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## GeoGebra Tutorial 5 – Discovering the Pythagorean Theorem This is the fifth tutorial in the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series. In this tutorial, we compare areas of squares formed on the sides of a right triangle. To construct this figure, we first construct a right triangle, and form three squares, each of which contains one of the three sides as shown below. Then we observe the relationship among the areas of the squares. Figure 1 – Squares formed containing the sides of a right triangle In this tutorial, we learn how to use the Regular Polygon tool. Most of the constructions that you will make here are review of the first four tutorials. Instructions 1.) Open GeoGebra and select Geometry from the Perspective panel. 2.) Select the Segment between Two Points tool and click two distinct places on the Graphics view to construct segment AB. 3.) If the labels of the points are not displayed, click the Move tool, right click each point and click Show label from the context menu. 4.) Next, we construct a line perpendicular to segment AB and passing through point B. To do this, choose the Perpendicular line tool, click on segment AB, then click on point B. 5.) Next, we create point C on the line. To do this, click the New point tool and click on the line. Be sure that the label of the third point is displayed. Figure 2 – Point C on the line passing through B You have to be sure that C is on the line passing through B. Be sure that you cannot drag point C out of the line. Otherwise, delete the point and create a new point C. 6.) Hide everything except the three points by right clicking tmem and unchecking the Show Object option. 7.) Next, we rename point B to point C and vice versa. To rename point B to C, right click point B, click Rename and then type the new name, in this case point C, in the Rename text box, then click the OK button. Now, rename B (or B1) to C. 8.) Next we construct a square with side AC. Click the Regular polygon tool, then click on point C and click on point A. 9.) In the Points text box of the Regular polygon tool, type 4. If the position of the square is displayed the wrong way (right hand side of AC) just undo button and reverse the order of the clicks when creating the polygon. Figure 3 – Square containing side AC 10. ) With the Polygon tool still active, click point B and click point C to create a square with side BC. Similarly, click point A, then click point B to create a square with side AB. After step 10, your drawing should look like the one shown below. Figure 4 – Squares containing sides of right triangle ABC 11.) Hide the label of the sides of the side of the squares. 12.) Rename the sides of the rectangle as shown below. Figure 5 – Triangle ABC wth side lengths a, b and c. 13.) Now, let us reveal the area of the three squares. Right click the interior of the square with side AC, then click Object Properties from the context menu to display the Preferences window. 14.) In the Basic tab of the Preferences window, check the Show Label check box and choose Value from the drop-down list box. Do this to the other two squares as well. Figure 6 – Properties of squares shown in the Preferences window. 15.) Move the vertices of the triangle. What do you observe about the area of the squares? 16.) You may have observed that the area of the biggest square is equal to the sum of the areas of the two smaller squares. To verify this, we can put a label in the GeoGebra window displaying the areas of the three squares. 17.) Suppose the side of the two smaller squares are a and b, and the side of the biggest square is c, what equation can you make to express the relationship of the of the three squares? 18.) What conjecture can you make based on your observation? ## GeoGebra Tutorial 4 – Graphs and Sliders This is the fourth tutorial in the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series first. In this tutorial, we use GeoGebra to investigate the effects of parameters of the equations of functions to the appearance of their graphs. First, we type equations $y = mx$ and then use a slider investigate the effects of $m$. Figure 1 – Sample Graphs created with GeoGebra Graphing Functions 1. Open GeoGebra and be sure the the Algebra and Graphics view is selected for the Perspective menu. 2. To graph $y = 2x$, type y = 2x in the input field. (The input field is the text box with the label Input located at the bottom of your GeoGebra window.) Press the ENTER key on your keyboard. 