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Select Page Welcome back to our series on Today we will see what happens when we throw negatives into our exponential expressions. We will explore both negative bases and negative exponents. First, the bases. The rule to remember for negative bases is that odd powers of negative bases are negative and even powers of negative bases are positive. This rule makes sense when you remember that exponents simply notate a number of multiplications by the base (and remember your rules about multiplication with negative factors). Multiplication with an odd number of negative factors yields a negative product: (-x)3 = -x * -x * -x (-x)3 < 0 Multiplication with an even number of negative factors yields a positive product: (-x)4 = -x * -x * -x * -x (-x)4 > 0 A note on notation: parentheses should always be used around a negative value as the base of an exponent. If they are not, then the order of operations dictates that the exponent be applied before the negative sign. To avoid confusion, whenever the negative is meant to be left out of the exponential operation, parentheses are used like this: -(x)4 to make the order clear. Please note that -(x)4 is less than 0, and (-x)4 is greater than 0. Now for negative exponents. Here is a simple rule is the best way to explain it: x-n = 1/xn A negative exponent indicates a value reciprocal to the value with a positive exponent. It’s good practice to “translate” any exponential expressions with negative exponents to their reciprocal positive forms. Seeing it both ways can help you make sense of problems. 5-4 = 1 / (54) = 1 / 625 17-2 = 1 / (172) = 1 / 289 (9 / 16)-2 = (16 / 9)2 = 256 / 81 Integer bases with negative exponents go under a numerator of 1; fractional bases with negative exponents simply flip. Let’s look at some “double negative” exponential expressions. (-6)-3 = 1 / (-6)3 = 1 / -216 (-2)-10 = 1 / (210) = 1 / 1024 (-4 / 3)-4 = (-¾)4 = 81 / 256 Let’s get into some official GMAT problems. Be careful with this first one! From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers? (A) (-10)20 (B) (-10)10 (C) 0 (D) -(10)19 (E) -(10)20 One incorrect answer is chosen far more often than any other on this problem: answer choice D. Trying to minimize the product, many people consider taking the maximum number (20) of the lowest value (-10). The common mistake is then thinking that -(10)20 (answer choice E) is actually a positive value since it involves an even number (20) of negative factors. Many people then take “the next best thing” in answer choice D, which shifts the exponent to the next odd number down from 20. In fact, -(10)20 does not involve an even number of negative factors, since the negative sign is excluded from the exponential expression by the parentheses. Answer E means “take 1020 and make it negative.” It is true that taking 20 negative tens and multiplying them all together produces a large positive value (the opposite of what we are aiming for on this problem), but this misguided idea is notated by answer choice A – not answer choice E. Remember that the answer choices notate the product of the 20 factors, not necessarily a condensed list of the 20 factors. It is possible to choose 20 integers from -10 to 10 inclusive that, when multiplied, yield a product of -(10)20 (answer choice E). The “least possible value” is obtained by finding the greatest absolute value (distance from 0) in negative form. So we want all 20 of our factors to be either 10 or -10 since this will maximize the absolute value (distance from 0) of the product. To ensure that the product is also negative, we simply need an odd number of negative tens. We can use nineteen negative 10s and 1 positive 10, 1 negative 10 and 19 positive 10s, or any odd combination in between. Any of these options will yield a product of -(10)20. Read the notation carefully! Let’s try another: The value of 2(-14)+2(-15)+2(-16)+2(-17)/5 is how many time the value of 2(-17)? (A) 3/2 (B) 5/2 (C) 3 (D) 4 (E) 5 This problem benefits from the skill of noticing patterns and “checking” them. You should see the pattern in the numerator and generalize it by saying “the negative exponent on the 2 keeps decreasing by 1.” Then you can see how this pattern “works” by checking a single case. 2-17 = 1 / 217 2-16 = 1 / 216 Since 217 = 2 * 216, (1 / 217) is half the value of (1 / 216). Or, to say it a more useful way, 2-16 = 2 * 2-17 This pattern will continue through the numerator. Since we are looking for how many copies of 2-17 we have in this expression, we can replace 2-17 with 1 and follow the pattern. (214 + 2-15 + 2-16 + 2-17) / 3 (8 + 4 + 2 + 1) / 3 15 / 3 = 3 And the correct answer is C. Here’s a final problem for today: a is a nonzero integer. Is a2greater than 1? • a < -1 • a is even. To evaluate statement 1, simply start by checking a = -2. (-2)-2 = 1 / (-2)2 = ¼. Moving on to a = -3, (-3)-3 = 1 / (-3)3 = 1 / -27. This time the value is negative, but the positive value from a = -2 is still less than 1. If you imagine continuing with a = -4, a = -5, etc., you will just keep making smaller and smaller fractions. Statement 1 alone is sufficient. Statement 2 on its own is easy to check since we already know from checking statement 1 that some even values for the variable a yield an aa with a value less than 1. And it shouldn’t be hard to imagine an even value for variable a where aa is greater than 1. For example, 22 = 4 and 44 = 256. So statement 2 alone is insufficient, and the correct answer is A. If you went with answer choice C, here’s what might have happened. Noticing that statement 1 tells you the base is negative, you might have seen next that statement 2 tells you the exponent is even. You might have thought that this “evenness” of the exponent makes the difference since it determines the positivity or negativity of the expression. Very often in DS problems with negative bases, the even/odd identity of the exponent really matters. But in this case, it’s a trap, because we were asked whether aa is greater than 1 (not 0), and the fact that the exponent is also negative means that it’s even/odd identity is irrelevant – the value is always less than 1. The rules governing negative exponents and negative bases are simple, but the GMAT and EA problems that employ these rules can catch you if you aren’t careful. Next time we will look at another tricky exponential scenario: when the base is between -1 and 1. Register now for a free consultation with one of our top tutors. Contributor: Elijah Mize (Apex GMAT Instructor)
Which graphs represent a function Value tables and function graphs bettermarks »Math book» Algebra and functions »Functions and their representations» Linear functions »Value tables and function graphs In these explanations you will learn the relationship between value tables and graphical representations of linear functions. Tables of values A function can be represented in different ways, one possibility is the value table. From the table of values to the graphic representation The graph of a linear function is a straight line. A straight line is clearly defined by two points. You can find the coordinates of these points in the table of values and enter them in a suitable coordinate system. Given is the table of values of a linear function f. Draw the graph of the function. Enter points in the coordinate system Tables of values and graphical representation for material tasks There are many situations in everyday life that you can describe with the help of linear functions. If you want to draw the graph (the straight line) for a factual situation with a linear connection, you first consider which coordinate system is suitable. The labeling and scaling of the coordinate axes and are important the choice of the quadrants of the coordinate system. In order to make the right decisions here, you are guided by the definition and value range of the function and what information you want to take from the graphical representation. Ms. Meier is going away for a few days. To prevent her plants from drying out on the windowsill, she places a 10-liter bucket with water next to it and attaches a woolen thread to each flower pot, one end of which is in the water and the other end of which is in the ground.The water now runs evenly from the bucket into the flower pots. After half a day there are still 8 liters in the bucket and after 2.5 days there are still 4 liters. How much water did Ms. Meier put in the bucket? After what time is the bucket only half full? After how many days is the bucket empty ? Enter points in the coordinate system and draw straight lines
### Theory: • Depending upon the lengths of the sides and diagonal, create a triangle of $$PQR$$ based on $$SSS$$ construction. • Make an arc (point $$S$$) at a certain distance from $$P$$. • Make an arc at a certain distance from point $$R$$ on the earlier arc on $$S$$. Name the two points intersection as $$S$$. • $$P$$ and $$R$$ join $$S$$. The quadrilateral $$PQRS$$ will be achieved. Example: Construct a quadrilateral $$ABCD$$ with the following measurements. $$AB =$$ $$4.5 cm$$, $$BC =$$ $$5.5 cm$$, $$CD = 4cm$$, $$AD = 6cm$$, $$AC = 7cm$$. Step 1:Draw side $$BC = 5.5 cm$$ and cut arcs above it from $$B$$ ($$4.5 cm$$) and $$C$$ ($$7 cm$$). Mark the intersection as $$A$$. Join $$AB$$ and $$AC$$. Step 2: Draw and arc from $$A$$ equal to $$6 cm$$ which is the length of $$AD$$. Step 3: Draw and arc from $$C$$ equal to $$4 cm$$ which is the length of $$CD$$. Mark the intersection as $$D$$ and join $$AD$$ and $$CD$$. Thus, the $$ABCD$$ is a required quadrilateral. Area of the quadrilateral $$ABCD$$ $$=$$ $$\frac {1}{2}$$ $$\times$$ d $$\times$$ $$h_1 + h_2$$ sq. units $$=$$ $$\frac{1}{2} × 10 (1.9 +2.3)$$ $$=$$ $$5\times 4.2$$ $$=$$ $$21 cm²$$.
# 2001 AMC 12 Problems/Problem 21 ## Problem Four positive integers $a$, $b$, $c$, and $d$ have a product of $8!$ and satisfy: $$\begin{array}{rl} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{array}$$ What is $a-d$? $\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$ ## Solution Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows: \begin{align} (a+1)(b+1) & = 525 \\ (b+1)(c+1) & = 147 \\ (c+1)(d+1) & = 105 \end{align} Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$. We get: \begin{align} ef & = 3\cdot 5\cdot 5\cdot 7 \\ fg & = 3\cdot 7\cdot 7 \\ gh & = 3\cdot 5\cdot 7 \end{align} Clearly $7^2$ divides $fg$. On the other hand, $7^2$ can not divide $f$, as it then would divide $ef$. Similarly, $7^2$ can not divide $g$. Hence $7$ divides both $f$ and $g$. This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$. The first case solves to $(e,f,g,h)=(75,7,21,5)$, which gives us $(a,b,c,d)=(74,6,20,4)$, but then $abcd \not= 8!$. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$. (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.) The second case solves to $(e,f,g,h)=(25,21,7,15)$, which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$, and we have $a-d=24-14 =\boxed{10}$.
## Search Here ### Two trains leave different cities heading toward each other at different speeds. When and where do they meet? Train A, traveling 70 miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away. At the same time Train B, traveling 60 mph, leaves Eastford heading toward Westford. When do the two trains meet? How far from each city do they meet? To solve this problem, we'll use the distance formula: Distance = Rate x Time Since an equation remains true as long as we perform the same operation on both sides, we can divide both sides by rate: Distance ---------- = Time Rate or by time: Distance ---------- = Rate Time So rate is defined as distance divided by time, which is a ratio. Speed is another word that is used for rate. When a problem says that a train is moving at a speed of 40 mph, you can understand this to mean that the train's rate is 40 mph, which means it will travel 40 miles in one hour. Here are two different ways to approach this problem. Let's start by listing the information given: Speed of Train A: 70 mph Speed of Train B: 60 mph Distance between Westford and Eastford: 260 miles Method I: We'll use the notion of relative speed 1 (or relative rate) in order to express the rates of the two trains in one number that can then be used in the distance formula. Imagine you're on Train A. You're going 70 mph, so your speed relative to the trees, houses, and other non-moving things outside the train is 70 mph. (All of those objects look as if they're going by at 70 mph.) Now imagine you're the engineer and you can see Train B coming toward you - not on the same track, of course! Since Train B is moving 60 mph, it will look as if it's approaching faster than if it were sitting still in the station - a lot faster than the trees and houses appear to be moving. The relative speed of the two trains is the sum of the speeds they are traveling. (If you're on either of the trains, this is the speed you appear to be moving when you see the other train.) In our problem, the relative speed of the two trains is 70 mph + 60 mph = 130 mph. What if the trains were traveling in the same direction? Then we'd need to subtract the speed of the slower train from the speed of the faster train, and their relative speed would be 10 mph. At this point we know two of the three unknowns: rate and distance, so we can solve the problem for time. Remember that time = distance/rate, the distance traveled is 260 miles, and the relative speed is 130 mph: t = 260 miles/130 mph t = 2 hrs. We find that the trains meet two hours after leaving their respective cities. Method II: Here we'll begin by noting that the distance between Westford and Eastford is 260 miles: this is the total distance the trains will travel. Using the distance formula (Distance = rate x time, or D = rt) we can express the distance traveled by each train: Train A moving at 70 mph in t hours will cover 70t miles Train B moving at 60 mph in t hours will cover 60t miles Together the two trains will cover the distance 70t + 60t Since we know that this distance is 260 miles, we can write the following algebraic equation to represent this information. 70t + 60t = 260 Solving this equation we find that: 130t = 260 t = 2 which tells us that the trains will meet in 2 hours. Now, where do the trains meet? We again use the distance formula to find how far each train has traveled in two hours: For Train A: 70 mph x 2 hrs = 140 miles For Train B: 60 mph x 2 hrs = 120 miles Thus the two trains meet at a point 140 miles from Westford and 120 miles from Eastford. To check this, we can add 140 to 120: the answer is 260, which was the given distance between the two cities. Let's look at a variation on this problem. Train A, traveling 40 mph, leaves Westford heading toward Eastford, 260 miles away. One hour after Train A leaves Westford, Train B, traveling 70 mph, leaves Eastford heading toward Westford. When do the two trains meet? Notice that in this problem, the two trains do not leave their respective cities at the same time. Method I Let's move the starting point for Train A so we can treat the problem as if the trains leave at the same time, which we already know how to do. We know that Train A is moving 40 mph, and will therefore travel 40 miles in the hour before Train B leaves Eastford. This means that by the time Train B starts moving, the two trains are only 260 - 40 = 220 miles apart. Now we can use the relative speed of the trains, which is 40 + 70 = 110 mph. Using the distance formula for time (time = distance/rate), we write: t = 220 miles/110 mph t = 2 hrs. Since t represents the time traveled by each train after Train A has already traveled for one hour, Train B travels 2 hours before meeting Train A. Adding the extra hour that Train A travels before Train B starts moving, we see that Train A must travel 3 hours before meeting Train B. Method II Let t represent the time that Train A travels. Since Train B leaves one hour after Train A, let t -1 represent the time that Train B travels. Again, the sum of the distances traveled by the two trains up until the time they meet is 260 miles: between the two of them, they cover all 260 miles of track. Using the distance formula: distance traveled by Train A = 40t distance traveled by Train B = 70(t -1) 40t + 70(t-1) = 260 miles t = 3 hrs. Since t represents the time that Train A has been traveling, this means that Train A travels 3 hours before meeting Train B. But how long has Train B been traveling? Train B travels t -1 hours, which means that it travels 2 hours before meeting Train A. Note: To help us make sense of relative speed and how it relates to distance traveled, let's think of it in terms of the distance the two trains travel toward each other in one hour. We'll again use the distance formula: Rate x Time = Distance Train A's rate is 70 mph, so: 70 mph x 1 hr = 70 miles Train B's rate is 60 mph, so: 60 mph x 1 hr = 60 miles If we add these two numbers together, we get 130 miles, which means that 130 miles of track will be covered in one hour. This is the same as saying that the two trains are going 130 mph relative to each other, so: 130 mph x 1 hr = 130 miles Either way, the numbers tell you that 130 miles of track will be covered in an hour.
# Perimeter of Triangle – Formula with Examples Welcome to Brighterly, where our mission is to make math exciting and enjoyable for children of all ages! Today, we’re exploring the fascinating world of geometry, focusing on the perimeter of a triangle. This fundamental concept is essential for students to grasp as it sets the foundation for more advanced mathematical topics. So, without further ado, let’s embark on an adventure into the realm of triangle perimeters and discover the secrets behind this geometric wonder! ## What is the Perimeter of a Triangle? The perimeter of any shape refers to the total length of its outer edges. In the case of a triangle, the perimeter is the sum of the lengths of its three sides. Triangles come in various types, including scalene, isosceles, equilateral, and right triangles. Depending on the type of triangle, the method of calculating its perimeter may vary slightly. Keep reading to learn how to find the perimeter of a triangle and the formulas for each type of triangle! ## How to Find The Perimeter of a Triangle? Finding the perimeter of a triangle is quite simple! All you need to do is add up the lengths of its three sides. However, if you’re missing some side lengths, you may need to use other information about the triangle to find the missing values. For instance, you could use the Pythagorean theorem for right triangles or apply the properties of isosceles and equilateral triangles. Below, we’ll discuss the formulas for finding the perimeter of various triangle types. ## Perimeter of a Triangle Formula The general formula for finding the perimeter of a triangle is: Perimeter = a + b + c where ‘a’, ‘b’, and ‘c’ represent the lengths of the triangle’s sides. This formula applies to all types of triangles, but some specific triangle types have unique properties that can simplify the calculation. Let’s explore the formulas for each triangle type! ## Perimeter of a Scalene Triangle A scalene triangle has three sides of different lengths. To find the perimeter of a scalene triangle, simply use the general formula: Perimeter = a + b + c ## Perimeter of an Isosceles Triangle An isosceles triangle has two sides of equal length. Since two sides are equal, the formula for its perimeter is: Perimeter = 2a + b where ‘a’ is the length of the equal sides and ‘b’ is the length of the remaining side. ## Perimeter of an Equilateral Triangle An equilateral triangle has three sides of equal length. Therefore, the formula for its perimeter is: Perimeter = 3a where ‘a’ is the length of each side. ## Perimeter of a Right Triangle A right triangle has one 90-degree angle. To find its perimeter, you can use the general formula: Perimeter = a + b + c However, if you only have the lengths of the two legs (the sides adjacent to the right angle), you can use the Pythagorean theorem to find the length of the hypotenuse (the side opposite the right angle): c² = a² + b² Then, you can calculate the perimeter using the general formula. ## Perimeter of Isosceles Right Triangle An isosceles right triangle has one 90-degree angle and two equal sides. Since the equal sides are adjacent to the right angle, you can use the Pythagorean theorem to find the hypotenuse: c² = a² + a² Then, calculate the perimeter using the general formula. ## Perimeter of an Isosceles, Equilateral and Scalene Triangle As a recap, here are the formulas for finding the perimeter of isosceles, equilateral, and scalene triangles: • Isosceles: Perimeter = 2a + b • Equilateral: Perimeter = 3a • Scalene: Perimeter = a + b + c Keep these formulas in mind as we explore some examples below! ## Perimeter of a Triangle Examples Let’s take a look at some examples to help you understand how to find the perimeter of a triangle: 1. Scalene Triangle: If a scalene triangle has sides with lengths of 5, 7, and 8, the perimeter is: Perimeter = 5 + 7 + 8 = 20 2. Isosceles Triangle: If an isosceles triangle has two equal sides with lengths of 6 and a remaining side with a length of 4, the perimeter is: Perimeter = 2(6) + 4 = 12 + 4 = 16 3. Equilateral Triangle: If an equilateral triangle has sides with lengths of 9, the perimeter is: Perimeter = 3(9) = 27 ## Practice Questions on Perimeter of Triangle 1. A scalene triangle has sides with lengths of 6, 10, and 8. What is its perimeter? 2. An isosceles triangle has two equal sides with lengths of 5 and a remaining side with a length of 2. What is its perimeter? 3. An equilateral triangle has sides with lengths of 7. What is its perimeter? ## Conclusion We hope this comprehensive guide on the perimeter of a triangle has not only helped you grasp the concept but also inspired a newfound appreciation for geometry. At Brighterly, we believe that practice is the key to unlocking your full potential. So, don’t hesitate to tackle more problems and apply the formulas for each triangle type. As you continue to hone your skills, you’ll soon become a master at calculating triangle perimeters and be ready to tackle even more complex mathematical challenges! Remember, learning should be a joyous journey, and at Brighterly, we’re dedicated to making that journey as enjoyable and engaging as possible. Keep exploring, stay curious, and most importantly, have fun! ## Frequently Asked Questions on Perimeter of Triangle ### What is the perimeter of a triangle? The perimeter of a triangle is the total length of its outer edges or the sum of the lengths of its three sides. Perimeter is an important concept in geometry and is used to measure the boundary of any shape. In the case of triangles, the perimeter helps us understand the size and scale of the triangle based on the lengths of its sides. ### How do you find the perimeter of a triangle? To find the perimeter of a triangle, you need to add the lengths of its three sides. The method of calculating the perimeter may vary slightly depending on the type of triangle (scalene, isosceles, equilateral, or right triangle). For each triangle type, there is a specific formula that can be used to calculate the perimeter: • Scalene: Perimeter = a + b + c • Isosceles: Perimeter = 2a + b • Equilateral: Perimeter = 3a • Right Triangle: Perimeter = a + b + c (use the Pythagorean theorem if necessary) ### What is the formula for the perimeter of a triangle? The general formula for the perimeter of a triangle is Perimeter = a + b + c, where ‘a’, ‘b’, and ‘c’ represent the lengths of the triangle’s sides. This formula applies to all types of triangles, but some specific triangle types have unique properties that can simplify the calculation. For example, an isosceles triangle has two equal sides, so its formula is Perimeter = 2a + b, and an equilateral triangle has three equal sides, so its formula is Perimeter = 3a. Information Sources
Word Problems on Simultaneous Linear Equations - Embibe # Word Problems on Simultaneous Linear Equations: Definition, Examples Word Problems on Simultaneous Linear Equations: When we have to find the values of two unknown quantities in a problem, we take the two unknown quantities to be x, y, or any two other algebraic symbols. Then, based on the stated condition or conditions, we build the equation and solve the two simultaneous equations to determine the values of the two unknown numbers. As a result, we can solve the problem. Simultaneous equations are ones that need simultaneous solutions. Word problems, also known as applied problems, are situations in which unknowns are described using words. For success with word (or applied) difficulties, practice and mastery of certain basic translation processes are essential. Solving word problems can be done in two ways. To begin, translate the problems’ wording into algebraic equations. The second step is to solve the resulting equations. We will go over word problems on simultaneous linear equations in-depth in this article. Learn Method of Substitution to Solve Simultaneous Linear Equations here ## What is a Word Problem? Mathematical issues rarely reveal themselves as $$2+3$$ or $$6-4$$ in real life. They are, in fact, a little more complicated than we believe. Authors of mathematics curricula occasionally use word problems to assist students to comprehend how their work relates to the actual world. Word problems frequently represent mathematics as it occurs in the real world. The difficulty of word problems can range from easy to complicated. Here are a few examples to get you some ideas: 1. Keerthi had $$5$$ chocolates. Her mother gave her $$7$$ more chocolates. Now, how many chocolates does Keerthi have altogether? 2. There were $$18$$ pens and $$9$$ erasers. How many more pens than erasers are there? 3. Kriti has one dozen bananas. Her friends ate $$4$$ for breakfast. Now, how many bananas are left with Kriti? 4. There are $$45$$ mangoes. Jyoti, Priya, and Preethu want to eat them in equal shares. How many mangoes will each of the friends get? As you can notice, word problems involve addition, subtraction, multiplication, division or even multiple operations. ## Simultaneous Linear Equations Simultaneous linear equations in two variables depict real-life situations by involving two unknown values. It aids in establishing a link between quantities, pricing, speed, time, distance, and other factors, which leads to a better knowledge of the issues. Simultaneous linear equations are used in our daily lives without us even realising them. ## Solving Word Problems No approach works in all instances due to the great range of word (or application) issues. The following broad ideas, however, might be helpful: (i) Carefully read and re-read the problem statement to discover what amounts must be found. (ii) Use letters to represent the unknown quantities. (iii) Write equations by determining whether expressions are equivalent. (iv) Solve the resulting equations. ## Necessary Steps for Forming and Solving Simultaneous Linear Equations Let us take a mathematical problem to indicate the necessary steps for forming simultaneous equations: In a stationery shop, the cost of $$3$$ pencil cutters exceeds the price of $$2$$ pens by $$â‚¹2.$$ Also, the total price of $$7$$ pencil cutters and $$3$$ pens is $$â‚¹43.$$ Step I: Identify the unknown variables; assume one of them as $$x$$ and the other as $$y$$ Here two unknown quantities (variables) are: Price of each pencil cutter$$=â‚¹x$$ Price of each pen$$=â‚¹y$$ Step II: Identify the relationship between the unknown quantities. Price of $$3$$ pencil cutters$$=â‚¹3x$$ Price of $$2$$ pens$$=â‚¹2y$$ Therefore, the first condition gives: $$3x-2y=2$$ Step III: Express the conditions of the problem in terms of $$x$$ and $$y$$ Again price of $$7$$ pencil cutters$$=â‚¹7x$$ Price of $$3$$ pens$$=â‚¹3y$$ Therefore, the second condition gives: $$7x+3y=43$$ Simultaneous equations formed from the problem is $$3x – 2y = 2 \ldots \ldots \ldots {\rm{(i)}}$$ $$7x – 3y = 43 \ldots \ldots \ldots {\rm{(ii)}}$$ ## Word Problems on Simultaneous Linear Equations Example: Twice one number minus three times a second equals $$2,$$ and the sum of these numbers is $$11.$$ Find the numbers. Solution: Let the two numbers be $$x, y.$$ According to the problem, $$3x – 3y = 2 \ldots ..{\rm{(i)}}$$ and $$x – y = 11 \ldots {\rm{(ii)}}$$ Multiplying both sides of $${\rm{(ii)}}$$ by $$3,$$ we get $$3x – 3y = 33 \ldots ..{\rm{(iii)}}$$ On adding $${\rm{(i)}}$$ and $${\rm{(iii)}},$$ we get $$5x=35â‡’x=7$$ Substituting this value of $$x$$ in $${\rm{(ii)}},$$ we get $$7+y=11â‡’y=11-7â‡’y=4$$ Hence, the required numbers are $$7$$ and $$4.$$ ## Solved Examples â€“ Word Problems on Simultaneous Linear Equations Q.1: A man buys postage stamps of denominations $$25$$ paise and $$50$$ paise for $$â‚¹10.$$ He buys $$28$$ stamps in all. Find the number of $$25$$ paise stamps bought by him. Ans: Let the number of $$25$$ paise stamps be $$x,$$ and the number of $$50$$ paise stamps be $$y.$$ According to the problem, $$x + y = 28 \ldots {\rm{(i)}}$$ and $$25x+50y=1000$$ $$[â‚¹10=1000]$$ paise i.e. $$x + 2y = 40 \ldots ({\rm{ii}})$$ Subtracting $${\rm{(i)}}$$ from $${\rm{(ii)}},$$ we get, $$y=12$$ Substituting this value of $$y$$ in $${\rm{(i)}},$$ we get $$x+12=28$$ $$â‡’x=28-12$$ $$â‡’x=16$$ Hence, the number of $$25$$ paise stamps is $$16.$$ Q.2: A chemist has one solution, $$50\%$$ acid, and a second, $$25\%$$ acid. How much of each should be mixed to make $${\rm{10}}\,{\rm{litres}}$$ of a $$40\%$$ acid solution? Ans: Let $$x$$ litres of $$50\%$$ and y litres of $$25\%$$ acid solutions be mixed. Then, according to the given conditions, we have $$x + y = 10 \ldots ({\rm{i}})$$ and $$25\%$$ of $$x + 50\%$$ of $$y = 40\%$$ of $$10$$ $$\Rightarrow \frac{{50}}{{100}}x + \frac{{25}}{{100}}y = \frac{{40}}{{100}} \times 10$$ $$\Rightarrow \frac{1}{2}x + \frac{1}{4}y = 4$$ $$\Rightarrow 2x + y = 16 \ldots {\rm{(ii)}}$$ Subtracting equation $${\rm{(i)}}$$ from equation $${\rm{(ii)}},$$ we get $$x=6$$ Substituting this value of $$x$$ in equation $${\rm{(i)}},$$ we get $$6+y=10â‡’y=4$$ Hence, $$6\,{\rm{litres}}$$ of $$50\%$$ and $$4\,{\rm{litres}}$$ of $$25\%$$ acid solutions be mixed to get $$10\,{\rm{litres}}$$ of $$40\%$$ acid solution. Q.3: A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is $$18$$ less than the original number. Find the number. Ans: Let $$x$$ be the digit at ten’s place, and y be the digit at unit’s place. Then the number is $$10x+y.$$ According to the first condition of the problem, $$10x+y=7(x+y)$$ $$â‡’10x+y=7x+7y$$ $$â‡’3x=6y$$ $$\Rightarrow x + 2y \ldots ({\rm{i}})$$ The number formed by reversing the digits is $$10y+x.$$ According to the second condition of the problem, $$10y+x=(10x+y)-18$$ $$â‡’10y-y=10x-x-18$$ $$â‡’9y=9x-18$$ $$â‡’y=x-2$$ $$â‡’y=2y-2$$ [Using $$({\rm{i}})$$] $$â‡’y=2$$ From $$({\rm{i}}),$$ $$x=2Ã—2=4$$ Hence, the required number is $$42.$$ Q.4: The number of heads and legs of deer and human visitors at a deer park was counted at one point, and it was found that there were $$41$$ heads and $$136$$ legs. Calculate the number of deer and people who have visited the park. Ans: Let there be $$x$$ deer and $$y$$, human visitors, in the park at the given time. As a deer and a human each has one head, we get $$x + y = 41 \ldots ({\rm{i}})$$ As a deer has $$4$$ legs and a human has $$2$$ legs, we get $$4x+2y=136$$ $$\Rightarrow 2x + y = 68 \ldots ({\rm{ii}})$$ Subtracting $${\rm{(i)}}$$ from $$({\rm{ii}}),$$ we get $$x=27$$ Substituting $$x=27$$ in $${\rm{(i)}},$$ we get $$27+y=41$$ $$â‡’y=14$$ Hence, there were $$27$$ deer and $$14$$ human visitors in the park. Q.5: Six years after a woman’s age will be three times her daughter’s age, and three years ago, she was nine times as old as her daughter. Find their present ages. Ans: Let the present age of the woman be $$x$$ years, and the present age of her daughter be $$y$$ years. $$6$$ years after, their ages will be $$(x+6)$$ years and $$(y+6)$$ years. According to the problem, $$x+6=3(y+6)$$ $$â‡’x+6=3y+18$$ $$\Rightarrow x + 3y = 12 \ldots ({\rm{i}})$$ 3 years ago, their ages were $$(x-3)$$ years and $$(y-3)$$ years. According to the problem, $$x-3=9(y-3)$$ $$â‡’x-3=9y-27$$ $$\Rightarrow x – 9y = – 24 \ldots ({\rm{ii}})$$ Subtracting $${\rm{(ii)}}$$ from $${\rm{(i)}},$$ we get $$6y=36â‡’y=6$$ Substituting this value of $$y$$ in $${\rm{(i)}},$$ we get $$x-3Ã—6=12$$ $$â‡’x-18=12$$ $$â‡’x=12+18=30$$ Hence, the present age of the woman is $$30$$ years, and that of her daughter is $$6$$ years. ## Summary In this article, we have discussed word problems and how to solve a word problem. Simultaneous equations are those that must be solved at the same time. Word problems or applied problems are real-life situations involving unknowns that are described in words. Also, we covered simultaneous linear equations and the necessary steps to solve word problems on simultaneous linear equations. And at last, we had some solved word problems on simultaneous linear equations. ## Frequently Asked Questions (FAQ) â€“ Word Problems on Simultaneous Linear Equations Q.1: Why is it called a simultaneous equation? Ans: Sometimes, you’ll come across two or more unknown quantities, as well as two or more equations that relate to them. Simultaneous equations are what they’re called. As a result, the only pair of numbers that answer both equations simultaneously may be found in this manner. As a result, these are referred to as simultaneous equations. Q.2: What is the difference between linear and simultaneous equations? Ans: A linear equation is one in which all the terms are either constants or the products of constants. Simultaneous equations are a collection of equations with several variables. Q.3: How do you create a simultaneous equation from a word problem? Ans: Step I: Identify the unknown variables; assume one of them as $$x$$ and the other as $$y$$ Step II: Identify the relationship between the unknown quantities. Step III: Express the conditions of the problem in terms of $$x$$ and $$y$$ Step IV: Solve for the values of $$x$$ and $$y$$ Q.4: What are linear equations in two variables? Ans: An equation is said to be a linear equation in two variables if it is written in the form of $$Ax+By+C=0,$$ where $$A, B\& C$$ are real numbers and the coefficients of $$x$$ and $$y,$$ and $$aâ‰ 0,bâ‰ 0$$ For example, $$6x+4y=3$$ and $$-3x+5y=12$$ are linear equations in two variables. Q.5: How many ways can you solve simultaneous equations? Ans: You can solve for both unknowns if you have two separate equations with the same two unknowns in each. The elimination, substitution, and graphical methods are the three most popular approaches for solving problems. Learn to Algebraically Solve Simultaneous Linear Equations here We hope you find this article on ‘Word Problems on Simultaneous Linear Equations helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them. 53 Views Achieve Your Best With 3D Learning, Book Practice, Tests & Doubt Resolutions at Embibe
# How do you differentiate f(x) =2cosx+sin2x ? Mar 15, 2018 $2 \cos \left(2 x\right) - 2 \sin \left(x\right)$ #### Explanation: the derivative of $\cos \left(x\right)$ is defined as $- \sin \left(x\right)$ therefore for the first term, the derivative of a constant multiplied by $\cos \left(x\right)$ gives that same constant multiplied by $- \sin \left(x\right)$ therefore derivative of the first term is $- 2 \sin \left(x\right)$ the derivative of the second term can be found by using The Chain Rule $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$ therefore, let $f \left(x\right) = \sin \left(x\right)$ and $g \left(x\right) = 2 x$ therefore, $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \sin \left(2 x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = \cos \left(2 x\right) \cdot 2$ $= 2 \cos \left(2 x\right)$ therefore, the entire derivative is, $- 2 \sin \left(x\right) + 2 \cos \left(2 x\right)$ =
Courses Courses for Kids Free study material Offline Centres More Store # One card is drawn from a well-shuffled deck of 52 cards. The probability that the card will not be an ace is: ${\text{A}}{\text{. }}\dfrac{1}{{13}} \\ {\text{B}}{\text{. }}\dfrac{1}{4} \\ {\text{C}}{\text{. }}\dfrac{{12}}{{13}} \\ {\text{D}}{\text{. }}\dfrac{3}{4} \\$ Last updated date: 13th Jun 2024 Total views: 411.9k Views today: 11.11k Verified 411.9k+ views Hint: In order to find the probability that the card will not be an ace, we first find out how many aces are there in a deck of 52 cards in order to compute its possibilities. Then we find the total number of possibilities and substitute these quantities in the formula of probability to determine the answer. Given Data, Deck of 52 cards Ace Firstly, a deck of 52 cards has four sets of 13 cards each representing four different symbols. Each symbol has an Ace in its 13 cards. So there are 4 aces in total in the deck. Now the probability of an event not happening = 1 – the probability of an event happening. As the total probability of an even is always equal to 1. Therefore the probability of the card will not be an ace = 1 – the probability of the card being an ace. The probability of the card being an ace is: Number of favorable outcomes = 4 (there are 4 aces in total) Total outcomes = 52 (total number of cards) We know the formula of probability is, ${\text{P = }}\dfrac{{{\text{favorable outcomes}}}}{{{\text{total outcomes}}}}$ Hence, P (A) = $\dfrac{4}{{52}}$ Therefore the probability of not getting an ace = 1 – P (A) $\Rightarrow {\text{P}}\left( {{\text{not ace}}} \right) = 1 - \dfrac{4}{{52}}$ $\Rightarrow {\text{P}}\left( {{\text{not ace}}} \right) = \dfrac{{48}}{{52}}$ $\Rightarrow {\text{P}}\left( {{\text{not ace}}} \right) = \dfrac{{12}}{{13}}$ So, the correct answer is “Option A”. Note: In order to solve this type of problems the key is to first have a good idea about how a deck of 52 cards is arranged and what constitutes it. Once we know this we can calculate the number of possibilities of each of the given in the question and substitute them in the formula of probability. The total probability of an event is always exactly equal to 1, neither more than 1 nor less than 1.
# AP Statistics Curriculum 2007 Normal Std (Difference between revisions) Revision as of 20:22, 5 November 2008 (view source)IvoDinov (Talk \| contribs) (→Standard Normal Distribution: added the CDF)← Older edit Revision as of 16:09, 18 May 2009 (view source)IvoDinov (Talk \| contribs) (fixed a typo)Newer edit → Line 3: Line 3: === Standard Normal Distribution=== === Standard Normal Distribution=== The Standard Normal Distribution is a continuous distribution with the following density: The Standard Normal Distribution is a continuous distribution with the following density: - * Standard Normal ''density'' function $f(x)= {e^{-x^2} \over \sqrt{2 \pi}}.$ + * Standard Normal ''density'' function $f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}.$ - * Standard Normal ''cumulative distribution'' function $\Phi(y)= \int_{-\infty}^{y}{{e^{-x^2} \over \sqrt{2 \pi}} dx}.$ + * Standard Normal ''cumulative distribution'' function $\Phi(y)= \int_{-\infty}^{y}{{e^{-x^2 \over 2} \over \sqrt{2 \pi}} dx}.$ Note that the following exact ''areas'' are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin: Note that the following exact ''areas'' are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin: ## General Advance-Placement (AP) Statistics Curriculum - Standard Normal Variables and Experiments ### Standard Normal Distribution The Standard Normal Distribution is a continuous distribution with the following density: • Standard Normal density function $f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}.$ • Standard Normal cumulative distribution function $\Phi(y)= \int_{-\infty}^{y}{{e^{-x^2 \over 2} \over \sqrt{2 \pi}} dx}.$ Note that the following exact areas are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin: • The area: -1 < x < 1 = 0.8413 - 0.1587 = 0.6826 • The area: -2.0 < x < 2.0 = 0.9772 - 0.0228 = 0.9544 • The area: -3.0 < x < 3.0 = 0.9987 - 0.0013 = 0.9974 • Note that the inflection points (f''(x) = 0)of the Standard Normal density function are $\pm$ 1. • The Standard Normal distribution is also a special case of the more general normal distribution where the mean is set to zero and a variance to one. The Standard Normal distribution is often called the bell curve because the graph of its probability density resembles a bell. ### Experiments Suppose we decide to test the state of 100 used batteries. To do that, we connect each battery to a volt-meter by randomly attaching the positive (+) and negative (-) battery terminals to the corresponding volt-meter's connections. Electrical current always flows from + to -, i.e., the current goes in the direction of the voltage drop. Depending upon which way the battery is connected to the volt-meter we can observe positive or negative voltage recordings (voltage is just a difference, which forces current to flow from higher to the lower voltage.) Denote Xi={measured voltage for battery i} - this is random variable with mean of 0 and unitary variance. Assume the distribution of all Xi is Standard Normal, $X_i \sim N(0,1)$. Use the Normal Distribution (with mean=0 and variance=1) in the SOCR Distribution applet to address the following questions. This Distributions help-page may be useful in understanding SOCR Distribution Applet. How many batteries, from the sample of 100, can we expect to have? • Absolute Voltage > 1? P(X>1) = 0.1586, thus we expect 15-16 batteries to have voltage exceeding 1. • \|Absolute Voltage\| > 1? P(\|X\|>1) = 1- 0.682689=0.3173, thus we expect 31-32 batteries to have absolute voltage exceeding 1. • Voltage < -2? P(X<-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than -2. • Voltage <= -2? P(X<=-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than or equal to -2. • -1.7537 < Voltage < 0.8465? P(-1.7537 < X < 0.8465) = 0.761622, thus we expect 76 batteries to have voltage in this range.
# High School Physics : Impulse and Momentum ## Example Questions ### Example Question #1 : Impulse And Momentum crate slides along the floor for before stopping. If it was initially moving with a velocity of , what is the force of friction? Explanation: The fastest way to solve a problem like this is with momentum. Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force. Using this transformation, we can see that momentum is also equal to force times time. can also be thought of as . Expand this equation to include our given values. Since the object is not moving at the end, its final velocity is zero. Plug in the given values and solve for the force. We would expect the answer to be negative because the force of friction acts in the direction opposite to the initial velocity. ### Example Question #2 : Impulse And Momentum crate slides along a floor with a starting velocity of . If the force due to friction is , how long will it take for the box to come to rest? Explanation: The fastest way to solve a problem like this is with momentum. Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force. Using this transformation, we can see that momentum is also equal to force times time. can also be thought of as . Expand this equation to include our given values. Since the box is not moving at the end, its final velocity is zero. Plug in the given values and solve for the time. ### Example Question #1 : Impulse And Momentum A man with a mass of m is painting a house. He stands on a tall ladder of height h. He leans over and falls straight down off the ladder. If he is in the air for s seconds, what will be his momentum right before he hits the ground? Explanation: The problem tells us he falls vertically off the ladder (straight down), so we don't need to worry about motion in the horizontal direction. The equation for momentum is: We can assume he falls from rest, which allows us to find the initial momentum. . From here, we can use the formula for impulse: We know his initial momentum is zero, so we can remove this variable from the equation. The problem tells us that his change in time is s seconds, so we can insert this in place of the time. The only force acting upon man is the force due to gravity, which will always be given by the equation . ### Example Question #4 : Impulse And Momentum If an egg is dropped on concrete, it usually breaks. If an egg is dropped on grass, it may not break. What conclusion explains this result? Grass has less mass than concrete Grass is a softer texture The egg is in contact with the grass for longer, so it absorbs less force The density of grass is less than the density of concrete The coefficient of friction of the grass on the egg is less than the coefficient of the concrete on the egg The egg is in contact with the grass for longer, so it absorbs less force Explanation: In both cases the egg starts with the same velocity, so it has the same initial momentum. In both cases the egg stops moving at the end of its fall, so it has the same final velocity. The only thing that changes is the time and force of the impact. The force is produced by the deceleration resulting from the time that the egg is in contact with its point of impact. As the time of contact increases, acceleration decreases and force decreases. As the time of contact decreases, acceleration increases and force increases. The grass will have less force on the egg, allowing for a lesser acceleration, due to a longer period of impact. ### Example Question #5 : Impulse And Momentum tennis ball strikes a racket, moving at . After striking the racket, it bounces back at a speed of . What is the change in momentum? Explanation: The change in momentum is the final momentum minus the initial momentum, or . Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounced back," it begins to move in the opposite direction, so its velocity at this point will be negative. Plug in our values to solve: ### Example Question #1 : Impulse And Momentum The area under the curve on a Force versus time (F versus t) graph represents Impulse Work Kinetic energy Momentum
# 2003 AMC 12A Problems/Problem 7 ## Problem How many non-congruent triangles with perimeter $7$ have integer side lengths? $\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$ ## Solution By the triangle inequality, no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$. Since all sides must be integers, the largest possible length of a side is $3$. Therefore, all such triangles must have all sides of length $1$, $2$, or $3$. Since $2+2+2=6<7$, at least one side must have a length of $3$. Thus, the remaining two sides have a combined length of $7-3=4$. So, the remaining sides must be either $3$ and $1$ or $2$ and $2$. Therefore, the number of triangles is $\boxed{\mathrm{(B)}\ 2}$.
# The first term in an arithmetic sequence is 9. The fourth term in the sequence is 24.the twentieth ter is 104. What is the common difference of this sequence? How do you find the nth term of the arithmetic sequence? The first term in an arithmetic sequence is 9. The fourth term in the sequence is 24.the twentieth ter is 104. What is the common difference of this sequence? How do you find the nth term of the arithmetic sequence? You can still ask an expert for help ## Want to know more about Polynomial arithmetic? • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it pierretteA Step 1 Given, The first term in an arithmetic sequence is 9. The fourth term in the sequence is 24. And the twentieth term is 104.We know that, The general term of the arithmetic sequence is given by ${a}_{n}=a+\left(n-1\right)$ where d is the common difference a is the first term n is the number of terms Step 2 Now, First term in an arithmetic sequence is 9. $a=9$ The fourth term in the sequence is 24 and the twentieth term is 104 $a+3d=24anda+19d=104$ Put $a=9$ then $⇒9+3d=24$ $⇒3d=24-9$ $⇒3d=15$ $⇒d=5$ The general term of the arithmetic sequence is ${a}_{n}=a+\left(n-1\right)d$ $⇒{a}_{n}=9+\left(n-1\right)5$ $⇒{a}_{n}=9+5n-5$ $⇒{a}_{n}=5n+4$ $\therefore$ The nth term of the arithmetic sequence is ${a}_{n}=5n+4$
# CONVERTING CUSTOMARY UNITS "Converting customary units" is the stuff which is required to the students who would like to practice problems on the customary units like length, weight, capacity and time in the united states. ## Converting customary units - Practice problems Problem 1 : Convert 2 feet into inches. Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 2 feet = 2 x 12 inches 2 feet = 24 inches Hence, 2 feet is equal to 24 inches Problem 2 : Convert 3.5 yards into inches. Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 3.5 yards = 3.5 x 36 inches 3.5 yards = 126 inches Hence, 3.5 yards is equal to 126 inches Problem 3 : Convert 30 inches into feet. Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 30 inches = 30 / 12 feet 30 inches = 2.5 feet Hence, 30 inches is equal to 2.5 feet Problem 4 : Convert 5280 yards into miles. Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 5280 yards = 5280 / 1760 miles 5280 yards = 3 miles Hence, 5280 yards is equal to 3 miles Problem 5 : Convert 24 feet into yards Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 24 feet = 24 / 3 yards 24 feet = 8 yards Hence, 24 feet is equal to 8 yards Problem 6 : Convert 2 pounds into ounces. Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 2 pounds = 2 x 16 ounces 2 pounds = 32 ounces Hence, 2 pounds is equal to 32 ounces. Problem 7 : Convert 3.5 tons into pounds . Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 3.5 tons = 3.5 x 2000 pounds 3.5 tons = 7000 pounds Hence, 3.5 tons is equal to 7000 pounds. Problem 8 : Convert 0.5 tons into ounces. Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 0.5 tons = 0.5 x 2000 pounds 0.5 tons = 1000 pounds 0.5 tons = 1000 x 16 ounces 0.5 tons = 16000 ounces Hence, 0.5 tons is equal to 16000 ounces. Problem 9 : Convert 48 ounces into pounds. Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 48 ounces = 48 / 16 pounds 48 ounces = 3 pounds Hence, 48 ounces is equal to 3 pounds. Problem 10 : Convert 5000 pounds into tons Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 5000 pounds = 5000 / 2000 tons 5000 pounds = 2.5 tons Hence, 5000 pounds is equal to 2.5 tons. Problem 11 : Convert 2 pints into cups. Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 2 pints = 2 x 2 cups 2 pints = 4 cups Hence, 2 pints is equal to 4 cups. Problem 12 : Convert 3.5 quarts into cups . Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 3.5 quarts = 3.5 x 4 cups 3.5 quarts = 14 cups Hence, 3.5 quarts is equal to 14 cups. Problem 13 : Convert 32 cups into quarts. Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 32 cups = 32 / 4 quarts 32 cups = 8 quarts Hence, 32 cups is equal to 8 quarts. Problem 14 : Convert 256 cups into gallons. Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 256 cups = 256 / 16 gallons 256 cups = 16 gallons Hence, 256 cups is equal to 16 gallons. Problem 15 : Convert 24 quarts into gallons Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 24 quarts = 24 / 4 gallons 24 quarts = 6 gallons Hence, 24 quarts is equal to 6 gallons. Problem 16 : Convert 2 minutes into seconds. Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 2 minutes = 2 x 60 seconds 2 minutes = 120 seconds Hence, 2 minutes is equal to 60 seconds. Problem 17 : Convert 3.5 hours into minutes . Solution : Here, we convert bigger unit into smaller unit. So we have to multiply. 3.5 hours = 3.5 x 60 minutes 3.5 hours = 210 minutes Hence, 3.5 hours is equal to 210 minutes. Problem 18 : Convert 3 days into minutes. Solution : Here, we convert bigger unit into smaller unit unit. So we have to multiply. 3 days = 3 x 24 hours 3 days = 72 hours 3 days = 72 x 60 minutes 3 days = 4320 minutes Hence, 3 days is equal to 4320 minutes. Problem 19 : Convert 480 seconds into minutes. Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 480 seconds = 480 / 60 minutes 480 seconds = 8 minutes Hence, 480 seconds is equal to 8 minutes. Problem 20 : Convert 112 days into weeks. Solution : Here, we convert smaller unit into bigger unit. So we have to divide. 112 days = 112 / 7 weeks 112 days = 16 weeks Hence, 112 days is equal to 16 weeks. ## Converting customary units - Word problems Problem 1 : David travels 60 miles in two hours. How many yards of distance will he cover in one minute ? Solution : Distance covered in 2 hours = 60 miles Distance covered in 1 hour = 30 miles We know that 1 hour = 60 minutes and 1 mile = 1760 yards 1 hour -----> 30 miles =====> 60 minutes -----> 30 x 1760 yards 60 minutes -----> 52800 yards So, distance covered in 60 minutes = 52800 yards Distance covered in one minute = 52800 / 60 yards = 880 yards Hence 880 yards of distance covered in 1 minute. Let us look at the next word problem on "Converting customary units" Problem 2 : Mark jogged 15840 feet in 45 minutes. Find the speed of Mark in feet per minute. Solution : Speed = Distance / Time Speed = 15840 / 45 Speed = 352 feet per minute Hence the speed of Mark is 352 feet per minute. Let us look at the next word problem on "Converting customary units" Problem 3 : Use a fraction to find the length in feet of a shoe that is 9 inches long. Solution : Here, we convert smaller unit (inches) into bigger unit (foot). So we have to divide. Since we divide, we have to use the fraction 1/12. Because, 1 foot = 12 inches 9 inches = 9 x 1/12 ft 9 inches = 3/4 ft Hence 9 inches is equal to 3/4 ft. Let us look at the next word problem on "Converting customary units"" Problem 4 : Kevin has a new television that is 24 inches tall. If Kevin sets the television on a 3-foot-tall stand, how far from the floor will the top of the television be (in inches) ? Solution : Height of the television = 24 inches Height of the stand = 3 feet = 3 x 12 = 36 inches Distance from the floor to the top of the television is = Height of the stand + Height of the television = 36 + 24 = 60 inches Hence the top of the television is 60 inches far from the floor. Let us look at the next word problem on "Converting customary units" Problem 5 : Becky and Keith each ran for exactly 20 minutes on a treadmill. Keith’s treadmill said he had run 10,000 feet. Becky’s treadmill said she had run 2 miles. Who ran farther, and how much farther? Solution : Both Becky and Keith took the same amount of time. That is 20 minutes. Distance covered by Becky and Keith are given in different units. (Miles and Feet) We have to make the units to be same. Let us convert miles in to feet. So, 2 miles = 2 x 5280 feet = 10560 feet Therefore, Distance covered by Becky = 10560 feet -----(1) Distance covered by Keith = 10000 feet -----(2) Difference between (1) and (2) ----> 10560 - 10000 = 560 feet Hence Becky ran farther by 560 feet. Let us look at the next word problem on "Converting customary units" Problem 6 : David prepares 24 pounds of metal in 1 hour 36 minutes. At the same rate, How many ounces of metal will he prepare in one minute ? Solution : 1 hour 36 minutes = 60 min + 36 min = 96 minutes 1 pound = 16 ounces 24 pounds = 24 x 16 ounces = 384 ounces 1 hour 36 min -----> 24 pounds ====> 96 minutes ----> 384 pounds So, no. of pounds prepared in 96 minutes = 384 ounces No. of ounces prepared in in one minute = 384 / 96 = 4 Hence 4 ounces of metal is prepared in one minute. Let us look at the next word problem on "Converting customary units" Problem 7 : Mark used 15840 ounces of metal to make an alloy in 45 minutes. Find the amount metal used in one minute (in ounces). Solution : No. of ounces used in 45 minutes = 15840 No. of ounces used in 1 minute = 15840 / 45 No. of ounces used in 1 minute = 352 Hence 352 ounces of metal used in 1 minute. Let us look at the next word problem on "Converting customary units" Problem 8 : Mrs. Moore handed out 4 ounces of almonds to each of her 22 students . How many pounds of almonds did Mrs. Moore hand out? Solution : Total no. of ounces of almonds handed out = 4 x 22 = 88 ounces Total no. of pounds of almonds handed out = 88 / 16 = 5.5 lb Hence, Mrs. Moore handed out 5.5 pounds of almonds. Let us look at the next word problem on "Converting customary units" Problem 9 : Tommy uses 4 ounces of cheese in each pizza he makes. How many pounds of cheese does Tommy need to make 28 pizzas ? Solution : 1 pizza -----> 4 ounces of cheese 28 pizzas -----> 28 x 4 ounces of cheese 28 pizzas ------> 112 ounces of cheese Hence Tommy needs 112 ounces of cheese to make 28 pizzas. Let us look at the next word problem on "Converting customary units" Problem 10 : A standard elevator in a mid rise building can hold a maximum weight of about 1.5 tons. Assuming an average adult weight of 150 pounds, what is the maximum number of adults who could safely ride the elevator ? Solution : First let us convert 1.5 tons into pounds 1.5 tons = 1.5 x 2000 = 3000 pounds So, the elevator can hold 3000 pounds of weight. If the average weight of an adult is 150 pounds, Maximum no. of adults could safely ride the elevator is = 3000/150 = 20 Hence, maximum number of adults who could safely ride the elevator is 20. Let us look at the next word problem on "Converting customary units" Problem 11 : David prepares 60 pints of juice in two hours. At the same rate, How many cups of juice will he prepare in one minute ? Solution : No. of pints prepared in 2 hours = 60 No. of pints prepared in 1 hour = 30 We know that 1 hour = 60 minutes and 1 pint = 2 cups 1 hour -----> 30 pints =====> 60 minutes -----> 30 x 2 cups 60 minutes -----> 60 cups So, no. of cups prepared in 60 minutes = 60 No. of cups prepared in in one minute = 60 / 60 = 1 cup Hence 1 cup of juice is prepared in 1 minute. Let us look at the next word problem on "Converting customary units" Problem 12 : Mark used 15840 cups of fuel in 45 minutes. Find the amount fuel used in one minute (in cups). Solution : No. of cups used in 45 minutes = 15840 No. of cups used in 1 minute = 15840 / 45 No. of cups used in 1 minute = 352 Hence 352 cups of fuel used in 1 minute. Let us look at the next word problem on "Converting customary units Problem 13 : Kemka's little sister needs to take a bubble bath. The package says to put in a drop of bubble bath for every half gallon of water in the bath tub. If bathtub has 12 gallons of water, how many drops can she put into the bath for her sister? Solution : Half gallon of water -------> 1 drop of bubble bath 1 gallon of water -------> 2 drops of bubble bath 12 gallons of water -------> 12 x 2 drops of bubble bath 12 gallons of water -------> 24 drops of bubble bath Hence, Kemka can put into 24 drops of bubble bath for her sister with 12 gallons of water. Let us look at the next word problem on "Converting customary units" Problem 14 : Ivan needs gas for his truck. He knows his truck holds 40 gallons of gas. If he is allowed to fill up 8 quarts of gas once in a time, how many times will he have to fill up his gas can to get his truck full of gas ? Solution : 1 gallon = 4 quarts 40 gallons = 40 x 4 quarts = 160 quarts So, he needs 160 quarts of gas to make his truck full of gas. Once in a time, he can fill up 8 quarts of gas. No. of times of filling to make the truck full of gas is = 160 / 8 = 20 Hence Ivan has to fill up his gas can 20 times to get his truck full of gas. Let us look at the next word problem on "Converting customary units" Problem 15 : A bath hols 83 gallons and a shower uses 34 gallons.Mrs. Hitchins has a bath. How much water will be saved if Mrs. Hitchins decides to have a shower ? Solution : No. of gallons used when Mrs. Hitchins has a bath = 83 ----(1) No. of gallons used when Mrs. Hitchins has a shower = 34 ----(2) Water saved = Dbetween (1) and (2) Water saved = 83 - 34 Water saved = 49 Hence, 49 gallons water will be saved if Mrs. Hitchins decides to have a shower. Let us look at the next word problem on "Converting customary units" Problem 16 : Mrs. Moore took 4 hours 30 minutes to complete a work. How many seconds will Mrs. Moore take to complete the same work ? Solution : 4 hours 30 minutes = 4x60 min + 30 min 4 hours 30 minutes = 240 min + 30 min 4 hours 30 minutes = 270 minutes 4 hours 30 minutes = 270 x 60 seconds 4 hours 30 minutes = 16200 seconds Hence, Mrs. Moore will take 16200 seconds to complete the same work. Let us look at the next word problem on "Converting customary units" Problem 17 : Tommy takes 10 minutes time for each pizza he makes. How many seconds will he take to make 4 pizzas ? Solution : 1 pizza -----> 10 minues 4 pizzas -----> 4 x 10 minutes 4 pizzas ------> 40 minutes 4 pizzas ------> 40 x 60 seconds 4 pizzas ------> 2400 seconds Hence Tommy will take 2400 seconds to make 4 pizzas. Let us look at the next word problem on "Converting customary units" Problem 18 : A piece of work can be done by Mr. David in 9 days working 10 hours per day. How many hours will be taken by Mr. David to complete another work which is 4 times the first one ? Solution : Time needed to complete the given work = 9 x 10 hours = 90 hours Time needed to complete another work which is 4 times the first first work is = 4 x 90 hours = 360 hours Hence, the required time is 360 hours. Let us look at the next word problem on "Converting customary units" Problem 19 : Jose needs 6 hours to complete a work. But Jacob need 3/4 of time taken by Jose to complete the same work. In how many minutes will Jose complete the work ? Solution : Time required for Jose = 6 hours Time required for Jacob = 3/4 of time taken Jose Time required for Jacob = (3/4) x 6 hours Time required for Jacob = (3/4) x 6 x 60 minutes Time required for Jacob = 270 minutes Hence, Jacob needs 270 minutes to complete the work. Let us look at the next word problem on "Converting customary units" Problem 20 : Who is driving faster, Lenin covers 6 miles in 2 minutes or Daniel covers 225 miles in 1.5 hours ? Solution : To compare the given measures, convert them in to unit rates in distance per hour. LeninDistance in 2 min = 6 miles Distance in 1 min = 3 miles 1 hour = 60 minutesDistance in 1hr = 60x3Distance in 1 hr = 180 miles DanielDistance in 1.5 hrs =225 miles Distance in 1 hr = 225 / 1.5 Distance in 1 hr = 150 miles From the above unit rates, Lenin covers more miles than Daniel per hour. Hence, Lenin is driving faster. After having gone through the problems explained above, we hope that the students would have understood the stuff given on "Converting customary units". If you need any other stuff in math, please use our google custom search here. 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# CBSE Class 7 Maths Revision Notes Chapter 4 ## CBSE Class 7 Mathematics Revision Notes Chapter 4 – Simple Equations The Class 7 Mathematics Chapter 4 Notes are designed for students preparing for the final examinations. Students should focus on subjects such as mathematics from the start, as they are considered building blocks for higher classes. Simple language with a proper explanation of each aspect is given in the CBSE revision notes. Candidates can rely on these notes as they are created according to the latest CBSE guidelines and NCERT books. Extramarks recommends that students read the textbook properly and solve the numerical questions to score well in the examination. In addition to the revision notes, one can access CBSE sample papers, CBSE previous year question papers, and CBSE extra questions, including many more from the official website Extramarks. ### Simple Equations A simple equation is a combination of variables, constants, and mathematical operations, such as division, multiplication, subtraction, and addition, that are balanced by an equal sign. The left side of the equation is known as the LHS, and the right side of the equation is called the RHS. Any letter from ‘A to Z’ can represent a variable in any form z. For example, x + 3 = 8, so the variable in this equation is x. The general form of a simple equation can be written as ax + b = c. Variables: A variable is any quantity, number or characteristic that is measurable and can be counted. A data item can also be termed a variable. Different examples of variables are vehicle type, eye colour, class grades, capital expenditure, place of birth, business expenditure, business income, sex, age, etc. Constant: The values will remain constant throughout the equation. We can also say that a symbol with a fixed numeric value is known as a constant. Equal to sign: An equal indicates a balanced status between the left and right sides of the equation. ### Solving Simple Equations Most of the equations to be solved require a simple rearrangement of the variables and constants to get the solution. To get a better understanding of the concept of rearrangement, we can refer to the simple steps given below. They are as follows: • To start with solving, one should always try to balance the equation using any of the mathematical operations. You should focus on preserving the relationship between the left and right-hand sides. • If you are doing simultaneous operations on both sides of the equations, then your equation will be correct. To find an accurate solution, you should always try to balance the equations properly. • You can also solve simple equations using the BODMAS rule. While solving the equation, try to remember the required formulas and do the calculations accurately to get the desired solutions. • Always try to simplify the equations properly through multiplication, division and addition or subtraction of the terms. ### Advantages of Revision Notes of CBSE Class 7 Chapter 4 One can get the following advantages from the Class 7 Chapter 4 Mathematics Notes. • Candidates will be able to revise the important questions and formulas accurately. • Students can have a quick revision of the important concepts which are essential from the examination point of view. • The CBSE revision notes are extremely important as they can help the students gain good marks in the final examination. Proper revision of the notes will help the students analyse their mistakes and rectify them properly. • All the topics are systemically covered in the Chapter 4 Mathematics Class 7 Notes, so candidates can rely on them for exam preparation. ### Conclusion The Class 7 Chapter 4 Mathematics Notes include all the important questions and topics required to score well in the examination. The revision notes are created using an easy-to-use language that will help the candidates understand the concepts. One should revise all the formulas before the examination, especially in subjects like mathematics. Hence, the important formulas and concepts related to it are accurately covered in the CBSE revision notes. The notes will provide the students with a summary that can be read before the examination for quick revision. ### 1. Is Chapter 4 Mathematics Class 7 Notes difficult to understand? Candidates should not worry about the difficulty level as easy-to-use language is used to prepare the revision notes for the students. Also, all the topics are covered which will help students in revising properly. ### 2. How many questions are there in chapter 4 of Mathematics? In the NCERT books, there are a total of four exercises and eighteen questions. A detailed overview is given below Exercise 4.1: Six questions with subdivisions in each of them. Exercise 4.2: Four questions with subdivisions in each question. Exercise 4.3: Four questions in total. Exercise 4.4: In total, there are four questions with further subdivisions. So, in total there are 18 questions.
