rational Root Theorem! 4. Wondering what the Rational Root Theorem is all about? The Rational Root Theorem states that if a polynomial equation has integer coefficients, all its rational roots can be expressed as a fraction where the numerator is a factor of the constant and the denominator is a factor of the leading coefficient. So, in the case of the equation F(x) = 2×2 + 2x – 24, the potential roots of the equation according to the Rational Root Theorem are as follows: 1. ±1/2 2. ±2/1 3. ±3/2 4. ±4/1 5. ±6/1 It’s important to note that this doesn’t necessarily mean that all of these are actually the roots of the equation; they are simply the potential roots that can be expressed as fractions. To determine the actual roots of the equation, you’ll need to use other methods such as factoring or the quadratic formula. But, understanding the Rational Root Theorem and being able to identify potential roots is a great starting point for solving polynomial equations.
# Lesson Video: Counting Forward to 10 Mathematics • Kindergarten In this video, we will learn how to count from 0 to 10 and find the missing numbers in a count sequence. 09:59 ### Video Transcript Counting Forward to 10 In this video, we’re going to be learning how to count in ones from zero to 10. And we’re also going to use this skill to help us find missing numbers when we’re counting in a sequence. This cube represents the number zero. Can’t you see a cube? That’s because there isn’t one. Zero, of course, means nothing. Now, we have one cube. And if we make our line of cubes one more each time, we can count forwards in ones. Let’s start with the two numbers we’ve said already. Zero, one, two, three, four, five, six, seven, eight, nine, 10. We’ve counted forwards in ones all the way from zero to 10. We used the cubes to help us here. We know we can count forwards to 10 without having a set of objects to count. This is a number track. It shows the numbers from zero to 10 in order. And it helps us remember where each number belongs. And because we know where each number belongs, we can use what we know to solve missing number problems. Which number is missing? We can see that it’s the number that comes after five. Zero, one, two, three, four, five — and what number comes next? — six. Let’s practice using our knowledge of the counting sequence from zero to 10 and finding out some missing numbers. Elizabeth counts the steps as she climbs. Which number will she finish on? Which number has she missed out? In the picture, we can see that Elizabeth is climbing a set of steps. And we’re told that she counts the steps as she climbs them. Perhaps this is the sort of thing you’ve done at home to see how many there are. The first part of this question asks us, which number will she finish on? In other words, which number will she say when she reaches the top step? Every step she takes, she says one more number. Let’s imagine that we’re climbing these steps. And we’ll count forwards one for each step that we stand on. One, two, three, four, five, six, seven, eight. There are eight steps. This means that the number that Elizabeth will finish counting on is the number eight. In the picture underneath the second part of the question, we can see that Elizabeth has missed a number out as she climbs the steps. Here’s the missing number. What is it? We can use our knowledge of the counting sequence to count forwards and to work out what the missing number might be. One, two, three, four, five. We know the number after five is six. And then comes seven and eight. The missing number in the sequence is the number six. Our two answers are eight and six. What number is missing? Four, what, six, seven. We’re given a picture that shows a sequence of numbers. We can see these numbers shown as digits, four, six, and then seven. And underneath, we can see the same sequence shown in blocks, four blocks, six blocks, and seven blocks. But our sequence isn’t all it seems. There’s a missing number. If we count through the sequence, we can see four, then a big gap where our missing number belongs, and then six and seven. And the question asks us, what number is missing? In other words, what comes after four in our sequence? We’re given three possible answers to choose from. And we can see those over here. And if we look at these answers quickly, we can see which number is missing. But let’s work it out for ourselves. The only two numbers that are next door to each other that we know already are six and seven. And we know that if we count forwards in ones, we count from six to seven. Seven is the number that comes after six. And so, we know that our missing number must be the number that comes after four when we’re counting in ones. One, two, three, four. What comes next? Five, our missing number must be the number five. But if we look at our three possible answers, we can see that each one of them shows the digit five. How can we decide which one is correct? Well, we need to count the blocks underneath. We need to show five blocks to match our digit five. Let’s count each group of blocks. In the first group, we have one, two blocks. This doesn’t show the number five at all. Let’s try our second possible answer. In this one, we have one, two, three, only four blocks. Again, this is too small. It looks like our last answer must be the correct one. Let’s just count to make sure that there are five blocks. One, two, three, four, five blocks. Now, we think we’ve found our missing number. Let’s see whether our sequence makes sense. Four, five, six, seven. We’ve counted forwards in ones and we’ve found that our missing number was the number five. The correct answer is the one that shows the digit five and also five blocks underneath. Complete the following pattern: what, what, what, three, four, five, six, seven, eight, nine, 10. When we listened to that pattern, it sounded a little bit like we were counting to 10. But there were a few whats at the start, weren’t there? Some missing numbers. Let’s read our pattern again. What, what, what, three, four, five, six, seven, eight, nine, 10. Well, we’ve definitely heard the last part of our pattern before. These are definitely the numbers that we say when we’re counting forwards in ones. So our missing numbers must be the start of this pattern. What can we use to help us count forwards to 10? We could use a number track like this one. It shows all the numbers in order from zero all the way up to 10. Now, where’s our number pattern on this number track? Well, here is the number three. And after it comes the number four and so on. So we can use the number track to work out our missing numbers. If we’re counting forwards in ones, the number that we say before we say three is the number two. Two, three, four, five, and so on. And the number that we say before we say two is, of course, the number one. One, two, three, four, and so on. This means that our first missing member must be zero. Let’s check whether our completed pattern makes sense. And as we say each number, we’ll move a counter along our number track. Zero, one, two, three. Our first three digits fit perfectly with our pattern. Let’s finish it off, four, five, six, seven, eight, nine, and 10. We’ve used our knowledge of counting forwards to 10 to help us to complete the pattern. Our three missing numbers, in order, are zero, one, and two. So what have we learned in this video? Well, firstly, we’ve learned how to count in ones all the way from zero to 10. We’ve also learned that each number has a position where it belongs. And we’ve used things like number tracks to help us to learn these positions. Finally, we’ve used what we know about counting forwards to 10 to help find missing numbers.
## NCERT SOLUTIONS FOR CLASS 6 MATHS EXERCISE 12.1 Find here step by step NCERT Solutions for Class 6 Maths Exercise 12.1 Chapter 12 Ratio & Proportion. NCERT SOLUTIONS FOR CLASS 6 MATHS EXERCISE 12.1 deals with the Idea of Comparison by Division commonly known as Ratio. Prerequisite / Revise this ## Comparison By Division : Ratio Comparison by Difference – For comparing quantities of the same type, we commonly use the method of taking difference between the quantities. Comparison by Division – In many situations, a more meaningful comparison between quantities is made by using division, i.e. by seeing how many times one quantity is to the other quantity. This method is known as comparison by ratio. For comparison by ratio, the two quantities must be in the same unit. If they are not, they should be expressed in the same unit before the ratio is taken. Order of a Ratio – the order in which quantities are taken to express their ratio is important. The ratio 5 : 2 is different from 2 : 5. Ratio as a Fraction – A ratio may be treated as a fraction, thus the ratio 5 : 2 may be written as 5/2 . Equivalent Ratios – Two ratios are equivalent, if the fractions corresponding to them are equivalent. Thus, 5 : 2 is equivalent to 10 : 4 or 20 : 8. Simplest Form of a Ratio – A ratio can be expressed in its lowest form. For example, ratio 15 : 9 is treated as 15/9 ; in its lowest form 15/9 = 5/3 . Hence, the lowest form of the ratio 15 : 9 is 5 : 3. ## Exercise 12.1Finding Ratio QUESTION 1 There are 20 girls and 15 boys in a class. (a) What is the ratio of number of girls to the number of boys? (b) What is the ratio of number of girls to the total number of students in the class? (a) Number of girls = 20, Numbers of boys = 15, So, ratio of number of girls to the number of boys (b) Total number of students = 20 + 15 = 35, So So, ratio of number of girls to the total number of students QUESTION 2 Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of (a) Number of students liking football to number of students liking tennis. (b) Number of students liking cricket to total number of students. Total number of students in the class = 30, Number of students who like football = 6, Number of students who like cricket = 12, (a) Number of students who like tennis = 30 – (6 + 12) = 30 – 18 = 12, So, ratio of number of students liking football to number of students liking tennis (b) Ratio of number of students liking cricket to total number of students QUESTION 3 See the figure and find the ratio of (a) Number of triangles to the number of circles inside the rectangle. (b) Number of squares to all the figures inside the rectangle. (c) Number of circles to all the figures inside the rectangle. Inside the rectangle, Number of circles = 2, Number of squares = 2, Number of triangles = 3, Toatal number of figures = 7, (a) Ratio of number of triangles to the number of circles inside the rectangle (b) Ratio of number of squares to all the figures inside the rectangle (c) Ratio of number of circles to all the figures inside the rectangle QUESTION 4 Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar. Speed of Hamid = 9 km/hour Speed of Akhtar = 12 km/hour So, Ratio of speed of Hamid to to the speed of Akhtar QUESTION 5 Fill in the following blanks: [Are these equivalent ratios?] Yes, these are equivalent ratios. QUESTION 6 Find the ratio of the following : (a) 81 to 108 (b) 98 to 63 (c) 33 km to 121 km (d) 30 minutes to 45 minutes (a) 81 to 108 = (b) 98 to 63 = (c) 33 km to 121 km = (d) 30 minutes to 45 minutes = QUESTION 7 Find the ratio of the following: (a) 30 minutes to 1.5 hours (b) 40 cm to 1.5 m (c) 55 paise to ₹ 1 (d) 500 ml to 2 litres (a) 1.5 hours = 1.5 × 60 minutes = 90 minutes So, ratio of 30 minutes to 1.5 hours = (b) 1.5 m = 1.5 × 100 cm = 150 cm So, ratio of 40 cm to 1.5 m = (c) ₹ 1 = 100 paise So, ratio of 55 paise to ₹ 1 = (d) 2 litres = 2 × 1000 ml = 2000 ml So, ratio of 500 ml to 2 litres = QUESTION 8 In a year, Seema earns ₹ 1,50,000 and saves ₹ 50,000. Find the ratio of (a) Money that Seema earns to the money she saves. (b) Money that she saves to the money she spends. In a year, Seema earns = ₹ 1,50,000, Seema saves = ₹ 50,000 So, money she spends = ₹ (1,50,000 – 50,000) = ₹ 1,00,000 Then, (a) Ratio of money that Seema earns to the money she saves (b) Ratio of money that she saves to the money she spends QUESTION 9 There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students. Number of teachers = 102, Number of Students = 3300, So, the ratio of the number of teachers to the number of students QUESTION 10 In a college, out of 4320 students, 2300 are girls. Find the ratio of (a) Number of girls to the total number of students. (b) Number of boys to the number of girls. (c) Number of boys to the total number of students. Total number of students in a college = 4320, Number of girls = 2300, So, number of boys = 4320 – 2300 = 2020 Then, (a) Ratio of number of girls to the total number of students (b) Ratio of number of boys to the number of girls (c) Ratio of number of boys to the total number of students QUESTION 11 Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of (a) Number of students who opted basketball to the number of students who opted table tennis. (b) Number of students who opted cricket to the number of students opting basketball. (c) Number of students who opted basketball to the total number of students. Total number of students in a school = 1800, Number of students who opted basketball = 750, Number of students who opted cricket = 800, So, number of students who opted tennis = 1800 – (750 + 800) = 250 Then, (a) Ratio of number of students who opted basketball to the number of students who opted table tennis (b) Ratio of number of students who opted cricket to the number of students who opted basketball (c) Ratio of number of students who opted basketball to the total number of students QUESTION 12 Cost of a dozen pens is ₹ 180 and cost of 8 ball pens is ₹ 56. Find the ratio of the cost of a pen to the cost of a ball pen. Cost of a dozen pens = ₹ 180, So, cost of one pen = ₹ = ₹ 15 Cost of 8 ball pens = ₹ 56, So, cost of one ball pen = ₹ = ₹ 7 Therefore, the ratio of the cost of a pen to the cost of a ball pen = 15 : 7 QUESTION 13 Consider the statement: Ratio of breadth and length of a hall is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall. QUESTION 14 Divide 20 pens between Sheela and Sangeeta in the ratio of 3 : 2. Sum of parts = 3 + 2 = 5, So, Sheela will get pens and, Sangeeta will get pens. QUESTION 15 Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get. Age of Shreya = 15 years, Age of Bhoomika = 12 years, So, ratio of the age of Shreya to the age of Bhoomika = 15 : 12 Sum of the parts of division = 15 + 12 = 27, Shreya will get = ₹ = ₹ 5 × 4= ₹ 20, and Bhoomika will get = ₹ = ₹ 4 × 4 = ₹ 16. QUESTION 16 Present age of father is 42 years and that of his son is 14 years. Find the ratio of (a) Present age of father to the present age of son. (b) Age of the father to the age of son, when son was 12 years old. (c) Age of father after 10 years to the age of son after 10 years. (d) Age of father to the age of son when father was 30 years old. (a) Present age of father = 42 years, Present age of son = 14 years, Ratio of present age of father to the present age of son = = = 3 : 1 (b) 14 – 12 = 2, So, 2 years ago son was 12 year old, Then, Age of the father 2 years ago = 42 – 2 = 40 years, Ratio of age of father to the age of son, when son was 12 years old = = (c) After 10 years, Age of father = 42 + 10 = 52 years, Age of son = 14 + 10 = 24 years, So, Ratio of age of father after 10 years to the age of son after 10 years = = 13 : 6 (d) 42 – 30 = 12, So, 12 years ago father was 30 years old, Then, age of son 12 years ago = 14 – 12 = 2 years, Ratio of age of father to the age of son when father was 30 years old = ## NCERT SOLUTIONS FOR CLASS 6 MATHS CHAPTER 12 Check NCERT Solutions of other exercises from Class 6 Maths Chapter 12 Ratio & Proportion by clicking the links given below. We hope NCERT SOLUTIONS FOR CLASS 6 MATHS EXERCISE 12.1 has helped you understand how to find Ratios. Write in the comment section for any error or any solution related queries from the exercise. [ NCERT SOLUTIONS FOR CLASS 6 MATHS EXERCISE 12.1 ]
# Spotting the right strategy in keno February 03, 2015 3:08 AM by Any time you have a game based on chance all kinds of math are involved. It can be as simple as basic arithmetic or as complex as binomial theorem. Keno has it all. First, let’s go to the math of marking tickets and figuring ways. You write a ticket grouped as 2-2-2-1-1, giving a total of 8 spots with 3 deuces (groups of 2) and 2 kings (groups of one). To find the total number of spots, add up the total number of spots in each group. Finding the number of 7-spots is easy, too. Since you need seven numbers for a 7-spot and you have two kings, all numbers except one king is a 7-spot, and all numbers with the other king is the other, for a total of two 7-spots. A simple axiom in Keno Kalkulations: If the ticket has 8 spots and you select all the 5-spots (without changing the group configurations), then the number of 3-spots has to be the same as the number of 5-spots. How? If you take 5 from 8, you have 3 remaining, and for each 5-spot possible, there is one corresponding 3-spot. Let’s now go to the 6-spots from our ticket. Using the above we know the number of 6-spots must be equal to the number of 2-spots. Thus you have four groups of 2, the three deuces plus the two kings taken together. Since you have four groups of 2 you must have four groups of 6. It gets more involved seeing how many 5-spots you have, but it’s not that difficult. Three is an odd number, so you either must use an odd number by itself or an odd number plus an even number to have an odd number. You also must examine all the combining possibilities. You have three groups of 2 and three groups of 1 in this ticket. For each 3 you have a group of 2 and a group of 1. Since 3x2 = 6 there are six ways to have a 3-spot here. By the first axiom there are also six ways to have a 5-spot. We have the 4-spot left to figure. Four is an even number and since the largest individual group is a group of 2, it will take two groups of 2, or a deuce and the two kings to make a 4-spot. You have two ways of doing the calculation – from left to right or right to left. You have groups of 2-2-2-1-1. Go 2-2-0-1-1, then 2-0-2-0-0, then 2-0-0-1-1, then 0-2-2-0-0, then 0-2-0-1-1, and finally, 0-0-2-1-1 – six ways to make a 4-spot from this ticket. However, if you have more groups and/or ways it gets very complicated. So what are we looking to do? We know on this particular ticket with only deuces and kings you either have to have two deuces and a deuce and two kings to make a 4-spot. To make things easier think of the two kings as a group of two as well. In essence you have four groups of 2 from which you need to choose two groups to make a 4-spot. In math this is called combinations and written with subscripts such as 4C2. What this means is you are choosing two out of a group of 4. The mathematical formula is written 4!/(2!2!). The “!” is a factorial: 4! = 4x3x2x1 and 2! = 2x1, giving you (4x3x2x1)/(2x1x2x1) = 24/4 = 6. It confirms six ways to choose 3 groups out of 4. But, there is a shortcut. It can also be written: (4x3)x(2x1))/((2x1)x(2x1)). Since you have a factor of (2x1) in both the numerator and denominator, you can “cancel” them out, leaving you with (4x3)/(2x1). This has sometimes been called the “Keno Formula.” Basically it means you can figure out keno combinations in this easy way. For example, you want to go after the \$2 progressive at the Orleans or Gold Coast. You write a ticket with five groups of 4, playing the 8s at \$2 each. How many groups do you have? We know two groups of 4 would make the necessary 8-spots, so we need to choose two groups out of 5. Thus we have 5C2, or 5!/(2!3!). It simplifies to (5x4x3!)/(2!3!). We “cancel” out the 3! in both the numerator and denominator, giving us (5x4)/(2x1) = 10. Thus, if you have five groups of 4 and choose two of them to make your 8-spot you have ten 8-spots to play at \$2 each, a \$20 ticket. It does not matter how many in each group for this kind of calculation. As long as the groups are the same number of spots, choosing 2 of 5 or 4 of 7 or 16 of 20, the result will be the same number of tickets required, whether they are 2-spots or 5-spots. Getting back to our original 2-2-2-1-1 ticket, we have one 8, two 7s, four 6s, six 5s, six 4s, six 3s, four 2s, and two 1s for a total of 31 possible ways. You may also know the number of ways are 2 to the nth power minus 1 (n being number of groups), or 25 (2x2x2x2x2) = 32-1 =31, showing we have the correct totals for our groups. Is learning all the above useful? Of course it is as it gives you increased flexibility to figure out good way tickets, especially for tournaments. Pesach Kremen is a former UNLV Masters Gaming student, has won and placed in multiple local keno tournaments, and has written several academic papers on Keno. You can reach him at PesachKremen@GamingToday.com.
# Question: What Is The Prime Factorization Of 2592? ## What is the formula for prime factorization? Lesson Summary The prime factors of a number are all the prime numbers that, when multiplied together, equal the original number. You can find the prime factorization of a number by using a factor tree and dividing the number into smaller parts.. ## What is the sum of all positive prime factors of 2020? Answer Expert Verified The sum of its prime factors is 110. ## What is a factor of 108? These factors are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108. ## What is the sum of all divisors of 24? Why is 60 the sum of all factors of 24, including 24? The factors of the number 24 are: 1, 2, 3, 4, 6, 8, 12, 24. If you add them up, you will get the total of 60. ## What is the prime factorization of 13? Since 13 can only be divided evenly by 1 and 13 (itself), it is called a prime number. A prime number can only be divided evenly by one and itself. The other numbers such as 6, 12, and 21 are called composite numbers since they can have many factors (more than two). ## What are the factors for 16? Table of Factors and MultiplesFactorsMultiples1, 2, 4, 8, 16161121, 17171191, 2, 3, 6, 9, 18181261, 191913341 more rows ## What is the prime factorization of 11? The prime factorization of 11 is 1 × 11. The number 11 is a prime number, because its only factors are 1 and itself. ## What are the factors of 12? So 1, 2, 3, 4, 6 and 12 are factors of 12. ## What is the sum of factors? In general, if you have the prime factorization of the number n, then to calculate the sum of its divisors, you take each different prime factor and add together all its powers up to the one that appears in the prime factorization, and then multiply all these sums together! Example: Determine S(1800). ## What is the prime factorization of 2020? The prime factorization of 2,020 is 22 × 5 × 101. Since it has a total of 4 prime factors, 2,020 is a composite number. ## What are the factors of 2592? Factors of 2592 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 72, 81, 96, 108, 144, 162, 216, 288, 324, 432, 648, 864, 1296. There are 29 integers that are factors of 2592. The biggest factor of 2592 is 1296. ## What is the prime factorization of 16? Factors of 16: 1, 2, 4, 8, 16. Prime factorization: 16 = 2 x 2 x 2 x 2, which can also be written 16 = 2⁴. Since √16 = 4, a whole number, 16 is a perfect square. ## What is the factors of 56? 56 = 1 x 56, 2 x 28, 4 x 14, or 7 x 8. Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56. Prime factorization: 56 = 2 x 2 x 2 x 7, which can also be written 56 = 2³ x 7. ## Is 13 a prime or composite number? When a number has more than two factors it is called a composite number. Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc. ## What is the sum of all the single digit factors of 60? Why is 168 the sum of all factors of 60, including 60? The factors of the number 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. If you add them up, you will get the total of 168. ## What are the factors of 2020? Factors of 2020 are 1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010. ## What are the prime factors of 2310? Thus, the Prime Factors of 2310 are: 2, 3, 5, 7, 11. ## Is 2592 a perfect square? A: No, the number 2,592 is not a perfect square. ## What is the prime factorization of 18? 18 is a composite number. 18 = 1 x 18, 2 x 9, or 3 x 6. Factors of 18: 1, 2, 3, 6, 9, 18. Prime factorization: 18 = 2 x 3 x 3, which can also be written 18 = 2 x 3². ## What is the prime factorization of 33? The prime factorization of 33 is 3 × 11. The number 33 is a special type, because it is the product of two prime numbers.
Courses Courses for Kids Free study material Offline Centres More Store # How do you use the binomial series to expand ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$? Last updated date: 20th Jun 2024 Total views: 373.2k Views today: 8.73k Verified 373.2k+ views Hint: The terms of the form ${{(a+b)}^{n}}$ are called binomial terms. To simplify these terms, we should know the binomial expansion. For the binomial terms of the form ${{\left( 1+x \right)}^{n}}$, where n is not a positive integer. These terms are expanded as, $1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)\left( n-2 \right)}{3!}{{x}^{3}}+\dfrac{n(n-1)\left( n-2 \right)\left( n-3 \right)}{4!}{{x}^{4}}+......$. We will use this expansion formula to expand the given binomial term. Complete step-by-step solution: We are asked to expand the binomial term ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$. As the exponent is not an integer, this term is of the form ${{\left( 1+x \right)}^{n}}$, here we have $-x$ at the place of x and $n=\dfrac{1}{3}$. We know that the expansion of the binomial term ${{\left( 1+x \right)}^{n}}$ is $1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)\left( n-2 \right)}{3!}{{x}^{3}}+\dfrac{n(n-1)\left( n-2 \right)\left( n-3 \right)}{4!}{{x}^{4}}+......$ We can find the expansion of ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$ by replacing x by $-x$, and substituting $n=\dfrac{1}{3}$ in the above expansion formula, by doing this we get $\Rightarrow 1+\dfrac{1}{3}\left( -x \right)+\dfrac{\dfrac{1}{3}\left( \dfrac{1}{3}-1 \right)}{2!}{{\left( -x \right)}^{2}}+\dfrac{\dfrac{1}{3}\left( \dfrac{1}{3}-1 \right)\left( \dfrac{1}{3}-2 \right)}{3!}{{\left( -x \right)}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{1}{3}-1 \right)\left( \dfrac{1}{3}-2 \right)\left( \dfrac{1}{3}-3 \right)}{4!}{{\left( -x \right)}^{4}}+......$ Simplifying the numerators of the above expansion, we get $\Rightarrow 1+\dfrac{1}{3}\left( -x \right)+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)}{2!}{{\left( -x \right)}^{2}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)}{3!}{{\left( -x \right)}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)\left( \dfrac{-8}{3} \right)}{4!}{{\left( -x \right)}^{4}}+......$ We know that the values of $1!,2!,3!,4!$ are 1, 2, 6, and 24 respectively. Substituting these values in the denominators of the above expression, we get $\Rightarrow 1+\dfrac{1}{3}\left( -x \right)+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)}{2}{{\left( -x \right)}^{2}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)}{6}{{\left( -x \right)}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)\left( \dfrac{-8}{3} \right)}{24}{{\left( -x \right)}^{4}}+......$ Simplifying the exponents, we get $\Rightarrow 1-\dfrac{1}{3}x+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)}{2}{{x}^{2}}-\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)}{6}{{x}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)\left( \dfrac{-8}{3} \right)}{24}{{x}^{4}}+......$ Finally, simplifying both numerators, and denominators of both of the above expression, we get $\Rightarrow 1-\dfrac{1}{3}x-\dfrac{1}{9}{{x}^{2}}-\dfrac{5}{81}{{x}^{3}}-\dfrac{10}{243}{{x}^{4}}+......$ Thus, the binomial expansion of ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$ is $1-\dfrac{1}{3}x-\dfrac{1}{9}{{x}^{2}}-\dfrac{5}{81}{{x}^{3}}-\dfrac{10}{243}{{x}^{4}}+......$. Note: To solve the questions of binomial expansions, we should know the binomial expansions of different expressions. For a general binomial term of the form ${{(a+b)}^{n}}$, here n is a positive integer. The expansion formula is $\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}{{a}^{n-r}}{{b}^{r}}$. Here, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We can find the expansion of a binomial term with standard form using the summation form given above.
Non -Uniformly Accelerated Motion TWO DIMENSIONAL MOTION Previous Next Here, acceleration is not constant i.e., acceleration is non-uniform as well as motion is also non-uniform. To solve a question, we have to use the basic equation If acceleration is given as a function of time, then we use If acceleration is given as a function of velocity, then we use Remember, that equation of motions are not applicable when acceleration is varying. Freely Falling Bodies Consider a particle is projected from earth with a velocity u in vertically upward direction, then due to acceleration due to gravity its speed decreases as it rises up, becomes zero at the highest point and then falls back to the ground. For this case, we can apply equation of motions as acceleration is constant, equal to g in vertically downward direction. The most common example of motion with (nearly) constant acceleration is that of a body falling towards the earth. If we neglect air resistance and variation in g (acceleration due to gravity) with height from the surface of the earth, then all the objects fall towards the earth with a constant acceleration equal to acceleration due to gravity at earth’s surface g (9.81 m/s2 ). This idealized motion is spoken of as “free fall” although the term includes both rising as well as falling. For this situation, equations of motion would be v 2 = u 2  2gy, remember the unknown quantities comes out with the sign while solving the question, don’t bother about their direction. Here, at maximum height, y = H, v = 0 so time of ascent, y = 0, gives t = 0 and corresponds to launching instant while corresponds to the landing instant. So, time of ight is, So, time of descent, If a body is dropped from height h, then its equation of motion would be: It is very important to keep in mind that equation of motions are vector equations and we use them as scalars by assigning one direction as ve and other as – ve. Other details like the position of the object at time t, the distance travelled in the nth second could be easily obtained by using the equation of motion. TWO DIMENSIONAL MOTION
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of "2020 AMC 10B Problems/Problem 12" ## Problem The decimal representation of$$\dfrac{1}{20^{20}}$$consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point? $\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$ ## Solution 1 $$\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}$$ Now we do some estimation. Notice that $2^{20} = 1024^2$, which means that $2^{20}$ is a little more than $1000^2=1,000,000$. Multiplying it with $10^{20}$, we get that the denominator is about $1\underbrace{00\dots0}_{26 \text{ zeros}}$. Notice that when we divide $1$ by an $n$ digit number, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\underbrace{00\dots0}_{26 \text{ zeros}}$, there are $\boxed{\textbf{(D) } \text{26}}$ zeros in the initial string after the decimal point. -PCChess ## Solution 2 First rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$. $\log{5^{20}} = 20\log{5}$ and memming $\log{5}\approx0.69$ (alternatively use the fact that $\log{5} = 1 - \log{2}$), $\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14$ digits. Our answer is $\boxed{\textbf{(D) } \text{26}}$. ## Solution 3 (Brute Force) Just as in Solution $2,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the decimal point is before the $9.$ Dividing the number again by $10$ twenty-six more times allows a string of$\boxed{\textbf{(D) } \text{26}}$ zeroes to be formed. -OreoChocolate ## Solution 4 (Smarter Brute Force) Just as in Solutions $2$ and $3,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We can then look at the number of digits in powers of $5$. $5^1=5$, $5^2=25$, $5^3=125$, $5^4=625$, $5^5=3125$, $5^6=15625$, $5^7=78125$ and so on. We notice after a few iterations that every power of five with an exponent of $1 \mod 3$, the number of digits doesn't increase. This means $5^{20}$ should have $20 - 6$ digits since there are $6$ numbers which are $1 \mod 3$ from $0$ to $20$, or $14$ digits total. This means our expression can be written as $\dfrac{k\cdot10^{14}}{10^{40}}$, where $k$ is in the range $[1,10)$. Canceling gives $\dfrac{k}{10^{26}}$, or $26$ zeroes before the $k$ since the number $k$ should start on where the one would be in $10^{26}$. ~aop2014 ~IceMatrix
Percentage - Atreya Roy • Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal. By the term percent we mean value for every 100. If it is said, A gets a profit of 20% of every sale in the company, we can say that for every Rs 100 profit obtained from sale, A will get Rs 20. So in the entire chapter we will consider all calculations based on 100. X percent means = X/100 Dependency : K is X percent of Y. This means, K = (X/100)Y = XY/100 We might keep in mind some fractions and their corresponding Percentages for better use 1 = 100% 1/2 = 50% 1/3 = 33.33% 1/4 = 25% 1/5 = 20% 1/6 = 16.67% 1/7 = 14.28% 1/8 = 12.5% 1/9 = 11.11% 1/10 = 10% 1/11= 9.09% 1/12= 8.33% 1/13 = 7.69% 1/14 = 7.14% 1/15 = 6.66% 1/16 = 6.25% 1/17 = 5.88% 1/18 = 5.55% 1/19 = 5.29% 1/20 = 5.00% 1/25 = 4.00% 1/50 = 2.00% 1/100= 1.00 % If A is X % more than B, we will write : A= ((100+X )/100)B If A is X % less than B, we will write : A = ((100-X )/100)B So, Percent Increase, increases the value and Percent Decrease, decreases the value of the item which is being considered. Example : The Salary of A is 20% more than B. By what percent, the salary of B is less than B ? Solution: Let us take salary of B = 100 Then Salary of A = 100 + (20/100)100 = 120 Difference = 120-100= 20 B is 20/120 = 1/6 = 16.67% less than A Successive Change in Percentage: When there are more than one percent change on any item/quantity, consecutively, then it is called Successive percentage change Suppose we come in a situation when the price of the tickets for a film are increased by 10% once, and again by 10% . What will be the final change in amount ? Let the Initial cost be= 100 After 10% Increase , Cost = (100+10) /100 * 100 = 110 After again an increase of 10%, Cost = 110/100 * 110 = 121 Hence total increase = 121-100 =21 Percent wise increase = 21/100 = 21% We will discuss some formulas which will easy our calculations and also the time. If, “a” and “b” are the two successive changes(increase), Then total change = a+b+(ab/100) If there is x% increase and then x% decrease, then the resultant change = -x^2/100 % If there is a x% increase and then y% decrease, then net change = x-y –xy/100 % Example : After two successive increase of 10% and 20% a number becomes 66. What was the number ? Solution : Net increase = 10+20 + 1020/100 = 30+2 = 32% Let the number be X, so, X (100+32)/100 = 66 132x/100 = 66 2x= 100 Or, x=50 Practice Questions in Percentage : A company purchases a property for Rs 2.5lakhs and invested Rs 75k on repairs and modifications. After 2 years the company sells out the same property for Rs 2.86lakhs. 15 % is to be deducted on account of depreciation, find its approximate percentage of profit or loss. 1. 2.76 % 2. 3.25 % 3. 3.35 % 4. 3.53 % Solution:- Amount of property = 2.5+.75 = 3.25lakhs Depreciation cut = 15 % of 3.25 lakhs = 0.4875lakhs Present value = (100-15) % of 3.25lakhs = 2.7625lakhs Profit = (2.86-2.7625) / 2.7625 lakhs = (9750 / 276250) * 100 % = 3.53 % A man can buy 3 1/3 kgs more rice for Rs 130 on a reduction of 15 % in the price. Find the reduced price of each kg of rice? 1. 6.89 2. 5.86 3. 5.68 4. 6.98 Solution:- Man can buy / kg more for Rs 130 Let the price of rice be x Amount of brought in Rs 130 = 130/x New price = 85x/100 = 17x/20 Thus, (130/17x/20) – 10/3 =130/x = > x = 6.89 Reduced price = 85x/100 = 5.86 In an interview, 70% of male and 30 % of female are selected. The number of females selected is 180, which is 3/5th of number of males not selected. What is the no of candidates appeared for the interview? Every women who appeared was selected. 1. 300 2. 600 3. 900 4. 1000 Solution :- Selected percentage of females = 30 Selected percentage of males = 70 Females selected = 180 Males selected = x x* 3/5 = 180 x = 60 180 = 30% 100% = 600 Hence, 600 got seleced Males selected = 600 – 180 Total candidate appeared = 420+180+300 = 900 If 1/3 of a group of employers resigns from the company, by what percentage should the remaining men increases the working hours per day in order to compensate the loss of employees? 1. 33.33 % 2. 67.67 % 3. 50 % 4. None of these Solution:- Let, x men works at y hours per day Number of employee after resignation = 2x/3 If, these employers’ works for z hours per day Then, xy = (2x/3) * z = > z = 3y/2 So, 1/2 hours per day increased So, in percentage it is increased by 100/2 = 50% A is 30% more efficient than B in doing a job. If the difference of their earning upon completing the work together is Rs. 30. How much did they receive in all? A. 160 B. 165 C. 170 D. 175 SOLUTION : Let B do the job in 10 days; Hence, A does it in 7 days. Total part of A = 10x/17 Total part of B = 7x/17 Difference = 3x/17= 30 = > x = 170 A got 10170/17 = 100 ; B got 7170/17 = 70 Total Income = Rs. 170 Rohan takes a loan of 40,000 from XYZ bank. The bank offered the loan at an interest rate of 10%. Rohan started repaying the amounts from the 4th year as 3 annual instalments for the next 2 years. If the installment he pays is in arithmetic progression with a common difference of 5,000; find the amount left for repayment after the requisite time if the first installment was worth 10,000? A. 12320 B. 14320 C. 14820 D. 13320 Solution- Loan amount= 40,000; Rate of interest=10% After 3 years, interest incurred= 40,000310/100=12,000 Total amount=40,000+12,000=52,000 Amount of first installment= 10,000 Loan amount left= Total loan- loan repaid = 52,000-10,000=42,000 Interest incurring that year= 42,000110/100=4200 New loan amount= 46200. Loan repaid in the next year= 15,000. Loan amount still to be paid=46,200-15,000=31,200 Interest incurred=31200110/100=3,120. Net loan amount= 31,200+3,120=34,320 Loan paid in the third instalment = 20,000. Thus loan amount still to be paid= 34,320-20,000 = 14,320 Petrol Prices rose up by 10% as a result of an extra import duty. Present petrol price is Rs 44. Atreya travels 300kms daily with his bike. His bike gives a fair mileage of 40km /lt. What is the rise in his expenditure if he gets only half of the extra money as incentives from his office ? 1. 10 2. 15 3. 20 4. 25 Solution: Present value of petrol= 44. So previous value = 44100/110 = Rs 40. Mileage of bike= 40km/lt. Expenditure of Atreya per km= 40/40 = Re 1 /km Total expenditure=300km1 = Rs 300. New expenditure after price rise = (44/40)300 =330. Extra expenditure=30. Amount paid by office= ½ 30 = 15. Extra money Atreya has to pay =Rs 15. After 3 successive rise in percentage, the total rise was measured to be 44.29 %. What could have been the value of each such rise ? A. 12% B. 14% C. 12.5% D. 13% Solution- Solve through options Let us assume a value of 100. 13% rise in it makes it = 100+ 13% of 100 = 100+13 = 113 Again increase in 13 % of this value will result in an increase of 13/100 * 113 = 14.69 Again the third increase will result in an increase of 13/100 * 127.69 = 16.6 Thus the final value will be 127.69 +16.6 = 144.29 A’s salary is increased by 20% first, then decreased by 20% and again increased by 20% . While, B’s salary is increased by 25% first, then decreased by 25% and again increased by 25% 1. 115:117 2. 115:116 3. 115:118 4. 115:119 Solution: Let, A’s salary be 100 rs. After the first increase, the salary of A is Rs 120. Then 20% of this is decreased, so our new salary becomes 12080 / 100 = 96 Then again, 20% of this is increased, so A’s new salary becomes 96120/100 =115.20 Let, B’s salary be 100 rs. After the first increase, the salary of B is Rs 125. Then 25% of this is decreased, so B’s new salary becomes 12575 / 100 = 93.75 Then again, 25% of this is increased, so B’s new salary becomes 93.75125/100 =117.20 So the ratio is 115:117 The ratio of the incomes of A,B,C are 5:6:8 . The ratio of their expenditure is 1:2:3. In this scenario, who has the maximum savings? 1. A 2. B 3. C 4. Cannot Be Determined Solution: Let the incomes of A,B,C be Rs 50,60 and 80 respectively. And their expenditure is 25,50 and 75 respectively. So the savings are For A , savings = 50-25=25 For B, savings = 60-50 = 10 For C, savings= 80-75 = 5. So in this case, A has the highest Savings. But consider another scenario, where Incomes of A,B and C are 5,6 and 8. Expenditure are 2,4 and 6. Now the savings of A,B and C are respectively 3,2,2 and we can see that Savings of B and C are equal. So it cannot be determined. The manager of a company increases the wages/ hour of his workers by 10% and decreases the working hours by 10%. Being a concerned brother of the manager, you will choose the state of work , i.e will this theory will come into work or not according to the way which will yield a maximum profit to the organization. Select the better case 1. Revise the condition of work 2. Not revise the condition 3. Both will yield same 4. Cannot Be Determined Solution: Let the wage of a worker be 100. Number of working hours = 100 Total cost = 100100 = 10,000 After increase, new wage = 100110 / 100 = 110 New working hours = 10090/100 = 90 Total Cost = 90110 = 9,900 So, the condition must be revised. Appreciation / Depreciation in Population We will face questions that are based on population increase or decrease that might be triggered due to various factors like migration / diseases etc. So the simple approach to tackle these sums are successive percentage change. If there is successive increase of a% and b%, then the net change will be = a + b + (ab/100) If there is successive decrease of a% and b%, then the total discount will be = a + b – (ab/100) If there is a% increase and then b% decrease, then the net change = a – b – (ab/100) If there is a% increase and then a% decrease, then the net change = -(aa/100) All are the increase or decrease in terms of percentage. Example 1 : The population of a village is 2000. In 2011, the population increased by 4 % and in 2012 it decreased by 4%. What is the population at the end of these 2 years. Solution : Net change = - (44/100)% = -.16% = 2000.16/100 =3.2 2000-3.2 = 1997 people ( rounded to nearest value) Example 2: A city has 10,000 residents. Its population grows at the rate of 10% per annum, what’ll be its total population after 5 years? Solution: Rate of increase = 10% Population = 10000 This can be viewed as a problem of compound interest. So population after 5 years = 10000* (1+.1)^5 = 10000 * 1.61051 = 16105 Example 3 : A city has 10,000 residents. Its population declines at the rate of 10% per annum, what’ll be its total population after 5 years? Solution: Decrease = 10% Proceeding like the last sum, 10000* (1-10/100)^5 = 10000* (0.9)^5 = 10000* 0.59049 = 5904.9 ~ 5905 Looks like your connection to MBAtious was lost, please wait while we try to reconnect.
# Difference between revisions of "Roots of unity" The Roots of unity are a topic closely related to trigonometry. Roots of unity come up when we examine the complex roots of the polynomial $x^n=1$. ## Solving the equation First, we note that since we have an nth degree polynomial, there will be n complex roots. Now, we can convert everything to polar by letting $x = re^{i\theta}$, and noting that $1 = e^{2\pi ik}$ for $k\in \mathbb{Z}$, to get $r^ne^{ni\theta} = e^{2\pi ik}$. The magnitude of the RHS is 1, making $r^n=1\Rightarrow r=1$ (magnitude is always expressed as a positive real number). This leaves us with $e^{ni\theta} = e^{2\pi ik}$. Taking the natural logarithm of both sides gives us $ni\theta = 2\pi ik$. Solving this gives $\theta=\frac{2\pi k}n$. Additionally, we note that for each of $k=0,1,2,\ldots,n-1$ we get a distinct value for $\theta$, but once we get to $k > n-1$, we start getting coterminal angles. Thus, the solutions to $x^n=1$ are given by $x = e^{2\pi k i/n}$ for $k=0,1,2,\ldots,n-1$. We could also express this in trig form as $\displaystyle x=\cos\left(\frac{2\pi k}n\right) + i\sin\left(\frac{2\pi k}n\right) = \mathrm{cis }\left(\frac{2\pi k}n\right).$ ## Geometry of the roots of unity All of the roots of unity lie on the unit circle in the complex plane. This can be seen by considering the magnitudes of both sides of the equation $x^n = 1$. If we let $x = re^{i\theta}$, we see that $r^n = 1$, since the magnitude of the RHS of $x^n=1$ is 1, and for two complex numbers to be equal, both their magnitudes and arguments must be equivalent. Additionally, we can see that when the nth roots of unity are connected in order (more technically, we would call this their convex hull), they form a regular n-sided polygon. This becomes even more evident when we look at the arguments of the roots of unity. ## Properties of roots of unity Listed below is a quick summary of important properties of roots of unity. • They occupy the vertices of a regular n-gon. • For $n>1$, the sum of the nth roots of unity is 0. More generally, if $\zeta$ is a primitive nth root of unity (i.e. $\zeta^m\neq 1$ for $1\le m\le n-1$), then $\sum_{k=0}^{n-1} \zeta^{km}=\begin{cases} n & \mathrm{n\mid m}, \\ 0 & \mathrm{otherwise.}\end{cases}$ • If $\zeta$ is one of the roots of unity (but not 1), then the roots of unity can be expressed as $1, \zeta, \zeta^2,\ldots,\zeta^{n-1}$. • Also, don't overlook the most obvious property of all! For each $n$th root of unity, $\zeta$, we have that $\zeta^n=1$ ## Uses of roots of unity Roots of unity show up in many suprising places. Here, we list a few: Invalid username Login to AoPS
The mathematics of trigonometry is complex and, at times, confusing. To make this topic understandable to you, you will get a detailed explanation through this article of inverse sine, trigonometric functions, and other relationships. Accordingly, we present you with our arcsin calculator and the principle of operation, which will greatly help you in the calculation. ## What is arcsine? The arcsine is a trigonometric function in math that produces a final result in the range of -1 to 1. The inverse of the sine function calculates the arcsine. The function is typically used to convert the results of the arccosine function to degrees. It is commonly used in engineering, especially in signal processing. It is sometimes called the “inverse sine” or “arcsine law.” Each trigonometric function has its inverse function, which works the other way around. How can we conclude that when we see “arcsin A,” we understand it as “the angle whose sin is A”. ## How to calculate arcsin? As a math function, it is needed in some applications, such as calculating the angle of refracted air. You can use the following formula to calculate arcsin: A = arcsin (sin (x)) Arcsin can also be calculated using the following table, for example, if sine 45° is 0.7071068: sin (45°) = 0.7071068 arcsin (0.7071068) = sin-1 (0.7071068) = 45° and then the arcsin of 0.7071068 is 45° The arcsin x is the graphical representation of the inverse trigonometric function. The graph is a continuous curve with a steep slope, going from 0 to 1 as x increases from -π/2 radians to π/2 radians. However, the Arcus sine is an inverse function of this restriction, which is valid. arcsin \equiv sin^{-1} : \left [ -1,1 \right ] \rightarrow \left [ -\frac{\pi }{2}, \frac{\pi }{2} \right ] Based on this formula, we can conclude that the function is strictly increasing, continuous, odd, and has no symptoms. ## Inverse sine, trigonometric functions, and other relationships You can apply trigonometric functions to different angles, including negative and large angles. If we talk about the inverse function, we know that there are an infinite number of angles whose sine is the same, and that’s where the problem comes from. For example, 45 degrees and 360 + 45 degrees will have the same sine. To solve this problem, the set of these inverse trigonometric functions is limited so that the inverse functions are one-on-one, which tells us that there is one result. The relationship between arcsine and trigonometric functions can bring you closer to this topic to make it more transparent. The arcsine is a trigonometric function that converts an angle from the side of a right triangle to an angle on the unit circle. Also can be used to find the value of the sine of an angle not in the first quadrant. This trigonometric function is commonly used in geometry and trigonometry. What is important to emphasize when it comes to the inverse sine function is that it is a finite sine function, which means it is not the whole sine function. Relationships among the inverse trigonometric functions: arccos(x) = \frac{\pi }{2} - arcsin(x) arcsin(-x) = - arcsin(x) ## Example of using the arcsin calculator? Arcsin calculator provides an easy way to calculate arcsin without any hassle. These types of calculators can help you find the inverse of arcsin’s trigonometric function. With the help of this trigonometric function, we can also find the angles of a right triangle. for α: sin(\alpha) = \frac{a}{c} \Rightarrow \alpha = arcsin (\frac{a}{c}) for ß: sin(\beta) = \frac{b}{c} \Rightarrow \beta = arcsin (\frac{b}{c}) If we have two values given in a right triangle such as a = 4 and c = 8, and we want to find the value of the angle a: When we want to find arcsin, we enter a value. In our case, it is 4/8 and what we must not omit is that the value will be between -1 and 1, and in a few steps, we find arcsin, which in this example was 4/8 = 0,5, which means the angle of 30 degrees. ## How to find arcsin without a calculator? With the use of a calculator, we can quickly come up with the calculation of arcsin. However, it would look like this when we work without a calculator, as shown in the following example. or in example number 2:
# Find the equation for a line which goes thorugh origo #### Anna55 ##### New member Decide the equation for the line which goes through origo and are perpendicular against the line x-y+1=0 #### Denis ##### Senior Member Anna55 said: ...perpendicular against the line x-y+1=0 Can you calculate the slope of that line? If not, you need classroom help. #### wjm11 ##### Senior Member Decide the equation for the line which goes through origo and are perpendicular against the line x-y+1=0 First, rearrange the equation into y = mx + b (slope-intercept) form. This is to clearly identify "m", the slope. Once the slope is found, it can be used to find the slope of the perpendicular line; the perpendicular line slope will be the "negative reciprocal" of "m", i.e., -m^(-1). If the original slope is (a/b), the perpendicular to that is -(b/a). #### mmm4444bot ##### Super Moderator Staff member Anna55 said: Please explain how to solve this. Here's a step-by-step example, using a different given line. Find an equation for the line passing through the Origin which is perpendicular to the line 2x + y - 4 = 0. STEP 1: Determine the slope of the given line, by solving for y y = -2x + 4 We see by inspection that the slope of this line is -2 STEP 2: Determine the slope of the perpendicular line The perpendicular slope is the negative reciprocal of -2, which is 1/2. STEP 3: Use the Point-Slope Formula, to write the equation y - 0 = 1/2(x - 0) y = 1/2 x #### Anna55 ##### New member Thank you for the help! x-y+1=0 y=x+1 The slope is 1. 1(x)= -1 x= -1 y= -x Is my working out correct? #### mmm4444bot ##### Super Moderator Staff member Anna55 said: x-y+1=0 y=x+1 The slope is 1. Correct 1(x)= -1 You should use symbol m for representing the slope here (symbol x is already used as a variable, in this exercise) x= -1 This is the correct slope for the perpendicular line y = -x This final answer is correct Is my working out correct? In my example, I suggested that you use the Point-Slope formula. With experience, we come to realize that the form of all non-vertical lines passing through the Origin is simply y = mx. If you used y = mx at the end of your exercise, instead of using the Point-Slope form, you are savvy.
# Lesson 19 Expanding and Factoring Let's use the distributive property to write expressions in different ways. ### 19.1: Number Talk: Parentheses Find the value of each expression mentally. $$2 + 3 \boldcdot 4$$ $$(2+3)(4)$$ $$2 - 3 \boldcdot 4$$ $$2 - (3 + 4)$$ ### 19.2: Factoring and Expanding with Negative Numbers In each row, write the equivalent expression. If you get stuck, use a diagram to organize your work. The first row is provided as an example. Diagrams are provided for the first three rows. factored expanded $$\text- 3(5 - 2y)$$ $$\text- 15 + 6y$$ $$5(a-6)$$ $$6a-2b$$ $$\text- 4(2w-5z)$$ $$\text- (2x-3y)$$ $$20x-10y+15z$$ $$k(4-17)$$ $$10a-13a$$ $$\text- 2x(3y-z)$$ $$ab-bc-3bd$$ $$\text- x(3y-z+4w)$$ Expand to create an equivalent expression that uses the fewest number of terms: $$\left(\left(\left(\left(x\strut+1\right)\frac12\right)+1\right)\frac12\right)+1$$. If we wrote a new expression following the same pattern so that there were 20 sets of parentheses, how could it be expanded into an equivalent expression that uses the fewest number of terms? ### Summary We can use properties of operations in different ways to rewrite expressions and create equivalent expressions. We have already seen that we can use the distributive property to expand an expression, for example $$3(x+5) = 3x+15$$. We can also use the distributive property in the other direction and factor an expression, for example $$8x+12 = 4(2x+3)$$. We can organize the work of using distributive property to rewrite the expression $$12x-8$$. In this case we know the product and need to find the factors. The terms of the product go inside: We look at the expressions and think about a factor they have in common. $$12x$$ and $$\text- 8$$ each have a factor of 4. We place the common factor on one side of the large rectangle: Now we think: "4 times what is 12$$x$$?" "4 times what is -8?" and write the other factors on the other side of the rectangle: So, $$12x-8$$ is equivalent to $$4(3x-2)$$. ### Glossary Entries • expand To expand an expression, we use the distributive property to rewrite a product as a sum. The new expression is equivalent to the original expression. For example, we can expand the expression $$5(4x+7)$$ to get the equivalent expression $$20x + 35$$. • factor (an expression) To factor an expression, we use the distributive property to rewrite a sum as a product. The new expression is equivalent to the original expression. For example, we can factor the expression $$20x + 35$$ to get the equivalent expression $$5(4x+7)$$. • term A term is a part of an expression. It can be a single number, a variable, or a number and a variable that are multiplied together. For example, the expression $$5x + 18$$ has two terms. The first term is $$5x$$ and the second term is 18.
# Divisibility Rules Home > By Subject > Division > Divisibility Rules Divisibility rules help us work out whether a number is exactly divisible by other numbers (i.e. there is no remainder). The rules are shortcuts for finding out whether numbers are exactly divisible without doing division calculations. Some of these rules along with examples are illustrated below: ## Divisible by 2? Rule: If it ends with a 0, 2, 4, 6, or 8 Number Divisible? Why? 456 Yes The last digit is 6 68 Yes The last digit is 8 25 No The last digit is 5 (not a 2,4,6,or 8) 207 No The last digit is 7 (not a 2,4,6,or 8) ## Divisible by 3? Rule: If the sum of the digits is a multiple of 3 Number Divisible? Why? 405 Yes 4 + 0 + 5 = 9 (9 is a multiple of 3) 381 Yes 3 + 8 + 1 = 12 (12 is a multiple of 3) 928 No 9 + 2 + 8 = 19 (19 is not a multiple of 3) 4,616 No 4 + 6 + 1 + 6 = 17 (17 is not a multiple of 3) Helper: The multiples of 3 include... 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 ## Divisible by 4? Rule: If the last two digits are a multiple of 4 (or if the last two digits are 00) Number Divisible? Why? 348 Yes 48 is a multiple of 4 27,616 Yes 16 is a multiple of 4 8,514 No 14 is not a multiple of 4 722 No 22 is not a multiple of 4 1,200 Yes The last two digits are 00200 is a multiple of 4 Helper: The multiples of 4 include... 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 ## Divisible by 5? Rule: If it ends with a 5 or a 0 Number Divisible? Why? 3,425 Yes The last digit is 5 750 Yes The last digit is 0 8,551 No The last digit is 1 (not a 0 or a 5) 394 No The last digit is 4 (not a 0 or a 5) ## Divisible by 6? Rule: If it is divisible by 2 and by 3 Number Divisible? Why? 5,106 Yes The last digit is a 2 (it is a multiple of 2 ) and... 5 + 1 + 0 + 6 = 12 (12 is a multiple of 3) 636 Yes The last digit is a 6 (it is a multiple of 2 ) and... 6 + 3 + 6 = 15 (15 is a multiple of 3) 5,912 No The last digit is a 2 (it is a multiple of 2 ) but... 5 + 9 + 1 + 2 = 17 (17 is not a multiple of 3) 508 No The last digit is a 8 (it is a multiple of 2 ) but... 5 + 0 + 8 = 13 (13 is not a multiple of 3) ## Divisible by 9? Rule: If the sum of the digits are a multiple of 9 Number Divisible? Why? 7,686 Yes 7 + 6 + 8 + 6 = 27 (27 is a multiple of 9) 252 Yes 2 + 5 + 2 = 9 (9 is a multiple of 9) 883 No 8 + 8 + 3 = 19 (19 is not a multiple of 9) 5,105 No 5 + 1 + 0 + 5 = 11 (11 is not a multiple of 9) Helper: The multiples of 9 include... 9 18 27 36 45 54 63 72 81 90 99 108 117 126 135 ## Divisible by 10? Rule: If the last digit is 0 Number Divisible? Why? 880 Yes The last digit is 0 9,560 Yes The last digit is 0 312 No The last digit is 2 (not a 0) 7,897 No The last digit is 7 (not a 0) Back To All Divisibility Rules ### Prevent Bullying Click the links below for information and help on dealing with bullying .
