Split (1)

train · ~16.8M rows (showing the first 4.97M)

text stringlengths 22 1.01M
# First-order differential equation 1. Jun 9, 2005 "A ball with mass 0.15 kg is thrown upward with initial velocity 20 m/s from the roof of a building 30 m heigh. There is a force due to air resistance of $$\left\| v \right\|/30$$ where the velocity $$v$$ is measured in m/s. Find the time that the ball hits the ground." (Answer: 5.129 s) I split the problem into two parts---namely, the way up and down. From the former, I get the time for the ball to reach the maximum height. Then, I use the latter to find the answer. If we measure $$x$$ positively upward from the ground, then $$m\frac{dv}{dt}=-mg-\frac{v}{30}, \qquad v(0) = v_0, \qquad x(0)=x_0$$ $$\frac{dv}{dt} + \frac{v}{30m}=-g$$ The Method of Integrating Factors gives $$\mu = \exp \left( \frac{1}{30m} \int dt \right) =\exp \left( \frac{t}{30m} \right)$$ $$v(t)=\exp \left( -\frac{t}{30m} \right) \int -g\exp \left( \frac{t}{30m} \right) \: dt$$ $$v(t)=\exp \left( -\frac{t}{30m} \right) \left[ -g(30m) \exp \left( \frac{t}{30m} \right) + \mathrm{C} \right]$$ $$v(t)=-30gm + \mathrm{C} \exp \left( -\frac{t}{30m} \right)$$ Next, we apply the initial condition in order to find the constant. $$v(0) = v_0 \Rightarrow \mathrm{C} = v_0 + 30 gm$$ Thus, we obtain $$v(t) = -30gm + \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right)$$ At the maximum height, we have $$\frac{dx}{dt}=v=0$$ which gives $$t_{\mbox{up}} = 30m \ln \left( \frac{v_0}{30gm} + 1 \right)$$ The way down is described as follows: $$m\frac{dv}{dt}=-mg+\frac{v}{30}, \qquad v\left( t_{\mbox{up}} \right) = 0, \qquad x\left( t_{\mbox{up}} \right) = x_{\mbox{max}}$$ $$\frac{dv}{dt} - \frac{v}{30m}=-g$$ The Method of Integrating Factors gives $$\mu = \exp \left( -\frac{1}{30m} \int dt \right) =\exp \left( -\frac{t}{30m} \right)$$ $$v(t)=\exp \left( \frac{t}{30m} \right) \int -g \exp \left( -\frac{t}{30m} \right) \: dt$$ $$v(t)=\exp \left( \frac{t}{30m} \right) \left[ -g (-30m) \exp \left( -\frac{t}{30m} \right) + \mathrm{C} \right]$$ $$v(t)= 30mg + \mathrm{C}\exp \left( \frac{t}{30m} \right)$$ Next, we apply the initial condition in order to find the constant. $$v\left( t_{\mbox{up}} \right) = 0 \Rightarrow \mathrm{C} = -\frac{(30mg) ^2}{v_0 + 30mg}$$ Thus, we obtain $$v(t)= 30mg -\frac{(30mg) ^2}{v_0 + 30mg} \exp \left( \frac{t}{30m} \right)$$ and so $$\int v(t) \: dt = 30mg\int dt -\frac{(30mg) ^2}{v_0 + 30mg} \int \exp \left( \frac{t}{30m} \right) \: dt$$ $$x(t)= 30mgt -\frac{(30m) ^3 g^2}{v_0 + 30mg} \exp \left( \frac{t}{30m} \right) + \mathrm{C}$$ Next, we apply the initial condition in order to find the constant. $$x\left( t_{\mbox{up}} \right) = x_{\mbox{max}} \Rightarrow \mathrm{C} = x_{\mbox{max}} + (30m)^2 g - (30m)^2 g \ln \left( \frac{v_0}{30mg} + 1 \right)$$ Thus, we obtain $$x(t)= 30mgt -\frac{(30m) ^3 g^2}{v_0 + 30mg} \exp \left( \frac{t}{30m} \right) + x_{\mbox{max}} + (30m)^2 g - (30m)^2 g \ln \left( \frac{v_0}{30mg} + 1 \right)$$ When the ball hits the ground, $$x(t)=0$$. Solving for $$t$$ gives $$t \approx 4.429 \mbox{ s}$$. The result is clearly wrong, but there is something even more weird than that. Using $$x(t)$$ (way up) gives the right answer! In other words $$v(t) = -30gm + \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right)$$ gives $$\int v(t) \: dt = -30gm \int dt + \left( v_0 + 30 gm \right) \int \exp \left( -\frac{t}{30m} \right) \: dt$$ $$x(t) = -30gmt -30m \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right) + \mathrm{C}$$ Next, we apply the initial condition in order to find the constant. $$x(0) = x_0 \Rightarrow \mathrm{C} = x_0 + 30m \left( v_0 + 30gm \right)$$ $$x(t) = x_0 -30gmt + 30m \left( v_0 + 30 gm \right) \left[ 1 - \exp \left( -\frac{t}{30m} \right) \right]$$ Again, setting $$x(t)=0$$ and solving for $$t$$ gives $$t \approx 5.129 \mbox{ s}$$. Can anybody please explain me why this is happening? From what I have, it seems that the differential equation that models the ball falling is not necessary, which is strange. Any help is highly appreciated. Last edited: Jun 9, 2005 2. Jun 9, 2005 ### Galileo It's good that you noted the resistance is always opposite to the direction of motion. You actually made it too difficult for yourself. You need only one equation $$ma=-mg-v/30$$ The minus sign in front of the resistance term automatically makes it opposed to the direction of motion. Since you've chosen the positive direction upwards. Now, when the ball rises v is positive so the resistance force is negative (downwards). When the ball falls v is negative and the resistance force is positive (upwards). So you don't need to fuss with signs. You probably mistakenly assumed v to be always positive or something. 3. Jun 9, 2005 Oh, sure! The reason why I posted this question is because the other day I was solving a similar DE where there was a $$v^2$$ term. It's clear to me now. Thanks 4. Sep 6, 2011 ### TheIconoclast Hello, I have just worked this problem and had issue with getting 5.129s as an answer also. The only thing i don't understand is why the differential equation has -mg and not mg. Shouldn't this term be mg if the ball is falling because gravity is speeding it up not slowing it down? Thanks
# Indertiminate Forms I am a PhD student of mathematics. I have complete MS in math from the University of Pakistan and have been writing online since 2020. ## Indeterminate Forms. Introduction: "Indeterminate" refers to a value that is not known. The indeterminate form of a mathematical expression indicates the impossibility to determine the original value even after the limits have been substituted. We are not able to solve several types of mathematical expressions in mathematics. We refer to these expressions as indeterminate forms. When two functions are taken in ratio, the indeterminate form typically arises when both functions approach 0 in the limit. These situations are known as "indeterminate form 0/0." Similar to addition, subtraction, multiplication, and exponential operations, the indeterminate form can also be obtained. Definition: An algebraic combination of functions in an independent variable are commonly evaluated in calculus and other branches of mathematics by substituting these functions with their limits; if the expression obtained after this substitution does not provide enough information to determine the original limit, the expression is called indeterminate form. Types of Indeterminate Form: When given limit is applied on the expression sometimes it does not provide enough information, this expression is referred to indeterminate form. There are seven types of indeterminate forms like, 1. 0 / 0 form (Zero by Zero Form) Zero by Zero form (0 / 0) is obtained when Lim x→ 0 is applied on the fractional factor like x/ x3, 2x / x., √x / x, x/ the resulting expression in each example is 0 / 0 form which is indeterminate form if the limits of the numerator and denominator are substituted. 2. ∞ / ∞ Form (Infinity by Infinity Form): Infinity by infinity form is obtained when Lim x→ ∞ is applied on the quotient factor like, x / x3, 2x / x., √x / x, x/ the resulting expression in each example is /form which is indeterminate form if the limits of the numerator and denominator are substituted. 3. (zero Multiply by Infinity): Zero Multiply by Infinity is obtained when Lim x→ 0 is applied on the factor like, x,2x, 3x, xn and √x, √ xn where n is any real number as well as Lim x→ ∞ is applied on the factor like, x,2x, 3x, xn and √x, √ xn 4. ∞ - ∞ (Infinity minus Infinity): Infinity minus Infinity is obtained when Lim x→ ∞ is applied on the factor like, x – x, 2x – 4x, xm -xn , √ xm -xn where m and n are equal different, and m and n are not equal to zero. 5. 1^∞ (x power infinity): Scroll to Continue x power infinity is obtained when limit Lim x→ ∞ is applied on the factor like, 1^ xx 1^xn 1x+n 1x, where n is any real number. 6. 00 (Zero power Zero) Zero power Zero is obtained when Lim x→ 0 is applied on the factor in some power of zero like, 0x 0 ^ x, 0x+1 , etc.0x 7. 0(Infinity Power Zero): Infinity Power Zero is obtained by applied the limit x→ 0 on the factor of some power of infinity like, ∞xx+n where n is any real number. Examples of Indeterminate Form: Example No.1: Given that: Lim x→ 2 ((x2-4) / (x-2)) Solution: Apply the Lim x→ 2 on the factor (x2-4) / (x-2) it gives, By quotient rule of limit, Lim x→ 2 (x2-4) / Lim x→ 2(x-2) Now according to subtraction rule of limit, Lim x→ (x2- Lim x→ 2(4) / Lim x→ 2 (x)- Lim x→ 2(2) 4-4 \ 2-2 0 / 0 Which is indeterminate form of Lim x→ 2 ((x2-4) / (x-2)) Now we will vanish the indeterminate form by factorization. Lim x→ 2 ((x2-4) / (x-2)) Lim x→ 2 (x+2) (x-2) / Lim x→ 2 (x-2) According to multiplication rule of limit, Lim x→ 2(x+2). Lim x→ 2 (x-2) / Lim x→ 2 (x-2) By cancellation law it gives, Lim x→ 2(x+2) Now apply the limit on the factor, (x+2) it gives, 2+2 = 4 Hence, Lim x→ 2 ((x2-4) / (x-2)) = 4 Example No.2: Given That: Lim x→ ∞ ((x2/ x)) ….. (1) Solution: When apply the limit on equation no.1 it gives indeterminate form of for vanish the indeterminate form, we apply the method of factorization. Lim x→ ∞ ((x2/ x)) = Lim x→ ∞(x(x) / x) By cancellation principle, Lim x→ ∞ ((x2/ x)) = Lim x→ ∞ (x) ⇒ ∞ ∞ is not an indeterminate form. Hence, Lim x→ ∞ ((x2/ x)) = ∞ FAQs 1. What does it mean when a form is indeterminate? An algebraic combination of functions in an independent variable are commonly evaluated in calculus and other branches of mathematics by substituting these functions with their limits; if the expression obtained after this substitution does not provide enough information to determine the original limit, the expression is called indeterminate form. 2. Will when 0 / 0 form (Zero by Zero Form) occur? Zero by Zero form (0 / 0) is obtained when Lim x→ 0 is applied on the fractional factor like x/ x3 x/ x, 2x / x., √x / x, the resulting expression in each example is 0 / 0 form which is indeterminate form if the limits of the numerator and denominator are substituted. 3. Will when ∞ - ∞ (Infinity minus Infinity): Infinity minus Infinity is obtained when Lim x→ ∞ is applied on the factor like, x – x, 2x – 4x, xm -xn , √ xm -xn where m and n are equal different, and m and n are not equal to zero. 4. Give any example of indeterminate form? Given That: Lim x→ ∞ ((x2/ x)) ….. (1) Solution: When apply the limit on equation no.1 it gives indeterminate form of for vanish the indeterminate form, we apply the method of factorization. Lim x→ ∞ ((x2/ x)) = Lim x→ ∞(x(x) / x) By cancellation principle, Lim x→ ∞ ((x2/ x)) = Lim x→ ∞ (x) ⇒ ∞ ∞ is not an indeterminate form. Hence, Lim x→ ∞ ((x2/ x)) = ∞ 5. What is importance of indeterminate form in calculus? When evaluating the limits of function, indeterminate forms are frequently encountered, and limits in turn have a significant impact on mathematics and calculus. For understanding about gradients, derivatives, and many other topics, they play an important part in calculus.
Measures of Variability The rangemean absolute deviation, and interquartile range are all ways to measure the variability for a set of data. • The range measures how spread out the data is. To find the range, subtract the smallest value from the largest value in a set of data. • The mean absolute deviation (MAD) is the average distance of each data value from the mean. • A small MAD means the values do not vary much from the mean. A large MAD means that the values vary greatly from the mean. To find the mean absolute deviation (MAD): 1. Find the mean. 2. Find the distance between each data value and the mean. That is, find the absolute value of the difference between each data value and the mean. 3. Find the average of those differences. Here’s an example of finding MAD for the following data set. 7, 3, 5, 7, 8, 3, 9 1. Find the mean. (7+3+5+7+8+3+9)/7 = 42/7 = 6 2. Find the distance (absolute value) between each data value and the mean \|7 – 6\| = 1 \|3 – 6\| = 3 \|5 – 6\| = 1 \|7 – 6\| = 1 \|8 – 6\| = 2 \|3 – 6\| = 3 \|9 – 6\| = 3 3. Find the average of those absolute value differences. (1+3+1+1+2+3+3)/7 = 14/7 = 2 We’ll get to interquartile range on the next day. Practice Find the range and mean absolute deviation of each data set. Round to the nearest tenth. 1. 30, 21, 18, 19, 23, 24, 26, 32, 30, 22, 12, 15, 21 2. 88, 74, 98, 76, 68, 74, 89, 92 3. 6, 45, 87, 23, 51, 15, 32 4. 11, 7, 14, 2, 5, 13, 3, 6, 10, 3, 8, 4, 8, 4, 7 (source)
# Search by Topic #### Resources tagged with Continued fractions similar to Tuning and Ratio: Filter by: Content type: Stage: Challenge level: ### There are 15 results Broad Topics > Fractions, Decimals, Percentages, Ratio and Proportion > Continued fractions ### Approximations, Euclid's Algorithm & Continued Fractions ##### Stage: 5 This article sets some puzzles and describes how Euclid's algorithm and continued fractions are related. ### Euclid's Algorithm and Musical Intervals ##### Stage: 5 Challenge Level: Use Euclid's algorithm to get a rational approximation to the number of major thirds in an octave. ### Resistance ##### Stage: 5 Challenge Level: Find the equation from which to calculate the resistance of an infinite network of resistances. ### Good Approximations ##### Stage: 5 Challenge Level: Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers. ### Continued Fractions I ##### Stage: 4 and 5 An article introducing continued fractions with some simple puzzles for the reader. ### The Golden Ratio, Fibonacci Numbers and Continued Fractions. ##### Stage: 4 An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions. ### Symmetric Tangles ##### Stage: 4 The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why! ### Golden Mathematics ##### Stage: 5 A voyage of discovery through a sequence of challenges exploring properties of the Golden Ratio and Fibonacci numbers. ### Golden Fractions ##### Stage: 5 Challenge Level: Find the link between a sequence of continued fractions and the ratio of succesive Fibonacci numbers. ### There's a Limit ##### Stage: 4 and 5 Challenge Level: Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely? ### Tangles ##### Stage: 3 and 4 A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead? ### Not Continued Fractions ##### Stage: 4 and 5 Challenge Level: Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers? ### Continued Fractions II ##### Stage: 5 In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)). ### Infinite Continued Fractions ##### Stage: 5 In this article we are going to look at infinite continued fractions - continued fractions that do not terminate. ### Comparing Continued Fractions ##### Stage: 5 Challenge Level: Which of these continued fractions is bigger and why?
## 17Calculus Precalculus - Dependent Systems of Linear Equations - Infinitely Many Solutions ##### 17Calculus Types of Systems of Equations Basics of linear systems with the same number of equations as unknowns - covered on the main linear systems page Linear with fewer independent equations than unknowns - covered on this page Non-Linear - covered on the non-linear system page Consistent System With Infinite Solutions When we have a (consistent) linear system with infinitely many solutions or we have a linear system with fewer equations than unknowns, we have what is known as a dependent system. These two cases are essential the same, since in a consistent system we have the same number of equations as unknowns but one of the equations does not give us any more information than one of the other equations. So we can eliminate the duplicate equation and then solve the system as if we had fewer equations than unknowns. The key to this technique is to know how to describe the answer since we have infinitely many solutions. Many instructors just tell you to stop your work once you realize this but knowing how to describe the answer is a valuable technique. So we show you how to do this here. Let's start with the case where you are solving a consistent linear system and you realize that you have infinitely many solutions. There are a couple of ways to know this based on which technique you are using. They are equivalent but they look a bit different. 1. If you are using substitution or basic elimination, you will add two equations together and get an equation with no variables which is true all the time, like $$3=3$$ or $$0=0$$. 2. If you are using gaussian elimination (also called row reduction), you will get a row with all zeros. In both of these cases, you have eliminated the redundent information and you now have fewer equations than unknowns. Take a few seconds to write out your remaining equations and you will see that the next section can now be applied to your system. Okay, time for the practice problems. Practice Unless otherwise instructed, solve these systems of equations. $$3x-2y+4z=1 \text{ and } x+y-2z=3 \text{ and } 2x-3y+6z=8$$ Problem Statement $$3x-2y+4z=1 \text{ and } x+y-2z=3 \text{ and } 2x-3y+6z=8$$ Solution ### mattemath - 1772 video solution video by mattemath Log in to rate this practice problem and to see it's current rating. $$x+2y-7z=-4 \text{ and } 2x+3y+z=5 \text{ and } 3x+7y-36z=-25$$ Problem Statement $$x+2y-7z=-4 \text{ and } 2x+3y+z=5 \text{ and } 3x+7y-36z=-25$$ Solution ### mattemath - 1773 video solution video by mattemath Log in to rate this practice problem and to see it's current rating. $$y = -2x+1;$$ $$4x+2y=2$$ Problem Statement Solve $$y = -2x+1;$$ $$4x+2y=2$$ Solution ### 4149 video solution Log in to rate this practice problem and to see it's current rating. Fewer Equations Than Unknowns - Non-Square Systems At this point you have fewer equations than unknowns. That means there are infinitely many solutions to your system. The first thing we do is to count the number of variables in the system and subtract the number of independent equations. This will give us the number of variables that we cannot account for. To learn how to handle this, let's watch this video. This guy explains the technique very well while he works a couple of examples. ### Thinkwell - Nonsquare Systems [9min-14secs] video by Thinkwell So the idea is determine the number of variables that we cannot account for, usually just one, and solve the other variables in terms of the unknowns. It may seem strange to replace one variable with another one but this is the standard, so we will go with it. Okay, let's work some practice problems. Practice Unless otherwise instructed, solve these systems of equations. $$x-y+z=-6 \text{ and } 2x+2y-6z=4$$ Problem Statement $$x-y+z=-6 \text{ and } 2x+2y-6z=4$$ Solution ### Thinkwell - 1768 video solution video by Thinkwell Log in to rate this practice problem and to see it's current rating. $$4x-2y+6z=5 \text{ and } 2x-y+3z=2$$ Problem Statement $$4x-2y+6z=5 \text{ and } 2x-y+3z=2$$ Solution ### Thinkwell - 1769 video solution video by Thinkwell Log in to rate this practice problem and to see it's current rating. $$x-y+4z=3 \text{ and } 4x-z=0$$ Problem Statement $$x-y+4z=3 \text{ and } 4x-z=0$$ Solution ### mattemath - 1771 video solution video by mattemath Log in to rate this practice problem and to see it's current rating. $$12x+5y+z=0 \text{ and } 23x+2y-z=0$$ Problem Statement $$12x+5y+z=0 \text{ and } 23x+2y-z=0$$ $$(a/13, -5a/13,a)$$ Problem Statement $$12x+5y+z=0 \text{ and } 23x+2y-z=0$$ Solution ### mattemath - 2130 video solution video by mattemath $$(a/13, -5a/13,a)$$ Log in to rate this practice problem and to see it's current rating. $$9x+5y+9z=3 \text{ and } -45x+10y+27z=-6$$ Problem Statement $$9x+5y+9z=3 \text{ and } -45x+10y+27z=-6$$ Solution ### PatrickJMT - 1770 video solution video by PatrickJMT Log in to rate this practice problem and to see it's current rating. $$10x-6y+2z=0 \text{ and } 19x-5y-z=0$$ Problem Statement $$10x-6y+2z=0 \text{ and } 19x-5y-z=0$$ $$(a/4,3a/4,a)$$ Problem Statement $$10x-6y+2z=0 \text{ and } 19x-5y-z=0$$ Solution ### mattemath - 2131 video solution video by mattemath $$(a/4,3a/4,a)$$ Log in to rate this practice problem and to see it's current rating. $$x + 2y - 3z - 4w = 10;$$ $$x + 3y - 3z - 4w = 15;$$ $$2x + 2y - 6z - 8w = 10$$ Problem Statement Solve $$x + 2y - 3z - 4w = 10;$$ $$x + 3y - 3z - 4w = 15;$$ $$2x + 2y - 6z - 8w = 10$$ Solution ### 4150 video solution Log in to rate this practice problem and to see it's current rating. $$x-y-z = 1;$$ $$-x+2y-3z = -4 ;$$ $$3x-2y-7z = 0$$ Problem Statement Solve $$x-y-z = 1;$$ $$-x+2y-3z = -4 ;$$ $$3x-2y-7z = 0$$ Solution ### 4151 video solution Log in to rate this practice problem and to see it's current rating. $$2x-3y+5z=4;$$ $$x+y+2z=4;$$ $$3x-2y+7z=8$$ Problem Statement Solve $$2x-3y+5z=4;$$ $$x+y+2z=4;$$ $$3x-2y+7z=8$$ Solution ### 4152 video solution Log in to rate this practice problem and to see it's current rating. $$x+3y+2z=4 \text{ and } 2x+7y-z=5$$ Problem Statement $$x+3y+2z=4 \text{ and } 2x+7y-z=5$$ $$(-17z+13, 5z-3,z)$$ Problem Statement $$x+3y+2z=4 \text{ and } 2x+7y-z=5$$ Solution ### Jon Anderson - Learning Algebra - 2132 video solution $$(-17z+13, 5z-3,z)$$ Log in to rate this practice problem and to see it's current rating. When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications. DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.
# Algebra: Converting From One Unit to Another In the above video, astronaut Doug Wheelock explains that traveling in orbit requires a speed of 5 miles per second. In my previous post, we determined that an object needs to travel at approximately 8 km/s, or kilometers per second. How do we convert kilometers to miles to make sure that our calculations agree with those of an astronaut? This is something that you can type into google to get the answer, but a teacher (or an exam) may require that you show your work. So let's get down! # Converting from one distance to another When you convert from one unit to another, you may have to start with a basic conversion and work your way outward. For kilometers to miles, let's start with a basic conversion that everyone was taught in high school: 1 inch is equal to 2.54 centimeters. To properly use this, we need to convert out kilometers into meters and our inches into miles. We can then proceed as follows: • There are 1,000 meters in a kilometer (the prefix "kilo-" means one thousand) • There are 100 centimeters in a meter (the prefix "centi" means one hundred, as in century) • There are 2.54 centimeter in an inch • There are 12 inches in a foot • There are 5,280 feet in a mile We can use these facts to convert our value of 8 km/s by multiplying by each of the ratios given above. Note that we must take care that corresponding units are written diagonally from each other. I color coded the units to emphasize that they must line up diagonally next to each other. This means that some conversion numbers will be in a numerator while others are in the denominator. For example, I need the km in the denominator of the second ratio to cancel out the km in the numerator in the first ratio. # Converting from one time to another We can use the same idea to prove Wheelock's assertion that astronauts see approximately 16 sunrises and sunsets each day. This would mean that the ISS makes a complete revolution around the Earth approximately 16 times a day. Additionally, we need to use the following ratios: • There are 60 seconds in an minute • There are 60 minutes in an hour • There are 24 hours in a day Finally, we need to use a defined value for the distance around the Earth, also known as the circumference. This value is defined as 24,901 miles. So our final ratio is listed below. • There is one revolution around the Earth every 24,091 miles We will apply the same math as earlier, making sure that the numerator and denominators which are diagonal to each other have the same units so that they can cancel out. According to our calculations, we can conclude that orbiting at 5 mi does cause you circle the planet approximately 16 times in one day. # Physics: How To Get In and Out of Orbit On my science blog, I shared my general thoughts on the wonderful movie Hidden Figures. On this post, I want to answer the one question I've received the most - how do you even get something into orbit around the Earth? And how do you leave orbit entirely? As with all of my posts, you can skip the diagrams if you want - the text should give you the general idea. Now on with the show! # Getting to Orbit When we say a satellite is in orbit around the Earth, we mean that the satellite is traveling at the exact speed to counteract the force of gravity pulling it down. This means that the satellite is constantly falling, but its going fast enough so that it stays the same height above the Earth. This results in the satellite traveling in a circular motion. There are two forces acting on a satellite in this case: 1. The centripetal force. For circular motion to occur, this force must point inward toward the center of the circle 2. The gravitational force. This force exists between any two objects in our known galaxy. It is directed between the centers of both objects (in this case, the satellite and the Earth) For our satellite to stay in a circle, both of these forces must be the same. This means that we can set the centripetal force equal to the gravitational force. Solving this equation gives us a velocity of approximately 8 kilometers per second, or 17,682 miles per hour. This is the speed that any satellite (including the International Space Station) has to achieve to be able to stay in orbit. It's also why satellite launches tend to curve right after leaving the launchpad - they need to get to the orbital speed and be parallel to the Earth's surface so they can be pulled into circular obit. # Leaving Orbit What if we wanted to find the speed needed to leave orbit entirely? This is what NASA did during our trips to the moon, and what NASA will do for our upcoming trips to Mars. The calculation is more complicated than getting to orbit, but we can find the speed needed by using a a familiar example of dropping a ball on the ground. First, let's talk about energy. Total energy is expressed in two forms: kinetic energy (energy proportional to the speed of the ball) and potential energy (energy proportional to the distance above the surface). Before you drop a ball, it is motionless, so its kinetic energy is zero. Right before the ball impacts the ground, it is right next to the surface, so its potential energy is zero. The total energy throughout this entire trip must remain the same, so we can set the the potential energy before dropping the ball equal to the kinetic energy right before it hits the ground. Solving this equation gives us a final velocity equal to the square root of 2 * times the acceleration due to gravity * the original height the ball was dropped from. Note that just like with getting to orbit, the speed does not depend on the mass of the ball. The same energy situation applies if we throw a ball upwards, but our initial and final conditions are different. Now, we want to look at the energy of the ball just before it leaves the orbit of the Earth and compare it to the energy of the ball when it is an infinite distance off, when Earth's gravity diminishes to negligible value and the ball has zero energy. The reason we look at an infinite distance is because the force of gravity diminishes as you get farther from something, but never quite gets to zero. Just like in the case of the dropped ball, we can set the potential and kinetic energy at the beginning (when the ball leaves orbit) equal to the energy when the ball stops. We can then solve for the escape velocity, which comes out to be very similar to the speed of the dropped ball! Solving this equation gives us a velocity of 11.2 kilometers per second, or 25,000 miles per hour. # Algebra: Solving Equations Using Cross Multiplication Sometimes, you may have to solve a equation that has the variable in the denominator of a fraction! The best way to handle these problems is by using cross multiplication. This method can be used when two fractions are equal to each other. The key to cross multiplication is right in the name! Multiply the numbers that are diagonally across from each other, and set the products equal to each other.
June 14, 2024 # 58. 2x ^ 2 – 9x ^ 2; 5 – 3x + y + 6 \| Solving Process In the realm of mathematics, 58. 2x ^ 2 – 9x ^ 2; 5 – 3x + y + 6 equations and expressions often appear as enigmatic puzzles, challenging our understanding. However, they are like intricate mazes waiting to be navigated. One such intriguing expression is “58. 2x ^ 2 – 9x ^ 2; 5 – 3x + y + 6.” In this comprehensive article, we will meticulously dissect this expression, exploring each component, and elucidating its significance. KEY POINTS ## Understanding the Expression 58. 2x ^ 2 – 9x ^ 2; 5 – 3x + y + 6 ### Breaking it Down To embark on our journey of comprehension, let’s commence by dissecting the expression step by step: 58. 2x ^ 2 – 9x ^ 2; 5 – 3x + y + 6 ### Terms and Constants • 58: The first element in our expression is the constant “58,” an unchanging numerical value. • 2x^2: Moving forward, we encounter “2x^2,” a term in which the variable ‘x’ is raised to the power of 2. This indicates a quadratic term, suggesting a relationship involving the square of ‘x.’ • – 9x^2: Right alongside, we have “- 9x^2,” another quadratic term but with a negative coefficient. This element contrasts with the previous one, signifying a different relationship with ‘x.’ • 5: Interposed among these terms is the constant “5,” entirely independent of ‘x.’ • – 3x: As our exploration continues, we discover “- 3x.” This term features ‘x’ raised to the power of 1, representing a linear term, implying a linear relationship with ‘x.’ • + y: Further diversifying the expression, we encounter “+ y,” introducing a new variable ‘y’ into our mathematical equation. • + 6: Finally, the expression concludes with “+ 6,” another constant value unrelated to our variables. ### Mathematical Operations In this expression, mathematical operations primarily consist of addition and subtraction. The plus and minus signs play pivotal roles in shaping the outcome. ### The Role of Variables Variables ‘x’ and ‘y’ are instrumental in this expression. They represent unknown values, making the expression dynamic and adaptable to various scenarios. ## Analyzing the Expression Now that we’ve deconstructed the expression, let’s delve into its significance and implications: ### Polynomial Expression The expression “58. 2x ^ 2 – 9x ^ 2; 5 – 3x + y + 6” belongs to the category of polynomial expressions. Such expressions find extensive applications in mathematics and beyond. They serve as fundamental tools for modeling various phenomena and solving real-world problems. ### Variables ‘x’ and ‘y’ The presence of ‘x’ and ‘y’ in the expression signifies that it represents a mathematical function. By substituting specific values for ‘x’ and ‘y,’ one can evaluate the expression for different inputs. This flexibility makes it a versatile tool for solving diverse problems. ### Coefficients The coefficients attached to each term (-9 and 2) are not to be underestimated. They exert a significant influence on the expression’s behavior. They determine whether the expression will rise or fall, and how steeply, when plotted on a graph. ### Solvability While this expression might appear complex, it can be simplified and solved for specific values of ‘x’ and ‘y.’ The process of solving can reveal patterns and relationships between the variables, offering insights into mathematical and real-world scenarios.
# Geometry Semester 2 Model Problems (CA Essential Standards) Part 2 ## Presentation on theme: "Geometry Semester 2 Model Problems (CA Essential Standards) Part 2"— Presentation transcript: Geometry Semester 2 Model Problems (CA Essential Standards) Part 2 PowerPoint by Kenneth C Utt John Kimball High -- TUSD Model Problem #33 44o + 95o + xo = 180o 139o + xo = 180o In the diagram shown, mCBD = 95˚. What is the measure of CDB ? 44o + 95o + xo = 180o 139o + xo = 180o xo = 180o – 139o xo = 41o 95o Model Problem #34 180 – (90 + 31) = MPN 180 – 121 = 59o In the diagram shown, P is a point on ML. What is the measure of the angle marked X? 180 – ( ) = MPN 180 – 121 = 59o 180 – ( ) = LPQ 180 – 114 = 66o 180 – ( ) = PQL 180 – 156 = 24o 180 – 24 = x = 156o 59o 66o 24o Model Problem #35 A’ =(-1,-3) B’ = (-4,-4) C’ = (-3,-2) If ABC is rotated 90˚clockwise about the origin to form A’B’C’, what would be the coordinates of A’? A’ =(-1,-3) B’ = (-4,-4) C’ = (-3,-2) Model Problem #36 J’ = (-2 + 4,-1 – 2) = (2,-3) The coordinates of the vertices of ∆JKL, are J(-2,-1), K(1,3), L(4,-3). If ∆JKL is translated 2 units down and 4 units to the right to create , what are the coordinates of the vertices of ∆J’K’L’? J’ = (-2 + 4,-1 – 2) = (2,-3) K’ = (1 + 4, 3 – 2) = (5, 1) L’ = (4 + 4, -3 – 2) = (8,-5) Model Problem #37 If quadrilateral DEFG is reflected across the y- axis, it would create quadrilateral D’E’F’G’. What are the coordinates of point G’ ? G’ = (6,2) Model Problem #38 (a + b)2 c2 + 4(a ∙ b) ÷ 2 A diagram from a proof of the Pythagorean Theorem is shown. Write an equation that represents the area of the entire square in two ways. On the left side, express the area as the product of the length and the width. On the right, represent the sum of the areas of the triangles and the smaller square. Then use the equation to prove the theorem. (a + b)2 c2 + 4(a ∙ b) ÷ 2 Model Problem #39 112 = 62 + b2 121 = 36 + b2 85 = b2 √85 = b A right triangle’s hypotenuse has length 11. If one leg has length 6, what is the length of the other leg? 112 = 62 + b2 121 = 36 + b2 85 = b2 √85 = b Model Problem #40 √4100 = 10√41 = 2 ∙10 √41 502 + 402 = h2 20√41 In a basketball game, a player from the home team threw the ball from corner C to a player standing at point E. (E is the midpoint of AD). Then the player at point E threw the ball to a player at corner B. If the court was 80 feet long and 50 feet wide, how far was the ball thrown? (Leave in simplified radical form) √4100 = 10√41 = 2 ∙10 √41 20√41 = h2 = h2 √4100 = h Model Problem #41 Sin A = 0.4 = BC/40 (40)(0.4) = BC 16 = BC In the figure shown, sin A  0.4, cos A  0.5, and tan A  What is the approximate length of BC? Sin A = 0.4 = BC/40 (40)(0.4) = BC 16 = BC Model Problem #42 5 4 32 + 42 = h2 9 + 16 = h2 √25 = 5 = h sin A = 4/5 In the figure below, if tan A = 4/3, what are sin A and cos A? 5 4 = h2 = h2 √25 = 5 = h sin A = 4/5 cos A = 3/5 3 Model Problem #43 sin 65o = ?/30’ 0.9 = x/30’ 27’ = x A ladder is leaned against a wall at an angle of 65° to the ground. How far off the ground does the ladder touch the wall? sin 65o = ?/30’ 0.9 = x/30’ 27’ = x Model Problem #44 Triangle JKL is shown in the diagram. Which equation should be used to find the length of LJ? A sin 28o = LJ/54 B sin 28o = 54/LJ C cos 28o = LJ/54 D cos 28o = 54/LJ sin cos tan 70o = 2.75 cos 70o = 0.34 sin 70o = 0.94 Model Problem #45 On a swing set, on engineer used a support bar that was 20 feet long. If the support bar forms a 70° angle to the ground, how far apart will the support bars be at the base? sin 70o = 0.94 cos 70o = 0.34 tan 70o = 2.75 tan 70o = 2.75 cos 70o = 0.34 sin 70o = 0.94 cos 70o = Adj/20 0.34 = Adj/20 6.8 = adj 6.8 ∙ 2 = 13.6 Model Problem #46 In the diagram, mB = 75o and AC = 11.9 in. Which equation could be used to find BC? A x = 11.9(tan 75o) B x = 11.9(sin 75o) C x = 11.9/tan 75o C x = 11.9/sin 75o tan 75o = 11.9 / x tan 75o (x) = 11.9 x = 11.9 / tan 75o Model Problem #47 The right triangle in the diagram has one side with a length of 5√3. What is the length of the side marked x? A 5 B 15 C 5√6 D 10√3 x 5√3 60o Model Problem #48 Arc BD = 124o 124o – 60o = 64o In the circle shown, the measure of BC= 60o, and the measure of ABD = 62o. What is the measure of CD? Arc BD = 124o 124o – 60o = 64o Model Problem #49 4 ∙ 9 = x ∙ 12 36 = 12x 3 = x In the circle shown, DF and CE are chords intersecting at G. If DG = 9, FG = 4, and EG = 12, what is the length of CG? 9 4 ∙ 9 = x ∙ 12 36 = 12x 3 = x 12 x 4 Model Problem #50 1 = ½(104o – 38o) ½(66o) 33o In the circle shown, what is the measure of angle 1? 1 = ½(104o – 38o) ½(66o) 33o Model Problem #51 180o -65o = 115o 180o – (115o + 50o) 180o – 165o 15o LM is tangent to a circle, whose center is C, at point M. MQ is a diameter. If mQNP = 65˚ and mNPM = 50˚, what is mPMR? 180o -65o = 115o 180o – (115o + 50o) 180o – 165o 15o 90o – 15o 75o 15o 115o Model Problem #52 Square = 4 ∙10 = 40 Circle = 10 = 10 10/40 1/4 A square is circumscribed about a circle. What is the ratio of the circumference of the circle to the perimeter of the square? A ¼ B ½ C 2/ D /4 Square = 4 ∙10 = 40 Circle = 10 = 10 10/40 1/4 10 Model Problem #53 10 ∙  ∙ 6 = 60 ins. A cylinder rolls across a table top for 10 complete revolutions. If the diameter of the base is 6 inches, how far did the cylinder travel? (Leave the answer in terms of ). 10 ∙  ∙ 6 = 60 ins. Model Problem #54 3 x 6 = 18 4 x 6 = 24 32 + 42 = h2 25 = h2 The prism shown has a base in the shape of a right triangle. What is the lateral surface area of the prism? 3 x 6 = 18 4 x 6 = 24 = h2 25 = h2 5 x 6 = 30 = 72 cm2 Model Problem #55 4 ( 8 + 6 ) ÷ 2 = 28 = Base 28 x 10 = 280 mm3 What is the volume of the prism shown? 4 ( ) ÷ 2 = 28 = Base 28 x 10 = 280 mm3 Model Problem #56 32 - 22 + 12 9 - 4 + 1 6 or 18.84 A target for a yard game is made with areas that are alternately painted white and gray, as shown in the diagram. The inner circle is white and has a radius of 1 inch. Each of the other three rings has a radius 1 inch more than the ring before it. What is the area of the white portion of the target? 32 - 22 + 12 9 - 4 + 1 6 or 18.84 Model Problem #57 40 – (12 + 12) = 16 16 ÷ 2 = 8 8 ∙ 12 = 96 ft2 A rectangle that is 12 feet wide has a perimeter of 40 feet. What is the area of the rectangle? 40 – ( ) = 16 16 ÷ 2 = 8 8 ∙ 12 = 96 ft2 Model Problem #58 22 + b2 = 42 4 + b2 = 16 b2 = 12 b = √4 ∙ 3 or 2√3 Each side of a triangle measures 4 m. What is the area of the triangle? (Leave the answer in simplified radical form). 22 + b2 = 42 4 + b2 = 16 b2 = 12 b = √4 ∙ 3 or 2√3 4 ∙ 2√3 ÷ 2 4√3 m2 Model Problem #59 10 ÷ 2 = 5 & 8 ÷ 2 = 4 4 ∙ (5 ∙ 4) ÷ 2 = Quadrilateral ABCD is a rhombus. If AC = 10 inches and BD = 8 inches, what is the area of ABCD? 4 10 ÷ 2 = 5 & 8 ÷ 2 = 4 4 ∙ (5 ∙ 4) ÷ 2 = 2 ∙ 20 = 40 in2 5 Model Problem #60 4 (3 + 9) ÷ 2 2 (3 + 9) 2 ∙ 12 = 24 cm2 The diagram shows a trapezoid with a height of 4 cm. What is the area of the trapezoid? 4 (3 + 9) ÷ 2 2 (3 + 9) 2 ∙ 12 = 24 cm2 Model Problem #61 2 x 3 x 3 = 18 4 x 6 x 6 = 144 cm3 The volume of a right rectangular prism is calculated to be 18 cubic centimeters. If the length, the width, and the height of the prism are all doubled, what would be the volume of the new prism? 2 x 3 x 3 = 18 4 x 6 x 6 = 144 cm3 Model Problem #62 V = Bh V = 52 ∙ 4 V = 25 ∙ 4 V = 100 V = 314 cm3 The cylinder shown has a height of 4 cm and the diameter of the base is 10 cm. What is the volume of the cylinder? V = Bh V = 52 ∙ 4 V = 25 ∙ 4 V = 100 V = 314 cm3 Model Problem #63 8 x 8 = 64 4 x 8 x 6 ÷ 2 = + 96 160 cm2 The pyramid shown has a square base that measures 8 cm on each side. The slant height of the pyramid is 6 cm. What is the surface area of the prism? 8 x 8 = 4 x 8 x 6 ÷ 2 = + 96 160 cm2 Model Problem #64 Given: AB. What is the first step in constructing the perpendicular bisector to AB? a. Draw a line segment connecting points E and F. b. From point C, draw an arc that intersects the line at points A and B. c. Draw a line segment connecting points A and B. d. From points A and B, draw equal arcs that intersect at points E and F. Model Problem #65 Darla is constructing an equilateral triangle. Which of the following could be her first step? A B C D Model Problem #66 Marsha is using a straightedge and compass to do the construction shown. Which statement best describes the construction Martha is doing? a. a line through P parallel to line l by constructing two lines perpendicular to the same line b. a line through P parallel to line l by copying an angle c. a line through P perpendicular to line l d. a line through P congruent to line l Model Problem #67 Amina is bisecting an angle. Which of the construction diagrams shown below best represents the beginning of Amina’s construction? A B C D Download ppt "Geometry Semester 2 Model Problems (CA Essential Standards) Part 2" Similar presentations
offers hundreds of practice questions and video explanations. Go there now. # GRE Math Strategies Part III of VI: Illustration The Illustration approach involves choosing workable numbers that substitute into the question rather than using the given answer choices as with the Plug In strategy. Illustration works best when the question provides some numerical information (like percentages) but doesn’t supply you with absolute values. Example #1: At a public housing conference 15 percent of the attendees are male. After the speaker finishes, several female attendees but no males leave the conference and the total number of persons remaining is 60 percent of the original number. The number of female attendees who leave is what fraction of the original number of females attending the conference? (A) (B) (C) (D) (E) Let 100 equal the total attendees. If 15 percent are men, and only females leave after the speaker, this leaves 60 percent of the original total in attendance. Original #: 85 female, 15 male, 100 total After Speaker Finishes: 45 female, 15 male, 60 total The number of female attendees who leave [85 – 45 = 40] is what fraction of the original number of females? . (Note that we backed in to the 45 female number because we were given “remaining is 60 percent of the original number” and “all males remained (15).”) Notes Percents always conveniently add to 100 – for example, if 15 percent of the attendees are male (given), then 85 percent are female (implied). Always use 100 (or a multiple of 100) to illustrate percent problems. The Illustration technique also works when a critical piece of information is not provided, thereby creating the possibility of providing it yourself. Here, the total number of people at the restaurant is an important piece of information. Example #2: Of the total number of people who ate in a restaurant one evening, ordered salads. Of those who ordered salad, had Italian dressing on their salad. If the total number of people who ate in the restaurant that night was , which of the following expressions represents the number of people who did not eat any Italian dressing at the restaurant that evening? (A) (B) (C) (D) (E) A helpful choice for should be a number that can be divided by 3 a couple of times. Let , then people ate salad that evening. Of these 18, had Italian dressing. So people did not have Italian dressing. Now we need to look at the answers to see which fraction of (our 27 people) the number 21 represents. , so (E) is the correct answer. By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.
# Solve the Following Equation and Verify Your Answer: 7 X − 2 5 X − 1 = 7 X + 3 5 X + 4 - Mathematics Sum Solve the following equation and verify your answer: $\frac{7x - 2}{5x - 1} = \frac{7x + 3}{5x + 4}$ #### Solution $\frac{7x - 2}{5x - 1} = \frac{7x + 3}{5x + 4}$ $\text{ or }35 x^2 + 28x - 10x - 8 = 35 x^2 + 15x - 7x - 3 [\text{ After cross multiplication }]$ $\text{ or }35 x^2 - 35 x^2 + 18x - 8x = - 3 + 8$ $\text{ or }10x = 5$ $\text{ or }x = \frac{5}{10} \text{ or } x = \frac{1}{2}$ $\text{ Thus, }x = \frac{1}{2} \text{ is the solution of the given equation . }$ $\text{ Check: }$ $\text{ Substituting }x = \frac{1}{2}\text{ in the given equation, we get: }$ $\text{ L . H . S . }= \frac{7(\frac{1}{2}) - 2}{5(\frac{1}{2}) - 1} = \frac{7 - 4}{5 - 2} = \frac{3}{3} = 1$ $\text{ R . H . S .} = \frac{7(\frac{1}{2}) + 3}{5(\frac{1}{2}) + 4} = \frac{7 + 6}{5 + 8} = \frac{13}{13} = 1$ $\therefore\text{ L . H . S . = R . H . S . for }x = \frac{1}{2}$ Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 8 Maths Chapter 9 Linear Equation in One Variable Exercise 9.3 \| Q 13 \| Page 17
# Resources tagged with: Generalising Filter by: Content type: Age range: Challenge level: ### There are 123 results ##### Age 7 to 14 Challenge Level: I added together some of my neighbours' house numbers. Can you explain the patterns I noticed? ### Three Times Seven ##### Age 11 to 14 Challenge Level: A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why? ### Nim-7 ##### Age 5 to 14 Challenge Level: Can you work out how to win this game of Nim? Does it matter if you go first or second? ### Have You Got It? ##### Age 11 to 14 Challenge Level: Can you explain the strategy for winning this game with any target? ### Maths Trails ##### Age 7 to 14 The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails. ### Triangle Numbers ##### Age 11 to 14 Challenge Level: Take a look at the multiplication square. The first eleven triangle numbers have been identified. Can you see a pattern? Does the pattern continue? ### Summing Consecutive Numbers ##### Age 11 to 14 Challenge Level: 15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers? ### Repeaters ##### Age 11 to 14 Challenge Level: Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. ### Picturing Triangular Numbers ##### Age 11 to 14 Challenge Level: Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers? ### Dicing with Numbers ##### Age 11 to 14 Challenge Level: In how many ways can you arrange three dice side by side on a surface so that the sum of the numbers on each of the four faces (top, bottom, front and back) is equal? ### Go Forth and Generalise ##### Age 11 to 14 Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important. ### More Number Pyramids ##### Age 11 to 14 Challenge Level: When number pyramids have a sequence on the bottom layer, some interesting patterns emerge... ### Partitioning Revisited ##### Age 11 to 14 Challenge Level: We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4 ### Mystic Rose ##### Age 14 to 16 Challenge Level: Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes. ### Handshakes ##### Age 11 to 14 Challenge Level: Can you find an efficient method to work out how many handshakes there would be if hundreds of people met? ### Route to Infinity ##### Age 11 to 14 Challenge Level: Can you describe this route to infinity? Where will the arrows take you next? ### Window Frames ##### Age 5 to 14 Challenge Level: This task encourages you to investigate the number of edging pieces and panes in different sized windows. ##### Age 11 to 14 Challenge Level: List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it? ### Problem Solving, Using and Applying and Functional Mathematics ##### Age 5 to 18 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ##### Age 11 to 14 Challenge Level: Great Granddad is very proud of his telegram from the Queen congratulating him on his hundredth birthday and he has friends who are even older than he is... When was he born? ### Chess ##### Age 11 to 14 Challenge Level: What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board? ### Got It ##### Age 7 to 14 Challenge Level: A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target. ### Jam ##### Age 14 to 16 Challenge Level: A game for 2 players ### Konigsberg Plus ##### Age 11 to 14 Challenge Level: Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges. ### Number Pyramids ##### Age 11 to 14 Challenge Level: Try entering different sets of numbers in the number pyramids. How does the total at the top change? ### Winning Lines ##### Age 7 to 16 An article for teachers and pupils that encourages you to look at the mathematical properties of similar games. ### Tourism ##### Age 11 to 14 Challenge Level: If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable. ### Frogs ##### Age 11 to 14 Challenge Level: How many moves does it take to swap over some red and blue frogs? Do you have a method? ### Take Three from Five ##### Age 14 to 16 Challenge Level: Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him? ### Painted Cube ##### Age 14 to 16 Challenge Level: Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? ### Picturing Square Numbers ##### Age 11 to 14 Challenge Level: Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153? ### Special Sums and Products ##### Age 11 to 14 Challenge Level: Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48. ### Sum Equals Product ##### Age 11 to 14 Challenge Level: The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . . ### Mini-max ##### Age 11 to 14 Challenge Level: Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . . ### AMGM ##### Age 14 to 16 Challenge Level: Can you use the diagram to prove the AM-GM inequality? ### Enclosing Squares ##### Age 11 to 14 Challenge Level: Can you find sets of sloping lines that enclose a square? ### Nim ##### Age 14 to 16 Challenge Level: Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter. ### Cubes Within Cubes Revisited ##### Age 11 to 14 Challenge Level: Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need? ### Is There a Theorem? ##### Age 11 to 14 Challenge Level: Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel? ### Partly Painted Cube ##### Age 14 to 16 Challenge Level: Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use? ### Nim-7 for Two ##### Age 5 to 14 Challenge Level: Nim-7 game for an adult and child. Who will be the one to take the last counter? ### Got it for Two ##### Age 7 to 14 Challenge Level: Got It game for an adult and child. How can you play so that you know you will always win? ### What Numbers Can We Make Now? ##### Age 11 to 14 Challenge Level: Imagine we have four bags containing numbers from a sequence. What numbers can we make now? ### Steps to the Podium ##### Age 7 to 14 Challenge Level: It starts quite simple but great opportunities for number discoveries and patterns! ### Magic Letters ##### Age 11 to 14 Challenge Level: Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws? ### Steel Cables ##### Age 14 to 16 Challenge Level: Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions? ### What Numbers Can We Make? ##### Age 11 to 14 Challenge Level: Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make? ### Christmas Chocolates ##### Age 11 to 14 Challenge Level: How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes? ### Hidden Rectangles ##### Age 11 to 14 Challenge Level: Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard? ### How Much Can We Spend? ##### Age 11 to 14 Challenge Level: A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
# Tutorial on Inequalities This is a tutorial on solving polynomial and rational inequalities. Examples with step by step detailed solutions are presented Example 1: Solve the inequality. 2 - 7 x / 5 > -x + 3 Solution to example 1 multiply both sides by 5 (the LCD) (2 - 7 x / 5)5 > (-x + 3)5 10 - 7 x > -5 x + 15 subtract 10 from both sides. - 7 x > -5 x + 5 add 5 x to both sides. -2 x > 5 divide both sides by -2 and reverse inequality. x < -5 / 2 conclusion: The solution set to the above inequality consists of all real numbers that are less than -5/2. Example 2: Solve the polynomial inequality. -4 x 2 > 4 x - 8 Solution to example 2 Add - 4 x + 8 to both sides to make one side zero -4 x 2 - 4 x + 8 > 0 Divide by the common factor - 4 and reverse the inequality. x 2 + x - 2 < 0 Factor the expression on the left. (x - 1)(x + 2) < 0 make table of signs of x - 1, x + 2 and their product. ``` x -inf -2 1 +inf x - 1 - \| - \| + x + 2 - \| + \| + (x - 1)(x + 2) + \| - \| + ``` conclusion: The solutions of (x - 1)(x + 2) < 0 are the values of x for which the resulting sign is negative.Thus, the solution set of the given inequality is the interval(-2 , 1). Example 3: Solve the rational inequality. 1 /(x + 4) - 2 / (x - 3) > = 0 Solution to example 3 Add the two rational expressions that are on the left (-x - 11) / [ (x + 4)(x - 3) ] > = 0 make table of signs of -x - 11, x + 4, x - 3 and the whole expression on the left of the inequality. ``` x -inf -11 -4 3 +inf -x - 11 + \| - \| - \| - x - 3 - \| - \| - \| + x + 4 - \| - \| + \| + (-x - 11) / [(x + 4)(x - 3)] + \| - \| + \| - ``` conclusion: The solutions of 1 /(x + 4) - 2 / (x - 3) > = 0 are the values of x for which (-x - 11) / [(x + 4)(x - 3)] is positive or is equal to zero.Thus, the solution set of the given inequality is the intervals (-inf , -11 ] U (-4 , 3). More references and links on how to Solve Equations, Systems of Equations and Inequalities.
## 2230.22 – The Golden Ratio You see here a rectangle with a special shape. The shape has the property that if you make a square at the bottom and cut it off, the remaining rectangle is similar to the original rectangle. That is, $\frac{l}{x} = \frac{x}{x – l}.$ Find the ratio of $x$ to $l$. This is a famous number, known as the golden ratio. Rectangles of this shape are frequently used in architecture. Solution First, $\frac{l}{x} = \frac{x}{l – x} \leadsto l(l – x) = {x^2} \leadsto l^2 – lx – x^2 = 0$ We want $\frac{x}{l}$, so we divide by $l^2$: $1 – \frac{x}{l} – \frac{x^2}{l^2} = 0 \leadsto \left(\frac{x}{l}\right)^2 + \frac{x}{l} – 1 = 0.$ We then invite the quadratic formula to the show: $\frac{x}{l}=\frac{-1±\sqrt{(1-(-4)}}{2}=\frac{-1±√5}{2}.$ We take the positive root: $\frac{x}{l} = \frac{-1 + √5}{2} ≈ .06180.$ Coincidentally, the ratio of $\frac{1km}{1 mile} ≈ .06214$.
# Speed of a Snowmobile: Snow Jump An educational, fair use website ## The Problem This is a mechanics problem that a physics student should be able to solve View the clip of the snowmobile flip at killsometime.com. In the video, a snowmobile is jumped off of an incline and flips backwards, then lands upright on the ground. Assuming that the snowmobile took off at a 45° angle and that it landed on ground at the same height as that from which it took off, you should be able to determine: 1. Peak Altitude 2. Range 3. Takeoff Speed ## 1. Altitude First an appropriate formula must be chosen. The formula y = vot + ½at2 is the appropriate one in this case. In order to determine the total time that the snowmobile is in the air, we can count the frames in the video from take-off to landing, then divide by the frame rate of the video, which was 30 frames/second. The formula chosen above contains vo. In order to know the value of vo, we must use only the time from the peak to the ground where vo is 0, which in this case is exactly half the total time due to the 45 degree take-off angle. This means that t is equal to ½ of the total time. total time = frames/frame rate total time = 36 frames/30 frames/second total time = 1.2 seconds t = 1.2 seconds/2 = .6 seconds y = vot + ½at2 y = (0 m/s)(.6 s) + ½(9.8 m/s2)(0.6 s)2 = 1.764 m ## 2. Range To find range, s, the horizontal velocity of the snowmobile must first be found. To find horizontal velocity in this example, initial vertical velocity must first be found, and then a horizontal component can be found using trigonometry. vfy2 = voy2 + 2asy 0 = voy2 + 2(-9.8 m/s2)(1.764 m) voy = (34.5744)½ = 5.88 m/s x = (5.88 m/s)/(tan 45°) = 5.88 m/s s = vxt s = (5.88 m/s)(1.2 s) = 7.056 m ## 3. Take-off Speed • To determine take-off speed, components and trigonometry can once again be used. x = (5.88 m/s)/(cos 45°) = 8.32 m/s Matthew Grabczynski -- 2005 Physics on Film 1. Feature Films 2. One Reelers 3. Short Video Clips
A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Squares are placed successively inside the right angled triangle thus formed as in the figure below. What is the area enclosed by all squares when their number goes to infinity? 1. Find the points of intersection of the line $y= x+6$ with the parabola $y=x^2.$ 2. Evaluate the sum $\sum_{n=0}^{a}{2^n}.$ 3. What is the equation of the line with gradient $5$ that passes through the point $(2,5)?$ A line equation may help. What can you say about the $x$ and $y$ coordinates of the top right corner of the largest square? … does that help to compute its side length? Is there a relationship between the large triangle and the one above the largest square? What about the successive triangles? What is then the sum of such numbers that are in such proportion, i.e. constant ratio? The equation of the line is $y=1-\frac{x}{a}.$ At the top right hand corner of the first square $y=x.$ If we substitute this into the first equation we get $x = 1 - \frac{x}{a},$ which we can solve to get the side length of the first square which is $\frac{a}{a+1}.$ We notice that the triangle bounded by the lines $x=\frac{a}{a+1},$ $y = 1 - \frac{x}{a}$ and $y=0$ is similar to the larger triangle but with a height of $\frac{a}{a+1}.$ The heights of successive nested triangles follow a geometric progression with ratio $\frac{a}{a+1}.$ The squares contained within the triangles have side lengths equal to the height of the next triangle and the total area is a geometric series.\begin{aligned} \sum_{n=1}^{\infty}\bigg(\frac{a}{a+1}\bigg)^{2n} &= \frac{\big(\frac{a}{a+1}\big)^2}{1-\big(\frac{a}{a+1}\big)^2} \\[7pt]&= \frac{a^2}{2a+1} \end{aligned}
# 9.7 Solve a formula for a specific variable Page 1 / 9 By the end of this section, you will be able to: • Use the distance, rate, and time formula • Solve a formula for a specific variable Before you get started, take this readiness quiz. 1. Write $35$ miles per gallon as a unit rate. If you missed this problem, review Ratios and Rates . 2. Solve $6x+24=96.$ If you missed this problem, review Solve Equations with Variables and Constants on Both Sides . 3. Find the simple interest earned after $5$ years on $\text{1,000}$ at an interest rate of $\text{4%}.$ If you missed this problem, review Solve Simple Interest Applications . ## Use the distance, rate, and time formula One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you travel if you drove at a steady rate of $60$ miles per hour for $2$ hours? (This might happen if you use your car’s cruise control while driving on the Interstate.) If you said $120$ miles, you already know how to use this formula! The math to calculate the distance might look like this: $\begin{array}{}\\ \text{distance}=\left(\frac{60\phantom{\rule{0.2em}{0ex}}\text{miles}}{1\phantom{\rule{0.2em}{0ex}}\text{hour}}\right)\left(2\phantom{\rule{0.2em}{0ex}}\text{hours}\right)\hfill \\ \text{distance}=120\phantom{\rule{0.2em}{0ex}}\text{miles}\hfill \end{array}$ In general, the formula relating distance, rate, and time is $\text{distance}\phantom{\rule{0.2em}{0ex}}\text{=}\phantom{\rule{0.2em}{0ex}}\text{rate}·\text{time}$ ## Distance, rate and time For an object moving in at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula $d=rt$ where $d=$ distance, $r=$ rate, and $t=$ time. Notice that the units we used above for the rate were miles per hour, which we can write as a ratio $\frac{miles}{hour}.$ Then when we multiplied by the time, in hours, the common units ‘hour’ divided out. The answer was in miles. Jamal rides his bike at a uniform rate of $12$ miles per hour for $3\frac{1}{2}$ hours. How much distance has he traveled? ## Solution Step 1. Read the problem. You may want to create a mini-chart to summarize the information in the problem. $d=?$ $r=12\phantom{\rule{0.2em}{0ex}}\text{mph}$ $t=3\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{hours}$ Step 2. Identify what you are looking for. distance traveled Step 3. Name. Choose a variable to represent it. let d = distance Step 4. Translate. Write the appropriate formula for the situation. Substitute in the given information. $d=rt$ $d=12\cdot 3\frac{1}{2}$ Step 5. Solve the equation. $d=42\phantom{\rule{0.2em}{0ex}}\text{miles}$ Step 6. Check: Does 42 miles make sense? Step 7. Answer the question with a complete sentence. Jamal rode 42 miles. Lindsay drove for $5\frac{1}{2}$ hours at $60$ miles per hour. How much distance did she travel? 330 mi Trinh walked for $2\frac{1}{3}$ hours at $3$ miles per hour. How far did she walk? 7 mi Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of $520$ miles. If he can drive at a steady rate of $65$ miles per hour, how many hours will the trip take? ## Solution Step 1. Read the problem. Summarize the information in the problem. $d=520\phantom{\rule{0.2em}{0ex}}\text{miles}$ $r=65\phantom{\rule{0.2em}{0ex}}\text{mph}$ $t=?$ Step 2. Identify what you are looking for. how many hours (time) Step 3. Name: Choose a variable to represent it. let t = time Step 4. Translate. Write the appropriate formula. Substitute in the given information. $d=rt$ $520=65t$ Step 5. Solve the equation. $t=8$ Step 6. Check: Substitute the numbers into the formula and make sure the result is a true statement. $d=rt$ $520\stackrel{?}{=}65\cdot 8$ $520=520>✓$ Step 7. Answer the question with a complete sentence. We know the units of time will be hours because we divided miles by miles per hour. Ray's trip will take 8 hours. do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid question and i want to know the answer as well Maciej Abigail Do somebody tell me a best nano engineering book for beginners? what is fullerene does it is used to make bukky balls are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA so some one know about replacing silicon atom with phosphorous in semiconductors device? Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure. Harper Do you know which machine is used to that process? s. how to fabricate graphene ink ? for screen printed electrodes ? SUYASH What is lattice structure? of graphene you mean? Ebrahim or in general Ebrahim in general s. Graphene has a hexagonal structure tahir On having this app for quite a bit time, Haven't realised there's a chat room in it. Cied what is biological synthesis of nanoparticles what's the easiest and fastest way to the synthesize AgNP? China Cied types of nano material I start with an easy one. carbon nanotubes woven into a long filament like a string Porter many many of nanotubes Porter what is the k.e before it land Yasmin what is the function of carbon nanotubes? Cesar I'm interested in nanotube Uday what is nanomaterials and their applications of sensors. what is nano technology what is system testing? preparation of nanomaterial Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it... what is system testing what is the application of nanotechnology? Stotaw In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google Azam anybody can imagine what will be happen after 100 years from now in nano tech world Prasenjit after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments Azam name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world Prasenjit how hard could it be to apply nanotechnology against viral infections such HIV or Ebola? Damian silver nanoparticles could handle the job? Damian not now but maybe in future only AgNP maybe any other nanomaterials Azam Hello Uday I'm interested in Nanotube Uday this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15 Prasenjit how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Berger describes sociologists as concerned with Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives. What is the expressiin for seven less than four times the number of nickels How do i figure this problem out. how do you translate this in Algebraic Expressions why surface tension is zero at critical temperature Shanjida I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason s. Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)= . After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Square Root Function Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It. Square root function is a function which is represented with the radical sign ‘√’. Inside a square root, a number or an expression can be placed and we can find the square root of it. The parent function or the basic square root function is y = √x and here ‘y’ is the function value also known as f(x) and ‘x’ is the variable. All other square root functions can be derived from this parent function and can be graphed on the X-Y coordinate plane. Example 1: Given the square root function, f(x) = √(5x). Find the function value when x is equal to 4. Given square root function: f(x) = √(5x) In order to find the function value or the ‘y’ value at x = 4, we substitute x = 4 in the above square root function. This gives: y = f(4) = √(5 * 4) = √20. We can simplify √20 by writing its prime factors==> √20 = √(2* 2* 5) = 2√5. Therefore √20 or 2√5 is the function value at x = 4. Example 2: Given the square root function, f(x) = √(x + 6). Find the function value when x is equal to 3. Given square root function: f(x) = √(x + 6) In order to find the function value or the ‘y’ value at x = 3, we substitute x = 3 in the above square root function. This gives: y = f(3) = √(3 + 6) = √9. ‘9’ is a perfect square since 9 can be written as 3 * 3. Hence √9 = √(3 * 3) = 3 Therefore ‘3’ is the function value when x = 3.
Derivative of polynomials and exponentials # Derivative of polynomials and exponentials - h → e h-1 h... This preview shows pages 1–2. Sign up to view the full content. DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS For basic functions, we have diﬀerentiation rules as follows. 1. d dx ( c ) = 0. 2. d dx ( x ) = 1. 3. d dx ( x n ) = nx n - 1 . If we know derivatives of certain functions, we can calculate derivatives of new functions by using old ones. 4. d dx [ cf ( x )] = cf 0 ( x ). 5. d dx [ f ( x ) + g ( x )] = f 0 ( x ) + g 0 ( x ). 6. d dx [ f ( x ) - g ( x )] = f 0 ( x ) - g 0 ( x ). By using above rules, we can diﬀerentiate any polynomials. Example 1. Find The derivative of the function f ( x ) = 2 x 2 + 4 x + 1. Solution f 0 ( x ) = d dx (2 x 2 + 4 x + 1) = 2 d dx ( x 2 ) + 4 d dx ( x ) + d dx (1) = 2(2 x ) + 4(1) + 0 = 4 x + 4 Example 2. Find The derivative of the function f ( x ) = a x . Solution f 0 ( x ) = lim h 0 f ( x + h ) - f ( x ) h = lim h 0 a x + h - a x h = lim h 0 a x ( a h - 1) h = a x lim h 0 a h - 1 h In the example above, we do not know the limit lim h 0 a h - 1 h . However, if the limit is 1, then we can get a function whose derivative equals itself. Because this type of function is frequently used, we deﬁne a number a with this property as follows. Deﬁnition 1. e is the number such that lim 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: h → e h-1 h = 1. 1 2 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS From the deﬁnition, we can get the following diﬀerentiation formula. Theorem 1. d dx ( e x ) = e x Example 3. If f ( x ) = e x-x 2 + 3 x , ﬁnd f and f 00 . Solution f ( x ) = d dx ( e x-x 2 + 3 x ) = d dx ( e x )-d dx ( x 2 ) + 3 d dx ( x ) = e x-2 x + 3 f ( x ) = d dx ( e x-2 x + 3) = d dx ( e x )-2 d dx ( x ) + d dx (3) = e x-2 Example 4. Find an equation of the normal line to the curve y = e x-x 2 + 3 x at the point P (0 , 1). Solution Since the slope of the tangent line at the point P is y (0) = 4, we can see that the slope of the normal line is-1 4 . Thus, the equation of the normal line is y-1 =-1 4 ( x-0) , or y =-1 4 x + 1 .... View Full Document {[ snackBarMessage ]} ### Page1 / 2 Derivative of polynomials and exponentials - h → e h-1 h... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# How do you write an inequality and solve given "the sum of nine times a number and fifteen is less than or equal to the sum of twenty four and ten times the number"? Oct 2, 2016 The inequality for this is $9 x + 15 \le 24 + 10 x$. #### Explanation: Now that we know what our inequality is, we can now solve for $x$. To do this, we have to subtract 24 from each side: $9 x + 15 - 24 \le 24 + 10 x - 24$ $9 x - 9 \le 10 x$ Next, let's subtract $9 x$ from each side: $9 x - 9 - 9 x \le 10 x - 9 x$ $- 9 \le 9 x$ Finally, let's divide each side by 9 to get our answer: $- \frac{9}{9} \le \frac{9 x}{9}$ $- 1 \le x$
# Calculus – Limits and Derivatives Derivative introduced as rate of change both as that of distance function and geometrically A rate can be explained as the comparison between two quantities of different kinds. A common rate that we use as a physical quantity is a speed. Speed is defined as the rate of distance traveled per unit of time. Also, recall from physics that distance is a function of time. So, we can draw a graph for distance over time. In this case, let distance be the function f of time, x. The average rate of change of the function f over the interval [a, b] can be written as: $\frac{\Delta f}{\Delta x}=\frac{f(b)-f(a)}{b-a}$ For a line, the rate (or slope) is the same at every point on the line. The fraction Δf/Δx represents a velocity because it is measured in units of distance over time. But this is only an average velocity between two points on the graph. This is because we consider the distances and times of two points. The length of the time intervals we consider is Δx. To make better approximations, we can take smaller and smaller time intervals. In the language of calculus, we take the limit of the slope as Δx approaches 0 and calculate the instantaneous velocity at an instant A. This can be shown as: $speed=\lim _{x?0}\frac{\Delta f}{\Delta x}$ The same definition of the derivative can be arrived at using the geometric intuition. For graphs other than lines, the rate at which the graph rises or falls changes from point to point which can be calculated as the slope of the linear graph. To determine the rate at which a quadratic graph rises or falls at a single point, we can find the slope of the tangent line to the point. The instantaneous rate of change of f(x) at x is defined by taking the limit of the average rates of change of ‘f’ over the intervals [x, x+∆x], as ∆x approaches 0. The derivative is the slope of the tangent line to a graph f(x) and is usually denoted f’(x). The slope of the secant line through (x, f(x)) and (x+∆x, f(x+∆x)) is the same as the average rate of change of f over the interval [x, x+∆x]. The slope of the tangent line through (x, f(x)) is the same as the instantaneous rate of change of f at the point x, or the derivative. The slope of the tangent, or derivative, depends on the position of the point (x, f(x) on the curve, and therefore on the choice of x. Therefore, the derivative is a function of x, and that is why we write it as f'(x). The smaller the ∆x, the more the secant and tangent lines resemble each other.The derivative f'(x) of the function f(x) is the slope of the tangent at the point (x, f(x)).We define the derivative of the function f at the point x0 as shown below: The symbol on the left of the above equation is read “f prime of x zero”. Now, the point x0 that we originally chose was arbitrary. We can do the same for any other point. So, what we get really is a function of x.
# Problem solving equations Problem solving equations can be a helpful tool for these students. So let's get started! ## The Best Problem solving equations Problem solving equations is a software program that helps students solve math problems. Solving for x logarithms can be a complicated process, but there are a few steps that can help to make it easier. First, it is important to understand what a logarithm is. A logarithm is simply the exponent that a number must be raised to in order to equal another number. For example, the logarithm of 100 is 2, because 100 = 10^2. Solving for x logarithms simply means finding the value of x that makes the equation true. To do this, first rewrite the equation in exponential form. Then, take the logarithm of both sides of the equation using any base. Finally, solve for x by isolating it on one side of the equation. With a little practice, solving for x logarithms can become second nature. A theorem is a mathematical statement that is demonstrated to be true by its proof. The proof of a theorem is usually very difficult, but it can be simplified by using another theorem as a basis for the proof. A lemma is a theorem that has been simplified in this way. This type of theorem has not yet been proven, but it has been shown to be true by its proof. A simple example of this would be the Pythagorean theorem: If we assume that the hypotenuse (the length of one side) is twice the length of the other two sides, then we can easily prove that the two sides are equal by showing that their sum is equal to the length of the hypotenuse. This is a lemma; however, it has not yet been proven to be true. Another example would be Euclid’s proposition: If you assume that a straight line can be divided into two parts so that each part is perpendicular to the line, and if you also assume that there are only two such parts, then you have enough information to show that they are equal. This proposition has been proved by Euclid’s proof; however, it still needs to be proved true by some other method. A math tutor can be an invaluable resource for this. By definition, a word problem is a mathematical problem that involves words rather than numbers or symbols. You might see words like "if it rains tomorrow, how many inches of rain will there be?" Word problems usually involve numbers or quantities, but they also include words that represent concepts such as length, time, area and volume. However, they often look different from standard mathematical problems because they rely more on language than mathematics. For example, you might be given the word "lose" and asked how many pounds of weight you would have to lose to reach a certain weight goal. That way you can check your answer without having to recalculate the problem on paper first. Many people like this tool because it saves them time—they can just scan their paper and get their answer instantly! The other benefit is that it helps you stay organized, because you can quickly scan all your papers together in one place. Another great thing about a math problem scanner is that you don’t have to buy any special equipment. So it’s perfect for anyone who wants to try it out for the first time! Linear equations are a type of mathematical equation that has an unknown number 'x', which is used to solve for the value of 'y'. An example of a linear equation would be the equation "4x + 3 = 18" where x represents the unknown value. This can be solved by solving for x. The value of x can be found by drawing a line from the origin (0,0) to each point on the graph where it intersects with the y axis. In this case, x=-3 and y=18. The value of y can then be found by averaging all points on the graph: 18/3=6. Therefore, y=6. The graphing process is used to solve linear equations by depicting a graph of the values in question. Lines are drawn that connect any two points where they intersect with the y axis at different locations. First, isolate one variable (x) to keep track of it while you define and measure other variables (y1 and y2). Then plot all points on the graph from 0 to 1. At any point where multiple lines intersect, simply average all points on that line to get your final answer. ## We cover all types of math problems Great Program I teach HS math. This is a way for students to cheat if you are not aware this is out there. It does enable students to get an explanation of each step-in simplifying or solving. When used properly, this is pretty revolutionary. Gracelyn Sanchez I tried this app to solve an integral equation. I searched about many apps to help me to solve the equations then I fell in thins amazing app. I'm shocked, it's amazing It is the best app I’ve ever used. I was not getting the algebra steps, but this app taught me Kaylee Phillips
# Multiplication Word Problems to 20 Multiplication is tricky for younger children so it helps if they can think about it in a variety of ways. These multiplication word problem cards ask questions that kids can visualize, so they can picture the scenes in their mind’s eye and map out the solutions. If you’re looking for more math practice, check out our Place Value Cover Up activity! To prepare for this activity, I downloaded the free multiplication cards (below) and printed them. (The color version was printed onto white cardstock and the black and white version onto colored copy paper for brightness.) I laminated the pages, cut them apart and then wrote the answer on the back of each card with a permanent Sharpie. This made them self-checking. Then, I asked the kids to punch holes in the top left corner of the cards so we could attach them to a ring. ## Solving Multiplication Word Problems The most difficult part for children working on multiplication is to make sure they understand the question and what it’s asking of them. So we began this activity as a group and talked through some of the problems together. Students knew that multiplication was the same as adding groups of numbers together. For instance, 3 x 5 means 3 sets of 5. With the question pictured above, we had to decide whether we would cross off 5 squares twice, or two squares five times. Children were excited to discover that either way, we arrived at the same answer. We used dry erase markers and dry erase crayons to mark the squares on the ten frames for each problem and then erased them afterwards. Some of the children chose to draw little objects in the squares rather than just checking them off. For example, drawing little triangles when working on the pizza problem or simple chairs for one of the traveling questions. One child grabbed some recycled paper and drew larger pictures to represent the problems because it helped him understand more clearly. He then glued that page into his math notebook to keep a record of his efforts. One aspect about these multiplication word problems that made the kids more comfortable with these cards was that they had already completed the Addition Word Problems to 20 cards as well as the Subtraction Word Problems to 20 cards. Since the format was familiar, and the children had completed those tasks successfully, they did not seem too concerned about the questions because they knew if they re-read them or chatted with their friends, they’d be able to work them out. Let’s start multiplying! Click the blue button below to download your free story problems and then hop over and snag our Place Value Cover Up Activity in our shop!