# AP Physics 1 : Motion in Two Dimensions ## Example Questions ### Example Question #11 : Motion An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent? 140m 420m 70m 280m 140m Explanation: The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use . Next we find how long it takes to reach the top of its trajectory using . t = 5.3s Finally, find how high the object goes with . ### Example Question #1 : Calculating Motion In Two Dimensions An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land? 62.5m 675m 1350m 250m 1350m Explanation: First, find the horizontal (x) and vertical (y) components of the velocity Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number. t = 6.25s Total time in the air is therefore 12.5s (twice this value). Finally, find distance traveled my multiplying horizontal velocity and time. ### Example Question #2 : Calculating Motion In Two Dimensions A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released. What is the velocity of the box just before it hits the ground? Explanation: We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp. Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation . We can plug these values into the following distance equation and solve for time. Now that we know the acceleration on the box and the time of travel, we can use the equation to solve for the velocity. ### Example Question #1 : Motion In Two Dimensions A ball gets pushed off a high table with a horizontal speed of . How far does the ball travel horizontally before hitting the ground? Explanation: This is a two step problem. The first step is to calculate the time it takes for the ball to reach the ground. To find this time, we use the following kinematic equation dealing with vertical motion. Choosing the ground to be the zero height, we have and . Also, knowing that the initial vertical velocity is zero, we know that . The kinematic equation simplifies using these values. Rearrange the equation to isolate time. We know that is the acceleration due to gravity: . Plug in the values to solve for time. We now have the time the ball is travelling before it hits the ground. Use this value to find the horizontal distance before it hits the ground with the kinematic equation . We know that and that . Using these values and the time, we can solve for the horizontal distance travelled. ### Example Question #1 : Calculating Motion In Two Dimensions A car drives north at for , then turns east and drives at for . What is the magnitude and direction of the average velocity for the trip? Explanation: First, determine how far the car travels in each direction: Now that we have the directional displacements, we can find the total displacement by using the Pythagorean Theorem. Find the average velocity by dividing the total displacement by the total time. Velocity is a vector, meaning it has both magnitude and direction. Now that we have the magnitude, we can find the direction by using trigonometry. Use the north and the east directional displacements to find the angle. Our final answer will be: ### Example Question #1 : Motion In Two Dimensions A ball is launched at an angle of above the horizontal with an initial velocity of . At what time is its vertical velocity ? Explanation: Any projectile has a vertical velocity of zero at the peak of its flight. To solve this question, we need to find the time that it takes the ball to reach this height. The easiest way is to solve for the initial vertical velocity using trigonometry, and then use the appropriate kinematics equation to determine the time. We know that the final vertical velocity will be zero. We can solve for the initial vertical velocity using the given angle and total velocity. Using this in our kinematics formula, we solve for the time. Keep in mind that this is only the vertical velocity. The total velocity at the peak is not zero, since the ball will still have horizontal velocity. ### Example Question #21 : Motion A 2kg box is at the top of a frictionless ramp at an angle of . The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released. When the box is released, how long will it take the box to reach the ground? Explanation: We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation . Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of . Since the angle is 60o, the acceleration on the box is Finally, we can plug these values into the following distance equation and solve for time. ### Example Question #1 : Motion In Two Dimensions A plane is traveling from Portland to Seattle, which is 100 miles due north of Portland. There is a constant wind traveling southeast at 30 mph. If the plane needs to get to Seattle in one hour while flying due north, at what speed (relative to the wind) and angle should the pilot fly? None of the other answers Explanation: First we need to find out at what the speed relative to the ground the plane needs to fly. The plane needs to cover 100 miles in 1 hour, so it's simply 100 mph due north. Now we need to calculate its speed relative to the wind and its angle. We know the sum of the wind vector and plane vector needs to equal 100 mph due north. Therefore, all east/west movement must cancel out and all north/south movement must add to 100mph. We can separate the wind velocity into its components: We can also represent the components of the plane's velocity: If the trig functions seem reversed, this is because the angle in question is between the y-coordinate and a vector pointing slightly west of north. We can also represent the components of the velocity relative to the ground: Since the horizontal velocity is equal to 0, we can set the x-components of the wind and plane vectors equal to each other: The sum of the y-components of the wind and plane vectors must equal 100: The wind vector is subtracted because it is in the opposite direction of the plane vector. Now we just need to isolate a variable and substitute one equation into another. We will isolate the total plane velocity in the first equation: Substituting this into the second equation, we get: Solving for theta, we get (west of north) We can plug this into the first equation to get: ### Example Question #9 : Motion In Two Dimensions A baseball is traveling with a velocity of at an angle of above horizontal. What is the velocity of the ball after two seconds? Explanation: To solve this problem, we first need to split the velocity into its vector components. Initial vertical velocity: Initial horizontal velocity: Since we are neglecting air resistance, horizontal velocity does not change over time. We only need to calculate the new vertical velocity after two seconds, using kinematics: The negative sign simply means that the vertical velocity has changed direction, and is now pointed downward. We can use the following equation to determine the total velocity, which will be the sum of the horizontal and vertical velocity vectors: ### Example Question #1 : Motion In Two Dimensions Suppose that a golf ball is struck such that it travels at a speed of at an angle to the horizontal. Neglecting air resistance, how long will the golf ball remain in the air before it touches the ground again? Explanation: We're told that the golf ball is starting its parabolic journey with a certain velocity at an angle to the ground. To solve for the time the golf ball stays airborne, we'll need to consider the x and y-components of the ball's trajectory. First, we'll need an expression that considers the ball's velocity in the x-direction. We'll also need an expression for the y-component of the velocity. We'll need an expression that can relate the vertical distance traveled with the time spent in the air. Since the only acceleration occurring in this scenario is due to gravity, the result is that acceleration is constant. Therefore, we can make use of some of the kinematics equations: It's also important to note that once the ball lands back on the ground, only its horizontal displacement will have changed, while its vertical displacement will remain unchanged. We also have to remember that in this case, the source of acceleration is from gravity, which points downwards. If we define up as the positive y direction, then down must be the negative y direction. Therefore, we can write: Plug in the vertical component of velocity and solve for time.
# Hearing Perturbation Theory In numerical linear algebra, we create ways for a computer to solve a linear system of equations $A\vec{x} = \vec{b}$. In doing so, we analyze how efficiently and accurately we can find the solution $\vec{x}$. Perturbation theory concerns how much error we incur in the solution $\vec{x}$ when we perturb (spoil) the data $A$ and $\vec{b}$. A classic statement tells us that the amount of error depends on the condition number of the matrix $A$. I will define and prove the statement, and help you understand it by “hearing” it. # 1. Problem statement Let’s consider the equation, $A\vec{x}_{true} = \vec{b}$, where $A \in \mathbb{R}^{n \times n}$ is nonsingular and $\vec{b} \neq \vec{0}$ (otherwise, we get the trivial case $\vec{x}_{true} = \vec{0}$). Note that $\vec{x}_{true}$ denotes the true solution. It’s what we hope to get. Suppose, instead, we solve the perturbed system, $(A + \delta\!A)\vec{x} = \vec{b} + \vec{\delta b}$. We seek to find a bound on the error $\vec{x} - \vec{x}_{true}$. We expect the error to be small when $\delta\!A$ and $\vec{\delta b}$ are small changes. Given an induced matrix norm $\|\|\cdot\|\|$, we can define the condition number of $A$: $\kappa(A) := \|\|A\|\| \cdot \|\|A^{-1}\|\|$. Note that we always have $\kappa(A) \geq 1$. You can check why, along with the definition and properties of an induced matrix norm, in Notes. We will show that, under a mild assumption $\displaystyle\kappa(A) \cdot \frac{\|\|\delta\!A\|\|}{\|\|A\|\|} < 1$, $\boxed{\displaystyle\frac{\|\|\vec{x} - \vec{x}_{true}\|\|}{\|\|\vec{x}_{true}\|\|} \leq \frac{\kappa(A)}{1 - \kappa(A)\frac{\|\|\delta\!A\|\|}{\|\|A\|\|}} \,\cdot \left(\frac{\|\|\delta\!A\|\|}{\|\|A\|\|} + \frac{\|\|\vec{\delta b}\|\|}{\|\|\vec{b}\|\|}\right)}$. The RHS is a product of terms, so we can conclude two things. (1) If $\kappa(A)$ is a small number, then small relative errors in $A$ and $\vec{b}$ will result in a small relative error in the solution $\vec{x}_{true}$. (2) If $\kappa(A)$ is a large number, however, we may get a large relative error in $\vec{x}_{true}$ even when the relative errors in $A$ and $\vec{b}$ are small. We see that the condition number influences how much change occurs in the output when we change the input of a system. As a result, we say that the matrix $A$ is well-conditioned if $\kappa(A)$ is small, and ill-conditioned if $\kappa(A)$ is large. # 2. Mathematical proof ## a. Step 1 For any matrix $X \in \mathbb{R}^{n \times n}$ with $\|\|X\|\| < 1$, the following statements hold: (i) $I - X$ is nonsingular, and its inverse is given by $\displaystyle(I - X)^{-1} = \sum_{i\,=\,0}^{\infty}\,X^{i} = I + X + X^{2} + \cdots$. (ii) $\displaystyle\|\|(I - X)^{-1}\|\| \leq \frac{1}{1 - \|\|X\|\|}$. To prove (i), we check that $(I - X)(I - X)^{-1} = I$ and $(I - X)^{-1}(I - X) = I$. We just need to check one of them because, in finite dimensions, injectivity and surjectivity occur together. We see that, \begin{aligned} (I - X)(I - X)^{-1} &= (I - X)(I + X + X^{2} + \cdots) \\[12pt] &= (I + X + X^{2} + \cdots) - (X + X^{2} + X^{3} + \cdots) \\[12pt] &= I. \end{aligned} Note, to be rigorous, we would first show that the infinite series $\sum_{i\,=\,0}^{\infty}\,X^{i}$ converges. The sequence of finite sums, $\left\{\sum_{i\,=\,0}^{k}\,X^{i}\right\}_{k\,=\,0}^{\infty}$, is a Cauchy sequence in the norm $\|\|\cdot\|\|$. Since $\mathbb{R}^{n \times n}$ is a Banach space, i.e. complete, the sequence converges. Let’s prove (ii). By triangle inequality and the properties of an induced matrix norm, \begin{aligned} \|\|(I - X)^{-1}\|\| &= \|\|I + X + X^{2} + \cdots\|\| \\[12pt] &\leq \|\|I\|\| + \|\|X\|\| + \|\|X^{2}\|\| + \cdots \\[12pt] &\leq \|\|I\|\| + \|\|X\|\| + \|\|X\|\|^{2} + \cdots. \end{aligned} Next, we use the infinite geometric sum formula: For any real number $r \in (-1,\,1)$, $\displaystyle\sum_{i\,=\,0}^{\infty}\,r^{i} = \frac{1}{1 - r}$. Since $\|\|X\|\| \in [0,\,1)$, we conclude that, $\displaystyle\|\|(I - X)^{-1}\|\| \leq \frac{1}{1 - \|\|X\|\|}$. ## b. Step 2 Recall that $A$ is nonsingular. Assume further that, $\displaystyle\kappa(A) \cdot \frac{\|\|\delta\!A\|\|}{\|\|A\|\|} < 1$. Then, $A + \delta\!A$ is also nonsingular. Since $A$ is nonsingular, we can write $A + \delta\!A$ as a product of two terms: $A + \delta\!A = A(I + A^{-1}\delta\!A)$. Hence, $A + \delta\!A$ is nonsingular, if and only if, $I + A^{-1}\delta\!A$ is nonsingular. Let’s show that the latter is true by applying (i) from Step 1. We find that, \displaystyle\begin{aligned} \|\|A^{-1}\delta\!A\|\| &\leq \|\|A^{-1}\|\| \cdot \|\|\delta\!A\|\| \\[12pt] &= \kappa(A)\frac{\|\|\delta\!A\|\|}{\|\|A\|\|} \\[12pt] &< 1. \end{aligned} Hence, $I + A^{-1}\delta\!A$ is nonsingular. Before we move on, let’s read the statement in Step 2 again. It tells us that, if a matrix is nonsingular, then there are infinitely many matrices nearby that are nonsingular, too. (And infinitely many around them, and so on.) This supports the fact that there are far more nonsingular matrices than singular ones. Isn’t that a marvel? ## c. Step 3 Show that, $\displaystyle \frac{\|\|\vec{x} - \vec{x}_{true}\|\|}{\|\|\vec{x}_{true}\|\|} \leq \|\|(A + \delta\!A)^{-1}\|\| \,\cdot \left(\|\|\delta\!A\|\| + \frac{\|\|\vec{\delta b}\|\|}{\|\|\vec{x}_{true}\|\|}\right)$. We’re almost there! Recall the original problems: $\begin{array}{rcl} A\vec{x}_{true} & = & \vec{b} \\[12pt] (A + \delta\!A)\vec{x} & = & \vec{b} + \vec{\delta b}. \end{array}$ Subtract the two equations to get, $\begin{array}{ll} & (A + \delta\!A)\vec{x} - A\vec{x}_{true} = \vec{\delta b} \\[12pt] \Rightarrow & (A + \delta\!A)(\vec{x} - \vec{x}_{true}) = \vec{\delta b} - \delta\!A\vec{x}_{true} \\[12pt] \Rightarrow & \vec{x} - \vec{x}_{true} = (A + \delta\!A)^{-1}(\vec{b} - \delta\!A\vec{x}_{true}). \end{array}$ Take the vector norm to both sides: \begin{aligned} \|\|\vec{x} - \vec{x}_{true}\|\| &= \|\|(A + \delta\!A)^{-1}(\vec{b} - \delta\!A\vec{x}_{true})\|\| \\[12pt] &\leq \|\|(A + \delta\!A)^{-1}\|\| \cdot \|\|\vec{b} - \delta\!A\vec{x}_{true}\|\| \\[12pt] & \leq \|\|(A + \delta\!A)^{-1}\|\| \cdot \Bigl(\|\|\vec{b}\|\| + \|\|\delta\!A\|\| \cdot \|\|\vec{x}_{true}\|\|\Bigr). \end{aligned} Finally, we divide both sides by $\|\|\vec{x}_{true}\|\|\,(\neq 0)$. ## d. Step 4 Finally, prove that, $\displaystyle\frac{\|\|\vec{x} - \vec{x}_{true}\|\|}{\|\|\vec{x}_{true}\|\|} \leq \frac{\kappa(A)}{1 - \kappa(A)\frac{\|\|\delta\!A\|\|}{\|\|A\|\|}} \,\cdot \left(\frac{\|\|\delta\!A\|\|}{\|\|A\|\|} + \frac{\|\|\vec{\delta b}\|\|}{\|\|\vec{b}\|\|}\right)$. Let’s find an upper bound for $\|\|(A + \delta\!A)^{-1}\|\|$ in Step 3. We use (ii) from Step 1: \begin{aligned} \|\|(A + \delta\!A)^{-1}\|\| &\leq \|\|(I + A^{-1}\delta\!A)^{-1}\|\| \cdot \|\|A^{-1}\|\| \\[12pt] &\leq \frac{\|\|A^{-1}\|\|}{1 - \|\|A^{-1}\delta\!A\|\|} \\[12pt] &\leq \frac{\|\|A^{-1}\|\|}{1 - \kappa(A)\frac{\|\|\delta\!A\|\|}{\|\|A\|\|}}. \end{aligned} Hence, \begin{aligned} \frac{\|\|\vec{x} - \vec{x}_{true}\|\|}{\|\|\vec{x}_{true}\|\|} &\leq \frac{\|\|A\|\| \cdot \|\|A^{-1}\|\|}{1 - \kappa(A)\frac{\|\|\delta\!A\|\|}{\|\|A\|\|}} \,\cdot \left(\frac{\|\|\delta\!A\|\|}{\|\|A\|\|} + \frac{\|\|\vec{\delta b}\|\|}{\|\|A\|\| \cdot \|\|\vec{x}_{true}\|\|}\right) \\[12pt] &\leq \frac{\kappa(A)}{1 - \kappa(A)\frac{\|\|\delta\!A\|\|}{\|\|A\|\|}} \,\cdot \left(\frac{\|\|\delta\!A\|\|}{\|\|A\|\|} + \frac{\|\|\vec{\delta b}\|\|}{\|\|\vec{b}\|\|}\right). \end{aligned} # 3. Application We can represent an audio as a vector $\vec{x}_{true} \in \mathbb{R}^{n}$ of frequencies. The provided audio file, williams.wav, lends to $n = 189930$. At a sampling rate of 44100 Hz, the audio is 4.3068 seconds long. For computational efficiency, we will write $\vec{x}_{true}$ as a matrix $X_{true} \in \mathbb{R}^{10 \times 18993}$, by storing the first 10 entries of $\vec{x}_{true}$ as the first column of $X_{true}$, the next 10 entries as the second, and so on. Next, we apply a nonsingular matrix $A \in \mathbb{R}^{10 \times 10}$ to $X_{true}$ to get the encrypted audio $B \in \mathbb{R}^{10 \times 18993}$. We can always vectorize $B$ into $\vec{b} \in \mathbb{R}^{189930}$ by reversing the storage process described above. From now on, we assume that we only have $A$ and $B$. We seek to decrypt the audio, i.e. solve the equation $AX_{true} = B$, when there are perturbations in the matrix $A$ and/or in the input $B$: $(A + \delta\!A)X = (B + \delta\!B)$. ## a. Generating matrices To generate a well-conditioned $A$, we find the QR factorization of a random $10 \times 10$ matrix and set $A = Q$. We know that the 2-norm of an orthogonal matrix is always 1. Hence, $\kappa_{2}(Q) = \|\|Q\|\|_{2} \cdot \|\|Q^{T}\|\|_{2} = 1$. For an ill-conditioned $A$, we use the Hilbert matrix $H$: $H = \left[\begin{array}{ccccc} 1 & \frac{1}{2} & \cdots & \frac{1}{9} & \frac{1}{10} \\[12pt] \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{10} & \frac{1}{11} \\[12pt] \vdots & \vdots & \ddots & \vdots & \vdots \\[12pt] \frac{1}{9} & \frac{1}{10} & \cdots & \frac{1}{17} & \frac{1}{18} \\[12pt] \frac{1}{10} & \frac{1}{11} & \cdots & \frac{1}{18} & \frac{1}{19} \end{array}\right]$. Then, $\kappa_{2}(H) \approx 1.6025 \times 10^{13}$. Note that both $Q$ and $H$ are nonsingular. ## b. Generating perturbations The entries of perturbations $\delta\!A$ and $\delta\!B$ are randomly generated from a scaled normal distribution. For convenience, we will assume that $A + \delta\!A$ is (most likely) nonsingular because $\delta\!A$ is randomly chosen. We are more interested in ensuring that the values of $\|\|\delta\!A\|\|_{2}$ and $\|\|\vec{\delta b}\|\|_{2}$ stay about the same each time we run the program. ## c. Simulations We can run perturbation_theory.m under 6 different cases: $\begin{tabular}{c\|c\|c\|c} & Perturb A only & Perturb B only & Perturb A and B \\[8pt]\hline Well-conditioned A & (1,\,1) & (1,\,2) & (1,\,3) \\[8pt]\hline Ill-conditioned A & (2,\,1) & (2,\,2) & (2,\,3) \end{tabular}$ The table lists the input parameters for each case. First, let’s consider what happens when $A$ is well-conditioned. This slideshow requires JavaScript. These plots show that the obtained solution $\vec{x}$ matches the true solution $\vec{x}_{true}$ well. The fidelity of the obtained audios remains largely pristine. We do hear some added noise. In the 100 ms sample, the relative error stays around 2% (median) when we perturb only the matrix $A$. When we perturb the encrypted audio $B$, the relative error jumps to about 10%. Download and listen to the obtained audios: • audio_obtained11.wav (well-conditioned $A$, perturbation to $A$ only) • audio_obtained12.wav (well-conditioned $A$, perturbation to $B$ only) • audio_obtained13.wav (well-conditioned $A$, perturbation to both $A$ and $B$) Next, let’s consider what happens when $A$ is ill-conditioned. This slideshow requires JavaScript. This time, the obtained solution hardly matches the true solution. You can hear much more noise in the obtained audios, and listening to them becomes almost unbearable. The relative error is orders of magnitude larger. It’s interesting how perturbing $B$ only results in a completely garbled audio, but perturbing $A$ in addition reconstructs some of the original audio. Two wrongs do make a right, it seems. Download and listen to the obtained audios: • audio_obtained21.wav (ill-conditioned $A$, perturbation to $A$ only) • audio_obtained22.wav (ill-conditioned $A$, perturbation to $B$ only) • audio_obtained23.wav (ill-conditioned $A$, perturbation to both $A$ and $B$) We can also check that our matrices satisfy the statement that we painstakingly proved. I will leave writing the extra code to you. # Notes Given a vector norm $\|\|\cdot\|\|$ for the vector space $\mathbb{R}^{n}$, we can always create a matrix norm $\|\|\cdot\|\|$ for the vector space $\mathbb{R}^{n \times n}$. As a result, we call this an “induced” matrix norm. For simplicity, I use the double bar notation for both types of norms. It is clear from context whether we are looking at the vector norm or the induced matrix norm. The induced matrix norm of $A \in \mathbb{R}^{n \times n}$ is defined as, $\displaystyle\|\|A\|\| := \max\limits_{\vec{x}\,\neq\,\vec{0}}\,\frac{\|\|A\vec{x}\|\|}{\|\|\vec{x}\|\|} = \max\limits_{\|\|\vec{x}\|\|\,=\,1}\,\|\|A\vec{x}\|\|$. Hence, the induced matrix norm measures how far out we can map a vector relative to its original size. We can also generalize the definition to rectangular matrices. Because of its definition, the induced matrix norm satisfies these useful properties: (i) For any $A \in \mathbb{R}^{n \times n}$ and $\vec{x} \in \mathbb{R}^{n}$, $\|\|A\vec{x}\|\| \leq \|\|A\|\| \cdot \|\|\vec{x}\|\|$. (ii) For any $A,\,B \in \mathbb{R}^{n \times n}$, $\|\|AB\|\| \leq \|\|A\|\| \cdot \|\|B\|\|$. In particular, property (ii) implies that the condition number of a nonsingular matrix is always at least 1. $AA^{-1} = I \,\,\Rightarrow\,\, \|\|AA^{-1}\|\| = 1 \,\,\Rightarrow\,\, \|\|A\|\| \cdot \|\|A^{-1}\|\| \geq 1$. You can find the code and audio files in their entirety here: Download from GitHub
# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### 3/5 + 7/10 = 13/10 = 1 3/10 = 1.3 Spelled result in words is thirteen tenths (or one and three tenths). ### How do you solve fractions step by step? 1. Add: 3/5 + 7/10 = 3 · 2/5 · 2 + 7/10 = 6/10 + 7/10 = 6 + 7/10 = 13/10 For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(5, 10) = 10. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 5 × 10 = 50. In the next intermediate step, the fraction result cannot be further simplified by canceling. In words - three fifths plus seven tenths = thirteen tenths. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Pears There were pears in the basket, I took two-fifths of them, and left six in the basket. How many pears did I take? • The tap For one day flows 148 l of water out of the tap. How much water will flow out for 3/4 day? • UN 1 If we add to an unknown number his quarter, we get 210. Identify unknown number. Shade the area on the grid that shows 5/8 x 2/4 • Fruits Amy bought a basket of fruits 1/5 of them were apples,1/4 were oranges, and the rest were 33 bananas. How many fruits did she buy in all? • Trees 3/5 trees are apples, cherries are 1/3. 5 trees are pear. How many is the total number of trees? • Number I think the number. If I add to its third seven I get same as when to its quarter add 8. Which is the number? • Expressions Let k represent an unknown number, express the following expressions: 1. The sum of the number n and two 2. The quotient of the numbers n and nine 3. Twice the number n 4. The difference between nine and the number n 5. Nine less than the number n • Vehicle tank A vehicle tank was 3/5 full of petrol. When 21 liters of fuel was added it was 5/6 full. How many liters of petrol can the tank hold? • Red diplomas The numbers of students with honors in 2013 and 2014 are in ratio 40:49. How big is the year-on-year percentage increase? • Calories 2 Ben eats approximately 2400 calories per day. His wife Sarah eats 5/8 as much. How many calories does Sarah eat per day? • The cube The cube has an edge of 12 dm. The second cube has an edge exactly 20% longer. How many % is more water in the second cube than in the first cube, if the first cube is full to 3/4 and the second to 3/8? • Excavation Excavation for the base of the cottage has dimensions 4.5 m x 3.24 m x 60 cm. The excavated soil will increase its volume by one-quarter. Calculate the volume of excavated soil.
# 11a.5. Ratio theorem Derivation of ratio theorem Let $A,B$ and $C$ be points with position vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ respectively. Let $C$ divide $AB$ internally such that $AC:CB = \lambda : \mu$. $\overrightarrow{AB} = \mathbf{b}-\mathbf{a}$ Since the ratio $AC:CB= \lambda:\mu$, the ratio $AC:AB = \lambda:\lambda + \mu$. Hence $\overrightarrow{AC} = \frac{\lambda}{\lambda+\mu} (\mathbf{b} - \mathbf{a})$. $\mathbf{c} - \mathbf{a} = \frac{\lambda}{\lambda+\mu} (\mathbf{b} - \mathbf{a})$. $\mathbf{c} = \frac{\lambda}{\lambda+\mu} (\mathbf{b} - \mathbf{a}) + \mathbf{a}$. $\displaystyle \mathbf{c} = \frac{\lambda \mathbf{b} - \lambda \mathbf{a} + \lambda \mathbf{a} + \mu \mathbf{a}}{\lambda+\mu}$ Hence ### Solution to examples Applying ratio theorem, $\overrightarrow{OP} = \frac{(-2\mathbf{i}+5\mathbf{k})+2(\mathbf{i}+2\mathbf{j}-3\mathbf{k})}{1+2} =$ Since $AB:AQ=2:5$, $B$ divides $AQ$ internally in the ratio $2:3$. $\overrightarrow{OB} = \frac{2 \overrightarrow{OQ} + 3 \overrightarrow{OA} }{2+3}$. $\overrightarrow{OQ} = \frac{5 \overrightarrow{OB} - 3 \overrightarrow{OA}}{2} = \frac{1}{2} (-13\mathbf{i}-6\mathbf{j}+34\mathbf{k})$. Hence coordinates of $Q$:
# What is a cubic polynomial function in standard form with zeros 3, -4, and 5? Feb 16, 2018 $y = {x}^{3} - 4 {x}^{2} - 17 x + 60$ #### Explanation: . The standard form of a cubic polynomial function is: $y = a {x}^{3} + b {x}^{2} + c x + d$ If we know the zeros as being $\alpha$, $\beta$, and $\theta$ then we know its factored form to be: $y = \left(x - \alpha\right) \left(x - \beta\right) \left(x - \theta\right)$ simply because you would have had to set the expression within each set of parentheses equal to $0$ and solve for a root. Therefore, we can write the polynomial in this problem as: $y = \left(x - 3\right) \left(x + 4\right) \left(x - 5\right)$ Now, if we multiply these through and remove the parentheses we will have the standard form of it: $y = \left(x - 3\right) \left({x}^{2} - x - 20\right)$ $y = {x}^{3} - {x}^{2} - 20 x - 3 {x}^{2} + 3 x + 60$ $y = {x}^{3} - 4 {x}^{2} - 17 x + 60$ Feb 16, 2018 ${x}^{3} - 4 {x}^{2} - 17 x + 60$ #### Explanation: There is only one cubic polynomial in standard form (i.e. monic) with three given distinct zeros, $a , b$ and $c$. It is given by the product: $\left(x - a\right) \left(x - b\right) \left(x - c\right)$ $\left(x - 3\right) \left(x + 4\right) \left(x - 5\right) = \left({x}^{2} + x - 12\right) \left(x - 5\right)$ $= {x}^{3} - 4 {x}^{2} - 17 x + 60$
of 24 /24 1 Sampling Sampling Distributions Distributions Presentation 2 • Sampling Distribution of sample proportion • Sampling Distribution of sample means • Author melvin-pope • Category ## Documents • view 222 3 TAGS: Embed Size (px) ### Transcript of 1 Sampling Distributions Presentation 2 Sampling Distribution of sample proportions Sampling... • Slide 1 • 1 Sampling Distributions Presentation 2 Sampling Distribution of sample proportions Sampling Distribution of sample means • Slide 2 • 2 Statistics VS Parameters Statistic is a numerical value computed from a sample. Parameter is a numerical value associated with a population. Essentially, we would like to know the parameter. But in most cases it is hard to know the parameter since the population is too large. So we have to estimate the parameter by some proper statistics computed from the sample. • Slide 3 • 3 Some Notation p = population proportion p = population proportion = sample proportion = sample proportion = population mean = population mean = sample mean = sample mean = standard deviation = standard deviation s = sample standard deviation s = sample standard deviation • Slide 4 • 4 A.Sampling Distribution of the Sample Proportion Situation 1: A survey is undertaken to determine the proportion of PSU students who engage in under-age drinking. The survey asks 200 random under-age students (assume no problems with bias). Suppose the true population proportion of those who drink is 60%. Thus, p = 0.6 and is the proportion in the sample who drink. Thus, p = 0.6 and is the proportion in the sample who drink. • Slide 5 • 5 Repeated Samples Imagine repeating this survey many times, and each time we record the sample proportion of those who have engaged in under-age drinking. What would the sampling distribution of look like? Sample (n=200) Sample Proportion Sample Proportion 1 1 2 2 3 3 4 4 5 5 150,000 150,000 150,000 is a random variable assigning a value to each sample! • Slide 6 • 6 Histogram of for 150 000 samples. • Slide 7 • 7 Sampling Distribution of Let X be the number of respondents who say they engage in under age drinking. Let X be the number of respondents who say they engage in under age drinking. X is binomial with n =200 and p =0.6. X is binomial with n =200 and p =0.6. So, we can calculate the probability of X for each possible outcome (0-200). The PDF is plotted below: So, we can calculate the probability of X for each possible outcome (0-200). The PDF is plotted below: • Slide 8 • 8 Sampling Distribution of Since X ~Bin (n =200, p =0.6), the sampling distribution of is the same as that of the binomial distribution divided by n. Since X ~Bin (n =200, p =0.6), the sampling distribution of is the same as that of the binomial distribution divided by n. Therefore we have Therefore we have • Slide 9 • 9 Sampling Distribution of - Cont. Using the Normal approximation to the binomial distribution we have that the sampling distribution of is approximately Normal with mean p and std. dev. Using the Normal approximation to the binomial distribution we have that the sampling distribution of is approximately Normal with mean p and std. dev. i.e. i.e. The conditions for this approximation to be valid are: The conditions for this approximation to be valid are: 1. The sample selected from the population is random. 2. The sample must be large enough, np and n(1-p) MUST be greater than 5, and should be greater than 10. • Slide 10 • 10 Example: Recent studies have shown that about 20% of American adults fit the medical definition of being obese. Recent studies have shown that about 20% of American adults fit the medical definition of being obese. A large medical clinic would like to estimate what percent of their patients are obese, so they take a random sample of 100 patients and find that 18 percent are obese. A large medical clinic would like to estimate what percent of their patients are obese, so they take a random sample of 100 patients and find that 18 percent are obese. Suppose in truth, the same percentage holds for the patients of the medical clinic as for the general population, 20%. Suppose in truth, the same percentage holds for the patients of the medical clinic as for the general population, 20%. Give notation and the numerical value for the following. Give notation and the numerical value for the following. • Slide 11 • 11 Problem - Cont. a.The population proportion of obese patients in the medical clinic: b.The proportion of obese patients in the sample of 100 patients: c.The mean of the sampling distribution of : d.The standard deviation of the sampling distribution of : e.The variance of the sampling distribution of : • Slide 12 • 12 B. Sampling Distribution of the Sample Mean Situation 2: The mean height of women age 20 to 30, X, is normally distributed (bell-shaped) with a mean of 65 inches and a standard deviation of 3 inches. i.e. X ~N(65,9) A random sample of 200 women was taken and the sample mean recorded. A random sample of 200 women was taken and the sample mean recorded. Now IMAGINE taking MANY samples of size 200 from the population of women. For each sample we record the. What is the sampling distribution of ? • Slide 13 • 13 Histograms for the Distribution of X and X -Bar Original Population of Women: X= height of random woman Distribution of Sample Means: X-bar = mean of random sample of size 200. • Slide 14 • 14 Normal Data Consider a Normal random variable X with mean and standard deviation , Consider a Normal random variable X with mean and standard deviation , X ~N( , 2 ). The sampling distribution of the sample mean of X for a sample of size n is Normal with The sampling distribution of the sample mean of X for a sample of size n is Normal withi.e. • Slide 15 • 15 Skewed or Non-Normal Data Situation 3: In a college survey, students were asked to report the number of cds they own. Clearly CDs is a right skewed data set. Suppose our population looked something like this, let us take repeated samples from this population and see what the sample mean looks like. • Slide 16 • 16 Suppose we take repeated samples of size n = 4, 8, 16, 32 n = 4 n = 32n = 16 n = 8 • Slide 17 • 17 Statistics From Skewed Data Using that CD sample as the population, Using that CD sample as the population, = 87.6, = 87.8 = 87.6, = 87.8 The sample means from the previous slide had the following summary statistics: The sample means from the previous slide had the following summary statistics: Sample Size Mean of X-bar Std. Dev. of X-bar n = 4 86.643.2 n = 8 86.830.9 n = 16 n = 1686.721.9 n = 32 n = 3286.615.6 Note: that the mean remains constant, and the std. deviation decreases as the sample size increases! • Slide 18 • 18 Central Limit Theorem For non-normal data coming from a population with mean and standard deviation the sampling distribution of the sample mean is approximately normal with For non-normal data coming from a population with mean and standard deviation the sampling distribution of the sample mean is approximately normal with Conditions: The above is true if the sample size is large enough, usually n > 30 is sufficient. • Slide 19 • 19 What next? We have shown that both the sampling distribution of the sample proportion, and the sampling distribution of the sample mean are both normal under certain conditions. We have shown that both the sampling distribution of the sample proportion, and the sampling distribution of the sample mean are both normal under certain conditions. Now we can use what we know about normal distributions to make conclusions about and ! Now we can use what we know about normal distributions to make conclusions about and ! In the following we will see how to use the values of the statistics (p-hat, x-bar) to make inferences about the parameters (p, ). In the following we will see how to use the values of the statistics (p-hat, x-bar) to make inferences about the parameters (p, ). • Slide 20 • 20 Exercise 1 The population proportion is 0.30. Consider the following questions. The population proportion is 0.30. Consider the following questions. 1. Find the sampling distribution of p-hat for each of the following sample sizes n=100, n=200, n=1000 2. What is the probability that a sample proportion will be within .04 of the population proportion for each of these sample sizes? 3. What is the advantage of larger sample size? • Slide 21 • 21 Exercise 2 A certain antibiotic in known to cure 85% of strep bacteria infections. A scientist wants to make sure the drug does not lose its potency over time. He treats 100 strep patients with a 1 year old supply of the antibiotic. Let be the proportion of individuals who are cured. A certain antibiotic in known to cure 85% of strep bacteria infections. A scientist wants to make sure the drug does not lose its potency over time. He treats 100 strep patients with a 1 year old supply of the antibiotic. Let be the proportion of individuals who are cured. ASSUME the drug has NOT lost potency, answer the following questions 1.What is the sampling distribution of ? Draw a picture 2.If we repeated this study many times we would expect 95% of to fall within what interval? 3.What is the probability that more than 90% in the sample are cured? • Slide 22 • 22 Exercise 3 A newspaper conducts a poll to determine the proportion of adults who favor a certain candidate. They ask a random sample of 800 people whether or not they favor that candidate (Assume no bias!). Suppose the true proportion of adults who favor the candidate is 58%. A newspaper conducts a poll to determine the proportion of adults who favor a certain candidate. They ask a random sample of 800 people whether or not they favor that candidate (Assume no bias!). Suppose the true proportion of adults who favor the candidate is 58%. 1. The newspaper records the sample proportion who favor the candidate. What is the sampling distribution of the sample proportion? Draw a picture of its PDF (center it correctly and include the appropriate scale). 2. What is the probability that the newspaper would have recorded a sample proportion greater than 62%? 3. What is the probability that less than 50% of the newspaper respondents would support this candidate? 4. What is the probability that a randomly selected individual favors this candidate? • Slide 23 • 23 Exercise 4 Suppose the number of calories FIT students consume in a day is normally distributed with mean 2000 and standard deviation 300. Suppose the number of calories FIT students consume in a day is normally distributed with mean 2000 and standard deviation 300. 1. About 95% of PSU students have a daily caloric intake between what two values? 2. What is the probability that a randomly selected individual consumed between 1800 and 2100 calories yesterday? 3. Suppose I take a random sample of 36 students and recorded the number of calories each consumed on a given day. Describe the sampling distribution of the sample mean. 4. Draw a picture of the sampling distribution of the sample mean (center it correctly and include the appropriate scale). 5. If I take a sample of size 36 from the student body, what is the probability that the sample mean will be less than 2050? • Slide 24 • 24 Exercise 5 Assume the length of trout living in the Susquehanna River is normally distributed with mean of 14 inches and standard deviation of 2 inches. A random sample of 16 trout is taken from the river. Assume the length of trout living in the Susquehanna River is normally distributed with mean of 14 inches and standard deviation of 2 inches. A random sample of 16 trout is taken from the river. 1. What is the sampling distribution of the average trout length (i) in a sample of size 16 (ii) in a sample of size 100? 2. What happens to the sampling distribution of the sample mean as the sample size increases? (Draw a picture) 3. What is the probability that a random sample of 16 trout will provide a sample mean within one in of the population mean? 4. What is the probability that a random sample of 100 trout will provide a sample mean within one in of the population mean? 5. What is the advantage of a larger sample size when one is attempting to estimate the population mean?
# Construction of Tangent to a Circle A tangent of a circle is a line that starts from outside the circle and intersects the plane of the circle at its periphery at one exact point. The point where a tangent intersects the circle is called the point of tangency (See Figure 1). Construction of Tangent is one of the most basic parts of geometry. Let’s study more. ## Tangents A common tangent can be any line, a ray or a segment that can be a tangent to more than one circle at the same time. There can be one to four number of common tangents to two circles. The point to be noted is that a tangent of a circle touches the circle but never enters it. Also, multiple tangents to the two circles can be an outer tangent or an inner one to each other’s circles. Figure 1: Circle with a centre O and tangent as PQ. X is the point of tangency Note: There can only be two tangents drawn from one point outside the circle. In this module, we’ll learn the art of constructing two tangents to a given circle from a point outside the circle without the help of a scale. ## Constructing Tangent of a Circle We can come across two cases while constructing tangent of a circle. These are described as follows: ### Case I: When the centre of a circle is known Let’s say that we are given a circle with centre O and a point P outside it, and we have to construct two tangents from the point P to the circle without the help of a ruler. Follow the steps of construction as below: 1. Join OP and bisect it. To bisect OP, take a compass, and open it slightly more than half of the length of the line segment. 2. From point P, mark a minimal arc above and below the line segment. Repeat the similar step from point O keeping the opening of the compass as same as it was from point P. Two points will be created where the two arcs, produced from point O and point P, meet. 3. Joint these two points with a line segment using a scale. This line segment bisects the OP. Let’s consider H as the mid-point of PO. 4. Taking the point H as a centre and HO as a radius, draw a circle. Let it intersect the given circle at the points, T and T’. 5. Join PT and PT’. Then PT and PT’ are the required two tangents. Figure 2: Drawing tangents PT and PT’ (marked in red) to the circle ### Proof and Rationale To check if the construction is correct: • Join OT. Then ∠PTO is an angle in the semicircle and, therefore, ∠PTO = 90° [ angle in a semicircle is always 90°] • Check with the 180° protractor ensuring that ∠PTO = 90°. Since OT is the radius of the circle with centre O and ∠PTO = 90°, therefore, PT has to be a tangent to the circle. This is because of the property of the circle which states that the radius from the centre of the circle to the point of tangency is perpendicular to the tangent line. Similarly, PT’ also becomes a tangent to the circle. ### Case II: When the centre of the given circle is not known In that case: 1. Draw any two non-parallel chords in the given circle 2. Draw perpendicular bisectors to both of the chords (as described in steps 2 and 3 above in Case I) (See Figure 3) 3. Find the point of intersection of two perpendicular bisectors which will be the centre of the given circle. 4. Follow the same steps from 1 to 5 as described above in Case I. Figure 3: Intersection of two perpendicular bisectors is the centre point O ## Solved Examples for You Question 1: Given a circle without a centre point and a point P outside the circle. Find the centre point of the circle and draw two tangents from the point P to the circle. Describe the steps. 1. Draw any two non-parallel chords (CD and EF) in the given circle 2. Draw perpendicular bisectors to both of the chords. 1. To bisect CD, take a compass, and open it slightly more than half of the length of the line segment. 2. From point D, mark a minimal arc above and below the line segment. Repeat the similar step from point C keeping the opening of the compass as same as it was from point D. Two points will be created where the two arcs, produced from point C and point D, meet. 3. Joint these two points with a line segment using a protractor. This line segment bisects the CD. 4. Repeat steps from 2 a to 2 c for drawing a perpendicular bisector for EF. 3. Find the point of intersection of two perpendicular bisectors which will be the centre of the given circle, say point O. (See Figure A) 4. Join the centre point O to the given point P outside the circle. 5. Bisect OP by following steps from 2 a to 2 c. Taking the point G as a centre and OG as a radius, draw a circle. Let it intersect the given circle at the points, T and T’. 6. Join PT and PT’. Then PT and PT’ are the required two tangents. Question 2: What does it mean if a line is tangent to a circle? Answer: A tangent line refers to a line that is intersecting a circle at one point. We call such a line the tangent to that circle. Moreover, the point at which the circle and the line intersect is referred to as the point of tangency. In other words, for any tangent line, there will be a perpendicular radius Question 3: How many tangents can a circle have? Answer: A circle has 0 tangents. We can draw exactly 0 tangents through a circle. A tangent will intersect a circle in only one point so that it does not enter the area of the circle. Moreover, we can draw an infinite number of tangents to a certain circle as there won’t be any limits to the points for a tangent to intersect. Question 4: Is a tangent line perpendicular? Answer: A tangent to a circle is a line that intersects the circle at precisely one point. It is the point of tangency or tangency point. A significant result is that the radius from the centre of the circle to the point of tangency will be perpendicular to the tangent line Question 5: What does tangent mean? Answer: In geometry, the tangent line or sometimes simply called the tangent, is a plane curve at a certain point is the straight line which “just touches” the curve at that point. Leibniz stated that it is the line through a pair of infinitely close points on the curve. Moreover, the word “tangent” is derived from the Latin ‘tangere’ which means ‘to touch’. Figure A: Tangents PT and PT’ from the given point P outside the given circle (highlighted) 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 Subscribe Notify of
2 2 9 9 7 6 # AGES SET-3 Q.1 Total age of some 7 years old & some 5 years old children is 60 yrs. If a team is to be selected from these children such that their total age is 48 years, in how many ways can it be done ? A) 2 B) 3 C) 4 D) 1 Ans. D 74+54=28+20 =48 Q.2 At present ages of a father & his son are in the ratio of 7:3 & they will be in the ratio of 2 : 1 after 10 years. Then the present age of father is : A) 77 B) 56 C) 70 D) 42 Ans. C Shortcut: Father’s age = xt (a-b)/ difference of cross product Or x =7 t=10 a=2 b=1 therefore, 710(2-1)/1 =70 Q.3 The present ages of A and B are in the ratio 2 : 3 and the ages of A and C are in the ratio 3 : 4. If the sum of the present ages of A, B and C is 92 years, how many years will it take by B and C to get their ages in the ratio 10 : 9? A) 2 B) 4 C) 5 D) 6 Ans. B A : B = 2 : 3 = 6 : 9 (Note: In this since A is common that’s why we have multiplied “2 with 3” & “3 with 9”) A : C = 3 : 4 = 6 : 8 (Multiplied “3 with 2” & “4 with 2”) ? A : B : C = 6 : 9 : 8 ? A age = × 92 = 24 years B age = × 92 = 36 years C = × 92 = 32 years After x years, = ? x = 4 years Q.4 The sum of ages of a son & father is 56 years. After 4 years, the age of father will be three times that of the son. Their ages respectively are A) 12 & 44 yrs B) 16 & 42 yrs C) 16 & 48 yrs D) 18 & 36 yrs Ans. A Shortcut: Son’s age=S1+t2(N-1)/N+1 [When age is in years ago] But in this question use this formula because [It is given after years] Son’s age= S1-t2(N-1)/N+1 Therefore, Son’s age is 56-4*2/4=12 where N= 3 as it is given three times Now, for father’s age= It is given in Question , the sum of ages of a son & father is 56 years Then,12 + father’s age = 56 Father’s age = 56-12= 44 Other explanation: Suppose the age of father be a yrs Then the age of the son = (56-a) After four yrs the age of the father would be (a+4) yrs & the age of the son would (56-a+4) Now (a+4) = 3(56-a+4) (a+4) = 180 – 3a a = 44 yrs So, the age of father = 44 yrs & the age of son = 56-44 = 12 yrs Q.5 The average age of three girls is 16 years. If their ages are in the ratio 4 : 5 : 7, the age of the youngest girl is : A) 21 B) 15 C) 16 D) 12 Ans. C Let the ages of three girls be 4x, 5x, 7x (ratio 4 : 5 : 7 is given) Average age of the three girls = 16 years Average age = = 16 16x = 16 3 x = 3. Age of the youngest girl = 4x = 4 3 = 12 years. Q.6 The ratio of the ages of two persons (A and B) is 2 : 3 and that of (B and C) is 4 : 5. If the average age of all the three is 35 years. What is B’s age? A) 24 B) 36 C) 40 D) 30 Ans. B Total age of A + B + C = 35 × 3 = 105 = ? A = = ? C = ? + B + = 105 ? B = 36 years.
# How many inches are in 1 2/5 groups of 1 2/3 2297961 Created Rating 0 ## 1 2/5 = 7/5 1 2/3 = 5/3 7/5 * 5/3 = 7/3 = 2 1/3 2297963 Created Rating 0 ## To find the number of inches in 1 2/5 groups of 1 2/3, we need to perform the following steps: Step 1: Convert 1 2/5 into an improper fraction. To convert 1 2/5 into an improper fraction, multiply the whole number (1) by the denominator (5) and add the numerator (2). This gives us (15 + 2)/5 = 7/5. Step 2: Convert 1 2/3 into an improper fraction. To convert 1 2/3 into an improper fraction, multiply the whole number (1) by the denominator (3) and add the numerator (2). This gives us (13 + 2)/3 = 5/3. Step 3: Multiply the two fractions. To multiply fractions, multiply the numerators together and multiply the denominators together. This gives us (7/5) * (5/3) = (75) / (53) = 35/15. Step 4: Simplify the fraction, if possible. To simplify the fraction, find the greatest common divisor (GCD) of the numerator and denominator, and divide both by it. In this case, the GCD of 35 and 15 is 5. Dividing both by 5 gives us 7/3. So, there are 7/3 inches in 1 2/5 groups of 1 2/3. 3438569 Created Rating 0 ## To determine how many inches are in 1 2/5 groups of 1 2/3, we need to perform some mathematical operations. Let's break it down step by step: 1. Begin by converting the mixed numbers to improper fractions. 1 2/5 = (5 * 1 + 2) / 5 = 7 / 5 1 2/3 = (3 * 1 + 2) / 3 = 5 / 3 2. Next, multiply the two fractions. To multiply fractions, you simply multiply the numerators together and the denominators together. (7/5) * (5/3) = (7 * 5) / (5 * 3) = 35 / 15 3. Simplify the resulting fraction if possible. In this case, we can simplify by dividing both the numerator and denominator by their greatest common divisor, which is 5. (35/5) / (15/5) = 7/3 4. Finally, convert the improper fraction back to a mixed number. 7/3 = 2 1/3 Therefore, 1 2/5 groups of 1 2/3 is equivalent to 2 1/3 inches.