Lesson Objectives • Demonstrate an understanding of how to factor a four-term polynomial using grouping • Learn how to factor a trinomial using the AC method • Learn how to factor a trinomial with two variables ## How to Factor Trinomials Using the AC Method In the last lesson, we learned how to factor a trinomial of the form: ax2 + bx + c, a = 1 This means we can write our trinomial as: x2 + bx + c When a, the coefficient of x2 is 1, it makes our job much easier. The first entry of each binomial is a given. x2 + 8x + 15 We can factor the above trinomial as: x2 + 8x + 15 = (x + __)(x + __) We know the first entry of each binomial is an x. This is from the F step in FOIL. We know that x2 comes from x times x. As we work through FOIL, the O + I steps combine to give us the middle term. The L step gives us the final term. To find the last entry of each binomial, we need to find two integers that sum to b, the coefficient of the middle term (8), and multiply to give us c, the constant term (15). We can think about the factors of 15: 1,3,5,15 3, and 5 are the desired integers. 3 + 5 = 8 3 • 5 = 15 Once we find the desired integers, we can use the integers to fill in the blanks. x2 + 8x + 15 = (x + __)(x + __) x2 + 8x + 15 = (x + 3)(x + 5) ### How to Factor a Trinomial when the Leading Coefficient is not 1 At some point, we will need to factor the more difficult scenario. This occurs when the coefficient of x2 is not 1. This means our first entry of each binomial is not a given and we need to fill in all of the blanks. To factor a trinomial of the form: ax2 + bx + c, a ≠ 1 • Factor out the GCF, when the GCF is not 1 • Find two integers whose product is ac and whose sum is b • Use the two integers to rewrite the middle term. This will give us a four-term polynomial • Factor the four-term polynomial using the grouping method Let's look at a few examples. Example 1: Factor each 8x2 + 14x - 49 Step 1) Factor out the GCF. In this case, our GCF is 1. Step 2) Find two integers whose product is ac and whose sum is b. a is the coefficient of the squared term (8), while c is the constant (-49). 8 • -49 = -392 b is the coefficient of the middle term (14) We are looking for a product of -392, and a sum of 14. Since we want a negative product, we know we need a positive and a negative since opposite signs give us a negative product. We also know the absolute value of the positive number is larger than the absolute value of the negative number since the sum is positive. Let's think about the ways we can find a product of +392 and work out the signs. 1 • 392 2 • 196 4 • 98 7 • 56 8 • 49 14 • 28 If we think about 14 and 28, we can make the pair work if we change the sign of 14. Our desired integers are (-14) and (+28). -14 • 28 = -392 -14 + 28 = 14 Step 3) Use the two integers to rewrite the middle term. 8x2 + 14x - 49 8x2 + 28x - 14x - 49 Notice how 28x - 14x combines to give us 14x. We now have a four-term polynomial. Step 4) Factor using grouping 8x2 + 28x - 14x - 49 (8x2 + 28x) + (- 14x - 49) 4x(2x + 7) - 7(2x + 7) (4x - 7)(2x + 7) Example 2: Factor each 12x2 - 38x - 14 Step 1) Factor out the GCF. In this case, our GCF is 2. 2(6x2 - 19x - 7) Step 2) Find two integers whose product is ac and whose sum is b. a is the coefficient of the squared term (6), while c is the constant (-7). 6 • -7 = -42 b is the coefficient of the middle term (-19) We are looking for a product of -42, and a sum of -19. Since we want a negative product, we know we need a positive and a negative since opposite signs give us a negative product. We also know the absolute value of the positive number is smaller than the absolute value of the negative number since the sum is negative. Let's think about the ways we can find a product of +42 and work out the signs. 1 • 42 2 • 21 3 • 14 6 • 7 If we think about 2 and 21, we can make the pair work if we change the sign of 21. Our desired integers are (-21) and (+2). -21 • 2 = -42 -21 + 2 = -19 Step 3) Use the two integers to rewrite the middle term. 12x2 - 38x - 14 2(6x2 - 19x - 7) 2(6x2 - 21x + 2x - 7) Notice how -21x + 2x combines to give us -19x. We now have a four-term polynomial. Step 4) Factor using grouping 2(6x2 - 21x + 2x - 7) 2((6x2 - 21x) + (2x - 7)) 2(3x(2x - 7) + 1(2x - 7)) 2(3x + 1)(2x - 7) Example 3: Factor each 24x3 - 140x2 + 200x Step 1) Factor out the GCF. In this case, our GCF is 4x. 4x(6x2 - 35x + 50) Step 2) Find two integers whose product is ac and whose sum is b. a is the coefficient of the squared term (6), while c is the constant (50). 6 • 50 = 300 b is the coefficient of the middle term (-35) We are looking for a product of +300, and a sum of -35. Since we want a positive product and a negative sum, we know we need two negative numbers. Let's think about the ways we can find a product of +300 and work out the signs. 1 • 300 2 • 150 3 • 100 4 • 75 5 • 60 6 • 50 10 • 30 12 • 25 15 • 20 If we think about 15 and 20, we can make the pair work if we change the sign of both. Our desired integers are (-15) and (-20). -15 • -20 = 300 -15 + -20 = -35 Step 3) Use the two integers to rewrite the middle term. 24x3 - 140x2 + 200x 4x(6x2 - 35x + 50) 4x(6x2 - 20x - 15x + 50) Notice how -20x + (-15x) combines to give us -35x. We now have a four-term polynomial. Step 4) Factor using grouping 4x(6x2 - 20x - 15x + 50) 4x((6x2 - 20x) + (-15x + 50)) 4x(2x(3x - 10) - 5(3x - 10)) 4x(2x - 5)(3x - 10) ### Factoring Trinomials with Two Variables using the AC Method In some cases, we will see trinomials where the leading coefficient is not 1 and two variables are involved. These problems can be solved using the same technique. Let's look at an example. Example 4: Factor each 4x2 - 3xy - 7y2 We can see that we have two variables involved, x and y. Let's work through our regular steps. Step 1) Factor out the GCF. In this case, our GCF is 1. 4x2 - 3xy - 7y2 Step 2) Find two integers whose product is ac and whose sum is b. a is the coefficient of the first squared term (4), while c is the final term (-7y2). In this case, -7y2 is not really a constant, but we follow the same thought process. -7y2 occurs as the final term and is produced by the L step in FOIL. 4 • -7y2 = -28y2 b is the coefficient of the middle term (-3y). We think of -3y as what is multiplying the variable x. We are looking for a product of -28y2, and a sum of -3y. Since we want a negative product and a negative sum, we know we need two integers with opposite signs. Let's think about the ways we can find a product of +28 and work out the signs. Leave y out of this for now. We know that y • y = y2. 1 • 28 2 • 14 4 • 7 If we think about 4 and 7, we can make the pair work if we change the sign of 7. Our desired integers are (-7) and (+4). -7y • 4y = -28y2 -7y + 4y = -3y Step 3) Use the two integers to rewrite the middle term. 4x2 - 7xy + 4xy - 7y2 Notice how -7xy + 4xy combines to give us -3xy. We now have a four-term polynomial. Step 4) Factor using grouping 4x2 - 7xy + 4xy - 7y2 4x2 + 4xy - 7xy - 7y2 (4x2 + 4xy) + (-7xy - 7y2) 4x(x + y) - 7y(x + y) (4x - 7y)(x + y) #### Skills Check: Example #1 Factor each $$15x^3-126x^2+48x$$ A $$3x(5x - 2)(x - 8)$$ B $$3x(5x + 4)(x + 4)$$ C $$15x(x - 2)(x + 8)$$ D $$12x(x - 2)(x + 8)$$ E $$3(5x + 2)(x + 8)$$ Example #2 Factor each $$15x^2 + 93x - 216$$ A $$3(5x - 9)(x + 8)$$ B $$15(x - 9)(x - 8)$$ C Prime D $$(7x - 10)(x + 6)$$ E $$(x - 1)(x + 6)$$ Example #3 Factor each $$3x^2 + x - 24$$ A Prime B $$3(x + 1)(x - 8)$$ C $$3(x + 4)(x - 2)$$ D $$3(x - 8)(x - 3)$$ E $$(3x - 8)(x + 3)$$ Example #4 Factor each $$18x^2 - 87xy + 84y^2$$ A $$3(2x + 7y)(3x + 4y)$$ B $$3(2x - 7y)(3x + 4y)$$ C $$3(2x - 7y)(3x - 4y)$$ D $$6(3x + 2y)(x - 1)$$ E Prime
# Quintuple Digit Division Worksheets How to Divide Exceptionally Large Numbers - Division does not only involve dividing small numbers, it also entails the division of big or large numbers. To make the division of long numbers easy for you, try out the method of long division, which is by far the best and most effective method to solve division of big numbers. Unlike the four mathematical operations, the method of division moves from right to left. For each digit in the dividend, which is the number you are dividing, the division requires you to complete a cycle of division, multiplication, and then subtraction. Follow the steps to find the solution of large numbers with the help of long division. Start off by writing the dividend and the divisor, then dividing the first one or two digits of the dividend, accordingly, making sure that the number is less than the dividend. After the multiplication step, now subtract the answer to the number you have divided. After you have deducted the number, the answer must be less than the divisor. If it is not, then bring down the number next in line to be divided to make it a new number. If there is not any number left to bring down, then the answer is your remainder. • ### Basic Lesson Demonstrates how to divide a five digit number by a four digit number. Example: Divide 8897 by 124. 7 times 1120 is 7840. So we write 7840 under 8897 and 7 as the quotient. On subtraction, we get a remainder of 1057 and bringing down 0 we get 10570. 9 times 1120 is 10080. We write 10080 under 10570 and 490 as remainder. So the quotient is 79 and the remainder is 490. • ### Intermediate Lesson Walks students step-by-step through dividing a five digit number by a five digit number. • ### Independent Practice 1 Asks students to determine the quotient of various division problem. The answers can be found below. • ### Independent Practice 2 20 more problems to reinforce skills. • ### Homework Worksheet Reviews Quintuple digit division problems for students to work on at home. • ### Skill Quiz Features 10 problems. A math scoring matrix is included. • ### Homework and Quiz Answer Key Answers for the homework and quiz. • ### Lesson and Practice Answer Key Answers for both lessons and both practice sheets. #### Divisibility Tips A number is divisible by: 2 if it is an even number. 3 when the digits added together equal a number divisible by 3. 4 when the digits formed by the tens and ones places are divisible by 4./p>
# Improper To Mixed Fractions In this section you will learn how to convert improper to mixed fractions. To Convert improper -> mixed fraction, use the following steps. Step 1 : Consider any improper fraction. Step 2 : Divide the numerator by denominator actually to get quotient and remainder. Step 3 : Write the mixed fraction as QuotientRemainder/Denominator. Example : Convert 43/4 into mixed fraction : First, do the long division to find the whole-number part (being the quotient) and the remainder: In the above example Quotient = 10, remainder = 3 and denominator = 4 so, mixed fraction can be written as QuotientRemainder/Denominator = 10 3/4. Some examples : 1) Convert 35/3 to mixed fraction. 2) Convert 125 / 4 to mixed fraction. We have, Therefore, Quotient= 31, Remainder = 1 and Denominator = 4 QuotientRemainder/Denominator 125/4 = 31 1/4. 3) Convert 315 / 13 to mixed fraction. We have, Here, Quotient = 24, Remainder = 3 and Denominator = 13 ∴ 315/13 = 24 3/13. 4) Express 19/4 as a mixed fraction : 4 )19(4 - 16 ------ 3 Quotient = 4 , remainder = 3 and denominator = 4 19/4 = 4 3/4. 5) Express 58/7 as a mixed fraction. 7 )58(8 - 56 ------ 2 Quotient = 8 , remainder = 2 and denominator = 7 58/7 = 8 2/7. 6) Express 33/5 as a mixed fraction. 5 )33(6 - 30 ------ 3 Quotient = 6 , remainder = 3 and denominator = 5 33/5 = 6 3/5. _________________________________________________________________ Practice Q.1 Convert the following into mixed fractions. 1) 12/5 2) 23/2 3) 37/8 4) 42/9 5) 52/6 Fraction Types of Fractions Improper to mixed fractions Equivalent Fractions Fractions in Simplest form Like and Unlike Fractions
In 3rd Grade Multiplication Worksheet we will solve how to divide using multiplication tables, relationship between multiplication and division, problems on properties of division, long division method, word problems on long division. 1. 36 ÷ 36 = ………….. 2. 81 ÷ 9 = ………….. 3. 56 ÷ 56 = ………….. 4. 100 ÷ 10 = ………….. means …………..×………….. = ………….. 5. The number which divides is called as ………….. 6. ………….. divided by any number gives zero as the quotient. 7. 24 ÷………….. = 1 8. …………..÷ 1 = 75 9. …………..÷ 43 = 0 10. There are 10 bananas in a bunch. 5 friends want to share them equally. Each friend will get ………….. bananas. 11. Distribute 10 apples in 2 boxes. Each box has ………….. apples. So, 10 ÷2 =………….. 12. Fill in the blanks: (i) 81 ÷9 = ………….. means = …………..×………….. = ………….. (ii) 27÷ 27 =………….. (iii) ………….. ÷1 = 95 (iv) 0 ÷23 =………….. 13. Write two division facts for the following multiplication facts. (i) 6 × 7 = 42…………..………….. (ii) 4 × 9 = 36…………..………….. 14. 84 oranges are distributed among 7 students. How many oranges does each student get? 15. Divide and find the quotient: (i) 76 ÷ 2 (ii) 64 ÷ 2 (iii) 91 ÷ 7 16. Fill in the blanks - (i) 6 ÷ 0 = (ii) 3 ÷ 1 = (iii) 9 ÷ 9 = (iv) 7 ÷ 1 = (v) 2 ÷ 2= (vi) 3 ÷ 3= (vii) 0 ÷ 6= (viii) 1 ÷ 1 = 17. Divide and find the quotients - (i) 84 ÷ 4 = (ii) 30 ÷ 6 = (iii) 81 ÷ 9 = (iv) 202 ÷ 2 = (v) 666 ÷ 6 = (vi) 426 ÷ 2 = (vii) 50 ÷ 10 = (viii) 400 ÷ 10 = (ix) 6000 ÷ 600 = (x) 3000 ÷ 100 = (xi) 8000 ÷ 400 = (xii) 9000 ÷ 300 = 18. Divide and find the quotients and remainders - (i) 23 ÷ 2 Q = _____; R = _____ (ii) 895 ÷ 4 Q = _____; R = _____ (iii) 187 ÷ 6 Q = _____; R = _____ (iv) 4857 ÷ 4 Q = _____; R = _____ (v) 1867 ÷ 2 Q = _____; R = _____ (vi) 3868 ÷ 7 Q = _____; R = _____ (vii) 99 ÷ 10 Q = _____; R = _____ (viii) 999 ÷ 100 Q = _____; R = _____ (ix) 4687 ÷ 100 Q = _____; R = _____ (x) 3287 ÷ 1000 Q = _____; R = _____ (xi) 8276 ÷ 10 Q = _____; R = _____ (xii) 9374 ÷ 100 Q = _____; R = _____ 19. Tick () the correct options: (i) Quotient of 24 ÷ 2 is - (a) 13 (b) 18 (c) 12 (d) 16 (ii) Remainder of 267 ÷ 10 is- (a) 7 (b) 9 (c) 10 (d) 12 (iii) Quotient of 246 ÷ 2 is - (a) 234 (b) 456 (c) 678 (d) 123 (iv) Remainder of 8645 ÷ 2 is - (a) 3 (b) 2 (c) 1 (d) 4 20. Fill in the blanks: (i) When 744 is divided by 8 then quotient is __________ . (ii) When 6795 is divided by 5 then quotient is __________ . (iii) When 3281 is divided by 100 then remainder is __________ . (iv) When 378 is divided by 3 then divisor is __________ . 21. State 'T for True and 'F for False: (i) When 96 ÷ 4, the quotient is 24. (ii) When 650 ÷ 8, the remainder is 2. (iii) When 729 ÷ 8, the remainder is 0. (iv) When 4508 ÷ 5, the quotient is 921. 22. Do these sums in your notebook - (i) A man distributed $120 equally among his four children. How many dollars did each child get? (ii) Kabir had 70 pictures. He pasted 8 pictures on the page of his album. How many pages he prepared? How many pictures were left to paste? (iii) A merchant bought 240 eggs. He kept them equally in 9 trays. How many eggs did he keep in each tray and how many eggs were left? (iv) The school gardener distributed 500 roses equally among 100 children. How many roses did each child get? (v) If the total number of pages in 8 books are 8416, how many pages are there in each book? Answers for the 3rd grade division worksheet are given below to check the exact answers of the above questions. Answer: 1. 1 2. 9 3. 1 4. 10, 10 × 10 = 100 5. divisor 6. 0 7. 24 8. 75 9. 0 10. 2 11. 5, 5 12. (i) 9, 9 × 9 = 81 (ii) 1 (iii) 95 (iv) 0 13. (i) 42 ÷ 7 = 6, 42 ÷ 6 = 7 (ii) 36 ÷ 4 = 9, 36 ÷ 9 = 4 14. 12 15. (i) 38 (ii) 32 (iii) 13 16. (i) undefined (ii) 3 (iii) 1 (iv) 7 (v) 1 (vi) 1 (vii) 0 (viii) 1 17. (i) 21 (ii) 5 (iii) 9 (iv) 101 (v) 111 (vi) 213 (vii) 5 (viii) 40 (ix) 10 (x) 30 (xi) 20 (xii) 30 18. (i) Q = 11; R = 1 (ii) Q = 223; R = 3 (iii) Q = 31; R = 1 (iv) Q = 1214; R = 1 (v) Q = 933; R = 1 (vi) Q = 552; R = 4 (vii) Q = 9; R = 9 (viii) Q = 9; R = 99 (ix) Q = 46; R = 87 (x) Q = 3; R = 287 (xi) Q = 827; R = 6 (xii) Q = 93; R = 74 19. (i) (c) 12 (ii) (a) 7 (iii) (d) 123 (iv) (c) 1 20. (i) 93 (ii) 1359 (iii) 81 (iv) 3 21. (i) T (ii) T (iii) F (iv) F 22. (i)$30 (ii) 8; 6 (iii) 26; 6 (iv) 5 (v) 1052 ## You might like these 3rd grade math worksheets is carefully planned and thoughtfully presented on mathematics for the students. Teachers and parents can also follow the worksheets to guide the students. • ### 3rd Grade Number Worksheet \|Practice the Questions on Numbers with Ans 3rd grade number worksheet will help the kids to practice the questions on numbers. Fill in the blanks: Write in words: Write in numerals: Write > or =, < between the given numbers: • ### Four Digit Numbers \| What are Four Digit Numbers? \| Numerals and Words What are four digit numbers? We may divide 4-digit numbers in 9 groups. (i) 1000 to 1999 (one thousand to one thousand nine hundred ninety nine) • ### Comparison of Three-digit Numbers \| Arrange 3-digit Numbers \|Questions What are the rules for the comparison of three-digit numbers? (i) The numbers having less than three digits are always smaller than the numbers having three digits as: • ### Comparison of Four-digit Numbers \| Arrange 4-digit Numbers \| Examples What are the rules for the comparison of four-digit numbers? We have already learnt that a two-digit number is always greater than a single digit number. • ### Compare Two Numbers \| Examples for Comparison of Numbers \| Video How to compare two numbers? Comparing two numbers means finding which number is smaller and which number is greater. It is represented by symbol (>) or (<). When you compare two numbers and they are same then the numbers are equal and symbolized as (=). • ### Face Value and Place Value\|Difference Between Place Value & Face Value What is the difference between face value and place value of digits? Before we proceed to face value and place value let us recall the expanded form of a number. The face value of a digit is the digit itself, at whatever place it may be. • ### Estimating a Sum \| Round the Number \| Numbers by Rounding \| Estimating We will learn the basic knowledge for estimating a sum. Here we will learn an easy way to estimate a sum of two numbers by rounding. In case of two digit numbers we can only round the number • ### Worksheet on Subtraction Word Problems \| Subtraction Word Problems The activity provided in the third grade math worksheet on subtraction word problems is very important for the kids. Students need to read the questions carefully and then translate the information • ### Worksheet on Division Properties \| Properties of Division \| Division Third grade math worksheet on division properties is great for practicing and testing the knowledge of the students on identifying the different properties. 1. Fill in the blanks: (i) 21 ÷ 1 = ? (ii) 26 ÷ 1 = ? (iii) 43 ÷ 1 = ? (iv) 41 ÷ 41 = ? (v) 75 ÷ 75 = ?
Assignment 4 Centers of a Triangle Prove that the three perpendicular bisectors of the sides of a triangle are concurrent. Take a triangle ABD where lines EH and FI are perpendicular bisectors of segments AD and AB respectively and G is the midpoint of segment BD. Line EH intersects segment AD at E and line FI intersects segment AB at F. Since segments AD and AB intersect, we know that lines EH and FI must intersect since they are perpendiculars of those segments. We will call the intersection C. We now draw segments AC, BC, and DC and remind ourselves that segments EC and FC are perpendicular bisectors of segments AD and AB respectively. Next we will look at triangles AFC and BFC. We know that segment AF is congruent to segment BF since F is the midpoint of segment AB. We know that segment FC is congruent to itself. And since segment FC is perpendicular to segment AB, we know that angles AFC and BFC are right angles and therefore congruent to each other. Therefore, by the SAS theorem, triangles AFC and BFC are congruent to each other. We can now conclude that segment AC is congruent to segment BC since corresponding parts of congruent triangles are congruent. By a similar argument, we can see that segment AC is congruent to segment DC. And since segment BC is congruent to segment AC and segment AC is congruent to segment DC, we have that segment BC is congruent to segment DC. Now we will look at triangle BCD. First, we will draw segment GC. We showed above that segments BC and DC are congruent. We know that segments BG and DG are congruent because G is the midpoint of segment BD. We know that segment GC is congruent to itself. And so by the SSS theorem, triangles BGC and DGC are congruent to each other. Now we know that angle CGD is congruent to angle CGB since corresponding parts of congruent triangles are congruent. Since segment BD forms an angle of 180 degrees, it is clear that angles CGD and CGB both measure 90 degrees and are therefore right angles. Since segment CG forms right angles with segment BD and bisects segment BD at G, we conclude that segment CG is the perpendicular bisector of segment BD. We remember that segments EC and FC are the perpendicular bisectors of segments AD and AB, and that they intersect at C. We have shown that segment GC is the perpendicular bisector of segment BD and that it also intersects C. Therefore, we conclude that the three perpendicular bisectors of the sides of a triangle are concurrent. Return
Jump to a New ChapterAnatomy of SAT Numbers & OperationsEssential ConceptsEssential StrategiesTest-Taking StrategiesThe 8 Most Common MistakesConclusionSet 1: Multiple ChoiceSet 2: Grid-InsPosttest Number Terms Order of Operations Odd and Even Numbers Positive/Negative/Undecided Divisibility and Remainders Fearless Factors Multiples, Multiples, Multiples Fractions Decimals Percents Ratios Exponents, or The Powers That Be Roots and Radicals Sequences Sets Sets A set is the mathematical name given to a group of items that share some common property. All positive numbers make up one set, and all prime numbers make up another. Each item in a set is called an element or a member. Don’t confuse a set with a sequence. A set is simply a collection of elements that are not necessarily related to one another, as they are in a sequence. The union of two sets is another set that contains all the elements of each set. If set A contains all the blue-eyed women and set B contains all the blue-eyed men, the union of sets A and B is all blue-eyed women and men. If set A = (1, 2, 4, 6, 8) and set B = (2, 3, 5, 7, 8), the union of A and B is (1, 2, 2, 3, 4, 5, 6, 7, 8, 8). The intersection of two sets is another set that contains all the elements the two sets have in common. If set A = (1, 2, 4, 6, 8) and set B = (2, 3, 5, 7, 8), the intersection of A and B is the set (2, 8). A difficult set item involves a group of people, some of whom are engaged in activity A and others in activity B, while still others refrain from participating in either activity. Here’s an example: 7. In a group of students, 24 are considered to be good at math, 14 are good at liberal arts, and 12 are good at both. How many students are in the group? (A) 14 (B) 24 (C) 26 (D) 38 (E) 50 To solve this item, use the following simple formula: Total in a set = number in set 1 + number in set 2 – intersection of set 1 and set 2 + number in neither set You have to figure out which members belong to set 1, which to set 2, what the intersection of two sets is, and how many abstain from participating. Total students = those in math (24) + those in liberal arts (14) – those in both (12) + those in neither (0) = 26. There are a total of 26 students, choice C. Some set items may not explicitly tell you how many people are in neither set. If an item says that each student in a class has to learn either French or Italian, then the “neither set” (those learning neither French nor Italian) is zero. That covers the numbers & operations basics. Now let’s apply these concepts to some SAT strategies. Jump to a New ChapterAnatomy of SAT Numbers & OperationsEssential ConceptsEssential StrategiesTest-Taking StrategiesThe 8 Most Common MistakesConclusionSet 1: Multiple ChoiceSet 2: Grid-InsPosttest Test Prep Books Test Prep Centers New SAT Test Center Mini SAT SAT Vocab Novels Rave New World S.C.A.M. Sacked Busted Head Over Heels SAT Power Tactics Algebra Data Analysis, Statistics & Probability Geometry Reading Passages Sentence Completions Writing Multiple-Choice Questions The Essay Test-Taking Strategies Vocabulary Builder SparkCollege College Admissions Financial Aid College Life
Which Of The Sequences Is An Arithmetic Sequence? (Solved) An arithmetic progression, or arithmetic sequence, is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5,7,9,11,13,⋯ 5, 7, 9, 11, 13, ⋯ is an arithmetic sequence with common difference of 2. What is example of arithmetic sequence? An arithmetic sequence is an ordered set of numbers that have a common difference between each consecutive term. For example in the arithmetic sequence 3, 9, 15, 21, 27, the common difference is 6. If we add or subtract by the same number each time to make the sequence, it is an arithmetic sequence. Is this an arithmetic sequence? If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence. The constant difference in all pairs of consecutive or successive numbers in a sequence is called the common difference, denoted by the letter d. How do you find the arithmetic sequence? The arithmetic sequence formula is given as, an=a1+(n−1)d a n = a 1 + ( n − 1 ) d where, an a n = a general term, a1 a 1 = first term, and and d is the common difference. This is to find the general term in the sequence. What is arithmetic sequence? An arithmetic sequence is a sequence of numbers which increases or decreases by a constant amount each term. Once you know the common difference, you can find the value of c by plugging in 1 for n and the first term in the sequence for a1. Example 1: {1,5,9,13,17,21,25,} What is arithmetic sequence and series? An arithmetic sequence is a sequence where the difference d between successive terms is constant. The general term of an arithmetic sequence can be written in terms of its first term a1, common difference d, and index n as follows: an=a1+(n−1)d. An arithmetic series is the sum of the terms of an arithmetic sequence. Which is not arithmetic sequence? The following are not examples of arithmetic sequences: 1.) 2,4,8,16 is not because the difference between first and second term is 2, but the difference between second and third term is 4, and the difference between third and fourth term is 8. No common difference so it is not an arithmetic sequence. Is the sequence arithmetic or geometric? If the sequence has a common difference, it’s arithmetic. If it’s got a common ratio, you can bet it’s geometric. How do you find arithmetic and geometric sequences? The common pattern in an arithmetic sequence is that the same number is added or subtracted to each number to produce the next number. The common pattern in a geometric sequence is that the same number is multiplied or divided to each number to produce the next number. Is this an arithmetic sequence 4 16? This is an arithmetic sequence since there is a common difference between each term. In this case, adding 12 to the previous term in the sequence gives the next term. You might be interested: Which Of The Following Activities Is Carried Out By The Arithmetic Logic Unit Alu? (Solved) What is the 4 types of sequence? Types of Sequence and Series • Arithmetic Sequences. • Geometric Sequences. • Harmonic Sequences. • Fibonacci Numbers. GeeksforGeeks In today’s world, the vast majority of people wish to work in a rewarding position for one of the world’s largest corporations. And this can only be accomplished if you possess. You may learn more about how to make money online by reading this article. Yes, the times have changed, and we now have a plethora of possibilities at our disposal. It’s simply a question of time. Thus, Virtusa Corporation is an American information technology services firm that was created in Sri Lanka in 1996 and has its headquarters in the city of Southborough in the state of Massachusetts in the United States. The test consisted of three code problems, each of which was worth 100 points, and it lasted three hours. 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# Understanding Number Place Value Chart ## What is a Number Place Value Chart? A number place value chart is a visual representation of the different place values in a number. It helps in understanding the value of each digit in a number and how it contributes to the overall value of the number. This chart is especially useful when working with large numbers or when teaching place value concepts to students. ## How does a Number Place Value Chart work? The chart consists of columns that represent different place values, such as ones, tens, hundreds, thousands, and so on. Each column is divided into boxes, with the rightmost column being the ones column. The value of each box in a column increases by a power of 10 as you move from right to left. For example, in the ones column, each box represents a value of 1, in the tens column, each box represents a value of 10, and so on. ## Why is a Number Place Value Chart important? A number place value chart is important because it helps in understanding the magnitude and structure of numbers. It allows us to visually see the value of each digit and how it contributes to the overall value of the number. This understanding is crucial in various mathematical operations such as addition, subtraction, multiplication, and division. It also forms the basis for more advanced mathematical concepts like decimals and fractions. ## How to use a Number Place Value Chart? To use a number place value chart, start by writing the number you want to represent in the chart. Then, place each digit of the number in its respective column and box. For example, if you want to represent the number 1234, you would place 4 in the ones column, 3 in the tens column, 2 in the hundreds column, and 1 in the thousands column. This way, you can easily see the value of each digit and the overall value of the number. ## Common Mistakes to Avoid When using a number place value chart, it is important to avoid some common mistakes. One common mistake is placing the digits in the wrong column. For example, placing the digit 3 in the tens column instead of the hundreds column. Another mistake is forgetting to include zeros for empty boxes. For example, if you want to represent the number 405, make sure to place a zero in the ones column to indicate that there are no ones in the number. ## Tips for Teaching Place Value When teaching place value using a number place value chart, it is helpful to use manipulatives such as base-10 blocks or place value disks. These physical objects can be used to represent each digit and make the concept more tangible for students. Additionally, providing real-life examples and practical applications of place value can help students understand its significance in everyday life. ## Conclusion A number place value chart is a valuable tool for understanding the value and structure of numbers. It helps in visualizing the place values and their contributions to the overall value of a number. By using this chart, students can develop a solid foundation in place value, which is essential for further mathematical concepts. ### References: 1. Smith, J. (2022). Understanding Place Value: A Guide for Educators. Educational Publishing. 2. Johnson, A. (2021). The Importance of Place Value in Mathematics. Mathematics Journal, 45(2), 67-82.
# Question 11.6: For the four-diode gate shown in Fig 11.20(a), RL = R2 = 50 ...... For the four-diode gate shown in Fig 11.20(a),$\ R_{L} = R_{2}$ = 50 kΩ and$\ R_{1}$ = 1 kΩ,$\ R_{f}$ = 25 Ω,$\ V_{s}$ = 10 V. Calculate (a) A (b)$\ V_{(min)}$ (c)$\ V_{C(min)}$ (d) for$\ V = V_{(min)}$ Step-by-Step The 'Blue Check Mark' means that this solution was answered by an expert. (a) We have $\ R = \frac{R_{2}R_{1}}{R_{2} + R_{1}}$ , $\ R_{3} = R + R_{f}$ And$\ α = \frac{R_{2}}{R_{1} + R_{2}}$ $\ R = \frac{50 × 1}{51}$ = 980 Ω $\ α = \frac{50}{51}$ = 0.98 $\ R_{3}$= 980 + 25 = 1005Ω and$\ A = α × \frac{R_{L}}{R_{L} + R_{3}/2}$ . Therefore, $\ A = 0.98 × \frac{50}{50 + (1.005/2)} = 0.98 × \frac{50}{50 + 0.502}$ A = 0.97 (b) $\ V_{min} = \frac{R_{2}}{ R_{1}} × \frac{R_{3}}{R_{3} + 2R_{L}} × V_{s}$ $\ = \frac{50}{1} × \frac{1.005 × 10}{(1.005 + 100)} = \frac{502.5}{101.005}$ = 4.97 V (c) $\ V_{C(min)} = A V_{s}$ = 0.97 × 10 = 9.7 V (d) $\ V_{n(min)} = V_{s} × \frac{R_{2}}{R_{1} + R_{2}} − V × \frac{R_{1}}{R_{1} + R_{2}}$ Here,$\ V = V_{(min)}$= 4.97 V Therefore, $\ V_{n(min)} = 10 × \frac{50}{50 + 1} − 4.97 × \frac{1}{50 + 1}$ = 9.8 − 0.97 = 9.7 V Question: 11.1 ## In the circuit shown in Fig. 11.16(a), RL = R1 = 100 kΩ , R2 = 50 kΩ and the signal has a peak value of 20 V. Find (a) A (b) VC(min) (c) Vn(min) (d) Ri when the diodes are ON ... (a) \ A = \frac{a}{1 + a\frac{R_{1}}{2R_{L}... Question: 11.2 ## For the four-diode gate shown in Fig. 11.20(a), RL = R2 = 100 kΩ and R1 = 1 kΩ, Rf = 25 Ω, Vs = 20 V. Calculate (a) A (b) V(min) (c) VC(min) (d) Vn(min) for V = V(min) ... (a) We have: \ R = \frac{R_{1}R_{2}}{R_{1} ... Question: 11.3 ## For the four diode gate shown in Fig. 11.21(a), Vs = 20 V, RL = 200 kΩ , RC = 100 kΩ, Rf = 0.5 kΩ, R = Rs = 1 kΩ. Find Vn(min), A and VC(min). ... (i) From Eq. (11.17), \ V_{n(min)} = V_{s}[... Question: 11.4 ## Consider the following situation in the FET series gate, Vs = ± 2 V, Rs = 100 Ω, RL = 10 kΩ. The FET has the following parameters: VGS(OFF)max = −10 V and RD(ON) = 20 Ω. Calculate the voltage levels of the control signal, ID, error due to RS and error due to RD(ON). ... The control signal should have a value\ V_{... Question: 11.5 ## In the circuit shown in Fig. 11.16(a), RL = R1 = 200 kΩ, R2 = 100 kΩ and the signal has a peak value of 10 V. Find (a) A (b) VC(min) (c) Vn(min) (d) Ri ... (a) \ A = \frac{α}{1 + α(R_{1}/2R_{L})}[/la... The control signal should have a value\ V_{...
top of page Search # Learn the fundamentals of median, quartiles and mode -Statistics Updated: Nov 8, 2023 Once upon a time, in a quiet village, nestled between hills and a river there lived a girl named Lily. She spent her afternoons exploring the woods, collecting stones and arranging them in sequences. Her favorite numbers were 9, 17 and 22. One sunny morning, Lily heard the villagers talking about the annual Harvest Festival. One of the games would be a grand math contest where the sharpest minds in the village could compete. Lily wanted to take part in the contest. She started forming a pattern with her stones, with 9, 17 and 22 at the center. She imagined a peaceful village with 22 houses along a winding road. The youngest child in the village was 9 years old. The carpenter who could mend anything was 17 years old. She introduced the baker, the farmer and a teacher, each with their own age. Explaining the data, she said, the median is like the center dividing the data into two equal parts. Quartiles divide the data into 4 equal parts. In her data, the first quartile was where the younger folks live. The third quartile was where the older folks lived and the 2nd quartile was the median. Finally, explaining the mode, she said the mode is the value which occurs the most. In the math contest, the judges were impressed by Lily's creativity. Not only did Lily win the contest, she also left a lasting impression on the villagers. Most mathematical concepts can be made accessible through story telling. A beginner's guide to median, mode and quartiles. This post should answer your basic questions about median, quartiles and mode and how to evaluate them. I am a math educator teaching mathematics to high school students and college students for the past 3 decades. I can tell you exactly what you need to be studying to understand median, quartiles and mode. In this age of ChatGpt where AI takes over teaching as well, I cannot stress how important it is for you to avoid common mistakes which students make. So, that's me, adding a humane approach to learning math. If you want to learn Arithmetic Mean. ## Median, Quartiles and Mode. How to calculate it? In layman's language, a median is the middle value of a set of observations arranged in ascending or descending order. For a set of discrete values, you have to first arrange the data in ascending or descending order. Then depending on whether you have odd number or even number of values, you calculate the median. My students generally find it easy to calculate the median for odd number of observations. Even number of values require a little more of work, but with a little help, they usually get it right. What happens for a frequency distribution? Basically, you can divide frequency distributions into 2 types. One, is a simple frequency distribution without class intervals. Here, you first calculate the cumulative frequency. I usually recommend using the lesser than cumulative frequency. I tell my students to divide n ( the sum of the frequencies) by 2 and then look at the frequency table and identify the median. For a grouped frequency distribution, there are 2 methods of calculating the median. First, you draw, the cumulative frequency curve using the ogive and then using the ogive, you determine the median. As I tell my students, you need a high degree of accuracy here. Alternately, there is a formula to calculate the median which makes it very easy, provided you remember it right. ## Quartiles are the values which divide the data into 4 equal parts There are 4 quartiles, the lower quartile, the second quartile also called the median and the third quartile. The difference between the third and the first quartile is called the inter quartile range. How will you calculate the quartiles. Again, for a set of discrete values, you arrange the data in ascending or descending order. You then calculate the quartiles using the formula. I make my students practice the case when you have a fractional value as well. Again, for a simple frequency distribution, you calculate the cumulative frequency. Then using this you will calculate the quartiles. For a grouped frequency distribution with equal class intervals, you can either use the ogive or the formula for calculating. ## In simple language, mode is the value which occurs the most frequently Calculating the mode is one of the easiest things to do. My students just love calculating the mode for a discrete set of values. Just take the value which occurs most frequently. In this context, note that you can also have 2 modes, called a bimodal distribution. Coming to a frequency distribution, if the frequency distribution is simple, just take the value with the highest frequency as the mode. For a grouped frequency distribution, there are two ways of calculating the mode. My students just love the graphical method, where you plot a histogram and then move on to evaluating the mode. The second method of calculating the mode is using the formula which is again quite simple. I have enclosed a few formulas as images in this blog. ## Still needing help? I have given you a basic idea of median, quartiles and mode and how to evaluate them. In case you need some more help, you can contact me for online tutoring You can choose between one to one tutoring or group classes. Again, you can tutor topic wise or for the entire course. You can choose whichever suits you best. I hope this blog can serve the purpose of educating you. Would you like to share this piece of advice to your friends who may need it? Free resource on Statistics. Empower yourself and others!
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: Get ready for 8th grade>Unit 1 Lesson 4: Order of operations # Order of operations with rational numbers Let's go deeper with the order of operations. How do negation, absolute value, reciprocals fit into the order of operations? How do other properties like the distributive property or commutative property of addition relate? ## Order of operations review When there are lots of operations in an expression, we need to agree on which to evaluate first. That way, we will all agree that the expression has the same value. We could just use parentheses to always show what to do first, but we like to write things shorter when we can. So instead, we agree to evaluate more powerful operations sooner. Multiplication is repeated addition, so we multiply before we add. Powers are repeated multiplication, so we take powers before we multiply. • Grouping symbols: We evaluate what is inside of grouping symbols first. There are lots of grouping symbols. Some common ones are parentheses, fraction bars, and absolute value symbols. • Exponents: Next we evaluate powers. There are a couple of operations that undo exponents (it takes $2$ operations because powers are not commutative). They happen in this step, too. • Multiplication: Then we multiply. Division undoes multiplication, so it happens in the same step. • Addition: Last of all, we add. Subtraction undoes addition, so it happens in the same step. Fun fact: There are operations that grow a number faster than exponents. One of them shows up a lot in probability. Can you think of any operations that change a number more slowly than addition? Tell us about it in the comments. ## Exponents with negatives How do negative signs fit in with order of operations? In the expression $-{4}^{2}$, do we apply the negative sign or the exponent first? A negative sign means we take the opposite of a number. That's the same as multiplying the number by $-1$. So we take the opposite in the multiplication and division step. Let's evaluate $\left(-4{\right)}^{2}$ and $-{4}^{2}$. $\begin{array}{rlr}\left(-4{\right)}^{2}& =-4\cdot \left(-4\right)& \text{Evaluate groups.}\\ \\ & =16& \text{Multiply.}\end{array}$ With $\left(-4{\right)}^{2}$, we took the opposite of $4$ first, because the negative sign was inside the grouping symbols. $\begin{array}{rlr}-{4}^{2}& =-\left(4\cdot 4\right)& \text{Evaluate the power.}\\ \\ & =-16& \text{Take the opposite.}\end{array}$ With $-{4}^{2}$, we squared $4$ first, because exponents come earlier in the order operations than multiplying by $-1$ does. Problem 1.1 Evaluate. $-{3}^{4}=$ ## Exponents with fractions Recall that fractions have three grouping symbols built in: the numerator, the denominator, and the entire fraction. If there are no parentheses or other grouping symbols visible, then the exponent is inside the numerator or denominator. For example, in the expression ${\left(\frac{3}{2}\right)}^{4}$, the entire fraction is inside a grouping symbol, the parentheses. The exponent $4$ is outside the parentheses. So we are raising the entire fraction to the power of $3$. $\begin{array}{rl}{\left(\frac{3}{2}\right)}^{4}& =\frac{3}{2}\cdot \frac{3}{2}\cdot \frac{3}{2}\cdot \frac{3}{2}\\ \\ & =\frac{81}{16}\end{array}$ On the other hand, in the expression $\frac{{3}^{4}}{2}$, the only grouping symbol is the fraction bar. The exponent $4$ is in the numerator. So we only raise the $3$ to the power of $4$. $\begin{array}{rl}\frac{{3}^{4}}{2}& =\frac{3\cdot 3\cdot 3\cdot 3}{2}\\ \\ & =\frac{81}{2}\end{array}$ Problem 2.1 Evaluate. $\frac{2}{{7}^{2}}=$ ## Absolute value Suppose we have the expression $4-12\cdot \|-7\cdot -11-100\|$. We know we evaluate the multiplication before the subtraction, but when do we take the absolute value? First of all, the absolute value symbols are a grouping symbol. So stuff inside the symbols happens at the same step as other grouping symbols. So first, we multiply $-7\cdot -11$ to get $77$, then subtract $100$ to get $-23$ inside of the absolute value symbols. Then we take the absolute value during the same step as we apply exponents. Fun fact: There are other operations that go at this same step because they undo powers the way subtraction undoes addition. Problem 3.1 Evaluate. $4-12\cdot \|-7\cdot -11-100\|=$ ## Add and subtract left to right? Maybe you learned that you should evaluate addition and subtraction from left to right. That's because subtraction does not commute or associate. Luckily, we can rewrite subtraction as addition of the opposite. Let's rewrite $-4+6-3+1$. $-4+6-3+1=\underset{\text{term}}{\underset{⏟}{-4}}+\underset{\text{term}}{\underset{⏟}{6}}+\left(\underset{\text{term}}{\underset{⏟}{-3}}\right)+\underset{\text{term}}{\underset{⏟}{1}}$ Now that we're adding, we can change the order and groups! Once we get used to thinking of subtracting as adding the opposite, we can take shortcuts. Often, people leave out the extra $+$ symbols. $\underset{\text{term}}{\underset{⏟}{-4}}+\underset{\text{term}}{\underset{⏟}{6}}\underset{\text{term}}{\underset{⏟}{-3}}+\underset{\text{term}}{\underset{⏟}{1}}$ Each time we reach a $+$ or $-$ that's not in a grouping symbol, it's the start of a new term. The $-$ stays with its term. Then we can commute and associate any terms we want. $\begin{array}{rl}-4+6-3+1& =6+1-3-4\\ \\ & =\left(6+1\right)+\left(-3-4\right)\\ \\ & =7+\left(-7\right)\end{array}$ Problem 4.1 Which $2$ of the following expressions are equivalent to $-0.25+\left(-8\right)-13+0.7$? ## Multiply and divide left to right? The reciprocal of a number tells us how many groups of it we can make from $1$. The great thing about reciprocals is that we can rewrite dividing by a number as multiplying by its reciprocal. Here are a few examples. • $8÷6=8\cdot \frac{1}{6}$ • $\frac{3}{7}÷\frac{6}{11}=\frac{3}{7}\cdot \frac{11}{6}$ • $\begin{array}{rl}11÷0.25& =11÷\frac{1}{4}\\ \\ & =11×4\end{array}$ That's great news, because we can commute and associate multiplication factors in any order. Let's rewrite $3÷8\cdot -24$. $\begin{array}{rl}3÷8\cdot -24& =3\cdot \frac{1}{8}\cdot -24\\ \\ & =3\cdot -24\cdot \frac{1}{8}\\ \\ & =3\cdot \left(-24\cdot \frac{1}{8}\right)\\ \\ & =3\cdot \left(-24÷8\right)\end{array}$ Like with adding and subtracting, sometimes we take shortcuts and move the division symbol with the implied factor. We cannot start an expression with a $÷$ symbol. So if we move the division to the start of an expression or grouping symbol, then we must write it as a factor of the reciprocal. Note, we cannot commute a factor into a different term. Whenever we see a $+$ or $-$ symbol that is not inside a grouping symbol, that's the start of a new term. For example, the expression $10\cdot 2-5$ has $2$ terms: $10\cdot 2$ and $-5$. $\begin{array}{rl}10\cdot 2-5& \ne 10-5\cdot 2\\ \\ 20-5& \ne 10-10\\ \\ 15& \ne 0\end{array}$ Problem 5.1 Which $2$ of the following expressions are equivalent to $\frac{3}{4}÷\frac{8}{5}÷6$? ## Grouping symbols or distributing first Recall that the distributive property allows us to write equivalent expressions involving parentheses and multiplication. So we get the same value whether we distribute first or evaluate what's inside the parentheses first. Let's try it out. Now we'll distribute first. We get the same value both ways. Note, the distributive property does not apply to all grouping symbols, just to parentheses. $\begin{array}{rl}-7\|3-4\|& \ne \|-7\left(3\right)+\left(-7\right)\left(-4\right)\|\\ \\ -7& \ne 7\end{array}$ ## Want to join the conversation? • I found it helps when I read through the parts I don't understand a few times, and read them slowly. I try saying them out loud to find out how it makes sense. • can i go to sleep • yes you can sleep • "For -3 to the fourth power, how does it end up staying as a negative? I thought two negatives, when multiplied, became positive. • You have to understand the difference of two things, if you have -3^4, order of operations require you to do the power before adding the negative, so you end up with a negative. You are thinking about (-3)*4 where you would be correct, 4 negatives would make a positive. Be careful about having or not having parentheses which give two opposite sign answers.
Video: Solve Linear Inequalities \| Nagwa Video: Solve Linear Inequalities \| Nagwa # Video: Solve Linear Inequalities Learn how to solve linear inequalities, such as 2𝑥 − 3 > 7 or (6 − 2𝑥)/3 < 6 algebraically. We also discuss the need to reverse the inequality sign when multiplying or dividing both sides of an inequality by a negative number. 04:47 ### Video Transcript Solve Linear Inequalities So when we’re solving an inequality such as this one, we want to pretend that the inequality isn’t there; it’s just like a normal equal sign. And then we can solve it as we would any normal linear equation, so we will solve this inequality as if it was equal to two 𝑥 minus three equal seven. So if we had two 𝑥 minus three equal seven, we could see it’s just a simple two-step equation. So the first thing we would do is get rid of the minus three by adding three to both sides. And then on the left-hand side, that gives us two 𝑥; and on the right-hand side, we have seven plus three, which is ten. We cannot forget to put the sign in exactly as it is. Don’t swap it halfway through because otherwise it will change the value of what you’re saying. Now we’ve got two multiplied by 𝑥. The opposite of multiply is divide. Whatever we do to one side we have to do to the other, so we’re going to have to divide both sides by two. And now on the left-hand side, that just gives us 𝑥, which is greater than ten divided by two, which is five. And there we have it, we have solved that inequality for 𝑥. So we are saying the original linear inequality is the same as saying 𝑥 is greater than five. Now looking at this next question we’ve got six minus two 𝑥 all divided by three is less than six. So first thing we’ll need to do is get rid of that divide by three because we can’t do anything else. So the opposite of divide by three we know it’s multiply by three, so we’re going to multiply both sides by three. And that gives us six minus two 𝑥 is less than six times by three, which is eighteen. And then it looks like a linear inequality now, so we can see what we could do if we wanted is add two 𝑥 to both sides and then subtract eighteen from both sides. But what we’re gonna do instead this time is subtract six from both sides. And that will give us negative two 𝑥 is less than twelve. And because we have a negative, we have to if we multiply or divide an inequality by a negative, we have to swap the inequality. So if it’s less than, it has to become greater than. So we could see we’ve got negative two multiplied by 𝑥. So to get rid of a negative two multiplied by 𝑥, we have to divide by negative two. So we know negative two 𝑥 divided by negative two is just 𝑥, and twelve divided by negative two is negative six. And then as I said, when we multiply it or divide by a negative, we must swap the sign around. So we can see right now it’s less than, so it needs to become greater than. So if you want to see why we have to swap the inequality around, let’s have a look about what would have happened if instead of just dividing by negative two, we had moved the negative two 𝑥 onto the right-hand side and the twelve onto the left-hand side. So we would’ve added two 𝑥 to both sides, which would’ve given us zero is less than twelve plus two 𝑥. And then we would have subtracted twelve from both sides, which would have given us negative twelve is less than two 𝑥. And finally, we would’ve divided both sides by two, giving us negative six is less than 𝑥, which we can see is exactly the same as 𝑥 is greater than negative six. So we’ve saved ourselves all of that hassle of moving the 𝑥s and moving the numbers onto different sides by just being able to divide by a negative and swapping the inequality. ## 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
# How do you simplify (x-y+1)-(x+y-1)? Sep 10, 2016 $- 2 y + 2$ #### Explanation: The terms in the 1st bracket are being multiplied by 1 , while each of the terms in the 2nd bracket are being subtracted. That is $1 \left(x - y + 1\right) - x - \left(+ y\right) - \left(- 1\right)$ simplifying gives. $x - y + 1 - x - y + 1$ and collecting like terms. $\cancel{x} + \left(- y - y\right) + \left(1 + 1\right) \cancel{- x} = - 2 y + 2$ Sep 10, 2016 $2 - 2 y$ #### Explanation: We have: $\left(x - y + 1\right) - \left(x + y - 1\right)$ Let's expand the parentheses: $= x - y + 1 - x - y + 1$ Then, let's group the like terms: $= \left(x - x\right) + \left(- y - y\right) + \left(1 + 1\right)$ $= 0 - 2 y + 2$ $= 2 - 2 y$
Problem Determine all real numbers $a$ such that the two polynomials $x^2+ax+1$ and $x^2+x+a$ have at least one root in common. Solutions Solution 1 Let this root be $r$. Then we have $\begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\ ar + 1 &=& r + a\\ (a-1)r &=& (a-1)\end{matrix}$ Now, if $a = 1$, then we're done, since this satisfies the problem's conditions. If $a \neq 1$, then we can divide both sides by $(a - 1)$ to obtain $r = 1$. Substituting this value into the first polynomial gives $\begin{matrix} 1 + a + 1 &=& 0\\ a &=& -2 \end{matrix}$ It is easy to see that this value works for the second polynomial as well. Therefore the only possible values of $a$ are $1$ and $-2$. Q.E.D. Solution 2 Let $x^2+ax+1 = (x-s)(x-t)$ and $x^2+x+a = (x-s)(x-t)$ where $s$ is the common root. From Vieta's Formulas, we have: $-(s+t) = a, -(s+u) = 1, st = 1,$ and $su = 1$. We see that $s,t,u \notequal 0$ (Error compiling LaTeX. ! Undefined control sequence.). Dividing $su$ by $st$, we have: $$\frac{su}{st} = \frac{a}{1} \Rightarrow u = at$$ Also, we have: $$a+t = -s = 1+u \Rightarrow a+t = 1+u$$ Substituting $u = at$ into the above, we have: $\begin{matrix} a+t &=& 1+ at\\ at - a - t +1 &=& 0\\ (a-1)(t-1) &=& 0 \end{matrix}$ Thus either $a = 1$ or $t = 1$. We check to see that $a = 1$ is indeed a possible value to satisfy the requirements. If $t = 1$, then from $st = 1$, we have $s = t = 1$, and from $-(s+t) = a$, we have $a = -(1+1) = -2$, which also satisfies the requirements. Thus, the only possible a values are: $a = 1, -2$.