3. Type the following equations: y = 3x, y = 4x , y = -8x, and press the ENTER key after each equation. 4. Type more equations of the form $y = mx$ where $m$ is any real number. 5. How does the value of $m$affect the appearance of the graph of the function $y = mx$? Using Sliders To avoid typing over and over again for varying values $m$, we use the slider tool. A slider is a visual representation of a number. This time, we add the parameter b. This means that we will explore the graph of the function of $y = mx + b$, where $m$ and$b$ are real numbers. Instructions 1.) Open the GeoGebra and be sure that Algebra View is selected from the Perspective menu. You can also do this by selecting Algebra from the view menu. 2.) Select the Slider tool and click on the Graphics view to display the Slider dialog box. 3.) In the Slider dialog box, change the slider’s name to m, change the interval to $-10$ to $10$ as shown below.Leave the other values as is and click the Apply button to finish. 4.) Adjust the position of your slider if necessary. Move the small black circle on your slider. What do you observe? 5.) To graph the function $y = mx$, type y = mx in the Input box located at the bottom of the window. If the Input bar is not displayed, you can display it using the View menu. 6.) Now, Move the small circle on your slider. What do you observe? 7.) Now, create a new Slider and name it $b$. Set the interval to $-10$ and $10$ and leave the other values as they are. 8.) To change the function to $y = mx + b$, double click the graph to display the Redefine window, and then type y = mx + b. 9.) Now, move your $b$ slider. What do you observe? 10. ) How does the value of $b$ affect the appearance of the graph of the function $y = mx + b$? 11.) As an exercise, graph the function $y = ax^2 + b$. Explain the effects of the parameters $a$ and $b$ to the graph of the function. Notes: 1. The ^ symbol is used for exponentiation. Hence, we write $ax^2 + c$ as a*x^2 + c. 2. Instead of using $y$ in writing equations of functions, you can also use $f(x)$ for the function $f$. Last Update: November 12, 2014 for GeoGebra 5.0. ## GeoGebra Tutorial 3 – Constructing a Square This is the third tutorial of the GeoGebra Intermediate Tutorial Series and the fourth part of the GeoGebra Basic Construction Series. If this is your first time to use GeoGebra, I strongly suggest that you read the GeoGebra Essentials Series. This tutorial, answers the following problem using GeoGebra. Problem: How will you draw an equilateral triangle without using the Regular polygon tool? In this tutorial, we mimic compass and straightedge construction using the Circle tool, the Parallel Line tool, the Perpendicular Line tool of GeoGebra to construct a square instead of using the Regular Polygon tool. We will also reinforce the use of the Angle tool, this time, learn how to use it to measure angle using three points. » Read more 1 67 68 69 70
# Finding the Normal Equations A normal curve is a straight line passing through the point where the tangent touches the curve and is perpendicular (at right angles) to the tangent at that point. The gradient of the tangent to a curve is $m$, then the gradient of the normal is $\displaystyle -\dfrac{1}{m}$, as the product of the gradients of $2$ perpendicular lines equals to $-1$. The gradient of the tangent at $x=a$ is $f^{\prime}(a)$. Therefore the gradient of the normal is $\displaystyle -\dfrac{1}{f^{\prime}(a)}$. The equation of the normal is: $$y-f(a) = -\dfrac{1}{f^{\prime}(a)}(x-a)$$ ### Example 1 Find the gradient of the normal to the curve $f(x)=2x^3-x^2+1$ at $x=1$. \begin{align} \displaystyle \require{AMSsymbols} \require{color} f^{\prime}(x) &= 6x^2-2x \\ f^{\prime}(1) &= 6 \times 1^2-2 \times 1 &\color{red} \text{gradient of tangent} \\ &= 4 \\ \end{align} Therefore, the normal gradient is $\displaystyle -\dfrac{1}{4}$. ### Example 2 Find the equation of the normal to $f(x)=x^3-2x+3$ at the point $(1,2)$. \begin{align} \displaystyle \require{AMSsymbols} \require{color} f^{\prime}(x) &= 3x^2-2 \\ f^{\prime}(1) &= 3 \times 1^2-2 &\color{red} \text{gradient of tangent} \\ &= 1 \end{align} Therefore the gradient of the normal is $-1$. The equation of the normal is: \begin{align} \displaystyle \require{color} y-2 &= -1(x-1) \\ \therefore y &= -x+3 \end{align} ### Example 3 Find the equation of the normal to $f(x)=3x^2-1$ at $x=1$. \begin{align} \displaystyle \require{AMSsymbols} \require{color} f(1) &= 3 \times 1^2-1 \\ &= 2 &\color{red} y\text{-coordinate of the point} \\ f^{\prime}(x) &= 6x \\ f^{\prime}(1) &= 6 \times 1 &\color{red} \text{gradient of tangent} \\ &= 6 \end{align} Therefore, the normal gradient is $\displaystyle -\dfrac{1}{6}$. The equation of the normal is: \begin{align} \displaystyle \require{AMSsymbols} \require{color} y-2 &= -\dfrac{1}{6}(x-1) &\color{red} y-f(1) = -\dfrac{1}{f^{\prime}(1)}(x-1) \\ 6y-12 &= -x+1 \\ \therefore x+6y-13 &= 0 \end{align} Discover more enlightening videos by visiting our YouTube channel! ## Mastering Integration by Parts: The Ultimate Guide Welcome to the ultimate guide on mastering integration by parts. If you’re a student of calculus, you’ve likely encountered integration problems that seem insurmountable. That’s… ## High School Math for Life: Making Sense of Earnings Salary Salary refers to the fixed amount of money that an employer pays an employee at regular intervals, typically on a monthly or biweekly basis,… ## The Best Practices for Using Two-Way Tables in Probability Welcome to a comprehensive guide on mastering probability through the lens of two-way tables. If you’ve ever found probability challenging, fear not. We’ll break it… ## Binomial Expansions The sum $a+b$ is called a binomial as it contains two terms.Any expression of the form $(a+b)^n$ is called a power of a binomial. All… ## How to Find the Gradient of a Straight Line Understanding Straight-Line Gradients: A Practical Guide Straight-line gradients are a fundamental concept in mathematics and physics, crucial for understanding the steepness, incline, or slope of…
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2. Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 9 Chapter Name Algebraic Expressions and Identities Exercise Ex 9.2 Number of Questions Solved 5 Category NCERT Solutions NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Question 1. Find the product of the following pairs of monomials: (i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq (iv) $$4{ p }^{ 3, },\quad -3p$$ (v) 4p, 0. Solution. Question 2. Find the areas of rectangles with the following pairs of mononials as their lengths and breadths respectively: (i) (p, q); (ii) (10m, 5n); (iii) ($$20{ x }^{ 2 },\quad 5{ y }^{ 2 }$$); (iv) ($$4x,\quad 3{ x }^{ 2 }$$); (v) (3mn, 4np). Solution. (i) (p, q) Length = p ∴ Area of the rectangle = pxq = pq (ii) (10m, 5n) Length = 10 m ∴ Area of the rectangle = (10m) x (5n) = (10 x 5) x (m x n) = 50 x (mn) = 50 mn (iii) ($$20{ x }^{ 2 },\quad 5{ y }^{ 2 }$$) Length = $$20{ x }^{ 2 }$$ Breadth = $$5{ y }^{ 2 }$$ ∴ Area of the rectangle = ($$20{ x }^{ 2 }$$) x ($$5{ y }^{ 2 }$$) = (20 x 5) x ($${ x }^{ 2 }\times { y }^{ 2 }$$) = 100 x ($${ x }^{ 2 }{ y }^{ 2 }$$) = 100$${ x }^{ 2 }{ y }^{ 2 }$$ (iv) (4x, 3xP) Length = 4.x Breadth = $$3{ x }^{ 2 }$$ ∴ Area of the rectangle (4x) x ($$3{ x }^{ 2 }$$) = (4 x 3) x ($$x\times { x }^{ 2 }$$) = 12 x $${ x }^{ 3 }$$ = 12×3 (v) (3mn, 4np) Length = 3 mn ∴ Area of the rectangle = (3mn) x (4np) = (3 x 4) x (mn) x (np) = 12 x m x (n x n) x p = 12$$m{ n }^{ 2 }p$$ Question 3. Complete the table of products. Solution. Question 4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively: (i) $$5a,\quad 3{ a }^{ 2 },\quad 7{ a }^{ 4 }$$ (ii) 2p, 4q, 8r (iii) $$xy,\quad 2{ x }^{ 2 }y,\quad 2x{ y }^{ 2 }$$ (iv) a, 2b, 3c Solution. (i) $$5a,\quad 3{ a }^{ 2 },\quad 7{ a }^{ 4 }$$ (ii) 2p, 4q, 8r (iii) $$xy,\quad 2{ x }^{ 2 }y,\quad 2x{ y }^{ 2 }$$ (iv) a, 2b, 3c Question 5. Obtain the product of (i) xy, yz, zx (ii) $$a,\quad -{ a }^{ 2 },\quad { a }^{ 3 }$$ (iii) $$2,\quad 4y,\quad 8{ y }^{ 2 },\quad 16{ y }^{ 3 }$$ (iv) a, 2b, 3c, 6abc (v) m, – mn, mnp. Solution. (i) xy, yz, zx Required product = (xy) x (yz) x (zx) (ii) $$a,\quad -{ a }^{ 2 },\quad { a }^{ 3 }$$ (iii) $$2,\quad 4y,\quad 8{ y }^{ 2 },\quad 16{ y }^{ 3 }$$ (iv) a, 2b, 3c, 6abc (v) m, – mn, mnp. We hope the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2, drop a comment below and we will get back to you at the earliest.