# Section 1.4 If-Then Statements and Postulates ## Presentation on theme: "Section 1.4 If-Then Statements and Postulates"— Presentation transcript: Section 1.4 If-Then Statements and Postulates 4/6/2017 Geometry Objectives-What we’ll learn Recognize and analyze a conditional statement Write postulates about points, lines, and planes using conditional statements 4/6/2017 Geometry Postulate vs. Theorem A postulate is a statement that is assumed true without proof. A theorem is a true statement that can be proven. Conditional Statement A conditional statement has two parts, a hypothesis and a conclusion. When conditional statements are written in if-then form, the part after the “if” is the hypothesis, and the part after the “then” is the conclusion. p → q represents “if p then q” 4/6/2017 Geometry Examples If you are 13 years old, then you are a teenager. Hypothesis: Conclusion: You are a teenager 4/6/2017 Geometry Rewrite in the if-then form (Conditional Statement) All mammals breathe oxygen If an animal is a mammal, then it breathes oxygen. A number divisible by 9 is also divisible by 3 If a number is divisible by 9, then it is divisible by 3. 4/6/2017 Geometry Rewrite in the if-then form (Conditional Statement) Two lines intersect at a point. If two lines intersect, then they intersect at a point. Three non-collinear points determine a plane. If there are three non-collinear points, then they determine a plane. 4/6/2017 Geometry Writing a Counterexample Write a counterexample to show that the following conditional statement is false If x2 = 16, then x = 4. As a counterexample, let x = -4. The hypothesis is true, but the conclusion is false. Therefore the conditional statement is false. 4/6/2017 Geometry Converse The converse of a conditional statement is formed by switching the hypothesis and the conclusion. The converse of p → q is q → p 4/6/2017 Geometry Rewrite in the Converse form. If you are 13 years old, then you are a teenager. If you are a teenager, then you are 13 years old. If a number divisible by 9, then it is also divisible by 3 If a number is divisible by 3, then it is divisible by 9. Rewrite in the Converse form. If two angles are vertical angles, then they are congruent. If two angles are congruent, then they are vertical angles. If a quadrilateral has 4 right angles, then it is a rectangle. If a quadrilateral is a rectangle, then it has 4 right angles. Point, Line, and Plane Postulates Postulate 1-1: Through any two points there exists exactly one line Postulate 1-2: Through any three noncollinear points there exists exactly one plane Postulate 1-3: A line contains at least two points Postulate 1-4: A plane contains at least three points not on the same line 4/6/2017 Geometry Postulate 2-5: If two points lie in a plane, then the line containing them lies in the plane Postulate 2-6: If two planes intersect, then their intersection is a line 4/6/2017 Geometry Similar presentations
Question Video: Finding the Limit of a Function from Its Graph at a Point of Removable Discontinuity If the Limit Exists \| Nagwa Question Video: Finding the Limit of a Function from Its Graph at a Point of Removable Discontinuity If the Limit Exists \| Nagwa # Question Video: Finding the Limit of a Function from Its Graph at a Point of Removable Discontinuity If the Limit Exists Mathematics • Second Year of Secondary School ## Join Nagwa Classes Determine the limit as π₯ βΆ β1 of the function represented by the graph. 03:03 ### Video Transcript Determine the limit as π₯ approaches negative one of the function represented by the graph. Weβre given the graph of a function. And we need to determine the limit as π₯ approaches negative one of this function by using the graph. To start, we can see that our π¦-axis is labeled π of π₯. So weβll call our function π of π₯. Letβs now recall what we mean by the limit as π₯ approaches negative one of a function π of π₯. The first thing we need to recall is the notation which represents this. We represent this by saying the limit as π₯ approaches negative one of π of π₯. And what this means is, itβs the value that π of π₯ approaches as π₯ tends to negative one. In other words, as our input values of π₯ are getting closer and closer to negative one, we want to see what happens to our output values π of π₯. Remember that our input values of π₯ will be on the π₯-axis. Since we want to know what happens as π₯ gets closer and closer to negative one, letβs mark this on our π₯-axis. And now we can see something interesting. We can see that our function π of π₯ is not defined when π₯ is equal to negative one. This is represented by the hollow circle in our curve. We might think this is a problem. However, remember, weβre only interested in what happens to our outputs as π₯ tends to negative one. This means weβre only interested in what happens to our outputs as π₯ gets closer and closer to negative one. Our value of π₯ will never be equal to negative one. We want to know what happens around this value. Letβs start by seeing what happens as our values of π₯ approach negative one from the left. In other words, our inputs will all be less than negative one. To start, we can see if we input a value of π₯ is equal to negative six, then our function π of π₯ outputs negative five. In other words, π of negative six is negative five. We can do the same when we input negative four. We can see that our function will output negative three. In other words, π of negative four is equal to negative three. And we can continue doing this, getting closer. When we input negative two, our function outputs negative one. Well, we want to know what happens as our inputs are getting closer and closer to negative one. And if we continue doing this, we can see that our outputs are getting closer and closer to zero. But this is only one side of the story. What happens when our values of π₯ get closer and closer to negative one from the right? In other words, our inputs will all be bigger than negative one. And we can answer this in exactly the same way. First, if we input six, we can see that our function outputs seven. Next, when our input value of π₯ is equal to four, we can see that our function outputs five. And we can keep doing this. When we input π₯ is equal to two, our function outputs three. And when we input π₯ is equal to zero, our function outputs one. And if we keep getting closer and closer to π₯ is equal to negative one, we can see that once again our outputs are approaching zero. So in both cases, our outputs were getting closer and closer to zero. And because in both cases we were getting closer and closer to zero, we can conclude that this limit must be equal to zero. Therefore, we were able to show the limit as π₯ approaches negative one of the function π of π₯ given to us in the graph is equal to zero. ## 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
###### Pedal Triangles Alex Szatkowski This exploration will take a look at pedal triangles and what happens when the pedal point is the centroid, incenter, orthocenter, circumcenter, as well as the center of the 9 point circle of the original triangle. First let's take a look at what a pedal triangle looks like. A pedal triangle can be easily constructed using GSP. A pedal triangle is created by joining the intersections of lines perpindicular to each of the three sides of the triangle. Lets see what this looks like: Pedal point outside triangle Pedal point inside triangle # 1. Lets take a look at when the pedal point is also the centroid of the triangle. To begin we need to look at the three different types of triangles; acute, obtuse and right to see if there are any differences. As we can see from the pictures above, when the pedal point is also the centroid of the triangle, the pedal triangle is always completely inside the original triangle. # 2. Lets take a look at when the pedal point is also the incenter of the triangle. We will look at the same three types of triangles, acute, obtuse and right. Similar to when the pedal point was also the centroid, when the pedal point is also the incenter, the pedal triangle is always completely inside the original triangle. # 3. Lets take a look at when the pedal point is also the orthocenter of the triangle. Unlike the centroid and incenter, the orthocenter of a triangle is not always inside the triangle so we need to make sure we look at that case. We can make three unique observations: a. When the triangle is acute, the orthocenter is inside as well as the entire pedal triangle. b. When the triangle is obtuse, the orthocenter is outside and the pedal triangle overlaps the triangle. c. When the triangle is right, the orthocenter is at the right angle and the pedal triangle appears as a line. This line is called the Simson Line and is where the three vertices of the pedal triangle are collinear. This is a potential extension to this exploration. # 4. Lets take a look at when the pedal point is also the circumcenter of the triangle. Similar to the orthocenter the circumcenter can be both inside and outside the triangle. Lets see if we get similar results. Some observations: a. Unlike with the orthocenter, when the pedal point was also the circumcenter the pedal triangle was always inside the original triangle. b. When the triangle was acute, the circumcenter could be inside or outside the triangle as we can see in the top two triangles, however both times the pedal triangle remains inside. c. In the last picture, the right triangle, the circumcenter, also fell on one of the vertices of our pedal triangle. # 5. Lets take a look at when the pedal point is also the center of the 9 point circle of the triangle. We will take a look at an acute, obtuse and right triangle. Some observations: a. For both the acute and right triangles the pedal triangle is completely inside the original triangle. b. The obtuse triangle does not have a 9-point circle, therefore we cannot investigate what the pedal triangle would look like when the pedal point is the same as the center of the 9 point circle. # One final thought regarding the triangles chosen to investigate is that one might wonder why we did not look at an equilateral triangle. To begin, equilateral triangles were included, however they all appeared the same due to the fact that the centers of an equilateral triangle are coincident. Therefore the pedal point is the same no matter what center we are looking at. The pedal triangle is always inside the original triangle. Return
# What is difference between ASA and AAS congruence criteria? ## What is difference between ASA and AAS congruence criteria? ASA stands for “angle, side, angle” and means that we have two triangles where we know two angles and the included side are equal. And AAS stands for “angle, angle, side” and means that we have two triangles where we know two angles and the non-included side are equal. ## What is ASA congruence explain with an example? ASA (Angle-Side- Angle) If any two angles and the side included between the angles of one triangle are equivalent to the corresponding two angles and side included between the angles of the second triangle, then the two triangles are said to be congruent by ASA rule. Which answer choice best explains the difference between ASA and AAS? AAS is when two angles and a non-included side are congruent, but in ASA the side is included between the two angles. They are really the same since they both begin with an A for “angle”. AAS is when you check the congruence in a clockwise direction, but in a counterclockwise direction it would match ASA. ### Is AAS congruence? The AAS Theorem says: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. Notice how it says “non-included side,” meaning you take two consecutive angles and then move on to the next side (in either direction). ### Is aas a congruence rule? Whereas the Angle-Angle-Side Postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent. What is AAS triangle congruence? ## Is SAA and AAS the same? A variation on ASA is AAS, which is Angle-Angle-Side. Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent. ## Does ASA prove congruence? Angle-Side-Angle (ASA) Rule Angle-side-angle is a rule used to prove whether a given set of triangles are congruent. The ASA rule states that: If two angles and the included side of one triangle are equal to two angles and included side of another triangle, then the triangles are congruent. What is the difference between ASA and AAS in geometry? If two triangles are congruent, all three corresponding sides are congruent and all three corresponding angles are congruent. This shortcut is known as angle-side-angle (ASA). Another shortcut is angle-angle-side (AAS), where two pairs of angles and the non-included side are known to be congruent. ### Can AAS prove congruence? Angle-Angle-Side (AAS) Rule. Angle-angle-side is a rule used to prove whether a given set of triangles are congruent. The AAS rule states that. If two angles and a non-included side of one triangle are equal to two angles and a non-included side of another triangle, then the triangles are congruent. ### What is AAS and Asa? ASA and AAS – Concept. If two pairs of corresponding angles and the side between them are known to be congruent, the triangles are congruent. This shortcut is known as angle-side-angle (ASA). Another shortcut is angle-angle-side (AAS), where two pairs of angles and the non-included side are known to be congruent. What is SSS SAS ASA AAS? There are five ways to find if two triangles are congruent: SSS, SAS, ASA, AAS and HL. SSS stands for “side, side, side” and means that we have two triangles with all three sides equal. For example: (See Solving SSS Triangles to find out more) If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent. ## Is AAS a congruence theorem? AAS Theorem of Congruence (Angle-Angle-Side) Definition: If two angles and the side opposite one of them in a triangle are congruent to the corresponding parts in another triangle , then the triangles are congruent.
# How do you find an equation of the circle that passes through the pts (0,0) and (1,1) and has its center on the line y=2? Jan 20, 2016 ${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 5$ #### Explanation: If the center of the circle is on $y = 2$ then the center of the circle is at $\left({x}_{c} , 2\right)$ for some value ${x}_{c}$ The radius of the circle, $r$, is • The distance from $\left({x}_{c} , 2\right)$ to $\left(0 , 0\right)$ $\textcolor{w h i t e}{\text{XXX}} = \sqrt{{\left({x}_{c} - 0\right)}^{2} + {\left(2 - 0\right)}^{2}} = \sqrt{{x}_{c}^{2} + 4}$ • The distance from $\left({x}_{c} , 2\right)$ to $\left(1 , 1\right)$ $\textcolor{w h i t e}{\text{XXX}} = \sqrt{{\left({x}_{c} - 1\right)}^{2} + {\left(2 - 1\right)}^{2}} = \sqrt{{x}_{c}^{2} - 2 {x}_{c} + 1 + 1}$ Therefore $\textcolor{w h i t e}{\text{XXX}} {x}_{c}^{2} + 4 = {x}_{c}^{2} - 2 {x}_{c} + 2$ $\textcolor{w h i t e}{\text{XXX}} - 2 {x}_{c} = - 2$ $\textcolor{w h i t e}{\text{XXX}} {x}_{c} = - 1$ and since $r = \sqrt{{x}_{c}^{2} + 4}$ $\textcolor{w h i t e}{\text{XXX}} r = \sqrt{5}$ The general equation of a circle with center $\left(a , b\right)$ and radius $r$ is $\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ By substituting $\left(- 1 , 2\right)$ for $\left(a , b\right)$ and $\sqrt{5}$ for $r$, we get $\textcolor{w h i t e}{\text{XXX}} {\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 5$ graph{(x+1)^2+(y-2)^2=5 [-5.8, 5.3, -0.764, 4.783]}
# 初级代数: 最大公因数 1. $$12,30$$ 答案 2. $$20,30,100$$ 答案 3. $$16,8,36$$ 答案 4. $$4,8,14$$ 答案 5. $$8,12$$ 答案 6. $$15,25,40$$ 答案 # Greatest Common Factor - Introduction All whole numbers have factors — numbers that can be multiplied together to yield the original value. For example, both $$2$$ and $$3$$ are factors of $$6$$ , because $$2\times 3=6$$ . Some numbers, such as prime numbers, have only two factors: The number itself and $$1$$ . Consider the case of $$17$$ — no whole numbers multiply together to give this answer, except for $$17$$ and $$1$$ . Many numbers have common factors, which are factors that appear in both of their lists of factors. Let’s look at the factors of $$8$$ and $$10$$ , and see if they have any factors in common. Factors of $$8$$ $$1$$ , $$2$$ , $$4$$ Factors of $$10$$ $$1$$ , $$2$$ , $$5$$ Looking at the lists, we can see that their common factors are $$1$$ and $$2$$ . # Finding the Greatest Common Factor Listing the factors also helps us find the greatest common factor (GCF) — the largest whole number factor that the given numbers have in common. This is especially useful when simplifying fractions. Because, once we found the GCF, we can divide both the numerator and denominator by the GCF, which will result in a simpler fraction. How do you find the greatest common factor? Below are two methods. ## Method 1: Listing Factors Let’s find the GCF of these numbers: $$16$$ , $$28$$ and $$52$$ . Factors of $$16$$ $$1$$ , $$2$$ , $$4$$ , $$8$$ Factors of $$28$$ $$1$$ , $$2$$ , $$4$$ , $$7$$ , $$14$$ Factors of $$52$$ $$1$$ , $$2$$ , $$4$$ , $$13$$ , $$26$$ Looking at the list, we can see that the greatest common factor is $$4$$ . ## Method 2: Prime Factors You can also find the GCF using prime factors. Let’s try it for $$12$$ and $$16$$ . Prime factors of $$12$$ $$1$$ , $$2$$ , $$2$$ , $$3$$ Prime factors of $$16$$ $$1$$ , $$2$$ , $$2$$ , $$2$$ , $$2$$ , $$2$$ Find the “intersection” of these primes — the factors both numbers have in common — to get your answer. In this case, that’s $$2\times 2=4$$ , which is the GCF of $$12$$ and $$16$$ . # What's Next Once you’re familiar with finding the GCF, you can use it to quickly simplify algebra equations and reduce complicated fractions. Need a quick answer for more complex factor questions? Try our greatest common factor calculator. Ready to dive in and do more practice problems yourself? Start with Cymath’s free online practice problems or join Cymath Plus for an ad-free experience complete with more in-depth math assistance.
# Website that answers math word problems Best of all, Website that answers math word problems is free to use, so there's no sense not to give it a try! Our website can solving math problem. ## The Best Website that answers math word problems In this blog post, we discuss how Website that answers math word problems can help students learn Algebra. The common factors of 3 and 4 are 1 and 3, so we can cancel out the 3 in both the numerator and denominator, leaving us with the simplified fraction 1/4. In general, it's helpful to start by finding any common factors in the numerator and denominator that are larger than 1. Once you've cancelled out as many factors as possible, you can then multiply both the numerator and denominator by any remaining factors in order to further simplify the fraction. Just be careful not to cancel out any essential parts of the fraction (like 2 in ¾). If you do, you'll end up with an incorrect answer! Solving for exponents can be a tricky business, but there are a few basic rules that can help to make the process a bit easier. First, it is important to remember that any number raised to the power of zero is equal to one. This means that when solving for an exponent, you can simply ignore anyterms that have a zero exponent. For example, if you are solving for x in the equation x^5 = 25, you can rewrite the equation as x^5 = 5^3. Next, remember that any number raised to the power of one is equal to itself. So, in the same equation, you could also rewrite it as x^5 = 5^5. Finally, when solving for an exponent, it is often helpful to use logs. For instance, if you are trying to find x in the equation 2^x = 8, you can take the log of both sides to get Log2(8) = x. By using these simple rules, solving for exponents can be a breeze. How to solve perfect square trinomial. First, identify a, b, and c. Second, determine if a is positive or negative. Third, find two factors of ac that add to b. Fourth, write as the square of a binomial. Fifth, expand the binomial. Sixth, simplify the perfect square trinomial 7 eighth, graph the function to check for extraneous solutions. How to solve perfect square trinomial is an algebraic way to set up and solve equations that end in a squared term. The steps are simple and easy to follow so that you will be able to confidently solve equations on your own! With a good generator, you can input the parameters of the problem you want students to solve, and it will spit out a variety of different problems that meet those criteria. This can be a valuable tool for teachers who want to give their students some extra practice on a specific concept or for those who are looking for some fresh material to spice up their lesson plans. There are a number of different math problem generators available online, so take some time to explore and find one that meets your needs. In other words, x would be equal to two (2). However, if x represented one third of a cup of coffee, then solving for x would mean finding the value of the whole cup. In this case, x would be equal to three (3). The key is to remember that, no matter what the size of the fraction, solving for x always means finding the value of the whole. With a little practice, solving for x with fractions can become second nature. ## We cover all types of math issues I absolutely love this app and highly recommend it for students who are struggling in Math at school. It's such a lifesaver! If you need to show your work then this is the perfect app for you! And the answers are always correct. I just love it! Haniya Rogers Great experience overall, but for some reason I am unable to subscribe to the app plus. I click on the app plus where it should prompt me to purchase a subscription but it just loads forever and doesn't open an option for me to subscribe. I really love the app and would love to use plus as well, if only I could access it. Gracelyn Campbell Math generator Secant line solver Equations with variables on both sides solver App that scans math problems Combining like terms solver
2012 AMC 12A Problems/Problem 24 Problem Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$, and in general, $$a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}$$ Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$. What is the sum of all integers $k$, $1\le k \le 2011$, such that $a_k=b_k?$ $\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012$ Solution 1 First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$(decreasing exponential function), and $g(x) = x^k$ for $k > 0$(increasing exponential function for positive $x$). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant. We will now examine the first few terms. Comparing $a_1$ and $a_2$, $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$. Therefore, $0 < a_1 < a_2 < 1$. Comparing $a_2$ and $a_3$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$. Comparing $a_1$ and $a_3$, $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$. Therefore, $0 < a_1 < a_3 < a_2 < 1$. Comparing $a_3$ and $a_4$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$. Comparing $a_2$ and $a_4$, $0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$. Therefore, $0 < a_1 < a_3 < a_4 < a_2 < 1$. Continuing in this manner, it is easy to see a pattern(see Note 1). Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$. Solving gives $\boxed{\textbf{(C)} 1341}$. Note 1: We claim that $0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1$. We can use induction to prove this statement. (not necessary for AMC): Base Case: We have already shown the base case above, where $0 < a_1 < a_2 < 1$. Inductive Step: Rearranging in decreasing order gives $1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b_{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} = a_1 > 0$. Solution 2 Start by looking at the first few terms and comparing them to each other. We can see that $a_1 < a_2$, and that $a_1 < a_3 < a_2$, and that $a_3 < a_4 < a_2$, and that $a_3 < a_5 < a_4$ ... From this, we find the pattern that $a_k-1 < a_k+1 <$a_k$. Examining this relationship, we see that every new number$ (Error compiling LaTeX. ! Missing $inserted.)a_k$will be between the previous two terms,$a_k-1$and$a_k-2$. Therefore, we can see that$a_1$is the smallest number,$a_2$is the largest number, and that all odd terms are less than even terms. Furthermore, we can see that for every odd k,$a_k < a_k+2$, and for every even k,$a_k > a_k+2$This means that rearranging the terms in descending order will first have all the even terms from$a_2$to$a_2012$, in that order, and then all odd terms from$a_2011$to$a_1$, in that order (so$\{b_k\}_{k=1}^{2011} = {a_2, a_4, a_6, ... a_2008, a_2010, a_2011, a_2009, ... a_5, a_3, a_1}$). We can clearly see that there will be no solution k where k is even, as the$(Error compiling LaTeX. ! Missing$ inserted.)k$th term in$\{a_k\}_{k=1}^{2011}$will appear in the same position in its sequence as the$2k$th term does in$\{a_k\}_{k=1}^{2011}$, where k is even. Therefore, we only have to look at the odd terms of$a_k$, which occur in the latter part of$b_k$. Looking at the back of both sequences, we see that term k in$ (Error compiling LaTeX. ! Missing $inserted.)\{a_k\}_{k=1}^{2011}$progresses backwards in the equation$2012 - k$, and that term k in$\{a_k\}_{k=1}^{2011}$progresses backwards in the equation$2k - 1$. Setting these two expressions equal to each other, we get$671$. However, remember that we started counting from the back of both sequences. So, plugging$(Error compiling LaTeX. ! Missing$ inserted.)671$back into either side of the equation from earlier, we get our answer of$\boxed{\textbf{(C)} 1341}\$. Sorry for the sloppy explanation. It's been two years since I've tried to give a solution to a problem, but I think this solution takes a different approach than the one above. ~dolphin7
# How do you use the distributive property to multiply 5(7.1)? ##### 1 Answer May 1, 2017 $5 \left(7.1\right) = 35.5$ #### Explanation: The distributive property comes in two forms: Left distributive property: $a \left(b + c\right) = a b + a c$ Right distributive property: $\left(a + b\right) c = a c + b c$ We can use the left distributive property to find: $5 \left(7.1\right) = 5 \left(7 + 0.1\right) = \left(5 \cdot 7\right) + \left(5 \cdot 0.1\right) = 35 + 0.5 = 35.5$
What is 51/331 as a decimal? Solution and how to convert 51 / 331 into a decimal 51 / 331 = 0.154 51/331 or 0.154 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). Now, let's solve for how we convert 51/331 into a decimal. 51/331 is 51 divided by 331 The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 51 is being divided into 331. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 331. We use this as our equation: numerator(51) / denominator (331) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is how we look at our fraction as an equation: Numerator: 51 • Numerators are the portion of total parts, showed at the top of the fraction. Any value greater than fifty will be more difficult to covert to a decimal. 51 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. Time to evaluate 331 at the bottom of our fraction. Denominator: 331 • Denominators represent the total parts, located at the bottom of the fraction. Larger values over fifty like 331 makes conversion to decimals tougher. But 331 is an odd number. Having an odd denominator like 331 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. Now let's dive into how we convert into decimal format. Converting 51/331 to 0.154 Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 331 \enclose{longdiv}{ 51 }$$ Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once. Step 2: Extend your division problem $$\require{enclose} 00. \\ 331 \enclose{longdiv}{ 51.0 }$$ Uh oh. 331 cannot be divided into 51. Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 331 into 510. Step 3: Solve for how many whole groups you can divide 331 into 510 $$\require{enclose} 00.1 \\ 331 \enclose{longdiv}{ 51.0 }$$ How many whole groups of 331 can you pull from 510? 331 Multiple this number by our furthest left number, 331, (remember, left-to-right long division) to get our first number to our conversion. Step 4: Subtract the remainder $$\require{enclose} 00.1 \\ 331 \enclose{longdiv}{ 51.0 } \\ \underline{ 331 \phantom{00} } \\ 179 \phantom{0}$$ If there is no remainder, you’re done! If you have a remainder over 331, go back. Your solution will need a bit of adjustment. If you have a number less than 331, continue! Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value. Why should you convert between fractions, decimals, and percentages? Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. And the same is true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples: When you should convert 51/331 into a decimal Dining - We don't give a tip of 51/331 of the bill (technically we do, but that sounds weird doesn't it?). We give a 15% tip or 0.154 of the entire bill. When to convert 0.154 to 51/331 as a fraction Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches. Practice Decimal Conversion with your Classroom • If 51/331 = 0.154 what would it be as a percentage? • What is 1 + 51/331 in decimal form? • What is 1 - 51/331 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.154 + 1/2?
# How do you graph y=8/(x^2-x-6) using asymptotes, intercepts, end behavior? Oct 15, 2017 Vertical asymptote: x=3,-2 Horizontal asymptote: $y = 0$ $x$- intercept: none $y$-intercept: $- \frac{4}{3}$ End behaviour: As x rarr -∞,y rarr ∞ As x rarr ∞, y rarr ∞ #### Explanation: Denote the function as $\frac{n \left(x\right)}{d \left(x\right)}$ To find the vertical asymptote, Find $d \left(x\right) = 0$ $\Rightarrow {x}^{2} - x - 6 = 0$ $\left(x - 3\right) \left(x + 2\right) = 0$ $\therefore$ The vertical asymptotes are at $x = 3 , - 2$ To find the $x$-intercept, plug in $0$ for $y$ and solve for $x$. $0 = \frac{8}{{x}^{2} - x - 6}$ $\therefore$ There are no $x$-intercepts. To find the $y$-intercept, plug in $0$ for $x$ and solve for $y$. y=(8)/(0^2-0-6 $y = - \frac{4}{3}$ $\therefore$ The $y$-intercept is at $- \frac{4}{3}$. To find the horizontal asymptote, Compare the leading degrees of the numerator and denominator. For $n \left(x\right)$, the leading degree is $0$, since $8 \cdot {x}^{0}$ would give $8$. Denote this as $\textcolor{\pi n k}{m}$. For $d \left(x\right)$, the leading degree is $2$. Denote this as $\textcolor{b r o w n}{n}$. If $\textcolor{\pi n k}{n} < \textcolor{b r o w n}{m}$, then the horizontal asymptote is $y = 0$. To find the end behaviour of the graph, plot it in a graphic calculator. As x rarr -∞,y rarr ∞ As x rarr ∞, y rarr ∞
Courses Courses for Kids Free study material Offline Centres More Store # To draw parallel lines we use, A. Compass B. Divider C. Scale and set squares D. Scale and protractor Last updated date: 20th Jun 2024 Total views: 394.8k Views today: 5.94k Verified 394.8k+ views Hint: We recall the definition uses of scale, divider, compass, protractor and set squares. We recall the definitions of parallel lines and demonstrate how we can use edges of set squares to draw a parallel line of $\overleftrightarrow {AB}$ passing through a point P. Complete step-by-step solution We know that in our geometry box different types of instruments are used. The scale or ruler is an instrument used to measure the length in cm or inch of distance between two points and also to draw straight lines. A divider otherwise known as a compass is used to locate points while a compass is used to draw arc and circle. A protractor is an instrument to measure angles with the unit degree. The set squares are a type of drawing instrument whose two edges are perpendicular to each other. There are two types of set squares. The smaller one is ${{45}^{\circ }}-{{45}^{\circ }}$ set square whose edges of equal length and both the acute angles are of measure ${{45}^{\circ }}$.The larger one is ${{30}^{\circ }}-{{60}^{\circ }}$ set square whose edges are not of equal length and the acute angles are of measure ${{30}^{\circ }},{{60}^{\circ }}$. Parallel lines are two straight lines that never intersect each other. We draw the rough figure of two straight lines $\overleftrightarrow {AB},\overleftrightarrow {PQ}$ which are parallel below . We draw parallel lines with the help of a scale and set squares, for that we first draw the line $\overleftrightarrow {AB}$ with a scale and locate the point P. We align the hypotenuse of ${{45}^{\circ }}-{{45}^{\circ }}$ set square (shown in red colour below)with it and then align ${{30}^{\circ }}-{{60}^{\circ }}$ set square with the side of ${{45}^{\circ }}-{{45}^{\circ }}$ show in green colour below. We fix the ${{30}^{\circ }}-{{60}^{\circ }}$ with our left hand and move the ${{45}^{\circ }}-{{45}^{\circ }}$ set square so that the hypotenuse passes through P. We use to draw the line through P and extend with a scale to get the parallel line of $\overleftrightarrow {AB}$ passing through P as $\overleftrightarrow {PQ}$ So the correct option is C. Note: We note that the distance between two parallel lines is always equal. This method of drawing parallel lines is also used to construct trapezium. If the lines are not parallel then they are intersecting. We can also draw parallel lines using scale and compass taking arcs of equal width.
Polynomial Identities MathBitsNotebook.com Terms of Use Contact Person: Donna Roberts In Quadratics, we dealt with quadratic identities. Let's extend our study to all polynomials. An equation that is true for every value of the variable is called an identity. To show that an equation is an identity: Start with either side of the equation and show that it can algebraically be changed into the other side. Or start with both sides of the equation and show that they both can be changed into the same algebraic expression. You may be asked to verify (show, prove) that a known "rule" is an identity (always true). You may, or may not, be asked to supply "justification" for your work. 1. Show, with justification, that (x2 + y2)2 = (x2 - y2)2 + (2xy)2 is an identity. x4 + 2x2y2 +y4 = x4 - 2x2y2 +y4 + (4x2y2) expand by multiplication and distributive property. x4 + 2x2y2 +y4 = x4 + 2x2y2 +y4 combine like terms. (x2 + y2)2 = (x2 - y2)2 + (2xy)2 Identity! (always true) This identity is used to create Pythagorean Triples. For example, substituting x = 3 and y = 2 will give 132 = 52 + 122 which is the Pythagorean Triple 5, 12, 13. You may be asked to verify (show, prove) that any algebraic equation is an identity. NOTE: If you are asked to show that an equation is an identity, you may assume that the equation IS an identity. If you are asked to determine whether an equation is an identity, be careful as the answer may be that it is not an identity. 2. Show that x3 + y3 = (x + y)(x2 - xy + y2) is an identity. Starting with the right hand side: (distribute) (x + y)(x2 - xy + y2) = x3 - x2y + xy2 + x2y - xy2 + y3 (simplify) = x3 + y3 This is an identity. We have just shown that the formula for factoring the sum of two cubes is true. 3. Determine whether (2a2 + b3)(2a2 - b3) = 4a2 - b3 is an identity. Starting with the left hand side: (2a2 + b3)(2a2 - b3) = 4a4 + 2a2b3 - 2a2b3 - b6 = 4a4 - b6 4a4 - b6 ≠ 4a2 - b3 (FALSE) This is not an identity. 4. Determine whether (a + b)4 + (a - b)4 = 2a4 + 12a2b2 + 2b4 is an identity. First, expand ( a + b )4 ( a + b )( a + b )( a + b )( a + b) = (a2 + 2ab + b2)(a2 + 2ab + b2) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 Now, expand ( a - b )4 ( a - b )( a - b )( a - b )( a - b) = (a2 - 2ab + b2)(a2 - 2ab + b2) = a4 - 4a3b + 6a2b2 - 4ab3 + b4 Add the results. (a4 + 4a3b + 6a2b2 + 4ab3 + b4) added to (a4 - 4a3b + 6a2b2 - 4ab3 + b4 ) gives 2a4+ 12a2b2 + 2b4 This is an identity. You may be asked to "check" someone's work to see if they properly showed the existence of an identity. 5. Riley gave the following justification that n3 - n = n(n + 1)(n - 1) is an identity. Is this correct? If not, where is the error? n(n + 1)(n - 1) Right side of equation. Step 1: = (n2 + n)(n - 1) Multiply first two terms. Step2: = n3 + n2 - n2- n Multiplication (distribution). Step 3: = n3 - n Combining like terms. n3 - n = n(n + 1)(n - 1) is an identity. Riley's solution is correct! NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use".
# Base quantity A quantity that splits another quantity as its factors is called the base. ## Introduction A quantity can be expressed in product form on the basis of another quantity and the product form expression can be written simply in exponential notation as per exponentiation. The quantity that is considered to split a quantity as a product of its terms is called the base of exponential form. ### Example $81$ is a number and express it in product form on the basis of another number $3$ by factorization. $81 \,=\, 3 \times 3 \times 3 \times 3$ According to exponentiation, write the product form expression in exponential notation. $81 \,=\, 3^4$ On the basis of number $3$, the number $81$ is expressed as factors of $3$. Hence, the number $3$ is called the base of the exponential notation for the number $81$. #### More Examples Observe the following examples to understand what exactly the base is in exponential notation. $(1)\,\,\,\,\,\,\,$ $8 = 2 \times 2 \times 2 = 2^3$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ In this example, $2$ is the base of the exponential term. $(2)\,\,\,\,\,\,\,$ $16 = 4 \times 4 = 4^2$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ In this example, $4$ is the base of the exponential term. $(3)\,\,\,\,\,\,\,$ $3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ In this example, $5$ is the base of the exponential term. $(4)\,\,\,\,\,\,\,$ $343 = 7 \times 7 \times 7 = 7^3$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ In this example, $7$ is the base of the exponential term. $(5)\,\,\,\,\,\,\,$ $28561 = 13 \times 13 \times 13 \times 13 = 13^4$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ In this example, $13$ is the base of the exponential term. #### Algebraic form $m$ is literal and it represents a quantity. Take, the quantity $m$ is divided as factors on the basis of the quantity $b$. The total number of factors of $b$ is $n$ by the factorisation. $m$ $\,=\,$ $\underbrace{b \times b \times b \times … \times b}_{\displaystyle n factors}$ $\,=\,$ $b^n$ Therefore, the number $b$ is called the base of the exponential notation $b^n$. ###### Math Questions The math problems with solutions to learn how to solve a problem. Learn solutions Practice now ###### Math Videos The math videos tutorials with visual graphics to learn every concept. Watch now ###### Subscribe us Get the latest math updates from the Math Doubts by subscribing us.
How do you solve x^2-2x-5<=0 using a sign chart? Oct 30, 2016 The answer is $1 - \sqrt{6} \le x \le 1 + \sqrt{6}$ Explanation: First we start by solving $y = {x}^{2} - 2 x - 5 = 0$ We calculate $\Delta = {b}^{2} - 4 a c = 4 + 20 = 24$ So the roots are $= \frac{2 \pm \sqrt{24}}{2} = \frac{2 \pm 2 \sqrt{6}}{2}$ so the roots are $1 + \sqrt{6}$ and $1 - \sqrt{6}$ We can do the sign chart $x$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a}$$1 - \sqrt{6}$$\textcolor{w h i t e}{a a a}$$1 + \sqrt{6}$$\textcolor{w h i t e}{a a a}$$+ \infty$ $1 - \sqrt{6}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$ $1 + \sqrt{6}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a a a a}$$+$ $y$$\textcolor{w h i t e}{a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$ So ${x}^{2} - 2 x - 5 \le 0$ when $1 - \sqrt{6} \le x \le 1 + \sqrt{6}$
# Algebra 1 : Functions and Lines ## Example Questions ### Example Question #1 : Midpoint Formula Find the midpoint of the line segment with these endpoints: and . Explanation: The midpoint formula is . Add the two values together and divide the total by two to find the value of the midpoint. Add the two values together and divide the total by two to find the value of the midpoint. ### Example Question #861 : Functions And Lines What is the midpoint of the points and ? Explanation: The midpoint between two points can be found by the equation . Substiuting in the equation becomes . If you simplfy it the the midpoint is . ### Example Question #4148 : Algebra 1 What would be the midpoint of a line with endpoints and ? Explanation: To find the midpoint, we just need to find the mean of the x-coordinates, and then the mean of the y-coordinates. To find the mean, add the two numbers together and then divide by 2. Our x-coordinates are -3 and 15. Their mean is . Our y-coordinates are 6 and 4. Their mean is . This gives us a new coordinate pair of . ### Example Question #861 : Functions And Lines The endpoints of a line segment are and . What is the midpoint? Explanation: In order to find the midpoint, we have to find the mean of the x-coordinates of the endpoints, and then find the mean of the y-coordinates of the endpoints. To find the mean, add the two numbers together then divide by 2. The x-coordinates are 0 and -20. Their mean is . The y-coordinates are 0 and 4. Their mean is . This gives us a new x-, y-coordinate pair of . ### Example Question #862 : Functions And Lines Find the midpoint of a line containing the points and . Explanation: Use the midpoint formula: Remember points are written in the following format: Substitute. Simplify. Our midpoint is . ### Example Question #862 : Functions And Lines What is the midpoint of the coordinates and ? Explanation: The questions is asking for the midpoint of a line between two coordinates. The midpoint formula is . Now, fill in the variables to find and ### Example Question #861 : Functions And Lines Find the midpoint between the given points: Explanation: Using the midpoint formula: Plugging in our points we can find the midpoint. ### Example Question #41 : Midpoint Formula Find the midpoint of the line segment that has endpoints at . Explanation: To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates. ### Example Question #42 : Midpoint Formula Find the midpoint of the line segment that has endpoints at . Explanation: To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates. ### Example Question #862 : Functions And Lines Find the midpoint of a line segment that has endpoints at and .
<meta http-equiv="refresh" content="1; url=/nojavascript/"> You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 7 Go to the latest version. # 7.11: Two-Step Equations from Verbal Models Difficulty Level: At Grade Created by: CK-12 % Progress Practice Two-Step Equations from Verbal Models Progress % Have you ever tried to earn money babysitting? Kara worked on earning money for her trip to Boston. For each babysitting job she took on, Kara charged$2 for bus fare plus an additional$6 for each hour she worked. On Saturday, Kara earned $26 for the one babysitting job. Write an equation to represent $h$ , the total number of hours that Manuel babysat during his$26 job. Do you know how to do this? This Concept will teach you how to write two -step equations from verbal models. You will know how to do this by the end of the Concept. ### Guidance In the last few Concepts, you worked on solving single variable equations. Here is a single variable equation. Here is a problem like the ones you worked on in the last two Concepts. $3x=15$ Now if you think about this equation, you only had to perform one operation to solve it. We used an inverse or opposite operation to solve it. Because this is a multiplication equation, we used division to solve for the unknown variable. We could say that we performed one-step to solve this equation. The one-step was to divide. We call an equation where one operation is needed to solve for the unknown variable a one-step equation. What happens when there is more than one operation needed to solve an equation? When this happens, we have two-step equation . A two-step equation has more than one operation in it. Let’s look at a few two-step equations. $3x+5 &= 20\\\frac{x}{3}-2 &= 5$ If you look at both of these equations, you will see that there are two operations present in each. The first equation has multiplication and addition. The second equation has division and subtraction. You will see other variations of two-step equations too, but this gives you an idea of a few different situations. Before we begin solving equations, let’s look at how we can write a two-step equation from a verbal model. First, let’s think about some of the words that mean addition, subtraction, multiplication and division. Identifying these key words is going to assist us when writing equations from verbal models. Remember that a verbal model uses words. Sum Altogether In all Plus and Difference Less than More than Take away subtract Product Times Groups of Quotient Split Up Divided Take a few minutes to write down these key words in your notebook. Now apply this information. Six times a number plus five is forty-one. First, identify any key words that identify operations. Six times a number plus five is forty-one. Next, begin to translate the words into an equation. Six = 6 Times $=x$ A number = variable Plus = + Five = 5 Is means = Forty-one is 41 Next, we put it altogether. $6x+5=41$ Here is another one. Four less than two times a number is equal to eight. First, identify any key words that identify operations. Four less than two times a number is equal to eight. Now we can translate each part. Four becomes 4 Less than means subtraction Two becomes 2 Times $= x$ A number = variable Is equal to means = Eight = 8 Notice that the one tricky part is in the words “less than” because it is less than two times a number, the two times a number needs to come first in the equation. Then we can subtract. Put it altogether. $2x-4=8$ Now it's time for you to try a few on your own. Write an equation for each situation. #### Example A The product of five and a number plus three is twenty-three. Solution: $5x+3=23$ #### Example B Six times a number minus four is thirty-two. Solution: $6y-4=32$ #### Example C A number, $y$ , divided by 3 plus seven is ten. Solution: $\frac{y}{3}+7=10$ Here is the original problem once again. Kara worked on earning money for her trip to Boston. For each babysitting job she took on, Kara charged $2 for bus fare plus an additional$6 for each hour she worked. On Saturday, Kara earned $26 for the one babysitting job. Write an equation to represent $h$ , the total number of hours that Manuel babysat during his$26 job. Do you know how to do this? To begin, Kara worked an unknown amount of hours. This is our variable. We can call this $h$ She earns six dollars an hour. We multiply that times the unknown number of hours. $6h$ Kara also charges \$2.00 for bus fare. $6h + 2$ She earned twenty -six dollars. $6h + 2 = 26$ This is our equation. ### Vocabulary Here are the vocabulary words in this Concept. One-Step Equation an equation with one operation in it Two-Step Equation an equation with two operations in it ### Guided Practice Here is one for you to try on your own. Write an equation for this statement. A number divided by two and six is equal to fourteen. First, look for key words that identify operations. A number divided by two and six is equal to fourteen. Next, translate each word into an equation. A number means variable Divided means $\div$ By two means 2 is our divisor Six means 6 Is means = Fourteen means 14 Put it altogether. $\frac{x}{2}+6=14$ ### Practice Directions: Write the following two-step equations from verbal models. 1. Two times a number plus seven is nineteen. 2. Three times a number and five is twenty. 3. Six times a number and ten is forty-six. 4. Seven less than two times a number is twenty-one. 5. Eight less than three times a number is sixteen. 6. A number divided by two plus seven is ten. 7. A number divided by three and six is eleven. 8. Two less than a number divided by four is ten. 9. Four times a number and eight is twenty. 10. Five times a number take away three is twelve. 11. Two times a number and seven is twenty - nine. 12. Four times a number and two is twenty- six. 13. Negative three time a number take a way four is equal to negative ten. 14. Negative two times a number and eight is equal to negative twelve. 15. Negative five times a number minus eight is equal to seventeen. ### Vocabulary Language: English One-Step Equation One-Step Equation A one-step equation is an algebraic equation with one operation in it that requires one step to solve. Two-Step Equation Two-Step Equation A two-step equation is an algebraic equation with two operations in it that requires two steps to solve. Nov 30, 2012 May 09, 2015
## How do you find the equation of a line given two points? We know two points: point “A” is (6,4) (at x is 6, y is 4) point “B” is (2,3) (at x is 2, y is 3)Step 1: Find the Slope (or Gradient) from 2 Pointssubtract the Y values,subtract the X values.then divide. ## What is the formula of a line? The equation of a straight line is usually written this way: y = mx + b. ## What are the three equations of a line? There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form. We review all three in this article. There are three main forms of linear equations. ## What is the general equation for a vertical line? Vertical lines help determine if a relation is a function in math. The equation of a vertical line always takes the form x = k, where k is any number and k is also the x-intercept . ## What is an equation for a vertical line? Vertical Lines Similarly, in the graph of a vertical line, x only takes one value. Thus, the equation for a vertical line is x = a, where a is the value that x takes. Example 3: Write an equation for the following line: Graph of a Line Since x always takes the value 2 = , the equation for the line is x = . ## What is the equation of a vertical line through (- 5 2? Any vertical line has the equation x = k, where k is the x-coordinate of any point the line goes through. x = -5. ## How do you convert an equation into standard form? The standard form of such an equation is Ax + By + C = 0 or Ax + By = C. When you rearrange this equation to get y by itself on the left side, it takes the form y = mx +b. This is called slope intercept form because m is equal to the slope of the line, and b is the value of y when x = 0, which makes it the y-intercept. ## What is equation standard form? The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations. ## How do you solve a line equation? The slope-intercept formula of a line is written as y = mx+b, where m is the slope and b is the y-intercept (the point on the y-axis where the line crosses it). Plug the number you found for your slope in place of m. In our example, the formula would read y = 1x+b or y = x+b when you replace the slope value. ## What is B in the equation of a line? The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis. ### Releated #### Energy equation physics What is the formula for energy? The formula that links energy and power is: Energy = Power x Time. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second. How do you solve for energy in physics? In classical mechanics, kinetic energy (KE) […] #### Equation for exponential growth How do you calculate exponential growth? To calculate exponential growth, use the formula y(t) = a__ekt, where a is the value at the start, k is the rate of growth or decay, t is time and y(t) is the population’s value at time t. What is an exponential growth function? An exponential function can describe […]
# 2’s complement If you know what is 1’s complement, you should know 2’s complement. The basic concept to use either 1’s complement or 2’s complement is to use it as a negative number. Subtraction operation is carried as addition of the negative number, ie, the complement of the number. For example, 47-47=0 means 47+(-47)=0 and therefore a number X adds to its complement will become ZERO. Now it depends on whether you are using 1’s complement as negative number or 2’s complement. Firstly we will use 1’s complement as negative number so you will see the following addition will give you: Therefore we can easily say that if “1111″ represents negative ZERO then 1001 is the 1’s complement of 0110 or negative of 0110. To summarize the idea, say 0110 is +6 as the 4th bit is 0, 1001 is -6 as the 4th bit is 1. 0110-0110=0110+1001=1111=-0. In decimal, +6-6=+6+(-6)=-0. Try +6-5, in binary we have 0110-0101=0110+1010(1’s complement)=10000, move the overflow bit around to have 1+0000=0001 to get the proper result of 6-5=1. Try +6-4, in binary we have 0110-0100=0110+1011(1’s complement)=10001, add the overflow bit to have 1+0001=0010 to get the proper result of 6-4=2. You see how the numbers are manipulated to get the result of subtration using 1’s complement. Now we will use 2’s complement as negative number so you will see the following addition will give you: 0110+1010(2’s complement=1’s complement +1)=10000 in binary addition Therefore we can easily say that if “10000″ is “0000″ which represents ZERO and the 5th bit is 1 and it is just the overflow. Then 1010 is called the 2’s complement of 0110 or negative of 0110. To summarize the idea, say 0110 is +6 (positive) as the 4th bit is 0, 1010 is -6 as the 4th bit is 1. 0110-0110=0110+1010(1001+1)=10000=0. In decimal, +6-6=+6+(-6)=0. Try +6-5, in binary we have 0110-0101=0110+(1010+1)(2’s complement)=10001, drop the overflow bit to have 1\|0001=0001 to get the proper result of 6-5=1. Try +6-4, in binary we have 0110-0100=0110+(1011+1)(2’s complement)=10010, drop the overflow bit to have 1\|0010=0010 to get the proper result of 6-4=2. You see how the numbers are manipulated to get the result of subtration using 2’s complement.
Jazz asked in Science & MathematicsMathematics · 7 years ago # Help me with probability please? How would I solve these questions?! Suppose that you roll a pair of fair dice. a. What is the probability that the sum is at least 8? b. What is the probability that the sum is not 7? c. Given that you roll a sum of 7 on the first roll, what is the probability that you roll a sum of 7 on the second roll? d. What is the probability that you roll a sum of 7 two times in a row? Relevance • 7 years ago These are all relative frequency probability problems. First, you need to know that there are 36 possible outcomes. Imagine you roll a red die and a green one. For each value of the red die, there are 6 possible ones for the green one. Since the red die has 6 possible values, 6*6=36. Here are the number of ways you can roll each number with 2 dice: 2, 12: 1 way (1,1 or 6,6) 3, 11: 2 ways (1,2; 2,1 or 5,6; 6,5) Remember that the red and green dice are different. 4, 10: 3 ways 5, 9 : 4 ways 6, 8 : 5 ways 7: 6 ways. So the probability of rolling a 7 is 6/36 = 1/6. The probability of rolling a 6 or a 7 is (6+5)/36 = 11/36 And so on. 1. Sum is at least 8: Sum is 8,9,10,11,or12. Probability is (5+4+3+2+1)/36 = 15/36 = 5/12 2. Probability of 7 is 1/6, so probability of complement is 1-(1/6) = 5/6 3. 1/6: The two rolls are independent. The value of one does not effect the probability of any outcome ON THE SECOND ROLL. It will affect the probability of the outcome of the sum. For example, if you roll 2 on the first roll, there is 0 probability of rolling 15 as the total of two rolls. 4. Since the rolls are independent, the probability of them BOTH occurring is the product of their probabilities. VERY IMPORTANT: this is only true when the events are independent. so 1/6 on the first roll, 1/6 on the second, so 1/36 all together.
# How do you divide ( -i-8) / (-i +7 ) in trigonometric form? Dec 17, 2015 $\frac{- i - 8}{- i + 7} = \sqrt{\frac{65}{50}} {e}^{\arccos \left(- \frac{8}{\sqrt{65}}\right) - \arccos \left(- \frac{7}{\sqrt{50}}\right)}$ #### Explanation: Usually I always simplify this kind of fraction by using the formula $\frac{1}{z} = \frac{z \overline{z}}{\left\mid z \right\mid} ^ 2$ so I'm not sure what I'm going to tell you works but this is how I'd solve the problem if I only wanted to use trigonometric form. $\left\mid - i - 8 \right\mid = \sqrt{64 + 1} = \sqrt{65}$ and $\left\mid - i + 7 \right\mid = \sqrt{50}$. Hence the following results : $- i - 8 = \sqrt{65} \left(- \frac{8}{\sqrt{65}} - \frac{i}{\sqrt{65}}\right)$ and $- i + 7 = \sqrt{50} \left(\frac{7}{\sqrt{50}} - \frac{i}{\sqrt{50}}\right)$ You can find $\alpha , \beta \in \mathbb{R}$ such that $\cos \left(\alpha\right) = - \frac{8}{\sqrt{65}}$, $\sin \left(\alpha\right) = - \frac{1}{\sqrt{65}}$, $\cos \left(\beta\right) = \frac{7}{\sqrt{50}}$ and $\sin \left(\beta\right) = - \frac{1}{\sqrt{50}}$. So $\alpha = \arccos \left(- \frac{8}{\sqrt{65}}\right) = \arcsin \left(- \frac{1}{\sqrt{65}}\right)$ and $\beta = \arccos \left(- \frac{7}{\sqrt{50}}\right) = \arcsin \left(- \frac{1}{\sqrt{50}}\right)$, and we can now say that $- i - 8 = \sqrt{65} {e}^{\arccos} \left(- \frac{8}{\sqrt{65}}\right)$ and $- i + 7 = \sqrt{50} {e}^{\arccos} \left(- \frac{7}{\sqrt{50}}\right)$.
# HSPT Math : How to do other word problems ## Example Questions ← Previous 1 3 4 5 6 7 8 9 10 ### Example Question #1 : How To Do Other Word Problems What is the prime factorization of 48? Explanation: We can find the prime factorization of a number by dividing 48 into smaller numbers. 2 is prime 2 is prime 2 is prime Both 2 and 3 are prime. So we have four 2's and one 3. Multiply these numbers to get 48. ### Example Question #1 : How To Do Other Word Problems What is the least common multiple of 6 and 8? Explanation: One easy way to find the least common multiple is to take the largest of the numbers given to you, and find its multiples. 8: 8, 16, 24, 32, 40, 48... Now start with 8, and see if any of these numbers is divisible by 6. The first number that is divisible by 6 is the least common multiple of 6 and 8. 8: No 16: No 24: Yes 24 is the least common multiple of 6 and 8. ### Example Question #1 : How To Do Other Word Problems How many whole numbers are between and ? Explanation: Convert each fraction to a mixed number. So how many whole numbers are between and ? 4, 5, 6, and 7 would fall between these two numbers. 3 would not. So there are 4 total whole numbers between them. ### Example Question #4 : How To Do Other Word Problems What is the next number in this progression? 18, 10, 2, -6 Explanation: This is an arithmetic progression. Therefore you must add or subtract a number to the number before it to arrive at the next number. We must find the difference between each number and apply it to the final number to find the next number in the series. To start, take the second number and subtract the first number. In this case it yields Then take the third number and subtract the second number which results in If the resulting number is the same you now know the difference between each number in the sequence and can apply the difference to each resulting number. In this case the difference is So we take the final number and add the difference to it to result with ### Example Question #1 : How To Do Other Word Problems How many inches are in a foot? Explanation: This is a conversion question. There are 12 inches in a foot. ### Example Question #42 : Word Problems What is Explanation: When multiplying monomials with the same base, you add the powers of each monomial together. So in this case ### Example Question #43 : Word Problems Explanation: This tests your knowledge of scientific notation. To solve you must move the decimal places to the right to the power of the ten. In this case you must move the decimal place to the right 7 spaces from its original spot. The solution is ### Example Question #44 : Word Problems You have 5 types of meat, 3 different types of vegetables, and 8 different types of bread. How many different sandwiches can you make using only one of each ingredient? Explanation: This is a combination question. To find the total number of sandwiches using only one ingredient we create three spaces for each ingredient and place each number of possible ingredients into each space. We then multiply each number in each space together to find the answer ### Example Question #2 : How To Do Other Word Problems The square of what number is 75 divided by 3? Explanation: To answer this question we must first divide the original number by the number following it to get Once we have this number we then square root it to find the answer ### Example Question #482 : Hspt Mathematics What is the absolute value of -57?
In this lesson, children will place shoes in order by size. ### Math Lesson for: Toddlers/Preschoolers (See Step 5: Adapt lesson for toddlers or preschoolers.) Algebra Measurement ### Learning Goals: This lesson will help toddlers and preschoolers meet the following educational standards: • Understand patterns, relations and functions • Understand measurable attributes of objects and the units, systems and processes of measurement ### Learning Targets: After this lesson, toddlers and preschoolers should be more proficient at: • Comparing and ordering objects by length • Sorting, classifying and ordering objects by size, number and other properties ### Lesson plan for toddlers/preschoolers #### Step 1: Gather materials. • Shoes • Paper • Markers • Tape measure Note: Small parts pose a choking hazard and are not appropriate for children age five or under. Be sure to choose lesson materials that meet safety requirements. #### Step 2: Introduce activity. 1. Say: “Put your feet out in front of you, so that everyone in the circle can see the bottoms of your shoes.” 2. Ask: “Who do you think has the longest shoe? Who do you think has the shortest shoe?” 3. Say: “We are going to see who has a longer shoe.” #### Step 3: Engage children in lesson activities. 1. Ask two children come to the front of the circle. Try to choose two children with obviously different shoe sizes. 2. Ask: “Who do you think is going to have the longer shoe?” 3. Tell the children to put their feet side by side. 4. Ask the children if they can tell which shoe is longer and which is shorter. 5. Using blank paper, trace around each child’s shoe. Hold up the tracing and ask: “Which one is longer? Which is shorter?” 6. Tell the children to compare the two tracings and ask them to point to the one that is longer. 7. Continue to pair up the children to compare their shoe lengths until all of the children have had their turn. #### Step 4: Math vocabulary. • Long, Longer, Longest: Comparison words for length • Short, Shorter, Shortest: Comparison words for length • Compare: To view at least two things and identify the similarities and differences #### Step 5: Adapt lesson for toddlers or preschoolers. ###### Toddlers may: • Watch as you measure their shoes • Listen as you compare the lengths • Decorate their shoe tracings ###### Home child care providers may: • Use the vocabulary terms “longer” and “shorter” when describing the child’s shoes • Show the children the shoes side by side and point out which one is the longest and which one is the shortest ###### Preschoolers may: • Make comparisons • Measure their shoes • Use a tape measure to compare lengths of shoes ###### Home child care providers may: • Write the measurements down on a piece of paper or on a white board or chalk board • Help the children graph the shoe patterns on the wall by hanging the shoes from the shortest to the longest • Use the vocabulary terms “shortest” and “longest” ### Suggested Books • The Best Bug Parade by Stuart Murphy (New York: Harper Trophy, 1996) • The Giant Cabbage: An Alaskan Folktale by Cherie Stihler (Seattle: Sasquatch, 2003) ### Music and Movement Finger play: Little, Bigger, Biggest A little ball, (Make ball with your finger and thumb.) A bigger ball, (Make ball with your two hands.) And a great big ball I see. (Make a big ball shape with your arms.) Now help me count them, One, two, three! (Repeat gestures for each size.) Variations: • Use cutout paper shapes or real objects to show the three sizes. • Create additional verses with other shapes of different sizes. ### Outdoor Connections Go for a walk and pick up sticks. Identify who found the shortest stick and who found the longest stick.