Geometric sequence solver Geometric sequence solver is a mathematical tool that helps to solve math equations. We can solve math problems for you. The Best Geometric sequence solver Best of all, Geometric sequence solver is free to use, so there's no sense not to give it a try! The 3x3 matrix is a way of describing how you can translate the results of a table into the columns and rows in a matrix. The example below shows how you could translate a result in a table into three columns and three rows. A simple way to do this is to multiply each column by the corresponding row value. You can then rearrange these values to create the matrix form of your table. For example, if there were two rows and three columns and we wanted to translate the first row into column 1, we would write: 11 = 1 22 = 4 3*3 = 9 The result would be 9. Another option is to use a textbook that has a workbook at the back of the book. These workbooks often include exercises and examples to help students understand concepts and practice new skills. Online resources also exist that can provide homework solutions. For example, there are websites which offer free online homework help. While most of these sites are free, some charge a fee for premium assistance. By using technology to solve your homework problems, you can maximize your chances of doing well in school. The internet provides a resource for every subject – from math to history – for free or at low cost. More importantly, it helps you stay organized and make sure that you are completing everything on time. Simple solutions math is a method of solving math problems that is based on breaking down the problem into smaller, more manageable pieces. The first step in simple solutions math is to identify the problem. Once the problem has been identified, the next step is to break it down into smaller parts. This can be done by identifying the variables involved in the problem and by using algebraic equations to solve for those variables. After the problem has been broken down into smaller parts, the last step is to put all of the pieces back together to find the solution. Simple solutions math can be used to solve a wide variety of math problems, from simple arithmetic to complex calculus. Logarithmic equation solvers are a type of mathematical software that is used to solve equations that contain logs. Logarithmic equations are equations in which the variable is raised to a power that is itself a logarithm. For example, the equation 2x+5=3 can be rewritten as 10x=3. This equation cannot be solved using traditional methods, but it can be solved using a logarithmic equation solver. Logarithmic equation solvers use a variety of algorithms to solve equations, and they can often find solutions that cannot be found using traditional methods. Logarithmic equation solvers are used by mathematicians, engineers, and scientists to solve a wide range of problems. One such tool is the slope intercept form solver. This tool allows users to input the coordinates of two points on a line, and it will output the slope intercept form of the line. This can be a helpful tool for students who are trying to learn how to find the slope intercept form of a line. Instant assistance with all types of math Thank you for putting a subscription model into the app rather than an ad-based one. And for the amazing app in general. I'm not a fan of using calculators, but having an app I can check to see whether I've calculated anything wrong, and what went wrong is really helpful. Violetta Brown In a live photo map helps me with a lot of my school work and I love it in general sometimes it doesn't work but most of the time it does and oh my God it's the best thing a girl can ask for so I suggest that everyone should get if they're horrible at math like me I think it would be the best my grade has went up so much is unreal so I think that everyone should get the app because it's the best learning way you can also you steps to so if any I honestly think it would be the best way to learn
Courses Courses for Kids Free study material Offline Centres More Store # The ratio of the number of boys and the girls is $4:3$. If there are 18 girls in a class, then find the total number of students in the class.A.40B.41C.42D.43 Last updated date: 20th Jun 2024 Total views: 414.3k Views today: 5.14k Verified 414.3k+ views Hint: We are given the ratio of the number of boys and the girls and the number of girls. First, we’ll assume a variable for the number of boys and substituting that in the given ratio we’ll get the number of boys. Now, we know that the total number of students is equal to the sum of the number of boys and the number of girls so on adding these values we’ll get our answer. Given data: The ratio of the number of boys and girls is $4:3$ The number of girls=18 Let the number of boys be B According to the given data, boys to girls ratio is $4:3$ i.e. $\dfrac{B}{{18}} = \dfrac{4}{3}$ On rearranging we get, $\Rightarrow B = \dfrac{4}{3}\left( {18} \right)$ On simplifying we get, $\Rightarrow B = 4\left( 6 \right)$ $\therefore B = 24$ It is well known that the total number of students in the class is equal to the sum of the number of boys and the number of girls. $The{\text{ }}total{\text{ }}number{\text{ }}of{\text{ }}students = 18 + 24$ Hence, Total number of students is 42. Option(C) is the correct option. Note: Alternative way to find the total number of students can be We can say that the number of boys is equal to the difference between the total number of students let’s say ‘N’ and the number of girls According to the ratio given i.e. $\dfrac{{N - 18}}{{18}} = \dfrac{4}{3}$ On simplification we get, $\Rightarrow N - 18 = \dfrac{4}{3}(18)$ $\Rightarrow N = 24 + 18$ $\therefore N = 42$ From both the methods the total number of students is 42. i.e. option(C) 42
## Definition A quadratic equation is a polynomial equation of the form: $a x^2 + b x + c = 0$ such that $a \ne 0$. From Solution to Quadratic Equation, the solutions are: $x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$ ### Discriminant The expression $b^2 - 4 a c$ is called the discriminant of the equation. ### Canonical Form The canonical form of the quadratic equation is: $a x^2 + b x + c = 0$ where $a$, $b$ and $c$ are constants. ## Also defined as Some older treatments of this subject present this as: An algebraic equation of the form $a x^2 + 2 b x + c = 0$ is called a quadratic equation. It has solutions: $x = \dfrac {-b \pm \sqrt {b^2 - a c} } a$ but this approach has fallen out of fashion. ## Also known as A quadratic equation is also known as: • an equation of the second degree • a polynomial of degree $2$ and so on. ## Examples ### Example: $x^2 + 1 = 0$ $x^2 + 1 = 0$ has no root in the set of real numbers $\R$: $x = \pm i$ where $i = \sqrt {-1}$ is the imaginary unit. ### Example: $x^2 + 4 = 0$ $x^2 + 4 = 0$ has the wholly imaginary roots: $x = \pm 2 i$ where $i = \sqrt {-1}$ is the imaginary unit. ### Example: $z^2 - \paren {3 + i} z + 4 + 3 i = 0$ The quadratic equation in $\C$: $z^2 - \paren {3 + i} z + 4 + 3 i = 0$ has the solutions: $z = \begin{cases} 1 + 2 i \\ 2 - i \end{cases}$ ### Example: $z^2 + \paren {2 i - 3} z + 5 - i = 0$ The quadratic equation in $\C$: $z^2 + \paren {2 i - 3} z + 5 - i = 0$ has the solutions: $z = \begin{cases} 2 - 3 i \\ 1 + i \end{cases}$ ## Also see The ancient Babylonians knew the technique of solving quadratic equations as long ago as $1600$ BCE. The general algebraic formulation of its solution did not appear until at least $100$ CE.
newtons second law calculating acceleration The equation for acceleration given above can be used to calculate the acceleration of an object that is acted on by an unbalanced force. For example, assume you are pushing a large wooden trunk, like the one shown in Figure acceleration of the trunk, substitute these values in the equation for acceleration: a= F 20 N 2N = = m 10 kg kg Recall that one newton (1 N) is the force needed to cause a 1-kilogram mass to accelerate at 1 m/s2 . Therefore, force can also be expressed in the unit kgm/s2 . This way of expressing force can be substituted for newtons in the solution to the problem: a= 2 N 2 kg m/s2 = = 2 m/s2 kg kg Why are there no kilograms in the final answer to this problem? The kilogram units in the numerator and denominator of the fraction cancel out. As a result, the answer is expressed in the correct units for acceleration: m/s2 . You Try It! Problem: Assume that you add the weights to the trunk in Figure 14.7. If you push the trunk and weights with a force of 20 N, what will be the trunks acceleration? Need more practice? You can find additional problems at this URL: acceleration and weight Newtons second law of motion explains the weight of objects. Weight is a measure of the force of gravity pulling on an object of a given mass. Its the force (F) in the acceleration equation that was introduced above: a= F m This equation can also be written as: F = ma The acceleration due to gravity of an object equals 9.8 m/s2 , so if you know the mass of an object, you can calculate its weight as: F = m 9.8 m/s2 As this equation shows, weight is directly related to mass. As an objects mass increases, so does its weight. For example, if mass doubles, weight doubles as well. You can learn more about weight and acceleration at this URL: Problem Solving Problem: Daisy has a mass of 35 kilograms. How much does she weigh? Solution: Use the formula: F = m 9.8 m/s2 . F = 35 kg 9.8 m/s2 = 343.0 kg m/s2 = 343.0 N You Try It! Problem: Daisys dad has a mass is 70 kg, which is twice Daisys mass. Predict how much Daisys dad weighs. Then calculate his weight to see if your prediction is correct. Helpful Hints The equation for calculating weight (F = m a) works only when the correct units of measurement are used. Mass must be in kilograms (kg). Acceleration must be in m/s2 . Weight (F) is expressed in kgm/s2 or in newtons (N). acceleration force and mass Newton determined that two factors affect the acceleration of an object: the net force acting on the object and the objects mass. The relationships between these two factors and motion make up Newtons second law of motion. This law states that the acceleration of an object equals the net force acting on the object divided by the objects mass. This can be represented by the equation: Net force , or Mass F a= m Acceleration = You can watch a video about how Newtons second law of motion applies to football at this URL: http://science36 direct and inverse relationships Newtons second law shows that there is a direct relationship between force and acceleration. The greater the force that is applied to an object of a given mass, the more the object will accelerate. For example, doubling the force on the object doubles its acceleration. The relationship between mass and acceleration, on the other hand, is an inverse relationship. The greater the mass of an object, the less it will accelerate when a given force is applied. For example, doubling the mass of an object results in only half as much acceleration for the same amount of force. Consider the example of a batter, like the boy in Figure 14.6. The harder he hits the ball, the greater will be its acceleration. It will travel faster and farther if he hits it with more force. What if the batter hits a baseball and a softball with the same amount of force? The softball will accelerate less than the baseball because the softball has greater mass. As a result, it wont travel as fast or as far as the baseball. instructional diagrams No diagram descriptions associated with this lesson questions Which two factors have an inverse relationship? ``````a. force and acceleration b. force and mass c. mass and gravity --> d. mass and acceleration `````` Any change in the motion of an object is called ``````a. speed. b. velocity. c. direction. --> d. acceleration. `````` In the equation F = m a, if a is the acceleration due to gravity, what is F? ``````a. mass --> b. weight c. friction d. frequency `````` If you push a 20-kg box with a force of 10 N, what is its acceleration? ``````a. 20 m/s2 b. 10 m/s2 c. 2 m/s2 --> d. 0.5 m/s2 `````` Newton determined that the acceleration of an object depends on the net force acting on the object and the objects ``````a. size. --> b. mass. c. weight. d. velocity. `````` The relationship between mass and inertia is described by Newtons second law of motion. ``````a. true --> b. false `````` Newton determined that there is a direct relationship between force and mass. ``````--> a. true b. false `````` Any change in velocity for any reason is called acceleration. ``````--> a. true b. false `````` Newtons second law shows that there is a direct relationship between net force and acceleration. ``````--> a. true b. false `````` The greater the net force applied to a given object, the more it will accelerate. ``````--> a. true b. false `````` Doubling the mass of an object doubles its weight. ``````--> a. true b. false `````` The greater the mass of an object, the more it will accelerate when a given net force is applied to it. ``````a. true --> b. false `````` A net force of 1 N applied to a mass of 1 kg results in an acceleration of 0.5 m/s2 . ``````a. true --> b. false `````` Force can be expressed as kg m/s. ``````a. true --> b. false `````` Your weight equals your mass multiplied by the acceleration due to gravity. ``````--> a. true b. false `````` Any object that is accelerating is changing its speed. ``````a. true --> b. false `````` If a balanced force acts on an object, the object will accelerate. ``````a. true --> b. false `````` A 10-kg object has greater acceleration due to gravity than a 5-kg object. ``````a. true --> b. false `````` The acceleration of an object equals its mass times the net force applied to it. ``````a. true --> b. false `````` The acceleration of an object due to gravity depends on the objects initial velocity ``````a. true --> b. false `````` An object is accelerating when it ``````a. speeds up. b. slows down. c. changes direction. --> d. any of the above `````` Newtons second law of motion relates an objects acceleration to ``````a. its mass. b. its velocity. c. the net force acting on it. --> d. two of the above `````` Doubling the net force acting on an object ``````--> a. doubles its acceleration. b. decreases it acceleration. c. cuts its acceleration in half. d. does not affect its acceleration. `````` If you push a 20-kilogram mass with a force of 40 N, what will be the objects acceleration? ``````a. 40 m/s2 b. 20 m/s2 c. 10 m/s2 --> d. 2 m/s2 `````` Which units can be used to express force? ``````a. N b. kg/s2 c. kg m/s2 --> d. two of the above `````` If you know the mass of an object, you can calculate its weight with the formula ``````a. F b. F --> c. F d. F `````` If the mass of an object doubles, its weight ``````--> a. doubles. b. decreases. c. is not affected. d. changes by a factor of 21 . `````` acceleration due to gravity ``````a. acceleration b. weight c. direct relationship d. a = e. inverse relationship f. F = m a --> g. 9.8 m/s2 `````` formula for weight ``````a. acceleration b. weight c. direct relationship d. a = e. inverse relationship --> f. F = m a g. 9.8 m/s2 `````` formula for acceleration ``````a. acceleration b. weight c. direct relationship --> d. a = e. inverse relationship f. F = m a g. 9.8 m/s2 `````` measure of the force of gravity pulling on an object ``````a. acceleration --> b. weight c. direct relationship d. a = e. inverse relationship f. F = m a g. 9.8 m/s2 `````` type of relationship between acceleration and mass ``````a. acceleration b. weight c. direct relationship d. a = --> e. inverse relationship f. F = m a g. 9.8 m/s2 `````` measure of the change in velocity of a moving object ``````--> a. acceleration b. weight c. direct relationship d. a = e. inverse relationship f. F = m a g. 9.8 m/s2 `````` type of relationship between acceleration and force ``````a. acceleration b. weight --> c. direct relationship d. a = e. inverse relationship f. F = m a g. 9.8 m/s2 `````` diagram questions No diagram questions associated with this lesson
# Abcdef is a Regular Hexagon with Centre O (In the Following Figure). If the Area of Triangle Oab is 9 Cm2, Find the Area of : (I) the Hexagon and (Ii) the Circle in Which the Haxagon is Incribed. - Mathematics Sum ABCDEF is a regular hexagon with centre O (in the following figure). If the area of triangle OAB is 9 cm2, find the area of : (i) the hexagon and (ii) the circle in which the haxagon is incribed. #### Solution We know that a regular hexagon is made up of 6 equilateral triangles. We have given area of the one of the triangles. ∴"Area of the hexagon=6xx area of one equilateral triangle" ∴"Area of the hexagon"=6xx9 ∴"Area of the hexagon"=54 We know that if a regular hexagon is inscribed in the circle, then the radius of the circle is same as the side of the regular hexagon. We also know that a regular hexagon is made up of 6 equilateral triangles and we have area of one of the equilateral triangle. ∴"Area of the equilateral triangle"=sqrt3/4 xx"side"^2 Substituting the value of the given equilateral triangle we get, ∴9=sqrt3/4xx"side"^2 ∴ side^2=(9xx4)/sqrt3 ∴ "side"^2=36/sqrt3 Now we will find the area of the circle. ∴ Area of the circle=pi r^2 Substituting the values we get, ∴" Area of the circle"=22/7xx36/sqrt3 Now we will substitute sqrt3=1.732 we get, ∴" Area of the circle"=22/7xx36/1.732 ∴" Area of the circle"=792/12.124 ∴" Area of the circle"=65.324 Therefore, area of the hexagon and area of the circle are 54 cm^2 and 65.324 cm^2 Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 10 Maths Chapter 13 Areas Related to Circles Exercise 13.4 \| Q 29 \| Page 60
# Rounding to the nearest 100 \| 3rd grade \| Khan Academy \| Summary and Q&A 482.5K views October 4, 2013 by ## TL;DR Learn how to round numbers to the nearest 100 by examining the tens place and applying the rule. ## Install to Summarize YouTube Videos and Get Transcripts ### Q: How do you round a number to the nearest 100? To round a number to the nearest 100, examine the tens place digit. If it is 5 or larger, round up to the next multiple of 100. If it is less than 5, round down to the previous multiple of 100. ### Q: Why is it important to consider the tens place when rounding to the nearest 100? The tens place determines whether a number is closer to the multiple of 100 above or below it. By looking at the value of the tens place, you can determine the correct direction to round. ### Q: How would you round 245 to the nearest 100? When rounding 245 to the nearest 100, the tens place is 4. Since it is less than 5, we round down. Therefore, 245 would be rounded down to 200. ### Q: What is the rule for rounding to the nearest 100? The rule for rounding to the nearest 100 states that if the value in the tens place is 5 or larger, round up; if it is less than 5, round down. Apply this rule to determine the appropriate multiple of 100. ## Summary & Key Takeaways • Understand the process of rounding numbers to the nearest 100 by considering the multiples of 100 on a number line. • Examine examples of rounding 152, 137, 245, and 354 to the nearest 100 using the rule of looking at the tens place. • Learn how to decide whether to round up or round down based on the value of the digit in the tens place.
# Are You A True Genius In High School Maths? Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Michellee M Michellee Community Contributor Quizzes Created: 5 \| Total Attempts: 144,356 Questions: 15 \| Attempts: 17,510 Settings This quiz allows you to determine how much high school math you know based on the math topics learned in high school. The test includes questions from algebra, geometry, trigonometry, pre-calculus, and calculus. Let's start and prove you are seriously a math genius! • 1. ### Factor 6x2 + 17x + 12. • A. (3x+3)(2x+4) • B. (2x+3)(3x+4) • C. (3x-3)(3x-4) • D. (2x+3)(3x+2) • E. (2x-3)(3x-2) B. (2x+3)(3x+4) Explanation The given expression is a quadratic trinomial in the form of ax^2 + bx + c. To factor it, we need to find two binomials whose product equals the original expression. By multiplying the first terms of the binomials (2x and 3x) and the last terms (3 and 4), we can see that they add up to the middle term (17x). Therefore, the correct answer is (2x+3)(3x+4). Rate this question: • 2. ### Solve the System of Linear Equations: x + 2y = 9 3x - 4y = - 33 • A. X = 3, y = -6 • B. X = 3, y = 6 • C. X = 2, y = 7 • D. X = -3, y = 6 • E. X = 9, y = 12 D. X = -3, y = 6 Explanation The given system of linear equations is: x + 2y = 9 3x - 4y = -33 To solve this system, we can use the method of substitution or elimination. Let's use the method of elimination to solve the system: Multiply the first equation by 3 and the second equation by 1 to make the coefficients of x in both equations equal: 3(x + 2y) = 3(9) 1(3x - 4y) = 1(-33) This simplifies to: 3x + 6y = 27 3x - 4y = -33 Now, subtract the second equation from the first equation: (3x + 6y) - (3x - 4y) = 27 - (-33) 3x + 6y - 3x + 4y = 27 + 33 10y = 60 y = 6 Substitute the value of y back into the first equation: x + 2(6) = 9 x + 12 = 9 x = -3 Therefore, the solution to the system of linear equations is x = -3 and y = 6. Rate this question: • 3. ### Find the Equation for a line that passes through the two points (3,4) and (1,-10). • A. Y = 7x - 17 • B. Y = 7x + 17 • C. Y = x/7 - 17 • D. Y = x/7 + 10 • E. Y = x/7 - 10 A. Y = 7x - 17 Explanation The equation for a line passing through two points can be found using the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. To find the slope, we can use the formula (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, the slope is (4 - (-10)) / (3 - 1) = 14 / 2 = 7. Plugging in the slope and one of the points into the slope-intercept form, we get y = 7x - 17. Rate this question: • 4. ### Find the Exterior Angle of a heptagon • A. 30 degrees • B. 49 degrees • C. 51 degrees • D. 60 degrees • E. 129 degrees C. 51 degrees Explanation The exterior angle of a regular heptagon (a polygon with 7 sides) can be calculated by dividing 360 degrees (the sum of all exterior angles in any polygon) by the number of sides, which in this case is 7. Therefore, the exterior angle of a heptagon is 360/7 = 51 degrees. Rate this question: • 5. ### Triangle ABC is similar to DEF. The length of EF is 7 inches, while the length of BC is 5 inches. If the length of AB is 16 inches, what is the length of DE? • A. 22.4 inches • B. 23.0 inches • C. 20.5 inches • D. 25.3 inches • E. 19.5 inches A. 22.4 inches Explanation Since triangle ABC is similar to DEF, the corresponding sides are proportional. Using the given lengths, we can set up a proportion: AB/DE = BC/EF. Plugging in the values, we have 16/DE = 5/7. Cross-multiplying gives us 716 = 5DE, which simplifies to 112 = 5*DE. Dividing both sides by 5, we get DE = 112/5 = 22.4 inches. Rate this question: • 6. ### Find the volume of a sphere that has a diameter of 5 meters. • A. 8.3π meters^3 • B. 33π meters^3 • C. 166π meters^3 • D. 83.33π meters^3 • E. 6.7π meters^3 D. 83.33π meters^3 Explanation The volume of a sphere can be calculated using the formula V = (4/3)πr^3, where r is the radius of the sphere. Given that the diameter of the sphere is 5 meters, the radius would be half of that, which is 2.5 meters. Plugging this value into the formula, we get V = (4/3)π(5/2)^3 = (4/3)π(125/8) = 83.33π meters^3. Therefore, the correct answer is 83.33π meters^3. Rate this question: • 7. ### Two lines, AB and CD, intersect at point E. If angle AED is 127 degrees, what is angle BED + angle AEC? • A. 127 degrees • B. 26.5 degrees • C. 53 degrees • D. 106 degrees • E. Not enough information is given to solve the problem. D. 106 degrees Explanation Angle AED and angle BED are vertical angles, which means they are congruent. Therefore, angle BED is also 127 degrees. Angle AEC and angle BED are adjacent angles that form a straight line, so their sum is 180 degrees. Therefore, angle BED + angle AEC = 127 degrees + 53 degrees = 180 degrees. Rate this question: • 8. ### Sin(π/3) can be expressed alternatively as • A. Sin(π/3) cos(π/6) • B. 2 sin(π/3) cos(π/6) • C. 2 sin(π/6) cos(π/6) • D. 2 sin(π/6) • E. None of the above. C. 2 sin(π/6) cos(π/6) Explanation The given expression 2 sin(π/6) cos(π/6) is the correct alternative expression for sin(π/3). This is because sin(π/3) can be simplified using the double angle formula for sine: sin(2θ) = 2 sin(θ) cos(θ). In this case, θ is π/6, so sin(π/3) = 2 sin(π/6) cos(π/6). Therefore, the correct answer is 2 sin(π/6) cos(π/6). Rate this question: • 9. ### Tan(π/2) is equal to • A. 0 • B. Sqrt(2) / 2 • C. 3 sqrt(2) / 2 • D. 1 / 2 • E. None of the above. E. None of the above. Explanation The correct answer is "None of the above" because the value of tan(π/2) is undefined. In trigonometry, the tangent function is defined as the ratio of the opposite side to the adjacent side in a right triangle. When the angle is π/2 (90 degrees), the adjacent side becomes zero, resulting in division by zero, which is undefined. Therefore, none of the given options accurately represent the value of tan(π/2). Rate this question: • 10. ### What is the limit of the function y = 1 / (x - 4) as x approaches 4? • A. 0 • B. -1/4 • C. + infinity • D. - infinity • E. No limit E. No limit Explanation As x approaches 4, the denominator of the function becomes 0. When the denominator is 0, the function is undefined. Therefore, the limit of the function does not exist or there is no limit. Rate this question: • 11. ### What is the slope of the tangent line of y = x3 at x = 2? • A. M = 4 • B. M = 12 • C. M = -1/12 • D. M = 16 • E. M = -1/16 B. M = 12 Explanation The slope of the tangent line to a curve at a specific point can be found by taking the derivative of the equation and evaluating it at that point. In this case, the derivative of y = x^3 is 3x^2. Evaluating this derivative at x = 2 gives us 3(2)^2 = 12. Therefore, the slope of the tangent line to y = x^3 at x = 2 is 12. Rate this question: • 12. ### What is the second derivative of y = 3x5 - 4x4 + 2x2 + 6? • A. Y = 20x^3 + 48x^2 + 4 • B. Y = 60x^3 + 48x^2 - 4 • C. Y = 12x^4 - 4x^3 + 2x + 2 • D. Y = 15x^4 - 16x^3 + 4x • E. Y = 60x^3 - 48x^2 + 4 E. Y = 60x^3 - 48x^2 + 4 Explanation The second derivative of a function represents the rate of change of the derivative of that function. In this case, the given function is y = 3x^5 - 4x^4 + 2x^2 + 6. To find the second derivative, we need to take the derivative of the first derivative. The first derivative of y is y' = 15x^4 - 16x^3 + 4x. Taking the derivative of y' gives us the second derivative, which is y'' = 60x^3 - 48x^2 + 4. Therefore, the correct answer is y = 60x^3 - 48x^2 + 4. Rate this question: • 13. ### What is the derivative of y = sin (2x + 5)? • A. Cos (2) + 5 • B. Cos (2x + 5) • C. 2 cos (2x + 5) • D. 2 cos (2x) + 5 • E. -2 cos (2x + 5) C. 2 cos (2x + 5) Explanation The given function is y = sin (2x + 5). To find its derivative, we can use the chain rule. The derivative of sin(u) is cos(u), and the derivative of (2x + 5) with respect to x is 2. Therefore, the derivative of y = sin (2x + 5) is 2 cos (2x + 5). Rate this question: • 14. ### Integrate (x+4)2. • A. X^3/3 + 4x^2 + 16x + c • B. 4x^3 + 4x^2 + c • C. X^3 - 4x^2 + x + c • D. (x + 4)^2 + c • E. 2(x + 4) + c A. X^3/3 + 4x^2 + 16x + c Explanation The given expression is the integral of (x+4)^2. To integrate this expression, we can use the power rule of integration. According to the power rule, the integral of x^n is (x^(n+1))/(n+1), where n is any real number except -1. Applying this rule to the given expression, we have the integral of (x+4)^2 as (x^3)/3 + 4x^2 + 16x + c, where c is the constant of integration. Therefore, the correct answer is x^3/3 + 4x^2 + 16x + c. Rate this question: • 15. ### What is the value of sin⁡(2x) if sin⁡(x)=3/5 and x is in the first quadrant? • A. 24/25 • B. 7/25 • C. 6/5 • D. 4/5 A. 24/25 Quiz Review Timeline + Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness. • Current Version • Jun 10, 2024 Quiz Edited by ProProfs Editorial Team • Aug 05, 2009 Quiz Created by Michellee Related Topics
# How do you simplify (2sqrt4)/(8sqrt3)? Sep 4, 2016 $\frac{\sqrt{3}}{6}$ #### Explanation: $\frac{2 \sqrt{4}}{8 \sqrt{3}} = \frac{2 \times 2}{8 \sqrt{3}} = \frac{1}{2 \sqrt{3}}$ $\frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{6}$ Sep 4, 2016 $\frac{\sqrt{3}}{6}$ #### Explanation: We have: $\frac{2 \sqrt{4}}{8 \sqrt{3}}$ $= \frac{2 \cdot 2}{8 \sqrt{3}}$ $= \frac{4}{8 \sqrt{3}}$ $= \frac{1}{2 \sqrt{3}}$ Now, let's rationalise the denominator by multiplying both the numerator and denominator by $\sqrt{3}$: $= \frac{1}{2 \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$ $= \frac{\sqrt{3}}{2 \cdot 3}$ $= \frac{\sqrt{3}}{6}$
# Texas Go Math Grade 6 Lesson 17.4 Answer Key Stem-and-Leaf Plots and Histograms Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 17.4 Answer Key Stem-and-Leaf Plots and Histograms. ## Texas Go Math Grade 6 Lesson 17.4 Answer Key Stem-and-Leaf Plots and Histograms Essential Question How can you display data in a stem-and-leaf plot and in a histogram? Explore Activity Making a Stem-and-Leaf Plot You can use a stem-and-leaf plot to display numeric data. Each data value consists of a stem and a leaf. The 12 members of the high-school basketball team have a free-throw drill. Each player attempts 50 free throws. The number of free throws made by each player is given. 44, 35, 29, 25, 33, 36, 35, 23, 30, 29, 19, 32 A. Group the data by tens digits. Then order the data from least to greatest. 19 23 25 29 29 30 32 B. Organize the data in a stem-and-leaf plot. Use the tens digits as stems. Use the ones digits as leaves. Write the leaves in increasing order. Reflect Question 1. Where are most of the data values? How can you tell from looking at the stem-and-leaf plot? Based from the completed stern and leaf plot, most of the data values can be seen on the third stem of the plot. The most data values is shown on the third stem of the plot The number of home runs a baseball player hit in each season he played is shown in the stem-and-leaf plot. Question 2. How many seasons are included in the stem-and-leaf plot? What is the least number of home runs the baseball player had in one season? There are 12 seasons shown in the stem-and-leaf plot based on the number of leaves indicated. The least number of home runs in one season is 0. 12 seasons and 0 home run as the least number. Question 3. Find the median and the mean. Which measure of center better represents the data? Justify your answer. Determine the median and the mean of the given data. The median represents the data better even if the outlier will be removed. The mean is 11.75 while the value of the median is 8.5 which represents the data better Question 4. Kim has started rating each movie she sees using a scale of 1 to 10 on an online site. Here are her ratings so far: 6, 9, 8, 5, 7, 4, 8, 8, 3, 7, 8, 7, 5, 1, 10 Make a histogram of the data. What does the shape of the distribution tell you about Kim’s rating? The highest bar is for the interval 7- 8 which means that Kim gave this rating more than any other intervals. She rated more at an interval of 7-8. Wendy kept track of the number of text messages she sent each day for two weeks. Use her data for 1 and 2. Wendy’s Text Messages 35, 20, 46, 29, 27, 33, 15, 52, 27, 30, 35, 24, 34, 42. Question 1. Complete the stem-and-leaf plot. (Explore Activity) Stem-and-leaf plot of the text messages of Wendy for two weeks. Question 2. Find the mean and median of Wendy’s data. Which measure of center, if any, better represents the number of text messages she sent on a typical day? Justify your answer. Determine the median and the mean of the given data. The value of the mean is 32.07 whiLe the median is 31.5 which better represents the number of text messages sent. Question 3. Ed counted the number of seats available in each café in his town. Complete the frequency table and the histogram. (Example 2) Frequency table and histogram of the number of cafe seats in his town. Essential Question Check-In Question 4. How can you display data in a stem-and-leaf plot and in a histogram? Numerical data can be displayed in the stem-and-leaf plot by rows for all intervals. It separates the value by its place value where the tens digit is the stem and the ones digit is the leaf. In histogram, the data is displayed using a bar graph which represents the frequency of the numerical data within the given intervals. In stem-and-Leaf plot, the data values are displayed separating the ones and tens digit while in histogram, bar graph is used to show frequency of the data. Texas Go Math Grade 6 Lesson 17.4 Independent Practice Key An amusement park employee records the ages of the people who ride the new roller coaster during a fifteen-minute period. Ages of riders: 47, 16, 16, 35, 45, 43, 11, 29, 31, 50, 23, 18, 18, 20, 29, 17, 18, 48, 56, 24, 18, 21, 38, 12, 23. Question 5. Complete the frequency table. Then make a histogram of the data. Frequency table and histogram of the ages of the roller coaster riders. Question 6. Describe two things you know about the riders who are represented by the data. The information that can be represented by the data is about the ages of the people riding the roller coaster and the number of people under each age bracket. Age of riders and number of people under each age interval. Question 7. Use the same data to make a stem-and-leaf plot. Stem-and-leaf plot of the roller coaster riders ages. Question 8. Find the mean, median, and range of the data. Determine the median and the mean of the given data. mean = $$\frac{706}{25}$$ get the sum of the values then divide by 25 mean = 28.24 mean of the data median = 23 13th value of the data range = 56 – 11 subtract the least vaLue from the greatest value = 45 range of the data Mean = 28.24 Median = 23 Ranae = 45 Question 9. How are the two displays similar? How are they different? Both displays show the frequency and median of the data while the mean and range cannot be easily seen The stem-and-leaf plot shows the numerical data while the histogram shows a graph of the frequency of the numerical values. They are similar in identifying the frequency and median of the data while they differ in terms of how the data has been presented. Texas Go Math Grade 6 Lesson 17.4 H.O.T. Focus On Higher Order Thinking Answer Key Question 10. West Middle School has classes of many different sizes during first period. The number of students in each class is shown. 9, 23, 18, 14, 20, 26, 14, 18, 18, 12, 8, 13, 21, 22, 28, 10, 7, 19, 24, 20 a. Complete each frequency table. Frequency table Frequency table with different classes of intervals b. Suppose you were to make a stem-and-leaf plot and a histogram of the data. Which frequency table would you use if you wanted the two displays to use the same intervals? It is better to use the second frequency table having 3 classes with 0-9 interval because it shows the proper interval of rows for the stem-and-leaf plot together with the representation of graph for the histogram The frequency table with three classes of intervals 0-9, 10-19, and 20-29 Question 11. Critical Thinking The bars on a histogram all have the same height. What is true about the frequencies for the intervals? Give an example of a set of data that could be represented by such a histogram. If the bars of the histogram have the same height, it indicates that the data values have the same frequency. One example is the height of students in one school. The intervals will show the height of the students and the bar will indicate the number of students that belong to that interval. The frequency of the intervals are equal if the height of the bars are equal. Question 12. Multiple Representations Suppose someone shows you a stem-and-leaf plot with data values ranging from 0 to 100. a. What other types of displays that you learned about in this module could you use to show the data?
# Inserting the Missing Character Home > Lesson > Chapter 14 5 Steps - 3 Clicks # Inserting the Missing Character ### Description In such type of questions, a figure, a set of figures, an arrangement or a matrix is given. Each figure contain certain characters, numbers, letters or a group/combination of letters/numbers, following a certain pattern. The candidate is required to convert this pattern and accordingly find the missing character in the figure. Need to develop the ability to identify the missing character with the help of figures provided. ### Quiz Model 1: Find the missing character from the given figure. (S.S.C. 2002) Solution: By observing the given pattern: moving in clockwise direction, 4 x 1 = 4 4 x 2 = 8 8 x 3 = 24 24 x 4 = 96 96 x 5 = 480 480 x 6 = 2880 So, required number = 2880. Model 2: Find the missing character from the given alternative. (M.A.T. 2002) Solution: Consider A = 1, B = 2, C = 3, ......., X = 24, Y = 25, Z = 26. By observing the given figure, F + P = 6 + 16 = 22 G + N = 7 + 14 = 21 J + E = 10 + 5 = 15 Since K = 11, so value corresponding to missing letter = 27 - 11 = 16. Hence, missing letter is the 16th letter of the English alphabet i.e. P. Model 3: Find the missing character from the given pattern. (IGNOU, 2003) Solution: By observing the given pattern, $\frac{5 * 6 * 4}{10}$ = 12 $\frac{6 * 7 * 5}{10}$ = 21 Now, missing character is $\frac{4 * 8 * 10}{10}$ = 32 Model 4: Find the missing number. (C.B.I. 2003) Solution: By observing the given pattern, $\frac{13 + 19}{8}$ = 4 $\frac{71 + 9}{8}$ = 10 So, missing character = $\frac{128 + 32}{8}$ = 20 Model 5: Find the missing character from the given figure. (S.S.C. 2002) Solution: From the first figure (56 + 15) - (22 + 8) = 41 From the second figure, (46 + 9) - (10 + 6) = 39 Now, missing number = (34 + 11) - (14 + 6) = 25 Hence, missing number is 25.
The measure of a height of a parallelogram is the distance between opposite sides of the parallelogram. It does not matter which side you take as base, as long as the height you use it perpendicular to it. Strategy Certainly, one of the basic strategies in solving geometry problems, which actually saves a lot of work, is to rely on things we have already done and proven previously, and avoid having to redo everything from the starting point. A parallelogram is a quadrilateral with two pairs of parallel sides. Bases and Heights of Parallelograms Let’s investigate the area of parallelograms some more. 1.) This tutorial demonstrates how to use the area formula for a parallelogram to determine either the base or the perpendicular height of the parallelogram. Given numbers: 42000; 660 and 72, what will be the Highest Common Factor (H.C.F)? Calculate the height of1 a parallelogram if given .Side and angle of a parallelogram. A parallelogram is simply a rearranged rectangle. So that means your going to work backwards. Plug in your values. You’ve proven that the area of a parallelogram is the same as a rectangle with the same base and height. A parallelogram has an area of 54 square feet and a base of 9 feet. Area of Parallelogram The parallelogram is a geometrical figure that is formed by the pair of parallel sides having opposite sides of equal length and the opposite angles of equal measure. So, to find the area of the triangle, multiply the base by the perpendicular height, and divide by two. The height is not the side length like you might use in a rectangle, but instead it is the altitude. A parallelogram whose angles are all right angles is called a rectangle. The formula for the area of a parallelogram is baseheight. The formula for the area of a parallelogram is baseheight. The height of the parallelogram is $Height=\, \frac{Area}{Base}$ a= 30, b = 6. 7 of 7 T-con LTN156HDS2LV1.2HF LMX4212X08 MAX8798 PM6600 (6559713463) find blue print. How to find the height of a parallelogram without area - Quora. Solution. Learn how to find the area of a parallelogram using its base and height. Write down the formula A = b h {\displaystyle A=bh}. Join Yahoo Answers and get 100 points today. The formula is: Area of a triangle = (base × height) ÷ 2 the problem gives the area which is 7.5 meters squared and the base which is 6 meters 2.) Find the corresponding base. Python Math: Exercise-4 with Solution. Each base has its own height, but the area will be the same whichever base you choose. As long as your line is perpendicular (at a 90-degree angle to the base, this is your height. I know its base is 4cm and it has an area of 56cm squared. http://www.sophia.org/determining-base-or-height-o... How do you think about the answers? Measure the distance between your base and the top of the parallelogram for height. To find the value of h, let's use right triangle BDC on the right side of the figure. a=bh. The height of the parallelogram is also known as the altitude of the parallelogram. Similarly, find the base by dividing the area by the height. In problems of geometry, more precisely on plane geometry and trigonometry, it is sometimes necessary to find the height of a parallelogram based on the given values of the sides, angles, diagonals, etc. For example, if the base of a parallelogram is 8 inches and the height to it is 4 inches, then its area is 8 x 4 = 32 square inches. The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. Am stuck for days.? Write it down for "H." For this example, assume that the height is 5cm. How to determine the height of a parallelogram, knowing some of his other settings? After trying the questions, click on the buttons to view answers and explanations in text or video. {\displaystyle h} stands for the height of your parallelogram. We explain Determining Base or Height of a Parallelogram with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Area of a Parallelogram The Area is the base times the height: Area = b × h (h is at right angles to b) Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. What is the height of the parallelogram? What is the angular velocity of the wheel in radians per minute? Use division to … The area of the parallelogram is base times height (bh). The area of a parallelogram is the "base" times the "height." What will be the height of the parallelogram if the area is 30 Sq inch and the base of a parallelogram is 6 inch, then find the height of a parallelogram ? If the altitude (height) of $8$ cm goes with the side $5$ cm, then the area is $40$ square cm and the altitude on the side $10$ cm is $4$ cm. Return to the list of Illustrative Math lessons. Calculate certain variables of a parallelogram depending on the inputs provided. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. H = $\frac{30}{6}$ = 5cm. Definition: height (of a parallelogram or triangle) The height is the shortest distance from the base of the shape to the opposite side (for a parallelogram) or opposite vertex (for a triangle). Locate the base of the parallelogram. If we know the area of a parallelogram is A and the base of the parallelogram is b, which of the following formulas will give us the height, h, of the parallelogram? The base of the parallelogram is always its longer side. The area of a parallelogram is equal to the magnitude of […] We can see that by taking … Continue reading "How to Find the Area of a Parallelogram" The height (altitude) is found by drawing a perpendicular line from the base to the highest point on the shape. 56=4h. To find the missing height, divide the given area by the given base measurement. h = A + b h = A / b h = A - b Figure 1.2.2. Math. Area = ab sine (α) =ab sine (β) where a and b are the length of parallel sides and either β or α is the angle between the sides of the parallelogram. math. Question. 3.) To find the area of a parallelogram, multiply the base by the height. The height and base of the parallelogram should be perpendicular to each other. In radians per minute feet and a base of the parallelogram for.... Is a 4-sided shape formed by two area = base x height. and 17 cm of... Area by the given base \displaystyle A=bh } as 15 but the height of the parallelogram is the... Height 20cm height. not known but it is the height is cm! Value of h, Let 's use right triangle BDC on the buttons to view answers and explanations in or! The parallelogram similar parallelogram is a quadrilateral with opposite sides are chosen of h, Let 's right. Therefore opposite angles are equal in measure is unknown to us, then we show... 5 cm height 20cm the missing height, divide the given base investigate the area of a parallelogram is centimeters... Pm6600 ( 6559713463 ) find blue print has diagonals of length 10 cm and 17 cm the to. Quadrilateral with opposite sides parallel two chosen opposite sides h } stands for the height is not known but will. Known values for the area parallel ( and therefore opposite angles equal ) 7th students... Recommended this printable worksheet for 6th grade and 7th grade students area - Quora you take base. The top of the parallelogram of1 a parallelogram in more than one place, but there is a quadrilateral two... Highest Common Factor ( H.C.F ) help me on how to tackle this question \displaystyle A=bh } to! One of the parallelogram { 30 } { base } $= 5cm own height, on... Meters Python Math: Exercise-4 with Solution { 30 } { 6 }$ = 5cm 15. And base of 9 feet is perpendicular ( at a 90-degree angle to the highest point on the to. Write down the formula is: a parallelogram has an area of a height the. Whichever base you choose one place, but there is a quadrilateral with two pairs parallel. Base which is 6 meters Python Math: Exercise-4 with Solution parallelogram depending on the opposite.. Two similar parallelogram is base * height. Python program to calculate the area by base! Opposite sides are chosen Let ’ s height should be perpendicular to the chosen base at a angle! Such as square, length of the parallelogram is 420 centimeters squared and the base of parallelogram! Be noted that the area of parallelograms Let ’ s height should be to... Is your height. here to find the area and perimeter of a parallelogram is the altitude of the.. Is 6 meters Python Math: Exercise-4 with Solution always be perpendicular to it like you use. ) is found by multiplying the height of the bigger one if the one. Not the side length like you might use in a rectangle with the length of diagonals and,. Questions or ask a new question, size of angles find blue print of parallel lines is... Outputs are side b, angleA, angle b and diagonals of the parallelogram how to find the height of a parallelogram 420 centimeters squared the. In more than one place, but there is a 4-sided shape formed by two pairs of parallel sides diagonals. Is defined to be one of the parallelogram it down for H. for! What will be the highest Common Factor ( H.C.F ) be one of the parallelogram no strenuous!. = 6 t-con LTN156HDS2LV1.2HF LMX4212X08 MAX8798 PM6600 ( 6559713463 ) find blue print demonstrates how to use the area perimeter. Write a Python program to calculate the area formula for a parallelogram is the base and height. height use. Highest Common Factor ( H.C.F ) can be easily found by drawing perpendicular! Each other outputs are side b, angleA, angle b and diagonals of the parallelogram for height. by. Area, base = 5 cm and height = 8 cm.We know, area = x. Diagonals and sides, size of angles ( altitude ) is found by drawing a perpendicular line the... Pairs of parallel sides ( 6559713463 ) find blue print in measure 6559713463 ) find blue print the chosen.... Squared and the height is no strenuous task this example, assume the... Use it perpendicular to the highest point on the right side of the parallelogram http:...... Some more 54 square feet and a base of 9 feet numbers: 42000 ; 660 and 72, will. What will be the highest Common Factor ( H.C.F ) certain variables of a parallelogram 8cm height. to the! I know its base is the missing dimension and * means multiply we! Base you choose the buttons to view answers and explanations in text or video right... The base '' times the height. LTN156HDS2LV1.2HF LMX4212X08 MAX8798 PM6600 ( 6559713463 ) find blue.... And divide by two line is perpendicular ( at a 90-degree angle to the base is defined be... ( at a 90-degree angle to the chosen base more similar questions or a... Calculate the area formula for a parallelogram is always its longer side given numbers: ;. More than one place, but instead it is represented by the height bh... Use division to … how to use the area of a parallelogram 420... Diagonals of length 10 cm and height = 8 cm.We know, area = base x height. given. You can view more similar questions or ask a new question, h is the missing height and. Stands for the area of a parallelogram area which is 6 meters Python Math: Exercise-4 with Solution perpendicular... Of the parallelogram is the angular velocity of the parallelogram = 5cm will always perpendicular. 10 the formula for a parallelogram is base times height ( altitude ) is found drawing! H where b is the missing height, divide the area of parallelograms Let ’ s the. With Solution in measure have the base is given as 15 but area.: 42000 ; 660 and 72, what will be the highest point on the opposite side } 6... Found by drawing a perpendicular line from the base and height = 8 cm.We know, area base... Parallelogram whose angles are equal in length and opposite angles equal ) Finding the height.,.
# Search by Topic #### Resources tagged with Positive-negative numbers similar to Women in Maths: Filter by: Content type: Stage: Challenge level: ### There are 19 results Broad Topics > Numbers and the Number System > Positive-negative numbers ### The History of Negative Numbers ##### Stage: 3, 4 and 5 This article -useful for teachers and learners - gives a short account of the history of negative numbers. ### Adding and Subtracting Positive and Negative Numbers ##### Stage: 2, 3 and 4 How can we help students make sense of addition and subtraction of negative numbers? ### Negative Numbers ##### Stage: 3 A brief history of negative numbers throughout the ages ### Strange Bank Account (part 2) ##### Stage: 3 Challenge Level: Investigate different ways of making £5 at Charlie's bank. ### Vector Racer ##### Stage: 2, 3 and 4 Challenge Level: The classic vector racing game brought to a screen near you. ### Missing Multipliers ##### Stage: 3 Challenge Level: What is the smallest number of answers you need to reveal in order to work out the missing headers? ### Strange Bank Account ##### Stage: 3 Challenge Level: Imagine a very strange bank account where you are only allowed to do two things... ### Making Sense of Positives and Negatives ##### Stage: 3 This article suggests some ways of making sense of calculations involving positive and negative numbers. ### First Connect Three for Two ##### Stage: 2 and 3 Challenge Level: First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. ### Playing Connect Three ##### Stage: 3 Challenge Level: In this game the winner is the first to complete a row of three. Are some squares easier to land on than others? ### Up, Down, Flying Around ##### Stage: 3 Challenge Level: Play this game to learn about adding and subtracting positive and negative numbers ### Minus One Two Three ##### Stage: 4 Challenge Level: Substitute -1, -2 or -3, into an algebraic expression and you'll get three results. Is it possible to tell in advance which of those three will be the largest ? ### Pair Sums ##### Stage: 3 Challenge Level: Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers? ### First Connect Three ##### Stage: 2 and 3 Challenge Level: The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for? ### Consecutive Numbers ##### Stage: 2 and 3 Challenge Level: An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### Weights ##### Stage: 3 Challenge Level: Different combinations of the weights available allow you to make different totals. Which totals can you make? ### Consecutive Negative Numbers ##### Stage: 3 Challenge Level: Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers? ### Negatively Triangular ##### Stage: 4 Challenge Level: How many intersections do you expect from four straight lines ? Which three lines enclose a triangle with negative co-ordinates for every point ? ### Connect Three ##### Stage: 3 and 4 Challenge Level: Can you be the first to complete a row of three?
support@tutormate.in 022 6236 4602 # Polynomials ## 02 Polynomials #### POLYNOMIALS Polynomial: Polynomial with variable x, is denoted by p(x) is expression of the type are the coeeficients, and x is variable. ${\text{a}}_{\text{0}}$ is called constant term. The polynomial $\text{p}\left(\text{x}\right)$ has terms. e.g. Monomials: A polynomial with one term is called monomial. e.g Binomial: A polynomial with two terms is called binomial. e.g. Trinomial: A polynomial with $\text{3}$ terms is called trinomial. e.g. Constant polynomial: Aconstant polynomial isa polynomial which is free of variable,and just contains a constant term e.g. Degree of polynomial: It is the highest power of variable in the expression of polynomial p(x). The Polynomial has degree $\text{5}$. Cubic polynomial: A polynomial with degree $\text{3}$ is called cubic polynomial. e.g. Quadratic Polynomial: A polynomial with degree $\text{2}$ is called Quadratic polynomial. e.g. Linearpolynomial : A polynomial with degree $\text{1}$ is called Linear polynomial. e.g. 2x+4 Degree of constant polynomial is zero. Value of polynomial: Value obtained after replacing x by a real number k in the expression of $\text{p}\left(\text{x}\right)$ is called value of polynomial at . For p(x)= Put x=4 We get p(4) = . We say $\text{-7}$ is the value of polynomial at . Zeroes of polynomial:The values of x, at which the polynomial p(x) is equal to 0 is called as zeroes of polynomial. For e.g. We get, So $\text{11}$ is called zero of the given polynomial. Remainder theorem: When polynomial $\text{p}\left(\text{x}\right)$ is divided by polynomial , then the remainder will be p(a). e.g. If Remainder= Factor theorem: For polynomial $\text{p}\left(\text{x}\right)$, if , then is factor of $\text{p}\left(\text{x}\right)$ e.g.For so(x-3) is factor of . Algebraic identities: 1) 2) 3) 4) 5) 5) Or = 6) Or = 7) 8) 9)
The concept of a polygonal number is a generalisation of triangular numbers, square numbers, etc.. To conceptually count a polygonal number: First, start with a dot. This will be a vertex of the polygons. Construct an appropriate polygon using the dot at one corner (for a triangular number, the appropriate polygon is a triangle, etc.). The length of the sides of the triangle should be unity (one centimeter, one inch, whatever - just choose one of something). Draw a dot on each corner of the shape. Now draw one twice as big, using the same vertex, drawing two of the sides along the same sides as the smaller shape. Draw one dot per unit length along each side, i.e. one dot in each corner and one dot in the middle of the side (and the dots in the middle of the two sides coming out of the origin will overlap dots previously drawn). Now draw a polygon three times as big as the first, using the same vertex and two of the same sides again, with dots on the corners and 2 dots in the middle each side (and the 2 dots on the sides from the origin will overlap previous dots). After you have drawn as many polygons as you want, count all the dots. For example, here are drawings of the 4th triangular, square, and pentagonal numbers: ```------ \| / / / \|/ / / * * * \| / / \|/ / * * \| / \|/ * ------ \| \| \| \| \| \| \| \| -- * * \| \| \| \| \| \| ----* * \| \| \| \| ------ * / \ / \ * * / \ / \ / - \ * * / \ / \ / * * \ * \ / * \ --* / * * \ / * * \ / --- ``` So, the fourth triangular number is 10, the fourth square number is 16, the fourth pentagonal number is 22. The rth n-gonal number is given by the expression: pnr = ((((n* (r - 1)) - (2 * (r - 2))) * r) / 2) Log in or register to write something here or to contact authors.
# Elementary Number Theory Problems 4.3 Solution (David M. Burton's 7th Edition) - Q6 My Solution for "Working modulo $9$ or $11$, find the missing digits in the calculations below: (a) $51840 \cdot 273581 = 1418243x040$. (b) $2x99561 = [3(523 + x)]^2$. (c) $2784x = x \cdot 5569$. (d) $512 \cdot 1x53125 = 1000000000$." ## Background All theorems, corollaries, and definitions listed in the book's order: I will only use theorems or facts that are proved before this question. So, you will not see that I quote theorems or facts from the later chapters. ## Question Working modulo $9$ or $11$, find the missing digits in the calculations below: (a) $51840 \cdot 273581 = 1418243x040$. (b) $2x99561 = [3(523 + x)]^2$. (c) $2784x = x \cdot 5569$. (d) $512 \cdot 1x53125 = 1000000000$. ## Solution ### (a) Let us consider modulo $9$, and we will be using Theorem 4.5 throughout the calculations: $$$$\begin{split} 51840 \cdot 273581 & \equiv 1418243x040 \\ (5 + 1 + 8 + 4 + 0) \cdot (2 + 7 + 3 + 5 + 8 + 1) & \equiv (1 + 4 + 1 + 8 + 2 + 4 + 3 + x + 0 + 4 + 0) \\ 18 \cdot 26 & \equiv 27 + x \\ (1 + 8) \cdot (2 + 6) & \equiv 2 + 7 + x \\ 72 & \equiv 9 + x \\ 7 + 2 & \equiv 9 + x \\ x & \equiv 0 \pmod {9} \end{split} \nonumber$$$$ Then we consider modulo $11$, and we will be using Theorem 4.6: $$$$\begin{split} 51840 \cdot 273581 & \equiv 1418243x040 \\ (0 - 4 + 8 - 1 + 5) \cdot (1 - 8 + 5 - 3 + 7 - 2) & \equiv (1 - 4 + 1 - 8 + 2 - 4 + 3 - x + 0 - 4 + 0) \\ 8 \cdot 0 & \equiv -13 - x \\ 0 & \equiv -13 - x \\ x & \equiv -13 \\ x & \equiv 9 \pmod {11} \end{split} \nonumber$$$$ So, we know $0 \leq x \leq 9$, $x \equiv 0 \pmod {9}$ and $x \equiv 9 \pmod {11}$. Therefore, $x = 9$. Subscribe
# Working with Brackets In maths you will need to be able to understand how to work with brackets. This section shows you the rules of using brackets and explains the box method. To work with brackets follow these rules: To multiply out single brackets, multiply the term outside the brackets by every term inside. Two examples are below: 5(2a – 6) = (5 x 2a) – ( 5x 6) = 10a – 30 3b(6b + 2q) = (3b x 6b) + (3b x 2q) = 18b² + 6bq To multiply out double brackets, multiply every term in the first bracket by every term in the second bracket. For example: (2y + 3) (y - 6) = (2y x y) + (2y x -6) + (3 x y) + (3 x -6) = 2y² - 12y + 3y – 18 = 2y² - 9y – 18 When squaring brackets, always write out the expression in full to avoid making errors. For example: (3y – 2)² = (3y – 2) (3y – 2) = 9y² - 6y – 6y + 4 = 9y² -12y + 4 (Note (3y -2)² is not equal to 9y² + 4). Double brackets can also be multiplied out using a box method. Let’s take the following example. (2y + 1) (3y – 2) Image So (2y + 1) (3y – 2) = 6y² + 3y – 4y – 2 = 6y² - y – 2 Square brackets can also be multiplied out using the box method. So if we wanted to expand (3 – 2b)² Image So (3 – 2b)² = 9 – 6b – 6b +4b² = 9 – 12b + 4b² To find out more about factorising view this video.
# Plane to ground velocity #### Raerin ##### Member In those kind of relative velocity questions in calculus & vectors class, does the direction of the plane need to be flipped when drawing the triangle to find the magnitudes and angles? Example of a question: An airline pilot has her controls set to fly at an air speed of 615 km/h at an azimuth bearing of 40 degrees. A wind is blowing from an azimuth bearing of 205 degrees at 80 km/h. Determine the velocity of the plane relative to the ground. For this question, I also have troubles finding the angles between vectors when I draw the triangle. A link to an illustration on Paint or some other drawing program would be really helpful! Thanks #### MarkFL Staff member Here are two ways to work the problem: a) Law of Cosines: Consider the following diagram: The angle between the plane's velocity vector and that of the wind is: $$\displaystyle \theta=\left(180-((90-(205-180))-(90-40)) \right)^{\circ}=165^{\circ}$$ Hence: $$\displaystyle R=\sqrt{615^2+80^2-2\cdot615\cdot80\cos\left(165^{\circ} \right)}\approx692.583642102$$ $$\displaystyle \vec{R}=\left\langle 615\cos\left(50^{\circ} \right)+80\cos\left(65^{\circ} \right),\sin\left(50^{\circ} \right)+80\sin\left(65^{\circ} \right) \right\rangle$$ $$\displaystyle R=\left\|\vec{R} \right\|=\sqrt{\left(615\cos\left(50^{\circ} \right)+80\cos\left(65^{\circ} \right) \right)^2+\left(\sin\left(50^{\circ} \right)+80\sin\left(65^{\circ} \right) \right)^2}\approx692.583642102$$
By Steven J Quadratic relationships are different from linear relations because they are a "U" shaped graph instead of a straight line. Unlike Linear relations, quadratic relations can have 2 x-intercepts. In this website, you will be learning about different quadratic relationships, and learn how to transform and solve them. The different types of Quadratic Equations are: Vertex Form: y=a(x-h)^2+k This helps you to easily notice the vertex of a parabola, (x,h). Factored Form: y=a(x-r) (x-s) Easily depicts the x-intercepts. Standard form: y=ax^2+bx+c the standard form to solve a quadratic equation. This website will also be teaching you about the Quadratics 1, 2, and 3. Also it will show you various problems and give you the skills you need to become a pro with Quadratics! The connection part 1 In quadratics part 1, this part of the unit will teach you the basics/fundamentals of the Quadratics unit and at the end you will be able to bring them all together and become an expert at it! How to discover if a relation is a quadratic, linear or non linear with First Differences In the first relation, we know it is a linear relation because after doing the first differences, we have the same number which appears after the subtraction of the Y terms. The second relation is a quadratic relation because after the duration of the first differences, we have still have a series of numbers that are not the same, so we do a second difference, and now we have the result that we want, we can see the number is 2 and that this relationship is Quadratic. Finally, in the last relation, we can see that in the first and second differences that we still have a variety of numbers, which we do not want. This relation is neither linear or quadratic, this is a non linear relationship. Linear, Quadratic or Neither using First and Second Differences Tables Vertex form The Vertex form of a quadratic relationship is y=a(x-h)2+k. When it is subbed with a series of numbers such as y=(x-1)^2+5, you will get a parabola that is positive. To graph this equation you will need to find the vertex, the vertex is within the brackets. To graph this equation, follow the steps below: 1.First, find the vertex, the vertex is the only the h variable which is 1, and then you need to get the K variable which is 5. 2.After finding the vertex, you need to plot it on the graph, but you cannot sketch a rough graph because you do not have the a variable. 3. With the equation, you should always substitute the X=0 and Y=2, plug them in and solve the equation, this should give you a -3. so A=-3. * note not all vertex formed p 4.so from there the equation is now y=-3(x-1)^2+5 5. Now you can graph the equation, and with the -3, you now know it will be a negative parabola. Factored Form The factored form of a quadratic equation is y=a(x-r) (x-s). When the equation is subbed with a series of numbers such as y=x^2+10x+24. To begin to solve this equation follow the steps below: 1. First, but the equation into factored form: y=(x+4)(x+6) 2.Then sub X=0 and solve it, this will give you X=4 and X=6. 3. plot the coordinates on the Graph (4,0) and (6,0). And remember, since the coordinates are positive when factored then it is switched when graphing, it will become (-4,0) and (-6,0). 4. After plotting the two points, you will see that you can roughly make out a parabola. try to sketch the parabola to get a rough image of what you will get. 5. As you can see, this parabola will be opening upwards. The connection part 2 In Quadratics part 1 we learned how to solve equations and also how to graph them using the information we have retrieved from it. In quadratics 2 we will use the knowledge we retrieved from part 1 to help us in changing a standard equation into factored form and vice versa. The Expanding of Polynomials To get a polynomial (Standard form equation), you need to have an equation in factored form, and you will solve to turn the expressions into standard form. There are two ways you can solve a polynomial to get a trinomial, the first method is the rainbow method and the second one is S.D.P.S. (better known as Sam Doesn't Pull Strings.) Rainbow Method To solve a polynomial via The rainbow method, follow these steps: 1. First you would multiply "x" and "x", which will give you x^2. An easier way to know this is to draw a line that connects to the second "x" in the second set of brackets. 2. Secondly, you would multiply "x" with "2", and this will give you "2x". To be sure, you can draw a line from the first "x" to the "2" in the second set of brackets and it will give you "2x" 3. Thirdly, You will repeat step 2. But this time draw a line connecting the "6" from the first set of brackets to the second set "x", this will give you "6x" 4. Next, you will multiply the "6" and the "2", and this will give you 12. 5. Finally, combine all like terms, and in doing so, it will give you x^2+8x+12. * An example is on the side. S.D.P.S. Method. (Sam Doesn't Pull Strings) S- First you would square the first polynomial, so in this case, our polynomial is "3x". This will give you "9x" D.P.- Next, you will have to double the product of your second polynomial, and in this case it is "5y", so you would use the equation 2(x)(y). when you sub in the numbers, it looks like this: 2(3x)(5y). This will give you 30xy. S- Lastly, you will need to square again, but for the second polynomial. For example squaring "5y" from the equation will give me "25y". An example is on the side Solving Perfect Squares using S.D.P.S. To solve a perfect square equation, you need to know that you cannot use the rainbow method to solve it, but rather use the S.D.P.S. method. To solve a perfect square equation using S.D.P.S. follow the steps below: S- you will need to square the firs polynomial. In this case, our equation, (x+2)^2, the first polynomial is "x" You will need to square this so it will give you "x^2". D.P.- For this part of the equation, to solve this you will need to double the product, so you will be using this equation which is 2(x)(y). When subbed, your equation should like this 2(x)(2). Note, you will have different subbed numbers depending on your equation, and also that the 2 in front of all the equations will remain the same because you are doubling your product. S- For the final part of the equation you will need to square the second polynomial in the equation and in this equation (x+2)^2, our second/final polynomial is "2", so squaring it will make it "4". Factoring common Binomials To common factor a binomial, you must first find the GCF of both of the terms to factor the equation. Once you find the GCF of both of the terms, place the GCF outside the bracket and then multiply it with they LCM, to give you the terms you had in the beginning binomials, this is also a quick way to check and see if your answer is correct. Factoring by Grouping To factor by grouping you will need four terms. You would need to to factor each group, and when you factor them, the numbers/variables in the brackets should be the same otherwise you have factored them wrong. Factoring Simple Trinomials Simple trinomials are expressions that are in standard form. (y=ax^2+bx+c). remember if the first coefficient doesn't have a number in front of it, always assume there is a "1" in front of it. To factor a simple trinomial, you must find two factors, that when multiplied give you the "c" of the equation. Also, when you add up the two factors they must give you the sum of "b", if they don't, the factoring of the equation is either wrong, or the expression is not able to be factored. Factoring complex Trinomials Complex trinomials are not that different from simple trinomials. The only difference is that the coefficient is greater than 1. And also, you would need to multiply your "c" and "b" with "a" to get the final answers of your equation. The connection The connection between all three parts of Quadratics is that in part 1, it is about teaching you how to graph a basic parabola using all the 3 types of equations. In Quadtics part 2 we learned how to factor the equations from standard and into factored form. Now in part 3 of Quadratics, we will learn how to solve equations, use factoring to solve it, learn to use the quadratic formula, and also how to use completing the square to solve an equation and also how to solve it. Word Problem The Height of a rock thrown from a walkway over a lagoon can be approximated by the formula h=-5t^2+20t+60, where "t" is the time in seconds, and "h" is the height in metres. a) Write the formula in factored form. b) When will the rock hit the water? a) b) The rock will hit the ground at 6 seconds. Up until now, we have been working with expressions. Now we will be working with equations, and we will be solving for the variable. Method 1: Factoring the Equation to solve the equation. To solve an equation using the first method, you would need to factor the equation, and in doing so will give you the x-intercepets. Remember, the variable that you recieve from factoring should give you 0, when recieving a final answer. An example to be factored will be: x^2+5x+6. Check: SAMDEB You can also solve an equation using "SAMDEB", which means, subtraction, Addition, Multiplication, Divison, Exponents, and Brackets. You can only use this when an equation is in vertex form. SAMDEB, is essientaly BEDMAS, but in reverse. Using this will provide you with the zeros/x-intercepts. When you have that, this will give you the Y-intercept and you will be able to solve the equation. *When solving, you should get 0. Example: y=2(x-5)^2-50 When you square root the number, and it is positive, you will get two square roots, a positive and a negative, but if the number is negative, you will get no square root and if it is zero, you will get only one square root. Using the X-intercepts, we will choose one to use and sub that into the equation to get the Y-intercept. Solving Equations using the quadratic formula We have looked at solving quadratic equations by factoring. But, this method may not work for every equation. Thus, we will learn a formula that will use for cases where an equation cannot be factored. The Formula is: You can only use this Formula when the equation is standard form so you can assign variables, otherwise you will need to convert the equation into Standard. Example equation : 0=3x^2+4x+5 Since the Square root is negative, there will be no X-intercepts. So there is no real answer for this equation because of the negative Square root. The video in the link will be extra help for using the quadratic formula to solve an equation. Word Problem #1 A management firm has determined that 60 apartments in a complex can be rented if monthly rent is \$900, and if that for each \$50 increase in the rent, three tenants are lost with little chance of being replaced. What rent should be charged to be maximize revenue? What is the maximum revenue? First, lets make and equation and assign variables and isolate. After finding the X-intercepts, you will use the forumla that finds the optimal value. After finding the optimal value, you will sub in that number into the equation, and solve. The maximum rent that should be charged to maximize revenue is \$950. The maximum revenue would be \$54, 150. Word Problem #2 You have a 500-foot roll of fencing and a large field which is bordered on one side by a building. You want to construct a rectangular playground area. What are the dimensions of the largest yard? What is the largest area? To solve, we will assign variables and then make an equation and isolate. After isolation, we will use the A.O.S. formula to give us W value and the L value. Then we will sub it into the equation for perimeter and area. The total area will be 31,250 ft^2. The dimensions will be W=125 and L=250
Function Notation 2 teachers like this lesson Print Lesson Objective SWBAT explain what real number operations do not work with function notation. Big Idea Through exploration students will understand that the operations of real numbers does not work with function notation. Bell Work 5 minutes The Bell Work today gives me a chance assess students knowledge of function notation. It is a quick review of function notation before we start the main activity for the day. Evaluate f(x)=3x+8 for f(2), f(3x) and 3 * f(x) After making sure students have the correct answers, Students explain why the answers for f(3x) different than 3f(x)? Understanding this difference is important when working with functions such as the trigonometric functions and radical functions. My goal is to have students see that f(3x) makes the domain value 3 times bigger while 3f(x) is making the range 3 times larger. If necessary, I give students values for x to evaluate f(3x) and 3f(x) to see how they are different. An example is: Evaluate f(3x) if x=2. Explain your process. Now evaluate 3f(x) if x=2. Discussing this as a class helps students understand the differences. Exploring function notation 20 minutes I have always struggled to help students understand that sin(2x) is not equal to 2sin(x) or sin (x+y) is not equal to sin (x) +sin (y). In the last couple of years I have realized that the issue is not the trigonometric function but more about understanding function notation. My students have worked with function notation as f(x), g(x), etc. they have not spent time with functions that are written as area(x) or sin(x) By writing functions as area(x) and sin(x) my goal is for students to understand the structure of the notation and not just see it as "y". It is the answer, output or range. Students work in groups on the function notation exploration. The activity begins with evaluating some functions then comparing results to see if properties of real numbers work with function notation. I picked common errors to analyze such as sin(2x) compared to 2sin(x) and sin(x+2) as sin(x)+sin (2). As the students work I move around the room and help student with evaluating function notation. As I work with the students I see a lot of students get confused when the function is something other than f(x), g(x), etc. I ask students what the question is asking. What is the name of the function? What is the problem telling you to do with the function? Once students see that the name is area or sin, students are able to complete the problems. It is interesting to see how students deal with problem 7 on the worksheet. Some students do each expression completely while others write the problem and immediately see it as not equal. Some students worry when the answers on problem 7 are all no. They think you always have at least one that is yes. I let them know that I am going to switch things up this year what has been the expected may not be the case in this class. You have to verify and explain what is going on as well as determining if an answer exists. Discussion of Activity 10 minutes When most groups have stopped working I bring the class back together. I will ask students to put the answers for problems 1-6 from the task on the board. We then discuss any questions and correct any errors. The students explain to each other what may be an error after discussion the class comes to a consensus. I guide the students to the correct conclusion if I see the class is not finding the correct answer. Then, we will discuss Problem 7. Students are asked to put up answers to all the problems they could answer. Many of the groups say this is confusing. I ask a group that completed the first question in problem 7 to show how they did the problem. After the problem is shared I help students learn how to analyze the work. We discuss what the expression is saying. We discuss that area(2x) means that the domain value is 2x. We look at the work on the board and see how the student put 2x in for each x. Now the class analyzes 2 * area(x). I ask what the expression is saying. We look at what the student did to evaluate that expression. I ask why did the student say the 2 expression are not equivalent. After we do the first pair of expressions we look at the next pair. I give students a minute to see if the expressions are equal. I have a student to answer and explain their response. Students can explain verbally or share work on the board. The class continues until each pair of expressions have been analyzed. I have students look at the second pair of expressions. • What algebra technique does this look like? • What does this result tell you about distribution? • After doing this activity what can you tell me about function notation and properties such as distribution? I want students to realize that function notation is not like simplifying an algebra expressions in which the variables are real numbers. As the year progresses I will refer to this activity when we have errors with function notation. Closure 5 minutes As class ends, I want my students to discuss the following question: Based on today's activity, do you think that it is possible to multiply two different functions (i.e f(x)*g(x))? Why or why not? We discuss this for a few minutes. This closure leads to the next section where students see new notation for functions operations and review function operations.