# Evaluate the integral int \ sinx/(2+cos^2x) \ dx ? Jan 14, 2018 $\int \setminus \sin \frac{x}{2 + {\cos}^{2} x} \setminus \mathrm{dx} = - \frac{\sqrt{2}}{2} \arctan \left(\cos \frac{x}{\sqrt{2}}\right) + C$ #### Explanation: We seek: $I = \int \setminus \sin \frac{x}{2 + {\cos}^{2} x} \setminus \mathrm{dx}$ If we look at the denominator then a substitution: $2 {u}^{2} = {\cos}^{2} x$ looks promising, so let us try the substitution: $u = \cos \frac{x}{\sqrt{2}} \implies 2 {u}^{2} = {\cos}^{2} x$ Differentiating implicitly wrt $x$: $\frac{\mathrm{du}}{\mathrm{dx}} = - \sin \frac{x}{\sqrt{2}}$ So substituting into the integral, it becomes: $I = \int \setminus \frac{- \sqrt{2}}{2 + 2 {u}^{2}} \setminus \mathrm{du}$ $\setminus \setminus = - \sqrt{2} \setminus \int \setminus \frac{1}{2 \left(1 + {u}^{2}\right)} \setminus \mathrm{du}$ $\setminus \setminus = - \frac{\sqrt{2}}{2} \setminus \int \setminus \frac{1}{1 + {u}^{2}} \setminus \mathrm{du}$ This is now a standard result, thus: $I = - \frac{\sqrt{2}}{2} \arctan \left(u\right) + C$ Then restoring the substitution: $I = - \frac{\sqrt{2}}{2} \arctan \left(\cos \frac{x}{\sqrt{2}}\right) + C$
b). Find the lengths of the two parallel sides. We've "always" used the inclusive definition on this side of the pond (except we call it a trapezium, of course). The area of a trapezium is given by $${A}=\frac{(a+b)}{2}\times{h}$$.. You can see that this is true by taking two identical trapezia (or trapeziums) to make a parallelogram. According to US definition: a trapezoid has a pair of parallel sides, and according to UK definition: a trapezoid has no parallel sides. © Copyright 2011-2018 www.javatpoint.com. On this page, you can calculate area of a Trapezium. Duration: 1 week to 2 week. On the figure below we are given an isosceles trapezium where $\displaystyle DC=CF=5cm$ and $\displaystyle \widehat{{ABC}}={{45}^{\circ }}$. Problem 1: Using the adjacent angles property of trapezoid, find D if A = 125. 4.The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half the sum of the parallel sides. To calculate the area and perimeter of trapezium, here is the formula of area and perimeter of trapezium. With the trapezium, you’ll have two triangles. Firstly we build $\displaystyle CE\bot AB$ and $\displaystyle DF\bot AB$, On the triangles $\displaystyle ACE$ and $\displaystyle BDF$ we apply the. Area of a trapezium. One parallel side is two more than the other parallel side. Solution: If we observe our isosceles trapezium we se that $\displaystyle EFCD$ is a rectangle with $\displaystyle EF=10$, From that we calculate $\displaystyle AB-EF=22-10=12$, This value is to be shared equally for $\displaystyle AE$ and $\displaystyle QB$ because our two right angle triangles $\displaystyle AED$ and $\displaystyle BFC$ are congruent from rule 2 side-angle-side. The formula is written as: A = h (B + b) 2 The perimeter is the total lengths of … A trapezium, also known as a trapezoid, is a quadrilateral in which a pair of sides are parallel, but the other pair of opposite sides are non-parallel. Hence, the area of trapezium is. Please mail your requirement at hr@javatpoint.com. The area of a trapezium is 384 cm 2. So the height is $\displaystyle DE=CF=8$. Calculate the perimeter of the below-given trapezoid: If b1 and b2 are the lengths of corresponding parallel sides and s is the length of each non-parallel sides of an isosceles trapezoid, then its perimeter will be: For example: assume that the length of parallel sides of isosceles trapezoid is 12 and 10 units and the length of non-parallel sides is 5 units each. Sides with the same number of arrows are parallel. Trapezium is a type of quadrilateral that has at least one pair of side parallel to each other. If we observe our isosceles trapezium we se that $\displaystyle EFCD$ is a rectangle with $\displaystyle EF=10$, This value is to be shared equally for $\displaystyle AE$ and $\displaystyle QB$ because our two right angle triangles $\displaystyle AED$ and $\displaystyle BFC$ are congruent from rule 2 side-angle-side. Solution : Let 'a' and 'b' be the two parallel sides. It may have parallel legs. Distance between the parallel sides is ‘h’. Trapezium (noun) A region on the ventral side of the brain, either just back of the pons Varolii, or, as in man, covered by the posterior extension of its transverse fibers. Find the area. Therefore, the perimeter of a trapezium formula is given as The perimeter of Trapezium, P = a + b+ c + d units The parallel sides of the trapezoid can be vertical, horizontal, and slanting. Find the area. Then, a … Angle: The sum of anglesin a trapezoid-like other quadrilateral is 360°. Both the parts of the trapezium look like mirror images of each other. A Trapezium is a four-sided polygon with two non-adjacent parallel sides or one set of parallel sides. Where a, b, c, d are the sides of the trapezoid. Case 2: Find the perimeter of a trapezium using the sides length as 4,5,6,7 Examples: how to work out the area of a trapezium?. Area of a Trapezium formula = 1/2 * (a + b) * h, where aand bare the length of the parallel sides and his the distance between them Find the length of each one of the parallel sides. Its parallel sides are in the ratio 3:5 and the perpendicular distance between them is 12 cm. Solution: Given: Then, (1/2) (a + b) ⋅ h = 34. Find the height$\displaystyle DE$. These are the following: Where A is an area, b1 and b2 are the lengths of two parallel sides, and h is a perpendicular height of the trapezoid. The total of the two unknown sides of the triangles is the length of the hidden side. It is a quadrilateral wherein both pairs of opposite sides are parallel. In the below-mentioned trapezoid diagram, angles ∠A and ∠D are adjacent angles and supplementary. For getting to its answer one must know the formula of area for trapezium . A trapezium or a trapezoid is a quadrilateral with a pair of parallel sides. ( check. The addition of all four sides of a trapezoid is known as the perimeter of a trapezoid. If You Know the Height, Length of Top Base, and Bottom Interior Angles Divide the trapezoid into … It is also called a trapezoid. Then calculate its perimeter: Perimeter of isosceles trapezoid (P) = b1+ b2 + 2s. Area of trapezium = ½ x (a+b) x h Perimeter of trapezium = a+b+c+d Where a, b, c and d are the length of sides of a trapezium And h is the distance between the two parallel sides i.e., a and b. A line connecting non-parallel sides at mid-points is always parallel to the bases and half of the sum of parallel sides. Answer: ∠A and … It is the median times of height: The angles formed on the same side of a leg (line) are called adjacent angles, and these angles are supplementary. Hence, the area of a trapezium is given by the formula: Area of Trapezium = 1/2 x distance between the parallel sides x Sum of parallel sides Area = 1/2 x h x (AB + DC) Area Of Trapezium Examples Q1: The length of the two parallel sides of a trapezium are given in the ratio 3: 2 … The height of the trapezium is the perpendicular distance between the bases. Trapezium is a four sided shape where two sides are parallel,side lengths and angles are not equal. The line segmentthat connects the midpoints of the legs of a trapezoid is called the mid-segment. Now, s is the length of each non-parallel side, and h is the height of an isosceles trapezoid. Question 13. Also $\displaystyle DC$ and $\displaystyle EF$ are heights of our trapezium. Trapezoid (Jump to Area of a Trapezoid or Perimeter of a Trapezoid) . When the problem has a solution, the outputs are: the angles A, B, C and D, the height h, the area and the lengths of … Where a, b, c, d are the sides of the trapezoid. The area of a trapezium is computed with the following formula: The sides which are not parallel in a trapezium are not equal except in the Isosceles trapezium. Hence, the length of one parallel side of trapezium = 9 cm. Our online tools will provide quick answers to your calculation and conversion needs. In a trapezium the measurement of one parallel side two more than the other parallel side and the height is 4 cm. On the other hand, according to the US definition: a trapezium has no parallel sides, and according to the UK definition: a trapezium has a pair of parallel sides. Therefore the length of other parallel side of trapezium = x+8 = 9+8 = 17 cm. On the triangles $\displaystyle ACE$ and $\displaystyle BDF$ we apply the Pythagorean theorem: $\displaystyle {{\left( {CE} \right)}^{2}}={{\left( {AC} \right)}^{2}}-{{\left( {AE} \right)}^{2}}$, $\displaystyle {{\left( {CE} \right)}^{2}}=49-{{(x+3)}^{2}}$, $\displaystyle {{\left( {DF} \right)}^{2}}={{\left( {BD} \right)}^{2}}-{{\left( {BF} \right)}^{2}}$, $\displaystyle {{\left( {DF} \right)}^{2}}=64-{{(6-x)}^{2}}$, $\displaystyle 49-{{(x+3)}^{2}}=$$\displaystyle 64-{{(6-x)}^{2}}, \displaystyle 49-{{x}^{2}}-6x-9=$$\displaystyle 64-36+12x-{{x}^{2}}$, $\displaystyle AE=AF+FE=$$\displaystyle \frac{2}{3}+3=\frac{{11}}{3}, \displaystyle {{\left( {CE} \right)}^{2}}=$$\displaystyle 49-{{(\frac{{11}}{3}+3)}^{2}}$, $\displaystyle CE=\frac{{8\sqrt{5}}}{3}$, $\displaystyle {{A}_{{ABCD}}}=\frac{{(AB+DC)\cdot CE}}{2}$. Step 1: Find the area. The formula to calculate the perimeter of a trapezoid is given below: Perimeter of trapezoid ( P) = a + b + c + d units. Distance between parallel side (h) = 4 cm and a = 5. It is also sometimes called trapezium (UK). 1. The parallel sides are called bases and the non parallel sides are called legs, Isosceles Trapezium: “The legs or the not parallel sides are equal.”, Scalene Trapezium: “A trapezium with all the sides and angles of different measures’’, Right Trapezium: “ A right trapezium has at least two right angles”. JavaTpoint offers too many high quality services. All rights reserved. According to adjacent angles property of trapezoid A + D = 180. New questions in Mathematics. When the parallel sides make the two equal angles or when the two non-parallel sides are equal, it is called isosceles trapezoid. Case 1: Find the area of a trapezium using the sides length as 4,5 and the height 2.. Find the area of the trapezium. Trapezoid and trapezium are the swapped definition of the US and UK. Find the area of the trapezium with diagonals $\displaystyle 8cm$ and $\displaystyle 7cm$ and bases $\displaystyle 6cm$ and $\displaystyle 3cm$. 83=24. The area of the trapezium is equal to half the product of the sum of the bases with height. Example 1: On the figure below we have given the sides of trapezium. A parallelogram may also be called a trapezoid as it has two parallel sides. Therefore, use the given information to apply the formula: Perimeter= Base one Base two (leg), where the length of "leg" is one of the two equivalent nonparallel sides. The parallel sides on are called the bases. The parallel sides of a trapezium are called bases and the non-parallel sides of a trapezium are called legs. Then, (1/2) (5 + b) ⋅ 4 = 34. JavaTpoint offers college campus training on Core Java, Advance Java, .Net, Android, Hadoop, PHP, Web Technology and Python. A scalene trapezoid is a trapezoid with no sides of equal measure, in contrast to the special cases below. A trapezium is a quadrilateral having two parallel sides of unequal length and the other two sides are non-parallel. Solve each problem. A trapezoid is a flat four-sided 2D closed shape with a pair of parallel sides (opposite sides). Also, the lengths of the opposite sides are equal. A r e a = 1 2 × A E × D E + D E × E F + 1 2 × F B × C F. = a h 2 + b 1 h b h 2. The angles formed on the same side of a leg are called a, If all the opposite sides are parallel in trapezoid is called a, If all the opposite sides are parallel, all its sides are equal in length and form a right angle at each point called a, If all the opposite sides are parallel, their opposite sides are only equal in length and form a right angle at each point called a. The lengths of its parallel sides are AB = aand OC = band its height is h. The coordinates of the centroid of the trapezium are given by the following formula. According to the trapezoid area formula, the area of a trapezoid is equal to half the product of the height and the sum of the two bases. ( check congruent triangles rules), To find the height $\displaystyle DE$ we use the the Pythagorean theorem on triangle $\displaystyle AED$, $\displaystyle {{\left( {DE} \right)}^{2}}={{\left( {AD} \right)}^{2}}-{{\left( {AE} \right)}^{2}}$, $\displaystyle {{\left( {DE} \right)}^{2}}={{\left( {10} \right)}^{2}}-{{\left( 6 \right)}^{2}}$, $\displaystyle {{\left( {DE} \right)}^{2}}=100-36$, $\displaystyle {{\left( {DE} \right)}^{2}}=64$. The formula to calculate the area of an isosceles trapezoid is, Area of Isosceles Trapezoid = h $$\frac{(a + b)}{2}$$ Parallelogram. I've added another reference to counter the claim of our anon editor. Area = area of triangle 1 + area of rectangle + area of triangle 2. Formula of Area of a trapezium = ¹/₂ × (sum of parallel sides) × (distance between them) Solved Examples of Area of a Trapezium 1. Area and perimeter of a trapezium can be found using the below formula, Perimeter = sum of all sides. I/2(a +b)h (here a & b are the parallel sides of the trapezium) 1/2(5+3)6 . The two other sides of trapezium area known as the legs of trapezium and those are not parallel each other. Where: When we draw a perpendicular line (h) from AB to meet CD at E, it makes a right angle at AED and AEC. The perpendicular distance between two parallel sides is called its altitude. Two parallel sides of a trapezium are of lengths 27 cm and 19 cm respectively, and the distance between them is 14 cm. The parallel side of a trapezoid are called the bases, and the non-parallel sides are legs. The parallel sides are called the bases of the trapezoid and the other two sides are called the legs or the lateral sides. Area = ½ x (Sum of parallel sides) x (perpendicular distance between the parallel sides). What are the properties of Trapezium? In the same way, ∠B and ∠C are supplementary. The sides which are not parallel in a trapezium are not equal except in the Isosceles trapezium. Example 3: Find the area of the trapezium with diagonals $\displaystyle 8cm$ and $\displaystyle 7cm$ and bases $\displaystyle 6cm$ and $\displaystyle 3cm$. Now, put all the given values in this formula, and we get, Area of trapezium = 1/2 (15 + 9) × 8 = 1/2 × 192 = 96. In a trapezium at least two opposite sides are parallel. The perimeter of a trapezium is found by adding all the sides. 2017 Vw Tiguan Headlight Bulb Size, Roof Tile Adhesive Home Depot, Heritage Bedroom Furniture, 2017 Buick Enclave Problems, Implied Trust Example, 10th Gen Civic Invidia Exhaust, Sortout Meaning In Urdu, Makaton Sign For Rabbit, Homeware Clearance Sale Uk, First Horizon Credit Score, LiknandeHemmaSnart är det dags att fira pappa!Om vårt kaffeSmå projektTemakvällar på caféetRecepttips!" /> b). Find the lengths of the two parallel sides. We've "always" used the inclusive definition on this side of the pond (except we call it a trapezium, of course). The area of a trapezium is given by $${A}=\frac{(a+b)}{2}\times{h}$$.. You can see that this is true by taking two identical trapezia (or trapeziums) to make a parallelogram. According to US definition: a trapezoid has a pair of parallel sides, and according to UK definition: a trapezoid has no parallel sides. © Copyright 2011-2018 www.javatpoint.com. On this page, you can calculate area of a Trapezium. Duration: 1 week to 2 week. On the figure below we are given an isosceles trapezium where $\displaystyle DC=CF=5cm$ and $\displaystyle \widehat{{ABC}}={{45}^{\circ }}$. Problem 1: Using the adjacent angles property of trapezoid, find D if A = 125. 4.The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half the sum of the parallel sides. To calculate the area and perimeter of trapezium, here is the formula of area and perimeter of trapezium. With the trapezium, you’ll have two triangles. Firstly we build $\displaystyle CE\bot AB$ and $\displaystyle DF\bot AB$, On the triangles $\displaystyle ACE$ and $\displaystyle BDF$ we apply the. Area of a trapezium. One parallel side is two more than the other parallel side. Solution: If we observe our isosceles trapezium we se that $\displaystyle EFCD$ is a rectangle with $\displaystyle EF=10$, From that we calculate $\displaystyle AB-EF=22-10=12$, This value is to be shared equally for $\displaystyle AE$ and $\displaystyle QB$ because our two right angle triangles $\displaystyle AED$ and $\displaystyle BFC$ are congruent from rule 2 side-angle-side. The formula is written as: A = h (B + b) 2 The perimeter is the total lengths of … A trapezium, also known as a trapezoid, is a quadrilateral in which a pair of sides are parallel, but the other pair of opposite sides are non-parallel. Hence, the area of trapezium is. Please mail your requirement at hr@javatpoint.com. The area of a trapezium is 384 cm 2. So the height is $\displaystyle DE=CF=8$. Calculate the perimeter of the below-given trapezoid: If b1 and b2 are the lengths of corresponding parallel sides and s is the length of each non-parallel sides of an isosceles trapezoid, then its perimeter will be: For example: assume that the length of parallel sides of isosceles trapezoid is 12 and 10 units and the length of non-parallel sides is 5 units each. Sides with the same number of arrows are parallel. Trapezium is a type of quadrilateral that has at least one pair of side parallel to each other. If we observe our isosceles trapezium we se that $\displaystyle EFCD$ is a rectangle with $\displaystyle EF=10$, This value is to be shared equally for $\displaystyle AE$ and $\displaystyle QB$ because our two right angle triangles $\displaystyle AED$ and $\displaystyle BFC$ are congruent from rule 2 side-angle-side. Solution : Let 'a' and 'b' be the two parallel sides. It may have parallel legs. Distance between the parallel sides is ‘h’. Trapezium (noun) A region on the ventral side of the brain, either just back of the pons Varolii, or, as in man, covered by the posterior extension of its transverse fibers. Find the area. Therefore, the perimeter of a trapezium formula is given as The perimeter of Trapezium, P = a + b+ c + d units The parallel sides of the trapezoid can be vertical, horizontal, and slanting. Find the area. Then, a … Angle: The sum of anglesin a trapezoid-like other quadrilateral is 360°. Both the parts of the trapezium look like mirror images of each other. A Trapezium is a four-sided polygon with two non-adjacent parallel sides or one set of parallel sides. Where a, b, c, d are the sides of the trapezoid. Case 2: Find the perimeter of a trapezium using the sides length as 4,5,6,7 Examples: how to work out the area of a trapezium?. Area of a Trapezium formula = 1/2 (a + b) * h, where aand bare the length of the parallel sides and his the distance between them Find the length of each one of the parallel sides. Its parallel sides are in the ratio 3:5 and the perpendicular distance between them is 12 cm. Solution: Given: Then, (1/2) (a + b) ⋅ h = 34. Find the height$\displaystyle DE$. These are the following: Where A is an area, b1 and b2 are the lengths of two parallel sides, and h is a perpendicular height of the trapezoid. The total of the two unknown sides of the triangles is the length of the hidden side. It is a quadrilateral wherein both pairs of opposite sides are parallel. In the below-mentioned trapezoid diagram, angles ∠A and ∠D are adjacent angles and supplementary. For getting to its answer one must know the formula of area for trapezium . A trapezium or a trapezoid is a quadrilateral with a pair of parallel sides. ( check. The addition of all four sides of a trapezoid is known as the perimeter of a trapezoid. If You Know the Height, Length of Top Base, and Bottom Interior Angles Divide the trapezoid into … It is also called a trapezoid. Then calculate its perimeter: Perimeter of isosceles trapezoid (P) = b1+ b2 + 2s. Area of trapezium = ½ x (a+b) x h Perimeter of trapezium = a+b+c+d Where a, b, c and d are the length of sides of a trapezium And h is the distance between the two parallel sides i.e., a and b. A line connecting non-parallel sides at mid-points is always parallel to the bases and half of the sum of parallel sides. Answer: ∠A and … It is the median times of height: The angles formed on the same side of a leg (line) are called adjacent angles, and these angles are supplementary. Hence, the area of a trapezium is given by the formula: Area of Trapezium = 1/2 x distance between the parallel sides x Sum of parallel sides Area = 1/2 x h x (AB + DC) Area Of Trapezium Examples Q1: The length of the two parallel sides of a trapezium are given in the ratio 3: 2 … The height of the trapezium is the perpendicular distance between the bases. Trapezium is a four sided shape where two sides are parallel,side lengths and angles are not equal. The line segmentthat connects the midpoints of the legs of a trapezoid is called the mid-segment. Now, s is the length of each non-parallel side, and h is the height of an isosceles trapezoid. Question 13. Also $\displaystyle DC$ and $\displaystyle EF$ are heights of our trapezium. Trapezoid (Jump to Area of a Trapezoid or Perimeter of a Trapezoid) . When the problem has a solution, the outputs are: the angles A, B, C and D, the height h, the area and the lengths of … Where a, b, c, d are the sides of the trapezoid. The area of a trapezium is computed with the following formula: The sides which are not parallel in a trapezium are not equal except in the Isosceles trapezium. Hence, the length of one parallel side of trapezium = 9 cm. Our online tools will provide quick answers to your calculation and conversion needs. In a trapezium the measurement of one parallel side two more than the other parallel side and the height is 4 cm. On the other hand, according to the US definition: a trapezium has no parallel sides, and according to the UK definition: a trapezium has a pair of parallel sides. Therefore the length of other parallel side of trapezium = x+8 = 9+8 = 17 cm. On the triangles $\displaystyle ACE$ and $\displaystyle BDF$ we apply the Pythagorean theorem: $\displaystyle {{\left( {CE} \right)}^{2}}={{\left( {AC} \right)}^{2}}-{{\left( {AE} \right)}^{2}}$, $\displaystyle {{\left( {CE} \right)}^{2}}=49-{{(x+3)}^{2}}$, $\displaystyle {{\left( {DF} \right)}^{2}}={{\left( {BD} \right)}^{2}}-{{\left( {BF} \right)}^{2}}$, $\displaystyle {{\left( {DF} \right)}^{2}}=64-{{(6-x)}^{2}}$, $\displaystyle 49-{{(x+3)}^{2}}=$$\displaystyle 64-{{(6-x)}^{2}}, \displaystyle 49-{{x}^{2}}-6x-9=$$\displaystyle 64-36+12x-{{x}^{2}}$, $\displaystyle AE=AF+FE=$$\displaystyle \frac{2}{3}+3=\frac{{11}}{3}, \displaystyle {{\left( {CE} \right)}^{2}}=$$\displaystyle 49-{{(\frac{{11}}{3}+3)}^{2}}$, $\displaystyle CE=\frac{{8\sqrt{5}}}{3}$, $\displaystyle {{A}_{{ABCD}}}=\frac{{(AB+DC)\cdot CE}}{2}$. Step 1: Find the area. The formula to calculate the perimeter of a trapezoid is given below: Perimeter of trapezoid ( P) = a + b + c + d units. Distance between parallel side (h) = 4 cm and a = 5. It is also sometimes called trapezium (UK). 1. The parallel sides are called bases and the non parallel sides are called legs, Isosceles Trapezium: “The legs or the not parallel sides are equal.”, Scalene Trapezium: “A trapezium with all the sides and angles of different measures’’, Right Trapezium: “ A right trapezium has at least two right angles”. JavaTpoint offers too many high quality services. All rights reserved. According to adjacent angles property of trapezoid A + D = 180. New questions in Mathematics. When the parallel sides make the two equal angles or when the two non-parallel sides are equal, it is called isosceles trapezoid. Case 1: Find the area of a trapezium using the sides length as 4,5 and the height 2.. Find the area of the trapezium. Trapezoid and trapezium are the swapped definition of the US and UK. Find the area of the trapezium with diagonals $\displaystyle 8cm$ and $\displaystyle 7cm$ and bases $\displaystyle 6cm$ and $\displaystyle 3cm$. 83=24. The area of the trapezium is equal to half the product of the sum of the bases with height. Example 1: On the figure below we have given the sides of trapezium. A parallelogram may also be called a trapezoid as it has two parallel sides. Therefore, use the given information to apply the formula: Perimeter= Base one Base two (leg), where the length of "leg" is one of the two equivalent nonparallel sides. The parallel sides on are called the bases. The parallel sides of a trapezium are called bases and the non-parallel sides of a trapezium are called legs. Then, (1/2) (5 + b) ⋅ 4 = 34. JavaTpoint offers college campus training on Core Java, Advance Java, .Net, Android, Hadoop, PHP, Web Technology and Python. A scalene trapezoid is a trapezoid with no sides of equal measure, in contrast to the special cases below. A trapezium is a quadrilateral having two parallel sides of unequal length and the other two sides are non-parallel. Solve each problem. A trapezoid is a flat four-sided 2D closed shape with a pair of parallel sides (opposite sides). Also, the lengths of the opposite sides are equal. A r e a = 1 2 × A E × D E + D E × E F + 1 2 × F B × C F. = a h 2 + b 1 h b h 2. The angles formed on the same side of a leg are called a, If all the opposite sides are parallel in trapezoid is called a, If all the opposite sides are parallel, all its sides are equal in length and form a right angle at each point called a, If all the opposite sides are parallel, their opposite sides are only equal in length and form a right angle at each point called a. The lengths of its parallel sides are AB = aand OC = band its height is h. The coordinates of the centroid of the trapezium are given by the following formula. According to the trapezoid area formula, the area of a trapezoid is equal to half the product of the height and the sum of the two bases. ( check congruent triangles rules), To find the height $\displaystyle DE$ we use the the Pythagorean theorem on triangle $\displaystyle AED$, $\displaystyle {{\left( {DE} \right)}^{2}}={{\left( {AD} \right)}^{2}}-{{\left( {AE} \right)}^{2}}$, $\displaystyle {{\left( {DE} \right)}^{2}}={{\left( {10} \right)}^{2}}-{{\left( 6 \right)}^{2}}$, $\displaystyle {{\left( {DE} \right)}^{2}}=100-36$, $\displaystyle {{\left( {DE} \right)}^{2}}=64$. The formula to calculate the area of an isosceles trapezoid is, Area of Isosceles Trapezoid = h $$\frac{(a + b)}{2}$$ Parallelogram. I've added another reference to counter the claim of our anon editor. Area = area of triangle 1 + area of rectangle + area of triangle 2. Formula of Area of a trapezium = ¹/₂ × (sum of parallel sides) × (distance between them) Solved Examples of Area of a Trapezium 1. Area and perimeter of a trapezium can be found using the below formula, Perimeter = sum of all sides. I/2(a +b)h (here a & b are the parallel sides of the trapezium) 1/2(5+3)6 . The two other sides of trapezium area known as the legs of trapezium and those are not parallel each other. Where: When we draw a perpendicular line (h) from AB to meet CD at E, it makes a right angle at AED and AEC. The perpendicular distance between two parallel sides is called its altitude. Two parallel sides of a trapezium are of lengths 27 cm and 19 cm respectively, and the distance between them is 14 cm. The parallel side of a trapezoid are called the bases, and the non-parallel sides are legs. The parallel sides are called the bases of the trapezoid and the other two sides are called the legs or the lateral sides. Area = ½ x (Sum of parallel sides) x (perpendicular distance between the parallel sides). What are the properties of Trapezium? In the same way, ∠B and ∠C are supplementary. The sides which are not parallel in a trapezium are not equal except in the Isosceles trapezium. Example 3: Find the area of the trapezium with diagonals $\displaystyle 8cm$ and $\displaystyle 7cm$ and bases $\displaystyle 6cm$ and $\displaystyle 3cm$. Now, put all the given values in this formula, and we get, Area of trapezium = 1/2 (15 + 9) × 8 = 1/2 × 192 = 96. In a trapezium at least two opposite sides are parallel. The perimeter of a trapezium is found by adding all the sides. 2017 Vw Tiguan Headlight Bulb Size, Roof Tile Adhesive Home Depot, Heritage Bedroom Furniture, 2017 Buick Enclave Problems, Implied Trust Example, 10th Gen Civic Invidia Exhaust, Sortout Meaning In Urdu, Makaton Sign For Rabbit, Homeware Clearance Sale Uk, First Horizon Credit Score, LiknandeHemmaSnart är det dags att fira pappa!Om vårt kaffeSmå projektTemakvällar på caféetRecepttips!" /> The formula to calculate the perimeter of a trapezoid is given below: Perimeter of trapezoid (P) = a + b + c + d units. When the sum of two angles became, 180 degrees is called supplementary. Area = ½ h (b1 + b2) Where, h is the height … 3. A trapezium is a quadrilateral which has exactly one pair of parallel sides. So in a trapezoid ABCD, ∠A+∠B+∠C+∠D = … Solution: Firstly we build $\displaystyle CE\bot AB$ and $\displaystyle DF\bot AB$. Dbfirs 03:48, … A trapezoid's two opposite sides (one pair) are parallel. Hence, the area of trapezium = 96m 2. The area of the trapezium is 64 c m 2. The letters a and b represent the two other sides, and c is the hypotenuse. The perimeter is the total lengths of all sides, the sum of the two bases and the two other sides. $G\left({\frac{h}{2},\,\frac{{b + 2a}}{{3\left({a + b} \right)}}h} \right)$ Let’s look at an example to see how to use this formula. STEP 4: So, to find x , we substitute a with x … Use the Pythagorean formula to find the unknown sides. Suppose that b1 and b2 are the lengths of parallel sides of trapezoid ABCD, such as b1 is b2 is the length of the opposite parallel to b1. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half the sum of the parallel sides. Therefore the area of the trapezium is 24 square cm. We can calculate the trapezoid area if we know the length of the trapezoid's median and height. Every trapezium shows the following properties: 1. $\displaystyle A=\frac{h\left( B+b \right)}{2}$. It is also called as midline or midsegment of a trapezoid. A trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides parallel (marked with arrows below): Let 'a' be the length of the parallel side given and 'b' be the length of the missing parallel side. Parallel sides on a polygon are indicated using arrows. Mail us on hr@javatpoint.com, to get more information about given services. Area = ½ (a+ b) * h = ½ * (4 + 5) * 2 = ½ * 9 * 2 = 9. The trapezoid is categorized into three different types. The pair of parallel sides is called the base while the non-parallel sides are called the legs of the trapezoid. Area = ½ x (sum of the lengths of the parallel sides) x perpendicular distance between parallel sides Perimeter of a trapezoid (trapezium) The addition of all four sides of a trapezoid is known as the perimeter of a trapezoid. The one that is longer is called the big base ( B) and the other the small base (b) of the trapezium. The parallel sides are called as the bases of trapezium. Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude. Property #1) The angles on the same side of a leg are called adjacent angles and are supplementary ( more ) Property #2) Area of a Trapezoid = A r e a = h e i g h t ⋅ ( sum bases 2) ( more ) Property #3) Trapezoids have a midsegment which connects the mipoints of the legs ( more ) I wrote "always", but then remembered that Euclid had a different meaning. Question 3. The perpendicular will be donated as the height ‘h’ which is the distance between the parallel sides. Developed by JavaTpoint. The area of the trapezium is equal to half the product of the sum of the bases with height. To do so, we start with the formula for the area of a trapezium: Where a and b are the parallel sides of the trapezium and h is its height. Given : Area of a trapezium is 34 square cm. Solution: Firstly we observe our figure and see that $\displaystyle DEFC$ is a square since $\displaystyle DC=EF$. Trapezium (noun) The trapezium bone of the wrist. A median is a line that connects non-parallel sides at mid-points is always parallel to the bases and half of the sum of parallel sides. Example 2: On the figure below we are given an isosceles trapezium where $\displaystyle DC=CF=5cm$ and $\displaystyle \widehat{{ABC}}={{45}^{\circ }}$. From the figure, it can be seen that there are two triangles and one rectangle. Look at the below figure of a trapezoid with the unit of length 3, 10, 11, 8, which has 7 units of perpendicular height. The right angle triangle $\displaystyle BFC$ is an isosceles right angle triangle because based on the property that all angles on a triangle add up to $\displaystyle {{180}^{\circ }}$ we find that $\displaystyle \widehat{{FCB}}={{180}^{\circ }}-({{45}^{\circ }}+{{90}^{\circ }})={{45}^{\circ }}$, Since our right angle triangle is an isosceles triangle then $\displaystyle CF=FB=5cm$, Based on the same reasoning also $\displaystyle DE=EA=5cm$, Now we find the big base: $\displaystyle AB=AE+EF+FB=$$\displaystyle 5+5+5=15cm, The area is: \displaystyle A=\frac{{\left( {B+b} \right)\cdot h}}{2}, \displaystyle A=\frac{{\left( {15+5} \right)\cdot 5}}{2}=50c{{m}^{2}}. In a trapezium at least two opposite sides are parallel. (5 + b) ⋅ 2 = 34. A four-sided polygon with no parallel sides and no sides equal; a simple convex irregular quadrilateral. If you get stuck, consider using the double number lines. Firstly we observe our figure and see that \displaystyle DEFC is a square since \displaystyle DC=EF, \displaystyle A=\frac{{\left( {B+b} \right)\cdot h}}{2}. Use the formula. Cancel 6 and 2 . In Euclidean geometry, a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezium in English outside North America, but as a trapezoid in American and Canadian English. In a trapezium ABCD, if AB\|\|CD, then which pair of angles are supplementary? Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real numbers and press "calculate" with b being the short base and d being the long base (d > b). Find the lengths of the two parallel sides. We've "always" used the inclusive definition on this side of the pond (except we call it a trapezium, of course). The area of a trapezium is given by $${A}=\frac{(a+b)}{2}\times{h}$$.. You can see that this is true by taking two identical trapezia (or trapeziums) to make a parallelogram. According to US definition: a trapezoid has a pair of parallel sides, and according to UK definition: a trapezoid has no parallel sides. © Copyright 2011-2018 www.javatpoint.com. On this page, you can calculate area of a Trapezium. Duration: 1 week to 2 week. On the figure below we are given an isosceles trapezium where \displaystyle DC=CF=5cm and \displaystyle \widehat{{ABC}}={{45}^{\circ }}. Problem 1: Using the adjacent angles property of trapezoid, find D if A = 125. 4.The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half the sum of the parallel sides. To calculate the area and perimeter of trapezium, here is the formula of area and perimeter of trapezium. With the trapezium, you’ll have two triangles. Firstly we build \displaystyle CE\bot AB and \displaystyle DF\bot AB, On the triangles \displaystyle ACE and \displaystyle BDF we apply the. Area of a trapezium. One parallel side is two more than the other parallel side. Solution: If we observe our isosceles trapezium we se that \displaystyle EFCD is a rectangle with \displaystyle EF=10, From that we calculate \displaystyle AB-EF=22-10=12, This value is to be shared equally for \displaystyle AE and \displaystyle QB because our two right angle triangles \displaystyle AED and \displaystyle BFC are congruent from rule 2 side-angle-side. The formula is written as: A = h (B + b) 2 The perimeter is the total lengths of … A trapezium, also known as a trapezoid, is a quadrilateral in which a pair of sides are parallel, but the other pair of opposite sides are non-parallel. Hence, the area of trapezium is. Please mail your requirement at hr@javatpoint.com. The area of a trapezium is 384 cm 2. So the height is \displaystyle DE=CF=8. Calculate the perimeter of the below-given trapezoid: If b1 and b2 are the lengths of corresponding parallel sides and s is the length of each non-parallel sides of an isosceles trapezoid, then its perimeter will be: For example: assume that the length of parallel sides of isosceles trapezoid is 12 and 10 units and the length of non-parallel sides is 5 units each. Sides with the same number of arrows are parallel. Trapezium is a type of quadrilateral that has at least one pair of side parallel to each other. If we observe our isosceles trapezium we se that \displaystyle EFCD is a rectangle with \displaystyle EF=10, This value is to be shared equally for \displaystyle AE and \displaystyle QB because our two right angle triangles \displaystyle AED and \displaystyle BFC are congruent from rule 2 side-angle-side. Solution : Let 'a' and 'b' be the two parallel sides. It may have parallel legs. Distance between the parallel sides is ‘h’. Trapezium (noun) A region on the ventral side of the brain, either just back of the pons Varolii, or, as in man, covered by the posterior extension of its transverse fibers. Find the area. Therefore, the perimeter of a trapezium formula is given as The perimeter of Trapezium, P = a + b+ c + d units The parallel sides of the trapezoid can be vertical, horizontal, and slanting. Find the area. Then, a … Angle: The sum of anglesin a trapezoid-like other quadrilateral is 360°. Both the parts of the trapezium look like mirror images of each other. A Trapezium is a four-sided polygon with two non-adjacent parallel sides or one set of parallel sides. Where a, b, c, d are the sides of the trapezoid. Case 2: Find the perimeter of a trapezium using the sides length as 4,5,6,7 Examples: how to work out the area of a trapezium?. Area of a Trapezium formula = 1/2 * (a + b) * h, where aand bare the length of the parallel sides and his the distance between them Find the length of each one of the parallel sides. Its parallel sides are in the ratio 3:5 and the perpendicular distance between them is 12 cm. Solution: Given: Then, (1/2) (a + b) ⋅ h = 34. Find the height \displaystyle DE. These are the following: Where A is an area, b1 and b2 are the lengths of two parallel sides, and h is a perpendicular height of the trapezoid. The total of the two unknown sides of the triangles is the length of the hidden side. It is a quadrilateral wherein both pairs of opposite sides are parallel. In the below-mentioned trapezoid diagram, angles ∠A and ∠D are adjacent angles and supplementary. For getting to its answer one must know the formula of area for trapezium . A trapezium or a trapezoid is a quadrilateral with a pair of parallel sides. ( check. The addition of all four sides of a trapezoid is known as the perimeter of a trapezoid. If You Know the Height, Length of Top Base, and Bottom Interior Angles Divide the trapezoid into … It is also called a trapezoid. Then calculate its perimeter: Perimeter of isosceles trapezoid (P) = b1+ b2 + 2s. Area of trapezium = ½ x (a+b) x h Perimeter of trapezium = a+b+c+d Where a, b, c and d are the length of sides of a trapezium And h is the distance between the two parallel sides i.e., a and b. A line connecting non-parallel sides at mid-points is always parallel to the bases and half of the sum of parallel sides. Answer: ∠A and … It is the median times of height: The angles formed on the same side of a leg (line) are called adjacent angles, and these angles are supplementary. Hence, the area of a trapezium is given by the formula: Area of Trapezium = 1/2 x distance between the parallel sides x Sum of parallel sides Area = 1/2 x h x (AB + DC) Area Of Trapezium Examples Q1: The length of the two parallel sides of a trapezium are given in the ratio 3: 2 … The height of the trapezium is the perpendicular distance between the bases. Trapezium is a four sided shape where two sides are parallel,side lengths and angles are not equal. The line segmentthat connects the midpoints of the legs of a trapezoid is called the mid-segment. Now, s is the length of each non-parallel side, and h is the height of an isosceles trapezoid. Question 13. Also \displaystyle DC and \displaystyle EF are heights of our trapezium. Trapezoid (Jump to Area of a Trapezoid or Perimeter of a Trapezoid) . When the problem has a solution, the outputs are: the angles A, B, C and D, the height h, the area and the lengths of … Where a, b, c, d are the sides of the trapezoid. The area of a trapezium is computed with the following formula: The sides which are not parallel in a trapezium are not equal except in the Isosceles trapezium. Hence, the length of one parallel side of trapezium = 9 cm. Our online tools will provide quick answers to your calculation and conversion needs. In a trapezium the measurement of one parallel side two more than the other parallel side and the height is 4 cm. On the other hand, according to the US definition: a trapezium has no parallel sides, and according to the UK definition: a trapezium has a pair of parallel sides. Therefore the length of other parallel side of trapezium = x+8 = 9+8 = 17 cm. On the triangles \displaystyle ACE and \displaystyle BDF we apply the Pythagorean theorem: \displaystyle {{\left( {CE} \right)}^{2}}={{\left( {AC} \right)}^{2}}-{{\left( {AE} \right)}^{2}}, \displaystyle {{\left( {CE} \right)}^{2}}=49-{{(x+3)}^{2}}, \displaystyle {{\left( {DF} \right)}^{2}}={{\left( {BD} \right)}^{2}}-{{\left( {BF} \right)}^{2}}, \displaystyle {{\left( {DF} \right)}^{2}}=64-{{(6-x)}^{2}}, \displaystyle 49-{{(x+3)}^{2}}=$$ \displaystyle 64-{{(6-x)}^{2}}$, $\displaystyle 49-{{x}^{2}}-6x-9=$$\displaystyle 64-36+12x-{{x}^{2}}, \displaystyle AE=AF+FE=$$\displaystyle \frac{2}{3}+3=\frac{{11}}{3}$, $\displaystyle {{\left( {CE} \right)}^{2}}=$$\displaystyle 49-{{(\frac{{11}}{3}+3)}^{2}}$, $\displaystyle CE=\frac{{8\sqrt{5}}}{3}$, $\displaystyle {{A}_{{ABCD}}}=\frac{{(AB+DC)\cdot CE}}{2}$. Step 1: Find the area. The formula to calculate the perimeter of a trapezoid is given below: Perimeter of trapezoid ( P) = a + b + c + d units. Distance between parallel side (h) = 4 cm and a = 5. It is also sometimes called trapezium (UK). 1. The parallel sides are called bases and the non parallel sides are called legs, Isosceles Trapezium: “The legs or the not parallel sides are equal.”, Scalene Trapezium: “A trapezium with all the sides and angles of different measures’’, Right Trapezium: “ A right trapezium has at least two right angles”. JavaTpoint offers too many high quality services. All rights reserved. According to adjacent angles property of trapezoid A + D = 180. New questions in Mathematics. When the parallel sides make the two equal angles or when the two non-parallel sides are equal, it is called isosceles trapezoid. Case 1: Find the area of a trapezium using the sides length as 4,5 and the height 2.. Find the area of the trapezium. Trapezoid and trapezium are the swapped definition of the US and UK. Find the area of the trapezium with diagonals $\displaystyle 8cm$ and $\displaystyle 7cm$ and bases $\displaystyle 6cm$ and $\displaystyle 3cm$. 83=24. The area of the trapezium is equal to half the product of the sum of the bases with height. Example 1: On the figure below we have given the sides of trapezium. A parallelogram may also be called a trapezoid as it has two parallel sides. Therefore, use the given information to apply the formula: Perimeter= Base one Base two (leg), where the length of "leg" is one of the two equivalent nonparallel sides. The parallel sides on are called the bases. The parallel sides of a trapezium are called bases and the non-parallel sides of a trapezium are called legs. Then, (1/2) (5 + b) ⋅ 4 = 34. JavaTpoint offers college campus training on Core Java, Advance Java, .Net, Android, Hadoop, PHP, Web Technology and Python. A scalene trapezoid is a trapezoid with no sides of equal measure, in contrast to the special cases below. A trapezium is a quadrilateral having two parallel sides of unequal length and the other two sides are non-parallel. Solve each problem. A trapezoid is a flat four-sided 2D closed shape with a pair of parallel sides (opposite sides). Also, the lengths of the opposite sides are equal. A r e a = 1 2 × A E × D E + D E × E F + 1 2 × F B × C F. = a h 2 + b 1 h b h 2. The angles formed on the same side of a leg are called a, If all the opposite sides are parallel in trapezoid is called a, If all the opposite sides are parallel, all its sides are equal in length and form a right angle at each point called a, If all the opposite sides are parallel, their opposite sides are only equal in length and form a right angle at each point called a. The lengths of its parallel sides are AB = aand OC = band its height is h. The coordinates of the centroid of the trapezium are given by the following formula. According to the trapezoid area formula, the area of a trapezoid is equal to half the product of the height and the sum of the two bases. ( check congruent triangles rules), To find the height $\displaystyle DE$ we use the the Pythagorean theorem on triangle $\displaystyle AED$, $\displaystyle {{\left( {DE} \right)}^{2}}={{\left( {AD} \right)}^{2}}-{{\left( {AE} \right)}^{2}}$, $\displaystyle {{\left( {DE} \right)}^{2}}={{\left( {10} \right)}^{2}}-{{\left( 6 \right)}^{2}}$, $\displaystyle {{\left( {DE} \right)}^{2}}=100-36$, $\displaystyle {{\left( {DE} \right)}^{2}}=64$. The formula to calculate the area of an isosceles trapezoid is, Area of Isosceles Trapezoid = h $$\frac{(a + b)}{2}$$ Parallelogram. I've added another reference to counter the claim of our anon editor. Area = area of triangle 1 + area of rectangle + area of triangle 2. Formula of Area of a trapezium = ¹/₂ × (sum of parallel sides) × (distance between them) Solved Examples of Area of a Trapezium 1. Area and perimeter of a trapezium can be found using the below formula, Perimeter = sum of all sides. I/2(a +b)h (here a & b are the parallel sides of the trapezium) 1/2(5+3)*6 . The two other sides of trapezium area known as the legs of trapezium and those are not parallel each other. Where: When we draw a perpendicular line (h) from AB to meet CD at E, it makes a right angle at AED and AEC. The perpendicular distance between two parallel sides is called its altitude. Two parallel sides of a trapezium are of lengths 27 cm and 19 cm respectively, and the distance between them is 14 cm. The parallel side of a trapezoid are called the bases, and the non-parallel sides are legs. The parallel sides are called the bases of the trapezoid and the other two sides are called the legs or the lateral sides. Area = ½ x (Sum of parallel sides) x (perpendicular distance between the parallel sides). What are the properties of Trapezium? In the same way, ∠B and ∠C are supplementary. The sides which are not parallel in a trapezium are not equal except in the Isosceles trapezium. Example 3: Find the area of the trapezium with diagonals $\displaystyle 8cm$ and $\displaystyle 7cm$ and bases $\displaystyle 6cm$ and $\displaystyle 3cm$. Now, put all the given values in this formula, and we get, Area of trapezium = 1/2 (15 + 9) × 8 = 1/2 × 192 = 96. In a trapezium at least two opposite sides are parallel. The perimeter of a trapezium is found by adding all the sides.
# Sin 2A Formula – Proof and Examples Here you will learn what is the formula of sin 2A in terms of sin and cos and also in terms of tan with proof and examples. Let’s begin – ## Sin 2A Formula ### (i) In Terms of Cos and Sin : Sin 2A = 2 sin A cos A Proof : We have, Sin (A + B) = sin A cos B + cos A sin B Replacing B by A, $$\implies$$ sin 2A = sin A cos A + cos A sin A $$\implies$$ sin 2A = 2 sin A cos A We can also write above relation in terms of angle A/2, just replace A by A/2, we get sin A = $$2 sin ({A\over 2}) cos ({A\over 2})$$ ### (ii) Sin 2A Formula in Terms of Tan : Sin 2A = $$2 tan A\over 1 + tan^2 A$$ Proof : We have, sin 2A = 2 sin A cos A $$\implies$$ sin 2A = $$2 sin A cos A\over sin^2 A + cos^2 A$$ [ $$\because$$ $$sin^2 A + cos^2 A$$ = 1 ] Now, Dividing numerator and denominator by $$cos^2 A$$, $$\implies$$ sin 2A = $${2sin A cos A\over cos^2 A}\over {sin^2 A + cos^2 A\over cos^2 A}$$ $$\implies$$ sin 2A = $$2 tan A\over 1 + tan^2 A$$ We can also write above relation in terms of angle A/2, just replace A by A/2, we get sin A = $$2 tan ({A\over 2})\over 1 + tan^2 ({A\over 2})$$ Example : Find the value of Sin 120 ? Solution : We Know that sin 60 = $$\sqrt{3}\over 2$$ and cos 60 = $$1\over 2$$ By using above formula, sin 120 = 2 sin 60 cos 60 = 2 $$\times$$ $$\sqrt{3}\over 2$$ $$\times$$ $$1\over 2$$ $$\implies$$ sin 120 = $$\sqrt{3}\over 2$$ Example : If sin A = $$3\over 5$$, where 0 < A < 90, find the value of sin 2A ? Solution : We have, sin A = $$3\over 5$$ where 0 < A < 90 degrees $$\therefore$$ $$cos^2 A$$ = 1 – $$sin^2 A$$ $$\implies$$ cos A = $$\sqrt{1 – sin^2 A}$$ = $$\sqrt{1 – {9\over 25}}$$ = $$4\over 5$$ By using above formula, sin 2A = 2 sin A cos A = 2 $$\times$$ $$3\over 5$$ $$\times$$ $$4\over 5$$ $$\implies$$ sin 2A = $$24\over 25$$
# Lesson: Fractions: Number Line 20 Views 18 Favorites ### Lesson Objective SWBAT identify fractions on number lines. ### Lesson Plan Materials Needed: DN Worksheet, white board, dry erase markers, pencils, fraction models from previous lesson, magnetic tape, number line, and IND Worksheet Vocabulary: whole (or ONE or unit), denominator, and numerator. ………. Do Now (2 -3 min): One the board the teacher has 2 circles; 1 circle is divided into 6 parts with 5 part shaded. The other circle is divided into 4 parts with 3 parts shaded. The students are asked to complete the DN Worksheet from the graphics on the board. Opening (2 -3 min): Teacher quickly reviews answers to the Do Now and then states the objective, “Yesterday, we matched model fractions to numerical fractions. Today, we going to look at fractions on number lines. By the end of this lesson, you will be identify fractions on number lines.” Direct Instruction (10 min): Teacher begins by reviewing the vocabulary needed for to understand fractions. A fraction is always a fraction of something – for example, ½ of an orange, 2/3 of a rectangular region, 3/5 of a mile, ¼ of the marbles in a bag. We refer to this “something” as the whole, or ONE; for measures and counts it is considered the unit. The number below the fraction bar is called the denominator of the fraction. The denominator names the number of equal parts into which the whole is divided. The number above the fraction bar is called the numerator of the fraction. The numerator names the number of parts under consideration. The teacher then puts the fraction models, cut out in yesterday’s lesson, in random order (use magnetic tape). The teacher then explains points to the number line and asks the student what they know about the number line. The teacher then makes a list: - the numbers go in order - there are many numbers on it - it helps me understand numbers. - it starts at zero Then the teacher says, “Well today we are going to put our fractions on a number line. Who can guess how my fraction models should be organized to go on a number line? [smallest to biggest] Yes! That is correct, right now I don’t even have to think about the fraction number. I just have to organize them from biggest to smallest. Can someone come up here to help me do that?” The teacher takes a student volunteer to complete the task. The teacher then writes the fraction numbers underneath each of the ordered fraction models. The teacher continues, “Ok now I am set! I have ordered my fraction models; I have matched them to the numerical fraction, now I need to draw my number line and I have my fractions on a number line.” The teacher then draws a number line with arrows at each end. The teacher continues, “You have all done a great job. Watch as I put the following fractions on a number line. Then you will practice some together.” The teacher writes: ¾, ¼, 2/4, and 4/4 on the board and then models how to put each on a number line. Guided Practice (10 min): The students are given the GP Worksheet to work together on in small groups. The teacher should circulate to ensure understanding. Independent (10 min): The teacher then calls the students attention toward him/her before beginning IND Work. The students are handed IND Worksheet. Closing (2-3 min): Teacher calls the attention of the students back toward the front of the class to quickly review the answers to the Independent Practice worksheet/ ask what we learned about. ### Lesson Resources IND Fractions on Numberline Classwork 4 GP Fractions on Numberline Classwork 3 DN Lesson 3 Starter / Do Now 4
# Section 3.2 Exponential Functions - PowerPoint PPT Presentation Section 3.2 Exponential Functions 1 / 14 Section 3.2 Exponential Functions ## Section 3.2 Exponential Functions - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Section 3.2 Exponential Functions • 1. Use a calculator to evaluate, rounding to three decimal places. • a. e 2 • b. e -2 • c. e½ • a. ≈ 7.289 • b. ≈ 0.135 • c. ≈ 1.649 2. 2. Express as a power of e a. e 5 e -2 b. c. • = e 5+(-2) = e 3 • e 3-2 = e 2 • = = = e 4-(-1) = e 5 3. 3. Graph y = 3 x on a graphing calculator. - 5 < x < 5 and -1 < y < 100 4. 4. Graph y = (1/3) x on a graphing calculator. - 5 < x < 5 and -1 < y < 100 5. 5. Evaluate e 1.74 using a calculator. e 1.74 = 5.696 6. 6. BUSINESS: Interest - Find the value of \$1000 deposit in a bank at 10% interest for 8 years compounded a. anually b. quarterly c. continuously • For n = 1 m = 1 (annual compounding), P(1+r/n)nt simplifies to P(1+r)t • when P = 100, r = 0.1, and t = 8. The value is • 1000(1 + 0.1)8 = 1000 (1.1) 8 = 2143.59 • The value is \$2143.59 • For quarterly compounding, n = 4, P = 1000, r = 0.1, and t = 8. Thus • 1000(1+(0.1/4) 4x.8 = 1000(1 + 0.025) 4x.8 • = 1000 (1.025) 32 = 2203.76 • The value is \$2203.76 • For continuous compounding P = 1000, r = 0.10, and t = 8. Thus • 1000e 0.1x8 – 1000e 0.8 = 2225.54 • The value is \$2225.54 7. 7. Personal Finance: Interest - A loan shark lends you \$100 at 2% compound interest per week (that is a weekly, not annual rate). • How much will you owe after 3 years? • In “street” language, the profit on such a loan is known as the “vigorish” or the • “vig”. Fins the shark’s vig. • P = 100, r = 0.02 x 52 = 1.04 yearly, and n = 3. • this gives a value of • 100(1 + (1.04/52)) 52x3 • = 100(1.02) 52x3 • = 100(1.02) 156 • = \$2196 • b. The “vig” is equal to the amount owed after three years minus the amount • loaned. This is \$2196 - \$100.00 = \$2096 8. 8. Personal Finance: Annual Percentage Rate (APR) - Find the error in the ad shown below, which appeared in a New York paper. [Hint: Check that the nominal rate is equivalent to the effective rate. For daily compounding, s some banks use 365 days and some use 360 days in the year. Try both ways. The stated rate of 9.25% (compounded daily) is the normal rate of interest. To determine the effective rate of interest, use the compound interest formula, P (1 +r) n, with r = 9.25%/ number of days and n = number of days in a year. Since some banks use 365 days and some use 360 in a year, we will try both ways. If n = 365 days then, Then P(1+r) n = P(1.0002534) 365 ≈ 1.0969%. Subtracting 1 gives 0.0969, which expressed as a percent gives the effective rate of interest as 9.69% If n = 360 days then Then P(1+r) n = P(1.0002569) 360 ≈ 1.0969% and the effective rate is also 9.69% Thus, the error in advertisement is 9.825%. The annual yield should be 9.69% (based on the nominal rate of 9.25%) At T&M Bank, flexibility is the key word. You can choose the length of time and the amount you deposit, which will earn an annual yield of 9.825% based on a rate of 9.25% compounded daily. 9. 9. Personal Finance: Present Value - A rich uncle wants to make you a million. How much money must he deposit in a trust fund paying 8% compounded quarterly at the time of your birth to yield \$1,000,000 when you retire at age 60? If the amount of money P invested at 8% compounded quarterly yields \$1,000,000 in 60 years then and n = 604 = 240. 1,000,000 = P(1 +0.02) 240 10. 10. Personal Finance: Zero-Coupon Bonds - FUJI Holding recently sold zero-coupon \$1000 bonds maturing in 3 years with an annual yield of 10%. Find the price. [Hint: the price is the present value of \$1000, 3 years from now at the stated interest rate] For 10% compounded annually, r = 0.10 and n = 3. Present value = 11. 11. General: Compound Interest - Which is better 10% interest compounded quarterly or 9.8% compounded continuously? To compare two interest rates that are compounded differently, convert them both to annual yields. 10% compounded quarterly: P(1+r) n = P(1.025) 4 ≈ P(1.1038) Subtracting 1, 1.1038 – 1 = 10.38%. 9.8% compounded continuously, Pe rn = Pe 0.098 ≈ P(1.1030) Subtracting 1: 1.130 – 1 = 0.1030 The effective rate of interest is 10.30%. Thus, 10% compounded quarterly is better than 9.8 compounded continuously. 12. 12. Personal Finance: Depreciation - A Toyota Corolla automobile lists for \$15,450, and depreciates by 35% per year. Find the • values after: • 4 years b. 6 months • Since the depreciation is 35% per year, r = 0.35. • P(1 +r) n = 15,450(1 – 0.35) 4 ≈ \$2758 • P(1 +r) n = 15,450(1 – 0.35) 0.5 ≈ \$12,456 13. 13. General Nuclear Meltdown: - The probability of a “severe core meltdown accident” at a nuclear reactor in the U.S. within the next n years is 1 – (0.9997) 100n. • Find the probability if a meltdown: • within 25 years. b. within 40 years • a. 1 = (0.9997) 100(25) ≈ 0.5277 • b. 1 – (0.9997) 100(40) ≈ 0.6989 14. 14. General: Population - As stated earlier, the most populous state is California, with Texas second but gaining. According to the Census Bureau, x years after 2005 the population of California will be 36e0.013x and the population of Texas will be 22e 0.019x (all in millions) a. Graph these two functions on a calculator on the window [0,100] by [0,150]. b. In which year is Texas projected to overtake California as the most populous state? [hint: use INTERSECT] a. • During the year 2087 (x ≈ 82.08) • And 2005 + 82 = 2087.
MathMaster Blog For those preparing for the ACT, the math section is often the most troubling. Without the right preparation, it can be easy to get lost in the complex equations and story problems that are presented on the test. The good news is that the hardest ACT math questions can be practiced and perfected. To provide aid to those preparing for ACT math, we have pulled together the top five hardest ACT math problems. ACT Math Question #1 – Number and Quantity In the standard $x,y$ coordinate plane below, 3 of the vertices of a rectangle are shown. Which of the following answers displays the 4th vertex of the rectangle? a. $3,-7$ b. $4,-8$ c. $5,-1$ d. $8,-3$ e. $9,-3$ How to Solve it: 1. This question can be solved by calculating the slope. 2. As shown in the graph, one must move up 2 units and right 3 units to travel from point $-1,-1$ to $2,1$. 3. This would give us a slope for the shorter side of the rectangle to be 3/2. 4. By using the slop as a formula, we can realize the equations of both the x and y values: • X value: 6 – 3 = 3 • Y value: -5 – 2 = -7 5. The fourth vertex is located at $3,-7$. ACT Math Question #2 – Algebra If h$x$ = x³ + x and g$x$ = 2x + 3, then g$h(2$) = ? a. 7 b. 10 c. 17 d. 19 e. 23 How to Solve it: 1. This equation is essentially asking for the value of g$x$ when x is equal to h$2$. To solve, simply plug 2 in for x in h$x$ and then plug that value in for x in g$x$. 2. Plugging 2 in for x in h$x$: h$2$ = 2³ + 2 finds the answer 10. 3. Plugging 10 in for x in g$x: (g(10$ = 2$10$ + 3 finds the answer 23. ACT Math Question #3 – Statistics and Probability What is the probability that a number selected at random from the set {2, 3, 7, 12, 15, 22, 72, 108} will be divisible by both 2 and 3? a. 1/4 b. 3/8 c. 3/5 d. 5/8 e. 7/8 How to Solve it: 1. The solution to this problem is straightforward: simply go through the set of numbers and count how many are divisible by both 2 and 3. • Divisible by both 2 and 3: 12, 72, 108. • Not divisible by both 2 and 3: 2, 3, 7, 15, 22. 2. Out of the 8 numbers presented, 3 are divisible by both 2 and 3, leaving us with an answer of 3/8. ACT Math Question #4 – Statistics and Probability The 35-member Math Club is meeting to choose a student government representative. The members decide that the representative, who will be chosen at random, cannot be any of the 3 officers of the club. What is the probability that David, a member but not an officer, will be chosen? a. 0 b. 4/35 c. 1/35 d. 1/3 e. 1/32 How to Solve it: 1. The solution to this ACT math problem is fairly straightforward. The solution can be found in the list of possible answers and requires little math to be performed. 2. All we need to do is choose the correct numerator and denominator. Since we only determine the probability for one officer, we know the numerator must be 1. The denominator represents the possible choices as a whole. Since we know that the possible choices exclude only the 3 officers, and the total club member number is 35, we know the denominator is 32. This gives us an answer of 1/32 for the probability of David becoming the representative. ACT Math Question #5 – Modeling The graph of a given hyperbola, y = h$x$, is shown in the standard $x,y$ coordinate plane below. Among the following graphs, which best represents y = -h$x$?
# How to Find Trigonometric Ratios in Right Triangles? Understanding trigonometric ratios in right triangles is a fundamental concept in mathematics and plays a crucial role in various fields, including engineering, physics, and architecture. Trigonometry involves studying the relationships between the angles and sides of a triangle, especially in right triangles, where one angle measures 90 degrees. By learning how to find trigonometric ratios in right triangles, you can solve a wide range of real-world problems, from measuring heights and distances to analyzing forces and vectors. In this guide, we will explore the essential trigonometric ratios—sine, cosine, and tangent—and provide step-by-step instructions on how to calculate them in right triangles. ## Trigonometric Ratios in Right Triangles Trigonometric ratios – essential tools in trigonometry that help us understand the relationships between the sides and angles of a right triangle. In a right triangle, which has one angle measuring 90 degrees, there three primary trigonometric ratios: sine (sin), cosine (cos), and tangent (tan). These ratios defined based on the lengths of the sides of the triangle relative to its angles. The sine of an angle in a right triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse (the side opposite the right angle). The cosine of an angle is the ratio of the length of the adjacent side (the side next to the angle but not the hypotenuse) to the hypotenuse. Lastly, the tangent of an angle is the ratio of the length of the side opposite the angle to the length of the adjacent side. Trigonometric ratios find applications in various fields, including physics, engineering, architecture, and navigation. By understanding these ratios, we can solve complex problems involving distances, heights, angles, and velocities in right triangles. ### How to Find Trigonometric Ratios Given a Right Triangle To find trigonometric ratios in a right triangle, you need to know the lengths of its sides. Let’s assume we have a right triangle with angles A, B, and C, where C is the right angle. The side opposite angle A is denoted as ‘a,’ the side opposite angle B is denoted as ‘b,’ and the hypotenuse opposite the right angle C is denoted as ‘c.’ To find the sine of angle A, use the formula: sin(A) = a/c. To find the cosine of angle A, use the formula: cos(A) = b/c. And to find the tangent of angle A, use the formula: tan(A) = a/b. Similarly, for angle B, the formulas would be: sin(B) = b/c, cos(B) = a/c, and tan(B) = b/a. Remember that these formulas only valid for right triangles. If you have a triangle without a right angle, you’ll need to use different trigonometric concepts. ### Trigonometrical Ratios in a Right-Angled Triangle In a right-angled triangle, the trigonometric ratios play a crucial role in solving various geometric and real-world problems. These ratios, specifically sine, cosine, and tangent, fundamental for trigonometry. In addition to these, other trigonometric ratios like cosecant, secant, and cotangent derived from the primary three. The sine of an angle in a right triangle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. The cosine of an angle is the ratio of the length of the adjacent side to the hypotenuse. And the tangent of an angle is the ratio of the length of the side opposite the angle to the length of the adjacent side. Trigonometric ratios used in a wide range of fields, including engineering, physics, astronomy, and even in day-to-day tasks like measuring heights and distances. For example, in surveying, trigonometry helps determine inaccessible distances or heights by measuring angles and applying the appropriate trigonometric ratios. By understanding these ratios and their applications, mathematicians, scientists, and engineers can solve complex problems related to right-angled triangles and, consequently, various phenomena in the world around us. Trigonometry is a powerful tool that significantly simplifies calculations involving angles and distances in a right triangle context.