# (1 - 15%) × 2,200 = ? Percentage Decrease Change: Decrement the Number 2,200 by 15% (Percent) of its Value and Calculate the Actual Change (the Absolute Difference, the Variation). How Do You Decrease the Number by a Percentage of Its Value? ## Decrease numbers by percentages of their values and calculate the absolute percentage change (difference). Examples ### How to calculate the new value of a number by decreasing it by percentage of its value and the absolute percentage change (difference) • Percentage decreased number = Initial value - Absolute percentage change (difference) • Absolute percentage change (difference) = Percentage decrease % × Initial value • Percentage decreased number = Initial value - Percentage decrease % × Initial value = (1 - Percentage decrease %) × Initial value • In conclusion: ### Four examples of calculations below. • How to decrease a number by percentage of its value • How to calculate the absolute percentage change: #### Example 1: • If Percentage decrease % = 20%, (1 - Percentage decrease %) = 1 - 20% = 100/100 - 20/100 = 80/100 = 0.8 => • Percentage decreased number = 0.8 × Initial value, and Initial value = Percentage decreased number / 0.8; Absolute percentage change (difference) = New value - Initial value; #### Example 2: • If Percentage decrease % = 5%, (1 - Percentage decrease %) = 1 - 5% = 100/100 - 5/100 = 95/100 = 0.95 => • Percentage decreased number = 0.95 × Initial value, and Initial value = Percentage decreased number / 0.95; Absolute percentage change (difference) = New value - Initial value; #### Example 3: • If Percentage decrease % = 120%, (1 - Percentage decrease %) = 1 - 120% = 100/100 - 120/100 = -20/100 = -0.2 => • Percentage decreased number = -0.2 × Initial value, and Initial value = Percentage decreased number / -0.2; Absolute percentage change (difference) = New value - Initial value; #### Example 4: • If Percentage decrease % = -30%, (1 - Percentage decrease %) = 1 - (-30%) = 1 + 30% = 100/100 + 30/100 = 130/100 = 1.3 => • Percentage decreased number = 1.3 × Initial value, and Initial value = Percentage decreased number / 1.3; this is in fact a percentage increase of the initial number; Absolute percentage change (difference) = New value - Initial value.
# Math in Focus Grade 5 Chapter 15 Practice 4 Answer Key Understanding and Measuring Volume This handy Math in Focus Grade 5 Workbook Answer Key Chapter 15 Practice 4 Understanding and Measuring Volume provides detailed solutions for the textbook questions. ## Math in Focus Grade 5 Chapter 15 Practice 4 Answer Key Understanding and Measuring Volume These solids are formed by stacking unit cubes in the corner of a room. Find the volume of each solid. Question 1. Volume = _________ cubic units Volume of give cube has 27 cubic units, Explanation: As we know volume of solid is l X w X h, so given cube has 3 unit X 3 unit X 3 unit = 27 cubic units. Question 2. Volume = _________ cubic units Volume of give cube has 32 cubic units, Explanation: As we know the volume of solid is l X w X h, so the given cube has 4 units X 4 units X 2 units = 32 cubic units. Question 3. Volume = _________ cubic units Volume of given cube has 16 cubic units, Explanation: As we know the volume of solid is l X w X h, Given cube contains 2 fewer small unit cubes so first we calculate the total surface and subtract missing cubic units, Total surface area has 3 units X 2 units X 3 units = 18 cubic units. the surface area of small unit cubes is 1 unit X 1 unit X 2 units = 2 cubic units, therefore the volume of the given cube is 18 cubic units – 2 cubic units = 16 cubic units. Question 4. Volume = _________ cubic units Volume of 9 cubic units, Explanation: Given solid cube has 9 unit cubes with 1 unit X 1 unit X 1 unit each, So the volume of given cube is 9 X (1 unit X 1 unit X 1 unit) = 9 cubic units. These solids are formed by stacking 1-centimeter cubes in the corner of a room. Find the volume of each solid. Question 5. Volume = ______ cm3 Volume 11 cm3, Explanation: Given solid cube has 11 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 11 X (1 cm X 1 cm X 1 cm) = 11 cm3. Question 6. Volume = _________ cm3 Volume 8 cm3, Explanation: Given solid cube has 8 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 8 X (1 cm X 1 cm X 1 cm) = 8 cm3. Question 7. Volume = _________ cm3 Volume 10 cm3, Explanation: Given solid cube has 10 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 10 X (1 cm X 1 cm X 1 cm) = 10 cm3. Question 8. Volume = ______ cm3 Volume 12 cm3, Explanation: Given solid cube has 12 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 12 X (1 cm X 1 cm X 1 cm) = 12 cm3. Question 9. Volume = _________ cm3 Volume 7 cm3, Explanation: Given solid cube has 7 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 7 X (1 cm X 1 cm X 1 cm) = 7 cm3. Question 10. Volume = _________ cm3 Volume 11 cm3, Explanation: Given solid cube has 11 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 11 X (1 cm X 1 cm X 1 cm) = 11 cm3. These solids are built using 1-centimeter cubes. Find the volume of each solid. Then compare their volumes and fill in the blanks. Question 11. Solid _________ has a greater volume than solid _________. Solid B has a greater volume than solid A, Explanation: Given solid cube A has 5 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 5 X (1 cm X 1 cm X 1 cm) = 5 cm3, and given solid cube B has 8 units cms with 1 cm X 1 cm X 1 cm each, So the volume of given cubes is 8 X (1 cm X 1 cm X 1 cm) = 8 cm3. Solid B has a greater volume than solid A. These solids are built using 1-meter cubes. Find the volume of each solid. Then compare their volumes and fill in the blanks. Question 12. Solid _________ has less volume than solid _________. Solid 8 m3 has less volume that solid 11 m3, Explanation: Given solid cube C has 8 units m’s with 1 m X 1 m X 1 m each, So the volume of given cubes is 8 X (1 m X 1 m X 1 m) = 8 m3, and given solid cube D has 11 units m’s with 1 m X 1 m X 1 m each, So the volume of given cubes is 11 X (1 m X 1 m X 1 m) = 11 m3. Solid 8 m3 has less volume that solid 11 m3. These solids are built using 1-inch cubes. Find the volume of each solid. Then compare their volumes and f411 in the blanks. Question 13. Length = __________ in. Width = __________ in. Height = _________ in. Volume = _________ in.3 Length = __________ in. Width = __________ in. Height = _________ in. Volume = _________ in.3 Solid _________ has less volume than solid _________. Solid E has less volume than solid F, Explanation: Given the solid cube of E which has a length of 2 in, width 2 in and height 3 in, so the volume of cube E is 12 in.3, Given solid cube of F which has a length of 4 in, width of 2 in, and height of 2 in, so volume of cube F is 16 in.3. Solid E has less volume than solid F. These solids are built using 1-foot cubes. Find the volumes of each solid. Then compare their volumes and fill in the blanks. Question 14. Length = _____2_____ ft. Width = ______2____ ft. Height = _____2____ ft. Volume = _____8____ ft.3 Length = ____4______ ft. Width = _____4_____ ft. Height = _____4____ ft. Volume = _____64____ ft.3 Solid _________ has a greater volume than solid _________.
# How do you differentiate y =sqrtln(x^2-3x) using the chain rule? dy/dx=(2x-3)/(2(x^2-3x)sqrt(ln (x^2-3x)) #### Explanation: $y = \sqrt{\ln \left({x}^{2} - 3 x\right)}$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\sqrt{\ln \left({x}^{2} - 3 x\right)}\right)$ using formula: $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\ln \left({x}^{2} - 3 x\right)}} \cdot \frac{d}{\mathrm{dx}} \left(\ln \left({x}^{2} - 3 x\right)\right)$ using now the formula: $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\ln \left({x}^{2} - 3 x\right)}} \cdot \frac{1}{{x}^{2} - 3 x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\ln \left({x}^{2} - 3 x\right)}} \cdot \frac{1}{{x}^{2} - 3 x} \left(2 x - 3\right)$ After simplification, the final answer is dy/dx=(2x-3)/(2(x^2-3x)sqrt(ln (x^2-3x))
# Differentiate the following functions with respect to x : Question: Differentiate the following functions with respect to $x$ : $\tan ^{-1}\left\{\frac{\mathrm{x}}{\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}\right\}, \mathrm{a}<\mathrm{x}<\mathrm{a}$ Solution: $y=\tan ^{-1}\left\{\frac{x}{\sqrt{a^{2}-x^{2}}}\right\}$ Let $x=a \sin \theta$ Now $y=\tan ^{-1}\left\{\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right\}$ Using $\sin ^{2} \theta+\cos ^{2} \theta=1$ $y=\tan ^{-1}\left\{\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}}\right\}$ $y=\tan ^{-1}\left\{\frac{\sin \theta}{\cos \theta}\right\}$ $y=\tan ^{-1}(\tan \theta)$ Considering the limits, $-a$-a $-1<\sin \theta<1$ $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ Now, $y=\tan ^{-1}(\tan \theta)$ $y=\theta$ $y=\sin ^{-1}\left(\frac{x}{a}\right)$ Differentiating w.r.t $\mathrm{x}$, we get $\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1}\left(\frac{x}{a}\right)\right)$ $\frac{d y}{d x}=\frac{a}{\sqrt{a^{2}-x^{2}}} \times \frac{1}{a}$ $\frac{d y}{d x}=\frac{1}{\sqrt{a^{2}-x^{2}}}$
# if Question: If $e^{y}+x y=e$, the ordered pair $\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)$ at $x=0$ is equal to : 1. (1) $\left(\frac{1}{e},-\frac{1}{e^{2}}\right)$ 2. (2) $\left(-\frac{1}{e}, \frac{1}{e^{2}}\right)$ 3. (3) $\left(\frac{1}{e}, \frac{1}{e^{2}}\right)$ 4. (4) $\left(-\frac{1}{e},-\frac{1}{e^{2}}\right)$ Correct Option: , 2 Solution: Given, $e^{y}+x y=e$ ........$\ldots$ (i) Putting $x=0$ in (i), $\Rightarrow e^{y}=e \Rightarrow y=1$ On differentiating (i) w. r. to $x$ $e^{y} \frac{d y}{d x}+x \frac{d y}{d x}+y=0$.....(ii) Putting $y=1$ and $x=0$ in (ii), $e \frac{d y}{d x}+0+1=0 \Rightarrow \frac{d y}{d x}=-\frac{1}{e}$ On differentiating (ii) w. r. to $x$, $e^{y} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \cdot e^{y} \cdot \frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{d y}{d x}=0$..(iii) Putting $y=1, x=0$ and $\frac{d y}{d x}=-\frac{1}{e}$ in (iii), $e \frac{d^{2} y}{d x^{2}}+\frac{1}{e}-\frac{2}{e}=0 \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{1}{e^{2}}$ Hence, $\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right) \equiv\left(-\frac{1}{e}, \frac{1}{e^{2}}\right)$
# Subtract the numbers: 3,964 - 9,471 = ? Calculate the numbers difference and learn how to do the subtraction, column subtracting method, from right to left ## The operation to perform: 3,964 - 9,471 ### Stack the numbers on top of each other. #### And so on... 9 4 7 1 - 3 9 6 4 ? ## Subtract column by column; start from the column on the right ### Subtract the digits in the ones column: #### When borrowing, 1 ten = 10 ones. Add 10 to the top digit in the column of the ones: 10 + 1 = 11. 6 9 4 7 11 - 3 9 6 4 #### After borrowing, the subtraction has become: 11 - 4 = 10 + 1 - 4 = 10 + 1 - 4 = 1 + (10 - 4) = 1 + 6 = 7. 7 is the ones digit. Write it down at the base of the ones column. 6 9 4 7 11 - 3 9 6 4 7 ### Subtract the digits in the tens column: #### 76 - 6 = 0. 0 is the tens digit. Write it down at the base of the tens column. 6 9 4 7 11 - 3 9 6 4 0 7 ### Subtract the digits in the hundreds column: #### When borrowing, 1 thousand = 10 hundreds. Add 10 to the top digit in the column of the hundreds: 10 + 4 = 14. 8 6 9 14 7 11 - 3 9 6 4 0 7 #### After borrowing, the subtraction has become: 14 - 9 = 10 + 4 - 9 = 10 + 4 - 9 = 4 + (10 - 9) = 4 + 1 = 5. 5 is the hundreds digit. Write it down at the base of the hundreds column. 8 6 9 14 7 11 - 3 9 6 4 5 0 7 ### Subtract the digits in the thousands column: #### 98 - 3 = 5. 5 is the thousands digit. Write it down at the base of the thousands column. 8 6 9 14 7 11 - 3 9 6 4 5 5 0 7 ## The latest numbers that were subtracted 3,964 - 9,471 = ? Oct 01 11:55 UTC (GMT) - 526 + 209 - 248 - 215 + 228 + 214 + 296 = ? Oct 01 11:55 UTC (GMT) - 403 + 478 - 360 + 632 + 505 = ? Oct 01 11:54 UTC (GMT) - 2,397 - 9,783 = ? Oct 01 11:54 UTC (GMT) 932 + 991 - 879 + 668 + 297 - 467 = ? Oct 01 11:54 UTC (GMT) - 637 + 4,289 - 846 + 2,224 - 770 - 762 = ? Oct 01 11:54 UTC (GMT) 3,825 + 464 - 437 + 1,969 + 97 = ? Oct 01 11:54 UTC (GMT) 639 + 284 + 356 - 271 + 320 - 294 - 399 = ? Oct 01 11:54 UTC (GMT) - 2,065 + 1,986 = ? Oct 01 11:54 UTC (GMT) 419 - 434 - 393 + 612 - 541 = ? Oct 01 11:54 UTC (GMT) - 100,004 + 98 = ? Oct 01 11:53 UTC (GMT) - 861 + 335 = ? Oct 01 11:53 UTC (GMT) 921 - 435 = ? Oct 01 11:53 UTC (GMT) All the numbers that were subtracted by users
# If 5+9+13+... to n terms7+9+11+...to (n+1) terms=1716, then n = Question: If $\frac{5+9+13+\ldots \text { to } n \text { terms }}{7+9+11+\ldots \text { to }(n+1) \text { terms }}=\frac{17}{16}$, then $n=$ (a) 8 (b) 7 (c) 10 (d) 11 Solution: Here, we are given, $\frac{5+9+13+\ldots \text { to } n \text { terms }}{7+9+11+\ldots \text { to }(n+1) \text { terms }}=\frac{17}{16}$ .......(1) We need to find n. So, first let us find out the sum of n terms of the A.P. given in the numerator. Here we use the following formula for the sum of n terms of an A.P., $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ Where; a = first term for the given A.P. d = common difference of the given A.P. = number of terms Here, Common difference of the A.P. (d) = $=9-5$ $=4$ Number of terms (n) = n First term for the given A.P. (a) = So, using the formula we get, $S_{n}=\frac{n}{2}[2(5)+(n-1)(4)]$ $=\left(\frac{n}{2}\right)[10+(4 n-4)]$ $=\left(\frac{n}{2}\right)(6+4 n)$ $=n(3+2 n)$ ...........(2) Similarly, we find out the sum of $(n+1)$ terms of the A.P. given in the denominator $(7+9+11+\ldots)$. Here, Common difference of the A.P. $(d)=a_{2}-a_{1}$ $=9-7$ $=2$ Number of terms (n) = n First term for the given A.P. (a) = So, using the formula we get, $S_{n+1}=\frac{n+1}{2}[2(7)+[(n+1)-1](2)]$ $=\left(\frac{n+1}{2}\right)[14+(n)(2)]$ $=(n+1)(7+n)$ $=7 n+7+n^{2}+n$ $=n^{2}+8 n+7$$\ldots(3)$ Now substituting the values of (2) and (3) in equation (1), we get, $\frac{2 n^{2}+3 n}{n^{2}+8 n+7}=\frac{17}{16}$ $16\left(2 n^{2}+3 n\right)=17\left(n^{2}+8 n+7\right)$ $32 n^{2}+48 n=17 n^{2}+136 n+119$ $32 n^{2}-17 n^{2}+48 n-136 n-119=0$ $15 n^{2}-88 n-119=0$ Further solving the quadratic equation for n by splitting the middle term, we get, $15 n^{2}-88 n-119=0$ $15 n^{2}-105 n+17 n-119=0$ $15 n(n-7)+17(n-7)=0$ $(15 n+17)(n-7)=0$ So, we get $15 n+17=0$ $15 n=-17$ $n=\frac{-17}{15}$ Or $n-7=0$ $n=7$ Since is a whole number, it cannot be a fraction. So, Therefore, the correct option is (b).
The "3M" Game I well remember my wonder that I experienced when I learned about logarithms in Advanced Algebra. The thing that most caught my attention was that you could "multiply two numbers" through a not-so- simple addition process. To do this, you first find the logarithm of each of the two numbers, add them, then reconvert that back to a number that would be the product required. Sounds complicated, but it works! (For those readers who have not reached this level of math yet, do not turn away from this page until you've read a little more.) In algebra, we write the basic logarithmic property this way: log (a × b) = log a + log b Of course, this is just a "taste" of a much larger and grander topic in mathematics, but it is meant to give a connection to what I want to show you this time, namely how you can "MULTPLY by ADDING", using a much simpler, arithmetic basis. Triangular Numbers Before we can proceed, we must introduce what to some persons may be a new topic, that of Triangular Numbers. Square numbers have been discussed frequently in this website, but now we are going to look at a family of numbers that form triangles. Observe: 10 is a triangular number, because 10 things can be arranged in a triangular array like this: * * * * * * * * * * From this sort of picture it is easy to form and determine many other triangular numbers. * * * * * * * * * * Here we see that 1, 3, and 6 are the first three triangular numbers. And OF COURSE with these few examples we can see a short cut for finding other triangular numbers. 10 = 1 + 2 + 3 + 4 6 = 1 + 2 + 3 3 = 1 + 2 1 = 1 So going in the higher direction, we have 15, 21, 28, 36, 45, and 55, thus giving us the first ten triangular numbers. To make discussing a particular triangular number a bit more convenient, we will use the following notation [using the 4th one as an example]: T4 = 10 The First 50 Triangular Numbers T1 = 1 T11 = 66 T21 = 231 T31 = 496 T41 = 861 T2 = 3 T12 = 78 T22 = 253 T32 = 528 T42 = 903 T3 = 6 T13 = 91 T23 = 276 T33 = 561 T43 = 946 T4 = 10 T14 = 105 T24 = 300 T34 = 595 T44 = 990 T5 = 15 T15 = 120 T25 = 325 T35 = 630 T45 = 1035 T6 = 21 T16 = 136 T26 = 351 T36 = 666 T46 = 1081 T7 = 28 T17 = 153 T27 = 378 T37 = 703 T47 = 1128 T8 = 36 T18 = 171 T28 = 406 T38 = 741 T48 = 1176 T9 = 45 T19 = 190 T29 = 435 T39 = 780 T49 = 1225 T10 = 55 T20 = 210 T30 = 465 T40 = 820 T50 = 1275 Magical Multiplying Method Finally we're ready to show you the magical way to multiply without multiplying anything. I call it the "Magical Multiplying Method" (or the "3M" way). First, I must confess we will do a little subtracting, too. Second, you will need a copy of the table above. I hope you don't mind too much. Let's take as an example 15 × 9. 1. Take the larger factor 15 and find T15 in the table above. It is 120. 2. Subtract 1 from 9, the smaller factor, getting 8. Find T8 in the table. It is 36. 3. Subtract the two factors, 15 - 9; that's 6. Find T6. It is 21. 4. Add the results of Steps #1 & #2, then subtract the result from Step #3. That's your product! 120 + 36 - 21 = 135 [Well, I never said it was going to be easier, shorter, or anything like that. It's just an interesting idea by itself, don't you agree? And with practice, it's not too difficult.] We can summarize this by a little algebra. This is the formula for the steps given above. a × b = Ta + Tb-1 - Ta-b Doesn't look so bad now, or does it? Rationale The reasons why I am presenting this activity are various: 1. It's a different way to do a common thing. 2. It's a good experience in "following steps" to achieve a certain goal. 3. It shows a connection between advanced math (logarithms) and simple arithmetic, both of which are being used to multiply two numbers. 4. It provides an experience in looking up information in tables to use for a certain purpose. 5. It show how an algebraic formula can show the various steps in compact style. Extension Activity For students who are, or have been, in Algebra I, it would be a good challenge to prove that the formula indeed works for ALL numbers a and b. To do so, however, one needs to know the general formula for the triangular numbers. I will not show how it is derived, rather just state it here.
## Precalculus (6th Edition) Blitzer $2$ ; $0.6+1.9i$ ; $-1.6+1.2i$ ; $-1.6-1.2i$ ; $0.6-1.9i$ Roots of a complex number represented in polar form can be found by the application of DeMoivre’s theorem. DeMoivre’s theorem for finding complex roots states that every complex number has two distinct complex square roots, three distinct complex cube roots, four distinct complex fourth roots, and so on. Each root has the same modulus. \begin{align} & r=z_{k}^{n} \\ & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\,\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \end{align} where $k$ is the number of distinct $n\text{th}$ root and $\theta$ is in radians. There are exactly five fifth roots of $32$ as per DeMoivre’s theorem. First write the provided number in rectangular form as follows: $32=32+0i$ Convert the above expression into polar form as follows: \begin{align} & 32=r\left( \cos \,\theta +i\,\sin \,\theta \right) \\ & =32\left( \cos \,0+i\,\sin \,0 \right) \end{align} Apply De Moivre’s theorem to find all fifth roots of $32$. ${{z}_{k}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi k}{5} \right)+i\,\sin \left( \frac{0+2\pi k}{5} \right) \right]$ Here, $k=0,1,2,3,4$. Insert the above values of $k$ to find five distinct fifth roots of the provided number. For $k=0$, \begin{align} & {{z}_{0}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 0 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 0 \right)}{5} \right) \right] \\ & =2\left[ \cos \,0+i\,\sin \,0 \right] \\ & =2 \end{align} For $k=1$, \begin{align} & {{z}_{1}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 1 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 1 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{2\pi }{5} \right)+i\,\sin \left( \frac{2\pi }{5} \right) \right] \end{align} Substitute the values of $\cos \left( \frac{2\pi }{5} \right)$ and $\sin \left( \frac{2\pi }{5} \right)$: \begin{align} & {{z}_{1}}=2\left[ 0.3090+i0.9511 \right] \\ & =0.6180+i1.9021 \\ & \approx 0.6+1.9i \end{align} For $k=2$, \begin{align} & {{z}_{2}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 2 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 2 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{4\pi }{5} \right)+i\,\sin \left( \frac{4\pi }{5} \right) \right] \end{align} Substitute the values of $\cos \left( \frac{4\pi }{5} \right)$ and $\sin \left( \frac{4\pi }{5} \right)$: \begin{align} & {{z}_{2}}=2\left[ -0.8090+i0.5877 \right] \\ & =-1.618+i1.1755 \\ & \approx -1.6+1.2i \end{align} For $k=3$, \begin{align} & {{z}_{3}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 3 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 3 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{6\pi }{5} \right)+i\,\sin \left( \frac{6\pi }{5} \right) \right] \end{align} Substitute the values of $\cos \left( \frac{6\pi }{5} \right)$ and $\sin \left( \frac{6\pi }{5} \right)$: \begin{align} & {{z}_{3}}=2\left[ -0.8090+i\left( -0.5878 \right) \right] \\ & =-1.6180-i1.1756 \\ & \approx -1.6-1.2i \end{align} For $k=4$, \begin{align} & {{z}_{4}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 4 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 4 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{8\pi }{5} \right)+i\,\sin \left( \frac{8\pi }{5} \right) \right] \end{align} Substitute the values of $\cos \left( \frac{8\pi }{5} \right)$ and $\sin \left( \frac{8\pi }{5} \right)$: \begin{align} & {{z}_{4}}=2\left[ 0.3090+i\left( -0.9511 \right) \right] \\ & =0.6180-i1.9021 \\ & \approx 0.6-1.9i \end{align}
# LINEAR INEQUALITY WORD PROBLEMS Linear inequality word problems : In this section, we will learn, how to solve word problems on linear inequalities. ## Linear Inequality Word Problems - Steps When we solve word problems on linear inequalities, we have to follow the steps given below. Step 1 : Read and understand the information carefully and translate the statements into linear inequalities. Step 2 : Solve for the variable using basic operations like addition, subtraction, multiplication and division. Step 3 : Find the solution set and obtain some of the possible solutions. Apart from the above steps, we have to make the following changes, when we multiply or divide each side of the inequality by a negative value. • If we have <, then change it as > • If we have >, then change it as < • If we have , then change it as • If we have , then change it as ## Linear Inequality Word Problems - Examples Example 1 : A and B are working on similar jobs, but their annual salaries differ by more than \$6000. If the annual salary of A is more than B and B earns \$27000 per year, then what are the possible values for the annual salary of A? Solution : Let x be the annual salary of A. Given : The annual salaries of A and B differ by more than \$6000. x - 27000 > 6000 x > 33000 So, the annual salary of A must be greater than \$33000. Example 2 : A worker can be paid according to the following schemes : In the first scheme he will be paid \$500 plus \$70 per hour, and in the second scheme he will paid \$120 per hour. If he works for x hours, then for what value of x does the first scheme gives better wages? Solution : Let x be the number of working hours. Amount of money, he gets in the first scheme : = 500 + 70x -----(1) Amount of money, he gets in the second scheme : = 120x -----(2) If the first scheme gives better wages, then we have (1) > (2) 500 + 70x > 120x Subtract 70x from each side. 500 > 50x Divide each side by 50. 10 > x x < 10 When he works for less than 10 hours (x < 10), the first scheme gives better wages. Example 3 : Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40. Solution : Let x and (x + 2) be the two required consecutive odd numbers. Given : The sum of the numbers is 40. x + (x + 2) < 40 2x + 2 < 40 Subtract 2 from each side. 2x < 38 x < 19 Given : The numbers are greater than 10. So, the required two consecutive odd numbers must be greater than 10 and less than 19. Therefore, the possible pairs of two consecutive odd numbers are (11, 13), (13, 15) and (15, 17) Example 4 : A model rocket is launched from the ground. The height h reached by the rocket after t seconds from lift off is given by h(t) = - 5t+ 100t, 0 ≤ t ≤ 20. If t is an integer, at what time the rocket is 495 feet above the ground ? Solution : According to the question, we have to find the value of t, when h(t) > 495 feet - 5t+ 100t > 495 Subtract 495 from each side. - 5t+ 100t - 495 > 0 Divide each side by (-5). t- 20t + 99 > 0 Solve the above quadratic inequality by factoring. t- 11t - 9t + 99 > 0 t(t - 11) - 9(t - 11) > 0 (t - 11)(t - 9) > 0 If we assume (t - 11)(t - 9) = 0, we will get the following values for t. t - 11 = 0t = 11 t - 9 = 0t = 9 Mark the values 9 and 11 on the number. From the number line above, we can get the following three intervals. (0, 9), (9, 11) and (11, 20) When t ∊ (0, 9), h(t) < 495 When t ∊ (9, 11), h(t) > 495 When t ∊ (11, 20), h(t) < 495 We get h(t) > 495, when ∊ (9, 11). Because t is an integer, the possible value we have for t in the interval is 10. So, the rocket is 495 feet above the ground, after 10 seconds, it is launched from the ground. After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems using linear inequalities. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. You can also visit our following web pages on different stuff in math. 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# What Is The Value Of X In The Equation (X + 12) = (X + 14) – 3 How do you find the value of x in an equation?, Isolate “x” on one side of the algebraic equation by multiplying the number that appears on the same side of the equation as part of an “x” fractional component. For example, in the equationx / 2 = 3, rewrite the equation as “x = 2 x 3″ and solve forx.” The solution is “x = 6.” Furthermore, What is the value of x enter your answer in the box X?, x is equal to seven. Finally, What is the value of x in the equation 221 17x?, If 221 is divided by 17, 13 is the answer. ### What is the value of x enter your answer in the box 39 102? The value of X is 39 degrees. You have to add 102 degrees and 39 degrees to get 141 degrees. ### What is the value of x enter your answer in the box 3x 50 6x 10? The value of x is 20. Step-by-step explanation: Given two vertical angles (3x+50)° and (6x10)°. ### What is the value of x in the equation 221 17x? If 221 is divided by 17, 13 is the answer. ### What is the value for Y enter your answer in the box Y? Answer: The value of y is 28. ### How do I calculate X? Just divide 3x and 9 by 3, the x term coefficient, to solve for x. 3x/3 = x and 3/3 = 1, so you’re left with x = 1. Check your work. To check your work, just plug x back in to the original equation to make sure that it works. ### How do you find X in a equation? Isolate “x” on one side of the algebraic equation by subtracting the sum that appears on the same side of the equation as the “x.” For example, in the equationx + 5 = 12″, rewrite the equation as “x = 12 – 5″ and solve for “x.” The solution is “x = 7.” ### How do you find X of a number? 1. How to calculate percentage of a number. Use the percentage formula: P% * X = Y 1. Convert the problem to an equation using the percentage formula: P% * X = Y. 2. P is 10%, X is 150, so the equation is 10% * 150 = Y. 3. Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10. ### What is the value for Y enter your answer in the box? Answer: The value of y is 28. ### What is the value for y 5y 10? The sum of angles in a triangle is 180 degrees, so to find 5y + 10, you do 180 – 50 – 50 = 80 degrees. So 5y + 10 = 80 degrees, and here you can solve. ### What is Y value? The vertical value in a pair of coordinates. How far up or down the point is. The Y Coordinate is always written second in an ordered pair of coordinates (x,y) such as (12,5). In this example, the value “5” is the Y Coordinate. ### What is the value of x enter your answer in the box X a triangle with angle measures thirty nine degrees and one hundred two degrees the third angle has an unknown measure X degrees? The value of X is 39 degrees. ### What is the value of x enter your answer in the box x Brainly? x is equal to seven. ### What is the value of x enter your answer in the box X a triangle with angle measures thirty nine degrees and one hundred two degrees the third angle has an unknown measure X degrees? The value of X is 39 degrees. ### What is the value of x enter your answer in the box 3x 50 6x 10? The value of x is 20. Step-by-step explanation: Given two vertical angles (3x+50)° and (6x10)°. ### What is the value of x enter your answer in the box X note image not drawn to scale triangle GEH with segment ED such that D is on Segment G? The value of x is 68 ft. ### What is the value of x in the equation 221 17x? If 221 is divided by 17, 13 is the answer. ### How do I calculate X? Just divide 3x and 9 by 3, the x term coefficient, to solve for x. 3x/3 = x and 3/3 = 1, so you’re left with x = 1. Check your work. To check your work, just plug x back in to the original equation to make sure that it works.
Question joanna has 7 rolls of coins worth 23 dollars. She has rolls of dimes worth 5 dollars each, and rolls of nicklets worth 2 dollars each. how many rolls of each kind does she have? 1. thugiang If She has rolls of dimes worth 5 dollars each, and rolls of nicklets worth 2 dollars each. how many rolls of each kind does she have is: 3 rolls of dimes and 4 rolls of nickels ### Number of rolls of dimes and nickels Number of rolls of dimes = x Number of rolls of nickels = y Total number of rolls = 7 x + y = 7 …… Equation 1 Total amount = \$ 23 5x + 2y = 23 ——– Equation 2 Solving the two equation x + y = 7 x=7-y Substitute into equation 2 5x + 2y = 23 5(7-y)+2y=23 35-5y+2y=23 Collect like terms -3y=23-35 -3y=-12 Divide both side by -3y y=-12/-3 y=4 nickels Substitution x+y=7 Where y=4 x+4=7 x=7-4 x=3 dimes Therefore If She has rolls of dimes worth 5 dollars each, and rolls of nicklets worth 2 dollars each. how many rolls of each kind does she have is: 3 rolls of dimes and 4 rolls of nickels
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> Chapter 1: Expressions, Equations, and Functions Difficulty Level: Basic Created by: CK-12 Introduction The study of expressions, equations, and functions is the basis of mathematics. Each mathematical subject requires knowledge of manipulating equations to solve for a variable. Careers such as automobile accident investigators, quality control engineers, and insurance originators use equations to determine the value of variables. Functions are methods of explaining relationships and can be represented as a rule, a graph, a table, or in words. The amount of money in a savings account, how many miles run in a year, or the number of trout in a pond are all described using functions. Throughout this chapter, you will learn how to choose the best variables to describe a situation, simplify an expression using the Order of Operations, describe functions in various ways, write equations, and solve problems using a systematic approach. Chapter Outline Summary This chapter first deals with expressions and how to evaluate them by using the correct order of operations. Tips on using a calculator are also given. It then builds on this knowledge by moving on to talk about equations and inequalities and the methods used to solve them. Next, functions are discussed in detail, with instruction given on using the proper notation, determining a function's domain and range, and graphing a function. How to determine whether or not a relation is a function is also covered. Finally, the chapter concludes by highlighting some general problem-solving strategies. Expressions, Equations, and Functions Review Define the following words: 1. Domain 2. Range 3. Solution 4. Evaluate 5. Substitute 6. Operation 7. Variable 8. Algebraic expression 9. Equation 10. Algebraic inequality 11. Function 12. Independent variable Evaluate the following expressions. 1. 3y(7(zy))\begin{align}3y (7 - (z - y))\end{align}; use y=7\begin{align}y = -7\end{align} and z=2\begin{align}z = 2\end{align} 2. m+3np4\begin{align}\frac{m + 3n - p}{4}\end{align}; use m=9, n=7,\begin{align}m = 9, \ n = 7,\end{align} and p=2\begin{align}p = 2\end{align} 3. \|p\|(n2)3\begin{align}\|p\| - \left (\frac{n}{2} \right )^3\end{align}; use n=2\begin{align}n = 2\end{align} and p=3\begin{align}p = 3\end{align} 4. \|v21\|\begin{align}\|v-21\|\end{align}; use v=70\begin{align}v=-70\end{align} Choose an appropriate variable to describe the situation. 1. The number of candies you can eat in a day 2. The number of tomatoes a plant can grow 3. The number of cats at a humane society 4. The amount of snow on the ground 5. The number of water skiers on a lake 6. The number of geese migrating south 7. The number of people at a trade show The surface area of a sphere is found by the formula A=4πr2\begin{align}A = 4 \pi r^2\end{align}. Determine the surface area for the following radii/diameters. 1. radius=10 inches\begin{align}radius=10 \ inches\end{align} 2. radius=2.4 cm\begin{align}radius=2.4 \ cm\end{align} 3. diameter=19 meters\begin{align}diameter=19 \ meters\end{align} 4. radius=0.98 mm\begin{align}radius=0.98 \ mm\end{align} 5. diameter=5.5 inches\begin{align}diameter=5.5 \ inches\end{align} Insert parentheses to make a true equation. 1. 1+23+4=15\begin{align}1 + 2 \cdot 3 + 4 = 15\end{align} 2. 532+6=35\begin{align}5 \cdot 3 - 2 + 6 = 35\end{align} 3. 3+172297=24\begin{align}3 + 1 \cdot 7 - 2^2 \cdot 9 - 7 = 24\end{align} 4. 4+6253=40\begin{align}4 + 6 \cdot 2 \cdot 5 - 3 = 40\end{align} 5. 32+274=33\begin{align}3^2 + 2 \cdot 7 - 4 = 33\end{align} Translate the following into an algebraic expression, equation, or inequality. 1. Thirty-seven more than a number is 612. 2. The product of u\begin{align}u\end{align} and –7 equals 343. 3. The quotient of k\begin{align}k\end{align} and 18 4. Eleven less than a number is 43. 5. A number divided by –9 is –78. 6. The difference between 8 and h\begin{align}h\end{align} is 25. 7. The product of 8, –2, and r\begin{align}r\end{align} 8. Four plus m\begin{align}m\end{align} is less than or equal to 19. 9. Six is less than c\begin{align}c\end{align}. 10. Forty-two less than y\begin{align}y\end{align} is greater than 57. Write the pattern shown in the table with words and with an algebraic equation. 1. Movies watchedTotal time0011.52334.54657.5\begin{align}&\text{Movies watched} && 0 && 1 && 2 && 3 && 4 && 5\\ &\text{Total time} && 0 && 1.5 && 3 && 4.5 && 6 && 7.5\end{align} 2. A case of donuts is sold by the half-dozen. Suppose 168 people purchase cases of donuts. How many individual donuts have been sold? 3. Write an inequality to represent the situation: Peter’s Lawn Mowing Service charges $10 per mowing job and$35 per landscaping job. Peter earns at least 8,600 each summer. Check that the given number is a solution to the given equation or inequality. 1. t=0.9, 547(9t+5)\begin{align}t=0.9, \ 54 \le 7(9t+5)\end{align} 2. f=2; f+2+5f=14\begin{align}f=2; \ f+2+5f = 14\end{align} 3. p=6; 4p5p5\begin{align}p=-6; \ 4p-5p \le 5\end{align} 4. Logan has a cell phone service that charges18 dollars per month and $0.05 per text message. Represent Logan’s monthly cost as a function of the number of texts he sends per month. 5. An online video club charges$14.99 per month. Represent the total cost of the video club as a function of the number of months that someone has been a member. 6. What is the domain and range for the following graph? 7. Henry invested $5,100 in a vending machine service. Each machine pays him$128. How many machines does Henry need to install to break even? 8. Is the following relation a function? Solve the following questions using the 4-step problem-solving plan. 1. Together, the Raccoons and the Pelicans won 38 games. If the Raccoons won 13 games, how many games did the Pelicans win? 2. Elmville has 250 fewer people than Maplewood. Elmville has 900 people. How many people live in Maplewood? 3. The cell phone Bonus Plan gives you 4 times as many minutes as the Basic Plan. The Bonus Plan gives you a total of 1200 minutes. How many minutes does the Basic Plan give? 4. Margarite exercised for 24 minutes each day for a week. How many total minutes did Margarite exercise? 5. The downtown theater costs $1.50 less than the mall theater. Each ticket at the downtown theater costs$8. How much do tickets at the mall theater cost? 6. Mega Tape has 75 more feet of tape than everyday tape. A roll of Mega Tape has 225 feet of tape. How many feet does everyday tape have? 7. In bowling DeWayne got 3.5 times as many strikes as Junior. If DeWayne got 28 strikes, how many strikes did Junior get? Expressions, Equations, and Functions Test 1. Write the following as an algebraic equation and determine its value. On the stock market, Global First hit a price of $255 on Wednesday. This was$59 greater than the price on Tuesday. What was the price on Tuesday? 2. The oak tree is 40 feet taller than the maple. Write an expression that represents the height of the oak. 3. Graph the following ordered pairs: (1, 2), (2, 3), (3, 4), (4, 5) (5, 6), (6, 7). 4. Determine the domain and range of the following function: 5. Is the following relation a function? Explain your answer. {(3,2),(3,4),(5,6),(7,8)}\begin{align}\left \{(3, 2), (3, 4), (5, 6), (7, 8) \right \}\end{align} 6. Evaluate the expression (5bc)a\begin{align}(5bc)-a\end{align} if a=2, b=3, and c=4\begin{align}a=2,\ b=3, \ \text{and } \ c=4\end{align}. 7. Simplify: 3[36÷(3+6)]\begin{align}3[36 \div (3+6)]\end{align}. 8. Translate the following into an algebraic equation and find the value of the variable: One-eighth of a pizza costs \$1.09. How much was the entire pizza? 9. Use the 4-step problem-solving method to determine the solution to the following: The freshman class has 17 more girls than boys. There are 561 freshmen. How many are girls? 10. Underline the math verb in this sentence: 8 divided by y\begin{align}y\end{align} is 48. 11. Jesse packs 16 boxes per hour. Complete the table to represent this situation. HoursBoxes02458101214\begin{align}&\text{Hours} && 0 && 2 && 4 && 5 && 8 && 10 && 12 && 14\\ &\text{Boxes}\end{align} 1. A group of students are in a room. After 18 leave, it is found that 78\begin{align}\frac{7}{8}\end{align} of the original number of students remain. How many students were in the room in the beginning? 2. What are the domain and range of the following relation: {(2,3),(4,5),(6,7),(2,3),(3,4)}?\begin{align}\left \{(2, 3), (4, 5), (6, 7), (-2, -3), (-3, -4) \right \}?\end{align} 3. Write a function rule for the following table: Time in hours, xDistance in miles, y00160212031804240\begin{align}& \text{Time in hours,}\ x && 0 && 1 && 2 && 3 && 4\\ & \text{Distance in miles,}\ y && 0 && 60 && 120 && 180 && 240\end{align} 1. Determine if the given number is a solution to the following inequality: 6yy>8; y=6\begin{align}\frac{6 - y}{y} > -8; \ y = 6\end{align} Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9611. 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# Distance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is: Save this PDF as: Size: px Start display at page: Download "Distance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is:" ## Transcription 1 Chapter 1 Kinematics 1.1 Basic ideas r(t) is the position of a particle; r = r is the distance to the origin. If r = x i + y j + z k = (x, y, z), then r = r = x 2 + y 2 + z 2. v(t) is the velocity; v = v is the speed. a(t) is the acceleration of the particle. In one-dimensional motion, we have Average speed during a time interval = Distance travelled, time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is: In more than one dimension, and the instantaneous velocity is v = ds dt = ṡ. r(t) = x(t) i + y(t) j + z(t) k = (x, y, z), v = dr dt A particle is at rest if v = 0. = ṙ = ẋ i + ẏ j + ż k = (ẋ, ẏ, ż). (1) 1.2 Acceleration The acceleration of a particle in one or more dimension is: a = dv dt = v = d2 r = r = (ẍ, ÿ, z). (2) dt2 1 2 2 1.3 Relative velocity In one dimension, let the position of a particle be r = s(t) i, with speed v = ṡ and acceleration a = v = s. Then a = dv dt = dv ds ds dt = v dv ds = d ( ) 1 ds 2 v2. (3) If we assume that the acceleration is uniform (a constant), and that at time t = 0, the particle is at position s = 0 and is travelling with speed u, then its speed at time t is: v = u + at, (4) its position at time t is s = ut at2, (5) and we can eliminate t from the two equations above to get: v 2 = u 2 + 2as. (6) With non-uniform acceleration in one or more dimensions, use (1) and (2) to calculate v and a from r (or a from v) by differentiation, and integrate to find r from v, or r and v from a. 1.3 Relative velocity The position of a particle P relative to a particle Q is r P Q = r P r Q. (7) Differentiating, we get the relative velocity v P Q = v P v Q, (8) and the relative acceleration a P Q = a P a Q. (9) 1.4 Angular speed Let θ be the angle that the line OP makes with a fixed reference line OA, so θ = POA, where O and A are fixed and P is moving. Then the instantaneous angular speed of P is ω = dθ dt = θ. (10) If the particle is moving in a circle of radius R, then the instantaneous speed of the particle is v = Rω. (11) 3 Chapter 2 Dynamics: Newton s laws of motion and gravitation 2.1 Newton s laws of motion A particle is a point mass, having no internal structure. We will not consider rigid body dynamics, where rotation of the object would have to be included (e.g., the spin on a tennis ball). All objects that we consider (stones, balls, people, trains, cars, boats etc) will be treated as particles. I: Newton s First Law of Motion: A particle moves at a constant velocity (perhaps zero) unless it is acted up on by a force. The momentum of a particle is its mass times its velocity: p = mv (12) II: Newton s Second Law of Motion: The rate of change of momentum of a particle is equal to the total force acting on the particle: F = dp dt = d dm (mv) = dt dt v + mdv = ṁv + m v. dt The total force F is the vector sum of all forces acting on the particle. If the mass of the particle is constant (ṁ = 0), then F = m dv dt = m v = ma (13) III: Newton s Third Law of Motion: If a particle A exerts a force F on particle B, then particle B exerts an equal and opposite force F on particle A. One consequence of Newton s Third Law of Motion is the Principle of Conservation of Momentum: the total momentum of two particles interacting though equal and opposite forces is constant. 3 4 4 2.2 Newton s Law of Gravitation 2.2 Newton s Law of Gravitation Two particles of masses M and m a distance r apart exert a mutually attractive force on each other: Gravitational Force = GMm r 2 (14) where G is the Universal Gravitational Constant G = m 3 /kg/s 2 in S.I. units. For an object close to the surface of a planet of mass M and radius R, the weight of the object is a force directed towards the centre of the planet, of magnitude Weight = GM GM m = mg where g = (15) R2 R 2 5 Chapter 3 Projectiles, lifts, ropes and pulleys, friction, collisions, circular motion 3.1 Projectiles A projectile starting from the origin at time t = 0 and with an initial speed U 0, launched with an angle α above the horizontal, moves under the influence of gravity: a = g j, where the x-direction is horizontal and the y-direction is vertically upwards, and g is the gravitational acceleration. We integrate to get the velocity: v = U 0 cos α i + (U 0 sin α gt) j, (using the given initial velocity), and integrate again to get the position: r = x i + y j = (U 0 cos α) t i + ((U 0 sin α) t 1 ) 2 gt2 j, (using the given initial position). Putting this together, let x(t) be the horizontal position, let y(t) be the vertical position, and let v = ẏ be the vertical component of the velocity (note that v is not the magnitude of v in this case), we have: x = (U 0 cos α) t (16) v = U 0 sin α gt (17) y = (U 0 sin α) t 1 2 gt2 (18) v 2 = (U 0 sin α) 2 2gy (19) Time t can be eliminated by writing t in terms of x using (16) and then substituting into (18). With a different initial velocity of postion, the expressions may be different, but the idea is the same. The range can be found by solving y = 0 (or whatever condition specifies when the particle comes back to the ground. 5 6 6 3.2 Lifts 3.2 Lifts If a lift (elevator) accelerates upwards with an acceleration a, then the apparent weight of an object of mass m in the lift is m(g + a). 3.3 Ropes and pulleys We will treat only massless inextensible ropes and strings, and the tension in a rope is the force that rope applies to the particles that are connected to either end of the rope. We consider only massless frictionless pulleys, so the tension in the rope on either side of the pulley is the same. Ropes can only pull particles, not push. If a rope connects two particles, then the speeds and accelerations of the two particles are the same (thought the directions will be different). 3.4 Friction If an object of mass m is moving across a rough horizontal surface, then there is a vertical normal force N that balances the object s weight mg, and there is an opposing frictional resistance R: R = µn, (20) where µ is called the coefficient of friction. If the object is not moving, but some horizontal forces are being applied, there may still be an opposing frictional force R < µn. 3.5 Impulse and collisions Newton s Second Law (II) is: F = m dv dt If we integrate this from time 0 to t, we get: t 0 F (t) dt = t 0 = m v = ma; m dv dt = mv(t) mv(0) dt In a collision, where the force is non-zero only for a very short time, from 0 to t, the Impulse is defined to be: I = t 0 F (t) dt = mv 1 mv 0, (21) where v 0 is the velocity before the collision, and v 1 is the velocity after the collision. Thus, the Impulse acting on a particle in a collision is equal to the change in the particle s momentum during the collision. The units of Impulse are N s (Newton seconds), and it is a vector. 7 Chapter 3 Projectiles, lifts, ropes and pulleys, friction, collisions, circular motion Inelastic collisions Suppose two particles collide and stick together. Then the total momentum before the collision is equal to the total momentum after the collision. This is called Conservation of momentum. In the case of one-dimensional motion, consider two particles of masses m A and m B moving to the right with speeds u A and u B respectively. If one of the particles is moving to the left, take its speed to be negative. The total momentum before any collision is therefore m A u A + m B u B. The particles stick together in an inelastic collision, then the new particle has mass m A + m B, and travels to the right with speed v, and so has total momentum (m A + m B )v. Equating these, we find v: Elastic collisions v = m Au A + m B u B m A + m B (22) If two particles collide and then bounce apart, then the total momentum is conserved: the total momentum before the collision is equal to the total momentum after the collision. In the case of one-dimensional motion, consider two particles of masses m A and m B moving to the right with speeds u A and u B respectively. If one of the particles is moving to the left, take its speed to be negative. After the collision, the particles have speeds v A and v B respectively, where again positive speeds mean that the particle is travelling to the right. The total momentum is the same before and after: m A u A + m B u B = m A v A + m B v B (23) In addition, we have Newton s Elastic Law (NEL), which says that when two particles of the same material collide directly, then their relative velocity after impact is in direct proportion to their relative velocity before impact, but in the opposite direction: NEL: v B v A = e(u B u A ), or v B v A u B u A = e (24) 3.6 Circular motion A particle P whose position vector is r(t) = R cos(ωt) i + R sin(ωt) j = x(t) i + y(t) j is moving in a circle of radius R, and the line OP makes an angle θ with the x-axis, where tan θ = y(t)/x(t) = tan(ωt), so θ = ωt. The particle has angular velocity ω see equation (10). The period of the oscillation is T = 2π/ω The velocity and acceleration of the particle are: v = ṙ = Rω sin(ωt) i+rω cos(ωt) j and a = v = Rω 2 cos(ωt) i Rω 2 sin(ωt) j = ω 2 r. 8 8 3.6 Circular motion The velocity v is tangent to the circle (since r v = 0) and the acceleration a is directed towards the centre of the circle (since a is a negative constant times r). The magnitudes of these vectors are constant: r = r = R, v = v = Rω and a = a = Rω 2 = v2 R. (25) The force required to keep the particle moving in a circle is the Centripetal Force. From Newton s second law, we have F = ma = mω 2 r, and F = F = mrω 2 = mv2 R. (26) This force could be provided from a variety of sources: gravity, friction, a string... 9 Chapter 4 Work, energy and power 4.1 Work and Kinetic Energy A particle moving in one dimension obeys Newton s Second Law (II). Using equation (3), we get: F = ma = m dv dt = mdv ds ds dt dv = mv ds = m d ( ) 1 ds 2 v2. If we integrate F ds from time 0 to t, where the particle moves from 0 to s in this time, we get: s s F ds = m d ( ) ds 2 v2 ds = 1 2 mv mv2 0, (27) where v 0 is the velocity at time 0 and v 1 is the velocity at time t. We define the Kinetic Energy (K.E.) of a particle of mass m, travelling at speed v, to be K.E. = 1 2 mv2 (28) and the work done by a force acting on a particle to be: Work done = F ds (29) so equation (27) can also be written as Work done on a particle = increase in the particle s Kinetic Energy. The units of work and Kinetic Energy are Newton-metres (Nm), or Joules (J). In the case where the force is a constant, we get Constant force: Work done = F s = Force distance, (30) where force and distance are measured in the same sense. 4.2 Gravitational Potential Energy Consider a particle moving under the influence of gravity, so the force acting on it is F = mg (negative because downwards). If the particle falls from h 1 to h 2 (with 9 10 Power h 1 > h 2 ), the work done by gravity is Work done = h2 h 1 mg ds = mg(h 2 h 1 ) = mg(h 1 h 2 ). This is equal to the increase in Kinetic Energy: mg(h 1 h 2 ) = 1 2 mv mv2 1, where v 1 is the initial speed (at height h 1 ) and v 2 is the final speed (at height h 2 ). Rearrange this to give the Principal of conservation of Energy : mgh mv2 1 = mgh mv2 2. (31) The sum of either side of this equation is called the total mechanical energy, and the first part of the sum is defined to be the Gravitational Potential Energy: the Potential Energy (P.E.) of a particle of mass m a distance h above the ground is: P.E. = mgh. (32) Mechanical energy is lost to friction and in collisions, but otherwise the total mechanical energy (Potential Energy + Kinetic Energy) is conserved. 4.3 Power Power is the rate at which work is done. 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Name: Name: 3834-1 - Page 1 1) If a woman runs 100 meters north and then 70 meters south, her total displacement is A) 170 m south B) 170 m north C) 30 m south D) 30 m north 2) The graph below represents the More information ### = o + t = ot + ½ t 2 = o + 2 Chapters 8-9 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the More information ### (1) (3) 1. This question is about momentum, energy and power. (a) In his Principia Mathematica Newton expressed his third law of motion as to every action there is always opposed an equal reaction. 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Backbone of Physics FORCES Forces Isaac Newton stated 3 laws that deal with forces and describe motion. Backbone of Physics Inertia Tendency of an object to remain in the same state of motion. Resists a change in motion. More information ### PHYSICS 025 FINAL EXAMINATION Friday, 2003 April 11 Print Name: Student No. A 1-10 A 11-20 A 21-30 B1 B2 B3 B4 C1 C2 C3 Total PHYSICS 025 FINAL EXAMINATION Friday, 2003 April 11 Time: 7:00 10:00 PM 1. This is a closed book test. You may use only a calculator, More information ### SIR MICHELANGELO REFALO CENTRE FOR FURTHER STUDIES VICTORIA GOZO SIR MICHELANGELO REFALO CENTRE FOR FURTHER STUDIES VICTORIA GOZO Half-Yearly Exam 2013 Subject: Physics Level: Advanced Time: 3hrs Name: Course: Year: 1st This paper carries 200 marks which are 80% of More information ### PHYS 101 Previous Exam Problems. Force & Motion I PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0-kg block is lowered with a downward More information ### AP Physics 1 Chapter 7 Circular Motion and Gravitation AP Physics 1 Chapter 7 Circular Motion and Gravitation Chapter 7: Circular Motion and Angular Measure Gravitation Angular Speed and Velocity Uniform Circular Motion and Centripetal Acceleration Angular More information ### Newton s 3 Laws of Motion Newton s 3 Laws of Motion 1. If F = 0 No change in motion 2. = ma Change in motion Fnet 3. F = F 1 on 2 2 on 1 Newton s First Law (Law of Inertia) An object will remain at rest or in a constant state of More information ### 1. Which one of the following situations is an example of an object with a non-zero kinetic energy? Name: Date: 1. Which one of the following situations is an example of an object with a non-zero kinetic energy? A) a drum of diesel fuel on a parked truck B) a stationary pendulum C) a satellite in geosynchronous More information ### Simple and Physical Pendulums Challenge Problem Solutions Simple and Physical Pendulums Challenge Problem Solutions Problem 1 Solutions: For this problem, the answers to parts a) through d) will rely on an analysis of the pendulum motion. There are two conventional More information ### Chapter 6 Work, Energy, and Power. Copyright 2010 Pearson Education, Inc. Chapter 6 Work, Energy, and Power What Is Physics All About? Matter Energy Force Work Done by a Constant Force The definition of work, when the force is parallel to the displacement: W = Fs SI unit: newton-meter More information
Courses Courses for Kids Free study material Offline Centres More Store If the ratio of the sum of n terms of two APs is $(7n\text{ }+\text{ }1):\left( 4n\text{ }+\text{ }27 \right)$, then the ratio of their ${{m}^{th}}$ terms is. Last updated date: 14th Jun 2024 Total views: 402.3k Views today: 12.02k Verified 402.3k+ views Hint: We start solving the problem by assigning variables for the first term and common difference for the two of given arithmetic progressions. We compare the given ratio of the sum of n terms using the sum of n terms of A.P ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$. We then make substitution for the value of n in order to get ${{m}^{th}}$ terms in both numerator and denominator by comparing with ${{n}^{th}}\ term=a+(n-1)d$ to get the final result. As mentioned in the question, the ratio of the sum of first n terms of two different arithmetic progressions is as follows For 1st arithmetic progression, let the first term and common difference be ‘a’ and ‘d’ respectively and for the 2nd arithmetic progression, let the first term and common difference be ‘A’ and ‘D’. Now, we can write as following using the formula given in the hint, $\dfrac{{{S}_{n}}}{{{S}_{n}}\grave{\ }}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}$ Now, we can write this ratio as follows $\dfrac{\dfrac{n}{2}\left( 2a+(n-1)d \right)}{\dfrac{n}{2}\left( 2A+(n-1)D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}$. $\dfrac{\left( 2a+(n-1)d \right)}{\left( 2A+(n-1)D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}$. $\dfrac{\left( a+\dfrac{(n-1)}{2}d \right)}{\left( A+\dfrac{(n-1)}{2}D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}$ ---(a). We know that the ${{n}^{th}}$ term of an A.P (Arithmetic progression) is defined as ${{T}_{n}}=a+\left( n-1 \right)d$---(1), where a is first term and d is the common difference. Now, we are asked the ratio of the ${{m}^{th}}$ term of these two series and let us assume it as ‘m’. Using equation (1) we get, $\dfrac{a+\left( m-1 \right)d}{A+\left( m-1 \right)D}=x\ \ \ \ \ ...(b)$ We need to make a substitution in place of n of the equation (a) to get equation (b). So, we compare both numerators (or denominators) to find the value that needs to substitute in n. So, we have $\dfrac{(n-1)}{2}=\left( m-1 \right)$. $n-1=2\left( m-1 \right)$. $n-1=2m-2$. $n=2m-2+1$. $n=2m-1$ ---(c). We now substitute the result obtained from equation (c) in equation (a). $\dfrac{\left( a+\dfrac{(2m-1-1)}{2}d \right)}{\left( A+\dfrac{(2m-1-1)}{2}D \right)}=\dfrac{\left( 7\times \left( 2m-1 \right)+1 \right)}{\left( 4\times \left( 2m-1 \right)+27 \right)}$. $\dfrac{\left( a+\dfrac{(2m-2)}{2}d \right)}{\left( A+\dfrac{(2m-2)}{2}D \right)}=\dfrac{\left( 14m-7+1 \right)}{\left( 8m-4+27 \right)}$. $\dfrac{\left( a+\left( m-1 \right)d \right)}{\left( A+\left( m-1 \right)D \right)}=\dfrac{\left( 14m-6 \right)}{\left( 8m+23 \right)}$. On comparing numerator and denominator with equation (1), we get $\dfrac{{{T}_{m}}}{T_{m}^{'}}=\dfrac{\left( 14m-6 \right)}{\left( 8m+23 \right)}$. We have found the ratio of the ${{m}^{th}}$ terms of two progressions as $\dfrac{14m-6}{8m+23}$. Note: We should not take the same first term and common difference for both progressions as it will make us confused and give wrong results. Whenever we get this type of problem, we need to first take the ratio and make a substitution that will be fit in order to get the ratio of required terms. We should not confuse the general terms of geometric, arithmetic and harmonic progressions.