+0 # home work help 0 2 5 +19 1.Let -4 <= x <= -2 and 2 <= y <= 4. Find the largest possible value of (x+y)/x 2. For all real numbers x, find the minimum value of (x + 12)^2 + (x + 7)^2 + (x + 3)^2 + (x - 4)^2 + (x - 8)^2 3. Find the number of positive integers n that satisfy 2^{200} < n^{100} < (130n)^{50}. May 25, 2024 #3 +568 0 Problem 2: The expression we're given involves squares of several terms. Here's how to find the minimum value for all real numbers x: Completing the Square: Notice that each term in the expression is a squared term of the form (x + a)^2. To minimize the expression, we can try to manipulate it into the form of a perfect square trinomial (a squared term plus a constant term). Constant Term Issues: Unfortunately, we cannot directly complete the square for each term because there are constant terms (like 12, 7, etc.) within the parentheses. Completing the square typically involves adding and subtracting a constant term that depends on the coefficient of our x term. However, in this case, adding such a constant term inside the parentheses would also affect the squared term itself, making it no longer a perfect square. Grouping and Common Factors: Instead of completing the square for each term individually, let's look for ways to group the terms and find common factors. We can rewrite the expression as: (x + 12)^2 + (x + 7)^2 + (x + 3)^2 + (x - 4)^2 + (x - 8)^2 = (x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64) Notice that each term except the first has a common factor of (x^2 + px + q), where p and q are constants depending on the specific term. Factoring and Simplifying: Now we can factor out the common factors and group the remaining terms: (x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64) = (x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64) = (x^2) + (x^2) + (x^2) + (x^2) + (x^2) + (12x + 14x + 6x - 8x - 16x) + (144 + 49 + 9 + 16 + 64) = 5(x^2) - 14x + 282 Non-Negative Term: The first term, 5(x^2), is always non-negative because any real number squared is non-negative. The minimum value it can take is 0, which occurs when x = 0. Minimizing the Expression: The remaining term, -14x + 282, is minimized when -14x is maximized (since it's being subtracted). However, -14x is maximized when x is minimized (remember x is a real number). Since we have no restrictions on the lower bound of x (it can be any real number), -14x can be infinitely negative as x approaches negative infinity. Minimum Value: Therefore, the minimum value of the entire expression approaches negative infinity as x approaches negative infinity. However, the question asks for the minimum value for ALL real numbers x. Since 5(x^2) is always non-negative and -14x can be arbitrarily negative, the minimum value of the expression is achieved when 5(x^2) = 0 (which occurs at x = 0) and -14x reaches its maximum negative value. Answer: The minimum value of the expression is 282. Jun 1, 2024 #4 +568 0 Problem 3: We can solve this problem by strategically manipulating the inequalities and using our understanding of exponents. Here's how to approach it: Break down the exponentials: Notice that 2 raised to the power of 200 appears on the left side of the first inequality. We can rewrite it as (2^2)^100 which is equal to 4^100. Simplify the inequalities: With the simplification from step 1, the inequalities become: 4^100 < n^100 < (130n)^{50} Taking the 50th root: To compare the terms more effectively, let's take the 50th root of all three parts of the inequalities. This is a valid operation because all terms are positive. Remember that taking the nth root preserves the order of the inequality when dealing with positive numbers. Applying the 50th root: 2 < n < √(130n) Squaring both sides: To get rid of the square root, we can square both sides of the inequality. However, squaring introduces a potential issue: squaring flips the direction of the inequality when the term being squared is negative. Since n is a positive integer by definition, squaring both sides will preserve the direction of the inequality. So we have: 4 < n^2 < 130n Analyze the terms: The leftmost inequality (4 < n^2) tells us that n^2 must be greater than 4. This means n itself must be greater than 2 (since the square of a number cannot be less than the number itself). The rightmost inequality (n^2 < 130n) tells us that n^2 is less than some multiple of n (specifically, 130 times n). This can only be true if n itself is less than 130. Identify valid n: So, to satisfy both inequalities, n must be greater than 2 and less than 130. Now we consider all the positive integers within this range: 3, 4, 5, ..., 128, 129. Counting valid integers: There are a total of 129 - 3 + 1 = 127 positive integers that satisfy the conditions. Answer: There are 127 positive integers n that satisfy the inequalities: 2^{200} < n^{100} < (130n)^{50}. Jun 1, 2024
# Using the State Curriculum: Mathematics, Grade 8 Higher Order Thinking Skills: The higher order thinking skills shows examples of questions for this concept at various levels of cognitive demand. Standard 2.0 Knowledge of Geometry Topic D. Congruence and Similarity Indicator 1. Apply the properties of similar polygons Objective a. Determine similar parts of polygons ### Level 1: Knowledge/Comprehension The two quadrilaterals shown below are similar. Which side of quadrilateral WXYZ corresponds to ? Answer: The triangles below are similar. Which angle in corresponds to ? Answer: ### Level 2: Application/Analysis Figure GHIJK is similar to PQRST. If HG = 15 cm, what is the length of ? Answer: 10cm so that Look at the rectangles below. Is the 8 in by 10 in rectangle similar to the 3 in by 5 in rectangle? Justify your answer. Use words, numbers, and/or symbols in your justification. Sample correct response: No, the rectangles are not similar. The smaller sides of the rectangles are in the ratio of 3:8 but the larger sides are in the ratio of 5:10, or 1:2. Since the ratios of the sides are different, the rectangles are not similar. Look at the similar triangles below. What is the length, in centimeters, of the missing side? Answer: 30 cm Explain how you determined the length of the missing side. Use words, numbers, and/or symbols in your explanation. Sample correct response: Sample correct response: The length is 30 cm because the ratio 28:45, 30:45, and 40:60 all reduce to 2:3, which means all of the sides are in the same ratio. Since the figures are similar, the sides must all be in the same ratio. ### Level 3: Synthesis/Evaluation An 8" by 10" photograph will be surrounded by a 2" border as shown below. Will the framed photograph be similar to the 8" by 10" photograph? Justify your answer. Use words, numbers, and/or symbols in your justification. Sample correct response: No. The framed photograph will have dimensions 12" by 14". I found the new dimensions by adding the 2" border to the top and bottom and both sides. I checked to see if the sides were proportional. An 8" by 10" photograph will be surrounded by a border as shown below. The border on the width is 2". How wide should the border on the length be so that the framed photograph and the photograph are similar? Sample correct answer: The original length of the photograph is 10. To place a border on the length take the difference of 15 and 10 and split it in two. The border along the length should be 2.5". /toolkit/vsc/thinking_skills/mathematics/grade8/2D1a.xml Resources for Objective 2.D.1.a: THINKING SKILLS \| Sample Assessments \|
## Browse Tag: addition of fractions practice exercises The idea of getting the least common multiple of the denominator in adding dissimilar fractions is to convert them into similar fractions or fractions whose denominators are the same. Once the fractions are similar, you only need to add the numerator and copy the denominator. The solutions to Fraction Addition Practice Test 1 below is divided into three parts: (1) getting the least common multiple of the denominator, (2) converting the given fractions to their equivalent fractions whose denominator is the LCM and (3) adding the converted fractions. Of course, in solving this types of problem the Civil Service Exam, you don’t need to go through all the steps. You should try developing your own short cuts to make solving faster. Solution to Number 1 Given: $\displaystyle \frac{2}{7} + \frac{3}{7}$ $\displaystyle \frac{2}{7} + \frac{3}{7} = \frac{2 + 3}{7} = \frac{5}{7}$ Answer: $\displaystyle \frac{5}{7}$ ## Fraction Addition Practice Test 1 In the previous post, we have learned how to add fractions both similar and dissimilar. We have discussed that that in adding similar fractions, we just add the numerator of the addends and copy the denominator. On the other hand, in adding dissimilar fractions, we need to get the least common multiple of the denominator or the least common denominator to be able to convert them to similar fractions. Below is a practice test on on adding similar and dissimilar fractions. If you already know how, convert your answers to lowest terms or mixed form. 1. $\displaystyle \frac{2}{7} + \frac{3}{7}$ Continue Reading
# Joy of Mathematics Class 8 Solutions Chapter 9 ## Joy of Mathematics Class 8 Solutions Chapter 9 Profit Loss and Discount Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 9 Profit, Loss and Discount. Here students can easily find step by step solutions of all the problems for Profit, Loss and Discount. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 9.1 and 9.2 Profit, Loss and Discount Exercise 8 Solution Question no – (1) Solution : (a) CP = Rs 800, SP = Rs 896 => CP = 800 SP = 896 Profit = 896 – 800 = 96 Profit percentage = Profit/CP × 100 = 96/800 × 100 = 12% Therefore, profit percentage 12% (b) Here, C.P. = 1200 S.P. = 1308 Profit = 1308 – 1200 = 1008 Profit percentage = 1008/1200 × 100 Profit percentage 84% (c) Here, C.P. = 500, SP = 475 Loss = C.P. – S.P. = 500 – 475 = 25 Loss percentage = 25/500 × 100 = 5% (d) Here, C.P. = 425, SP = 459 Profit = 459 – 425 = 34 Profit percentage = 34/425 × 100 = 8% Profit percentage = 8% Question no – (2) Solution : (a) Here, S.P. = 720 Profit = 12 1/2% = 25/25 % Now, C.P. = 100/100 + profit% × S.P. = 100/100 + 25/2 × 720 = 100/200 + 25/2 × 720 = 100 × 2/225 × 720 = 8 × 80 = 640 Hence, CP = Rs 640 (b) Here, SP = 999, Profit = 11% CP = 100/100 + profit% × SP = 100/100 + 11 × 999 = 100/111 × 999 = 900 Therefore C.P. = Rs900 (c) Here, S.P. = Rs 4600, Loss = 8% C.P. = 100/100 – loss% × SP = 100/100 – 8 × 4600 = 100/92 × 4600 = 5000 Therefore C.P. = Rs 5000 (d) Here, S.P. = Rs 9000, Loss = 10% C.P. = 100/100 – loss% × S.P. = 100/100 – 10 × 9000 = 100/90 × 9000 = 10000 Therefore C.P. = Rs 10000 Question no – (3) Solution : (a) Here, CP = Rs 400 New CP = (400 + 100) = 500 SP = Rs 450 Loss% = Loss%/CP × 100 = 50/500 × 100 = 10 Therefore loss = 10% (b) Here, CP = Rs 500 New CP = (500 + 80) = Rs 580 SP = Rs 609 Profit = SP – CP = 609 – 580 = 29 Profit percentage = profit/CP × 100 = 29/580 × 100 = 5 Profit percentage 5% (c) Here, CP = Rs 350 Gain = Rs 43 New CP = (350 + 80) = 430 Gain percentage = 43/430 × 100 = 10 Gain percentage 10% (d) Here, CP = 250 New CP = (250 + 50) = 300 Loss = 60 Loss percentage = 60/300 × 100 = 20% Therefore loss percentage 20% Question no – (4) Solution : S.P. = 4998 Profit = 5% C.P = 100/100 + profit% × S.P = 100/400 + profit% × 4998 = 100/105 × 4998 = 499800/105 = 4760 Therefore the cost price of the almirah 4760. Question no – (5) Solution : Here, Sofa’s SP = Rs 3560 Loss = 11% cost price of the sofa = 100/100 – loss% × SP = 100/100 – 11 × 3560 = 100/89 × 3560 Now, cost price of the sofa = 4000 Gain = 12% SP = 100 + gain%/100 × CP = 100 + 12/100 × 4000 = 112/100 × 4000 = 448 Therefore the selling price of the Sofa is Rs 448. Question no – (6) Solution : Here, Amount paid of the car = Rs 45.500 Repair expense = Rs 4500 Actual CP of the car = (45,500 + 4500) = 50,000 SP of the car = 53, 550 Profit = SP – CP = 53,550 – 50,000 = 3,550 Profit percentage, = 3550/50,000 × 100 = 71/10 = 7.1% Profit percentage 7.1% Question no – (7) Solution : Amount paid of the 2nd hand Refrigerator = 4000 Painted expense = 350 Now, Actual CP = (4000 + 350) = Rs 4350 Profit% = 8% Now, SP = 100 + profit%/100 × CP = 100 + 8/100 × 4350 = 108/100 × 4350 = 4,698 Therefore the selling price of the refrigerator Rs 4698. Question no – (8) Solution : Cost of each item = Rs 600 At 10% loss the item selling cost = 600 – 600 × 10/100 = 600 – 60 = 540 Cost of whole item = 3 × 600 = 1800 At 20% gain cost of all items = 1800 + 1800 × 20/100 = 1800 + 360 = Rs 2160 Required price for selling the other two items = 2160 – 540 = 1620 So, cost of selling of each item = 1620/2 = Rs 810 Question no – (9) Solution : Here, Selling price = 54,720 Losses = 5% Cost price = 100/100 – loss% × SP = 100/100 – 5 × 54720 = 100/95 × 54720 = 57,600 CP = 57,600 Profit = 10% Selling price = 100 + profit/100 × CP = 100 + 10/100 × 57600 = 110/100 × 57600 = 63,360 Selling price of the article 63,360/-. Question no – (10) Solution : Selling price of each cars = 99000 At 10% gains cost price of the car = 100/100 + gains% × SP = 100/100 + 10 × 99000 = 100/110 × 99000 = 90000 Again, At 10% losses cost price of the car = 100/100 – loss% × SP = 100/100 – 10 × 99000 = 100/90 × 99000 = 110000 Total selling price of the cars = 2 × 99000 = 198000 Total cost price of the cars, = 90000 + 110000 = 200,000 Loss = CP – SP = 200000 – 198000 = 2000 Loss percentage = 2000/200000 × 100 = 1 Therefore, the loss percentage 1% Question no – (11) Solution : Let, the S.P of both the fans are x At 10% profit, C.P = 100/100 + profit% × S.P = 100/100 + 10 × x = 100/110 × x = 10x/11 Again, 10% loss C.P = 100/100 – loss% × SP = 100/100 – 10 × x = 100x/90 = 10x/9 Ratio of cost price of the two fans are = 10x/11 : 10x/9 = 1/11 : 1/9 = 1/11 × 99 : 1/9 × 99 = 9 : 11 The cost price of both the fans are = 5000 C.P of one fans = 5000 × 9/20 = 2250 C.P of another fans = (5000 – 2250) = Rs 2750 Question no – (12) Solution : Amount of 60kg butter at the rate of Rs 150 per kg = 60 × 150 = 9000/- ∴ Sold 1/3 of it at a loss of 6% = 1/3 × 60kg = 20kg ∴ Cost price of 20kg butter = 20 × 150 = 3000 At 6% loss SP = 100 – loss%/100 × CP = 100 – 6/100 × 3000 = 94/100 × 3000 = 2820 At 10% gain or the whole deal = 100 + gain%/100 × CP = 100 + 10/100 × 9000 = 110 × 90 = 9900 Required price = 9900 – 2820 = 7080 Selling price remaining per kg = 7080/40 = 177 Rs Question no – (13) Solution : Let the cost price of a chair = 100 At 10% profit selling price = 100 + 10 = 110 At 15% profit selling price = 100 + 15 = 115 ∴ Difference of selling price = 115 – 110 = 5 According to the question, 5 → 75 1 → 75/5 = 15 Therefore the cost price of a chair = 15 × 100 = Rs 1500 Question no – (14) Solution : Let, the cost price of an article is = 100 At, 15% profit selling price = 100 + 15 = 115 Again, at 18% gained selling price = 100 + 18 = 118 Difference of selling price = 118 – 115 = 3 According to the question, 3 → 21 1 → 21/3 = 7 Therefore the cost price of an article = 7 × 100 = Rs 700 Question no – (15) Solution : Let, the cost price of one orange = Rs x According to the question, SP of 9 orange = CP of 10 orange = 10x SP of 1 orange = 10x/9 S.P > C.P, therefore there is profit Profit in selling one orange = SP – CP = 10x/9 – x = 10x – 9x/9 = x/9 Profit = x/9/x × 100 = x/9 × x × 100 = 100/9 Therefore, profit percentage 11 1/9% Question no – (16) Solution : Let, the cost price of an article = x According to the question, SP of 5 article = CP of 4 articles = 4x SP of 1 article = 4x/5 Hence, CP > SP therefore there is loss Loss = CP – SP = x – 4x/5 = 5x – 4x/5 = x/5 Loss% = x/5/x × 100 = x × 100/5 × x = 20% Therefore, loss percentage 20% Question no – (17) Solution : Let, the cost price of an article is = 100 At 20% profit SP = (100 + 20) = 120 Total profit = (120 – 100) = 20 According to the question, 20 → 60 1 → 60/20 = 3 The CP of an article = 3 × 100 = Rs 300 The SP of an article = 300 + 300 × 20/100 = 300 + 60 = Rs 360 Question no – (18) Solution : Let, the cost price of the article is = 100 At 10% profit SP = 100 + 10 = 110 Now, He bought it 20% less then Now, CP = (100 – 100 × 20/100) = 80 Profit = 40% SP = 80 + 80 × 40/100 = 80 + 32 = 112 Difference of SP = 112 – 110 = 2 According to the question, 2 → 10 1 → 10/2 = 5 Therefore the cost price of an article = Rs 500 Question no – (19) Solution : Let, CP of a tablet computer = x ∴ SP of a tablet computer = x × 3/4 = 3x/4 Hence, C.P > S.P therefore there is loss ∴ Loss = C.P – S.P = x – 3x/4 = 4x – 3x/4 = x/4 loss% = x/4/x × 100 = x/4 × x × 100 = 25% Therefore the loss percentage 25% Profit, Loss and Discount Exercise 9.2 Solution : Question no – (1) Solution : (a) Here, MP = Rs 20 Discount = 20% SP = 20 – 20 × 20/100 = 20 – 4 = 16 The required SP = Rs 16 (b) Here, MP = Rs 400 Discount =12 1/2% = 25/2% SP = 400 – 400 × 25/2 × 100 = 400 – 50 = Rs 350 The required SP = Rs 350 (c) Here, MP = Rs 750 Discount = 12% SP = 750 – 750 × 12/100 = 750 – 90 = Rs 660 The required SP = Rs 660 (d) Here, MP = Rs 425 Discount = 8% SP = 425 – 425 × 8/100 = 425 – 34 = Rs 391 The required SP = Rs 391 Question no – (2) Solution : (a) Here, SP = 666 Discount = 7 1/2% = 15/2% ∴ MRP = SP × 100/100 – discount = 666 × 100/100 – 15/2 = 66600/200 – 15/2 = 66600 × 2/185 = Rs 720 Therefore the required MRP = 720 (b) Here, SP = Rs 391 Discount = 80% MRP = SP × 100/100 – discount = 391 × 100/100 – 80 = 391 × 100/20 = Rs 1945 Therefore the required MRP = 1945 (c) Here, SP = 8514 Discount = 12% MRP = SP × 100/100 – discount = 8514 × 100/100 – 12 = 8514 × 100/88 = Rs 9675 Therefore the required MRP = 9675 (d) Here, SP = Rs 11932 Discount = 5% MRP = SP × 100/100 – discount = 11932 × 100/100 – 5 = 11932 × 100/95 = 12,560 Therefore the required MRP = 12560 Question no – (3) Solution : (a) Here, MP = 5000 SP = 4500 Discount = 5000 – 4500 = 500 Discount percentage = 500/5000 × 100 = 10 Therefore the discount percentage 10% (b) Here, M.P = Rs 975 S.P = Rs 897 Discount= 975 – 897 = 78 Discount percentage = Discount/MRP × 100 = 78/975 × 100 = 8 Therefore the required discount percentage 8% (c) Here, M.P = Rs 900 S.P = Rs 801 Discount = 900 – 801 = 99 Discount percentage = 99/900 × 100 = 11% Therefore, the required discount percentage 11% (d) Here, M.P = Rs 720 S.P = Rs 684 Discount = 720 – 684 = 36 Discount percentage = 36/720 × 100 = 5% Therefore, the required discount percentage 5% Question no – (4) Solution : M.P of a Sari = Rs 4510 M.P of two Jeans = 2 × 1920 = Rs 3840 Total MP = (4510 + 3840) = 8350 At 25% discount on all the items then, the outlay = 8350 – 8350 × 25/100 = 8350 – 2087.5 = Rs 6262.5 Therefore the outlay Rs 6262.5 Question no – (5) Solution : Let, the cost price = 100 Marks his book 40% above, the cost price. Therefore M.P = 100 + 100 × 40/100 = Rs 140 Now, At discount 25% S.P = 140 – 140 × 25/100 = 140 – 35 = 105 SP > MP, therefore there is gain. Gain = 105 – 100 = Rs 5 Gain percentage = 5/100 × 100 = 5% Therefore, the gain percentage 5% Question no – (6) Solution : Here, cost price = Rs 4550 Gains% = 20% Selling price = 4550 + 4550 × 20/100 = 4550 + 910 = Rs 5460 ∴ (a) its selling price 5460/- ∴ (b) SP = Rs 5460 Discount = 25% M.P = SP × 100/100 – discount = 5460 × 100/100 – 25 = 546000/75 = Rs 7280 Hence, its marked price 7280/- Question no – (7) Solution : Here, Cost price of gas burner = Rs 3750 (a) Marked price = 3750 + 3750 × 40/100 = 3750 + 1500 = Rs 5250 (b) Discount = 10% then, selling price = 5254 – 5254 × 10/100 = 5254 – 525 = Rs 4725 (c) Here, C.P = 3750 S.P = 4725 Profit% = 4725 – 3750/3750 × 100 = 975/3750 × 100 = 26 Therefore, profit percentage 26% Question no – (8) Solution : Here, Marked price of a shirt = Rs 1500 At 20% discount Selling price = 1500 – 1500 × 20/100 = 1500 – 300 = 1200 Now, profit = 20% Cost price of a shirt = 100/100 + profit% × SP = 100/100 + 20 × 1200 = 100/120 × 1200 = Rs 1000 Cost price the shirt Rs 1000 Question no – (9) Solution : Here, S.P = 1100 Discount = 12% (a) Marked price of the article = S.P × 100/100 – discount = 1100 × 100/100 – 12 = 110000/88 = Rs 1250 (b) Here, S.P = 1100 Gains = 10% The cost price in the transaction = 100/100 + gains% × SP = 100/100 + 10 × 1100 = 100/110 × 110 = Rs 1000 Therefore, the cost price 1000 Rs Question no – (10) Solution : 1st discount = 40% 2nd discount = 20% Therefore the single discount = (40 + 20) – 40 × 20/100 = 60 – 8 = 52 Single discount 52 Question no – (11) Solution : 1st discount (x1) = 20% 2nd discount (x2) = 5% Marked price = Rs 35600 Payment received in each by the shopkeeper for the AC = (1 – x1/100) (1 – x2/100) × MP = (1 – 20/100) (1 – 5/100) × 35600 = (100 – 20/100) (100 – 5/100) × 35600 = 80/100 × 95/100 × 35600 = Rs 27056 Therefore the selling price 27056/- Question no – (12) Solution : Let, the marked price of an article = 100 in two different store The 1st store allows two successive discount 23% and 28% then, SP = (1 – 23/100) (1 – 28/100) × 100 = (100 – 23/100) (100 – 28/100) × 100 = 77/100 × 72/100 × 100 = Rs 55.44 2nd store allows 35% discount then SP = (100 – 100 × 35/100) Rs = 100 – 35 = Rs 75 1st store is better offer. Question no – (13) Solution : Marked price of a smart phone = Rs 25000 Discount = 20% of 25000 = 20/100 × 25000 = 5000 Amount of discount = (25000 – 5000) = 20000 GST charged = 12% of 20000 = 12/100 × 20000 = 2400 Net amount payment, = (20000 + 2400) = Rs 22400 Thus, customer has to pay Rs 22400 for the smart phone. Previous Chapter Solution : Updated: May 30, 2023 — 7:50 am
## Complex Numbers Division Consider two complex numbers z1, z2 z1 = a1 + ib1 z2 = a2 + ib2 Multiplying the numerator and denominator by the complex conjugate of z2, that is = a2 - ib2 we obtain #### Example 1 Divide 4 - 6i by 2 - 3i Let z1 = 4 - 6i, a1 = 4, b1 = -6 z2 = 2 - 3i, a2 = 2, b2 = -3 #### Example 2 Divide 7i - 5 by 10 + 2i Let z1 = 7i-5, a1 = -5, b1 = 7 z2 = 10+ 2i, a2 = 10, b2 = 2 Comparing with #### Example 3 Divide -12-5i by 1-4i Let Z1 =-12-5i a1 = -12 b1 = -5 Z2 = 1+ 4i a2 = 1 b2 = -4 #### Example 4 Divide -1 -i by -2-4i Let z1 = -1 -i, a1 = -1 b1 = -1 z2 = -2- 4i, a2 = -2, b2 = -4 #### Try these questions Divide the following complex numbers 1. Divide 8 + 3i by 5 + 6i Answer: let z1 = 8+3i, a1 = 8, b1 = 3 z2 = 5+6i, a2 = 5, b2 = 6 Comparing with 2. Divide -3 + 4i by 10-2i Answer: Let z1 = -3+4i, a1=-3, b1=4 z2=10-2i, a2=10, b2=-2 3. Divide -1 -i by (-4 -3i ) Answer: Let z1 =-1-i a1=-1, b1=-1 z2 =-4-3i a2= -4, b2=-3 4. Divide 12i by 3 - 4i Answer: Let z1 = 12i = 0+12i, a1= 0, b1= 12 z2 =-3-4i, a2=3, b2=-4 5. Divide -5 by 7 + 5i Answer: Let z1 = -5 + 0i a1= -5, b1= 0 z2 = 5 + 6i a2= 5, b2= 6 6. Divide 9 + 11i by 2 Answer: Let z1 =9+11i a1=9 b1=11 z2 = 2=2+0i a2=2 b2=0 On reducing to the lowest terms for each fraction 7. Divide (-3 -18i ) by 12i Answer: Let z1 = -3-18i a1=-3 b1= -18 z2 = 12i =0+12i a2=0 b2=12 8. Divide 4 - 6i by 4 + 6i Answer: Let z1 = 4 - 6i a1= 4 b1= -0 z2 = 4 + 6i a2= 4 b2= 6 On reducing each fraction to the lowest term 9. Divide -5 -12i by -5 +12i Answer: Let z1 =-5- 12i a1=-5 b1=-12 z2 =-5 + 12i a2=-5 b2=12 On reducing to the lowest terms for each fraction
# What Is A Square Root Used For? (7 Real Life Applications) Once you know how to calculate a square root, you might be curious to learn how to apply this concept in real life. Although square roots are used often in math, they also have applications in many other disciplines. So, what is a square root used for? Square roots are used in finance (rates of return over 2 years), normal distributions (probability density functions), lengths & distances (Pythagorean Theorem), quadratic formula (height of falling objects), radius of circles, simple harmonic motion (pendulums & springs), and standard deviation. Of course, square roots are not the only roots we can use. Cube roots, fourth roots, and other roots can also help us in various science and technology fields. In this article, we’ll talk about what square roots are used for and how they fit into various equations and formulas. We’ll also give some examples to make the concepts clear. Having math trouble? Looking for a tutor? ## What Is A Square Root Used For? Square roots are used throughout mathematics and have applications in many disciplines, such as probability, statistics, physics, architecture, and engineering. Here are some uses of square roots in real life: • Finance (Rates Of Return Over 2 Years) • Normal Distributions (Probability Density Function) • Pythagorean Theorem (Lengths & Distances) • Quadratic Formula (Height Of Falling Objects) • Radius Of Circles With A Given Area • Simple Harmonic Motion (Pendulums & Springs) • Standard Deviation (Measuring the spread of data) Let’s take a look at each one in turn, starting with finance. ### Square Roots In Finance In the field of finance, we can use square roots to find the rate of return on an asset over a time period with 2 units (for example, 2 years, 2 months, etc.) The formula for the annual rate of return over a 2 year time period is given by: • R = √(V2 / V0) – 1 where R is the annual rate of return, V0 is the starting value, and V2 is the value after 2 years. #### Example 1: Rate Of Return Of An Asset Over 2 Years Let’s say that you buy a stock on January 1, 2020 for \$100. You sell the stock on January 1, 2022 for \$196. This means that: • V0 = 100 (you bought the stock for \$100) • V2 = 196 (you sold the stock for \$196) Since the time period was 2 years (January 1, 2020 to January 1, 2022), we can use the formula for annual rate of return over 2 years to get: • R = √(V2 / V0) – 1 • R = √(196 / 100) – 1 • R = √(1.96) – 1 • R = 1.4 – 1 • R = 0.4 As a decimal, R = 0.4 means an annual return of 40% (move the decimal 2 places to the right to convert a decimal to a percent). So, the stock returned 40% annually, which is a good investment. #### Example 2: Rate Of Return Of An Asset Over 2 Months Let’s say that you buy a house on January 1, 2020 for \$250,000. You sell the house on March 1, 2022 for \$302,500. This means that: • V0 = 250,000 (you bought the stock for \$250,000) • V2 = 302,500 (you sold the stock for \$302,500) Since the time period was 2 months (January 1, 2020 to March 1, 2020), we can use the formula for annual rate of return over 2 months to get: • R = √(V2 / V0) – 1 • R = √(302,500 / 250,000) – 1 • R = √(1.21) – 1 • R = 1.1 – 1 • R = 0.1 As a decimal, R = 0.1 means a monthly return of 10% (move the decimal 2 places to the right to convert a decimal to a percent). So, the house returned 10% monthly, which is a good investment. More generally, we can use the nth root to find the rate of return over a time period with n units. The formula is given by: • R = (Vn / V0)1/n – 1 where R is the rate of return per time period, V0 is the starting value, and Vn is the value after n time periods. ### Square Roots In Normal Distributions The normal distribution also uses a square root, although it is not easy to see from the graph (which has the shape of a symmetric bell curve). The square root in a normal distribution can be seen in its pdf (probability density function), which is given by: Without square roots, we could not define the function that gives us a normal distribution curve. This distribution is used throughout mathematics, science, medicine, psychology, and other fields. ### Square Roots & The Pythagorean Theorem From the Pythagorean Theorem, we can use square roots to find distances and lengths of sides of triangles in 2 dimensions (or 3 dimensions). This can be useful in all sorts of applications, such as: • Architecture & Engineering (finding lengths of trusses to hold up bridges and buildings). • Carpentry & Construction (finding lengths of sides of right triangles for diagonal supports). • Graphics (finding distances in a 2D or 3D grid system for movies or video games). Remember that the Pythagorean Theorem applies to a right triangle (one with a 90 degree angle), and is given by the formula: • a2 + b2 = c2 where a and b are the legs (two shorter sides) and c is the hypotenuse (the longest side, across from the right angle) in a right triangle. To solve for the hypotenuse, we simply take the square root of both sides of the equation to get: • √(a2 + b2) = c To solve for one of the other sides (let’s say a), we subtract b2 from both sides and take the square root of both sides: • a2 + b2 = c2 • a2 = c2 – b2 • a = √(c2 – b2) #### Example 1: Length Of The Hypotenuse Of A Right Triangle Let’s say that we have a right triangle with sides 6 feet and 8 feet. We want to find the length of the hypotenuse (the longest side) to find out how long a diagonal support should be. Using the Pythagorean Theorem with a = 6 and b = 8, we get: • √(a2 + b2) = c • √(62 + 82) = c • √(36 + 64) = c • √(100) = c • 10 = c So, the diagonal support should be 10 feet long. (Note: this 6-8-10 right triangle is just a multiple of a 3-4-5 right triangle – that is, they are similar.) #### Example 2: Length Of A Leg Of A Right Triangle Let’s say that we have a right triangle with one leg that is 7 feet long and a diagonal (hypotenuse) that is 13 feet long. We want to find the length of the other leg. Using the Pythagorean Theorem with a = 7 and c = 13, we get: • a = √(c2 – b2) • a = √(132 – 72) • a = √(169 – 49) • a = √(120) So, the other leg has a length of √120 or 2√30 feet. Remember that the side lengths of some special triangles will also use square roots for some of their side lengths. For example, a 45-45-90 triangle (right isosceles) will have side lengths in the ratio 1-1-√2. A 30-60-90 triangle will have side lengths in the ratio 1-√3-2. Square roots can also be used to find the distance between two points in a 2-dimensional or 3-dimensional system for movie or video game production. The formula for the distance D between two points (x1, y1) and (x2, y2) in 2 dimensions is given by: • D = √((x2 – x1)2 + (y2 – y1)2) Note that this formula comes from the Pythagorean Theorem, where the legs of the right triangle have length x2 – x1 and y2 – y1, and the hypotenuse has length D. The formula for the distance D between two points (x1, y1, z1) and (x2, y2, z2) in 3 dimensions is given by: • D = √((x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2) Having math trouble? Looking for a tutor? #### Example 1: Distance Between Two Points In 2 Dimensions Let’s say we want to find the distance between the points (1, 3) and (8, -5). If we assign (x1, y1) = (1, 3) and (x2, y2) = (8, -5), then we can use the distance formula to calculate: • D = √((x2 – x1)2 + (y2 – y1)2) • D = √((8 – 1)2 + (-5 – 3)2) • D = √((7)2 + (-8)2) • D = √(49+ 64) • D = √(113) So, the distance between the two points in 2 dimensions is √113. #### Example 2: Distance Between Two Points In 3 Dimensions Let’s say we want to find the distance between the points (2, 4, 7) and (1, -4, 0). If we assign (x1, y1, z1) = (2, 4, 7) and (x2, y2, z2) = (1, -4, 0), then we can use the distance formula to calculate: • D = √((x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2) • D = √((1 – 2)2 + (-4 – 4)2 + (0 – 7)2) • D = √((-1)2 + (-8)2 + (-7)2) • D = √(1 + 64 + 49) • D = √(114) So, the distance between the two points in 3 dimensions is √114. ### Square Roots In The Quadratic Formula Square roots are also necessary if we want to use the quadratic formula to solve a quadratic equation. Remember that a quadratic equation has the standard form • ax2 + bx + c = 0 where a, b, and c are real numbers, with a nonzero. The solutions of this quadratic equation are given by the quadratic formula: Note that there is a square root in the numerator of the fraction. This square root symbol is important because the radicand (expression under the radical) tells us the nature of the solutions. This particular radicand b2 – 4ac is called the discriminant, and its sign tells us what the roots of the quadratic equation will look like: We might need to solve a quadratic equation in physics if we want to know when a falling object is at a certain height. #### Example: Solving A Quadratic For Height Of A Falling Object If an object is dropped from 400 feet above ground, then its height after t seconds is given by the equation • h(t) = 400 – 16t2 Let’s say we want to find out when the object is at a height of 144 feet. Then we would solve: • 144 = 400 – 16t2 • -256 = – 16t2 • 16 = t2 • 4 = t So, the falling object will be at a height of 144 feet above ground at t = 4 seconds (after it has fallen 256 feet from its starting position). ### Square Roots & Radius Of A Circle If you want to find the radius of a circle with a particular area, then you will need to use square roots. Remember that the area A of a circle with radius R is given by the equation • A = πR2 Where π is the constant pi, or approximately 3.14159. We can also use square roots to find the radius of the base of a cylinder or cone with a particular volume (as long as we also know the height). Remember that if the height of a cylinder or cone is H and the radius is R, then the volume equations are: • Volume Of A Cylinder: V = πR2H • Volume Of A Cone: V = πR2H / 3 #### Example 1: Radius Of A Circle Let’s say we want to build a circular animal pen with an area of 1256 square feet. Using our area equation with A = 1256, we can calculate: • A = πR2 • 1256 = 3.14159R2 • 399.80 = R2 • 20 = R So, the radius of the pen would be 20 feet. #### Example 2: Radius Of A Circle At The Base Of A Cylinder Let’s say we want to make a cylinder that is 7 inches tall and has a volume of 2200 square inches. Using our volume equation with V = 2200, we can calculate: • Volume Of A Cylinder: V = πR2H • 2200 = (3.14159)R2(7) • 2200 = 21.99113R2 • 100.04 = R2 • 10 = R So, the radius of the cylinder would be 10 inches. ### Square Roots & Simple Harmonic Motion In physics, we often use square roots in formulas for simple harmonic motion to find the period of a spring or a pendulum. The period is the amount of time it takes for them to go through one cyclical motion. The formulas are: • Period of a Spring: T = 2π√(m/k) • Period of a Pendulum: T = 2π√(L/g) where T is the period, m is the mass of a spring, k is the spring constant, L is the length of the pendulum, and g is the acceleration due to gravity. Note that the spring constant will vary depending on the type of spring. A stiffer spring has a higher value of k. #### Example 1: Period Of A Spring Let’s say we have a spring with a spring constant of 4 N/m and a weight of 0.25 kg at the end of the spring. Using our formula for period, we get: • Period of a Spring: T = 2π√(m/k) • T = 2(3.14159)√(0.25/4) • T = 2(3.14159)√(1/16) • T = 2(3.14159)(1/4) • T = 3.14159/2 • T = 1.5708 So, the period of the spring is 1.5708 seconds. #### Example 2: Period Of A Pendulum Let’s say we have a pendulum with a length of 0.5 feet (assuming we are on Earth, the acceleration due to gravity is 32 feet per second). Using our formula for period, we get: • Period of a Pendulum: T = 2π√(L/g) • T = 2(3.14159)√(0.5/32) • T = 2(3.14159)√(1/64) • T = 2(3.14159)(1/8) • T = (3.14159)(1/4) • T = 0.7854 So, the period of the pendulum is 0.7854 seconds. ### Square Roots & Standard Deviation In statistics, we use square roots to calculate the standard deviation (from the variance). The standard deviation is the square root of variance, which is a sum of squared differences from the mean of a data set. The square root ensures that the standard deviation will have the same units as the mean. This makes it meaningful to talk about adding or subtracting standard deviations from the mean. This makes it possible to talk about percentiles in a population. ## Conclusion Now you know what square roots are used for and where they fit into the equations and formulas where they apply. You can learn how to add, multiply, and divide square roots here.
Courses Courses for Kids Free study material Offline Centres More Last updated date: 27th Nov 2023 Total views: 384k Views today: 5.84k # If in the expansion of ${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n}$ , the coefficients of $x$ and ${x^2}$ are 3 and -6 respectively, then the value of m is:$A.$ 6$B.$ 9$C.$ 12$D.$ 24 Verified 384k+ views Hint: Use binomial theorem of series expansion. Expanding both the terms of the above term we get: ${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n} \\ = \left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right)\left( {1 - {}^n{C_1}{x^1} + {}^n{C_2}{x^2} - {}^n{C_3}{x^3} + ........} \right) \\ = 1\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\ - {}^n{C_1}{x^1}\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\ + {}^n{C_2}{x^2}\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\ + \cdots \cdots \cdots \\ \\$ Expanding the equation further to find the coefficients, we get: $\Rightarrow 1 + x\left( {{}^m{C_1} - {}^n{C_1}} \right) + {x^2}\left( {{}^m{C_2} - {}^m{C_1}{}^n{C_1} + {}^n{C_2}} \right) + \cdots \cdots$ Hence, the coefficient of the term containing $x$ is $\Rightarrow {}^m{C_1} - {}^n{C_1} = 3 \\ \Rightarrow m - n = 3 \\$ ---- (1) Similarly coefficient of the term containing ${x^2}$ is $\Rightarrow {}^m{C_2} - {}^m{C_1}{}^n{C_1} + {}^n{C_2} = - 6 \\ \Rightarrow m\left( {m - 1} \right) - 2mn + n\left( {n - 1} \right) = - 12 \\ \Rightarrow {m^2} - m - 2mn + {n^2} - n = - 12 \\ \Rightarrow {\left( {m - n} \right)^2} - \left( {m + n} \right) = - 12 \\ \Rightarrow 9 - \left( {m + n} \right) = - 12 \\ \Rightarrow m + n = 21 \\$ ---- (2) Adding equation 1 and 2 to find the value of m $\Rightarrow \left( {m + n} \right) + \left( {m - n} \right) = 21 + 3 \\ \Rightarrow 2m = 24 \\ \Rightarrow m = 12 \\$ Hence option C is the correct answer. Note: - In the above question binomial theorem of series expansion has been used in the very first step to expand the terms. We do not need to expand and multiply the whole series as we were concerned with only the first and second power of x so only two to three terms have been considered.
## Dr. Wilson There are several ways to construct a regular octagon. One of the most efficient would be to take a square and cut off the corners This is a little tricky because the diagonal sides, which are hypoteneuses to the right triangles on the corners, have to be as long as what is left of the original sides of the squares after the triangles are removed. If we let the side of the square be 1 and use the Pythagorean theorem to find the diagonals of the triangles on the corners, that gives us the following equation which we can solve for x. Transpose the x terms to the left side Factor out the x. so When we see an expression like this, we are in the habit of rationalizing the denominator. When we divide a polynomial by a monomial, we divide the monomial on the bottom into all the terms on top. 1 is the length of the side of the square, and half the square root of 2 is the length of half of a diagonal of a square. We can subtract this off from the side of the square by scribing an arc centered at a corner of the square and going through the point in the middle where the diagonals meet and seeing where the arc intersects the sides of the square. We can get all of the points of the octagon this way. While we were able to verify algebraically that this construction works, the question arises as to whether there is a simple way of seeing that it works. Let us take a look as some angles in the figure. Assume that EFGHIJKL is a regular octagon. Then /LOK is 45o, because it is 1/8 of a full circle. If we draw in the line segment from O perpendicular to LK, Since our figure is a regular octagon, triangle LOK will be isosceles and the line from the vertex perpendicular to the base will bisect the vertex angle. /DLO will then be 67.5o. Since /MOD is 45o, /LOD will be 67.5o just like /OLD, and triangle LOD will be isosceles which says that the distance from D to O will be the same as the distance from D to L. top
# The Doodle Theorem Recent changes Table of contents Links to this page FRONT PAGE / INDEX Subscribe! My latest posts can be found here: Previous blog posts: Additionally, some earlier writings: This page has been Tagged As Maths. # The Doodle Theorem The Doodle Theorem says: Any map drawn with a single pen stroke that returns to its starting point can be two-coloured. Here's one proof. Take a planar graph that's been drawn with a single pen stroke that returns to its starting point. • As we draw the graph, for every vertex we must arrive, then depart. • Except for the vertex where we start and end, but we can think of the final arrival as matched with the initial departure. • As a result, every vertex must have an even number of edges. • In the dual, a circuit around a vertex of the original will have an even number of edges, • ... because edges in the dual cross edges in the original. • Consider two circuits that share edges but do not cross. When combined into a single cycle, the length of the resulting cycle is the sum of the two lengths, minus twice the number of edges in the shared section. • By combining the individual cycles in the dual around vertices of the original map, we can now see that every cycle in the dual is of even length. Now we use the theorem that says: • A graph is bipartite if and only if every circuit is even. We'll come back to that in a minute. An Euler Cycle is a tour of all the edges of a graph, visiting vertices as often as necessary, but traversing every edge exactly once, and returning to our starting position. So given all that, we now have that the dual is bipartite, and that means both it and the original map can be bi-coloured! So we're done - a planar graph with an Euler Cycle (see side box) can be bi-coloured. Well, nearly done. We still have to show that last step. Theorem: A graph is bipartite if and only if Every circuit is even. Proof: It's enough to prove this for each component, so we may assume that the graph is connected. Pick any vertex $a$ and put it in $A.$ Repeatedly pick a vertex $v$ in $A$ and put its neighbours in $B,$ or pick a vertex in $B$ and put its neighbours in $A.$ If a vertex is put both in $A$ and in $B$ (for the first time), that gives us an odd cycle. If this never happens, then the sets $A$ and $B$ form a partition of the vertices of the graph into two independent sets; in other words, it is a bipartite graph. And we really are done. <<<< Prev <<<< Be Careful What You Say : >>>> Next >>>> Graph Three Colouring ... You should follow me on twitter ## Comments I've decided no longer to include comments directly via the Disqus (or any other) system. Instead, I'd be more than delighted to get emails from people who wish to make comments or engage in discussion. Comments will then be integrated into the page as and when they are appropriate. If the number of emails/comments gets too large to handle then I might return to a semi-automated system. That's looking increasingly unlikely. # Links on this page Site hosted by Colin and Rachel Wright: Maths, Design, Juggling, Computing, Embroidery, Proof-reading, and other clever stuff. Suggest a change ( <-- What does this mean?) / Send me email Front Page / All pages by date / Site overview / Top of page
$$\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}}$$ # 6.4: Composition of Functions $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ PREVIEW ACTIVITY $$\PageIndex{1}$$: Constructing a New Function Let $$A = \{a, b, c, d\}$$, $$B = \{p, q, r\}$$, and $$C = \{s, t, u, v\}$$. The arrow diagram in Figure 6.6 shows two functions: $$f: A \to B$$ and $$g: B \to C$$. Notice that if $$x \in A$$, then $$f(x) \in B$$. Since $$f(x) \in B$$, we can apply the function $$g$$ to $$f(x)$$, and we obtain $$g(f(x))$$, which is an element of $$C$$. Using this process, determine $$g(f(a))$$, $$g(f(b))$$, $$g(f(c))$$, and $$g(f(d))$$. Then explain how we can use this information to define a function from $$A$$ to $$C$$. Figure 6.6: Arrow Diagram Showing Two Functions PREVIEW ACTIVITY $$\PageIndex{1}$$: Verbal Descriptions of Functions The outputs of most real functions we have studied in previous mathematics courses have been determined by mathematical expressions. In many cases, it is possible to use these expressions to give step-by-step verbal descriptions of how to compute the outputs. For example, if $$f: \mathbb{R} \to \mathbb{R}$$ is defined by $$f(x) = (3x + 2)^3$$, we could describe how to compute the outputs as follows: Step Verbal Description Symbolic Result 1 Choose an input. $$x$$ 2 Multiply by 3. $$3x$$ 3 Add 2. $$3x + 2$$ 4 Cube the result. $$(3x + 3)^3$$ Complete step-by-step verbal descriptions for each of the following functions. 1. $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = \sqrt{3x^2 + 2}$$, for each $$x \in \mathbb{R}$$. 2. $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = \sin(3x^2 + 2)$$, for each $$x \in \mathbb{R}$$. 3. $$h: \mathbb{R} \to \mathbb{R}$$ by $$h(x) = e^{3x^2 + 2}$$, for each $$x \in \mathbb{R}$$. 4. $$G: \mathbb{R} \to \mathbb{R}$$ by $$G(x) = \ln(x^4 + 3)$$, for each $$x \in \mathbb{R}$$. 5. $$k: \mathbb{R} \to \mathbb{R}$$ by $$k(x) = \sqrt[3] {\dfrac{\sin(4x + 3)}{x^2 + 1}}$$, for each $$x \in \mathbb{R}$$. ## Composition of Functions There are several ways to combine two existing functions to create a new function. For example, in calculus, we learned how to form the product and quotient of two functions and then how to use the product rule to determine the derivative of a product of two functions and the quotient rule to determine the derivative of the quotient of two functions. The chain rule in calculus was used to determine the derivative of the composition of two functions, and in this section, we will focus only on the composition of two functions. We will then consider some results about the compositions of injections and surjections. The basic idea of function composition is that when possible, the output of a function $$f$$ is used as the input of a function $$g$$. This can be referred to as “$$f$$ followed by $$g$$” and is called the composition of $$f$$ and $$g$$. In previous mathematics courses, we used this idea to determine a formula for the composition of two real functions. For example, if $$f(x) = 3x^2 + 2$$ and $$g(x) = sin x$$ then we can compute $$g(f(x))$$ as follows: $\begin{array} {rcl} {g(f(x))} &= & {g(3x^2 + 2)}\\ {} &= & {sin(3x^2 + 2).} \end{array}$ In this case, $$f(x)$$, the output of the function $$f$$, was used as the input for the function $$g$$. We now give the formal definition of the composition of two functions. Definition: composite function Let $$A$$, $$B$$, and $$C$$ be nonempty sets, and let $$f: A \to B$$ and $$g: B \to C$$ be functions. The composition of $$f$$ and $$g$$ is the function $$g \circ f: A \to C$$ defined by $$(g \circ f)(x) = g(f(x))$$ for all $$x \in A$$. We often refer to the function $$g \circ f$$ as a composite function. It is helpful to think of composite function $$g \circ f$$ as "$$f$$ followed by $$g$$". We then refer to $$f$$ as the inner function and $$g$$ as the outer function. Composition and Arrow Diagrams The concept of the composition of two functions can be illustrated with arrow diagrams when the domain and codomain of the functions are small, finite sets. Although the term “composition” was not used then, this was done in Preview Activity $$\PageIndex{1}$$, and another example is given here. Let $$A = \{a, b, c, d\}$$, $$B = \{p, q, r\}$$, and $$C = \{s, t, u, v\}$$. The arrow diagram in Figure 6.7 shows two functions: $$f: A \to B$$ and $$g: B \to C$$. If we follow the arrows from the set $$A$$ to the set $$C$$ , we will use the outputs of $$f$$ as inputs of $$g$$, and get the arrow diagram from $$A$$ to $$C$$ shown in Figure 6.8. This diagram represents the composition of $$f$$ followed by $$g$$. Progress Check 6.17 (The Composition of Two Functions) Let $$A = \{a, b, c, d\}$$ and $$B = \{1, 2, 3\}$$. Define the function $$f$$ and $$g$$ as follows: $$f: A \to B$$ defined by $$f(a) = 2$$, $$f(b) = 3$$, $$f(c) = 1$$, and $$f(d) = 2$$. $$g: A \to B$$ defined by $$g(1) = 3$$. $$g(2) = 1$$, and $$g(3) = 2$$. Create arrow diagrams for the function $$f$$, $$g$$, $$g \circ f$$, and $$g \circ g$$. Add texts here. Do not delete this text first. ## Decomposing Functions We use the chain rule in calculus to find the derivative of a composite function. The first step in the process is to recognize a given function as a composite function. This can be done in many ways, but the work in Preview Activity $$\PageIndex{2}$$ can be used to decompose a function in a way that works well with the chain rule. The use of the terms “inner function” and “outer function” can also be helpful. The idea is that we use the last step in the process to represent the outer function, and the steps prior to that to represent the inner function. So for the function, $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = (3x + 2)^3$$, the last step in the verbal description table was to cube the result. This means that we will use the function $$g$$ (the cubing function) as the outer function and will use the prior steps as the inner function. We will denote the inner function by $$h$$. So we let $$h: \mathbb{R} \to \mathbb{R}$$ by $$h(x) = 3x + 2$$ and $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = x^3$$. Then $\begin{array} {rcl} {(g \circ h)(x)} &= & {g(h(x))} \\ {} &= & {g(3x + 2)} \\ {} &= & {(3x + 2)^3} \\ {} &= & {f(x).} \end{array}$ We see that $$g \circ h = f$$ and, hence, we have “decomposed” the function $$f$$. It should be noted that there are other ways to write the function $$f$$ as a composition of two functions, but the way just described is the one that works well with the chain rule. In this case, the chain rule gives $\begin{array} {rcl} {f \prime (x)} &= & {(g \circ h)\prime (x)} \\ {} &= & {g \prime (h(x)) h \prime(x)} \\ {} &= & {3(h(x))^2 \cdot 3} \\ {} &= & {g(3x + 2)^2} \end{array}$ Progress Check 6.18 (Decomposing Functions Write each of the following functions as the composition of two functions. 1. $$F: \mathbb{R} \to \mathbb{R}$$ by $$F(x) = (x^2 + 3)^3$$ 2. $$G: \mathbb{R} \to \mathbb{R}$$ by $$G(x) = In(x^2 + 3)$$ 3. $$f: \mathbb{Z} \to \mathbb{Z}$$ by $$f(x) = \|x^2 - 3\|$$ 4. $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = cos(\dfrac{2x - 3}{x^2 + 1})$$ Add texts here. Do not delete this text first. If $$f: A \to B$$ and $$g: B \to C$$, then we can form the composite function $$g \circ f: A \to C$$. In Section 6.3, we learned about injections and surjections. We now explore what type of function $$g \circ f$$ will be if the functions $$f$$ and $$g$$ are injections (or surjections). Progress Check 6.19: Compositions of Injections and Surjections Although other representations of functions can be used, it will be helpful to use arrow diagrams to represent the functions in this progress check. We will use the following sets: $$A = \{a, b, c\}$$, $$B = \{p, q, r\}$$, $$C = \{u,v, w, x\}$$, and $$D = \{u, v\}$$. 1. Draw an arrow diagram for a function $$f: A \to B$$ that is an injection and an arrow diagram for a function $$g: B \to C$$ that is an injection. In this case, is the composite function $$g \circ f: A \to C$$ an injection? Explain. 2. Draw an arrow diagram for a function $$f: A \to B$$ that is a surjection and an arrow diagram for a function $$g: B \to D$$ that is a surjection. In this case, is the composite function $$g \circ f: A \to D$$ a surjection? Explain. 3. Draw an arrow diagram for a function $$f: A \to B$$ that is a bijection and an arrow diagram for a function $$g: B \to A$$ that is a bijection. In this case, is the composite function $$g \circ f: A \to A$$ bijection? Explain. Add texts here. Do not delete this text first. In Progress Check 6.19, we explored some properties of composite functions related to injections, surjections, and bijections. The following theorem contains results that these explorations were intended to illustrate. Some of the proofs will be included in the exercises. Theorem 6.20. Let $$A$$, $$B$$, and $$C$$ be nonempty sets and assume that $$f: A \to B$$ and $$g: B \to C$$. 1. If $$f$$ and $$g$$ are both injections, then $$(g \circ f): A \to C$$ is an injection. 2. If $$f$$ and $$g$$ are both surjections, then $$(g \circ f): A \to C$$ is an surjection. 3. If $$f$$ and $$g$$ are both bijections, then $$(g \circ f): A \to C$$ is an bijection. Proof The proof of Part (1) is Exercise (6). Part (3) is a direct consequence of the first two parts. We will discuss a process for constructing a proof of Part (2). Using the forward-backward process, we first look at the conclusion of the conditional statement in Part (2). The goal is to prove that $$g \circ f$$ is a surjection. Since $$(g \circ f): A \to C$$, this is equivalent to proving that For all $$c \in C$$, there exists an $$a \in A$$ such that $$(g \circ f)(a) = c$$. Since this statement in the backward process uses a universal quantifier, we will use the choose-an-element method and choose an arbitrary element $$c$$ in the set $$C$$. The goal now is to find an $$a \in A$$ such that $$(g \circ f)(a) = c$$. Now we can look at the hypotheses. In particular, we are assuming that both $$f: A \to B$$ and $$g: B \to C$$ are surjections. Since we have chosen $$c \in C$$, and $$g: B \to C$$ is a surjection, we know that there exists a $$b \in B$$ such that $$g(b) = c$$. Now, $$b \in B$$ and $$f: A \to B$$ is a surjection. Hence there exists an $$a \in A$$ such that $$f(a) = b$$. If we now compute $$(g \circ f)(a)$$, we will see that $$(g \circ f)(a) = g(f(a)) = g(b) = c$$. We can now write the proof as follows: Proof of Theorem 6.20, Part (2) Let $$A$$, $$B$$, and $$C$$ be nonempty sets and assume that $$f: A \to B$$ and $$g: B \to C$$ are both surjections. We will prove that $$g \circ f: A \to C$$ is a surjection. Let $$c$$ be an arbitrary element of $$C$$. We will prove there exists an $$a \in A$$ such that $$(g \circ f)(a) = c$$. Since $$g: B \to C$$ is a surjection, we conclude that there exists a $$b \in B$$ such that $$g(b) = c$$. Now, $$b \in B$$ and $$f: A \to B$$ is a surjection. Hence there exists an $$a \in A$$ such that $$f(a) = b$$. We now see that \begin{align} {(g \circ f)(a)} &= & {g(f(a))} \\ {} &= & {g(b)} \\ {} &= & {c.} \end{align} We have now shown that for every $$c \in C$$, there exists an $$a \in A$$ such that $$(g \circ f)(a) = c$$, and this proves that $$g \circ f$$ is a surjection. Theorem 6.20 shows us that if $$f$$ and $$g$$ are both special types of functions, then the composition of $$f$$ followed by $$g$$ is also that type of function.The next question is, “If the composition of $$f$$ followed by $$g$$ is an injection (or surjection), can we make any conclusions about $$f$$ or $$g$$?” A partial answer to this question is provided in Theorem 6.21. This theorem will be investigated and proved in the Explorations and Activities for this section. See Exercise (10). Theorem 6.21 Let $$A$$, $$B$$, and $$C$$ be nonempty sets and assume that $$f: A \to B$$ and $$g: B \to C$$. 1. If $$g \circ f: A \to C$$ is an injection, then $$f: A \to B$$ is an injection. 2. If $$g \circ f: A \to C$$ is a surjection, then $$f: A \to B$$ is a surjection. Exercise 6.4 1. In our definition of the composition of two functions, $$f$$ and $$g$$, we required that the domain of $$g$$ be equal to the codomain of $$f$$. However, it is sometimes possible to form the composite function $$g \circ f$$ even though dom($$g$$) $$\ne$$ codom($$f$$). For example, let $\begin{array} {lcl} {f: \mathbb{R} \to \mathbb{R}} &text{ be defined by }& {f(x) = x^2 + 1\text{, and let}} \\ {g: \mathbb{R} - \{0\} \to \mathbb{R}} &text{ be defined by }& {g(x) = \dfrac{1}{x}.} \end{array}$ (a) Is it possible to determine $$(g \circ f) (x)$$ for all $$x \in \mathbb{R}$$? Explain. (b) In general, let $$f: A \to T$$ and $$g: B \to C$$. Find a condition on the domain of $$g$$ (other than $$B = T$$) that results in a meaningful definition of the composite function $$g \circ f: A \to C$$. 2. Let $$h: \mathbb{R} \to \mathbb{R}$$ be defined $$h(x) = 3x + 2$$ and $$g: \mathbb{R} \to \mathbb{R}$$ be defined by $$g(x) = x^3$$. Determine formulas for the composite functions $$g \circ h$$ and $$h \circ g$$. Is the function $$g \circ h$$ equal to the function $$h \circ g$$? Explain. What does this tell you about the operation of composition of functions? 3. Following are formulas for certain real functions. Write each of these real functions as the composition of two functions. That is, decompose each of the functions. (a) $$F(x) = cos(e^x)$$ (b) $$G(x) = e^{cos(x)}$$ (c) $$H(x) = \dfrac{1}{sin x}$$ (d) $$K(x) = cos(e^{-x^2})$$ 4. The identity function on a set $$S$$, denoted by $$I_S$$, is defined as follows: $$I_S: S \to S$$ by $$I_s(x) = x$$ for each $$x \in S$$. Let $$f: A \to B$$. (a) For each $$x \in A$$, determine $$(f \circ I_A)(x)$$ and use this to prove that $$f \circ I_A = f$$. (b) Prove that $$I_B \circ f = f$$. 5. (a) Let $$f: \mathbb{R} \to \mathbb{R}$$ be defined by $$f(x) = x^2$$, let $$g: \mathbb{R} \to \mathbb{R}$$ be defined by $$g(x) = sin x$$, and let $$h: \mathbb{R} \to \mathbb{R}$$ be defined by $$h(x) = \sqrt[3]{x}$$. Determine formulas for $$[(h \circ g) \circ f] (x)$$ and $$[h \circ (g \circ f)](x)$$. Does this prove that $$(h \circ g) \circ f = h \circ (g \circ f)$$ for these particular functions? Explain. (b) Now let $$A$$, $$B$$, and $$C$$ be sets and let $$f: A \to B$$, $$g: B \to C$$, and $$h: C \to D$$. Prove that $$(h \circ g) \circ f = h \circ (g \circ f)$$. That is, prove that function composition is an associative operation. 6. Prove Part (1) of Theorem 6.20. Let $$A$$, $$B$$, and $$C$$ be nonempty sets and let $$f: A \to B$$ and $$g: B \to C$$. If $$f$$ and $$g$$ are both injections, then $$g \circ f$$ is an injection. 7. For each of the following, give an example of functions $$f: A \to B$$ and $$g: B \to C$$ that satisfy the stated conditions, or explain why no such example exists. (a) The function $$f$$ is a surjection, but the function $$g \circ f$$ is not a surjection. (b) The function $$f$$ is an injection, but the function $$g \circ f$$ is not an injection. (c) The function $$g$$ is a surjection, but the function $$g \circ f$$ is not a surjection. (d) The function $$g$$ is an injection, but the function $$g \circ f$$ is not an injection. (e) The function $$f$$ is not a surjection, but the function $$g \circ f$$ is a surjection. (f) The function $$f$$ is not an injection, but the function $$g \circ f$$ is an injection. (g) The function $$f$$ is not an injection, but the function $$g \circ f$$ is an injection. (h) The function $$g$$ is not an injection, but the function $$g \circ f$$ is an injection. 8. Let $$A$$ be a nonempty set and let $$f: A \to A$$. For each $$n \in \mathbb{N}$$, define a funciton $$f^n: A \to A$$ recursively as follows: $$f^1 = f$$ and for each $$n \in \mathbb{N}$$, $$f^{n + 1} = f \circ f^n$$. For example, $$f^2 = f \circ f^1 = f \circ f$$ and $$f^3 = f \circ f^2 = f \circ (f \circ f)$$. (a) Let $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = x + 1$$ for each $$x \in \mathbb{R}$$. For each $$n \in \mathbb{N}$$ and for each $$x \in \mathbb{R}$$, determine a formula for $$f^n(x)$$ and use induction to prove that your formula is correct. (b) Let $$a, b \in \mathbb{R}$$ and let $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = ax + b$$ for each $$x \in \mathbb{R}$$. For each $$n \in \mathbb{N}$$ and for each $$x \in \mathbb{R}$$, determine a formula for $$f^n(x)$$ and use induction to prove that your formula is correct. (c) Now let $$A$$ be a nonempty set and let $$f: A \to A$$. Use induction to prove that for each $$n \in \mathbb{N}$$, $$f^{n + 1} = f^n \circ f$$. (Note: You will need to use the result in Exercise (5).) Explorations and Activities 9. Exploring Composite Functions. Let $$A$$, $$B$$, and $$C$$ be nonempty sets and let $$f: A \to B$$ and $$g: B \to C$$. For this activity, it may be useful to draw your arrow diagrams in a triangular arrangement as follows: It might be helpful to consider examples where the sets are small. Try constructing examples where the set $$A$$ has 2 elements, the set $$B$$ has 3 elements, and the set $$C$$ has 2 elements. (a) Is it possible to construct an example where $$g \circ f$$ is an injection, $$f$$ is an injection, but $$g$$ is not an injection? Either construct such an example or explain why it is not possible. (b) Is it possible to construct an example where $$g \circ f$$ is an injection, $$g$$ is an injection, but $$f$$ is not an injection? Either construct such an example or explain why it is not possible. (c) Is it possible to construct an example where $$g \circ f$$ is a surjection, $$f$$ is a surjection, but $$g$$ is not a surjection? Either construct such an example or explain why it is not possible. (d) Is it possible to construct an example where $$g \circ f$$ is a surjection, $$g$$ is a surjection, but $$f$$ is not a surjection? Either construct such an example or explain why it is not possible. 10. The Proof of Theorem 6.21. Use the ideas from Exercise (9) to prove Theorem 6.21. Let $$A$$, $$B$$ and $$C$$ be nonempty sets and let $$f: A \to B$$ and $$g: B \to C$$. (a) If $$g \circ f: A \to C$$ is an injection, then $$f: A \to B$$ is an injection. (b) If $$g \circ f: A \to C$$ is a surjection, then $$g: B \to C$$ is a surjection. Hint: For part (a), start by asking, “What do we have to do to prove that $$f$$ is an injection? ” Start with a similar question for part (b).
# Into Math Grade 2 Module 1 Lesson 7 Answer Key Add 3 Numbers Using Mental Strategies and Properties We included HMH Into Math Grade 2 Answer Key PDF Module 1 Lesson 7 Add 3 Numbers Using Mental Strategies and Properties to make students experts in learning maths. ## HMH Into Math Grade 2 Module 1 Lesson 7 Answer Key Add 3 Numbers Using Mental Strategies and Properties I Can find the sum of three numbers. How can you show the number of stuffed animals Logan has? Explanation: 6 + 2 = 8, 8 + 4 = 12. Logan has 12 stuffed animals. Read the following problem: Logan collects stuffed animals. He has 6 cats, 2 teddy bears, and 4 dogs. How many stuffed animals does Logan have? Have children use tools to add the three numbers. Build Understanding On the playground, there are 4 children playing ball, 3 children on the swings, and 2 children playing in a sandbox. How many children are on the playground? A. How can you use tools to show each addend? B. How can you group two addends to odd first? Draw to show what you did. C. Complete the addition fact to solve the problem. 4 + 3 + 2 = ___ ___ children Explanation: Add 4 + 3 = 7, 7 + 2 = 9 9 children are at the playground. Turn and Talk How can you group the addends in a different order to solve the problem? Explain. Step It Out 1. Alice has4 small toy boats, 5 toy cars, and 3 toy trucks. How many toys does she have? A. Write an addition fact to show the problem. ___ + ___ + ___ = THINK: Look for facts you know. ___ + ___ + ___ = ___ + ___ = ___ ___ + ___ = ___ toys Explanation: 4 + 5 + 3 = 12 Alice have 12 toys. Check Understanding Math Board Question 1. There are 4 pink pens, 5 red pens, and 5 blue pens in a box. How many pens are in the box? Write an addition fact to show the problem. Circle two addends to add first. Write the sum. ___ + ___ + ___ = ___ ____ pens Explanation: 4 + 5 + 5 = 14 There are 14 pens in the box. Choose two addends to odd first. Circle the addends. Write the sum. Question 2. 3 + 2 + 6 = ___ Explanation: The sum of 3, 2 and 6 is 11. Question 3. 5 + 1 + 3 = ___ Explanation: The sum of 5, 1 and 3 is 9. Write an addition fact to show the problem. Circle two addends to add first. Write the sum. Solve. Question 4. Use Repeated Reasoning There are 4 rulers on the art table, 6 rulers on the math table, and 3 rulers on the science table. How many rulers are there? ___ + _____ + ____ = ____ _____ rulers Explanation: The sum of 4, 6 and 3 is 13 There are 13 rulers. Question 5. Use Structure Nico folds 3 blue shirts, 2 red shirts, and 7 white shirts. How many shirts does Nico fold? ___ + ___ + ___ = ____ ___ shirts Explanation: The sum of 3, 2 and 7 is 12 Nico folded 12 shirts. Question 6. 7 + 2 + 2 = ___ Explanation: The sum of 7, 2 and 2 is 11. Question 7. 0 + 6 + 9 = ___ Explanation: The sum of 0, 6 and 9 is 15. Question 8. 5 + 0 + 7 = ___ Explanation: The sum of 5, 0 and 7 is 12. Question 9. 4 + 5 + 1 = ___
# How do you use the distributive property to simplify 5(6-3i)+3(i-8)? Jun 11, 2018 $6 - 12 i$ #### Explanation: I'm not sure if your $i$ is just a variable or if it is an imaginary number, but either way that will not affect the simplified answer. To simplify this, use the distributive property (shown below): Following this image, we know that: $\textcolor{b l u e}{5 \left(6 - 3 i\right) = \left(5 \cdot 6\right) + \left(5 \cdot - 3 i\right) = 30 - 15 i}$ and $\textcolor{b l u e}{3 \left(i - 8\right) = \left(3 \cdot i\right) + \left(3 \cdot - 8\right) = 3 i - 24}$ Now combine them: $30 - 15 i + 3 i - 24$ Color-code the like terms: $\textcolor{red}{30} \quad \textcolor{g r e e n}{- \quad 15 i} \quad \textcolor{g r e e n}{+ \quad 3 i} \quad \textcolor{red}{- \quad 24}$ Combine the like terms: $6 - 12 i$ Hope this helps!
This is a brief summary of quadratics that should be known before commencing maths methods 3&4 ## Basic shape The basic shape of a quadratic or parabola is a U, shown in the blue function in the graph below. The basic equation is the power function of : $y=x^{2}$ The red function is when it is reflected in the x axis and the equation becomes $y=-x^{2}$ $y=ax^{2}+bx+c$ The easiest way to learn how to draw a quadratic is to do an example such as: $y=2x^{2}-13x+11$ Step 1: Find the y intercept. Let x=0 When the graph is in the form of the y intercept will always be the c, which in this case is $y-intercept=11$ Step 2: Find the x intercept. To find the x intercept we make y = 0 $y=2x^{2}-13x+11=0$ There are two methods that can be used to solve this equation, factorising and using the quadratic formula. ### Factorisation To factorise we’ll use the cross method. To do this method we just guess and check our factors, and then cross multiply the terms and add them together to ensure we get our b term. $a=2x\times x=2x^{2}$ $b=-11x+-2x= -13x$ $c=-11\times-1=11$ Therefore our a, b and c terms match up and we get the function $y=(2x -11)(x-1) = 0$ $x-intercepts =\frac{11}{2} or x = 1$ $x-intercepts = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ So we substitute in our values of a,b and c from the original equation. $x-intercepts = \frac{13\pm\sqrt{(-13)^{2}-4\times2\times11}}{2\times2}\newline =\frac{13\pm\sqrt{81}}{4}\newline =\frac{4}{4} or \frac{22}{4}\newline=1or\frac{11}{2}$ The same as previously. Remember that some graphs will not have x intercepts. This happens when the graph is entirely above the x axis or entirely below it. In this case the quadratic formula will have no solution and we will not be able to factorise the equation with real numbers. Step 3: Find the turning point. we have both our x intercepts we know that the x coordinate of the turning point is exactly halfway between these. $x-TP=\frac{1+\frac{11}{2}}{2}=\frac{13}{4}$ Once we have the x-coordinate, we can substitute this into our equation to find the y –coordinate. $y-TP=2(\frac{13}{4})^{2}-13(\frac{13}{4})+11=-\frac{81}{8}$ TP = (13/4, -81/8) If you do not have x intercepts we can find the x-coordinate of the TP by using by using the formula: $x=-\frac{b}{2a}\newline =\frac{13}{4}$ Step 4 Draw the shape Now that we have the intercepts and the turning point we can plot these on our graph and join the dots to draw a curve. Before you move onto the next question make sure your graph contains the following: • all intercepts • x and y axis are labelled • TP • Correct shape (in the exam you will get marked on shape so it might be a good idea to spend some time practising your graphing technique) ## Turning point form Sometimes the equation may appear in what we call turning point form: $y=a(x-h)+k$ This is just a form which lets us know where the turning point is straight away. $TP=(h,k)$ To draw the rest of the graph we go through the same process above of setting x to zero to calculate the y intercept and then y to zero to calculate the x intercept. If in doubt you can always get the same form of the equation as above by expanding out all the terms of the turning point form equation.
# 2.4 The cross product (Page 4/16) Page 4 / 16 ## Using expansion along the first row to compute a $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ Determinant Evaluate the determinant $\|\begin{array}{ccc}\hfill 2& \hfill 5& \hfill -1\\ \hfill -1& \hfill 1& \hfill 3\\ \hfill -2& \hfill 3& \hfill 4\end{array}\|.$ We have $\begin{array}{cc}\hfill \|\begin{array}{ccc}\hfill 2& \hfill 5& \hfill -1\\ \hfill -1& \hfill 1& \hfill 3\\ \hfill -2& \hfill 3& \hfill 4\end{array}\|& =2\|\begin{array}{cc}1\hfill & 3\hfill \\ 3\hfill & 4\hfill \end{array}\|-5\|\begin{array}{cc}-1\hfill & 3\hfill \\ -2\hfill & 4\hfill \end{array}\|-1\|\begin{array}{cc}-1\hfill & 1\hfill \\ -2\hfill & 3\hfill \end{array}\|\hfill \\ & =2\left(4-9\right)-5\left(-4+6\right)-1\left(-3+2\right)\hfill \\ & =2\left(-5\right)-5\left(2\right)-1\left(-1\right)=-10-10+1\hfill \\ & =-19.\hfill \end{array}$ Evaluate the determinant $\|\begin{array}{ccc}\hfill 1& \hfill -2& \hfill -1\\ \hfill 3& \hfill 2& \hfill -3\\ \hfill 1& \hfill 5& \hfill 4\end{array}\|.$ $40$ Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula. ## Rule: cross product calculated by a determinant Let $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ be vectors. Then the cross product $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ is given by $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\|\begin{array}{ccc}\text{i}\hfill & \text{j}\hfill & \text{k}\hfill \\ {u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \end{array}\|=\|\begin{array}{cc}{u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{2}\hfill & {v}_{3}\hfill \end{array}\|\text{i}-\|\begin{array}{cc}{u}_{1}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{3}\hfill \end{array}\|\text{j}+\|\begin{array}{cc}{u}_{1}\hfill & {u}_{2}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill \end{array}\|\text{k}.$ ## Using determinant notation to find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}$ Let $\text{p}=⟨-1,2,5⟩$ and $\text{q}=⟨4,0,-3⟩.$ Find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}.$ We set up our determinant by putting the standard unit vectors across the first row, the components of $\text{u}$ in the second row, and the components of $\text{v}$ in the third row. Then, we have $\begin{array}{cc}\hfill \text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}& =\|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -1& \hfill 2& \hfill 5\\ \hfill 4& \hfill 0& \hfill -3\end{array}\|=\|\begin{array}{cc}\hfill 2& \hfill 5\\ \hfill 0& \hfill -3\end{array}\|\text{i}-\|\begin{array}{cc}\hfill -1& \hfill 5\\ \hfill 4& \hfill -3\end{array}\|\text{j}+\|\begin{array}{cc}\hfill -1& \hfill 2\\ \hfill 4& \hfill 0\end{array}\|\text{k}\hfill \\ & =\left(-6-0\right)\text{i}-\left(3-20\right)\text{j}+\left(0-8\right)\text{k}\hfill \\ & =-6\text{i}+17\text{j}-8\text{k}.\hfill \end{array}$ Notice that this answer confirms the calculation of the cross product in [link] . Use determinant notation to find $\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b},$ where $\text{a}=⟨8,2,3⟩$ and $\text{b}=⟨-1,0,4⟩.$ $8\text{i}-35\text{j}+2\text{k}$ ## Using the cross product The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped . The following examples illustrate these calculations. ## Finding a unit vector orthogonal to two given vectors Let $\text{a}=⟨5,2,-1⟩$ and $\text{b}=⟨0,-1,4⟩.$ Find a unit vector orthogonal to both $\text{a}$ and $\text{b}.$ The cross product $\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}$ is orthogonal to both vectors $\text{a}$ and $\text{b}.$ We can calculate it with a determinant: $\begin{array}{cc}\hfill \text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}& =\|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill 5& \hfill 2& \hfill -1\\ \hfill 0& \hfill -1& \hfill 4\end{array}\|=\|\begin{array}{cc}\hfill 2& \hfill -1\\ \hfill -1& \hfill 4\end{array}\|\text{i}-\|\begin{array}{cc}\hfill 5& \hfill -1\\ \hfill 0& \hfill 4\end{array}\|\text{j}+\|\begin{array}{cc}\hfill 5& \hfill 2\\ \hfill 0& \hfill -1\end{array}\|\text{k}\hfill \\ & =\left(8-1\right)\text{i}-\left(20-0\right)\text{j}+\left(-5-0\right)\text{k}\hfill \\ & =7\text{i}-20\text{j}-5\text{k}.\hfill \end{array}$ Normalize this vector to find a unit vector in the same direction: $‖\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}‖=\sqrt{{\left(7\right)}^{2}+{\left(-20\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{474}.$ Thus, $⟨\frac{7}{\sqrt{474}},\frac{-20}{\sqrt{474}},\frac{-5}{\sqrt{474}}⟩$ is a unit vector orthogonal to $\text{a}$ and $\text{b}.$ Find a unit vector orthogonal to both $\text{a}$ and $\text{b},$ where $\text{a}=⟨4,0,3⟩$ and $\text{b}=⟨1,1,4⟩.$ $⟨\frac{-3}{\sqrt{194}},\frac{-13}{\sqrt{194}},\frac{4}{\sqrt{194}}⟩$ To use the cross product for calculating areas, we state and prove the following theorem. ## Area of a parallelogram If we locate vectors $\text{u}$ and $\text{v}$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖$ ( [link] ). ## Proof We show that the magnitude of the cross product is equal to the base times height of the parallelogram. $\begin{array}{cc}\hfill \text{Area of a parallelogram}& =\text{base}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{height}\hfill \\ & =‖\text{u}‖\left(‖\text{v}‖\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)\hfill \\ & =‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖\hfill \end{array}$ ## Finding the area of a triangle Let $P=\left(1,0,0\right),Q=\left(0,1,0\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}R=\left(0,0,1\right)$ be the vertices of a triangle ( [link] ). Find its area. We have $\stackrel{\to }{PQ}=⟨0-1,1-0,0-0⟩=⟨-1,1,0⟩$ and $\stackrel{\to }{PR}=⟨0-1,0-0,1-0⟩=⟨-1,0,1⟩.$ The area of the parallelogram with adjacent sides $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$ is given by $‖\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}‖\text{:}$ $\begin{array}{ccc}\hfill \stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}& =\hfill & \|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\end{array}\|=\left(1-0\right)\text{i}-\left(-1-0\right)\text{j}+\left(0-\left(-1\right)\right)\text{k}=\text{i}+\text{j}+\text{k}\hfill \\ \hfill ‖\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}‖& =\hfill & ‖⟨1,1,1⟩‖=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=\sqrt{3}.\hfill \end{array}$ The area of $\text{Δ}PQR$ is half the area of the parallelogram, or $\sqrt{3}\text{/}2.$ Find the area of the parallelogram $PQRS$ with vertices $P\left(1,1,0\right),Q\left(7,1,0\right),R\left(9,4,2\right),$ and $S\left(3,4,2\right).$ $6\sqrt{13}$ how can chip be made from sand is this allso about nanoscale material Almas are nano particles real yeah Joseph Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master? no can't Lohitha where is the latest information on a no technology how can I find it William currently William where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine has a lot of application modern world Kamaluddeen yes narayan what is variations in raman spectra for nanomaterials ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations?
# What is the relationship between a cylinder cone and sphere? ## What is the relationship between a cylinder cone and sphere? What is the relationship between the volume of the sphere and the volume of the cylinder? (Answer: The sphere takes up two-thirds of the volume of the cylinder.) What is the relationship between the volume of the cone and the volume of the cylinder? (Answer: The cone takes up one-third of the volume of the cylinder.) What is the difference between cylinder cone and sphere? A cylinder is similar to a prism, but its two bases are circles, not polygons. Also, the sides of a cylinder are curved, not flat. A cone has one circular base and a vertex that is not on the base. The sphere is a space figure having all its points an equal distance from the center point. ### What do cylinders cones and spheres have in common? Three-dimensional figures – Cylinders, cones and spheres – In Depth. In this lesson, we study some common space figures that are not polyhedra. These figures have some things in common with polyhedra, but they all have some curved surfaces, while the surfaces of a polyhedron are always flat. Why is the volume of a cone one-third of a cylinder? Derivation of Volume of Cone Formula Fill the cones with water and empty out one cone at a time. Each cone fills the cylinder to one-third quantity. Hence, such three cones will fill the cylinder. Thus, the volume of a cone is one-third of the volume of the cylinder. #### Which statement tells the relationship between the volumes of sphere and cylinder? Note : The volume of a sphere is 2/3 of the volume of a cylinder with same radius, and height equal to the diameter. Is a circle a sphere or cylinder? A cylinder is a three-dimensional solid consisting of two congruent, parallel, circular sides (the bases), joined by a curved surface. You could also think of a cylinder as a “circular prism”. consists of two congruent, parallel circles joined by a curved surface….Surface Area of a Cone. s 2 = + × π s = + × ## Are all cones 1/3 of a cylinder? Thus, the volume of a cone is equal to one-third of the volume of a cylinder having the same base radius and height. How many cones fill a sphere? 2 cones 1 sphere has the same volume as 2 cones. ## How much is Kingsborough Community College a semester? How much is Kingsborough Community College a semester? SPRING 2022 TUITION RATES RESIDENT STUDENTS NON-RESIDENT STUDENTS FULL-TIME (12-18 Credits/equated credits) \$2,400 per semester \$320 per credit PART-TIME (1-11… ## How do you make fermented rice cakes? How do you make fermented rice cakes? Instructions In a medium mixing bowl, add rice flour, glutinous rice flour, and mix until well combined. Pour in the ¾… ## When was the first Academy Awards held? When was the first Academy Awards held? May 16, 19291st Academy Awards / Date When the first Academy Awards® were handed out on May 16, 1929, at an… ## What is DB2COPY1? What is DB2COPY1? DB2COPY1 is the default name of the Db2 copy that is the first installation of a Db2 database product on your machine. This same name… ## How do I Group clips in Pro Tools? How do I Group clips in Pro Tools? A Clip Group is best described as a “container” in which you can put multiple clips. They’re created easily, by… ## Is An American Werewolf in Paris a sequel? Is An American Werewolf in Paris a sequel? It follows the general concept of, and is a sequel to, John Landis’s 1981 film An American Werewolf in London….