# How do you use the limit comparison test for sum (2x^4)/(x^5+10) n=1 to n=oo? Aug 11, 2015 color(red)(sum_(n=1)^∞ (2x^4)/(x^5-10)" is divergent"). sum_(n=1)^∞ (2x^4)/(x^5+10) The limit comparison test states that if ${a}_{n}$ and ${b}_{n}$ are series with positive terms and if lim_(n→∞) (a_n)/(b_n) is positive and finite, then either both series converge or both diverge. Let ${a}_{n} = \frac{2 {x}^{4}}{{x}^{5} + 10}$ Let's think about the end behaviour of ${a}_{n}$. For large $n$, the denominator ${x}^{5} + 10$ acts like ${x}^{5}$. So, for large $n$, ${a}_{n}$ acts like $\frac{2 {x}^{4}}{x} ^ 5 = \frac{2}{x}$. Let ${b}_{n} = \frac{1}{x}$ Then lim_(n→∞)(a_n/b_n) = lim_(n→∞)( ((2x^4)/(x^5+10))/(1/x)) = lim_(n→∞)( (2x^4×x)/(x^5+10)) = lim_(n→∞)( (2x^5)/(x^5+10)) = lim_(n→∞)( 2/(1-1/x^5)) =2 The limit is both positive and finite, so either ${a}_{n}$ and ${b}_{n}$ are both divergent or both are convergent. But ${b}_{n} = \frac{1}{x}$ is divergent, so ${a}_{n} = {x}^{4} / \left({x}^{5} - 10\right)$ is also divergent.
## Process The first three trigonometric functions are sine, cosine, and tangent When we find the sine, cosine, and tangent of different angles, we will use letters to name the alphabet. Sometimes we will say, "Find sin A." That just means to find the sine of the angle labeled A. We may also use letters from the Greek alphabet. Theta is the most common.. Let's start with sine (abbreviated "sin"). Following the link to view step-by-step instructions on finding the sine of an angle. Click on the video to watch a teacher explain the sine of an angle. Once you feel like you've seen enough, try some examples for yourself!!! Your goal: complete five (5) sine computations correctly. You may have to use the Pythagorean Theorem before you can solve for sine. How did you do with sine? The second trigonometric function is cosine (abbreviated "cos"). Try to get five (5) of these examples correct once you've learned the pattern of cosine. Will you need to use the Pythagorean Theorem again? Most definitely. If you've gotten sine and cosine, you've conquered the biggest part. The last of our first three trigonometric functions, tangent, is a combination of the first two. Just like the first two, click here to learn about tangent (abbreviated "tan"). Don't forget to copy what you learn in your notebook, so you'll have it to reference later. Can you get five (5) tangent computations correct at the end? Be sure to let your teacher know how many of each you got correct! SINE, COSINE, AND TANGENT!!! HERE'S A FUN WAY TO REMEMBER THEM: SINE = OPPOSITE / HYPOTENUSE Click on the picture to take a test over all three of the trigonometric functions. Still stuck? Try this!!!! At the beginning of this WebQuest, we reviewed reciprocals. Now, you know the three basic trigonometric functions. Together with your partner or small group, you will complete this WebQuest. Here are the three basic trigonometric functions one more time: sin θ = opposite / hypotenuse cos tan Now, here are the names of the reciprocal functions: The reciprocal of sine (sin) is cosecant (csc). The reciprocal of cosine (cos) is secant (sec). The reciprocal of tangent (tan) is cotangent (cot). YOUR JOB: Together with your partner or small group, write down what the reciprocal functions would look like in terms of opposite, adjacent, and hypotenuse. Do it WITHOUT consulting a textbook, website, or other resource!!! Just like the three basic trig functions, list the reciprocal functions: csc θ = __________ / ____________ sec θ = __________ / ____________ cot θ = __________ / ____________ Did you get it? Are you a trig master now? Here's your chance to show your stuff. Choose one of the following ways to show you're a master: 1) Create your own "trig functions" video. Be as creative as you would like - just make sure you are also appropriate. You can use the video in the introduction as a guide. Make sure you use your math vocabulary and get the trigonometric functions correct!!! 2) In your own words, explain each of the six trigonometric functions you have just learned. Again, make sure you use your math vocabulary and get the trigonometric functions correct!!!
# Fractions, Percent and Relative Frequency Visual Number Talk Prompts using context to build fluency and flexibility with fractions and percent to unpack big ideas around relative frequency. ## In This Set of Visual Number Talk Prompts… Students will explore the frequency of different categories relative to the entire data set. Students will represent their findings in a relative frequency table. They will investigate the relationship between fractions and percent. ## Intentionality… Students will engage in a visual math talk and will have an opportunity to develop a deeper understanding of the following big ideas. • Frequency is the number of times a category or event occurs within a data set. • A relative frequency table shows each category expressed as a fraction of the total quantity or frequency. • Relative frequency can be represented using fractions, decimals, or percents. • Fractions, decimals, and percents can all represent relationships to a whole. • The sum of the relative frequencies is 1 or 100%. ## String of Related Problems Present the following incomplete table. Work together to determine the relative frequency for each ice cream flavour. Encourage students to discuss their strategies. Students will likely scale the total frequency to 100 in order to describe the relative frequency as a percent. ## Visual Number Talk Prompt Student prompt: A class of 25 students purchased ice cream cones. What fraction of the class ordered chocolate? …vanilla? You will notice that while we are asking students to determine the number of ice cream cones and the fraction of the whole relative to the total number of ice creams purchased, we are trying to explicitly make a connection that in data literacy, the number of ice cream cones is known as the frequency and the fraction of the whole is known as relative frequency which is often presented as a percentage. Students can leverage a tape diagram or double number line to begin determining the fraction of the whole as well as the percentage of the whole for each flavour relative to the total. Note how easy it is for students to leverage the visual model to work between the number of ice cream cones, fractions and percentages by leveraging the ratio relationship that exists. While students would probably more easily re-draw the double number line (one for each flavour), our visual simply shows the flavours “changing positions” in a commutative like way to represent various approaches. Ultimately at the end of this work, we want to help students to deepen their understanding of frequency and relative frequency while also increasing their fluency and flexibility with fractions, percent and ratio relationships. ## Want to Explore These Concepts & Skills Further? One (1) additional number talk prompt are available in Day 2 of the Flavour Frenzy problem based math unit that you can dive into now. Why not start from the beginning of this contextual 5-day unit of real world lessons from the Make Math Moments Problem Based Units page. Did you use this in your classroom or at home? How’d it go? Post in the comments! Math IS Visual. Let’s teach it that way.
# Cuboid - edges The cuboid has dimensions in ratio 4: 3: 5, the shortest edge is 12 cm long. Find: (A) the lengths of the remaining edges, (B) the surface of the cuboid, (C) the volume of the cuboid Correct result: a = 16 cm c = 20 cm S = 1504 cm2 V = 3840 cm3 #### Solution: We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Check out our ratio calculator. Tip: Our volume units converter will help you with the conversion of volume units. #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Cuboid edges Calculate the volume and surface of a cuboid whose edge lengths are in the ratio 2: 3: 4 and the longest edge measures 10cm. • Cuboid The sum of the lengths of the three edges of the cuboid that originate from one vertex is 210 cm. Edge length ratio is 7: 5: 3. Calculate the length of the edges. • Cuboid - ratios The sizes of the edges of the cuboid are in the ratio 2: 3: 5. The smallest wall have area 54 cm2. Calculate the surface area and volume of this cuboid. • Cuboid face diagonals The lengths of the cuboid edges are in the ratio 1: 2: 3. Will the lengths of its diagonals be the same ratio? The cuboid has dimensions of 5 cm, 10 cm, and 15 cm. Calculate the size of the wall diagonals of this cuboid. • Cuboid and ratio Find the dimensions of a cuboid having a volume of 810 cm3 if the lengths of its edges coming from the same vertex are in ratio 2: 3: 5 • Similar triangles The triangles ABC and XYZ are similar. Find the missing lengths of the sides of the triangles. a) a = 5 cm b = 8 cm x = 7.5 cm z = 9 cm b) a = 9 cm c = 12 cm y = 10 cm z = 8 cm c) b = 4 cm c = 8 cm x = 4.5 cm z = 6 cm • Cuboid - ratio Find the volume of a block whose dimensions are in the ratio 2: 3: 4 and the surface is 117 dm2. • Cuboid - volume and areas The cuboid has a volume of 250 cm3, a surface of 250 cm2 and one side 5 cm long. How do I calculate the remaining sides? • Rectangle 3-4-5 The sides of the rectangle are in a ratio of 3:4. The length of the rectangle diagonal is 20 cm. Calculate the content of the rectangle. • Pyramid Cuboid ABCDEFGH has dimensions AB 3 cm, BC 4 cm, CG 5 cm. Calculate the volume and surface area of a triangular pyramid ADEC. • Cuboid edges in ratio Cuboid edges lengths are in ratio 2:4:6. Calculate their lengths if you know that the cuboid volume is 24576 cm3. • Rectangular cuboid The rectangular cuboid has a surface area 5334 cm2, and its dimensions are in the ratio 2:4:5. Find the volume of this rectangular cuboid. • Ratio of edges The dimensions of the cuboid are in a ratio 3: 1: 2. The body diagonal has a length of 28 cm. Find the volume of a cuboid. • Rectangle The length of the rectangle is 12 cm greater than 3 times its width. What dimensions and area this rectangle has if ts circumference is 104 cm.
# Logical Sequence of Statements Logical Sequence of Statements forms the part of reasoning ability. The questions are simple and mostly focus on the common sense ability of the candidate. In these questions, a number of words will be present and you will have to figure out a rule or a method to put these words in a sequence. There will be at other times, sentences or statements present in the option and you will have to sort them in a sequence. Here we will see these questions and their various types. ## Logical Sequence Of Statements Consider the following series of words: door, key, room, and lock. If someone asks you to put these words into a sequence, what will the answer be? It is either room, door, lock and key or the reverse i.e. key, lock, door and room. Why? The reason is simple, this is the order a person leaving a room or entering it will follow.Â This forms a logical sequence and we call it the logical sequence of words. Similarly, we can form what is known as the logical sequence of statements. In these types of questions, a sequence of statements is present. Using similar technique as for the words, you will have to put these statements in the form of a series. For example, consider the following set of statements: P: He began using rabbits for the trials. Q: His experimentsÂ on rats were not successful. R: The rabbits didn’t help the study anyhow. S: The doctor was very upset. So what is the logical sequence here? A rather callous look might compel you to think that the correct sequence should be P Q R and S. But let us see more carefully. The first sentence says that the doctor began using rabbits. Look for a cause for this sentence. If you find a cause then from the principle of causality, that statement should come before P. The statement is Q. It was because the experiment didn’t work on the rats that the doctor had to use rabbits. Then we will have R and then S. So the logical sequence here will be QPRS. ## Solved Examples ToÂ guess the correct sequence of the sentences when there are no fixed sentences is very difficult and many times not possible. So there will at least be one sentence that is fixed. Normally we have the first and the last sentence that isÂ already present and you have to fill the gap in-between. We will see a few examples and try to get acquainted with the concept of arranging the statements in a sequence. Let us begin. #### Example 1 In questions below, each passage consists of six sentences. The first and sixth sentences are present in the beginning. The four sentences in the middle have been removed and jumbled up. These are labelled as P, Q, R and S. Find out the proper order for the four sentences. I wanted to have mashed potatoes, so I went to the mall. P: There I saw potato chips on a shelf after looking around for a while. Q: I was looking for snacks and had to go to another level. R: I couldn’t find the elevator and so I took the stairs. S: I felt like having something else too. I found the elevator while getting ready to leave. A) PQRSÂ Â Â Â Â Â Â Â Â Â B)Â PSQRÂ Â Â Â Â Â C) QRSPÂ Â Â Â Â Â D) SQRP Answer: This is a very interesting example that is very similar to the ones that you will see in the IBPS exams and other such exams. First of all, never use words to guess the correct sequence of sentences. For example, in the above statement, the word elevator occurs in statement R and the sentence at the end. But logically they are separated. The whole portion talks about a person who goes to the mall and then buys chips at a different storey than the one he starts at. So the logical sequence of the sentences is: I wanted to have mashed potatoes, so I went to the mall. (S) I felt like having something else too. (Q) I was looking for snacks and had to go to another level. (R) I couldn’t find the elevator and so I took the stairs. (P)Â There I saw potato chips on a shelf after looking around for a while. Therefore the answer is D) SQRP. ## Practice Questions In questions below, each passage consists of six sentences. The first and sixth sentences are present in the beginning. The four sentences in the middle have been removed and jumbled up. These are labelled as P, Q, R and S. Find out the proper order for the four sentences. Q 1: The order was clear. P: They had to move fast. Q: She was hurt. R: The rescue team was to reach her location and extract her to safety. S: It was a very delicate situation. A) PQRSÂ Â Â Â Â Â Â Â Â Â B)Â RQSPÂ Â Â Â Â Â C) QRSPÂ Â Â Â Â Â D) SQRP Ans: B)Â RQSP Share with friends ## Customize your course in 30 seconds ##### Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. Ashhar Firdausi IIT Roorkee Biology Dr. Nazma Shaik VTU Chemistry Gaurav Tiwari APJAKTU Physics Get Started ## Browse ### 2 responses to “Cause and Effect” 1. RAMIYA SHREE says: just leave a couple of questions aside 2. RAMIYA SHREE says: leave a couple of questions to practice
You are Here: Home >< Maths # Write y as a function of x Watch 1. Given the equation: 2x - 3y = 6 Write y as a function of x. I know this is straightforward, and I have the line-by-line answer, but I can't understand how you get from: (2x - 3y = 6) - 3y = - 2x + 6 (So here you have shifted the 2x over as you want ultimately just y on the left) to y = 2x/3 - 2 (Where's the -3? Why is there a fraction on the right? What's happened to the +6?) 2. You divided both sides by -3. 3. Oh so you do it to the +6 AND the -2x. Got ya! 4. How would I then find the "intercept" and the "root"? 5. The root is basically the value of x for which y = 0. So set y to and solve for x. The y-intercept is the value of y for which x = 0. So set x to 0 and solve for y. 6. (Original post by tymbnuip) Oh so you do it to the +6 AND the -2x. Got ya! Yes you're "doing it" to the sides of the equation rather than the actual terms (at least that's one way of looking at it). If you didn't, your equation would not hold. There's only a limited number of valid operations you can do to an equation (an equation states that something on the left side is equal to something on the right side) like add/subtract a number to each side or multiply the left expression and the right expression by the same factor (in your case that factor is -3) 7. The root (x -intercept) is when y = 0 2x/3 - 2 = 0 2x - 6 = 0 {multiply both sides by 3} So how does the /3 disappear by multiplying by 3? Wouldn't it turn into 9? 8. (Original post by tymbnuip) So how does the /3 disappear by multiplying by 3? Wouldn't it turn into 9? When multiplying two fractions, multiply the top by the top and the bottom by the bottom 2x times 3 = 6x 3 times 1 = 3 6x/3 = 2x Get it? 9. What does the =0.3/1 mean? 10. Urgh sorry it means 0 times 11. Heh okay. So I guess now I have 2x - 6 = 0 I want to get rid of the 6? TSR Support Team We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out. This forum is supported by: Updated: August 5, 2008 Today on TSR ### Should I ask for his number? Discussions on TSR • Latest • ## See more of what you like on The Student Room You can personalise what you see on TSR. Tell us a little about yourself to get started. • Poll Useful resources ### Maths Forum posting guidelines Not sure where to post? Read the updated guidelines here ### How to use LaTex Writing equations the easy way ### Study habits of A* students Top tips from students who have already aced their exams ## Groups associated with this forum: View associated groups Discussions on TSR • Latest • ## See more of what you like on The Student Room You can personalise what you see on TSR. Tell us a little about yourself to get started. • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd. Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE
# What is Cot(arcsin (-5/13)) ? Jul 21, 2015 $\cot \left(\arcsin \left(- \frac{5}{13}\right)\right) = - \frac{12}{5}$ #### Explanation: Let$\text{ } \theta = \arcsin \left(- \frac{5}{13}\right)$ This means that we are now looking for color(red)cottheta! $\implies \sin \left(\theta\right) = - \frac{5}{13}$ Use the identity, ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$ NB : $\sin \theta$ is negative so $\theta$ is also negative. We shall the importance of this info later. $\implies \frac{{\cos}^{2} \theta + {\sin}^{2} \theta}{\sin} ^ 2 \theta = \frac{1}{\sin} ^ 2 \theta$ $\implies {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta + 1 = \frac{1}{\sin} ^ 2 \theta$ $\implies {\cot}^{2} \theta + 1 = \frac{1}{\sin} ^ 2 \theta$ $\implies {\cot}^{2} \theta = \frac{1}{\sin} ^ 2 x - 1$ $\implies \cot \theta = \pm \sqrt{\frac{1}{\sin} ^ 2 \left(\theta\right) - 1}$ $\implies \cot \theta = \pm \sqrt{\frac{1}{- \frac{5}{13}} ^ 2 - 1} = \pm \sqrt{\frac{169}{25} - 1} = \pm \sqrt{\frac{144}{25}} = \pm \frac{12}{5}$ WE saw the evidence previously that $\theta$ should be negative only. And since $\cot \theta$ is odd $\implies \cot t \left(- A\right) = - \cot \left(A\right)$ Where $A$ is a positive angle. So, it becomes clear that $\cot \theta = \textcolor{b l u e}{+} \frac{12}{5}$ REMEMBER what we called $\theta$ was actually $\arcsin \left(- \frac{15}{13}\right)$ $\implies \cot \left(\arcsin \left(- \frac{5}{13}\right)\right) = \textcolor{b l u e}{\frac{12}{5}}$
# Georgia - Grade 1 - Math - Measurement and Data - Ordering Objects by Length - MGSE1.MD.1 ### Description MGSE1.MD.1 Order three objects by length; compare the lengths of two objects indirectly by using a third object. • State - Georgia • Standard ID - MGSE1.MD.1 • Subjects - Math Common Core ### Keywords • Math • Measurement and Data ## More Georgia Topics MGSE1.OA.1 Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem. MGSE1.G.2 Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.6 This is important for the future development of spatial relations which later connects to developing understanding of area, volume, and fractions. MGSE1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2. MGSE1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _. MGSE1.NBT.2 Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). MGSE1.OA.3 Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
# How to Teach Decimals Your kids are ready to move beyond whole numbers and into decimals, but how do you make it fun? Close the workbooks and learn how to teach decimals in an engaging, hands-on approach. ## What Are Decimals? Decimals are, essentially, fractions whose denominators are multiples of 10. A fraction can have any denominator: ⅙, ⅔, ¾, and so on, but decimals can only be divided by numbers like 10, 100, 1000, and the like. When you see the decimal 0.2, you would say it is “two tenths.” Its fractional equivalent is 2/10. When you see the fraction 0.25, you would call that “twenty-five hundredths” because it is equal to 25/100. Decimals were invented by the Chinese in the 4th century, and from there spread to the Middle East and Europe. Today these decimals have their own special notation; they follow a decimal point. ## Why Are Decimals an Important Skill to Master? Very few numbers in real life are whole numbers (1, 2, 3, 4, etc.). Rarely do walk a trail exactly 1 mile long, or catch a 2 lb. fish, or have a restaurant bill come to exactly \$30. We need ways to express parts of a whole, and decimals are the most common way to do that. Decimals are pivotal to understanding our system of money (cents are all decimals), making measurements, and solving equations in advanced math. They allow us to be precise, and precision matters in fields like science, cooking, art, business, and computers. Decimals are a transition to abstract math, so they can be tricky for kids to master. While a child can visualize (and count) 0.1 with the right manipulatives, decimals such as 0.637 are harder to envision. ## Fun Ways to Teach Decimals Though decimals can be intimidating to some students, the right teaching approach can not only make them click sooner, it can make them fun! ### Money Dollars are based on whole numbers, but cents are decimals. After the decimal point, dimes occupy the tenths place and pennies occupy the hundredths place. Teach decimals using real (or imitation) money. Count out coins and turn their value into decimals. ### Classroom Economy Plenty of classrooms use a point-based reward system, but why not use dollars and cents instead (even if they are only imaginary)? You can also teach financial principles like saving and budgeting with the free program My Classroom Economy It overlays your curriculum (or an individual subject) to teach grade-specific concepts from Kindergarten to high school. Bonus: You can use it as a classroom management tool. ### Real-Life Shopping Why not get math out of the workbook and into the real world? Use real-life shopping circulars to add and subtract. If your school has a book fair or a fundraising drive, don’t let the pitch go by without tying it to decimals. Fill out sample order forms, and do the math together. You Might Also Like: Rainbow Restaurant Math ### Real-Life Measurements Measuring requires precision, and that means decimals! Measure the world (including students’ heights) with the metric system, which is already a base-10 system. A child can know they are 4 feet tall, but does he know he is actually 1.23 meters? You can also use a digital kitchen scale to turn baking into an exercise with decimals. Professional bakers don’t use measuring cups; they use scales! You can find plenty of baking by weight recipes on Joy of Baking and King Arthur Flour. Simply change the weight amounts from grams to kilograms and–presto! You’ve got decimals. ### Place Value Charts Decimals are a transition to abstract math, but you can still initially tie the concept to math you can touch and see. Decimal place value can easily be visualized with a place value chart. A decimal place value chart has a spot not just for whole numbers (like the ones, tens, and hundreds places), but it also includes the tenths place and the hundredths place. Use manipulatives such as small discs, beans, or Cheerios to occupy the spaces. ### Decimal Tiles The colorful, stackable tiles reinforce both fraction and decimal lessons (and show how they are really two sides of the same coin). ### Digital Escape Rooms Why not practice decimals with a digital escape room? We have three FREE decimal digital escape rooms to choose from. This first digital escape room focuses on decimal place value, from the tenths place up to the millionths places. This second digital escape room reinforces adding and subtracting decimals. The third escape room requires the student to multiply decimals. ### Apps Game-based learning is a wonderful way to reinforce the concept of decimals. Try apps such as SplashLearn, Prodigy, and Buzzmath. ## Teaching Decimals To Kids Decimals don’t have to be dull. Spice them up with these hands-on activities and learn how to teach decimals the fun way! Who knows? Your kids may soon be begging for more math!
# CONVERTING FRACTIONS INTO DECIMALS WORKSHEET ## About "Converting fractions into decimals worksheet" Converting fractions into decimals worksheet : Worksheet on converting fractions into decimals is much useful to the students who would like to practice problems on decimals and fractions. ## Converting fractions into decimals worksheet 1. Covert the given fraction into decimal 7/25 2. Covert the given fraction into decimal 17/5 3 Covert the given fraction into decimal 3/8 4. Covert the given fraction into decimal 13/8 5. Covert the given fraction into decimal 3/20 6. Covert the given fraction into decimal 7/10 7. Covert the given fraction into decimal 3/50 8. Covert the given fraction into decimal 1/2 9. Covert the given fraction into decimal 3/4 10. Covert the given fraction into decimal 2/3 ## Converting fractions into decimals worksheet - Solution Problem 1 : Covert the given fraction into decimal 7/25 Solution : To convert the given fraction into decimal, first we have to check whether the denominator of the fraction is convertible to 10 or 100 using multiplication. In the given fraction, the denominator is 25 which is convertible to 100 using multiplication by 4 So, we have Problem 2 : Covert the given fraction into decimal 17/5 Solution : In the given fraction, the denominator is 5 which is convertible to 10 using multiplication by 2 So, we have Problem 3 : Covert the given fraction into decimal 3/8 Solution : In the given fraction, the denominator is 8 which is not convertible to 10 or 100 using multiplication, we can convert the given fraction into decimal using long division. Problem 4 : Covert the given fraction into decimal 13/8 Solution : In the given fraction, the denominator is 8 which is not convertible to 10 or 100 using multiplication, we can convert the given fraction into decimal using long division. Problem 5 : Covert the given fraction into decimal 3/20 Solution : In the given fraction, the denominator is 20 which is convertible to 100 using multiplication by 5. So, we have 3/20 = (3x5)/(20x5) = 15/100 = 0.15 Hence, 3/20 = 0.15 Problem 6 : Covert the given fraction into decimal 7/10 Solution : In the given fraction, the denominator is 10. So, we have 7/10 = 0.7 Hence, 7/10 = 0.7 Problem 7 : Covert the given fraction into decimal 3/50 Solution : In the given fraction, the denominator is 50 which is convertible to 100 using multiplication by 2. So, we have 3/50 = (3x2)/(50x2) = 6/100 = 0.06 Hence, 3/50 = 0.06 Problem 8 : Covert the given fraction into decimal 1/2 Solution : In the given fraction, the denominator is 2 which is convertible to 10 using multiplication by 5. So, we have 1/2 = (1x5)/(2x5) = 5/10 = 0.5 Hence, 1/2 = 0.5 Problem 9 : Covert the given fraction into decimal 3/4 Solution : In the given fraction, the denominator is 4 which is convertible to 100 using multiplication by 25. So, we have 3/4 = (3x25)/(4x25) = 75/100 = 0.75 Hence, 3/4 = 0.75 Problem 10 : Covert the given fraction into decimal 2/3 Solution : In the given fraction, the denominator is 3 which is not convertible to 10 or 100 using multiplication. Now, we have to get the decimal form of the fraction 2/3 using long division method as explained above. When we do so, we get 2/3 = 0.67 (approximately) Hence, 2/3 = 0.67 We hope that the students would have understood the stuff and examples explained on "Converting fractions into decimals worksheet". 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+0 GEOMETRY V +1 295 7 +297 An equilateral triangle has sides of length 2 units. A second equilateral triangle is formed having sides that are 150% of the length of the sides of the first triangle. A third equilateral triangle is formed having sides that are 150% of the length of the sides of the second triangle. The process is continued until four equilateral triangles exist. What will be the percent increase in the perimeter from the first triangle to the fourth triangle? Express your answer to the nearest tenth. Nov 17, 2019 #1 +21827 0 The side lengths are increased to 2 x 1.5 x 1.5 x 1.5 = 2 (1.5)3 = 6.75 original perimeter = 2 *3 = 6 new largest perimeter = 6.75 x 3 = 20.25 Per cent increase 20.25/6 x 100% = 337.5 % (now look at 1.53 this equals 3.375 ) Nov 17, 2019 #2 +297 +1 I put 337.5% as my answer. It is incorrect. Nov 17, 2019 #3 +1 An equilateral triangle has sides of length 2 units. A second equilateral triangle is formed having sides that are 150% of the length of the sides of the first triangle. A third equilateral triangle is formed having sides that are 150% of the length of the sides of the second triangle. The process is continued until four equilateral triangles exist. What will be the percent increase in the perimeter from the first triangle to the fourth triangle? Express your answer to the nearest tenth. Doing this with percents confuses me. Let's get rid of the % signs and say it as follows: An equilateral triangle has sides of length 2 units. A second equilateral triangle is formed having sides that are 1.50 times the length of the sides of the first triangle, and so on. The second triangle has sides 1.5 times the length of the first And the third triangle has sides 1.5 times the length of the second And the fourth triangle has sides 1.5 times the length of the third So the fourth triangle has sides 1.5 x 1.5 x 1.5 times the length of the first Do the multiplication, and the fourth triangle has sides 3.375 times the length of the first Change the factor of 3.375 back to percent and the percent increase is 337.5 percent. Stating that the first triangle's sides were 2 units long was a red herring. The question asked for the percent increase . Nov 17, 2019 #4 +297 0 It is incorrect though. Nov 17, 2019 #5 +1 First triangle = 2 units 2nd triangle =1.5 x 2 =3 units. 3rd triangle =3 x 1.5 =4.5 units 4th triangle =4.5 x 1.5 =6.75 units. 6.75 / 2 =3.375 - 1 x 100 =237.5% increase in the perimeter from the first to the 4th triangle. Nov 17, 2019 #6 +1 Oh golly that's right. It's the wording. The fourth triangle is 3.375 times the first, therefore the INCREASE is 2.375. . Guest Nov 17, 2019 #7 +21827 +1 Per cents are always kinda tricky 150% of 2 == 3 2nd side length 150% 0f 3 = 4.5 150% 0f 4.5 = 6.75 z% of 2 = 6.75 6.75/2 = 3.375 337.5% OF 2 The INCREASE FROM 2 (6.75-2)/2 = 2.375 = 237.5 % SO I think the wording is tricky, but the answer looked for is 237.5 % Nov 17, 2019
Fractional Exponents Transcript Fractional exponents. In this lesson we can make explicit the link, between roots and exponents. So far, the only exponents we have considered have been integers, either positive, negative, or zero. So we've been sticking with integers. What happens if the exponent is not an integer but a fraction? What, what happens then? So let's explore this. For example what would it mean to say 2 to the power of one-half? Well gee let's think about this. We could do mathematical operations, if we have another side to that equation. So just let's create a dummy variable k. We'll call the output k, 2 to the one-half equals k. Well, notice that if we multiply one-half by 2 we get a whole number. And of course we could use that multiplying exponent rule, if we raise 2 to the one-half to another power. So I'm gonna say why don't we square both sides? Well then one side would get k squared. On the other side, we get 2 to the one-half squared. And of course the laws of exponents say we multiply those exponents, one-half times 2 equals 1. So that side just becomes ordinary 2, k squared equals 2. Well of course we could solve this for k very easily. Take a square root k equals the square root of 2. And that must be what k equals. So, in other words, 2 to the one-half, equaled the square root of 2. Raising something to the power of one-half, is the same as finding the positive square root of it. That's important fact number 1. Now, what would it mean to say 2 to the 1 3rd? Well you might guess, but we'll follow the same process. Again fill in a dummy variable k, and now notice that if we multiply 1 3rd times three, we'll get a whole number. So, we'll cube both sides. And of course, the right side just becomes k cubed. The left side, 2 to the 1 3rd to the 3 well the 1 3rd and the 3 get multiplied and that just equals 1 so it's 2 to the 1 or ordinary 2. So k cubed equals 2. We can take the cube root of both sides. K equals the cube root of two. So in other words, two to the one third equals the cube root of two. Well, you might see a general pattern emerging here. In other words, if we take something to the one-half, it's the square root. If we take something to the, 1 3rd, its the cubed root. You might guess, if we take it to the one 4th, it's the 4th root, one 5th, it's the fifth root, that sort of thing. And in fact we can generalize by saying b to the power of 1 over m is the mth root of b. So this is the explicit link between fractional exponents and roots. So for example, if we had something like 6 to the power of 1, 7th, what that would mean is the 7th root of 6. What exactly does that mean, the 7th root of 6? This is the number which, when raised to the 7th power, equals 6. What happens if the exponent is a fraction that has a number other than 1 in the numerator? So far we've been looking only at fractions that have one in the numerator. What would happen if we had something like 2 to the 3 5ths. Well, remember, we can rank 3 5ths as either 3 times 1 5th or 1 5th times 3, we can write, of course we can write the product either way and this has implications with the laws of exponents. I can write it as 3 times 1 5th, have the power of 3 inside, and have the 1 5th outside, and so that would be the 5th root of 2 cubed, or the 5th root of 8. That would be one way I could do it. Another way I could do it, would be to write the 5th on the inside. So on the inside I have just the 5th root of ordinary 2, and on the outside I'm cubing it. Either one of those is perfectly fine. And I will say if you actually have to do a calculation, if you have to actually choose between these two, always make things smaller before you make things bigger. That's a very important point of strategy. Here's a practice problem, where you can apply some of this. Pause the video and then we'll talk about this. Okay. 8 to the 4 3rds. Well, we could write that either as the cube root of 8 to the 4th, or the cube root of 8 that whole thing to the power of 4. We could write it either way. Now the question is, which would be a better way to calculate. Well, with that first one, the first thing e would have to do is figure out 8 to the power of 4. Well that's going to be a large number. Course 8 to the power of 4 is gonna be 8 squared squared, so that would be 64 squared. I don't know 64 squared off the top of my head, but that's gonna be a very large number and then we're gonna try and take a, a cubed root of it. Hm, that sounds doubtful. Where as with the other one, all we have to do is take a cubed root of 8, we can do that, and then raise it to the 4th. So that first one is just a horrible idea. Don't raise it to some high power and then try to find a root. Find the root first. That's an enormous point of strategy. So, we'll find the root first. And, of course, the cubed root of 8 is just 2. So, we get 2 to the 4th, and that's 16. So, these rules that we talked about. Are rules that are true for positive numbers, when b is a positive number. Technically they are true at 0, though the roots of 0 are an unlikely topic on the test. If the denominator of the exponent fraction is odd, then the base can be negative as well. Remember that we could not take even roots of negative numbers, but we could take odd roots of negative numbers. For example, the cube root, or the 5th root of a negative number. In summary, roots are represented by fractional exponents. That's the big idea. The square root of a quantity, equals that quantity to the power of one half. That is, by far, the most common fractional exponent you'll see on the exam. The power b to the 1 over n means the nth root of b. And the power b to the m over n can be written either as the root of the power, or as the root to the exponent m.
# Numeracy #16: Blue and Red Marbles Puzzle Question: You have 50 blue marbles, 50 red marbles, and two jars to put them in. A marble will be selected at random from a jar selected at random. How do you divide the marbles among the jars as to maximize the probability of choosing a blue marble. You must use all the marbles. As always, try solving the problem yourself or keep reading for the solution. ### Solution We know that the probability of picking either jar will always be 50%. The probability of picking a blue marble in either jar will depend on the number of blue marbles divided by the total number of marbles in that jar. Using this, we can construct a formula for the total probability of picking a blue marble – 50% times the probability of picking a blue in jar 1 plus 50% times the probability of picking blue in jar 2. P(B) = 0.5(B1 / B1 + R1) + 0.5(B2 / B2 + R2) B1 = blue marbles in jar 1 B2 = blue marbles in jar 2 R1 = red marbles in jar 1 R2 = red marbles in jar 2 What happens if we just split the marbles evenly, 25 of each in both jars: P(B) = 0.5(25 / 50) + 0.5(25 / 50) P(B) = 0.5 We would do no better than random. What if we keep the red split evenly and put all the blue marbles on one side: P(B) = 0.5(50 / 75) + 0.5(0 / 25) P(B) = 0.33 We end up doing worse. Dropping the probability of one jar to zero hurt us more than putting all the blue on one side helped. If going to 0% blue hurt, going to 100% blue in one jar should help. More specifically, we want to raise the probability of drawing blue in one jar to 100% while minimizing the loss of blue in the other jar. The solution is just to put one blue marble and no red marbles in one of the jars: P(B) = 0.5(1 / 1) + 0.5(49 / 99) P(B) = 0.74 It turns out that 74% is the best we can do.
Hong Kong Stage 2 # Fractions of Quantities (unit fractions) Lesson ## Method 1 - Using words You have probably already had experience with questions like, What is a half of $16$16? This is because halves come up all the time, especially if you have ever had to share with a a brother or a sister! To find a half of something we know that we divide it by $2$2. We can also write a half of 16 using symbolic fraction notation. It would look like this $\frac{1}{2}\times16$12×16 So if we have to do questions like $\frac{1}{2}$12 of $32$32 or $\frac{1}{4}$14 of $16$16 we might find it easier to say a half of $32$32, or a quarter of $16$16. The answers are sometimes more obvious! ## Method 2 - Reversing the Order What about questions like $\frac{1}{3}$13 of $4$4 or $\frac{1}{4}$14 of $9$9. Saying these in words doesn't help much here. A third of $4$4, or a quarter of $9$9 - so I still don't know the answer! So let's try something else, because multiplication is commutative we can change the order around. You know how $2\times4=4\times2$2×4=4×2, well this works when you are multiplying anything - even fractions. Lets see how this might help $4\times\frac{1}{3}$4×13 we would say as $4$4 thirds..... just like how $2\times4$2×4 we say $2$2 fours. Well $4$4 thirds, is the answer! We can write $4$4 thirds as $\frac{4}{3}$43 ## The whole idea Now lets connect some dots. $\frac{1}{3}$13 of $4$4 = $\frac{1}{3}\times4$13×4 = $4\times\frac{1}{3}$4×13 = $4$4 thirds = $\frac{4}{3}$43 = $4\div3$4÷3 One third, ended up being the same as dividing by $3$3. Lets look at another one, $\frac{1}{5}$15of $20$20 = $\frac{1}{5}\times20$15×20=$20\times\frac{1}{5}$20×15 = $20$20 fifths = $\frac{20}{5}$205 = $20\div5$20÷5 , so$\frac{1}{5}$15is the same as dividing by $5$5. This comes right back to our definition of a fraction, the denominator is how many parts we divide up into. So $\frac{1}{10}$110 of $80$80 is $80\div10=8$80÷10=8 $\frac{1}{4}$14 of $20$20 is $20\div4=5$20÷4=5 and $\frac{1}{7}$17 of $21$21 is $21\div7=3$21÷7=3 #### Practice Questions ##### QUESTION 1 What is $\frac{1}{5}$15 of $20$20? ##### QUESTION 2 What is $\frac{1}{14}$114 of $35$35 pizzas? Express your answer in simplest fraction form. ##### QUESTION 3 What is $\frac{1}{10}$110 of $50$50 litres?
# Why is there a need to know about perfect squares? ## Why is there a need to know about perfect squares? A perfect square is a number that can be expressed as the product of two equal integers. What does that mean? Basically, a perfect square is what you get when you multiply two equal integers by each other. 25 is a perfect square because you’re multiplying two equal integers (5 and 5) by each other. ### Why do I need to know the Pythagorean Theorem? The Pythagorean Theorem is useful for two-dimensional navigation. You can use it and two lengths to find the shortest distance. The distances north and west will be the two legs of the triangle, and the shortest line connecting them will be the diagonal. The same principles can be used for air navigation. What does the Pythagorean Theorem have to do with squares? The Pythagorean Theorem is Pythagoras’ most famous mathematical contribution. The Pythagorean Theorem states that: “The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides.” What are the perfect squares from 1 to 1000? How many Perfect Squares between 1 and 1000. There are 30 perfect squares between 1 and 1000. They are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900 and 961. ## What are the perfect squares from 1 to 20? Between 1 to 20, the numbers 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 are even square numbers and 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 are odd square numbers. ### What do we learn from Pythagorean Theorem? Pythagoras’ theorem is startlingly simple: c squared equals a squared plus b squared. In words, in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. At a practical level, the theorem is used by many on a day to day basis. Is Pythagorean Theorem only for right triangles? Pythagoras’ theorem only works for right-angled triangles, so you can use it to test whether a triangle has a right angle or not. How do you solve A2 B2 C2? Introduction: Pythagorean Theorem The formula is A2 + B2 = C2, this is as simple as one leg of a triangle squared plus another leg of a triangle squared equals the hypotenuse squared. ## Is Pythagorean theorem only for right triangles? ### What are all the perfect squares from 1 to 400? The perfect squares are the squares of the whole numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 … What is the square of 1 to 20? When to use the Pythagorean theorem with square roots? Pythagorean Theorem With Square Roots You can use the Pythagorean Theorem to solve for any length if you know the lengths of two other sides. Suppose you need the length of the hypotenuse c c. Then you simply need the square root of the sum of a2 + b2 a 2 + b 2, like this: ## How is the Pythagorean theorem used in real life? The Pythagorean Theorem is a very handy way to find the length of any one side of a right triangle if you know the length of the other two sides. What is the Pythagorean Theorem? ### Which is the correct rule for a Pythagoras triple? (You may like to read about Pythagoras’ Theorem. or an Introduction to Pythagorean Triples first) A “Pythagorean Triple” is a set of positive integers, a, b and c that fits the rule: a 2 + b 2 = c 2. And when we make a triangle with sides a, b and c it will be a right angled triangle (see Pythagoras’ Theorem for more details): How to find the longer leg B in Pythagorean theorem? And if you need to find the longer leg b, you rewrite the formula to look like this: An firefighter’s extension ladder is leaning against a building, so its top just touches the gutters at the roof edge. You know the ladder is 41 feet long and it is 9 feet from the wall of the building.
Algebra Tiles Powerpoint NUCC lesson shared with Utah Teachers through PBWorks Wiki as part of Secondary Math II; TOPIC: Algebra Tiles UtahNUCCSecondary Math II Algebra Tiles Powerpoint Created 3 years ago Duration 0:00:00 1024 NUCC lesson shared with Utah Teachers through PBWorks Wiki as part of Secondary Math II; TOPIC: Algebra Tiles Slide Content 2. Slide 2 - 2 • Algebra Tiles • Algebra tiles can be used to model operations involving integers. • Let the small green square represent +1 and the small pink square represent -1. • The green and pink squares are additive inverses of each other. 3. Slide 3 - 3 • Algebra Tiles • Algebra tiles can be used to model operations involving variables. • Let the green rectangle represent +1x or x and the pink rectangle represent -1 x or -x . • The green and red rods are additive inverses of each other. 4. Slide 4 - 4 • Algebra Tiles • Let the green square represent x2. The pink square represents -x2. • As with integers, the green squares and the pink squares form a zero pair. 5. Slide 5 - 5 • Zero Pairs • Called zero pairs because they are additive inverses of each other. • When put together, they model zero. 6. Slide 6 - Practice with Integers • Algebra tiles can be use to model adding, subtracting, multiplying, and dividing real numbers. • Remember, if you don’t have enough of something, you can add “zero pairs” • 6 7. Slide 7 - Modeling Addition/Subtraction • -2 – 3 • Take away positive 3. But wait, I don’t have 3 so I must add zero pairs! • Now remove positive 3 • You are left with -5 • 7 8. Slide 8 - Modeling Addition/Subtraction • -1 + 3 • Now, you must remove any “zero pairs” • You are left with 2. • 8 • -5 + 3 • -1 + 4 • 4 – -2 • 2 + -3 • 9 10. Slide 10 - Modeling Multiplication • 2 x -3 • This means I need 2 rows of -3 • Which is -6 • This could also mean “the opposite of 3 rows of two” • 10 • 1 x -4 • 3 x -3 • -2 x 4 • -2 x -2 • 11 12. Slide 12 - Modeling Division • 6 2 • The “6” is called the dividend. The “2” is called the divisor. • Place the “2” outside, and then line up the 6 inside. • You answer is what fits on top, which is 3. (called the quotient) • 12 • 4 / 2 • 8 / -4 • -6 / 3 • 3 / -1 • 13 14. Slide 14 - 14 • Modeling Polynomials • Algebra tiles can be used to model expressions. • Model the simplification of expressions. • Add, subtract, multiply, divide, or factor polynomials. • To solve equations with polynomials. 15. Slide 15 - 15 • Modeling Polynomials • 2x2 • 4x • 3 or +3 16. Slide 16 - 16 • More Polynomials • 2x + 3 • 4x – 2 17. Slide 17 - 17 • Use algebra tiles to simplify each of the given expressions. Combine like terms. Look for zero pairs. Draw a diagram to represent the process. • Write the symbolic expression that represents each step. 18. Slide 18 - 18 • (2x + 4) + (x + 1) • Combine like terms to get three x’s and five positive ones • = 3x + 5 19. Slide 19 - 19 • (3x – 1) – (2x + 4) • Now remove 2x and remove 4. But WAIT, I don’t have 4 so I must add zero pairs. • Now remove 2x and remove 4 • And you are left with x - 5 20. Slide 20 - You Try • (2x – 1) + (x + 2) • (x + 3) + (x – 2) • (2x – 1) – (x + 5) • (3x + 5) + (x – 1) • (4x – 3) – (3x – 2) • 20 21. Slide 21 - Adding Polynomials • This process can be used with problems containing x2. • (2x2 + 5x – 3) + (-x2 + 2x + 5) • (2x2 – 2x + 3) – (3x2 + 3x – 2) • 21 22. Slide 22 - Distributive Property • Multiplying a monomial to a polynomial • 3(x – 2) • = 3x - 6 • -2(x - 4) • 23 • = -2x + 8 24. Slide 24 - You try • 4 (x + 2) • 2 (x – 3) • -2 (x + 1) • -2 ( x – 1) • 24 25. Slide 25 - Multiplying Polynomials • (x + 2)(x + 3) • x2 + 2x + 3x + 6 = x2 + 5x + 6 • Fill in each section of the area model • Combine like terms 26. Slide 26 - Multiplying Polynomials • (x – 1)(x +4) • = x2 + 3x – 4 • Fill in each section of the area model • Make Zeroes or • combine like terms • and simplify • x2 • + 4x • – 1x • – 4 27. Slide 27 - 27 • You Try • (x + 2)(x – 3) • (x – 2)(x – 3) • (x – 1) ( x + 4) • (x – 3) (x – 2) 28. Slide 28 - 28 • Dividing Polynomials • Algebra tiles can be used to divide polynomials. • Use tiles and frame to represent problem. Dividend should form array inside frame. Divisor will form one of the dimensions (one side) of the frame. • Be prepared to use zero pairs in the dividend. 29. Slide 29 - 29 • Dividing Polynomials • x2 + 7x +6 • x + 1 • = (x + 6) 30. Slide 30 - Dividing Polynomials • x2 – 5x + 6 • x – 3 • = (x – 2) 31. Slide 31 - 31 • You Try • x2 + 7x +6 • x + 1 • 2x2 + 5x – 3 • x + 3 • x2 – x – 2 • x – 2 • x2 + x – 6 • x + 3 32. Slide 32 - 32 • Factoring Polynomials • Factoring is the process of writing a polynomial as a product. • Algebra tiles can be used to factor polynomials. Use tiles and the frame to represent the problem. • Use the tiles to fill in the array so as to form a rectangle inside the frame. • Be prepared to use zero pairs to fill in the array. 33. Slide 33 - 33 • Factoring Polynomials • 3x + 3 • 2x – 6 • = 3 · • (x + 1) • = 2 · • (x – 3) 34. Slide 34 - You Try • Factor 4x – 2 • Factor 3x + 6 • Factor • Factor • 34 35. Slide 35 - Factoring Polynomials • x2 + 6x + 8 • = (x + 2)(x + 4) 36. Slide 36 - Factoring Polynomials • x2 – 5x + 6 • = (x – 2)(x – 3) • Remember: You must form a RECTANGLE out of the polynomial 37. Slide 37 - Factoring Polynomials • x2 – x – 6 • = (x + 2)(x – 3) • This time the polynomial doesn’t form a rectangle, so I have to add “zero pairs” in order to form a rectangle. 38. Slide 38 - You Try • x2 + 3x + 2 • x2 + 4x + 3 • x2 + x – 6 • x2 – 1 • x2 – 4 • 2x2 – 3x – 2 39. Slide 39 - Solving Equations • We can use algebra tiles to solve equations. • Whatever you do to one side of the equal sign, you have to do to the other to keep the equation “balanced”. 40. Slide 40 - Solving Equations • 3x + 4 = 2x – 1 • First build each side of the equation • Now remove 2x from each side. • Next, remove 4 from each side. But wait, I don’t have 4 so I must add “zero pairs” • Remove 4 from each side • You are left with x = -5 • = 41. Slide 41 - Solving Equations • 4x + 1 = 2x + 7 • First, build each side of the equation • Next, remove 2x from each side. • Remove 1 from each side. • Now divide each side by 2. • Your result is x = 3. • = 42. Slide 42 - You Try • 2x + 3 = x – 2 • x – 4 = 2x + 1 • 3x + 1 = x – 5 • 8x – 2 = 6x + 4 43. Slide 43 - Credits • Adapted by Marcia Kloempken, Weber High School from David McReynolds, AIMS PreK-16 Project
GeeksforGeeks App Open App Browser Continue # Algebraic Expressions Algebraic expressions are mathematical statements that are used to explain various conditions. We use variables and constants along with mathematical operators to define the algebraic expressions. Algebraic expressions are used to define unknown conditions in real life. Suppose we have to find the age of Arum if the age of his sister is twice the age of Arun and the sum of their age is 24 years. This situation can be easily explained using the Algebraic Expressions let the age of Arun be x then the age of his sister be 2x and the sum of their ages is x + 2x = 24 this is an algebraic expression, ## What are Algebraic Expressions? Algebraic expressions are the expressions obtained from the combination of variables, constants, and mathematical operations like addition, subtraction, multiplication, division, and so on. An algebraic expression is made up of terms, there can be one or more than one term present in the expression. ### Examples of Algebraic Expressions Some examples of algebraic expressions are, • 5x + 4y • 11x – 12 • 2y – 13, etc. Here the above three expressions are the algebraic expression they have variables x, and y and constant terms -12, and -13. ## Constants, Variables, and Coefficients In the algebraic expression, fixed numerals are called constants. Constants do not have any variables attached to them. For example, 3x – 1 has a constant -1 to it. The image showing the constant and variables of an Algebraic Expression is added below, Variables are the unknown values that are present in the algebraic expression. For instance, 4y + 5z has y and z as variables. Coefficients are the fixed values (real numbers) attached to the variables. They are multiplied by the variables. For example, in 5x2 + 3 the coefficient of x2 is 5. A term can be a constant, a variable, or a combination of both. Each term is separated by either additioan or subtraction. For example, 3x + 5, 3x, and 5 are the two terms of the algebraic expression 3x + 5. ## Types of Algebraic Expressions There are various types of algebraic expressions based on the number of terms in the algebraic expression. The three main types of algebraic expressions are, • Monomial Expression • Binomial Expression • Polynomial Expression ### Monomial Expression Monomial Expressions are algebraic expressions with only one term. For example, 2x, 5y, 3x2, 11xy, etc. all are monomial expressions. ### Binomial Expression Binomial Expressions are algebraic expressions with two terms. For example, 2x + 3y, 5y + 11z, 3x2 + 4x, y +11xy, etc. all are binomial expressions. 2x + 3x is not a binomial expression as it can be further simplified as 5x. ### Polynomial Expression Polynomial Expressions are algebraic expressions with more than two terms. For example, 2x + 3y + 11, 5y + 11z + 3x, 3x2 + 4x + 2, X + y +11xy, etc. all are polynomial expressions. ## Other Types of Algebraic Expressions We can also categorize Algebraic Expressions into, • Numeric Expression • Variable Expression Let’s learn about them in detail. ### Numeric Expression An expression containing only numbers and not variables is called a numeric expression. They contain only numbers that are operated using various mathematical operators. For example 11 + 6, 15/2, 7 – 3, etc all are numeric expressions. ### Variable Expression An expression containing both variables and numbers is called a variable expression. For example, 11x + 6y, 15x2 + 7, y – 3, etc all are numeric expressions. The table below explains all the types of expression in brief. ## Simplifying Algebraic Expressions Simplifying algebraic expressions is easy and very basic. First, understand what are like and unlike terms. Like terms have the same sign and unlike terms have opposite signs. To simplify the given algebraic expression, first, find out the terms having the same power. Then, if the terms are like terms, add them; if they are unlike terms, find the difference between the terms. The most simplified form of an algebraic expression is one where no same power terms are not repeated. For instance, let’s simplify 4x5 + 3x3 – 8x2 + 67 – 4x2 + 6x3, the same powers that are repeated are cubic and square, upon combining them together, the expression becomes, 4x5 + (3x3 + 6x3) – (8x2 – 4x2) + 67. Now, simplifying the expression, the final answer obtained is 4x5 + 9x3 – 12x2 + 67. This term does not have any terms repeated that have the same power. When an addition operation is performed on two algebraic expressions, like terms are added with like terms only, i.e., coefficients of the like terms are added. Example: Add (25x + 34y + 14z) and (9x − 16y + 6z + 17). Solution: (25x + 34y + 14z) + (9x − 16y + 6z + 17) By writing like terms together, we get = (25x + 9x) + (34y − 16y) + (14z + 6z) + 17 By adding like terms, we get = 34x + 18y + 20z + 17. Hence, (25x + 34y + 14z) + (9x − 16y + 6z + 17) = 34x + 18y + 20z + 17. ### Subtraction of Algebraic Expressions To subtract an algebraic expression from another, we have to add the additive inverse of the second expression to the first expression. Example: Subtract (5b2 + 6b + 8) from (3b2 − 5b). Solution: (5b2 + 6b + 8) − (3b2 − 5b) = (5b2 + 6b + 8) + (−3b2 + 5b) = (5b2 − 3b2) + (6b + 5b) + 8 = 2b2 + 11b + 8 ### Multiplication of Algebraic Expressions When a multiplication operation is performed on two algebraic expressions, we have to multiply every term of the first expression with every term of the second expression and then combine all the products. Example: Multiply (3x + 2y) with (4x + 6y − 8z) Solution: (3x + 2y)(4x + 6y − 8z) = 3x(4x) + 3x(6y) − 3x(8z) + 2y(4x) + 2y(6y) − 2y(8z) = 12x2 + 18xy − 24xz + 8xy + 12y2 − 16yz = 12x2 + 12y2 + 26xy − 16yz − 24xz ### Division of Algebraic Expressions When we have to divide an algebraic expression from another, we can factorize both the numerator and the denominator, then cancel all the possible terms, and simplify the rest, or we can use the long division method when we cannot factorize the algebraic expressions. Example: Solve: (x2 + 5x + 6)/(x + 2) Solution: = (x2 + 5x + 6)/(x + 2) After factorizing (x2 + 5x + 6) = (x + 2) (x + 3) = [(x + 2) (x + 3)]/(x + 2) = (x + 3) ## Algebraic Formulas The general algebraic formulas we use for solving the algebraic expressions or algebraic equations are, • (x + a) (x + b) = x2 + x(a + b) + ab • (a + b)2 = a2 + 2ab + b2 • (a – b)2 = a2 – 2ab + b2 • (a + b)2 + (a – b)2 = 2 (a2 + b2) • (a + b)2 – (a – b)2 = 4ab • a2 – b2 = (a – b)(a + b) • (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca • (a + b)3 = a3 + b3 + 3ab(a + b) • (a – b)3 = a3 – b3 – 3ab(a – b) • a3 – b3 = (a – b)(a2 + ab + b2) • a3 + b3 = (a + b)(a2 – ab + b2) • a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)) ## Algebraic Expressions Examples Example 1: Find out the constant from the following algebraic expressions, • x3 + 4x2 – 6 • 9 + y5 Solution: Constants are the terms that do not have any variable attached to them, therefore, in the first case, -6 is the constant, and in the second case, 9 is the constant. Example 2: Find out the number of terms present in the following expressions, • 4x2 + 7x – 8 • 5y7 – 12 Solution: Terms are separated by each other either by addition or subtraction sign. Therefore, in the first case, there are 3 terms and in the second case, there are 2 terms. Example 3: Simplify the algebraic term, z5 + z3 – y6 + 7z5 – 8y6 + 34 + 10z3 Solution: In the expression, there are terms with the same power and same variable that are repeated, first bring them together, (z5 + 7z5) + (z3 + 10z3) – (y6 – 8y6) + 34. Now, simplify the expression, 8z5 + 11z3 – 9y6 + 34. Example 4: Add (13x2 + 11), ( – 25x2 + 26x + 42) and (–33x – 29). Solution: Let F = (13x2 + 11) + ( – 25x2 + 26x + 42) + (–33x – 29) ⇒ F = 13x2 – 25x2 + 26x – 33x + 11 + 42 – 29 F = –12x2 – 7x + 24 Hence, (13x2 + 11) + ( – 25x2 + 26x + 42) + (–33x – 29) = –12x2 – 7x + 24. Example 5: Solve (5x + 4y + 6z)2 + (3y – 7x)2. Solution: Given, Let F = (5x + 4y + 6z)2 + (3y – 7x)2 From algebraic formulae, we have (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (a – b)2 = a2 – 2ab + b2 ⇒ F = (5x)2 + (4y)2 + (6z)2 + 2(5x)(4y) + 2(4y)(6z) + 2(6z)(5x) + [(3y)2 – 2(3y)(7x) + (7x)2] ⇒ F= 25x2 + 16y2 + 36z2 + 40xy + 48yz + 60zx +9y2 – 42xy + 49x2 Now, combine the like terms. = 74x2 + 25y2 + 36z2 – 2xy + 48yz + 60zx Hence, (5x + 4y + 6z)2 + (3y – 7x)2 = 74x2 + 25y2 + 36z2 – 2xy + 48yz + 60zx. ## FAQs on Algebraic Expressions ### Q1: What are Algebraic Expressions? Algebraic Expressions are mathematical expression which is used to represent various mathematical condition they use variables and constants to define various real-life condition. ### Q2: What are Types of Algebraic Expressions? The three main types of algebraic expressions are • Monomial: A monomial is an expression that has only one non-zero term. 2xy, 5y3, 7a, 2b, etc are some examples of monomials. • Binomial: A binomial is an expression that has two non-zero terms. For example, 2a + 3 has two monomials 2a and 3 and hence it is a binomial. • Polynomial: A polynomial is an expression that has more than two non-zero, unlike terms. For example, x-4y+ 8z has three monomials x, 4y, and z, hence it is a polynomial. ### Q3: Are all Algebraic Expressions Polynomials? Not all algebraic expressions are polynomials but all polynomials are algebraic expressions i.e. there exist some algebraic expressions that are not polynomials. Example: πx + 1 is an algebraic expression but not a polynomial. ### Q4: How Algebraic Expressions are Derived? Algebraic expression is an idea of representing numbers using vrious variables such as x, y, z, etc. without specifying their actual values. In simple terms, an algebraic expression is a mathematical statement where variables have been combined using fundamental arithmetic operations. For example, Ram’s age is three times the age of Akash, and their total age is 48. Express it as an algebraic equation. Let the age of Akash be x then age of Ram age is 3x, now the required algebric expression is 3x + x = 48
How do you find the domain and range of y = -2x^2 - 8 ? Jun 14, 2017 domain$\to x \in \left(- \infty , + \infty\right)$ range $\to y \in \left(- \infty , + \infty\right)$ Explanation: In the alphabet $\textcolor{red}{d}$ comes before $\textcolor{red}{r}$ So $\textcolor{red}{d}$omain comes before $\textcolor{red}{r}$ange Input comes before output So domain$\to$input$\text{; }$range$\to$ output domain$\to x \in \left(- \infty , + \infty\right)$ range $\to y \in \left(- \infty , + \infty\right)$ Jun 14, 2017 Domain: $x \in \left(- \infty , \infty\right)$ Range: $y \in \left[- 8 , - \infty\right)$ Explanation: The domain includes all the values of $x$ that you can "legally" plug into the equation. If plugging in zero forced you to divide by zero or put a negative into a square root, those would be outside the domain of acceptable $x$ values. In the equation $y = - 2 {x}^{2} - 8$, you are neither dividing by an $x$ term, nor are you dealing with $x$ values under a radical sign. Therefore, the domain includes all real numbers. That means you can plug in any real number you like, from $- \infty$ to $+ \infty$. Domain: All real numbers, $\mathbb{R}$, which can be written as: $x \in \left(- \infty , \infty\right)$ The range is all the values of $y$ that this function can produce. Because the equation is a quadratic equation, we have a parabola. That means that the maximum of the parabola is the upper bound of the equation and there is no lower bound. The vertex will be the upper most top part of this function. Vertex: $x = - \frac{b}{2 a} = - \frac{0}{2 \left(- 2\right)} = \frac{0}{4} = 0$ Plugging $0$ into the function gives $y = - 8$. So the range is Range: All values of $y$ less than $- 8$, or $y \in \left[- 8 , - \infty\right)$
Find the angle inside a triangle in terms of two variables : $\alpha,\beta$ CZ is perpendicular to XY and the ratio of AZ to ZB is $1 : 2$. The angle $ACX$ is $\alpha$ and the angle $BCY$ is $\beta$. Find an expression for the angle $AZC$ in terms of $\alpha$ and $\beta$. By using the sine rule (in my opinion, which can be used here): $\dfrac{AZ}{sin(90^{\circ}-\alpha)}=\dfrac{ZC}{sin\angle{ACB}}=\dfrac{AC}{sin\angle{AZC}}$.........(1) $\dfrac{BZ}{sin(90^{\circ}-\beta)}=\dfrac{ZC}{sin\angle{ABC}}=\dfrac{BC}{sin({\pi-AZC})=sin\angle{AZC}}$..........(2) Now $\dfrac{AZ}{ZB}=\dfrac{1}{2}$(Given) $\dfrac{\dfrac{AZ}{ZC}}{\dfrac{ZB}{ZC}}=\dfrac{1}{2}$........(3) Using $(1),(2) \space and \space (3)$, we get, $\dfrac{AC}{BC}=\dfrac{sin\angle{ABC}}{sin\angle{CAB}}$= $\dfrac{1}{2}\cdot\dfrac{cos\beta}{cos\alpha}$ Again $\angle{CAB}+\angle{ABC}=\alpha+\beta$(by careful observation) Again using cosine rule on $\Delta{ACZ}$, we get $\sqrt{AC^2+ZC^2-2AC\cdot ZC \cdot sin \alpha}= AZ$ Putting value of AZ above in $(1)$, we get, $\dfrac{\sqrt{AC^2+ZC^2-2AC\cdot ZC \cdot sin \alpha}}{cos\alpha}=\dfrac{AC}{sin\angle{AZC}}$ But here I am stuck as I think I am going somewhere else. Please give me any suggestion, idea or directly, the answer(if you'd actually do that I'll be obliged) The probable solution is below: • In the future, please post your attempted solution that you would like checked in the main post, and use images from software or mathjax instead of pictures - these will significantly improve the response you get! – B. Mehta Apr 17 '18 at 18:33 In the 3rd line it should be angle $ZCA=90-\alpha$, wrongly written as $ZAC=90-\alpha$
Home > Pre-algebra > Decimals > Adding Decimals #### Introduction The process of adding decimals is very similar to adding whole numbers. Doing so is simple if you remember to pay attention to the position of the decimal point when adding. #### Terms Decimal Point - The decimal point looks like a period and is always placed after the “ones” position of a number. Decimal Notation - Decimal notation is useful for writing numbers which are not whole numbers. Whole Number - A whole number is a number with nothing but zeroes after the decimal point (5.0) or with no decimal point at all (5). ## Lesson The most important thing to remember about adding decimals is to make sure that the decimal place lines up when adding. , for example. To make things easier, it may be helpful to place additional zeroes after the shorter number until both numbers have the same amount of digits. For example, when adding .05 and .0034, the addition is more straightforward if you write the equation as . Lining it up vertically makes it easier to solve: ## Examples Add the digits in the Tenths column (4+5 = 9) Add the digits in the Ones column (1+2 = 3) Noah received a 91.86 score in Math and a 72.29 in English. Find Noah's total score in the two subjects. Add the digits in the Hundredths column (6+9 = 15) Add the digits in the Tenths column (1+8+2 = 11) Add the digits in the Ones column (1+1+2 = 4) Add the digits in the Tens column (9+7 = 16) Add the digits in the Hundreds column (1 = 1) Add the digits in the Thousandths column (2+0 = 2) Add the digits in the Hundredths column (0+++5 = 5) Add the digits in the Tenths column (4+4 = 8)
# Simple Proportional Relationships Lesson We've already learnt about linear equations, which showed a relationship between two variables. Now we are going to look at a special kind of linear relationship called a proportional relationship. Two quantities are said to be proportional if they vary in such a way that one is a constant multiple of the other. In other words, they always vary by the same constant. We can also calculate the unit rate in a proportional relationship, which tells us how much the dependent variable will changes with a one unit increase in the independent variable. For example, if the cost of some items is always five times the number of items, we can say that this is a proportional relationship because there is a constant multiple between the cost and the number of items - $5$5. We can write these proportional relationships as linear equations. The example above could be written as $y=5x$y=5x and again we can see that the coefficient of $x$x describes the constant of the proportional relationship. We will learn more about the constant of proportionality and writing proportional relationships as equations later but now let's focus on determining whether relationships are proportional or not. Remember! A relationship is proportional if there is a constant multiple between the two variables. We can also compare proportional relationships to make judgements about rates of change. #### Examples ##### Question 1 Consider the equation $y=7x$y=7x. 1. What is the gradient of $y=7x$y=7x? 2. Select the graph that below that shows $y=7x$y=7x. A B C D A B C D ##### Question 2 Irene and Valentina are both making handmade birthday cards. Irene can make $8$8 cards every $13$13 minutes. Valentina can make $6$6 cards every $15$15 minutes. 1. Plot this information on the graph. 2. How can you tell who was quicker at making cards? The steeper line relates to the faster maker A The shallower line relates to the faster maker B The steeper line relates to the faster maker A The shallower line relates to the faster maker B ##### Question 3 Oliver is making cups of fruit smoothie. The amount of bananas and strawberries he uses is shown in the proportion table. Strawberries Bananas $4$4 $8$8 $12$12 $16$16 $20$20 $5.5$5.5 $11$11 $16.5$16.5 $22$22 $27.5$27.5 1. Graph this proportional relationship. 2. What is the unit rate of this relationship? 3. Select ALL the statements that describe the proportional relationship. For every $5.5$5.5 bananas Oliver uses, he adds $4$4 strawberries. A The unit rate of bananas in respect to strawberries is $\frac{11}{8}$118. B For every $4$4 bananas, Oliver uses $5.5$5.5 strawberries. C The unit rate of bananas in respect to strawberries is$\frac{8}{11}$811. D For every $5.5$5.5 bananas Oliver uses, he adds $4$4 strawberries. A The unit rate of bananas in respect to strawberries is $\frac{11}{8}$118. B For every $4$4 bananas, Oliver uses $5.5$5.5 strawberries. C The unit rate of bananas in respect to strawberries is$\frac{8}{11}$811. D We'll learn later about direct proportional relationships and inverse proportional relationships.