# Differentiation from First Principles Differentiation is the process of finding the gradient of a variable function. A variable function is a polynomial function that takes the shape of a curve, so is therefore a function that has an always-changing gradient. #### Create learning materials about Differentiation from First Principles with our free learning app! • Flashcards, notes, mock-exams and more • Everything you need to ace your exams There is a traditional method to differentiate functions, however, we will be concentrating on finding the gradient still through differentiation but from first principles. This means using standard Straight Line Graphs methods of $$\frac{\Delta y}{\Delta x}$$ to find the gradient of a function. ## How does differentiation from first principles work? Differentiation from first principles involves using $$\frac{\Delta y}{\Delta x}$$ to calculate the gradient of a function. We will have a closer look to the step-by-step process below: STEP 1: Let $$y = f(x)$$ be a function. Pick two points x and $$x+h$$. The coordinates of x will be $$(x, f(x))$$ and the coordinates of $$x+h$$ will be ($$x+h, f(x + h)$$). STEP 2: Find $$\Delta y$$ and $$\Delta x$$. $$\Delta y = f(x+h) - f(x); \Delta x = x+h-x = h$$STEP 3: Complete $$\frac{\Delta y}{\Delta x}$$.$$\frac{\Delta y}{\Delta x} = \frac{f(x+h) - f(x)}{h}$$STEP 4: Take a limit: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$. ## The formula for differentiation from first principles The formula below is often found in the formula booklets that are given to students to learn differentiation from first principles: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ ## Derivative of sin(x) using first principles To find out the derivative of sin(x) using first principles, we need to use the formula for first principles we saw above: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ Here we will substitute f(x) with our function, sin(x): $f'(x) = \lim_{h\to 0} \frac{\sin(x+h) - \sin (x)}{h}$ For the next step, we need to remember the trigonometric identity: $$\sin(a + b) = \sin a \cos b + \sin b \cos a$$ Using the trigonometric identity, we can come up with the following formula, equivalent to the one above: $f'(x) = \lim_{h\to 0} \frac{(\sin x \cos h + \sin h \cos x) - \sin x}{h}$ We can now factor out the $$\sin x$$ term: \begin{align} f'(x) &= \lim_{h\to 0} \frac{\sin x(\cos h -1) + \sin h\cos x}{h} \\ &= \lim_{h \to 0}(\frac{\sin x (\cos h -1)}{h} + \frac{\sin h \cos x}{h}) \\ &= \lim_{h \to 0} \frac{\sin x (\cos h - 1)}{h} + lim_{h \to 0} \frac{\sin h \cos x}{h} \\ &=(\sin x) \lim_{h \to 0} \frac{\cos h - 1}{h} + (\cos x) \lim_{h \to 0} \frac{\sin h}{h} \end{align} here we need to use some standard limits: $$\lim_{h \to 0} \frac{\sin h}{h} = 1$$, and $$\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$$. Using these, we get to: $f'(x) = 0 + (\cos x) (1) = \cos x$ And so: $\frac{d}{dx} \sin x = \cos x$ ## Derivative of cos(x) using first principles To find out the derivative of cos(x) using first principles, we need to use the formula for first principles we saw above: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ Here we will substitute f(x) with our function, cos(x): $f'(x) = \lim_{h\to 0} \frac{\cos(x+h) - \cos (x)}{h}$ For the next step, we need to remember the trigonometric identity: $$cos(a +b) = \cos a \cdot \cos b - \sin a \cdot \sin b$$. Using the trigonometric identity, we can come up with the following formula, equivalent to the one above: $f'(x) = \lim_{h\to 0} \frac{(\cos x\cdot \cos h - \sin x \cdot \sin h) - \cos x}{h}$ We can now factor out the $$\cos x$$ term: $f'(x) = \lim_{h\to 0} \frac{\cos x(\cos h - 1) - \sin x \cdot \sin h}{h} = \lim_{h\to 0} \frac{\cos x(\cos h - 1)}{h} - \frac{\sin x \cdot \sin h}{h}$. Now we need to change factors in the equation above to simplify the limit later. For this, you'll need to recognise formulas that you can easily resolve. The equations that will be useful here are: $$\lim_{x \to 0} \frac{\sin x}{x} = 1; and \lim_{x_to 0} \frac{\cos x - 1}{x} = 0$$ If we substitute the equations in the hint above, we get: $\lim_{h\to 0} \frac{\cos x(\cos h - 1)}{h} - \frac{\sin x \cdot \sin h}{h} \rightarrow \lim_{h \to 0} \cos x (\frac{\cos h -1 }{h}) - \sin x (\frac{\sin h}{h}) \rightarrow \lim_{h \to 0} \cos x(0) - \sin x (1)$ Finally, we can get to: $\lim_{h \to 0} \cos x(0) - \sin x (1) = \lim_{h \to 0} (-\sin x)$ Since there are no more h variables in the equation above, we can drop the $$\lim_{h \to 0}$$, and with that we get the final equation of: $\frac{d}{dx} (\cos x) = -\sin x$ ## Worked examples of differentiation from first principles Let's look at two examples, one easy and one a little more difficult. Differentiate from first principles $$y = f(x) = x^3$$. SOLUTION: Steps Worked out example STEP 1: Let $$y = f(x)$$ be a function. Pick two points x and x + h. Coordinates are $$(x, x^3)$$ and $$(x+h, (x+h)^3)$$.We can simplify$$(x+h)^3 = x^3 + 3x^2 h+3h^2x+ h^3$$ STEP 2: Find $$\Delta y$$ and $$\Delta x$$. $$\Delta y = (x+h)^3 - x = x^3 + 3x^2h + 3h^2x+h^3 - x^3 = 3x^2h + 3h^2x + h^3; \\ \Delta x = x+ h- x = h$$ STEP 3:Complete $$\frac{\Delta y}{\Delta x}$$ $$\frac{\Delta y}{\Delta x} = \frac{3x^2h+3h^2x+h^3}{h} = 3x^2 + 3hx+h^2$$ STEP 4: Take a limit. $$f'(x) = \lim_{h \to 0} 3x^2 + 3h^2x + h^2 = 3x^2$$ ANSWER $$3x^2$$ however the entire proof is a differentiation from first principles. So differentiation can be seen as taking a limit of a gradient between two points of a function. You will see that these final answers are the same as taking derivatives. Let's look at another example to try and really understand the concept. This time we are using an exponential function. Differentiate from first principles $$f(x) = e^x$$. SOLUTION: Steps Worked out example STEP 1: Let y = f(x) be a function. Pick two points x and x + h. Co-ordinates are $$(x, e^x)$$ and $$(x+h, e^{x+h})$$. STEP 2: Find $$\Delta y$$ and $$\Delta x$$ $$\Delta y = e^{x+h} -e^x = e^xe^h-e^x = e^x(e^h-1)$$$$\Delta x = (x+h) - x= h$$ STEP 3:Complete $$\frac{\Delta y}{\Delta x}$$ $$\frac{\Delta y}{\Delta x} = \frac{e^x(e^h-1)}{h}$$ STEP 4: Take a limit. $$f'(x) = \lim_{h \to 0} \frac{e^x(e^h-1)}{h} = e^x(1) = e^x$$Because $$\lim_{h \to 0} \frac{(e^h-1)}{h} = 1$$ ANSWER $$e^x$$, but of course, the entire proof is an answer as this is differentiation from first principles. ## Differentiation from First Principles - Key takeaways • Differentiation is the process of finding the gradient of a curve. • The gradient of a curve changes at all points. • Differentiation can be treated as a limit tending to zero. • The formula to differentiate from first principles is found in the formula booklet and is $$f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$ #### Flashcards in Differentiation from First Principles 2 ###### Learn with 2 Differentiation from First Principles flashcards in the free StudySmarter app We have 14,000 flashcards about Dynamic Landscapes. How do we differentiate from first principles? We take the gradient of a function using any two points on the function (normally x and x+h). What is the differentiation from the first principles formula? The formula is: limh->0((f(x+h)-f(x))/h) How do we differentiate a quadratic from first principles? We simply use the formula and cancel out an h from the numerator. This should leave us with a linear function. How do we differentiate a trigonometric function from first principles? We use addition formulae to simplify the numerator of the formula and any identities to help us find out what happens to the function when h tends to 0. StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance. ##### StudySmarter Editorial Team Team Math Teachers • Checked by StudySmarter Editorial Team ## Join over 22 million students in learning with our StudySmarter App The first learning app that truly has everything you need to ace your exams in one place • Flashcards & Quizzes • AI Study Assistant • Study Planner • Mock-Exams • Smart Note-Taking
### Theory: Multiplication: When more than $$2$$ numbers are multiplied, depending upon the number of negative numbers involved, the sign of the answer varies. We know that, $$(-1) × (1) = -1$$ $$(-1) × (-1) = +1$$ Now, $$(-1) × (-2) × (-3) = -6$$ $$(-1) × (-2) × (-3) × (-4) = +24$$ $$(-1) × (-2) × (-3) × (-4) × (-5) = -120$$ Inference: When $$2$$ negative numbers are multiplied, the result will be a positive number. When $$3$$ negative numbers are multiplied, the result will be a negative number. When $$4$$ negative numbers are multiplied, the result will be a positive number. Conclusion: From the above, we can conclude that, Number of negative integers in multiplication Sign of the result Even $$+$$ odd $$-$$ Division: Let's learn, how to deal with the division of more than $$2$$ numbers, $$a,b,c$$ are three numbers, $$a/b/c$$ should be evaluated as $$(a×c)/b$$. Example: $\begin{array}{l}1.\phantom{\rule{0.147em}{0ex}}\frac{1/2}{3}=\frac{1×3}{2}=\frac{3}{2}\\ \\ 2.\phantom{\rule{0.147em}{0ex}}\frac{2/6}{5}=\frac{2×5}{6}=\frac{10}{6}\end{array}$ $$a,b,c,d$$ are four numbers, a/b/c/d should be evaluated as $$(a×d)/(b×c)$$ Example: $\begin{array}{l}1.\phantom{\rule{0.147em}{0ex}}\frac{5/2}{3/6}=\frac{5×6}{3×2}=\frac{30}{6}\\ \\ 2.\phantom{\rule{0.147em}{0ex}}\frac{2/6}{5/2}=\frac{2×2}{5×6}=\frac{4}{30}\\ \\ 3.\phantom{\rule{0.147em}{0ex}}\frac{2/4}{6/2}=\frac{2×2}{6×4}=\frac{4}{24}\end{array}$ Similar to multiplication, when more than $$2$$ numbers are involved, depending upon the number of negative numbers involved, the sign of the answer varies. Number of negative integers in division Sign of the result Even $$+$$ odd $$-$$ Example: $\begin{array}{l}1.\phantom{\rule{0.147em}{0ex}}\frac{-5/-2}{-2/8}=\frac{-5×8}{-2×-2}=\frac{-40}{4}\\ \\ 2.\phantom{\rule{0.147em}{0ex}}\frac{-8/-6}{2/4}=\frac{-8×4}{-6×2}=\frac{-32}{-12}=\frac{32}{12}\\ \\ 3.\phantom{\rule{0.147em}{0ex}}\frac{-9/1}{-1/6}=\frac{-9×6}{-1×1}=\frac{-54}{-1}=\frac{54}{1}\\ \\ 4.\phantom{\rule{0.147em}{0ex}}\frac{-1/2}{1/2}=\frac{-1×2}{1×2}=\frac{-2}{2}=-1\\ \\ 5.\phantom{\rule{0.147em}{0ex}}\frac{\frac{-1}{2}}{2}=\frac{-1×2}{2×1}=-1\end{array}$
FutureStarr 5 1 4 to improper fraction ## 5 1 4 to improper fraction At fractional practions, 5 1 4 doesn’t become 5/4. In this article, we'll talk about how to make fractions easy for learners. ### fraction via GIPHY An improper fraction is a fraction whose numerator is greater than or equal to its denominator. For example, 9/4, 4/3 are improper fractions. Numerically, an improper fraction always equals to or greater than 1. On the other hand, a mixed fraction is a fraction that is written as a combination of a natural number and a proper fraction. It is a simplified form of an improper fraction. For example, $$21\dfrac{4}{5}, 16\dfrac{2}{3}$$ are mixed fractions. Numerically, a mixed fraction is always greater than 1. Also, any mixed fraction can be written as an improper fraction. For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(5, 5) = 5. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 5 × 5 = 25. In the following intermediate step, it cannot further simplify the fraction result by canceling. The calculator performs basic and advanced operations with mixed numbers, fractions, integers, decimals. Mixed fractions are also called mixed numbers. A mixed fraction is a whole number and a proper fraction combined, i.e. one and three-quarters. The calculator evaluates the expression or solves the equation with step-by-step calculation progress information. Solve problems with two or more mixed numbers fractions in one expression.This is called a mixed fraction. Thus, an improper fraction can be expressed as a mixed fraction, where quotient represents the whole number, remainder becomes the numerator and divisor is the denominator. A fraction, where the numerator is less than the denominator is called the proper fraction for example, $$\frac{2}{3}$$, $$\frac{5}{7}$$, $$\frac{3}{5}$$ are proper fractions. A fraction with numerator 1 is called a unit fraction. (Source: www.math-only-math.com) ## Related Articles • #### A 22 Out of 25 As a Percentage: June 26, 2022 \| Abid Ali • #### How Many 18x18 Tiles Do I Need June 26, 2022 \| sheraz naseer • #### 12 13 As a Percentage ORR June 26, 2022 \| Bilal Saleem • #### A 4 Is What Percent of 14 June 26, 2022 \| Shaveez Haider • #### Calc: June 26, 2022 \| Abid Ali • #### Less Than or Equal to Calculator, June 26, 2022 \| Jamshaid Aslam • #### How Much Should My Car Lease Payment Be June 26, 2022 \| sheraz naseer • #### How to Find 20 Percent of a Number in Excel orr June 26, 2022 \| Bushra Tufail • #### Inch Calculator Online June 26, 2022 \| Future Starr • #### A Interpret Stem and Leaf Plots Calculator June 26, 2022 \| Shaveez Haider • #### A 23 25 As a Percent: June 26, 2022 \| Abid Ali • #### What Percent Is 18 Out of 30 OR June 26, 2022 \| Jamshaid Aslam • #### How many teaspoons in a tablespoon June 26, 2022 \| Future Starr • #### 3 in Fraction Form OR June 26, 2022 \| Jamshaid Aslam • #### 740 Area Code June 26, 2022 \| sajjad ghulam hussain
# complex - MATH 135 COMPLEX NUMBERS(WINTER 2008 The... This preview shows pages 1–3. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: MATH 135: COMPLEX NUMBERS (WINTER, 2008) The introduction of complex numbers in the 16th century made it possible to solve the equation x 2 + 1 = 0. These notes 1 present one way of defining complex numbers. 1. The Complex Plane A complex number z is given by a pair of real numbers x and y and is written in the form z = x + iy , where i satisfies i 2 =- 1. The complex numbers may be represented as points in the plane (sometimes called the Argand diagram). The real number 1 is represented by the point (1 , 0), and the complex number i is represented by the point (0 , 1). The x-axis is called the “real axis”, and the y-axis is called the “imaginary axis”. For example, the complex numbers 3 + 4 i and 3- 4 i are illustrated in Fig 1a . Fig 1a r 3 + 4 i 7 r 3- 4 i S S S S Sw Fig 1b : 3 r r r 4 + i 2 + 3 i 6 + 4 i Complex numbers are added in a natural way: If z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 , then z 1 + z 2 = ( x 1 + x 2 ) + i ( y 1 + y 2 ) (1) Fig 1b illustrates the addition (4 + i ) + (2 + 3 i ) = (6 + 4 i ). Multiplication is given by z 1 z 2 = ( x 1 x 2- y 1 y 2 ) + i ( x 1 y 2 + x 2 y 1 ) Note that the product behaves exactly like the product of any two algebraic expressions, keeping in mind that i 2 =- 1. Thus, (2 + i )(- 2 + 4 i ) = 2(- 2) + 8 i- 2 i + 4 i 2 =- 8 + 6 i We call x the real part of z and y the imaginary part , and we write x = Re z , y = Im z . ( Remem- ber : Im z is a real number.) The term “imaginary” is an historical holdover; it took mathematicians Date : November 26, 2007. 1 These notes are based on notes written by Bob Phelps, with modifications by Tom Duchamp. 1 2 (WINTER, 2008) some time to accept the fact that i (for “imaginary”, naturally) was a perfectly good mathematical object. Electrical engineers (who make heavy use of complex numbers) reserve the letter i to denote electric current and they use j for √- 1. There is only one way we can have z 1 = z 2 , namely, if x 1 = x 2 and y 1 = y 2 . An equivalent statement (one that is important to keep in mind) is that z = 0 if and only if Re z = 0 and Im z = 0. If a is a real number and z = x + iy is complex, then az = ax + iay (which is exactly what we would get from the multiplication rule above if z 2 were of the form z 2 = a + i 0). Division is more complicated (although we will show later that the polar representation of complex numbers makes it easy). To find z 1 /z 2 it suffices to find 1 /z 2 and then multiply by z 1 . The rule for finding the reciprocal of z = x + iy is given by: 1 x + iy = 1 x + iy · x- iy x- iy = x- iy ( x + iy )( x- iy ) = x- iy x 2 + y 2 (2) The expression x- iy appears so often and is so useful that it is given a name. It is called the complex conjugate of z = x + iy and a shorthand notation for it is z ; that is, if z = x + iy , then z = x- iy . For example, 3 + 4 i = 3- 4 i , as illustrated in the Fig 1a . Note that z = z and z 1 + z 2 = z 1 + z 2 . Exercise (3b) is to show that z 1 z 2 =... View Full Document {[ snackBarMessage ]} ### Page1 / 8 complex - MATH 135 COMPLEX NUMBERS(WINTER 2008 The... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# What is the slope of any line perpendicular to the line passing through (8,12) and (5,-2)? Feb 16, 2016 The slope $= - \frac{3}{14}$ #### Explanation: Consider the points : (x_1, y_1) = color(blue)((8,12) (x_2, y_2) = color(blue)((5,-2) The slope joining the pair of points is calculated as : $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ $= \frac{- 2 - 12}{5 - 8}$ $= \frac{- 14}{- 3}$ $= \frac{14}{3}$ The product of slopes of two lines perpendicular to each other is $- 1$. Hence slope of line perpendicular to the line passing through $\left(8 , 12\right)$ and $\left(5 , - 2\right)$ will be $- \frac{1}{\frac{14}{3}}$ or $- \frac{3}{14}$.
# Classifications of Triangles with properties \| Triangle Area Formulas ## Types of Triangles With examples \| Properties of Triangle A simple closed figure bounded by three lines segments is called triangle. The triangle can be defined as “A two dimensional plane figure with three straight sides and three angles”. ### Terminology and Formulas of the Triangles: #### Triangle Sides : The three lines segments that form the triangle area called sides of triangle. AB = c, BC = a, CA = b are the sides of the ΔABC ( triangle ABC). The Sum of any two sides of a triangle is always greater than the third side. i.e a + b > c , b + c > a , c + a > b #### Vertex : Two adjacent sides of a triangle intersect at a point. That point is called vertex. The points A, B, &C area called vertices or angular of ΔABC ( triangle ABC) . #### Angle : Two lines segments with the same end point will determine one angle. Triangle has 3 angles namely ∠ABC, ∠BCA, ∠CAB and they also denoted as ∠B, ∠C, ∠A respectively. Always the sum of the angles of a triangle is i.e ∟A + ∟B + ∟C = #### Perimeter of a Triangle : The sum of the measures of the three sides of the a triangle is called the perimeter of the triangle. The sum of the sides AB = c , BC = a & CA= b is equal to perimeter of ΔABC ( triangle ABC). i.e perimeter of ΔABC = AB + BC + CA = a +b + c #### Triangular Area or Region : The interior of a triangle together with its boundary is called the triangular region or triangular area. Area of the triangle = (1/2) x Base x Height Area of the triangle ABC = (1/2) x a x h #### Median of the triangle : The line joining the midpoint of a side of a triangle to the positive vertex is called the median of the triangle. Here AR, BQ, CP are the medians of the triangle ΔABC. The median of a triangle divides the triangle into two triangles with equal areas. i.e Area of the ΔARB = Area of the ΔARC Area of the ΔBQA= Area of the ΔBQC Area of the ΔCPB = Area of the ΔCPA #### Triangle centroid: The point where the three medians of a triangle meet is called centroid of that triangle. The point ” s” is the centroid of the triangle ΔABC. The triangle centroid divides each the median of the triangle into the ratio segments. i.e As : sR = 2 : 1 Bs : sQ = 2 : 1 Cs : sP = 2 : 1 Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle. Area of the triangle ABC = 1/4 area of the triangle PQR ( area of ΔABC = 0.25 x area of ΔPQR ) ### Classification of the triangles: Triangles can be classified into three kinds according to measures of their sides and three kinds according to the measures of angles. #### Classification according to sides: According to the sides the triangles classified as Equilateral, isosceles and Scalene ##### Equilateral triangles : A triangle having all the three sides of the same length is called an equilateral triangle. All three angles of equilateral triangles are equal. i.e PQ = QR = RP = a ∠P, ∠Q, ∠R = 60° Formulas for equilateral triangle : Perimeter of the equilateral triangle = 3 a Area of the equilateral triangle = (√3/4) x a2 Radius of incircle of an equilateral triangle = a / (2 √3) Radius of circumference circle of an equilateral triangle = a / √3 ##### Isosceles Triangles : A triangle having two sides of equal length is called an isosceles triangle. The angle opposite to the equal sides of an isosceles triangle area equal. Here PQ = PR = a and ∠Q = ∠R ##### Scalene Triangles : If the three sides of a triangle are of different length then the triangle is called scalene triangle. If PQ ≠ QR ≠ R or ∠P ≠ ∠Q ≠ ∠R then ΔPQR is called scalene triangle. #### Classification according to angle: According to their angles divided three kinds of triangles. They are acute angled triangle, Right angled triangle and obtuse angled triangle ##### Acute Triangle : All angles of this triangles area having less than 90°. So all equilateral triangles area acute triangles. If ∠P < 90° , ∠Q < 90° & ∠R < 90° than ΔPQR is called Acute triangle. ##### Right Triangle : If triangle having one 90° angle than it is called right angled triangle. If triangle having a right angle (i.e 90°) and also two equal angles ( i.e 45° & 45° ) then it is called Right Isosceles Triangle Here ∠Q = 90° So ΔPQR is called right angle triangle. If ∠P = ∠R = 45° then ΔPQR is called Right Isosceles Triangle. ##### Obtuse Triangle : If triangle having one more than 90° angle than it is called Obtuse angled triangle. Here ∠P > 90° So ΔPQR is called Obtuse angled triangle. ### Properties of the Triangles: 1. The sum of the three angles of the a triangle is two right angles (i.e 180° ). 2. If one sides of the triangle is produced , the exterior angle formed is equal to the sum of the interior opposite angles. Here ∠P = ∠Q = ∠PRS. 3. The internal bisectors of ∠Q and ∠R of ΔPQR intersect at A. Then ∠A = 180° + (∠P / 2 ) ( Note : Bisectors means ∠AQP = ∠AQR and ∠ARP = ∠ARQ ) 4. If any two angles and a sides of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent. 5. If two medians of triangle are equal then the triangle must be an Isosceles triangles. ### Geometry Math Two dimensional shapes formulas. Properties of circle in math \| Arc, Perimeter, Segment of circle Quadrilateral Properties \| Trapezium, parallelogram, Rhombus Topics in Quantitative aptitude math for all types of exams Hi friends Thanks for reading. I Hope you liked it. Give feed back, comments and please don’t forget to share it. ## 2 thoughts on “Classifications of Triangles with properties \| Triangle Area Formulas” (October 17, 2017 - 5:48 pm) […] Classifications of Triangles with properties \| Triangle Area Formulas […] #### Two dimensional shapes formulas of area and perimeter calculation (October 21, 2017 - 4:04 pm) […] Classification according to angle and according to sides like Equilateral, isosceles, Scalene, acute… […]
Application Center - Maplesoft # Calculus I: Lesson 11: Implicit Functions and Implicit Differentiation You can switch back to the summary page by clicking here. L11-implicitDifferentiation.mws Calculus I Lesson 11: Implicit Functions and Implicit Differentiation Usually when we speak of functions, we are talking about explicit functions of the form y = f(x). Sometimes we have instead an equation in x and y, for example, x + y + xy = sin(x + y), which may not be solvable for y. The solutions to this equation are a set of points {(x,y)} which implicitly define a relation between x and y which we will call an implicit function . Implicit functions are often not actually functions in the strict definition of the word, because they often have multiple y values for a single x value. They do have graphs and derivatives however. You are probably familiar with some linear implicit functions from algebra. Lines in point-slope and standard form (such as y -3 = 2(x + 5) and 7x + 9y = 63) are both examples of implicit functions since they are not explicitly solved for y in terms of x. These can be solved for y easily enough and converted into explicit functions. However, many implicit functions are difficult or impossible to convert to explicit functions and must be left in implicit form. Our first task will be, given an implicit defined function, can we plot some individual points on the graph. Lets start with , and find all points where x has the value 2. > restart; with(plots): ```Warning, the name changecoords has been redefined ``` > imp_fun := -4x + 10(x^2) * (y^(-2)) + y^2 = 11; Our method will consist of two steps : first, substitute for the value of x, then solve the resulting equation in one variable for y. We substitute x= 2 into the implicit equation. This gives us an equation with only one unknown. We solve this equation and get exact radical solutions. We can convert these to decimal answers in this way. > subs( x = 2, imp_fun ); > s := solve( % ); > s := evalf( % ); GRAPHING IMPLICIT FUNCTIONS Graphing implicit functions opens up an exciting new world of graphing possibilities. Unfortunately, its quite time-consuming to graph most implicit functions by hand - often rendering them effectively ungraphable without technological aids. Fortunately, Maple graphs these functions effectively.The results are amazing and somewhat a kin to mathematical tea leaves with many intricacies not usually seen in explicit functions. Instead of using the plot command, we use the implicitplot command. The first thing you may notice about implicit functions is that most of these functions fail the vertical line test.... miserably! That failure is one of the reasons they are so interesting. Here is a tilted ellipse. The implicit equation , an x range, and a y range. > implicitplot( x^2 + xy + y^2 = 16, x = -5..5, y = -5..5, grid=[50,50] ); Some of the shapes are remarkably simple. > implicitplot( x^4 + 8(x^3) + y^4 = 16, x = -12..10, y = -10..10, grid=[50,50],thickness = 2, color = brown, scaling = constrained); On the other hand, some functions can be remarkably complicated. In fact, this one is so complicated that our first attempt just gives a hint as to the complexity. By increasing the number of points used in the production of the graph, a great amount of detail and subtlety become visible. > implicitplot( xycos(x^2 + y^2) = 1, x = -10..10, y = -10..10, grid=[30,30] ); > implicitplot( xycos(x^2 + y^2) = 1, x = -10..10, y = -10..10, grid=[80,80] ); Many of these graphs are symmetrical in various ways. These are symmetric to the y axis. > implicitplot( x^2 + 1.5yx^2 + y^2 = 1, x = -10..10, y = -10..10, grid=[50,50] ); > implicitplot( (x^2 + y^2 -2) = (.5 + yx^2)^2 , x = -5..5, y = -5..5, grid=[100,100] ); This one is symmetrical to the x axis > implicitplot( -4x + 10(x^2) + (y^(-2)) + y^2 = 11, x = -5..5, y = -5..5, grid=[100,100] ); And this one is symmetrical to both the x and y axes. > implicitplot( x^2 - y^2 = xysin(xy), x = -4..4, y = -3..3, grid=[100,100] ); This graph is apparently symmetrical to the line y = -x. > implicitplot( x - y + sin(2.5xy) = sin(x) - sin(y) + sin(xy), x = -5..5, y = -5..5, grid=[100,100] ); Some of the graphs have symmetries and patterns which are not as easy to describe. > implicitplot( (xy)sin(y) = xcos(x-y), x = -10..10, y = -10..10, grid=[100,100]); This one has more complex pattern which hints as a more complex type of symmetry. > implicitplot( ln( (x + 7sin(y))^2 ) = exp(y + 2cos(x)) , x = -9..9, y = -12..3, grid=[100,100]); This graph creates a Tesslation-like pattern in the plane. > implicitplot( sin(x + 2sin(y) ) = cos( y + 3cos(x)), x = -10..10, y = -10..10, grid=[100,100]); C IMPLICIT DERIVATIVES Just as we use the implicitplot command rather than the plot command to graph these functions, we use the implicitdiff command instead of the diff command to find the derivatives of implicit functions. > implicitdiff( -4x + 10(x^2)(y^(-2) ) + y^2 = 11, y, x); The format of this command is an equation which implicitly defines a function, then the dependent and independent variables y, and x. Note that the result of taking an implicit derivative is a function in both x and y. Since an implicit function often has multiple y values for a single x value, there are also multiple tangent lines. In this example, we will go through several steps to construct all of the tangent lines for the value of x = 2. Lets start with the same function we used above. > imp_fun := -4x + 10(x^2)(y^(-2)) + y^2 =11; Step 1 : Choose an x value (c) > c := 2; Step 2 : Find the Corresponding y Values > s := evalf( solve( subs( x = c, imp_fun))); Step 3 : Fiind the slope for each point - remember the implicit derivative is a function of both x and y. For x = c, there are four different values of y, and thus 4 different slopes. > m1 := evalf( subs ( { x = c, y = s[1] }, implicitdiff( imp_fun, y,x))); > m2 := evalf( subs ( { x = c, y = s[2] }, implicitdiff( imp_fun, y,x))); > m3 := evalf( subs ( { x = c, y = s[3] }, implicitdiff( imp_fun, y,x))); > m4 := evalf( subs ( { x = c, y = s[4] }, implicitdiff( imp_fun, y,x))); We are finding the value of the slope for each of the 4 y values by substituting x = c, and y into the implicit derivative for each case. Step 4 : Plot the function and its 4 tangent lines > implicitplot( {y - s[1] = m1(x-c), y - s[2] = m2(x-c), y - s[3] = m3(x-c), y - s[4] = m4(x-c), imp_fun }, x = -5..5, y = -5..5, grid=[100,100]);
Is Every Rational Number Is a natural Number Every natural number is a rational number but a rational number need not to be a natural number. We know that 1/1 = 1, 2 = 2/1, 3 = 3/1, 4 = 4/1 and so on….. In other words we can say that every natural number can […] Equivalent Rational Numbers – Definition – Examples Rational Numbers A rational number can be represented in the form of p/q, where q is not equal to zero. Equivalent Rational Numbers Two rational numbers are said to be equivalent, if they have same value but can be represented in different forms. We know that if we […] Division of Two Rational Numbers In this tutorial, we will learn, division of rational numbers. Rational numbers are written as fractions, therefore to divide a given rational number by another rational number, we have to multiply the given rational number by the reciprocal of the second rational number. Let a/b is a rational number then […] Rational Numbers – Multiplication In this tutorial, we will learn, multiplication of rational numbers. Multiplication of two rational numbers is just like multiplication of two integers. Product of two or more rational numbers is founded by multiplying the corresponding numerators and denominators of the numbers and writing them in the standard form. Product of numerators […] Arithmetical Operations – Rational Numbers Rational numbers are numbers which can be represented in the form of p/q, where p and q are any two integers and q is not equal to zero(q ≠ 0). A rational number p/q is said to be in its standard form if p and q do not have any […] Rational Numbers on Number Line Rational Numbers Rational numbers are numbers which can be represented in the form of p/q, where p and q are any two integers and q is not equal to zero(q ≠ 0). Rational Numbers on Number Line Rational numbers can easily represented on number line by following some simple steps. […] Property of Rational Numbers – Additive Identity and Additive Inverse There are two basic additive properties of rational numbers. (1) Additive Identity Property (2) Additive Inverse Property (1) Additive Identity Property: Additive identity of rational numbers states that the sum of any rational number (a/b) and zero is the rational number itself. Suppose a/b is […]
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 9.6: Solving Trigonometric Equations $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ Skills to Develop • Solve linear trigonometric equations in sine and cosine. • Solve equations involving a single trigonometric function. • Solve trigonometric equations using a calculator. • Solve trigonometric equations that are quadratic in form. • Solve trigonometric equations using fundamental identities. • Solve trigonometric equations with multiple angles. • Solve right triangle problems. Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids. ## Solving Linear Trigonometric Equations in Sine and Cosine Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is $$2\pi$$. In other words, every $$2\pi$$ units, the y-values repeat. If we need to find all possible solutions, then we must add $$2\pi k$$,where $$k$$ is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is $$2\pi$$: $\sin \theta=\sin(\theta \pm 2k\pi)$ There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections. Example $$\PageIndex{1A}$$: Solving a Linear Trigonometric Equation Involving the Cosine Function Find all possible exact solutions for the equation $$\cos \theta=\dfrac{1}{2}$$. Solution From the unit circle, we know that \begin{align} \cos \theta &=\dfrac{1}{2} \\[4pt] \theta &=\dfrac{\pi}{3},\space \dfrac{5\pi}{3} \end{align} These are the solutions in the interval $$[ 0,2\pi ]$$. All possible solutions are given by $\theta=\dfrac{\pi}{3} \pm 2k\pi \quad \text{and} \quad \theta=\dfrac{5\pi}{3} \pm 2k\pi \nonumber$ where $$k$$ is an integer. Example $$\PageIndex{1B}$$: Solving a Linear Equation Involving the Sine Function Find all possible exact solutions for the equation $$\sin t=\dfrac{1}{2}$$. Solution Solving for all possible values of $$t$$ means that solutions include angles beyond the period of $$2\pi$$. From the section on Sum and Difference Identities, we can see that the solutions are $$t=\dfrac{\pi}{6}$$ and $$t=\dfrac{5\pi}{6}$$. But the problem is asking for all possible values that solve the equation. Therefore, the answer is $t=\dfrac{\pi}{6}\pm 2\pi k \quad \text{and} \quad t=\dfrac{5\pi}{6}\pm 2\pi k \nonumber$ where $$k$$ is an integer. Howto: Given a trigonometric equation, solve using algebra 1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity. 2. Substitute the trigonometric expression with a single variable, such as $$x$$ or $$u$$. 3. Solve the equation the same way an algebraic equation would be solved. 4. Substitute the trigonometric expression back in for the variable in the resulting expressions. 5. Solve for the angle. Example $$\PageIndex{2}$$: Solve the Linear Trigonometric Equation Solve the equation exactly: $$2 \cos \theta−3=−5$$, $$0≤\theta<2\pi$$. Solution Use algebraic techniques to solve the equation. \begin{align} 2 \cos \theta-3&= -5\\ 2 \cos \theta&= -2\\ \cos \theta&= -1\\ \theta&= \pi \end{align} Exercise $$\PageIndex{2}$$ Solve exactly the following linear equation on the interval $$[0,2\pi)$$: $$2 \sin x+1=0$$. $$x=\dfrac{7\pi}{6},\space \dfrac{11\pi}{6}$$ ## Solving Equations Involving a Single Trigonometric Function When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link]). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is $$\pi$$,not $$2\pi$$. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of $$\dfrac{\pi}{2}$$,unless, of course, a problem places its own restrictions on the domain. Example $$\PageIndex{3A}$$: Solving a Problem Involving a Single Trigonometric Function Solve the problem exactly: $$2 {\sin}^2 \theta−1=0$$, $$0≤\theta<2\pi$$. Solution As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate $$\sin \theta$$. Then we will find the angles. \begin{align} 2 {\sin}^2 \theta-1&= 0\\ 2 {\sin}^2 \theta&= 1\\ {\sin}^2 \theta&= \dfrac{1}{2}\\ \sqrt{ {\sin}^2 \theta }&= \pm \sqrt{ \dfrac{1}{2} }\\ \sin \theta&= \pm \dfrac{1}{\sqrt{2}}\\ &= \pm \dfrac{\sqrt{2}}{2}\\ \theta&= \dfrac{\pi}{4}, \space \dfrac{3\pi}{4},\space \dfrac{5\pi}{4}, \space \dfrac{7\pi}{4} \end{align} Example $$\PageIndex{3B}$$: Solving a Trigonometric Equation Involving Cosecant Solve the following equation exactly: $$\csc \theta=−2$$, $$0≤\theta<4\pi$$. Solution We want all values of $$\theta$$ for which $$\csc \theta=−2$$ over the interval $$0≤\theta<4\pi$$. \begin{align} \csc \theta&= -2\\ \dfrac{1}{\sin \theta}&= -2\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\space \dfrac{11\pi}{6},\space \dfrac{19\pi}{6}, \space \dfrac{23\pi}{6} \end{align} Analysis As $$\sin \theta=−\dfrac{1}{2}$$, notice that all four solutions are in the third and fourth quadrants. Example $$\PageIndex{3C}$$: Solving an Equation Involving Tangent Solve the equation exactly: $$\tan\left(\theta−\dfrac{\pi}{2}\right)=1$$, $$0≤\theta<2\pi$$. Solution Recall that the tangent function has a period of $$\pi$$. On the interval $$[ 0,\pi )$$,and at the angle of $$\dfrac{\pi}{4}$$,the tangent has a value of $$1$$. However, the angle we want is $$\left(\theta−\dfrac{\pi}{2}\right)$$. Thus, if $$\tan\left(\dfrac{\pi}{4}\right)=1$$,then \begin{align} \theta-\dfrac{\pi}{2}&= \dfrac{\pi}{4}\\ \theta&= \dfrac{3\pi}{4} \pm k\pi \end{align} Over the interval $$[ 0,2\pi )$$,we have two solutions: $$\theta=\dfrac{3\pi}{4}$$ and $$\theta=\dfrac{3\pi}{4}+\pi=\dfrac{7\pi}{4}$$ Exercise $$\PageIndex{3}$$ Find all solutions for $$\tan x=\sqrt{3}$$. $$\dfrac{\pi}{3}\pm \pi k$$ Example $$\PageIndex{4}$$: Identify all Solutions to the Equation Involving Tangent Identify all exact solutions to the equation $$2(\tan x+3)=5+\tan x$$, $$0≤x<2\pi$$. Solution We can solve this equation using only algebra. Isolate the expression $$\tan x$$ on the left side of the equals sign. \begin{align} 2(\tan x)+2(3)&= 5+\tan x\\ 2\tan x+6&= 5+\tan x\\ 2\tan x-\tan x&= 5-6\\ \tan x&= -1 \end{align} There are two angles on the unit circle that have a tangent value of $$−1$$: $$\theta=\dfrac{3\pi}{4}$$ and $$\theta=\dfrac{7\pi}{4}$$. ## Solve Trigonometric Equations Using a Calculator Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem. Example $$\PageIndex{5A}$$: Using a Calculator to Solve a Trigonometric Equation Involving Sine Use a calculator to solve the equation $$\sin \theta=0.8$$,where $$\theta$$ is in radians. Solution Make sure mode is set to radians. To find $$\theta$$, use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the $${\sin}^{−1}$$ function. What is shown on the screen is $${\sin}^{−1}$$ .The calculator is ready for the input within the parentheses. For this problem, we enter $${\sin}^{−1}(0.8)$$, and press ENTER. Thus, to four decimals places, $${\sin}^{−1}(0.8)≈0.9273$$ The solution is $$\theta≈0.9273\pm 2\pi k$$ The angle measurement in degrees is \begin{align} \theta&\approx 53.1^{\circ}\\ \theta&\approx 180^{\circ}-53.1^{\circ}\\ &\approx 126.9^{\circ} \end{align} Analysis Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using $$\pi−\theta$$. Example $$\PageIndex{5B}$$: Using a Calculator to Solve a Trigonometric Equation Involving Secant Use a calculator to solve the equation $$\sec θ=−4,$$ giving your answer in radians. Solution We can begin with some algebra. \begin{align} \sec \theta&= -4\\ \dfrac{1}{\cos \theta}&= -4\\ \cos \theta&= -\dfrac{1}{4} \end{align} Check that the MODE is in radians. Now use the inverse cosine function \begin{align}{\cos}^{-1}\left(-\dfrac{1}{4}\right)&\approx 1.8235\\ \theta&\approx 1.8235+2\pi k \end{align} Since $$\dfrac{\pi}{2}≈1.57$$ and $$\pi≈3.14$$,$$1.8235$$ is between these two numbers, thus $$\theta≈1.8235$$ is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure $$\PageIndex{2}$$. So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is $$\theta '≈\pi−1.8235≈1.3181$$. The other solution in quadrant III is $$\theta '≈\pi+1.3181≈4.4597$$. The solutions are $$\theta≈1.8235\pm 2\pi k$$ and $$\theta≈4.4597\pm 2\pi k$$. Exercise $$\PageIndex{5}$$ Solve $$\cos \theta=−0.2$$. $$\theta≈1.7722\pm 2\pi k$$ and $$\theta≈4.5110\pm 2\pi k$$ ## Solving Trigonometric Equations in Quadratic Form Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as $$x$$ or $$u$$. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations. Example $$\PageIndex{6A}$$: Solving a Trigonometric Equation in Quadratic Form Solve the equation exactly: $${\cos}^2 \theta+3 \cos \theta−1=0$$, $$0≤\theta<2\pi$$. Solution We begin by using substitution and replacing $$\cos \theta$$ with $$x$$. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let $$\cos \theta=x$$. We have $$x^2+3x−1=0$$ The equation cannot be factored, so we will use the quadratic formula: $$x=\dfrac{−b\pm \sqrt{b^2−4ac}}{2a}$$. \begin{align} x&= \dfrac{ -3\pm \sqrt{ {(-3)}^2-4 (1) (-1) } }{2}\\ &= \dfrac{-3\pm \sqrt{13}}{2}\end{align} Replace $$x$$ with $$\cos \theta$$ and solve. \begin{align} \cos \theta&= \dfrac{-3\pm \sqrt{13}}{2}\\ \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right) \end{align} Note that only the + sign is used. This is because we get an error when we solve $$\theta={\cos}^{−1}\left(\dfrac{−3−\sqrt{13}}{2}\right)$$ on a calculator, since the domain of the inverse cosine function is $$[ −1,1 ]$$. However, there is a second solution: \begin{align} \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 1.26 \end{align} This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is \begin{align} \theta&= 2\pi-{\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 5.02 \end{align} Example $$\PageIndex{6B}$$: Solving a Trigonometric Equation in Quadratic Form by Factoring Solve the equation exactly: $$2 {\sin}^2 \theta−5 \sin \theta+3=0$$, $$0≤\theta≤2\pi$$. Solution Using grouping, this quadratic can be factored. Either make the real substitution, $$\sin \theta=u$$,or imagine it, as we factor: \begin{align} 2 {\sin}^2 \theta-5 \sin \theta+3&= 0\\ (2 \sin \theta-3)(\sin \theta-1)&= 0 \qquad \text {Now set each factor equal to zero.}\\ 2 \sin \theta-3&= 0\\ 2 \sin \theta&= 3\\ \sin \theta&= \dfrac{3}{2}\\ \sin \theta-1&= 0\\ \sin \theta&= 1 \end{align} Next solve for $$\theta$$: $$\sin \theta≠\dfrac{3}{2}$$, as the range of the sine function is $$[ −1,1 ]$$. However, $$\sin \theta=1$$, giving the solution $$\theta=\dfrac{\pi}{2}$$. Analysis Make sure to check all solutions on the given domain as some factors have no solution. Exercise $$\PageIndex{6}$$ Solve $${\sin}^2 \theta=2 \cos \theta+2$$, $$0≤\theta≤2\pi$$. [Hint: Make a substitution to express the equation only in terms of cosine.] $$\cos \theta=−1$$, $$\theta=\pi$$ Example $$\PageIndex{7A}$$: Solving a Trigonometric Equation Using Algebra Solve exactly: $$2 {\sin}^2 \theta+\sin \theta=0;\space 0≤\theta<2\pi$$ Solution This problem should appear familiar as it is similar to a quadratic. Let $$\sin \theta=x$$. The equation becomes $$2x^2+x=0$$. We begin by factoring: \begin{align} 2x^2+x&= 0\\ x(2x+1)&= 0\qquad \text {Set each factor equal to zero.}\\ x&= 0\\ 2x+1&= 0\\ x&= -\dfrac{1}{2} \end{align} Then, substitute back into the equation the original expression $$\sin \theta$$ for $$x$$ . Thus, \begin{align} \sin \theta&= 0\\ \theta&= 0,\pi\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align} The solutions within the domain $$0≤\theta<2\pi$$ are $$\theta=0,\pi,\dfrac{7\pi}{6},\dfrac{11\pi}{6}$$. If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero. \begin{align} {\sin}^2 \theta+\sin \theta&= 0\\ \sin \theta(2\sin \theta+1)&= 0\\ \sin \theta&= 0\\ \theta&= 0,\pi\\ 2 \sin \theta+1&= 0\\ 2\sin \theta&= -1\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align} Analysis We can see the solutions on the graph in Figure $$\PageIndex{3}$$. On the interval $$0≤\theta<2\pi$$,the graph crosses the $$x$$-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value. We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well. Example $$\PageIndex{7B}$$: Solving a Trigonometric Equation Quadratic in Form Solve the equation quadratic in form exactly: $$2 {\sin}^2 \theta−3 \sin \theta+1=0$$, $$0≤\theta<2\pi$$. Solution We can factor using grouping. Solution values of $$\theta$$ can be found on the unit circle. \begin{align} (2 \sin \theta-1)(\sin \theta-1)&= 0\\ 2 \sin \theta-1&= 0\\ \sin \theta&= \dfrac{1}{2}\\ \theta&= \dfrac{\pi}{6}, \dfrac{5\pi}{6}\\ \sin \theta&= 1\\ \theta&= \dfrac{\pi}{2} \end{align} Exercise $$\PageIndex{7}$$ Solve the quadratic equation $$2{\cos}^2 \theta+\cos \theta=0$$. $$\dfrac{\pi}{2}, \space \dfrac{2\pi}{3}, \space \dfrac{4\pi}{3}, \space \dfrac{3\pi}{2}$$ ## Solving Trigonometric Equations Using Fundamental Identities While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation. Example $$\PageIndex{8A}$$: Use Identities to Solve an Equation Use identities to solve exactly the trigonometric equation over the interval $$0≤x<2\pi$$. $$\cos x \cos(2x)+\sin x \sin(2x)=\dfrac{\sqrt{3}}{2}$$ Solution Notice that the left side of the equation is the difference formula for cosine. \begin{align} \cos x \cos(2x)+\sin x \sin(2x)&= \dfrac{\sqrt{3}}{2}\\ \cos(x-2x)&= \dfrac{\sqrt{3}}{2}\qquad \text{Difference formula for cosine}\\ \cos(-x)&= \dfrac{\sqrt{3}}{2}\qquad \text{Use the negative angle identity.}\\ \cos x&= \dfrac{\sqrt{3}}{2} \end{align} From the unit circle in the section on Sum and Difference Identities, we see that $$\cos x=\dfrac{\sqrt{3}}{2}$$ when $$x=\dfrac{\pi}{6},\space \dfrac{11\pi}{6}$$. Example $$\PageIndex{8B}$$: Solving the Equation Using a Double-Angle Formula Solve the equation exactly using a double-angle formula: $$\cos(2\theta)=\cos \theta$$. Solution We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine: \begin{align} \cos(2\theta)&= \cos \theta\\ 2{\cos}^2 \theta-1&= \cos \theta\\ 2 {\cos}^2 \theta-\cos \theta-1&= 0\\ (2 \cos \theta+1)(\cos \theta-1)&= 0\\ 2 \cos \theta+1&= 0\\ \cos \theta&= -\dfrac{1}{2}\\ \cos \theta-1&= 0\\ \cos \theta&= 1 \end{align} So, if $$\cos \theta=−\dfrac{1}{2}$$,then $$\theta=\dfrac{2\pi}{3}\pm 2\pi k$$ and $$\theta=\dfrac{4\pi}{3}\pm 2\pi k$$; if $$\cos \theta=1$$,then $$\theta=0\pm 2\pi k$$. Example $$\PageIndex{8C}$$: Solving an Equation Using an Identity Solve the equation exactly using an identity: $$3 \cos \theta+3=2 {\sin}^2 \theta$$, $$0≤\theta<2\pi$$. Solution If we rewrite the right side, we can write the equation in terms of cosine: \begin{align} 3 \cos \theta+3&= 2 {\sin}^2 \theta\\ 3 \cos \theta+3&= 2(1-{\cos}^2 \theta)\\ 3 \cos \theta+3&= 2-2{\cos}^2 \theta\\ 2 {\cos}^2 \theta+3 \cos \theta+1&= 0\\ (2 \cos \theta+1)(\cos \theta+1)&= 0\\ 2 \cos \theta+1&= 0\\ \cos \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{2\pi}{3},\space \dfrac{4\pi}{3}\\ \cos \theta+1&= 0\\ \cos \theta&= -1\\ \theta&= \pi\\ \end{align} Our solutions are $$\theta=\dfrac{2\pi}{3},\space \dfrac{4\pi}{3},\space \pi$$. ## Solving Trigonometric Equations with Multiple Angles Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as $$\sin(2x)$$ or $$\cos(3x)$$. When confronted with these equations, recall that $$y=\sin(2x)$$ is a horizontal compression by a factor of 2 of the function $$y=\sin x$$. On an interval of $$2\pi$$,we can graph two periods of $$y=\sin(2x)$$,as opposed to one cycle of $$y=\sin x$$. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to $$\sin(2x)=0$$ compared to $$\sin x=0$$. This information will help us solve the equation. Example $$\PageIndex{9}$$: Solving a Multiple Angle Trigonometric Equation Solve exactly: $$\cos(2x)=\dfrac{1}{2}$$ on $$[ 0,2\pi )$$. Solution We can see that this equation is the standard equation with a multiple of an angle. If $$\cos(\alpha)=\dfrac{1}{2}$$,we know $$\alpha$$ is in quadrants I and IV. While $$\theta={\cos}^{−1} \dfrac{1}{2}$$ will only yield solutions in quadrants I and II, we recognize that the solutions to the equation $$\cos \theta=\dfrac{1}{2}$$ will be in quadrants I and IV. Therefore, the possible angles are $$\theta=\dfrac{\pi}{3}$$ and $$\theta=\dfrac{5\pi}{3}$$. So, $$2x=\dfrac{\pi}{3}$$ or $$2x=\dfrac{5\pi}{3}$$, which means that $$x=\dfrac{\pi}{6}$$ or $$x=\dfrac{5\pi}{6}$$. Does this make sense? Yes, because $$\cos\left(2\left(\dfrac{\pi}{6}\right)\right)=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$. Are there any other possible answers? Let us return to our first step. In quadrant I, $$2x=\dfrac{\pi}{3}$$, so $$x=\dfrac{\pi}{6}$$ as noted. Let us revolve around the circle again: \begin{align} 2x&= \dfrac{\pi}{3}+2\pi\\ &= \dfrac{\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{7\pi}{3}\\ x&= \dfrac{7\pi}{6}\\ \text {One more rotation yields}\\ 2x&= \dfrac{\pi}{3}+4\pi\\ &= \dfrac{\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{13\pi}{3}\\ \end{align} $$x=\dfrac{13\pi}{6}>2\pi$$, so this value for $$x$$ is larger than $$2\pi$$, so it is not a solution on $$[ 0,2\pi )$$. In quadrant IV, $$2x=\dfrac{5\pi}{3}$$, so $$x=\dfrac{5\pi}{6}$$ as noted. Let us revolve around the circle again: \begin{align} 2x&= \dfrac{5\pi}{3}+2\pi\\ &= \dfrac{5\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{11\pi}{3} \end{align} so $$x=\dfrac{11\pi}{6}$$. One more rotation yields \begin{align} 2x&= \dfrac{5\pi}{3}+4\pi\\ &= \dfrac{5\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{17\pi}{3} \end{align} $$x=\dfrac{17\pi}{6}>2\pi$$,so this value for $$x$$ is larger than $$2\pi$$,so it is not a solution on $$[ 0,2\pi )$$. Our solutions are $$x=\dfrac{\pi}{6}, \space \dfrac{5\pi}{6}, \space \dfrac{7\pi}{6}$$, and $$\dfrac{11\pi}{6}$$. Note that whenever we solve a problem in the form of $$sin(nx)=c$$, we must go around the unit circle $$n$$ times. ## Solving Right Triangle Problems We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, $a^2+b^2=c^2 \label{Pythagorean}$ and model an equation to fit a situation. Example $$\PageIndex{10A}$$: Using the Pythagorean Theorem to Model an Equation One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is $$69.5$$ meters above the ground, and the second anchor on the ground is $$23$$ meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure $$\PageIndex{4}$$. Solution Use the Pythagorean Theorem (Equation \ref{Pythagorean}) and the properties of right triangles to model an equation that fits the problem. Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem. \begin{align} a^2+b^2&= c^2\\ {(23)}^2+{(69.5)}^2&\approx 5359\\ \sqrt{5359}&\approx 73.2\space m \end{align} The angle of elevation is $$\theta$$,formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places. \begin{align} \tan \theta&= 69.523\\ {\tan}^{-1}(69.523)&\approx 1.2522\\ &\approx 71.69^{\circ} \end{align} The angle of elevation is approximately $$71.7°$$, and the length of the cable is $$73.2$$ meters. Example $$\PageIndex{10B}$$: Using the Pythagorean Theorem to Model an Abstract Problem OSHA safety regulations require that the base of a ladder be placed $$1$$ foot from the wall for every $$4$$ feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall. Solution For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be $$4a$$ feet. See Figure $$\PageIndex{5}$$. The side adjacent to $$\theta$$ is $$a$$ and the hypotenuse is $$4a$$. Thus, \begin{align} \cos \theta&= \dfrac{a}{4a}\\ &= \dfrac{1}{4}\\ {\cos}^{-1}\left (\dfrac{1}{4}\right )&\approx 75.5^{\circ} \end{align} The elevation of the ladder forms an angle of $$75.5°$$ with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem: \begin{align} a^2+b^2&= {(4a)}^2\\ b^2&= {(4a)}^2-a^2\\ b^2&= 16a^2-a^2\\ b^2&= 15a^2\\ b&= a\sqrt{15} \end{align} Thus, the ladder touches the wall at $$a\sqrt{15}$$ feet from the ground. Media Access these online resources for additional instruction and practice with solving trigonometric equations. ## Key Concepts • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example $$\PageIndex{1}$$, Example $$\PageIndex{2}$$, and Example $$\PageIndex{3}$$. • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example $$\PageIndex{4}$$, Example $$\PageIndex{5}$$, and Example $$\PageIndex{6}$$, and Example $$\PageIndex{7}$$. • We can also solve trigonometric equations using a graphing calculator. See Example $$\PageIndex{8}$$ and Example $$\PageIndex{9}$$. • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example $$\PageIndex{10}$$, Example $$\PageIndex{11}$$, Example $$\PageIndex{12}$$, and Example $$\PageIndex{13}$$. • We can also use the identities to solve trigonometric equation. See Example $$\PageIndex{14}$$, Example $$\PageIndex{15}$$, and Example $$\PageIndex{16}$$. • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example $$\PageIndex{17}$$. • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example $$\PageIndex{18}$$.