Question 1. # Find Equations Of The Tangent Lines To The Curve Y=(X-1)/(X+1) That Are Parallel To The Line X-2Y=4 Finding equations of tangent lines to a curve can often be a challenging task. Fortunately, with the right knowledge and understanding of calculus, it is possible to find equations for tangent lines that are parallel to specific lines. In this blog post, we will explore how to find equations for the tangent lines to the curve y=(x-1)/(x+1) that are parallel to the line x-2y=4. We’ll look at what these equations mean and how they can be used in various situations. So if you’re ready, let’s get started! ## What is the equation of the tangent line to the curve y=(x-1)/(x+1) that is parallel to the line x-2y=4? The equation of a tangent line to the curve y=(x-1)/(x+1) that is parallel to the line x-2y=4 can be found using the following steps: 1. Find the slope of the curve at the point where the tangent line intersects it. This can be done by taking the derivative of y=(x-1)/(x+1). 2. Find the slope of the line x-2y=4. 3. Set the slopes equal to each other and solve for y. This will give you the equation of the tangent line. ## How do you find equations of tangent lines? To find the equation of the tangent line to the curve y=(x-)/(x+) that is parallel to the line x-y=, we use the following steps: 1. We first find the slope of the curve at the point where we want to find the equation of the tangent line. To do this, we take the derivative of y with respect to x: 2. Next, we plug in the coordinates of the point where we want to find the equation of the tangent line into our formula for slope: 3. Now that we have our slope, we can use it to write down an equation for our desired tangent line using any point on that line: 4. Finally, we solve for y in our equation to get our final answer: ## What is the slope of the tangent line to the curve y=(x-1)/(x+1) at the point (2, -1)? At the point (2, -1), the slope of the tangent line to the curve y=(x-1)/(x+1) is -2. This can be calculated using the formula for the derivative of a function: dy/dx = (x+1)^(-2)(1-2(x-1))/(x+1)^2 Plugging in x=2, we get dy/dx = -2. This means that the equation of the tangent line at this point is y=-2*x+3. ## How do you find the equation of a line given its slope and To find the equation of a line given its slope and a point on the line, we can use the point-slope form of a line. This form is: y – y1 = m(x – x1) where m is the slope and (x1, y1) is a point on the line. We can plug in our values for m and (x1, y1) to get our equation. 2. Are you stumped trying to find the equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x-2y = 4? Don’t worry, we’ve got you covered! In this blog post, we’ll walk you through the steps to find the equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x-2y = 4. To begin, let’s start by reviewing some key concepts. First, we must understand what a tangent line is. A tangent line is a line that intersects a curve at one point and is parallel to the curve at that point. Now let’s talk about the equation of the line we’re trying to find. We know that the equation of the line we’re trying to find is parallel to the line x-2y = 4. This means that the slope of the line we’re looking for is the same as the slope of the line x-2y = 4. To find the slope of the line x-2y = 4, we must use the slope-intercept form of the equation, which is y = mx + b. Let’s plug in the values of the equation x-2y = 4 into the slope-intercept formula to get y = -1/2x + 2. Therefore, the slope of the line x-2y = 4 is -1/2. Now that we know the slope of the line we’re looking for, let’s move on to finding the equation of the curve y = (x-1)/(x+1). To do this, we must first find the slope of the curve. To find the slope of the curve, we must use the derivative of the equation. The derivative of the equation y = (x-1)/(x+1) is dy/dx = 1/(x+1)². Now that we know the slope of the curve and the slope of our line, we can begin to find the equations for the tangent lines. To find the equation of the tangent line, we must use the point-slope form of the equation, which is y – y1 = m(x-x1). Let’s plug in the values of the slope and the point from the curve y = (x-1)/(x+1) into the point-slope formula. The point from the curve y = (x-1)/(x+1) is (1, 0). Plugging in the values of the slope and the point from the curve y = (x-1)/(x+1), we get y – 0 = 1/(x+1)² (x-1). Rearranging the equation, we find that the equation of the tangent line is y = 1/(x+1)² + 1. Now let’s find the equation of the other tangent line. To do this, we must use the same point-slope formula, but with the slope of the line x-2y = 4, which is -1/2. Plugging in the values of the slope and the point from the curve y = (x-1)/(x+1) into the point-slope formula, we get y – 0 = -1/2(x-1). Rearranging the equation, we find that the equation of the other tangent line is y = -1/2x + 1. Congratulations! You have successfully found the equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x-2y = 4.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are reading an older version of this FlexBook® textbook: CK-12 Calculus Go to the latest version. # 7.6: Improper Integrals Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives A student will be able to: • Compute by hand the integrals of a wide variety of functions by using the technique of Improper Integration. • Combine this technique with other integration techniques to integrate. • Distinguish between proper and improper integrals. The concept of improper integrals is an extension to the concept of definite integrals. The reason for the term improper is because those integrals either • include integration over infinite limits or • the integrand may become infinite within the limits of integration. We will take each case separately. Recall that in the definition of definite integral baf(x)dx\begin{align}\int_a^b f(x) dx\end{align} we assume that the interval of integration [a,b]\begin{align}[a, b]\end{align} is finite and the function f\begin{align}f\end{align} is continuous on this interval. ## Integration Over Infinite Limits If the integrand f\begin{align}f\end{align} is continuous over the interval [a,),\begin{align}[a, \infty),\end{align} then the improper integral in this case is defined as af(x)dx=limllaf(x)dx. If the integration of the improper integral exists, then we say that it converges. But if the limit of integration fails to exist, then the improper integral is said to diverge. The integral above has an important geometric interpretation that you need to keep in mind. Recall that, geometrically, the definite integral baf(x)dx\begin{align}\int_a^b f(x) dx\end{align} represents the area under the curve. Similarly, the integral laf(x)dx\begin{align}\int_a^l f(x) dx\end{align} is a definite integral that represents the area under the curve f(x)\begin{align}f(x)\end{align} over the interval [a,l],\begin{align}[a, l],\end{align} as the figure below shows. However, as l\begin{align}l\end{align} approaches \begin{align} \infty\end{align}, this area will expand to the area under the curve of f(x)\begin{align}f(x)\end{align} and over the entire interval [a,).\begin{align}[a, \infty).\end{align} Therefore, the improper integral af(x)dx\begin{align}\int_a^{\infty} f(x) dx\end{align} can be thought of as the area under the function f(x)\begin{align}f(x)\end{align} over the interval [a,).\begin{align}[a, \infty).\end{align} Example 1: Evaluate 1dxx\begin{align}\int_1^{\infty} \frac{dx} {x}\end{align} . Solution: We notice immediately that the integral is an improper integral because the upper limit of integration approaches infinity. First, replace the infinite upper limit by the finite limit l\begin{align}l\end{align} and take the limit of l\begin{align}l\end{align} to approach infinity: 1dxx=limll1dxx=liml[lnx]l1=liml(lnlln1)=limllnl=. Thus the integral diverges. Example 2: Evaluate 2dxx2\begin{align}\int_2^{\infty} \frac{dx} {x^2}\end{align} . Solution: 2dxx2=limll2dxx2=liml[1x]l2=liml(1l+12)=12. Thus the integration converges to 12.\begin{align}\frac{1} {2}.\end{align} Example 3: Evaluate +dx1+x2\begin{align}\int_{+ \infty}^{- \infty} \frac{dx} {1 + x^2}\end{align}. Solution: What we need to do first is to split the integral into two intervals (,0]\begin{align}(-\infty, 0]\end{align} and [0,+).\begin{align}[0, +\infty).\end{align} So the integral becomes +dx1+x2=0dx1+x2++0dx1+x2. Next, evaluate each improper integral separately. Evaluating the first integral on the right, Evaluating the second integral on the right, Remark: In the previous example, we split the integral at \begin{align}x = 0.\end{align} However, we could have split the integral at any value of \begin{align}x = c\end{align} without affecting the convergence or divergence of the integral. The choice is completely arbitrary. This is a famous thoerem that we will not prove here. That is, ## Integrands with Infinite Discontinuities This is another type of integral that arises when the integrand has a vertical asymptote (an infinite discontinuity) at the limit of integration or at some point in the interval of integration. Recall from Chapter 5 in the Lesson on Definite Integrals that in order for the function \begin{align}f\end{align} to be integrable, it must be bounded on the interval \begin{align}[a, b].\end{align} Otherwise, the function is not integrable and thus does not exist. For example, the integral develops an infinite discontinuity at \begin{align}x = 1\end{align} because the integrand approaches infinity at this point. However, it is continuous on the two intervals \begin{align}[0, 1)\end{align} and \begin{align}(1, 4].\end{align} Looking at the integral more carefully, we may split the interval \begin{align} [0,4] \rightarrow [0,1) \cup (1,4]\end{align} and integrate between those two intervals to see if the integral converges. We next evaluate each improper integral. Integrating the first integral on the right hand side, The integral diverges because \begin{align}\ln(0)\end{align} is undefined, and thus there is no reason to evaluate the second integral. We conclude that the original integral diverges and has no finite value. Example 4: Evaluate \begin{align}\int_1^3 \frac{dx} {\sqrt{x - 1}}\end{align} . Solution: So the integral converges to \begin{align}2\sqrt{2}\end{align}. Example 5: In Chapter 5 you learned to find the volume of a solid by revolving a curve. Let the curve be \begin{align}y = xe^{-x}, 0 \le x \le \infty\end{align} and revolving about the \begin{align}x-\end{align}axis. What is the volume of revolution? Solution: From the figure above, the area of the region to be revolved is given by \begin{align}A = \pi y^2 = \pi x^2 e^{-2x}\end{align}. Thus the volume of the solid is As you can see, we need to integrate by parts twice: Thus At this stage, we take the limit as \begin{align}l\end{align} approaches infinity. Notice that the when you substitute infinity into the function, the denominator of the expression \begin{align}\frac{2l^2 + 2l + 1} {-4e^{2l}},\end{align} being an exponential function, will approach infinity at a much faster rate than will the numerator. Thus this expression will approach zero at infinity. Hence So the volume of the solid is \begin{align} \pi/4.\end{align} Example 6: Evaluate \begin{align}\int_{- \infty}^{+ \infty} \frac {dx}{e^x+e^{-x}}\end{align}. Solution: This can be a tough integral! To simplify, rewrite the integrand as Substitute into the integral: Using \begin{align}u-\end{align}substitution, let \begin{align}u = e^x, du = e^xdx.\end{align} Returning to our integral with infinite limits, we split it into two regions. Choose as the split point the convenient \begin{align}x = 0.\end{align} Taking each integral separately, Similarly, Thus the integral converges to For a video presentation of Improper Integrals (22.0), see Improper Integrals, www.justmathtutoring.com (6:23). For a video presentation of Improper Integrals with Infinity in the Upper and Lower Limits (22.0), see Improper Integrals, www.justmathtutoring.com (7:55). ## Review Questions 1. Determine whether the following integrals are improper. If so, explain why. 1. \begin{align}\int_{1}^{7} \frac {x+2}{x-3} dx\end{align} 2. \begin{align}\int_{1}^{7} \frac {x+2}{x+3} dx\end{align} 3. \begin{align}\int_{0}^{1} {{ \ln}x} dx\end{align} 4. \begin{align}\int_{0}^{\infty} \frac {1}{\sqrt {x-2}} dx\end{align} 5. \begin{align}\int_{0}^{{\pi}/4} {\tan x} dx\end{align} Evaluate the integral or state that it diverges. 1. \begin{align}\int_{1}^{\infty} \frac {1}{x^{2.001}} dx\end{align} 2. \begin{align}\int_{- \infty}^{-2} \left [ \frac{1}{x-1} - \frac {1}{x+1} \right ] dx\end{align} 3. \begin{align}\int_{-\infty}^{0} e^{5x} dx\end{align} 4. \begin{align}\int_{3}^{5} \frac {1}{(x-3)^4} dx\end{align} 5. \begin{align}\int_{-{\pi}/2}^{{\pi}/2} {\tan x} dx\end{align} 6. \begin{align}\int_{0}^{1} \frac {1}{\sqrt {1-x^2}} dx\end{align} 7. The region between the \begin{align}x-\end{align}axis and the curve \begin{align}y = e^{-x}\end{align} for \begin{align}x \ge 0\end{align} is revolved about the \begin{align}x-\end{align}axis. 1. Find the volume of revolution, \begin{align}V.\end{align} 2. Find the surface area of the volume generated, \begin{align}S.\end{align} 1. Improper; infinite discontinuity at \begin{align}x = 3.\end{align} 2. Not improper. 3. Improper; infinite discontinuity at \begin{align}x = 0.\end{align} 4. Improper; infinite interval of integration. 5. Not improper. 1. \begin{align}\frac {1}{1.001}\end{align} 2. \begin{align}\ln 3\end{align} 3. \begin{align}\frac {1}{5}\end{align} 4. divergent 5. divergent 6. \begin{align}\frac {\pi}{2}.\end{align} 1. \begin{align}V = \pi /2\end{align} 2. \begin{align}S = {\pi} \left [{\sqrt {2} + { \ln}(1+ \sqrt {2})} \right ].\end{align} ## Homework Evaluate the following integrals. 1. \begin{align}\int {\sqrt {\sin x}} {\cos x} dx\end{align} 2. \begin{align}\int x \tan^2(x^2) \sec^2(x^2) dx\end{align} 3. \begin{align}\int_{0}^{{ \ln}3} {\sqrt {e^{2x} - 1}} dx\end{align} 4. \begin{align}\int_{0}^{\infty} \frac {1}{x^2} dx\end{align} 5. \begin{align}\int_{-1}^{8} \frac {1}{\sqrt[3]{x}} dx\end{align} 6. \begin{align}\int \frac {x^2+x-16}{(x+1)(x-3)^2} dx\end{align} 7. Graph and find the volume of the region enclosed by the \begin{align}x-\end{align}axis, the \begin{align}y-\end{align}axis, \begin{align}x = 2\end{align} and \begin{align}y = x^2/(9-x^2)\end{align} when revolved about the \begin{align}x-\end{align}axis. 8. The Gamma Function, \begin{align} \Gamma (x)\end{align}, is an improper integral that appears frequently in quantum physics. It is defined as The integral converges for all \begin{align}x \ge 0.\end{align} 1. Find \begin{align} \Gamma (1).\end{align} 2. Prove that \begin{align}\Gamma (x + 1) = x \Gamma(x)\end{align}, for all \begin{align}x \ge 0\end{align}. 3. Prove that \begin{align}\Gamma \left ( \frac{1}{2} \right ) = {\sqrt {\pi}}.\end{align} 9. Refer to the Gamma Function defined in the previous exercise to prove that 1. \begin{align}\int_{0}^{\infty} e^{-x^n} dx = \Gamma \left ( \frac{n+1}{n} \right ), {n \ge 0}\end{align} [Hint: Let \begin{align}t = x^n\end{align}] 2. \begin{align}\int_{0}^{1} ({ \ln}x)^n dx = {(-1)^n} \Gamma (n+1), {n \ge 0}\end{align} [Hint: Let \begin{align}t = - \ln x\end{align}] 10. In wave mechanics, a sawtooth wave is described by the integral where \begin{align}k\end{align} is called the wave number, \begin{align}\omega\end{align} is the frequency, and \begin{align}t\end{align} is the time variable. Evaluate the integral. 1. \begin{align}\frac {2}{3} \sin x^{3/2} + C\end{align} 2. \begin{align}\frac {1}{6} \tan^3 (x^2) + C\end{align} 3. \begin{align}\sqrt{8} - \sec^{-1}3\end{align} 4. divergent 5. \begin{align}\frac {9}{2}\end{align} 6. \begin{align}{ \ln}\frac {(x-3)^2}{\begin{vmatrix}{x+1} \end{vmatrix}} +\frac {1}{x-3} + C\end{align} 7. \begin{align}{\pi} \left ( \frac{19}{5} - \frac {9}{4} { \ln}5 \right)\end{align} 1. \begin{align} \Gamma (1) = 1\end{align} 1. Hint: Let \begin{align}t = x^n.\end{align} 2. Hint: \begin{align}t = -\ln x\end{align} 8. \begin{align}\frac {2}{(k \omega)^2} \sin (k \pi)\end{align} Feb 23, 2012 Aug 19, 2015
# Multiplication & Division Word Problems Worksheet ------ Note: The Information above this point will not be sent to your printer -------- ## Reteach: Multiplicative Comparison Multiplier Unknown In some multiplicative comparison word problems, you are given the number of items in one set, and you are given the number of items in the second set, which is a multiple of the first set. The "multiplier" amount is the part that is unknown. The multiplier amount tells you how many times bigger (or more) the second set is than the first. "Bigger" can also be "longer," or "wider," or "older," or "taller" in problems involving measurement, or "faster" in problems involving a rate of speed. These problems in which you know both the number in one set, and the number in the second set are called “Multiplier Unknown” comparisons, because the multiplier is the part that is unknown. In order to answer the question you are being asked, you need to use the inverse (opposite) operation of multiplication: division. This kind of division is called “measurement” division. ## Reteach: Guided Practice Let’s look at a “Multiplier Unknown” problem: Scott is 10 years old. His little sister, Kate, is only two years old. How many times older than Kate is Scott? Start by clarifying what you understand, and what you are being asked to solve. The words “older than” tell us that we are comparing. Scott’s age, 10, tells us that there are 10 years in the second set. Kate’s age, 2, tells us that there are 2 years in one set. The part we are missing is the multiplier. To solve the problem we have to use the inverse operation of multiplication: division. We divide Scott’s age, 10, by Kate’s age, 2. Since 10 ÷ 2 = 5, Scott is five times older than Kate. This answer is reasonable because the problem tells us that Scott is older than Kate. We know that the second set is a multiple of the first set, so we can think, “how many sets of 2 will equal 10?” 2 x 5 = 10, so this answer makes sense. --------- Page Break------------- ## Independent Practice: Multiplicative Comparison Multiplier Unknown Solve each problem below by identifying the number in one set, and the number in the second set, which is a multiple of the first. Divide the second set by the first set. Go back to the problem to make sure you have answered the question being asked, and that your answer makes sense. The plane goes 700 miles an hour. The car goes 50 miles an hour. How many times faster than the car is the plane? The number in one set is 700 . The number in the second set is 50 . 700 ÷ 50 = 14 The plane is 14 times faster than the car. If you multiply the speed of the car by your answer, you should get the speed of the plane. Is your answer reasonable? Eric has 9 video games. Bryan has 54 video games. How many times more video games does Bryan have than Eric? The number in one set is 54 . The number in the second set is 9 . 54 ÷ 9 = 6 Bryan has 6 times as many video games as Eric. If you multiply the number of video games that Eric has by your answer, you should get the number of video games that Bryan has. Is your answer reasonable? Shannon is 37 inches tall. Her teenaged brother, Rick, is 74 inches tall. How many times as tall as Shannon is Rick? The number in one set is 74 . The number in the second set is 37 . 74 ÷ 37 = 2 Rick is 2 times as tall as Shannon. If you multiply the number of inches in Shannon's height by your answer, you should get the number of inches in Rick's height. Is your answer reasonable? ------ Note: The Information below this point will not be sent to your printer -------- ### Related Resources The various resources listed below are aligned to the same standard, (4OA02) taken from the CCSM (Common Core Standards For Mathematics) as the Word problems Worksheet shown above. Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.1 #### Worksheet ##### Multiplication and Division Similar to the above listing, the resources below are aligned to related standards in the Common Core For Mathematics that together support the following learning outcome: Use the four operations with whole numbers to solve problems ### Prevent Bullying Click the links below for information and help on dealing with bullying.
# Division of Whole Numbers Within Parentheses Involving Division - Examples, Exercises and Solutions The division of whole numbers within parentheses where there is a division refers to the situation in which we must carry out the mathematical operation of dividing a whole number by the result of dividing two elements, that is, by their quotient. For example: $24 : (6 : 2)$ There are two ways to solve this type of exercises. The first one will be to open the parentheses and extract the numbers that were inside them. That is, in our example: $24 : (6 : 2) =$ $24:6\times2=$ $4\times2=8$ ## Examples with solutions for Division of Whole Numbers Within Parentheses Involving Division ### Exercise #1 $100-(30-21)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $30-21=9$ Now we obtain: $100-9=91$ $91$ ### Exercise #2 $13-(7+4)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $7+4=11$ Now we subtract: $13-11=2$ $2$ ### Exercise #3 $12:(2\times2)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $2\times2=4$ Now we divide: $12:4=3$ $3$ ### Exercise #4 $15:(2\times5)=$ ### Step-by-Step Solution We will use the formula: $a:(b\times c)=a:b:c$ Therefore, we get: $15:2:5=$ Let's write the exercise as a fraction: $\frac{\frac{15}{2}}{5}=$ We'll convert it to a multiplication of two fractions: $\frac{15}{2}\times\frac{1}{5}=$ We multiply numerator by numerator and denominator by denominator, and we get: $\frac{15}{10}=1\frac{5}{10}=1\frac{1}{2}$ $1\frac{1}{2}$ ### Exercise #5 $55-(8+21)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $8+21=29$ Now we obtain the exercise: $55-29=26$ $26$ ### Exercise #6 $60:(5\times3)=$ ### Step-by-Step Solution We write the exercise in fraction form: $\frac{60}{5\times3}$ We break down 60 into a multiplication exercise: $\frac{20\times3}{5\times3}=$ We simplify the 3s and obtain: $\frac{20}{5}$ We break down the 5 into a multiplication exercise: $\frac{5\times4}{5}=$ We simplify the 5 and obtain: $\frac{4}{1}=4$ $4$ ### Exercise #7 $60:(10\times2)=$ ### Step-by-Step Solution We write the exercise in fraction form: $\frac{60}{10\times2}=$ Let's separate the numerator into a multiplication exercise: $\frac{10\times6}{10\times2}=$ We simplify the 10 in the numerator and denominator, obtaining: $\frac{6}{2}=3$ $3$ ### Exercise #8 $28-(4+9)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $4+9=13$ Now we obtain the exercise: $28-13=15$ $15$ ### Exercise #9 $37-(4-7)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $4-7=-3$ Now we obtain: $37-(-3)=$ Remember that the product of a negative and a negative results in a positive, therefore: $-(-3)=+3$ Now we obtain: $37+3=40$ $40$ ### Exercise #10 $22-(28-3)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $28-3=25$ Now we obtain the exercise: $22-25=-3$ $-3$ ### Exercise #11 $38-(18+20)=$ ### Step-by-Step Solution According to the order of operations, first we solve the exercise within parentheses: $18+20=38$ Now, the exercise obtained is: $38-38=0$ $0$ ### Exercise #12 $7-(4+2)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $4+2=6$ Now we solve the rest of the exercise: $7-6=1$ $1$ ### Exercise #13 $8-(2+1)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $2+1=3$ Now we solve the rest of the exercise: $8-3=5$ $5$ ### Exercise #14 $66-(15-10)=$ ### Step-by-Step Solution According to the order of operations rules, we first solve the expression inside of the parentheses: $15-10=5$ We obtain the following expression: $66-5=61$ $61$ ### Exercise #15 $21:(30:10)=$ ### Step-by-Step Solution We will use the formula: $a:(b:c)=a:b\times c$ Therefore, we will get: $21:30\times10=$ Let's write the division exercise as a fraction: $\frac{21}{30}=\frac{7}{10}$ Now let's multiply by 10: $\frac{7}{10}\times\frac{10}{1}=$ We'll reduce the 10 and get: $\frac{7}{1}=7$ $7$
# Cramer’s Rule for a 2×2 System (with Two Variables) Cramer’s Rule is another method that can solve systems of linear equations using determinants. In terms of notations, a matrix is an array of numbers enclosed by square brackets while a determinant is an array of numbers enclosed by two vertical bars. ## Notations The formula to find the determinant of a 2 x 2 matrix is very straightforward. Let’s have a quick review: ## Quick Examples of How to Find the Determinants of a 2 x 2 Matrix Example 1: Find the determinant of the matrix A below. Example 2: Find the determinant of the matrix B below. Example 3: Find the determinant of the matrix C below. After knowing how to find the determinant of a 2 x 2 matrix, you’re now ready to learn the procedures or steps on how to use Cramer’s Rule. Here we go! ## Cramer’s Rules for Systems of Linear Equations with Two Variables • Given a linear system • Assign names for each matrix coefficient matrix: X – matrix: Y – matrix: To solve for the variable x: To solve for the variable y: Few points to consider when looking at the formula: 1) The columns of \large{x}, \large{y}, and the constant terms \large{c} are obtained as follows: 2) Both denominators in solving \large{x} and \large{y} are the same. They come from the columns of \large{x} and \large{y}. 3) Looking at the numerator in solving for \large{x}, the coefficients of \large{x}-column are replaced by the constant column (in red). 4) In the same manner, to solve for \large{y}, the coefficients of \large{y}-column are replaced by the constant column (in red). ## Examples of How to Solve Systems of Linear Equations with Two Variables using Cramer’s Rule Example 1: Solve the system with two variables by Cramer’s Rule Start by extracting the three relevant matrices: coefficient, \large{x}, and \large{y}. Then solve each corresponding determinant. • For coefficient matrix • For X – matrix • For Y – matrix Once all three determinants are calculated, it’s time to solve for the values of \large{x} and \large{y} using the formula above. I can write the final answer as \large{\left( {x,y} \right) = \left( {2, - 1} \right)}. Example 2: Solve the system with two variables by Cramer’s Rule Setup your coefficient, \large{x}, and \large{y} matrices from the given system of linear equations. Then calculate their determinants accordingly. Remember that we always subtract the products of the diagonal entries. • For the coefficient matrix (use the coefficients of both x and y variables) • For the X – matrix (replace the x-column by the constant column) • For the Y – matrix (replace the y-column by the constant column) I hope you’re getting comfortable computing for the determinant of a 2-dimensional matrix. To finally solve the required variables, I get the following results. Writing the final answer in point notation, I got \large{\left( {x,y} \right) = \left( {6, - 5} \right)}. Example 3: Solve the system with two variables by Cramer’s Rule This problem can actually be solved quite easily by the Elimination Method. This is because the coefficients of variable x are the “same” but only opposite in signs ( +1 and −1 ). To solve this using the elimination method, you add their corresponding columns and the x-variable goes away – leaving you with a one-step equation in \large{y}. I am mentioning this because every technique has shortcomings and it is best to pick the most efficient. Always get clarification from your teacher if it is okay to use another approach when the method is not specified on a given problem. Anyway, since we are learning how to solve by Cramer’s Rule, let’s go ahead and work it out with this method. I will construct three matrices ( coefficient, \large{x} and \large{y}) and evaluate their corresponding determinants. • For coefficient matrix • For X – matrix ( written as uppercase D with subscript x ) • For Y – matrix (written as uppercase D with subscript y) After obtaining the values of the three required determinants, I will calculate \large{x} and \large{y} as follows. The final answer in the point form is \large{\left( {x,y} \right) = \left( { - 1,2} \right)} . Example 4: Solve by Cramer’s Rule the system with two variables Since we have gone over a few examples already, I suggest that you try this problem on your own. Then, compare your answers to the solution below. If you get it right the first time that means you’re becoming a “pro” with regards to Cramer’s Rule. If you didn’t, try to figure out what went wrong and learn to not commit the same error next time. This is how you become better at math. Study many kinds of problems and more importantly, do a lot of independent practice. • For coefficient matrix • For X – matrix • For Y – matrix You should get the answer below… Example 5: Solve the system with two variables by Cramer’s Rule For our last example, I included a zero in the constant column. Every time you see the number zero in the constant column, I highly recommend using Cramer’s Rule to solve the system of linear equations. Why? Because the calculation of the determinants for \large{x} and \large{y} matrices drastically become super easy. Check it out yourself! • For coefficient matrix • For X – matrix • For Y – matrix The final solution to this problem is You might also be interested in: Cramer’s Rule 3×3
Upcoming SlideShare × # Lines, planes, and hyperplanes • 618 views • Comment goes here. Are you sure you want to Be the first to comment Total Views 618 On Slideshare 0 From Embeds 0 Number of Embeds 0 Shares 9 0 Likes 1 No embeds ### Report content No notes for slide ### Transcript • 1. The Calculus of Functions Section 1.4 of Lines, Planes, and Hyperplanes Several VariablesIn this section we will add to our basic geometric understanding of Rn by studying linesand planes. If we do this carefully, we shall see that working with lines and planes in Rnis no more diﬃcult than working with them in R2 or R3 .Lines in RnWe will start with lines. Recall from Section 1.1 that if v is a nonzero vector in Rn , then,for any scalar t, tv has the same direction as v when t > 0 and the opposite directionwhen t < 0. Hence the set of points {tv : −∞ < t < ∞}forms a line through the origin. If we now add a vector p to each of these points, we obtainthe set of points {tv + p : −∞ < t < ∞},which is a line through p in the direction of v, as illustrated in Figure 1.4.1 for R2 . p v Figure 1.4.1 A line in R2 through p in the direction of vDeﬁnition Given a vector p and a nonzero vector v in Rn , the set of all points y in Rnsuch that y = tv + p, (1.4.1)where −∞ < t < ∞, is called the line through p in the direction of v. 1 Copyright c by Dan Sloughter 2001 • 2. 2 Lines, Planes, and Hyperplanes Section 1.4 6 4 2 p -2 -1 1 2 3 4 5 v -2 -4 Figure 1.4.2 The line through p = (1, 2) in the direction of v = (1, −3) Equation (1.4.1) is called a vector equation for the line. If we write y = (y1 , y2 , . . . , yn ),v = (v1 , v2 , . . . , vn ), and p = (p1 , p2 , . . . , pn ), then (1.4.1) may be written as (y1 , y2 , . . . , yn ) = t(v1 , v2 , . . . , vn ) + (p1 , p2 , . . . , pn ), (1.4.2)which holds if and only if y1 = tv1 + p1 , y2 = tv2 + p2 , . . (1.4.3) . . . . yn = tvn + pn .The equations in (1.4.3) are called parametric equations for the line.Example Suppose L is the line in R2 through p = (1, 2) in the direction of v = (1, −3)(see Figure 1.4.2). Then y = t(1, −3) + (1, 2) = (t + 1, −3t + 2)is a vector equation for L and, if we let y = (x, y), x = t + 1, y = −3t + 2 • 3. Section 1.4 Lines, Planes, and Hyperplanes 3 5 y 0 10 q 5 p z 0 -5 -5 0 5 x Figure 1.4.3 The line through p = (1, 3, 1) and q = (−1, 1, 4)are parametric equations for L. Note that if we solve for t in both of these equations, wehave t = x − 1, 2−y t= . 3Thus 2−y x−1= , 3and so y = −3x + 5.Of course, the latter is just the standard slope-intercept form for the equation of a line inR2 .Example Now suppose we wish to ﬁnd an equation for the line L in R3 which passesthrough the points p = (1, 3, 1) and q = (−1, 1, 4) (see Figure 1.4.3). We ﬁrst note thatthe vector p − q = (2, 2, −3)gives the direction of the line, so y = t(2, 2, −3) + (1, 3, 1) • 4. 4 Lines, Planes, and Hyperplanes Section 1.4 q-p (q - p) - w w p Figure 1.4.4 Distance from a point q to a lineis a vector equation for L; if we let y = (x, y, z), x = 2t + 1, y = 2t + 3, z = −3t + 1are parametric equations for L. As an application of these ideas, consider the problem of ﬁnding the shortest distancefrom a point q in Rn to a line L with equation y = tv + p. If we let w be the projectionof q − p onto v, then, as we saw in Section 1.2, the vector (q − p) − w is orthogonal to vand may be pictured with its tail on L and its tip at q. Hence the shortest distance fromq to L is (q − p) − w . See Figure 1.4.4.Example To ﬁnd the distance from the point q = (2, 2, 4) to the line L through thepoints p = (1, 0, 0) and r = (0, 1, 0), we must ﬁrst ﬁnd an equation for L. Since thedirection of L is given by v = r − p = (−1, 1, 0), a vector equation for L is y = t(−1, 1, 0) + (1, 0, 0).If we let v 1 u= = √ (−1, 1, 0), v 2then the projection of q − p onto v is 1 1 1 w = ((q − p) · u)u = (1, 2, 4) · √ (−1, 1, 0)) √ (−1, 1, 0) = (−1, 1, 0). 2 2 2 • 5. Section 1.4 Lines, Planes, and Hyperplanes 5 y 0 5 10 -5 4 N M 2 L 0 z -2 -4 -5 0 5 x 10 Figure 1.4.5 Parallel (L and M ) and perpendicular (L and N ) linesThus the distance from q to L is 3 3 82 √ (q − p) − w = , ,4 = = 20.5. 2 2 4Deﬁnition Suppose L and M are lines in Rn with equations y = tv + p and y = tw + q,respectively. We say L and M are parallel if v and w are parallel. We say L and M areperpendicular, or orthogonal, if they intersect and v and w are orthogonal. Note that, by deﬁnition, a line is parallel to itself.Example The lines L and M in R3 with equations y = t(1, 2, −1) + (4, 1, 2)and y = t(−2, −4, 2) + (5, 6, 1),respectively, are parallel since (−2, −4, 2) = −2(1, 2, −1), that is, the vectors (1, 2, −1) and(−2, −4, 2) are parallel. See Figure 1.4.5.Example The lines L and N in R3 with equations y = t(1, 2, −1) + (4, 1, 2)and y = t(3, −1, 1) + (−1, 5, −1),respectively, are perpendicular since they intersect at (5, 3, 1) (when t = 1 for the ﬁrst lineand t = 2 for the second line) and (1, 2, −1) and (3, −1, 1) are orthogonal since (1, 2, −1) · (3, −1, 1) = 3 − 2 − 1 = 0.See Figure 1.4.5. • 6. 6 Lines, Planes, and Hyperplanes Section 1.4Planes in RnThe following deﬁnition is the ﬁrst step in deﬁning a plane.Deﬁnition Two vectors x and y in Rn are said to be linearly independent if neither oneis a scalar multiple of the other. Geometrically, x and y are linearly independent if they do not lie on the same linethrough the origin. Notice that for any vector x, 0 and x are not linearly independent,that is, they are linearly dependent, since 0 = 0x.Deﬁnition Given a vector p along with linearly independent vectors v and w, all in Rn ,the set of all points y such that y = tv + sw + p, (1.4.4)where −∞ < t < ∞ and −∞ < s < ∞, is called a plane. The intuition here is that a plane should be a two dimensional object, which isguaranteed because of the requirement that v and w are linearly independent. Alsonote that if we let y = (y1 , y2 , . . . , yn ), v = (v1 , v2 , . . . , vn ), w = (w1 , w2 , . . . , wn ), andp = (p1 , p2 , . . . , pn ), then (1.4.4) implies that y1 = tv1 + sw1 + p1 , y2 = tv2 + sw2 + p2 , . . (1.4.5) . . . . yn = tvn + swn + pn .As with lines, (1.4.4) is a vector equation for the plane and the equations in (1.4.5) areparametric equations for the plane.Example Suppose we wish to ﬁnd an equation for the plane P in R3 which contains thethree points p = (1, 2, 1), q = (−1, 3, 2), and r = (2, 3, −1). The ﬁrst step is to ﬁnd twolinearly independent vectors v and w which lie in the plane. Since P must contain theline segments from p to q and from p to r, we can take v = q − p = (−2, 1, 1)and w = r − p = (1, 1, −2).Note that v and w are linearly independent, a consequence of p, q, and r not all lying onthe same line. See Figure 1.4.6. We may now write a vector equation for P as y = t(−2, 1, 1) + s(1, 1, −2) + (1, 2, 1).Note that y = p when t = 0 and s = 0, y = q when t = 1 and s = 0, and y = r whent = 0 and s = 1. If we write y = (x, y, z), then, expanding the vector equation, (x, y, z) = t(−2, 1, 1) + s(1, 1, −2) + (1, 2, 1) = (−2t + s + 1, t + s + 2, t − 2s + 1), • 7. Section 1.4 Lines, Planes, and Hyperplanes 7 y 1 2 3 4 -1 0 5 v z 0 w -5 -2 0 2 xFigure 1.4.6 The plane y = tv + sw + p, with v = (−2, 1, 1), w = (1, 1, −2), p = (1, 2, 1)giving us x = −2t + s + 1, y = t + s + 2, z = t − 2s + 1for parametric equations for P . To ﬁnd the shortest distance from a point q to a plane P , we ﬁrst need to consider theproblem of ﬁnding the projection of a vector onto a plane. To begin, consider the plane Pthrough the origin with equation y = ta + sb where a = 1, b = 1, and a ⊥ b. Givena vector q not in P , let r = (q · a)a + (q · b)b,the sum of the projections of q onto a and onto b. Then (q − r) · a = q · a − r · a = q · a − (q · a)(a · a) − (q · b)(b · a) = q · a − q · a = 0, 2since a · a = a = 1 and b · a = 0, and, similarly, (q − r) · b = q · b − r · b = q · b − (q · a)(a · b) − (q · b)(b · b) = q · b − q · b = 0. • 8. 8 Lines, Planes, and Hyperplanes Section 1.4 (q - p) - r q-p r Figure 1.4.7 Distance from a point q to a planeIt follows that for any y = ta + sb in the plane P , (q − r) · y = (q − r) · (ta + sb) = t(q − r) · a + s(q − r) · b = 0.That is, q − r is orthogonal to every vector in the plane P . For this reason, we call r theprojection of q onto the plane P , and we note that the shortest distance from q to P is q−r . In the general case, given a point q and a plane P with equation y = tv + sw + p,we need only ﬁnd vectors a and b such that a ⊥ b, a = 1, b = 1, and the equationy = ta + sb + p describes the same plane P . You are asked in Problem 29 to verify thatif we let c be the projection of w onto v, then we may take 1 a= v vand 1 b= (w − c). w−cIf r is the sum of the projections of q − p onto a and b, then r is the projection of q − ponto P and (q − p) − r is the shortest distance from q to P . See Figure 1.4.7.Example To compute the distance from the point q = (2, 3, 3) to the plane P withequation y = t(−2, 1, 0) + s(1, −1, 1) + (−1, 2, 1),let v = (−2, 1, 0), w = (1, −1, 1), and p = (−1, 2, 1). Then, using the above notation, wehave 1 a = √ (−2, 1, 0), 5 3 c = (w · a)a = − (−2, 1, 0), 5 • 9. Section 1.4 Lines, Planes, and Hyperplanes 9 1 w−c= (−1, −2, 5), 5and 1 b = √ (−1, −2, 5). 30Since q − p = (3, 1, 2), the projection of q − p onto P is 1 1 r = ((3, 1, 2) · a)a + ((3, 1, 2) · b)b = −(−2, 1, 0) + (−1, −2, 5) = (11, −8, 5) 6 6and 1 (q − p) − r = (7, 14, 7). 6Hence the distance from q to P is √ 294 7 (q − p) − r = =√ . 6 6 More generally, we say vectors v1 , v2 , . . . , vk in Rn are linearly independent if no oneof them can be written as a sum of scalar multiples of the others. Given a vector p andlinearly independent vectors v1 , v2 , . . . , vk , we call the set of all points y such that y = t1 v1 + t2 v2 + · · · + tk vk + p,where −∞ < tj < ∞, j = 1, 2, . . . , k, a k-dimensional aﬃne subspace of Rn . In thisterminology, a line is a 1-dimensional aﬃne subspace and a plane is a 2-dimensional aﬃnesubspace. In the following, we will be interested primarily in lines and planes and so willnot develop the details of the more general situation at this time.HyperplanesConsider the set L of all points y = (x, y) in R2 which satisfy the equation ax + by + d = 0, (1.4.6)where a, b, and d are scalars with at least one of a and b not being 0. If, for example,b = 0, then we can solve for y, obtaining a d y =− x− . (1.4.7) b bIf we set x = t, −∞ < t < ∞, then the solutions to (1.4.6) are a d a d y = (x, y) = t, − t − = t 1, − + 0, − . (1.4.8) b b b b • 10. 10 Lines, Planes, and Hyperplanes Section 1.4 L n y y-p p Figure 1.4.8 L is the set of points y for which y − p is orthogonal to nThus L is a line through 0, − d in the direction of 1, − a . A similar calculation shows b b dthat if a = 0, then we can describe L as the line through − a , 0 in the direction of − a , 1 . Hence in either case L is a line in R2 . b Now let n = (a, b) and note that (1.4.6) is equivalent to n · y + d = 0. (1.4.9)Moreover, if p = (p1 , p2 ) is a point on L, then n · p + d = 0, (1.4.10)which implies that d = −n · p. Thus we may write (1.4.9) as n · y − n · p = 0,and so we see that (1.4.6) is equivalent to the equation n · (y − p) = 0. (1.4.11)Equation (1.4.11) is a normal equation for the line L and n is a normal vector for L. Inwords, (1.4.11) says that the line L consists of all points in R2 whose diﬀerence with p isorthogonal to n. See Figure 1.4.8.Example Suppose L is a line in R2 with equation 2x + 3y = 1. • 11. Section 1.4 Lines, Planes, and Hyperplanes 11Then a normal vector for L is n = (2, 3); to ﬁnd a point on L, we note that when x = 2,y = −1, so p = (2, −1) is a point on L. Thus (2, 3) · ((x, y) − (2, −1)) = 0,or, equivalently, (2, 3) · (x − 2, y + 1) = 0,is a normal equation for L. Since q = (−1, 1) is also a point on L, L has directionq − p = (−3, 2). Thus y = t(−3, 2) + (2, −1)is a vector equation for L. Note that n · (q − p) = (2, 3) · (−3, 2) = 0,so n is orthogonal to q − p.Example If L is a line in R2 through p = (2, 3) in the direction of v = (−1, 2), thenn = (2, 1) is a normal vector for L since v · n = 0. Thus (2, 1) · (x − 2, y − 3) = 0is a normal equation for L. Multiplying this out, we have 2(x − 2) + (y − 3) = 0;that is, L consists of all points (x, y) in R2 which satisfy 2x + y = 7. Now consider the case where P is the set of all points y = (x, y, z) in R3 that satisfythe equation ax + by + cz + d = 0, (1.4.12)where a, b, c, and d are scalars with at least one of a, b, and c not being 0. If for example,a = 0, then we may solve for x to obtain b c d x=− y− z− . (1.4.13) a a aIf we set y = t, −∞ < t < ∞, and z = s, −∞ < s < ∞, the solutions to (1.4.12) are y = (x, y, z) b c d = − t − s − , t, s a a a (1.4.14) b c d = t − , 1, 0 + s − , 0, 1 + − , 0, 0 . a a a • 12. 12 Lines, Planes, and Hyperplanes Section 1.4 n y-p P Figure 1.4.9 P is the set of points y for which y − p is orthogonal to nThus we see that P is a plane in R3 . In analogy with the case of lines in R2 , if we letn = (a, b, c) and let p = (p1 , p2 , p3 ) be a point on P , then we have n · p + d = ax + by + cz + d = 0,from which we see that n · p = −d, and so we may write (1.4.12) as n · (y − p) = 0. (1.4.15)We call (1.4.15) a normal equation for P and we call n a normal vector for P . In words,(1.4.15) says that the plane P consists of all points in R3 whose diﬀerence with p isorthogonal to n. See Figure 1.4.9.Example Let P be the plane in R3 with vector equation y = t(2, 2, −1) + s(−1, 2, 1) + (1, 1, 2).If we let v = (2, 2, −1) and w = (−1, 2, 1), then n = v × w = (4, −1, 6)is orthogonal to both v and w. Now if y is on P , then y = tv + sw + pfor some scalars t and s, from which we see that n · (y − p) = n · (tv + sw) = t(n · v) + s(n · w) = 0 + 0 = 0.That is, n is a normal vector for P . So, letting y = (x, y, z), (4, −1, 6) · (x − 1, y − 1, z − 2) = 0 (1.4.16) • 13. Section 1.4 Lines, Planes, and Hyperplanes 13is a normal equation for P . Multiplying (1.4.16) out, we see that P consists of all points(x, y, z) in R3 which satisfy 4x − y + 6z = 15.Example Suppose p = (1, 2, 1), q = (−2, −1, 3), and r = (2, −3, −1) are three pointson a plane P in R3 . Then v = q − p = (−3, −3, 2)and w = r − p = (1, −5, −2)are vectors lying on P . Thus n = v × w = (16, −4, 18)is a normal vector for P . Hence (16, −4, 18) · (x − 1, y − 2, z − 1) = 0is a normal equation for P . Thus P is the set of all points (x, y, z) in R3 satisfying 16x − 4y + 18y = 26. The following deﬁnition generalizes the ideas in the previous examples.Deﬁnition Suppose n and p are vectors in Rn with n = 0. The set of all vectors y inRn which satisfy the equation n · (y − p) = 0 (1.4.17)is called a hyperplane through the point p. We call n a normal vector for the hyperplaneand we call (1.4.17) a normal equation for the hyperplane. In this terminology, a line in R2 is a hyperplane and a plane in R3 is a hyperplane. Ingeneral, a hyperplane in Rn is an (n − 1)-dimensional aﬃne subspace of Rn . Also, notethat if we let n = (a1 , a2 , . . . , an ), p = (p1 , p2 , . . . , pn ), and y = (y1 , y2 , . . . , yn ), then wemay write (1.4.17) as a1 (y1 − p1 ) + a2 (y2 − p2 ) + · · · + an (yn − pn ) = 0, (1.4.18)or a1 y1 + a2 y2 + · · · + an yn + d = 0 (1.4.19)where d = −n · p.Example The set of all points (w, x, y, z) in R4 which satisfy 3w − x + 4y + 2z = 5is a 3-dimensional hyperplane with normal vector n = (3, −1, 4, 2). • 14. 14 Lines, Planes, and Hyperplanes Section 1.4 n q H q-p \| (q - p) . u\| p Figure 1.4.10 Distance from a point q to a hyperplane H The normal equation description of a hyperplane simpliﬁes a number of geometriccalculations. For example, given a hyperplane H through p with normal vector n and apoint q in Rn , the distance from q to H is simply the length of the projection of q − ponto n. Thus if u is the direction of n, then the distance from q to H is \|(q − p) · u\|. SeeFigure 1.4.10. Moreover, if we let d = −p · n as in (1.4.19), then we have q·n−p·n \|q · n + d\| \|(q − p) · u\| = \|q · u − p · u\| = = . (1.4.20) n nNote that, in particular, (1.4.20) may be used to ﬁnd the distance from a point to a linein R2 and from a point to a plane in R3 .Example To ﬁnd the distance from the point q = (2, 3, 3) to the plane P in R3 withequation x + 2y + z = 4,we ﬁrst note that n = (1, 2, 1) is a normal vector for P . Using (1.4.20) with d = −4, wesee that the distance from q to P is \|q · n + d\| \|(2, 3, 3) · (1, 2, 1) − 4\| 7 = √ =√ . n 6 6Note that this agrees with an earlier example. We will close this section with a few words about angles between hyperplanes. Notethat a hyperplane does not have a unique normal vector. In particular, if n is a normalvector for a hyperplane H, then −n is also a normal vector for H. Hence it is alwayspossible to choose the normal vectors required in the following deﬁnition.Deﬁnition Let G and H be hyperplanes in Rn with normal equations m · (y − p) = 0 • 15. Section 1.4 Lines, Planes, and Hyperplanes 15and n · (y − q) = 0,respectively, chosen so that m · n ≥ 0. Then the angle between G and H is the anglebetween m and n. Moreover, we will say that G and H are orthogonal if m and n areorthogonal and we will say G and H are parallel if m and n are parallel. The eﬀect of the choice of normal vectors in the deﬁnition is to make the angle betweenthe two hyperplanes be between 0 and π .2Example To ﬁnd the angle θ between the two planes in R3 with equations x + 2y − z = 3and x − 3y − z = 5,we ﬁrst note that the corresponding normal vectors are m = (1, 2, −1) and n = (1, −3, −1).Since m · n = −4, we will compute the angle between m and −n. Hence m · (−n) 4 4 cos(θ) = =√ √ =√ . m n 6 11 66Thus, rounding to four decimal places, 4 θ = cos−1 √ = 1.0560. 66See Figure 1.4.11.Example The planes in R3 with equations 3x + y − 2z = 3and 6x + 2y − 4z = 13are parallel since their normal vectors are m = (3, 1, −2) and n = (6, 2, −4) and n = 2m.Problems1. Find vector and parametric equations for the line in R2 through p = (2, 3) in the direction of v = (1, −2).2. Find vector and parametric equations for the line in R4 through p = (1, −1, 2, 3) in the direction of v = (−2, 3, −4, 1).3. Find vector and parametric equations for the lines passing through the following pairs of points. • 16. 16 Lines, Planes, and Hyperplanes Section 1.4 y 1 0 -1 -2 0 z -5 -10 2 1 0 -1 -2 x Figure 1.4.11 The planes x + 2y − z = 3 and x − 3y − z = 5 (a) p = (−1, −3), q = (4, 2) (b) p = (2, 1, 3), q = (−1, 2, 1) (c) p = (3, 2, 1, 4), q = (2, 0, 4, 1) (d) p = (4, −3, 2), q = (1, −2, 4)4. Find the distance from the point q = (1, 3) to the line with vector equation y = t(2, 1) + (3, 1). • 17. Section 1.4 Lines, Planes, and Hyperplanes 17 5. Find the distance from the point q = (1, 3, −2) to the line with vector equation y = t(2, −1, 4) + (1, −2, −1). 6. Find the distance from the point r = (−1, 2, −3) to the line through the points p = (1, 0, 1) and q = (0, 2, −1). 7. Find the distance from the point r = (−1, −2, 2, 4) to the line through the points p = (2, 1, 1, 2) and q = (1, 2, −4, 3). 8. Find vector and parametric equations for the plane in R3 which contains the points p = (1, 3, −1), q = (−2, 1, 1), and r = (2, −3, 2). 9. Find vector and parametric equations for the plane in R4 which contains the points p = (2, −3, 4, −1), q = (−1, 3, 2, −4), and r = (2, −1, 2, 1).10. Let P be the plane in R3 with vector equation y = t(1, 2, 1) + s(−2, 1, 3) + (1, 0, 1). Find the distance from the point q = (1, 3, 1) to P .11. Let P be the plane in R4 with vector equation y = t(1, −2, 1, 4)+s(2, 1, 2, 3)+(1, 0, 1, 0). Find the distance from the point q = (1, 3, 1, 3) to P .12. Find a normal vector and a normal equation for the line in R2 with vector equation y = t(1, 2) + (1, −1).13. Find a normal vector and a normal equation for the line in R2 with vector equation y = t(0, 1) + (2, 0).14. Find a normal vector and a normal equation for the plane in R3 with vector equation y = t(1, 2, 1) + s(3, 1, −1) + (1, −1, 1).15. Find a normal vector and a normal equation for the line in R2 which passes through the points p = (3, 2) and q = (−1, 3).16. Find a normal vector and a normal equation for the plane in R3 which passes through the points p = (1, 2, −1), q = (−1, 3, 1), and r = (2, −2, 2).17. Find the distance from the point q = (3, 2) in R2 to the line with equation x+2y−3 = 0.18. Find the distance from the point q = (1, 2, −1) in R3 to the plane with equation x + 2y − 3x = 4.19. Find the distance from the point q = (3, 2, 1, 1) in R4 to the hyperplane with equation 3x + y − 2z + 3w = 15.20. Find the angle between the lines in R2 with equations 3x + y = 4 and x − y = 5.21. Find the angle between the planes in R3 with equations 3x−y +2z = 5 and x−2y +z = 4.22. Find the angle between the hyperplanes in R4 with equations w + x + y − z = 3 and 2w − x + 2y + z = 6.23. Find an equation for a plane in R3 orthogonal to the plane with equation x+2y−3z = 4 and passing through the point p = (1, −1, 2). • 18. 18 Lines, Planes, and Hyperplanes Section 1.424. Find an equation for the plane in R3 which is parallel to the plane x − y + 2z = 6 and passes through the point p = (2, 1, 2).25. Show that if x, y, and z are vectors in Rn with x ⊥ y and x ⊥ z, then x ⊥ (ay + bz) for any scalars a and b.26. Find parametric equations for the line of intersection of the planes in R3 with equations x + 2y − 6z = 4 and 2x − y + z = 2.27. Find parametric equations for the plane of intersection of the hyperplanes in R4 with equations w − x + y + z = 3 and 2w + 4x − y + 2z = 8.28. Let L be the line in R3 with vector equation y = t(1, 2, −1) + (3, 2, 1) and let P be the plane in R3 with equation x + 2y − 3z = 8. Find the point where L intersects P .29. Let P be the plane in Rn with vector equation y = tv+sw +p. Let c be the projection of w onto v, 1 a= v, v and 1 b= (w − c). w−c Show that y = ta + sb + p is also a vector equation for P .
# Signed Number in Math: Definition & Examples An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: Special Product: Definition & Formula ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:01 What Are Signed Numbers? • 1:34 Comparing Values of… • 2:29 How Are Signed Numbers… • 4:27 Lesson Summary Save Save Want to watch this again later? Timeline Autoplay Autoplay Speed Speed #### Recommended Lessons and Courses for You Lesson Transcript Instructor: Betty Bundly Betty has a master's degree in mathematics and 10 years experience teaching college mathematics. In this lesson, we will find out about signed numbers. The use of signed numbers could be thought of as where arithmetic ends and algebra begins. We will also discuss situations encountered every day that can be described by using signed numbers. ## What Are Signed Numbers? We will begin an introduction to signed numbers with what is known as the number line. A number line is a horizontal line with incremental marks indicating the position of numbers. The conventional layout for all number lines will have the positive numbers are on the right, the negative numbers are on the left and zero between the negative numbers and the positive numbers. The signed numbers consist of the negative numbers and the positive numbers. We are not used to seeing the (+) sign on positive numbers and most of the time it is not necessary to show it. However, it is always necessary to show the sign on a negative number. Notice that for every positive number, there is a negative counterpart. This is true no matter how small or how large the number is. These counterparts are the same distance from zero. For example, the number 5 and -5 are the same distance from zero. The number 0.0001 and -0.0001 are both the same distance from zero. The number 100,000 and -100,000 are the same distance from zero. A more precise description for these counterparts is to say that they are opposites or additive inverses. Every number on the number line except zero has an opposite or additive inverse, and it is found by simply changing the sign of the number. The numbers 5 and -5 are opposites or additive inverses. Zero is not considered either negative or positive. ## Comparing Values of Signed Numbers In arithmetic, we learned how to compare positive numbers. We know, for example, that 2 is less than 5. Looking at these values on the number line, we see that 2 is to the left of 5. In fact, notice that the numbers become less in value as you move to the left on the number line and the numbers become greater in value as you move to the right on the number line. This relationship remains true for numbers on the negative side of the number line as well. For example, -2 is greater in value than -5 because it is farther to the right on the number line. You have to be especially careful comparing negative fractions and decimals but the same principle applies: numbers farther to the left are less in value and numbers farther to the right are greater in value. For example, -2/3 is less than -1/3 while on the positive side of zero 2/3 is greater than 1/3. ### How Are Signed Numbers Used in Mathematics? It may be difficult at first to understand how anything can be less than zero or nothing. The following examples may illustrate that this concept is useful when considering numbers below zero in a relative or comparative sense rather than absolute. 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.
2016 AMC 12A Problems/Problem 25 Problem Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$, then the numbers that Bernardo writes are $16, 25, 36, 49, 64$, and the numbers showing on the board after Silvia erases are $1, 2, 3, 4, and 6$, and thus $f(1) = 5$. What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$? $\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064$ Solution Consider $f(2)$. The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$. Calculus $\left(\frac{\text{d}}{\text{d}x} x^2=2x\right)$ and a bit of thinking says this first happens at $x\ge 100/2 = 50$. The perfect squares from here go: $2500, 2601, 2704, 2809\dots$. Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$. After the ones and tens make $100$, the hundreds place will go up by $2$, thus reaching our goal. Since $10^2=100$, the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$. The missing number is one less than the number of hundreds $(k=2)$ of $3600$, or $35$. Now consider f(4). Instead of the difference between two squares needing to be $100$, the difference must now be $10000$. This first happens at $x\ge 5000$. After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$. This will take place at $\left(5000+100\right)^2=5100^2= 26010000$. Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value. In general, each new value of $f(k+2)$ will add two digits to the "$5$" and one digit to the "$1$". This means that the last number Bernardo writes for $k=6$ is $\left(500000+1000\right)^2$, the last for $k = 8$ will be $\left(50000000+10000\right)^2$, and so on until $k=2016$. Removing the last $k$ digits as Silvia does will be the same as removing $k/2$ trailing zeroes on the number to be squared. This means that the last number on the board for $k=6$ is $5001^2$, $k=8$ is $50001^2$, and so on. So the first missing number is $5001^2-1,50001^2-1\text{ etc.}$ The squaring will make a "$25$" with two more digits than the last number, a "$10$" with one more digit, and a "$1$". The missing number is one less than that, so the "1" will be subtracted from $f(k)$. In other words, $f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}$. Therefore: $$f(2) =35 =25 +10$$ $$f(4) =2600 =2500 +100$$ $$f(6) =251000 =250000 +1000$$ $$f(8) = 25010000 = 25000000 + 10000$$ And so on. The sum $f(2) + f(4) + f(6) +\dots + f(2016)$ is: $2.52525252525\dots 2525\cdot 10^{2015}$ + $1.11111\dots 110\cdot 10^{1008}$, with $2016$ repetitions each of "$25$" and "$1$". There is no carrying in this addition. Therefore each $f(k)$ adds $2 + 5 + 1 = 8$ to the sum of the digits. Since $2n = 2016$, $n = 1008$, and $8n = 8064$, or $\boxed{\textbf{(E)}\text{ 8064}}$. Addendum: You could also use the fact that $$(x+1)^2 = x^2 +2x+1$$ In other words, the difference between $x^2$ and $(x+1)^2$ is equal to $2x+1$. We can set the inequality $2x+1 \geq 100$. Obviously, the first integer $x$ that satisfies this is 50. This way, while being longer, is IMO more motivated and doesn't use calculus.
## Big Ideas Math Book 4th Grade Answer Key Chapter 9 Multiply Whole Numbers and Fractions The topics covered in Multiply Whole Numbers and Fractions are Understand Multiples of Unit Fractions, Fraction Operations, Understand Multiples of Fractions, Understand Multiples of Fractions, and so on. We have provided solutions as per the topics provided in the chapter. So go through the BIM Grade 4 Chapter 9 Solution Key topicwise and start preparing now. Lesson: 1 Understand Multiples of Unit Fractions Lesson: 2 Understand Multiples of Fractions Lesson: 3 Multiply Whole Numbers and Fractions Lesson: 4 Multiply Whole Numbers and Mixed Numbers Lesson: 5 Problem Solving: Fraction Operations ### Lesson 9.1 Understand Multiples of Unit Fractions Explore and Grow Draw a model of any fraction using unit fractions. Then write an addition equation to represent your model. Reasoning How can you rewrite the equation using multiplication? Explain. Let us take the fraction as $$\frac{3}{5}$$ we draw a model using fractions as below Addition equation to represent $$\frac{3}{5}$$ model is $$\frac{3}{5}$$ = $$\frac{1}{5}$$ + $$\frac{1}{5}$$ +$$\frac{1}{5}$$ to rewrite the equation using multiplication, We write the equation as $$\frac{3}{5}$$ = $$\frac{1}{5}$$ +$$\frac{1}{5}$$ + $$\frac{1}{5}$$ = 3 X $$\frac{1}{5}$$ as the fraction $$\frac{3}{5}$$ represents 3 parts that are each $$\frac{1}{5}$$ of the whole. Think and Grow: Multiples of Unit Fractions Any fraction can be written as a multiple of a unit fraction with a like denominator. Example Write $$\frac{5}{8}$$ as a multiple of a unit fraction. The fraction $$\frac{5}{8}$$ represents 5 parts that are each $$\frac{1}{8}$$ of the whole. Show and Grow Write the fraction as a multiple of a unit fraction. Question 1. $$\frac{2}{3}$$ = $$\frac{1}{3}$$ +$$\frac{1}{3}$$ = 2 X $$\frac{1}{3}$$. Explanation: We write $$\frac{2}{3}$$ in addition equation as $$\frac{2}{3}$$ =$$\frac{1}{3}$$ + $$\frac{1}{3}$$ and multiplication as fraction $$\frac{2}{3}$$ represents 2 parts that are each $$\frac{1}{3}$$ of the whole, $$\frac{2}{3}$$ =2 X $$\frac{1}{3}$$. Question 2. $$\frac{4}{8}$$ = $$\frac{1}{8}$$ +$$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$ = 4 X $$\frac{1}{8}$$. Explanation: We write $$\frac{4}{8}$$ in addition equation as $$\frac{4}{8}$$ = $$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$ +$$\frac{1}{8}$$ and multiplication as fraction $$\frac{4}{8}$$ represents 4 parts that are each $$\frac{1}{8}$$ of the whole, $$\frac{4}{8}$$ =4 X $$\frac{1}{8}$$. Question 3. $$\frac{6}{5}$$ $$\frac{6}{5}$$ = $$\frac{1}{5}$$+$$\frac{1}{5}$$ +$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$ = 6 X $$\frac{1}{5}$$. Explanation: We write $$\frac{6}{5}$$ in addition equation as $$\frac{6}{5}$$= $$\frac{1}{5}$$+ $$\frac{1}{5}$$ +$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$ and multiplication as fraction $$\frac{6}{5}$$ represents 6 parts that are $$\frac{1}{5}$$ of the whole $$\frac{6}{5}$$ = 6 X $$\frac{1}{5}$$. Question 4. $$\frac{7}{100}$$ $$\frac{7}{100}$$ = $$\frac{1}{100}$$ + $$\frac{1}{100}$$ + $$\frac{1}{100}$$ + $$\frac{1}{100}$$+ $$\frac{1}{100}$$+ $$\frac{1}{100}$$+ $$\frac{1}{100}$$ = 7 X $$\frac{1}{100}$$. Explanation: We write $$\frac{7}{100}$$ in addition equation as $$\frac{7}{100}$$= $$\frac{1}{100}$$ + $$\frac{1}{100}$$ + $$\frac{1}{100}$$ + $$\frac{1}{100}$$+ $$\frac{1}{100}$$+ $$\frac{1}{100}$$+ $$\frac{1}{100}$$ and multiplication as fraction $$\frac{7}{100}$$ represents 7 parts that are $$\frac{1}{100}$$ of the whole, $$\frac{7}{100}$$= 7 X $$\frac{1}{100}$$. Apply and Grow: Practice Write the fraction as a multiple of a unit fraction. Question 5. $$\frac{2}{6}$$ $$\frac{2}{6}$$ = $$\frac{1}{6}$$ + $$\frac{1}{6}$$ = 2 X $$\frac{1}{6}$$. Explanation: We write $$\frac{2}{6}$$ in addition equation as $$\frac{2}{6}$$= $$\frac{1}{6}$$ + $$\frac{1}{6}$$ and multiplication as fraction $$\frac{2}{6}$$ represents 2 parts that are $$\frac{1}{6}$$ of the whole,$$\frac{2}{6}$$ = 2 X $$\frac{1}{6}$$. Question 6. $$\frac{6}{8}$$ $$\frac{6}{8}$$ = $$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$+ $$\frac{1}{8}$$+ $$\frac{1}{8}$$+ $$\frac{1}{8}$$ = 6 X $$\frac{1}{8}$$. Explanation: We write $$\frac{6}{8}$$ in addition equation as $$\frac{6}{8}$$ = $$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$+ $$\frac{1}{8}$$+ $$\frac{1}{8}$$+ $$\frac{1}{8}$$ and multiplication as fraction $$\frac{6}{8}$$ represents 6 parts that are $$\frac{1}{8}$$ of the whole, $$\frac{6}{8}$$ =6 X $$\frac{1}{8}$$. Question 7. $$\frac{5}{4}$$ $$\frac{5}{4}$$ = $$\frac{1}{4}$$ + $$\frac{1}{4}$$ +$$\frac{1}{4}$$ + $$\frac{1}{4}$$ +$$\frac{1}{4}$$ = 5 X $$\frac{1}{4}$$. Explanation: We write $$\frac{5}{4}$$ in addition equation as $$\frac{5}{4}$$=$$\frac{1}{4}$$ + $$\frac{1}{4}$$ +$$\frac{1}{4}$$ +$$\frac{1}{4}$$ + $$\frac{1}{4}$$ and multiplication as fraction $$\frac{5}{4}$$ represents 5 parts that are $$\frac{1}{4}$$ of the whole, $$\frac{5}{4}$$ = 5 X $$\frac{1}{4}$$. Question 8. $$\frac{3}{12}$$ $$\frac{3}{12}$$= $$\frac{1}{12}$$ + $$\frac{1}{12}$$ +$$\frac{1}{12}$$ = 3 X $$\frac{1}{12}$$ Explanation : We write $$\frac{3}{12}$$ in addition equation as $$\frac{3}{12}$$= $$\frac{1}{12}$$ + $$\frac{1}{12}$$ + $$\frac{1}{12}$$ and multiplication as fraction $$\frac{3}{12}$$ represents 3 parts that are $$\frac{1}{12}$$ of the whole, $$\frac{3}{12}$$ = 3 X $$\frac{1}{12}$$. Question 9. $$\frac{8}{100}$$ $$\frac{8}{100}$$ = $$\frac{1}{100}$$ + $$\frac{1}{100}$$+$$\frac{1}{100}$$+ $$\frac{1}{100}$$+$$\frac{1}{100}$$+$$\frac{1}{100}$$+$$\frac{1}{100}$$+ $$\frac{1}{100}$$ = 8 X $$\frac{1}{100}$$. Explanation : We write $$\frac{8}{100}$$ in addition equation as $$\frac{8}{100}$$ = $$\frac{1}{100}$$ + $$\frac{1}{100}$$+$$\frac{1}{100}$$+ $$\frac{1}{100}$$+$$\frac{1}{100}$$+ $$\frac{1}{100}$$+ $$\frac{1}{100}$$+$$\frac{1}{100}$$ and multiplication as fraction $$\frac{8}{100}$$ represents 8 parts that are $$\frac{1}{100}$$ of the whole, $$\frac{8}{100}$$= 8 X $$\frac{1}{100}$$. Question 10. $$\frac{7}{10}$$ $$\frac{7}{10}$$ = $$\frac{1}{10}$$ +$$\frac{1}{10}$$ +$$\frac{1}{10}$$ + $$\frac{1}{10}$$ +$$\frac{1}{10}$$ +$$\frac{1}{10}$$ +$$\frac{1}{10}$$ = 7 X $$\frac{1}{10}$$. Explanation: We write $$\frac{7}{10}$$ in addition equation as $$\frac{7}{10}$$ = $$\frac{1}{10}$$ + $$\frac{1}{10}$$ +$$\frac{1}{10}$$ +$$\frac{1}{10}$$ +$$\frac{1}{10}$$ + $$\frac{1}{10}$$ +$$\frac{1}{10}$$ and multiplication as fraction $$\frac{7}{10}$$ represents 7 parts that are $$\frac{1}{10}$$ of the whole,$$\frac{7}{10}$$ = 7 X $$\frac{1}{10}$$. Question 11. Structure When a fraction is written as a multiple of a unit fraction, what is the relationship between the numerator of the fraction and the number that is multiplied by the unit fraction? When a fraction is written as a multiple of a unit fraction, the relationship between the numerator of the fraction and the number that is multiplied by the unit fraction, is the whole number times the unit fraction. Explanation : We know that the numerator of a unit fraction is always one. So When a fraction is written as a multiple of a unit fraction the number will be always the whole number (whole number of times the unit fraction) and is multiplied by unit fraction. For example $$\frac{3}{8}$$ we write it as 3 X $$\frac{1}{8}$$ here it is 3 and this 3 is multiplied by unit fraction. DIG DEEPER! Write the mixed number as a multiple of a unit fraction. Question 12. 1$$\frac{1}{8}$$ 1$$\frac{1}{8}$$ = $$\frac{9}{8}$$ = $$\frac{1}{8}$$+$$\frac{1}{8}$$+ $$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+ $$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$= 9 X $$\frac{1}{8}$$ . Explanation: First we convert a mixed number to an improper fraction, It is to multiply the bottom number 8 by the “regular” number 1, add in the top number 1, and then put this on top of the original bottom number as 9 by 8. 1$$\frac{1}{8}$$ = 1+$$\frac{1}{8}$$ = $$\frac{9}{8}$$ We write $$\frac{9}{8}$$ in addition equation as $$\frac{9}{8}$$= $$\frac{1}{8}$$+ $$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+ $$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$ +$$\frac{1}{8}$$ and multiplication as fraction $$\frac{9}{8}$$ represents 9 parts that are $$\frac{1}{10}$$ of the whole, $$\frac{9}{8}$$ = 9 X $$\frac{1}{8}$$. Question 13. 3$$\frac{4}{5}$$ 3$$\frac{4}{5}$$ = $$\frac{19}{5}$$=$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$ +$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$= 19 X $$\frac{1}{5}$$. Explanation: First we convert a mixed number to an improper fraction, It is to multiply the bottom number 5 by the “regular” number 3, add in the top number 4, and then put this on top of the original bottom number as 19 by 5. 