## Section8.2Another look at distributing apples or folders A recurring problem so far in this book has been to consider problems that ask about distributing indistinguishable objects (say apples) to distinct entities (say children). We started in Chapter 2 by asking how many ways there were to distribute $$40$$ apples to $$5$$ children so that each child is guaranteed to get at least one apple and saw that the answer was $$C(39,4)\text{.}$$ We even saw how to restrict the situation so that one of the children was limited and could receive at most $$10$$ apples. In Chapter 7, we learned how to extend the restrictions so that more than one child had restrictions on the number of apples allowed by taking advantage of the Principle of Inclusion-Exclusion. Before moving on to see how generating functions can allow us to get even more creative with our restrictions, let's take a moment to see how generating functions would allow us to solve the most basic problem at hand. ###### Example8.4. We already know that the number of ways to distribute $$n$$ apples to $$5$$ children so that each child gets at least one apple is $$C(n-1,4)\text{,}$$ but it will be instructive to see how we can derive this result using generating functions. Let's start with an even simpler problem: how many ways are there to distribute $$n$$ apples to one child so that each child receives at least one apple? Well, this isn't too hard, there's only one way to do it—give all the apples to the lucky kid! Thus the sequence that enumerates the number of ways to do this is $$\{a_n\colon n\geq 1\}$$ with $$a_n=1$$ for all $$n\geq 1\text{.}$$ Then the generating function for this sequence is \begin{equation} x+x^2+x^3+\cdots = x(1+x+x^2+x^3+\cdots) = \frac{x}{1-x}. \end{equation} How can we get from this fact to the question of five children? Notice what happens when we multiply \begin{equation} (x+x^2+\cdots)(x+x^2+\cdots)(x+x^2+\cdots)(x+x^2+\cdots) (x+x^2+\cdots). \end{equation} To see what this product represents, first consider how many ways can we get an $$x^6\text{?}$$ We could use the $$x^2$$ from the first factor and $$x$$ from each of the other four, or $$x^2$$ from the second factor and $$x$$ from each of the other four, etc., meaning that the coefficient on $$x^6$$ is $$5 = C(5,4)\text{.}$$ More generally, what's the coefficient on $$x^n$$ in the product? In the expansion, we get an $$x^n$$ for every product of the form $$x^{k_1}x^{k_2}x^{k_3}x^{k_4}x^{k_5}$$ where $$k_1+k_2+k_3+k_4+k_5 = n\text{.}$$ Returning to the general question here, we're really dealing with distributing $$n$$ apples to $$5$$ children, and since $$k_i> 0$$ for $$i=1,2,\dots,5\text{,}$$ we also have the guarantee that each child receives at least one apple, so the product of the generating function for one child gives the generating function for five children. Let's pretend for a minute that we didn't know that the coefficients must be $$C(n-1,4)\text{.}$$ How could we figure out the coefficients just from the generating function? The generating function we're interested in is $$x^5/(1-x)^5\text{,}$$ which you should be able to pretty quickly see satisfies \begin{align} \frac{x^5}{(1-x)^5} \amp = \frac{x^5}{4!}\frac{d^4}{dx^4}\left(\frac{1}{1-x}\right) = \frac{x^5}{4!}\sum_{n=0}^\infty n(n-1)(n-2)(n-3)x^{n-4}\\ \amp =\sum_{n=0}^\infty \frac{n(n-1)(n-2)(n-3)}{4!}x^{n+1} = \sum_{n=0}^\infty \binom{n}{4}x^{n+1}. \end{align} The coefficient on $$x^n$$ in this series $$C(n-1,4)\text{,}$$ just as we expected. We could revisit an example from Chapter 7 to see that if we wanted to limit a child to receive at most $$4$$ apples, we would use $$(x+x^2+x^3+x^4)$$ as its generating function instead of $$x/(1-x)\text{,}$$ but rather than belabor that here, let's try something a bit more exotic. ###### Example8.5. A grocery store is preparing holiday fruit baskets for sale. Each fruit basket will have $$20$$ pieces of fruit in it, chosen from apples, pears, oranges, and grapefruit. How many different ways can such a basket be prepared if there must be at least one apple in a basket, a basket cannot contain more than three pears, and the number of oranges must be a multiple of four? Solution. In order to get at the number of baskets consisting of $$20$$ pieces of fruit, let's solve the more general problem where each basket has $$n$$ pieces of fruit. Our method is simple: find the generating function for how to do this with each type of fruit individually and then multiply them. As in the previous example, the product will contain the term $$x^n$$ for every way of assembling a basket of $$n$$ pieces of fruit subject to our restrictions. The apple generating function is $$x/(1-x)\text{,}$$ since we only want positive powers of $$x$$ (corresponding to ensuring at least one apple). The generating function for pears is $$(1+x+x^2+x^3)\text{,}$$ since we can have only zero, one, two, or three pears in basket. For oranges we have $$1/(1-x^4) = 1+x^4+x^8+\cdots\text{,}$$ and the unrestricted grapefruit give us a factor of $$1/(1-x)\text{.}$$ Multiplying, we have \begin{equation} \frac{x}{1-x} (1+x+x^2+x^3) \frac{1}{1-x^4} \frac{1}{1-x} = \frac{x}{(1-x)^2(1-x^4)} (1+x+x^2+x^3). \end{equation} Now we want to make use of the fact that $$(1+x+x^2+x^3) =(1-x^4)/(1-x)$$ (by (8.1.1)) to see that our generating function is \begin{align} \frac{x}{(1-x)^3} \amp= \frac{x}{2}\sum_{n=0}^\infty n(n-1)x^{n-2} = \sum_{n=0}^\infty\frac{n(n-1)}{2} x^{n-1} \\ \amp=\sum_{n=0}^\infty\binom{n}{2} x^{n-1} = \sum_{n=0}^\infty\binom{n+1}{2} x^n. \end{align} Thus, there are $$C(n+1,2)$$ possible fruit baskets containing $$n$$ pieces of fruit, meaning that the answer to the question we originally asked is $$C(21,2) = 210\text{.}$$ The compact form of the solution to Example 8.5 suggests that perhaps there is a way to come up with this answer without the use of generating functions. Thinking about such an approach would be a good way to solidify your understanding of a variety of the enumerative topics we have already covered. ###### Example8.6. Find the number of integer solutions to the equation \begin{equation} x_1 + x_2 + x_3 = n \end{equation} ($$n\geq 0$$ an integer) with $$x_1 \geq 0$$ even, $$x_2\geq 0\text{,}$$ and $$0\leq x_3\leq 2\text{.}$$ Solution. Again, we want to look at the generating function we would have if each variable existed individually and take their product. For $$x_1\text{,}$$ we get a factor of $$1/(1-x^2)\text{;}$$ for $$x_2\text{,}$$ we have $$1/(1-x)\text{;}$$ and for $$x_3$$ our factor is $$(1+x+x^2)\text{.}$$ Therefore, the generating function for the number of solutions to the equation above is \begin{equation} \frac{1+x+x^2}{(1-x)(1-x^2)} = \frac{1+x+x^2}{(1+x)(1-x)^2}. \end{equation} In calculus, when we wanted to integrate a rational function of this form, we would use the method of partial fractions to write it as a sum of “simpler” rational functions whose antiderivatives we recognized. Here, our technique is the same, as we can readily recognize the formal power series for many rational functions. Our goal is to write \begin{equation} \frac{1+x+x^2}{(1+x)(1-x)^2} = \frac{A}{1+x} + \frac{B}{1-x} + \frac{C}{(1-x)^2} \end{equation} for appropriate constants, $$A\text{,}$$ $$B\text{,}$$ and $$C\text{.}$$ To find the constants, we clear the denominators, giving \begin{equation} 1+x+x^2 = A(1-x)^2 + B(1-x^2) + C(1+x). \end{equation} Equating coefficients on terms of equal degree, we have: \begin{align} 1 \amp = A+B+C\\ 1 \amp = -2A + C\\ 1 \amp = A - B \end{align} Solving the system, we find $$A=1/4\text{,}$$ $$B=-3/4\text{,}$$ and $$C=3/2\text{.}$$ Therefore, our generating function is \begin{align} \amp\frac{1}{4}\frac{1}{1+x} -\frac{3}{4} \frac{1}{1-x} +\frac{3}{2} \frac{1}{(1-x)^2}\\ =\amp \frac{1}{4}\sum_{n=0}^\infty (-1)^n x^n - \frac{3}{4} \sum_{n=0}^\infty x^n + \frac{3}{2}\sum_{n=0}^\infty n x^{n-1}\text{.