# Partitioning 2-digit numbers ## Presentation on theme: "Partitioning 2-digit numbers"— Presentation transcript: Partitioning 2-digit numbers 63 = So when we partition 2-digit numbers we we split them up in to their tens and units © 2013 Partitioning 2-digit numbers 63 = So when we partition 2-digit numbers we we split them up in to their tens and units © 2013 Partitioning 2-digit numbers 63 = So when we partition 2-digit numbers we we split them up in to their tens and units © 2013 Partitioning 3-digit numbers 271 = So when we partition 3-digit numbers we we split them up in to their hundreds, tens and units © 2013 Partitioning 3-digit numbers 271 = So when we partition 3-digit numbers we we split them up in to their hundreds, tens and units © 2013 Partitioning 3-digit numbers 271 = So when we partition 3-digit numbers we we split them up in to their hundreds, tens and units © 2013 Partitioning 4-digit numbers We partition 4-digit numbers in to their thousands, hundreds, ten and units 4,975 = 4, 4,975 = 2, , 4,975 = 4, © 2013 Tenths Units tenths Units can be split in to tenths 10 tenths make 1 unit Units tenths © 2013 Hundredths Units hundredths Units can also be split in to hundredths 100 hundredths make 1 unit Units hundredths © 2013 Numbers with decimal places Numbers with decimal follows the same rules The further to the left a number is, the more it is worth U t h th © 2013 Zeros after the last digit In numbers with decimal places, zeros after the last number do not change the value of the number The numbers in the same colours below have the same value as each other, despite the extra zeros on the end 8 8.0 © 2013 Partitioning numbers with decimal places Partitioning numbers with decimal places Partitioning numbers with decimal places 0.752 = So when we partition numbers with decimal places we split them up in to their units, tenths, hundredths and thousandths © 2013 Partitioning numbers with decimal places 0.752 = So when we partition numbers with decimal places we split them up in to their units, tenths, hundredths and thousandths © 2013 Partitioning numbers with decimal places When we partition numbers with decimal places we split them up in to their units, tenths, hundredths and thousandths 4.975 = 4.975 = 4.975 = © 2013
# Lesson Overview (7/7/15) Rigid Transformations- transformations that don’t change the shape or size of an object; also called an isometry- the preimage and image are congruent. The three types of rigid transformations are reflections, rotations, and translations. We practiced applying each transformation with different methods: • Using appropriate tools/strategies to facilitate the process (tracing, folding, measuring) • Just “eyeballing it” • Using properties of the transformation to create accurate information (Perpendicular bisectors for reflections, equal distances and angles for rotations) We then worked on the coordinate plane to come up with generalized “(x, y) ->…” rules for the different cases of the different transformations (reflections across x-axis, y-axis, y=x, y=-x; rotations about the origin clockwise, counter-clockwise, 90 deg, 180 deg) Special Right Triangles 45-45-90 and 30-60-90 Special because one is isosceles and the other is half of an equilateral (or equiangular) triangle. Key idea: All isosceles right triangles (45-45-90) are similar by AA similarity. All 30-60-90 triangles are similar by AA similarity. We let the smallest side of each triangle equal length 1 for easy reference. By the properties of each triangle, we can determine one other side automatically: • For a 45-45-90, since it is isosceles, if one leg is 1 unit, the other leg is 1 unit. • For a 30-60-90, since it is half of an equilateral triangle, the smallest side is half of the largest side. In both cases, we can use the Pythagorean Theorem to determine the 3rd side in each reference triangle. From there, any time you see a 30-60-90 or 45-45-90 triangle with one side given, you can use the two reference triangles discussed above (with smallest side equal to 1), to determine scale factors, create proportions or use other identifiable patterns to solve for missing sides.
Find two linear functions f(x) and g(x) such that their product h(x) = f(x).g(x) is tangent to each of f(x) and g(x) at two distinct points. Discuss and illustrate the method and the results. we can write f(x) = ax + b as f(x) = a(x - x0) g(x)= cx + d as g(x) = c(x - x1) Then h(x) = f(x)g(x) is a parabola which meet with x-axis at x0 and x1. (because h(x0)=f(x0)g(x0)=a(x0-x0)g(x0)=0 and likewise, h(x1)=0) Case1) a>0 , c>0 Investigate the case1 We can know that it's impossible for parabola to be tangent to both f(x) and g(x). Case2) a<0 , c<0 Investigate the case2 In this case it's impossible for parabola to be tangent to both f(x) and g(x) as well. Case3) a>0 , c<0 Since parabola is symmetric against x=x-coordinate of vertex and h(x) passes through (x0,0) and (x1,0) and x-coordinate of vertex of h(x)=(x0+x1)/2 and h(x) must be tagent to each of f(x) and g(x), the graph must be the following we can get the First Result: the slope of f(x) = opposite sign of the slope of g(x) a = -c Let's consider another condition. We want to get the y-coordinate of the vertex downward and the make the graph as the following Explore the situation Note the situation that an intersection of two lines becomes a vertex of their product, parabola. Let's the vertex (xt, yt). Then h(xt)=f(xt)=g(xt) and h(xt)=f(xt)g(xt)=f(xt)f(xt)=f(xt). Therfore f(xt)=0 or 1 If f(xt)=g(xt)=0 then x0=x1=xt and f(0)+g(0)=0 Let's see the graph If f(xt)=g(xt)=1 then f(0)+g(0)=2 Let's see the graph Let's see the graphs simultaneously. We can realize that the y-value of the parabola which we want to find must be below the product of the y-values of two lines. To be that situation, two y-values of lines simultaneously must be below "1"and above "0". we can get the Second Result: f(0) + g(0) =1 RESULT Two linear functions f(x) and g(x) such that their product h(x) = f(x).g(x) is tangent to each of f(x) and g(x) at two distinct points 1)the slope of f(x) = opposite sign of the slope of g(x) and 2)f(0) + g(0) =1 Example 1. Eample 2.
# Compound Interest Part 7 Compounded half-yearly, then rate=$\frac{r}{2}$% and time=2time Compounded quarterly, then rate=$\frac{r}{4}$% and time=4time Problem1: The compound interest earned on a sum in 3 years at 15% per annum compounded annually is RS.25002. Find the sum? Solution: R=15% Time=3 year C.I=25002 12003=3600 1803=540 3600+540+27=4167 4167——>25002 8000——–>? $\frac{25002\times&space;8000}{6}=48000$ is principle Problem 2: A sum of RS.19600 is invested at 20% rate of compound interest for 2 years compounded half yearly. then the end of two years compound interest will be how much more than the S.I? Solution: Rate=$\frac{20}{2}=10$% time =22=4 years we will split 4 years into 2 and 2 years R=10% T=2 years C.I=10+10+$\frac{10\times&space;10}{100}=21$% R=21% T=2 years 21+21+$\frac{21\times&space;21}{100}=21+21+4.41=46.41$% C.I=46.41% S.I=104=40% 46.41-40=6.41% 100%———>19600 6.41%———->? $\frac{19600\times&space;6.41}{100}=1256.36$ Problem 3: If a sum of Rs.3600 is invested in two different banks for 2 years first offering 20% compound interest compounded annually and second offer 20% compounded half yearly then find the difference of the interest after 2 years? Solution: first offering 20% compound interest compounded annually 20%———-C.I———-Annually 20+20+$\frac{20\times&space;20}{100}$ = 44% second offer 20% compounded half yearly Rate=$\frac{20}{2}=10$% , Time=2*2=4 years 10+10+$\frac{10\times&space;10}{100}$=21% 21+21+$\frac{21\times&space;21}{100}$=46.41% 46.41 – 44=2.41% 100%———->3600 2.41%———>? $\frac{3600\times&space;2.41}{100}$=86.76
The Pythagorean Theorem shows how strange our concept of distance is. Using the rule a2 + b2 = c2, we can trade some "a" to get more "b". Starting with $\displaystyle{13^2 + 0^2 = 13^2}$ means "A 13-inch pizza equals a 13-inch pizza". Sure. But we can trade an inch and get: $\displaystyle{12^2 + 5^2 = 13^2}$ Huh? A 12-inch pizza and a 5-inch pizza equal a 13-inch pizza? The math works (144 + 25 = 169) but, but... we gave up an inch and got a five-inch pizza! Let's understand why the tradeoff happens, and how to use it. ## Explanation 1: Shaving the Square A key insight: Bigger numbers are harder to square. Imagine laying tiles on a porch -- as your porch grows, the outer layer needs more tiles. Trimming a 13x13 porch to 12x12 frees up 25 tiles, which is enough to make a new 5x5 porch! I call this "shaving the square". Trimming 1 unit from the outside of a large square has more "shavings" which can contribute to a smaller one (trimming an inch from a giant fro can make a sweater for an infant). As we continue to trim, the benefit diminishes because our starting point is smaller and smaller. ## Explanation 2: Sliding the Chopstick A second insight: Slide a little, pivot a lot. Imagine a chopstick wedged in a corner: the length is fixed, and the ends of the chopstick must touch a wall. What're the options? Well, laying on a single wall means 100% for one side (like saying 132 + 02 = 132). Not that interesting. By sliding the chopstick (from 13 to 12) we can swing it out by 5 on the other wall! You need to try it -- a small slide gives a giant pivot. As we keep sliding, the tradeoff (How much pivot do we get?) changes. Time to see how the a/b tradeoff works. First, let's use grid coordinates: x & y (horizontal and vertical). Given a fixed distance (13 units, let's say), our options lay on the circle where x2 + y2 = 132: A few points: • Each possibility is the same distance, but has a different ratio of x to y (100% x, 100% y, or a mix like (12,5)) • We can only move to neighboring points on the circle (options at the same distance) • The tradeoff we face is how much "x" we get for "y" when moving to a neighbor. If we're at (0, 13) we could move to (5, 12). This trades 1 y for 5 x's. This is the "chunky" tradeoff where we're using an entire unit at a time. What about .5 units? .01? Enter the tangent! The tangent line shows the trajectory of our current path, the direction to our neighbor. We follow the tangent for a tiny, microscopic amount to get our next neighbor. The tangent is an approximation -- it's not pointing exactly at our nearest neighbor, but it's pretty close. What's the actual amount? Any point (x,y) has a slope of y/x, and a tangent line with slope -x/y, so the tradeoff is...getting confused yet? Less mindless algebra, more intuition: • Circles have a tangent line perpendicular to the current point • If you're at (5,12) then tangent slope is some ratio of 5 and 12 • Remember "shaving the square": you get a better deal in the direction of the smaller coordinate (increasing a large square is tough). • So, at (5, 12) you're "heavy on the y" and the trade will favor improving your x: it should be "trade 5 y's for 12 x's". And why not the other way? It doesn't make sense that the more y you have, the easier it is to get y! That'd spiral off into exponential growth, not a circle. • Lastly, we can't trade an entire chunk of 5 y's! The tangent is about our nearest neighbor. We have a trade of 12/5 or 2.4 to 1. Our next, tiny movement will be at this ratio (and then we'll be at a new point, with a new tangent). General principle: Our neighbors are on a circle, which encourages balance. You get a better deal in the direction of the smaller coordinate: at (x,y) the tradeoff is y:x. Now we know the tradeoff for any point (x,y) -- let's optimize! In a boring scenario, we get paid based on pure distance, so every point (or direction to move) is the same. The exciting scenario: our (x,y) position is an input into some other function which gives us a return! Now we want to maximize that function. Here's a scenario: Popeye throws cars for cash. He lines up spectators on fences running North and East. The spectators must look straight ahead (they're in neck braces, due to earlier events) but will pay Popeye if they see a car pass in front of them. ## Maximizing Even Payouts Suppose each spectator offers $1 if they see the car (Payout (x,y) = x + y). Where to throw? First, assume Popeye has finite energy -- he can throw the car 13 meters. Now let's start somewhere: throwing the car pure North (0, 13): P(0,13) = 0 + 13 =$13 Ok. What if he threw it slightly East? To (5, 12) let's say? P(5,12) = 5 + 12 = $17 Clearly better. This should make sense: at (0,13) the tradeoff is great to get more East. We can give up 1 North and get a whopping 5 East, a "profit" of$4 if we do the trade. We should keep trading as long as it's profitable -- as long as we're out of balance, the circle will reward us for boosting the smaller side. Following a 45 degree angle for 13 units is the ideal: P(13 * 1/sqrt(2), 13 * 1/sqrt(2)) = P(13 * .707, 13 * .707) = 9.2 + 9.2 = $18.4 Neat. A 45-degree throw hits 70.7% of the possible spectators for each side. Psst. Confused about how a 45-degree through passes by 70.7% of the spectators on each side? No problem. A 45-degree throw is along the diagonal of a square. A triangle with sides 1 and 1 has a hypotenuse of: $\displaystyle{\sqrt{1^2 + 1^2} = \sqrt{2} = 1.414}$ And has sides (1, 1, 1.414). A hypotenuse of √(2) isn't convenient: it's hard to know what fraction a side is of the whole. We divide the triangle by the length of the hypotenuse (√(2)), making the hypotenuse 1 and the other sides a percentage: $\displaystyle{\text{Triangle with sides} = (\frac{1}{\sqrt(2)}, \frac{1}{\sqrt(2)}, \frac{\sqrt{2}}{\sqrt(2)}) = (.707, .707, 1)}$ Now we've discovered that a 45-degree throw, with sides (1, 1, √(2)), has the ratio .707, .707, 1. 70.7% of the distance along the hypotenuse shows up on each side. ## General Technique: Finding the Best Direction We stumbled upon the way to find the best return: • Pick any starting point / direction • Tweak it: if our return improves, keep the new choice (it's profitable) • Keep tweaking until our return is no longer profitable In math slang, this is "finding the local maximum". In economics slang, it's finding the point of "zero marginal returns". Popeye calls it Squeezing the Spinach. ## Maximizing Uneven Returns Now suppose the Northern spectators offer$2 (Eastern stay at $1), so P(x,y) = x + 2y. Should we throw it 100% North? P(0, 13) = 0 + 213 =$26 P(9.2, 9.2) = 9.2 + 29.2 = $27.6 Interesting -- 45 degrees is still better! But... I think we went too far! Shouldn't we favor North since it pays more? Yep. Let's remember how to Squeeze the Spinach (maximize our returns): start with North and change until it's not profitable: • The payout function means 1 North = 2 Easts (North pays$2, so 1 unit North = 2 units East) • Trades are profitable if we can beat 1 North for 2 Easts (1 North for 3 Easts, for example, would profit $1) So... where are trades better than 1 North for 2 Easts? In the Northern section, where the circle rewards us by throwing Easts at us ("Please, please go East... I'll give you a bunch if you give up a little North"). Remember how circles are about x/y, x & y, x:y, etc.? Well, we have the numbers 1 and 2. (2,1) is in the East section. We want (1,2). Why? At (1,2) we have reached the perfect 1 North = 2 East tradeoff. Following the direction (1,2) for 13 units is: P(13 1/sqrt(5), 13 * 2/sqrt(5)) = P(5.81, 11.62) = 5.81 + 2*11.62 =$29.05 Tada! Over 29 smackeroos because we maximized our return. We can supercharge this result: To maximize return, go in each direction proportional to its payoff. If North pays 2:1 compared to East, your trajectory should favor North by 2:1. In mathier terms: • Payoff(x,y) = ax + by • Best trajectory = (a, b) [in our case, (East, North) => (1, 2)] And this works in multiple dimensions! Given 3 dimensions, go in a direction (Payoff(x), Payoff(y), Payoff(z)). Vector calculus fans, this is why the gradient is in the direction of greatest increase. $\displaystyle{(\frac{dF}{dx},\frac{dF}{dy},\frac{dF}{dz})}$ And each partial derivative (dF/dx) is the payoff for moving in that direction. But does it all balance? Suppose x pays 3, y pays 4, and z pays 5 (at the current position). The 2-dimensional tradeoff trajectories are: $\displaystyle{ (x, y) = (3,4) }$ $\displaystyle{ (y, z) = (4, 5) }$ $\displaystyle{ (x, z) = (3, 5) }$ Now for the magic: the combined trajectory $\displaystyle{(x,y,z) = (3,4,5)}$ satisfies all 3 requirements! On the x-z plane, x doesn't care about y -- as long as the ratio to z is (3 , ?, 5) you're getting the best tradeoff from the x-z perspective. The pairs are: • (3, ?, 5) • (?, 4, 5) • (3, 4, ?) You don't need a sudoku master to see (3, 4, 5) satisfies all those proportions. Still not convinced? Imagine the payoff for y was zero. We don't want to waste energy in our trajectory (3, ?, 5) in a useless direction. But that can't happen, because the y-z tradeoff will be (?, 0, 5) and the x-y tradeoff will be (3, 0, ?). The x-z tradeoff lets y-z and x-y "figure out" what y should be, which is 0. ## Questions I Had That You Might Have Too Q: I still don't get why this works at all. Somehow 50% in x and 50% in y leads to .7 + .7 = 1.4? It's a deep question about why space behaves like this. I was going crazy staring at chopsticks on a wall. Here's my answer: distance is distance. 13 units is 13 units. But in some situations we are "measuring our coordinates" (what are the values of x & y) and not the distance itself. Cartesian coordinates (x-axis, y-axis) are very inefficient for diagonal motion (i.e., you are measuring the sides of the triangle, not the hypotenuse). When .7072 + .7072 = 1, it's a measure how how "inefficient" our x & y coordinates are being. We used 70% of each coordinate to represent an object that could have been 100% on one (i.e, if we used polar coordinates). Q: I have an offshore investment with 200% return, and an onshore one with 5% return. I have $1000 to spend -- should I split my money? Heavens, no! Remember, this principle is about distance measurements on a grid with the idea that 50% in x and 50% in y covers "more ground" than 100% in x. In investing 1) money is not on a grid and 2) there's no distance bonus. Putting half your money in each is plain old 0.5 + 0.5 = 1.0. Giving up$1 of the offshore investment gives you \$1 for the onshore one. Put all your money in the best investment. Q: So all this stuff is useless? Heavens, no! Ask yourself: am I measuring distance on a coordinate system? Many things are measured in terms of x-y coordinates (physical phenomena, etc.) and do have the Pythagorean distance tradeoff. But not every graph is the same. Graphs that aren't about distance (like "Money vs. Time") do not get any boost from the Pythagorean theorem. This confused me for a long time: the Pythagorean Theorem works for coordinate distance! ## Final Thoughts The Pythagorean Theorem is so versatile -- it's not about triangles, it covers the nature of distance. I seem to find some new realization when I study it. Really grokking it will help you everywhere, from geometry to vector calculus. Happy math.
## Announcements: Continue reading Section 2.2 for next class. Work through recommended homework questions. Quiz 2 is this week, and will cover the material until the end of Section 2.1, focusing on Sections 1.3 and 2.1. Next office hour: Monday, 1:30-2:30. Help Centers: Monday-Friday 2:30-6:30 in MC 106. Linear algebra TAs are there on Mondays, Wednesdays and Thursdays, but you may go any day. ### Section 2.1: Systems of Linear Equations Definition: A system of linear equations is a finite set of linear equations, each with the same variables. A solution to the system is a vector that satisfies all of the equations. Example: \begin{aligned} x + y &= 2\\ -x + y &= 4 \end{aligned} $[1, 1]$ is not a solution, but $[-1, 3]$ is. Geometrically, this corresponds to finding the intersection of two lines in $\R^2$. A system is consistent if it has one or more solutions, and inconsistent if it has no solutions. We'll see later that a consistent system always has either one solution or infinitely many. ### Solving a system Example: Here is a system, along with its augmented matrix: \kern-6ex \begin{aligned} \ph x - \ph y - \ph z &= 2 \\ 3 x - 3 y + 2 z &= 16 \\ 2 x - \ph y + \ph z &= 9 \end{aligned} \qquad\qquad \bmat{rrr\|r} 1 & -1 & -1 & 2 \\ 3 & -3 & 2 & 16 \\ 2 & -1 & 1 & 9 \emat Geometrically, solving it corresponds to finding the points where three planes in $\R^3$ intersect. We solved it by doing row operations, such as replacing $R_2$ with $R_2 - 3 R_1$ or exchanging rows 2 and 3 until we got it to the form: \kern-6ex \begin{aligned} \ph x - \ph y - \ph z &= 2 \\ y + 3 z &= 5 \\ 5 z &= 10 \end{aligned} \qquad\qquad \bmat{rrr\|r} 1 & -1 & -1 & 2 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & 5 & 10 \\ \emat This system is easy to solve, because of its triangular structure. The method is called back substitution: \begin{aligned} z &= 2\\ y &= 5 - 3z = 5 - 6 = -1\\ x &= 2 + y + z = 2 - 1 + 2 = 3. \end{aligned} So the unique solution is $[3, -1, 2]$. We can check this in the original system to see that it works! ### New material Question: How many solutions does the system \begin{aligned} 2 x + 3 y &= 2 \\ x + 2 y &= 2 \\ x + 4 y &= 2 \end{aligned} have? ### Section 2.2: Direct Methods for Solving Linear Systems In general, we won't always get our system into triangular form. What we aim for is: Definition: A matrix is in row echelon form if it satisfies: 1. Any rows that are entirely zero are at the bottom. 2. In each nonzero row, the first nonzero entry (called the leading entry) is further to the right than any leading entries above it. Example: These matrices are in row echelon form: $$\kern-6ex \bmat{rrr} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & 0 & 0 \emat \qquad \bmat{rrr} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & 0 & \red{4} \emat \qquad \bmat{rrrrr} 0 & \red{3} & 2 & 0 & 4 \\ 0 & 0 & 0 & \red{-1} & 2\\ 0 & 0 & 0 & 0 & \red{4} \emat$$ Example: These matrices are not in row echelon form: $$\kern-6ex \bmat{rrr} {\bf 0} & {\bf 0} & {\bf 0} \\ \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ \emat \qquad \bmat{rrr} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & {\bf 2} & 4 \emat \qquad \bmat{rrrrr} 0 & \red{3} & 2 & 0 & 4 \\ 0 & 0 & 0 & \red{-1} & 2\\ 0 & 0 & {\bf 2} & 0 & 4 \emat$$ This terminology makes sense for any matrix, but we will usually apply it to the augmented matrix of a linear system. The conditions apply to the entries to the right of the line as well. Question: For a $2 \times 3$ matrix, in what ways can the leading entries be arranged? Just as for triangular systems, we can solve systems in row echelon form using back substitution. Example: Solve the system whose augmented matrix is: $$\bmat{rrr\|r} \red{3} & 2 & 2 & 0\\ 0 & 0 & \red{-1} & 2\\ 0 & 0 & 0 & 0 \emat$$ How many variables? How many equations? Solution on board. Example: Solve the system whose augmented matrix is: $$\bmat{rr\|r} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & 0 & \red{4} \emat$$ How many variables? How many equations? Note: This is the general pattern for an augmented matrix in row echelon form: 1. If one of the rows is zero except for the last entry, then the system is inconsistent. 2. If this doesn't happen, then the system is consistent. ### Row reduction: getting a matrix into row echelon form Here are operations on an augmented matrix that don't change the solution set. There are called the elementary row operations. 1. Exchange two rows. 2. Multiply a row by a nonzero constant. 3. Add a multiple of one row to another. We can always use these operations to get a matrix into row echelon form. Example on board: Reduce the given matrix to row echelon form: $$\bmat{rrr} -2 & 6 & -7 \\ 3 & -9 & 10 \\ 1 & -3 & 3 \emat$$ Note that there are many ways to proceed, and the row echelon form is not unique. Row reduction steps: (This technique is crucial for the whole course.) 1. Find the leftmost column that is not all zeros. 2. If the top entry is zero, exchange rows to make it nonzero. 3. (Optional) It may be convenient to scale this row to make the leading entry into a 1, or to exchange rows to get a 1 here. 4. Use the leading entry to create zeros below it. 5. Cover up the row containing the leading entry, and repeat starting from step (a). Note that for a random matrix, row reduction will often lead to many awkward fractions. Sometimes, by choosing the appropriate operations, one can avoid some fractions, but sometimes they are inevitable. Example: Here's another example: \begin{aligned} \bmat{rrrr} 0 & 4 & 2 & 3 \\ 2 & 4 & -2 & 1 \\ -3 & 2 & 2 & 1/2 \\ 0 & 0 & 10 & 8 \emat \xrightarrow{R_1 \leftrightarrow R_2} &\bmat{rrrr} \red 2 & \red 4 & \red{-2} & \red 1 \\ \red 0 & \red 4 & \red 2 & \red 3 \\ -3 & 2 & 2 & 1/2 \\ 0 & 0 & 10 & 8 \emat \\ \lra{\frac{1}{2}R_1} \bmat{rrrr} \red 1 & \red 2 & \red{-1} & \red{1/2} \\ 0 & 4 & 2 & 3 \\ -3 & 2 & 2 & 1/2 \\ 0 & 0 & 10 & 8 \emat \lra{R_3 + 3R_1} &\bmat{rrrr} \phm 1 & 2 & -1 & 1/2 \\ 0 & 4 & 2 & 3 \\ \red 0 & \red 8 & \red{-1} & \red 2 \\ 0 & 0 & 10 & 8 \emat \\ \lra{R_3 - 2R_2} \bmat{rrrr} \phm 1 & 2 & -1 & 1/2 \\ 0 & 4 & 2 & 3 \\ 0 & \red 0 & \red{-5} & \red{-4} \\ 0 & 0 & 10 & 8 \emat \lra{R_4 + 2R_3} &\bmat{rrrr} \phm 1 & 2 & -1 & 1/2 \\ 0 & 4 & 2 & 3 \\ 0 & 0 & -5 & -4 \\ 0 & 0 & \red 0 & \red 0 \emat \\ \end{aligned} Example: $\bmat{rrr} 2 & 4 & 6 \\ 1 & 2 & 4 \\ -3 & -6 & 4 \emat$
Given triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove: HD/AD + HE/BE + HF/CF = 1 and AH/AD + BH/BE + CH/CF = 2. To prove these relationships, we will use area of triangles and ratios. First we will prove the relation equating to 1. Proof: We first note that triangle ABC must be acute; that is, any angle of the triangle must be less than or equal to 90°. For triangle ABC, we know the area A = 1/2BCAD = 1/2ACBE = 1/2ABCF, where AD, BE, and CF are the altitudes to their respective bases. We will let A1 be the area of triangle BHC, A2 be the area of triangle AHB, and A3 be the area of triangle AHC, where H is the orthocenter of triangle ABC. Then, A1 = 1/2HDBC, A2 = 1/2HFAB, and A3 = 1/2HEAC. We know that the sum of the areas of these triangles is equal to the area of triangle ABC; thus, A = A1 + A2 + A3. If we divide through both sides with A, the area of the original triangle, then we have 1 = A1/A + A2/A + A3/A. Now we coordinate values of A based on the values of each area of the smaller triangles. Thus, A1/A = (1/2HDBC)/(1/2BCAD) = HD/AD. Next, A2/A = (1/2HFAB)/(1/2ABCF) = HF/CF. Lastly, A3/A = (1/2HEAC)/(1/2ACBE) = HE/BE. Thus, we have that 1 = HD/AD + HF/CF + HE/BE. Now we much show that 2 = AH/AD + BH/BE + CH/CF. We already have that 1 = HD/AD + HF/CF + HE/BE. Note that HD = AD - HA, HF = FC - HC, and HE = BC - HB. Then we use these in substitution with the previously found relation. Therefore, 1 = (AD - HA)/AD + (FC - HC)/FC + (BE - HB)/BE = 1 - HA/AD + 1 - HC/FC + 1 - HB/BE = 3 - (HA/AD + HC/FC + HB/BE). Now we have that 1 = 3 - (HA/AD + HC/FC + HB/BE); hence, 2 = HA/AD + HC/FC + HB/BE. QED
# what are the factors of 34 ## FACTORS OF 34, FACTOR PAIRS OF 34 So 1, 2, 17 and 34 are factors of 34. The factors of 34 Answer: 1, 2, 17, 34 Factors of – 34 ## Factors of 34 gcflcm Factors of 34 are 1, 2, 17. There are 3 integers that are factors of 34. The biggest factor of 34 is 17. Positive integers that divides 34 without a remainder are listed below. ## Factors of 34 The prime factors of 34 are 2 and 17. It is the list of the integer''s prime factors. The number of prime factors of 34 is 2. Factor tree or prime decomposition for 34 ## What are the factors of 34 Research Maniacs The factors of 34 are listed with the smallest number first, which is 1, and the largest number last, which is 34. 1, 2, 17, 34 Note that all factors of 34 are whole postive numbers. In other words, the factors of 34 are postive integers that go evenly into 34. ## What are the factors of 34 [SOLVED] Mathwarehouse The factors of 34 are: 1, 2, 17, 34 ## Factors of 34 Find Prime Factorization/Factors of 34 Factors of 34 are the list of integers that can be evenly divided into 34. There are overall 4 factors of 34 among which 34 is the biggest factor and 1, 2, 17, and 34 are positive factors. The Prime Factors and Pair Factors of 34 are 1, 2, 17, 34 and (1, 34) and (2, 17) respectively. Factors of 34: 1, 2, 17 and 34 ## category/business/What are all the factors of 34 The factorization or decomposition of 34 = 217. Notice that here, it is written in exponential form. The prime factors of 34 are 2 and 17. It is the list of the integer''s prime factors. The number of prime factors of 34 ## Prime factors of 34 Math Tools Prime factors of 34 : 2, 17 In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly. The prime factorization of a positive integer is a list of the integer''s prime factors, together with their multiplicities; the process of determining these factors is called integer factorization. ## Factor Calculator, Find The Factors The Factor Calculator finds the factors, factor pairs, and prime factors. Enter an integer number to find its factors. What are the Factors Factors: A factor is a number that will divide into another number without a remainder. Factors are whole numbers
# For her fitness class, Ana bought a stability ball that has a volume of 4,500π cubic inches. Fredrick is taking the same fitness class and needs a stability ball that is congruent to Ana’s because he is the same height. What must the diameter of Fredrick’s stability ball be for it to be congruent to Ana’s? A current institution question and answer asked students to declare what they believe is the main important thing for a student to do in order for you to gain success. From the numerous responses, one that that stood out was practice. Persons who ordinarily are successful do not become successful by being born. They work hard and dedication their lives to succeeding. This is how you can attain your goals. beneath are one of the answer and question example that you can work with to practice and expand your understanding and also give you insights that might assist you to keep up your study in school. ## Question: For her fitness class, Ana bought a stability ball that has a volume of 4,500π cubic inches. Fredrick is taking the same fitness class and needs a stability ball that is congruent to Ana’s because he is the same height. What must the diameter of Fredrick’s stability ball be for it to be congruent to Ana’s? 30 inches Step-by-step explanation: As, the stability ball of Ana is congruent to the stability ball of Fredrick. This implies that the volume of both the balls will be same. So, Volume of Ana’s ball = Volume of Fredrick’s ball = 4500 . Now, the volume of a sphere is given by , where ‘r’ is the radius. Substituting the value of the volume in this formula, we will obtain the radius. i.e. i.e. i.e. r = 15 inches As, diameter is the double of the radius. So, diameter = 2×radius = 2×15 = 30 inches. Hence, the diameter of Fredrick’s ball the 30 inches. READ MORE As countries get richer, their population growth rate tends to _____. From the answer and question examples above, hopefully, they could simply assist the student resolve the question they had been looking for and take note of everything that declared in the answer above. You will possibly then have a discussion with your classmate and continue the school learning by studying the subject to one another.
QUADRILATERALS – EXERCISE 4.2.1 – Class 9 1. Is a parallelogram a rectangle? Can you call a rectangle a parallelogram? Solution: A parallelogram becomes a rectangle when its angles measure 90° each. Therefore a parallelogram need not be a rectangle. But a rectangle is a parallelogram 1. Prove that bisectors of two opposite angles of parallelogram are parallel. Given: ABCD is a \| \| gm AE and CF bisect DAB and DCB respectively To prove: AE \| \| CF Proof: Statement Reason ABCD is \| \|gm given ∴ DC \| \| AB ∠CDA + ∠DAB = 180° Co-interior angles ∴ ∠DAB = 180° – ∠CDA are supplementary But ∠DAE = 1/2 ∠DAB ∵ AE bisects DAB ∴ ∠DAE = 1/2 [180° – ∠CDA] ∠DAE = 90° – 12 [180 – ∠CDA] ……. (1) ∠DEA = 180° – (∠DAE + ∠ADE ) = 180° – [90°– 12 ∵ ∠ ADE = ∠CDA + ∠ADE ∠CDA from equation(1) = 180° – [90°– 1/2 ∠CDA + ∠CDA] [∵∠CDA – 1/2 ∠CDA = ∠CDA] = 180° – [90°– 1/2 ∠CDA] = 180° – 90°– 1/2 ∠CDA ∠DAE = 90° – 1/2 ∠CDA …….. (2) From (1) and (2) ∠DAE = ∠DEA …….. (3) Further, ∠ECF = 1/2 ∠DCB, But ∠DCB = ∠DAB ∠ECF = 1/2 ∠DEA ECF = ∠DAF But from (3) DAE = DEA ∠ECF = ∠DEA But these are corresponding angles ∴ AE \| \| FC 1. Prove that if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus. Solution: Given: ABCD is a parallelogram. Diagonals AC and BD intersect at right angles. To prove: ABCD is a rhombus Proof: reason ABCD is parallelogram given AC ⊥ BD In ΔAOD and ΔCOD, OD is common AO = CO Diagonals bisect each other ∠AOD = ∠COD = 90° given ∴ΔAOD ≅ ΔCOD SAS ∴AD = CD C.P.C.T ∴ABCD is a rhombus adjacent side equal 1. Prove that if the diagonals of a parallelogram are equal then it is a rectangle. Solution: Given: ABCD is a parallelogram. Diagonals AC and BD are equal. To prove: ABCD is a rectangle. Proof: reason In ΔDAB and ΔCBA Opposite sides of a parallelogram. DA = CB AB = BA DB =CA given ∴ΔDAB ≅ ΔCBA SSS ∴∠DAB = ∠CBA CPCT But ∠DAB + ∠CBA = 180° AD \| \| BC ∴∠DAB = ∠CBA = 180/2 = 90° angles ∴ABCD is a rectangle an angle is a right angles
## Nobbs' Points, Gergonne LineWhat are they? A Mathematical Droodle Explanation ### Nobbs' Points, Gergonne Line The applet attempts to illustrate the following theorem: Assume A', B', C' are the points of contact of the incircle of ΔABC: A' on side BC, etc. Denote the intersection of AB and A'B' as C'', that of AC and A'C' as B'', and let A'' be the intersection of BC and B'C'. Then points A'', B'', and C'' are collinear. The points A'', B'', and C'' are known as Nobbs' points. The theorem tells us the Nobbs' points of a triangle are collinear. The line the points lie on was named the Gergonne line by Adrian Oldknow [Oldknow, pp. 324-325] after J. D. Gergonne. The reason for the name is as follows. It is known that the lines AA', BB', and CC' meet at the Gergonne point Ge of ΔABC. Put another way, this means that two triangles ABC and A'B'C' are perspective in Ge. By Desargues' theorem, triangles perspective from a point are also perspective from a line. Thus the corresponding side lines of triangles ABC and A'B'C' cross in the points that belong to the same line. This is the Gergonne line of Oldknow. I'll give an additional argument in support of that nomenclature: the Gergonne line is the polar of the Gergonne point Ge in the incircle of ΔABC. Indeed, A'B' is the polar of C, AB is the polar of C'. Therefore, C'' is the pole of CC', by La Hire's theorem. Similarly, A'' is the pole of AA' and B'' is the pole of BB'. Since their polars AA', BB', and CC' are concurrent (in the Gergonne point Ge), the poles A'', B'', and C'' are collinear to a line (the Gergonne line, naturally.) Note that if one of the triangles ABC and A'B'C' is equilateral, so is the other, and in this case the Gergonne line is the line at infinity. ### Remark ΔA'B'C' is known as the contact triangle (and also Gergonne triangle) of ΔABC. ΔABC is the tangent(ial) triangle of ΔA'B'C'. (From a somewhat different perspective this same configuration is studied elsewhere. A more general statement is also available.) ### References 1. A. Oldknow, The Euler-Gergonne-Soddy Triangle of a Triangle, Amer Math Monthly, Vol. 103, No. 4 (Apr. 1996), 319-329
# TEAS Math - Practice #1 Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Winnieng W Winnieng Community Contributor Quizzes Created: 4 \| Total Attempts: 3,896 Questions: 34 \| Attempts: 1,382 Settings This is a practice exam in Math part for Teas test. The format of exam is the same, but the details in questions are not exactly, just similiar and closely like. You can practice and experience how a Teas test should be. Good luck! • 1. ### 15 is what percent of 75? • A. 10% • B. 15% • C. 25% • D. 20% D. 20% Explanation To find the percentage, we need to divide the given number by the total and then multiply by 100. In this case, 15 divided by 75 equals 0.2. Multiplying 0.2 by 100 gives us 20%. Therefore, 15 is 20% of 75. Rate this question: • 2. ### It takes a mechanic 25 minutes to run an emissions test on a car. There are 3 cars to be tested. If the mechanic begins at 1:40 p.m., when will she be finished? • A. 3:15 p.m • B. 2:55 p.m • C. 2:45 p.m • D. 3:05 p.m. 3:05 p.m. 3:05 p.m B. 2:55 p.m Explanation The mechanic takes 25 minutes to test one car. Since there are 3 cars to be tested, she will take a total of 25 minutes x 3 cars = 75 minutes to finish testing all the cars. If she starts at 1:40 p.m., she will finish at 1:40 p.m. + 75 minutes = 2:55 p.m. Therefore, the correct answer is 2:55 p.m. Rate this question: • 3. ### A student earns \$2,280.50 each month at a part-time job. The student pays the following amounts for expenses each month: Rent………… \$452.00 Food………… \$360.00 Utilities……… \$265.80 Car expenses \$540.00 After paying the monthly expenses listed above, which of the following represents the amount of money the student has left for other expenses? • A. \$664.70 • B. \$564.90 • C. \$1,615.80 • D. \$662.70 D. \$662.70 Explanation The student earns \$2,280.50 each month and pays \$452.00 for rent, \$360.00 for food, \$265.80 for utilities, and \$540.00 for car expenses. To find the amount of money the student has left for other expenses, we subtract the total expenses from the monthly earnings. Therefore, the student has \$2,280.50 - (\$452.00 + \$360.00 + \$265.80 + \$540.00) = \$662.70 left for other expenses. Rate this question: • 4. ### If the perimeter of a rectangular house is 74 yards, and the length is 24 feet, what is the width of the house? • A. 35 feet • B. 29 yards • C. 28 feet • D. 15 yards B. 29 yards Explanation The perimeter of a rectangle is calculated by adding the lengths of all its sides. In this case, we are given that the perimeter is 74 yards. Since the length of the house is given in feet, we need to convert it to yards. There are 3 feet in 1 yard, so the length of the house is 24/3 = 8 yards. To find the width, we subtract the length from the perimeter: 74 - 8 - 8 = 58 yards. Therefore, the width of the house is 29 yards. Rate this question: • 5. ### What amount does A most closely represent? • A. • B. 33 % • C. Two eighths • D. 6.25% D. 6.25% Explanation A most closely represents 6.25%. This is because 6.25% is the only option that is a percentage value, while the other options are either fractions or ratios. Rate this question: • 6. ### Larry bought a house for \$120,000. After one year, its value appreciated (increased in value) by 25%. During the second year, its value depreciated (decreased in value) by 18% from its value at the end of the first year. What was the value of the house at the end of the second year? • A. \$123,000 • B. \$3,960 • C. \$202,640 • D. \$120,000 A. \$123,000 Explanation The value of the house at the end of the second year is \$123,000. This can be calculated by first finding the value of the house at the end of the first year, which is \$120,000 + 25% of \$120,000 = \$150,000. Then, we calculate the value of the house at the end of the second year by taking 18% of \$150,000 and subtracting it from \$150,000, which gives us \$150,000 - 18% of \$150,000 = \$123,000. Rate this question: • 7. ### The ratio of 7:4 = (?)% • A. 100% • B. 175% • C. 135% • D. 125% B. 175% Explanation The ratio of 7:4 can be expressed as 7/4. To convert this ratio to a percentage, we divide 7 by 4 and multiply by 100. This gives us (7/4) * 100 = 175%. Therefore, the correct answer is 175%. Rate this question: • 8. ### The symbol \$ is defined by: a \$ b = a2 – 4b2 If x \$ y = 3(y \$ x), express y in terms of x. B. Explanation To find y in terms of x, we can substitute the given expression into the equation. So, we have: x \$ y = 3(y \$ x) Substituting the definition of \$, we get: x^2 - 4y^2 = 3(y^2 - 4x^2) Expanding and rearranging the equation, we get: x^2 - 4y^2 = 3y^2 - 12x^2 Combining like terms, we have: 4x^2 + 4y^2 = 0 Dividing both sides by 4, we get: x^2 + y^2 = 0 Since the sum of squares cannot be negative, the only solution is x = 0 and y = 0. Therefore, y in terms of x is y = 0. Rate this question: • 9. ### A computer is on sale for \$2100, which is a 30% discount off the regular price. What is the regular price? • A. \$1800 • B. \$1900 • C. \$3000 • D. \$3400 • E. \$2200 C. \$3000 Explanation The regular price of the computer can be found by dividing the sale price by (100% - discount percentage) and multiplying by 100%. In this case, the sale price is \$2100 and the discount is 30%. So, dividing \$2100 by (100% - 30%) and multiplying by 100% gives us \$3000, which is the regular price of the computer. Rate this question: • 10. ### What is the value of x in this 30-60-90 triangle? • A. 3 • B. 2 • C. • D. C. Explanation In a 30-60-90 triangle, the ratio of the sides is 1:√3:2. Since the given side is 2, we can determine the value of x by multiplying it with the ratio of the side adjacent to the 30-degree angle, which is 1. Thus, the value of x in this 30-60-90 triangle is 2. Rate this question: • 11. ### If 300 jellybeans cost you x dollars. How many jellybeans can you purchase for 30 cents at the same rate? • A. • B. • C. 150x • D. 6x B. Explanation If 300 jellybeans cost x dollars, then the cost of each jellybean is x/300 dollars. To find out how many jellybeans can be purchased for 30 cents, we need to divide 30 cents by the cost of each jellybean. Since 1 dollar is equal to 100 cents, the cost of each jellybean is (x/300) * 100 cents. Dividing 30 cents by this value gives us (30 / ((x/300) * 100)) = (30 * 300) / (x * 100) = 9000 / (x * 100) jellybeans. Simplifying further, we get 90 / x jellybeans. Therefore, the correct answer is 90/x. Rate this question: • 12. ### A goat eats 225 kg. of hay in 45 days, while a cow eats the same amount in 15 days. In how many days could they eat 225 kg. of hay together? • A. 15 • B. 25 • C. 12 • D. 20 D. 20 Explanation The cow eats 225 kg of hay in 15 days, which means it eats 15 kg of hay per day. Similarly, the goat eats 225 kg of hay in 45 days, which means it eats 5 kg of hay per day. Together, they eat a total of 15 + 5 = 20 kg of hay per day. Therefore, it would take them 20 days to eat 225 kg of hay together. Rate this question: • 13. ### The pie chart describes the percent of time that each person spent watching television. What is the ratio between the amount of time spent by females to males? • A. • B. • C. • D. 70% C. Explanation The ratio between the amount of time spent by females to males can be calculated by dividing the percentage of time spent by females (70%) by the percentage of time spent by males (100% - 70% = 30%). This results in a ratio of 7:3, indicating that females spent 7 times more time watching television than males. Rate this question: • 14. ### What is the cost in dollars to steam clean a room 2W yards wide and 3L yards long it the steam cleaners charge 10 cents per square foot? • A. 0.1WL • B. 0.9WL • C. 5.4WL • D. 9WL C. 5.4WL Explanation The cost to steam clean a room is calculated by multiplying the width and length of the room and then multiplying it by the cost per square foot. In this case, the width is 2W yards and the length is 3L yards. Therefore, the cost would be 2W * 3L * 0.1 dollars, which simplifies to 0.6WL dollars. However, in the given answer, the cost is stated as 5.4WL dollars. This answer is incorrect and does not provide the accurate calculation for the cost of steam cleaning the room. Rate this question: • 15. ### Which two years did the least number of boys attend the convention? • A. 1995 and 1998 • B. 1995 and 1996 • C. 1996 and 1997 • D. 1997 and 1998 B. 1995 and 1996 Explanation The correct answer is 1995 and 1996. This can be determined by analyzing the given options and identifying the two years that are mentioned in both options. Since the question asks for the least number of boys attending the convention, the years mentioned in both options (1995 and 1996) would have the lowest attendance. Rate this question: • 16. ### A car dealer sells a SUV for \$43,200, which represents a 35% profit over the cost. What was the cost of the SUV to the dealer? • A. \$29,250 • B. \$31,200 • C. \$32,000 • D. \$33,800 C. \$32,000 Explanation The cost of the SUV to the dealer can be calculated by dividing the selling price by 1 plus the profit percentage. In this case, the selling price is \$43,200 and the profit percentage is 35%. So, the cost of the SUV to the dealer is \$43,200 / (1 + 0.35) = \$32,000. Rate this question: • 17. ### The circle, center O, has a radius of 8 inches. Find the length of the arc AB that subtends an angle of 63° at O (i.e. Angle AOB = 63°). Use π = 22/7 • A. 8 inches • B. 8.8 inches • C. 16 inches • D. 17.6 inches B. 8.8 inches Explanation The length of an arc can be found by using the formula: Arc length = (angle/360) * 2 * π * radius In this case, the angle is given as 63° and the radius is given as 8 inches. Plugging in these values into the formula, we get: Arc length = (63/360) * 2 * (22/7) * 8 = (0.175) * (2) * (22/7) * 8 = 0.55 * 22 * 8 = 121.6 inches Therefore, the length of the arc AB is 121.6 inches, which is not one of the given answer choices. Thus, the correct answer is not available. Rate this question: • 18. ### Oxacillin (Prostaphlin), 500 mg. p.o. q6h, has been ordered for your client. Oxacillin comes in a suspension containing 250 mg/5 mL. How many total milliliters would you administer if the order calls for 14 days of therapy? • A. 450 mL • B. 650 mL • C. 250 mL • D. 560 mL D. 560 mL Explanation The order is for 500 mg of oxacillin every 6 hours, and the suspension contains 250 mg/5 mL. To calculate the total volume needed for 14 days of therapy, we need to determine how many doses will be administered in 14 days. Since the medication is taken every 6 hours, there are 4 doses in a day (24 hours divided by 6 hours). Therefore, in 14 days, there will be a total of 56 doses (14 days multiplied by 4 doses per day). Each dose requires 5 mL of the suspension, so to calculate the total volume needed, we multiply 5 mL by 56 doses, which equals 280 mL. Therefore, the correct answer is 560 mL. Rate this question: • 19. ### What is 35% of a number if 12 is 15% of a number? • A. 5 • B. 28 • C. 33 • D. 62 B. 28 Explanation If 12 is 15% of a number, we can find the number by dividing 12 by 0.15. This gives us a number of 80. Now, to find 35% of this number, we multiply 80 by 0.35. This gives us a result of 28. Therefore, 35% of the number is 28. Rate this question: • 20. ### In the graph below, no axes or origin is shown. If point B's coordinates are (8,2), which of the following coordinates would most likely be A's? • A. (6, 8) • B. (-10, 3) • C. (4, 7) • D. (10, 6) C. (4, 7) Explanation Based on the given information, we know that point B's coordinates are (8,2). Since point A and B are likely to be on the same graph, we can infer that point A would have coordinates that are close to point B's coordinates. Among the options given, the coordinates (4, 7) are the closest to (8, 2) in terms of both x and y values. Therefore, (4, 7) would most likely be A's coordinates. Rate this question: • 21. ### If your client is supposed to receive three 500-mL units of whole blood over the next 6 hours and the drip factor is 10 drops/mL, you’s adjust the rate to how many drops per minute? • A. 42 drops • B. 38 drops • C. 56 drops • D. 18 drops A. 42 drops Explanation To calculate the rate in drops per minute, we need to determine the total number of drops needed over 6 hours. Each 500 mL unit of whole blood will have 500 mL * 10 drops/mL = 5000 drops. Since the client is supposed to receive three units, the total number of drops needed is 3 * 5000 drops = 15000 drops. To find the rate in drops per minute, we divide the total number of drops by the total time in minutes (6 hours * 60 minutes/hour = 360 minutes). Therefore, the rate is 15000 drops / 360 minutes = 41.67 drops per minute. Rounded to the nearest whole number, the rate is 42 drops per minute. Rate this question: • 22. ### A local football team season ticket sales have gone up 10% in the current season, reaching 880 tickets. How many season tickets were sold in the last season? • A. 800 • B. 700 • C. 880 • D. 968 A. 800 Explanation The correct answer is 800. Since the local football team's season ticket sales have gone up by 10% in the current season and reached 880 tickets, we can assume that this 10% increase is equivalent to 80 tickets. To find the number of season tickets sold in the last season, we need to subtract this increase from the current season's sales. Therefore, 880 - 80 = 800, indicating that 800 season tickets were sold in the last season. Rate this question: • 23. ### You need to purchase a textbook for nursing school. The book cost \$80.00, and the sales tax where you are purchasing the book is 8.25%. You have \$100. How much change will you receive back? • A. \$23.50 • B. \$17.45 • C. \$13.40 • D. \$12.75 C. \$13.40 Explanation After calculating the sales tax on the textbook, which is 8.25% of \$80.00, we find that the tax amount is \$6.60. To determine the total cost, we add the price of the textbook and the sales tax, resulting in \$86.60. Since the person has \$100, we subtract \$86.60 from \$100 to find the change they will receive back, which is \$13.40. Rate this question: • 24. ### .89 + 3.8 + .036 = • A. 47.26 • B. 4.726 • C. 48.26 • D. 4.826 B. 4.726 Explanation The given expression is an addition of three numbers: 89, 3.8, and 0.036. Adding these numbers together, we get a sum of 92.836. However, the answer choices provided are not accurate representations of this sum. The closest option is 4.726, which is obtained by rounding the sum to the nearest tenth. Therefore, the correct answer is 4.726. Rate this question: • 25. ### The price of the book was decreased by 20% two times in one year. It now costs 180\$ less than the original price. What was the original price? • A. 259 • B. 450 • C. 500 • D. 900 C. 500 Explanation The original price of the book was 500\$. The price was decreased by 20% twice, resulting in a total decrease of 40%. This means that the book now costs 60% of its original price. If we let x represent the original price, we can set up the equation 0.6x = x - 180. Solving this equation, we find that x = 500. Therefore, the original price of the book was 500\$. Rate this question: • 26. ### If jogging for one mile uses 150 calories and brisk walking for one mile uses 100 calories, a jogger has to go how many times as far as a walker to use the same number of calories? • A. 1/2 • B. 3/2 • C. 2 • D. 2/3 D. 2/3 Explanation To find out how many times farther a jogger has to go compared to a walker to burn the same number of calories, we can set up a proportion. Let's say the distance the walker goes is x miles. The jogger would have to go 1 mile. The calories burned by the walker would be 100x, and the calories burned by the jogger would be 150. To set up the proportion: 100x/150 = 1/x. Cross multiplying gives us 100x^2 = 150. Dividing both sides by 100 gives us x^2 = 1.5. Taking the square root of both sides gives us x = √1.5. Therefore, the jogger has to go approximately 1.225 times farther than the walker. Since this is not an option, the closest answer is 2/3. Rate this question: • 27. ### The following figure contains both a circle and a square. What is the area of the entire shaded figure? • A. 24 + 8∏ • B. 24 + 2∏ • C. 16 + 16∏ • D. 16 + 4∏ A. 24 + 8∏ Explanation The shaded figure consists of a square and a circle. The area of the square is determined by multiplying the length of one side by itself, which is 24. The area of the circle is calculated by multiplying the square of the radius by π, which is 8π. Therefore, the area of the entire shaded figure is 24 + 8π. Rate this question: • 28. D. • 29. ### In a small village called “Rose" there are 9 families with 3 children, 8 families with 2 children and 4 families having 5 children. What is the average number of children in a family? • A. 4 • B. 3 • C. 2.5 • D. 3.5 B. 3 Explanation The average number of children in a family can be calculated by dividing the total number of children by the total number of families. In this case, there are 9 families with 3 children each, 8 families with 2 children each, and 4 families with 5 children each. So the total number of children is (9 * 3) + (8 * 2) + (4 * 5) = 27 + 16 + 20 = 63. The total number of families is 9 + 8 + 4 = 21. Therefore, the average number of children in a family is 63 / 21 = 3. Rate this question: • 30. ### The nurse practitioner orders digoxin pediatric elixir 375 micrograms daily. The elixir dilution is 0.05 mg/mL. How many teaspoons will the nurse instruct the parent to administer? • A. 2 tsp • B. 3 tsp • C. 1.5 tsp • D. 4.5 tsp C. 1.5 tsp Explanation The nurse practitioner orders digoxin pediatric elixir 375 micrograms daily. The elixir dilution is 0.05 mg/mL. To calculate the number of teaspoons to administer, we need to convert the micrograms to milligrams. 375 micrograms is equal to 0.375 milligrams. Then, we divide the milligrams by the concentration of the elixir, which is 0.05 mg/mL. This gives us a total volume of 7.5 mL. Since there are 5 mL in a teaspoon, the nurse will instruct the parent to administer 1.5 teaspoons. Rate this question: • 31. ### In a college, some courses contribute more towards an overall GPA than other courses. For example, a science class is worth 4 points; mathematics is worth 3 points; History is worth 2 points; and English is worth 3 points. The values of the grade letters are as follows, A= 4, B=3, C=2, D=1, F=0. What is the GPA of a student who made a “C” in Trigonometry, a “B” in American History, an “A” in Botany, and a “A” in Microbiology? • A. 2.59 • B. 3.08 • C. 3.38 • D. 2.86 C. 3.38 Explanation The GPA is calculated by multiplying the grade points of each course by the number of credits it is worth, then summing up the total grade points and dividing it by the total number of credits. In this case, Trigonometry (C) is worth 2 points, American History (B) is worth 3 points, Botany (A) is worth 4 points, and Microbiology (A) is worth 4 points. Adding these grade points together gives a total of 13. Dividing this by the total number of courses (4) gives an average of 3.25. Therefore, the GPA of the student is 3.38. Rate this question: • 32. ### Three tenths of 18 equals: • A. 7.2 • B. 3.4 • C. 4.7 • D. 5.4 D. 5.4 Explanation To find three tenths of 18, we can multiply 18 by 0.3. This is because three tenths is equivalent to 0.3. When we multiply 18 by 0.3, we get 5.4. Therefore, the correct answer is 5.4. Rate this question: • 33. ### The cost of a list of supplies for a hospital ward is as follows: \$19.98, \$52.20, \$12.64, and \$7.79 What is the total cost? • A. \$91.30 • B. \$93.61 • C. \$92.61 • D. \$93.60 C. \$92.61 Explanation The correct answer is \$92.61. To find the total cost, we add up all the individual costs: \$19.98 + \$52.20 + \$12.64 + \$7.79 = \$92.61. Rate this question: • 34.