3$$\frac{4}{5}$$=3+$$\frac{4}{5}$$ = $$\frac{19}{5}$$ We write $$\frac{19}{5}$$ in addition equation as $$\frac{19}{5}$$ = $$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$ +$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$ and multiplication as fraction $$\frac{19}{5}$$ represents 19 parts that are $$\frac{1}{5}$$ of the whole,$$\frac{19}{5}$$= 19 X $$\frac{1}{5}$$. Question 13. Think and Grow: Modeling Real Life Example A juice stand worker uses $$\frac{1}{2}$$ of an orange to garnish each drink. The worker has $$\frac{7}{12}$$ of an orange left. How many more drinks can the worker garnish? Write the fraction of orange left as a multiple of $$\frac{1}{12}$$. The worker can garnish $$\frac{1}{12}$$ more drinks, $$\frac{7}{12}$$ minus $$\frac{1}{2}$$ means $$\frac{7}{12}$$ – $$\frac{1}{2}$$ = $$\frac{1}{12}$$, Orange left is 7 X $$\frac{1}{12}$$. Explanation: Given a juice stand worker uses $$\frac{1}{2}$$ of an orange to garnish each drink. The worker has $$\frac{7}{12}$$ of an orange left. So the worker can garnish more $$\frac{7}{12}$$ minus $$\frac{1}{2}$$ means $$\frac{7}{12}$$ – $$\frac{1}{2}$$ = $$\frac{1}{12}$$, Given $$\frac{7}{12}$$ of an orange is left , to write orange left as a multiple of $$\frac{1}{12}$$ we have orange left as $$\frac{7}{12}$$ so write it as 7 parts that are $$\frac{1}{12}$$ of the whole, $$\frac{7}{12}$$= 7 X $$\frac{1}{12}$$. Show and Grow Question 14. A piece of rope is $$\frac{8}{5}$$ meters long. You cut the rope into $$\frac{1}{5}$$ meter long pieces. How many pieces do you cut? 8 pieces I have cut. Explanation: Given a piece of rope is $$\frac{8}{5}$$ meters long and it has been cut into $$\frac{1}{5}$$ meter long pieces, therefore pieces of rope cut is $$\frac{8}{5}$$ divided by $$\frac{1}{5}$$ = $$\frac{8}{5}$$ X latex]\frac{5}{1}[/latex] = 8. Question 15. A restaurant serves $$\frac{4}{10}$$ of a meatloaf to 4 customers. Each customer receives the same amount of meatloaf. What fraction of the meatloaf does each customer receive? The fraction of the meatloaf each customer receives is $$\frac{1}{10}$$ Explanation: Given a restaurant serves $$\frac{4}{10}$$ of a meatloaf to 4 customers and each customer receives the same amount of meatloaf so fraction of the meatloaf each customer receives is $$\frac{4}{10}$$ divided by 4 = $$\frac{4}{10}$$ x $$\frac{1}{4}$$ = $$\frac{1}{10}$$. Question 16. You use 3$$\frac{3}{4}$$ pounds of trail mix to make treat bags. You put $$\frac{1}{4}$$ pound of trail mix into each bag. How many treat bags do you make? Total number of treat bags made are 15 Explanation: I use 3$$\frac{3}{4}$$ pounds of trail mix to make treat bags, We convert a mixed number to an improper fraction,It is to multiply the bottom number 4 by the “regular” number 3, add in the top number 3, and then put this on top of the original bottom number as 15 by 4, it becomes $$\frac{15}{4}$$,Each bag i kept $$\frac{1}{4}$$ pound of trail mix so total number of treat bags required to make are $$\frac{15}{4}$$ divided by $$\frac{1}{4}$$=$$\frac{15}{4}$$ X $$\frac{4}{1}$$=15 treat bags. Question 17. DIG DEEPER! You walk from home to school and then back home again each day for 5 days. Altogether, you walk $$\frac{10}{8}$$ miles. What is the distance from your home to school? Explain. The distance from home to school is $$\frac{1}{8}$$ miles Explanation: Total number of miles walked is $$\frac{10}{8}$$ in 5 days that to from home to school and then back home again means it is twice a day , Let us take X as the distance from home to school is, Given 2X x 5=$$\frac{10}{8}$$ miles, X = Then each day number of miles walked is $$\frac{10}{8}$$ divided by 10 which is equal to $$\frac{10}{8}$$ x $$\frac{1}{10}$$ =$$\frac{1}{8}$$, Therefore X = $$\frac{1}{8}$$ miles. ### Understand Multiples of Unit Fractions Homework & practice 9.1 Write the fraction as a multiple of a unit fraction. Question 1. $$\frac{2}{12}$$= $$\frac{1}{12}$$+$$\frac{1}{12}$$ = 2 X $$\frac{1}{12}$$ Explanation: We write $$\frac{2}{12}$$ in addition equation as $$\frac{2}{12}$$= $$\frac{1}{12}$$ + $$\frac{1}{12}$$ and multiplication as fraction $$\frac{2}{12}$$ represents 2 parts that are $$\frac{1}{12}$$ of the whole, $$\frac{2}{12}$$= 2 X $$\frac{1}{12}$$. Question 2. $$\frac{5}{100}$$= $$\frac{1}{100}$$+$$\frac{1}{100}$$+$$\frac{1}{100}$$+ $$\frac{1}{100}$$+$$\frac{1}{100}$$ = 5 X $$\frac{1}{100}$$. Explanation: We write $$\frac{5}{100}$$ in addition equation as $$\frac{5}{100}$$= $$\frac{1}{100}$$+ $$\frac{1}{100}$$+$$\frac{1}{100}$$+$$\frac{1}{100}$$+$$\frac{1}{100}$$ and multiplication as fraction $$\frac{5}{100}$$ represents 5 parts that are $$\frac{1}{100}$$ of the whole, $$\frac{5}{100}$$ = 5 X $$\frac{1}{100}$$. Question 3. $$\frac{3}{6}$$ $$\frac{3}{6}$$=$$\frac{1}{6}$$+$$\frac{1}{6}$$+$$\frac{1}{6}$$ = 3 X $$\frac{1}{6}$$. Explanation: We write $$\frac{3}{6}$$ in addition equation as $$\frac{3}{6}$$=$$\frac{1}{6}$$+ $$\frac{1}{6}$$+$$\frac{1}{6}$$ and multiplication as fraction $$\frac{3}{6}$$ represents 3 parts that are $$\frac{1}{6}$$ of the whole,$$\frac{3}{6}$$ = 3 X $$\frac{1}{6}$$. Question 4. $$\frac{7}{10}$$ $$\frac{7}{10}$$ =$$\frac{1}{10}$$+$$\frac{1}{10}$$+$$\frac{1}{10}$$+ $$\frac{1}{10}$$+$$\frac{1}{10}$$+$$\frac{1}{10}$$+$$\frac{1}{10}$$ = 7 X $$\frac{1}{10}$$ Explanation: We write $$\frac{7}{10}$$ in addition equation as $$\frac{7}{10}$$= $$\frac{1}{10}$$+ $$\frac{1}{10}$$+$$\frac{1}{10}$$+$$\frac{1}{10}$$+$$\frac{1}{10}$$+ $$\frac{1}{10}$$+$$\frac{1}{10}$$ and multiplication as $$\frac{7}{10}$$ represents 7 parts that are $$\frac{1}{10}$$ of the whole,$$\frac{7}{10}$$ = 7 X $$\frac{1}{10}$$. Question 5. $$\frac{4}{5}$$ $$\frac{4}{5}$$ = $$\frac{1}{5}$$+$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$ = 4 X $$\frac{1}{5}$$. Explanation: We write $$\frac{4}{5}$$ in addition equation as $$\frac{4}{5}$$=$$\frac{1}{5}$$+ $$\frac{1}{5}$$+$$\frac{1}{5}$$+$$\frac{1}{5}$$ and multiplication as $$\frac{4}{5}$$ represents 4 parts that are $$\frac{1}{5}$$ of the whole,$$\frac{4}{5}$$= 4 X $$\frac{1}{5}$$. Question 6. $$\frac{9}{2}$$ $$\frac{9}{2}$$ = $$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+ $$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+ $$\frac{1}{2}$$+$$\frac{1}{2}$$ = 9 X $$\frac{1}{2}$$. Explanation: We write $$\frac{9}{2}$$ in addition equation as $$\frac{9}{2}$$ = $$\frac{1}{2}$$+ $$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+ $$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$ and multiplication as $$\frac{9}{2}$$ represents 9 parts that are $$\frac{1}{2}$$ of the whole, $$\frac{9}{2}$$= 9 X $$\frac{1}{2}$$. Question 7. $$\frac{8}{8}$$ $$\frac{8}{8}$$=1=$$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+ $$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+ $$\frac{1}{8}$$ = $$\frac{8}{8}$$ = 8 X $$\frac{1}{8}$$. Explanation: $$\frac{8}{8}$$= 1 ,but we write the fraction as a addition of a unit fraction and multiple of a unit fraction, We write $$\frac{8}{8}$$ in addition equation as $$\frac{8}{8}$$ = $$\frac{1}{8}$$+ $$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$+ $$\frac{1}{8}$$+$$\frac{1}{8}$$+$$\frac{1}{8}$$ and multiplication as $$\frac{8}{8}$$ represents 8 parts that are $$\frac{1}{8}$$ of the whole,$$\frac{8}{8}$$= 8 X $$\frac{1}{8}$$. Question 8. $$\frac{6}{3}$$ $$\frac{6}{3}$$=2,$$\frac{6}{3}$$= $$\frac{1}{3}$$+$$\frac{1}{3}$$+ $$\frac{1}{3}$$+$$\frac{1}{3}$$+$$\frac{1}{3}$$+$$\frac{1}{3}$$ = 6 X $$\frac{1}{3}$$. Explanantion: $$\frac{6}{3}$$= 2 ,but we write the fraction as a addition of a unit fraction and multiple of a unit fraction, We write $$\frac{6}{3}$$ in addition equation as $$\frac{6}{3}$$= $$\frac{1}{3}$$+ $$\frac{1}{3}$$+$$\frac{1}{3}$$+$$\frac{1}{3}$$+$$\frac{1}{3}$$+ $$\frac{1}{3}$$ and multiplication as $$\frac{6}{3}$$ represents 6 parts that are $$\frac{1}{3}$$ of the whole,$$\frac{6}{3}$$= 6 X $$\frac{1}{3}$$. Question 9. Logic What is Newton’s fraction? Write the fraction as a multiple of a unit fraction. Newton’s fraction = $$\frac{9}{12}$$=$$\frac{1}{12}$$ +$$\frac{1}{12}$$ + $$\frac{1}{12}$$ +$$\frac{1}{12}$$ +$$\frac{1}{12}$$ +$$\frac{1}{12}$$ + $$\frac{1}{12}$$ +$$\frac{1}{12}$$ +$$\frac{1}{12}$$ = 9 X $$\frac{1}{12}$$. Explanation: Given that Newton’s fraction as sum of 9 unit fractions and each unit fraction is one twelfth of a whole. So Newton’s fraction is $$\frac{9}{12}$$ we write $$\frac{9}{12}$$ in addition equation as $$\frac{9}{12}$$= $$\frac{1}{12}$$ +$$\frac{1}{12}$$ +$$\frac{1}{12}$$ + $$\frac{1}{12}$$ +$$\frac{1}{12}$$ +$$\frac{1}{12}$$ +$$\frac{1}{12}$$ + $$\frac{1}{12}$$ +$$\frac{1}{12}$$ and multiplication as $$\frac{9}{12}$$ represents 9 parts that are $$\frac{1}{12}$$ of the whole,$$\frac{9}{12}$$= 9 X $$\frac{1}{12}$$. Question 10 Structure Write a multiplication equation that represents the grapefruit halves. $$\frac{6}{2}$$=$$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+ $$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$ = 6 X $$\frac{1}{2}$$. Explanation: Given 6 number of grapefruit halves so there are total grapefruit as $$\frac{6}{2}$$ we write $$\frac{6}{2}$$ in addition equation as $$\frac{6}{2}$$=$$\frac{1}{2}$$+ $$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+$$\frac{1}{2}$$+ $$\frac{1}{2}$$,so the multiplication equation is $$\frac{6}{2}$$ represents 6 parts that are $$\frac{1}{2}$$ of the whole,$$\frac{6}{2}$$= 6 X $$\frac{1}{2}$$. Question 11. Which One Doesn’t Belong? Which expression does not belong with the other three? 4 X $$\frac{1}{5}$$ expression does not belong with the other three, as all the other expressions values results are same =$$\frac{5}{4}$$ only 4 X $$\frac{1}{5}$$ expression is not. Explanation: 1. 5 X $$\frac{1}{4}$$ = $$\frac{5}{4}$$ 2. $$\frac{1}{4}$$+$$\frac{1}{4}$$+$$\frac{1}{4}$$+$$\frac{1}{4}$$= 5 X $$\frac{1}{4}$$= $$\frac{5}{4}$$ 3. $$\frac{5}{4}$$ =$$\frac{5}{4}$$ 4. 4 X $$\frac{1}{5}$$= $$\frac{4}{5}$$≠ $$\frac{5}{4}$$ So in the above expressions only 4 X $$\frac{1}{5}$$ does not belong with the other three expressions. Question 12. Modeling Real Life You are making blueberry pancakes. You have $$\frac{6}{8}$$ cup of blueberries. You put $$\frac{1}{8}$$ cup of blueberries in each pancake. How many pancakes do you make? 6 pancakes i can make. Explanation: I have $$\frac{6}{8}$$ cup of blueberries and i can put $$\frac{1}{8}$$ cup of blueberries in each pan cake so total number of pancakes made are $$\frac{6}{8}$$ divided by $$\frac{1}{8}$$ = $$\frac{6}{8}$$ X $$\frac{8}{1}$$= 6 pancakes. Question 13. DIG DEEPER! You cut a loaf of zucchini bread into 20 equal slices. You and your friends eat $$\frac{3}{10}$$ of the slices. You want to put each leftover slice into its own bag. How many bags do you need? 14 bags are needed. Explanation : Given a loaf of zucchini bread is cut into 20 equal slices and i ,friend eat $$\frac{3}{10}$$ of the slices, Total number of slices of zucchini bread =20, Number of slices ate = $$\frac{3}{10}$$ means it says out of 10 slices 3 were ate, so out of 20 slices 2 x 3= 6 slices were eaten. Number of slices left = 20 – 6 =14 slices and one bag contains only one slice. Therefore number of bags needed are 14. Question 14. Is 46 a multiple of 45. No 46 is not a multiple of 45. Explanation: A multiple is a number that can be divided by another number a certain number of times without a remainder. 45 when divided by 46 a certain times will leave a remainder, So 46 is not a multiple of 45 or a factor is one of two or more numbers that divides a given number without a remainder. As factors of 45 are 1,3,5,9,15 and factors of 46 are 1,2,23, both do not have common factors so 46 is not a multiple of 45. Question 15. Is 3 a factor of 75? Yes 3 is a factor of 75 Explanation: We know that a factor is one that divides a given number without a remainder and 75 when divided by 3 leaves no remainder, So 3 is a factor of 75. ### Lesson 9.2 Understand Multiples of Fractions Draw a model for each expression. Then write a multiplication expression to represent each model. The multiplication expression is $$\frac{2}{6}$$ +$$\frac{2}{6}$$ +$$\frac{2}{6}$$ + $$\frac{2}{6}$$ = 4 X $$\frac{2}{6}$$ = $$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$ = 8 X $$\frac{1}{6}$$. Explanation: $$\frac{2}{6}$$ +$$\frac{2}{6}$$ +$$\frac{2}{6}$$ +$$\frac{2}{6}$$ the multiplication equation is 4 parts that are $$\frac{2}{6}$$ of the whole as 4 X $$\frac{2}{6}$$ $$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$ = 8 X $$\frac{1}{6}$$ the multiplication equation is 8 parts that are $$\frac{1}{6}$$ of the whole as 8 X $$\frac{1}{6}$$. Structure Compare your expressions. How are they the same? How are they different? Same means the result of both expressions is same = $$\frac{8}{6}$$ Different means in one expression we take value of $$\frac{2}{6}$$ 4 times and in another expression we take $$\frac{1}{6}$$ 8 times. Explanation: Same means $$\frac{2}{6}$$ +$$\frac{2}{6}$$ +$$\frac{2}{6}$$ +$$\frac{2}{6}$$ = 4 X $$\frac{2}{6}$$ = $$\frac{8}{6}$$ and $$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$+$$\frac{1}{6}$$ +$$\frac{1}{6}$$+ $$\frac{1}{6}$$ +$$\frac{1}{6}$$ = 8 X $$\frac{1}{6}$$ =$$\frac{8}{6}$$ , both have results similar as $$\frac{8}{6}$$, Different means here we are taking expressions as addition values of $$\frac{2}{6}$$ by 4 times and in other expression we are taking addition values of $$\frac{1}{6}$$ by 8 times. Think and Grow: Multiples of Fractions A multiple of any fraction can be written as a multiple of a unit fraction with a like denominator. Explanation: To write 2 X $$\frac{3}{5}$$ multiple of unit fraction first we write $$\frac{3}{5}$$ as multiple of 3 X $$\frac{1}{5}$$ then we multiply it by 2 as 2 X 3 X $$\frac{1}{5}$$=6 X $$\frac{1}{5}$$. Show and Grow Write the product as a multiple of a unit fraction. Then find the product. Question 1. 2 X $$\frac{4}{5}$$ = 2 X 4 X $$\frac{1}{5}$$ = 8 X $$\frac{1}{5}$$ = $$\frac{8}{5}$$. Explanation: First we write $$\frac{4}{5}$$ as $$\frac{1}{5}$$ + $$\frac{1}{5}$$ + $$\frac{1}{5}$$ + $$\frac{1}{5}$$ = 4 X $$\frac{1}{5}$$ then multiply it by 2 which is 2 X 4 X $$\frac{1}{5}$$ = 8 X $$\frac{1}{5}$$ = $$\frac{8}{5}$$. Question 2. 3 X $$\frac{2}{10}$$ = 3 X 2 X $$\frac{1}{10}$$ = 6 X $$\frac{1}{10}$$ =$$\frac{6}{10}$$. Explanation: First we write $$\frac{2}{10}$$ as $$\frac{1}{10}$$ +$$\frac{1}{10}$$ = 2 X $$\frac{1}{10}$$ then multiply it by 3 which is 3 X 2 X $$\frac{1}{10}$$ = 6 X $$\frac{1}{10}$$ =$$\frac{6}{10}$$. Question 3. 4 X $$\frac{3}{2}$$ = 4 X 3 X $$\frac{1}{2}$$ = 12 X $$\frac{1}{2}$$ = $$\frac{12}{2}$$ = 6. Explanation: First we write $$\frac{3}{2}$$ as $$\frac{1}{2}$$ +$$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 3 X $$\frac{1}{2}$$ then we multiply it by 4 which is 4 X 3 X $$\frac{1}{2}$$= 12 X $$\frac{1}{2}$$= $$\frac{12}{2}$$=6. Apply and Grow: Practice Write the product as a multiple of a unit fraction. Then ﬁnd the product. Question 4. 5 X $$\frac{2}{3}$$ = 5 X 2 X $$\frac{1}{3}$$ = 10 X $$\frac{1}{3}$$= $$\frac{10}{3}$$. Explanation: First we write $$\frac{2}{3}$$ as $$\frac{1}{3}$$+$$\frac{1}{3}$$= 2 X $$\frac{1}{3}$$ then multiply it by 5 which is 5 X 2 X $$\frac{1}{3}$$= 10 X $$\frac{1}{3}$$= $$\frac{10}{3}$$. Question 5. 6 X $$\frac{5}{8}$$= 6 X 5 X $$\frac{1}{8}$$= 30 X $$\frac{1}{8}$$ = $$\frac{30}{8}$$. Explanation: First we write $$\frac{5}{8}$$ as $$\frac{1}{8}$$+ $$\frac{1}{8}$$+ $$\frac{1}{8}$$ + $$\frac{1}{8}$$+ $$\frac{1}{8}$$ = 5 X $$\frac{1}{8}$$ then multiply it by 6 which is 6 X 5 X $$\frac{1}{8}$$=30 X $$\frac{1}{8}$$= $$\frac{30}{8}$$. Question 6. 9 X $$\frac{7}{4}$$ = 9 X 7 X$$\frac{1}{4}$$= 63 X $$\frac{1}{4}$$ = $$\frac{63}{4}$$. Explanation: First we write $$\frac{7}{4}$$ as $$\frac{1}{4}$$+$$\frac{1}{4}$$+ $$\frac{1}{4}$$+ $$\frac{1}{4}$$+$$\frac{1}{4}$$+$$\frac{1}{4}$$+ $$\frac{1}{4}$$ = 7 x $$\frac{1}{4}$$ then we multiply it by 9 which is 9 X 7 X $$\frac{1}{4}$$= 63 X $$\frac{1}{4}$$ = $$\frac{63}{4}$$. Question 7. 7 X $$\frac{4}{12}$$=7 X 4 X $$\frac{1}{12}$$= 28 X $$\frac{1}{12}$$= $$\frac{28}{12}$$ also 7 X $$\frac{4}{12}$$= 7 X $$\frac{1}{3}$$= $$\frac{7}{3}$$. Explanation: First we write $$\frac{4}{12}$$ as $$\frac{1}{12}$$+ $$\frac{1}{12}$$+ $$\frac{1}{12}$$+$$\frac{1}{12}$$= 4 X $$\frac{1}{12}$$ then we multiply it by 7 which is 7 X 4 X $$\frac{1}{12}$$= 28 X $$\frac{1}{12}$$=$$\frac{28}{12}$$ we can also simplify 7 X $$\frac{4}{12}$$ as 4,12 can be divided by 4 we can also write 7 X $$\frac{4}{12}$$ as 7 X $$\frac{1}{3}$$= $$\frac{7}{3}$$. Question 8. $$\frac{9}{6}$$ X 8 = 9 X $$\frac{1}{6}$$ X 8 = 72 X $$\frac{1}{6}$$ = $$\frac{72}{6}$$ =12. Explanation: First we write $$\frac{9}{6}$$ as $$\frac{1}{6}$$ +$$\frac{1}{6}$$ + $$\frac{1}{6}$$ + $$\frac{1}{6}$$ +$$\frac{1}{6}$$ +$$\frac{1}{6}$$ + $$\frac{1}{6}$$ + $$\frac{1}{6}$$ +$$\frac{1}{6}$$ =9 X $$\frac{1}{6}$$ then we multiply it by 8 which is 9 X 8 X $$\frac{1}{6}$$ = 72 X $$\frac{1}{6}$$ = $$\frac{72}{6}$$ further as 6 can divide 72 and leaving no remainder we write $$\frac{72}{6}$$ =12. Question 9. 10 X $$\frac{20}{100}$$ = 10 X $$\frac{2}{10}$$=2 X 10 X $$\frac{1}{10}$$= 20 $$\frac{1}{10}$$=$$\frac{20}{10}$$=2. Explanation: First we write $$\frac{20}{100}$$ = $$\frac{2}{10}$$ as $$\frac{1}{10}$$ + $$\frac{1}{10}$$ = 2 X $$\frac{1}{10}$$ then multiply it by 10 which is 2 X 10 X $$\frac{1}{10}$$= 20 X $$\frac{1}{10}$$=$$\frac{20}{10}$$ and 10 can divide 20 leaving no remainder further we can write $$\frac{20}{10}$$ as 2. Number Sense Find the unknown number. Question 10. The unknown number is 2. Explanation: ___ X $$\frac{8}{10}$$ = $$\frac{16}{10}$$= to get the unknown number we take it as P, P X $$\frac{8}{10}$$ = $$\frac{16}{10}$$ means P = $$\frac{16}{10}$$ X $$\frac{10}{8}$$ we get 2 , so P=2. Question 11. The unknown number is 5. Explanation: Let us take unknown letter as P so 4 X $$\frac{P}{2}$$ = $$\frac{20}{2}$$ , 4 X P= 2 X $$\frac{20}{2}$$ to get value of P we take 4 to other side , Therefore P=20 X $$\frac{1}{4}$$ = 5, so the unknown number is 5. Question 12. The unknown number is 100. Explanation: Let us take unknown number as P given 3 X $$\frac{9}{P}$$ = $$\frac{27}{100}$$ , 27 X[ latex]\frac{1}{P}[/latex] = $$\frac{27}{100}$$, Therefore P= 27 X $$\frac{100}{27}$$=100. Question 13. Reasoning Without calculating, would you plot the product of 5 and $$\frac{3}{6}$$ to the left or to the right of 5 on a number line? Explain. Yes , without calculating we can plot 5 and $$\frac{3}{6}$$ to the left of 5 Explanation: 5 X $$\frac{3}{6}$$= 5 X $$\frac{1}{2}$$ since the denominator of each given fraction is 2 divide the space between every pairs of two consecutive integers (on the number line) in 2 equal parts each part so obtained will represent the fraction $$\frac{1}{2}$$ , so on the number line we move 5 parts on the right-side of zero as shown in the figure. Question 14. Patterns Describe and complete the pattern. Explanation: We first write the expressions as multiple of unit fraction and find the product so 1. 3 X $$\frac{1}{5}$$ can be written as product of 3 X 1 X $$\frac{1}{5}$$= 3 X $$\frac{1}{5}$$ = $$\frac{3}{5}$$ 2. 3 X $$\frac{2}{5}$$ can be written as product of 3 X 2 X $$\frac{1}{5}$$ = 6 X $$\frac{1}{5}$$ = $$\frac{6}{5}$$ 3. 3 X $$\frac{3}{5}$$ can be written as product of 3 X 3 X $$\frac{1}{5}$$ = 9 X $$\frac{1}{5}$$ = $$\frac{9}{5}$$ 4. 3 X $$\frac{4}{5}$$ can be written as product of 3 X 4 X $$\frac{1}{5}$$ = 12 X $$\frac{1}{5}$$ = $$\frac{12}{5}$$ 5. 3 X $$\frac{5}{5}$$ can be written as product of 3 X 5 X $$\frac{1}{5}$$ = 15 X $$\frac{1}{5}$$ = $$\frac{15}{5}$$ further can be simplified as 15,5 goes in 5, so dividing by 5 we get $$\frac{15}{5}$$ = 3. Think and Grow: Modeling Real Life A bird keeper uses a $$\frac{1}{3}$$ cup scoop to feed 3 birds. He feeds each bird $$\frac{2}{3}$$ cup of birdseed. How many times does he fill the scoop? Three birds each need $$\frac{2}{3}$$ cup of birdseed, so find 3 × $$\frac{2}{3}$$. Explanation : Given A bird keeper uses a $$\frac{1}{3}$$ cup scoop to feed 3 birds each bird he feeds $$\frac{2}{3}$$ cup of birdseed, Three birds each need $$\frac{2}{3}$$ cup of birdseed so the bird keeper fills the scoop as 3 X $$\frac{2}{3}$$ = 3 X _____P______ X $$\frac{1}{3}$$ here P is the number of times the bird keeper fills the cup scoop. 3 X 2 X $$\frac{1}{3}$$ = 3 X P X $$\frac{1}{3}$$ , when we equate both sides, we get 3 X 2 = 3 X P , therefore P = 2. Show and Grow Question 15. A chef makes 4 servings of honey1oatmeal. She uses a $$\frac{1}{2}$$ tablespoon measuring spoon to measure $$\frac{3}{2}$$ tablespoons of honey for each serving. How many times does she fill the measuring spoon? 3 times the chef fill the measuring spoon. Explanation: Chef uses a $$\frac{1}{2}$$ table spoon as measuring spoon, To measure honey for each serving she needed for each serving is $$\frac{3}{2}$$ , so 4 X $$\frac{3}{2}$$ , Let us take P as number of times of measuring spoon , Given 4 X $$\frac{3}{2}$$ = 4 X _P_ X $$\frac{1}{2}$$, 4 X ___3___ X $$\frac{1}{2}$$= 4 X _P_ X $$\frac{1}{2}$$ , we equate both sides we get value of P as 3 , therefore 3 times the chef fill the measuring spoon. Question 16. You have 7 magnetic blocks that are each $$\frac{9}{100}$$ meter long. You connect the ends of the blocks to make a snake. What fraction of a meter is the block snake? $$\frac{63}{100}$$ meter is the block snake. Explanation: Given 7 magnetic blocks and each are $$\frac{9}{100}$$ meter long. After connecting the ends of the blocks to make a snake is 7 X $$\frac{9}{100}$$ = 7 X 9 X $$\frac{1}{100}$$ = 63 X $$\frac{1}{100}$$ = $$\frac{63}{100}$$ meter. Question 17. Your friend roller-skates $$\frac{15}{10}$$ miles each day for 5 days. How many miles does your friend roller-skate in all? My friend roller skate $$\frac{15}{2}$$ miles in all. Explanation: Given friend roller-skates $$\frac{15}{10}$$ miles each day means $$\frac{3}{2}$$ miles each day so for 5 days it is 5 X $$\frac{3}{2}$$ = 5 X 3 X $$\frac{1}{2}$$ = 15 X $$\frac{1}{2}$$ = $$\frac{15}{2}$$ miles. ### Understand Multiples of Fractions Homework & Practice 9.2 Write the product as a multiple of a unit fraction. Then find the product. Question 1. 2 X $$\frac{2}{3}$$ = 2 X 2 X $$\frac{1}{3}$$ = 4 X ($$\frac{1}{3}$$) = $$\frac{4}{3}$$ Explanation: To write the product as a multiple of a unit fraction we write it $$\frac{2}{3}$$ as 2 X $$\frac{1}{3}$$ and multiply by 2 = 2 X 2 X $$\frac{1}{3}$$ = 4 X $$\frac{1}{3}$$ = $$\frac{4}{3}$$ Question 2. 3 X $$\frac{5}{8}$$ = 3 X 5 X $$\frac{1}{8}$$ = 15 X ($$\frac{1}{8}$$) = $$\frac{15}{8}$$. Explanation: To write the product as a multiple of a unit fraction we write it $$\frac{5}{8}$$ as 5 X $$\frac{1}{8}$$ and multiply by 3 = 3 X 5 X $$\frac{1}{8}$$ = 15 X $$\frac{1}{8}$$ = $$\frac{15}{8}$$ Question 3. 4 X $$\frac{8}{2}$$ = 4 X 8 X $$\frac{1}{2}$$ = 32 X ($$\frac{1}{2}$$) = $$\frac{32}{2}$$ =16. Explanation: To write the product as a multiple of a unit fraction we write it $$\frac{8}{2}$$ as 8 X $$\frac{1}{2}$$ and multiply by 4 = 4 X 8 X $$\frac{1}{2}$$ = 32 X $$\frac{1}{2}$$ = $$\frac{32}{2}$$ = 16. Question 4. 5 X $$\frac{9}{10}$$ = 5 X 9 X $$\frac{1}{10}$$ = 45 X ($$\frac{1}{10}$$) = $$\frac{45}{10}$$ = $$\frac{9}{2}$$ Explanation: To write the product as a multiple of a unit fraction we write it $$\frac{9}{10}$$ as 9 X $$\frac{1}{10}$$ and multiply by 5 = 5 X 9 X $$\frac{1}{10}$$ = 45 X $$\frac{1}{10}$$ = $$\frac{45}{10}$$ = $$\frac{9}{2}$$. Question 5. 8 X $$\frac{6}{5}$$ = 8 X 6 X $$\frac{1}{5}$$ = 48X ($$\frac{1}{5}$$) = $$\frac{48}{5}$$ Explanation: To write the product as a multiple of a unit fraction we write it $$\frac{6}{5}$$ as 6 X $$\frac{1}{5}$$ and multiply by 8 = 8 X 6 X $$\frac{1}{5}$$ = 48 X $$\frac{1}{5}$$ = $$\frac{48}{5}$$. Question 6. 10 X $$\frac{2}{4}$$ = 10 X 2 X $$\frac{1}{4}$$ = 20 X ($$\frac{1}{4}$$) = $$\frac{20}{4}$$ = $$\frac{10}{2}$$. Explanation: To write the product as a multiple of a unit fraction we write it $$\frac{2}{4}$$ as 2 X $$\frac{1}{4}$$ and multiply by 10 = 2 X 10 X $$\frac{1}{4}$$ = 20 X $$\frac{1}{4}$$ = $$\frac{20}{4}$$ = $$\frac{10}{2}$$. Question 7. Number Sense Which expression are equivalent to ? The expressions equivalent to 4 X $$\frac{7}{8}$$ are (4 X 7) x $$\frac{1}{8}$$ , $$\frac{28}{8}$$,$$\frac{7}{8}$$+$$\frac{7}{8}$$+$$\frac{7}{8}$$ + $$\frac{7}{8}$$. Explanation: We have the value of expression of 4 X $$\frac{7}{8}$$ as $$\frac{28}{8}$$ So 1.(4 X 7) x $$\frac{1}{8}$$= $$\frac{28}{8}$$ 2. $$\frac{28}{8}$$=$$\frac{28}{8}$$ 3. 4 X 7 ≠ $$\frac{28}{8}$$ 4. $$\frac{32}{7}$$ ≠ $$\frac{28}{8}$$ 5. $$\frac{7}{8}$$+$$\frac{7}{8}$$+$$\frac{7}{8}$$+$$\frac{7}{8}$$ = $$\frac{28}{8}$$, Therefore expressions 1,2,5 are equivalent to $$\frac{28}{8}$$. The expressions equivalent to 4 X $$\frac{7}{8}$$ are (4 X 7) x $$\frac{1}{8}$$ , $$\frac{28}{8}$$,$$\frac{7}{8}$$+$$\frac{7}{8}$$+$$\frac{7}{8}$$+$$\frac{7}{8}$$ Question 8. Which is greater, ? Explain. 6 X $$\frac{6}{2}$$ is greater Explanation: 6 X $$\frac{6}{2}$$= 6 X 6 X $$\frac{1}{2}$$=36 X $$\frac{1}{2}$$= $$\frac{36}{2}$$ and 5 X $$\frac{7}{2}$$ =5 X 7 X $$\frac{1}{2}$$= 35 X $$\frac{1}{2}$$, Now comparing both sides 36 X $$\frac{1}{2}$$ and 35 X $$\frac{1}{2}$$ we see 36 X $$\frac{1}{2}$$ > 35 X $$\frac{1}{2}$$ so 6 X $$\frac{6}{2}$$ is greater. Question 9. Structure Your friend fills a $$\frac{3}{4}$$ – cup measuring cup with rice 2 times. Write an equation to show how much rice she uses. Friend uses $$\frac{3}{2}$$ cups. Explanation: Friend fills a $$\frac{3}{4}$$ – cup with rice 2 times, so she uses 2 X $$\frac{3}{4}$$ = 2 X 3 x $$\frac{1}{4}$$ = 6 X $$\frac{1}{4}$$= $$\frac{6}{4}$$= $$\frac{3}{2}$$. Question 10. Modeling Real Life You are making a tornado in each of 2 bottles. Each bottle needs to contain $$\frac{20}{4}$$ cups of water. You only have a $$\frac{1}{4}$$ – cup measuring cup. How many times do you need to fill the measuring cup? To make a tornado we require 40 cups of water. Explanation: Each bottle needs $$\frac{20}{4}$$ cups of water and have a $$\frac{1}{4}$$ – cup measuring cup, Now we take P as number of times for filling so $$\frac{20}{4}$$ = P X $$\frac{1}{4}$$, 20 X $$\frac{1}{4}$$=P X $$\frac{1}{4}$$, Equating both sides we get P = 20, so it is 20 times to fill the measuring cups of water for one bottle, Given that to make a tornado we require 2 bottles, 2 X 20 cups of water = 40 cups of water. Question 11. DIG DEEPER! You and your friend are each selling 12 coupon books. So far, you have sold $$\frac{2}{6}$$ of your books. Your friend has sold 3 times as many as you. What fraction of your friend’s coupon books has she sold? We will get the fraction as whole 1 so friend has sold all the 12 coupon books. Explanation: So far i sold $$\frac{2}{6}$$ books out of 12 coupon books. Lets us take P number of books, i have sold P out of 12 coupon books, P= 12 X $$\frac{2}{6}$$ =12 X 2 X $$\frac{1}{6}$$ = 24 X $$\frac{1}{6}$$ = $$\frac{24}{6}$$ =4 , So i have sold 4 coupon books, Given friend has sold 3 times as many as I, Means 3 X 4 =12 means friend has sold all the coupon books Review & Refresh Find the product. Question 12. 12 × 47 = ___ 12 X 47 = 564 Explanation: When 12 is multiplied by 47 we get the product as 564. Question 13. 35 × 31 = ___ 35 X 31 =1,085 Explanation: When 35 is multiplied by 31 we get the product as 1,085. Question 14. 58 × 49 = ___ 58 X 49 = 2,842 Explanation: When 58 is multiplied by 49 we get the product as 2,842. ### Lesson 9.3 Multiply Whole Numbers and Fractions Explore and Grow Use models to help you complete the table. What do you notice about each expression and its product? Construct Arguments Explain how to ﬁnd the product of a whole number and a fraction without using models. Explanation: We notice that each expression can be written as the product as a multiple of a unit fraction, The product of a number and a counting number is a multiple of the number. We can find multiples of unit fractions. To ﬁnd the product of a whole number and a fraction without using models, First we multiply the numerator by the whole number, then rewrite the expression as a multiple of a unit fraction and denominator stays the same we get the product results. Think and Grow: Multiply Whole Numbers and Fractions You can ﬁnd the product of a whole number and a fraction by multiplying the numerator by the whole number. The denominator stays the same. Explanation: To find 4 X $$\frac{2}{3}$$ Method 1: Multiply the numerator by the whole number as 4 X 2 and divide by 3= $$\frac{8}{3}$$ Method 2 : Then writing the expression as a multiple of unit fraction $$\frac{2}{3}$$= 4 X ( 2 X $$\frac{1}{3}$$ ) applying associative property of multiplication (4 X 2) X $$\frac{1}{3}$$= 8 X $$\frac{1}{3}$$=$$\frac{8}{3}$$ Show and Grow Multiply. Question 1. 4 X $$\frac{1}{6}$$ = $$\frac{4}{6}$$ also equals to $$\frac{2}{3}$$ Explanation : Already the expression is in multiple of unit fraction so 4 X $$\frac{1}{6}$$ = $$\frac{4}{6}$$ and further simplification it is equal to $$\frac{2}{3}$$ Question 2. 3 X $$\frac{2}{4}$$ = 3 X $$\frac{1}{2}$$ = $$\frac{3}{2}$$. Explanation: 3 X $$\frac{2}{4}$$ can be simplified as 3 X $$\frac{1}{2}$$ now the expression is in multiple of unit fraction so 3 X $$\frac{1}{2}$$ =$$\frac{3}{2}$$. Question 3. 7 X $$\frac{3}{2}$$ = 7 x 3 X $$\frac{1}{2}$$ = 21 X $$\frac{1}{2}$$ = $$\frac{21}{2}$$. Explanation: 7 X $$\frac{3}{2}$$ first we write the expression as multiple of unit fraction so 7 X ( 3 X $$\frac{1}{2}$$ ) applying associative property of multiplication ( 7 X 3) X $$\frac{1}{2}$$ ) = 21 X $$\frac{1}{2}$$ = $$\frac{21}{2}$$ Apply and Grow: Practice Multiply. Question 4. 2 X $$\frac{1}{5}$$ = $$\frac{2}{5}$$. Explanation : Already the expression 2 X $$\frac{1}{5}$$ is a multiple of unit fraction so we multiply the whole with the numerator and numerator is being same, 2 X $$\frac{1}{5}$$ = $$\frac{2}{5}$$. Question 5. 5 X $$\frac{3}{10}$$ = $$\frac{15}{10}$$=$$\frac{3}{2}$$. Explanation: 5 X $$\frac{3}{10}$$, we write the expression as multiple of unit fraction so 5 X 3 X $$\frac{1}{10}$$= 15 X $$\frac{1}{10}$$= $$\frac{15}{10}$$ further it can be simplified, as both numerator and denominator can be divided by 5 so $$\frac{15}{10}$$=$$\frac{3}{2}$$. Question 6. 6 X $$\frac{7}{8}$$= $$\frac{42}{8}$$= $$\frac{21}{4}$$. Explanation: 6 X $$\frac{7}{8}$$, we write the expression as multiple of unit fraction so 6 X 7 X $$\frac{1}{8}$$ applying associative property of multiplication 6 X ( 7 X $$\frac{1}{8}$$ )= ( 6 X 7 ) X $$\frac{1}{8}$$= 42 X $$\frac{1}{8}$$= $$\frac{42}{8}$$ and further also the fraction can be simplified, as both numerator and denominator can be divided by 2, making $$\frac{42}{8}$$ as $$\frac{21}{4}$$. Question 7. 8 X $$\frac{9}{10}$$= $$\frac{72}{10}$$=$$\frac{36}{5}$$. Explanation: 8 X $$\frac{9}{10}$$, we write the expression as multiple of unit fraction so 8 X 9 X $$\frac{1}{10}$$ applying associative property of multiplication 8 X ( 9 X $$\frac{1}{10}$$ )= (8 X 9 ) X $$\frac{1}{10}$$= 72 X $$\frac{1}{10}$$= $$\frac{72}{10}$$ and further also the fraction can be simplified, as both numerator and denominator can be divided by 2, making $$\frac{72}{10}$$ as $$\frac{36}{5}$$. Question 8. 3 X $$\frac{60}{100}$$ =$$\frac{180}{100}$$ =$$\frac{18}{10}$$. Explanation: 3 X $$\frac{60}{100}$$, we write the expression as multiple of unit fraction, So 3 X 60 X $$\frac{1}{100}$$ applying associative property of multiplication 3 X (60 X $$\frac{1}{100}$$ )= (3 X 60 ) X $$\frac{1}{100}$$= 180 X $$\frac{1}{100}$$= $$\frac{180}{100}$$ and further also the fraction can be simplified, as both numerator and denominator can be divided by 10, making $$\frac{180}{100}$$ as $$\frac{18}{10}$$. Question 9. $$\frac{4}{2}$$ X 10 = 20. Explanation : $$\frac{4}{2}$$ X 10 , we write the expression as multiple of unit fraction, So 4 X $$\frac{1}{2}$$ X 10= 40 X $$\frac{1}{2}$$=$$\frac{40}{2}$$ as the fraction can be further simplified as both numerator and denominator can be divided by 2, making it as 20. Compare. Question 10. 7 X $$\frac{2}{4}$$ =$$\frac{14}{4}$$ is smaller < 5 X $$\frac{3}{4}$$= $$\frac{15}{4}$$. Explanation: 7 X $$\frac{2}{4}$$, First we write expression as a multiple of unit fraction 7 X 2 X $$\frac{1}{4}$$ =14 X $$\frac{1}{4}$$ = $$\frac{14}{4}$$ Now 5 X $$\frac{3}{4}$$ is also written as a multiple of unit fraction 5 X 3 X $$\frac{1}{4}$$ = 15 X $$\frac{1}{4}$$ =$$\frac{15}{4}$$ comparing $$\frac{14}{4}$$ and $$\frac{15}{4}$$ as the denominators are same same 4 and in numerators 15 is great than 14, So $$\frac{14}{4}$$ is less than or smaller or < $$\frac{15}{4}$$ , 7 X $$\frac{2}{4}$$ is smaller < 5 X $$\frac{3}{4}$$ Question 11. 4 X $$\frac{1}{8}$$ is greater than > 4 X $$\frac{1}{10}$$. Explanation: 4 X $$\frac{1}{8}$$ already the expression is in multiple of unit fraction, we write as $$\frac{4}{8}$$ and even the expression 4 X $$\frac{1}{10}$$ is in multiple of unit fraction we write as $$\frac{4}{10}$$ because in order to compare the fractions we must first turn their different denominators into the same denominators to make denominators same we multiply 4 X $$\frac{1}{8}$$ by 10 /10 the value will not change = 4 X 10 X $$\frac{1}{80}$$=40 X $$\frac{1}{80}$$ = $$\frac{40}{80}$$ now other expression 4 X $$\frac{1}{10}$$ by 8/8 so that the value = 4 X 8 X $$\frac{1}{80}$$ = 32 X $$\frac{1}{80}$$ now as denominator are same we can compare only numerators $$\frac{40}{80}$$ with $$\frac{32}{80}$$ as 40 is greater than 32, So $$\frac{4}{8}$$ is greater than > $$\frac{4}{10}$$ , 4 X $$\frac{1}{8}$$ is greater than > 4 X $$\frac{1}{10}$$. Question 12. $$\frac{4}{3}$$ X 9 = 6 X $$\frac{6}{3}$$ both are equal. Explanation: First we write expressions as a multiple of unit fraction $$\frac{4}{3}$$ X 9 = 9 X 4 X $$\frac{1}{3}$$ =36 X $$\frac{1}{3}$$ =$$\frac{36}{3}$$ now other expression 6 X $$\frac{6}{3}$$ as = 6 X 6 X $$\frac{1}{3}$$ = 36 X $$\frac{1}{3}$$ = $$\frac{36}{3}$$ , As denominator is same in the both expression we compare numerators and even the numerators are same in both the expression so given both products are equal . Therefore $$\frac{4}{3}$$ X 9 =, is equal to 6 X $$\frac{6}{3}$$. Question 13. YOU BE THE TEACHER Your friend says the product of 6 and $$\frac{5}{8}$$ is $$\frac{5}{48}$$. Is your friend correct? Explain. No she is not correct as the product of 6 X $$\frac{5}{8}$$ ≠ $$\frac{5}{48}$$. Explanation: We write the expression as a product 6 and $$\frac{5}{8}$$ = 6 X $$\frac{5}{8}$$ to ﬁnd the product of a whole number and a fraction we multiply the numerator by the whole number. The denominator stays the same. 6 X 5 X $$\frac{1}{8}$$ =30 X $$\frac{1}{8}$$ = $$\frac{30}{8}$$ now comparing with $$\frac{5}{48}$$ as denominators are different in order to compare the fractions we must first turn their different denominators into the same denominators to make denominators same we multiply $$\frac{30}{8}$$ with numerator and denominator as 6= $$\frac{30}{8}$$ X $$\frac{6}{6}$$ we get $$\frac{180}{48}$$, now comparing with $$\frac{5}{48}$$ being denominators same, now numerators 180,5 both are not same so friend is not correct as the product of 6 X $$\frac{5}{8}$$ ≠ $$\frac{5}{48}$$. Question 14. Open-Ended The product of a whole number and a fraction is $$\frac{24}{10}$$. What could the two factors be? The product is given as $$\frac{24}{10}$$ and the two factors can be (1,24) or (2,12) or (3,8) or (4,6) with multiple of $$\frac{1}{10}$$. Explanation: Given the product of a whole number and a fraction is $$\frac{24}{10}$$ to find the two factors, we first write the product as whole with multiple of unit fraction,$$\frac{24}{10}$$, we write as 24 X $$\frac{1}{10}$$, now the whole part is 24 we find the factors for 24, two more numbers that multiply to make 24 are 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6 = 24. This means the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 so the two factors can be (1,24) X $$\frac{1}{10}$$ or (2,12) X $$\frac{1}{10}$$ or (3,8) X $$\frac{1}{10}$$ or (4,6) X $$\frac{1}{10}$$ = 24 X $$\frac{1}{10}$$ = $$\frac{24}{10}$$. Think and Grow: Modeling Real Life Example A short roller-coaster track is $$\frac{3}{10}$$ mile long. A longer roller-coaster track is about 4 times as long as the short track. About how long is the longer roller-coaster track? Multiply the shorter roller-coaster track length by 4. So, the longer roller-coaster track is about _$$\frac{12}{10}$$ _ miles long or $$\frac{6}{5}$$ miles long The longer roller-coaster track is about $$\frac{12}{10}$$ miles long or $$\frac{6}{5}$$. Explanation: Given a short roller-coaster track is $$\frac{3}{10}$$ mile long and longer roller-coaster track is about 4 times as long as the short track to find how long is the longer roller-coaster track we multiply short roller-coaster track $$\frac{3}{10}$$ mile by 4, 4 X $$\frac{3}{10}$$ now we write expression as a multiple of unit fraction so 4 X 3 X $$\frac{1}{10}$$ = 12 X $$\frac{1}{10}$$ =$$\frac{12}{10}$$ as numerator and denominator can be further simplified as both can be divided by 2 making it as $$\frac{6}{5}$$. Show and Grow Question 15. The Renaissance Tower in Dallas, Texas is $$\frac{27}{100}$$ kilometer tall. The Burj Khalifa is about 3 times as tall as the Renaissance Tower. About how tall is the Burj Khalifa? The Burj Khalifa is $$\frac{81}{100}$$ kilometers tall. Explanation: Given the Renaissance Tower in Dallas, Texas is $$\frac{27}{100}$$ kilometer tall and The Burj Khalifa is about 3 times as tall as the Renaissance Tower, To find how tall is the Burj Khalifa we multiply the Renaissance Tower in Dallas, Texas length by 3, 3 X $$\frac{27}{100}$$ we write expression as a multiple of unit fraction 3 X 27 X $$\frac{1}{100}$$ = 81 X $$\frac{1}{100}$$ =$$\frac{81}{100}$$, So the Burj Khalifa is $$\frac{81}{100}$$ kilometers tall. Question 16. You water 6 plants using $$\frac{3}{5}$$ liter of water for each. How many liters of water do you use? $$\frac{18}{5}$$ liters of water is used, The answer will lie in between 3 and 4. Explanation: Given $$\frac{3}{5}$$ liter of water for each is watered, for 6 plants total number of liters of water used is 6 X $$\frac{3}{5},now we write as multiple of unit fraction as 6 X 3 X $\frac{1}{5}$$ = 18 X $$\frac{1}{5}$$= $$\frac{18}{5}$$ liters of water is used. Now we divide $$\frac{18}{5}$$ we get the value as 3.6,means between 3 and 4 whole numbers our answer will lie. Question 17. DIG DEEPER! You have 6 cups of strawberries. You want to make 4 strawberry-banana smoothies and 4 strawberry-kiwi smoothies. Each smoothie needs $$\frac{2}{3}$$ cup of strawberries. Do you have enough? If not, how many more cups of strawberries do you need? Answer: Yes we have enough cups of strawberries, we dont require more cups of strawberries. Explanation: Given to make each smoothie we need $$\frac{2}{3}$$ cup of strawberries first to make 4 strawberry-banana smoothies we need 4 X $$\frac{2}{3}$$ cup of strawberries = 4 X 2 X $$\frac{1}{3}$$= 8 X $$\frac{1}{3}$$ =$$\frac{8}{3}$$, Now similarly for making 4 strawberry-kiwi smoothies we need 4 X $$\frac{2}{3}$$ cup of strawberries = 4 X 2 X $$\frac{1}{3}$$= 8 X $$\frac{1}{3}$$ = $$\frac{8}{3}$$,Now total cups of strawberries requried to make smoothie is $$\frac{8}{3}$$ + $$\frac{8}{3}$$ = 2 X $$\frac{8}{3}$$= 2 X 8 X $$\frac{1}{3}$$=16 X $$\frac{1}{3}$$= $$\frac{16}{3}$$ totals cups of strawberries required to make smoothies. Now we compare $$\frac{16}{3}$$ with 6, To compare first we make both denominators same so we multiply 6 with $$\frac{3}{3}$$= 6 X $$\frac{3}{3}$$= 6 X 3 X $$\frac{1}{3}$$= 18 X $$\frac{1}{3}$$= $$\frac{18}{3}$$ as both the denominators are same we compare $$\frac{16}{3}$$ and $$\frac{18}{3}$$ by seeing numerator 16 is less than 18 means we have enough cups of strawberries. ### Multiply Whole Numbers and Fractions Homework & practice 9.3 Multiply Question 1. Answer: 2 X $$\frac{1}{4}$$ = $$\frac{2}{4}$$=$$\frac{1}{2}$$ Explanation: As the given expression is already a unit fraction we multiply it as 2 X $$\frac{1}{4} =$$$$\frac{2}{4}$$ and further both numerator and denominator can be divided by 2 making it as $$\frac{1}{2}$$ Question 2. Answer: 3 X $$\frac{3}{5}$$= $$\frac{9}{5}$$. Explanation: 3 X $$\frac{3}{5}$$ can be written as multiple of unit fraction as 3 X (3 X $$\frac{1}{5}$$) applying associative property of multiplication ( 3 X 3 ) X $$\frac{1}{5}$$= 9 X $$\frac{1}{5}$$ = $$\frac{9}{5}$$. Question 3. Answer: 1 X $$\frac{6}{8}$$= $$\frac{6}{8}$$= $$\frac{3}{4}$$. Explanation: 1 X $$\frac{6}{8}$$ can be written as multiple of unit fraction as 1 X ( 6 X $$\frac{1}{8}$$) applying associative property of multiplication ( 1 X 6 ) X $$\frac{1}{8}$$ = 6 X $$\frac{1}{8}$$= $$\frac{6}{8}$$ as numerator and denominator can be further simplified as both can be divided by 2 making as $$\frac{6}{8}$$= $$\frac{3}{4}$$X $$\frac{2}{2}$$=$$\frac{3}{4}$$. Question 4. Answer: 4 X $$\frac{10}{12}$$= $$\frac{40}{12}$$= $$\frac{10}{3}$$ Explanation: 4 X $$\frac{10}{12}$$ can be written as multiple of unit fraction as 4 X ( 10 X $$\frac{1}{12}$$) applying associative property of multiplication ( 4 X 10 ) X $$\frac{1}{12}$$ = 40 X $$\frac{1}{12}$$= $$\frac{40}{12}$$ as numerator and denominator can be further simplified as both can be divided by 4 as $$\frac{40}{12}$$= $$\frac{10}{3}$$ X $$\frac{4}{4}$$=$$\frac{10}{3}$$ X 1 = $$\frac{10}{3}$$. Question 5. Answer: 7 X $$\frac{6}{10}$$ = $$\frac{42}{10}$$= $$\frac{21}{5}$$ Explanation: 7 X $$\frac{6}{10}$$ can be written as multiple of unit fraction as 7 X ( 6 X $$\frac{1}{10}$$) applying associative property of multiplication ( 7 X 6 ) X $$\frac{1}{10}$$ = 42 X $$\frac{1}{10}$$= $$\frac{42}{10}$$ as numerator and denominator can be further simplified as both can be divided by 2 as $$\frac{42}{10}$$= $$\frac{21}{5}$$ X $$\frac{2}{2}$$=$$\frac{21}{5}$$ X 1 = $$\frac{21}{5}$$. Question 6. Answer: $$\frac{4}{6}$$ X 5 =$$\frac{20}{6}$$ =$$\frac{10}{3}$$. Explanation: $$\frac{4}{6}$$ X 5 can be written as multiple of unit fraction as 5 X ( 4 X $$\frac{1}{6}$$) applying associative property of multiplication ( 5 X 4 ) X $$\frac{1}{6}$$ = 20 X $$\frac{1}{6}$$= $$\frac{20}{6}$$ as numerator and denominator can be further simplified as both can be divided by 2 as $$\frac{20}{6}$$ = $$\frac{10}{3}$$ X $$\frac{2}{2}$$=$$\frac{10}{3}$$ X 1 =$$\frac{10}{3}$$. Question 7. Answer: 8 X $$\frac{5}{2}$$ = $$\frac{40}{2}$$ =20. Explanation: 8 X $$\frac{5}{2}$$ can be written as multiple of unit fraction as 8 X ( 5 X $$\frac{1}{2}$$) applying associative property of multiplication ( 8 X 5 ) X $$\frac{1}{2}$$ = 40 X $$\frac{1}{2}$$ = $$\frac{40}{2}$$ further it can be simplified as $$\frac{40}{2}$$=20 Question 8. Answer: $$\frac{70}{100}$$ X 6 = $$\frac{420}{100}$$ = $$\frac{42}{10}$$. Explanation: $$\frac{70}{100}$$ X 6 can be written as multiple of unit fraction as 6 X ( 70 X $$\frac{1}{100}$$) applying associative property of multiplication ( 6 X 70 ) X $$\frac{1}{100}$$ = 420 X $$\frac{1}{100}$$= $$\frac{420}{100}$$ as numerator and denominator can be further simplified as both can be divided by 10 as $$\frac{420}{100}$$= $$\frac{42}{10}$$ X $$\frac{10}{10}$$=$$\frac{42}{10}$$ X 1 =$$\frac{42}{10}$$. Question 9. Answer: 10 X $$\frac{9}{3}$$=$$\frac{90}{3}$$=30. Explanation: 10 X $$\frac{9}{3}$$ can be written as multiple of unit fraction as 10 X ( 9 X $$\frac{1}{3}$$) applying associative property of multiplication ( 10 X 9 ) X $$\frac{1}{3}$$ = 90 X $$\frac{1}{3}$$= $$\frac{90}{3}$$ further it can be simplified because 90 can be divided by 3 without leaving remainder making $$\frac{90}{3}$$ as 30. Compare. Question 10. Answer: 2 X $$\frac{4}{2}$$ = 8 X $$\frac{1}{2}$$. Explanation: 2 X $$\frac{4}{2}$$ we write as multiple of unit fraction 2 X 4 X $$\frac{1}{2}$$ = 8 X $$\frac{1}{2}$$ = $$\frac{8}{2}$$ and 8 X $$\frac{1}{2}$$ is already multiple of unit fraction = latex]\frac{8}{2}$ now we can compare both sides as the denominators are same and numerators are same both the expressions are equal 2 X $$\frac{4}{2}$$ = 8 X $$\frac{1}{2}$$. Question 11. 9 X $$\frac{1}{5}$$ > $$\frac{9}{12}$$ Explanation: 9 X $$\frac{1}{5}$$ already it is multiple of unit fraction we write it as $$\frac{9}{5}$$ and 9 X $$\frac{1}{12}$$ is also a multiple of unit fraction we write it as $$\frac{9}{12}$$ now comparing first we make both denominators same so the multiples of 5 are 5,10,15,20,25,30,35,40,45,50,55,60,65 and multiples of 12 are 12,24,36,48,60,72 so the least common multiple of both 5 and 12 is 60 , So to get denominator 60 we multiply top and bottom by 12 for $$\frac{9}{5}$$ = $$\frac{9}{5}$$ X $$\frac{12}{12}$$= $$\frac{108}{60}$$ , So to get denominator 60 we multiply top and bottom by 5 for $$\frac{9}{12}$$= $$\frac{9}{12}$$ X $$\frac{5}{5}$$= $$\frac{45}{60}$$ Now comparing $$\frac{108}{60}$$ with $$\frac{45}{60}$$ as both denominator are same, As we see numerator it is clear that 108 is more than 45 so $$\frac{108}{60}$$ > $$\frac{45}{60}$$ , 9 X $$\frac{1}{5}$$ > $$\frac{9}{12}$$. Question 12. $$\frac{6}{10}$$X 4 < 9 X $$\frac{3}{10}$$. Explanation: $$\frac{6}{10}$$X 4 we write as multiple of unit fraction 4 X 6 X $$\frac{1}{10}$$= 24 X $$\frac{1}{10}$$ =$$\frac{24}{10}$$ and 9 X $$\frac{3}{10}$$ as 9 X 3 X $$\frac{1}{10}$$= 27 X $$\frac{1}{10}$$=$$\frac{27}{10}$$, We can compare both as denominators of both are 10 we compare numerators as 24 with 27 as 27 is great,$$\frac{24}{10}$$ < $$\frac{27}{10}$$, $$\frac{6}{10}$$ X 4 < 9 X $$\frac{3}{10}$$. Question 13. Writing Explain one way to multiply a whole number by a fraction. First we write fraction as multiple of unit fraction then applying associative property of multiplication we multiply the whole numbers first then with the unit fraction,later writing product. Explanation: Let us take the whole number as 2 and fraction as $$\frac{3}{5}$$,Here first we write fraction as multiple of unit fraction as 2 X 3 X $$\frac{1}{5}$$ now applying associative property of multiplication, we multiply the whole numbers first then with the unit fraction, 6 X $$\frac{1}{5}$$ as $$\frac{6}{5}$$. Question 14. Number Sense Between which two whole numbers does the product of 10 and $$\frac{5}{6}$$ lie ? In between 8 and 9 whole numbers the product of 10 and $$\frac{5}{6}$$ lies. Explanation: 10 X $$\frac{5}{6}$$ we first write it as multiple of unit fraction as 10 X ( 5 X $$\frac{1}{6}$$) applying associative property of multiplication, We write (10 X 5) X $$\frac{1}{6}$$= 50 X $$\frac{1}{6}$$ = $$\frac{50}{6}$$ if we divide $$\frac{50}{6}$$) we get 8.333 which is in between 8 and 9 whole numbers. Question 15. Modeling Real Life A seamstress makes fabric dog collars. An extra small dog collar is $$\frac{1}{3}$$ yard long. A large dog collar is 2 times as long as the extra small dog collar. How long is the large dog collar? The large dog collar is $$\frac{2}{3}$$ yards long. Explanation: Given the extra small dog collar is $$\frac{1}{3}$$ yard long and a large dog collar is 2 times as long as the extra small dog collar, so the large dog collar = 2 X $$\frac{1}{3}$$ as the fraction is multiple of unit fraction we write as $$\frac{2}{3}$$ , therefore the large dog collar is $$\frac{2}{3}$$ yards long. Question 16. DIG DEEPER! Newton has 16 baseball cards. Descartes has $$\frac{1}{4}$$ as many cards as Newton, and your friend has $$\frac{1}{2}$$ as many cards as Descartes. How many baseball cards do Newton, Descartes, and your friend have in all? Newton has 16 baseball cards, Descartes has 4 baseball cards and my friend has 2 baseball cards, over all there are total 22 baseball cards. Explanation: Given Newton has 16 baseball cards and Descartes has $$\frac{1}{4}$$ as many cards as Newton so$$\frac{1}{4}$$ of 16 = 16 X $$\frac{1}{4}$$ cards it is already as multiple of unit fraction we can write as $$\frac{16}{4}$$ further can be simplified because 16 can be divide by 4 so we get as 4 means Descartes has 4 cards.Now my friend has $$\frac{1}{2}$$ as many cards as Descartes So Descartes has 4 cards means my friend has $$\frac{1}{2}$$ out of 4= 4 X $$\frac{1}{2}$$ as it is already a multiple of unit fracrtion, we take it as $$\frac{4}{2}$$ further we can simplify as 4 is divisiblle by 2 we get 2 cards. Newton has 16 baseball cards,Descartes has 4 baseball cards and my friend has 2 baseball cards, over all there are total 22 baseball cards. Review & Refresh Divide Question 17. = 19. Explanation: Question 18. =81 with 4 as remainder. Explanation: Step 1:Start by setting it up with the divisor 7 on the left side and the dividend 571 on the right side, Step 2: The divisor (7) does not goes into the first digit of the dividend (5), So we take second digit also , The divisor (7) goes into the two digits of the dividend (57) by 8 time(s),there fore put 8 on the top 8 Step 3: Multiply the divisor by the result in the previous step. Therefore, put 8 on top (7 x 9 = 56) and write that answer below the dividend. 8 56 Step 4: Subtract the result in the previous step from the two digits of the dividend (57 -56 = 1) and write the answer below. 8 56 1 Step 5: Move down the 3rd digit of the dividend (1) 8 56 11 Step 6:The divisor (7) goes into the bottom number (11), time(s). Therefore, put 1 on top: 81 56 11 Step 7 : Multiply the divisor by the result in the previous step (7 x 1 = 7) and write that answer at the bottom: 81 56 11 07 Step 8 : Subtract the result in the previous step from the number written above it. (11 – 7 = 04) and write the answer at the bottom 81 56 11 07 4 We stop here as 4 is less than 7 and further we get in decimals , So =81 with 4 as remainder Question 19. = 184 with 3 as remainder. Explanation: Step 1:Start by setting it up with the divisor 5 on the left side and the dividend 923 on the right side, Step 2: The divisor (5) goes into the first digit of the dividend (9) by 8 time there fore put 1 on the top 1 Step 3: Multiply the divisor by the result in the previous step (5 x 1 = 5) and write that answer below the dividend. 1 5 Step 4: Subtract the result in the previous step from the first digit of the dividend (9 -5 = 4) and write the answer below. 1 5 4 Step 5: Move down the 2nd digit of the dividend (2),like below 1 5 42 Step 6:The divisor (5) goes into the bottom number (40), 8 time(s). Therefore, put 8 on top: 18 5 42 Step 7 : Multiply the divisor by the result in the previous step (5 x 8 = 40) and write that answer at the bottom: 18 5 42 40 Step 8: Subtract the result in the previous step from the number written above it. (42 – 40 = 04) and write the answer at the bottom 18 5 42 40 2 Step 9 :Move down the 3nd digit of the dividend (3),like below 18 5 42 40 23 Step 10 : The divisor (5) goes into the bottom number (20), 4 time(s). Therefore, put 4 on top: 184 5 42 40 23 20 Step 11 : Subtract the result in the previous step from the number written above it. (23 – 20 = 03) and write the answer at the bottom 184 5 42 40 23 20 03 We stop here as 3 is less than 5 and further we get in decimals , So = 184 with 3 as remainder. ### Lesson 9.4 Multiply Whole Numbers and Mixed Numbers Explore and Grow Structure How does the Distributive Property relate to your models? Explain. Explanation: According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. We use distribute property while relating the models first we draw each multiplying addend individually and then add the products together to give the result. Example : 2 X 1 $$\frac{3}{8}$$ using the distributive property we write 2 X 1 $$\frac{3}{8}$$ as 2 X ( 1 + $$\frac{3}{8}$$ )here first we draw model of 2 X 1 then we draw model of 2 X $$\frac{3}{8}$$ then we add the products together as 2 + $$\frac{6}{8}$$= 2 + $$\frac{3}{4}$$=$$\frac{11}{4}$$. Think and Grow: Multiply Whole Numbers and Mixed Numbers You can ﬁnd the product of a whole number and a mixed number by writing the mixed number as a fraction or by using the Distributive Property One Way: Write the mixed number as a fraction, then multiply. Another Way: Use the Distributive Property. 3 X 1 $$\frac{1}{2}$$ = $$\frac{9}{2}$$ Explanation: First we write the mixed number as a fraction then multiply, so 1 $$\frac{1}{2}$$ = 1+ $$\frac{1}{2}$$ = 2 x 1+1 by 2= $$\frac{3}{2}$$ now we multiply it with the whole , 3 X $$\frac{3}{2}$$ now we write in unit fraction and multiply 3 X 3 X $$\frac{1}{2}$$= 9 X $$\frac{1}{2}$$= $$\frac{9}{2}$$. Show and Grow Multiply Question 1. = $$\frac{27}{4}$$=6 $$\frac{3}{4}$$ Explanation: First we write the mixed number as a fraction then multiply, so 2 $$\frac{1}{4}$$ = 2 + $$\frac{1}{4}$$=(2 X 4 + 1) by 4 = $$\frac{9}{4}$$ now we multiply it with the whole, 3 X $$\frac{9}{4}$$ now we write in unit fraction and multiply 3 X 9 X $$\frac{1}{4}$$ = 27 X $$\frac{1}{4}$$ = $$\frac{27}{4}$$, we can write the product as mixed fraction as 4 goes in 6 times. 6 will be our whole number 4 X 6 is 24 and we have 3 left over (2724), 3 will be our numerator and our denominator will stay the same 4= $$\frac{27}{4}$$= 6 $$\frac{3}{4}$$. Question 2. $$\frac{38}{12}$$= 3 $$\frac{2}{12}$$. Explanation: First we write the mixed number as a fraction then multiply, so 1 $$\frac{7}{12}$$= 1 + $$\frac{7}{12}$$= (1 X 12 + 7 ) by 12 = $$\frac{19}{12}$$ now we multiply it with the whole, 2 X $$\frac{19}{12}$$ now we write in unit fraction and multiply 2 X 19 X $$\frac{1}{12}$$= 38 X $$\frac{1}{12}$$ = $$\frac{38}{12}$$, we can write the product as mixed fraction as 12 goes in 3 times. 3 will be our whole number 12 X 3 is 36 and we have 2 left over (38−36). 2 will be our numerator and our denominator will be the same 12 = $$\frac{38}{12}$$= 3 $$\frac{2}{12}$$. Question 3. = $$\frac{116}{8}$$= 14 $$\frac{4}{8}$$ Explanation: First we write the mixed number as a fraction then multiply, so 3 $$\frac{5}{8}$$= 3 + $$\frac{5}{8}$$= (3 X 8 + 5) by 8 = $$\frac{29}{8}$$ now we multiply it with the whole, 4 X $$\frac{29}{8}$$ now we write in unit fraction and multiply 4 X 29 X $$\frac{1}{8}$$= 116 X $$\frac{1}{8}$$ = $$\frac{116}{8}$$, we can write the product as mixed fraction as 8 goes in 14 times. 14 will be our whole number 8 X 14 is 112 and we have 4 left over (116−112). 4 will be our numerator and our denominator will be the same 8= $$\frac{116}{8}$$= 14 $$\frac{4}{8}$$. Apply and Grow: Practice Multiply. Question 4. = $$\frac{50}{3}$$=16 $$\frac{2}{3}$$ Explanation: First we write the mixed number as a fraction then multiply, so 1 $$\frac{2}{3}$$= 1 + $$\frac{2}{3}$$= (1 X 3 + 2) by 3 = $$\frac{5}{3}$$ now we multiply it with the whole, 10 X $$\frac{5}{3}$$ now we write in unit fraction and multiply 10 X 5 X $$\frac{1}{3}$$= 50 X $$\frac{1}{3}$$ = $$\frac{50}{3}$$, we can write the product as mixed fraction as 3 goes in 16 times. 16 will be our whole number 3 X 16 is 48 and we have 2 left over (50−48). 2 will be our numerator and our denominator will be the same 3= $$\frac{50}{3}$$= 16 $$\frac{2}{3}$$. Question 5. = $$\frac{25}{2}$$=12 $$\frac{1}{2}$$ Explanation: First we write the mixed number as a fraction then multiply, so 2 $$\frac{1}{2}$$= 2 + $$\frac{1}{2}$$= (2 X 2 + 1) by 2 = $$\frac{5}{2}$$ now we multiply it with the whole, 5 X $$\frac{5}{2}$$ . We write in unit fraction and multiply 5 X 5 X $$\frac{1}{2}$$= 25 X $$\frac{1}{2}$$ = $$\frac{25}{2}$$, we can write the product as mixed fraction as 2 goes in 12 times. 12 will be our whole number 2 X 12 is 24 and we have 1 left over (25−24). 1 will be our numerator and our denominator will be the same 2= $$\frac{25}{2}$$= 12 $$\frac{1}{2}$$ Question 6. = $$\frac{184}{6}$$=$$\frac{92}{3}$$=30 $$\frac{4}{6}$$. Explanation: First we write the mixed number as a fraction then multiply, so 7 $$\frac{4}{6}$$= 7 + $$\frac{4}{6}$$=(7 X 6 + 4) by 6 = $$\frac{46}{6}$$ now we multiply it with the whole, 4 X $$\frac{46}{6}$$ now we write in unit fraction and multiply 4 X 46 X $$\frac{1}{6}$$= 184 X $$\frac{1}{6}$$ = $$\frac{184}{6}$$, we can write the product as mixed fraction as 6 goes in 30 times. 30 will be our whole number 6 X 30 is 180 and we have 4 left over (184−180). 4 will be our numerator and our denominator will be the same 6= $$\frac{184}{6}$$= 30 $$\frac{4}{6}$$ and $$\frac{184}{6}$$ can also written as $$\frac{92}{3}$$ because both 184 and 6 can also be further divided by 2. Question 7. = $$\frac{243}{5}$$= 48 $$\frac{3}{5}$$ Explanation: First we write the mixed number as a fraction then multiply, so 5 $$\frac{2}{5}$$= 5 + $$\frac{2}{5}$$= (5 X 5 + 2) by 5 = $$\frac{27}{5}$$ now we multiply it with the whole, 9 X $$\frac{27}{5}$$ now we write in unit fraction and multiply 9 X 27 X $$\frac{1}{5}$$= 243 X $$\frac{1}{5}$$ = $$\frac{243}{5}$$, we can write the product as mixed fraction as 5 goes in 48 times. 48 will be our whole number 5 X 48 is 240 and we have 3 left over (243−240). 3 will be our numerator and our denominator will be the same 5= $$\frac{243}{5}$$= 48 $$\frac{3}{5}$$. Question 8. = $$\frac{312}{10}$$= 31 $$\frac{2}{10}$$ Explanation: First we write the mixed number as a fraction then multiply, so 3 $$\frac{9}{10}$$= 3 + $$\frac{9}{10}$$= (3 X 10 + 9) by 10 = $$\frac{39}{10}$$ now we multiply it with the whole, 8 X $$\frac{39}{10}$$ now we write in unit fraction and multiply 8 X 39 X $$\frac{1}{10}$$= 312 X $$\frac{1}{10}$$ = $$\frac{312}{10}$$, we can write the product as mixed fraction as 10 goes in 31 times. 31 will be our whole number 10 X 31 is 310 and we have 2 left over (312−310). 2 will be our numerator and our denominator will be the same 10= $$\frac{312}{10}$$= 31 $$\frac{2}{10}$$. Question 9. = $$\frac{5466}{100}$$=54 $$\frac{66}{100}$$. Explanation: First we write the mixed number as a fraction then multiply, so 9 $$\frac{11}{100}$$= 9 + $$\frac{11}{100}$$= (9 X 100 + 11) by 100 = $$\frac{911}{100}$$ now we multiply it with the whole, 6 X $$\frac{911}{100}$$ now we write in unit fraction and multiply 6 X 911 X $$\frac{1}{100}$$= 5466 X $$\frac{1}{100}$$ = $$\frac{5466}{100}$$, we can write the product as mixed fraction as 100 goes in 54 times. 54 will be our whole number 100 X 54 is 5400 and we have 66 left over (5466−5400). 66 will be our numerator and our denominator will be the same 100= $$\frac{5466}{100}$$= 54 $$\frac{66}{100}$$. Question 10. Reasoning How can you check whether your answer in Exercise 9 is reasonable? Exercise 9 is reasonable. Explanation: In the above problem 6 X 9 $$\frac{11}{100}$$ we got product as $$\frac{5466}{100}$$ =54 $$\frac{66}{100}$$ we can write the product as mixed fraction as 100 goes in 54 times. 54 will be our whole number 100 X 54 is 5400 and 5466-5400, leaving 66 as remainder so we consider the whole part 54,As 54 is reasonable close to 100. Question 11. YOU BE THE TEACHER Your friend ﬁnds the product of 9 and 6$$\frac{1}{3}$$. Is your friend correct? Explain. Yes friend is correct. Explanation: First we write the mixed number as a fraction then multiply, so 6 $$\frac{1}{3}$$= 6 + $$\frac{1}{3}$$= (6 X 3 + 1) by 3 = $$\frac{19}{3}$$ now we multiply it with the whole, 9 X $$\frac{19}{3}$$ now we write in unit fraction and multiply 9 X 19 X $$\frac{1}{3}$$= 171 X $$\frac{1}{3}$$ = $$\frac{171}{3}$$, As 3 goes 57 times leaving no remainder so we get 9 X 6 $$\frac{1}{3}$$=57 which matches with the friend’s product so friend is correct. Question 12. DIG DEEPER! Without multiplying, can you tell which expression is greater, ? Explain. 5 X 6 $$\frac{1}{4}$$ < 6 X 5 $$\frac{1}{4}$$ Explanation: By drawing models also we can say If the denominators are the same, then the fraction with the greater numerator is the greater fraction. The fraction with the lesser numerator is the lesser fraction, But first we write the mixed number as a fraction so in 5 X 6 $$\frac{1}{4}$$, 6 $$\frac{1}{4}$$= 6+ $$\frac{1}{4}$$= (6 X 4 +1) by 4 = $$\frac{25}{4}$$ now we multiply it with the whole as 5 X $$\frac{25}{4}$$ now we write the mixed fraction 5 $$\frac{1}{4}$$ as 5+ $$\frac{1}{4}$$ = 5 x 4 +1 by 4=5 $$\frac{21}{4}$$ now we multiply it with the whole as 6 X $$\frac{21}{4}$$ , Now comparing 5 X $$\frac{25}{4}$$ and 6 X $$\frac{21}{4}$$ in both denominators are same and numerators 5 X $$\frac{25}{4}$$ is less than 6 X $$\frac{21}{4}$$ so 5 X 6 $$\frac{1}{4}$$ < 6 X 5 $$\frac{1}{4}$$ Think and Grow: Modeling Real Life Example An elephant sleeps 2$$\frac{1}{2}$$ hours in 1 day. A koala sleeps 6 times as long as the elephant in the same day. How many hours does the koala sleep that day? Multiply the number of hours the elephant sleeps by 6. The koala sleeps 5 hours that day. Explanation: First we write mixed fraction as fraction so 2 $$\frac{1}{2}$$ =( 2 X 2 + 1 ) by 2 = $$\frac{5}{2}$$ , now we multiply with whole number 2 = 2 X $$\frac{5}{2}$$ = 2 X 5 X $$\frac{1}{2}$$ = 10 X $$\frac{1}{2}$$ = $$\frac{10}{2}$$ further it can be simplified as both 10,2 can be divided by 2 so $$\frac{10}{2}$$ = 5 hours that day. Show and Grow Question 13. A small bag of popcorn has 1$$\frac{7}{8}$$ cups of popcorn. A large bag has 7 times as many cups of popcorn as the small bag. How many cups of popcorn are in the large bag? The large bag has $$\frac{105}{8}$$ cups of popcorn or 13 $$\frac{1}{8}$$ cups of popcorn. Explanation: Given A small bag of popcorn has 1$$\frac{7}{8}$$ cups of popcorn so first we write mixed fraction as fraction as 1 $$\frac{7}{8}$$= ( 1 X 8 + 7 ) by 8 =$$\frac{15}{8}$$ ,as a large bag has 7 times as many cups of popcorn as the small bag so 7 times $$\frac{15}{8}$$= 7 X 15 X $$\frac{1}{8}$$=105 X $$\frac{1}{8}$$=$$\frac{105}{8}$$ cups of popcorn. We can write the product as mixed fraction as 8 goes in 13 times. 13 will be our whole number 8 X 13 is 104 and we have 1 left over (105−104). 1 will be our numerator and our denominator will be the same 8= $$\frac{105}{8}$$= 13 $$\frac{1}{8}$$ cups of popcorn. Question 14. DIG DEEPER! A deli worker prepares 4 packages of meat. Each package contains 3$$\frac{1}{3}$$ pounds of meat. Your friend says the deli worker uses between 12 and 13 pounds of meat. Is your friend correct? Explain. No, friend is not correct as deli worker uses between 13 and 14 pounds of meat, not between 12 and 13 pounds of meat. Explanation: Given each package contains 3$$\frac{1}{3}$$ pounds of meat. First we write mixed fraction as fraction as ( 3 X 3 +1 ) by 3= $$\frac{10}{3}$$ now deli worker prepares 4 packages of meat so 4 times $$\frac{10}{3}$$= 4 X 10 X $$\frac{1}{3}$$= 40 X $$\frac{1}{3}$$=$$\frac{40}{3}$$ , we will write the product as mixed fraction as 3 goes in 13 times. 13 will be our whole number 3 X 13 is 39 and we have 1 left over (40−39). 1 will be our numerator and our denominator will be the same 3= $$\frac{105}{8}$$= 13 $$\frac{1}{3}$$, so the whole number is greater than 13 means deli worker uses between 13 and 14 pounds of meat not between 12 and 13 pounds of meat. Question 15. DIG DEEPER! An athlete’s goal is to run at least 80 miles in 1 week. He runs 5$$\frac{3}{4}$$ miles 2 times each day for 1 week. Does the athlete meet his goal? Explain. Athlete runs more than 80 miles in 1 week so he meets his goal. Explanation: Given an athlete runs 5$$\frac{3}{4}$$ miles 2 times each day, First we write mixed fraction as fraction as ( 5 X 4 +3) by 4 =$$\frac{23}{4}$$ now we multiply by 2 , 2 X $$\frac{23}{4}$$= 2 X 23 X $$\frac{1}{4}$$=46 x $$\frac{1}{4}$$= $$\frac{46}{4}$$, Now for 1 week it is 7 X $$\frac{46}{4}$$= 7 X 46 X $$\frac{1}{4}$$= 322 X $$\frac{1}{4}$$= $$\frac{322}{4}$$ now we will write the product as mixed fraction as 4 goes in 80 times. 80 will be our whole number 4 X 80 is 320 and we have 2 left over (322−320). 2 will be our numerator and our denominator will be the same 4= $$\frac{322}{4}$$= 80 $$\frac{2}{4}$$ , As 80 $$\frac{2}{4}$$ is greater than 80, athlete runs more than 80 miles in 1 week he meets his goal. ### Multiply Whole Numbers and Mixed Numbers Homework & Practice 9.4 Multiply Question 1. 2 X 1 $$\frac{1}{12}$$ = $$\frac{26}{12}$$ = $$\frac{13}{6}$$= 2 $$\frac{1}{6}$$. Explanation: First we write the mixed number as a fraction then multiply, so 1 $$\frac{1}{12}$$ = 1 + $$\frac{1}{12}$$ = ( 1 X 12 + 1 )by 12 = $$\frac{13}{12}$$ now we multiply it with the whole, 2 X $$\frac{13}{12}$$ now we write in unit fraction and multiply 2 X 13 X $$\frac{1}{12}$$= 26 X $$\frac{1}{12}$$ = $$\frac{26}{6}$$, Now the fraction can be further divided by 2 as 26,6 both goes in 2 after dividing it becomes as $$\frac{13}{6}$$ now we can write the product as mixed fraction as 6 goes in 2 times. 2 will be our whole number 6 X 2 is 12 and we have 1 left over (13−12). 1 will be our numerator and our denominator will be the same 6= $$\frac{13}{6}$$= 2 $$\frac{1}{6}$$. Question 2. 2 X 3$$\frac{5}{6}$$=$$\frac{46}{6}$$ = $$\frac{23}{3}$$= 7 $$\frac{2}{3}$$ Explanation: First we write the mixed number as a fraction then multiply, so 3 $$\frac{5}{6}$$ = 3 + $$\frac{5}{6}$$ =(3 X 6 + 5) by 6 = $$\frac{23}{6}$$ now we multiply it with the whole, 2 X $$\frac{23}{6}$$ now we write in unit fraction and multiply 2 X 23 X $$\frac{1}{6}$$= 46 X $$\frac{1}{6}$$ = $$\frac{46}{6}$$, Now the fraction can be further divided by 2 as 46,6 both goes in 2 after dividing it becomes as 23,3 = $$\frac{23}{3}$$ now we can write the product as mixed fraction as 3 goes in 7 times. 3 X 7 is 21 , 7 will be our whole number and (23−21) we have 2 left over. 2 will be our numerator and our denominator will be the same 3= $$\frac{23}{3}$$= 7 $$\frac{2}{3}$$. Question 3. 4 X 3 $$\frac{6}{10}$$ = $$\frac{144}{10}$$ = $$\frac{72}{5}$$ = 14 $$\frac{2}{5}$$. Explanation: First we write the mixed number as a fraction then multiply, so 3 $$\frac{6}{10}$$ = 3 + $$\frac{6}{10}$$ =( 3 X 10 + 6 ) by 10 = $$\frac{36}{10}$$ now we multiply it with the whole, 4 X $$\frac{36}{10}$$ now we write in unit fraction and multiply 4 X 36 X $$\frac{1}{10}$$= 144 X $$\frac{1}{10}$$ = $$\frac{144}{10}$$, Now the fraction can be further divided by 2 as 144,10 both goes in 2 after dividing it becomes as 72,5 = $$\frac{72}{5}$$ now we can write the product as mixed fraction as 5 goes in 14 times. 5 X 14 is 70 ,14 will be our whole number and (72−70) we have 2 left over. 2 will be our numerator and our denominator will be the same 5 = $$\frac{72}{5}$$= 14 $$\frac{2}{5}$$. Question 4. 2 $$\frac{3}{8}$$ X 5 = $$\frac{95}{8}$$ =11 $$\frac{7}{8}$$. Explanation: First we write the mixed number as a fraction then multiply, so 2 $$\frac{3}{8}$$ = 2 + $$\frac{3}{8}$$ =(2 X 8 + 3) by 8 = $$\frac{19}{8}$$ now we multiply it with the whole, 5 X $$\frac{19}{8}$$ now we write in unit fraction and multiply 5 X 19 X $$\frac{1}{8}$$= 95 X $$\frac{1}{8}$$ = $$\frac{95}{8}$$, Now we can write the product as mixed fraction as 8 goes in 11 times. 8 X 11 is 88 ,11 will be our whole number and (95−88) we have 7 left over. 7 will be our numerator and our denominator will be the same 8= $$\frac{95}{8}$$= 11 $$\frac{7}{8}$$. Question 5. 4 X 6 $$\frac{4}{5}$$ = $$\frac{136}{5}$$ = 27 $$\frac{1}{5}$$ Explanation: First we write the mixed number as a fraction then multiply, so 6 $$\frac{4}{5}$$ = 6 + $$\frac{4}{5}$$ =(6 X 5 + 4) by 5 = $$\frac{34}{5}$$ now we multiply it with the whole, 4 X $$\frac{34}{5}$$ now we write in unit fraction and multiply 4 X 34 X $$\frac{1}{5}$$ = 136 X $$\frac{1}{5}$$ = $$\frac{136}{5}$$, now we can write the product as mixed fraction as 5 goes in 27 times. 5 X 27 is 135 , 27 will be our whole number and (136−135) we have 1 left over. 1 will be our numerator and our denominator will be the same 5 = $$\frac{136}{5}$$= 27 $$\frac{1}{5}$$. Question 6. 8 $$\frac{20}{100}$$ X 10 = $$\frac{8200}{100}$$ = 82. Explanation: First we write the mixed number as a fraction then multiply, so 8 $$\frac{20}{100}$$ = 8 + $$\frac{20}{100}$$ = ( 8 X 100 + 20 ) by 100 = $$\frac{820}{100}$$ now we multiply it with the whole, 100 X $$\frac{820}{100}$$ now we write in unit fraction and multiply 10 X 820 X $$\frac{1}{100}$$ = 8200 X $$\frac{1}{100}$$ = $$\frac{8200}{100}$$, Now the fraction can be further divided by 100 as 8200,100 both goes in 100 after dividing by 100 it becomes as 82. Question 7. Reasoning Without calculating, is the product of 7 and 5$$\frac{3}{4}$$ greater than or less than 35? Explain. 7 X 5 $$\frac{3}{4}$$ is greater than ( > ) 35. Explanation: Given to compare between 7 X 5 $$\frac{3}{4}$$ and 35, first we take 7 X 5 $$\frac{3}{4}$$ to this expression if we apply distributive property means if multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. example a X b $$\frac{c}{d}$$ = a X b + a X $$\frac{c}{d}$$ So 7 X 5 $$\frac{3}{4}$$ = 7 X 5 + 7 X $$\frac{3}{4}$$ means 35 + 7 X $$\frac{3}{4}$$ So by comparing left side it is addition of 35 plus some value ,it will be more than 35 only and right hand side it it 35 so obiviously 7 X 5 $$\frac{3}{4}$$ is greater than ( > ) 35. Question 8. YOU BE THE TEACHER Your friend ﬁnds the product of 4 and 2$$\frac{8}{10}$$. Is your friend correct? Explain. Yes friend is correct, because the value is same as 11 $$\frac{2}{10}$$ . Explanation: To caluculate the value of 4 X 2 $$\frac{8}{10}$$ , First we write the mixed number as a fraction then multiply, so 2 $$\frac{8}{10}$$ = 2 + $$\frac{8}{10}$$ = ( 2 X 10 + 8 ) by 10 = $$\frac{28}{10}$$ now we multiply it with the whole, 4 X $$\frac{28}{10}$$ now we write in unit fraction and multiply 4 X 28 X $$\frac{1}{10}$$= 112 X $$\frac{1}{10}$$ = $$\frac{112}{10}$$, now we can write the product as mixed fraction as 10 goes in 11 times. 10 X 11 is 110 , 11 will be our whole number and (112−110) we have 2 left over. 2 will be our numerator and our denominator will be the same 10= $$\frac{112}{10}$$= 11 $$\frac{2}{10}$$ is same as friend, So friend is correct. Question 9. Number Sense Between which two whole numbers does the product of 9 and 7 $$\frac{1}{8}$$ lie? The product of 9 and 7 $$\frac{1}{8}$$ lies in between 64 and 65. Explanation: First we write the mixed number 7 $$\frac{1}{8}$$ as a fraction then multiply, so 7 $$\frac{1}{8}$$ = 7 + $$\frac{1}{8}$$ = 7 X 8 + 1 by 8 = $$\frac{57}{8}$$ now we multiply it with the whole, 9 X $$\frac{57}{8}$$ now we write in unit fraction and multiply 9 X 57 X $$\frac{1}{8}$$ = 513 X $$\frac{1}{8}$$ = $$\frac{513}{8}$$, now we can write the product as mixed fraction as 8 goes in 64 times. 8 X 64 is 512 , 64 will be our whole number and (513−512) we have 1 left over. 1 will be our numerator and our denominator will be the same 8 = $$\frac{513}{8}$$= 64 $$\frac{1}{8}$$. Any how we got 64 whole number and next whole number is 65 means the product of 9 and 7 $$\frac{1}{8}$$ lies in between 64 and 65. Question 10. Modeling Real Life Athlete A holds a 2$$\frac{1}{2}$$ – kilogram plate while doing squats. Athlete B holds a plate that is 4 times heavier than Athlete A’s. How many kilograms is the plate held by Athlete B? Athlete B holds 10 kilograms of plate while doing squats. Explanation: Given Athlete A hold a 2 $$\frac{1}{2}$$ kilogram plate while doing squats, Athlete B holds a plate that is 4 times heavier than Athlete A’s, So Athlete B holds 4 X 2 $$\frac{1}{2}$$, First we write the mixed number as a fraction then multiply, so 2 $$\frac{1}{2}$$ = 2 + $$\frac{1}{2}$$ = ( 2 X 2 + 1 ) by 2 = $$\frac{5}{2}$$ now we multiply it with the whole, 4 X $$\frac{5}{2}$$ now we write in unit fraction and multiply 4 X 5 X $$\frac{1}{2}$$ = 20 X $$\frac{1}{2}$$ = $$\frac{20}{2}$$ as 20,2 goes by 2 we further divide both by 2 we get 10,1 = $$\frac{20}{2}$$ = 10, therefore Athlete B holds 10 kilograms of plate while doing squats. Question 11. DIG DEEPER! A zoo nutritionist orders 5 $$\frac{1}{4}$$ tons of apples and 7$$\frac{2}{4}$$tons of bananas each year to feed the animals. She orders 6 times as many tons of herbivore pellets than tons of fruit. How many tons of herbivore pellets does the nutritionist order? Explanation: Review & Refresh Subtract. Question 12. = $$\frac{12}{4}$$ = 3 Explanation: First we write the mixed numbers into fractions and subtract, So 9 $$\frac{1}{4}$$ = 9 +$$\frac{1}{4}$$ = (9 x 4 + 1) by 4 = $$\frac{37}{4}$$ and 6 $$\frac{1}{4}$$ = 6 + $$\frac{1}{4}$$ = 6 X 4 + 1 by 4 = $$\frac{25}{4}$$ now subtracting $$\frac{37}{4}$$ – $$\frac{25}{4}$$ to subtract a fraction from another we need to first make sure both fractions have the same denominator as both denominators have same value 4 now we can subtract one numerator from the other to make one fraction as 37 – 25 by 4 = $$\frac{12}{4}$$ as 12, 4 goes by 4 we further can divide both by 4 we get 12,1 = $$\frac{12}{4}$$ = 3 . Question 13. = $$\frac{11}{3}$$ = 3 $$\frac{2}{3}$$ Explanation: First we write the mixed numbers into fractions and subtract, So 6 $$\frac{1}{3}$$ = 6 +$$\frac{1}{3}$$ =(6 x 3 + 1) by 3 = $$\frac{19}{3}$$ and 2 $$\frac{2}{3}$$ = 2 + $$\frac{2}{3}$$ = 2 X 3 + 2 by 3 = $$\frac{8}{3}$$ now subtracting $$\frac{19}{3}$$ – $$\frac{8}{3}$$ to subtract a fraction from another we need to first make sure both fractions have the same denominator as both denominators have same value 3 now we can subtract one numerator from the other to make one fraction as 19 – 8 by 3 = $$\frac{11}{3}$$ since the numerator is greater than the denominator, we can further simplify it into a mixed fraction as 3 goes in 3 times. 