} \end{align} The solution to our question is thus the coefficient on $$x^n$$ in the above generating function, which is \begin{equation} \frac{(-1)^n}{4} - \frac{3}{4} + \frac{3(n+1)}{2}, \end{equation} a surprising answer that would not be too easy to come up with via other methods! The invocation of partial fractions in Example 8.6 is powerful, but solving the necessary system of equations and then hoping that the resulting formal power series have expansions we immediately recognize can be a challenge. If Example 8.6 had not asked about the general case with $$n$$ on the right-hand side of the equation but instead asked specifically about $$n=30\text{,}$$ you might be wondering if it would just be faster to write some Python code to generate all the solutions or more interesting to huddle up and devise some clever strategy to count them. Fortunately, technology can help us out when working with generating functions. In SageMath, we can use the series() method to get the power series expansion of a given function. The two arguments to series are the variable and the degree of the terms you want to truncate. In the cell below, we ask SageMath to expand the generating function from Example 8.6 by giving us all the terms of degree at most 30 and then collapsing the rest of the series into its form of big-Oh notation, which we discard by storing the output from series() in a polynomial f(x). If all we really want is the coefficient on a specific term, we can use the list() method to turn the polyomial into a list of its coefficients and then index into that list using standard SageMath or Python syntax: Let's see that the answer agrees with what our formula in the solution to Example 8.6 gives us for $$n=30\text{:}$$ That's a relief, and so long as we only need a single coefficient, we're now in good shape. But what if we really need a formula for the coefficient on $$x^n$$ in general? Let's see how we can use SageMath to help us with some of the other steps in Example 8.6. The first thing we'll want is the partial_fraction() method: If you don't like the way that looks, the pretty_print() function can make it easier to read: Up to the location of a minus sign, this is what we got by hand, but we get it much faster! From this stage, it's frequently possible to use our knowledge of certain fundamental power series that appear when doing the partial fractions expansion to come up with the general form for the coefficient on an arbitrary term of the power series. To facilitate this, we close this section with an example that illustrates how we can use solutions to counting problems we have already studied in order to figure out the coefficients on generating functions. ###### Example8.7. Let $$n$$ be a positive integer. What is the coefficient on $$x^k$$ in the generating function \begin{equation} \frac{1}{(1-x)^n}\text{?} \end{equation} Solution. We have already encountered the case $$n=5$$ in the midst of working on Example 8.4, but there we appealed to calculus. Let's take a look at this from the perspective of just counting. The generating function $$1/(1-x) = 1+x+x^2+\cdots$$ encodes the sequence for the number of ways to distribute $$n$$ apples to one child. There's only one way to do that task: give the lucky kid all the apples. Multiplying together a bunch of copies of $$1/(1-x)$$ then serves to increase the number of children to whom the apples are being distributed, and since each power series being multiplied starts with $$1\text{,}$$ we are in the situation where the number of apples each child receives must be nonnegative. This is therefore a problem from Section 2.5. We have $$n$$ children and the coefficient on $$x^k$$ is the number of ways of distributing $$k$$ apples to them. This requires $$n$$ artificial apples, so we distribute $$k+n$$ apples, which determine $$k+n-1$$ gaps and we must choose $$n-1$$ of them as the locations for dividers. Therefore, we can conclude that \begin{equation} \frac{1}{(1-x)^n} = \sum_{k=0}^\infty \binom{k+n-1}{n-1}x^k = \sum_{k=0}^\infty \binom{k+n-1}{n}x^k\text{.} \end{equation} It's possible to arrive at this conclusion using techniques from calculus, but there are a lot of factorials and $$-1$$s to monitor, so this combinatorial approach may be less error prone!