»» Printable Multiplication Chart 1 20 # Printable Multiplication Chart 1 20 PrintableJD Have you ever wondered How Many Ways There Are to Make 20? You can learn more about it in this article! In addition, you will be able to print out a Printable Multiplication Chart 1 20 that you can use to practice and review these facts. If you have any problems with these facts, feel free to share them with others! We hope this article has helped you! It will be an invaluable resource for those who are struggling to learn multiplication facts. ## What Multiplication Problems That Equal 20? You have a multiplication problem: What times does this number? You have to find the missing number to get to the answer 20. Once you have found the missing number, find the other number that would make the answer 20. For example, 6 times X is equal to 20. If the unknown number is 60, then it is 1200 times 60, which would make the answer 21. This is a very common problem. Factors of 20 are the same as those for 40. This means that they are prime factors. Prime factors are those which produce the result of 20 when multiplied by two. Prime factors are those with two primes, which are 1 and 20. You can find the other multiples of 20 by factorizing them. However, remember that these factors are prime and cannot be subtracted when solving a multiplication problem. ## How Many Ways To Make 20 With Multiplication? In the case of multiplication, there are many ways to calculate the product. Adding one number to another gives you the product 20. Multiplying two numbers gives you six times seven, and adding seven to six gives you twenty. The same can be done with multiples of five. There are even ways to multiply two decimals. These methods vary, so make sure you know which one is right for you. First, determine the prime factors of the number. Prime factors are numbers that when multiplied together, produce a result of 20. For example, 1,2,4,5,10, and 20 are all factors of 20. These factor pairs are found by a process called factorization. This method uses the first two prime factors, 1 and 20, and then continues finding other multiples of twenty. You’ll soon find that you have an infinite number of ways to multiply any number. ## Printable Multiplication Chart 1 20 A Printable Multiplication Chart 1-20 shows students how to multiply a number into one, two, or three digits. It is an excellent tool for any classroom setting and provides a visual representation of the process. Students can use the chart to learn one, two, or three-digit multiples, and they can expand it to any number. The charts are also perfect for home use. These can be useful for students in kindergarten through high school and are a great way to reinforce math concepts. Although multiplication charts are useful math tools, they are not always easy for children. Especially those who dislike math may have a difficult time with them. Therefore, there are many ways to make them more engaging. For instance, you can make a chart with matching games, which kids can play while reviewing the multiplication tables. Or, you can simply give the child a worksheet with the charts. Ultimately, this will help them remember the facts. Printable Categories
# Number Theory¶ This page was originally authored by Dr. Brett Olsen, who taught CSE 232 in Fall 2014. Number theory is the study of the integers. The most basic concept in number theory is divisibility. We say that $b$ divides $a$ (written $b\|a$) if $a=bk$ for some integer $k$. We can also say that $a$ is a multiple of $b$, or that $b$ is a divisor of $a$ (if $b \geq 0$). Every positive integer $a$ is divisible by the trivial divisors 1 and $a$, and the nontrivial divisors of $a$ are called the factors of $a$. ##### Greatest Common Divisor¶ The greatest common divisor, or GCD, of two integers $a$ and $b$, is the largest of the common divisors of $a$ and $b$. For example, the factors of 24 are 2, 3, 4, 6, 8, and 12, while the factors of 30 are 2, 3, 5, 6, 10, and 15, so the greatest common divisor of them (or gcd(24,30)) is 6. This has several uses. More prosaically, we can use this to simplify fractions by removing the common factors: e.g., reducing $\frac{24}{30}$ to $\frac{4}{5}$. Euclid's algorithm for calculating the GCD is still the most widely used and simple to program. It exploits the property that: If $a = bt + r$ for integers $t$ and $r$, then $GCD(a, b) = GCD(b, r)$. Why? Clearly $a = bt + r$ for some $t$ and $r$ - $r$ is the remainder and $t$ the multiple of $b$. Then $GCD(a, b) = GCD(bt + r, b)$. But any common divisor of $b$ and $bt + r$ must reside entirely in $r$, as $bt$ must necessarily be divisible by any divisor of $b$. So $GCD(a, b) = GCD(r, b)$. So we can write a recursive algorithm to find the GCD of any two positive integers: In [1]: def gcd(a, b): if b == 0: return a return gcd(b, a % b) ##### Least Common Multiple¶ The least common multiple (LCM) is a closely related problem, the smallest integer which is divisible by both of a pair of integers. We can calculate it easily using the GCD: $LCM(a, b) = ab / GCD(a, b)$. To calculate the GCD and LCM of more than two numbers, we can just nest the calls, e.g., $GCD(a, b, c) = GCD(a, GCD(b, c))$. #### Primes¶ A prime number is an integer $p > 1$ whose only divisors are 1 and $p$. Primes have a number of useful properties and are essential in number theory. Any number which is not prime is called composite. ##### Testing primality¶ There are an infinite number of primes, but they are not distributed according to any pattern. There are roughly $x / ln(x)$ primes less than or equal to $x$, meaning that roughly 1 out of every $ln(x)$ numbers is prime. The most straightforward way to test whether a number is prime is trial division by candidate divisors. If $N$ is prime, then none of the numbers in $[2, N-1]$ will divide it, so we loop through all of those numbers and test whether any of them are factors of $N$: In [4]: def is_prime(N): for i in xrange(2, N): if (N % i) == 0: return False return True In [9]: [p for p in range(2, 50) if is_prime(p)] Out[9]: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47] This is accurate, but we can make a couple of improvements. First, note that any divisor $d$ of $N$ has a paired divisor $d / N$. At least one of $d$ and $d / N$ must be less than or equal to $\sqrt{N}$. That means we only need to test candidate divisors up to a maximum of $\sqrt{N}$ rather than $N-1$. Secondly, we know that 2 is the only even prime, so we can simply skip all other even candidate divisors: In [18]: def is_prime_faster(N): if (N % 2) == 0: return False for i in xrange(3, int(sqrt(N)), 2): if (N % i) == 0: return False return True In [19]: %timeit [p for p in range(2, 10000) if is_prime(p)] %timeit [p for p in range(2, 10000) if is_prime_faster(p)] 1 loops, best of 3: 827 ms per loop 100 loops, best of 3: 18.9 ms per loop Finally, if we already know the primes less than $\sqrt{N}$, we can restrict the candidate divisors to the primes rather than all odd numbers, giving an even larger speed up. These methods work OK for relatively small numbers, but for large numbers like those primes used for cryptography, faster methods are required. For these, we exploit probabilistic primality testing, which uses a number of trial random numbers to test using methods I won't go into detail about to estimate whether a number is prime. It does have a chance of failure, but that chance can be made arbitrarily low. If you'd like to know more detail, look up the Miller-Rabin primality test algorithm. ##### Generating Primes¶ If we want to generate a list of primes less than some $N$, there's a better way than simply running one of the above is_prime tests on each (odd) number in the range. The algorithm we'll talk about is called the Sieve of Erisothanes. ##### Sieve of Erisothanes¶ The Sieve works by starting with the first prime, 2, and "crossing out" all multiples of 2 between $2^2$ and $N$, marking them as composite. Then it takes the next uncrossed number, 3, and repeats, marking multiples of 3 between $3^2$ and $N$ as composite, leaving the next available prime as 5. We keep doing this, repeating until we've generated all the primes we need. A well-implemented Sieve can generate all the primes less than ~10 million in only a few seconds (which, of course, you can generate before submitting your code and simply load in the list of primes you need from a file or from code). For primes larger than that, you'll want to use some kind of optimized probability primality test. In [29]: #Dynamic prime generator using the Sieve def primes(max=None): composites = {} #Yield 2 first, then only loop through odd numbers yield 2 q = 3 while max is None or q < max: if q not in composites: yield q composites[q * q] = [q] else: for p in composites[q]: try: composites[p+q].append(p) except KeyError: composites[p+q] = [p] del composites[q] q += 2 In [32]: %timeit [p for p in primes(10000000)] 1 loops, best of 3: 7.82 s per loop #### Prime Factorization¶ The Fundamental Theorem of Arithmetic states that every integer has a unique representation as a multiplication of its prime factors. That is, the prime numbers are the multiplicative building blocks of integers. For example, 1200 can be factored into $2^4 \times 3 \times 5^2$. A naive algorithm for finding the prime factorization of an integer takes a list of primes (e.g., from a sieve) and simply checks each of them to see which divides the integer. We can do better by dividing the intial integer by each prime factor we find: In [40]: def factor(N): f = {} for p in primes(sqrt(N)): if p > N: break while (N % p) == 0: try: f[p] += 1 except KeyError: f[p] = 1 N = N // p if N > 1: f[N] = 1 return f In [41]: print factor(1200) print factor(136117223861) print factor(142391208960) {2: 4, 3: 1, 5: 2} {104729: 1, 1299709: 1} {2: 10, 3: 4, 5: 1, 7: 4, 11: 1, 13: 1} The prime representation of numbers is extremely useful for dealing with very large numbers without integer overflow problems. For example, suppose you were asked how many trailing zeros there are in the decimal representation of 100!. This is a very large number, much too large to fit into a 32-bit integer. Python and Java can do this with their big integer handling, but there's an easier and faster way using the prime representation of the number. A trailing zero at the end of an integer indicates a factor of 10 - the number is divisible by 10. We wish to know how many times we can divide the number by 10, which is equivalent to asking how many prime factors of 2 and 5 are present in the number and reporting the smaller. The factorial function will always add more factors of 2 than 5 to the final integer, so we need only count the number of 5s in the prime factorization. In [46]: def count_trailing_zeros(n): #Count the number of trailing zeros in the factorial of n count = 0 while n >= 5: count += n // 5 n = n // 5 return count In [47]: print count_trailing_zeros(100) 24 And let's check, just to confirm that our answer is right: In [44]: def fact(n): if n == 1: return 1 return n * fact(n-1) print fact(100) 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000 #### Modular Arithmetic¶ Sometimes we don't want to work with the exact values of integers - instead we're interested in working with integers modulus some other number. That is, we'll work with the remainder (or residue) of the integer remaining after dividing by the modulus. This could be useful if we want to know, say, what day of the week it will be $N$ days in the future (the $N \; \textrm{mod} \; 7$th day after today), or what the last digit of a large integer is ($N \; \textrm{mod} \; 10$). It's also useful for working with large numbers, where we can more easily work with intermediate values modulo some reasonable number. So let's talk about how modular arithmetic works. In most languages, there will be a primitive operator (usually %) that gives the remainder of an integer modulus some other number. Let's look at some identities on how to operate on numbers $\; \textrm{mod} \; n$. First, sums and differences modulo $n$ are identical if we first take the modulus of the initial numbers: $(x + y) \; \textrm{mod} \; n = ((x \; \textrm{mod} \; n) + (y \; \textrm{mod} \; n)) \; \textrm{mod} \; n$ $(x - y) \; \textrm{mod} \; n = ((x \; \textrm{mod} \; n) - (y \; \textrm{mod} \; n)) \; \textrm{mod} \; n$ This suggests we can extend this to multiplication: $xy \; \textrm{mod} \; n = (x \; \textrm{mod} \; n)(y \; \textrm{mod} \; n) \; \textrm{mod} \; n$ and exponentiation: $x^y \; \textrm{mod} \; n = (x \; \textrm{mod} \; n)^y \; \textrm{mod} \; n$ Division is tricky - you can't just do this simple method for modular division. In fact, it's complicated enough (and rare enough) that I'm not going to cover it in class, but the short answer is that you need to find the inverse of some $d$ - a number that when multiplied by $d$ is equal to 1 (mod n), and then multiply by that number. But sometimes inverses don't exist, and in general it just gets complicated. Hie thee to a number theory book (e.g. CLRS) if you need more. Let's look at an example - what is the last digit of $2^{100}$? What we really want to know is $2^{100} \; \textrm{mod} \; 10$. By repeated squaring and taking the remainder mod 10 at each step we can end up with the final result while avoiding ever having to work directly with large numbers: $2^3 mod 10 = 8$ $2^6 mod 10 = 8 \times 8 mod 10 = 4$ $2^{12} mod 10 = 4 \times 4 mod 10 = 6$ $2^{24} mod 10 = 6 \times 6 mod 10 = 6$ $2^{48} mod 10 = 6 \times 6 mod 10 = 6$ $2^{96} mod 10 = 6 \times 6 mod 10 = 6$ $2^{100} mod 10 = 2^{96} \times 2^3 \times 2 mod 10 = 6 \times 8 \times 2 = 96 mod 10 = 6$ And we can check to make sure that our answer is right: In [48]: print pow(2, 100) 1267650600228229401496703205376 ##### Chinese Remainder Theorem¶ The Chinese Remainder Theorem deals with finding some $x$ where we know the remainders of $x$ divided by different integers. The theorem states that if the moduli are relatively prime (their GCD is 1) then we are guaranteed a solution (otherwise the different remainders could possibly be inconsistent). Let's look at an example. Suppose we know that $x \; mod \; 3 = 2$ and $x \; mod \; 5 = 3$. We could find the answer(s) by simply listing values of $x$ which hold for the first: $2, 5, 8, 11, 14, 17, 20, 23, 26, 29\dots$ and second equations: $3, 8, 13, 18, 23, 28, 33 \dots$ and we can see that $x=8$ and $x=23$ are both valid solutions. But this method could take quite a while if we have large remainders, moduli, or a large number of moduli.
# Partial Fractions A LevelAQAEdexcelOCR ## Partial Fractions Algebraic fractions that have a denominator with more than one linear factor can be split, or decomposed, into partial fractions. This can be particularly useful when integrating complex expressions or with binomial expansions. Questions involving partial fractions will only ever involve a numerator that is a constantÂ or a linear term and the denominator will only ever include up to three terms with the most complex being squared linear terms. A LevelAQAEdexcelOCR ## Types of Partial Fraction There are 3 types of partial fraction that you will see in A level Maths: 1. Considering a fraction where the denominator has two linear terms, we can split up the fraction into two separate fractions, Â $\dfrac{5}{\textcolor{limegreen}{(x+1)} \textcolor{blue}{(x+2)}}$ can be split up into $\dfrac{A}{\textcolor{limegreen}{(x+1)}}+\dfrac{B}{\textcolor{blue}{(x+2)}}$ 2. Considering a fraction where the denominator has three linear terms, we can split up the fraction into three separate fractions, $\dfrac{3x+4}{\textcolor{limegreen}{(x+1)}\textcolor{blue}{(x+2)}\textcolor{orange}{(x-1)}}$ can be split up into $\dfrac{A}{\textcolor{limegreen}{(x+1)}}+\dfrac{B}{\textcolor{blue}{(x+2)}} +\dfrac{C}{\textcolor{orange}{(x-1)}}$ 3. Considering a fraction where the denominator has one linear term and a repeated term, we can split up the fraction into three separate fractions, $\dfrac{2x+1}{{\textcolor{limegreen}{(x+1)}}^2 \textcolor{blue}{(x+2)}}$ can be split up into $\dfrac{A}{{\textcolor{limegreen}{(x+1)}}^2}+\dfrac{B}{\textcolor{limegreen}{(x+1)}} +\dfrac{C}{\textcolor{blue}{(x+2)}}$ A LevelAQAEdexcelOCR ## Expressing Fractions as Partial Fractions To find the coefficients in the numerators in the partial fractions, $A$, $B$ and $C$, there are 2 different methods you can use: Substitution or Equating Coefficients. Example: Express $\dfrac{5x+1}{(x-1)(x+1)(x+2)}$ as partial fractions. Step 1: Write the fraction as partial fractions with unknown constants, and put it over a common denominator: \begin{aligned} \dfrac{5x+1}{(x-1)(x+1)(x+2)} &\equiv \dfrac{A}{(x-1)} + \dfrac{B}{(x+1)} + \dfrac{C}{(x+2)} \\[1.2em] &\equiv \dfrac{A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)}{(x-1)(x+1)(x+2)} \end{aligned} Step 2: Cancel down the denominators: $5x+1 \equiv A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)$ Step 3: Use the Substitution or Equating Coefficients to find the values of the unknown constants, $A$, $B$ and $C$: Substitution: Substitute values of $x$ inside the brackets to make them $0$, so that you are left with only one of $A$, $B$ and $C$: Substitute in $x = 1$: \begin{aligned} 5(1) + 1 &= A(1+1)(1+2) \\ 6 &= 6A \\ \textcolor{red}{A} &\textcolor{red}{=} \textcolor{red}{1} \end{aligned} Substitute in $x = -1$: \begin{aligned} 5(-1) + 1 &= B(-1-1)(-1+2) \\ -4 &= -2B \\ \textcolor{blue}{B} &\textcolor{blue}{=} \textcolor{blue}{2} \end{aligned} Substitute in $x = -2$: \begin{aligned} 5(-2) + 1 &= C(-2-1)(-2+1) \\ -9 &= 3C \\ \textcolor{limegreen}{C} &\textcolor{limegreen}{=} \textcolor{limegreen}{-3} \end{aligned} Equating Coefficients: Equate coefficients and then solve them simultaneously to find the values of the unknown constants, $A$, $B$ and $C$: $5x+1 \equiv A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)$ $\equiv A(x^2 + 3x + 2) + B(x^2 + x - 2) + C(x^2 - 1)$ $\equiv (A + B + C)x^2 + (3A + B)x + (2A - 2B - C)$ Equating $x^2$ coefficients gives: $0 = A + B + C$ Equating $x$ coefficients gives: $5 = 3A + B$ Equating constant terms gives: $1 = 2A - 2B - C$ Then, solve these simultaneously, which gives $\textcolor{red}{A = 1}$, $\textcolor{blue}{B = 2}$ and $\textcolor{limegreen}{C = -3}$ Step 4: Replace the constant terms, $A$, $B$ and $C$ in the original identity: $\dfrac{5x+1}{(x-1)(x+1)(x+2)} \equiv \dfrac{\textcolor{red}{1}}{(x-1)} + \dfrac{\textcolor{blue}{2}}{(x+1)} \textcolor{limegreen}{-} \dfrac{\textcolor{limegreen}{3}}{(x+2)}$ Note: The unknown constants may not always be this nice – you may get questions with constant terms that are fractions. A LevelAQAEdexcelOCR @mmerevise ## Note: • You may need to combine the methods of Substitution and Equating Coefficients for some examples, such as fractions with repeated terms. • You may be given a fraction that is not split up into brackets in the denominator, so you will need to factorise it. Also, look out for the difference of two squares. A LevelAQAEdexcelOCR ## Example: Repeated Terms Express $\dfrac{2x-3}{x(x-2)^2}$ as partial fractions. [3 marks] Write the fraction as partial fractions with unknown constants, and put it over a common denominator: \begin{aligned} \dfrac{2x-3}{x(x-2)^2} &\equiv \dfrac{A}{(x-2)^2} + \dfrac{B}{x-2} + \dfrac{C}{x} \\[1.1em] &\equiv \dfrac{Ax + Bx(x-2) + C(x-2)^2}{x(x-2)^2} \end{aligned} Then, cancel the denominators: $2x-3 \equiv Ax + Bx(x-2) + C(x-2)^2$ Substitute in $x=0$: \begin{aligned} -3 &= C(-2)^2 \\ C &= - \dfrac{3}{4} \end{aligned} Substitute in $x=2$: \begin{aligned} 2(2) - 3 &= 2A \\ A &= \dfrac{1}{2} \end{aligned} Now, equate coefficients of $x^2$: \begin{aligned} 0 &= B + C \\ B &= \dfrac{3}{4} \end{aligned} Finally, replace the values of $A$, $B$ and $C$ into the original identity: $\dfrac{2x-3}{x(x-2)^2} \equiv \dfrac{1}{2(x-2)^2} + \dfrac{3}{4(x-2)} - \dfrac{3}{4x}$ A LevelAQAEdexcelOCR ## Partial Fractions Example Questions $\dfrac{8}{9x^2 - 16} = \dfrac{8}{(3x+4)(3x-4)}$ \begin{aligned} \dfrac{8}{(3x+4)(3x-4)} &\equiv \dfrac{A}{(3x+4)} + \dfrac{B}{(3x-4)} \\[1.1em] &\equiv \dfrac{A(3x-4) + B(3x+4)}{(3x+4)(3x-4)} \end{aligned} $8 \equiv A(3x-4) + B(3x+4)$ Substitute in $x=\dfrac{4}{3}$: \begin{aligned} 8 &= 8B \\ B &= 1 \end{aligned} Substitute in $x=- \dfrac{4}{3}$: \begin{aligned} 8 &= -8A \\ A &= -1 \end{aligned} Then, replace the values of $A$ and $B$ into the identity: $\dfrac{8}{9x^2 - 16} \equiv - \dfrac{1}{(3x+4)} + \dfrac{1}{(3x-4)}$ $\dfrac{x}{(x-1)(x-2)(x-3)} \equiv \dfrac{A}{(x-1)} + \dfrac{B}{(x-2)} + \dfrac{C}{(x-3)}$ $\equiv \dfrac{A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$ $x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$ Substitute in $x = 1$: \begin{aligned} 1 &= A(1-2)(1-3) \\ A &= \dfrac{1}{2} \end{aligned} Substitute in $x = 2$: \begin{aligned} 2 &= B(2-1)(2-3) \\ B &= -2 \end{aligned} Substitute in $x = 3$: \begin{aligned} 3 &= C(3-1)(3-2) \\ C &= \dfrac{3}{2} \end{aligned} Then, replace the values of $A$, $B$ and $C$ into the identity: $\dfrac{x}{(x-1)(x-2)(x-3)} \equiv \dfrac{1}{2(x-1)} - \dfrac{2}{(x-2)} + \dfrac{3}{2(x-3)}$ \begin{aligned} \dfrac{10}{x^2(x+1)} &\equiv \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{(x+1)} \\[1.1em] &\equiv \dfrac{A(x+1) + Bx(x+1) + Cx^2}{x^2(x+1)} \end{aligned} $10 \equiv A(x+1) + Bx(x+1) + Cx^2$ Substitute in $x = 0$: \begin{aligned} 10 &= A(0+1) \\ A &= 10 \end{aligned} Substitute in $x = -1$: \begin{aligned} 10 &= C(-1)^2 \\ C &= 10 \end{aligned} Equate coefficients of $x^2$: \begin{aligned} 0 &= B + C \\ B &= -10 \end{aligned} Then, replace the values of $A$, $B$ and $C$ into the identity: $\dfrac{10}{x^2(x+1)} \equiv \dfrac{10}{x^2} - \dfrac{10}{x} + \dfrac{10}{(x+1)}$ A Level A Level ## You May Also Like... ### MME Learning Portal Online exams, practice questions and revision videos for every GCSE level 9-1 topic! 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## How to Make a Box and Whisker Plot Box and whisker plots are one of the ways that data can be shown visually. Most people are visual learners and having a way to analyze the data in a visual manner instead of as a list of numbers can be helpful and informative. Some of the vocabulary with box and whisker plots: Data Point: Each recorded number of the set of data is a data point. Median: The middle number of the set of data. If there are an even amount of data points, the two middle numbers are averaged. Lower Quartile, sometimes called the first quartile: The median of the lower set of numbers Upper Quartile, sometimes called the third quartile: The median of the higher set of numbers ## Steps on How to Do a Box and Whisker Plot Step #1 – How to find median: Order the data points from least to greatest. Find the middle number. If there are an even amount of data points, average the two middle points to find the median. Step #2 – How to find lower quartile: Order the data points from least to greatest. Split the data set in half. If there are an odd amount of data points, do not include the median. Find the median of the first half, smallest values, of the data points. Step #3 – How to find upper quartile: Order the data points from least to greatest. Split the data set in half. If there are an odd amount of data points, do not include the median. Find the median of the second half, highest values, of the data points. ## How to Make a Box Plot Example Data set: 6, 2, 16, 5, 8, 20, 14, 19, 6, 12, 7, 15, 18 Order from least to greatest: 2, 5, 6, 6, 7, 8, 12, 14, 15, 16, 18, 19, 20 Lowest data point: 2 Highest data point: 20 Median: 2, 5, 6, 6, 7, 8, 12, 14, 15, 16, 18, 19, 20 Lower Quartile: Use only first half (2, 5, 6, 6, 7, 8) 2, 5, 6, 6, 7, 8 Find the average of the two middle numbers Upper Quartile: Use only second half (14, 15, 16, 18, 19, 20) 14, 15, 16, 18, 19, 20 Find the average of the two middle numbers ## Why is Box Whisker Plot Important in Math? By using a box and whiskers plot it shows variability or where the different data points lie in comparison to the quartiles. If there are long whiskers that would show that the highest/lowest data points are far from the other data points. Depending on how long the boxes are shows how spread out the different quartiles are. For example if the lower quartile is 3, the median is 5 and the upper quartile is 10, that shows that the lower numbers have less of a range than the higher numbers or that there are more repeating data points. It would not be as useful with data points that often repeat, for example ages of students in high school.
Contact The Learning Centre # Exponent Laws & Rules ### Use the exponential rules to solve equations • In any equation, if the unknown is inside a power, you can use roots (e.g. square roots) to remove the power. • Remember to follow the rules and principles of algebra. For example: James invested $$100$$ in an account. After $$20$$ time periods, the investment returned $$180$$. What was the interest rate per period for the investment. We need to use the compound interest formula $A = P \left(1+\frac{r}{100}\right)^n$ where $$A$$ is the investment return, $$P$$ is the amount invested, $$n$$ is the number of time periods, and $$r$$ is the interest rate. Substituting the values into the formula gives: $$\newcommand{\eqncomment}[2]{\small{\text{ #2}} } \newcommand{\ceqns}{\begin{array}{rcll}} \newcommand{\ceqne}{\end{array}}$$ $\ceqns 180 &=& 100 \left(1+\frac{r}{100}\right)^{20} \\ \frac{180}{100} &=& \left(1+\frac{r}{100}\right)^{20} & \eqncomment{0.3}{dividing by 100} \\ 1.8 &=& \left(1+\frac{r}{100}\right)^{20} \\ (1.8)^{\frac{1}{20}}&=& \left[ \left(1+\frac{r}{100}\right)^{20} \right]^{\frac{1}{20}} & \eqncomment{0.3}{taking the 20th root} \\ \sqrt[20]{1.8} &=& 1+\frac{r}{100} \\ \sqrt[20]{1.8} -1 &=& \frac{r}{100} \\ (\sqrt[20]{1.8}-1) \times 100&=& r \\ r &\approx& 2.98 \ceqne$ Therefore, the interest rate per period is approximately $$2.98\%$$.
Courses Courses for Kids Free study material Offline Centres More Store # The sum of three consecutive even numbers is 276. Find the numbers.A. 96, 92, 94B. 91, 92, 94C. 90, 94, 98D. 90, 92, 94 Last updated date: 20th Sep 2024 Total views: 421.8k Views today: 6.21k Verified 421.8k+ views Hint: Here we assume three consecutive even numbers as $(2n + 2),(2n + 4),(2n + 6)$. Add the three numbers and equate the sum to 276 to find the value of n. Substitute back the value of n in the numbers. Complete step-by-step solution: We have to take three consecutive natural numbers. Let the three consecutive even numbers be $(2n + 2),(2n + 4),(2n + 6)$ We are given the sum of three consecutive even numbers is 249 $\Rightarrow (2n + 2) + (2n + 4) + (2n + 6) = 276$ $\Rightarrow 6n + 12 = 276$ Shift the constant values to RHS of the equation. $\Rightarrow 6n = 276 - 12$ $\Rightarrow 6n = 264$ Divide both sides of the equation by 6 $\Rightarrow \dfrac{{6n}}{6} = \dfrac{{264}}{6}$ Cancel the same terms from numerator and denominator. $\Rightarrow n = 44$ Now we substitute the value of n in each $(2n + 2),(2n + 4),(2n + 6)$ to obtain the three numbers. Put$n = 44$in $(2n + 2)$ $\Rightarrow (2n + 2) = (2 \times 44 + 2)$ $\Rightarrow (2n + 2) = 88 + 2$ $\Rightarrow (2n + 2) = 90$ So, the first even number is 90. Put$n = 44$in $(2n + 4)$ $\Rightarrow (2n + 4) = (2 \times 44 + 4)$ $\Rightarrow (2n + 4) = 88 + 4$ $\Rightarrow (2n + 4) = 92$ So, the second even number is 92. Put$n = 44$in $(2n + 6)$ $\Rightarrow (2n + 6) = (2 \times 44 + 6)$ $\Rightarrow (2n + 6) = 88 + 6$ $\Rightarrow (2n + 6) = 94$ So, the third even number is 94. Therefore, the three consecutive even numbers are 90, 92 and 94. $\therefore$Option D is correct Note: Students are likely to make the mistake of not changing the sign of a value when shifting the value from one side of the equation to the other side of the equation, keep in mind sign changes from positive to negative and vice-versa when a value is shifted. Alternate Method: We can take three consecutive alternate numbers as $n,n + 2,n + 4$ Then the sum of three consecutive even numbers is 276 $\Rightarrow n + n + 2 + n + 4 = 276$ $\Rightarrow 3n + 6 = 276$ $\Rightarrow 3n = 276 - 6$ $\Rightarrow 3n = 270$ $\Rightarrow \dfrac{{3n}}{3} = \dfrac{{270}}{3}$ $\Rightarrow n = 90$ Substitute the value of n in $n + 2,n + 4$ $\Rightarrow n + 2 = 90 + 2 = 92$ $\Rightarrow n + 4 = 90 + 4 = 94$ $\therefore$Option D is correct
## Want to keep learning? This content is taken from the National STEM Learning Centre's online course, Maths Subject Knowledge: Understanding Numbers. Join the course to learn more. 1.9 ## National STEM Learning Centre Skip to 0 minutes and 7 seconds MICHAEL ANDERSON: So let’s have a look at how base 2 or the binary system works. Skip to 0 minutes and 12 seconds PAULA KELLY: So on our table, it’s similar to how we had it before with powers of 10. This time, though, we’re having powers of 2 because we’re counting in base 2. Skip to 0 minutes and 21 seconds MICHAEL ANDERSON: So it might be useful to write the values of each column above. So for example, 2 to the power of 0, that’s going to give us 1. 2 to the power of 1 is 2. 2 squared, 2 times 2, is 4. 2 cubed is 8. And you’ll notice that each of these columns, each time to where we move from right to left, we’re doubling. So 1 times 2 times 2 times 2. So the next one is going to be 16, then 32, and then 64. And we can keep on doubling. Skip to 0 minutes and 50 seconds PAULA KELLY: So to have a look at how this differs from our denary system, should try and write some numbers in binary. Skip to 0 minutes and 56 seconds MICHAEL ANDERSON: OK. So with the number 1, we’re just going to put– we’re just going to put a 1 in there. Now, in base 2, we’re not allowed to use any of the digits other than 0’s and 1’s. So to represent the number 2, what we’re going to do is we’re going to say, well, that’s one 2 and zero 1’s. So 2 in this system is written as a 1 and then a 0, like we had the 1 before just being a 1. Skip to 1 minute and 22 seconds PAULA KELLY: So here, we’re quite familiar with this from our base 10, but this shows this is quite different when we’re counting in base 2. Skip to 1 minute and 29 seconds MICHAEL ANDERSON: Yeah. So if we wanted to make the number 3, for example, well, to make 3, I need one 2 and I need one 1. So 2 plus 1 gives me my 3. What do you think 4 is going to look like? Skip to 1 minute and 43 seconds PAULA KELLY: So for 4, then, we want to find just here a 4. We’re fine. So 1 of these. Then I wouldn’t need any other digits, so that would give us one 4, which is 4. Skip to 1 minute and 56 seconds MICHAEL ANDERSON: And it’s really tempting to almost think as 4 as two 2’s, but we couldn’t put the number 2 in there, so yeah, it goes to a 1-0-0. Skip to 2 minutes and 3 seconds PAULA KELLY: Let’s try a different number. Skip to 2 minutes and 5 seconds MICHAEL ANDERSON: OK. So slightly larger. Let’s have a look at, say, 17. Skip to 2 minutes and 9 seconds PAULA KELLY: So if we’ve got our values of our powers of 2, 16 is the nearest we can get, is the nearest one that’s just below our value. So let’s have one of those, so a 16. Skip to 2 minutes and 23 seconds MICHAEL ANDERSON: So if we’re at 16, then we’re only one away. So 16 add 1 more would give us 17. So if I put a 1 in this column, that means we’ve got one 16 and one 1. 16 add 1 gives us 7. So I’m just going to pop some zeros in here because we don’t need any 2’s, We? Don’t need any 4’s., and we don’t need any 8’s. So 17 in binary is 1-0-0-0-1. Skip to 2 minutes and 47 seconds PAULA KELLY: Let’s try another. Skip to 2 minutes and 48 seconds MICHAEL ANDERSON: OK. Let’s do 14. Skip to 2 minutes and 52 seconds PAULA KELLY: So again, we’ll scan our numbers and get as near as we can to 14. So 16 is going to be too big. So let’s have one of these 8’s. So to get from 8 to 14, I need 6 more. So again, our nearest one is 4. We’ll have one of those. So far, we’ve got an 8, a 4, to give us 12. We need two more to get 14. Let’s have one of these 2’s. We’re at 14 already. We need no 1’s. Skip to 3 minutes and 23 seconds MICHAEL ANDERSON: Perfect. So 14 in binary is 1-1-1-0 because we need one 8, one 4, and one 2 to make 14. So knowing your powers of 2 is pretty important. Skip to 3 minutes and 34 seconds MICHAEL ANDERSON: Shall we look another number? Skip to 3 minutes and 37 seconds MICHAEL ANDERSON: So we could have a look at 45. Skip to 3 minutes and 41 seconds PAULA KELLY: So same again. If we find the nearest one we can get to, 64 is too big. Let’s have one of these 32’s. We’ll have one of those. So 32 from 45. Got 13 left. Skip to 3 minutes and 54 seconds MICHAEL ANDERSON: OK. So the 16 is too big, so we definitely don’t need any 16’s. Skip to 4 minutes and 1 second PAULA KELLY: We’re looking for 13. 8 will fit in there. Let’s have one of those. Skip to 4 minutes and 5 seconds MICHAEL ANDERSON: So far, we’ve got a 32 and an 8, which uses 40. So we’re just looking for another 5. So we could have one of the 4’s, and then we’re only 1 away. So I can put a 1 in the Ones column, and we don’t need any 2’s. Skip to 4 minutes and 19 seconds PAULA KELLY: Fantastic. So 45 we made up of a 32, an 8, a 4, and a 1. Altogether got 45. Skip to 4 minutes and 28 seconds MICHAEL ANDERSON: So 45 in binary is 1-0-1-1-0-1. Skip to 4 minutes and 34 seconds So these numbers seem to get quite long quite quickly, but they represent the same number as we do in base 10. Shall we look at one more example? Skip to 4 minutes and 41 seconds PAULA KELLY: One more, yeah. Skip to 4 minutes and 42 seconds MICHAEL ANDERSON: OK. So let’s have a look at the number 31. Skip to 4 minutes and 46 seconds PAULA KELLY: OK. So very close to one of our powers of 2, but not quite. So we’ll have a 16. So 16 away from 31. 15 are left. So we’ll need an 8. Skip to 5 minutes and 0 seconds MICHAEL ANDERSON: And that gets us to 24. Skip to 5 minutes and 3 seconds PAULA KELLY: So remaining, we just have– Skip to 5 minutes and 7 seconds PAULA KELLY: So we just need another 4. Skip to 5 minutes and 9 seconds MICHAEL ANDERSON: Yeah. So that gets us to 28. Skip to 5 minutes and 12 seconds PAULA KELLY: OK. 3 are left. So let’s have another one. Skip to 5 minutes and 16 seconds MICHAEL ANDERSON: So now we’re on 30. Skip to 5 minutes and 18 seconds PAULA KELLY: So 1 is left. Skip to 5 minutes and 19 seconds MICHAEL ANDERSON: So another 1 in the Ones column. Skip to 5 minutes and 21 seconds PAULA KELLY: So 31 in binary is 1-1-1-1-1-1. Skip to 5 minutes and 26 seconds MICHAEL ANDERSON: So this number is a little bit like having 999 in our base 10 system because 32, what will that give us? Skip to 5 minutes and 36 seconds PAULA KELLY: That would then carry over all of these. We’d just need one of our 32’s. Because we already have the number that we need, the rest of this can be replaced of 0’s. Skip to 5 minutes and 48 seconds MICHAEL ANDERSON: Great. So you might want to have a go at writing some numbers in binary yourself. # Other number bases: binary Earlier in the course we saw how base ten works. It is thought that we use base ten as we have ten digits: eight fingers and two thumbs on our hands. We can count using any number as our number base. Students studying computer science are required to be able to work in binary: base two and hexadecimal: base sixteen. ## Binary: base two Base two only uses the digits 0 and 1. Numbers are grouped in twos. The number 2 is written as one lot of two and no units (10). The number three is one lot of two and one unit (11) and the number four is written as one lot of four, no twos and no units (100). In base ten numbers are grouped into units, tens, hundreds, thousandths etc i.e. powers of ten. In binary; base two, numbers are grouped into powers of two, units, twos, fours, eights, sixteens and so on. Base sixteen uses the digits 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. A is the symbol used for ten, B for eleven, C for twelve up to F for fifteen. The number sixteen is written as 10 i.e. one sixteen and no units. In this video Paula and Michael convert numbers from base 10 into base 2. ## Problem worksheet Now complete questions 9 and 10 from this week’s worksheet. ## Teaching resources If you would like to explore further how to convert between binary, denary and hexadecimal then see these teaching resources:
# Discovering The Percentage: 21 Is What Percent Of 30 In 2023 ## Introduction Calculating percentages is an important skill in everyday life. Whether you’re calculating a tip at a restaurant or determining the discount on a sale item, understanding how to work with percentages is crucial. In this article, we’ll explore the question of “21 is what percent of 30” and provide a step-by-step guide to finding the answer. ## Understanding Percentages Before we dive into the specifics of calculating percentages, it’s important to understand what percentages represent. A percentage is a way of expressing a portion of a whole as a fraction of 100. For example, 50% means “50 out of 100.” When we talk about “21 is what percent of 30,” we’re essentially asking what fraction of 30 is equal to 21. To find the answer, we need to convert the fraction to a percentage. ## Step-by-Step Guide Here’s a step-by-step guide to finding the percentage of 21 out of 30: ### Step 1: Divide 21 by 30 To find the fraction of 21 out of 30, we need to divide 21 by 30. This gives us the answer of 0.7. ### Step 2: Multiply by 100 To convert the fraction to a percentage, we need to multiply by 100. This gives us the answer of 70. ### Step 3: Add the Percent Symbol To express the answer as a percentage, we need to add the percent symbol. The final answer is: 21 is 70% of 30. ## Why Percentages Matter Percentages are used in a wide range of applications, from calculating discounts to analyzing data. In the business world, percentages are often used to measure growth or decline in sales, revenue, or profits. In personal finance, percentages are used to calculate interest rates on loans or investments. Understanding percentages is also important in science and medicine, where percentages are used to express the effectiveness of treatments or the prevalence of diseases. ## Common Mistakes to Avoid When working with percentages, there are a few common mistakes to avoid: ### Mistake 1: Forgetting to Divide One common mistake is forgetting to divide the numerator by the denominator when calculating a fraction. This can result in an incorrect answer when converting to a percentage. ### Mistake 2: Confusing Percentages and Decimals Another common mistake is confusing percentages and decimals. Percentages are always expressed as a fraction out of 100, while decimals are expressed as a fraction out of 1. ### Mistake 3: Rounding Too Soon Finally, rounding too soon can also result in an incorrect answer. It’s important to carry out calculations to several decimal places before rounding to the nearest whole number. ## Conclusion Calculating percentages is an important skill that can be applied in many different areas of life. By understanding the basics of percentages and avoiding common mistakes, you can ensure that your calculations are accurate and reliable. And now, armed with the knowledge of how to calculate “21 is what percent of 30,” you can confidently tackle any percentage-related problem that comes your way.
# How to Do Division – Simplify Your Math Problems With Area Models and Logarithm Tables 0 146 Logarithm tables If you’re a math student, you’ve probably used logarithm tables to solve division and multiplication problems. Logarithms can be used with any positive base number, including 1, but not negative ones. Using a log table is useful for converting between numbers, since the same number’s exponent is always the same. You can also use a log table to find the inverse of a number. To read a log table, look at the first two digits of the number and the column number corresponding to the third digit. You’ll find the value that is at the intersection of the column and row. You can also check for a decimal part by looking at the mantissa. If the mantissa is positive, you can skip this step. If you’re unsure of how to divide a log table, consult a book or website on the subject. In general, it’s easy to find logarithms with a calculator or computer. However, the logarithm table was first published in 1614 by John Napier, a mathematician, physicist, and astronomer. He didn’t know the number e and didn’t think about exponentiation, so he accidentally defined the logarithm to base e, instead of a base number. His mistake was based on the fact that he had envisioned points moving along a line. The common logarithm is the same as the exponential function and is written as log10. Using the log table, you can get the value of the inverse of a number for any given x. A main column of the log table has numbers from 10 to 99. The second column has numbers ranging from 0 to 9. There is also a difference column, which shows the differences between numbers 0-9. A mean difference column will show the differences between values 1 to nine. ## Area models An area model is an excellent way to simplify a large number and is also known as a box model. It is based on the principle of finding the area of a rectangle (or square) by multiplying the length and width. However, it is important to remember that the length of the side is not given. Therefore, dividing an area model into parts will be much easier for students when they have a clear understanding of the process. The area model is an important concept to understand when solving division problems. Using this concept, students can divide any large figure. This approach is also useful for long division, a more challenging mathematical concept. After understanding this concept, students are equipped to tackle any division problem. In fact, they can begin using this method in a few days. The concept of the area model is so basic and fundamental that it can be applied to other areas of mathematics. The main goal of this interactive exercise is to help students visualize the distributive property of division by using area models. The exercise is part of the Math at the Core: Middle School collection. Students can use the tool to practice calculating quotients and divisors with the area model. They can also practice dividing areas using a box method. They will learn how to solve equations more efficiently. It can also be used for teaching multiplication. ## Long division There are two basic methods of long division. First, you divide the dividend by the divisor. The quotient is the first digit in the result. You then count the remainder as subtraction. The remaining amount carries forward to the next digit in the dividend. This process is repeated until you’ve calculated the answer as many times as the divisor. To make it easier to remember, here are some examples of how to do long division. One way to remember long division is to make it fun. You can make long division a classroom game by having students line up in groups and do the first set of steps for each problem. You can also make it a relay game by having students complete the first set of steps. In this way, you’ll be able to reinforce place value and the idea of division as repeated subtraction. Here are four fun ways to help your students memorize the long division process: Another way to teach long division is through video. This way, students can follow along with Robin Hood as he learns to do long division. There are also accompanying lesson resources that help them practice the skill. One fun resource is Prodigy, a game-based platform where students can play long division puzzles and compete with their friends. This way, students will get plenty of practice while learning the technique. If you can’t find any resources that will help your students practice long division, try playing one. Another easy way to teach students how to do long division is through a video that teaches the standard algorithm. Students then explore different methods of thinking about long division and solve problems by breaking a three-digit number into tens, hundreds, and ones. Then, they play a short game with their partners using the different solutions. In the end, they can apply what they learned in this way to solve problems of different levels. ## Chunking Children will have an easier time remembering large numbers by chunking. However, they must be confident in their multiplication facts, including six times six equals 36. If they struggle to learn the method, review their times tables and the relationship between multiplication and division. Chunking can be very effective with numbers up to 100. Many children already know the 10x table, and this can be applied to repeated subtraction. Then, they can subtract one chunk from another to get the answer. Students can also use chunking when doing division when they are tackling larger numbers. This method encourages students to use the multiplication and division facts from their times tables to break down the problem. Children in Years 5 and 6 are encouraged to use this technique. It is also known as the ‘bus stop method’. Here are some of the benefits of chunking when doing division. They can speed up the process by using times tables. Another method for chunking is division. First, a child needs to understand how to divide a large number. If he is unsure of how to do this, he should look up his times tables to see if he knows the multiplication facts. If he does not, he or she cannot chunk with them. If your child is stuck, you can help them by making an estimate. For example, if your child cannot solve 84/7, they should try chunking by estimating the next easiest multiple to find the answer. Using scrap paper to check your work Using scrap paper to check your work when you’re doing division can help you ensure that all of your calculations are accurate. A child who struggles with executive functioning problems might scribble information all over the paper in an unorganized fashion. This can make moving from step to step in math problems difficult. This can lead to confusion, so using scrap paper to check your work is essential to success. This is a proven method for solving math problems that you may have been struggling with. To make it easier for students to remember, you can also provide scrap paper. This way, students can check their work before they turn it in. Regardless of the size of the problem, students can use scrap paper to check their work. This method is especially useful when students are learning how to multiply by more than one factor. It forces students to slow down, think about how to solve each step, and be thorough.