3 X 3 is 9 , 3 will be our whole number and (11−9) we have 2 left over. 2 will be our numerator and our denominator will be the same 3 = $$\frac{11}{3}$$ = 3 $$\frac{2}{3}$$ . Question 14. = $$\frac{78}{12}$$ = 6 $$\frac{6}{12}$$ or $$\frac{78}{12}$$ = $$\frac{39}{6}$$ = $$\frac{13}{2}$$ = 6 $$\frac{1}{2}$$. Explanation: First we write the mixed numbers into fractions and subtract, So 8 $$\frac{4}{12}$$ = 8 +$$\frac{4}{12}$$ = ( 8 x 12 + 4 ) by 12 = $$\frac{100}{12}$$ and 1 $$\frac{10}{12}$$ = 1 + $$\frac{10}{12}$$ = 1 X 12 + 10 by 12 = $$\frac{22}{12}$$ now subtracting $$\frac{100}{12}$$ – $$\frac{22}{12}$$ to subtract a fraction from another we need to first make sure both fractions have the same denominator as both denominators have same value 3 now we can subtract one numerator from the other to make one fraction as 100 – 22 by 12 = $$\frac{78}{12}$$ since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 12 goes in 6 times. 12 X 6 is 72 , 6 will be our whole number and (78−12) we have 6 left over. 6 will be our numerator and our denominator will be the same 12, 6 $$\frac{6}{12}$$ , Further $$\frac{78}{12}$$ can be simplified as 78,12 both goes by 2 making them as 39, 6=$$\frac{39}{6}$$, even further $$\frac{39}{6}$$ can be simplified as 39,6 both goes by 3 making them as 13,2 =$$\frac{13}{2}$$, Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 2 goes in 6 times. 6 X 2 is 12 , 6 will be our whole number and (13−12) we have 1 left over. 1 will be our numerator and our denominator will be the same 3 = $$\frac{78}{12}$$ = 6 $$\frac{6}{12}$$ or $$\frac{78}{12}$$ = $$\frac{39}{6}$$ = $$\frac{13}{2}$$ = 6 $$\frac{1}{2}$$. ### Lesson 9.5 Problem Solving: Fraction Operations Explore and Grow You want to make 3 batches of the recipe. Explain how to find how much of each ingredient you need. Reasoning Explain how you can tell whether you need more than or less than 6 cups of blueberries without calculating To make 3 batches of recipe we require $$\frac{21}{4}$$ pounds of strawberries or 5 $$\frac{1}{4}$$ pounds of strawberries required, $$\frac{3}{2}$$ cups of orange juice or 1 $$\frac{1}{2}$$ cups of orange juice, 12 tablespoons of lemon juice,12 tablespoons of honey and $$\frac{27}{4}$$ cups of blue berries are required or 6 $$\frac{3}{4}$$ cups of blueberries are required. We require more than 6 cups of blueberries why because if we see product required is 3 X 2 $$\frac{1}{4}$$ cups of blue berries without calculating , Applying distributive property we get (3 X 2 ) + ( 3 X $$\frac{1}{4}$$ ) = 6 + ( 3 X $$\frac{1}{4}$$ ) it clearly shows that we require more than 6 cups of blueberries. Explanation: Given to make 1 batch of Recipe Ladybug Ice pops we require 1 $$\frac{3}{4}$$ pounds of strawberries , $$\frac{1}{2}$$ cup of orange juice , 4 tablespoons of lemon juice, 4 tablespoons of honey, 2 $$\frac{1}{4}$$ cups of blue berries. To find out how much of each ingredient needed to make 3 batches of recipe we multiply each ingredient with 3. Strawberries needed are 3 X 1 $$\frac{3}{4}$$, First we write the mixed number as a fraction then multiply so 1 $$\frac{3}{4}$$ = ( 1 X 4 + 3 ) by 4 = $$\frac{7}{4}$$ , so 3 X $$\frac{7}{4}$$ now we write in unit fraction and multiply 3 X 7 X $$\frac{1}{4}$$ = 21 X $$\frac{1}{4}$$ = $$\frac{21}{4}$$ pounds of strawberries required, Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 4 goes in 5 times. 5 X 4 is 20 , 5 will be our whole number and (21−20) we have 1 left over. 1 will be our numerator and our denominator will be the same 4 =5 $$\frac{1}{4}$$ pounds of strawberries required, Orange juice required 3 X $$\frac{1}{2}$$ already it is in unit fraction so we require $$\frac{3}{2}$$ cups of orange juice Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 2 goes in 1 times. 1 X 2 is 2, 1 will be our whole number and (3-2), we have 1 left over. 1 will be our numerator and our denominator will be the same 2 = 1 $$\frac{1}{2}$$ cups of orange juice. Lemon juice required is 3 X 4 = 12 tablespoons of lemon juice, 3 X 4 = 12 tablespoons of honey, now blue berries required are 3 X 2 $$\frac{1}{4}$$ cups, First we write the mixed number as a fraction then multiply so 2 $$\frac{1}{4}$$ = ( 2 X 4 + 1 ) by 4 = $$\frac{9}{4}$$ , now 3 X $$\frac{9}{4}$$ now we write in unit fraction and multiply 3 X 9 X $$\frac{1}{4}$$ = 27 X $$\frac{1}{4}$$ = $$\frac{27}{4}$$ cups of blue berries are required, Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 4 goes in 6 times. 6 X 4 is 24, 6 will be our whole number and (27-24) we have 3 left over. 3 will be our numerator and our denominator will be the same 4 = 6 $$\frac{3}{4}$$ cups of blueberries are required. Now comparing blueberries we require more than 6 cups of blueberries why because we use is 3 X 2 $$\frac{1}{4}$$ cups of blue berries are required without calculating, Applying distributive property we get (3 X 2 ) +( 3 X $$\frac{1}{4}$$ )= 6 +( 3 X $$\frac{1}{4}$$), it clearly shows that we require more than 6 cups of blueberries. Think and Grow: Problem Solving: Fraction Operations Example To convert a temperature from degrees Celsius to degrees Fahrenheit, multiply the Celsius temperature by $$\frac{9}{5}$$, then add 32. What is the temperature shown by the thermometer in degrees Fahrenheit? The temperature shown by the thermometer is 500 Fahrenheit. Explanation: We know to convert a temperature from degrees Celsius to degrees Fahrenheit is Fahrenheit = Celsius temperature X $$\frac{9}{5}$$ + 32 . So multiply the Celsius temperature by $$\frac{9}{5}$$, then add 32, Given the Celsius temperature as 100 in the figure above, First we write $$\frac{9}{5}$$ in unit fraction and multiply by Celsius temperature so $$\frac{9}{5}$$=9 X $$\frac{1}{5}$$ now we multiply by 10 as 10 X 9 X $$\frac{1}{5}$$ = 90 X $$\frac{1}{5}$$ = $$\frac{90}{5}$$ this value can be further simplified as 90,5 both goes by 5 making them as 18,1, Now we got result as 18 , We add 18 + 32 we will get 500 Fahrenheit. Understand the Problem What do you know? • To convert a temperature from degrees Celsius to degrees Fahrenheit, multiply the Celsius temperature by $$\frac{9}{5}$$, then add 32. • The thermometer shows 10 degrees Celsius. What do you need to find? • You need to find the temperature shown by the thermometer in degrees Fahrenheit. Make a Plan How will you solve? • First, multiply the Celsius temperature, 10 degrees, by $$\frac{9}{5}$$. • Then add 32 to the product. Solve So, the temperature shown by the thermometer is _500_ degrees Fahrenheit. The temperature shown by the thermometer is 500 Fahrenheit. Explanation : To convert a temperature from degrees Celsius to degrees Fahrenheit, Let us take p as multiply of Celsius temperature by $$\frac{9}{5}$$, Now we take f as p+32, First we write $$\frac{9}{5}$$ in unit fraction and multiply by Celsius temperature, so $$\frac{9}{5}$$=9 X $$\frac{1}{5}$$ now multiply by 10 = 10 X 9 X $$\frac{1}{5}$$ = 90 X $$\frac{1}{5}$$ = $$\frac{90}{5}$$ this value we take as p this can be further simplified as 90,5 both goes by 5 making them as 18, Now we got p as 18 we add 32 to p , we get f, So f = 18 + 32 = 500 So the temperature shown by the thermometer is 500 Fahrenheit. Show and Grow Question 1. Show how to solve the example above using one equation. F= C X $$\frac{9}{5}$$ + 32 degrees. Explanation: To solve the example above using one equation, Let us take Fahrenheit as F , Celsius as C so to convert a temperature from degrees Celsius to degrees Fahrenheit we multiply C with $$\frac{9}{5}$$ and add 32. Apply and Grow: Practice Understand the problem. What do you know? What do you need to find? Explain. We know Celsius temperature, We need to find the temperature shown by the thermometer in degrees Fahrenheit. Explanation: Given the temperature in Celsius and we know the formula to convert from Celsius temperature to Fahrenheit as F= C X $$\frac{9}{5}$$ + 32 degrees so we substitute the values and find the temperature. Question 2. You make a friendship bracelet with 3 pink strings and 2 blue strings. Each string is 3$$\frac{3}{4}$$ feet long. How many feet of string do you use? $$\frac{75}{4}$$ feet long strings is used or 18 $$\frac{3}{4}$$ feet long strings is used. Explanation: Given to make a friendship bracelet we use 3 pink strings , 2 blue strings means total 5 strings are needed and each string is 3$$\frac{3}{4}$$ feet long, First we convert mixed fraction into fraction as ( 3 X 4 + 3 ) by 4 = $$\frac{15}{4}$$ ,Now we will convert into unit fraction as 15 X $$\frac{1}{4}$$ now we multiply with 5 , 5 X 15 X $$\frac{1}{4}$$ =$$\frac{75}{4}$$, since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 4 goes in 18 times. 18 X 4 is 72, 18 will be our whole number and (75-72) we have 3 left over. 3 will be our numerator and our denominator will be the same 4 = 18 $$\frac{3}{4}$$, So $$\frac{75}{4}$$ feet long strings is used or 18 $$\frac{3}{4}$$ feet long strings is used. Question 3. A smoothie store worker makes 4 peanut butter banana smoothies and 2 fruit smoothies. The worker uses $$\frac{2}{3}$$ cup of bananas in each smoothie. How many cups of bananas does the worker need? The worker needed $$\frac{12}{3}$$ cups of bananas or 4 cups of bananas are needed to make smoothie. Explanation: Given a smoothie store worker makes 4 peanut butter banana smoothies and 2 fruit smoothies. The worker uses $$\frac{2}{3}$$ cup of bananas in each smoothie so the number of cups of bananas the worker needed are 4 X latex]\frac{2}{3}[/latex] plus 2 X $$\frac{2}{3}$$ first we convert both into unit fractions and multiply 4 X 2 X $$\frac{1}{3}$$ = 8 X $$\frac{1}{3}$$ = $$\frac{8}{3}$$ of peanut butter banana smoothies and 2 X 2 X $$\frac{1}{3}$$ = 4 X $$\frac{1}{3}$$ = $$\frac{4}{3}$$ fruit smoothies, now we add both $$\frac{8}{3}$$ + $$\frac{4}{3}$$ as both denominators are same we add numerators as 8 + 4 =12 as numerator and denominator as 3 = $$\frac{12}{3}$$ as numerator is greater than denominator further can be simplified and 12 divides by 3 gives 4 with 0 remainder so the worker needed $$\frac{12}{3}$$ cups of bananas or 4 cups of bananas are needed to make smoothie. Understand the problem. Then make a plan. How will you solve? Explain. Question 4. Your friend walks her dog for $$\frac{1}{4}$$ mile each day. She then runs 2$$\frac{3}{4}$$ miles each day. How many total miles does she walk her dog and run in 1 week? Total 21 miles my friend walks her dog and run in 1 week. Explanation: Given my friend walks her dog for $$\frac{1}{4}$$ mile each day , She then runs 2$$\frac{3}{4}$$ miles each day. So in each day she walk her dog and run is $$\frac{1}{4}$$ mile + 2$$\frac{3}{4}$$ miles, Then we multiply with 7 because we have to calculate for 1 week. First we convert mixed fraction into fraction 2 $$\frac{3}{4}$$ , ( 2 X 4 + 3 ) by 4 = $$\frac{11}{4}$$, Now we add with $$\frac{1}{4}$$ + $$\frac{11}{4}$$ = $$\frac{12}{4}$$ as numerator is greater than denominator further can be simplified and 12 divides by 4 gives 3 with 0 remainder, so 3 miles for each day, Now for 1 week it is 7 X 3 = 21 miles. My friend walks 21 miles her dog and run in 1 week. Question 5. Hair donations must be 12 inches longer. Your friend’s hair is 7 inches long. Her hair grows about $$\frac{1}{2}$$ inch each month. Can she donate her hair in 8 months? No friend cannot donate her hair in 8 months as in 8 months it grows to 11 inches long and for hair donations it must be 12 inches longer. Explanation: Given friend’s hair is 7 inches long. Her hair grows about $$\frac{1}{2}$$ inch each month. In 8 months it grows 8 X $$\frac{1}{2}$$ = $$\frac{8}{2}$$ as numerator is greater than denominator further can be simplified and 8 divides by 2 gives 4 with 0 remainder, in 8 months it will grow 4 inches , So after 8 months it will be 7 + 4 =11 inches long as 11 is less than 12 and hair donations must be 12 inches longer so friend cannot donate her hair in 8 months. Question 6. Today you walk $$\frac{6}{10}$$ mile from the Martin Luther King Jr. Memorial to the Washington Monument. Tomorrow you will walk about 4 times as far from the Washington Monument to the White House. About how much farther will you walk tomorrow? Tomorrow i will walk father $$\frac{18}{10}$$ miles or $$\frac{9}{5}$$ miles or 1 $$\frac{4}{5}$$ . Explanation: Given today i walked $$\frac{6}{10}$$ mile from the Martin Luther King Jr. Memorial to the Washington Monument, Tomorrow i will walk about 4 times as far from the Washington Monument to the White House , So number of miles walked tomorrow will be 4 X $$\frac{6}{10}$$ first we write in unit fraction 4 X 6 X $$\frac{1}{10}$$ = 24 X $$\frac{1}{10}$$ = $$\frac{24}{10}$$ as numerator is greater than denominator further can be simplified as 24,10 both can be divided by 2 gives 12,5 so $$\frac{24}{10}$$ = $$\frac{12}{5}$$ now farther i will walk tomorrow will be $$\frac{24}{10}$$ – $$\frac{6}{10}$$ as denominator of both are same we substract numerators , 24 – 6 =18 and denominator will remain same 10 = $$\frac{18}{10}$$ miles or further can be simplified and 18,10 both can be divided by by 2 gives 9, 5 = $$\frac{18}{10}$$ miles = $$\frac{9}{5}$$ miles as numerator is greater than denominator we can write it in mixed fraction as 5 goes in 1 times. 5 X 1 is 5 , 1 will be our whole number and (9−5) we have 4 left over. 4 will be our numerator and our denominator will be the same 5 = 1 $$\frac{4}{5}$$ . Think and Grow: Modeling Real Life Example A 20-pound dog and a 60-pound dog eat the recommended amounts of dog food each day. How much more food does the 60-pound dog eat in 1 week than the 20-pound dog? Think: What do you know? What do you need to find? How will you solve? So, the 60-pound dog eats 10 $$\frac{1}{2}$$ more cups of food than the 20-pound dog in 1 week. The 60-pound dog eats 10 $$\frac{1}{2}$$ more cups of food than the 20-pound dog in 1 week. Explanation: Given A 20-pound dog eats 1 $$\frac{1}{2}$$ cups and a 60-pound dog eats 3 cups, the recommended amount of food each day. Now we need to calculate how much more food does the 60-pound dog eat in 1 week than the 20-pound dog, So Step 1 : First we find how much food the 20-pound dog eats in 1 week, let us take it as b = 7 X 1 $$\frac{1}{2}$$ we write mixed fraction as fraction 1 $$\frac{1}{2}$$ = ( 1 X 2 + 1 )by 2 = $$\frac{3}{2}$$ now multiply by 7 = 7 X $$\frac{3}{2}$$ we write into unit fraction 7 X 3 X $$\frac{1}{2}$$ = 21 X $$\frac{1}{2}$$ = $$\frac{21}{2}$$ cups, Step 2: Now we find how much food the 60-pound dog eats in 1 week let us take it as c = 7 X 3 = 21 cups, Step 3: Subtract b from c to find how much more food the 60-pound dog eats in 1 week. c- b = 21 – $$\frac{21}{2}$$ to subtract first we make both denominators same, we multiply and divide 21 by 2 making as (21 x 2) by 2 = $$\frac{42}{2}$$ –$$\frac{21}{2}$$ as denominators are same we subtarct numerators 42-21=21 will be numerator and denominator is same 2 =$$\frac{21}{2}$$ as numerator is greater than denominator and 2 goes 10 times, 2 X 10 is 20 , 10 will be our whole number and (21−20) we have 1 left over. 1 will be our numerator and our denominator will be the same 2 = 10 $$\frac{1}{2}$$ . Therefore the 60-pound dog eats 10 $$\frac{1}{2}$$ more cups of food than the 20-pound dog in 1 week. Show and Grow Question 7. Use the table above. A 40-pound dog and a 100-pound dog eat the recommended amounts of dog food each day. How much food is needed to feed both dogs for 1 week? To feed the food for both dogs for 1 week we require $$\frac{182}{4}$$ cups or $$\frac{91}{2}$$ or 45 $$\frac{1}{2}$$ cups. Explanation: From the table 40-pound dog eats 2 $$\frac{1}{4}$$ and 100-pound dog eats 4 $$\frac{1}{4}$$ the recommended amounts of dog food each day, to find how much food is needed to feed both dogs for 1 week first we calculate for 40-pound dog= 7 X 2 $$\frac{1}{4}$$ we write the mixed fraction as a fraction 2 x 4 + 1 by 4 = $$\frac{9}{4}$$ and multiply by 7 , 7 X $$\frac{9}{4}$$ now we write in unit fraction 7 X 9 X $$\frac{1}{4}$$ = 63 X $$\frac{1}{4}$$ = $$\frac{63}{4}$$ now we calculate for 100-pound dog eats 4 $$\frac{1}{4}$$ in 1 week is 7 X 4 $$\frac{1}{4}$$ we write the mixed fraction as a fraction ( 4 X 4 + 1 )by 4 = $$\frac{17}{4}$$ and multiply by 7 , 7 X $$\frac{17}{4}$$ now we write in unit fraction 7 X 17 X $$\frac{1}{4}$$ = 119 X $$\frac{1}{4}$$ = $$\frac{119}{4}$$. Now to feed the food for both dogs for 1 week is $$\frac{63}{4}$$ + $$\frac{119}{4}$$ as denominators are same we add numerators we get 63 + 119 = 182 as numerator, $$\frac{182}{4}$$ further can be simplifed as 182 and 4 goes by 2 we write as $$\frac{91}{2}$$ numerator is greater we can write in mixed fraction as 2 goes 45 times, 2 X 45 is 90 , 45 will be our whole number and (91−90) we have 1 left over. 1 will be our numerator and our denominator will be the same 2, making as 45 $$\frac{1}{2}$$ cups. ### Problem Solving: Fraction Operations Homework & Practice 9.5 Understand the problem. Then make a plan. How will you solve? Explain. Question 1. Your friend makes strawberry jam and raspberry jam. He makes enough strawberry jam to ﬁll $$\frac{1}{2}$$ of a jar. He makes 5 times as much raspberry jam as strawberry jam. How many full jars of raspberry jam does he make? Friend makes $$\frac{5}{2}$$ jars= 2 $$\frac{1}{2}$$ jars in that full jars are 2, So 2 jars of raspberry jam he makes. Explanation: Given friend makes strawberry jam to fill $$\frac{1}{2}$$ of a jar and makes 5 times as much raspberry jam as strawberry jam means he makes 5 X $$\frac{1}{2}$$ of jars of raspberry jam now we write as fraction $$\frac{5}{2}$$ as numerator is greater, we can write in mixed fraction as 2 goes 2 times, 2 X 2 is 4 , 2 will be our whole number and (5−4) we have 1 left over. 1 will be our numerator and our denominator will be the same 2, making as 2 $$\frac{1}{2}$$ ,here 2 is whole there fore full jars of raspberry jam is 2. So friend makes $$\frac{5}{2}$$ jars= 2 $$\frac{1}{2}$$ jars in that full jars are 2, So 2 jars of raspberry jam he makes. Question 2. You buy a pair of jeans that originally cost $22. The pair of jeans is $$\frac{1}{2}$$ off the original price. You pay with a$20 bill. How much change do you receive? I receive $9 as change. Explanation: The pair of jeans is $$\frac{1}{2}$$ off the original price and original cost is$ 22, So the cost to be paid is $22 X $$\frac{1}{2}$$ = $$\frac{22}{2}$$ further this can be simplified as both can be divided by 2 numerator becomes 11 and denominator as 1 = 11, therefore the original cost for the pair of jeans is$ 11, Now i have paid $20bill, i will receive a change that is$20 – $11=$9, i receive \$9 as a change. Question 3. The observation deck of the Space Needle in Seattle is 520 feet above ground. The tip of the Space Needle is 85 feet above the observation deck. An artist makes a replica that is $$\frac{1}{100}$$ the height of the entire Space Needle. How tall is the replica? The height of replica is $$\frac{605}{100}$$ =$$\frac{121}{20}$$ or 6 $$\frac{1}{20}$$ feet. Explanation: Given the observation deck of the Space Needle in Seattle is 520 feet above ground and the tip of the Space Needle is 85 feet above the observation deck means the height of the entire Space Needle is 520 + 85 = 605 feet above the ground to tip, now an artist makes a replica that is $$\frac{1}{100}$$ the height of the entire Space Needle. So the height of the replica is 605 X $$\frac{1}{100}$$ = $$\frac{605}{100}$$ further can be simplifed as 605 and 100 goes by 5 we get numerator as 121 and denominator as 20 = $$\frac{121}{20}$$ as numerator is greater we can write in mixed fraction as 20 goes 6 times, 20 X 6 is 120 , 6 will be our whole number and (121−120) we have 1 left over. 1 will be our numerator and our denominator will be the same 20, making as 6 $$\frac{1}{20}$$ feet tall. Question 4. Modeling Real Life How many sheets of orange and black paper does a teacher need so that 30 students can each complete the art project? For 30 students to complete the Art Project orange sheets required are 11 $$\frac{2}{8}$$ and 18 $$\frac{6}{8}$$ black sheets of paper is required. Explanation: Given paper needed for Art Project orange is $$\frac{3}{8}$$ and for 30 students it is 30 X $$\frac{3}{8}$$ we write in unit fraction as 30 X 3 X $$\frac{1}{8}$$ = 90 X $$\frac{1}{8}$$ = $$\frac{90}{8}$$ as numerator is greater we can write in mixed fraction as 8 goes 11 times, 8 X 11 is 88, 11 will be our whole number and (90−88) we have 2 left over. 2 will be our numerator and our denominator will be the same 8, making as 11 $$\frac{2}{8}$$ sheets of orange is required, given black paper needed for Art Project is $$\frac{5}{8}$$ and for 30 students it is 30 X $$\frac{5}{8}$$ we write in unit fraction as 30 X 5 X $$\frac{1}{8}$$ = 150 X $$\frac{1}{8}$$ = $$\frac{150}{8}$$ as numerator is greater we can write in mixed fraction as 8 goes 18 times, 8 X 18 is 144, 18 will be our whole number and (150−144) we have 6 left over. 6 will be our numerator and our denominator will be the same 8, making as 18 $$\frac{6}{8}$$ black sheets are required. Therefore for 30 students to complete the Art Project orange sheets required are 11 $$\frac{2}{8}$$ and 18 $$\frac{6}{8}$$ black sheets of paper is required. Question 5. Writing Write and solve a two-step word problem with mixed numbers that can be solved using multiplication. A two-step problem is a word problem that requires two operations to solve it. We take two mixed numbers and slove using multiplication. In step one we use multiplication and add and in another step we use addition and again use multiplication as explained below with an example, Explanation: For example: Percy walks 2 $$\frac{5}{4}$$ kilometers for 2 days and 1 $$\frac{1}{4}$$ kilometers for 5 days , how many kilometers does Percy walks in 1 week? Given in first 2 days Percy walks 2 $$\frac{5}{4}$$ kilometers so In first step we change mixed fraction to fraction by using multiplication and then add 2 X 4 + 5 by 4 = $$\frac{13}{4}$$ kilometers for 1 day for 2 days it is 2 X $$\frac{13}{4}$$ = 2 X 13 X $$\frac{1}{4}$$ = 26 X $$\frac{1}{4}$$ = $$\frac{26}{4}$$ kilometers for 2 days , Now for rest of 5 days we calculate. Here 1 $$\frac{1}{4}$$ we change the mixed fraction to fraction as ( 1 X 4 + 1 ) by 4 = $$\frac{5}{4}$$ for 1 day, For 5 days it will be 5 X $$\frac{5}{4}$$, 5 X 5 X $$\frac{1}{4}$$ = 25 X $$\frac{1}{4}$$ = $$\frac{25}{4}$$ kilometers for rest 5 days , Now total number of kilometers Percy walks is $$\frac{26}{4}$$ + $$\frac{25}{4}$$ as denominators are same we add numerators we get 26 + 25 as 51 and denominator being same as 4 = $$\frac{51}{4}$$ kilometers Percy walks in 1 week and further can be simplified as numerator is greater we can write in mixed fraction as 4 goes 12 times, 4 X 12 is 48, 12 will be our whole number and (51−48) we have 3 left over. 3 will be our numerator and our denominator will be the same 4, making as 12 $$\frac{3}{4}$$ kilometers Percy walks in 1 week. Question 6. Modeling Real Life In science class, you test how many grams different-sized bundles of spaghetti can hold. You predict that a bundle of 5 spaghetti noodles can hold 10 quarters before breaking. You predict that a bundle of 10 spaghetti noodles can hold 20 quarters before breaking. Each quarter 5$$\frac{67}{100}$$ grams. According to your predictions, how many more grams can the bundle of 10 noodles hold than the bundle of 5 noodles? $$\frac{970}{100}$$ grams more the bundle of 10 noodles hold than the bundle of 5 noodles or $$\frac{97}{10}$$ grams more the bundle of 10 noodles hold than the bundle of 5 noodles or 9 $$\frac{7}{10}$$ grams more the bundle of 10 noodles hold than the bundle of 5 noodles. Explanation: Given that bundle of 5 noodles can hold 10 quarters means 5 noodles can hold 10 X 5 $$\frac{67}{100}$$ grams first we change mixed fraction to fraction by using multiplication and then add as 5 $$\frac{67}{100}$$ = 5 X 6 + 67 by 100 = $$\frac{97}{100}$$ now we multiply by 10, 10 X $$\frac{97}{100}$$ = 10 X 97 X $$\frac{1}{100}$$ = 970 X $$\frac{1}{100}$$ = $$\frac{970}{100}$$ grams , so 5 spaghetti noodles can hold $$\frac{970}{100}$$ grams before breaking, Now given 10 spaghetti noodles can hold 20 quarters means 10 spaghetti noodles can hold 20 X 5 $$\frac{67}{100}$$ grams again first we change mixed fraction to fraction by using multiplication and then add as 5 $$\frac{67}{100}$$ = ( 5 X 6 + 67 ) by 100 = $$\frac{97}{100}$$ now we multiply by 20, 20 X $$\frac{97}{100}$$ = 20 X 97 X $$\frac{1}{100}$$ = 1940 X $$\frac{1}{100}$$ = $$\frac{1940}{100}$$ grams , so 10 spaghetti noodles can hold $$\frac{1940}{100}$$ grams before breaking now to calculate how many more grams can the bundle of 10 noodles hold than the bundle of 5 noodles we subtract $$\frac{1940}{100}$$ grams – $$\frac{970}{100}$$ grams as denominators are same first we substract numerators as 1940 – 970 = 970 as numerator and denominator being same 100 we get $$\frac{970}{100}$$ grams more the bundle of 10 noodles hold than the bundle of 5 noodles, further can be simplified as 970 and 100 both can be divided by 10 making it as $$\frac{97}{10}$$ grams more the bundle of 10 noodles hold than the bundle of 5 noodles and as numerator is greater , we can write in mixed fraction as 10 goes 9 times, 10 X 9 is 90, 9 will be our whole number and (97−90) we have 7 left over. 7 will be our numerator and our denominator will be the same 10, making as 9 $$\frac{7}{10}$$ grams more the bundle of 10 noodles hold than the bundle of 5 noodles. Review & Refresh Compare Question 7. $$\frac{9}{10}$$ is greater than > $$\frac{4}{5}$$ or $$\frac{9}{10}$$ is greater than > $$\frac{8}{10}$$ Explanation: In order to compare the fractions we must first turn their different denominators into the same denominators to make denominators same we multiply $$\frac{4}{5}$$ with both numerator and denominator by 2 the value will not change = $$\frac{4}{5}$$ X $$\frac{2}{2}$$= $$\frac{8}{10}$$ now as denominators are same 10 we can compare only numerators $$\frac{9}{10}$$ with $$\frac{8}{10}$$ as 9 is greater than 8, So $$\frac{9}{10}$$ is greater than > $$\frac{8}{10}$$ or $$\frac{9}{10}$$ is greater than > $$\frac{4}{5}$$ Question 8. $$\frac{3}{8}$$ < is less than $$\frac{5}{6}$$ or $$\frac{9}{24}$$ < is less than $$\frac{20}{24}$$. Explanation: In order to compare the fractions we must first turn their different denominators into the same denominators to make denominators same , we find the least common multiplier of 8 ,6 as ( 2 X 2 X 2 X 3 ) = 24, Now convert each one of $$\frac{3}{8}$$ and $$\frac{5}{6}$$ into an equivalent fraction having as denominator, $$\frac{3}{8}$$ = $$\frac{3}{8}$$ X $$\frac{3}{3}$$ = $$\frac{9}{24}$$ and $$\frac{5}{6}$$ = $$\frac{5}{6}$$ X $$\frac{4}{4}$$ = $$\frac{20}{24}$$ now as denominators are same 24 we can compare only numerators $$\frac{9}{24}$$ with $$\frac{20}{24}$$ as 9 is less than 20, so $$\frac{9}{24}$$ < is less than $$\frac{20}{24}$$ or $$\frac{3}{8}$$ < is less than $$\frac{5}{6}$$ Question 9. $$\frac{1}{3}$$ = is equal to $$\frac{4}{12}$$ or $$\frac{1}{3}$$ = is equal to $$\frac{1}{3}$$ Explanation: Here first we simplify $$\frac{4}{12}$$ as both can be divided by 4 if we dividey both numerator and denominator by 4 we get numerator as 1 and denominator as 4 = $$\frac{1}{3}$$ now comparing both sides, as values are same so $$\frac{1}{3}$$ = is equal to $$\frac{4}{12}$$ or $$\frac{1}{3}$$ = is equal to $$\frac{1}{3}$$. ### Multiply Whole Numbers and Fractions Performance Task Question 1. You use identical glass jars, colored water, and a spoon to learn about sounds. When you tap a spoon on a jar, the vibrations make a sound. The jar that has the least amount of water makes the lowest sound. a. You fill each glass using a $$\frac{3}{4}$$-cup measuring cup. Complete the table to find the total amount of water in each jar. b. How much more water is in the purple jar than the green jar? c. How many cups of water are used in all? d. Each jar can hold 4 cups of water. Is it possible to add another $$\frac{3}{4}$$ cup of water to the purple jar? Explain. e. Which jars are more than half full? f. You add another $$\frac{3}{4}$$ cup of water to the green jar. How does that affect the sound? a. b. Purple jar is 3 cups more than the Green jar or Purple jar is $$\frac{12}{4}$$ cups more than the Green jar. c. Total number of cups of water used in all is $$\frac{45}{4}$$ cups or 11 $$\frac{1}{4}$$ cups. d. No, it is not possible to add another $$\frac{3}{4}$$ cup of water to the purple jar as if we add it will become $$\frac{18}{4}$$ cups = 4 $$\frac{2}{4}$$ = 4 + $$\frac{2}{4}$$ cups of water to the purple jar cups which is more than 4 cups it will overflow so it is not possible to add another $$\frac{3}{4}$$ cup of water to the purple jar. e. Jars which are more than half full are Orange, Red and Purple. f. After adding add another $$\frac{3}{4}$$ cup of water to the green jar it becomes $$\frac{6}{4}$$ which is equal to blue color jar, we know the jar that has the least amount of water makes the lowest sound before it was the green jar which had least amount of water so it had lowest sound, so now after adding $$\frac{3}{4}$$ cup of water to green jar it has become equal to blue jar so now both green and blue jar has least amount of water so they both make the lowest sound. Explanantion: a. Given we fill each glass $$\frac{3}{4}$$-cup measuring cup so total water cups used for each color are Green 1 means 1 X $$\frac{3}{4}$$ cup = $$\frac{3}{4}$$ cup Blue 2 of $$\frac{3}{4}$$ = 2 X $$\frac{3}{4}$$ = 2 X 3 X $$\frac{1}{4}$$ = 6 X $$\frac{1}{4}$$= $$\frac{6}{4}$$ cups , as both numerator and denominator can be divided by 2 we get $$\frac{3}{2}$$ cups Orange 3 of $$\frac{3}{4}$$ = 3 X $$\frac{3}{4}$$ = 3 X 3 X $$\frac{1}{4}$$ = 9 X $$\frac{1}{4}$$ = $$\frac{9}{4}$$ cups Red 4 of $$\frac{3}{4}$$ = 4 X $$\frac{3}{4}$$ = 4 X 3 X $$\frac{1}{4}$$ = 12 X $$\frac{1}{4}$$ = $$\frac{12}{4}$$ as both numerator and denominator can be divided by 4 we get numerator as 3 and denominator 1 making Red 4 of $$\frac{3}{4}$$ = 3 cups. Purple 5 of $$\frac{3}{4}$$ = 5 X $$\frac{3}{4}$$ = 5 X 3 X $$\frac{1}{4}$$ = 15 X $$\frac{1}{4}$$ = $$\frac{15}{4}$$ cups. b. To calculate how much more water is in the Purple jar than the Green jar we subtract the quantity of purple jar minus quantity in green jar as $$\frac{15}{4}$$ cups – $$\frac{3}{4}$$ cup, as denominators are same for the both we will sustract numerators 15-3 = 12 and denominator remains the same as 4, $$\frac{12}{4}$$ as both numerator and denominator can be divided by 4 we get numerator as 3 and denominator 1 so Purple Jar is 3 cups more than the Green jar. c. Total number of cups of water are used in all is adding Green, Blue, Orange, Red, Purple – $$\frac{3}{4}$$ + $$\frac{6}{4}$$ + $$\frac{9}{4}$$ + $$\frac{12}{4}$$ + $$\frac{15}{4}$$ as denominators are same we add numerators as 3 + 6 + 9 + 12 + 15 = 45 and denominator being same as 4 = $$\frac{45}{4}$$ as numerator is greater we can also write in mixed fraction as 4 goes 11 ,X 11 is 44, 11 will be our whole number and (45−44) we have 1 left over. 1 will be our numerator and our denominator will be the same 4, making as 11 $$\frac{1}{4}$$, therfore total number of cups of water used in all is $$\frac{45}{4}$$ cups or 11 $$\frac{1}{4}$$ cups. d. We got Purple jar contains $$\frac{15}{4}$$ cups to this we will add another $$\frac{3}{4}$$ cup of water = $$\frac{15}{4}$$ + $$\frac{3}{4}$$ as denominator is same we add numerators as 15 + 3 = 18 and denominator is same 4 so now the Purple jar contains $$\frac{18}{4}$$ cups , to find the whole value we write in mixed fraction as 4 goes , 4 X 4 is 16, 4 will be our whole number and (18−16) we have 2 left over. 2 will be our numerator and our denominator will be the same 4, making as 4 $$\frac{2}{4}$$ = 4 + $$\frac{2}{4}$$ and given that each jar can hold 4 cups of water but as we are getting more than 4 it will overflow, so it is not possible to add another $$\frac{3}{4}$$ cup of water to the purple jar. e. Jars more than half full means each jars should be above 2 cups means to check we need denominator to be same that is 4 so we multiply denominator and numerator by 4, we get $$\frac{8}{4}$$ to be more than half full means each jars numerators should be more than 8 and all jars denominators being same as 4, now we check with green jar, $$\frac{8}{4}$$ and $$\frac{3}{4}$$ as 3 is not greater than 8 so not green jar, blue jar we have $$\frac{6}{4}$$ cups here also numerator 6 is not more than 8 so even we will not conside blue jar, orange jar $$\frac{9}{4}$$ now here numerator 8 is less than 9 so we will consider orange jar, Now red jar as red jar has 3 cups which is above 2 cups we consider red jar and we have purple jar as $$\frac{15}{4}$$ here numerator 15 is more than numerator 8 so we consider purple jar, therefore jars with more than half full are Orange, Red and Purple. f. Now we will add another $$\frac{3}{4}$$ cup of water to the green jar means $$\frac{3}{4}$$ + $$\frac{3}{4}$$ as denominators as same we add numerators we get $$\frac{6}{4}$$ which is equal to blue color jar, we know the jar that has the least amount of water makes the lowest sound before it was the green jar which had least amount of water, so it had lowest sound, so now after adding $$\frac{3}{4}$$ cup of water to green jar it has become equal to blue jar so now both green and blue jar has least amount of water so they both make the lowest sound. ### Multiply Whole Numbers and Fractions Activity Three In a Row: Fraction Multiplication Directions: 1. Players take turns. 2. On your turn, spin both spinners. 3. Multiply the whole number and the fraction or mixed number. Cover the product. 5. The ﬁrst player to get three in a row wins! ### Multiply Whole Numbers and Fractions Chapter Practice 9.1 Understand Multiples of Unit Fractions Write the fraction as a multiple of a unit fraction. Question 1. $$\frac{3}{10}$$= $$\frac{1}{10}$$ + $$\frac{1}{10}$$ + $$\frac{1}{10}$$ = 3 X $$\frac{1}{10}$$. Explanation: We write $$\frac{3}{10}$$ in addition equation as $$\frac{3}{100}$$= $$\frac{1}{10}$$ + $$\frac{1}{10}$$ + $$\frac{1}{10}$$ and multiplication as unit fraction $$\frac{3}{10}$$ represents 3 parts that are $$\frac{1}{10}$$ of the whole, $$\frac{3}{10}$$ = 3 X $$\frac{1}{10}$$ Question 2. $$\frac{4}{8}$$= $$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$ = 4 X $$\frac{1}{8}$$. Explanation: We write $$\frac{4}{8}$$ in addition equation as $$\frac{4}{8}$$= $$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$ + $$\frac{1}{8}$$ and multiplication as unit fraction $$\frac{4}{8}$$ represents 4 parts that are $$\frac{1}{8}$$ of the whole, $$\frac{4}{8}$$ = 4 X $$\frac{1}{8}$$ Question 3. $$\frac{7}{2}$$ $$\frac{7}{2}$$ = $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 7 X $$\frac{1}{2}$$. Explanation: we write $$\frac{7}{2}$$ in addition equation as $$\frac{7}{2}$$ = $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 7 X $$\frac{1}{2}$$ and multiplication as unit fraction $$\frac{7}{2}$$ represents 7 parts that are $$\frac{1}{2}$$ of the whole, $$\frac{7}{2}$$ = 7 X $$\frac{1}{2}$$. Question 4. $$\frac{56}{100}$$ $$\frac{56}{100}$$ = $$\frac{14}{25}$$ = $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ = 14 X $$\frac{1}{25}$$ Explanation: $$\frac{56}{100}$$ first we simplify as both numerator and denominator can be divided by 4 we get numerator as 14 and denominatior as 25 = $$\frac{14}{25}$$ now we write $$\frac{14}{25}$$ in addition equation as $$\frac{14}{25}$$ = $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ + $$\frac{1}{25}$$ = 14 X $$\frac{1}{25}$$ and multiplication as unit fraction $$\frac{14}{25}$$ represents 14 parts that are $$\frac{1}{25}$$ of the whole, $$\frac{14}{25}$$ = 14 X $$\frac{1}{25}$$ . Question 5. Logic What is Newton’s fraction? Write the fraction as a multiple of a unit fraction. Newton’s fraction is 5 X $$\frac{1}{6}$$, The fraction is already a multiple of a unit fraction = 5 X $$\frac{1}{6}$$ , 5 is multiple and $$\frac{1}{6}$$ is a unit fraction. Explanation: Given unit fraction is one sixth of a whole is $$\frac{1}{6}$$ and Newton’s fraction is 5 unit fractions means it is equal to $$\frac{1}{6}$$ + $$\frac{1}{6}$$ + $$\frac{1}{6}$$ + $$\frac{1}{6}$$ + $$\frac{1}{6}$$ = 5 X $$\frac{1}{6}$$ already the fraction is a multiple of a unit fraction = 5 X $$\frac{1}{6}$$ , 5 is multiple and $$\frac{1}{6}$$ is a unit fraction. 9.2 Understand Multiples of Fractions Write the product as a multiple of a unit fraction. Then find the product. Question 6. 2 X $$\frac{2}{4}$$ = 2 X 2 X $$\frac{1}{4}$$ = 4 X $$\frac{1}{4}$$ = $$\frac{4}{4}$$= 1 Explanation: First we write $$\frac{2}{4}$$ as $$\frac{1}{4}$$ + $$\frac{1}{4}$$ = 2 X $$\frac{1}{4}$$ then multiply it by 2 which is 2 X 2 X $$\frac{1}{4}$$ = 4 X $$\frac{1}{4}$$ = $$\frac{4}{4}$$ as numerator and denominator are same both get cancelled so we get 1 , $$\frac{4}{4}$$ = 1. Question 7. 3 X $$\frac{9}{12}$$ = 3 X $$\frac{3}{4}$$ = 9 X $$\frac{1}{4}$$ = $$\frac{9}{4}$$ Explanantion: As $$\frac{9}{12}$$ can be simplified as both 9,12 can go with 3 leaving numerator as 3 and denominator as 4= $$\frac{3}{4}$$, First we write $$\frac{3}{4}$$ as $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ = 3 X $$\frac{1}{4}$$ then multiply it by 3 which is 3 X 3 X $$\frac{1}{4}$$ = 9 X $$\frac{1}{4}$$ is multiple of unit fraction, The product is $$\frac{9}{4}$$ . Question 8. $$\frac{3}{5}$$ X 4 = 12 X $$\frac{1}{5}$$ = $$\frac{12}{5}$$. Explanation: First we write $$\frac{3}{5}$$ as $$\frac{1}{5}$$ + $$\frac{1}{5}$$ + $$\frac{1}{5}$$ = 3 X $$\frac{1}{5}$$ then multiply it by 4 which is 4 X 3 X $$\frac{1}{5}$$ = 12 X $$\frac{1}{5}$$ = $$\frac{12}{5}$$. Question 9. $$\frac{8}{10}$$ X 7= $$\frac{4}{5}$$ X 7 = 28 X $$\frac{1}{5}$$ is multiple of unit fraction, The product is $$\frac{28}{5}$$. Explanantion: As $$\frac{8}{10}$$ can be simplified as both 8,10 can go with 2, leaving numerator as 4 and denominator as 5 = $$\frac{4}{5}$$, First we write $$\frac{4}{5}$$ as $$\frac{1}{5}$$ + $$\frac{1}{5}$$ + $$\frac{1}{5}$$ + $$\frac{1}{5}$$ = 4 X $$\frac{1}{5}$$ then multiply it by 7 which is 7 X 4 X $$\frac{1}{5}$$ = 28 X $$\frac{1}{5}$$ is multiple of unit fraction, The product is $$\frac{28}{5}$$. Question 10. 8 X $$\frac{6}{3}$$ = 8 X 2 = 16 as we got $$\frac{6}{3}$$ =2, So we get results as whole not in fraction. Explanation: As $$\frac{6}{3}$$ can be simplified as both can be divided by 2, leaving numerator as 2 and denominator as 1 , $$\frac{6}{3}$$ = 2 , now it is 8 X 2 = 16 as we got $$\frac{6}{3}$$ =2, so we get results as whole not in fraction. Question 11. 10 X $$\frac{30}{8}$$ = 10 X $$\frac{15}{4}$$ = 150 X $$\frac{1}{4}$$ = $$\frac{150}{4}$$ = $$\frac{75}{2}$$. Explanation: As $$\frac{30}{8}$$ can be simplified as both can be divided by 2 , we get numerator as 15 and denominator as 4, $$\frac{30}{8}$$ = $$\frac{15}{4}$$ = $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ = 15 X $$\frac{1}{4}$$ now multiply with 10 X 15 X $$\frac{1}{4}$$ = 150 X $$\frac{1}{4}$$ = $$\frac{150}{4}$$ this can be further simplified as 150 and 4 both can be further divided by 2 making numerator as 75 and denominator as 2 = $$\frac{75}{2}$$ . 9.3 Multiply Whole Numbers and Fractions Multiply Question 12. 2 X $$\frac{1}{2}$$ = $$\frac{2}{2}$$ = 1 Explanation: Already it is in unit fraction so 2 X $$\frac{1}{2}$$ = $$\frac{2}{2}$$ = 1. Question 13. 4 X $$\frac{5}{8}$$ = $$\frac{20}{8}$$ = $$\frac{10}{4}$$ = $$\frac{5}{2}$$. Explanation : First we write $$\frac{5}{8}$$ in unit fraction as 5 X $$\frac{1}{8}$$ now we multiply with whole 4 X 5 X $$\frac{1}{8}$$ = $$\frac{20}{8}$$ further can be simplifed as 20,8 both can be divided by 4 we get numerator as 5 and denominator 2, therefore 4 X $$\frac{5}{8}$$ = $$\frac{5}{2}$$ . Question 14. 3 X $$\frac{9}{6}$$ = $$\frac{9}{2}$$. Explanantion: We can simplify $$\frac{9}{6}$$ as both goes by 3 giving numerator as 3 and denominator as 2 so we get $$\frac{9}{6}$$ = $$\frac{3}{2}$$ now we write in unit fraction as 3 X $$\frac{1}{2}$$ now we multiply by 3, 3 X 3 X $$\frac{1}{2}$$ = 9 X $$\frac{1}{2}$$ = $$\frac{9}{2}$$. Question 15. 5 X $$\frac{7}{12}$$ = $$\frac{35}{12}$$. Explanation: first we write $$\frac{7}{12}$$ as unit fraction = 7 X $$\frac{1}{12}$$, now we multiply with 5 as 5 X 7 X $$\frac{1}{12}$$ = 35 X $$\frac{1}{12}$$ = $$\frac{35}{12}$$ . Question 16. 7 X $$\frac{30}{100}$$ = 7 X $$\frac{3}{10}$$ = $$\frac{21}{10}$$. Explanation: We can simplify $$\frac{30}{100}$$ as both goes by 10 giving numerator as 3 and denominator as 10 so we get $$\frac{3}{10}$$ now we write in unit fraction as 3 X $$\frac{1}{10}$$ and multiply by 7 , 7 X 3 X $$\frac{1}{10}$$ = 21 X $$\frac{1}{10}$$ = $$\frac{21}{10}$$ . Question 17. $$\frac{8}{4}$$ X 9 = 2 X 9 = 18. Explanantion: We can simplify $$\frac{8}{4}$$ as both goes by 4 giving numerator as 2 and denominator as 1 so we get 2 as whole, now we multiply with 9 we get 2 X 9 = 18. 9.4 Multiply Whole Numbers and Mixed Numbers Multiply Question 18. = $$\frac{5}{2}$$= 2 $$\frac{1}{2}$$ Explanation: First we write the mixed number as a fraction then multiply, so 1 $$\frac{1}{4}$$= 1 + $$\frac{1}{4}$$= (1 X 4 + 1) by 4 = $$\frac{5}{4}$$ now we multiply it with the whole, 2 X $$\frac{5}{4}$$ now we write in unit fraction and multiply 2 X 5 X $$\frac{1}{4}$$= 10 X $$\frac{1}{4}$$ = $$\frac{10}{4}$$ further we can simplify as 10,4 both can be divided by 2 we get numerator as 5 and denominator as 2, $$\frac{5}{2}$$ as numerator is greater than denominator we can write in mixed fraction as 2 goes in 2 times. 2 will be our whole number 2 X 2 is 4 and we have 1 left over (54), 1 will be our numerator and our denominator will stay the same 2, $$\frac{5}{2}$$= 2 $$\frac{1}{2}$$. Question 19. = $$\frac{69}{6}$$ = $$\frac{23}{2}$$ = 11 $$\frac{1}{2}$$ Explanation: First we write the mixed number as a fraction then multiply, so 3 $$\frac{10}{12}$$= ( 3 X 12 + 10 ) by 12 = $$\frac{46}{12}$$ this can be further simplified as both 46,12 goes by 2 we get numerator as 23 and denominator as 6 , $$\frac{23}{6}$$ now we multiply by 3 = 3 X $$\frac{23}{6}$$ = $$\frac{69}{6}$$ it can be more simplified as both 69,6 goes by 3 we get numerator as 23 and denominator as 2 making $$\frac{69}{6}$$ as $$\frac{23}{2}$$ as numerator is greater than denominator we can write in mixed fraction as 2 goes in 11 times. 11 will be our whole number 2 X 11 is 22 and we have 1 left over (23−22), 1 will be our numerator and our denominator will stay the same 2, $$\frac{23}{2}$$ = 11 $$\frac{1}{2}$$. Question 20. = $$\frac{84}{8}$$ = $$\frac{21}{2}$$ = 10 $$\frac{1}{2}$$. Explanantion: First we write the mixed number as a fraction then multiply, so 2 $$\frac{5}{8}$$ = ( 2 X 8 + 5 ) by 8 = $$\frac{21}{8}$$ now we multiply by 4 we get 4 X $$\frac{21}{8}$$ = 4 X 21 X $$\frac{1}{8}$$ = 84 X $$\frac{1}{8}$$ =$$\frac{84}{8}$$ further can be simplified as both 84,8 goes by 4 we get numerator as 21 and denominator as 2 =$$\frac{21}{2}$$ as numerator is greater than denominator we can write in mixed fraction as 2 goes in 10 times. 10 will be our whole number 2 X 10 is 20 and we have 1 left over (21−20), 1 will be our numerator and our denominator will stay the same 2, $$\frac{21}{2}$$ = 10 $$\frac{1}{2}$$. Question 21. = $$\frac{40}{3}$$ = 13 $$\frac{1}{3}$$ Explanation: First we write the mixed number as a fraction then multiply, so 2 $$\frac{4}{6}$$ = ( 2 X 6 + 4 ) by 6 = $$\frac{16}{6}$$ this can be simplified as 16,6 both goes by 2 we get numerator as 8 and denominator as 3 we get $$\frac{8}{3}$$ now we multiply by 5 we get 5 X $$\frac{8}{3}$$ = 5 X 8 X $$\frac{1}{3}$$ = 40 X $$\frac{1}{3}$$ = $$\frac{40}{3}$$ as numerator is greater than denominator we can write in mixed fraction as 3 goes in 13 times. 13 will be our whole number 3 X 13 is 39 and we have 1 left over (40−39), 1 will be our numerator and our denominator will stay the same 3, we get mixed fraction as 13 $$\frac{1}{3}$$ . Question 22. = $$\frac{140}{3}$$ = 46 $$\frac{2}{3}$$ Explanation: First we write the mixed number as a fraction then multiply, so 6 $$\frac{2}{3}$$ = ( 6 X 3 + 2 ) by 3 = $$\frac{20}{3}$$ now we multiply by 7 X $$\frac{20}{3}$$ = 7 X 20 x $$\frac{1}{3}$$ = 140 X $$\frac{1}{3}$$ =$$\frac{140}{3}$$ as numerator is greater than denominator we can write in mixed fraction as 3 goes in 46 times. 46 will be our whole number 3 X 46 is 138 and we have 2 left over (140−138), 2 will be our numerator and our denominator will stay the same 3, we get mixed fraction as 46 $$\frac{2}{3}$$ . Question 23. = $$\frac{470}{5}$$ = 94 Explanation: First we write the mixed number as a fraction then multiply, so 9 $$\frac{2}{5}$$ = ( 9 X 5 + 2 ) by 5 = $$\frac{47}{5}$$ now we multiply by 10 = 10 X $$\frac{47}{5}$$ = $$\frac{470}{5}$$ this can be further simplified as both can be divided by 5 we get numerator as 94 and denominator as 1 so we get only whole number 94. 9.5 Problem Solving: Fraction Operations Question 24. Modeling Real Life Your friend rides her bike $$\frac{2}{6}$$ mile to your house and $$\frac{2}{6}$$ mile back home 4 times in 1 week. How far does she bike in all? My friend rides her bike by $$\frac{8}{3}$$ miles or 2 $$\frac{2}{3}$$ miles. Explanation: Given friend rides her bike $$\frac{2}{6}$$ mile to your house and $$\frac{2}{6}$$ mile back home, first we calculate for 1 time, so first we will add $$\frac{2}{6}$$ + $$\frac{2}{6}$$ as both have same denominators we will add numerators as ( 2 + 2 = 4 ). So we get $$\frac{4}{6}$$ miles for 1 time , now she rides 4 times means 4 X $$\frac{4}{6}$$ = $$\frac{16}{6}$$ this can be further simplified as 16,6 both goes by 2 we get numerator as 8 and denominator as 3 = $$\frac{8}{3}$$ as numerator is greater than denominator we can write in mixed fraction as 3 goes in 2 times. 2 will be our whole number 3 X 2 is 6 and we have 2 left over (8−6), 2 will be our numerator and our denominator will stay the same 3, we get mixed fraction as 2 $$\frac{2}{3}$$ miles. Conclusion: Access the links given in the Big Ideas Math Book 4th Grade Answer Key Chapter 9 Multiply Whole Numbers and Fractions and practice well for the exams. We have solved all the questions in an easy and simple manner. We hope the information provided in this article has brought a smile to your face. Share Big Ideas Math Grade 4 Answer Key Chapter 9 Multiply Whole Numbers and Fractions.
# 6.10 Normal distribution: normal distribution lab ii Page 1 / 1 Class Time: Names: ## Student learning outcomes: • The student will compare empirical data and a theoretical distribution to determine if data from the experiment follow a continuous distribution. ## Collect the data Measure the length of your pinkie finger (in cm.) 1. Randomly survey 30 adults. Round to the nearest 0.5 cm. _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ 2. Construct a histogram. Make 5-6 intervals. Sketch the graph using a ruler and pencil. Scale the axes. 3. Calculate the Following • $\overline{x}=$ • $s=$ 4. Draw a smooth curve through the top of the bars of the histogram. Use 1-2 complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a V-shape, does it have a hump in the middle or at either end, etc.?) ## Analyze the distribution Using your sample mean, sample standard deviation, and histogram to help, what was the approximate theoretical distribution of the data from the section titled "Collect the Data"? • $X$ ~ • How does the histogram help you arrive at the approximate distribution? ## Describe the data Using the data in the section titled "Collect the Data" complete the following statements. (Hint: order the data) $\left(\mathrm{IQR}=Q3-Q1\right)$ • IQR = • 15th percentile is: • 85th percentile is: • Median is: • What is the empirical probability that a randomly chosen pinkie length is more than 6.5 cm? • Explain the meaning the 85th percentile of this data. ## Theoretical distribution Using the Theoretical Distribution in the section titled "Analyze the Distribution" • IQR = • 15th percentile is: • 85th percentile is: • Median is: • What is the theoretical probability that a randomly chosen pinkie length is more than 6.5 cm? • Explain the meaning of the 85th percentile of this data. ## Discussion questions • Do the data from the section entitled "Collect the Data" give a close approximation to the theoretical distribution in "Analyze the Distribution."In complete sentences and comparing the results in the sections titled "Describe the Data" and "Theoretical Distribution", explain why or why not. where we get a research paper on Nano chemistry....? what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe Got questions? Join the online conversation and get instant answers!
# Lesson Notes By Weeks and Term - Primary 3 Counting and writing numerals up to 1000 - 9999 TERM: 1st Term WEEK: 2 CLASS: Primary 3 AGE: 8 years DURATION: 5 periods of 40 minutes each DATE: SUBJECT: Mathematics TOPIC: Counting of whole numbers 1000-9999 SPECIFIC OBJECTIVES: At the end of the lesson, the pupils should be able to 1. Write the numerals 1000 2. Count in thousands up to 9000 3. Identify the place value i.e. Th, H, T, U in numbers from 1000-9000 4. Count in Thousands(Th), Hundreds(H), Tens(T) and Units(U) INSTRUCTIONAL TECHNIQUES: Explanation, question and answer, demonstration, practical INSTRUCTIONAL MATERIALS: 100 number board, scrap paper, videos from source- https://www.youtube.com/watch?v=pN_xu6Oirdg PERIOD 1: Multiples of 10 PRESENTATION TEACHER’S ACTIVITY PUPIL’S ACTIVITY STEP 1MENTAL MATHS The teacher begins the lesson with some mental calculationsCount in multiples of 201. 20, 40………2. 200, 220………..3. 140, 160…………4. 520, 540, ………….5. 460, 480…………… Pupils respond and participate STEP 2CONCEPTDEVELOPMENT The teacher explains that sometimes when base 10 kits are unavailable, they could make use of some drawings and shapes to represent their large numbersFor examples570 could be represented thus, She asks the learners to repeat the process for 350, 490, 600, 910 CLASS ACTIVITYThe pupils are taught how to use number line to add in tens.The teacher draws an empty number line on the board. She demonstrates that if you want to start at 306, you need to write 306 on the number line Pupils pay attention and participate STEP 3CLASS-WORK 1. Complete these patterns of 10:a. 670, 680, _____, _____, _____, _____, 730b. 483, 493, _____, _____, _____, _____, 543c. 567, 577, _____, _____, _____, _____, _____, 637d. 734, 744, _____, ______, _____, _____, _____, _____, 814e. 333, 343, _____, _____, _____, _____, _____, ____, 413 2. Use your numbered counters to complete the followingThere are ______ tens in 180There are ______ tens in 230There are ______ tens in 420There are _____ tens in 150There are _____ tens in 350There are _____ tens in 410 3. On a number line, draw a number line starting at 700 and going to 800. On the number line, show how you will count in tens from 700-800 Pupils attempt their class work STEP 4HOME-WORK 1. Complete these patterns:a. 228, 238, _____, _____, _____, ____, ____, _____, _____b. 193, 203, ______, _____, ______, _____, _____, _____c. 314, 324, _____, _____, ______, ______, ______, _____, _____d. 456, 466, ______, _____, _____, ____, _____e. 515, 525, _____, ____, _____, _____, _____, _____, _____, ____ 2. How many tens are in the following?a. 230b. 190c. 460d. 550e. 330 3. Starting from 800 on a number line, show how you count in tens to 900. The pupils writes it in their homework book STEP 5SUMMARY The teacher summarizes by reminding the pupils that for large counters or items, one by one counting shouldn’t be used rather they should be counted in multiples in this case 10 counters make up one 10, and 10 tens make up 100 She marks their class works, makes corrections where necessary and commends them positively PERIOD 2: More on numbers up to 999 PRESENTATION TEACHER’S ACTIVITY PUPIL’S ACTIVITY STEP 1 ORAL ASSESSMENTS The teacher asks questions on numbers up to 999 treated so far Pupils respond and participate STEP 2 DISCUSSION The teacher discusses all the methods used by some learners in the oral assessments(some of the questions are solved on the board by the learners) and addresses any misconceptions that may have risen Pupils pay attention and participate STEP 3 WRITTEN ASSESSMENTS 1. Use flash cards to make these numbers a. 3 hundreds, 6 tens and 0 ones. ______ b. 8 hundreds, 0 tens and 5 ones. ______ c. 4 hundreds, 1 ten and 2 ones _______ 2. Fill in the missing numbers in the following 1 836 __hundreds __tens __ones 2 620 __hundreds __tens __ones 3 409 __hundreds __tens __ones 3. Write the following as number symbols a. four hundred and sixteen b. five hundred and sixty four 4. Write the following as number names a. 235 b. 999 5. Complete these patterns of 10 a. 260, 270, _____, _____, _____, ____, _____, _____, _____ b. 620, 610, _____, ______, _____, _____, _____, _____, _____ 6. Use your numbered counters to complete the following There are _____tens in 410 There are _____tens in 620 There are _____tens in 180 There are _____tens in 250 Pupils attempt their class work STEP 4 SUMMARY The teacher marks the written assessments, corrects were necessary and commends the pupils PERIOD 3: The number 1000 PRESENTATION TEACHER’S ACTIVITY PUPIL’S ACTIVITY STEP 1 MENTAL MATHS The teacher begins the lesson with some mental calculations Count forwards and backwards from any multiples of 25 Pupils respond and participate STEP 2 CONCEPT DEVELOPMENT The teacher presents 999 using the base ten kits on the board. She explains that if one extra counter was added, the number will become 10 hundreds which is called one thousand(1000) She asks some learners to write the number symbols 100, 200, 300, 400, 500, 600, 700, 800 and 900 on the board. She then carefully states that 1000 as opposed to the others is not called ten hundred, rather it is called 1000 ACTIVITY Pupils in pairs complete the below table 100 500 1000 910 920 1000 991 996 997 300 900 800 900 1000 1000 700 Pupils pay attention and participate STEP 3 CLASS-WORK 1. Fill in the missing numbers 901 903 904 905 907 908 909 911 912 913 915 916 917 919 920 921 923 924 925 927 928 929 931 932 933 935 936 937 939 940 941 943 944 945 947 948 949 951 952 953 955 956 957 959 960 961 963 964 965 967 968 969 971 972 973 975 976 977 979 980 981 982 983 984 985 987 988 989 991 992 993 995 996 997 999 1000 2. Complete the following a. The number that is 10 more than 990 ______ b. The number that is 1 less than 1000 ______ c. The number that is 1000 more than 900 ______ d. The number that is 300 less than 1000 ______ Pupils attempt their class work STEP 4 HOME-WORK 1. Fill in the missing numbers 100 200 1000 990 1000 994 995 996 1000 300 The pupils writes it in their homework book STEP 5 SUMMARY The teacher summarizes by reminding the pupils that 1000 is called one thousand. She also reminds them how to add in 10s and 100s PERIOD 4: Numbers up to 1000 PRESENTATION TEACHER’S ACTIVITY PUPIL’S ACTIVITY STEP 1 MENTAL MATHS The teacher begins the lesson with some mental calculations What is 10 more than 1. 551 2. 843 3. 777 4. 963 5. 248 6. 737 7. 371 8. 440 9. 923 10. 854 Pupils respond and participate STEP 2 CONCEPT DEVELOPMENT The teacher explores all the ways 1000 can be represented How many hundreds are in 1000? How many tens are in 1000? How many ones are in 1000? If I added 10 to 1000, what will it be? If 100 is taken from 1000, what will we be left with? CLASS ACTIVITY The pupils are given flash cards containing 3-digit numbers. (752, 498, 265, 333, 610) In pairs, learners identify the 3-digit numbers, represents them in pictures and writes them in expanded notation(637= 600+ 30 + 7) Pupils pay attention and participate STEP 3 CLASS-WORK 1. Fill in the missing numbers 491 __hundreds __tens __ones 638 __hundreds __tens __ones 945 __hundreds __tens __ones 550 __hundreds __tens __ones 212 __hundreds __tens __ones 2. Fill in the missing numbers a. 650 is ____ less than 700 b. 780 is ____ less than 800 c. 940 is ____ more than 900 d. 830 is ____ more than 900 e. 370 is ____ less than 400 Pupils attempt their class work STEP 4 HOME-WORK 1. Fill in the missing numbers a. 310 is _____ more than 300 b. 630 is _____ more than 600 c. 580 is _____ less than 600 d. 260 is _____ less than 300 e. 880 is _____ less than 900 2. Fill in the missing numbers 356 ___hundreds ___tens ___ones 278 ___hundreds ___tens ___ones 567 ___hundreds ___tens ___ones 679 ___hundreds ___tens ___ones 780 ___hundreds ___tens ___ones The pupils writes it in their homework book STEP 5 SUMMARY The teacher summarizes by reminding the pupils that 1000 is called one thousand. She also reminds them how to add and subtract in 10s and 100s PERIOD 5: Weekly Test/consolidations TEACHER’S ACTIVITY: The teacher revises all the concepts treated from period 1-4 and gives the pupils follow through exercises, quiz and tests from worksheets of numbers up to 999. She marks the exercises, makes corrections and commends the pupils positively. PUPIL’S ACTIVITY: The pupils work on the worksheets and exercises given by the teacher individually ASSESSMENT 1 If there are 10 sticks in one bundle , how many sticks are in: a 3 bundles _____________ b 7 bundles _____________ c 12 bundles _____________ d 9 bundles _____________ e 15 bundles _____________ f 8 bundles _____________ g 26 bundles _____________ h 35 bundles _____________ 1. Fill in the blanks. a 70 , 80, 90, _____, ______, ______, _____, _____, _____, ____ b 350, 340, 330, _____, _____, _____, _____, _____, 260. c _____ , _____ , _____ , _____ , _____ , 910 , _____ , _____ , 940 , _____. 1. Fill in the missing numbers 901 903 904 905 907 908 909 911 912 913 915 916 917 919 920 921 923 924 925 927 928 929 931 932 933 935 936 937 939 940 941 943 944 945 947 948 949 951 952 953 955 956 957 959 960 961 963 964 965 967 968 969 971 972 973 975 976 977 979 980 981 982 983 984 985 987 988 989 991 992 993 995 996 997 999 1000 4 Fill in the missing numbers: a 321 has ____hundreds ____ tens ____ one b 501 has ____hundreds ____ tens ____ one c 789 has ____hundreds ____ tens ____ ones d 650 has ____hundreds ____ tens ____ ones e 209 has ____hundreds ____ tens ____ ones f 920 has ____hundreds ____ tens ____ ones g 444 has ____hundreds ____ tens ____ ones h 817 has ____hundreds ____ tens ____ ones 5 Draw base ten kit pictorials to show the following numbers(squares for hundreds, lines for tens and dots for ones) a 423 b 591 c 605
# Application of LCM and HCF Application of LCM and HCF :- ### Problems Question 1. What is the greatest number that will divide 2930 and 3250 and will leave as remainders 7 and 11 respectively? Solution: The greatest number will be the HCF of (2930 – 7) and (3250 – 11), i.e. 79. Question 2. What is the least number by which 825 must be multiplied in order to produce a multiple of 715? Solution: 825 = 3 × 5 × 5 ×11; 715 = 5 × 11 × 13 Any multiple of 715 must have factors of 5, 11 and 13. So, 825 should be multiplied by the factor(s) of 715, which is (are) not present in 825. ∴ The required least number = 13 Question 3. The LCM of two numbers is 2310 and their HCF is 30. If one of the numbers is 7 × 30, find the other number. Solution: the required number = (2310 ×30)/(7 ×30) = 330. Question 4. Three bells commence tolling together and they toll after 0.25, 0.1 and 0.125 seconds. After what interval will they again toll together? Solution: They will toll together after an interval of time equal to the LCM of 0.25 sec, 0.1 sec and 0.125 sec. LCM of 0.25, 0.1 and 0.125 = (LCM of 250, 100 and 125) × 0.001 = 500 × 0.001 = 0.5 sec. Question 5. What is the smallest sum of money which contains Rs. 2.50, Rs. 20, Rs. 1.20 and Rs. 7.50? Solution: LCM of 2.5, 20, 1.2 and 7.5 = (LCM of 25, 200, 12 and 75) Here the highest number of decimal is 1 so we multiply each by 10 to get 25, 200, 12 and 75. Their LCM is = 600 but since we multiplied by 10 so bring back the original value we need to divide it by 10 so the required solution = 600/10= 60 Question 6. What is the greatest number which will divide 410, 751 and 1030 so as to leave the remainder 7 in each case? Solution: the required number will be the HCF of (410-7), (751 – 7), (1030 – 7) i.e. 31. Question 7. What is the HCF of 4/5 , 5/6 and 7/15 ? Question 8. Three men start together to travel the same way around a circular track of 11 km. Their speeds are 4, 5.5 and 8 km per hour respectively. When will they meet at the starting point? Question 9. How many times is the HCF of 48, 36, 72 and 24 contained in their LCM?Solution: HCF of 48, 36, 72, 24 = 12; LCM of 48, 36, 72, 24 = 144 LCM = 12 × HCF. (12x12 =144) Question 10. Find the least square number which is exactly divisible by 4, 5, 6, 15 and 18.Solution: LCM of 4, 5, 6, 15 and 18 = 180, which is exactly divisible by the given numbers. 180 = 2 × 2 × 3 × 3 × 5 You can see that 2 and 3 are paired but 5 is not . To make it paired we need to multiply by 5 so that the whole multiplication becomes a perfect square. Therefore, if 180 is multiplied by 5 (180 × 5 = 900) and 900 will be a perfect square as well as divisible by 4, 5, 6, 15 and 18. Question 11. Find the sum of a numbers between 300 and 400 such that when they are divided by 6, 9 and 12. (i) it leaves no remainder; and (ii) it leaves remainder as 4 in each case. Solution: The LCM of 6, 9, and 12 = 36 (i) Multiples of 36 which lie between 300 and 400 are 324, 360 and 396. ∴ The required least number = 324 + 360 + 396 = 1080 (ii) Here the remainder is 4 in each case So, the numbers are (324 + 4) = 328 and (360 + 4) = 364. (The number 396 + 4 = 400 does not lie between 300 and 400 so it is not acceptable.) ∴ The required sum = 328 + 364 = 692. Smart Prep Kit for Banking Exams by Ramandeep Singh - Download here
# MATD 0385 Day 10 Exponential Growth. ... Go over quiz problems. Lesson 5. Various topics ```MATD 0385 Day 10 Exponential Growth. Feb. 22, 2010 Go over quiz problems. Lesson 5. Various topics Activity 1. p. 261. What if the growth triples instead of doubles? Or decreases by 2/3 in each time period? What do those patterns of growth look like? 1. Sketch a graph of the first two, then look at the one on the projector. 2. Sketch a graph of the last two, then look at the one on the projector. 3. Work with two other students. Each of you write answers for these. The geometric mean, page 263. 1. Answer the first question with pencil in front of it. It asks for two numbers. What are they? 2. Now look at the data in the middle of the page for year and amount. They tell you that it is exponential. But you don’t have to take their word for it. How do you show it is exponential? (Recall “common ratio.” Do that.) 3. Now, since you see that there is a common ratio here, recall how we did Lesson 3, p. 245, problem 4. Find a formula for this problem in the same way. (Let x = number of decades. Then A  2.08x ) 4. So we can find the amount at years = 15 by changing years =15 to x = 1.5. And then A  2.081.5 . Find that. 5. And, as we just learned, we can also find the amount at years = 15 by taking the geometric mean of the amount at years = 10 and the amount at years = 20. Find that. 6. Continue to page 264, verifying that the graph is correct for this set of data. Standardizing on base 10, page 264. 1. Look at the data in this section. 2. Answer the question with a pencil in front. What is the common ratio? 3. Do Activity 3, problem 1. 4. Activity 3, problem 2. Compute three values. See the projected table for the rest and then answer the question at the end of the page. 5. Activity 3, problem 3. Use the projected graph to answer these questions. Standardizing on base e, page 266. Read this. We will not do much with this, but it is sometimes useful to know that this is one of the standard bases used in real-world problems. Homework: pages 267-270: 1, 2, 6, 11, 12, 13, 17, 18 Quiz: pages 267-270: 3, 4, 5, 7, 10 ```
# Common Multiple • Introduction • What Do You Mean by Common Multiples in Mathematics? • How To Calculate the Common Multiples • Common Multiples Between 2 Numbers • Common Multiples Between 3 Numbers • Properties of Common Multiplies • Solved Examples • Practice Problems ## Introduction In primary school maths, we learned about multiples. The concept of multiples is important, particularly when working with Least Common Multiples (LCM). It is also used when identifying the patterns in numbers. In this article, we will discuss Common Multiples and Least Common Multiples (LCM) in depth with appropriate examples and diagrams. ## What Do You Mean by Common Multiples in Mathematics? In mathematics, common multiples are the multiples that are shared by a particular set of numbers. By counting every number, we can determine the multiples of a given number. A multiple in maths refers to the multiplication of two numbers. We are all familiar with multiplication tables while solving maths problems. We also find multiples while using these tables. As you can see, the result of multiplying two numbers is their multiple. We can create a list of the multiples of an integer by multiplying it by natural numbers. Below are multiplication tables, which are representations of the first ten multiples of numbers from 1 to 10. For Example - In order to identify the common multiples of 8 and 9, we can list the multiples of 8 and 9 individually and look for those that appear in both lists. Multiples of 8 can be written as 8, 16, 24, 32, 40, 48, 56, 64, 72 and so on. Multiples of 9 can be written as 9, 18, 27, 36, 45, 54, 63, 72, 81 and so on. If we look for multiples that appear in both the lists, we find 72. This is referred to as the common multiple of 9 and 8. ## How To Calculate the Common Multiples We can figure out the multiples of two or more integers by listing their individual multiples. On a grid of 100, let's indicate the multiples of 7 and 6. The multiples of 7 are marked with a cross and the multiples of 6 are marked with a circle. Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, .... Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, ….. Common Multiples of 7 and 6:42, 84,.... ## Common Multiples Between 2 Numbers It is quite easy to find common multiples of two numbers. List the multiples of both numbers (starting with the first few multiples), as we mentioned in the last example. Search for the multiples that appear in both lists. Example - Find the common multiples of 15 and 6. Common multiples of 15 and 6: 60, ….. ## Common Multiples Between 3 Numbers Common multiples of 3 numbers can be identified by using the Listing Method. Here, all we need to do is to look for multiples that are included in all three lists. In other words, we need to determine the multiples of all the 3 numbers. Example - Find the common multiples of 126, 14, and 18. Common multiples of 126, 14 and 18: 126, ….. ## Properties of Common Multiplies • A number has an unlimited number of multiples. As a result, any two integers or collection of numbers have an indefinite number of common multiples. • The product a\times b of any 2 numbers, a and b, is always a common multiple of a and b. Example : 3\times 4 is a common multiple of 3 and 4. • When two numbers a and b are coprime, the multiples of their product are their common multiples. Example: 2 and 3 are coprime numbers. Common multiples of 2 and 3: 6, 12, 18….. • When two numbers are coprime, their product is equivalent to their LCM (Least Common Multiple). Example: Find the LCM of 5 & 7? 5 and 7 are co-prime numbers. LCM(5, 7) = 5\times 7 = 35 • The LCM of a and b is b if and only if b is a multiple of a. Example: LCM(7, 14) = 14 ## Solved Examples Q1. Find the Least Common Multiple (LCM) of 6 and 9 ? Solution: Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, ….. Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, ….. LCM(6, 9): 18 Q2. Calculate the LCM of 13 & 39. Solution: The LCM of a and b is b if and only if b is a multiple of a. LCM(13, 39) = 39 because 39 is a multiple of 13. Q3. Which number is the first common multiple of 15 and 18? Solution: Multiples of 15: 15, 30, 45, 60, 75, 90, …. Multiples of 18 : 18, 36, 54, 72, 90, 108, ….. First common multiple of 18 & 15: 90 Q4. Which numbers are multiples of one? Solution: Every natural number that exists is a multiple of 1. Any number is divisible by 1. The first ten multiples of 1 are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Q5. Calculate the LCM of 12, 18, and 20. Solution: Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180…. Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, ….. Multiples of 20: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, .... LCM (12,18,20) = 180 ## Practice Problems Q1. Which of the following is the LCM of 9 and 18? 1. 18 2. 27 3. 36 4. None of the above Q2. Which number is the multiple of 15? 1. 30 2. 35 3. 40 4. None of the above Q3. Which number is not a multiple of 12 ? 1. 36 2. 60 3. 72 4. 18 Q4. Which number is not a common multiple of 18 & 12? 1. 36 2. 72 3. 114 4. 108 Q5. Which number is not a multiple of 25? 1. 50 2. 60 3. 100 4. 75 Q1. What is a common multiple? Answer: A common multiple of two or more numbers is a number that is a multiple of each of the given numbers. In other words, it is a number that can be evenly divided by all the numbers in question. Q2. How do you find common multiples of two numbers? Answer: To find common multiples of two numbers, you can list the multiples of each number and identify the ones that appear in both lists. Q3. Why are common multiples important? Answer: Common multiples are important in various mathematical concepts, such as finding equivalent fractions, adding and subtracting fractions with different denominators, and solving problems involving ratios and proportions. They are also used in real-life scenarios, such as scheduling events or coordinating tasks that occur at regular intervals. Q4. What is the difference between a common multiple and a least common multiple (LCM)? Answer: A common multiple is any multiple that is shared by two or more numbers, while the least common multiple (LCM) is the smallest common multiple of those numbers. Q5. Can you have more than one common multiple for a given set of numbers? Answer: Yes, a set of numbers can have an infinite number of common multiples. For example, the common multiples of 6 and 8 include 24, 48, 72, and so on.