### Home > CC3MN > Chapter 8 Unit 8 > Lesson CC3: 8.2.3 > Problem8-96 8-96. Simplify each expression. 1. $- \frac { 9 } { 5 } \cdot \frac { 8 } { 15 }$ Multiply the numerators together and the denominators together. $-\frac{9(8)}{5(15)} = -\frac{72}{75}$ $-\frac{24}{25}$ 1. $\frac { 1 } { 5 } + ( - \frac { 2 } { 15 } ) - ( - \frac { 4 } { 9 } )$ Find a common denominator. $\frac{9}{45}+\left(-\frac{6}{45}\right)-\left(-\frac{20}{45}\right)$ 1. $- \frac { 4 } { 8 } \cdot \frac { 3 } { 7 } \cdot ( - \frac { 2 } { 5 } )$ Multiply, see (a) for help. 1. $\frac { 3 } { 5 } \cdot ( - \frac { 2 } { 7 } ) + ( - \frac { 5 } { 7 } ) ( \frac { 3 } { 10 } )$ Multiply the terms, then find the common denominator. $-\frac{27}{70}$ 1. $- 8 \frac { 1 } { 9 } + 3 \frac { 5 } { 6 }$ Make fractions greater than one, then find the least common denominator 1. $2 \frac { 1 } { 2 } \cdot 4 \frac { 1 } { 5 }$ Convert the mixed numbers into fractions greater than one before multiplying. $10\frac{1}{2}$
Subscribe to our Youtube Channel - https://you.tube/teachoo 1. Chapter 8 Class 8 Comparing Quantities 2. Concept wise 3. Compound Interest compounded half yearly Transcript Ex 8.3, 6 Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1 1/2 years if the interest is (i) compounded annually. Given, Principal (P) = 80000 Rate (R) = 10% p.a Time (N) = 1 1/2 years Since n is in fraction, We use the formula Compound Interest for 1 1/2 Years = Compound Interest for 1 year + Simple Interest for 1/2 year. Compound Interest for 1 Year P = 80000 R = 10% n = 1 Amount = P (1+๐/100)^๐ = 80000 ร (1+10/100)^1 = 80000 ร (1+1/10) = 80000 ร ((10 + 1)/10) = 80000 ร 11/10 = 88000 Since, Amount = Principal + Interest 88000 = 80,000 + Interest 88000 โ 80000 = Interest 8000 = Interest Interest = 8000 โด Interest for 1st year = Rs 8000 & Amount after 1st Year = Rs 88000 Simple Interest for next ๐/๐ year Principal = Amount in Previous Year = 88000 Rate = 10% p.a Time = 1/2 year Interest = (๐ ร ๐ ร ๐)/100 = (88000 ร10 ร 1/2)/100 = (88000 ร 5)/100 = 880 ร 5 = 4400 Simple Interest for next 1/2 Year = Rs 4400 Now, Interest after 1 1/4 years = Compound interest for 1 year + Simple interest for next 1/2 year = 8000 + 4400 = 12400 Also, Amount = Principal + Interest = 80000 + 12400 = 92400 โด Amount = Rs 92,400 Ex 8.3, 6 Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1 1/2 years if the interest is (ii) compounded half yearly. Given, Principal (P) = 80000 Rate(R) = 10% per annum Compound Half Yearly = 10/2 % per half yearly = 5 % Per half yearly Time (N) = 1 1/2 Years = 3/2 years = 3/2 ร 2 half years = 3 Half years Now, Amount = P (1+๐/100)^๐ = 80000 ร (1+5/100)^3 = 80000 ร (1+1/20)^3 = 80000 ร ((20 + 1)/20)^3 = 80000 ร (21/20)^3 = 80,000 ร ((21 ร 21 ร 21)/(20 ร 20 ร 20)) = 80,000 ร ((441 ร 21)/(400 ร 20)) = 80000 ร (9261/8000) = 10 ร 9261 = 92610 โด Amount = Rs 92,610 Difference in Amounts = Amount when interest is compounded half yearly โ Amount when interest is compounded annually = 92610 โ 92400 = 210 โด Difference in Amounts = Rs 210 Compound Interest compounded half yearly
# NCERT Solutions for Class 8 Maths Chapter 3- Understanding Quadrilaterals Exercise 3.3 The NCERT solutions for Class 8 maths Chapter 3- Understanding Quadrilaterals includes the solutions of all the questions present in the NCERT textbook. The subject experts at BYJUâ€™S have solved each question of NCERT exercises with utmost care, in order to help the students in solving any question from the NCERT textbook. NCERT Class 8 Exercise 3.3 is based on the different types of quadrilaterals. Students can download the NCERT Solutions of Class 8 mathematics to furthermore understand the concept. Â Â ### Access other exercise solutions of Class 8 Maths Chapter 3- Understanding Quadrilaterals Exercise 3.1 Solutions 7 Questions (1 Long Answer Questions, 6 Short Answer Questions) Exercise 3.2 Solutions 6 Questions (6 Short Answer Questions) Exercise 3.4 Solutions 6 Questions (1 Long Answer Questions, 5 Short Answer Questions) ### Access Answers to Maths NCERT Class 8 Chapter 3- Understanding Quadrilaterals Exercise 3.3 Page Number 50 1. Given a parallelogram ABCD. Complete each statement along with the definition or property used. (i) AD = â€¦â€¦ (ii) âˆ DCB = â€¦â€¦ (iii) OC = â€¦â€¦ (iv) m âˆ DAB + m âˆ CDA = â€¦â€¦ Solution: (i) AD = BC (Opposite sides of a parallelogram are equal) (ii) âˆ DCB = âˆ DAB (Opposite angles of a parallelogram are equal) (iii) OC = OA (Diagonals of a parallelogram are equal) (iv) m âˆ DAB + m âˆ CDA = 180Â° 2. Consider the following parallelograms. Find the values of the unknown x, y, z Solution: (i) y = 100Â° (opposite angles of a parallelogram) x + 100Â° = 180Â° (Adjacent angles of a parallelogram) â‡’ x = 180Â° â€“ 100Â° = 80Â° x = z = 80Â° (opposite angles of a parallelogram) âˆ´, x = 80Â°, y = 100Â° and z = 80Â° (ii) 50Â° + x = 180Â° â‡’ x = 180Â° â€“ 50Â° = 130Â° (Adjacent angles of a parallelogram) x = y = 130Â° (opposite angles of a parallelogram) x = z = 130Â° (corresponding angle) (iii) x = 90Â° (vertical opposite angles) x + y + 30Â° = 180Â° (angle sum property of a triangle) â‡’ 90Â° + y + 30Â° = 180Â° â‡’ y = 180Â° â€“ 120Â° = 60Â° also, y = z = 60Â° (alternate angles) (iv) z = 80Â° (corresponding angle) z = y = 80Â° (alternate angles) x + y = 180Â° (adjacent angles) â‡’ x + 80Â° = 180Â° â‡’ x = 180Â° â€“ 80Â° = 100Â° (v) x=28o y = 112o z = 28o 3. Can a quadrilateral ABCD be a parallelogram if (i) âˆ D + âˆ B = 180Â°? Â (ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm? (iii)âˆ A = 70Â° and âˆ C = 65Â°? Solution: (i) Yes, a quadrilateral ABCD be a parallelogram if âˆ D + âˆ B = 180Â° but it should also fulfilled some conditions which are: (a) The sum of the adjacent angles should be 180Â°. (b) Opposite angles must be equal. (ii) No, opposite sides should be of same length. Here, AD â‰ BC (iii) No, opposite angles should be of same measures. âˆ A â‰ âˆ C 4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure. Â Solution: ABCD is a figure of quadrilateral that is not a parallelogram but has exactly two opposite angles that is âˆ B = âˆ D of equal measure. It is not a parallelogram because âˆ A â‰ âˆ C. 5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram. Â Solution: Let the measures of two adjacent angles âˆ A and âˆ B be 3x and 2x respectively in parallelogram ABCD. âˆ A + âˆ B = 180Â° â‡’ 3x + 2x = 180Â° â‡’ 5x = 180Â° â‡’ x = 36Â° We know that opposite sides of a parallelogram are equal. âˆ A = âˆ C = 3x = 3 Ã— 36Â° = 108Â° âˆ B = âˆ D = 2x = 2 Ã— 36Â° = 72Â° 6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram. Â Solution: Let ABCD be a parallelogram. Sum of adjacent angles of a parallelogram = 180Â° âˆ A + âˆ B = 180Â° â‡’ 2âˆ A = 180Â° â‡’ âˆ A = 90Â° also, 90Â° + âˆ B = 180Â° â‡’ âˆ B = 180Â° â€“ 90Â° = 90Â° âˆ A = âˆ C = 90Â° âˆ B = âˆ D = 90 Â° 7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them. Â Solution: y = 40Â° (alternate interior angle) âˆ P = 70Â° (alternate interior angle) âˆ P = âˆ H = 70Â° (opposite angles of a parallelogram) z = âˆ H â€“ 40Â°= 70Â° â€“ 40Â° = 30Â° âˆ H + x = 180Â° â‡’ 70Â° + x = 180Â° â‡’ x = 180Â° â€“ 70Â° = 110Â° 8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm) Â Solution: (i) SG = NU and SN = GU (opposite sides of a parallelogram are equal) 3x = 18 x = 18/3 â‡’ x =6 3y â€“ 1 = 26 an d, â‡’ 3y = 26 + 1 â‡’ y = 27/3=9 x = 6 and y = 9 (ii) 20 = y + 7 and 16 = x + y (diagonals of a parallelogram bisect each other) y + 7 = 20 â‡’ y = 20 â€“ 7 = 13 and, x + y = 16 â‡’ x + 13 = 16 â‡’ x = 16 â€“ 13 = 3 x = 3 and y = 13 9. In the above figure both RISK and CLUE are parallelograms. Find the value of x. Â Solution: âˆ K + âˆ R = 180Â° (adjacent angles of a parallelogram are supplementary) â‡’ 120Â° + âˆ R = 180Â° â‡’ âˆ R = 180Â° â€“ 120Â° = 60Â° also, âˆ R = âˆ SIL (corresponding angles) â‡’ âˆ SIL = 60Â° also, âˆ ECR = âˆ L = 70Â° (corresponding angles) x + 60Â° + 70Â° = 180Â° (angle sum of a triangle) â‡’ x + 130Â° = 180Â° â‡’ x = 180Â° â€“ 130Â° = 50Â° 10. Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.32) Â Solution: When a transversal line intersects two lines in such a way that the sum of the adjacent angles on the same side of transversal is 180Â° then the lines are parallel to each other. Here, âˆ M + âˆ L = 100Â° + 80Â° = 180Â° Thus, MN \|\| LK As the quadrilateral KLMN has one pair of parallel line therefore it is a trapezium. MN and LK are parallel lines. 11. Find mâˆ C in Fig 3.33 if AB \|\| DC ? Â Solution: mâˆ C + mâˆ B = 180Â° (angles on the same side of transversal) â‡’ mâˆ C + 120Â° = 180Â° â‡’ mâˆ C = 180Â°- 120Â° = 60Â° 12. Find the measure of âˆ P and âˆ S if SP \|\| RQ ? in Fig 3.34. (If you find mâˆ R, is there more than one Â method to find mâˆ P?) Solution: âˆ P + âˆ Q = 180Â° (angles on the same side of transversal) â‡’ âˆ P + 130Â° = 180Â° â‡’ âˆ P = 180Â° â€“ 130Â° = 50Â° also, âˆ R + âˆ S = 180Â° (angles on the same side of transversal) â‡’ 90Â° + âˆ S = 180Â° â‡’ âˆ S = 180Â° â€“ 90Â° = 90Â° Thus, âˆ P = 50Â° and âˆ S = 90Â° Yes, there are more than one method to find mâˆ P. PQRS is a quadrilateral. Sum of measures of all angles is 360Â°. Since, we know the measurement of âˆ Q, âˆ R and âˆ S. âˆ Q = 130Â°, âˆ R = 90Â° and âˆ S = 90Â° âˆ P + 130Â° + 90Â° + 90Â° = 360Â° â‡’ âˆ P + 310Â° = 360Â° â‡’ âˆ P = 360Â° â€“ 310Â° = 50Â° Exercise 3.3 of NCERT Solutions for Class 8 Maths Chapter 3- Understanding Quadrilaterals is based on the following topics:
Help solve math problems step by step for free can support pupils to understand the material and improve their grades. We can solve math problems for you. ## The Best Help solve math problems step by step for free Help solve math problems step by step for free is a mathematical tool that helps to solve math equations. Then, you can use your knowledge of the value of one variable to eliminate that variable from the system. Once you have eliminated one variable, you can then use your knowledge of the remaining variable to eliminate that second variable. This process is repeated until only one equation remains, and then you can solve for your solution. Using this method, you can eliminate any number of variables in a linear system at once, allowing you to solve complex problems more quickly than other methods. While you can't completely prevent tanning, there are ways to protect yourself when you're out in the sun. Wear a hat, apply sunscreen regularly and stay out of the midday sun between 10 a.m. and 4 p.m., when UV rays hit their peak strength. Keep your skin hydrated by drinking plenty of water throughout the day and applying a hydrating moisturizer after you shower. And if you see any signs of skin damage or abnormalities, see your doctor right away so you can start treatment early. It is pretty simple to solve a geometric sequence. If we have a sequence A, B, C... of numbers and it looks like AB, then we can simply start at A and work our way down the list. Once we reach C, we are done. In this example, we can easily see AB = BC = AC ... Therefore once we reach C, the solution is complete. Let's try some other examples: A = 1, B = 2, C = 4 AB = BC = AC = ACB ACAB = ABC ==> ABC + AC ==> AC + AB ==> AC + B CABACCA ==> CA + AB ==> CA + B + A ==> CA + (B+A) ==> CABABABABABA The solutions are CABABABABABA and finally ABC. Then, you'd isolate the D on the left side by multiplying both sides by -1. This gives you: You can now substitute this value for D into your original equation and solve for x. When done correctly, you're left with two equations that are equal and one solution. It's important to note that solving simultaneous equations isn't always easy. Because they require so much mental juggling, sometimes people give up before they get started. However, with some practice it can become second nature. And once you understand how they work, you'll be able to solve them in your sleep! A right triangle is a triangle with two right angles. By definition, it has one leg that's longer than the other. A right triangle has three sides. A right triangle has three sides: the hypotenuse (the longest side) and two shorter sides. These are called legs. The legs are always equal in length. They have equal lengths to each other and to the hypotenuse. The hypotenuse is the longest side of a right triangle and is therefore the opposite side from the one with the highest angle. It is also called the altimeter or longer leg. Right triangles always have an altimeter (the longest side). It is opposite to the hypotenuse and is also called the longer leg or hypotenuse. The other two sides of a right triangle are called legs or short sides. These are always equal in length to each other and to the longer leg of the triangle, which is called the hypotenuse. The sum of any two angles in a right triangle must be 180 degrees, because this is one full turn in any direction around a vertical line from vertex to vertex of an angle-triangle intersection. An angle-triangle intersection occurs when two lines that intersect at a common point meet back together at another point on their way down from both vertexes to that point where they intersected at first! it really helps when you don't understand what you're doing. I originally got it to do my homework quickly, but as the math at school got harder, the app became more of a Reacher and less of a quick fix ### Octavia Hughes Literally THE BEST app for students with just one click you can solve any math question, no matter how complicated it is, and not only that, you get a step-by-step detailed explanation of how to solve the question, Plus the app shows you relevant graphs of your functions. ### Romina Washington Numerical solver Circle solver How to solve ratios How to solve limits Precalculus problem solver Prealgebra solver
# What Are The Kinematic Formulas? Kinematic Equations: The goal of this first unit of The Physics Classroom has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations.There are a variety of quantities associated with the motion of objects – displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object’s motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object’s motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations) ## 4 Kinematic Equations Kinematics is the study of objects in motion and their inter-relationships. There are four (4) kinematic equations, which relate to displacement, D, velocity, v, time, t, and acceleration, a. a) D = vit + 1/2 at2 b) (vi +vf)/2 = D/t c) a = (vf – vi)/t d) vf2 = vi2 + 2aD D = displacement a = acceleration t = time vf = final velocity vi = initial velocity If we know three of these five kinematic variables— Δ x , t , v 0 , v , a \Delta x, t, v_0, v, a Δx,t,v0,v,adelta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a—for an object under constant acceleration, we can use a kinematic formula, see below, to solve for one of the unknown variables. four kinematic equations The four kinematic equations that describe an object’s motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. Kinematic equations can be used to calculate various aspects of motion such as velocity, acceleration, displacement, and time. ## Kinematic Equations List The kinematic formulas are a set of formulas that relate the five kinematic variables listed below. 1, point, space, v, equals, v, start subscript, 0, end subscript, plus, a, t 2, point, space, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t 3, point, space, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript 4, point, space, v, start superscript, 2, end superscript, equals, v, start subscript, 0, end subscript, start superscript, 2, end superscript, plus, 2, a, delta, x Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing. Also, the kinematic formulas assume all variables are referring to the same direction: horizontal x, vertical y, etc. ## Angular Kinematic Equations A freely flying object is defined as any object that is accelerating only due to the influence of gravity. We typically assume the effect of air resistance is small enough to ignore, which means any object that is dropped, thrown, or otherwise flying freely through the air is typically assumed to be a freely flying projectile with a constant downward acceleration of magnitude g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction. This is both strange and lucky if we think about it. It’s strange since this means that a large boulder will accelerate downwards with the same acceleration as a small pebble, and if dropped from the same height, they would strike the ground at the same time. [How can this be so?] It’s lucky since we don’t need to know the mass of the projectile when solving kinematic formulas since the freely flying object will have the same magnitude of acceleration, g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction, no matter what mass it has—as long as air resistance is negligible. Note that g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction is just the magnitude of the acceleration due to gravity. If upward is selected as positive, we must make the acceleration due to gravity negative a, start subscript, y, end subscript, equals, minus, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction for a projectile when we plug into the kinematic formulas. ## Physics Kinematic Equations The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object’s motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object’s motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object’s motion. The four kinematic equations that describe an object’s motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object’s motion. As such, they can be used to predict unknown information about an object’s motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. ## Basic Kinematic Equations Kinematics is the study of the motion of objects without concern for the forces causing the motion. These familiar equations allow students to analyze and predict the motion of objects, and students will continue to use these equations throughout their study of physics. A solid understanding of these equations and how to employ them to solve problems is essential for success in physics. This article is a purely mathematical exercise designed to provide a quick review of how the kinematics equations are derived using algebra. This exercise references the diagram in Fig. 1, in which the x axis represents time and the y axis represents velocity. The diagonal line represents the motion of an object, with velocity changing at a constant rate. The shaded area (A1 + A2) represents the displacement of the object during the time interval between t1 and t2, during which the object increased velocity from v1 to v2. This document will make use of the following variables: v = the magnitude of the velocity of the object (meters per second, m/s) v1 = the magnitude of the initial velocity (meters per second, m/s) (in some texts this is vi or v0) v2 = the magnitude of the final velocity (meters per second, m/s) (in some texts this is vf) a = the magnitude of the acceleration (in meters per second squared, m/s2) s = the displacement vector, the magnitude of the displacement is the distance, s = │s│ = d (vectors are indicated in bold; the same symbol not in bold represents the magnitude of the vector) Δ indicates change, for example Δv = (v2 –v1) t = time t1 = the initial time t2 = the final time
# Applying the Work-Energy Theorem to Falling Objects So far we’ve applied the Work-Energy Theorem to a flying object, namely, Santa’s sleigh, and a rolling object, namely, a car braking to avoid hitting a deer. Here we’ll apply the Theorem to a falling object, that coffee mug we’ve been following through this blog series. We’ll use the Theorem to find the force generated on the mug when it falls into a pan of kitty litter. This falling object scenario is one I frequently encounter in previous topics and it’s something I’ve got to consider when designing objects that must withstand impact forces if they are dropped. Applying the Work-Energy Theorem to Falling Objects Here’s the Work-Energy Theorem formula again, F × d = ½ × m × [v22 – v12] where F is the force applied to a moving object of mass m to get it to change from a velocity of v1 to v2 over a distance, d. As we follow our falling mug from its shelf, its mass, m, eventually comes into contact with an opposing force, F, which will alter its velocity when it hits the floor, or in this case a strategically placed pan of kitty litter. Upon hitting the litter, the force of the mug’s falling velocity, or speed, causes the mug to burrow into the litter to a depth of d. The mug’s speed the instant before it hits the ground is v1, and its final velocity when it comes to a full stop inside the litter is v2, or zero. Inserting these values into the Theorem, we get, F × d = ½ × m × [0 – v12] F × d = – ½ × m × v12 The right side of the equation represents the kinetic energy that the mug acquired while in freefall. This energy will be transformed into Gaspard Gustave de Coriolis’ definition of work, which produces a depression in the litter due to the force of the plummeting mug. Work is represented on the left side of the equal sign. Now a problem arises with using the equation if we’re unable to measure the mug’s initial velocity, v1. But there’s a way around that, which we’ll elaborate next time when we put the Law of Conservation of Energy to work for us to do just that.
# Quartiles and Box-and-Whisker Plots ```0732_hsm07a1_se_ny_1002.qxd 6/13/07 3:49 PM Page 732 Quartiles and Box-and-Whisker Plots NY-2 Learning Standards for Mathematics GO for Help Check Skills You’ll Need A.S.11 Find the percentile rank of an item in a data set and identify the point values for ﬁrst, second, and third quartiles. Find the range and the median. 1. 3 8 12 15 20 3. 0 2 7 10 –1 –4 –11 A.S.5, A.S.9, A.S.6 Construct, analyze, and interpret a box-andwhisker plot using the ﬁve number summary. Lesson 1-6 2. 2.1 3.3 –5.4 0.8 3.5 4. 64 16 23 57 14 22 New Vocabulary • ﬁrst quartile • third quartile • ﬁve number summary • minimum • maximum • percentiles • percentile rank • quartile • box-and-whisker plot • interquartile range 1 2 Part Graphing and y ≠ ax Percentiles 1 1 Quartiles In Lesson 1-6, you used measures of central tendency to describe data sets. In this lesson, you will use quartiles to separate data in four equal parts. The median separates the data into upper and lower halves. The ﬁrst quartile (Q1) is the median of the lower half of the data. The third quartile (Q3) is the median of the upper half of the data. minimum 1 EXAMPLE Q1 Q2 Q3 first quartile median of lower half median third quartile median of upper half maximum Finding Median, First, and Third Quartiles For the data below, ﬁnd the median (Q2), ﬁrst quartile (Q1), and third quartile (Q3). 39 27.5 26 31 42 26 37 30 22 Step 1 Arrange the data in order from least to greatest. Find the median. 22 26 26 27.5 30 31 37 39 42 The median (Q2) is the middle value 30. Step 2 Find the ﬁrst quartile and third quartile, which are the medians of the lower and upper halves. 22 26 \| 26 27.5 30 31 37 \| 39 42 first quartile (Q1) = 26 1 26 = 26 2 Quick Check NY 732 third quartile (Q3) = 37 1 39 = 38 2 1 For the following data, ﬁnd the median, the ﬁrst quartile, and the third quartile. 480 500 600 250 660 570 490 610 530 Chapter NY New York Additional Topics 0732_hsm07a1_se_ny_1002.qxd 6/13/07 3:49 PM Page 733 The ﬁve number summary of a set of data includes the minimum, the ﬁrst quartile, the median, the third quartile, and the maximum. The minimum of a data set is the least value. The maximum is the greatest value. 2 EXAMPLE Finding the Five Number Summary Find the ﬁve number summary for the data below. 125 80 140 135 126 140 350 Step 1 Arrange the data in order from least to greatest. 80 125 126 135 140 140 350 Step 2 Find the minimum, maximum, and median. 80 125 126 135 140 140 350 The minimum is 80. The maximum is 350. The median is 135. Step 3 Find the ﬁrst quartile and third quartile. ﬁrst quartile (Q1) = 125 1 126 = 125.5 2 third quartile (Q3) = 140 1 140 = 140 2 The ﬁve number summary is minimum = 80, Q1 = 125.5, median = 135, Q3 = 140, and maximum = 350. Quick Check 2 Find the ﬁve number summary for each set of data. a. 95 85 75 85 65 60 100 105 75 85 75 b. 11 19 7 5 21 53 Data sets can be separated into smaller equal parts. Percentiles separate data sets into 100 equal parts. The percentile rank of a score is the percentage of scores that are less than or equal to that score. 3 EXAMPLE Finding a Percentile Rank Of 25 test scores, eight are less than or equal to 75. What is the percentile rank of a test score of 75? Write a ratio of the number of scores less than or equal to 75 compared to the total number of test scores. 8 25 8 = 0.32 25 = 32% Number of test scores less than or equal to 75 Total number of test scores 8 Rewrite 25 as a percent. The percentile rank of 75 is 32. Quick Check 3 a. Of the 25 test scores, 15 scores are less than or equal to 85. What is the percentile rank of 85? b. Critical Thinking Is it possible to have a percentile rank of 0? Explain. Lesson NY-2 Quartiles and Box-and-Whisker Plots NY 733 0732_hsm07a1_se_ny_1002.qxd 2 1 6/13/07 3:49 PM Page 734 Box-and-Whisker Plots A box-and-whisker plot is a graph that summarizes a data set along a number line. The box represents the ﬁrst quartile, the median, and the third quartile of the data. The whiskers on the left and right sides represent the minimum and maximum values. Use the ﬁve number summary to construct a box-and-whisker plot. Box-and-Whisker Plot whisker minimum 4 EXAMPLE whisker box first quartile third maximum quartile median Real-World Problem Solving Agriculture The table below shows United States crops harvested from 1992 to 2004. Find the ﬁve number summary and construct a box-and-whisker plot of the data. Step 1 Arrange the data in order from least to greatest. Find the minimum, maximum, and the median. 308 314 316 317 321 321 321 324 325 326 326 327 332 Step 2 Find the ﬁrst quartile and third quartile. first quartile = 316 1 317 = 633 = 316.5 2 2 third quartile = 326 1 326 = 652 = 326 2 2 Crops Harvested Acres Acres Year (millions) Year (millions) 1992 1993 1994 1995 1996 1997 1998 317 308 321 314 326 332 326 1999 2000 2001 2002 2003 2004 327 325 321 316 324 321 SOURCE: Statistical Abstract of the United States. Step 3 Draw a number line. Construct the box-and-whisker plot. Mark the values of the minimum, ﬁrst quartile, median, third quartile, and maximum. Crops Harvested (millions of acres) 290 300 310 320 330 340 350 Quick Check 4 Construct a box-and-whisker plot for the following distances of bird migrations in thousands of miles. 5 2.5 6 8 9 2 1 4 6.2 18 NY 734 Chapter NY New York Additional Topics 0732_hsm07a1_se_ny_1002.qxd 6/13/07 3:49 PM 5 Page 735 EXAMPLE Using a Graphing Calculator Use a graphing calculator to make a box-and-whisker plot of the data below. Find the ﬁve number summary. 10 26 18 35 14 11 17 29 31 25 27 20 19 12 13 26 L1 L2 L3 1 25 27 20 19 12 13 26 1:L1 Plot1 Plot2 Plot3 On Off Type: Xlist: L1 Freq: 1 L1(16) = 26 Med=19.5 Step 1 Press 1: Edit . . . . Enter data under L1. (To clear L1, move the cursor over L1 and press , .) Step 2 Y= for Press STAT PLOTS. Enter 1. Highlight ON and also Type: box-and-whisker plot. Press . Step 3 Press and enter 9. Press and move the cursor across the box-and-whisker plot to ﬁnd the ﬁve number summary. The ﬁve number summary is minimum = 10, Q1 = 13.5, median = 19.5, Q3 = 26.5, maximum = 35. Quick Check 5 Use a graphing calculator to ﬁnd the ﬁve number summary for the following set of 98 92 76 84 93 82 74 68 85 91 77 83 94 97 72 88 70 84 87 82 The interquartile range is the difference between the ﬁrst and third quartiles. In a box-and-whisker plot, this is the width of the box. You can use a parallel box-and-whisker plot to compare two sets of data. 6 EXAMPLE Comparing Data Sets The box-and-whisker plot below shows average monthly rainfall for Miami and New Orleans. Which city shows the greater range in average monthly rainfall? Average Monthly Rainfall (millimeters) 20 40 60 80 100 120 140 160 180 200 220 Miami, Fla. New Orleans, La. Miami has the longer box-and-whisker plot, so it has the greater range. Quick Check 6 What do the interquartile ranges tell you about the average monthly rainfall for Miami and New Orleans? Lesson NY-2 Quartiles and Box-and-Whisker Plots NY 735 0732_hsm07a1_se_ny_1002.qxd 6/22/07 12:59 AM Page 736 EXERCISES For more exercises, see Extra Skill and Word Problem Practice. Practice and Problem Solving A Practice by Example Example 1 GO for Help (page NY 732) Example 2 (page NY 733) Example 3 (page NY 733) Example 4 (page NY 734) Example 5 (page NY 735) Find the median, the ﬁrst quartile, and the third quartile. 1. 12 10 11 7 9 10 5 2. 4.5 3.2 6.3 5.2 5 4.8 6 3.9 12 3. 55 53 67 52 50 49 51 52 52 4. 101 100 100 105 101 102 104 5. Find the ﬁve number summary for the set of data. 1 1 3 3 4 5 7 7 8 9 9 32 31 29 35 30 32 11 6. Of 20 scores, 19 are less than or equal to 10. Find the percentile rank of 10. 7. Gerry ranks 48th in a class of 120 students. What is his percentile rank? 8 Construct a box-and-whisker plot for the ages of the members of the art club. 13 18 16 14 15 17 15 16 16 15 16 14 14 17 19 9. Use a graphing calculator to draw a box-and-whisker plot for the data below. Then ﬁnd the interquartile range. 16 18 59 75 30 34 25 49 27 16 21 58 71 19 50 Example 6 10. Compare the box-and-whisker plots below. What can you conclude? (page NY 735) Ages of U.S. Olympic Soccer Team Players 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 Men Women B For Exercises 11–16, use the data below. Find each measure. 1.8 2.5 3.9 4.6 4.7 4.8 4.8 4.9 11. median 12. ﬁrst quartile 13. minimum 14. maximum 15. third quartile 16. percentile rank of 4.6 17. Use the box-and-whisker plot below. What can you conclude about the areas of state parks? Areas of State Parks (acres) 100 200 300 400 500 600 For Exercises 18–21, use the graph below showing box-and-whisker plots for two sets of data, A and B. 66 68 70 72 74 76 78 Data Set A Data Set B 18. Which set of data has the greater range? 19. Which set of data has the lesser median? NY 736 Chapter NY New York Additional Topics 80 82 84 86 0732_hsm07a1_se_ny_1002.qxd 6/13/07 3:49 PM Page 737 20. Which set of data has the greater interquartile range? 21. Which set of data has the lesser minimum? For Exercises 22–24, use the ﬁve number summary given below. Minimum = 10 Q1 = 20 Median = 30 22. Find the interquartile range. Q3 = 42 Maximum = 75 23. Find the range of the data. 24. Use the data to construct a box-and-whisker plot. 25. Social Studies The plots below compare the percents of the voting-age population who said they registered to vote in U.S. elections to the percents who said they voted. Which conclusion best reﬂects the data collected? Percents of Population Who Registered and Voted, 1990–2000 40 50 60 70 Registered Voted A. The percent who voted was about 15% less than the percent who registered. B. The percent who voted was about half the percent who registered. C. The percent who voted was equal to the percent who registered. D. The percent who voted was about 15% more than the percent who registered. 26. Error Analysis In a class of 250 students, Emily had the tenth highest grade average. She computed her percentile rank as 4. What was her error? C Challenge 27. Reasoning Can you ﬁnd the mean, median, and mode of a set of data by looking at a box-and-whisker plot? Explain. REGENTS Test Prep Multiple Choice 28. Find the ﬁrst quartile and third quartile in the following data set. 17 20 30 19 20 18 25 28 31 23 17 29 31 33 28 A. 19 and 30 B. 17 and 25 C. 19.5 and 30.5 D. 17 and 33 29. Of 20 test scores, sixteen are less than or equal to 80. What is the percentile rank of a test score of 80? F. 16th Short Response G. 85th H. 25th J. 80th 30. Describe how you could ﬁnd the 75th percentile score in a set of 20 scores. Mixed Review Lesson 1-1 Deﬁne variables and write an equation to model each situation. 31. The total cost of the number of CDs times \$5.00. 32. The perimeter of a square equals 4 times the length of the side. Lesson 1-3 Decide whether each statement is true or false. If the statement is false, give the counterexample. 33. All whole numbers are rational numbers. 34. The square root of a number is always smaller than the number. Lesson NY-2 Quartiles and Box-and-Whisker Plots NY 737 ```
1 / 12 Correlation Correlation. In this lesson you will cover:. How to measure and interpret correlation About the effects of scaling data on correlation. When two sets of random variables (bivariate data) are displayed on a scatter graph; we are used to describing the correlation but how do you measure it?. Télécharger la présentation Correlation E N D Presentation Transcript 1. Correlation 2. In this lesson you will cover: • How to measure and interpret correlation • About the effects of scaling data on correlation 3. When two sets of random variables (bivariate data) are displayed on a scatter graph; we are used to describing the correlation but how do you measure it? 4. Two sets of random variables (bivariate data) we can describe correlation but how do you measure it? x - = - y - = + - x + = + x - = + y - = + + x + = + x - = - y - = - - x - = + x - = + y - = - + x - = + 5. Covariance – how do you interpret it? • When the covariance is positive it suggests positive correlation • When covariance is negative it suggests negative correlation • When the covariance is close to zero it suggests no correlation. 6. Covariance – can you see any potential problems with this method alone? • When the covariance is positive it suggest positive correlation • When covariance is negative it suggest negative correlation • When the covariance is close to zero it suggests no correlation. • You guessed it: • (you don’t know the range) 7. Pearson Moment Correlation Coefficient • Is to standardise the covariance so that it can interpreted easily. It converts the covariance to a number between -1 to 1, where: • -1 is a perfect negative correlation • 1 is a perfect positive correlation • 0 is no correlation Karl Pearson 1857 - 1936 8. Pearson Moment Correlation Coefficient can be simplified to: This is the covariance This is the standard deviation of y This is the standard deviation of x 9. Task • Exercise A • Page 140 10. The effect of scaling • If you work out the correlation coefficient for sales of ice-cream & temperature (t) in Fahrenheit. Would you expect the correlation to change if you worked on the same data but in Celsius? • No – scaling has no effect on correlation. 11. Be aware of correlation claims • Some things may look like they are connected but they are not: • General knowledge and height: Children in a school from year 7 to year 13 are asked general knowledge questions. The correlation is worked out using height and their score. In your opinion does height have any effect on their score? If not can you suggest what is the explanatory factor that is connected to both? • Outliners • As all data items are used outliners will effect the correlation coefficient. When outliners are obvious it is worth ignoring them altogether. • Non-linear relationships. • Pearson's p.m.c.c. is only suitable for linear relationships 12. Task • Exercise C • Page 144 • Test yourself • Page 145 • HOMEWORK!!!!!!!!!!!! • Past papers • Past papers • Past papers • Go through each of the chapters: • PowerPoint's • revision notes • Unanswered questions More Related
# GRE Math : How to multiply exponents ## Example Questions 2 Next → ### Example Question #21 : Exponents Simplify: Explanation: Remember, we add exponents when their bases are multiplied, and multiply exponents when one is raised to the power of another. Negative exponents flip to the denominator (presuming they originally appear in the numerator). ### Example Question #471 : Algebra is a real number such that . Quantity A: Quantity B: The two quantities are equal. Quantity A is greater. Quantity B is greater. The relationship cannot be determined from the information given. The relationship cannot be determined from the information given. Explanation: (y2)(y4) = y2+4 = y6 Plug in two different values for y. Plug in y = 1: y8 = y Plug in y = 2: y8 > y Since the results differ, the relationship cannot be determined from the information given. ### Example Question #13 : How To Multiply Exponents Evaluate: Explanation: Can be simplified to: ### Example Question #14 : How To Multiply Exponents Simplify the following expression. Explanation: To solve this problem, we must first understand some of the basic concepts of exponents. When multiplying exponents with the same base, one would simply add the powers together with the same base to obtain the result. For example, . When raising exponents to a certain power, one would simply multiply the power the exponent is being raised to with the exponent itself to obtain the new exponent. The base also gets raised to the same power as it normally would and the new exponent gets put on afterward. For example . Now that we have covered the basic concepts of exponent manipulation, we can now solve the problem. The top part of the expression give to us is . As we have stated before, when an exponent is raised to a certain power, you simply multiply the power and the exponent together. This results in . The new expression becomes . Solving for the denominator of the equation, we previously stated that when multiplying exponents together with the same base, we simply add the exponents together and keep the same base. Therefore . The new expression becomes , however , therefore the expression becomes . 2 Next → Tired of practice problems? Try live online GRE prep today.