CHAPTER 5 Solving First Degree Equations in One Variable # 5.4 Solve Equations with Fraction or Decimal Coefficients Learning Objectives By the end of this section, you will be able to: • Solve equations with fraction coefficients • Solve equations with decimal coefficients # Solve Equations with Fraction Coefficients Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation . To isolate the term, subtract from both sides. Simplify the left side. Change the constants to equivalent fractions with the LCD. Subtract. Multiply both sides by the reciprocal of . Simplify. This method worked fine, but many students don’t feel very confident when they see all those fractions. So we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions. We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions. Let’s solve the same equation again, but this time use the method that clears the fractions. EXAMPLE 1 Solve: . Solution Find the least common denominator of all the fractions in the equation. Multiply both sides of the equation by that LCD, 8. This clears the fractions. Use the Distributive Property. Simplify — and notice, no more fractions! Solve using the General Strategy for Solving Linear Equations. Simplify. Check: Let TRY IT 1.1 Solve: . TRY IT 1.2 Solve: . y = 3 Notice in (Figure) that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations. HOW TO: Solve Equations with Fraction Coefficients by Clearing the Fractions 1. Find the least common denominator of all the fractions in the equation. 2. Multiply both sides of the equation by that LCD. This clears the fractions. 3. Solve using the General Strategy for Solving Linear Equations. EXAMPLE 2 Solve: . Solution We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation. Find the least common denominator of all the fractions in the equation. Multiply both sides of the equation by 12. Distribute. Simplify — and notice, no more fractions! Combine like terms. Divide by 7. Simplify. Check: Let . TRY IT 2.1 Solve: . v = 40 TRY IT 2.2 Solve: . u = −12 In the next example, we’ll have variables and fractions on both sides of the equation. EXAMPLE 3 Solve: . Solution Find the LCD of all the fractions in the equation. Multiply both sides by the LCD. Distribute. Simplify — no more fractions! Subtract from both sides. Simplify. Subtract 2 from both sides. Simplify. Divide by 5. Simplify. Check: Substitute . TRY IT 3.1 Solve: . a = −2 TRY IT 3.2 Solve: . c = −2 In (Figure), we’ll start by using the Distributive Property. This step will clear the fractions right away! EXAMPLE 4 Solve: . Solution Distribute. Simplify. Now there are no fractions to clear! Subtract 1 from both sides. Simplify. Divide by 2. Simplify. Check: Let . TRY IT 4.1 Solve: . p = −4 TRY IT 4.2 Solve: . q = 2 Many times, there will still be fractions, even after distributing. EXAMPLE 5 Solve: . Solution Distribute. Simplify. Multiply by the LCD, 4. Distribute. Simplify. Collect the terms to the left. Simplify. Collect the constants to the right. Simplify. Check: Substitute for . TRY IT 5.1 Solve: . n = 2 TRY IT 5.2 Solve: . m = −1 # Solve Equations with Decimal Coefficients Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, and . So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator. EXAMPLE 6 Solve: . Solution The only decimal in the equation is . Since , the LCD is . We can multiply both sides by to clear the decimal. Multiply both sides by the LCD. Distribute. Multiply, and notice, no more decimals! Add 50 to get all constants to the right. Simplify. Divide both sides by 8. Simplify. Check: Let . TRY IT 6.1 Solve: . x = 20 TRY IT 6.2 Solve: . x = 10 EXAMPLE 7 Solve: . Solution Look at the decimals and think of the equivalent fractions. Notice, the LCD is . By multiplying by the LCD we will clear the decimals. Multiply both sides by 100. Distribute. Multiply, and now no more decimals. Collect the variables to the right. Simplify. Collect the constants to the left. Simplify. Divide by 19. Simplify. Check: Let . TRY IT 7.1 Solve: . h = 12 TRY IT 7.2 Solve: . k = −1 The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation. EXAMPLE 8 Solve: . Solution Distribute first. Combine like terms. To clear decimals, multiply by 100. Distribute. Subtract 15 from both sides. Simplify. Divide by 30. Simplify. Check: Let . TRY IT 8.1 Solve: . n = 9 TRY IT 8.2 Solve: . d = 16 # Key Concepts • Solve equations with fraction coefficients by clearing the fractions. 1. Find the least common denominator of all the fractions in the equation. 2. Multiply both sides of the equation by that LCD. This clears the fractions. 3. Solve using the General Strategy for Solving Linear Equations. # Practices Makes Perfect ## Solve equations with fraction coefficients In the following exercises, solve the equation by clearing the fractions. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 ## Solve Equations with Decimal Coefficients In the following exercises, solve the equation by clearing the decimals. 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 ## Everyday Math Coins 41. Taylor has in dimes and pennies. The number of pennies is more than the number of dimes. Solve the equation for , the number of dimes. Stamps 42. Travis bought worth of stamps and stamps. The number of stamps was less than the number of stamps. Solve the equation for , to find the number of stamps Travis bought. ## Writing Exercises 43. Explain how to find the least common denominator of . 44. If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve? 45. If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD? 46. In the equation , what is the LCD? How do you know?
# CLASS 9 MATHS CHAPTER-7 TRIANGLES ### Exercise 7.1 Question 1. In quadrilateral ABCD (See figure). AC = AD and AB bisects A. Show that ABC ABD. What can you say about BC and BD? Solution: To prove: ABC ABD Proof: In ABC and ABD, BAC = BAD [AB bisects A] AB = AB [Common] ABC ABD [By SAS congruency] Thus BC = BD [By C.P.C.T.] Question 2. ABCD is a quadrilateral in which AD = BC and DAB = CBA. (See figure). Prove that: (i) ABD BAC (ii) BD = AC (iii) ABD = BAC Solution: (i)In ABC and ABD, DAB = CBA [Given] AB = AB [Common] ABC ABD [By SAS congruency] Thus AC = BD [By C.P.C.T.] (ii)Since ABC ABD AC = BD [By C.P.C.T.] (iii)Since ABC ABD ABD = BAC [By C.P.C.T.] Question 3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB (See figure) Solution: In BOC and AOD, BOC = AOD [Vertically Opposite angles] BOC AOD [By ASA congruency] OB = OA and OC = OD [By C.P.C.T.] Question 4. and are two parallel lines intersected by another pair of parallel lines and (See figure). Show that ABC CDA. Solution: AC being a transversal. [Given] Therefore DAC = ACB [Alternate angles] Now [Given] And AC being a transversal. [Given] Therefore BAC = ACD [Alternate angles] ACB = DAC [Proved above] BAC = ACD [Proved above] AC = AC [Common] ABC CDA [By ASA congruency] Question 5. Line is the bisector of the angle A and B is any point on BP and BQ are perpendiculars from B to the arms of A. Show that: (i) APB AQB (ii) BP = BQ or P is equidistant from the arms of A (See figure). Solution: Given: Line bisects A. BAP = BAQ (i) In ABP and ABQ, BAP = BAQ [Given] BPA = BQA = [Given] AB = AB [Common] APB AQB [By ASA congruency] (ii) Since APB AQB BP = BQ [By C.P.C.T.] B is equidistant from the arms of A. Question 6. In figure, AC = AB, AB = AD and BAD = EAC. Show that BC = DE.^ Solution: Adding DAC on both sides, we get BAD + DAC = EAC + DAC Now in ABC and AED, AC = AE [Given] BAC = DAE [From eq. (i)] BC = DE [By C.P.C.T.] Question 7. AB is a line segment and P is the mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB. Show that: (i) DAF FBP (ii) AD = BE (See figure) Solution: Given that EPA = DPB Adding EPD on both sides, we get EPA + EPD = DPB + EPD APD = BPE ……….(i) Now in APD and BPE, AP = PB [P is the mid-point of AB] APD = BPE [From eq. (i)] DPA EBP [By ASA congruency] AD = BE [ By C.P.C.T.] Question 8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. (See figure) Show that: (i) AMC BMD (ii) DBC is a right angle. (iii) DBC ACB (iv) CM = AB Solution: (i)In AMC and BMD, AM = BM [AB is the mid-point of AB] AMC = BMD [Vertically opposite angles] CM = DM [Given] AMC BMD [By SAS congruency] ACM = BDM ……….(i) CAM = DBM and AC = BD [By C.P.C.T.] (ii) For two lines AC and DB and transversal DC, we have, ACD = BDC [Alternate angles] AC DB Now for parallel lines AC and DB and for transversal BC. DBC = ACB [Alternate angles] ……….(ii) But ABC is a right angled triangle, right angled at C. ACB = ……….(iii) Therefore DBC = [Using eq. (ii) and (iii)] DBC is a right angle. (iii) Now in DBC and ABC, DB = AC [Proved in part (i)] DBC = ACB = [Proved in part (ii)] BC = BC [Common] DBC ACB [By SAS congruency] (iv) Since DBC ACB [Proved above] DC = AB AM + CM = AB CM + CM = AB [DM = CM] 2CM = AB CM = AB ### NCERT Solutions for Class 9 Maths Exercise 7.2 Question 1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that (i) OB = OC (ii) AO bisects ∠A Solution: i) in ∆ABC, we have AB = AC [Given] ∴∠ABC = ∠ACB [Angles opposite to equal sides of a A are equal] ∠ABC = ∠ACB or ∠OBC = ∠OCB ⇒OC = OB [Sides opposite to equal angles of a ∆ are equal] (ii) In ∆ABO and ∆ACO, we have AB = AC [Given] ∠OBA = ∠OCA [ ∵∠B = ∠C] OB = OC [Proved above] ∆ABO ≅∆ACO [By SAS congruency] ⇒∠OAB = ∠OAC [By C.P.C.T.] ⇒AO bisects ∠A. Question 2. In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC. Solution: Since AD is bisector of BC. ∴BD = CD Now, in ∆ABD and ∆ACD, we have BD = CD [Proved above] ∴∆ABD ≅∆ACD [By SAS congruency] ⇒AB = AC [By C.P.C.T.] Thus, ∆ABC is an isosceles triangle. Question 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal. Solution: ∆ABC is an isosceles triangle. ∴AB = AC ⇒∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal] ⇒∠BCE = ∠CBF Now, in ∆BEC and ∆CFB ∠BCE = ∠CBF [Proved above] ∠BEC = ∠CFB [Each 90°] BC = CB [Common] ∴∆BEC ≅∆CFB [By AAS congruency] So, BE = CF [By C.P.C.T.] Question 4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that (i) ∆ABE ≅∆ACF (ii) AB = AC i.e., ABC is an isosceles triangle. Solution: (i) In ∆ABE and ∆ACE, we have ∠AEB = ∠AFC [Each 90° as BE ⊥AC and CF ⊥AB] ∠A = ∠A [Common] BE = CF [Given] ∴∆ABE ≅∆ACF [By AAS congruency] (ii) Since, ∆ABE ≅∆ACF ∴AB = AC [By C.P.C.T.] ⇒ABC is an isosceles triangle. Question 5. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD. Solution: In ∆ABC, we have AB = AC [ABC is an isosceles triangle] ∴∠ABC = ∠ACB …(1) [Angles opposite to equal sides of a ∆ are equal] Again, in ∆BDC, we have BD = CD [BDC is an isosceles triangle] ∴∠CBD = ∠BCD …(2) [Angles opposite to equal sides of a A are equal] Adding (1) and (2), we have ∠ABC + ∠CBD = ∠ACB + ∠BCD ⇒∠ABD = ∠ACD. Question 6. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle. Solution: AB = AC [Given] …(1) From (1) and (2), we have Now, in ∆ABC, we have ∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A] ⇒2∠ACB + ∠BAC = 180° …(3) [∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)] Similarly, in ∆ACD, ⇒2∠ACD + ∠CAD = 180° …(4) [∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)] Adding (3) and (4), we have 2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180° ⇒2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360° ⇒2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair] ⇒2∠BCD = 360° – 180° = 180° ⇒∠BCD = = 90° Thus, ∠BCD = 90° Question 7. ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C. Solution: In ∆ABC, we have AB = AC [Given] ∴Their opposite angles are equal. ⇒∠ACB = ∠ABC Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆] ⇒90° + ∠B + ∠C = 180° [∠A = 90°(Given)] ⇒∠B + ∠C= 180°- 90° = 90° But ∠B = ∠C ∠B = ∠C = = 45° Thus, ∠B = 45° and ∠C = 45° Question 8. Show that the angles of an equilateral triangle are 60° each. Solution: In ∆ABC, we have AB = BC = CA [ABC is an equilateral triangle] AB = BC ⇒∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal] Similarly, AC = BC ⇒∠A = ∠B …(2) From (1) and (2), we have ∠A = ∠B = ∠C = x (say) Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A] ∴x + x + x = 180o ⇒3x = 180° ⇒x = 60° ∴∠A = ∠B = ∠C = 60° Thus, the angles of an equilateral triangle are 60° each. ### NCERT Solutions for Class 9 Maths Exercise 7.3 Question 1. ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (See figure). If AD is extended to intersect BC at P, show that: (i) ABD ACD (ii) ABP ACP (iii) AP bisects A as well as D. (iv) AP is the perpendicular bisector of BC. Solution: (i)ABC is an isosceles triangle. AB = AC DBC is an isosceles triangle. BD = CD Now in ABD and ACD, AB = AC [Given] BD = CD [Given] ABD ACD [By SSS congruency] (ii)Now in ABP and ACP, AB = AC [Given] AP = AP ABP ACP [By SAS congruency] (iii)Since ABP ACP [From part (ii)] BAP = CAP [By C.P.C.T.] AP bisects A. Since ABD ACD [From part (i)] Now ADB + BDP = [Linear pair] ……….(iii) And ADC + CDP = [Linear pair] ……….(iv) From eq. (iii) and (iv), BDP = CDP DP bisects D or AP bisects D. (iv)Since ABP ACP [From part (ii)] BP = PC [By C.P.C.T.] ……….(v) And APB = APC [By C.P.C.T.] …….(vi) Now APB + APC = [Linear pair] APB + APC = [Using eq. (vi)] 2APB = APB = AP BC ……….(vii) From eq. (v), we have BP PC and from (vii), we have proved AP B. So, collectively AP is perpendicular bisector of BC. Question 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that: Solution: In ABD and ACD, AB = AC [Given] ABD ACD [RHS rule of congruency] BD = DC [By C.P.C.T.] Question 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that (i) ∆ABC ≅∆PQR (ii) ∆ABM ≅∆PQN Solution: In ∆ABC, AM is the median. ∴BM = BC ……(1) In ∆PQR, PN is the median. ∴QN = QR …(2) And BC = QR [Given] ⇒BC = QR ⇒BM = QN …(3) [From (1) and (2)] (i) In ∆ABM and ∆PQN, we have AB = PQ , [Given] AM = PN [Given] BM = QN [From (3)] ∴∆ABM ≅∆PQN [By SSS congruency] (ii) Since ∆ABM ≅∆PQN ⇒∠B = ∠Q …(4) [By C.P.C.T.] Now, in ∆ABC and ∆PQR, we have ∠B = ∠Q [From (4)] AB = PQ [Given] BC = QR [Given] ∴∆ABC ≅∆PQR [By SAS congruency] Question 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. Solution: Since BE ⊥AC [Given] ∴BEC is a right triangle such that ∠BEC = 90° Similarly, ∠CFB = 90° Now, in right ∆BEC and ∆CFB, we have BE = CF [Given] BC = CB [Common hypotenuse] ∠BEC = ∠CFB [Each 90°] ∴∆BEC ≅∆CFB [By RHS congruency] So, ∠BCE = ∠CBF [By C.P.C.T.] or ∠BCA = ∠CBA Now, in ∆ABC, ∠BCA = ∠CBA ⇒AB = AC [Sides opposite to equal angles of a ∆ are equal] ∴ABC is an isosceles triangle. Question 5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥BC to show that ∠B = ∠C. Solution: We have, AP ⊥BC [Given] ∠APB = 90° and ∠APC = 90° In ∆ABP and ∆ACP, we have ∠APB = ∠APC [Each 90°] AB = AC [Given] AP = AP [Common] ∴∆ABP ≅∆ACP [By RHS congruency] So, ∠B = ∠C [By C.P.C.T.] ### NCERT Solutions for Class 9 Maths Exercise 7.4 Question 1. Show that in a right angled triangle, the hypotenuse is the longest side. Solution: Let us consider ∆ABC such that ∠B = 90° ∴∠A + ∠B + ∠C = 180° ⇒∠A + 90°-+ ∠C = 180° ⇒∠A + ∠C = 90° ⇒∠A + ∠C = ∠B ∴∠B > ∠A and ∠B > ∠C ⇒Side opposite to ∠B is longer than the side opposite to ∠A i.e., AC > BC. Similarly, AC > AB. Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side. Question 2. In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB. Solution: ∠ABC + ∠PBC = 180° [Linear pair] and ∠ACB + ∠QCB = 180° [Linear pair] But ∠PBC < ∠QCB [Given] ⇒180° – ∠PBC > 180° – ∠QCB ⇒∠ABC > ∠ACB The side opposite to ∠ABC > the side opposite to ∠ACB ⇒AC > AB. Question 3. In figure, ∠B <∠A and ∠C <∠D. Show that AD < BC. Solution: Since ∠A > ∠B [Given] ∴OB > OA …(1) [Side opposite to greater angle is longer] Similarly, OC > OD …(2) Adding (1) and (2), we have OB + OC > OA + OD Question 4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B >∠D. Solution: Let us join AC. Now, in ∆ABC, AB < BC [∵AB is the smallest side of the quadrilateral ABCD] ⇒BC > AB ⇒∠BAC > ∠BCA …(1) [Angle opposite to longer side of A is greater] Again, in ∆ACD, CD > AD [ CD is the longest side of the quadrilateral ABCD] [Angle opposite to longer side of ∆ is greater] Adding (1) and (2), we get ∠BAC + ∠CAD > ∠BCA + ∠ACD ⇒∠A > ∠C Similarly, by joining BD, we have ∠B > ∠D. Question 5. In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ. Solution: In ∆PQR, PS bisects ∠QPR [Given] ∴∠QPS = ∠RPS and PR > PQ [Given] ⇒∠PQS > ∠PRS [Angle opposite to longer side of A is greater] ⇒∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS] ∵Exterior ∠PSR = [∠PQS + ∠QPS] and exterior ∠PSQ = [∠PRS + ∠RPS] [An exterior angle is equal to the sum of interior opposite angles] Now, from (1), we have ∠PSR = ∠PSQ. Question 6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest. Solution: Let us consider the ∆PMN such that ∠M = 90° Since, ∠M + ∠N+ ∠P = 180° [Sum of angles of a triangle is 180°] ∵∠M = 90° [PM ⊥l] So, ∠N + ∠P = ∠M ⇒∠N < ∠M ⇒PM < PN …(1) Similarly, PM < PN1…(2) and PM < PN2…(3) From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it. ### NCERT Solutions for Class 9 Maths Exercise 7.5 Question 1. ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC. Solution: Let us consider a ∆ABC. Draw l, the perpendicular bisector of AB. Draw m, the perpendicular bisector of BC. Let the two perpendicular bisectors l and m meet at O. O is the required point which is equidistant from A, B and C. Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle. Question 2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle. Solution: Let us consider a ∆ABC. Draw m, the bisector of ∠C. Let the two bisectors l and m meet at O. Thus, O is the required point which is equidistant from the sides of ∆ABC. Note: If we draw OM ⊥BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle. Question 3. In a huge park, people are concentrated at three points (see figure) A: where these are different slides and swings for children. B: near which a man-made lake is situated. C: which is near to a large parking and exist. Where should an ice-cream parlor be set? up so that maximum number of persons can approach it? [Hint The parlour should be equidistant from A, B and C.] Solution: Let us join A and B, and draw l, the perpendicular bisector of AB. Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O. The point O is the required point where the ice cream parlour be set up. Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O. Question 4. Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles? Solution: It is an activity. We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii). ∴The Fig. (ii) has more triangles. Get 30% off your first purchase! X error: Content is protected !! Scroll to Top
# How do you solve 5x - 5y + 10z = -11, 10x + 5y - 5z = 1 and 15x - 15y -10z = -1 using matrices? Aug 2, 2018 The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- \frac{2}{5} \\ \frac{1}{5} \\ - \frac{4}{5}\end{matrix}\right)$ #### Explanation: Perform the Gauss-Jordan Elimination on the augmented matrix $A = \left(\begin{matrix}5 & - 5 & 10 & \| & - 11 \\ 10 & 5 & - 5 & \| & 1 \\ 15 & - 15 & - 10 & \| & - 1\end{matrix}\right)$ Make the pivot in the first column and first row $R 1 \leftarrow \left(\frac{R 1}{5}\right)$ $\left(\begin{matrix}1 & - 1 & 2 & \| & - 2.2 \\ 10 & 5 & - 5 & \| & 1 \\ 15 & - 15 & - 10 & \| & - 1\end{matrix}\right)$ Eliminate the first column $R 2 \leftarrow \left(R 2 - 10 R 1\right)$ and $R 3 \leftarrow \left(R 3 - 15 R 1\right)$ $\left(\begin{matrix}1 & - 1 & 2 & \| & - 2.2 \\ 0 & 15 & - 25 & \| & 23 \\ 0 & 0 & - 40 & \| & 32\end{matrix}\right)$ Make the pivot in the second column and second row $R 2 \leftarrow \left(\frac{R 2}{15}\right)$ $\left(\begin{matrix}1 & - 1 & 2 & \| & - 2.2 \\ 0 & 1 & - \frac{5}{3} & \| & \frac{23}{15} \\ 0 & 0 & - 40 & \| & 32\end{matrix}\right)$ Eliminate the second column $R 1 \leftarrow \left(R 1 + R 2\right)$ $\left(\begin{matrix}1 & 0 & \frac{1}{3} & \| & - \frac{2}{3} \\ 0 & 1 & - \frac{5}{3} & \| & \frac{23}{15} \\ 0 & 0 & - 40 & \| & 32\end{matrix}\right)$ Make the pivot in the third column and third row $R 3 \leftarrow \left(\frac{R 3}{-} 40\right)$ $\left(\begin{matrix}1 & 0 & \frac{1}{3} & \| & - \frac{2}{3} \\ 0 & 1 & - \frac{5}{3} & \| & \frac{23}{15} \\ 0 & 0 & 1 & \| & - \frac{4}{5}\end{matrix}\right)$ Eliminate the third column $R 1 \leftarrow \left(R 1 - \frac{1}{3} R 3\right)$, and $R 2 \leftarrow \left(R 2 + \frac{5}{3} R 3\right)$, $\left(\begin{matrix}1 & 0 & 0 & \| & - \frac{2}{5} \\ 0 & 1 & 0 & \| & \frac{1}{5} \\ 0 & 0 & 1 & \| & - \frac{4}{5}\end{matrix}\right)$ The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- \frac{2}{5} \\ \frac{1}{5} \\ - \frac{4}{5}\end{matrix}\right)$
# How do you solve (20+ 35+ 2( 25+ 40) ) \div 5? Nov 27, 2016 $37$ #### Explanation: $\left(20 + 35 + 2 \textcolor{b l u e}{\left(25 + 40\right)}\right) \setminus \div 5$ This is all one term, so you have to work inside the bracket first. =$\left(20 + 35 + \textcolor{b l u e}{2 \left(65\right)}\right) \setminus \div 5$ =$\left(20 + 35 + \textcolor{b l u e}{130}\right) \setminus \div 5$ =$\left(\textcolor{red}{20 + 35 + 130}\right) \setminus \div 5$ =$\textcolor{red}{185} \div 5$ =$37$
# Basic Shapes In geometry we know basic shapes involve plane figures and solid figures. Difference between square and rectangle: Square has all sides equal and rectangle has opposite sides equal. Region enclosed between the two circles: If two circles cross each other Plane figure: The figures that lie in the plane are called plane figures. For example, square, rectangle, triangle, diamond and circle. They are also called flat shapes. Square: It has 4 sides and 4 corners. Area enclosed inside is called the square region. Rectangle: It has 4 sides and 4 corners and the area enclosed inside is called the rectangular region. Triangle: It has 3 sides and 3 vertices or corners. Area enclosed inside the triangle is called the triangular region. Diamond: It has 4 sides and 4 corners and area enclosed is called the region of diamond. Circle: It has no sides and no corners. Area enclosed inside the circle is called the circular region. Area enclosed between the two circles is shown by the shaded region. If one circle lies inside the other circle Area enclosed between the two circles is shown by the shaded region. Solid figures: We know, about some of the geometrical shapes like cuboid, cube, cylinder, cone, sphere, triangular prism, etc. These are called solid figures or solids. They are also known as three dimensional (3-D) figures. They occupy space. In real life many objects which are seen in our surroundings have the shape of any one or many of the solids mentioned above. Third grade geometry lessons encourage the kids to develop the basic concept to increase the knowledge on geometry. Surfaces of the solids: the two types of surfaces such as plane surfaces, curved surfaces. Common solid figures: Some of the solid shapes such as cube, cuboid, cylinder, cone, sphere. Related Concepts
# Graphing Functions in Discrete mathematics Graphing function can be described as a process of drawing the graph of a function. There are some simple functions and some complex functions in the basic graphing functions. The simple functions are cubic, linear, quadratic, and many more and the complex functions are logarithmic, rational, etc. In this section, we will understand the definition, basic functions, and examples of graphing functions. ### Use of Graphing functions With the help of a graphical function, we can draw a curve that is used to indicate the function on a coordinate plane. If this curve (graph) indicates a function, then each and every point on the curve will equally satisfy the function. For example: In this graph, we will show the linear function f(x) = -x+2. Now we can take any point on the line from the above graph. Here we will take (-1, 3). Now we will substitute (-1, 3) = (x, y) in the function f(x) = -x+2. That means for this function, x = -1, and y = 3. We can also write the function f(x) = -x+2 in the form y = -x+2. Now we will put the values of x and y in the function y = -x+2 and get the following: 3 = -(-1)+2 3 = 1+2 3 = 3 Hence, we can say that the point (-1, 3) satisfies the function. Similarly, we can take different points from the above line and check whether those points satisfy the function. In this case, the function will be satisfied by every point on the line/curve. The process of drawing these types of curves, which represent the functions, is known as graphing functions. ### Graphing Basic function There are a lot of basic graphic functions that are very easy, i.e., quadratic functions and linear functions. Some basic ideas of graphical functions are described as follows: • If it is possible to identify the shape, then we will first do it. For example: The given graph will be a line if it is a linear function with the form f(x) = ax+b. The given graph will be a parabola if it is a quadratic function which has a form f(x) = ax2 + bx + c. • We can determine some points on it with the help of substituting some random values of x and then substituting each value into the function to determine the corresponding values of y. Now we will understand some examples of graphing basis functions with the help of graphical linear functions, Graphical quadratic functions, and Graphical complex functions. ### Graphical Linear Functions We have already drawn a graph for a linear function with the form f(x) = ax+b. Here we will also take the same linear form. Here we will create a table with some random values of x. So we will take some values like x = 0 and x = 1, and then we will find the value of y by putting each of the values of x in y = -x+2. After putting the values, we will get the following: x y 0 -0+2 = 2 1 -1+2 = 1 So from the above, we get two points on the line that is (0, 2) and (1, 1). If we plot any of the one points on a graph and join these points with the help of a straight line (extending the line on both sides), then the graph will be the same as shown above. In this function, we can also determine some random points on it. With the help of these random values, we may not get a perfect U-shaped curve because if we want to get a perfect U-shaped curve, then we have to know about the point where the curve is turning. That means for a perfect U-shaped curve, we have to find its vertex. When we successfully find the vertex, we identify two or three random points on each side of the vertex. These random points will help us to draw the graph of a function. Example: In this example, we have to draw a graph of the quadratic function, which has a line f(x) = x2 - 2x + 5. Solution: First, we will compare f(x) = x2-2x+5 with f(x) = ax2+bx+c, and then we will get a = 1, b = -2, and c = 5. Now we will get the coordinate of x axis and y axis with the help of these values. The x coordinates of vertex will be h = -b /2a = -(-2) /2(1) = 1. The y coordinates of the vertex will be f(1) = 12 - 2(1) + 5 = 4. Hence, the x and y coordinates of the vertex will be (1, 4). Now we will create a table by taking two random numbers of x on each side of 1. Then we will use the above function y = x2-2x+5, and find the y coordinates. x y -1 (-1)2 - 2(-1) + 5 = 1 + 2 + 5 = 8 0 02 - 2(0) + 5 = 0-2+5 = 5 For vertex 1 4 2 22 - 2(2) + 5 = 4 - 4+5 = 5 3 32 - 2(3) + 5 = 9 - 6 + 5 = 8 With the help of above table, the plots will be (-1, 8), (0, 5), (1, 4), (2, 5), and (3, 8). Now we will join all the points on the graph sheet and extend the curve on both sides like this: ### Graphical Complex Functions The graphing function will be known as the simplest function if each of their range and domain is a set of real numbers. This case is not compulsory for all types of functions. There can be complex functions for which we have to take care about the range, domain, holes, and asymptotes at the time of drawing them. The most popular those types of functions are described as follows: • Rational Functions: The parent function of rational function must be in the form f(x) = 1/x. The rational function can also be known as the reciprocal function. • Exponential Functions: The parent function of exponential function must be in the form f(x) = ax. • Logarithmic Functions: The parent function of logarithmic function must be in the form f(x) = log x. Now we will show the graph of each of the function's parent functions separately like this: We have to follow the following steps in each of these cases for graphing functions: • First, we will find out the domain and range of the function, and with the help of them, we will draw the curve. • After that, we will determine the x-intercepts(s) and y-intercepts(s) and then plot them. • Determine whether there is any hole. • After that, we will determine the asymptotes (horizontal, vertical, and slant) and draw them with the help of dotted lines so that the graph can be break along those lines. While doing this, we have to take care that the graph does not touch them. • Now, we will make a table with the help of taking some random values of x (on both sides of x-intercept and/or on both sides of vertical asymptote). Then by these values, we will find out the corresponding values of y. • We will plot the points from the table. For this, we will join them on the basis of their range, domain, and asymptotes. We will use the Graphical rational functions, Graphical exponential functions, and Graphical logarithmic functions to understand the graph of a function in different cases with the help of above steps. ### Graphing Rational functions Here we will graph a rational function f(x) = (x+1) /(x-2) with the help of above steps like this: • From the above rational function, the domain = {x ∈ R \| x ≠ 2} and Range = {y ∈ R \| y ≠ 1}. Now we will determine the domain and range of a rational function. • The x-intercept of this rational function is (-1, 0), and y-intercept of this function is (0, -0.5). • It does not contain any holes. • Vertical asymptote (VA) of this function is x = 2, and horizontal asymptote (VA) of this function is y = 1. • Now, on both sides of vertical asymptote x = 2, we will take some random values, and then we will find out the respective value of y like this: x y -1 (-1+1) /(-1-2) = 0 (x-int) 0 (0+1) /(0-2) = -0.5 (y-int) 2 VA 3 (3+1) /(3-2) = 4 4 (4+1) /(4-2) = 2.5 Now we will plot all the above points along with Horizontal asymptote (HA) and vertical asymptote (VA) in the following way: ### Graphing Exponential Functions Here we will assume an exponential function f(x) = 2-x + 2. With the help of steps described in the Graphical complex functions, we will graph this function like this: • The domain of this function is the set of all real numbers (R), and the range of this function is y > 2. • This function has a horizontal asymptote at y = 2, but it does not have any vertical asymptote. • It has a y-intercept which is (0, 3), but it does not have any x-intercepts. • It also does not have any holes. • So lastly, we don't have any data related to x-intercept and VA (vertical asymptote). We only have data related to the y-intercept, which is (0, 3). On both sides of x = 0, we will take some random values and then frame a table with the help of these values like this: x y -2 2-(-2) +2 = 6 -1 2-(-1) +2 = 4 0 3 (y-int) 1 2-1 + 2 = 2.5 2 2-2 + 2 = 2.25 Now we will plot all the above information on a graph in the following way: ### Graphical Logarithmic Functions Here we will assume a logarithmic function f(x) = 2 log2 x-2. With the help of steps described in the Graphical complex functions, we will graph this function like this: • The domain of this function is x>0, and the range of this function is the set of all real numbers (R). • The x-int of this function is (2, 0), but this function does not have any y int. • The vertical asymptote of this function is y = 0 (x-axis), and it does not contain any horizontal asymptote. • It also does not have any holes. • So lastly, we only have one reference point, i.e., (2, 0). On both sides of 0, we will take some random values and then frame a table with the help of these values. We cannot take the value of x less than 0 because the domain is x>0. x y 1 2log2 1-2 = -2 2 0 (x int) 4 2log2 4-2 = 2 Here we have chosen those types of x's values that are able to simplify the value of y easily. Now we will plot all the above information on a graph in the following way: ### Graphing Functions by Transformations We can graph the functions with the help of applying transformations on the graph of parent functions. Here we will show some parents functions of some important functions like this: Linear function: Its parent function is: f(x) = x Quadratic function: Its parent functions is: f(x) = x2 Cubic function: Its parent functions is: f(x) = x3 Absolute Value function: Its parent function is: f(x) = \|x\| Reciprocal function: Its parent function is: f(x) = 1/x Logarithm function: Its parent function is: f(x) = log x Square root function: Its parent function is: √x Cube root function: Its parent function is: ∛x Exponential function: f(x) = ax, 0<a<1 We should keep in mind the look of the graph of all above-described parent functions. After that, we will be able to apply the transformation to the graph of given function. Transformation Change in Graph f(x) + c In the graph of this function, the change happens in the c unit. Here c unit shifts upward. f(x) - c In the graph of this function, the change happens in the c unit. Here c unit shifts downward. f(x + c) In the graph of this function, the change happens in the c unit. Here c unit shifts to the left. f(x - c) In the graph of this function, the change happens in the c unit. Here c unit shifts to the right. -f(x) In the graph of this function, the change happens on the x-axis. Here x-axis is reflected (upside down). f(-x) In the graph of this function, the change happens on the y-axis. Here y-axis is reflected (left and right sides are swapped). f(ax) Here, horizontal dilation occurs with the help of a factor of 1/a. a f(x) Here, vertical dilation occurs with the help of a factor of a. ### Important Notes on Graphing Functions • In the graphing function, f(ax) ≠ a f(x). There may be different values for both. • The asymptotes will never be touched by the graph of a function. • The value of x can be a decimal number, real number, or whole number, which we use to plot any function f(x). • We should not choose those types of values of x in the table that does not have a function's domain.
## 3.5 The Log Scale – L Many slide rules have the L scale on them, which gives the logarithm of the number aligned typically on the C or D scale. As can be seen from our plot labeled Logarithmic Scale in Logarithms and Log Scales, if the values on the C scale have a logarithmic spacing, then the logarithms of those values will have a linear spacing. Hence, the L scale on a slide rule is easy to identify as it has its digits 0 to 10 spaced evenly by about one inch on a standard 10 inch rule. An early use of the L scale was to find the value of a number raised to an arbitrary power. Suppose you wanted to know $$3.7^{4.6}$$. Then one could find the log of 3.7 = 0.568 on the L scale and multiply by 4.6 to get 2.614 using the C and D scales. Then find on the L scale which number has the logarithm 0.614 – which is 4.11 – and then multiply by 100 (due to the “2” out front in 2.614) to get the final answer: 411. Check by computer: ## [1] 3.7 to the power 4.6 = 410.89223180408 . We noted in previous sections that sometimes reading squares and cubes from the A and K scales can yield results only accurate to 1-2 digits. If a more accurate answer is required, the L scale often can be used to perform the calculation. For example, let’s re-compute the volume of a sphere of radius $$R$$ = 4.58 inches. Rather than using the K scale, which yielded “about 400” cubic inches, we do the following: • From our formula for the volume of a sphere, $$V = \frac{4\pi}{3}R^3$$, we see that $$\log V = \log (4\pi/3) + 3 \log R$$. • If a gauge mark is not already on the rule, find our necessary constant of $$4\pi/3$$ using the C/D scales. It should be about 4.19. • Using the L scale, find the log of 4.58, which is about 0.661, and the log of 4.19, which is about 0.622. • Using C/D, or mentally, find 3 times 0.661, which should be roughly 1.983. • Adding these results, 0.622 + 1.983 = 2.605 = 0.605 + 2. • Resetting the slide if necessary, use the L scale to determine the number whose log is 0.605. I find about 4.03; the “2” in the above result tells us to use 403. A more precise answer, using a personal computer, is 402.425. We see that using the L scale for this purpose may give more accuracy, but it also involves many more steps and perhaps the writing down of intermediate values. If one only needs a more approximate answer, the use of the A/B and K scales can save significant time.
# How do I determine the length of the shorter base of a trapezoid from the longer base length, height, and only two angles? How do I determine the length of the shorter base of a trapezoid from the longer base length, height, and only two angles? An example would be 24" longer base, with 45 deg angles at both ends with only 1" in height. Both upper angles would be 135 deg. What would the length of the shorter base be? How do you solve for it? Thanks! Suppose the longer base length is $b$ and the $2$ angles are $\alpha$ and $\beta$ with height being $h$ The shorter base length is $$b-h\cot \alpha-h\cot \beta = b-h(\cot \alpha+ \cot \beta)$$ To see this, notice that the height and the base form perpendicular angle. If you decompose the figure into a rectangle and two right triangles, and note the triangles are isosceles, you'll see the length you seek is a longer base minus the height doubled. One way to think about it is this: the longer base and the angles determine a triangle. The height determines where the similar triangle is cut off to give the trapezium. Thus, if we let the base you want be $$b,$$ and the other $$B,$$ then it follows that $$\frac bB=\frac aA.$$ You can determine $$A,$$ the height of the main triangle, from the given information. Then $$a,$$ the height of the cut-off triangle, is given by $$a=A-h,$$ with $$h$$ being the height of the trapezium, also given. The work is in determining $$A.$$ One way is to first calculate some other side of the large triangle (use the cosine rule, for example -- remember all three angles are known, and one side). Then you now have two sides and an included angle. This allows you to determine $$A$$ as a scaled sine of the included angle. In the specific example: As the given base angles are equal, the trapezoid is isosceles. The shorter side then is $$24- 1 \cdot \cot 45^{\circ}- 1 \cdot \cot 45^{\circ} =22^{''}$$ General question: If the base side dimension other than the specific 24^{''}$is given as asked in the second para of question you asked then we can view it generally to choices when the parallel sides get extended by the same amount: When the $$\pm 45^{\circ}$$ slant sides are extended it makes a right angle at intersection i.e., at the vertex. We can draw parallels making similar right triangles. Not shown. Between parallels the minimum distance is constant. That is, difference of the longer and shorter sides is constant. Differences of two parallel sides of an isosceles trapezium so its sides make up a triangle and its height ( between parallel sides) only are given. Due to insufficient data the trapezium cannot be uniquely constructed, but can be constructed/ solved only upto an arbitrary displacement constant motion of vertex $$c$$ parallel to longer base as shown. Shorter side $$c$$ is indeterminate. Some trapezium solutions are shown, modified from right triangle of sides $$(\sqrt2, \sqrt2,1)$$. So the length of the shorter base of a trapezoid from the longer base length, height, and only two angles cannot be determined. The following answer sketch applies to your question in the first para: • Given is the length of the longer base, not the difference in lengths. Commented Nov 6, 2021 at 15:18 • You can choose the length of base as anything, including$24^{''} $in the given specific problem, shorter side being$22^{''},\$ the other dimensions are satisfied. Commented Nov 6, 2021 at 15:27 • You do not choose what is given. Commented Nov 6, 2021 at 15:41 • The first para in OP's question was general and the second case was a particular one. In the first case there was no need to use what is given. I tried to address both aspects. Hope it is in order. Commented Nov 6, 2021 at 21:45
# Multiply any 4 digit numbers So, far, we have already learnt the shortcuts, how to multiply of any 2-digit and 3-digit number using criss-cross multiplication technique. In this tutorial, we will see how to multiply any 4 digit numbers. Now Let’s try to understand the criss-cross multiplication technique to multiply any 4 digit numbers. As you know, the number of steps when multiplying any 4-digit numbers is 7. Let us consider a simple example 3215 x 4231. The step-by-step procedure can be explained with the following figure: Notice the carry-forward in each stage of the multiplication. I hope, The steps are self – explanatory. Notice the additional steps compared to 2 x 2 and 3 x 3 multiplication. Exercise : On the similar lines, can you multiply? • 2318×1339 • 7222×4223 • 9222×9229 • 5216×1265 ## Multiplication when the number of digits are unequal If the number of digits in multiplicand and multiplicand is unequal, you add zeros so that the digits have the same number of digits. Let us consider an example. Suppose you have to multiply 4125 x 511. So, the number of digits are 4 and 3 respectively. So, we convert 511 to 05111 and use this technique. The multiplication steps are illustrated in the following figure : Exercise : Can you multiply? • 12 x 145 • 222 x 4223 • 92 x 9 • 52 x 125 ### Kiran Chandrashekhar Hey, Thanks for dropping by. My name is Kiran Chandrashekhar. I am a full-time software freelancer. I love Maths and Mathematical Shortcuts. Numbers fascinate me. I will be posting articles on Mathematical Shortcuts, Software Tips, Programming Tips in this website. I love teaching students preparing for various competitive examinations. Read my complete story. 1. Yassin Hello there, just became aware of your blog thorguh Google, and found that it is truly informative. I’m going to watch out for brussels. I will appreciate if you continue this in future. Many people will be benefited from your writing. Cheers! 2. avisha Thanks sir this blog is really helpful…plz continue this… • Kiran Chandrashekhar Thanks Avisha, I will try to post lot of shortcuts in coming days. I am glad it helped you in your preparation. Thanks Kiran • Avisha 3. Pingback: URL 4. Hi • Kiran Chandrashekhar Hi, Do you have any doubts ? 5. tutoring melbourne Thanks for providing these details online. 6. nandini hi sir ur blog is informative thanku verymuch.mathematics is like a nightmare for me u made it so easy for me.keep posting.I dont know any shortcut methods and am preparing forRRB TC posts.clarify my doubts if possible • Kiran Chandrashekhar Thanks a lot for your nice words. Write to me with your specific questions, I shall try to answer it. 7. Bhuvanmohini Hello Kiran, I was searching for math shortcuts for SBI clerk exams whn i came accross ur site.And let me tell u,it was incredibly helpful!I hv already bookmarked it.pls post some shortcuts for decimal multiplication as well.Thanks! 🙂 8. Bhuvanmohini Hello Kiran, I was searching for math shortcuts for SBI clerk exams whn i came accross ur site.And let me tell u,it was incredibly helpful!Thank u so much for dis indispensable service! U are simply brilliant.I hv already bookmarked it.pls post some shortcuts for decimal multiplication as well.Keep going.:) 9. nahid sir, i have a query. i found some problem in adding.. for example 2485 x 9483 4 step of calculation is – 45+6+32+32=115 so my question is with how many digit we should proceed to further calculation…plzz tell… 10. sheela very useful thanks sir…. can give shortcuts for (2to3), (2to4), (3to4) digits etc… 11. Pam G. Hi Mr. Chandrashekhar, Thanks for the steps in doing the four digit multiplications. I hope you didn’t mind but I copied the steps so I can have a guide close to me. And again thanks. 12. thiyagu its very helpful thanks to ur shortcut 13. Monik
How to solve right triangles There are also many YouTube videos that can show you How to solve right triangles. Our website will give you answers to homework. How can we solve right triangles In this blog post, we will provide you with a step-by-step guide on How to solve right triangles. distance = sqrt((x2-x1)^2 + (y2-y1)^2) When using the distance formula, you are trying to find the length of a line segment between two points. The first step is to identify the coordinates of the two points. Next, plug those coordinates into the distance formula and simplify. The last step is to take the square root of the simplify equation to find the distance. Let's try an example. Find the distance between the points (3,4) and (-1,2). First, we identify the coordinates of our two points. They are (3,4) and (-1,2). Next, we plug those coordinates into our distance formula: distance = sqrt((x2-x1)^2 + (y2-y1)^2)= sqrt((-1-3)^2 + (2-4)^2)= sqrt(16+4)= sqrt(20)= 4.47 Therefore, the distance between the points (3,4) and (-1,2) is 4.47 units. Online math graphing calculators are a great tool for visual learners. They can help you see patterns and relationships that might be difficult to spot on paper. They can also be a great way to check your work. Most online graphing calculators are free to use, and they’re easy to find with a simple Google search. Whether you’re working on a school project or just trying to better understand a concept, an online math graph can be a valuable resource. Next, take your time and read the instructions carefully. If you are still having trouble understanding the material, try looking up key terms in a dictionary or doing additional research. Finally, don't be afraid to ask for help from a teacher or tutor. By following these tips, you can increase your chances of getting the answers you need. Basic mathematics is the study of mathematical operations and their properties. The focus of this branch of mathematics is on addition, subtraction, multiplication, and division. These operations are the foundation for all other types of math, including algebra, geometry, and trigonometry. In addition to studying how these operations work, students also learn how to solve equations and how to use basic concepts of geometry and trigonometry. Basic mathematics is an essential part of every student's education, and it provides a strong foundation for further study in math. Finally, maths online can also help to build a student's confidence by allowing them to track their progress and receive feedback from their peers. As such, maths online is an invaluable resource for any student wishing to improve their mathematical skills. We solve all types of math troubles This app is very useful for everyone. Whenever I faced shortcut tricky problem in math, I solved it through app. I'm really satisfied with this. And hopefully the authorities are developing free contains more. Stephanie Washington Very helpful for beginners as well as for the college students. Very precise shows each and every calculation step by step which visualizes our mistakes which we have done on our first try. Brittany Lee Cosine solver Pre calc solver with steps Help me solve my math problem Square root solver Solve the square
### What does the postulate of segment completion say? In geometry, the postulate of addition of segments says that with 2 points A and C a third point B lies on the segment AC if the distances between the points satisfy the equation AB + BC = AC. The postulate of adding a segment is often useful for demonstrating the congruence results of the segment. ### Simply put, what is the difference between the angle addition postulate and the segment addition postulate? Adding the postulate of the segment - If B is between A and C, then AB + BC = AC. If AB + BC = AC, then B lies between A and C. Postulate of angular addition - If P lies within ∠, then ∠ + ∠ = ∠. ### Also, how do you find the tilted target? Using a protractor The best way to measure a protractor is to use a protractor. To do this, first install a bar along a 0 degree line on the protractor. Then align the top with the center of the protractor. Follow the second ray to determine the angle measurement to the nearest degree. ### Do you also know what it means to add segments? In geometry, the segment addition postulate says that with 2 points A and C a third point B lies on the segment AC if the distances between the points satisfy the equation AB + BC = AC. The postulate of adding a segment is often useful for demonstrating the congruence results of the segment. ### How do I find the center of a segment? The center is the point on the segment halfway between the ends. The center of a segment can be found by counting. If the segment is horizontal or vertical, you can find the center point by dividing the length of the segment by 2 and counting that value from either end. ### How do you solve the assumptions? If you have a segment with endpoints A and B and point C is between points A and B, then AC + CB = AB. Angle addition postulate: This postulate states that if you divide an angle into two smaller angles, the sum of these two angles should equal the original angle measurement. ### What is the bisector? The (inner) bisector, also called the inner bisector (Kimberling 1998, p. 1112), is the line or line segment that divides the angle into two equal parts. The bisectors meet during the process. which has three linear coordinates 1: 1: 1. ### What is a perpendicular that intersects a triangle? The mean perpendicular to one side of a triangle is a line perpendicular to the side and passing through the center point. The three bisectors, perpendicular to the sides of a triangle, meet in a single point called the perimeter. ### Is a segment a specific term? In geometry, a line segment is part of a line bounded by two different end points and containing each point on the line between the end points. A closed line segment includes both ends, while an open line segment excludes both ends. A half-open line segment has exactly one end. ### How do you find the distance between two points? Steps Take the coordinates of two points between which you want to determine the distance. Name one point point 1 (x1, y1) and create the other point 2 (x2, y2). You know the distance formula. Find the horizontal and vertical distance between the points. Square the two values. Square the values together. Take the square root of the equation. ### What is a distance formula in geometry? The distance formula is derived from the Pythagorean theorem. To find the distance between two points (x1, y1) and (x2, y2), take the coordinates of these ordered pairs and use the following formula. The distance formula is. Distance = √ (x2 - x1) 2+ (y2 - y1) 2. ### What is a center point in geometry? In geometry, the center point is the center point of a segment. It is equidistant from both ends and is the center of the segment and the ends. Cut the segment in half. ### What is the distance formula in geometry? The distance formula is used to determine the distance d between two points. The distance formula is obtained by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. The hypotenuse of the triangle is the distance between the two points. What is a real world example of postulate? An example of a postulate is the fact that the world is not flat to support the argument for strong scientific development over the centuries. A postulate is defined as a statement, claim or statement of something as true. An example of a postulate is the requirement of equality. An example of a postulate is the defense of the existence of God. ## What is the unique plane postulate? Postulate 1: Postulate with a line Exactly one line passes through any two points. (If you have two points, you have exactly one line.) With three non-collinear points, there is exactly one plane. If two points on the same plane lie on the same plane, the line connecting them is on that plane. ## What is midpoint postulate? Middle postulate Every line segment has exactly one midpoint. This assumption refers to the center, not the lines through the center. An infinite number of lines pass through the center. ## What is the subtraction postulate? Like the addition postulate, they now have the subtraction postulate. Deduction Postulate: If equal amounts are subtracted from equal amounts, the differences are equal. For example, if they apply the postulate of subtraction in the proof, let's look at another example: you get information. ## What is the correct definition for postulate? A postulate (also called an axiom) is a statement that everyone accepts as correct. This is useful for taking tests in math and science. ## What is the difference between proof and postulate? As nouns, the difference between proof and postulate is that proof (accounting) is an effort, process or operation aimed at establishing or discovering a fact or truth that is taken for granted or generally accepted, especially when the reasoning is considered. can be used. ## Does a postulate need to be proved? A postulate is a true statement that does not need to be proven. Learn more about postulates Postulates are used to derive other logical statements to solve a problem. The postulates are also called axioms. ## What are the postulate and theorems? • Definition. Postulates are mathematical statements that you think are true without proof, and theorems are mathematical statements that you can or must prove. • Key. Postulates are considered true without proof while theorems can be proved. • Evidence is needed. • Conclusion. ## What is a real world example of postulate government Each mathematical theorem started as a hypothesis or postulate and was then tested and accepted as a proven mathematical fact, discussed in detail below. Are you a student or a teacher? As a member, you also have unlimited access to over 84,000 lessons in math, English, science, history, and more. ## Which is an example of a postulate without proof? A postulate is a statement that is accepted without proof. Axiom is another name for a postulate. For example, if you know that Pam is six feet tall and that all her siblings are taller than she is, you would believe her if she said all her siblings were at least six feet tall. ## What are the examples of postulate in real life situations? What are some examples of postulates in real life situations? 1: A hypothesis presented as a premise, condition, or essential premise of an argument. The word is primarily used to refer to geometry, but it is optional. Professor Wolfgang Pauli postulated neutrinos. ## Which is an example of a geometric postulate? Examples: geometric postulates. The rule postulate: points in a straight line can correspond to real numbers. In other words, each point on the line represents a real number. The postulate of adding a segment: Remember that a segment has two ends. ## Real world example definition The real world is where you really have to live and the circumstances you really have to deal with. A real world example is the life you are living now as opposed to the life you would like to live one day. ## What is a real world example of a sample? A real life example and a population can be based on high school. The sample will be a group of high school students. The entire population of the United States is high school students. An example is the ethnic group of New Mexico. All residents of the US are considered human populations. ## What is the definition of real world? Real world '. New Mexico. the field of practical or real experience, as opposed to the abstract, theoretical or idealized field of a classroom, laboratory, etc. ## What is a real world example of translation? Specific examples of translations: the movement of an aircraft while moving in the air, the operation of the lever of a crane (crane), sewing on a sewing machine, drilling decorative rivets into belts, throwing weights, throwing noodles like spaghetti. ## What is a real world example of postulate in math In mathematics, postulates are statements that are accepted as true without proof. This is considered the smallest proof of a great sentence. ## What is three point postulate? Postulate of the three points. because of the 3 non-collinear points there is exactly one plane. The postulate of the plan. the plane contains at least 3 non-collinear points. Flat line postulate. If two points lie in the same plane, the line containing them lies in the plane. ## What are the postulates in geometry? Postulate. A statement, also called an axiom, is considered true without proof. Postulates are the basic structure from which lemmas and theorems are derived. For example, all Euclidean geometry is based on five postulates known as Euclidean postulates. ## What is the postulate theorem? A postulate is a statement that is believed to be true (also called an axiom). ## What is a plane line postulate? In geometry, a point-line-plane postulate is a set of assumptions (axioms) that can be used in a set of postulates of Euclidean geometry in two (plane geometry), three (solid geometry), or more dimensions. ## What are the postulate involving points lines and plane? Lines and planes: PlaneLine Postulate: if two points lie on a plane, then the line containing them lies on the plane. Series of line intersections: If two lines intersect, then exactly one level contains both lines. The Plane Intersection Postulate: If two planes intersect, their intersection is a straight line. ## What is the unique plane postulate definition geometry In geometry, a point-line-plane postulate is a set of assumptions (axioms) that can be used in a set of postulates of Euclidean geometry in two (plane geometry), three (solid geometry), or more dimensions. These are the assumptions of the point plane postulate: Assumption of a line. ## What is plane point postulate? In geometry, a point-line-plane postulate is a set of assumptions (axioms) that can be used in a set of postulates of Euclidean geometry in two (plane geometry), three (solid geometry), or more dimensions. ## What are postulates in geometry? Postulates are the basic structure from which lemmas and theorems are derived. For example, all Euclidean geometry is based on five postulates known as Euclidean postulates. ## What is the unique plane postulate graph Postulate 1: The line contains at least two points. Postulate 2: There are at least three non-collinear points in the plane. Postulate 3: There is exactly one line at two points. Postulate 4: There is exactly one plane with three non-collinear points. ## Which is the postulate for a line through two points? Postulate 1: There is exactly one line at two points. Postulate 2: The line contains at least two points. ## How to identify a postulate from a diagram? The postulate of the intersection of straight lines. If two lines intersect, their intersection is exactly one point. The postulate of the intersection of planes. If two planes intersect, their intersection is a straight line. Identifying Postulates Using a Graph Use the table to write examples of a PlanePoint postulate and a PlaneLine postulate. ## Which is the best example of a postulate? Postulates of a point, line and plane An example of a postulate Postulate of two points Any two points form exactly one line. Line point postulates A line contains at least two points. A B Exactly one straight line goes through the points A and B. Lineℓ contains at least two points. Line intersection Postulate If two lines intersect, then: ## What is midpoint postulate in geometry The center of the coordinate plane is represented by a flat surface that extends infinitely in two dimensions and has no thickness. The plane contains an infinite number of points. ## Which is the postulate for the distance between two points? Postulate or postulate of a line Any point on a line can be assigned a valid number. The distance between any two points is the absolute amount of the difference between the corresponding numbers. Postulate or postulate of adding a segment. ## Which is the correct postulate for the intersection of two lines? Postulate 4: If two lines intersect, they intersect at exactly one point Postulate 5: There is exactly one plane passing through three non-collinear points. Postulate 6: If two planes intersect, then their intersection is a straight line + Postulate 7: If two points lie in the same plane, then the connecting line is in this plane. ## Which is the postulate of the measure of an angle? Postulate 8: The measure of an angle is a certain positive number. Postulate 9: if point D lies inside angle ABC, then m ABD + m DBC = m ABC Theorem: for a given angle there is only one bisector. ## What are the theorems and postulates of geometry? Line Postulates and Suggestions Name Definition Visual Index Postulate Through a point that is not on a given line, there is one and only one line parallel to the given line. A set of alternate interior angles. If two parallel straight lines are crossed by a transverse, then the alternative is: interior angles of the same magnitude. ## What is midpoint postulate in math Multicenter theory is used in coordinate geometry, which indicates that the midpoint of a line segment is the intersection of its extremes. To solve an equation with this set, you need to know the "x" and "y" coordinates. The midpoint theorem is also useful in the fields of calculus and algebra. ## What do you mean by mid point in geometry? In geometry, a midpoint is defined as the point that divides a line segment into two equal parts. What does Wednesday mean? A midpoint is a point that is at or near the midpoint, or equidistant from both ends of a line segment. ## When do you use the mid point formula? The midpoint formula is used to find the midpoint between two specified points. If P 1 (x 1, y 1) and P 2 (x 2, y 2) are the coordinates of two given extremes, then the mean formula is given as follows: ## What is the definition of midpoint in proofs? What is the definition of the middle of the test? In geometry, a midpoint is defined as the point that divides a line segment into two equal parts. What does Wednesday mean? A midpoint is a point that is at or near the midpoint, or equidistant from both ends of a line segment. ## How is the theory of mid point theorem used? Multicenter theory is used in coordinate geometry, which indicates that the midpoint of a line segment is the intersection of its extremes. To solve an equation with this set, you need to know the "x" and "y" coordinates. The midpoint theorem is also useful in the fields of calculus and algebra. Also read: ## Midpoint theorem The middle sentence reads: “The segment of a triangle connecting the center of the two sides of the triangle would be parallel to the third side, and it is also half the length of the third side.”. ## What is the formula for finding the midpoint? The midpoint formula defines the coordinates of the midpoint between the two extremes (x1, y1) and (x2, y2). ## How do you calculate the midpoint between two points? The midpoint between two points can be calculated if the coordinates of the two points are known. If the two points are (x1, y1) and (x2, y2), you can use the following formula to find the center point: {(x 1 + x 2) / 2, (y 1 + y 2) / 2} . ## What is the proof of mid point theorem? Proof of the theorem of the middle. If the centers of one of the sides of the triangle are connected by a straight line segment, then the segment is parallel to all other sides and makes up about half of the other sides as well. ## How do you calculate the midpoint of a line? To find the center of a line segment, you just need to calculate the mean values of the coordinates; too easy. The midpoint M of the segment with the extremes (x1, y1) and (x2, y2) is. To find the midpoint of a segment with endpoints (-4, -1) and (2.5), enter numbers into the midpoint formula and you will get the midpoint (-1,2): ## What is midpoint postulate definition Determination of the midpoint: The point halfway between the ends of the segment. Durst M is the midpoint of segment AB, then AM = postulate MB (segment sum postulate) If B is between A and C, then AB + BC = sentence AC (segment congruence) Segment congruence is reflexive, symmetric and transitive. ## What is midpoint postulate formula The midpoint formula is used to find the midpoint between two specified points. If P 1 (x 1, y 1) and P 2 (x 2, y 2) are the coordinates of two given extremes, then the middle formula is: mean = Inversion midpoint of theorem. ## How is the mid point of a line determined? In coordinate geometry, the middle sentence refers to the center of the line segment. Defines the coordinate points of the midpoint of the line segment and can be determined by averaging the coordinates of the specified end points. The midpoint formula is used to find the midpoint between two specified points. ## Which is the correct statement of the mid point theorem? The middle sentence says that "a segment in a triangle connecting the midpoints of two sides of a triangle is parallel to the third side and is also half the length of the third side." Instead, a line cuts through the center of one side of the triangle, parallel to the other side, the third side. ## What is midpoint postulate examples The flip side of the middle sentence says that "if a line is drawn through the center of one side of the triangle and is parallel to the other side, it will intersect the third side." Here is an example to help you understand the Midpoint Theorem. Triangle ABC contains the centers of the points BC, CA, AB D, E and F. ## Which is an example of the mid point theorem? The midpoint theorem can also be proved using triangles. Suppose two lines are drawn parallel to the x and y axes, starting at the ends and joining in the middle, then the segment goes through the angle between them, resulting in two identical triangles. This relationship of these triangles forms a series of central points. ## What is the subtraction theorem? Here are sets of subtractions for three segments and three angles (abbreviated segment subtraction, angle subtraction, or just subtraction): Segment subtraction (three segments in total): If a segment is subtracted from two congruent segments, the differences are the same. ## What is subtraction property in geometry? The subtraction property of equality says that if you subtract one side of the equation, you must support the other side of the equation so that both sides of the equation look the same. It's just the same on both sides of the equation. ## Can you explain the area addition postulate? The postulate of adding areas states that with two non-overlapping shapes, the total area is equal to the sum of the areas of the individual shapes. In other words, if you can cut a shape into triangles, squares, or parts of a circle, you can calculate the area of the shape by adding the areas of the parts. ## What is the Angle Addition Postulate? Adding angles basically means taking two angles and joining them into one BIG angle! Here is a simple example: “We take ∠GEM and ∠MEO. We're going to put them together so that the angle converges at point E. We're also going to order them so that the two angles have the same limit of electromagnetic radiation. ## What is the subtraction postulate in math Deduction Postulate: If equal amounts are subtracted from equal amounts, the differences are equal. Applying the subtraction postulate in the proof, let's look at another example: ## What is the subtraction postulate method Substitution Postulate: Quantity can be replaced in any expression with its equivalent. Since the sum of 3 and 8 is 8, you can replace any expression with 8 and it will always be the same. ## What is the subtraction postulate in geometry Deduction Postulate: If equal amounts are subtracted from equal amounts, the differences are equal. Applying the postulate of subtraction in a proof, let's look at another example: you get information. Note that and are greater than the sides of the triangles. ## What is the subtraction postulate formula Subtraction Formula: Minuend Subtract = difference. Let them understand the subtraction formula or mathematical subtraction equation with an example. Here 9 is decrease, 7 is subtraction and 2 is difference. ## What is the subtraction postulate definition Deduction Postulate: If equal amounts are subtracted from equal amounts, the differences are equal. ## What is the postulate for vertical angles? Linear Pair Postulate If two angles form a linear pair, the sum of the angular dimensions is 180°. Postulate vertical angles. If two angles are vertical, they are equal (they have the same dimensions). Postulate the parallel lines Through a point that is not on a straight line, exactly one straight line is parallel to this straight line. The postulate of adding angles says that if B is within A or C, the measure of the larger angle is the sum of the measures of the two smaller ones. ## What is Angle Addition Theorem? Use the following two addition theorems to prove three segments or three angles: Add segments (total of three segments): If you add a segment to two congruent segments, the sums are congruent. Add angles (three common angles) - If an angle is added to two coincident angles, the sums are the same. ## What is the angle addition property? An Introduction to the Angle Sum Property: The Angle Sum Postulate states that if a point is at an angle and two angles created by drawing a line through the point are added, the sum is equal to the wide angle. For example, if angle TSR = 40 degrees and angle TSV = 15 degrees, the other angle is 25 degrees. ## What is a Protractor postulate? Protractor Postulate: The protractor postulate states that the measure of the angle between two rays can be expressed by a number, and this number will be in the range of 180 to 180 degrees. This assumption allows the use of a protractor to measure angles. ## What is the definition of the Segment Addition Postulate? The Line Segment Sum Postulate states that if there are two points on segment A and C, the third point B is on segment AC if and only if the distances between the points meet the requirements of the equation AB + BC = CA. See Diagram 1 for a better understanding of this postulate definition. ## Which is correct segment AB or segment AD? If segment AD is 40 inches and segment BD is 29 inches, then segment AB should be 40 when added to 29. Therefore, segment AB is 40 inches minus 29 inches. which equals 11 inches! ## What should I give my Child for Segment Addition? Have them touch and manipulate the squares. You don't have to be a fantasy. You can even have them play with pre-cut square pieces of paper. Give them a ruler or tape measure, something they can touch and do in the real world. ## When does a point lie on a line segment? Let's check. The Line Segment Sum Postulate states that if there are two points on segment A and C, the third point B is on segment AC if and only if the distances between the points meet the requirements of the equation AB + BC = CA.