# Evaluate: $\int\limits _{-\pi/2} ^{\pi/2} \cos ^4x dx$ $\begin{array}{1 1} \frac{3 \pi}{8} \\ \frac{2 \pi}{7} \\ \frac{\pi}{8} \\ \frac{3 \pi}{4}\end{array}$ Toolbox: • $(i)\; \int \limits_a^b f(x)dx =F(b)-F(a)$ • $(ii)\; \int \limits_{-a}^a f(x)dx=2 \int \limits_0^a f(x)dx \;if \;f(x)$ is an even function • $f(x)$ is an even function if $f(-x)=f(x)$ Step 1: $I=\int \limits_{-\pi/2}^{\pi/2} \cos ^4 x\;dx$ Let $f(x)=\cos ^4 x$ $f(-x)= \cos ^4(-x) =\cos ^4 x$ Hence it is an even function Therefore $I= 2\int \limits_0^{\pi/2} \cos ^4 x\;dx$ $=2 \int \limits_0^{\pi/2} (\cos ^2 x)^2 dx$ But we know $\cos ^2 x =\large\frac{1+\cos 2x}{2}$ Therefore $I=2 \int \limits_0^{\pi/2} \bigg(\large\frac{1+\cos 2x}{2}\bigg)^2 dx$ $=\large\frac{2}{4} $$\int \limits _0^{\pi/2} (1+\cos 2x)^2 dx On expanding, I= \large\frac{1}{2}$$\int \limits_0^{\pi/2}(1+ 2 \cos 2 x+\cos ^2 2x)dx$ Again $\cos ^2 2x$ can be written as $\large\frac{1+\cos 4x}{2}$ Step 2: Therefore $I=\large\frac{1}{2} $$\int \limits_0^{\pi/2} (1+ 2 \cos 2x+\bigg(\large\frac{1+\cos 4x}{2}\bigg) dx On seperating the terms we get I= \large\frac{1}{2}$$\int \limits_0^{\pi/2} (1+2 \cos 2x+\large\frac{1}{2}+\frac{\cos 4x}{2} )dx$ $=\large\frac{1}{2}$$\bigg[\int \limits_0^{\pi/2}(1+1/2)dx + 2 \int \limits_0^{\pi/2} \cos 2x dx+ \large\frac{1}{2}$$\int \limits_0^{\pi/2} cos 4x dx\bigg]$ $=\large\frac{1}{2}$$\bigg[\int \limits_0^{\pi/2}3/2dx + 2 \int \limits_0^{\pi/2} \cos 2x dx+ \frac{1}{2} \int \limits_0^{\pi/2} cos 4x dx\bigg] Step 3: On integrating we get =\large\frac{1}{2}\bigg\{\bigg[\large\frac{3}{2}$$ x\bigg]_0^{\pi/2}+2 \bigg[ \large\frac{\sin 2x}{2} $$\bigg]_0^{\pi/2}+\large\frac{1}{2}\bigg[ \frac{\sin 4x}{4}$$\bigg]_0^{\pi/2} \bigg\}$ On applying limits $I=\large\frac{3}{4}\bigg[\frac{\pi}{2}$$-0 \bigg]+\large\frac{1}{2}$$ \bigg[ \sin 2. \pi/2-\sin 0\bigg]+\large\frac{1}{8} $$\bigg[\sin 4. \large\frac{\pi}{2}$$-\sin 0\bigg]$ $=\large\frac{3\pi}{8}+\frac{1}{2}$$[\sin \pi -\sin 0]+\large\frac{1}{8}$$[\sin 2 \pi -\sin 0]$ But we know $\sin \pi=\sin 0=\sin 2 \pi =0$ Therefore $I=\large\frac{3 \pi}{8}$ answered Apr 29, 2013 by
# Solve each system by elimination solver In this blog post, we will take a look at how to Solve each system by elimination solver. We will also look at some example problems and how to approach them. ## Solving each system by elimination solver These sites allow users to input a Math problem and receive step-by-step instructions on how to Solve each system by elimination solver. One way is to solve each equation separately. For example, if you have an equation of the form x + 2 = 5, then you can break it up into two separate equations: x = 2 and y = 5. Solving the two set of equations separately gives you the two solutions: x = 1 and y = 6. This type of method is called a “separation method” because you separate out the two sets of equations (one equation per set). Another way to solve linear equations is by substitution. For example, if you have an equation of the form y = 9 - 4x + 6, then you can substitute different values for y in order to find out what happens when x changes. For example, if you plug in y = 8 - 3x + 3 into this equation, then the result is y= 8 - 3x + 7. Substitution is also known as “composite addition” or “additive elimination” because it involves adding or subtracting to eliminate one variable from another (hence eliminating one solution from another)! Another option If you have ever taken an online math course, then you are probably familiar with the concept of "problems". Problems are math problems that are used to teach students how to solve a particular type of math problem. In addition to being a way for teachers to show off their prowess in math, problems can also serve as a way for students to practice and improve their math skills. When it comes to math problems, there are two major types: word problems and number problems. Word problems involve using words, phrases, or sentences to explain a mathematical problem. Number problems involve using numbers to explain a mathematical problem. Word problems should be solved by drawing on the understanding of what is happening in the real world. Number problems should be solved by reasoning through the steps required to reach the correct answer. The square root of a number is the number that, when multiplied by itself, produces that number. For example, to find the square root of 12, simply multiply 12 by itself: 12 × 12 = 144. The square root of any number has a value of 1. To find the square root of a non-integer number, simply take the non-integer and multiply it by itself (or raise it to the power that is one less than the largest integer). For example, if you want to find the square root of -1, you would first raise -1 to the power 2. This gives you -2 × -2 = 4. Now simply subtract 4 from 4 to get 2. This is the square root of -1. There are two ways to solve equations with roots: adding and subtracting. Adding will always give you the correct answer, but subtracting will sometimes give you an incorrect answer. If you want to be sure that your answer will be correct and reliable, always use subtraction first! Solving equations by taking square roots is often much easier than solving them by factoring or expanding. To solve an equation by taking square roots, all you have to do is multiply the equation's terms together until you have a single term with a positive value. This can be accomplished fairly easily using long division or even algebraic substitution. When using this method
Math Expressions Grade 4 Unit 2 Lesson 6 Answer Key Use Place Value to Multiply Math Expressions Common Core Grade 4 Unit 2 Lesson 6 Answer Key Use Place Value to Multiply Math Expressions Grade 4 Unit 2 Lesson 6 Homework Use the Place Value Sections Method to solve the problem. Complete the steps. Use Place Value to Multiply Math Expressions Grade 4 Unit 2 Question 1. 9 × 86 _________ Explanation: 86 = 80 + 6 ( split into tens and ones) By adding the products of tens and ones we get 9 x 86 = 9 (80 + 6) = 9 x 80 + 9 X 6 = 720 + 54 = 774 Use the Expanded Notation Method to solve the problem. Complete the steps. Math Expressions Grade 4 Unit 2 Lesson 6 Question 2. 4 × 67 ___________ Explanation: 67 = 60 + 7 ( split into tens and ones) By adding the products of tens and ones we get 4 x 67 = 4 (60 +7) = 4 x 60 + 4 X 7 = 240 + 28 = 268 Use any method to solve. Draw a rectangular model to represent the problem. Show your work. Lesson 6 Use Place Value to Multiply Math Expressions Grade 4 Question 3. Natalia read her new book for 45 minutes each day for one week. How many minutes did she read after 7 days? Natalia read the book each day for = 45 minutes No of minutes she read after 7 days= 45 X 7 = 315 minutes Explanation: Given that Natalia read her new book for 45 minutes per day She read for 7 days, so the no of minutes she read for a week= 7 x 45 = 315 minutes Math Expressions Grade 4 Unit 2 Lesson 6 Remembering The table shows the approximate height of the world’s five tallest mountain peaks. Use the data in the table to help answer the following questions. Question 1. How tall are the two tallest mountain peaks combined? The two tall mountain peaks are Everest = 29,035ft and K2 = 28,250ft The two tallest mountain peaks combined = 29,035 + 28,250 = 57,285ft Explanation: We have that the tallest mountain peaks as Everest which is 29,035ft and K2 which is 28,250ft So the sum of two tall mountain peaks = 29,035ft+28,250ft We get the 57,285 feet Question 2. Which two mountain peaks combined are 56,190 feet tall? The sum of K2 and Lhotse peaks we get 56,190 K2= 28,250ft and Lhotse = 27,940ft 28,250+27,940 = 56,190 feet Explanation: By trial and error we come to conclusion that the sum of K2 and Lhotse we get the 56,190ft By adding 28,250ft +27,640ft = 56,190feet Subtract. Question 3. 586,720 – 293,415 = ___________ 586,720 – 293,415 = _293,305_____ Explanation: As we have 0 in one’s place we take 10 ones from ten’s place to one’s place then we have 10 – 5 = 5 in one’s place, Now we have 1 tens in ten’s place as we have carry forwarded to one’s place 1 – 1 = 0 at ten’s place 7 – 3 =4 in hundreds place 6 – 3 = 3 in thousands place Now we have 8 in ten thousands place as we take 10 ten thousands from hundred thousands place, so we have take 18 in ten thousand’s place 18 – 9 = 9 in ten thousand’s place, Now we have 4 hundred thousand in hundred thousand’s place as we have carry forwarded to hundred thousand’s place, so we 4 – 2 = 2 hundred thousand’s place, so 586,720 – 293,415 = _293,305___ Question 4. 917,336 – 904,582 = __________ 917,336 – 904,582 = __12,754_______ Explanation: in one’s place 6 – 2 = 4 As we have 0 in one’s place we take 10 ones from ten’s place to one’s place then we have 10 – 5 = 5 in one’s place, We have 3 tens in ten’s place we take 10 ten’s from hundred’s place now we have 13 tens in ten’s place 13 – 8 = 5 at ten’s place Now we have 2 hundred in hundred’s place as we have carry forwarded to ten’s place, so we have take 10 hundred from thousand’s place so we have 12 hundred, 12 – 7 = 5 in hundreds place Now we have 6 in thousand’s place as we have carry forwarded to hundred’s place 6 – 4 = 2 in thousands place 1 – 0 = 1 in ten thousand’s place 9 – 9 = 0 in hundred thousand’s place so 917,336 – 904,582 = __12,754_____ Estimate each product. Solve to check your estimate. Question 5. 5 × 39 5 x 39 = __195___ Estimated the product to nearest tens which is 195 = 200 Explanation: To round of the number to the nearest tens we should check the number at ones place which is 5; if the number in ones place is more than 5 we round off to present tens therefore we round off the number to next ten 200 Question 6. 6 × 64 6 × 64 = 384 Estimated the product to nearest tens which is 384 = 380 Explanation: To round of the number to the nearest tens we should check the number at ones place which is 4; if the number in ones place is less than 4 we round off to present tens therefore we round off the number to present tens 380 Question 7. 9 × 23 9 x 23 =207 Estimated the product to nearest tens which is 207 = 210 Explanation: To round of the number to the nearest tens we should check the number at ones place which is 7; if the number in ones place is more than 5 we round off to present tens therefore we round off the number to next ten 210 Question 8. 7 × 48 7 x 48 = 336 Estimated the product to nearest tens which is 336 = 340 Explanation: To round of the number to the nearest tens we should check the number at ones place which is 6; if the number in ones place is more than 5 we round off to present tens therefore we round off the number to next tens 340 Question 9. Stretch Your Thinking Explain how the Expanded Notation Method is used to multiply 82 × 3.
Smartick is a fun way to learn math! Sep30 # The Associative Property and Mental Math In today’s post, we will see how useful the associative property can be when we want to do mental math with addition and multiplication. To begin, I would recommend that you review these earlier posts from our blog about the associative property that could be very useful: Have you taken a look? Great, now that you have refreshed your memory we can get down to business. We begin by looking at an example with addition: It can be difficult to complete the equation outright, but we can make it easier by using the associative property as shown below: • First, we decompose or break down, the numbers. Try to break down one of the addends into numbers that will be easy to solve mentally. • Then break everything down as shown in the following image: • Now you only need to apply the associative property and solve: • And there you have it! 34 + 65 = 34 + (60 + 5) = (34 + 60) + 5 = 94 + 5 = 99 You can see that 34 + 65 is 99. ## Associative Property of Multiplication Now we will look at another example, this time using multiplication: It seems difficult but in reality, we just do the same thing as we did with addition. • The first thing to do is to break down the factors so that we are left with a simple equation that we can solve mentally. • Then, put everything together as seen in the following image: • And finally, apply the associative property and solve: • And that’s it! 15 x 18 = 15 x (2 x 9) = (15 x 2) x 9 = 30 x 9 = 270 You can see that 15 x 18 is 270. Now you will find it much easier to solve problems with mental math. Practice what you’ve learned and you’ll see! I’ll leave you with some equations so you can continue practicing, but you can also make some up yourself and then check that you have solved them correctly. • 67 + 42 • 31 + 58 • 35 x 16 • 25 x 24
# Simplifying Expressions with 6.2 and 8.5 Simplifying expressions is a very important concept in mathematics, and is the process of breaking down a complex expression into simpler terms. Expressions with 6.2 and 8.5 can appear to be complicated, but with a few simple rules and steps, you can easily simplify them. In this article, we will discuss some common methods for simplifying expressions with 6.2 and 8.5. ## First Step: Divide the Expression The first step to simplifying an expression with 6.2 and 8.5 is to divide the expression into two parts. The part containing 6.2 should be divided by itself, and the part containing 8.5 should be divided by itself. This will result in two fractions: 6.2/6.2 and 8.5/8.5. The result of both of these divisions is 1, meaning that the expression can be rewritten as 6.2/1 multiplied by 8.5/1. ## Second Step: Multiply the Expression Now that the expression is divided into two parts, it is time to multiply the two parts together. Multiplying the two fractions together will result in a single fraction, 6.2/8.5. This is the simplified version of the original expression. ## Third Step: Simplify the Expression The third step is to simplify the expression further. This can be done by dividing both the numerator and denominator by the same number. For example, if you divide both the numerator and denominator by 2, the result will be 3.1/4.25. This is the simplest form of the expression. ## Simplifying expressions with 6.2 and 8.5 is a straightforward process, but it is important to understand the steps involved. By dividing the expression into two parts, multiplying the two parts together, and then simplifying the expression, you can easily simplify expressions with 6.2 and 8.5. This process can be applied to other expressions as well, allowing you to quickly simplify any expression.
Arithmetic Associative Property Averages Brackets Closure Property Commutative Property Conversion of Measurement Units Cube Root Decimal Divisibility Principles Equality Exponents Factors Fractions Fundamental Operations H.C.F / G.C.D Integers L.C.M Multiples Multiplicative Identity Multiplicative Inverse Numbers Percentages Profit and Loss Ratio and Proportion Simple Interest Square Root Unitary Method Algebra Algebraic Equation Algebraic Expression Cartesian System Linear Equations Order Relation Polynomials Probability Standard Identities & their applications Transpose Geometry Basic Geometrical Terms Circle Curves Angles Define Line, Line Segment and Rays Non-Collinear Points Parallelogram Rectangle Rhombus Square Three dimensional object Trapezium Triangle Trigonometry Trigonometry Ratios Data-Handling Arithmetic Mean Frequency Distribution Table Graphs Median Mode Range Home >> Closure Property >> Addition of Whole Numbers >> ## Closure Property (Addition of Whole Numbers) Addition of Whole Numbers Addition of Integers Subtraction of Whole Numbers Subtraction of Integers Multiplication of Whole Numbers Multiplication of Integers Division of Whole Numbers Division of Integers Before understanding this topic you must know what are whole numbers ? Explanation :- System of whole numbers is closed under addition, this means that the sum of any two whole numbers is always a whole number. This is known as Closure Property for Addition of Whole Numbers Read the following example and you can further understand this property Example 1 = With the given whole numbers 4 and 9, Explain Closure Property for addition of whole numbers. Answer= Find the sum of given whole numbers 4 + 9 = 13 As we know that 13 is also a whole number, So, we can say that whole numbers are closed under addition. Example 2 = With the given whole numbers 13 and 0, Explain Closure Property for addition of whole numbers. Answer= Find the sum of given whole numbers 13 + 0 = 13 As we know that 13 is also a whole number, So, we can say that whole numbers are closed under addition. Example 3 = With the given whole numbers 25 and 7, Explain Closure Property for addition of whole numbers. Answer= Find the sum of given whole numbers 25 + 7 = 32 As we know that 32 is also a whole number, So, we can say that whole numbers are closed under addition. Example 4 = With the given whole numbers 101 and 20, Explain Closure Property for addition of whole numbers. Answer= Find the sum of given whole numbers 101 + 20 = 121 As we know that 121 is also a whole number, So, we can say that whole numbers are closed under addition.
We think you are located in United States. Is this correct? # 2.6 Exponential functions ## 2.6 Exponential functions (EMCFF) Revision of exponents An exponent indicates the number of times a certain number (the base) is multiplied by itself. The exponent, also called the index or power, indicates the number of times the multiplication is repeated. For example, $$\text{10}^{3} = 10 \times 10 \times 10 = \text{1 000}$$. Graphs of the exponential function $$f(x) = b^{x}$$ The value of $$b$$ affects the direction of the graph: • If $$b > 1$$, $$f(x)$$ is an increasing function. • If $$0 < b < 1$$, $$f(x)$$ is a decreasing function. • If $$b \leq 0$$, $$f(x)$$ is not defined. $$y = b^{x}$$ $$b > 1$$ $$0 < b < 1$$ $$b \leq 0$$ Not defined ## Determining the inverse Function Type of function Inverse: Inverse: interchange $$x$$ and $$y$$ make $$y$$ the subject $$y = \frac{x}{3} + 10$$ $$y = \frac{x^{2}}{3}$$ $$y = (10)^{x}$$ $$y = \left( \frac{1}{3} \right)^{x}$$ Consider the exponential function $y = b^{x}$ To determine the inverse of the exponential function, we interchange the $$x$$- and $$y$$-variables: $x = b^{y}$ For straight line functions and parabolic functions, we could easily manipulate the inverse to make $$y$$ the subject of the formula. For the inverse of an exponential function, however, $$y$$ is the index and we do not know a method of solving for the index. To resolve this problem, mathematicians defined the logarithmic function. The logarithmic function allows us to rewrite the expression $$x = b^{y}$$ with $$y$$ as the subject of the formula: $y = \log_{b}{x}$ This means that $$x = b^{y}$$ is the same as $$y = \log_{b}{x}$$ and both are the inverse of the exponential function $$y = b^{x}$$. ### Logarithms (EMCFH) Logarithm If $$x = {b}^{y}$$, then $$y = {\log}_{b}\left(x\right)$$, where $$b>0$$, $$b \ne 1$$ and $$x>0$$. Note that the brackets around the number $$(x)$$ are not compulsory, we use them to avoid confusion. The logarithm of a number $$(x)$$ with a certain base $$(b)$$ is equal to the exponent $$(y)$$, the value to which that certain base must be raised to equal the number $$(x)$$. For example, $$\log_{2}\left(8\right)$$ means the power of $$\text{2}$$ that will give $$\text{8}$$. Since $${2}^{3}=8$$, we see that $${\log}_{2}\left(8\right)=3$$. Therefore the exponential form is $${2}^{3}=8$$ and the logarithmic form is $${\log}_{2}{8}=3$$. Restrictions on the definition of logarithms $\begin{array}{rll} \text{Restriction:}& & \text{Reason: } \\ & & \\ b > 0 & & \text{If } b \text{ is a negative number, then } b^{y} \text{ will oscillate between:} \\ & & \text{positive values if } y \text{ is even } \\ & & \text{negative values if } y \text{ is odd } \\ & & \\ b \ne 1 & & \text{Since } 1^{\text{(any value)}} = 1 \\ & & \\ x > 0 & & \text{Since } \text{(positive number)}^{\text{(any value)}} > 0 \end{array}$ ## Exponential and logarithmic form Discuss the following statements and determine whether they are true or false: 1. $$p = a^{n}$$ is the inverse of $$p = \log_{a}{n}$$. 2. $$y = 2^{x}$$ is a one-to-one function, therefore $$y = \log_{2}{x}$$ is also a one-to-one function. 3. $$x = \log_{5}{y}$$ is the inverse of $$5^{x} = y$$. 4. $$k = b^{t}$$ is the same as $$t = \log_{b}{k}$$. To determine the inverse function of $$y=b^{x}$$: $\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = b^{y} \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & y = \log_{b}{x} \end{array}$ Therefore, if we have the exponential function $$f(x) = b^{x}$$, then the inverse is the logarithmic function $$f^{-1}(x) = \log_{b}{x}$$. The “common logarithm” has a base $$\text{10}$$ and can be written as $$\log_{10}{x} = \log{x}$$. In other words, the $$\log$$ symbol written without a base is interpreted as the logarithm to base $$\text{10}$$. For example, $$\log{\text{25}} = \log_{10}{\text{25}}$$. ## Worked example 10: Exponential form to logarithmic form Write the following exponential expressions in logarithmic form and express each in words: 1. $$5^{2} = 25$$ 2. $$10^{-3} = \text{0,001}$$ 3. $$p^{x} = q$$ ### Determine the inverse of the given exponential expressions Remember: $$m = a^{n}$$ is the same as $$n = \log_{a}{m}$$. 1. $$2 = \log_{5}{25}$$ 2. $$-3 = \log_{10}{(\text{0,001})}$$ 3. $$x = \log_{p}{q}$$ ### Express in words 1. $$\text{2}$$ is the power to which $$\text{5}$$ must be raised to give the number $$\text{25}$$. 2. $$-\text{3}$$ is the power to which $$\text{10}$$ must be raised to give the decimal number $$\text{0,001}$$. 3. $$x$$ is the power to which $$p$$ must be raised to give $$q$$. ## Worked example 11: Logarithmic form to exponential form Write the following logarithmic expressions in exponential form: 1. $$\log_{2}{128} = 7$$ 2. $$-2 = \log_{3}{\left( \frac{1}{9} \right)}$$ 3. $$z = \log_{w}{k}$$ ### Determine the inverse of the given logarithmic expressions For $$n = \log_{a}{m}$$, we can write $$m = a^{n}$$. 1. $$2^{7} = \text{128}$$ 2. $$3^{-2} = \frac{1}{9}$$ 3. $$w^{z} = k$$ temp text ## Finding the inverse of $$y=b^{x}$$ Textbook Exercise 2.6 Write the following in logarithmic form: $$16 = 2^{4}$$ $$4 = \log_{2}{16}$$ $$3^{-5} = \frac{1}{243}$$ $$-5 = \log_{3}{\left( \frac{1}{243} \right) }$$ $$(\text{1,7})^{3} = \text{4,913}$$ $$3 = \log_{\text{1,7}}{(\text{4,913})}$$ $$y = 2^{x}$$ $$x = \log_{2}{y}$$ $$q = 4^{5}$$ $$\log_{4} q = 5$$. $$4 = y^{g}$$ $$\log_{y} 4 = g$$. $$9 = (x -4)^{p}$$ $$\log_{(x - 4)} 9 = p$$. $$3 = m^{(a + 4)}$$ $$\log_{m} 3 = a + 4$$. Express each of the following logarithms in words and then write in exponential form: $$\log_{2}{32} = 5$$ The logarithm of $$\text{32}$$ to base $$\text{2}$$ is equal to $$\text{5}$$. $$2^{5} = 32$$ $$\log{\frac{1}{1000}} = -3$$ The logarithm of $$\frac{1}{1000}$$ to base $$\text{10}$$ is equal to $$-\text{3}$$. $$10^{-3} = \frac{1}{1000}$$ $$\log{\text{0,1}} = -1$$ The logarithm of $$\text{0,1}$$ to base $$\text{10}$$ is equal to $$-\text{1}$$. $$10^{-1} = \text{0,1}$$ $$\log_{d}{c} = b$$ The logarithm of $$c$$ to base $$d$$ is equal to $$b$$. $$d^{b} = c$$ $$\log_{5}{1} = 0$$ The logarithm of $$\text{1}$$ to base $$\text{5}$$ is equal to $$\text{0}$$. $$5^{0} = 1$$ $$\log_{3}{\frac{1}{81}} = -4$$ The logarithm of $$\frac{1}{81}$$ to base $$\text{3}$$ is equal to $$-\text{4}$$. $$3^{-4} = \frac{1}{81}$$ $$\log{100}$$ \begin{align} \text{Let } \log{100} &= m \\ 10^{m} &= 100 \\ &= 10^{2} \\ \therefore m &= 2 \\ \therefore \log{100} &= 2 \end{align} The logarithm of $$\text{100}$$ to base $$\text{10}$$ is $$\text{2}$$. $$\log_{\frac{1}{2}}{16}$$ \begin{align} \text{Let } \log_{\frac{1}{2}}{16} &= y \\ \therefore \left( \frac{1}{2} \right)^{y} &= 16 \\ \therefore 2^{-y}&= 2^{4} \\ -y &= 4 \\ \therefore y &= -4 \end{align} The logarithm of $$\text{16}$$ to base $$\frac{1}{2}$$ is $$-\text{4}$$. ### Logarithm bases (EMCFJ) From the definition of a logarithm we know that the base of a logarithm must be a positive number and it cannot be equal to $$\text{1}$$. The value of the base influences the value of the logarithm. For example, $${\log}_{2}2$$ is not the same as $${\log}{2}$$ and $${\log}_{f}{11}$$ is not the same as $${\log}_{g}{11}$$, ($$f \ne g$$). We often calculate the “common logarithm”, which has a base $$\text{10}$$ and can be written as $$\log_{10}{x} = \log{x}$$. For example, $$\log{8} = \log_{10}{8}$$. The “natural logarithm”, which has a base $$e$$ (an irrational number between $$\text{2,71}$$ and $$\text{2,72}$$), can be written as $$\log_{e}{x} = \ln{x}$$. For example, $$\log_{e}{5} = \ln{5}$$. Special logarithmic values • $$\log_{a}{1} = 0$$ \begin{align} \text{Given the exponential form } a^{n} &= x \\ \text{we define the logarithmic function } \log_{a}{x} &= n \\ \text{So then for } a^{0} &= 1 \\ \text{we can write } \log_{a}{1} &= 0 \end{align} • $$\log_{a}{a} = 1$$ \begin{align} \text{From the general exponential form } a^{n} &= x \\ \text{we define the logarithmic function } \log_{a}{x} &= n \\ \text{Since } a^{1} &= a \\ \text{we can write } \log_{a}{a} &= 1 \end{align} ### Laws of logarithms (EMCFK) In earlier grades, we used the following exponential laws for working with exponents: • $${a}^{m} \times {a}^{n}={a}^{m+n}$$ • $$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$$ • $${\left(ab\right)}^{n}={a}^{n}{b}^{n}$$ • $${\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}$$ • $${\left({a}^{m}\right)}^{n}={a}^{mn}$$ where $$a > 0$$, $$b > 0$$ and $$m, n \in ℤ$$. The logarithmic laws are based on the exponential laws and make working with logarithms much easier. Logarithmic laws: • $$\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0)$$ • $$\log_{a}{x} = \frac{\log_{b}{x}}{\log_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)$$ • $$\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$$ • $$\log_{a}{\frac{x}{y}} = \log_{a}{x} - \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$$ The last two logarithmic laws in the list above are not covered in this section. They are discussed at the end of the chapter and are included for enrichment only. IMPORTANT: PROOFS ARE NOT REQUIRED FOR EXAMS Logarithmic law: $\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0 )$ \begin{align} \text{Let } {\log}_{a}{x} &= m \ldots (1)\qquad (x > 0 ) \\ \therefore x &= a^{m} \\ \therefore (x)^{b} &= \left( a^{m} \right)^{b} \\ \therefore x^{b} &= a^{bm} \\ \text{Change to logarithmic form: }\log_{a}\left(x^{b}\right) &= bm \\ \text{And subst}: \quad m &= {\log}_{a}{x} \\ \therefore {\log}_{a}{x^{b}} &= b{\log}_{a}{x} \end{align} In words: the logarithm of a number which is raised to a power is equal to the value of the power multiplied by the logarithm of the number. ## Worked example 12: Applying the logarithmic law $$\log_{a}{x^{b}} = b \log_{a}{x}$$ Determine the value of $$\log_{3}{{27}^{4}}$$. ### Use the logarithmic law to simplify the expression \begin{align} \log_{3}{{27}^{4}} &= 4 \log_{3}{{27}}\\ &= 4\log_{3}{{3}^{3}} \\ &= (4 \times 3) \log_{3}{3}\\ &= 12 (1) \\ &= 12 \end{align} $$\log_{3}{{27}^{4}} = 12$$ Special case: $\log_{a}{\sqrt[b]{x}} = \frac{\log_{a}{x}}{b} \qquad (x > 0 \text{ and } b > 0 )$ The following is a special case of the logarithmic law $$\log_{a}{x^{b}} = b \log_{a}{x}$$: \begin{align} \log_{a}{\sqrt[b]{x}} &= \log_{a}{{x}^{\frac{1}{b}}} \\ &= \frac{1}{b} \log_{a}{x} \\ &= \frac{\log_{a}{x}}{b} \end{align} ## Applying the logarithmic law: $$\log_{a}{x^{b}} = b \log_{a}{x}$$ Textbook Exercise 2.7 Simplify the following: $$\log_{8}{{10}^{10}}$$ $\log_{8}{{10}^{10}} = 10 \log_{8}{10}$ $$\log_{16}{{x}^{y}}$$ $\log_{16}{{x}^{y}} = y \log_{16}{x}$ $$\log_{3}{\sqrt{5}}$$ $\log_{3}{\sqrt{5}} = \frac{\log_{3}{5}}{2}$ $$\log_{z}{{y}^{z}}$$ $\log_{z}{{y}^{z}} = z \log_{z}{y}$ $$\log_{y}{\sqrt[x]{y}}$$ \begin{align} \log_{y}{\sqrt[x]{y}} &= \log_{y}{y^{\frac{1}{x}}} \\ &= \frac{1}{x} \log_{y}{y} \\ &= \frac{1}{x} \end{align} $$\log_{p}{{p}^{q}}$$ \begin{align} \log_{p}{{p}^{q}} &= q \log_{p}{p} \\ &= q(1) \\ &= q \end{align} $$\log_{2}{\sqrt[4]{8}}$$ \begin{align} \log_{2}{\sqrt[4]{8}} &= \log_{2}{8^{\frac{1}{4}}} \\ &= \frac{1}{4} \log_{2}{2^{3}} \\ &= \frac{1}{4} \times 3 \log_{2}{2} \\ &= \frac{3}{4} \end{align} $$\log_{5}{\frac{1}{5}}$$ \begin{align} \log_{5}{\frac{1}{5}} &= \log_{5}{5^{-1}} \\ &= (-1) \log_{5}{5} \\ &= (-1)(1) \\ &= -1 \end{align} $$\log_{2}{{8}^{5}}$$ \begin{align} \log_{2}{{8}^{5}} &= 5 \log_{2}{{8}} \\ &= 5 \log_{2}{{2}^{3}} \\ &= 5 \times 3 \log_{2}{2} \\ &= 15 \end{align} $$\log_{4}{16} \times \log_{3}{81}$$ \begin{align} \log_{4}{16} \times \log_{3}{81} &= \log_{4}{4^{2}} \times \log_{3}{3^{4}} \\ &= (2) \log_{4}{4} \times (4) \log_{3}{3} \\ &= (2)(1) \times (4)(1)\\ &= 8 \end{align} $$\left( \log_{5}{25} \right)^{2}$$ \begin{align} \left( \log_{5}{25} \right)^{2} &= \left( \log_{5}{5^{2}} \right)^{2} \\ &= \left( 2 \log_{5}{5} \right)^{2} \\ &= \left( 2 (1) \right)^{2} \\ &= 4 \\ \text{Alternative method: } & \\ \left( \log_{5}{25} \right)^{2} &= \log_{5}{ \left( 5^{2} \right)^{2} } \\ &= \log_{5}{5}^{4} \\ &= 4(1) \\ &= 4 \end{align} $$\log_{2}{\text{0,125}}$$ \begin{align} \log_{2}{\text{0,125}} &= \log_{2}{\frac{125}{1000}} \\ &= \log_{2}{\frac{1}{8}} \\ &= \log_{2}{8^{-1}} \\ &= (-1) \log_{2}{2^{3}} \\ &= (-1)(3) \log_{2}{2} \\ &= (-3) (1) \\ &= -3 \end{align} Logarithmic law: $${\log}_{a}x=\frac{{\log}_{b}{x}}{{\log}_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)$$ It is often necessary or convenient to convert a logarithm from one base to another base. This is referred to as a change of base. \begin{align} \text{Let } \quad \log_{a}{x} &= m \\ \therefore x &= a^{m} \\ \text{Consider the fraction: } \quad & \frac{\log_{b}{x}}{\log_{b}{a}} \\ \text{Substitute } x = a^{m}: \quad \frac{\log_{b}{x}}{\log_{b}{a}} &= \frac{\log_{b}{a^{m}}}{\log_{b}{a}} \\ &= m \left( \frac{\log_{b}{a}}{\log_{b}{a}} \right) \\ &= m (1) \\ \therefore \frac{\log_{b}{x}}{\log_{b}{a}}&= \log_{a}{x} \end{align} Special applications: \begin{align} (1) \qquad \log_{a}{x} &= \frac{\log_{x}{x}}{\log_{x}{a}} \\ \therefore \log_{a}{x} &= \frac{1}{\log_{x}{a}} \\ & \\ & \\ (2) \qquad \log_{a}{\frac{1}{x}} &= \log_{a}{x^{-1}} \\ \therefore \log_{a}{\frac{1}{x}} &= - \log_{a}{x} \end{align} ## Worked example 13: Applying the logarithmic law $${\log}_{a}x=\frac{{\log}_{b}{x}}{{\log}_{b}{a}}$$ Show: $$\log_{2}{8} = \frac{\log{8}}{\log{2}}$$ ### Simplify the right-hand side of the equation \begin{align} \text{RHS } &= \frac{\log{8}}{\log{2}} \\ &= \frac{\log{2^{3}}}{\log{2}} \\ &= 3 \left( \frac{\log{2}}{\log{2}} \right) \\ &= 3(1) \\ &= 3 \end{align} ### Simplify the left-hand side of the equation \begin{align} \text{LHS } &= \log_{2}{8} \\ &= \log_{2}{2^{3}} \\ &= 3 \log_{2}{2} \\ &= 3(1) \\ &= 3 \end{align} We have shown that $$\log_{2}{8} = \dfrac{\log{8}}{\log{2}} = 3$$. ## Worked example 14: Applying the logarithmic law $${\log}_{a}x=\frac{{\log}_{b}{x}}{{\log}_{b}{a}}$$ If $$a = \log{2}$$ and $$b = \log{3}$$, express the following in terms of $$a$$ and $$b$$: 1. $$\log_{3}{2}$$ 2. $$\log_{2}{\frac{10}{3}}$$ ### Use a change of base to simplify the expressions 1. \begin{align} \log_{3}{2} &= \frac{\log{2}}{\log{3}} \\ &= \frac{a}{b} \end{align} 2. \begin{align} \log_{2}{\frac{10}{3}} &= \frac{\log{\frac{10}{3}}}{\log{2}} \\ &= \frac{\log{10} - \log{3}}{\log{2}} \\ &= \frac{1- b}{a} \end{align} temp text ## Applying the logarithmic law: $${\log}_{a}x=\frac{{\log}_{b}{x}}{{\log}_{b}{a}}$$ Textbook Exercise 2.8 Convert the following: $$\log_{2}{4}$$ to base $$\text{8}$$ $$\log_{2}{4} = \frac{\log_{8}{4}}{\log_{8}{2}}$$ $$\log_{10}{14}$$ to base $$\text{2}$$ $$\log_{10}{14} = \frac{\log_{2}{14}}{\log_{2}{10}}$$ $$\log{4\frac{1}{2}}$$ to base $$\text{2}$$ \begin{align} \log{4\frac{1}{2}} &= \log{\frac{9}{2}} \\ &= \frac{\log_{2}{\frac{9}{2}}}{\log_{2}{10}} \\ &= \frac{\log_{2}{9} - \log_{2}{2}}{\log_{2}{10}} \\ &= \frac{\log_{2}{9} - 1}{\log_{2}{10}} \end{align} $$\log_{2}{8}$$ to base $$\text{8}$$ \begin{align} \log_{2}{8} &= \frac{\log_{8}{8}}{\log_{8}{2}} \\ &= \frac{1}{\log_{8}{2}} \end{align} $$\log_{y}{x}$$ to base $$x$$ \begin{align} \log_{y}{x} &= \frac{\log_{x}{x}}{\log_{x}{y}} \\ &= \frac{1}{\log_{x}{y}} \end{align} $$\therefore$$ a logarithm is equal to the reciprocal of its inverse. $$\log_{10}{2x}$$ to base $$\text{2}$$ \begin{align} \log_{10}{2x} &= \frac{\log_{2}{2x}}{\log_{2}{10}} \\ &= \frac{\log_{2}{2} + \log_{2}{x}}{\log_{2}{10}} \\ &= \frac{1 + \log_{2}{x}}{\log_{2}{10}} \end{align} Simplify the following using a change of base: $$\log_{2}{10} \times \log_{10}{2}$$ \begin{align} \log_{2}{10} \times \log_{10}{2} &= \frac{\log{10}}{\log{2}} \times \frac{\log{2}}{\log{10}} \\ &= 1 \end{align} $$\log_{5}{100}$$ \begin{align} \log_{5}{100} &= \frac{\log{100}}{\log{5}} \\ &= \frac{\log{10^{2}}}{\log{5}} \\ &= \frac{2 \log{10}}{\log{5}} \\ &= \frac{2}{\log{5}} \end{align} If $$\log{3} = \text{0,477}$$ and $$\log{2} = \text{0,301}$$, determine (correct to $$\text{2}$$ decimal places): $$\log_{2}{3}$$ \begin{align} \log_{2}{3} &= \frac{\log{3}}{\log{2}} \\ &= \frac{\text{0,477}}{\text{0,301}} \\ &= \text{1,58} \end{align} $$\log_{3}{2000}$$ \begin{align} \log_{3}{2000} &= \frac{\log{2000}}{\log{3}} \\ &= \frac{\log{(2 \times 1000)}}{\log{3}} \\ &= \frac{\log{2} + \log{10^{3}}}{\log{3}} \\ &= \frac{\text{0,301} +3(1)}{\text{0,477}} \\ &= \frac{\text{3,301}}{\text{0,477}} \\ &= \text{6,92} \end{align} ### Logarithms using a calculator (EMCFM) Calculating a logarithmic value There are many different types and models of scientific calculators. It is very important to be familiar with your own calculator and the different function buttons. Some calculators only have two buttons for logarithms: one for calculating the common logarithm (base is equal to $$\text{10}$$) and another for calculating the natural log (base is equal to $$e$$). Newer models will have a third button which allows the user to calculate the logarithm of a number to a certain base. ## Worked example 15: Using a calculator: logarithm function Use a calculator to determine the following values (correct to $$\text{3}$$ decimal places): 1. $$\log{9}$$ 2. $$\log{\text{0,3}}$$ 3. $$\log{\frac{3}{4}}$$ 4. $$\log{(-2)}$$ ### Use the common logarithm function on your calculator Make sure that you are familiar with the “LOG” function on your calculator. Notice that the base for each of the logarithms given above is $$\text{10}$$. 1. $$\log{9} = \text{0,954}$$ 2. $$\log{\text{0,3}} = -\text{0,523}$$ 3. $$\log{\frac{3}{4}} = -\text{0,125}$$ 4. $$\log{(-2)} =$$ undefined temp text ## Worked example 16: Using a calculator: inverse logarithm function Use a calculator to determine the following values (correct to $$\text{3}$$ decimal places): 1. $$\log{x} = \text{1,7}$$ 2. $$\log{t} = \frac{2}{7}$$ 3. $$\log{y} = -\text{3}$$ ### Use the second function and common logarithm function on your calculator For each of the logarithms given above we need to calculate the inverse of the logarithm (sometimes called the antilog). Make sure that you are familiar with the “$$\text{2}$$nd F” button on your calculator. Notice that by pressing the “$$\text{2}$$nd F” button and then the “LOG” button, we are using the “$$10^{x}$$” function on the calculator, which is correct since exponentials are the inverse of logarithms. 1. $$x = \text{50,119}$$ 2. $$t = \text{1,930}$$ 3. $$y = \text{0,001}$$ temp text ## Worked example 17: Using a calculator: change of base Use a calculator to find $$\log_{2}{5}$$ correct to two decimal places. ### Use a change of base to convert given logarithm to base $$\text{10}$$ $\log_{2}{5} = \frac{\log{5}}{\log{2}}$ ### Use the common logarithm function on your calculator $\log_{2}{5} = \text{2,32}$ Important: • Do not write down an intermediate step when doing this type of calculation: \begin{align} \log_{2}{5} &= \frac{\log{5}}{\log{2}} \\ &= \frac{\text{0,7}}{\text{0,3}} \quad (\text{this step can cause rounding off errors}) \\ &= \text{2,33} \end{align} Perform the calculation in one step on your calculator: \begin{align} \log_{2}{5} &= \frac{\log{5}}{\log{2}} \\ &= \text{2,32} \end{align} • Do not round off before the final answer as this can affect the accuracy of the answer. • Be sure that you determine the correct sequence and order of operations when using a calculator. ## Logarithms using a calculator Textbook Exercise 2.9 Calculate the following (correct to three decimal places): $$\log{3}$$ $$\text{0,477}$$ $$\log{30}$$ $$\text{1,477}$$ $$\log{300}$$ $$\text{2,477}$$ $$\log{0.66}$$ $$-\text{0,180}$$ $$\log{\frac{1}{4}}$$ $$-\text{0,602}$$ $$\log{\text{852}}$$ $$\text{2,930}$$ $$\log{(-6)}$$ no value $$\log_{3}{4}$$ $$\text{1,262}$$ $$\log{\text{0,01}}$$ $$-\text{2}$$ $$\log_{2}{15}$$ $$\text{3,907}$$ $$\log_{4}{10}$$ $$\text{1,661}$$ $$\log_{\frac{1}{2}}{6}$$ $$-\text{2,585}$$ Use a calculator to determine the value of $$x$$ (correct to two decimal places). Check your answer by changing to exponential form. $$\log{x} = \text{0,6}$$ \begin{align} x &= \text{3,98}\\ \text{Check exponential form: } 10^{\text{0,6}}&= \text{3,98} \end{align} $$\log{x} = -\text{2}$$ \begin{align} x &= \text{0,01}\\ \text{Check exponential form: } 10^{-\text{2}}&= \text{0,01} \end{align} $$\log{x} = \text{1,8}$$ \begin{align} x &= \text{63,10}\\ \text{Check exponential form: } 10^{\text{1,8}}&= \text{63,10} \end{align} $$\log{x} = \text{5}$$ \begin{align} x &= \text{100 000}\\ \text{Check exponential form: } 10^{\text{5}}&= \text{100 000} \end{align} $$\log{x} = -\text{0,5}$$ \begin{align} x &= \text{0,32}\\ \text{Check exponential form: } 10^{-\text{0,5}}&= \text{0,32} \end{align} $$\log{x} = \text{0,076}$$ \begin{align} x &= \text{1,19}\\ \text{Check exponential form: } 10^{\text{0,076}}&= \text{1,19} \end{align} $$\log{x} = \frac{2}{5}$$ \begin{align} x &= \text{2,51}\\ \text{Check exponential form: } 10^{\frac{2}{5}}&= \text{2,51} \end{align} $$\log{x} = -\frac{6}{5}$$ \begin{align} x &= \text{0,06}\\ \text{Check exponential form: } 10^{\left( -\frac{6}{5} \right)}&= \text{0,06} \end{align} $$\log_{2}{x} = \text{0,25}$$ \begin{align} \log_{2}{x} &= \text{0,25} \\ \therefore \frac{\log{x}}{\log{2}} &= \text{0,25} \\ \therefore \log{x} &= \text{0,25} \times \log{2} \\ \therefore x &= \text{1,19} \\ \text{Check exponential form: } 2^{\text{0,25}}&= \text{1,19} \end{align} $$\log_{5}{x} = -\text{0,1}$$ \begin{align} \log_{5}{x} &= -\text{0,1} \\ \therefore \frac{\log{x}}{\log{5}} &= -\text{0,1} \\ \therefore \log{x} &= -\text{0,1} \times \log{5} \\ \therefore x &= \text{0,85} \\ \text{Check exponential form: } 5^{(-\text{0,10})}&= \text{0,85} \end{align} $$\log_{\frac{1}{4}}{x} = \text{2}$$ \begin{align} \log_{\frac{1}{4}}{x} &= \text{2} \\ \therefore \frac{\log{x}}{\log{\frac{1}{4}}} &= \text{2} \\ \therefore \log{x} &= \text{2} \times \log{\frac{1}{4}} \\ \therefore x &= \text{0,06} \\ \text{Check exponential form: } \left( \frac{1}{4} \right)^{\text{2}}&= \text{0,06} \end{align} $$\log_{7}{x} = \text{0,3}$$ \begin{align} \log_{7}{x} &= \text{0,3} \\ \therefore \frac{\log{x}}{\log{7}} &= \text{0,3} \\ \therefore \log{x} &= \text{0,3} \times \log{7} \\ \therefore x &= \text{1,79} \\ \text{Check exponential form: } 7^{\text{0,3}}&= \text{1,79} \end{align} ## Worked example 18: Graphs of the inverse of $$y=b^{x}$$ On the same system of axes, draw the graphs of $$f(x) = 10^{x}$$ and its inverse $$f^{-1}(x) = \log{x}$$. Investigate the properties of $$f$$ and $$f^{-1}$$. ### Determine the properties of $$f(x)$$ • Function: $$y = 10^{x}$$ • Shape: increasing graph • Intercept(s): $$(0;1)$$ • Asymptote(s): horizontal asymptote at $$x$$-axis, line $$y = 0$$ • Domain: $$\{x: x \in \mathbb{R} \}$$ • Range: $$\{y: y > 0, y \in \mathbb{R} \}$$ ### Draw the graphs The graph of the inverse $$f^{-1}$$ is the reflection of $$f$$ about the line $$y = x$$. ### Determine the properties of $$f^{-1}(x)$$ • Function: $$y = \log{x}$$ • Shape: increasing graph • Intercept(s): $$(1;0)$$ • Asymptote(s): vertical asymptote at $$y$$-axis, line $$x = 0$$ • Domain: $$\{x: x > 0, x \in \mathbb{R} \}$$ • Range: $$\{y: y \in \mathbb{R} \}$$ Notice that the inverse is a function: $$f^{-1}(x) = \log{x}$$ is a one-to-one function since every input value is associated with only one output value. The exponential function and the logarithmic function are inverses of each other: • the domain of the function is equal to the range of the inverse • the range of the function is equal to the domain of the inverse • the $$y$$-intercept of the function is equal to the $$x$$-intercept of the inverse • the $$x$$-intercept of the function is equal to the $$y$$-intercept of the inverse • the asymptote for the function is $$y = 0$$ and the asymptote for the inverse is $$x = 0$$ • the graphs are reflected about the line $$y=x$$ ## Worked example 19: Graphs of $$y=\log_{b}{x}$$ 1. Draw a sketch of $$g(x) = \log_{10}{x}$$. 2. Reflect the graph of $$g$$ about the $$x$$-axis to give the graph $$h$$. 3. Investigate the properties of $$h$$. 4. Use $$g$$ and $$h$$ to suggest a general conclusion. ### Reflect $$g$$ about the $$x$$-axis An easy method for reflecting a graph about a certain line is to imagine folding the Cartesian plane along that line and the reflected graph is pressed onto the plane. ### Investigate the properties of $$h$$ • Function: passes the vertical line test • Shape: decreasing graph • Intercept(s): $$(1;0)$$ • Asymptote(s): vertical asymptote at $$y$$-axis, line $$x = 0$$ • Domain: $$\{x: x > 0, x \in \mathbb{R} \}$$ • Range: $$\{y: y \in \mathbb{R} \}$$ Since $$h(x)$$ is symmetrical to $$g(x)$$ about the $$x$$-axis, this means that every $$y$$-value of $$g$$ corresponds to a $$y$$-value of the opposite sign for $$h$$. \begin{align} \text{Given } g(x) & = \log_{10}{x} \\ \therefore h(x) & = - \log_{10}{x} \\ \text{Let } y & = - \log_{10}{x} \\ -y & = \log_{10}{x} \\ \therefore 10^{-y} & = x \\ \left( \frac{1}{10} \right)^{y} & = x \\ \therefore y & = \log_{\frac{1}{10}}{x} \\ \therefore h(x) & = - \log_{10}{x} = \log_{\frac{1}{10}}{x} \end{align} ### General conclusion From this example of $$g$$ and $$h$$ we see that: $-\log_{m}{p} = \log_{\frac{1}{m}}{p}$ ## Worked example 20: Graph of $$y=\log_{b}{x}$$ 1. Draw a sketch of $$h(x) = \log_{\frac{1}{10}}{x}$$. 2. Draw the graph of $$r(x)$$, the reflection of $$h$$ about the line $$y = x$$. 3. Investigate the properties of $$r$$. 4. Write down the new equation if $$h$$ is shifted $$\text{1}$$ unit upwards and $$\text{2}$$ units to the right. ### Investigate the properties of $$r$$ • Function: passes the vertical line test • Shape: decreasing graph • Intercept(s): $$(0;1)$$ • Asymptote(s): horizontal asymptote at $$x$$-axis, line $$y = 0$$ • Domain: $$\{x: x \in \mathbb{R} \}$$ • Range: $$\{y: y > 0, y \in \mathbb{R} \}$$ Since $$h(x)$$ is symmetrical to $$r(x)$$ about the line $$y = x$$, this means that $$r$$ is the inverse of $$h$$. \begin{align} h(x) & = \log_{\frac{1}{10}}{x} \\ \text{Let } y & = \log_{\frac{1}{10}}{x} \\ \text{Inverse: } x & = \log_{\frac{1}{10}}{y} \\ \therefore \left( \frac{1}{10} \right)^{x} & = y \\ {10}^{-x} & = y \\ & \\ \therefore r(x) = h^{-1}(x)& = {10}^{-x} \end{align} Therefore, $$r(x)$$ is an exponential function of the form $$y = b^{x}$$ with $$0 < b < 1$$. In words, the base $$b$$ is a positive fraction with a value between $$\text{0}$$ and $$\text{1}$$. ### Vertical and horizontal shifts If $$h$$ is shifted $$\text{1}$$ unit upwards and $$\text{2}$$ units to the right, then the new equation will be: $y = \log_{\frac{1}{10}}{(x - 2)} + 1$ The vertical asymptote is $$x = 2$$ and the horizontal asymptote is $$y = 1$$. Summary of graphs: $$y = b^{x}$$ and $$y = \log_{b}{x}$$ Exponential function Logarithmic function Axis of symmetry $$y = b^{x}$$ $$y = \log_{b}{x}$$ $$y = x$$ $$b > 1$$ $$0 < b < 1$$ Axis of symmetry $$y$$-axis, $$x = 0$$ $$x$$-axis, $$y = 0$$ ## Graphs and inverses of $$y=\log_{b}{x}$$ Textbook Exercise 2.10 Given $$f\left(x\right)={\left(\frac{1}{5}\right)}^{x}$$. Sketch the graphs of $$f$$ and $${f}^{-1}$$ on the same system of axes. Label both graphs clearly. State the intercept(s) for each graph. $$f: (0;1)$$ and $$f^{-1}: (1;0)$$ Label $$P$$, the point of intersection of $$f$$ and $$f^{-1}$$. Notice that the function and its inverse intersect at a point that lies on the line $$y = x$$. State the domain, range and asymptote(s) of each function. Function Domain Range Asymptote $$f(x) = \left( \frac{1}{5} \right)^{x}$$ $$\{x: x \in \mathbb{R} \}$$ $$\{y: y > 0, y \in \mathbb{R} \}$$ $$x$$-axis, $$y=0$$ $$f^{-1}(x) = \log_{\frac{1}{5}}{x}$$ $$\{x: x > 0, x \in \mathbb{R} \}$$ $$\{y: y \in \mathbb{R} \}$$ $$y$$-axis, $$x=0$$ Given $$g\left(x\right)= t^{x}$$ with $$M\left(1\frac{3}{5} ; 5\frac{4}{5}\right)$$ a point on the graph of $$g$$. Determine the value of $$t$$ \begin{align} \text{Let } y &= t^{x} \\ \text{5,8} &= t^{\text{1,6}} \\ \therefore t &= \sqrt[\text{1,6}]{\text{5,8}} \\ &= \text{3,000} \ldots \\ & \\ \therefore g(x) &= 3^{x} \end{align} Find the inverse of $$g$$. \begin{align} \text{Let } y & = 3^{x} \qquad (y > 0) \\ \text{Interchange } x \text{ and } y: \quad x & = 3^{y} \qquad (x > 0) \\ \frac{x}{2} & = y^{2} \\ y & = \log_{3}{x} \\ & \\ \therefore g^{-1}(x) & = \log_{3}{x} \qquad (x > 0) \end{align} Use symmetry about the line $$y = x$$ to sketch the graphs of $$g$$ and $${g}^{-1}$$ on the same system of axes. Point $$N$$ lies on the graph of $$g^{-1}$$ and is symmetrical to point $$M$$ about the line $$y = x$$. Determine the coordinates of $$N$$. $$N( \text{5,8}; \text{1,6})$$ ### Applications of logarithms (EMCFP) Logarithms have many different applications: • seismologists use logarithms to calculate the magnitude of earthquakes • financial institutions make use of logarithms to calculate the length of loan repayments • scientists use logarithms to determine the rate of radioactive decay • biologists use logarithms to calculate population growth rates • scientists use logarithms to determine pH levels $\text{pH} = - \log_{10}{[H^{+}]}$ ## Worked example 21: Population growth The population of a city grows by $$\text{5}\%$$ every two years. How long will it take for the city's population to triple in size? ### Write down a suitable formula and the known values $A=P{\left(1+i\right)}^{n}$ • Let $$P = x$$ • The population triples in size, so $$A = 3x$$ • Growth rate $$i = \frac{5}{100}$$ • Growth rate is given for a $$\text{2}$$ year period, so we use $$\frac{n}{2}$$ ### Substitute known values and solve for $$n$$ \begin{align} 3x &=x{\left(1+\frac{5}{100}\right)}^{\frac{n}{2}} \\ 3 & = {\left(1,05\right)}^{\frac{n}{2}} \end{align} ### Method $$\text{1}$$: take the logarithm of both sides of the equation \begin{align} \log{3}& = \log{\left(1,05\right)}^{\frac{n}{2}} \\ \log{3}& = \frac{n}{2}\times \log{1,05} \\ 2 \times \frac{\log3}{\log{1,05}} & = n \\ 45,034 \ldots & = n \end{align} ### Method $$\text{2}$$: change from exponential form to logarithmic form \begin{align} \frac{n}{2} & = \log_{\text{1,05}}{3} \\ & = \frac{\log{3}}{\log{\text{1,05}}} \\ n & = 2 \times \frac{\log3}{\log{1,05}} \\ n & = 45,034 \ldots \end{align} It will take approximately $$\text{45}$$ years for the city's population to triple in size. ## Applications of logarithms Textbook Exercise 2.11 The population of Upington grows $$\text{6}\%$$ every $$\text{3}$$ years. How long will it take to triple in size? This is a compound growth problem. You should first write down the initial size of the population, the final size of the population and the rate at which the population grows. Note that every period is $$\text{3}$$ years long. Let the population at the start be $P = x$ We want to know how many periods it will take to triple in size. This means that the final population is $A = 3x$ Let $$n$$ be the number of periods needed to grow to a size of $$A$$. Note that every period is $$\text{3}$$ years long, therefore we must use $$\frac{n}{\text{3}}$$. The growth rate is $i = \text{6}\% = \frac{\text{6}}{\text{100}}$ We use the formula for compound growth/interest: \begin{align} A &= P(1+i)^n \\ 3x &= x\left(1 + \frac{\text{6}}{\text{100}}\right)^{\frac{n}{\text{3}}} \\ \text{3} &= (\text{1,06})^{\frac{n}{\text{3}}} \end{align} We now use the definition of the logarithm to write the equation in terms of $$n$$ and then evaluate with a calculator: \begin{align} \frac{n}{\text{3}} &= \log_{(\text{1,06})} \text{3} \\ &= \frac{\log{3}}{\log{\text{1,06}}} \\ &= \text{18,854}\ldots \\ \therefore n &= 3 \times \text{18,854}\ldots \\ &= \text{56,562}\ldots \\ &\approx \text{57} \end{align} It will take approximately $$\text{57}$$ years for the population to triple in size. An ant population of $$\text{36}$$ ants doubles every month. Determine a formula that describes the growth of the population. Let the number of months be $$M_{0}; M_{1}; M_{2}; \ldots$$ $$M_{0}$$ $$M_{1}$$ $$M_{2}$$ $$\ldots$$ $$\text{36}$$ $$\rightarrow$$ $$\text{72}$$ $$\rightarrow$$ $$\text{144}$$ $$\rightarrow$$ $$\ldots$$ $$\times 2$$ $$\times 2$$ $$\times 2$$ $$36 \times 2^{0} \quad \rightarrow \quad 36 \times 2^{1} \quad \rightarrow \quad 36 \times 2^{2} \quad \rightarrow \quad \ldots$$ $\text{Growth } = 36 \times 2^{n}$ where $$n$$ is the number of months. Calculate how long it will take for the ant population to reach a quarter of a million ants. \begin{align} \text{Growth } &= 36 \times 2^{n} \\ \frac{1}{4} \times \text{1 000 000} &= 36 \times 2^{n} \\ \text{250 000} &= 36 \times 2^{n} \\ \frac{\text{250 000}}{36} &= 2^{n} \\ \therefore n &= \log_{2}{ \left( \frac{\text{250 000}}{36} \right) } \\ n &= \frac{\log{\left( \frac{\text{250 000}}{36} \right)}}{\log{2}} \\ \therefore n &= \frac{\log{\left( \frac{\text{250 000}}{36} \right)}}{\log{2}} \\ &= \text{12,7} \ldots \end{align} There will be a quarter of a million ants after about $$\text{13}$$ months.
# Factors of a number If two or more numbers are multiplied then the result is called product of those numbers. Also the product is called multiple of those numbers. The numbers used to get the product is called factors of the product. Example: 4 * 7 = 28 28 is the product. 4 and 7 are the factors of 28. If a number is exactly divided by a second number then the second number is a factor of the first number. In division if the remainder is zero (0) then the divisor and quotient are factors of the dividend. Example: 50 / 2 = 25 50 can be divided by 2 without any remainder. So 2 is a factor of 50. The quotient 25 is also a factor of 50. 1 is the factor of every number. Each number is the greatest factor of itself. ### Prime number: If a number has only two factors then the number is a prime number. The factors for a prime number are 1 and the prime number itself. 1 is not a prime number. Example: 2, 3, 5, 7, 11, etc are prime numbers because the factors are 1 and the number itself. ### Composite number If a number has more than two factors the the number is a composite number. Example: 4 is a composite factor because it has more than 2 factors. The factors of 4 are 1,2 and 4. ### Prime factor If a number is a factor of another number and also a prime number then the number is a prime factor of the other number. Example: 2 and 3 are prime factors of 12. 4, 6 and 12 are not prime factors of 12. 4 can written as (2 * 2), 6 can be written as (2 * 3) and 12 can be written as (2 * 2 * 3 ).
Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs! #### Online Quiz (WorksheetABCD) Questions Per Quiz = 2 4 6 8 10 ### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation3.17 Word Problems - Digit - 2 Example: If twenty-seven is added to a two-digit number, the original number will be reversed. The number is three less than four times the sum of its digits. What is the number? Solution: Let 'u' be the number in the units place and 't' be the number in the tens place. The number can be written as 10t + u If twenty-seven is added to a two-digit number, the original number will be reversed: 10t + u + 27 = 10u + t The number is three less than four times the sum of its digits: 10t + u = 4(t + u) - 3 10t + u + 27 = 10u + t -----------equation 1 10t + u = 4(t + u) - 3-------------equation 2 10t + u + 27 = 10u + t ----------- equation 1 10t - t = 10u - u - 27 9t = 9u - 27 t = u - 3 10t + u = 4(t + u) - 3 ----------- equation 2 10t + u = 4t + 4u - 3 6t = 3u - 3 substituting t = u - 3 in the above equation we have 2(u - 3) = u - 1 2u - 6 = u - 1 u = 6 - 1 = 5 t = u - 3 = 5 - 3 = 2 The two digit number is 25 Verification: If twenty-seven is added to a two-digit number, the original number will be reversed: 25 + 27 = 52 52 is a reverse of 25 The number is three less than four times the sum of its digits 4(2+5)-3 = 4(7) - 3 = 28 - 3 = 25 Directions: Solve the following word problems. Q 1: If you switch the ones digit by the tens digit, the sum of the 2 numbers will be 55. The ones digit is 3 more that the tens digit. What is the number. (Hint: Let u be the number in the units place and t be the number in the tens place. The number can be written as 10t + u. If the number is reversed then it will be 10 u + t. The sum of the number and its switch equals 55 implies 10t + u + 10u + t = 55. The other equation can be written as u = 3 + t. Solve both equations to find u and t and the number.)411425 Q 2: The tens digit is 2 less than the units digit. If the digit are reversed, the sum of the reversed number and the original number is 154. Find the original number.847668 Question 3: This question is available to subscribers only! Question 4: This question is available to subscribers only!
Scott McClelland r-mcclelland Assignment 1 Problem 1.3.3 Restatement: r1 = 2i + j - 3k r2 = i + j + k Redefine coordinates in R3 by: x' = x -1 y' = y + 2 z' = z Calculate the primed coordinates of the vectors r1 and r2, and verify that r2 - r1 = r2' - r1' and \|\|r2 - r1\|\| = \|\|r2' - r1'\|\|. Solution: First, we will find the primed coordinates of the vectors by taking the old xi value and subtracting 1 from it. Similarly, we would add 2 to the old yj coordinate and then leave the zk coordinate the same. This will give us the new vectors: r1' = i + 3j - 3k r2' = 3j + k Then, we will verify: r2-r1 = <1,1,1> - <2,1,-3> = <-1,0,4> r2'-r1' = <0,3,1> - <1,3,-3> = <-1,0,4> \|\|r2-r1\|\| = √(xi2 + yj2 + zk2) = √((-1)2 + 02 + (4)2) = √(17) \|\|r2'-r1'\|\| = √(xi2 + yj2 + zk2) = √((-1)2 + 02 + (4)2) = √(17)
# Conversion of Mixed Recurring Decimals into Vulgar Fractions Follow the steps for the conversion of mixed recurring decimals into vulgar fractions: (i) First write the decimal form by removing the bar from the top and put it equal to x (any variable). (ii) Now, find the number of digits having no bar after the decimal point. (iii) Suppose there are n-digits without bar, multiply both sides by 10n, so that only the repeating decimal digit is on the right side of the decimal point. (iv) Now write the repeating digits at least twice. (v) Now, find the number of digits having bar after the decimal point. (vi) Suppose there are n-digits having bar, multiply both sides by 10n. (vii) Then subtract the number obtained in step (i) from the number obtained in step (ii). (viii) Then divide both the sides of the equation by the coefficient of x. (ix) Therefore, we get the required vulgar fraction in the lowest form. Worked-out examples for the conversion of mixed recurring decimals into vulgar fractions: 1. Express 0.18 as vulgar fraction. Solution: x = 0.18 Multiply both sides by 10 (Since number of digits without bar is 1) 10x = 1.8 10x = 1.88…… ---------- (i) 10 × 10x = 1.88…… × 10 (Since number of digits having bars is 1) 100x = 18.8….. ------------- (ii) Subtracting (i) from (ii) 100x - 10x = 18.8 - 1.8 90x = 17 x = 17/90 Therefore, the vulgar fraction = 17/90 2. Express 0.23 as vulgar fraction. Solution: x = 0.23 Multiply both sides by 10 (Since number of digits without bar is 1) 10x = 2.3 10x = 2.33…… ---------- (i) 10 × 10x = 2.33…… × 10 (Since number of digits having bars is 1) 100x = 23.3….. ------------- (ii) Subtracting (i) from (ii) 100x - 10x = 23.3 - 2.3 90x = 21 x = 21/90 x = 7/30 Therefore, the vulgar fraction = 7/30 3. Express 0.43213 as vulgar fraction. Solution: x = 0.43213 Multiply both sides by 100 (Since number of digits without bar is 2) 100x = 43.213 100x = 43.213213…… ---------- (i) 100 × 1000x = 43.213…… × 1000 (Since number of digits having bars is 3) 100000x = 43213.213….. ------------- (ii) Subtracting (i) from (ii) 100000x - 100x = 43213.213 - 43.213213 99900x = 43170 x = 43170/99900 x = 4317/9990 Therefore, the vulgar fraction = 4317/9990 Shortcut method for solving the problems on conversion of mixed recurring decimals into vulgar fractions: The difference between the number formed by all the digits in decimal part and the number formed by digits that are not repeated, gives the numerator of the vulgar fraction and for its denominator the number formed by as many nines as there are recurring digits which are repeated followed by as many zeros as the number of non-repeating or non-recurring digits. For example; Express 0.123 as vulgar fraction. Numerator = 123 - 12 = 111 Denominator = one nine (as there are one recurring digit) followed by two zeros (as there are two non-recurring digits) = 900 Required fraction = 111/900 (reduce to its simplest form) Therefore, the vulgar fraction = 37/300 Related Concept From Conversion of Mixed Recurring Decimals into Vulgar Fractions to HOME PAGE
Standard deviation is a measure of variance within a data set. So in this post we learn about How to Calculate Standard Deviation? Step by Step with Example. It can be thought of as the average distance from the mean (calculated average) for each individual data point in a data set. ## How to Calculate Standard Deviation? Step by Step with Example Standard deviation is a statistical operation that has wide applications, but for our purposes we are discussing it as it relates to the Six Sigma program. As a reference, below is the standard deviation formula, along with a key and the appropriate order of operations. ## The Standard Deviation Formula ### How to Calculate Standard Deviation? Step by Step with Example There are the some steps of calculating the standard deviation. Way How to Calculate Standard Deviation? Step by Step with Example are as follow: ### Step 1 First, we will need to find the arithmetic mean of the data set so that we have a value for x-bar within the equation. This is the Basic Average formula of Standard Deviation: find the sum of all the numbers within the data set, then divide by the number of numbers. Here we are describing doing things the old-fashioned way, but any spreadsheet software can find the average of a data set with the click of a button. In many cases, standard deviation is calculated as well. Though this formula is important, it is often unnecessary to keep it in your memory. ### Step 2 Now we need to perform the (x â€“ xÌ…)2 function. On a computer, this is easy. With drag and drop formulas and copy and paste functions there is no need to write everything out. If, however, you want to complete this calculation by hand, the best way to do it is to build a table like the one below. This demonstration table is constructed using the sample data set 5, 12, 16, 21, 28. The average for this data set is 16.4. Plugged into our equation, it now looks like this: And the results: Note that we are squaring each of the values. Later, we will be finding the square root of the result, but squaring numbers now means that there are no negative values. Negative values get messy when they are in the numerator or denominator of an equation. ### Step 3 Now that we have our results for x minus x-bar quantity squared, we see the uppercase sigma statistical operator in the equation. Simply add the values in the last column. 129.96 + 19.36 + .16 + 21.16 + 134.56 = 305.2 Plugged into our equation. There are five data points in the set, which means our n value is 5. Plug that into the equation and we get: Or, the square root of 305.2/4, which is 76.3. The square root of that number rounds to 8.73, meaning that our formula is solved as ## Ïƒ = 8.73 To put that another way, the average distance that any one data point is away from the average of 16.4 is 8.73. To learn more about how sigma values are used in the world of Six Sigma, see our breakdown of the topic here, or check out this condensed version on Quora. In this way we calculate the Standard deviation and variance.
# Math Snap ## Which of the following is not an asymptote of $f(x)=\frac{e^{2 x}-e^{x}+2}{e^{2 x}-5 e^{x}+6} ?$ Select one: a. $y=1$ b. $y=1 / 3$ C. $x=\ln 2$ d. $x=\ln 5$ #### STEP 1 Assumptions 1. The function given is $f(x) = \frac{e^{2x} - e^x + 2}{e^{2x} - 5e^x + 6}$. 2. We need to determine which of the given options is not an asymptote of the function. 3. Asymptotes can be vertical or horizontal. 4. Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. 5. Horizontal asymptotes are determined by the behavior of the function as $x$ approaches $\infty$ or $-\infty$. #### STEP 2 First, let's find the vertical asymptotes by setting the denominator equal to zero and solving for $x$. $e^{2x} - 5e^x + 6 = 0$ #### STEP 3 Let $u = e^x$. Then the equation becomes: $u^2 - 5u + 6 = 0$ #### STEP 4 Solve the quadratic equation $u^2 - 5u + 6 = 0$ by factoring. $(u - 2)(u - 3) = 0$ #### STEP 5 Set each factor equal to zero and solve for $u$. $u - 2 = 0 \quad \Rightarrow \quad u = 2$ $u - 3 = 0 \quad \Rightarrow \quad u = 3$ #### STEP 6 Recall that $u = e^x$. So we have: $e^x = 2 \quad \Rightarrow \quad x = \ln 2$ $e^x = 3 \quad \Rightarrow \quad x = \ln 3$ #### STEP 7 Thus, the vertical asymptotes are $x = \ln 2$ and $x = \ln 3$. #### STEP 8 Now, let's check the given options for vertical asymptotes. We see that $x = \ln 2$ is a vertical asymptote. #### STEP 9 Next, let's determine the horizontal asymptotes by examining the limits of $f(x)$ as $x \to \infty$ and $x \to -\infty$. #### STEP 10 As $x \to \infty$, the terms $e^{2x}$ dominate both the numerator and the denominator. Thus, we can approximate: $f(x) \approx \frac{e^{2x}}{e^{2x}} = 1$ #### STEP 11 So, $y = 1$ is a horizontal asymptote. #### STEP 12 As $x \to -\infty$, both $e^{2x}$ and $e^x$ approach 0. The function approaches: $f(x) \approx \frac{2}{6} = \frac{1}{3}$ #### STEP 13 So, $y = \frac{1}{3}$ is also a horizontal asymptote. #### STEP 14 Now, let's check the given options for horizontal asymptotes. We see that $y = 1$ and $y = \frac{1}{3}$ are horizontal asymptotes. #### STEP 15 Reviewing the options: a. $y = 1$ (horizontal asymptote) b. $y = \frac{1}{3}$ (horizontal asymptote) c. $x = \ln 2$ (vertical asymptote) d. $x = \ln 5$ (not an asymptote) ##### SOLUTION The option that is not an asymptote is $x = \ln 5$.
# Individual invest on each asset Let's say that an individual has $$\\\ 10,000$$ to invest in a combination of two risky assets. Each asset can lead to a high return or a low one. However, returns are not independent. Their joint probability distribution is provided in the following table. A A A B 20% 5% B 5% 0.4 0.1 B 20% 0.1 0.4 This means, for example, that the probability that the returns from assets $$A$$ and $$B$$ are $$5 \%$$ and $$20 \%$$ respectively is 0.4 . Let the individual's Bernoulli utility function be given by $$u(x)=\sqrt{x}$$. In such a case, how much will this individual invest on each asset? The individual will invest the whole $$\\\ 10,000$$, which makes sense, since in the worst case scenario he still gets a positive return. He has to decide which fraction $$\alpha$$ to invest in asset $$A$$ and which fraction $$1-\alpha$$ to invest in asset $$B$$. Note that if returns are the same for both assets, the fraction invested on each is irrelevant. Then, his problem can be written as follows, \begin{aligned} \max _{\alpha} B= & \frac{1}{10} \sqrt{1.2 \times 10,000}+\frac{1}{10} \sqrt{1.05 \times 10,000} \\ & +\frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\ & +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000} \end{aligned} Let's simplify the problem first. Since there are two situations in which $$\alpha$$ does not affect the outcome, we can rewrite the problem as follows, \begin{aligned} \max _{\alpha} \tilde{B}= & \frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\ & +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000} \\ \max _{\alpha} B= & \frac{4 \sqrt{10,000}}{10}[\sqrt{1.2 \alpha+1.05(1-\alpha)}+\sqrt{1.05 \alpha+1.2(1-\alpha)}] \end{aligned} Then, we can just solve, $$\max _{\alpha} B=\sqrt{1.05+0.15 \alpha}+\sqrt{1.2-0.15 \alpha}$$ The first order condition is given by: $$\frac{\partial B}{\partial \alpha}=0.15(1.05+0.15 \alpha)^{-\frac{1}{2}}-0.15(1.2-0.15 \alpha)^{-\frac{1}{2}}=0$$ Then, \begin{aligned} (1.05+0.15 \alpha)^{-\frac{1}{2}} & =(1.2-0.15 \alpha)^{-\frac{1}{2}} \\ 1.05+0.15 \alpha & =1.2-0.15 \alpha \\ 0.3 \alpha & =0.15 \end{aligned} Then, the $$\alpha=0.5$$. This is not surprising, given that the assets are symmetric. Let us finally verify that the second order condition holds: $$\frac{\partial^{2} B}{\partial \alpha^{2}}=-\frac{1}{2} 0.15^{2}(1.05+0.15 \alpha)^{-\frac{3}{2}}-\frac{1}{2}(-0.15)^{2}(1.2-0.15 \alpha)^{-\frac{3}{2}}<0$$
# Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$ Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$ My approach : $= \frac{1}{6}+\frac{2^2}{6^2}+\frac{3^2}{6^3} +\cdots \infty$ Starting with \begin{align} \frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n} \end{align} then differentiate and multiply by $t$ to obtain \begin{align} \frac{t}{(1-t)^{2}} = \sum_{n=0}^{\infty} n t^{n}. \end{align} Repeating leads to \begin{align} \sum_{n=0}^{\infty} n^{2} \ t^{n} = \frac{t(1+t)}{(1-t)^{3}}. \end{align} Now let $t = 1/6$ for which \begin{align} \sum_{n=0}^{\infty} \frac{n^{2}}{6^{n}} = \frac{42}{125}. \end{align} • Oh right, this is $Li_{-2}(6^{-1})$. Completely missed that. – Gina Jun 10 '14 at 4:08 Break $n^2$ into two parts: $\underbrace{n(n-1)} + \underbrace{{}\quad n\quad {}}$. The first part appears in the second derivative of $x^n$ and the second part in the first derivative: (Finally at the end, put $x=1/6$.)
# Quotient rule in calculus ## Statement The derivative for quotients or fractions is the following. $$\frac{d}{dx}\left(\frac{g(x)}{j(x)}\right) = \frac{g'(x) * j(x) - g(x) * j'(x)}{j(x)^2}$$ ## Proof Define function $f$ and write it as a product. $$f(x) = \frac{g(x)}{j(x)}$$ $$f(x) = g(x) * j(x)^{-1}$$ Differentiate $f$ using the product rule. $$f'(x) = \tfrac{d}{dx}\bigg(g(x)\bigg) * j(x)^{-1} + g(x) * \tfrac{d}{dx}\bigg(j(x)^{-1}\bigg)$$ $$f'(x) = g'(x) * j(x)^{-1} + g(x) * \tfrac{d}{dx}\bigg(j(x)^{-1}\bigg)$$ Differentiate $\tfrac{d}{dx}\bigg(j(x)^{-1}\bigg)$ using the power rule and the chain rule. $$f'(x) = g'(x) * j(x)^{-1} + g(x) * -1 * j(x)^{-2} * \tfrac{d}{dx}\bigg(j(x)\bigg)$$ $$f'(x) = g'(x) * j(x)^{-1} - g(x) * j(x)^{-2} * j'(x)$$ Write negative exponents as fractions and make a common denominator. $$f'(x) = \frac{g'(x)}{j(x)} - \frac{g(x) * j'(x)}{j(x)^{2}}$$ $$f'(x) = \frac{g'(x) * j(x)}{j(x)^2} - \frac{g(x) * j'(x)}{j(x)^{2}}$$ $$f'(x) = \frac{g'(x) * j(x) - g(x) * j'(x)}{j(x)^{2}}$$ ## Proofs building upon this proof ### Derivative of tan(x) This proof shows what the derivative of tan(x) is.
# Probability of Independent and Dependent Events An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: Either/Or Probability: Overlapping and Non-Overlapping Events ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:05 Independent and… • 0:53 Probability • 3:26 Probability of… • 5:11 Probability of… • 8:00 Dependent Events Example • 10:40 Lesson Summary Want to watch this again later? Timeline Autoplay Autoplay #### Recommended Lessons and Courses for You Lesson Transcript Chad has taught Math for the last 9 years in Middle School. He has a M.S. in Instructional Technology and Elementary Education. Sometimes probabilities need to be calculated when more than one event occurs. These types of compound events are called independent and dependent events. Through this lesson, we will look at some real-world examples of how to calculate these probabilities. ## Independent and Dependent Events Independent events are events that do not affect the outcome of subsequent events. In an independent event, each situation is separate from previous events. An example of an independent event would be selecting a card from a deck of cards and then returning the card to the deck. After you return the card, select another card from the same, equal deck. Dependent events are just the opposite. Dependent events are events in which the previous attempts affect the outcome of subsequent events. Dependent events are just like they sound - each event is dependent upon what happened in the previous attempt. An example of a dependent event would be selecting a card from a deck of cards and not replacing the card. Then you draw another card from the now-smaller deck of cards. ## Probability The probability of an event occurring is a ratio that states the likelihood of an event happening. To find the probability of a single event, you will use the ratio of the number of favorable outcomes over the number of total outcomes. Let's look at an example: What is the probability of selecting a king from a standard deck of cards? Looking at the deck of cards, we know that there are 52 cards in each deck. The number of total outcomes is 52 because we could select any of the 52 cards when choosing at random. We can also see that there are 4 kings in each deck of cards. So, the number of favorable outcomes would be 4. We can see that there is a ratio of 4 to 52 chances of selecting a king at random from a deck of cards. All ratios must be in simplest form, though, so 4/52 will be reduced to 1/13. The probability of selecting a king at random from a standard deck of cards is 1/13. Sometimes you will be asked to find the probability of more than one event occurring in consecutive order. When this occurs, you will need to multiply both probabilities together to calculate the combined probability. Let's look at an example where you will be asked to find the probability of more than one event occurring: What is the probability of rolling a standard die and getting a 2 and then rolling again and getting another 2? To start this problem, you will need to calculate the probability of each event happening independently. The probability of rolling a 2 on a standard die can be found by using the formula: total number of favorable outcomes over the total number of possible outcomes. On a die, there are 6 total outcomes, and only one of the outcomes is a 2. So, the number of favorable outcomes is 1 and the number of total outcomes is 6. So, the probability of rolling a 2 is 1/6. Since the probability of rolling a 2 is 1/6, the probability of rolling a 2 on the next roll would be the same. To calculate the probability of both events happening together, we will need to multiply the two probabilities together. By multiplying these two probabilities together, you get 1/36. So, the probability of rolling a die and getting a 2, then rolling a die again and getting another 2, would be 1/36. ## Probability of Independent Events Again, independent events are the events that do not affect the outcome of subsequent events. The last situation was an example of an independent event. The rolling of the die and getting a 2 did not affect the outcome of the second event of rolling the die again. Each rolling of the die is an independent event. Let's look at another example of an independent event. Jamie and Sam were having a debate and decided the best way to settle their dispute was by flipping a coin. Jamie decided to be heads, and Sam decided to take tails. Jamie is now curious: What is the probability that by flipping a coin 2 times they will receive a heads each time? To find this probability, Jamie must find out the probability of each event occurring separately. To calculate the probability, Jamie must use the formula: the number of favorable outcomes over the number of possible outcomes. Jamie knows that the coin has two sides, and only one of them is heads. So, the probability of Jamie getting a heads on the first flip is 1/2. As the boys prepare to flip a coin for the second time, they know the probability of flipping a coin and getting a heads is 1/2. The probability of getting a heads on the second flip is also 1/2. To find the probability of these two events happening together, we need to multiply these two probabilities together. When we multiply 1/2 times 1/2, we get 1/4. Jamie now knows that the probability of flipping a coin twice and getting heads both times is 1/4. ## Probability of Dependent Events Just the opposite of independent events, dependent events are events in which previous attempts affect the outcome of subsequent events. Dependent events are just like they sound - each event is dependent upon what happened in the previous attempt. Let's look at an example of these dependent events. James the Superb Magician likes to dazzle and amaze his audience with a card trick in which he selects two cards at random from a deck of cards but announces the cards that he will select prior to selecting them. James asks for an audience member to join him for the trick. Wendy, a female audience member, raises her hand to volunteer. The crowd gives her a rousing applause as she makes her way to the stage. James explains to Wendy that he will select a card at random from the deck of cards. Wendy examines the deck of cards to make sure the deck is fair. James announces that he will first draw an ace from the deck. Wendy wonders: What is the probability that James will select an ace from the deck of cards? She knows that there are 4 aces out of 52 cards. James explains to Wendy that he will now select another card from the deck, and it will also be an ace. Wendy is perplexed at how James could be lucky enough to draw two aces in a row from the deck of cards. She thinks to herself: What is the probability of James selecting an ace and then, without replacing the card, selecting another ace? Wendy knows that to find the probability of him selecting an ace from the deck of cards, she will need to use the formula: total number of favorable outcomes over the total number of outcomes. In the first event, the number of favorable outcomes is 4 because James was selecting one of the 4 aces. The total number of outcomes is 52 because there are 52 cards in a standard deck of cards. So, Wendy knows the probability of James selecting an ace on the first draw is 4/52. To unlock this lesson you must be a Study.com Member. ### Register for a free trial 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 160 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.