# How do you solve (q - 12) 3 <5q + 2? Jul 12, 2016 $q > - 19$ #### Explanation: Multiply out the bracket on the left hand side: $3 \cdot q - 3 \cdot 12 < 5 q + 2$ $3 q - 36 < 5 q + 2$ Subtract $3 q$ from both sides of the inequality: $- 36 < 2 q + 2$ Subtract 2 from both sides of the inequality: $- 38 < 2 q$ Divide both sides by 2 $- 19 < q$ So our solution is that $q > - 19$
# 2018 AMC 12B Problems/Problem 7 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem What is the value of $$\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?$$ $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$ ## Solution 1 From the Change of Base Formula, we have $$\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{\textbf{(C) } 6}.$$ ## Solution 2 Using the chain rule of logarithms $\log _{a} b \cdot \log _{b} c = \log _{a} c,$ we get \begin{align} \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27 &= (\log _{3} 7 \cdot \log _{7} 11 \cdots \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdots \log _{21} 25) \\ &= \log _{3} 27 \cdot \log _{5} 25 \\ &= 3 \cdot 2 \\ &= \boxed{\textbf{(C) } 6}. \end{align} ~ pi_is_3.14 ## Video Solution (HOW TO THINK CRITICALLY!!!) ~Education, the Study of Everything 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
# Using the Fibonacci numbers to represent whole numbers First we show how to use the Fibonacci numbers only to represent every whole number, how to use it to convert easily between miles and kilometres, that multiplication in this system is easy and a connection with the Rabbit sequence. All this is done from scratch and does not need lots of mathematical experience. ## Contents of This Page The icon means there is a Things to do investigation at the end of the section. The icon means there is an interactive calculator in this section. ## Our decimal system The way we write our numbers is based on a system of tens - the decimal system. Each column is worth ten times the one on its right so that the columns indicate powers of ten: ``` ... 1000 100 10 1 3 6 0 7 = three thousand, six hundred (no tens) and seven ``` Since each column is TEN times the one on its right, we need ten symbols to represent the ten values in each column: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, called digits. Each positive number has a unique representation in the decimal system. Why use 10? The reason is almost certainly that early writing systems were based on counting using the fingers. [Our word digit comes from the Latin for finger. ] Tally systems were ways of putting marks or notches in wooden sticks (tally sticks) and they can be read more easily if grouped in batches of 5 or 10 for convenience. ## Other bases What if we used another power or base rather than ten? ### Binary Using powers of 2, we have the binary system, used in almost all computers. Here the columns are labelled with the powers of 2, and there are just 2 binary digits, 0 and 1, called bits. ``` ... 16 8 4 2 1 1 0 1 1 = 8 + 2+1 = eleven ``` In order to distinguish 11 (eleven) from 11 in another base, we will put the base as a subscript (or sometimes in brackets) after the representation to avoid confusion. So 1011 in binary is 11 in base 10 is written as: 10112 = 1110 Note that the base numbers (here 2 and 10) are always ordinary decimal numbers. In the next section we will see that the binary system is used in musical notation. #### Musical Notation If a crotchet is taken as lasting for one beat, then the semibreve is 4 beats , the minim 2, a quaver 1/2, a semiquaver 1/4 and demi-semiquaver is 1/8. They are written in musical notation as shown here: NoteSymbolDuration (beats) Semi-breve4 Minim2 Crotchet1 Quaver1/2 Semiquaver1/4 Demisemiquaver1/8 A dot is placed after a note to add on one half of its value. So a dotted crotchet is a crotchet plus a quaver and has a duration of 1·5 time units; two dots after a crotchet give a duration of 1 + 1/2 + 1/4 = 1·75 units. F.J Budden in An Introduction to Number Scales and Computers, Longmans, 1965, page 65, says he thinks the record number of dots is 4 in Verdi's Requiem in the Rex Tremenda. It is useful when a long note is followed by a quick note and the next note is "on the beat". Binary fractions are written using column headings as follows: ` ... 8 4 2 1 · 1/2 1/4 1/8 ...` So 1/4 = 0·012 and 3/8 = 0·0112 since it is 1/4+1/8. In binary, a dot after a crotchet adds a one in a fractional column: crotchet = 1 crotchet dot = 1·12 crotchet dot dot = 1·112 crotchet dot dot dot = 1·1112 and so on. ### More bases Base 8 is called octal and is presumably used by intelligent octopuses (or should that be octopi)! It uses "digits" 0, 1, 2, 3, 4, 5, 6 and 7. Base 3 is ternary and uses only 0, 1 and 2. Here is one hundred expressed in all the bases from 2 to 9: 11001002 = 102013 = 12104 = 4005 = 2446 = 2027 = 1448 = 1219 = 10010 Base 2 is called binary, Base 3 is called ternary, Base 4 is called quaternary, Base 5 is called quinary, Base 6 is called senary, Base 7 is called septenary, Base 8 is called octonary or octal, Base 9 is called nonary, Base 10 is called denary or decimal, continued below ... What about Base 1? Numbers in base 1 have columns that are powers of 1, so they are all ones! You might think that this base is not very good, but it was, in fact, the earliest system of written numbers. In Base 1, 2 is 11, 3 is 111, 4 is 1111, etc. So we make marks (1s) to count the number. This was used in a Talley System which worked like this: Suppose I lent you 5 sheep to graze on your field and cut your grass, I would make a mark for each sheep on a stick. Then the stick would be broken in half lengthways (across the marks) so we each had a copy of the 5 marks. The two sticks could then be compared to see if they tallied (agreed) to prevent me adding a mark or you erasing a mark on the sticks. Our English word score for the number of points made by one side in a game also means "to cut". The system was widely used in England from the twelfth century until 1826 by the Exchequer (see Oystein Ore's book below for pictures). When the collection of old wooden tallies was burned in 1836, it caused a fire in the old wooden Parliament buildings in London and burnt them down. The current Houses of Parliament (Westminster Palace) was built to replace them. Number Theory and Its History O Ore, Dover (1988), 388 pages, is an excellent and classic book to get you into all aspects of the mathematics of whole numbers (or Number Theory). ### What about bases bigger than 10? There is no logical reason why we cannot use any integer bigger than zero for a base. The only problem is what to use to represent 10 or more in a single column? We need a single symbol for each value from 0 to B-1 in base B. Usually the capital letters, A, B, C, etc, are used which take us up to base 36 (using the 10 digits and the 26 letters) - after that, it's up to you! 1010 = A, 1110 = B, 1210 = C, and so on. Here is one hundred again, this time expressed in some bases bigger than ten: 10010 = 9111 = 8412 = 7913 = 7214 = 6A15 Base 11 is called undenary, Base 12 is called duodenary or duodecimal, Base 16 is called hexadecimal. Chad Lake at the University of Utah has a nice page on what he calls the Snake Algorithm for converting from one base to another on paper. It is a web page for a course he gave at Indiana University. C A L C U L A T O R in base to base R E S U L T S: ## Sumthing about Fibonacci Numbers Here is an investigation into representing numbers as sums of Fibonacci numbers. First, let's just use any Fibonacci number once in any of our sums to see what we get. For example: • 1 is a sum all on its own! So there is just one sum of Fibonacci numbers with a sum of 1! • 1+1=2 but we are not allowing this as a Fibonacci sum since we have used 1 more than once! • 2 is, however, a sum formed of Fibonacci numbers but with just one number all on its own. • 1+2=3 and 3 is also a Fibonacci number so there are two sums for 3. • 1+3=4 is the only way to make a total of 4 using only Fibonacci numbers. • 2+3=5 and again 5 is a Fibonacci number so there are two sums for 5 • Find two sums for 6. • Find the single sum for 7. • How many ways can you make 8? Did you remember that 8 is also a Fibonacci number in the last example above? If so, you will have found three ways to make 8 as a Fibonacci sum. 1 is the smallest number with just a single number in its Fibonacci sum 3 is the first number with two Fibonacci sums; 8 is the smallest number with three such sums: 1+2+5=3+5=8 ? What is the smallest number with 4 Fibonacci sums? (Hint: It is less than 20) ? What is the smallest number with 5 Fibonacci sums? (Hint: It is less than 30) ... Can you continue this series of smallest numbers with n Fibonacci sums? You might have noticed that we are assuming: Every number is the sum of some set of Fibonacci Numbers Note: By set here we mean the mathematical term for a collection of unique items, no item being repeated. The set {1,3,5} is allowed as a collection of Fibonacci numbers with a sum of 9 because each number in the collection appears no more than once (all the items are unique). But {2,2,5} is excluded as a set totalling 9 even though all the numbers are Fibonacci numbers because it has a repeated number in it. This result is indeed true but we do not prove it here. It forms the idea behind representing numbers using Fibonacci numbers that we will investigate now in different forms on the rest of this page. Our decimal system relies on the fact that Every number is the sum of some collection of Powers Of Ten where we can use each power of ten up to ten times. A013583 is Sloane's series of the smallest numbers that can be written as a sum of n unique Fibonacci numbers: use it to check your answers. You might expect that the more numbers we want in the Fibonacci sum, the larger will be the first number we find with such a sum, and, in general, you would be right. But notice that this series is not always increasing! You will therefore be able to find numbers a and b where a is less than b, but a will have more Fibonacci sums than the larger number b in this list! C A L C U L A T O R of unique Fibonacci numbers whose sum is R E S U L T S: #### Things to do 1. What about the number of Fibonacci sums for the Fibonacci numbers themselves? We have seen the number of Fibonacci sums for 1 and for 2 is 1; for 3 and for 5 is 2; for 8 it is 3. What about 13? and 21? What is the pattern here? Can you explain it? 2. What is special about the number of Fibonacci sums for these numbers: 1,2,4,7,12,20,... ? What are the next 2 numbers with the same property? Suggest a simple formula for the numbers in this series. Again, can you prove or explain your result? 3. Find the series of numbers that have just 2 Fibonacci sums. Quite a bit is now known about this series: the number of representations of n as a sum of unique Fibonacci numbers (Sloane's A000119) which is called R(n). Here is a graph of it up for n from 1 to 143. At first it looks fairly random, but look closely and you'll find several examples of a fractal pattern where a section of the pattern is repeated but surrounded before and afterwards by another pattern. If you have a nice explanation or a proof of your answers to the questions above or you find some other patterns, please email me using the address at the bottom of this page. Representations of N as a sum of distinct elements from special sequences D. A. Klarner, Fibonacci Quarterly, vol 4 (1966), pages 289-306 and 322. The smallest positive integer having Fk representations as sums of distinct Fibonacci numbers Marjorie Bicknell-Johnson in Applications of Fibonacci Numbers Vol 8 (Proceedings of the Eighth International Research Conference of Fibonacci Numbers and their Applications, Rochester, USA, June 1998, editor F T Howard, Kluwer Academic) pages 47-52. The above article solves the problem of a formula for R(n), the number of sets of Fibonacci numbers whose total is n, by giving a recursive definition for it. The article is followed by two others, the second of which is... Composing with Sequences:... but is it art? by John Bliss, in Applications of Fibonacci Numbers Vol 8 (Proceedings of the Eighth International Research Conference of Fibonacci Numbers and their Applications, Rochester, USA, June 1998, editor F T Howard, Kluwer Academic), pages 61-73, is about turning R(n) directly into music. ## The Fibonacci base system Going back to the decimal number system, what if we labelled the columns with the Fibonacci numbers instead of powers of 10? We follow the usual conventions of larger column sizes being on the LEFT: ... 13 8 5 3 2 1 We will show that a number is represented in this system by putting Fib after it: e.g.: 8 5 3 2 1 ten = 1 0 0 1 0 Fib = 8 + 2 which distinguishes it from ten thousand and ten (10010) in decimal. ### Digits in the Fibonacci system This time it is not clear what digits we should use in the columns. For instance, there are many ways to represent the value ten in this system as well as in the example above: 10 = 2 5 = 2000Fib = 5 + 3 + 2 = 1110Fib = 3 3 + 1 = 301Fib = 10 1 = AFib Usually a number representation system is most useful if it has a unique representation of every integer. If we use only the digits 0 and 1 then we have our Fibonacci Sums of the previous section. But we saw there that, although each number does have such a sum (i.e. it is representable), some numbers have more than one sum and so their representation is not unique. We can find a single way to write every number as a sum of Fibonacci numbers if we also have the rule that no two consecutive Fibonacci numbers can be used in the same sum. In terms of the base system with Fibonacci numbers as the headings, this means that no two ones can occur next to each other. This last condition is because the sum of any two consecutive Fibonacci numbers is just the following Fibonacci number, so we can always replace ..011.. by ..100.. . To convince yourself that every number can be represented in this system, write down the Fibonacci representations of all the numbers from 1 to 40. It starts as follows: DecimalFibonacci 00 11 210 3100 4101 51000 61001 71010 810000 910001 1010010 1110100 1210101 13100000 14100001 15100010 16100100 17100101 18101000 19101001 20101010 #### Historical Note We can also call this the Fibonaccimal system (pronounced fib-on-arch-i-mal) as Marijke van Gans does because decimal refers to Base 10. This system is also called the Zeckendorf representation of a number after Edouard Zeckendorf who wrote about it (in French) in 1972. He proved that each representation of a number n as a sum of distinct Fibonacci numbers, but where no two consecutive Fibonacci numbers are used (and there is onyl one column headed "1"), is unique. Earlier, Lekkerkerker had written about this representation in 1952 (in Dutch) showing that there is only one way to write a number in this system. Représentation des nombres naturels par une somme de nombres de Fibonacci ou de nombres de Lucas, E Zeckendorf, Bulletin de las Societe Royale des Science de Liege vol 41 (1972) pages 179-182. A Generalized Fibonacci Numeration E Zeckendorf Fibonacci Quarterly vol 10 (1972) page 365-372 with Errata Fibonacci Quarterly vol 11 (1973) page 524. Edouard Zeckendorf C Kimberling Fibonacci Quarterly 36 (1998), pages 416-418. C A L C U L A T O R R E S U L T S: #### Things to do 1. The Zeckendorf system uses the set with the fewest Fibonacci numbers in it. What about choosing that set with the most Fibonacci numbers with a sum of n, each Fibonacci number being used at most once? This is called the maximal Fibonacci bit representation. "Bit" means that the only digits in the representations are 0 and 1. Zeckendorf's is therefore the minimal Fibonacci bit representation. Make a table of the maximal bit representation for n from 1 to 25. 2. 4 has a Zeckendorf representation of 101Fib = 3+1. It is also the only set of Fibonacci numbers with a sum of 4. What other numbers have just a single set of Fibonacci numbers that sum to them? 3. Investigate the number of ones in the Zeckendorf representations. What patterns can you find? Can you express your patterns as mathematical formulae? 4. What about the size of the largest sets (the number of ones in the maximal bit representations)? Is there a formulae for this function (from n to the size of the largest set)? ## An Application of the Fibonacci Number Representation There are approximately 8 kilometres in 5 miles. Since both of these are Fibonacci numbers then there are approximately Phi (1.618..) kilometres in 1 mile and phi (0.618..) miles in 1 kilometres. The real figure is more like 1.6093.. kilometres in 1 mile. This comes from the precise definition of 1 inch equals 2.54 centimetres exactly, and 100,000 centimetres make 1 kilometre. In the imperial system, 36 inches are 1 yard and 1760 yards are 1 mile. Replacing each Fibonacci number by the one before it has the effect of reducing it by approximately 0.618 (phi) times (the ratio of a Fibonacci number to the one before it is nearly phi). So to convert 13 kilometres to miles, replace 13 by the previous Fibonacci number, 8, and 13 kilometres is about 8 miles. Similarly, 5 kilometres is about 3 miles and 2 kilometres is about 1 mile. Now suppose we want to convert 20 kilometres to miles where 20 is not a Fibonacci number. We can express 20 as a sum of Fibonacci numbers and convert each number separately and then add them up. Thus 20 = 13 + 5 + 2. Using to stand for approximately equals and replacing 13 by 8, 5 by 3 and 2 by 1, we have 20 kms = 13 + 5 + 2 kilometres 8 + 3 + 1 miles = 12 miles. To convert miles to kilometres, we write the number of miles as a sum of Fibonacci numbers and then replace each by the next larger Fibonacci number: 20 miles = 13 + 5 + 2 miles 21 + 8 + 3 kilometres = 32 kilometres. There is no need to use the Fibonacci Representation of a number, which uses the fewest Fibonacci numbers, but you can use any combination of numbers that add to the number you are converting. For instance, 40 kilometres is 2 20 and we have just seen that 20 kms is 12 miles. So 40 kms is 2 12 = 24 miles approximately. [With thanks to Paul V S Townsend for reminding me of this application.] #### Things to do 1. A few years ago, the speed limit in USA was 55 miles per hour (mph). What would that be in kilometres per hour (km/h)? 2. The speed limit on UK motorways is 70 mph. What is this in kilometres per hour? 3. The speed limit in built up areas is 30 mph in the UK. What is 30 mph in km/h? What do you think the "30" signs would be replaced by if road signs went metric, i.e. round your conversion to the nearest 5 km/h? 4. The current train speed record of 552 km/h was set on April 14 1999 in Japan. What is the equivalent speed in mph using the Fibonacci method? What is the equivalent speed in mph using the conversion factor of 1.6093 km per mile? #### Reference Concrete Mathematics (2nd edition) by Graham, Knuth and Patashnik, Addison-Wesley, section 6.6. ## An easy way to Multiply ### The Egyptian system - using Doubling... The Egyptians had an easy way to multiply two integers which involved only doubling numbers and adding - no multiplication tables to learn and no need for a calculator (except to do the addition). For example, 19 x 65. We write the two numbers at the head of two columns, choosing one column to keep doubling and the other to keep halving (ignoring remainders), until the halving column reaches 1: ``` halve double odd? 19 65 + 9 130 + 4 260 2 520 1 1040 + ``` Any row whose halving column entry was odd is marked (here with +) and we add the marked values from the doubling column. In our example 65+130+1040=1235 which is the product of 19 and 65. The method works because if we represent 19 in the binary system we have 16+2+1=10011(2) and so 19x65=(16+2+1)x65 which is 1665 + 265 + 1*65. ie, the 1st, 2nd and 5th values in the doubling column above. #### Things to do 1. Check that if you halve the 65 column and double the 19 column the method still works. 2. Try the Egyptian method on 32x65. 3. Try it on 31x65. ### The Fibonacci system A similar system uses the Fibonacci representation to replace each doubling of the Egyptian method with an addition. Let's take the same example: 19x65. This time we take just one number - say 65 - as the head of the right hand column, the left column starting with 1. The second row has 2 on the left and we double 65 to get 130 on the right. Now each successive row is the sum of the previous TWO entries above it, taking each column separately. So since we started with 1 and 2 on the left we will get 3,5,8,... that is, the Fibonacci numbers on the left hand side. Stop when we can find a Fibonacci number which is bigger than the other number in the product - here 19: ``` 1 65 + 2 130 3 195 5 325 + 8 520 13 845 + 21 ``` We mark the rows this time by finding those entries in the left column that add up to 19. There many be several ways to do this selection but any will do. Here we have chosen 13+5+1. If we add up the right hand entries on these rows we have: 65+325+845=1235 which is again 19x65. #### Things to do 1. Try it the other way round, starting with 19 and stop when the Fibonacci number exceeds 65. 2. Try the same multiplications as above: 32x65 and 31x65. 3. Look up the article where this idea was first presented: Fibonacci, Lucas and the Egyptians by S La Barbera in The Fibonacci Quarterly, Vol 9, 1971, pages 177-187. ## Patterns in the Fibonacci representations ### Patterns in the columns - the Rabbit sequence In base 10, if we list all the integers from 1, then there are patterns in the columns: #### Decimal patterns Column 1 (units) cycles through all the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 repeatedly; Column 2 (tens) cycles through all the digits but each digit occurs ten times; Column 3 (hundreds) is the same but each digits occurs 100 times; and so on. #### Fibonacci Representations patterns Is there a pattern in the columns of numbers in the Fibonacci base system? Yes there is! It is based on the Rabbit sequence which now includes the initial 0. The pattern in column one (the right-hand column) is derived from the rabbit sequence where every "1" in the rabbit sequence has been replaced by "10":- The rabbit sequence: `010110101101101011010... ` becomes: ``` 0 1 0 1 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 1 0 ... 0 10 0 10 10 0 10 0 10 10 0 10 10 0 10 0 10 10 0 10 0 ... ``` which is column 1 above, read downwards. [N.B. This is exactly the same as if we flipped the bits (1 changes to 0 and 0 to 1) in the Rabbit sequence (without its initial zero)!! However, there is a pattern in the other columns which is better seen with the description above.] What about column 2 of the Fibonacci representations? This is derived similarly: every "1" in the rabbit sequence is replaced by "100" and every "0" is replaced by "00". ``` 0 1 0 1 1 0 1 0 1 1 0 1 1 0 ... Rabbit Sequence 00 100 00 100 100 00 100 00 100 100 00 100 100 00 ... Column 2``` where column 2 in the Table of Fibonacci representations is read downwards. For column 3, replace "0" by "000" and "1" by "11000" For column 4, replace "0" by "00000" and "1" by "11100000" For column 5, replace "0" by "00000000" and "1" by "1111100000000" The same pattern follows for all the columns: Column i the just the rabbit sequence with "0" replaced by F(i) 0s and "1" replaced by F(i-1) 1s followed by F(i) 0s. DecimalFibonacci 00 11 210 3100 4101 51000 61001 71010 810000 910001 1010010 1110100 1210101 13100000 14100001 15100010 16100100 17100101 18101000 19101001 20101010 ### The number of 1s in a Fibonacci Representation What is the least number of Fibonacci numbers that sum to a given n? This is the number of 1s in the Fibonacci representation, since the description given above guarantees the least number of Fibonacci's and is also called the minimal Fibonacci representation. Here we repeat the Fibonacci Representation table from above but now include the number of 1's in each representation: nnFib1's 1 11 2 101 3 1001 4 1012 5 10001 6 10012 7 10102 8 100001 9 100012 10 100102 11 101002 12 101013 From the table, we can see that the number of numbers with a Fibonacci representation of a given length is a Fibonacci number: There is 1 of length 1, there is 1 of length 2, there are 2 of length 3, there are 3 of length 4, there are 5 of length 5,... Here is a more compact list of the number of 1s in the (minimal) Fibonacci representation of the first few whole numbers : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 ... 1 1 1 2 1 2 2 1 2 2 2 3 1 2 2 2 3 2 3 3 1 2 2 2 3 2 3 3 2 3 3 3 4 ... If we split this list into sublists corresponding to the different lengths of Fibonacci representations we have the following: 1=1Fib 1, 2=10Fib 1, 3=100Fib,4=101Fib 1,2 5=1000Fib, 6=1001Fib, 7=1010Fib 1,2,2 8, 9, 10, 11 and 12 1,2,2,2,3 13 to 20 1,2,2,2,3,2,3,3 21 to 33 1,2,2,2,3,2,3,3,2,3,3,3,4 34 to 54 1,2,2,2,3,2,3,3,2,3,3,3,4,2,3,3,3,4,3,4,4 ... ... It is quite easy to see where this pattern comes from: Each time we put a 1 at the start of our Fibonacci representations and then copy the earlier patterns. For example, 8, 9, 10, 11 and 12 are 8+0, 8+1, 8+2, 8+3 and 8+4. Can you see any patterns in these sequences? It seems that each sequence starts off the following sequence. Can you discover how the remainder of each is formed, that is, the part that follows (the copy of) the previous sequence? It is not quite the sequence before, but, one added to all the items of the sequence before: Start with 1 and 1. The next sequence is the preceding one followed by adding one to the sequence before the preceding one. Since each sequence in the list above starts off the following one, it defines a unique infinite sequence. ### Palindromic Fibonacci Representations A palindrome is a word or list that is the same when reversed, e.g. radar or 1001. Here is the start of the list of numbers which are palindromes in the Fibonacci representation: N Fib rep of N 11 4101 61001 910001 1210101 14100001 221000001 271001001 331010101 3510000001 5110100101 56100000001 64100010001 74100101001 80101000101 88101010101 901000000001 Is there a formula for the series 1,4,6,9,12,14,...? ## A more general Fibonacci System Suppose that, instead of finding just a set of Fibonacci numbers that sum to N, that is each Fibonacci number is included at most once in the representation, we allow multiples of any Fibonacci number. We will then have a multi-set or bag of Fibonacci numbers whose sum is N. NFib numbers whose sum is N allowing repeats Count 111 22, 1+12 33, 2+1, 1+1+13 43+1, 2+2, 2+1+1, 1+1+1+14 55, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+16 65+1, 3+3, 3+2+1, 3+1+1+1, 2+2+2, 2+2+1+1, 2+1+1+1+1, 1+1+1+1+1+18 Notice the following things about this table: 1. Since we can have just a single use of a Fibonacci number (no repeats) in this new system, all the Zeckendorf representations that we looked at above are also included in the table. Every number has a Zeckendorf representation which is given first. 2. There are more sets of Fibonacci numbers that sum to N (i.e. where no Fibonacci number is used more than once) than the Zeckendorf representation. In the Zeckendorf representation no two consecutive Fibonacci numbers are used. We can have any Fibonacci numbers in our bags and so this time we do allow neighbouring Fibonacci numbers in any collection (bag). 3. We are only interested in the collection of Fibonacci numbers that sum to n, where we allow repeated use of any Fibonacci number. For example 5 is 2+2+1 and this is the same collection or bag as 1+2+2 and 2+1+2, since each includes two 2's and one 1. If we had been interested in the order of the numbers, then we would be studying sequences or lists. Here is a count of the number of Fibonacci bags with a sum from 1 to 40: N No. ofreps N No. ofreps N No. ofreps N No. ofreps 1 2 3 4 5 6 7 8 9 10 1 2 3 4 6 8 10 14 17 22 11 12 13 14 15 16 17 18 19 20 27 33 41 49 59 71 83 99 115 134 21 22 23 24 25 26 27 28 29 30 157 180 208 239 272 312 353 400 453 509 31 32 33 34 35 36 37 38 39 40 573 642 717 803 892 993 1102 1219 1350 1489 C A L C U L A T O R of Fibonacci numbers (repetitions allowed) whose sum is R E S U L T S: ### Fibagonacci System Using bags of Fibonacci's that sum to n, and the existence of at least one bag for every number n, means we now have another system of representing numbers: the Fibagonacci Representation! The columns of this "Fibonacci base" system are again the Fibonacci numbers from 1 upwards (1 only occurring once). The entries in the columns are the number of copies of that Fibonacci number in a bag with a sum of n. This 13 is the sum of the Fibonacci bag 5,2,2,2,1,1 and so is written as ... 5 3 2 1 1 0 3 2 Fib = 13 ### How many Fibagonacci representations of N are there for a number N? Any number n is just the sum of n 1's and F(2)=1 so that's one bag of Fibonacci's that sum to n, no matter what n is. All the sets of Fibonacci numbers that sum to n are also included in the bags, since a set is a special kind of bag (with no items repeated in it). Neil Sloane's Online Encyclopedia of Integer Sequences lists this as sequence A003107. Advanced note: The generating function for this series is i = 2 1 = 1 + x + 2x2 + 3x3 + 4x4 + 6x5 + ... 1 – xFib(i) ## Some more representations using Fibonacci Numbers Here are some more research topics and ivestigations on other types of collections of Fibonacci numbers. #### Things to do 1. We can also look at other kinds of Fibonacci representations. For both sets and bags, it is the ityems in the collection (and hte number of them) that matters, the order in which they are lsited begin immaterial since 1,2,3 is the same set (and bag) as 3,1,2. If the order of elements in the collection now does matter we would be studying sequences of Fibonacci numbers whose sum is N also called compositions. List the number of compositions of n that contain only Fibonacci numbers. 2. Alternatively, we could look at sets but without the Zeckendorf restriction. Such sets (of unique items) could now contain consecutive Fibonacci numbers. On a later page we will investigate what happens if instead of using the Fibonacci numbers as column headers we use powers of Phi (1.61803..), ie base Phi. ## References A Primer for the Fibonacci Numbers: Part XII, V E Hoggatt Jr, N Cox, M Bicknell in Fibonacci Quarterly, vol 11 (1973), pages 317-331 is a useful introduction to results in this area, but for post-18 mathematics students. Représentation des nombres naturels par une sommme de nombres de Fibonacci ou de nombres de Lucas by E Zeckendorf in Bulletin de la Société Royale des Sciences de Liège vol 41, 1972, pages 179 - 182 is the article from which the name Zeckendorf Representation is derived. He says he first found the Theorem now named after him in 1939. This paper shows that we can use the Lucas numbers also as a representation system with some minor conditions. Voorstelling van natuurlijke getallen door een som van getallen van Fibonacci C. G. Lekkerkerker, Simon Stevin vol 29, 1952, pages 190 - 195 appears to be the first article in print that is about the Fibonacci Number System now usually called the Zeckendorf Representation. © 1996-2006 Dr Ron Knott updated 26 May 2004
## Saturday, April 24, 2010 ### Calculating Bacteria Population? Let's assume that bacteria can reproduce once every 20 minutes during binary fission. Assume further that his rate of reproduction can be maintained. Now, calculate how many bacteria there would be after 1-hour period if you started with a single bacterium Calculating Bacteria Population? Bacterial growth In this section we will return to the questions posed in the first section on exponential and logarithmic functions. Recall that we are studying a population of bacteria undergoing binary fission. In particular, the population doubles every three hours. We would like to know the following: How many bacteria are present after 51 hours if a culture is inoculated with 1 bacterium? With how many bacteria should a culture be inoculated if there are to be 81,920 bacteria present on hour 42? How long would it take for an initial population of 6 to reach a size of 12,288 bacteria? http://www.biology.arizona.edu/biomath/t... Exponential Population Growth How many bacteria are present after 51 hours if a culture is inoculated with 1 bacterium? Use the model, N(t) = Noe kt, and assume the population doubles every 3 hours. (N(t) is the population size at time t and k is a constant.) Tutorial Now that we have convinced ourselves that an exponential model is appropriate, but not perfect, we will use the general model N(t) = N0e k t (1) where N (t) represents the population size at time t (Note: you may also use the model, N (t) = N0a t ). The question asks for the population size when t = 51. In order to use this model, we will need to figure out the values of the constants, N0 and k. At time t = 0, there is a single bacterium, therefore N0 = 1. Substituting N0 = 1 into (1) gives, N(t) = 1 ⋅ e kt = e kt, (2) We now use the fact that when t = 3, N(3) = 2 (the population has doubled). Substituting the point (t, N(t)) = (3, 2) into (2), N(3) = 2 = e 3k (3) We can now solve for k in (3) by "undoing" the exponential using the natural logarithm, Using this value of k our model in (2) becomes, (4) Now that we have our model, we need to find the population size after 51 hours. Substituting t = 51 into (4) yields, Thus, we find that after 51 hours there are 131,072 bacteria. so if you start with 1 bacterium and replicates every 20 mins that means that in 1 hours there will be 3 replications therefore 2(3)=6 there will be 6 bacteria
# Tshapes Essay Research Paper Looking at the T-shapes Essay, Research Paper Looking at the 9-9 grid below and the T-shape drawn on it, The total number of the numbers on the inside of the T-shape is called the T-total T-shapes Essay, Research Paper Looking at the 9-9 grid below and the T-shape drawn on it, The total number of the numbers on the inside of the T-shape is called the T-total 123456789 101112131415161718 192021222324252627 282930313233343536 373839404142434445 464748495051525354 555657585960616263 646566676869707172 737475767778798081 828384858687888990 The t-total for this T-shape is: 1+2+3+11+20=37 So 37 = T-total The number at the bottom is the T-number, So the T-number for this shape is 20 Aims: 1)Investigate the relationship between the T-total and the T-number 2)Use the grids of different sizes. Translate the T-shape to different positions. Investigate relationships between the T-total and the T-number and the grid size. 3)Use grids of different sizes again, try other transformations and combinations of transformations. Investigate relationships between the T-total and the T-number and the grid size and the transformations. Aim 1- the solution 123456789 101112131415161718 192021222324252627 282930313233343536 373839404142434445 464748495051525354 555657585960616263 646566676869707172 737475767778798081 828384858687888990 T69= 50+51+52+60+69 =282 T22=3+4+5+13+22 =47 In the diagram below it shows the difference between the T-number and the other numbers. First is the T-shape in question: 123 11 20 This is the T-shape and here is the Difference T-shape: N-19N-18N-17 N-9 N This shows the difference N= T-number In the T on the previous page I have noticed that the first difference from N is 9 which is also the Width of the square. I?ll put that idea into another T. Note W= width number(9) N-(2W-1)N-2WN-(2W+1) N-W N This is the same thing as before but shown algebraically. The formula for the Value of the T-total now is shown as: 5N-7W=T-total Aim 2- different sizes and relationship I know this works for the grid 9 by 9 but I?m not sure if it?ll work for any other grids. Here is a test for a 10 by 10 grid 12345678910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899100 T22=1+2+3+12+22 =42 I notice this is 5 more than 9 by 9 T69=48+49+50+59+69 =275Obviously no pattern there. Method test? (695)-70=275 YES it worked My method seems to have worked out as it is logical and fairly straight forward to explain. 12345678910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899100 As there are 10 in each row it?s obvious that the row above will be 10 less than the row below. So 68 is 10 less than the T-number 78. If you calculate the whole T you realise that row 2 is 10 less than row 1 and row 3 is 20 less than row 1,but there are three relevant numbers in row 3 which are 19 less and 21 less than the T-number. These cancel out to form 20 each, So finally we get (110)+(610)=(710) =70 Or =7W 12 by 12 123456789101112 131415161718192021222324 252627282930313233343536 373839404142434445464748 495051525354555657585960 616263646566676869707172 737475767778798081828384 858687888990919293949596 979899100101102103104105106107108 109110111112113114115116117118119120 121122123124125126127128129130131132 133134135136137138139140141142143144 T62=37+38+39+50+62 also =(625)-(712) = 226 =226 T141=116+117+118+129+141 also =(1415)-(712) =621 =621 Aim3- Transformations stretches and there effects on the formula I?ll do this with a 12 by 12 first, as this will give me enough accuracy to start with. 123456789101112 131415161718192021222324 252627282930313233343536 373839404142434445464748 495051525354555657585960 616263646566676869707172 737475767778798081828384 858687888990919293949596 979899100101102103104105106107108 109110111112113114115116117118119120 121122123124125126127128129130131132 133134135136137138139140141142143144 Stretch A will be called ST64 as it starts at 64, it?s a stretch of 2 in both directions. St64=26+27+28+29+30+40+52+64 =296 I think I can work out the formula using my previous method so: 12+24+36+(436)=216 21612=18 This means the formula is: 8N-18W=T-total 8N= number of integers in the T-shape 18W=difference number calculated Conclusion: The size of the T-shape calculates the number before N in the formula and the grid size calculates the value of W. the number before W is calculated by looking at the rows and finding how many rows away from the T-number they are. If the T is regular then the W number is negative but if the T is flipped upside down the W number is positive. ## Похожие материалы ◯ Reflection Essay Essay Research Paper My Reflection ◯ Art 2 Essay Research Paper Essay IndexCustom ◯ Gambling Essay Research Paper Research Paper 2Gambling ◯ Gun Control Essay Research Paper Research Paper ◯ Art Research Paper Essay Research Paper Art ◯ Paper A Essay Research Paper Paper AHide ◯ MIT Essay Essay Research Paper I may ◯ Dante Essay Research Paper Essay IndexCustom WritingCollege ◯ Ethics Essay Essay Research Paper Essay Paper ◯ Iq Essay Research Paper The following essay
# How do you solve -6x+14<-28 or 9x+15<-12? Jun 20, 2017 $\left(- \infty , - 3\right) \cup \left(7 , \infty\right)$ #### Explanation: Solve each inequality separately, then combine them with the "or" operator, and then simplify if possible. $- 6 x + 14 < - 28$ $- 6 x < - 42$ $6 x > 42$ $x > 7$ Now, we solve the second inequality: $9 x + 15 < - 12$ $9 x < - 27$ $x < - 3$ Combining the two, we get: $x < - 3 \mathmr{and} x > 7$ This cannot be simplified, since the two solution regions do not overlap. The more formal way to write this (with interval notation) is: $\left(- \infty , - 3\right) \cup \left(7 , \infty\right)$ On a number line, the solution set looks like this:
# What Is 7/8 In Percent? ## Understanding the Basics of Percentages Before we can answer the question, “What is 7/8 in percent?”, it is important to understand the basics of percentages. A percentage is a value that expresses a fraction as a portion of a whole. It is used to represent a portion of a total and is often expressed as a fraction or a ratio. For example, if you have 8 apples and you eat 2 of them, you\’ve eaten 25% of your apples. This is because 2/8 is equal to 25%. In this case, 2 is the numerator, 8 is the denominator, and 25% is the percentage. ## Converting Fractions to Percentages Now that we understand the basics of percentages, we can answer the question, “What is 7/8 in percent?” To convert fractions to percentages, we must first convert the fraction to a decimal by dividing the numerator by the denominator. In this case, 7/8 is equal to 0.875. To convert this decimal to a percentage, we must multiply the decimal by 100. 0.875 x 100 = 87.5%. Therefore, 7/8 is equal to 87.5%. ## Understanding the Difference between Percentages and Decimals It is important to understand the difference between percentages and decimals. A decimal is a value expressed as a fraction of a whole. A percentage is also a value expressed as a fraction of a whole, but it is expressed in hundredths. Therefore, 7/8 is equal to 0.875 as a decimal, but it is equal to 87.5% as a percentage. ## Using Percentages in Everyday Life Percentages are used in everyday life to express a portion of a total. For example, if you have a pizza with 8 slices and you eat 2 of them, you\’ve eaten 25% of the pizza. This is because 2/8 is equal to 25%. Percentages are also used to calculate discounts and taxes. For example, if an item is on sale for 20% off, the price is reduced by 20%. Similarly, if an item is subject to a 10% tax, the price is increased by 10%. ## Using Percentages in the Workplace Percentages are also used in the workplace to measure performance. For example, if an employee has a goal of selling 100 items and they sell 85 items, they have achieved an 85% success rate. Similarly, if a company has a goal of increasing profits by 10% and they increase profits by 8%, they have achieved an 80% success rate. Percentages are also used to measure the return on investment (ROI) of a project or investment. ## Understanding Percentages in the Context of Statistics Percentages are also used in the context of statistics. For example, if a survey is conducted and the results show that 50% of the respondents are in favor of a certain policy, it means that 50% of the respondents are in favor of the policy while the other 50% are against it. Percentages can also be used to measure the probability of an event occurring. For example, if a coin is flipped and the probability of it landing on heads is 50%, it means that there is a 50% chance that the coin will land on heads. ## Conclusion To answer the question, “What is 7/8 in percent?”, it is important to understand the basics of percentages and how to convert fractions to percentages. 7/8 is equal to 0.875 as a decimal and 87.5% as a percentage. Percentages are used in everyday life to express a portion of a total, measure performance in the workplace, and measure the probability of an event occurring in the context of statistics.
# 3.4 Rational Functions I. A rational function is a function of the form Where p and q are polynomial functions and q is not the zero polynomial. The domain. ## Presentation on theme: "3.4 Rational Functions I. A rational function is a function of the form Where p and q are polynomial functions and q is not the zero polynomial. The domain."— Presentation transcript: 3.4 Rational Functions I A rational function is a function of the form Where p and q are polynomial functions and q is not the zero polynomial. The domain consists of all real numbers except those for which the denominator is 0. Find the domain of the following rational functions: All real numbers except -6 and-2. All real numbers except -4 and 4. All real numbers. y = L y = R(x) y x y = L y = R(x) y x Horizontal Asymptotes x = c y x y x Vertical Asymptotes If an asymptote is neither horizontal nor vertical it is called oblique. y x Theorem Locating Vertical Asymptotes A rational function In lowest terms, will have a vertical asymptote x = r, if x - r is a factor of the denominator q. Vertical asymptotes: x = -1 and x = 1 No vertical asymptotes Vertical asymptote: x = -4 Find the vertical asymptotes, if any, of the graph of each rational function. 1. If n < m, then y = 0 is a horizontal asymptote of the graph of R. 2. If n = m, then y = a n / b m is a horizontal asymptote of the graph of R. 3. If n = m + 1, then y = ax + b is an oblique asymptote of the graph of R. Found using long division. 4. If n > m + 1, the graph of R has neither a horizontal nor oblique asymptote. End behavior found using long division. Consider the rational function Horizontal asymptote: y = 0 Horizontal asymptote: y = 2/3 Find the horizontal and oblique asymptotes if any, of the graph of Oblique asymptote: y = x + 6 Similar presentations
# Factors of 28 ## What is the factor? Factors are integers you multiply together to get another integer. Also remember that the factors always include 1 and itself. NOTE: When finding the factors of a number, ask yourself “what numbers can be multiplied together to give me this number?. ## How to find the factors of 28? THINK: What pairs of numbers can be multiplied together to get 28? Step 1: 1 x 28 = 28, so put these numbers in the factor list. 1 … 28 Step 2: Take 2 and divide with 28. It gives 2 x 14 = 28. 1 2 14 28 Step 3:Take 3 and divide with 28. The remainder will be 1. But factors always give 0 remainder. This means 3 is NOT a factor of 28. Step 4: Take 4 and divide with 28. It gives 4 x 7 = 28. 1 2 4 7 14 28 Step 5: Take 5 and divide with 28. The remainder will be 3. But factors always give 0 remainder. This means 5 is NOT a factor of 28. Step 6: Take 6 and divide with 28. The remainder will be 4. This means 6 is NOT a factor of 28. Step 7: Take 7 and divide with 28. It gives 7 x 4 = 28 but factors 7 and 4 are already in the list. So you dont need to repeat the number. 1 2 4 7 14 28 Step 8: Take 8 and divide with 28. The remainder will be 4. This means 8 is NOT a factor of 28. Step 9: Take 9 and divide with 28. The remainder will be 1. This means 9 is NOT a factor of 28. Step 10: Take 10 and divide with 28. The remainder will be 8. This means 10 is NOT a factor of 28. So the factors of 28 are 1, 2, 4, 7, 14, and 28. TIPS: Try to keep tables (at least till 20) on your finger tips. Revise tables every day ## Is 28 a prime or composite number? 28 is a composite number not prime number because it has more than 2 factors. What is composite number? – The composite numbers are the positive integers which have more than two factors. What is prime number? – An integer number that is divisible by two distinct integers, 1 and itself. ## What is the prime factorization of 28 The prime factorization of 28 are 2 x 2 x 7 (i.e. 22 × 7) ## Factor pairs of 28 Factors are often given as pairs of numbers, which multiply together to give the original number. These are called factor pairs What is factor pairs: Factor pairs are combinations of two factors that multiply together to give the original number. So for 28, there are 6 factor pairs 1 x 28 = 28 2 x 14 = 28 4 x 7 = 28 7 x 4 = 28 14 x 2 = 28 28 x 1 = 28 ## Is 28 a square number? No, 28 is not a square number. The square root of 28 is 5.292. The square of 28 is 784. Hope you have learned to solve the factors of 28. Scroll to Top
# Lesson: Properties of Real Numbers ## Comment on Properties of Real Numbers ### This seems too basic for the This seems too basic for the GRE. ### While you might not see any While you might not see any questions on the test that directly test the concepts in this video, you still need to know those concepts, since they could play a role in a more complex problem. That said, the beauty of a self-paced course is that, if you already know the concepts taught in a certain lesson, you can always skip that lesson. Happy studying! Fair enough! ### I have a doubt here. The I have a doubt here. The reinforcement activity questions here- Are they related to real numbers? If yes then should we solve each reinforcement activity after completing that certain lesson or should we solve the reinforcement activity after finishing that whole module? ### The reinforcement activities The reinforcement activities are (somewhat) related to the video lesson. So, I recommend that you solve them once you watch the video. ### If 2,4,6 and 9 are the digits If 2,4,6 and 9 are the digits of two 2-digit integers, what is the least possible positive difference between the integers? ### Good question! Good question! I'm assuming that each digit may be used only ONCE (otherwise, we could just make both integers the same, like 29 and 29). The solution requires a bit of logic and a bit of fiddling. Since we're trying to MINIMIZE the difference, we want to create two 2-digit numbers that are as CLOSE AS POSSIBLE. So, we should start by making the two TENS digits as close as possible. We have two options: Option #1: Use 2 and 4 as tens digits (since they are only 2 apart) Option #2: Use 4 and 6 as tens digits (since they are only 2 apart as well) We want to place the remaining integers (6 and 9) in the units positions so that the difference between the 2 integers is minimized. We get 29 and 46 The difference = 46 - 29 = 17 Now place the remaining integers (2 and 9) in the units positions so that the difference is minimized. We get 49 and 62 The difference = 62 - 49 = 13 Looks like the least positive difference is 13 Cheers, Brent Thank you. Eke ### In a factory, machine A In a factory, machine A operates on a cycle of 20 hours of work followed by 4 hours of rest,and machine B operates on a cycle of 40 hours of work followed by 8 hours of rest. Last week, the two machines began their respective cycles at 12 noon on Monday and continued until 12 noon on the following Saturday. On which days during that time period was there a time when both machines were at rest? in this question the machine is stoping on both wednesday and friday . then why we have concluded it as friday? The official answer is C and E, which means both machines were at rest (at the same time) on Wednesday and on Friday. The question asks us to identify ALL days on which both machines were at rest. Does that help? Cheers, Brent ### Hello Brent, Hello Brent, I am trying to understand how to follow the book pages that you mentioned from the official guide, when I look for page 116 #3, which is currently a verbal part on my book. I am originally from Brazil, not quite sure where I am getting lost. Thank you in advance. ### I'm happy to help. I'm happy to help. In the Study Guide, I reference two different Official Guides for GRE Review: - The 3rd edition: https://www.amazon.com/Official-Guide-General-Test-Third/dp/1259862410 - The 2nd edition: https://www.amazon.com/Official-Guide-Revised-General-Test/dp/007179123X Are you referring to the correct edition? Also note that the GRE test-makers also have other resources that comprise of Quant-only and Verbal-only practice questions. Those two resources are not referenced in the Study Guide. That is, I don't assign specific questions from those Quant-only and Verbal-only books. Note: The practice questions are listed in the form page#(question#). For example, 118(2) refers to question #2 on page 118. Likewise, 262(13) refers to question #13 on page 262. Does that help? Cheers, Brent ### https://greprepclub.com/forum https://greprepclub.com/forum/term-after-the-first-term-is-equal-to-the-preceding-term-1828.html
# Lesson 18 • Let’s work to solve quadratic equations. ### 18.1: Math Talk: Operations with Roots Evaluate mentally: $$\sqrt{100}-15$$ $$\sqrt{125-10^2}$$ $$20-2\sqrt{49}$$ $$\sqrt{4^2+3^2}$$ ### 18.2: Checking Brother’s Work Priya's older brother is working on some higher-level math work and claims that $$x = 3$$ is a solution to the equation $$x^3 - 5x^2 -2x = \text{-}24$$. 1. Explain how she could check that his solution is correct using each of these tools. 1. A basic calculator 2. A graphing tool 2. When looking at his work, Priya sees that he has the equation $$(x-3)(x^2 -2x - 8) = 0$$. Knowing the zero product property holds, Priya recognizes that this equation means $$x-3 = 0$$ or $$x^2 -2x - 8 = 0$$ for this question. Find 2 other solutions to the original equation. Explain or show your reasoning. ### 18.3: Steps to Using the Quadratic Formula The quadratic formula solves equations of the form $$ax^2 + bx + c = 0$$ using the equation $$x=\frac{\text{-}b \pm \sqrt{b^2 - 4ac}}{2a}$$. Andre wants to use the quadratic formula to solve $$x^2 - 7x = \text{-}12$$. 1. What should Andre do first? 2. What values of $$a, b,$$ and $$c$$ should he use? 3. After substituting the values into the quadratic formula, what is the order he should use to calculate the solutions? 4. Use the quadratic formula to solve the equation.
# Aptitude Time and Work Test Paper 2 6) A, B, C can do a job in 10, 20 and 40 days respectively. In how many days A can complete the job if he is assisted by B and C on every third day? 1. 8 days 2. 7 days 3. 9 days 4. 6 days Explanation: A's one day work = B's one day work = C's one day work = (A+B+C)'s one day work = Work done in three days will be the sum of A's two-day work and (A+B+C)'s one day work. A's two-day work = Therefore, the work is done in three days = 3/8 part of the job is done in 3 days. The entire job will be done by A in = 3 x 7) If 5 men can colour 50-meter long cloth in 5 days, in many days 4 men can color a 40-meter long cloth? 1. 5 days 2. 6 days 3. 4 days 4. 3 days Explanation: 5 men colour 50-meter long cloth in 5 days. 1 man will colour= 10-meter long cloth in five days. So, in one day one man can colour= 2-meter long cloth. Now, as per question 4 men are supposed to colour the 40-meter long cloth. One man has to colour the = 10 meter cloth For 2-meter long cloth one man takes 1 day So, to paint 10-meter long cloth he will take = As, 4 men are working together so 40 m long cloth will be coloured in 5 days Quicker method: Apply formula: M1D1W2 = M2D2W1 5540 = 4D250 D2= 8) If 4 men can finish 4 times of a work in 4 days, in how many days 6 men can finish the 6 times of same work ? 1. 3 days 2. 4 days 3. 5 days 4. 6 days Explanation: 4 men can finish 4 times of work in four days. Therefore, one man can finish the one time of work in four days. So, 6 men will finish the six times of work in the same time (4 days) Quicker method: Apply formula: M1D1W2 = M2D2W1 Let the work be X Work done by 4 men, W1 = 4X Work done by 6 men, W2 = 6X 446X = 6D24X 96X = D2 *24X D2= 4 days 9) A can do a piece of work in 10 days. B is 50% more efficient than A. In how many days B alone can do the same job? 1. 6.2 days 2. 6.6 days 3. 7 days 4. 7.2 days Explanation: B is 50%more efficient than A so he will take less time to do a piece of work. Therefore, the ratio of the time taken by A and B = Let B takes X days to do the job. 3X =20 X = 6.6 days 10) A can do a job in 30 days. B alone can do the same job in 20 days. If A starts the work and joined by B after 10 days, in how many days the job will be done? 1. 15 days 2. 16 days 3. 17 days 4. 18 days Explanation: A's one day work = A's ten-day work = So the remaining work would be = B's one day work = A and B's one day work = of the job will be done by them in one day. So, the remaining job= 8 days Therefore, the total number of days required to do the job would be = 10 + 8 = 18 days Aptitude Time and Work Test Paper 1 Aptitude Time and Work Test Paper 3 Aptitude Time and Work Test Paper 4 Aptitude Time and Work Test Paper 5 Aptitude Time and Work Test Paper 6 Aptitude Time and Work Test Paper 7 Aptitude Time and Work Test Paper 8 Aptitude Time and Work Test Paper 9 Time and Work Concepts
## Algebra 1 $\frac{55}{8}$ The order of operations states that first we perform operations inside grouping symbols, such as parentheses, brackets, and fraction bars. Then, we simplify powers. Then, we multiply and divide from left to right. Finally, we add and subtract from left to right. We use these rules to simplify the expression. Start with the given expression: $\frac{22+1^3+(3^4-7^2)}{2^3}$ Since we have multiple layers of grouping (fraction bar and parentheses) we start with the inner most grouping and simplify on the inside of the parentheses: We simplify the exponents first: $\frac{22+1^3+(81-49)}{2^3}$ Then, we subtract inside the parentheses: $\frac{22+1^3+32}{2^3}$ Next, we simplify the numerator, which is above the fraction bar: We simplify the exponents: $\frac{22+1+32}{2^3}$ We add from left to right: $\frac{23+32}{2^3}=\frac{55}{2^3}$ We simplify the denominator by simplify the power: $\frac{55}{8}$
How do you multiply 5/8 div 2/5? May 4, 2018 Actually you don't this is division problem treat it as a division problem Explanation: This problem can be set up as a fraction divided by a fraction. $\frac{\frac{5}{8}}{\frac{2}{5}}$ To simply and solve multiply both the top and the bottom fraction by the inverse of the bottom fractions. This will cause the bottom fraction to become 1. $\frac{2}{5} \times \frac{5}{2} = 1$ $\left(\frac{5}{8}\right) \times \frac{5}{2} / \left(\frac{2}{5}\right) \times \frac{5}{2}$ $\frac{\frac{5}{2}}{\frac{5}{2}} = 1$ so multiplying both the top and bottom by 5/2 is the same as multiplying by 1 # (5/8) xx 5/2 /(2/5) xx 5/2 = ( 5/8 ) xx ( 5/2) $\frac{5}{8} \times \frac{5}{2} = \frac{25}{16}$ There are no common factors but the improper fraction can be changed into a mixed number $\frac{25}{16} = 1 \frac{9}{16}$
How to Multiply by 6 Instructor: Mary Grace Miller Mary Grace has taught first grade for 8 years and has a Bachelor's degree in Elementary Education and is licensed in ESL. It's your birthday! You have invited 6 friends to your party, and you're working on their party bags. You want to give them each 1 whistle, 2 bracelets, 10 pieces of gum, and lots of other favors! The fastest and easiest way to decide how much to buy is to multiply by 6, as you'll learn in this lesson. What is Multiplication? Multiplication is a quick way to solve a repeated addition problem. For example, if you need to solve the problem 3 + 3 + 3 + 3 + 3 + 3 = 18, you could just multiply the 6 groups of 3 to get 18: 6 x 3 = 18. In multiplication, the 'x' mean 'times' or 'multiply'. For example, you'd read 3 x 6 as '3 times 6.' In a multiplication sentence, the numbers you are multiplying, in this case 3 and 6, are called factors. The number you get when you multiply the factors, 18, is called the product. Multiplication will be really helpful when you make the party favor bags for your birthday. If you want to give each of your 6 friends 3 ring pops, you know that you need to buy 18 ring pops since 6 x 3 = 18. The order of factors doesn't matter; you can change the order and still get the same product. Multiplying by 6 Using a Times Table The best way to learn your multiplication facts is to memorize them using a times table. When you use a times table, the numbers on the top row and the left column (the yellow numbers on this chart) are the factors. You just choose your factors using the top row and the left column, and the product is where they meet. In this times table, your 6 facts are highlighted in blue to help you find them. If you want to find 6 x 6, you find the 6 on the top row, the 6 on the left row, and the number where they meet, or 36. That means that 6 x 6 = 36. Tips and Tricks If you have already memorized your 3 times tables, try doubling your number and then multiplying it by 3. For example, if you were solving 6 x 5, you could double 5 to make 10, and then multiply 10 x 3 to get 30. To unlock this lesson you must be a Study.com Member. Register to view this lesson Are you a student or a teacher? See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Back What teachers are saying about Study.com Earning College Credit Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
# Distance of a Point from the Origin We will discuss here how to find the distance of a point from the origin. The distance of a point A (x, y) from the origin O (0, 0) is given by OA = $$\sqrt{(x - 0)^{2} + (y - 0)^{2}}$$ i.e., OP = $$\sqrt{x^{2} + y^{2}}$$ Consider some of the following examples: 1. Find the distance of the point (6, -6) from the origin. Solution: Let M (6, -6) be the given point and O (0, 0) be the origin. The distance from M to O = OM = $$\sqrt{(6 - 0)^{2} + (-6 - 0)^{2}}$$ = $$\sqrt{(6)^{2} + (-6)^{2}}$$ = $$\sqrt{36 + 36}$$ = $$\sqrt{72}$$ = $$\sqrt{2 × 2 × 2 × 3 × 3}$$ = 6$$\sqrt{2}$$ units. 2. Find the distance between the point (-12, 5) and the origin. Solution: Let M (-12, 5) be the given point and O (0, 0) be the origin. The distance from M to O = OM = $$\sqrt{(-12 - 0)^{2} + (5 - 0)^{2}}$$ = $$\sqrt{(-12)^{2} + (5)^{2}}$$ = $$\sqrt{144 + 25}$$ = $$\sqrt{169}$$ = $$\sqrt{13 × 13}$$ = 13 units. 3. Find the distance between the point (15, -8) and the origin. Solution: Let M (15, 8) be the given point and O (0, 0) be the origin. The distance from M to O = OM = $$\sqrt{(15 - 0)^{2} + (-8 - 0)^{2}}$$ = $$\sqrt{(15)^{2} + (-8)^{2}}$$ = $$\sqrt{225 + 64}$$ = $$\sqrt{289}$$ = $$\sqrt{17 × 17}$$ = 17 units.