# What is the derivative of e^(9x)? Sep 4, 2016 $9 {e}^{9 x}$ #### Explanation: We have: ${e}^{9 x}$ This expression can be differentiated using the "chain rule". Let $u = 9 x \implies u ' = 9$ and $v = {e}^{u} \implies v ' = {e}^{u}$: $\implies \frac{d}{\mathrm{dx}} \left({e}^{9 x}\right) = 9 \cdot {e}^{u}$ $\implies \frac{d}{\mathrm{dx}} \left({e}^{9 x}\right) = 9 {e}^{u}$ We can now replace $u$ with $9 x$: $\implies \frac{d}{\mathrm{dx}} \left({e}^{9 x}\right) = 9 {e}^{9 x}$
Deviation from the Goal Page 1 2 3 By doing the graphical method to the problem found in Page 1, you would have found that the solution to the problem is L = 4 and E = 0. This means 400 gallons of latex paint is produced and 0 gallons of enamel paint Lets look back at each of our goals. Goal 1 (Overtime Labour) 10 L + 15 E ≤ 40 10 L + 15 E + Y1- – Y1+ = 40 Thus if we evaluate our goal we have 10(4) + 15 (0) = 40 hours Therefore, our goal has been met exactly. We have not used any overtime labour. There is no deviation from our goal, i.e. Y1- = Y1+= 0 Inserting these values we have: 10(4) + 15 (0) + 0 + 0 = 40 hours Therefore, there is no penalty associated with this goal Goal 2 (Profit) 100 L + 100 E ≥ 1000 100 L + 100 E + Y2- – Y2+ = 1000 Evaluating our goal 100(4) + 100(0) = 400 Our goal is a \$1000, but we have only achieved \$400. Thus, we have fallen short of our goal, by \$600 (i.e. \$1000 - \$400 = \$600). Since, we have not exceeded our goal, then Y2+ = 0. The corresponding variable for fallen short of our goal is our slack deviation variable, and thus Y2- = \$600. Therefore inserting these values we have: 100 (4) + 100 (0) + 600 – 0 = 1000 Thus, since Y2- is \$600, our penalty associated with this goal is thus (\$600) × (\$1) = \$600 Goal 3 (Enamel Paint Production) E ≥ 7 E + Y3- – Y3+ = 7 Thus if we evaluate our goal: Goal 3 = 0 Therefore our goal is 700 gallons of enamel paint, however we are producing 0 gallons of enamel paint. This means we have fallen short of our goal by 700 (i.e. 700 - 0 = 700 enamel paint). We have not exceeded our goal, therefore, Y3+ = 0. We have however, fallen short of goal, which corresponds to the slack variable, Y3- = 7. Remember that the deviation variables are measured in 100 gallons of paint. Therefore inserting these values we have 0 + 7 – 0 = 7 Since Y3- is 7, then we have fallen short of our goal, and our penalty associated with this goal is (700) × (\$0.50) = \$350 Objective function If we look back at our objective function Min 3Y1++ 100 Y2- + 50 Y3- Let us insert the values we found for the deviation variables 3(0) + 100 (600) + 50 (7) = \$950 Thus the total penalties we have suffered for not achieving our goals 2 and 3 is \$950. Constraints Let us take a look at our constraints Constraint 1 (Labour) 10 L + 15 E ≤ 70 Substituting the values 10 (4) + 15 (0) = 40 Therefore the constraint is non-binding and there is a slack of 70 40 = 30 hrs. The 30 hrs corresponds the overtime which we did not use (as found in Goal 1). Constraint 2 (Production Constraint) L E ≥ 0 Substituting E and L values 4 – 0 = 4 Therefore this constraint is non-binding and there is an excess of 400 gallons of paint (i.e. 4 0 = 4 100-gallons of paint ). Note: Remember the minimum required was for E and L to be equal or in other words L – E = 0
# 2014 IMO Problems/Problem 4 ## Problem Points $P$ and $Q$ lie on side $BC$ of acute-angled $\triangle{ABC}$ so that $\angle{PAB}=\angle{BCA}$ and $\angle{CAQ}=\angle{ABC}$. Points $M$ and $N$ lie on lines $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove that lines $BM$ and $CN$ intersect on the circumcircle of $\triangle{ABC}$. ## Solution Sorry guys I'm new to AOPS so I don't know how to insert equations and stuff. Please help if you can. Thanks. We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. Let the intersection of BM and AN be point E, and the intersection of AM and CN be point F. Let us assume (angle BDC) + (angle BAC) = 180. Note: This is circular reasoning. If angle BDC plus angle BAC is 180, then angle BAC should be equal to angles BDN and CDM. We can quickly prove that the triangles ABC, APB, and AQC are similar, so angles BAC = AQC = APB. We also see that angles AQC = BQN = APB = CPF. Also because angles BEQ and NED, MFD and CFP are equal, the triangles BEQ and NED, MDF and FCP must be two pairs of similar triangles. Therefore we must prove angles CBM and ANC, AMB and BCN are equal. We have angles BQA = APC = NQC = BPM. We also have AQ = QN, AP = PM. Because the triangles ABP and ACQ are similar, we have EC/EN = BF/FM, so triangles BFM and NEC are similar. So the angles CBM and ANC, BCN and AMB are equal and we are done. ## Solution 2 Let $L$ be the midpoint of $BC$. Easy angle chasing gives $\angle{AQP} = \angle{APQ} = \angle{BAC}$. Because $P$ is the midpoint of $AM$, the cotangent rule applied on triangle $MBA$ gives us $$\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.