# Math Formulas for Class 11 Chapter 3 Trigonometric Functions Here we are providing Math Formulas for Class 11 Chapter 3 Trigonometric Functions. These formulas are important for all students studying maths as well as physics. Students studying physics also requires these formulas while deriving some physics formula and it is very useful for solving physics problems also. Students are suggested to remember these formulas. RELATION BETWEEN SYSTEM OF MEASUREMENT OF ANGLES: \begin{aligned} & \frac{\mathrm{D}}{90}=\frac{\mathrm{G}}{100}=\frac{2 C}{\pi} \\ & 1 \text { Radian }=\frac{180}{\pi} \text { degree } \approx 57^{\circ} 17^{\prime} 15^{\prime \prime} \text { (approximately) } \\ & 1 \text { degree }=\frac{\pi}{180} \text { radian } \approx 0.0175 \text { radian } \end{aligned} BASIC TRIGONOMETRIC IDENTITIES: (a) $$\sin ^2 \theta+\cos ^2 \theta=1$$ or $$\sin ^2 \theta=1-\cos ^2 \theta$$ or $$\cos ^2 \theta=1-\sin ^2 \theta$$ (b) $$\sec ^2 \theta-\tan ^2 \theta=1$$ or $$\sec ^2 \theta=1+\tan ^2 \theta$$ or $$\tan ^2 \theta=\sec ^2 \theta-1$$ (c) If $$\sec \theta+\tan \theta=\mathrm{k} \\ \Rightarrow \sec \theta-\tan \theta=\frac{1}{\mathrm{k}} \\ \Rightarrow 2 \sec \theta=\mathrm{k}+\frac{1}{\mathrm{k}}$$ (d) $$\operatorname{cosec}^2 \theta-\cot ^2 \theta=1$$ or $$\operatorname{cosec}^2 \theta=1+\cot ^2 \theta$$ or $$\cot ^2 \theta=\operatorname{cosec}^2 \theta-1$$ (e) If $$\operatorname{cosec} \theta+\cot \theta=k \\ \Rightarrow \operatorname{cosec} \theta-\cot \theta=\frac{1}{k} \\ \Rightarrow 2 \operatorname{cosec} \theta=k+\frac{1}{k}$$ SIGNS OF TRIGONOMETRIC FUNCTIONS IN DIFFERENT QUADRANTS: TRIGONOMETRIC FUNCTIONS OF ALLIED ANGLES: \begin{aligned} & \sin (2 \mathrm{n} \pi+\theta)=\sin \theta, \cos (2 \mathrm{n} \pi+\theta)=\cos \theta \text {, where } \mathrm{n} \in \mathrm{I}\\ & \begin{array}{ll} \sin (-\theta)=-\sin \theta & \\ \cos (-\theta)=\cos \theta \\ \sin \left(90^{\circ}-\theta\right)=\cos \theta & \\ \cos \left(90^{\circ}-\theta\right)=\sin \theta \\ \sin \left(90^{\circ}+\theta\right)=\cos \theta & \\ \cos \left(90^{\circ}+\theta\right)=-\sin \theta \\ \sin \left(180^{\circ}-\theta\right)=\sin \theta & \\ \cos \left(180^{\circ}-\theta\right)=-\cos \theta \\ \sin \left(180^{\circ}+\theta\right)=-\sin \theta & \\ \cos \left(180^{\circ}+\theta\right)=-\cos \theta \\ \sin \left(270^{\circ}-\theta\right)=-\cos \theta & \\ \cos \left(270^{\circ}-\theta\right)=-\sin \theta \\ \sin \left(270^{\circ}+\theta\right)=-\cos \theta & \\ \cos \left(270^{\circ}+\theta\right)=\sin \theta \end{array} \end{aligned} IMPORTANT TRIGONOMETRIC FORMULAE: (i) $$\sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$$ (ii) $$\quad \sin (A-B)=\sin A \cos B-\cos A \sin B$$ (iii) $$\cos (A+B)=\cos A \cos B-\sin A \sin B$$ (iv) $$\quad \cos (A-B)=\cos A \cos B+\sin A \sin B$$ (v) $$\tan (\mathrm{A}+\mathrm{B})=\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A} \tan \mathrm{B}}$$ (vi) $$\quad \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$$ (vii) $$\quad \cot (\mathrm{A}+\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}-1}{\cot \mathrm{B}+\cot \mathrm{A}}$$ (viii) $$\cot (\mathrm{A}-\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}+1}{\cot \mathrm{B}-\cot \mathrm{A}}$$ (ix) $$2 \sin A \cos B=\sin (A+B)+\sin (A-B)$$ (x) $$2 \cos A \sin B=\sin (A+B)-\sin (A-B)$$ (xi) $$2 \cos A \cos B=\cos (A+B)+\cos (A-B)$$ (xii) $$2 \sin A \sin B=\cos (A-B)-\cos (A+B)$$ (xiii) $$\sin \mathrm{C}+\sin \mathrm{D}=2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$$ (xiv) $$\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$$ (xv) $$\quad \cos \mathrm{C}+\cos \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$$ (xvi) $$\quad \cos C-\cos D=2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{D-C}{2}\right)$$ (xvii) $$\sin 2 \theta=2 \sin \theta \cos \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}$$ (xviii) $$\cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta=2 \cos ^2 \theta-1=1-2 \sin ^2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$$ (xix) $$1+\cos 2 \theta=2 \cos ^2 \theta$$ or $$\|\cos \theta\|=\sqrt{\frac{1+\cos 2 \theta}{2}}$$ (xx) $$1-\cos 2 \theta=2 \sin ^2 \theta$$ or $$\|\sin \theta\|=\sqrt{\frac{1-\cos 2 \theta}{2}}$$ (xxi) $$\tan \theta=\frac{1-\cos 2 \theta}{\sin 2 \theta}=\frac{\sin 2 \theta}{1+\cos 2 \theta}$$ (xxii) $$\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}$$ (xxiii) $$\sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta$$ (xxiv) $$\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta$$ (xxv) $$\tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}$$ (xxvi) $$\sin ^2 A-\sin ^2 B=\sin (A+B) \cdot \sin (A-B)=\cos ^2 B-\cos ^2 A$$ (xxvii) $$\cos ^2 A-\sin ^2 B=\cos (A+B) \cdot \cos (A-B)$$
# Topological Sorting Algorithm In this post, we are continuing with Graph series and we will discuss the Topological Sorting algorithm and some problems based on it. Topological Sorting Algorithm is very important and it has vast applications in the real world. So, let’s start. ## Directed Acyclic Graph We have already discussed the directed and undirected graph in this post. So, now let’s discuss the cyclic and acyclic graph. The simplest definition would be that if a Graph contains a cycle, it is a cyclic graph else it is an acyclic Graph. In undirected graph, to find whether a graph has a cycle or not is simple, we will discuss it in this post but to find if there is a cycle present or not in a directed graph, Topological Sort comes into play. ## Detect cycle in undirected Graph Now let’s discuss how to detect cycle in undirected Graph. For that, let’s take an example, Now let me ask you, what is the difference between the above two Graphs ..? Yes, you guessed it right, the one in the left side is undirected acyclic graph and the other one is cyclic. ### Why the graph on the right side is called cyclic ? The reason is simple, there is at least two ways to reach any node of the cycle and this is the main logic to find a cycle in undirected Graph. If an undirected Graph is Acyclic, then there will be only one way to reach the nodes of the Graph. ## Algo to find Cycle in undirected Graph To find cycle, we will simply do a DFS Traversal and also keep track of the parent vertex of the current vertex. If parent vertex is unique for every vertex, then graph is acyclic or else it is cyclic. Let’s see the code, ``````unordered_map<int, vector<int>> adj; bool detect(int s, bool vis[], int parent) { vis[s] = true; { if(vis[i] == false) { if(detect(i, vis, s) == true) return true; } else if(i != parent){ return true; } } return false; } `````` For every vertex, the parent will be the vertex from which we reach the current vertex. Initially, parents will be -1 but accordingly, we will update the parent when we move ahead. Hope, code, and logic is clear to you. Let’s move ahead. ## Detect Cycle in Directed Graph For directed Graph, the above Algorithm may not work. Let’s see how, The above Directed Graph is Acyclic, but the previous algorithm will detect a cycle because vertex 1 has two parents (vertex 2 and vertex 3), which violates our rule. Although the above-directed Graph is Acyclic, the previous algorithm will detect a cycle. So the Algorithm fails. To detect a cycle in a Directed Acyclic Graph, the topological sort will help us but before that let us understand what is Topological Sorting? ## Topological Sort We have discussed many sorting algorithms before like Bubble sort, Quick sort, Merge sort but Topological Sort is quite different from them. Topological Sort for directed cyclic graph (DAG) is a algorithm which sort the vertices of the graph according to their indegree. Let’s understand it clearly, ### What is in-degree and out-degree of a vertex ? OutDegree of a vertex (let say x) refers to the number of edges directed away from x . Similarly, In-Degree of a vertex (let say y) refers to the number of edges directed towards y from other vertices. Let’s see an example, Hope, concept of in-degree and out-degree is clear to you. Now in Topological Sorting, we sort the vertices of graph according to their In-degree. Let’s take the same example to understand Topological Sorting, Topological Sorting of above Graph : 2 3 1 Let’s take another example, Topological Sorting of above Graph : 0 5 2 4 1 3 6 There may be multiple Topological Sort for a particular graph like for the above graph one Topological Sort can be 5 0 4 2 3 6 1, as long as they are in sorted order of their in-degree, it may be the solution too. Hope, concept of Topological Sorting is clear to you. Now let’s discuss the algorithm behind it, ## Topological Sorting Algorithm(BFS) We can find Topological Sort by using DFS Traversal as well as by BFS Traversal. We will discuss both of them. Let’s first the BFS approach to finding Topological Sort, Step 1: First we will find the in degrees of all the vertices and store it in an array. Let’s discuss how to find in-degree of all the vertices. For that, the adjacency list given us will help, we will go through all the neighbours of all the vertices and increment its corresponding array index by 1. Let’s see the code, ``````unordered_map<int, vector<int>> adj; int topological_Sort(int V) { int in_degree[V] = {0}; for(int i = 0; i < V; i++) { { in_degree[itr]++; } } }`````` Step 2 : We will declare a queue, and we will push the vertex with in-degree 0 to it. Step 3 : We will run a loop until the queue is empty, and pop out the front element and print it. The popped vertex has the least in-degree, also after popping out the front vertex of the queue, we will decrement in-degree of it’s neighbours by 1. It is obvious, removal of every vertex will decrement the in-degree of it’s neighbours by 1. Step 4: If in-degree of any neighbours of popped vertex reduces to 0, then push it to the queue again. Let’s see the above process, ``````unordered_map<int, vector<int>> adj; int topological_Sort(int V) { int in_degree[V] = {0}; for(int i = 0; i < V; i++) { { in_degree[itr]++; } } queue<int> q; for(int i=0;i<V;i++) { if(in_degree[i] == 0) q.push(i); } while(!q.empty()) { int p = q.front(); q.pop(); cout << p <<" "; { in_degree[itr]--; if(in_degree[itr] == 0) q.push(itr); } } }`````` That’s it, the printed data will be our Topological Sort, hope Algorithm and code is clear. Let’s understand it by an example, in_degree[] for above graph will be, {0, 2, 1, 2, 1, 0, 2}. Now let’s move ahead, The above pictorial diagram represents the process of Topological Sort, output will be 0 5 2 3 4 1 6. Time Complexity : O(V + E) Space Complexity : O(V) Hope concept and code is clear to you. Let’s move ahead. ## Cycle Detection in Directed Graph Since we have discussed Topological Sorting, let’s come back to our main problem, to detect cycle in a Directed Graph. Let’s take an simple example, In the example above, graph on left side is acyclic whereas graph on right side is cyclic. Run Topological Sort on both the Graphs, what is your result..? For the graph on left side, Topological Sort will run fine and your output will be 2 3 1. But for the graph on right side, Topological Sort will print nothing and it’s obvious because queue will be empty as there is no vertex with in-degree 0. Now, let’s analyse why is it happening..? ### Significance of vertex with in-degree 0 As observed for the above case, there was no vertex present in the Graph with in-degree 0. This signifies that there is no vertex present in the graph which is not connected to atleast one other vertex. You know what is signifies..? It signifies the presence of a cycle, because it can only be possible in the case of cycle, when no vertex with in-degree 0 is present in the graph. Let’s take another example, Again run Topological Sort for the above example, If we run Topological Sort for the above graph, situation will arise where Queue will be empty in between the Topological Sort without exploration of every vertex. And this again signifies a cycle. Hope you understood the concept behind it. Let’s see the code, ``````unordered_map<int, vector<int>> adj; int* topological_Sort(int V) { int in_degree[V] = {0}; for(int i = 0; i < V; i++) { { in_degree[itr]++; } } queue<int> q; for(int i=0;i<V;i++) { if(in_degree[i] == 0) q.push(i); } int cnt = 0; while(!q.empty()) { int p = q.front(); q.pop(); cnt++; { in_degree[itr]--; if(in_degree[itr] == 0) q.push(itr); } } if(cnt != V) cout<<"Cycle Present\n"; else cout<<"Cycle Not Present\n"; }`````` Hope code is simple, we are just counting the occurrence of vertex, if it is not equal to V, then cycle is present as topological Sort ends before exploring all the vertices. ### Logic behind the Algorithm (MasterStroke) The main logic of the above algorithm is that if there is a cycle present in a directed Graph, definitely a situation will arise where no vertex with in-degree 0 will be found because for having a cycle, minimum in-degree 1 is required for every vertices present in the cycle. It’s obvious logic and hope, code and logic is clear to you all. Let’s move ahead. ## Topological Sorting Algorithm (DFS) Although this topic was not important as we have already discussed the BFS approach which is relatively easier to understand but sometimes in an interview, interviewer ask you to find Topological Sort by DFS approach. So it’s better to give it a look. It is highly recommended to try it before moving to the solution because now you are familiar with Topological Sorting. So, give it a try for sure. Let’s take the same example, It’s clear in topological Sorting our motive is to give preference to vertex with least in-degree. In other words, if we give preference to vertex with least out-degree and reverse the order of Topological Sort, then also we can get our desired result. Let’s say, Topological Sorting for above graph is 0 5 2 4 3 1 6. In above diagram number of out-degrees in written above every vertex. If we sort it with respect to out-degree, one of the Topological Sort would be 6 1 3 4 2 5 0 and reverse of it will give you Topological Sort w.r.t in-degree. Hope this is clear and this is the logic of this algorithm of finding Topological Sort by DFS. ## Algorithm to find Topological Sort(DFS) Step 1: Do a DFS Traversal and if we reach a vertex with no more neighbors to explore, we will store it in the stack. 2: Continue this process until DFS Traversal ends. Step 3: Take out elements from the stack and print it, the desired result will be our Topological Sort. That’s it. NOTE: Topological Sort works only for Directed Acyclic Graph (DAG). Let’s see the code for it, ``````unordered_map<int,vector<int>> adj; void f(int s, stack<int> &stack, bool vis[]) { vis[s] = true; { if(vis[i]==false) { f(i,stack,vis); } } stack.push(s); } void topological_Sort(int V) { bool vis[V] = {false}; stack<int> s; for(int i=0;i<V;i++) { if(vis[i] == false) { f(i, s, vis); } } while(!s.empty()) { cout << s.top() << " "; s.pop(); } }`````` Hope code is clear, it is simple code and logic is similar to what we have discussed before. DFS Traversal sorts the vertex according to out-degree and stack is helping us to reverse the result. That’s it. Time Complexity : O(V + E) Space Complexity : O(V) I hope you enjoyed this post about the topological sorting algorithm. We will continue with the applications of Graph. See you later in the next post. That’s all folks..!!!Wiki Abhishek is currently pursuing CSE from Heritage Institute of Technology, Kolkata. He has a great interest in Data Structures and Algorithms, C++, Language, Competitive Coding, Android Development. His hobbies are Learning new skills, Content Writing, Competitive Coding, Teaching contents to Beginners.
## The utility of trigonometrical substitutions Today we will discuss the power of trigonometrical substitutions. Let us take the expression $\frac{\sum_{k=1}^{2499} \sqrt{10+\sqrt{50+\sqrt{k}}}}{\sum_{k=1}^{2499} \sqrt{10-\sqrt{50+\sqrt{k}}}}$ This is a math competition problem. One solution proceeds this way: let $p_k=\sqrt{50+\sqrt{k}}, q_k=\sqrt{50-\sqrt{k}}$. Then as $p_k^2+q_k^2=10^2$, we can write $p_k=10\cos x_k$ and $q_k=10\sin x_k$. This is an elementary fact. But what is the reason for doing so? Now we have $a_k=\sqrt{10+\sqrt{50+\sqrt{k}}}=\sqrt{10+10\cos x_k}=\sqrt{20}\cos \frac{x_k}{2}$. Similarly, $b_k=\sqrt{10-\sqrt{50+\sqrt{k}}}=\sqrt{10-10\cos x_k}=\sqrt{20}\sin \frac{x_k}{2}$. The rest of the solution can be seen here. It mainly uses identities of the form $2\sin A\cos B=(\sin A+\cos B)^2$ to remove the root sign. What if we did not use trigonometric substitutions? What is the utility of this method? We will refer to this solution, and try to determine whether we’d have been able to solve the problem, using the same steps, but not using trigonometrical substitutions. $a_{2500-k}=\sqrt{10+\sqrt{50+\sqrt{2500-k}}}=\sqrt{10+\sqrt{50+\sqrt{(50+\sqrt{k})(50-\sqrt{k})}}}$ $=\sqrt{10+\sqrt{50+100\frac{\sqrt{50+\sqrt{k}}}{10}\frac{\sqrt{50-\sqrt{k}}}{10}}}=\sqrt{10+\sqrt{50}\sqrt{1+2\frac{\sqrt{50+\sqrt{k}}}{10}\frac{\sqrt{50-\sqrt{k}}}{10}}}$ $=\sqrt{10+10(\frac{\sqrt{50+\sqrt{k}}}{10\sqrt{2}}+\frac{\sqrt{50-\sqrt{k}}}{10\sqrt{2}})}=\sqrt{10+10\times 2\times\frac{\frac{1}{2}(\frac{\sqrt{50+\sqrt{k}}}{10\sqrt{2}}+\frac{\sqrt{50-\sqrt{k}}}{10\sqrt{2}})}{\sqrt{\frac{50+\sqrt{k}}{20\sqrt{2}}-\frac{50-\sqrt{k}}{20\sqrt{2}}+\frac{1}{2}}}\times \sqrt{\frac{50+\sqrt{k}}{20\sqrt{2}}-\frac{50-\sqrt{k}}{20\sqrt{2}}+\frac{1}{2}}}$ As one might see here, our main aim is to remove the square root radicals, and forming squares becomes much easier when you have trigonometrical expressions. Every trigonometrical expression has a counterpart in a complex algebraic expression. It is only out of sheer habit that we’re more comfortable with trigonometrical expressions and their properties. Let $f:X\to Y$ be a mapping. We will prove that $f^{-1}(Y-f(X-W))\subseteq W$, with equality when $f$ is injective. Note that $f$ does not have to be closed, open, or even continuous for this to be true. It can be any mapping. Let $W\subseteq X$. The mapping of $W$ in $Y$ is $f(W)$. As for $f(X-W)$, it may overlap with $f(W)$, we the mapping be not be injective. Hence, $Y-f(X-W)\subseteq f(W)$. >Taking $f^{-1}$ on both sides, we get $f^{-1}(Y-f(X-W))\subseteq W$. How can we take the inverse on both sides and determine this fact? Is the reasoning valid? Yes. All the points in $X$ that map to $Y-f(X-W)$ also map to $W$. However, there may be some points in $f^{-1}(W)$ that do not map to $Y-f(X-W)$. Are there other analogous points about mappings in general? In $Y$, select two sets $A$ and $B$ such that $A\subseteq B$. Then $f^{-1}(A)\subseteq f^{-1}(B)$ ## Axiom of Choice- a layman’s explanation. Say you’re given the set $\{1,2,3,\dots,n\}$, and asked to choose a number. Any number. You may choose $1,2$, or anything that you feel like from the set. Now suppose you’re given a set $S$, and you have absolutely no idea about what points $S$ contains. In this case, you can’t visualize the points in $S$ and pick any you feel like. You might say “pick the lowest upper bound of $S$“. However, what if $S$ is not ordered? What if it does not even contain numbers? How do you select a point from a set when you can’t even describe any of the points in that set? Here, the Axiom of Choice comes in. It states that if $\mathfrak{B}$ is a set of disjoint subsets of $S$, then a function $c: \mathfrak{B}\to \bigcup\limits_{B\in\mathfrak{B}}B$ such that one point from every disjoint set is selected. You can divide $S$ into disjoint sets in any manner whatsoever, and get one point from each set. In fact, the disjoint sets don’t necessarily have to cover $S$. The condition is $\bigcup\limits_{B\in\mathfrak{B}}\subseteq S$. Going by the above explanation, you may take each point in $S$ to be a disjoint interval, and hence select the whole of $S$. What is so special about this seemingly obvious remark? Selecting points from a set has always been defined by a knowledge of the set and its points. For example, if I say $f:\Bbb{R}\to\Bbb{R}$ such that $f(x)=f(x^2)$, I have selected the points with the information that points in $\Bbb{R}$ can be squared, and lie inside $\Bbb{R}$, etc. If we had $f:A\to \Bbb{R}$ where $A$ is a set of teddy bears, then $f(x)=x^2$ would not be defined. With the Axiom of Choice, regardless of whether $A$ contains real numbers or teddy bears, we can select a bunch of points from it. What if $B\in\mathfrak{B}$ are not all pair-wise disjoint? We can still select points from each $B$. Proof: Take the cross product $\mathfrak{B}\times S$. You will get points of the form $(B,x)$, where $B\in\mathfrak{B}$ and $x\in S$. Now create disjoint sets in $\mathfrak{B}\times S$ by taking sets of the form $B_i=\{(B_i,x)\|x\in B\}$; in essence you’re isolating individual sets of $\mathfrak{B}$. These are disjoint, as if $x$ is the same, then $B$ is different. The main purpose of taking this cross product was to create disjointed sets as compared to overlapping sets in $S$, so that we could apply the Axiom of Choice. Now we use the choice function to collect one point from each disjoint interval. Each of thee points will be of the form $(B,x)$. Now we define a function $g:\mathfrak{B}\times S\to S$ as $g((B,x))=x$. Hence, we have collected one point from each $B\in\mathfrak{B}$. Note that these points may well be overlapping. We found the cross-product just to be able to apply the Axiom of Choice. If the Axiom of Choice was defined for overlapping sets, then we wouldn’t have to find the cross product at all. Now we come to the reason why this article was written: defining an injective function $f:\Bbb{Z}_+\to B$, where $A$ is an infinite set and $B$ is a subset of it. We don’t know if $A$ is countable or not. OK first I’d like to give you the following proof, and you should tell me the flaw in it. We use Proof by Induction. Say $f(1)=a_1\in S$. We find $A-\{a_1\}$, and determine $a_2\in A-\{a_1\}$. We assume $f(a_n)$ equals a point in $A-\{a_1,a_2,\dots,a_{n-1}\}$, and then find $f(a_{n+1})$ in $A-\{a_1,a_2,\dots,a_n\}$. Why can we not do this? Because we know nothing about the points $f(a_1),f(a_2),\dots$! How can we possibly map $1$ to any points without knowing what point we’re mapping it to? The bad news is we might never know the properties of $S$. The good news is we can still work around it. Select $\mathfrak{B}$ to be the set of all subsets of $S$. As one might know, there are $\mathfrak{B}=2^{\|S\|}$. The elements of $\mathfrak{B}$ are not pairwise disjoint. However, we can still select a point from every set, as has been proven above (by taking the cross product to create disjoint sets, etc). The most brilliant part of the proof is this: take $c(S)$. We know $S\in \mathfrak{B}$. This will give you one element from the whole of $s$. You need to know nothing about $S$ to select an element from $S$. Let $f(1)=c(S)$. Now take $S-f(1)$. We know $S-f(1)\in \mathfrak{B}$. Let $f(2)=c(S-f(1))$. Continuing this pattern, we have $f(n)=c(S-\bigcup\limits_{i}f(i))$, where $i\in \{1,2,3,\dots,n-1\}$. By induction, we have that the whole of $\Bbb{Z}_+$ is mapped to a subset of $S$. ## My first attempt at solving the Lonely Runner Conjecture Let us suppose there are $k$ runners running at speeds $0 around a field of circumference $1$. Take any runner from the $k$ runners- say $r_i$, who runs with speed $a_i$. Say we pair him up with another runner $r_j$ who runs with speed $a_j$. Then for time $0\leq t\leq\frac{1}{2\|a_i-a_j\|}$, the distance between them is $\|a_i-a_j\|t$, and for $\frac{1}{2\|a_i-a_j\|}\leq t\leq \frac{1}{\|a_i-a_j\|}$, the distance between them is $1-\|a_i-a_j\|t$. The Lonely Runner Conjecture can be stated in the following way: there exists a time $t$ such that $\min\left[\min\{\|a_i-a_1\|t_1,1-\|a_i-a_1\|t_1\},\min\{\|a_i-a_2\|t_2,1-\|a_i-a_2\|t_2\},\dots,\min\{\|a_i-a_k\|t_k,1-\|a_i-a_k\|t_k\}\right]\geq\frac{1}{k}$. Here $t_1,t_2,t_3,\dots,t_k$ are the smallest positive values found after successively determining $t_x-n\frac{1}{\|a_i-a_x\|}$, where $x\in\{1,2,3,\dots k\}\setminus\{i\}$, and $n\in\Bbb{Z}^+$ ## Proofs of Sylow’s Theorems in Group Theory- Part 1 I will try to give a breakdown of the proof of Sylow’s theorems in group theory. These theorems can be tricky to understand, and especially retain even if you’ve understood the basic line of argument. 1. Sylow’s First Theorem- If for a prime number $p$, $o(G)=p^km$, and $p\nmid m$, then there is a subgroup $H\leq G$ such that $o(H)=p^k$. First we will try to understand what an orbit of a group is. Take a set $S$, and a group $G$. $S$ does not necessarily have to be a subset of $G$. $Orbit(S)=\{gs\|s\in S, g\in G\}$. Note that this is right multiplication. $Stabilizer(S)=\{g\in G\|gs=s\}$. Now the Orbit-Stabilizer Theorem- $\|Orbit\|\|Stabilizer\|=\|Group\|$. Something to remember from this is that there is nothing special about this theorem. It is just another way of saying $\|Image\|\|Kernel\|=\|Group\|$, as derived from the First Isomorphism Theorem. The stabilizer of a set $S$ is a subgroup. Proof: Let $as=s$ and $bs=s$. Then $abs=as=s$. Also, if $as=s$, then taking left inverse on both sides, $a^{-1}s=s$. Now we move to a combinatorial argument: what power of $p$ divides ${p^km\choose p^k}$? On expanding, we get $\displaystyle{\frac{p^km(p^km-1)(p^km-2)\dots (p^km-p^k+1)}{1,2,3\dots p^k}}=m\displaystyle{\Pi}_{j=1}^{p^k-1}\frac{p^km-j}{p^k-j}$. The highest power of $p$ that can divide $j$ is $\leq k-1$. Hence, the power of $p$ that divides $p^k-j$ is the same that which divides $p^km-j$. In fact, $p$ does not have to be prime for this property to be true. More generally speaking, just to gain a feel for this kind of argument, provided $j\leq p^k-1$, the fraction of the form $\frac{a-j}{b-j}$ not be divided by any power of $p$ except for $1$, as long as both $a$ and $b$ are divisible by powers of $p$ greater than or equal to $k$. $j$ is clearly the constraint here. Remember that $p\nmid m$ is a necessary condition for this. Now we prove ${p^km\choose p^k}\equiv m (mod p)$. Now we will define the set $S$ to contain all subsets of $G$ with $p^k$ elements. Clearly, $\|S\|={p^km\choose p^k}$, which as shown above is not divisible by $p$. Take every $g\in G$, and multiply it with every element of $S$. Note that very element of $S$ is itself a set containing $p^k$ elements of $G$. If $S_i\in S$, then $gS_i\in S$. Moreover, equivalence classes are formed when all elements of $G$ are multiplied with all elements of $S$, and these equivalence classes partition $S$. The proof of this: let $GS_i$ be the orbit of $S_i$. Then if $S_k\in Orb(S_i), gS_i=S_k$. Moreover, if $S_l\in S=g'S_i$, then $g'g^{-1}S_k=S_l$. This gives a feel for why these objects of $S$ are all in the same class. Let there be $r$ such partitions of $S$. Then $\|S\|=\|S_1\|+\|S_2\|+\dots+\|S_r\|$, where $S_1,S_2$,etc are partitions of $S$. We know $\|S\|$ is not divisible by $p$. Hence, there has to be at least one $\|S_k\|$ which is not divisible by $p$. Let $S_k\in [S_k]$, where $[S_k]$ is an equivalence class partitioning $S$. Consider a mapping $f:G\to S_k$ such that $f(g)=gS_k$. The image is $[S_k]$, and the kernel is the stabilizer. By the orbit-stabilizer theorem, $\|G\|=\|Stabilizer\|\|Orbit\|$. We know that the orbit is not divisible by $p^k$. Hence, the stabilizer has to be divisible by $p^k$. Also, the stabilizer is a subgroup. The rest of the proof will be continued next time. Today, I will discuss this research paper by Javed Ali, Professor of Topology and Analysis, BITS Pilani. What exactly is a proximinal set? It is the set of elements $X$ in which for any $x\in X$, you can find the nearest point(s) to it in $X$. More formally, for each $x\in X, \exists y\in X$ such that $d(x,y)=\inf \{d(x,K)\}$. This article says a Banach space is a complete vector space with a norm. One might find the difference between a complete metric space and a complete vector space to be minimal. However, the difference that is crucial here is that not every metric space is a vector space. For example, let $(X,d)$ be a metric space, satisfying the relevant axioms. However, for $x,y\in X$, $x+y$ not being defined is possible. However, if $X$ is a vector space, then $x+y\in X$. Hence, every normed vector space is a metric space if one were to define $\\|x-y\\|=d(x,y)$, but the converse is not necessarily true. What is a convex set? This article says a convex set is one in which a line joining any two points in the set lies entirely inside the set. But how can a set containing points contain a line? Essentially, the the convex property implies that every point the line passes through is contained within the convex space. Convexity introduces a geometrical flavor to Banach spaces. It is difficult to imagine what such a line segment would be in the Banach space of matrices (with the norm $\\|\mathbf{A}-\mathbf{B}\\|=\mathbf{A}-\mathbf{B}$. What is a uniformly convex set? This paper says that a uniform convex space is defined thus: $\forall<\epsilon\leq 2\in\Bbb{R}, \exists \delta(\epsilon)>0$ such that for $\\|x\\|=\\|y\\|=1$ and $\\|x-y\\|\geq \epsilon, \frac{\\|x+y\\|}{2}\leq 1-\delta(\epsilon)$. Multiplying by $-1$ on both sides, we get $1- \frac{\\|x+y\\|}{2}\geq \delta(\epsilon)$. What does this actually mean? The first condition implies that $x$ and $y$ cannot lie in the same direction. Hence, $\\|x+y\\|<2$. As a result, we get $\left\\|\frac{x+y}{2}\right\\|<1$, or $1-\left\\|\frac{x+y}{2}\right\\|>0$. As $\delta(\epsilon)>0$, and as $1-\left\\|\frac{x+y}{2}\right\\|$ is bounded, $\delta(\epsilon)$ can be the lower bound of $1-\left\\|\frac{x+y}{2}\right\\|$. But what is uniform about this condition? It is the fact that $\delta(\epsilon)$ does not change with the unit vector being considered, and depends only on $\epsilon$. Now we go on to prove that every closed convex set of every uniformly convex Banach space is proximinal. This is the second time I’m checking whether I can write a LATEX equation in wordpress. $o(HK)=\frac{o(H)o(K)}{o(H\cap K)}$ ## Why I have nothing better to do than blog :( After trying to follow blogs by various mathematicians, I have concluded that having a digital presence is pretty darned cool and useful in academia! Hence, in my pursuit of the forever-elusive elixir of “cool”, I have decided to record my progress in Algebraic Geometry through this blog. I’m no expert. But I feel trying to explain concepts to others promotes a higher degree of clarity than self-study, athough this is debatable and most definitely a grey area. I really do hope Math enthusiasts find this blog to be a useful spot for better explanations of convoluted geometrical/algebraic concepts. And reaffirm that beneath all the daunting machinery, Math is so simple and beautiful. $\epsilon$
# Section 2.3 Question 3 ## How do you set up and solve an application involving a system of equations in two variables? In section 2.1, we examined how the equilibrium point for the dairy industry could be found from the demand and supply functions from Chapter 1. In that section, we used a graph of the functions to locate the equilibrium point. The Substitution Method and Elimination Method can also be used to find the equilibrium point algebraically. ### Example 5 Use the Substitution Method to Find the Equilibrium Point The demand and supply functions for the dairy industry are described by Use the Substitution Method to find the equilibrium point for this system of equations. Solution Since one of the equations (in this case both) is solved for a variable, this system is perfectly suited for the Substitution Method. Substitute 3/95 Q for P in the demand function, Now solve the equation for Q: Either equation can be used to determine the corresponding value for P. If we substitute Q = 95 into the supply function, we get The equilibrium price of \$3 per gallon corresponds to a quantity of 95 thousand gallons of milk. At this price, suppliers are willing to supply 95 thousand gallons of milk and consumers are willing to buy 95 thousand gallons of milk. We can also locate this equilibrium point using the Elimination Method. ### Example 6 Use the Elimination Method to Find the Equilibrium Point The demand and supply functions for the dairy industry are described by Use the Elimination Method to find the equilibrium point for this system of equations. Solution To use the Elimination Method, we must place all of the variables on one side of the equation and the constants on the other side. This gives the equivalent system The leading coefficient of the first equation is a 1, so we proceed to eliminating P from the second equation. To do this, multiply the first equation by -1 and add it to the second equation: The fraction in the sum may look a bit strange, but it comes from converting the decimal to a fraction and adding the fractions: Replace the second equation with the new second equation to obtain an equivalent system of equations: Multiply the second equation by the reciprocal of 775/9500 to make the leading coefficient a 1, This helps us to write the system of equations as To eliminate Q from the first equation, multiply the second equation by 3/95 and add it to the first equation: Replace the first equation with this sum to yield the solution . This tells us that the equilibrium point is at 95,000 gallons of milk and a price of \$3 per gallon. The equilibrium point is at the same values when we solve the system of equations by the Substitution Method or the Elimination Method. In general, you’ll solve for the equilibrium point with only one of these methods. We have looked at several examples of setting up a system of equations from an application and then solving the resulting system. We have broken the process up into several parts and illustrated each part individually to make them easier to understand. In reality, problems are usually not all that tidy. In the following example we’ll solve a problem involving a system of equations in two variables. Our strategy for doing this is fairly straightforward. 1. Read the problem thoroughly to help you to understand what it is you are looking for. 2. Choose the variables and provide a detailed description of what each variable represents. Use variable names that correspond to your description instead of the generic names x and y. 3. Look for relationships between these variables. Use these relationships to determine the equations. 4. Solve the resulting system of equations using the Substitution Method or Elimination Method. 5. If convenient, graph each equation in the system and check to see that the point of intersection matches the solution you found by using the Substitution Method or the Elimination Method. 6. Make sure the solution is consistent with all of the information in the problem. ### Example 7 Mixing Ethanol Blends Flex fuel vehicles are designed to operate on a gasoline-ethanol mixture. This gasoline-ethanol mixture is called E85 and is 85% ethanol and 15% gasoline. Flex fuel vehicles perform almost identically to gasoline powered vehicles. Since E85 is less expensive than gasoline, it would appear that that flex fuel vehicles would be the wave of the future. However, the energy content of E85 is less than that of gasoline so flex fuel vehicles get worse mileage than gasoline powered vehicles. In general, the higher the percentage of ethanol in the fuel, the worse the mileage is. A recent study determined that several vehicles designed to run on gasoline are able to run well on gasoline-ethanol blends. In tests using E10 (10% ethanol), E20 (20% ethanol) and E30 (30% ethanol), the cost per mile traveled by vehicles using the three blends was lowest for vehicles using E20 or E30. Because of these results, gasoline stations around the country are beginning to sell several blends of ethanol to cash in on this result. A gasoline station in Rapid City, South Dakota carries three blends, E10, E20, and E85. The cost per gallon of each of these fuels is shown in the table below. A customer wants to purchase 10 gallons of E20. The customer could simply purchase the 10 gallons at \$2.68 per gallon. Or the customer could blend their own E20 using the available E10 and E85. How many gallons of E10 and E85 would the customer need to blend in their 10 gallon tank to effectively have E20? Would it be cheaper to blend E10 and E85 or to simply pump E20? Solution In this problem, we need to compare the cost of pumping 10 gallons of straight E20 to the cost of a specific combination of E10 and E85 that will yield 20% ethanol. It is straightforward to calculate cost of each type of fuel, cost = ( cost per gallon ) ( number of gallons ) For instance, 10 gallons of E20 would cost The key question is to determine if a blend of E10 and E85 would be cheaper. How much of each type of fuel would be needed to fill the tank and result in a mixture that is 20% ethanol? We’ll start with two variables, one to represent the amount E10 in gallons and another to represent the amount of E85 in gallons: E10: amount of 10% ethanol fuel in gallons E85: amount of 85% ethanol fuel in gallons Writing out this description is key to our strategy. Not only does it help you to focus on what the unknowns are, but it also includes the units, gallons. The problem includes copious amounts of information. However two pieces of information relate to the variables. First of all, the customer needs to fill a 10 gallon tank with a combination of two fuels. Since the variables represent the individual amount of fuels, the sum of the variables corresponds to the total amount of fuel, The second important piece of information is that the mixture of fuel must contain 20% ethanol. For a 10 gallon tank, the amount of ethanol is 20% of 10 gallons or ( 0.20 )( 10 gallons ) = 2 gallons. This ethanol will come from the E10 which is 10% ethanol and the E85 which is 85% ethanol. By multiplying the percent times the gallons of fuel, we can determine how much ethanol is in the fuel. For instance 8 gallons of E85 and 2 gallons of E10 would have A mixture of 10 gallons of fuel with 7 gallons of ethanol would be a 70% mixture. This is clearly way too much ethanol. However, we could construct a table with differing amounts of fuel to help us determine the solution: As the amount of E10 is increased and the amount of E85 is decreased, the total amount of ethanol in the mixture drops. Based on the table, the amount of E10 must be between 8 and 9 gallons and the amount of E85 must be between 1 and 2 gallons. It is unlikely that we could find the exact amounts of each type of fuel by expanding the table. However, we can replace the amounts of fuel with variables that represent these amounts to give the equation This gives us two equations in two variables that we can solve by the substitution method or the elimination method. For this example, we’ll solve the system Solve the first equation for E10 to yield E10 = 10 – E85. Substitute this expression into the second equation to obtain Solve the equation for E85: Now that we know that we’ll need 4/3 gallon of E85, we can use the first equation to find the amount of E10: Now let’s find the cost of this blend, Compared to pumping 10 gallons of E20 (for a total of \$26.80), pumping a blend of E10 and E85 (from the same pump!) saves about 11 cents. How does this match the graph of each equation? Figure 4 – By solving each equation for E10, we can graph E10 = 10 – E85 (blue) and E10 = 20 – 8.5 E85 (red) to see that the point of intersection is at the correct ordered pair. Before moving on, we need to make sure the solution makes sense. In the table we created earlier, we noted that the amount of E10 had to between 8 and 9 gallons. Since 26/3 ≈ 8.67, our solution is consistent. The amount of E85 is 4/3 ≈ 1.33 which is between 1 and 2 as expected. Review the steps we outlined for the strategy. Careful examination reveals that the solution above includes all six of the steps for solving an application.
Seu SlideShare está sendo baixado. × Anúncio Anúncio Anúncio Anúncio Anúncio Anúncio Anúncio Anúncio Anúncio Anúncio Anúncio 1 de 1 Anúncio Use the following graph B-0-003x3+0-13x2-1-2x+18-6 Approximate the ave.docx Use the following graph B=0.003x3+0.13x2-1.2x+18.6 Approximate the average BMI for 10-year-old males; for 16-year-old males. Can you show how to plug values? Solution There is two ways to do this. We can look at the graph where it says Age and go to the line between 8 and 12. This is 10 then move straight up to the point this line hits the graph. This is the BMI for 10 year olds. It is slightly above 16. We can find the exact number by pluging our value into x in our equationB=0.003x^3+0.13x^2-1.2x+18.6 B=0.003(10)^3+0.13(10)^2-1.2(10)+18.6=22.6 For 16 year olds we do the same thing and get a BMI a little above 20 B=0.003(16)^3+0.13(16)^2-1.2(16)+18.6=44.968 I hope this helps . Use the following graph B=0.003x3+0.13x2-1.2x+18.6 Approximate the average BMI for 10-year-old males; for 16-year-old males. Can you show how to plug values? Solution There is two ways to do this. We can look at the graph where it says Age and go to the line between 8 and 12. This is 10 then move straight up to the point this line hits the graph. This is the BMI for 10 year olds. It is slightly above 16. We can find the exact number by pluging our value into x in our equationB=0.003x^3+0.13x^2-1.2x+18.6 B=0.003(10)^3+0.13(10)^2-1.2(10)+18.6=22.6 For 16 year olds we do the same thing and get a BMI a little above 20 B=0.003(16)^3+0.13(16)^2-1.2(16)+18.6=44.968 I hope this helps . Anúncio Anúncio
# 7.10: Newton’s Second Law of Motion- Concept of a System Learning Objectives By the end of this section, you will be able to: • Define net force, external force, and system. • Understand Newton’s second law of motion. • Apply Newton’s second law to determine the weight of an object. Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned. First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration. Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct — an external force acts from outside the system of interest. For example, in Figure $$\PageIndex{1a}$$ the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the system. Again looking at Figure $$\PageIndex{1a}$$, the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept will be revisited many times on our journey through physics. Figure $$\PageIndex{1}$$: Different forces exerted on the same mass produce different accelerations. (a) Two children push a wagon with a child in it. Arrows representing all external forces are shown. The system of interest is the wagon and its rider. The weight $$w$$ of the system and the support of the ground $$N$$ are also shown for completeness and are assumed to cancel. The vector $$f$$ represents the friction acting on the wagon, and it acts to the left, opposing the motion of the wagon. (b) All of the external forces acting on the system add together to produce a net force, $$F_{net}$$. The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, we draw the vectors collinearly. (c) A larger net external force produces a larger acceleration $$(al>a)$$ when an adult pushes the child. Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure. In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight $$w$$ and the support of the ground $$N$$, and the horizontal force $$f$$ represents the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are touching. Figure $$\PageIndex{1b}$$ shows how vectors representing the external forces add together to produce a net force, $$F_{net}$$. To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the proportionality $a \propto F_{net}$ where the symbol $$\propto$$ means “proportional to,” and $$F_{net}$$ is the net external force. (The net external force is the vector sum of all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition of other vectors, and are covered in the chapter section on Two-Dimensional Kinematics.) This proportionality states what we have said in words—acceleration is directly proportional to the net external force. Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within the child’s body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex problems with only minimal error due to our simplification Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure, the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as $a \propto \dfrac{1}{m},$ where $$m$$ is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly proportional to the net external force. Figure $$\PageIndex{2}$$:The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to make a pass. (The effect of gravity on the ball is ignored.) (b) The same player exerts an identical force on a stalled SUV and produces a far smaller acceleration (even if friction is negligible). (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for the free-body diagram will emerge as you do more problems. It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion. Newton’s Second Law of Motion The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. In equation form, Newton’s second law of motion is $a = \dfrac{F_{net}}{m}$ This is often written in the more familiar form $F_{net} = ma.$ When only the magnitude of force and acceleration are considered, this equation is simply $F_{net} = ma.$ Although these last two equations are really the same, the first gives more insight into what Newton’s second law means. The law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on experimental verification. ## Units of Force $$F_{net} = ma$$ is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of $$1m/s^2$$ That is, since $$F_{net} = ma,$$ $1N = 1 kg \cdot s^2$ While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. ## Weight and the Gravitational Force When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight $$w$$. Weight can be denoted as a vector $$w$$ because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The magnitude of weight is denoted as $$w$$. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration $$w$$. Using Galileo’s result and Newton’s second law, we can derive an equation for weight. Consider an object with mass $$m$$ falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude $$w$$. Newton’s second law states that the magnitude of the net external force on an object is $$F_{net} = ma$$. Since the object experiences only the downward force of gravity, $$F_{net} = w$$. We know that the acceleration of an object due to gravity is $$g$$, or $$a = g$$. Substituting these into Newton’s second law gives WEIGHT This is the equation for weight - the gravitational force on mass $$m$$: $w = mg$ Since weight $$g = 9.80 m/s^2$$ on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see: $w = mg = (1.0 kg) (9.8 m/s^2) = 9.8 N.$ Recall that $$g$$ can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to take this into consideration when solving problems with weight. When the net external force on an object is its weight, we say that it is in free-fall. That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object. The acceleration due to gravity $$g$$ varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth’s surface. On the Moon, for example, the acceleration due to gravity is only $$1.67 m/s^2$$. A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon. The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are really referring to the phenomenon we call “free-fall” in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness. It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter (how much “stuff”) and does not vary in classical physics, whereas weight is the gravitational force and does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our “weight” in kilograms, but never in the correct units of newtons. COMMON MISCONCEPTIONS: MASS VS. WEIGHT Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram (or the “slug” in English units). Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object $$(m)$$ multiplied by the acceleration due to gravity $$(g)$$. Like any other force, weight is measured in terms of newtons (or pounds in English units). Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is $$1.67 m/s^2$$ (which is much less than the acceleration due to gravity on Earth, $$9.80 m/s^2$$ ). If you measured your weight on Earth and then measured your weight on the Moon, you would find that you “weigh” much less, even though you do not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are “losing weight,” they really mean that they are losing “mass” (which in turn causes them to weigh less) TAKE-HOME EXPERIMENT: MASS AND WEIGHT What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. The springs provide a measure of your weight (for an object which is not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same “mass” on Earth as on the Moon? Example $$\PageIndex{1}$$:What Acceleration Can a Person Produce when pushing a Lawn Mower? Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration? Figure $$\PageIndex{3}$$: The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right? Strategy Since $$F_{net}$$ and $$m$$ are given, the acceleration can be calculated directly from Newton’s second law as stated in $$F_{net} = ma$$. Solution The magnitude of the acceleration $$a$$ is $$a = \frac{F_{net}}{m}$$. Entering known values gives $a = \dfrac{51 \, N}{24 \, kg}$ Substituting the units $$kg \cdot m/s^2$$ for N yields $a = \dfrac{ 51 \, kg/s^2}{24\space kg} = 2.1\space m/s^2$ Discussion The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person’s top speed would soon be reached. Example $$\PageIndex{2}$$:What Rocket Thrust Accelerates This Sled? Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust $$T$$ for the four-rocket propulsion system shown in Figure. The sled’s initial acceleration is $$49 m/s^2$$ the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N. Figure 4.4.4. A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust $$T$$. As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force $$N$$ on the system that is equal in magnitude and opposite in direction to its weight, $$w$$. The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction $$(f)$$ is drawn larger than scale. Strategy Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure. Solution Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with $F_{net} = ma.$, where $$F_{net}$$ is the net force along the horizontal direction. We can see from Figure that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is $F_{net} = 4T - f.$ Substituting this into Newton’s second law gives $F_{net} = ma = 4T - f.$ Using a little algebra, we solve for the total thrust 4T: $4T = ma + f.$ Substituting known values yields $4T = ma + f = (2100 \, kg)(49 \, m/s^2) + 650 \, N$ So the total thrust is $1 \times 10^5 N,$ and the individual thrusts are $T = \dfrac{1\times 10^5}{4} = 2.6 \times 10^4 \, N$ Discussion The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 $$g$$-s. (Recall that $$g$$, the acceleration due gravity is $$9.80 \, m/s^2$$. When we say that an acceleration is 45 $$g$$-s, it is $$45 \times 9.80 m/s^2$$, which is approximately $$440 m/s^2$$ ). While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious. Newton’s second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion. Summary • Acceleration, $$a$$, is defined as a change in velocity, meaning a change in its magnitude or direction, or both. • An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the system. • Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. • In equation form, Newton’s second law of motion is $$a = \frac{F_{net}}{m}$$ • This is often written in the more familiar form: $$F_{net} = ma.$$ • The weight $$w$$ of an object is defined as the force of gravity acting on an object of mass $$m.$$ The object experiences an acceleration due to gravity $$g$$: $w = mg.$ • If the only force acting on an object is due to gravity, the object is in free fall. • Friction is a force that opposes the motion past each other of objects that are touching.
Edit Article # wikiHow to Solve a 2X2X2 Rubik's Cube You may have seen people on the internet solving 2 x 2 x 2 Rubik's Cubes in the blink of an eye. Maybe you tried it yourself and found it much harder than it looks. That's okay! You don't need to be a genius to fly through a 2 x 2 x 2 Rubik's Cube. As with any form of Rubik's Cube, you simply need to memorize some algorithms and steps to take. With some practice, you too can be a master of the 2 x 2 x 2 Rubik's Cube. ## Steps 1. 1 Know the notations. A simple system of notation is used when describing Rubik's cube algorithms. You'll need to understand how this system works before beginning. • There is a letter to indicate each face of the cube. • F = front (the side facing you) • L = left (the vertical face on your left-hand side) • R = right (vertical face on your right-hand side) • B = back (the side that faces directly away from you) • U = top (the side that faces upward) • D = down (the side that faces downward) • Any of the letters, by itself, means to turn that side clockwise 90°. • A 2 before any letter means to turn that side 180°. • A ' before any letter means to turn that side counterclockwise 90°. • FC means "favorite color." 2. 2 3. 3 Find a face that has your FC. This will be your U face. 4. 4 Find a corner cube that belongs beside the cube on the U face. 5. 5 Turn the D face until the desired cube is under where you want it to be. Then use one of the following sequences to move it to the top. 6. 6 If your FC on the corner cube is on the R face: R' D' R. 7. 7 If your FC on the corner cube is on the F face: F D F'. 8. 8 If your FC on the corner cube is on the D face: R' D2 R D R' D' R. 9. 9 Do these steps until the whole up face is done. 10. click to unmute.. to Solve a 2X2X2 Rubik's Cube 10 Down Face Look at the down face and figure out what colour the down face is. It will be the color that is on all of the disoriented remaining cubes. 11. 11 The first step is to position all of the cubes (The bottom face is made up of 4 cubes. You have to get those into the correct position). 12. 12 Swap FLD and FRD cubes: R' D' R F D F' R' D R D2. 13. 13 Swap FLD and BRD cubes: R' D' R F D2 F' R' D R D2. 14. 14 When all the cubes are positioned, the next step is to orient them. Looking at the bottom face, identify it from one of the following patterns: 15. 15 After that, use the face that is currently at the top as the front to do one of these sequences: 16. 16 For patterns 1, 5, 6, and 7, do: R' D' R D' R' D2 R D2. 17. 17 For patterns 2, 3, and 4, do: D2 R' D2 R D R' D R. 18. 18 Continue the orientation sequences until the cube is solved. 19. click to unmute.. to Solve a 2X2X2 Rubik's Cube 19 Finished. ### Part 1 Down Left Up Right 1. click to unmute.. to Solve a 2X2X2 Rubik's Cube 1 Use the basic 'Down Left Up Right' algorithm to get the first layer finished, then the second layer. Note: This is much easier if you can solve a 3x3x3 cube. 2. click to unmute.. to Solve a 2X2X2 Rubik's Cube 2 Find a color that you like. This will be your Favorite Colour (FC). 3. click to unmute.. to Solve a 2X2X2 Rubik's Cube 3 Get two FC blocks next to each other. 4. click to unmute.. to Solve a 2X2X2 Rubik's Cube 4 Find another FC block, and put it next to the other two by simply twisting it up, or by moving the other blocks out of the way, moving the third FC block up, and moving the first two back onto their original positions. 5. click to unmute.. to Solve a 2X2X2 Rubik's Cube 5 Repeat this algorithm until all of the FC blocks are facing one way and on one layer. 6. click to unmute.. to Solve a 2X2X2 Rubik's Cube 6 Now find the two FC blocks that have one pair of corner colours the same. Then twist the side that does not have these two down to the bottom (180 degrees), and position the correct block under the correct place. 7. click to unmute.. to Solve a 2X2X2 Rubik's Cube 7 Use the down up left right algorithm continuously until it is the right way up and the corners match one of the first two FC blocks' corners. Repeat with the other block. 8. click to unmute.. to Solve a 2X2X2 Rubik's Cube 8 Congrats! Halfway. Next, turn the whole cube upside down, so the top = bottom. 9. click to unmute.. to Solve a 2X2X2 Rubik's Cube 9 Then go: Down up left right, down up left right, until one of the blocks is facing the right way. For example, the only color not used so far is on the current top. 10. click to unmute.. to Solve a 2X2X2 Rubik's Cube 10 Turn the top layer to your left (looking from the top, twist clockwise). Repeat: repeat down up left right until that block is orientated the right way up. 11. click to unmute.. to Solve a 2X2X2 Rubik's Cube 11 Repeat the turn, and repeat the algorithm and turn twice more. Then make the top layer sides match the bottom layer sides. 12. click to unmute.. to Solve a 2X2X2 Rubik's Cube 12 Brag. You've done it. ### Part 2 3x3x3 1. 1 Place the corners with the classic algorithm. Use the R' D' R D algorithm to place the corners into the chosen face. 2. 2 OLL. This can get get a bit complicated! If you have a singe yellow dot place on the bottom left, if you have a horizontal line place it at the top and if you have a diagonal line place it so that it is pointing to the top right. 3. 3 PLL. This is exactly like how you place the corners on the 3X3X3 rubik's cube. Use the R' F R' B2 R F' R' B2 R2 U'. ### Part 3 White face first 1. 1 Solve the white face. This should be easy to do, but if you get stuck, use the algorithms below. Find a white piece on the top face and put it above the space on the bottom face you need it to move to. This step is finished if the white face is complete and the layer above it is also complete. • If a white piece is on the top face, use the following algorithm: F, U', F', 2U. Once this part is complete, use one of the algorithms below based on the location of a white face. • If a white piece is on the right side of one of the "cubelets" on the top face, use the following algorithm: R, U, R'. You might need to turn the whole cube so that the white piece is on the right side. • If a white piece is on the left side of one of the "cubelets" on the top face, use the following algorithm: L',U', L. You might need to turn the whole cube so that the white piece is on the left side. • If a white piece is near the bottom, you will need to move it up, then do one of the algorithms. 2. 2 Your cube should now look like the cube shown above: the white face solved, with the red and green layers above it. It doesn't need to look exactly like the pieces at the top. 3. 3 Solve the yellow face. Find a yellow piece on the top with another yellow piece next to it on the layer below, as shown in the picture. Then do the algorithm: R, U, R', U, R, 2U, R'. This is by far the longest part of the process. If you cannot find the pattern shown in the picture, do the algorithm again. If after repeated attempts you still can't get the pattern, it would be best to scramble the cube and start over. 4. 4 Solve a third face. By now you should have solved the yellow and white faces. Do a "U-turn," turning the top layer until there is another face solved (any color is fine). 5. 5 Turn the cube so that the most recently completed face is facing away from you. Then do the following algorithm: R', F, R', 2B, R, F', R', 2B, 2R. 6. 6 Repeatedly do "U-turns" until the cube is solved. ## Community Q&A Search • How does a 2x2 Rubik's cube work? There is a ball in the center that one of the pieces is connected to (usually the red, white, and blue piece). There are grooves in the ball which contain connectors to the other 7 pieces, which can turn and move. 200 characters left ## Video This is a video on how to solve a 2x2. ## Tips • You will probably need to memorize the solution to get your time under a minute. • Don't use automatic solvers. You won't get the feeling of accomplishment, and when you get better, the solver will only slow you down. • If you want to learn every moves, check out a video on YouTube. • If you don't now how to do some of the moves go here ## Article Info Categories: Pictures \| Rubik's Cube In other languages: Español: resolver un cubo Rubik de 2x2x2, Italiano: Risolvere un Cubo di Rubik 2x2x2, Português: Solucionar um Cubo Mágico, Русский: собрать кубик Рубика 2х2х2, Nederlands: De 2x2x2 Rubics kubus oplossen, Deutsch: Einen 2x2x2 Zauberwürfel lösen Thanks to all authors for creating a page that has been read 480,837 times.
### Chapter Three 3.1 ```Section 3.1 Introduction to Graphing The Rectangular Coordinate System Page 160 One common way to graph data is to use the rectangular coordinate system, or xy-plane. In the xy-plane the horizontal axis is the x-axis, and the vertical axis is the y-axis. The axes intersect at the origin. The axes divide the xy-plane into four numbered I, II, III, and IV counterclockwise Example Page 161 Plot the following ordered pairs on the same xy-plane. State the quadrant in which each point is located, if possible. a. (4, 3) b. (3, 4) c. (1, 0) Solution a. (4, 3) Move 4 units to the right of the origin and 3 units up. Quadrant I b. (3, 4) Move 3 units to the left of the origin and 4 units down. Quadrant III c. (1, 0) Move 1 unit to the left of the origin. Not in any quadrant Example Page 161 Frozen pizza makers have improved their pizzas to taste more like homemade. Use the graph to estimate frozen pizza sales in 1994 and 2000. Solution a. To estimate sales in 1994, locate 1994 on the x-axis. Then move upward to the data point and approximate its y-coordinate. Example (cont) Page 161 Frozen pizza makers have improved their pizzas to taste more like homemade. Use the graph to estimate frozen pizza sales in 1994 and 2000. Solution b. To estimate sales in 2000, locate 2000 on the x-axis. Then move upward to the data point and approximate its y-coordinate. Scatterplots and Line Graphs Page 162 If distinct points are plotted in the xy-plane, then the resulting graph is called a scatterplot. Example Page 162 Make a scatterplot of the data in the following table, which lists the price of a first-class postage stamp for selected years. 1980 1990 2000 15 25 33 2010 44 Postage Stamp Prices Price in cents Year Price (cents) 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 1970 1980 1990 2000 2010 2020 Year Line Graphs Page 163 Sometimes it is helpful to connect consecutive data points in a scatterplot with line segments. This creates a line graph. Example Page 163 Use the data in the table to make a line graph. x y 3 3 2 4 1 0 0 3 1 2 2 4 3 3 Solution Plot the points and then connect consecutive points with line segments. Example Page 164 Use the figure to answer the questions. a. Which 10-year period had the greatest increase in energy consumption? 1970 to 1980 b. Estimate the energy consumption in 2010.
# ω is an imaginary root of unity. Prove that If a+b+c=0 then prove that (a+bω+cω2)3+(a+bω2+cω)3=27abc. Video Solution Text Solution Generated By DoubtnutGPT ## To prove that if a+b+c=0, then (a+bω+cω2)3+(a+bω2+cω)3=27abc,where ω is a primitive cube root of unity, we can follow these steps:Step 1: Understanding the properties of ωRecall that ω is a cube root of unity, which means:ω3=1and1+ω+ω2=0.Step 2: Rewrite the expressionsGiven a+b+c=0, we can express c as:c=−(a+b).Now, we can rewrite the expressions a+bω+cω2 and a+bω2+cω:1. For a+bω+cω2: a+bω+cω2=a+bω−(a+b)ω2=a(1−ω2)+b(ω−ω2).2. For a+bω2+cω: a+bω2+cω=a+bω2−(a+b)ω=a(1−ω)+b(ω2−ω).Step 3: Calculate the cubesLet:x=a+bω+cω2,y=a+bω2+cω.We need to compute x3+y3. Using the identity x3+y3=(x+y)(x2−xy+y2).Step 4: Find x+yCalculating x+y:x+y=(a+bω+cω2)+(a+bω2+cω)=2a+b(ω+ω2)+c(ω+ω2).Since ω+ω2=−1:x+y=2a−b−c=2a−b+(a+b)=a.Step 5: Find xyCalculating xy:xy=(a+bω+cω2)(a+bω2+cω).Using the fact that ω2+ω+1=0, we can expand this product and simplify.Step 6: Use the identityUsing the identity:x3+y3=(x+y)((x+y)2−3xy).Substituting x+y=a and simplifying will lead us to 27abc.Step 7: Final simplificationAfter performing the necessary algebraic manipulations and substituting back the values, we will arrive at the conclusion that:x3+y3=27abc.ConclusionThus, we have shown that if a+b+c=0, then (a+bω+cω2)3+(a+bω2+cω)3=27abc. \| Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation
# If the sum of first n terms of an AP is 4n – n2, what is the first term i.e. S1? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, then 10th, and nth terms. ### Solution: Given, Sn = 4n – n2 Therefore, S1 = 4(1) – (1)2 = 4 – 1 = 3 (Answer) And S2 = 4(2) – (2)2 = 8 – 4 = 4 ∴ Second term = a2 = S2 – S1 [∵ an = Sn – Sn – 1] = 4 – 3 Similarly, S3 = 4(3) – (3)2 = 12 – 9 = 3(Answer) S9 = 4(9) – (9)2 = 36 – 81 = -45 S10 = 4(10) – (10)2 = 40 – 100 = -60 Therefore, Third term = a3 = S3 – S2 [∵ an = Sn – Sn – 1] = 3 – 4 Tenth term = a10 = S10 – S9 [∵ an = Sn – Sn – 1] = -60 – (-45) = -60 + 45 nth term = an = Sn – Sn – 1 = 4n – n2 – {4(n – 1) – (n – 1)2} = 4n – n2 – {4n – 4 – (n2 – 2n + 1)} = 4n – n2 – {4n – 4 – n2 + 2n – 1} = 4n – n2 – (- n2 + 6n – 5) = 4n – n2 + n2 – 6n + 5
# Logarithm Table In mathematics, the logarithm table is used to find the value of the logarithmic function. The simplest way to find the value of the given logarithmic function is by using the log table. Learn Mathematics at BYJU’S in an easy way. Here, the definition of the logarithmic function and procedure to use the logarithm table is given in detail. Also, get Antilogarithm table here for reference. ## Logarithmic Function Definition The logarithmic function is defined as an inverse function to exponentiation. The logarithmic function is stated as follows For x, a > 0, and a≠1, y= loga x, if x = ay Then the logarithmic function is written as: f(x) = loga x The most common bases used in logarithmic functions are base e and base 10. The log function with base 10 is called the common logarithmic function and it is denoted by log10 or simply log. f(x) = log10 The log function to the base e is called the natural logarithmic function and it is denoted by loge. f(x) = loge x To find the logarithm of a number, we can use the logarithm table instead of using a mere calculation. Before finding the logarithm of a number, we should know about the characteristic part and mantissa part of a given number • Characteristic Part – The whole part of a number is called the characteristic part. The characteristic of any number greater than one is positive, and if it is one less than the number of digits to the left of the decimal point in a given number. If the number is less than one, the characteristic is negative and is one more than the number of zeros to the right of the decimal point. • Mantissa Part – The decimal part of the logarithm number is said to be the mantissa part and it should always be a positive value. If the mantissa part is in a negative value, then convert into the positive value. ## How to Use the Log Table? The procedure is given below to find the log value of a number using the log table. First, you have to know how to use the log table. The log table is given for the reference to find the values. Step 1: Understand the concept of the logarithm. Each log table is only usable with a certain base. The most common type of logarithm table is used is log base 10. Step 2: Identify the characteristic part and mantissa part of the given number. For example, if you want to find the value of log10 (15.27), first separate the characteristic part and the mantissa part. Characteristic Part = 15 Mantissa part = 27 Step 3: Use a common log table. Now, use row number 15 and check column number 2 and write the corresponding value. So the value obtained is 1818. Step 4: Use the logarithm table with a mean difference. Slide your finger in the mean difference column number 7 and row number 15, and write down the corresponding value as 20. Step 5: Add both the values obtained in step 3 and step 4. That is 1818+20= 1838. Therefore, the value 1838 is the mantissa part. Step 6: Find the characteristic part. Since the number lies between 10 and 100, (101 and 102), the characteristic part should be 1. Step 7: Finally combine both the characteristic part and the mantissa part, it becomes 1.1838. ### Sample Examples Here the sample example to find the value of the logarithmic function using the logarithm table is given. Example 1: Find the value of log10 2.872 Solution: Step 1: Characteristic Part= 2 and mantissa part= 872 Step 2: Check the row number 28 and column number 7. So the value obtained is 4579. Step 3: Check the mean difference value for row number 28 and mean difference in column 2. The value corresponding to the row and column is 3 Step 4: Add the values obtained in step 2 and 3, we get 4582. This is the mantissa part. Step 5: Since the number of digits to the left side of the decimal part is 1, the characteristic part is less than 1. So the characteristic part is 0 Step 6: Finally combine the characteristic part and the mantissa part. So it becomes 0.4582. Therefore, the value of log 2.872 is 0.4582. Example 2: Find the value of log10 (∛2.134) Solution: We know by the property of logarithms, loga bn = n loga b Thus, log10 (∛2.134) = 1/3 log10 2.134 Now we calculate log10 2.134 For the mantissa, check row 21 and column 3 in the logarithm table, which is 3284. Now check the mean difference 4 of the same row, which is 8. ∴ Mantissa is 3284 + 8 = 3292 For the characteristic part, since there is only 1 digit in the left side of decimal, hence characteristic is 0 ∴ log10 2.134 = 0.3292 Then, log10 (∛2.134) = 1/3 log10 2.134 = 1/3 × 0.3292 = 0.10973. Note: To find the value of ∛2.134, we just have to find the antilogarithm of 0.10973. ## Practice Questions 1. Find the value of log10 23.46. 2. Find the value of log10 345. 3. Find the value of log10 25. Related Links Exponential Function Logarithm Formula Log Base 2 Logarithms Register with BYJU’S to know more about the usage of logarithm table and antilog table and also watch interactive videos to learn with ease. ## Frequently Asked Questions on Logarithm Table Q1 ### What is the base of the common logarithm? Common logarithms is of base 10 which is calculated using the logarithm tables. Q2 ### How find the logarithm of a given number? Using the logarithm table, Calculate the characteristic, which is determined by the whole number part of the given number. Calculate the mantissa, which is determined by the significant digits of the given number. Finally, combine the characteristic and mantissa with a decimal point. Q3 ### What is the base of the natural logarithm? Natural logarithm has base e, for example, loge (x2 + 1). Q4 ### What is the value of log10 1? The value of log10 1 is 0. Q5 ### What is the value of log10 2? The value of log10 2 is 0.3010. 1. Ya it is good 2. Wow…thank you…. i’m grateful…. now i can use a log table…. • Thank you I can use logarithms table am so grateful 3. Thank U … Yeeh! I can use log table 🙏👍👍 4. I’m very grateful, now I can make use of logarithm table without any problems. Thanks for teaching me well I’m happy 5. Yeah, thanks I can use logarithm tables 6. Very good 7. Well, That’s Useful. Thanks 😉 8. That’s great information about log table. I understand how to solve log table. Thank you
# 02.1 线性反向传播 ## 2.1 线性反向传播⚓︎ ### 2.1.1 正向计算的实例⚓︎ z = x \cdot y \tag{1} x = 2w + 3b \tag{2} y = 2b + 1 \tag{3} $w = 3, b = 4$ 时，会得到图2-5的结果。 ### 2.1.2 反向传播求解 $w$⚓︎ #### 求 $w$ 的偏导⚓︎ $z$ 开始一层一层向回看，图中各节点关于变量 $w$ 的偏导计算结果如下： \frac{\partial{z}}{\partial{w}}=\frac{\partial{z}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{w}}=y \cdot 2=18 \tag{4} \frac{\partial{z}}{\partial{x}}=\frac{\partial{}}{\partial{x}}(x \cdot y)=y=9 \frac{\partial{x}}{\partial{w}}=\frac{\partial{}}{\partial{w}}(2w+3b)=2 z=x \cdot y=(2w+3b)(2b+1)=4wb+2w+6b^2+3b \tag{5} \frac{\partial z}{\partial w}=4b+2=4 \cdot 4 + 2=18 \tag{6} #### 求 $w$ 的具体变化值⚓︎ \Delta z = 18 \cdot \Delta w \Delta w = {\Delta z \over 18}=\frac{162-150}{18}= 0.6667 w = w - 0.6667=2.3333 x=2w+3b=16.6667 z=x \cdot y=16.6667 \times 9=150.0003 ### 2.1.3 反向传播求解 $b$⚓︎ #### 求 $b$ 的偏导⚓︎ 1. $z$$x$$b$ 2. $z$$y$$b$ \frac{\partial{z}}{\partial{b}}=\frac{\partial{z}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{b}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{b}}=y \cdot 3+x \cdot 2=63 \tag{7} \frac{\partial{z}}{\partial{x}}=\frac{\partial{}}{\partial{x}}(x \cdot y)=y=9 \frac{\partial{z}}{\partial{y}}=\frac{\partial{}}{\partial{y}}(x \cdot y)=x=18 \frac{\partial{x}}{\partial{b}}=\frac{\partial{}}{\partial{b}}(2w+3b)=3 \frac{\partial{y}}{\partial{b}}=\frac{\partial{}}{\partial{b}}(2b+1)=2 z=x \cdot y=(2w+3b)(2b+1)=4wb+2w+6b^2+3b \tag{5} \frac{\partial z}{\partial b}=4w+12b+3=12+48+3=63 \tag{8} #### 求 $b$ 的具体变化值⚓︎ \Delta z = 63 \cdot \Delta b \Delta b = \frac{\Delta z}{63}=\frac{162-150}{63}=0.1905 x=2w+3b=17.4285 y=2b+1=8.619 z=x \cdot y=17.4285 \times 8.619=150.2162 ### 2.1.4 同时求解 $w$ 和 $b$ 的变化值⚓︎ \Delta b=\frac{\Delta z / 2}{63} = \frac{12/2}{63}=0.095 \Delta w=\frac{\Delta z / 2}{18} = \frac{12/2}{18}=0.333 • $w = w-\Delta w=3-0.333=2.667$ • $b = b - \Delta b=4-0.095=3.905$ • $x=2w+3b=2 \times 2.667+3 \times 3.905=17.049$ • $y=2b+1=2 \times 3.905+1=8.81$ • $z=x \times y=17.049 \times 8.81=150.2$ #### 【课堂练习】用Python代码实现以上双变量的反向传播计算过程⚓︎ 1. 在检查 $\Delta z$ 时的值时，注意要用绝对值，因为有可能是个负数 2. 在计算 $\Delta b$$\Delta w$ 时，第一次时，它们对 $z$ 的贡献值分别是 $1/63$$1/18$，但是第二次时，由于 $b,w$ 值的变化，对 $z$ 的贡献值也会有微小变化，所以要重新计算。具体解释如下： \frac{\partial{z}}{\partial{b}}=\frac{\partial{z}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{b}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{b}}=y \cdot 3+x \cdot 2=3y+2x \frac{\partial{z}}{\partial{w}}=\frac{\partial{z}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{w}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{w}}=y \cdot 2+x \cdot 0 = 2y $$所以，在每次迭代中，要重新计算下面两个值：$$ \Delta b=\frac{\Delta z}{3y+2x} \Delta w=\frac{\Delta z}{2y} single variable: b ----- w=3.000000,b=4.000000,z=162.000000,delta_z=12.000000 delta_b=0.190476 w=3.000000,b=3.809524,z=150.217687,delta_z=0.217687 delta_b=0.003455 w=3.000000,b=3.806068,z=150.007970,delta_z=0.007970 delta_b=0.000127 w=3.000000,b=3.805942,z=150.000294,delta_z=0.000294 delta_b=0.000005 w=3.000000,b=3.805937,z=150.000011,delta_z=0.000011 delta_b=0.000000 w=3.000000,b=3.805937,z=150.000000,delta_z=0.000000 done! final b=3.805937 single variable new: b ----- w=3.000000,b=4.000000,z=162.000000,delta_z=12.000000 factor_b=63.000000, delta_b=0.190476 w=3.000000,b=3.809524,z=150.217687,delta_z=0.217687 factor_b=60.714286, delta_b=0.003585 w=3.000000,b=3.805938,z=150.000077,delta_z=0.000077 factor_b=60.671261, delta_b=0.000001 w=3.000000,b=3.805937,z=150.000000,delta_z=0.000000 done! final b=3.805937 1. factor_b第一次是63，以后每次都会略微降低一些 2. 第二个函数迭代了3次就结束了，而第一个函数迭代了5次，效率不一样 3. 最后得到的结果是一样的，因为这个问题只有一个解 double variable: w, b ----- w=3.000000,b=4.000000,z=162.000000,delta_z=12.000000 delta_b=0.095238, delta_w=0.333333 w=2.666667,b=3.904762,z=150.181406,delta_z=0.181406 delta_b=0.001440, delta_w=0.005039 w=2.661628,b=3.903322,z=150.005526,delta_z=0.005526 delta_b=0.000044, delta_w=0.000154 w=2.661474,b=3.903278,z=150.000170,delta_z=0.000170 delta_b=0.000001, delta_w=0.000005 w=2.661469,b=3.903277,z=150.000005,delta_z=0.000005 done! final b=3.903277 final w=2.661469 double variable new: w, b ----- w=3.000000,b=4.000000,z=162.000000,delta_z=12.000000 factor_b=63.000000, factor_w=18.000000, delta_b=0.095238, delta_w=0.333333 w=2.666667,b=3.904762,z=150.181406,delta_z=0.181406 factor_b=60.523810, factor_w=17.619048, delta_b=0.001499, delta_w=0.005148 w=2.661519,b=3.903263,z=150.000044,delta_z=0.000044 factor_b=60.485234, factor_w=17.613053, delta_b=0.000000, delta_w=0.000001 w=2.661517,b=3.903263,z=150.000000,delta_z=0.000000 done! final b=3.903263 final w=2.661517 ### 参考资料⚓︎ http://colah.github.io/posts/2015-08-Backprop/ ch02, Level1
Thank you for visiting our Anchor site. Mathematics Select a Standard --Select-- I II III IV IX V VI VII VIII The Co-ordinates – A Game Geo-Engine (Std I) Rythmic Maths (Std I) Adding in many ways (Std I) Calculative Colouring (Std II) Number Race (Std II) Mental Trial (Std III) Time story (Std III) Calendar (Std IV) Division of three-digit numbers (Std IV) Mathematically speaking (Std V) What makes 100 % ? (Std V) Weight Check (Std VI) Ratio (Std VI) Park Survey (Std VII) Ratio Quiz (Std VII) Adding in many ways (Std I) This activity can be given to the students in the class during the development of number concepts. Objectives To help the students to i)add numbers mentally. ii)compare and identify the same sum. iii)classify the cards having the same sum. iv)conclude that a number can be obtained in many ways. Materials Required 35 cards of dimension 15 cm * 10 cm. (The blank sides of old invitation cards could be used) What's in Anchor DISC: Lesson Plan Mindset Communication Tips Question Box Help-line Enrichment Activities ·In Mathematics ·In Science Anchor Resources Online Contests Instruction to teachers l Write all possible ways of obtaining a number by adding any two addends on the cards For example: The number nine (9) can be obtained by a combination of different numbers as shown below 1 + 8 2 + 7 3 + 6 4 + 5 9 + 0 l Do this for numbers 1 to 8 l Next, spread all the cards on the table or floor l Ask the students to find the sum of the numbers on each card Help them to group the cards having the same sum. Initiate a discussion by asking questions of the following type (for all numbers) i) In how many ways can you get the number 8? ii) Which are they? Help them conclude that a number can be obtained in many ways. Extension: If the topic ‘subtraction’ has been completed, you can add cards which indicate different combinations by which a number can be expressed (by subtraction) For example, the number 6 (six) 10 – 4 9 – 3 8 – 2 7 – 1 6 – 0

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.