# Go Math Grade 4 Answer Key Chapter 5 Factors, Multiples, and Patterns Go Math Grade 4 Answer Key Chapter 5 includes topics like Factors, Common factors, Divisibilities and Review tests, etc. that aid students to solve the homework and assessment tests. Also, it is the best and ultimate guide for exam preparation. You will find every question was explained in a simplistic way so that you are able to understand the concepts easily. Go Math Grade 4 Answer Key Chapter 5 Factors, Multiples, and Patterns pdf links are available here for each and every lesson. So, kickstart your preparation and score good grades in the exams. ## Go Math Grade 4 Answer Key Chapter 5 Factors, Multiples, and Patterns Improve your Problem-Solving Skills utilizing the Go Math Grade 4 Answer Key Chapter 5 Factors, Multiples, and Patterns. Start practicing the question covered in the Go Math 4th grade Solution Key and Cross Check the Solutions of Chapter 5 Factors, Multiples, and Patterns from here. So that you can easily rectify your mistakes and fill up the knowledge gap. Take the help from the direct links available below and solve the problems covered in Go Math Grade 4 Answer Key. ### Common Core – Model Factors – Page No. 283 Model Factors Use tiles to find all the factors of the product. Record the arrays on grid paper and write the factors shown. Question 1. Question 2. Write the factors of: 30 The Factors Of 30 are: 1,2,3,5,6,10,15,30. Explanation: Factors are the numbers which divides the original number completely. Here, we can see the numbers which gives the result as 30 when multiplied together.So the factors of 30 are 1,2,3.5,6,10,15,30. 1×30=30 2×15=30 3×10=30 5×6=30 6×5=30 10×3=30 15×2=30 30×1=30 Question 3. Write the factors of: 45 Answer: The Factors Of 45 are:1,3,5,9,15,45. Explanation: Factors are the numbers which divides the original number completely. Here, we can see the numbers which gives the result as 45 when multiplied together.So the factors of 45 are:1,3,5,9,15,45. 1×45=45 3×15=45 5×9=45 9×5=45 15×3=45 45×1=45 Question 4. Write the factors of: 19 Answer: The Factors Of 19 are:1,19. Explanation: Since 19 is a Prime number that means it is divisible by 1 and itself. So the factors of 19 are 1,19. 1×19=19 19×1=19. Question 5. Write the factors of: 40 Answer: The Factors Of 40 are:1,2,4,5,8,10,20,40. Explanation:Factors are the numbers which divides the original number completely. The Factors Of 40 are:1,2,4,5,8,10,20,40. 1×40=40 2×20=40 4×10=40 5×8=40 8×5=40 10×4=40 20×2=40 40×1=40 Question 6. Write the factors of: 36 Answer: The Factors Of 36 are:1,2,3,4,6,9,12,18,36. Explanation: Factors are the numbers which divides the original number completely. The factors of 36 are:1,2,3,4,6,9,12,18,36. 1×36=36 2×18=36 3×12=36 4×9=36 6×6=36 9×4=36 12×3=36 18×3=36 36×1=36. Question 7. Write the factors of: 22 Answer: The Factors Of 22 are:1,2,11,22. Explanation: Factors are the numbers which divides the original number completely. The factors of 22 are:1,2,11,22. 1×22=22 2×11=22 11×2=22 22×1=22. Question 8. Write the factors of: 4 Answer: The Factors Of 4 are:1,2,4. Explanation: Factors are the numbers which divides the original number completely. The Factors Of 4 are:1,2,4. 1×4=4 2×2=4 4×1=4. Question 9. Write the factors of: 26 Answer: The Factors Of 26 are:1,2,13,26. Explanation: Factors are the numbers which divides the original number completely. Here, we can see the numbers which gives the result as 26 when multiplied together.So the factors of 26 are:1,2,13,26. 1×26=26 2×13=26 13×2=26 26×1=26. Question 10. Write the factors of: 49 Answer: The Factors Of 49 are:1,7,49. Explanation: Factors are the numbers which divides the original number completely. The Factors Of 49 are:1,7,49. 1×49=49 7×7=49 49×1=49. Question 11. Write the factors of: 32 Answer: The Factors Of 32 are:1,2,4,8,16,32. Explanation: Factors are the numbers that divide the original number completely. Here, we can see the numbers which give the result as 32 when multiplied together.So the factors of 32 are:1,2,4,8,16,32. 1×32=32 2×16=32 4×8=32 8×4=32 16×2=32 32×1=32. Question 12. Write the factors of 23 Answer: The Factors Of 23 are:1,23. Explanation: Since 23 is a Prime number that means it is divisible by 1 and itself. So the factors of 23 are 1,23. 1×23=23 23×1=23. Question 13. Brooke has to set up 70 chairs in equal rows for the class talent show. But, there is not room for more than 20 rows. What are the possible number of rows that Brooke could set up? Explanation: Let the possible no.of rows be X, As there is no room for more than 20 rows so there should not be more than 20 rows.X should be less than or equal to 20(X<=20). As Brooke has 70 chairs to set up in equal rows we will find the factors of 70 and in that we must pick up the numbers which are less than equal to 20.Therefore the factors of 70 are 2,5,7,10,14. Question 14. Eduardo thinks of a number between 1 and 20 that has exactly 5 factors. What number is he thinking of? Explanation: If find factors for 1 to 20 we don’t get exactly 5 factors for any number except 16. So the answer is 16. ### Common Core – Factors – Page No. 284 Lesson Check Question 1. Which of the following lists all the factors of 24? Options: a. 1, 4, 6, 24 b. 1, 3, 8, 24 c. 3, 4, 6, 8 d. 1, 2, 3, 4, 6, 8, 12, 24 Answer: d(1, 2, 3, 4, 6, 8, 12, 24) Explanation:Factors are the numbers which divides the original number completely. Here, we can see the numbers which gives the result as 24 when multiplied together.So the factors of 24 are:1, 2, 3, 4, 6, 8, 12, 24. 1×24=24 2×12=24 3×8=24 4×6=24 6×4=4 8×3=24 12×2=24 24×1=24 Question 2. Natalia has 48 tiles. Which of the following shows a factor pair for the number 48? Options: a. 4 and 8 b. 6 and 8 c. 2 and 12 d. 3 and 24 Explanation: 6 and 8 are factor pair for 48 because 6×8=48. Spiral Review Question 3. The Pumpkin Patch is open every day. If it sells 2,750 pounds of pumpkins each day, about how many pounds does it sell in 7 days? Options: a. 210 pounds b. 2,100 pounds c. 14,000 pounds d. 21,000 pounds Explanation: Let’s round off 2750 pounds to 3000 pounds. In one day 3000 pounds pumpkins were sold out, and in 7 days?? —- 3000×7= 21,000 pounds. Question 4. What is the remainder in the division problem modeled below? Options: a. 2 b. 3 c. 5 d. 17 Explanation: We can see in the above figure 3 circles with 5 sub circles inside it and a pair of sub circles. Here total sub circles are (3×5)+2=17. If we divide 17 with 3 then we will get reminder as 2. So answer is 2. Question 5. Which number sentence is represented by the following array? Options: a. 4 × 5 = 20 b. 4 × 4 = 16 c. 5 × 2 = 10 d. 5 × 5 = 25 Explanation: As we can see 4 rows and 5 squares, So 4 × 5 = 20. Question 6. Channing jogs 10 miles a week. How many miles will she jog in 52 weeks? Options: a. 30 miles b. 120 miles c. 200 miles d. 520 miles Explanation: No.of weeks = 52. So 1 week = 10 miles, then 52 weeks =????? 52×10=520 miles. ### Page No. 287 Question 1. Is 4 a factor of 28? Draw a model to help. Think: Can you make a rectangle with 28 squares in 4 equal rows? 4 ______ a factor of 28. Type below: __________ Is 5 a factor of the number? Write yes or no. Question 2. 27 Explanation: Factors of 27 are 1,3,9,27. So the answer is No. Question 3. 30 Explanation: As the last digit is 0 which is divisible 5. Question 4. 36 Explanation: 36 is not divisible by 5, So the answer is no Question 5. 53 Explanation: Factors of 53 are 1, 53. So the answer is No. Is 9 a factor of the number? Write yes or no. Question 6. 54 Explanation: As 54 is divisible by 9. Question 7. 63 Explanation: 63 is divisible by 9, So the answer is Yes Question 8. 67 Explanation: 67 is a prime number that means it is divisible by 1 and itself. So the answer is No. Question 9. 93 Explanation: The factors of 93 are 1,3,31 and 93. So the answer is No. List all the factor pairs in the table. Question 10. 1×24=24 1,24 2×12=24 2,12 3×8=24 3,8 4×6=24 4,6 Explanation: Factors of 24. Question 11. 1×39=39 1,39 3×13=39. 3,13 Explanation: Factors of 39. Practice: Copy and Solve List all the factor pairs for the number. Make a table to help. Question 12. 56 1×56=56 1,56 2×23=56 2,23 4×14=56 4,14 7×8=56 7,8 8×7=56 8,7 Explanation: Factors of 56. Question 13. 64 1×64=64 1,64 2×32=64 2,32 4×16=64 4,16 8×8=64 8,8 Explanation: Factors of 64 and factor pair for 64 is 8,8. ### Page No. 288 Use the table to solve 14–15. Question 14. Dirk bought a set of stamps. The number of stamps in the set he bought is divisible by 2, 3, 5, 6, and 9. Which set is it? Explanation: 90 is divisible by all numbers 2,3,5,6, and 9. So the answer is 90. Question 15. Geri wants to put 6 stamps on some pages in her stamp book and 9 stamps on other pages. Explain how she could do this with the stamp set for Sweden. Answer: 10 pages with 6 stamps and 2 pages with 9 stamps. Explanation: Geri could break 78 into 60+18, As 60 is divisible by 6, and 18 is divisible by 9. Then she could make 10 pages with 6 stamps as 60÷6=10 and 2 pages with 9 stamps as 18÷9=2. Question 16. Use Counterexamples George said if 2 and 4 are factors of a number, then 8 is a factor of the number. Is he correct? Explain. Explanation: Because if we 12 as an example, 2 and 4 are factors of 12 but not 8. Question 17. Classify the numbers. Some numbers may belong in more than one box. Divisible by 5 and 9 — 45 Divisible by 3 and 9 — 27,45,54,72,81 Divisible by 2 and 6 — 54,72,84. ### Common Core – Factors and Divisibility – Page No. 289 Is 6 a factor of the number? Write yes or no. Question 1. Question 2. 56 Explanation: 56 is not divisible by 6. So the answer is No. Question 3. 42 Explanation: Since 42 is divisible by 6. Question 4. 66 Explanation: 66 is divisible by 6. Is 5 a factor of the number? Write yes or no. Question 5. 38 Explanation: If the end is 0 or 5 then the number is divisible by 5. As the number is 38 the answer is No Question 6. 45 Explanation: 45 is divisible by 5. So the answer is Yes Question 7. 60 Explanation: 60 is a factor of 5 because 60 is divisible by 5. Question 8. 39 Explanation: As 39 is not divisible by 5. So the answer is No. List all the factor pairs Question 9. Factors of 12 1 × 12 = 12; ( 1 , 12 ) 2 × 6 = 12; ( 2, 6 ) 3 × 4 = 12; ( 3 , 4 ) Question 10. Factors of 25 1 ×25 = 25; ( 1 , 25 ) 5 × 5 = 25; ( 5 , 5 ) Question 11. List all the factor pairs for 48. Answer: Factor pairs of 48 are (1,48),(2,24),(3,16),(4,12),(6,8),(12,2),(6,3),(24,2),(48,1). Explanation: Factor pairs are the pairs when we multiplied both numbers will get the result. Here factor pairs for 48 are 1×48=48 (1,48) 2×24=48 (2,24) 3×16=48 (3,16) 4×12=48 (4,12) 6×8 =48 (6,8) Problem Solving Question 12. Bryson buys a bag of 64 plastic miniature dinosaurs. Could he distribute them equally into six storage containers and not have any left over? Explanation: 64 is not divisible by 6, So he cannot distribute them equally into six storage containers. Question 13. Lori wants to distribute 35 peaches equally into baskets. She will use more than 1 but fewer than 10 baskets. How many baskets does Lori need? Explanation: First we need to know the factors of 35. The factors of 35 are 1,5,7,35. As Lori uses more than 1 but fewer than 10, the answer is 5 or 7. Lori can distribute 35 peaches equally in 5 or 7 baskets. ### Common Core – Factors – Page No. 290 Lesson Check Question 1. Which of the following numbers has 9 as a factor? Options: a. 28 b. 30 c. 39 d. 45 Explanation: 45 is divisible 9. So the answer is 45. Question 2. Which of the following numbers does NOT have 5 as a factor? Options: a. 15 b. 28 c. 30 d. 45 Explanation: 28 is not divisible by 5. So 28 is not a factor of 5. Spiral Review Question 3. Which of the following shows a strategy to use to find 4 × 275? Options: a. (4 × 300) + (4 × 25) b. (4 × 300) – (4 × 25) c. (4 × 275) – 100 d. (4 × 200) + 75 Explanation: First we must replace 300-25 in the place of 275 then it becomes 4×(300-25), Now we must use the distributive property of multiplication then (4×300)-(4×25). So the answer is b. Question 4. Jack broke apart 5 × 216 as (5 × 200) + (5 × 16) to multiply mentally. What strategy did Jack use? Options: a. the Commutative Property b. the Associative Property c. halving and doubling d. the Distributive Property Explanation: Distributive property means if we multiply a sum by a number is same as multiplying each addend by the number and adding the products. This is the strategy Jack used. Question 5. Jordan has $55. She earns$67 by doing chores. How much money does Jordan have now? Options: a. $122 b.$130 c. $112 d.$12 Explanation: Jordan has $55, she earns by doing chores is$67. So total money is $55+$67=$122. Question 6. Trina has 72 collector’s stamps. She puts 43 of the stamps into a stamp book. How many stamps are left? Options: a. 29 b. 31 c. 39 d. 115 Answer: a Explanation: Stamps left are 72-43=29. ### Page No. 293 Question 1. Lucy has 40 bean plants, 32 tomato plants, and 16 pepper plants. She wants to put the plants in rows with only one type of plant in each row. All rows will have the same number of plants. How many plants can Lucy put in each row? First, read the problem and think about what you need to find. What information will you use? How will you use the information? Answer: We will find common factors for 40,32 and 16. Question 1. Next, make a list. Find the factors for each number in the problem. Answer: Factors of 40 are — 1,2,4,5,8,10,20,40 Factors of 32 are — 1,2,4,8,16,32 Factors of 16 are — 1,2,4,8,16 Question 1. Finally, use the list. Circle the common factors. So, Lucy can put ___ , ___ , ___ , or ___ plants in each row. Answer: 1,2,4,8 Explanation: Because 1,2,4,8, are common factors in 40,32,16. Question 2. What if Lucy has 64 bean plants instead of 40 bean plants? How many plants can Lucy put in each row? Answer: 1,2,4,8,16 Explanation: Here we need to find the factors of 64,32 and 16. We get common factors as 1,2,4,8,16. Question 3. One common factor of two numbers is 40. Another common factor is 10. If both numbers are less than 100, what are the two numbers? ______ and ______ Answer: 40 and 80. Explanation: As the next multiple of 40 is 80. So both 40 and 80 are less than 100 and has a common factor as 10. Question 4. The sum of two numbers is 136. One number is 51. What is the other number? What are the common factors of these two numbers? Answer: 85. Common Factors are 1,17. Explanation: As 136-51= 85 Factors of 51 are 1,3,17,51 Factors of 85 are 1,5,17,85. ### Page No. 294 Question 5. Analyze A number is called a perfect number if it equals the sum of all of its factors except itself. For instance, 6 is a perfect number because its factors are 1, 2, 3, and 6, and 1 + 2 + 3 = 6. What is the next greater perfect number? Answer: 28 Explanation: The factors of 28 are 1,2,4,7,14 and 28. If we add 1+2+4+7+14 we will get 28. So 28 is a perfect number. Question 6. Sona knits 10 squares a day for 7 days. Can she sew together the squares to make 5 equal-sized blankets? Explain. Answer: Yes Explanation: As 10×7= 70 which is a factor of 5. Question 7. Julianne earned$296 working at a grocery store last week. She earns $8 per hour. How many hours did Julianne work? Answer: 37 hours Explanation: Julianne earned$296 in last week. Per hour she earns $8, So total no.of hours did she worked is 296÷8= 37 hours. Question 8. There are 266 students watching a play in the auditorium. There are 10 rows with 20 students in each row and 5 rows with 8 students in each row. How many students are sitting in each of the 2 remaining rows if each of those rows has an equal number of students? Answer: 13 Students Explanation: Total number of students is 266. In which 10 rows were filled with 20 students that means 10×20=200 students, and 5 rows were filled with 8 students which means 5×8= 40 students. The total students filled are 240. And to know how many students filled in the remaining 2 rows we need to subtract 266-240=26, As students are filled in 2 rows 26÷2= 13. Question 9. Ben is planting a garden with 36 zinnias, 18 marigolds, and 24 petunias. Each row will have only one type of plant. Ben says he can put 9 plants in each row. He listed the common factors of 36, 18 and 24 below to support his reasoning. 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 18: 1, 2, 3, 6, 8, 9, 18 24: 1, 2, 3, 4, 6, 8, 9, 12, 24 Is he correct? Explain your answer. If his reasoning is incorrect, explain how he should have found the answer. Answer: No Explanation: The factors of 18 and 24 are incorrect which he listed. And the common factors for 36,24 and 18 are 1,2,3 and 6. So he can put 1,2,3 and 6 plants in a row. ### Common Core – Common Factors – Page No. 295 Problem Solving Common Factors Solve each problem. Question 1. Grace is preparing grab bags for her store’s open house. She has 24 candles, 16 pens, and 40 figurines. Each grab bag will have the same number of items, and all the items in a bag will be the same. How many items can Grace put in each bag? Question 2. Simon is making wreaths to sell. He has 60 bows, 36 silk roses, and 48 silk carnations. He wants to put the same number of items on each wreath. All the items on a wreath will be the same type. How many items can Simon put on each wreath? Answer:1,2,3,4,6 or 12 items Simon puts on each wreath. Explanation: First we will find the common factors of 36,48,60 factors of 36 are: 1,2,3,4,6,9,12,18,36. factors of 48 are: 1,2,3,4,6,8,12,16,24,48 factors of 60 are: 1,2,3,4,5,6,10,12,15,20,30,60. The common factors of 36,48,60 are 1,2,3,4,6,12. So Simon can put 1,2,3,4,6 or 12 items on each wreath. Question 3. Justin has 20 pencils, 25 erasers, and 40 paper clips. He organizes them into groups with the same number of items in each group. All the items in a group will be the same type. How many items can he put in each group? Answer: Justin can put 1 or 5 items in each group. Explanation: We will find common factors of 20,25,40. factors of 20 are: 1,2,4,5,10,20. factors of 25 are: 1,5,25. factors of 40 are: 1,2,4,5,8,10,20,40 So common factors are 1 and 5. Question 4. A food bank has 50 cans of vegetables, 30 loaves of bread, and 100 bottles of water. The volunteers will put the items into boxes. Each box will have the same number of food items and all the items in the box will be the same type. How many items can they put in each box? Answer: 1,2,5, or 10. Explanation: 1,2,5 or 10 are the common factors of 30,50 and 100. factors for 30 are: 1,2,3,5,6,10,15,30 factors for 50 are: 1,2,5,10,25,50 factors of 100 are: 1,2,4,5,10,20,25,50,100 So answer is 1,2,5,10. Question 5. A debate competition has participants from three different schools: 15 from James Elementary, 18 from George Washington School, and 12 from the MLK Jr. Academy. All teams must have the same number of students. Each team can have only students from the same school. How many students can be on each team? Answer: 3 Explanation: Lets find the common factors of 12,15,18 factors of 12 are: 1,2,3,4,6,12 factors of 15 are: 1,3,5,15 factors of 18 are: 1,2,3,6,9,18 3 is the common factor for 12,15,18 ### Common Core – Common Factors – Page No. 296 Lesson Check Question 1. What are all the common factors of 24, 64, and 88? Options: a. 1 and 4 b. 1, 4, and 8 c. 1, 4, 8, and 12 d. 1, 4, 8, and 44 Answer: b Explanation: factors of 24 are: 1,2,3,4,8,12,24 factors of 64 are: 1,2,4,8,16,32,64 factors of 88 are: 1,2,4,8,11,22,44,88 Question 2. Which number is NOT a common factor of 15, 45, and 90? Options: a. 3 b. 5 c. 10 d. 15 Answer: c Explanation: As 15 and 45 are not divisible by 10. Spiral Review Question 3. Dan puts$11 of his allowance in his savings account every week. How much money will he have after 15 weeks? Options: a. $165 b.$132 c. $110 d.$26 Explanation: Dan puts $11 in his savings account every week, So after 15 weeks it will be 15×11=165. The total money he will have after 15 weeks is$165. Question 4. James is reading a book that is 1,400 pages. He will read the same number of pages each day. If he reads the book in 7 days, how many pages will he read each day? Options: a. 20 b. 50 c. 140 d. 200 Explanation: Total no.of.pages is 1400, no.of pages James read each day is 1400÷7= 200 Question 5. Emma volunteered at an animal shelter for a total of 119 hours over 6 weeks. Which is the best estimate of the number of hours she volunteered each week? Options: a. 10 hours b. 20 hours c. 120 hours d. 714 hours Explanation: Total hours Emma volunteered is 119 hours over 6 weeks, how much she volunteered each week is 119÷6= 19.833 i.e 20 hours. We must round off to the nearest one i.e 20 hours. Question 6. Which strategy can be used to multiply 6 × 198 mentally? Options: a. 6 × 198 = (6 × 19) + (6 × 8) b. 6 × 198 = (6 × 200) + (6 × 2) c. 6 × 198 = (6 × 200) – (6 × 2) d. 6 × 198 = (6 + 200) × (6 + 2) Explanation: By Distributive property of multiplication 6×198 can be written as (6 × 200) – (6 × 2). ### Page No. 297 Choose the best term from the box. Question 1. A number that is multiplied by another number to find a product is called a Question 2. A number is _________ by another number if the quotient is a counting number and the remainder is zero. List all the factors from least to greatest. Question 3. 8 Question 4. 14 Is 6 a factor of the number? Write yes or no. Question 5. 81 Explanation: 81 is not divisible by 6 Question 6. 45 Explanation: 45 is not divisible by 6 Question 7. 42 Explanation: 42 is divisible by 6 Question 8. 56 Explanation: 56 is not divisible by 6 List all the factor pairs in the table. Question 9. 1×64=64 1,64 2×32=64 2,32 4×16=64 4,16 8×8=64 8,8 Explanation: Factors of 64 Question 10. 1×44=44 1,44 2×22=44 2,22 11×4=44 11,4 List the common factors of the numbers. Question 11. 9 and 18 Explanation: Factors of 9 are: 1,3,9 Factors of 18 are: 1,2,3,9,18 Question 12. 20 and 50 Explanation: Factors of 20 are: 1,2,4,5,10,20 Factors of 50 are: 1,2,5,10,25,50 ### Page No. 298 Question 13. Sean places 28 tomato plants in rows. All rows contain the same number of plants. There are between 5 and 12 plants in each row. How many plants are in each row? Explanation: There are 28 tomato plants in a row. To find out how many plants in a row we will find the factors of 28 i.e 1,2,4,7,14,28. As there are between 5 and 12 plants 7 is the only number between 5 and 12. So 7 plants are planted in each row. Question 14. Ella bought some key chains and spent a total of $24. Each key chain costs the same whole-dollar amount. She bought between 7 and 11 key chains. How many key chains did Ella buy? Answer: 8 Explanation: Ella spent a total of$24. To find how many key chains first we will find the factors of 24. Factors of 24 are 1,2,3,4,6,8,12,24. As Ella bought between 7 and 11 key chains 8 is the only number between 7 and 11. So 8 key chains Ella bought. Question 15. Sandy has 16 roses, 8 daisies, and 32 tulips. She wants to arrange all the flowers in the bouquets. Each bouquet has the same number of flowers and the same type of flower. What is the greatest number of flowers that could be in a bouquet? Answer: 2 roses, 1 daisy, and 4 tulips in 8 bouquets. Explanation: First we must add all the flowers i.e 16+8+32= 56, Now we can divide 56 flowers equally in each bouquet. Like 2 roses, 1 daisy and 4 tulips in 8 bouquets or 8 roses in 2 bouquets, 8 daisies in 1 bouquet and 8 tulips in 4 bouquets. Question 16. Amir arranged 9 photos on a bulletin board. He put the photos in rows. Each row contains the same number of photos. How many photos could be in each row? Answer: 9 photos in a row and 3 photos in 3 rows or 9 photos in 1 row. Explanation: Factors of 9 are 1,3,9. So Amir can arrange 9 photos in a row and 3 photos in 3 rows or 9 photos in 1 row. ### Page No. 301 Question 1. Multiply to list the next five multiples of 4. 4 , _____ , _____ , _____ , _____ , _____ 1 × 4 4 , _____ , _____ , _____ , _____ , _____ 4 1×4 8 2×4 12 3×4 16 4×4 20 4×5 Explanation: Multiplies of 4 Is the number a factor of 6? Write yes or no. Question 2. 2 Explanation: 6 is divisible by 2. So 2 is the factor of 6. Question 3. 6 Explanation: 6 is divisible by 6. Question 4. 16 Explanation: 16 is not divisible by 6 Question 5. 18 Explanation: 18 is divisible by 6 Is the number a multiple of 6? Write yes or no. Question 6. 3 Explanation: Multiples of 6 are 6,12,18,24,30, etc. Question 7. 6 Explanation: 1×6= 6. So 6 is multiple of 6. Question 8. 16 Explanation: Multiples of 6 are 6,12,18,24,30, etc. Question 9. 18 Explanation: Multiples of 6 are 6,12,18,24,30, etc. Is the number a multiple of 3? Write yes or no. Question 10. 4 Explanation: Multiples of 3 are 3,6,9,12,15,etc. Question 11. 8 Explanation: Multiples of 3 are 3,6,9,12,15,etc. Question 12. 24 Explanation: Multiples of 3 are 3,6,9,12,15,etc. Question 13. 38 Explanation: Multiples of 3 are 3,6,9,12,15,18,21,24,27,30,33,36,39,42,etc. Question 14. List the next nine multiples of each number. Find the common multiples. Multiples of 2: 2, _________________ Multiples of 8: 8, _________________ Common multiples: _________________ Explanation: Multiples of 2: 2,4,6,8,10,12,14,16,18,20. Multiples of 8: 8,16,24,32,40,48,56,64,72,80. So common multiples are: 8,16 Generalize Algebra Find the unknown number. Question 15. 12, 24, 36, _____ Explanation: 12×1= 12 12×2= 24 12×3= 36 12×4= 28 Question 16. 25, 50, 75, 100, ______ Explanation: 25×1= 25 25×2= 50 25×3= 75 25×4= 100 25×5= 125 Tell whether 20 is a factor or multiple of the number. Write factor, multiple, or neither. Question 17. 10 Explanation: 2×10= 20. Question 18. 20 Explanation: 1×20= 20 20÷1= 20. Question 19. 30 Explanation: Factors of 30 are: 1,2,3,5,6,10,15,and 30. Multiples of 30 are: 30,60,90,etc. Write true or false. Explain. Question 20. Every whole number is a multiple of 1. Explanation: For every whole number which is multiplied with 1, the result will be that number. Question 21. Every whole number is a factor of 1. Explanation: Not every whole number is a factor of 1. Question 22. Julio wears a blue shirt every 3 days. Larry wears a blue shirt every 4 days. On April 12, both Julio and Larry wore a blue shirt. What is the next date that they will both wear a blue shirt? Explanation: As Julio wears a blue shirt every 3 days and another shirt in the remaining 4 days, So 4×3 days= 12 Larry wears a blue shirt every 4 days and another shirt in the remaining 3 days, So 3×4 days= 12 12+12= 24. So the next date will be April 24. ### Page No. 302 Complete the Venn diagram. Then use it to solve 23–25. Question 23. What multiples of 4 are not factors of 48? Explanation: Multiples of 4 are 4,8,12,16,20,24,28,32,36,40,44,48. Not a factors of 48 are 20,28,32,36,40,44. Question 24. What factors of 48 are multiples of 4? Explanation: Multiples of 4 are: 4,8,12,16,20,24,28,32,36,40,44,48. Factors of 48 are: 1,2,4,8,12,16,24,48. Question 25. Pose a Problem Look back at Problem 24. Write a similar problem by changing the numbers. Then solve. Answer: Let’s take factors of 64 are multiples of 8? 8,16,32,64. Explanation: Multiples of 8 are: 8,16,24,32,40,48,56,64,72,80 Factors of 64 are: 1,2,4,8,16,32,64 Question 26. Kia paid $10 for two charms. The price of each charm was a multiple of$2. What are the possible prices of the charms? Answer: $2,$8 and $4,$6. Explanation: Since the price was multiple of 2 and Kia paid $10 for two charms, So possible prices are$2+$8=$10 and $4+$6=$10. Question 27. Look for Structure The answer is 9, 18, 27, 36, 45. What is the question? Answer: Write the multiples of 9 Question 28. How do you know whether a number is a multiple of another number? Answer: When the number is divisible by the number then that number is multiple of another number. Explanation: For example, if we take a number i.e 8 which is divisible by 2 and 8 is a multiple of 2. Question 29. For numbers 29a–29e, select True or False for each statement. a. The number 45 is a multiple of 9. i. True ii. False Answer: True Explanation: As 9×5= 45, So 45 is multiple of 9. Question 29. b. The number 4 is a multiple of 16. i. True ii. False Answer: False. Explanation: As 16 is divisible by 4 and not a multiple of 16. Multiple of 16 are : 16,32,48,64,80. Question 29. c. The number 28 is a multiple of 4. i. True ii. False Answer: True. Explanation: 4×7=28. Question 29. d. The number 4 is a factor of 28. i. True ii. False Answer: True. Explanation: Factors of 28 are: 1,2,4,7,14,28. Question 29. e. The number 32 is a factor of 8. i. True ii. False Answer: Explanation: ### Common Core – Factors and Multiples – Page No. 303 Factors and Multiples Is the number a multiple of 8? Write yes or no. Question 1. Question 2. 8 Answer: Yes Explanation: Since 8×1=8, it is a multiple of 8 Question 3. 20 Answer: No Explanation: 20 is not a multiple of 8 Question 4. 40 Answer: Yes Explanation: 8×5=40, So 40 is multiple of 8 List the next nine multiples of each number. Find the common multiples. Question 5. Multiples of 4: Multiples of 7: Common multiples: Answer: Explanation: Multiples of 4: 4,8,12,16,20,24,28,32,36,40. Multiples of 7: 7,14,21,28,35,42,49,56,63,70. Common Multiples: 28, Question 6. Multiples of 3: Multiples of 9: Common multiples: Answer: 9,18,45,54,63, etc. Explanation: Multiples of 3: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63. Multiples of 9: 9,18,27,36,45,54,63,72,81,90. Common multiples: 9,18,45,54,63, etc. Question 7. Multiples of 6: Multiples of 8: Common multiples: Answer: 24,48,72. Explanation: Multiples of 6: 6,12,18,24,30,36,42,48,54,60,66,72,78. Multiples of 8: 8,16,24,32,40,48,56,64,72,80. Common multiples: 24,48,72. Tell whether 24 is a factor or multiple of the number. Write factor, multiple, or neither. Question 8. 6 Answer: Multiple Explanation: 6×4=24 Question 9. 36 Answer: Neither Explanation: 36 is not a factor or multiple of 24. Question 10. 48 Answer: Factor Explanation: 24×2= 48, So 48 is a factor of 24 Problem Solving Question 11. Ken paid$12 for two magazines. The cost of each magazine was a multiple of $3. What are the possible prices of the magazines? Answer:$3+$9=$12. Explanation: As each magazine cost was multiple of $3, The possible price for 2 magazines are$3+$9=$12, which is a multiple of 3 Question 12. Jodie bought some shirts for $6 each. Marge bought some shirts for$8 each. The girls spent the same amount of money on shirts. What is the least amount they could have spent? Answer: $24 Explanation: As they spent the same amount of money which means the number should be multiple of$6 and $8, So multiples of 6 are: 6,12,18,24,30,36,42 and multiples of 8 are: 8,16,24,32,40. The least amount they could spend is 24. As 24 is the least common multiple. ### Common Core – Factors and Multiples – Page No. 304 Lesson Check Question 1. Which list shows numbers that are all multiples of 4? Options: a. 2, 4, 6, 8 b. 3, 7, 11, 15, 19 c. 4, 14, 24, 34 d. 4, 8, 12, 16 Answer: d Explanation: Multiples of 4 are 4,8,12,16. Question 2. Which of the following numbers is a common multiple of 5 and 9? Options: a. 9 b. 14 c. 36 d. 45 Answer: 45 Explanation: 5×9= 45 Spiral Review Question 3. Jenny has 50 square tiles. She arranges the tiles into a rectangular array of 4 rows. How many tiles will be left over? Options: a. 0 b. 1 c. 2 d. 4 Answer: 2 Explanation: As Jenny arranges in 4 rows, each row contains 12 tiles. So 12×4= 48. The tiles left are 50-48=2. Question 4. Jerome added two numbers. The sum was 83. One of the numbers was 45. What was the other number? Options: a. 38 b. 48 c. 42 d. 128 Answer: a Explanation: The sum of two numbers is 83, in that one number is 45. To find another number we will do subtraction, i.e 83-45=38. Question 5. There are 18 rows of seats in the auditorium. There are 24 seats in each row. How many seats are in the auditorium in all? Options: a. 42 b. 108 c. 412 d. 432 Answer: d Explanation: No.of rows= 18, each row has 24 seats. So total no.of seats are 18×24= 432. Question 6. The population of Riverdale is 6,735. What is the value of the 7 in the number 6,735? Options: a. 7 b. 700 c. 735 d. 7,000 Answer: b Explanation: In 6,735 the 7 is in the Hundreds Place. So the answer is 7. ### Page No. 307 Question 1. Use the grid to model the factors of 18. Tell whether 18 is prime or composite. Factors of 18: ____ , ____ , ____ , ____ , ____ , ____ Think: 18 has more than two factors. So, 18 is _________ . Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 18 are: 1,2,3,6,9,18. Tell whether the number is prime or composite. Question 2. 11 Think: Does 11 have other factors besides 1 and itself? Answer: Prime number. Explanation: A Prime number is a number that is divisible 1 and itself. Question 3. 73 Answer: Prime number Explanation: A Prime number is a number that is divisible 1 and itself. Question 4. 69 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 69 are: 1,3,23,69. Question 5. 42 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 42 are: 1,2,3,6,7,21,42. Tell whether the number is prime or composite. Question 6. 18 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 18 are: 1,2,3,6,9,18. Question 7. 49 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 49 are 1,7,49. Question 8. 29 Answer: Prime number. Explanation: A Prime number is a number that is divisible 1 and itself. Question 9. 64 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 64 are: 1,2,4,8,32,64. Question 10. 33 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 33 are: 1,3,11,33. Question 11. 89 Answer: Prime number. Explanation: A Prime number is a number that is divisible 1 and itself. Question 12. 52 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 52 are: 1,2,4,13,26,52. Question 13. 76 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 76 are: 1,2,4,19,38,76. Write true or false for each statement. Explain or give an example to support your answer. Question 14. Only odd numbers are prime numbers. Answer: False. Explanation: Not all odd numbers are prime numbers. For example. 39 is an odd number but not a prime number because it is divisible by 3 and 13. Question 15. A composite number cannot have three factors. Answer: False Explanation: A Composite number is a number that has more than two factors. For example. 21 is a composite number and the factors of 21 are 1,3,7,21. Question 16. I am a number between 60 and 100. My ones digit is two less than my tens digit. I am a prime number. What number am I? Answer: 97 Explanation: Prime numbers between 60 to 100 are 61,67,71,73,79,83,89,97. 97 is the number which ones digit is two less than tens digit. Question 17. Name a 2-digit odd number that is prime. Name a 2-digit odd number that is composite. Answer: 2 digit Prime odd numbers are 11,13,17 etc. 2 digit Composite odd numbers are 15,21,39 Explanation: A Prime number is a number that is divisible 1 and itself. The number which has more than two factors is called composite numbers. Question 18. Choose the words that correctly complete the sentence. The number 9 is img 18 because it has img 19 two factors. Type below: __________ ### Page No. 308 The Sieve of Eratosthenes Eratosthenes was a Greek mathematician who lived more than 2,200 years ago. He invented a method of finding prime numbers, which is now called the Sieve of Eratosthenes. Question 19. Follow the steps below to circle all prime numbers less than 100. Then list the prime numbers. STEP 1 Cross out 1, since 1 is not prime. STEP 2 Circle 2, since it is prime. Cross out all other multiples of 2. STEP 3 Circle the next number that is not crossed out. This number is prime. Cross out all the multiples of this number. STEP 4 Repeat Step 3 until every number is either circled or crossed out. So, the prime numbers less than 100 are Answer: 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. Explanation: A Prime number is a number that is divisible 1 and itself. Question 20. Explain why the multiples of any number other than 1 are not prime numbers. Answer: ### Common Core – Prime and Composite Numbers – Page No. 309 Prime and Composite Numbers Tell whether the number is prime or composite Question 1. Question 2. 68 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 68 are: 1,2,4,17,34,69. Question 3. 52 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 52 are: 1,2,4,13,26,52. Question 4. 63 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 63 are: 1,2,3,7,9,21,63. Question 5. 75 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 75 are: 1,3,5,15,25,75 Question 6. 31 Answer: Prime number. Explanation: 31 is a prime number that means it is divisible by 1 and itself. Question 7. 77 Answer: Composite number. Explanation: The number which has more than two factors is called composite numbers. Factors of 77 are: 1,7,11,77. Question 8. 59 Answer: Prime number Explanation: 59 is a prime number that means it is divisible by 1 and itself. Question 9. 87 Answer: Composite Number. Explanation: The number which has more than two factors is called composite numbers. Factors of 87 are: 1,3,29,87. Question 10. 72 Answer: Composite Number. Explanation: The number which has more than two factors is called composite numbers. Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Question 11. 49 Answer: Composite Number. Explanation: The number which has more than two factors is called composite numbers. Factors of 49 are 1,7,49. Question 12. 73 Answer: Prime number. Explanation: A Prime number is a number that is divisible 1 and itself. Problem Solving Question 13. Kai wrote the number 85 on the board. Is 85 prime or composite? Answer: Composite number Explanation: The number which has more than two factors is called composite numbers. Factors of 85 are 1,5,17,85 Question 14. Lisa says that 43 is a 2-digit odd number that is composite. Is she correct? Answer: No Explanation: 43 is a prime number. A Prime number is a number that is divisible 1 and itself. ### Common Core – Prime and Composite Numbers – Page No. 310 Lesson Check Question 1. The number 5 is: Options: a. prime b. composite c. both prime and composite d. neither prime nor composite Answer: Prime number. Explanation: A Prime number is a number that is divisible 1 and itself. Question 2. The number 1 is: Options: a. prime b. composite c. both prime and composite d. neither prime nor composite Answer: d Explanation: A Prime number is a number that is divisible 1 and itself. So prime number should have two divisors but 1 has only one divisor. The number which has more than two factors is called composite numbers. So 1 doesn’t have more than two factors. So 1 is neither Prime nor Composite. Spiral Review Question 3. A recipe for a vegetable dish contains a total of 924 calories. The dish serves 6 people. How many calories are in each serving? Options: a. 134 calories b. 150 calories c. 154 calories d. 231 calories Answer: c Explanation: Total no.of calories are 924, which serves 6 people. To find each serving we will perform division 924÷6= 154 calories. Question 4. A store clerk has 45 shirts to pack in boxes. Each box holds 6 shirts. What is the fewest boxes the clerk will need to pack all the shirts? Options: a. 9 b. 8 c. 7 d. 6 Answer: b Explanation: As the box holds only 6 shirts, 42 shirts are packed in 7 boxes, and the remaining 3 shirts will be packed in another box. So the total number of boxes is 8. Question 5. Which number rounds to 200,000? Options: a. 289,005 b. 251,659 c. 152,909 d. 149,889 Answer: c Explanation: 152,909 is nearest to 200,000. Question 6. What is the word form of the number 602,107? Options: a. six hundred twenty thousand,seventeen b. six hundred two thousand, one hundred seven c. six hundred twenty-one thousand, seventeen d. six hundred two thousand, one hundred seventy Answer: b ### Page No. 313 Use the rule to write the numbers in the pattern. Question 1. Rule: Subtract 10. First term: 100 Answer: 100,90,80,70,60,.. Explanation: 100 100-10= 90 90-10= 80 80-10= 70 70-10= 60 Use the rule to write the numbers in the pattern. Describe another pattern in the numbers. Question 2. Rule: Multiply by 2. First term: 4 4 , _____ , _____ , _____ , _____ , ……. Answer: 4,8,16,32,64,… Explanation: 4 4×2= 8 8×2= 16 16×2= 32 32×2= 64 Question 3. Rule: Skip-count by 6. First term: 12 12 , _____ , _____ , _____ , _____ , ……. Answer: 12,18,24,30,36,… Explanation: 12 12+6= 18 18+6= 24 24+6= 30 30+6= 36 Use the rule to write the first twelve numbers in the pattern. Describe another pattern in the numbers. Question 4. Rule: Add 7. First term: 3 Answer: 3 3+7= 10 10+7= 17 17+7= 34 34+7= 41 41+7= 48 48+7= 55 55+7= 62 62+7= 69 69+7= 76 76+7= 83 83+7= 90. Explanation: Added 7 to the given term. Question 5. 5. Rule: Add 2, add 1. First term: 12 Answer: 12,14,15,17,19,21,22,24,25,27,28,30,31. Explanation: 12 12+2= 14 14+1= 15 15+2= 17 17+1= 19 19+2= 21 21+1= 22 22+2= 24 24+1= 25 25+2= 27 27+1= 28 28+2= 30 30+1= 31 Question 6. Use Patterns Marcie likes to collect stickers, but she also likes to give them away. Currently, Marcie has 87 stickers in her collection. If Marcie collects 5 new stickers each week and gives away 3 stickers each week, how many stickers will Marcie have in her collection after 5 weeks? _______ stickers Answer: 97 stickers Explanation: Marcie has 87 stickers, in 1st week she collects 5 stickers and gives away 3 stickers, that means 87+5-3= 89 2nd week 89+5-3= 91 3rd week 91+5-3= 93 4th week 93+5-3= 95 5th week 95+5-3= 97. ### Page No. 314 Question 7. John is saving for his trip to see the Alamo. He started with$24 in his savings account. Every week he earns $15 for baby-sitting. Out of that, he spends$8 and saves the rest. John uses the rule add 7 to find out how much money he has at the end of each week. What are the first 8 numbers in the pattern? Answer: $24,$31, $38,$45, $52,$59, $66,$73. Explanation: 24 24+7= 31 31+7= 38 38+7= 45 45+7= 52 52+7= 59 59+7= 66 66+7= 73. Question 8. Draw a check under the column that describes the number. Pose a Problem Question 9. An activity at the Math Fair shows two charts. Use at least two of the numbers and an operation from the charts to write a pattern problem. Include the first five terms of your pattern in the solution to your problem. Pose a problem. Solve your problem. Describe other patterns in the terms you wrote. 10-6= 4 Subtraction. 5×2= 10 Multiplication. ### Common Core – Number Patterns – Page No. 315 Number Patterns Use the rule to write the first twelve numbers in the pattern. Describe another pattern in the numbers. Question 1. Rule: Add 8. First term: 5 Question 2. Rule: Subtract 7. First term: 95 Explanation: 95 95-7= 88 88-7= 81 81-7= 74 74-7= 67 67-7= 60 60-7= 53 53-7= 46 46-7= 39 39-7= 32 32-7= 25 25-7= 18 18-7= 11 Question 3. Rule: Add 15, subtract 10. First term: 4 Explanation: 4 4+15= 19 19-10= 9 9+15= 24 24-10= 14 14+15= 29 29-10= 19 19+15= 34 34-10= 24 24+15= 39 39-10=29 29+15=44 44-10=34 Question 4. Rule: Add 1, multiply by 2. First term: 2 Explanation: 2 2+1= 2 2×2= 4 4+1= 5 5×2= 10 10+1= 11 11×2= 22 22+1= 23 23×2= 46 46+1= 47 47×2= 94 94+1= 95 95×2= 190. Problem Solving Question 5. Barb is making a bead necklace. She strings 1 white bead, then 3 blue beads, then 1 white bead, and so on. Write the numbers for the first eight beads that are white. What is the rule for the pattern? Explanation: 1 1+4= 5 5+4= 9 9+4= 13 13+4= 17 17+4= 21 21+4= 25 25+4=29 Question 6. An artist is arranging tiles in rows to decorate a wall. Each new row has 2 fewer tiles than the row below it. If the first row has 23 tiles, how many tiles will be in the seventh row? Explanation: 23 23-2= 21 21-2= 19 19-2= 17 17-2= 15 15-2= 13 13-2= 11 ### Common Core – Number Patterns – Page No. 316 Lesson Check Question 1. The rule for a pattern is add 6. The first term is 5. Which of the following numbers is a term in the pattern? Options: a. 6 b. 12 c. 17 d. 22 Explanation: 5 5+6= 11 11+6= 17 Question 2. What are the next two terms in the pattern 3, 6, 5, 10, 9, 18, 17, . . .? Options: a. 16, 15 b. 30, 31 c. 33, 34 d. 34, 33 Explanation: 3 3×2= 6 6-1= 5 5×2= 10 10-1= 9 9×2= 18 18-1= 17 17×2= 34 34-1= 33 Spiral Review Question 3. To win a game, Roger needs to score 2,000 points. So far, he has scored 837 points. How many more points does Roger need to score? Options: a. 1,163 points b. 1,173 points c. 1,237 points d. 2,837 points Explanation: Roger has scored 837 points, He needs to score 2000 points to win, So to know how much more points do Roger needs we need to subtract i.e 2,000-837= 1,163. Question 4. Sue wants to use mental math to find 7 × 53. Which expression could she use? Options: a. (7 × 5) + 3 b. (7 × 5) + (7 × 3) c. (7 × 50) + 3 d. (7 × 50) + (7 × 3) Explanation: Distributive property means if we multiply a sum by a number is the same as multiplying each addend by the number and adding the products. Question 5. Pat listed numbers that all have 15 as a multiple. Which of the following could be Pat’s list? Options: a. 1, 3, 5, 15 b. 1, 5, 10, 15 c. 1, 15, 30, 45 d. 15, 115, 215 Explanation: 1×15= 15 3×5= 15 5×3= 15 15×1= 15 Question 6. Which is a true statement about 7 and 14? Options: a. 7 is a multiple of 14. b. 14 is a factor of 7. c. 14 is a common multiple of 7 and 14. d. 21 is a common multiple of 7 and 14. Explanation: 7×2=14 14×1=14 ### Review/Test – Page No. 317 Question 1. List all the factors of the number. 14: ______ , ______ , ______ , ______ Explanation: Factors are the numbers that divide the original number completely. Here, we can see the numbers which give the result as 14 when multiplied together. So the factors of 14 are 1,2,7,14. Question 2. Select the numbers that have a factor of 5. Mark all that apply. Options: a. 15 b. 3 c. 45 d. 5 e. 50 f. 31 Explanation: Factors are the numbers that divide the original number completely. Question 3. Jackson was making a poster for his room. He arranged 50 trading cards in the shape of a rectangle on the poster. For 3a–3e, choose Yes or No to tell whether a possible arrangement of cards is shown. a. 5 rows of 10 cards i. yes ii. no Explanation: 5 rows of 10 cards that means 5×10= 50. So the answer is Yes. Question 3. b. 7 rows of 8 cards i. yes ii. no Explanation: 7×8= 56, There will be extra cards. So the answer is No. Question 3. c. 25 rows of 2 cards i. yes ii. no Explanation: 25×2=50. So the answer is Yes Question 3. d. 50 rows of 1 card i. yes ii. no Explanation: 50×1=50. So the answer is Yes. Question 3. e. 45 rows of 5 cards i. yes ii. no Explanation: 45×5= 225. Which is not equal to 50. So the answer No. Question 4. List all the factor pairs in the table. 1×48= 48 1,48 2×24= 48 2,24 3×16= 48 3,16 4×12= 48 4,12 6×8= 48 6,8 Explanation: Factors are the numbers that divide the original number completely. Here, we can see the numbers which give the result as 30 when multiplied together. ### Review/Test – Page No. 318 Question 5. Classify the numbers. Some numbers may belong in more than one box. Divisible by 5 and 9: 90 Divisible by 6 and 9: 54,72,90 Divisible by 2 and 6: 54,72,84,90,96 Question 6. James works in a flower shop. He will put 36 tulips in vases for a wedding. He must use the same number of tulips in each vase. The number of tulips in each vase must be greater than 1 and less than 10. How many tulips could be in each vase? Answer: 2, 3, 4, 6, 9. Explanation: Question 7. Brady has a card collection with 64 basketball cards, 32 football cards, and 24 baseball cards. He wants to arrange the cards in equal piles, with only one type of card in each pile. How many cards can he put in each pile? Mark all that apply. Options: a. 1 b. 2 c. 3 d. 4 e. 8 f. 32 Explanation: Factors of 64 are 1,2,4,8,16,32,64. Factors of 32 are 1,2,4,8,16,32. Factors of 24 are 1,2,3,4,6,8,12,24. Common factors are 1,2,4,8. Question 8. The Garden Club is designing a garden with 24 cosmos, 32 pansies, and 36 marigolds. Each row will have only one type of plant in each row. Ben says he can put 6 plants in each row. He listed the common factors of 24, 32, and 36 below to support his reasoning. 24: 1, 2, 3, 4, 6, 8, 12, 24 32: 1, 2, 4, 6, 9, 16, 32 36: 1, 2, 3, 4, 6, 8, 12, 18, 36 Is he correct? Explain your answer. If his reasoning is incorrect, explain how he should have found the answer. Answer: No. He can put 1,2,4 plants in each row Explanation: The factors of 32 are incorrect. He listed as 6 and 9 are factors of 32 which is wrong and 8 is not a factor of 36. Factors of 32 are 1,2,4,8,16,32. Factors of 36 are 1,2,3,4,6,9,18,36. Common factors of 24,32 and 36 are 1,2,4. So he can put 1,2,4 plants in each row. ### Review/Test – Page No. 319 Question 9. Part A The museum is hosting a show for July that features the oil paintings by different artists. All artists show the same number of paintings and each will show more than 1 painting. How many artists could be featured in the show? Explanation: Factors of 30 are 1,2,3,5,6,10,15,30. Question 9. Part B The museum wants to display all the art pieces in rows. Each row has the same number of pieces and the same type of pieces. How many pieces could be in each row? Explain how you found your answer. Explanation: Factors of 30 are 1,2,3,5,6,10,15,30. Factors of 24 are 1,2,3,4,6,8,12,24 Factors of 21 are 1,3,7,21 Common Factors are 1,3 Question 10. Charles was skip counting at the Math Club meeting. He started to count by 8s. He said 8, 16, 24, 32, 40, and 48. What number will he say next? Explanation: Multiples of 8 8×1= 8 8×2= 16 8×3= 24 8×4= 32 8×5= 40 8×6= 48 8×7= 56. Question 11. Jill wrote the number 40. If her rule is add 7, what is the fourth number in Jill’s pattern? How can you check your answer? Explanation: 40 40+7= 47 47+7= 54 54+7= 61, And the fourth number is 61 ### Review/Test – Page No. 320 Question 12. For numbers 12a–12e, select True or False for each statement. a. The number 36 is a multiple of 9. i. True ii. False Explanation: 9×4= 36. Question 12. b. The number 3 is a multiple of 9. i. True ii. False Explanation: Multiples of 9 are 9,18,27,36,45,54,63, etc. Question 12. c. The number 54 is a multiple of 9. i. True ii. False Explanation: 9×6= 54 Question 12. d. The number 3 is a factor of 9. i. True ii. False Explanation: Factors of 9 are 1,3,9. Question 12. e. The number 27 is a factor of 9. i. True ii. False Explanation: Factors of 27 are 1,3,9,27 Question 13. What multiple of 7 is also a factor of 7? Explanation: 7 is both multiple and a factor of 7. Question 14. Manny makes dinner using 1 box of pasta and 1 jar of sauce. If pasta is sold in packages of 6 boxes and sauce is sold in packages of 3 jars, what is the least number of dinners that Manny can make without any supplies leftover? Manny has 1 box of pasta and 1 jar of sauce and he sold in a package of 6 boxes of pasta and 3 jars of sauce. Let the packages of pasta be 6P and jars of sauce be 3s. As Manny sold without any leftover 3S=6P, If we take 1 package of pasta then P=1, And 3S=6×1, where S= 6/3 which is equal to 2, So for every package of pasta, we need 2 packages of sauce, So the minimum purchase is 2 packages of sauce and 1 package of pasta. Since pasta packages are 6 boxes the minimum number of meals is 6. Question 15. Serena has several packages of raisins. Each package contains 3 boxes of raisins. Which could be the number of boxes of raisins Serena has? Mark all that apply. Options: a. 9 b. 18 c. 23 d. 27 e. 32 Explanation: Factors of 3. Question 16. Choose the words that make the sentence true. The number 7 is because it has two factors. The number 7 is _________ because it has _________ two factors. Answer: The number 7 is a prime number because it has exactly two factors. Explanation: A Prime number is a number that is divisible 1 and itself. ### Review/Test – Page No. 321 Question 17. Winnie wrote the following riddle: I am a number between 60 and 100. My ones digit is two less than my tens digit. I am a prime number. Part A What number does Winnie’s riddle describe? Explain. Explanation: 97 is the number which ones digit is two less than tens digit. Question 17. Part B Winnie’s friend Marco guessed that her riddle was about the number 79. Why can’t 79 be the answer to Winnie’s riddle? Explain. Answer: It’s wrong because in Winnie’s riddle ones digit is two less than tens digit. But in 79 ones digit is two greater than tens digit. Explanation: In 79 ones digit is two greater than tens digit. So Marco guess was incorrect. Question 18. Classify the numbers as prime or composite. Answer: Prime numbers are 37, 71 Composite numbers are 65, 82 Explanation: A Composite number is a number that has more than two factors. A Prime number is a number that is divisible 1 and itself. Question 19. Erica knits 18 squares on Monday. She knits 7 more squares each day from Tuesday through Thursday. How many squares does Erica knit on Friday? Explanation: 18 18+7= 25 25+7= 32 32+7= 39 39+7= 46. Question 20. Use the rule to write the first five terms of the pattern. First term: 11 ______ ______ ______ ______ Explanation: 11 11+10= 21 21-5= 16 16+10= 26 26-5= 21 ### Review/Test – Page No. 322 Question 21. Elina had 10 tiles to arrange in a rectangular design. She drew a model of the rectangles she could make with the ten tiles. Part A How does Elina’s drawing show that the number 10 is a composite number? Answer: 10 is a composite number because it has more than two factors. Explanation: The number which has more than two factors is called composite numbers. Question 21. Part B Suppose Elina used 15 tiles to make the rectangular design. How many different rectangles could she make with the 15 tiles? Write a list or draw a picture to show the number and dimensions of the rectangles she could make. Explanation: one by 15 tiles and second by 3tiles in a row. Question 21. Part Cs Elina’s friend Luke said that he could make more rectangles with 24 tiles than with Elina’s 10 tiles. Do you agree with Luke? Explain. Explanation: As 24 has more factors than 10. ### Page No. 329 Use the model to write an equivalent fraction. Question 1. $$\frac{1}{5}$$ = $$\frac{□}{□}$$ Explanation: From the above figure we can see that there are 5 equal parts and in that 1 part is shaded. So the fraction of the shaded part is 1/5. Question 2. $$\frac{2}{3}$$ = $$\frac{□}{□}$$ Explanation: From the above figure we can see that there are 3 equal parts and in that 2 part is shaded. So the fraction of the shaded part is 2/3. Tell whether the fractions are equivalent. Write = or ≠. Question 3. $$\frac{1}{6}$$ _____ $$\frac{2}{12}$$ Explanation: The denominator and numerators are equal for both the fractions. So 1/6=2/12 are equal. Question 4. $$\frac{2}{5}$$ _____ $$\frac{6}{10}$$ Explanation: The denominator and numerators are not equal for both the fractions. Question 5. $$\frac{4}{12}$$ _____ $$\frac{1}{3}$$ Explanation: The denominator and numerators are equal for both the fractions. Question 6. $$\frac{5}{8}$$ _____ $$\frac{2}{4}$$ Explanation: The denominator and numerators are not equal for both the fractions. Question 7. $$\frac{5}{6}$$ _____ $$\frac{10}{12}$$ Explanation: The denominator and numerators are equal for both the fractions. Question 8. $$\frac{1}{2}$$ _____ $$\frac{5}{10}$$ Explanation: The denominator and numerators are equal for both the fractions. Question 9. Manny used 8 tenth-size parts to model $$\frac{8}{10}$$. Ana used fewer parts to model an equivalent fraction. How does the size of a part in Ana’s model compare to the size of a tenth-size part? What size part did Ana use? Answer: Larger than a tenth-size part. And she used the fifth-size part. Explanation: A part of Ana’s model is larger than a tenth-size part. And she used the fifth-size part. Question 10. Use a Concrete Model How many eighth-size parts do you need to model $$\frac{3}{4}$$? Explain. Explanation: Let the parts be X, then 1/8×X=3/4. By calculation, we will get X as 6. So we need 6 parts. ### Page No. 330 Question 11. Ben brought two pizzas to a party. He says that since 14_ of each pizza is left, the same amount of each pizza is left. What is his error? Answer: As the size of pizzas is not the same, So 1/4 of leftover pizza is not equal to another. Question 12. For numbers 12a–12d, tell whether the fractions are equivalent by selecting the correct symbol. a. $$\frac{3}{15}$$ _____ $$\frac{1}{6}$$ Question 12. b. $$\frac{3}{4}$$ _____ $$\frac{16}{20}$$ c. $$\frac{2}{3}$$ _____ $$\frac{8}{12}$$ d. $$\frac{4}{5}$$ _____ $$\frac{8}{10}$$
Courses Courses for Kids Free study material Offline Centres More Store # In the figure, ABC is a right triangle, right - angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5cm and AD = $\dfrac{{3\sqrt 5 }}{2}$ cm, find the length of CE. Last updated date: 20th Jun 2024 Total views: 449.4k Views today: 5.49k Verified 449.4k+ views Hint: Use Pythagoras theorem on the various right angle triangles to get the desired result. From the figure we can see that ABC is a right triangle, right angled at B and similarly $\vartriangle {\text{ABD}}$ is also a right triangle. Using Pythagoras theorem in $\vartriangle {\text{ABD}}$ ${\text{A}}{{\text{B}}^2} + B{D^2} = A{D^2} \\$ Now BC = 2BD as AD is a median (median divides a side in two equal parts) ${\text{A}}{{\text{B}}^2} + \dfrac{{B{C^2}}}{{{2^2}}} = A{D^2} \\$ So, ${\text{ 4A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2} = 4{\text{A}}{{\text{D}}^2}...........\left\{ 1 \right\} \\$ Now $\vartriangle {\text{BEC}}$ is also a right angled triangle so applying Pythagoras theorem ${\text{B}}{{\text{E}}^2} + {\text{B}}{{\text{C}}^2} = C{E^2} \\$ Now AB = 2BE as EC is a median (median divides a side in two equal parts) $\dfrac{{{\text{A}}{{\text{B}}^2}}}{{{2^2}}} + B{C^2} = C{E^2} \\$ So, ${\text{ 4B}}{C^2} + A{B^2} = 4C{E^2}..............\left\{ 2 \right\} \\$ Adding equation (1) and (2), we get ${\text{5A}}{{\text{B}}^2} + 5B{C^2} = 4\left( {A{D^2} + C{E^2}} \right) \\$ But ${\text{ A}}{{\text{B}}^2} + B{C^2} = A{C^2}$, so now we get, ${\text{5A}}{{\text{C}}^2} = 4\left( {A{D^2} + C{E^2}} \right) \\$ $AC = 5$ cm and $AD = \dfrac{{3\sqrt 5 }}{2}$ cm Substituting the values we get, ${\text{125 = 4}}\left( {\dfrac{{45}}{4} + C{E^2}} \right) \\$ On solving we get, $CE = 2\sqrt 5$ cm Note: Always remember that a median divides a side in two equal parts so while solving such questions collaborate this concept with pythagoras theorem to reach a solution.
# Operations With Rational Numbers Worksheet Kuta A Rational Figures Worksheet may help your kids be more knowledgeable about the ideas powering this proportion of integers. With this worksheet, individuals are able to fix 12 different issues associated with rational expression. They will likely learn to increase several numbers, group of people them in couples, and determine their products and services. They will likely also training simplifying logical expression. When they have mastered these methods, this worksheet will certainly be a valuable instrument for furthering their research. Operations With Rational Numbers Worksheet Kuta. ## Logical Numbers certainly are a percentage of integers There are two types of figures: rational and irrational. Rational numbers are understood to be entire amounts, in contrast to irrational numbers tend not to recurring, and get an infinite number of numbers. Irrational amounts are no-no, no-terminating decimals, and square origins that are not ideal squares. These types of numbers are not used often in everyday life, but they are often used in math applications. To determine a logical variety, you must know what a reasonable number is. An integer is actually a entire quantity, plus a realistic variety is actually a rate of two integers. The proportion of two integers is the number on the top divided up through the quantity on the bottom. For example, if two integers are two and five, this would be an integer. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction. ## They can be manufactured in a portion A reasonable number carries a numerator and denominator which are not absolutely nothing. Because of this they could be expressed as being a small fraction. Together with their integer numerators and denominators, reasonable amounts can also have a negative importance. The adverse worth ought to be located left of and its particular total value is its length from absolutely no. To easily simplify this instance, we are going to claim that .0333333 is really a small fraction that may be created like a 1/3. Along with unfavorable integers, a reasonable amount may also be manufactured right into a small fraction. For instance, /18,572 is actually a logical number, although -1/ will not be. Any small percentage comprised of integers is rational, as long as the denominator fails to consist of a and can be written as an integer. Likewise, a decimal that leads to a position is another rational variety. ## They make sensation Despite their label, rational numbers don’t make a lot perception. In math, these are one organizations with a exclusive span about the variety series. Which means that when we add up something, we could purchase the dimensions by its ratio to its authentic number. This retains correct regardless if you will find endless reasonable amounts involving two specific numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer. In real life, if we want to know the length of a string of pearls, we can use a rational number. To find the time period of a pearl, for example, we might count up its size. Just one pearl is twenty kgs, which is actually a reasonable quantity. Moreover, a pound’s excess weight equals 15 kgs. As a result, we should certainly separate a lb by ten, with out worry about the length of one particular pearl. ## They could be expressed as a decimal You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal quantity could be published like a numerous of two integers, so 4x 5 is equal to 8. The same difficulty necessitates the repeated portion 2/1, and either side should be divided up by 99 to find the appropriate response. But how do you create the transformation? Here are some examples. A logical variety will also be developed in great shape, including fractions plus a decimal. One way to represent a rational quantity inside a decimal is always to break down it into its fractional comparable. You can find three ways to separate a reasonable amount, and all these approaches produces its decimal equivalent. One of those techniques is to break down it into its fractional counterpart, and that’s what’s referred to as a terminating decimal.
## Lesson 7: Quadratic Equations (III) February 11, 2011 An equation is a quadratic equation in one variable (or unknown) when: • It has only one variable or unknown • The unknown is squared x2 at least once in the equation. For instance: 3x23x=x–1. Standard form If we bring all the terms over to the RHS (right hand side) and the LHS (left hand side) is equal to 0 we get: 3x2 – 4x + 1 = 0 which is the form we should always use to express quadratic equations to be able to solve them? Any quadratic equation in one variable can be expressed using its standard form: ax2+bx+c = 0 where a, b and c are known numbers and a≠0. NOTES: 1) If a quadratic equation has a = 0, then the equation is a linear equation; this is a first degree equation 2) If you look at the LHS you find a second degree polynomial in one variable. Because of it, these equations are also named second degree polynomial equations in one variable or second degree equations for shorting (it isn’t very precise). 3) In many cases once the equation has this form it can be simplified which is very useful. We saw that any quadratic equation can be expressed in its general form ax2+bx+c = 0 where a, b and c are known numbers and a≠0. We name incomplete quadratic equations to that quadratic equations that have or . They have the forms: ax2+c = 0 where b = 0 ax2+bx = 0 where c = 0. There are several methods you can use to solve a quadratic equation: • Factoring • Completing the Square • Graphing This course, we will practice the Quadratic Formula and a very elemental case of factoring for incomplete quadratic equations. First of all, we are going to study the easy cases, when b = 0 or c = 0. ### 7.1. Solving quadratic equations of the form ax2+c = 0 where b = 0 In this case you have to follow the same method as in linear equations: • Transpose the number to the LHS ax2 = c • Transpose the coefficient of the quadratic term to the RHS x2 = c/a • Isolate the unknown by squaring the RHS. x = ±√¯c/a Take into account that in a square root: • If the radicand is positive there are two roots of the same absolute value and opposite signs. • If the radicand is zero there one only root, zero. • If the radicand is negative there are no real roots. When solving quadratic equations of the form ax2+c = 0 where b = 0 • If –c/a > 0 , there are two solutions x1 = +√¯c/a and x2 = √¯c/a • If –c/a = 0 , there are two solutions x1 = x2 =0 • If –c/a < 0 , there are no solutions ### 7.2. Solving quadratic equations of the form ax2+bx = 0 where c = 0. In this case you have to follow the method of factoring the equation: In general, if we are solving an equation of the form • Factor out the common factor . x·(ax + b) = 0 • Apply the following property: “If the product of two numbers is zero then, one of the factors is zero”. x·(ax + b) = 0 → x = 0 or ax + b = 0 • Solve the resulting equations. x = 0 and ax + b = 0 → x= –b/a When solving quadratic equations of the form ax2+bx = 0 where c = 0.the solutions are x = 0 and x= –b/a ### 7.3. Solving equations in the standard form The quadratic formula uses the “a“, “b“, and “c” from “ax2 + bx + c“, where “a“, “b“, and “c” are just numbers; they are the “numerical coefficients”. The solutions of an equation of the form can be obtained by applying the quadratic formula: This is there are two solutions corresponding to each sign preceding the square root. We will denote them x1 and x2 . Practise plugging into the formula on this web page. Warning: Don ‘t panic if you find “complex numbers” roots with i. These number belong to a set of number you don’t know. If the radicand is negative there are no real solution. This is enough for you. The Discriminant Notice the expression under the radical b2 – 4ac in the quadratic formula is called the discriminant. From this number we can determine the nature of the solutions of a quadratic equation. Three possible cases: 1. Δ = b2 – 4ac = 0 Exactly one real number solution exists. 2. Δ = b2 – 4ac > 0 Two real number solutions exist. 3. Δ = b2 – 4ac < 0 There isn’t any real solution. Remark. The argument of the square root, the expression b2 – 4ac, is called the “discriminant” because, by using its value, you can discriminate between (tell the differences between) the number of solutions of a quadratic equation. No real number has a negative square. Practise finding the of a quadratic equation and deducing from it the number of solutions of the equation in this web page 1. When you are using the quadratic formula you’re just plugging into a formula. There are no “steps” to remember, and there are fewer opportunities for mistakes, but if you have an incomplete equations use the other techniques, are simpler. You can also use the formula, of course! 2. When using the formula, make sure you are careful not to omit the “±” sign, and be careful with the fraction line (don’t draw it as being only under the square root; it’s under the initial “–b” part, too). And, though many of your quadratic equations will start with “x2” so a = 1, don’t forget that the denominator of the Formula is “2a“, not just “2”; that is, when the leading term is something like “3x2“, you will need to remember to put the “a = 3″ value in the denominator. Finally here you are a video on how to use the discriminant to find out the number of solutions of a quadratic equation.

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