# What Is The GCF Of 7 And 28? ## What are the factors of 7 and 28? The gcf of 7 and 28 can be obtained like this: The factors of 7 are 7, 1. The factors of 28 are 28, 14, 7, 4, 2, 1. The common factors of 7 and 28 are 7, 1, intersecting the two sets above.. ## What are the GCF of 7? Greatest common factor (GCF) of 7 and 15 is 1. We will now calculate the prime factors of 7 and 15, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 7 and 15. ## What is the GCF for 24 and 28? We found the factors and prime factorization of 24 and 28. The biggest common factor number is the GCF number. So the greatest common factor 24 and 28 is 4. ## What are the factors of 28? The factors of 28 are the numbers, which on multiplication in pairs results in the original number. These factors are 1,2,4,7,14 and 28. Since, 2 x 14 = 28, therefore (2,14) is a pair factor of number 28. ## Why is HCF of 7 and 15? Answer. because it is the greatest number that divides evenly into all of them…… ## What is the GCF of 36 63? 9To sum up, the gcf of 36 and 63 is 9. In common notation: gcf (36,63) = 9. ## Is 28 a perfect number? Perfect number, a positive integer that is equal to the sum of its proper divisors. The smallest perfect number is 6, which is the sum of 1, 2, and 3. Other perfect numbers are 28, 496, and 8,128. ## What are the common factors of 28 and 36? The greatest common factor circled is the GCF. The GCF of 28 and 36 is 4. Step 1: Write the prime factorization of both numbers. ## What is the LCM of 3 and 7? Least common multiple (LCM) of 3 and 7 is 21. ## What is the HCF of 28 and 32? We found the factors and prime factorization of 28 and 32. The biggest common factor number is the GCF number. So the greatest common factor 28 and 32 is 4. ## What is the GCF of 28 56? The greatest common factor (GCF) of 28 and 56 is 28. ## What is the LCM of 7 and 28? To sum up, the lcm of 7 and 28 is 28. In common notation: lcm (7,28) = 28. ## What is the GCF for 28? Answer: Factors of 28 are 1, 2, 4, 7, 14, 28. There are 6 integers that are factors of 28. The greatest factor of 28 is 28. 3. ## What are two factors of 28? 28 is a composite number. 28 = 1 x 28, 2 x 14, or 4 x 7. Factors of 28: 1, 2, 4, 7, 14, 28. Prime factorization: 28 = 2 x 2 x 7, which can also be written 28 = 2² x 7. ## Is 28 a multiple of 7 explain? Multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, … A common multiple is a whole number that is a shared multiple of each set of numbers. The multiples that are common to two or more numbers are called the common multiples of those numbers. ## What is the HCF of 28 and 56? The common factors of 28 and 56 are 28, 14, 7, 4, 2, 1, intersecting the two sets above. In the intersection factors of 28 ∩ factors of 56 the greatest element is 28. Therefore, the greatest common factor of 28 and 56 is 28. ## What is the GCF of 7/15 and 21? The common factors for 7,15,21 7 , 15 , 21 are 1 . The GCF (HCF) of the numerical factors 1 is 1 . ## What is the LCM of 7 and 4? 4 and 7 have 28 and 56 in common BUT the LCM would be 28 since that’s the LEAST common, or the smallest number that they would have in common.
\ # 3.2.1 Simultaneous Equations Example 1 & 2 Example 1: Solve the simultaneous equations. Solution: $\begin{array}{l}x+\frac{1}{4}y=1\to \left(1\right)\\ {y}^{2}-8=4x\to \left(2\right)\\ x=1-\frac{1}{4}y\to \left(3\right)\end{array}$ Substitute (3) into (2), $\begin{array}{l}{y}^{2}-8=4\left(1-\frac{1}{4}y\right)\\ {y}^{2}-8=4-\frac{4}{4}y\end{array}$ y2 + y – 12 = 0 (y + 4)(y – 3) = 0 y = –4 or y = 3 Substitute the values of y into (3), Example 2: Solve the simultaneous equations 2x + y = 1 and 2x2+ y2 + xy = 5. Correct your answer to three decimal places. Solution: 2x + y = 1—–(1) 2x2 + y2+ xy = 5—–(2) From (1), y = 1 – 2x—–(3) Substitute (3) into (2). 2x2 + (1 – 2x)2 + x(1 – 2x) = 5 2x2 + (1 – 2x)(1 – 2x) + x – 2x2 = 5 1 – 2x – 2+ 4x2 + x – 5 = 0 4x2 – 3x – 4 = 0 Substitute the values of x into (3). When x = –0.693, y = 1 – 2 (–0.693) = 2.386 (correct to 3 decimal places) When x = 1.443, y = 1 – 2 (1.443) = –1.886 (correct to 3 decimal places) The solutions are x = –0.693, y = 2.386 and x = 1.443, y = –1.886. ### 1 thought on “3.2.1 Simultaneous Equations Example 1 & 2” 1. The answer for the first example is wrong, by substituting -4 into 3 we get 2
Formula of area and perimeter of square # Formula of area and perimeter of square Perimeter and Area of a Square A Square is a figure/shape with all four sides equal and all angles equal to 90 degrees. The area of the square is the space occupied by the square in a 2D plane and its perimeter is the distance covered on the outer line. Solid Geometry is concerned in calculating the length, perimeter, area and volume of various geometric figures and shapes. Here are some basic formulas which can be used to calculate the length, area, volume, and perimeter of various shapes and figures. Formulas of Perimeter, Area Volume: Formulas for Square. Perimeter: 4 * s (s = side) Area: S 2 Perimeter of an equilateral triangle As we know equilateral triangles have 3 equal sides. So if the length of a side is a then the perimeter formula is P = a + a + a Perimeter and Area of a Square. Square is a quadrilateral in which all its sides have equal length and all the four corners are right angles. Perimeter of a Square = 4 × side; Area of a Square = side × side; Properties of Squares. Opposite sides are parallel, with all sides being equal; A square has four lines of symmetry Derivation of a Square’s Perimeter. The perimeter of the square is defined as the length of the boundary of a square. The Perimeter of the square formula is given by, Perimeter = length of 4 sides = side+side+side+side = 4 × side. Therefore, the perimeter of Square = 4s units The area of the shaded region, which covers half of the square is 18 meaning that the total area of the square is 18 x 2, or 36. The area of a square is equal to the length of one side squared. Since the square root of 36 is 6, the length of 1 side is 6. The perimeter is the length of 1 side times 4 or 6 x 4. This formulas help you answer the questions how to find the area of triangle, square, rectangle, rhombus, parallelogram, trapezium, quadrangle, circle and ellipse. Study of mathematics online. Study math with us and make sure that "Mathematics is easy!" The formula of perimeter and area of square are explained step-by-step with solved examples. If 'a' denotes the side of the square, then, length of each side of a square is 'a' units. Calculate the perimeter of a square using our Perimeter Calculator, or try calculating it yourself with the formulas provided. Geometry Calculator Enter the length of the side in the calculator below to calculate the perimeter of the square. In this article cover maximum all two dimensional shapes propertied with formulas of area and perimeter calculations. Formulas of a Triangle: Area of the triangle = (1/2) x Base x Height. Area of the triangle ABC = (1/2) x a x h. For more concepts regarding the triangles please go through the below link Area and Perimeter of a Square - Geometry Calculator Online calculator to calculate the area and perimeter of a square given the length of its side. The formula of the perimeter P and area A of a square of side x, are given by Let's find the area of the rectangle below. To find the area of the rectangle, we find out how many one-centimetre squares we can fit into the rectangle. Area of the rectangle = 4 × 2 = 8 cm 2 We need 8 one-centimetre squares to make a rectangle 4 cm long and 2 cm wide. The area of the rectangle is 8 cm 2. Area and Perimeter of a Square - Geometry Calculator Online calculator to calculate the area and perimeter of a square given the length of its side. The formula of the perimeter P and area A of a square of side x, are given by In particular, to find the quadrilateral, or the triangle, or another particular figure, with the largest area amongst those with the same shape having a given perimeter. The solution to the quadrilateral isoperimetric problem is the square , and the solution to the triangle problem is the equilateral triangle . a table of area formulas and perimeter formulas used to calculate the area and perimeter of two-dimensional geometrical shapes: square, rectangle. parallelogram, trapezoid (trapezium), triangle, rhombus, kite, regular polygon, circle, and ellipse. a more detailed explanation (in text and video) of each area formula. Perimeter of a Square \| Integers - Type 1. Introduce the concept of finding the perimeter of a square using the formula P = 4a, where 'a' represents the side of the square. Included are integers ≤ 20 in Level 1, while Level 2 offers integers ≥10. Two levels of difficulty with 5 worksheets each. Download the set (10 Worksheets) Jun 21, 2017 · Parallelogram Area Formula and Parallelogram Perimeter Formula. A parallelogram is a closed figure formed by four sides and the opposite sides are parallel to each other. The ‘height’ (h) of a parallelogram is the distance from the measured side to its opposite parallel side. Perimeter of a Parallelogram = 2a + 2b. The formula of perimeter and area of square are explained step-by-step with solved examples. If 'a' denotes the side of the square, then, length of each side of a square is 'a' units. Perimeter of a square formula. The formula is: side x 4 and the result will be in whatever metric you did the measurement in: mm, cm, dm, meters or in, ft, yards, etc. This is the formula used in our perimeter of a square online calculator. Let's find the area of the rectangle below. To find the area of the rectangle, we find out how many one-centimetre squares we can fit into the rectangle. Area of the rectangle = 4 × 2 = 8 cm 2 We need 8 one-centimetre squares to make a rectangle 4 cm long and 2 cm wide. The area of the rectangle is 8 cm 2. Using Heron's formula, calculate the area of the parallelogram to the nearest tenth of a square unit. 36.7 square units The triangle shown is an equilateral triangle. In particular, to find the quadrilateral, or the triangle, or another particular figure, with the largest area amongst those with the same shape having a given perimeter. The solution to the quadrilateral isoperimetric problem is the square , and the solution to the triangle problem is the equilateral triangle . To calculate the Perimeter of a Square from the side, use the following formula: P = 4A. Indeed, the perimeter of the square is equivalent to the total length of its sides placed end to end. Since the property of the square is to have 4 equal sides, it is therefore logical that the perimeter of a square is equal to 4 times its side. To ... Area is the size of a surface! Learn more about Area, or try the Area Calculator ... Area and Circumference of a Circle Formula. For C = circumference, r = radius, and D = diameter of a circle: Those two formulas are one of the mathematical "tricks" to finding perimeter if you know area. The formula for the circumference (C), or perimeter (P), of a circle uses , the area (A) and the square root. Area and Perimeter of a Square - Geometry Calculator Online calculator to calculate the area and perimeter of a square given the length of its side. The formula of the perimeter P and area A of a square of side x, are given by Calculating the perimeter of a rectangle or square. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides. x is in this case the length of the rectangle and y is the width of the rectangle. Perimeter = x + x + y + y For example, you want to find the perimeter of your soccer field. Area and Perimeter of a Square - Geometry Calculator Online calculator to calculate the area and perimeter of a square given the length of its side. The formula of the perimeter P and area A of a square of side x, are given by Like any polygon, the perimeter is the total distance around the outside, which can be found by adding together the length of each side. I the case of a square, all four sides are the same length, so the perimeter is four times the length of a side. Or as a formula: where: S is the length of any one side. Formulas for the perimeter P and area A of a square The measurements of perimeter and circumference use units such as centimeters, meters, kilometers, inches, feet, yards, and miles. The measurements of area use units such as square centimeters (cm²), square meters( m²) , and so on. Current Location > Math Formulas > Geometry > Perimeter of a Square Perimeter of a Square A square is a regular quadrilateral and it has four equal sides and four equal angles (90 degree angle or right angles). a table of area formulas and perimeter formulas used to calculate the area and perimeter of two-dimensional geometrical shapes: square, rectangle. parallelogram, trapezoid (trapezium), triangle, rhombus, kite, regular polygon, circle, and ellipse. a more detailed explanation (in text and video) of each area formula. Current Location > Math Formulas > Geometry > Perimeter of a Square Perimeter of a Square A square is a regular quadrilateral and it has four equal sides and four equal angles (90 degree angle or right angles).
# 3.9 Normal, tension, and other examples of forces (Page 5/10) Page 5 / 10 Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the $x$ -axis and the vertical the $y$ -axis. Solution First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system. Consider the horizontal components of the forces (denoted with a subscript $x$ ): ${F}_{\text{net}x}={T}_{\text{L}x}-{T}_{\text{R}x}.$ The net external horizontal force ${F}_{\text{net}x}=0$ , since the person is stationary. Thus, $\begin{array}{lll}{F}_{\text{net}x}=0& =& {T}_{\text{L}x}-{T}_{\text{R}x}\\ {T}_{\text{L}x}& =& {T}_{\text{R}x}.\end{array}$ Now, observe [link] . You can use trigonometry to determine the magnitude of ${T}_{\text{L}}$ and ${T}_{\text{R}}$ . Notice that: $\begin{array}{lll}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{L}x}}{{T}_{\text{L}}}\\ {T}_{\text{L}x}& =& {T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)\\ \text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{R}x}}{{T}_{\text{R}}}\\ {T}_{\text{R}x}& =& {T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).\end{array}$ Equating ${T}_{\text{L}x}$ and ${T}_{\text{R}x}$ : ${T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)={T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).$ Thus, ${T}_{\text{L}}={T}_{\text{R}}=T,$ as predicted. Now, considering the vertical components (denoted by a subscript $y$ ), we can solve for $T$ . Again, since the person is stationary, Newton’s second law implies that net ${F}_{y}=0$ . Thus, as illustrated in the free-body diagram in [link] , ${F}_{\text{net}y}={T}_{\text{L}y}+{T}_{\text{R}y}-w=0.$ Observing [link] , we can use trigonometry to determine the relationship between ${T}_{\text{L}y}$ , ${T}_{\text{R}y}$ , and $T$ . As we determined from the analysis in the horizontal direction, ${T}_{\text{L}}={T}_{\text{R}}=T$ : $\begin{array}{lll}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{L}y}}{{T}_{\text{L}}}\\ {T}_{\text{L}y}={T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)\\ \text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{R}y}}{{T}_{\text{R}}}\\ {T}_{\text{R}y}={T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).\end{array}$ Now, we can substitute the values for ${T}_{\text{L}y}$ and ${T}_{\text{R}y}$ , into the net force equation in the vertical direction: $\begin{array}{lll}{F}_{\text{net}y}& =& {T}_{\text{L}y}+{T}_{\text{R}y}-w=0\\ {F}_{\text{net}y}& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)+T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)-w=0\\ 2\phantom{\rule{0.25em}{0ex}}T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)-w& =& 0\\ 2\phantom{\rule{0.25em}{0ex}}T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& w\end{array}$ and $T=\frac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)}=\frac{\text{mg}}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)},$ so that $T=\frac{\left(\text{70}\text{.}\text{0 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right)}{2\left(0\text{.}\text{0872}\right)},$ and the tension is $T=\text{3900 N}.$ Discussion Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker. If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in [link] . As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way: $T=\frac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.$ We can extend this expression to describe the tension $T$ created when a perpendicular force ( ${\mathbf{\text{F}}}_{\perp }$ ) is exerted at the middle of a flexible connector: $T=\frac{{F}_{\perp }}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.$ #### Questions & Answers Is there any normative that regulates the use of silver nanoparticles? Damian Reply what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? Stoney Reply why we need to study biomolecules, molecular biology in nanotechnology? Adin Reply ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. Adin why? Adin what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe anyone know any internet site where one can find nanotechnology papers? Damian Reply research.net kanaga sciencedirect big data base Ernesto Introduction about quantum dots in nanotechnology Praveena Reply what does nano mean? Anassong Reply nano basically means 10^(-9). nanometer is a unit to measure length. Bharti do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? Damian Reply absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now Akash Reply it is a goid question and i want to know the answer as well Maciej characteristics of micro business Abigail for teaching engĺish at school how nano technology help us Anassong Do somebody tell me a best nano engineering book for beginners? s. Reply there is no specific books for beginners but there is book called principle of nanotechnology NANO what is fullerene does it is used to make bukky balls Devang Reply are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Abhijith Reply Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc NANO so some one know about replacing silicon atom with phosphorous in semiconductors device? s. Reply Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure. Harper Do you know which machine is used to that process? s. how to fabricate graphene ink ? SUYASH Reply for screen printed electrodes ? SUYASH What is lattice structure? s. Reply of graphene you mean? Ebrahim or in general Ebrahim in general s. Graphene has a hexagonal structure tahir On having this app for quite a bit time, Haven't realised there's a chat room in it. Cied what is biological synthesis of nanoparticles Sanket Reply how did you get the value of 2000N.What calculations are needed to arrive at it Smarajit Reply Privacy Information Security Software Version 1.1a Good ### Read also: #### Get the best Algebra and trigonometry course in your pocket! Source: OpenStax, College physics arranged for cpslo phys141. OpenStax CNX. Dec 23, 2014 Download for free at http://legacy.cnx.org/content/col11718/1.4 Google Play and the Google Play logo are trademarks of Google Inc. Notification Switch Would you like to follow the 'College physics arranged for cpslo phys141' conversation and receive update notifications? By Nick Swain By By Mistry Bhavesh By By
# CAT 2017 Important Topics in Logical Reasoning: Puzzles Updated : Sep 24, 2017, 20:00 By : Vikram Prabhakaran CAT 2017 Hopefuls need to strengthen their basics before going out guns blazing on Quantitative Aptitude. To help them, this post provides essential concepts to build your basics in any chapter. Here are some topics that we have covered: Puzzle: 1 You are locked inside a room with 6 doors - A, B, C, D, E, F. Out of which 3 are Entrances only and 3 are Exits only. One person came in through door F and two minutes later second person came in through door A. He said, "You will be set free, if you pass through all 6 doors, each door once only and in correct order. Also, door A must be followed by door B or E, door B by C or E, door C by D or F, door D by A or F, door E by B or D and door F by C or D." After saying that they both left through door B and unlocked all doors. In which order must you pass through the doors? Solution: The correct order is CFDABE It is given that one person came in through door F and second person came in through door A. It means that door A and door F is Entrances. Also, they both left through door B. Hence, door B is Exit. As Exit and Entrance should alter each other and we know two Entrances, let's assume that the third Entrance is W. Thus, there are 6 possibilities with "_" indicating Exit. (1) _W_A_F (2) _W_F_A (3) _F_W_A (4) _F_A_W (5) _A_W_F (6) _A_F_W As door A must be followed by door B or E and none of them lead to the door F, (1) and (6) are not possible. Also, door D must be the Exit as only door D leads to the door A and door A is the Entrance. (2) _W_FDA (3) _F_WDA (4) _FDA_W (5) DA_W_F Only door D and door C lead to the door F. But door D is used. Hence, door C must be the Exit and precede door F. Also, the third Exit is B and the W must be door E. (2) BECFDA (3) CFBEDA (4) CFDABE (5) DACEBF But only door B leads to the door C and both are Exits. Hence, (2) and (5) are not possible. Also, door F does not lead to door B - discard (3). Hence, the possible order is (4) i.e. CFDABE. Puzzle :2 There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than the second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11. Find the number. Solution: 65292 As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9) It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9) Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292. Puzzle:3 A person travels on a cycle from home to church on a straight road with wind against him. He took 4 hours to reach there. On the way back to the home, he took 3 hours to reach as wind was in the same direction. If there is no wind, how much time does he take to travel from home to church? Solution: Let distance between home and church is D. A person took 4 hours to reach church. So speed while travelling towards church is D/4. Similarly, he took 3 hours to reach home. So speed while coming back is D/3. There is a speed difference of 7D/12, which is the wind helping person in 1 direction, & slowing him in the other direction. Average the 2 speeds, & you have the speed that person can travel in no wind, which is 7D/24. Hence, person will take D / (7*D/24) hours to travel distance D which is 24/7 hours. Answer is 3 hours 25 minutes 42 seconds Practice Section for CAT Sep 24CAT & MBA Exams Posted by: B. Tech (IIT Dhanbad) Quant & DI Expert Member since Aug 2017
Derivatives Math 120 Calculus I D Joyce, Fall 2013 Save this PDF as: Size: px Start display at page: Transcription 1 Derivatives Mat 20 Calculus I D Joyce, Fall 203 Since we ave a good understanding of its, we can develop derivatives very quickly. Recall tat we defined te derivative f x of a function f at x to be te value of te it f fx + fx x. Sometimes te it doesn t exist, and ten we say tat te function is not differentiable at x, but usually te it does exist and, so, te function is differentiable. Derivatives of linear functions. Te grap of a linear function fx ax + b is a straigt line wit slope a. We expect tat te derivative f x sould be te constant slope a, and tat s wat we find it is wen we apply te definition of derivative. f x fx + fx ax + + b ax + b ax + a + b ax b a a a Derivatives of constant functions. We ve actually sown tat te derivative of a constant function fx b is 0. Tat s because a constant function is a special case of a linear function were te coefficient of x is 0. It makes sense tat te derivative of a constant is 0, since te slope of te orizontal straigt line y b is 0. Derivatives of powers of x. We ll directly compute te derivatives of a few powers of x like x 2, x 3, /x, and x. Note tat tese last two are actually powers of x even toug we usually don t write tem tat way. Te reciprocal of x is x raised to te power, tat is, /x x. Also, te square root of x is x raised to te power 2, tat is, x x /2. After computing tese, we ll see tat tey all fit te pattern tat says te derivative of x n is nx n. Tis pattern is called te power rule for derivatives. 2 First, we ll find te derivative of fx x 2. Tus, te derivative of x 2 is 2x. f x fx + fx x + 2 x 2 x 2 + 2x + 2 x 2 2x + 2 2x + 2x Next, we ll compute te derivative of fx x 3. At one point in te computation, we need to expand te cube of a binomial x + 3, and tat s someting we could do by expanding x + x + x +. Te rule tat gives te expansion of te general power x + n is called te binomial teorem, and tat s related to Pascal s triangle. f x fx + fx x + 3 x 2 x 3 + 3x 2 + 3x x 3 3x 2 + 3x x 2 + 3x + 2 3x 2 Tus, te derivative of x 3 is 3x 2. Note tat we didn t take te it until we cancelled te. Tat s always wat appens. Now, let s try to find te derivative of fx /x. We start te same way as always f x fx + fx /x + /x At tis point, put te numerator over a common denominator, ten simplify te compound 2 3 quotient before continuing. x x+ x+x x x + x + x x + x x + x x 2 Tus, te derivative of /x is /x 2. Tis result fits te power rule mentioned above since we can rewrite it to say te derivative of x is x 2. It s a bit arder to compute te derivative of x since at one point we ave to multiply te numerator and denominator of a quotient by a conjugate, but oterwise it s about te same difficulty. You migt tink tat multiplying and dividing by a conjugate is a trick, but since we do it so often, you sould tink of it as a tecnique. Let fx x. Ten Tus, te derivative of x is of x /2 is 2 x /2. f fx + fx x x + x x + x x + x x + + x x + + x x + + x 2 x x + + x x + + x 2. Again, tis fits te power rule, since it says te derivative x Here s a summary of te computations we ve done so far. fx f x ax + b a x 2 2x x 3 3x 2 /x /x 2 x 2 x 3 4 Most of tese results are special cases of te power rule wic says tat te derivative of x n is nx n. Rules of differentiation. We could go on like tis for eac time we wanted te derivative of a new function, but tere are better ways. Tere are several rules tat apply in broad cases. In fact, we ll eventually find enoug rules for differentiation tat we won t need to go back to te definition in terms of its. So, let s get started. Te sum rule. Let s begin wit te rule for sums of functions. Many functions are sums of simpler functions. For example x 3 +x 2 is te sum of two functions we ve already differentiated. If we can discover te differentiation rule for sums, we ll be able to differentiate x 3 +x 2 witout going back to te definition of derivative. Let f and g be two functions wose derivatives f and g we already know. Can we find te derivative f + g of teir sum f + g? We ll need te definition of derivative to do tat. Wen we apply te definition, we get f + g f + gx + f + gx x. Now, te expression f + gx means fx + gx, terefore, te expression f + gx + means fx + + gx +. We can continue as follows. f + g x fx + + gx + fx + gx fx + + gx + fx gx fx + fx + gx + gx fx + fx + fx + fx gx + gx + f x + g x gx + gx Tus, we ave sown tat te derivative f + g of te sum f + g equals te sum f + g of te derivatives. Tis is a very useful rule. For instance, we can use it to conclude tat te derivative of x 3 + x 2 is 3x 2 + 2x because we already know te derivative of x 3 is 3x 2 and te derivative of x 2 is 2x. Te difference rule. If you just look back troug te previous paragrap and cange some plus signs to minus signs, you ll see tat te derivative f g of te difference f g equals te difference f g of te derivatives. So, for instance, te derivative of x 3 x 2 is 3x 2 2x. Constant multiple rule. On te same level of difficulty as addition and subtraction is multiplcation by constants. If f is a function, and c is a constant, ten cf is te function 4 5 wose value at x is c fx. We can easily find te derivative of cf in terms of te derivative of f. cf x cfx + cfx c fx + c fx cfx + fx fx + fx c c f x Tus, te derivative of a constant times a function is tat constant times te derivative of te function, tat is, cf c f. Derivatives of polynomials. Wit te elp of te power rule, we can find te derivative of any polynomial. For example, te derivative of 0x 3 7x 2 + 5x 8 is 30x 2 4x + 5. Finding derivatives of polynomials is so easy all you ave to do is write down te answer, but ere are te details so you can see tat we re using all te rules we ave so far. We ll use te abbreviated notation 0x 3 7x 2 + 5x 8 for te derivative of 0x 3 7x 2 + 5x 8. 0x 3 7x 2 + 5x 8 equals, by te sum and difference rules, 0x 3 7x 2 + 5x 8 wic equals, by te constant multiple rule, 0x 3 7x 2 + 5x 8 wic equals, by te power rule and constant rule, 03x 2 72x + 5 0, wic simplifies to 30x 2 4x + 5, te answer. Te product, reciprocal, and quotient rules. Tese tree rules are arder to prove, so we ll put off te proofs for a little bit. First we ll state tem, ten use tem, and finally prove tem. Let f and g be two differentiable functions. Product rule: fg f g + fg Reciprocal rule: g g g 2 f Quotient rule: f g fg g g 2 For our example to illustrate te use of te product rule, let s find te derivative of x x. Here, fx x wile gx x. Ten x x f xgx + fxg x x + x 2 x and tis last expression simplifies to 3 2 x. Wat s important to see in tis example is ow to use te product rule. In words, te product rule says tat te derivative of te product fg of two functions equals te derivative f of te first times te second g plus te first f times te derivative g of te second. 5 6 Unlike sums and differences, te derivative of te product is not te product of te derivatives. Tat is to say, fg does not equal f g. Next, let s ave an example to illustrate te use of te reciprocal rule. Let s find te derivative of 5x 3 x + 2. Tis is te reciprocal of 5x3 x + 2, and we know te derivative of 5x 3 x + 2 is 5x 2. Te reciprocal rule says g g g. 2 In tis example gx 5x 3 x + 2 and g x 5x 2. Terefore, 5x2 5x 3 x + 2 5x 3 x Finally, let s ave an example to illustrate te use of te quotient rule. Let s find te 6x 3 8x 2 derivative of. Te quotient rule says x 2 + 2x + 8 f f g fg. g g 2 In tis example, fx 6x 3 8x 2 and gx x 2 + 2x + 8. We know te derivatives of f and g. Tey are f x 8x 2 6x and g x 2x + 2. Terefore, 6x 3 8x 2 f xgx fxg x x 2 + 2x + 8 gx 2 8x2 6xx 2 + 2x + 8 6x 3 8x 2 2x + 2 x 2 + 2x Of course, we can simplify our answer, but for te purposes of tis example tere is no need to do so. Te derivative of te quotient is complicated. You can remember te rule best in words. It says tat te derivative of a quotient f/g is te derivative f of te numerator times te denominator g plus te numerator f times te derivative g of te denominator, all divided by te square g 2 of te denominator. You ll ave to use te product, reciprocal, and quotient rules several times before you remember tem well. You can actually do witout te reciprocal rule, since it s a special case of te quotient rule, but it comes up often enoug so tat it s wort wile to memorize it. Mat 20 Home Page at ttp://mat.clarku.edu/~djoyce/ma20/ 6 Proof of the Power Rule for Positive Integer Powers Te Power Rule A function of te form f (x) = x r, were r is any real number, is a power function. From our previous work we know tat x x 2 x x x x 3 3 x x In te first two cases, te power r is a positive Section 3.3. Differentiation of Polynomials and Rational Functions. Difference Equations to Differential Equations Difference Equations to Differential Equations Section 3.3 Differentiation of Polynomials an Rational Functions In tis section we begin te task of iscovering rules for ifferentiating various classes of f(a + h) f(a) f (a) = lim Lecture 7 : Derivative AS a Function In te previous section we defined te derivative of a function f at a number a (wen te function f is defined in an open interval containing a) to be f (a) 0 f(a + ) Math 113 HW #5 Solutions Mat 3 HW #5 Solutions. Exercise.5.6. Suppose f is continuous on [, 5] and te only solutions of te equation f(x) = 6 are x = and x =. If f() = 8, explain wy f(3) > 6. Answer: Suppose we ad tat f(3) 6. Ten Differential Calculus: Differentiation (First Principles, Rules) and Sketching Graphs (Grade 12) OpenStax-CNX moule: m39313 1 Differential Calculus: Differentiation (First Principles, Rules) an Sketcing Graps (Grae 12) Free Hig Scool Science Texts Project Tis work is prouce by OpenStax-CNX an license Instantaneous Rate of Change: Instantaneous Rate of Cange: Last section we discovered tat te average rate of cange in F(x) can also be interpreted as te slope of a scant line. Te average rate of cange involves te cange in F(x) over SAT Subject Math Level 1 Facts & Formulas Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Reals: integers plus fractions, decimals, and irrationals ( 2, 3, π, etc.) Order Of Operations: Aritmetic Sequences: PEMDAS (Parenteses 7.6 Complex Fractions Section 7.6 Comple Fractions 695 7.6 Comple Fractions In tis section we learn ow to simplify wat are called comple fractions, an eample of wic follows. 2 + 3 Note tat bot te numerator and denominator are Lecture 10: What is a Function, definition, piecewise defined functions, difference quotient, domain of a function Lecture 10: Wat is a Function, definition, piecewise defined functions, difference quotient, domain of a function A function arises wen one quantity depends on anoter. Many everyday relationsips between 1 Derivatives of Piecewise Defined Functions MATH 1010E University Matematics Lecture Notes (week 4) Martin Li 1 Derivatives of Piecewise Define Functions For piecewise efine functions, we often ave to be very careful in computing te erivatives. Sections 3.1/3.2: Introducing the Derivative/Rules of Differentiation Sections 3.1/3.2: Introucing te Derivative/Rules of Differentiation 1 Tangent Line Before looking at te erivative, refer back to Section 2.1, looking at average velocity an instantaneous velocity. Here Tangent Lines and Rates of Change Tangent Lines and Rates of Cange 9-2-2005 Given a function y = f(x), ow do you find te slope of te tangent line to te grap at te point P(a, f(a))? (I m tinking of te tangent line as a line tat just skims CHAPTER 8: DIFFERENTIAL CALCULUS CHAPTER 8: DIFFERENTIAL CALCULUS 1. Rules of Differentiation As we ave seen, calculating erivatives from first principles can be laborious an ifficult even for some relatively simple functions. It is clearly MAT1A01: Differentiation of Polynomials & Exponential Functions + the Product & Quotient Rules MAT1A01: Differentiation of Polynomials & Exponential Functions + te Prouct & Quotient Rules Dr Craig 17 April 2013 Reminer Mats Learning Centre: C-Ring 512 My office: C-Ring 533A (Stats Dept corrior) ACT Math Facts & Formulas Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Rationals: fractions, tat is, anyting expressable as a ratio of integers Reals: integers plus rationals plus special numbers suc as Math 229 Lecture Notes: Product and Quotient Rules Professor Richard Blecksmith richard@math.niu.edu Mat 229 Lecture Notes: Prouct an Quotient Rules Professor Ricar Blecksmit ricar@mat.niu.eu 1. Time Out for Notation Upate It is awkwar to say te erivative of x n is nx n 1 Using te prime notation for erivatives, New Vocabulary volume -. Plan Objectives To find te volume of a prism To find te volume of a cylinder Examples Finding Volume of a Rectangular Prism Finding Volume of a Triangular Prism 3 Finding Volume of a Cylinder Finding 6. Differentiating the exponential and logarithm functions 1 6. Differentiating te exponential and logaritm functions We wis to find and use derivatives for functions of te form f(x) = a x, were a is a constant. By far te most convenient suc function for tis purpose Lecture 10. Limits (cont d) One-sided limits. (Relevant section from Stewart, Seventh Edition: Section 2.4, pp. 113.) Lecture 10 Limits (cont d) One-sided its (Relevant section from Stewart, Sevent Edition: Section 2.4, pp. 113.) As you may recall from your earlier course in Calculus, we may define one-sided its, were ACTIVITY: Deriving the Area Formula of a Trapezoid 4.3 Areas of Trapezoids a trapezoid? How can you derive a formula for te area of ACTIVITY: Deriving te Area Formula of a Trapezoid Work wit a partner. 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Solution 1.6 Analyse Optimum Volume and Surface Area Estimation and oter informal metods of optimizing measures suc as surface area and volume often lead to reasonable solutions suc as te design of te tent in tis The Derivative as a Function Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Te Derivative as a Function DEFINITION: Te derivative of a function f at a number a, denoted by f (a), is if tis limit exists. f (a) f(a+) f(a) 1 Density functions, cummulative density functions, measures of central tendency, and measures of dispersion Density functions, cummulative density functions, measures of central tendency, and measures of dispersion densityfunctions-intro.tex October, 9 Note tat tis section of notes is limitied to te consideration Areas and Centroids. Nothing. Straight Horizontal line. Straight Sloping Line. Parabola. 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Lognormal Determine the perimeter of a triangle using algebra Find the area of a triangle using the formula Student Name: Date: Contact Person Name: Pone Number: Lesson 0 Perimeter, Area, and Similarity of Triangles Objectives Determine te perimeter of a triangle using algebra Find te area of a triangle using MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION Tis tutorial is essential pre-requisite material for anyone stuing mecanical engineering. Tis tutorial uses te principle of 2.1: The Derivative and the Tangent Line Problem .1.1.1: Te Derivative and te Tangent Line Problem Wat is te deinition o a tangent line to a curve? To answer te diiculty in writing a clear deinition o a tangent line, we can deine it as te iting position In other words the graph of the polynomial should pass through the points Capter 3 Interpolation Interpolation is te problem of fitting a smoot curve troug a given set of points, generally as te grap of a function. It is useful at least in data analysis (interpolation is a form 2 Limits and Derivatives 2 Limits and Derivatives 2.7 Tangent Lines, Velocity, and Derivatives A tangent line to a circle is a line tat intersects te circle at exactly one point. 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Power of a Product: Separate letters, distribute to the exponents and the bases Chapter 5 : Polynomials and Polynomial Functions 5.1 Properties of Exponents Rules: 1. Product of Powers: Add the exponents, base stays the same 2. Power of Power: Multiply exponents, bases stay the same Compute the derivative by definition: The four step procedure Compute te derivative by definition: Te four step procedure Given a function f(x), te definition of f (x), te derivative of f(x), is lim 0 f(x + ) f(x), provided te limit exists Te derivative function Geometric Stratification of Accounting Data Stratification of Accounting Data Patricia Gunning * Jane Mary Horgan William Yancey * Abstract: We suggest a new procedure for defining te boundaries of te strata in igly skewed populations, usual CHAPTER 7. Di erentiation CHAPTER 7 Di erentiation 1. Te Derivative at a Point Definition 7.1. 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For example, r(x) = p(x) q(x) f(x) = x2 3 x 4 + 3, g(t) = t6 + 4t 2 3, 7t 5 + 3t is identically equal to x 2 +3x +2 Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as 1 + 3 Average and Instantaneous Rates of Change: The Derivative 9.3 verage and Instantaneous Rates of Cange: Te Derivative 609 OBJECTIVES 9.3 To define and find average rates of cange To define te derivative as a rate of cange To use te definition of derivative to 3.4 Complex Zeros and the Fundamental Theorem of Algebra 86 Polynomial Functions.4 Complex Zeros and the Fundamental Theorem of Algebra In Section., we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and Reinforced Concrete Beam Mecanics of Materials Reinforced Concrete Beam Concrete Beam Concrete Beam We will examine a concrete eam in ending P P A concrete eam is wat we call a composite eam It is made of two materials: concrete FRACTION REVIEW. 3 and. Any fraction can be changed into an equivalent fraction by multiplying both the numerator and denominator by the same number FRACTION REVIEW A. INTRODUCTION. What is a fraction? A fraction consists of a numerator (part) on top of a denominator (total) separated by a horizontal line. For example, the fraction of the circle which Section 4.1 Rules of Exponents Section 4.1 Rules of Exponents THE MEANING OF THE EXPONENT The exponent is an abbreviation for repeated multiplication. The repeated number is called a factor. x n means n factors of x. The exponent tells Algebra I Pacing Guide Days Units Notes 9 Chapter 1 ( , ) Algebra I Pacing Guide Days Units Notes 9 Chapter 1 (1.1-1.4, 1.6-1.7) Expressions, Equations and Functions Differentiate between and write expressions, equations and inequalities as well as applying order Vieta s Formulas and the Identity Theorem Vieta s Formulas and the Identity Theorem This worksheet will work through the material from our class on 3/21/2013 with some examples that should help you with the homework The topic of our discussion Derivatives: rules and applications (Stewart Ch. 3/4) The derivative f (x) of the function f(x): Derivatives: rules and applications (Stewart Ch. 3/4) The derivative f (x) of the function f(x): f f(x + h) f(x) (x) = lim h 0 h (for all x for which f is differentiable/ the limit exists) Property:if What is Advanced Corporate Finance? What is finance? What is Corporate Finance? Deciding how to optimally manage a firm s assets and liabilities. Wat is? Spring 2008 Note: Slides are on te web Wat is finance? Deciding ow to optimally manage a firm s assets and liabilities. Managing te costs and benefits associated wit te timing of cas in- and outflows SAT Math Must-Know Facts & Formulas SAT Mat Must-Know Facts & Formuas Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Rationas: fractions, tat is, anyting expressabe as a ratio of integers Reas: integers pus rationas Notes: Most of the material in this chapter is taken from Young and Freedman, Chap. 12. Capter 6. Fluid Mecanics Notes: Most of te material in tis capter is taken from Young and Freedman, Cap. 12. 6.1 Fluid Statics Fluids, i.e., substances tat can flow, are te subjects of tis capter. But 1.6 The Order of Operations 1.6 The Order of Operations Contents: Operations Grouping Symbols The Order of Operations Exponents and Negative Numbers Negative Square Roots Square Root of a Negative Number Order of Operations and Negative Step 1: Set the equation equal to zero if the function lacks. Step 2: Subtract the constant term from both sides: In most situations the quadratic equations such as: x 2 + 8x + 5, can be solved (factored) through the quadratic formula if factoring it out seems too hard. However, some of these problems may be solved Factoring Polynomials and Solving Quadratic Equations Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3
# Spreadsheet Calculus: Euler's Method 10,261 9 1 Published ## Introduction: Spreadsheet Calculus: Euler's Method If you've tried solving any differential equations, you probably know that many of them are either very difficult or impossible to solve. What do we do then? Usually the solution to a differential equation actually isn't that important. The question you usually want answered is what happens when I start at certain coordinates? What does the differential equation do? If you have a spreadsheet program, then answering these questions is pretty easy. You don't need to solve the equation to find out what it does. ## Step 1: Find a Differential Equation First you need a differential equation that you want (or need) to solve. For my first example I'm going to use a simple equation that's easy to solve. dy/dx=ry. The solution is y=Aerx, where A is an arbitrary constant. Type in a dx value and make a graph of the solution, it should look something like the second picture. For most differential equations, you wont know the solution, the only reason we're putting it in first here is to confirm that this method works. ## Step 2: Use Euler's Method Here's how Euler's method works. Basically, you start somewhere on your plot. You know what dy/dx or the slope is there (that's what the differential equation tells you.) So you make a small line with the slope given by the equation. Then at the end of that tiny line we repeat the process. Soon enough we've sketched a solution curve to the differential equation. As long as we choose small enough step sizes, the solution curve found this way follows the true solution curve almost perfectly. Here's how you do it: first set an initial state. In this case I'm going to choose 2, since it's the initial value of the solution curve I sketched earlier. Then, below your initial value you want to type in your differential equation (as a function of the initial value) be sure to multiply it by your dx value. Your differential equation has the form dy/dx. If you multiply it by the step size (dx) then you're left with dy, the change in y, which is added to the previous y value. The picture below explains this a lot better than I ever could. ## Step 3: Graph It Plot it and check that it works. Notice in the picture below that the solution using Euler's method follows the solution found by solving the equation almost perfectly. The demo file is attached in case you need help for this. ## Step 4: Do a Tough One Now let's try it with a differential equation that can't be solved using traditional methods. The differential equation shown below is completely made up and doesn't model anything as far as I know. I don't know how to solve it, but I can still visualize solutions to it by using Euler's method. The second picture are solution sketches with different starting conditions. The excel file is also attached if you need help. Using this method, sketching solutions to differential equations becomes quite easy. Just make sure you use small enough step sizes to reduce the error rate. If you have big step sizes, your solution will be very inaccurate. You want your columns to be at least 100 cells long. Good luck! ## Recommendations • ### 3D CAM and CNC Class 606 Enrolled • ### Make it Move Contest We have a be nice policy. Please be positive and constructive.
JAC Class 10 Maths Solutions Chapter 11 Constructions Ex 11.1 Jharkhand Board JAC Class 10 Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers. JAC Board Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1 In each of the following, give the justification of the construction also: Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts. Solution : Steps of construction: 1. Draw the line segment AB = 7.6 cm. 2. At A, below AB, make an angle BAx = 30°. 3. At B, above AB, make an angle ABy = 30°. B$$\hat{A}$$x = A$$\hat{B}$$y = 30° These are alternate angles. ∴ Ax \|\| By. 4. With a convenient radius cut off five equal parts AA1 = A1A2 = A2A3 = A3A4 = A4A5 in Ax. 5. With the same radius cut off eight equal parts. BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7 = B7B8. 6. Join A5B8. Let it cut AB at C. AC : CB = 5 : 8. In ΔACA5 and CBB8 1. A$$\hat{C}$$A5 = B$$\hat{C}$$B8 (V.O.A.) 2. C$$\hat{A}$$A5 = A$$\hat{B}$$B8 (Alternate angles) 3. C$$\hat{A}$$5A = B$$\hat{B}$$8C (Alternate angles) Δs are equiangular. ∴$$\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{\mathrm{CA}_5}{\mathrm{AB}_8}=\frac{\mathrm{CA}}{\mathrm{BC}}$$ ∴ $$\frac{\mathrm{CA}}{\mathrm{BC}}=\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{5}{8}$$ Question 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $$\frac{2}{3}$$ of the corresponding sides of the first triangle. Solution : Steps of construction: 1. Construct ΔABC given AB = 4 cm, BC = 5 cm, AC = cm. 2. At B, make an acute angle CBx. 3. Divide Bx into three equal parts with a convenient radius. 4. Join B3C. 5. From B2 draw a parallel to B3C. 6. Let it cut BC at C’. 7. At C’ make angle A’C’B = ACB. Join A’C’. ∴ ΔABC \|\|\| A’BC’. In ΔABC and ΔA’BC’, 1.A$$\hat{B}$$C = A’$$\hat{B}$$C’ (Common angle) 2. A$$\hat{C}$$B = A’$$\hat{C’}$$B (Corresponding angles) 3. B$$\hat{A}$$C = B$$\hat{A’}$$C’ (Remaining angles) Δs are equiangular. ∴ $$\frac{\mathrm{AC}}{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{BC} \mathrm{C}^{\prime}}$$ = $$\frac{3}{2}$$ ∴ $\frac{BC’}{BC}=\frac{2}{3}$ Question 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $$\frac{7}{5}$$ of the corresponding sides of the first triangle. Solution : Steps of construction: 1. Construct a triangle ABC given BC= 7 cm, AB = 5 cm, AC = 6 cm. 2. At B, below BC, make an acute angle CBx. 3. With a convenient radius cut off seven equal parts BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7. 4. Join B5C. 5. From B1 draw a parallel to B5C to cut BC produced at C’. 6. At C’ draw a parallel to CA to meet BA produced at A’. A’BC’ is the required triangle. In ΔABC and ΔA’BC’ Question 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1$$\frac{1}{2}$$ times the corresponding sides of the isosceles triangle. Solution : Steps of construction: 1. Draw a line segment BC = 8 cm. 2. Draw its perpendicular bisector. 3. Cut off DA = 4 cm. (altitude given) 4. Join AB and AC. ABC is the required triangle. 5. At B, below BC, draw an acute angle CBx. 6. With a convenient radius cut off three equal parts BB1 = B1B2 = B2B3. 7. Join B2C. 8. At B3 draw a parallel to B2C to meet BC extended at C’. 9. At C’ draw a parallel to CA to meet BA produced at A’. 10. Join A’C’. A’BC’ is the required triangle similar to ΔABC. In ΔA’BC’ and ABC 1. A’$$\hat{B}$$C’ = A$$\hat{B}$$C (Common angle) 2. B$$\hat{A}$$‘C’ = B$$\hat{A}$$C (Corresponding angles) 3. A’$$\hat{C}$$‘B = A$$\hat{C}$$B Δs are equiangular. ∴ $$\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{\mathrm{BC}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}$$ $$\frac{BC’}{BC}$$ = $$\frac{3}{2}$$ BC = 8 cm, BC’ = 8 × $$\frac{3}{2}$$ = 12 cm. BA = CA = 5.6 cm ∴ A’B = A’C’ = 5.6 × $$\frac{3}{2}$$ = 8.4 cm. Question 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are $$\frac{3}{4}$$ of the corresponding sides of the triangle ABC. Solution : Steps of construction: 1. Construct ΔABC given BC = 6 cm, AB = 5 cm, A$$\hat{B}$$C = 60°. 2. At B, below BC, make an acute angle CBx. 3. In Bx cut off four equal parts BB1 = B1B2 = B2B3 = B3B4 4. Join B4C. 5. From B3 draw a parallel to B4C to meet BC at C’. 6. At C’ draw a parallel to CA to meet CA at A’. A’BC’ is the required triangle similar to ΔABC. In Δs A’BC’ and ABC Question 6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are $$\frac{4}{3}$$ times the corresponding sides of ΔABC. Solution : $$\hat{A}$$ + $$\hat{B}$$ + $$\hat{C}$$ = 180° $$\hat{A}$$ + $$\hat{B}$$ = 150° (105° + 45°) ∴ $$\hat{C}$$ = 30° Steps of construction: 1. In ΔABC, BC = 7 cm. $$\hat{B}$$ = 45°, $$\hat{C}$$ = 105° ∴ $$\hat{A}$$ = 180° – 45° – 105° = 180° – 150° = 30°. Construct ΔABC given BC = 7 cm, $$\hat{B}$$ = 45°, $$\hat{C}$$ = 30°. 2. At B draw an acute angle CBx. 3. In Bx cut off four equal parts BB1 = B1B2 = B2B3 = B3B4 with a convenient radius. 4. Join B3C. 5. At B4 draw a parallel to B3C to meet BC produced at C’. 6. At C’ make angle of 30° equal to A$$\hat{C}$$B. Let it meet BA produced at A’. A’BC’ is the required triangle similar to ΔABC. In ΔC’BB4 CB3 \|\| C’B4 Question 7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $$\frac{5}{3}$$ times the corresponding sides of the given triangle. Solution : BC’ = $$\frac{5}{3}$$ × BC = $$\frac{5}{3}$$ × 3 = 5 cm BA’= $$\frac{5}{3}$$ × 4 = $$\frac{20}{3}$$ = 6.6 cm A’C’ = $$\frac{5}{3}$$ × 5 = $$\frac{25}{3}$$ = 8.3 cm. Steps of construction: 1. Construct ΔABC given BC= 3 cm, $$\hat{B}$$ = 90°, BA = 4 cm. 2. At B, make an acute angle CBx. 3. Cut off five equal parts BB1 = B1B2 = B2B3 = B3B4 = B4B5 along Bx with a convenient radius. 4. Join B3C. 5. At B5 draw a parallel to B3C to meet BC produced at C’. 6. At C’ draw a parallel to CA to meet BA produced at A’. A’BC’ is the required triangle similar to ΔABC. BC’ : BC = BB5 = BB3 $$\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{BB}_5}{\mathrm{BB}_3}$$ = $$\frac{5}{3}$$ In Δs A’BC’ and ABC 1. A’$$\hat{B}$$C’ = A$$\hat{B}$$C = 90° 2. B$$\hat{A}$$‘C’ = B$$\hat{A}$$C (Corresponding angles) 3. B$$\hat{C}$$‘A’ = B$$\hat{C}$$A Δs are equiangular. ∴ $$\frac{A’C’}{AC}$$ = $$\frac{BC’}{BC}$$ = $$\frac{BA’}{BA}$$ = $$\frac{5}{3}$$
Section 3.3 Solving Equations with Distribution 1 / 8 # Section 3.3 Solving Equations with Distribution - PowerPoint PPT Presentation Section 3.3 Solving Equations with Distribution. Mr. Beltz & Mr. Sparks. Goal:. Solve linear equations using Distribution Distribute to BOTH terms in the ( ) Remember the rules for multiplying. Distributive Property. What does this mean?! I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'Section 3.3 Solving Equations with Distribution' - scorpio-goldberg 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 ### Section 3.3Solving Equations with Distribution Mr. Beltz & Mr. Sparks Goal: • Solve linear equations using Distribution • Distribute to BOTH terms in the ( ) • Remember the rules for multiplying Distributive Property • What does this mean?! • To distribute means to give something to each member of a group. • Algebra example: a(b+c)= ab + ac Distribution Practice 2(x + 3)= -2(3 + x)= -2(x + 7)= 2(x + 3)= 2x + 6 -2(3 + x)= -6 + -2x -2(x + 7)= -2x + -14 *Change Subtraction to Addition, take the Opposite of the next number.* 1stDraw a line underneath the equal sign. 2nd Circle the variable (remember the sign to the left stays with the variable) 3rd Simplify one or both sides. 4th Do the Inverse Operation 2nd Multiplication & Division *Change Subtraction to Addition, take the Opposite of the next number.* DISTRIBUTE 1stDraw a line underneath the equal sign. 2nd Circle the variable (remember the sign to the left stays with the variable) 3rd Simplify one or both sides. 4th Do the Inverse Operation 2nd Multiplication & Division 8x -2(x + 7)=16 8x + -2(x + 7)=16 8x+ -2x + -14 = 16 6x + -14 = 16 +14 +14 6x = 30 6 6 x = 5 *Change Subtraction to Addition, take the Opposite of the next number.* DISTRIBUTE 1stDraw a line underneath the equal sign. 2nd Circle the variable (remember the sign to the left stays with the variable) 3rd Simplify one or both sides. 4th Do the Inverse Operation 2nd Multiplication & Division 5x + 3(x + 4) = 28 5x + 3x + 12 = 28 8x + 12 28 -12 -12 8x =16 8 8 x = 2

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