$$ Hence, by the cotangent rule on $ABC$, we have $$\cot \angle{BAL} = 2\cot \angle{BAC} + \cot \angle{ABC} = \cot \angle{MBC}.$$ Because the period of cotangent is $180^\circ$, but angles are less than $180^\circ$, we have $\angle{BAL} = \angle{MBC}.$ Similarly, we have $\angle{LAC} = \angle{NCB}.$ Hence, if $BM$ and $CN$ intersect at $Z$, then $\angle{BZC} = 180^\circ - \angle{BAC}$ by the Angle Sum in a Triangle Theorem. Hence, $BACZ$ is cyclic, which is equivalent to the desired result. --Suli 23:27, 7 February 2015 (EST) ## Solution 3 Let $L$ be the midpoint of $BC$. By AA Similarity, triangles $BAP$ and $BCA$ are similar, so $\dfrac{BA}{AP} = \dfrac{BC}{CA}$ and $\angle{BPA} = \angle{BAC}$. Similarly, $\angle{CQA} = \angle{BAC}$, and so triangle $AQP$ is isosceles. Thus, $AQ = AP$, and so $\dfrac{BA}{AQ} = \dfrac{BC}{CA}$. Dividing both sides by 2, we have $\dfrac{BA}{AN} = \dfrac{BL}{AC}$, or $$\frac{BA}{BL} = \frac{AN}{AC}.$$ But we also have $\angle{ABL} = \angle{CAQ}$, so triangles $ABL$ and $NAC$ are similar by $SAS$ similarity. In particular, $\angle{ANC} = \angle{BAL}$. Similarly, $\angle{BMA} = \angle{CAL}$, so $\angle{ANC} + \angle{BMA} = \angle{BAC}$. In addition, angle sum in triangle $AQP$ gives $\angle{QAP} = 180^\circ - 2\angle{A}$. Therefore, if we let lines $BM$ and $CN$ intersect at $T$, by Angle Sum in quadrilateral concave $\angle{NTM} = 180^\circ + \angle{A}$, and so convex $\angle{BTC} = 180^\circ - \angle{A}$, which is enough to prove that $BACT$ is cyclic. This completes the proof. --Suli 10:38, 8 February 2015 (EST)
# NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials Exercise 2.4 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4 are provided here to aid students in their studies. These solutions are prepared by our subject experts in Maths to help the students prepare well for the Class 10 board exam. These experts create NCERT Solutions for Maths which would help students to solve the NCERT problems easily. They also make sure the concepts are easy to understand and that students can learn quickly. Exercise 2.4 is optional and is not given from the examination point of view. It consists of extra questions to practice from the chapter. Our experts provide a detailed solution for each answer to the questions given in Exercise 2.4 in the NCERT textbook for Class 10. The NCERT Solutions for Class 10 Maths Chapter 2 Polynomials are prepared by following NCERT guidelines and syllabus. These are very helpful in scoring well in the examinations. ### Access Other Exercise Solutions of Class 10 Maths Chapter 2 â€“ Polynomials Exercise 2.1 Solutions 1 Question Exercise 2.2 Solutions 2 Question (2 short) Exercise 2.3 Solutions 5 Questions (2 short, 3 long) ### Access Answers to NCERT Class 10 Maths Chapter 2 â€“ Polynomials Exercise 2.4 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: (i) 2x3+x2-5x+2; -1/2, 1, -2 Solution: Given, p(x) = 2x3+x2-5x+2 And zeroes for p(x) are = 1/2, 1, -2 Â âˆ´ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0 p(1) = 2(1)3+(1)2-5(1)+2 = 0 p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0 Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2. Now, comparing the given polynomial with the general expression, we get; âˆ´ ax3+bx2+cx+d = 2x3+x2-5x+2 a=2, b=1, c= -5 and d = 2 As we know, if Î±, Î², Î³ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then; Î± +Î²+Î³ = â€“b/a Î±Î²+Î²Î³+Î³Î± = c/a Î± Î²Î³ = â€“ d/a. Therefore, putting the values of zeroes of the polynomial, Î±+Î²+Î³ = Â½+1+(-2) = -1/2 = â€“b/a Î±Î²+Î²Î³+Î³Î± = (1/2Ã—1)+(1 Ã—-2)+(-2Ã—1/2) = -5/2 = c/a Î± Î² Î³ = Â½Ã—1Ã—(-2) = -2/2 = -d/a Hence, the relationship between the zeroes and the coefficients is satisfied. (ii) x3-4x2+5x-2 ;2, 1, 1 Solution: Given, p(x) = x3-4x2+5x-2 And zeroes for p(x) are 2,1,1. âˆ´ p(2)= 23-4(2)2+5(2)-2 = 0 p(1) = 13-(4Ã—12 )+(5Ã—1)-2 = 0 Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2 Now, comparing the given polynomial with the general expression, we get; âˆ´ ax3+bx2+cx+d = x3-4x2+5x-2 a = 1, b = -4, c = 5 and d = -2 As we know, if Î±, Î², Î³ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then; Î± + Î² + Î³ = â€“b/a Î±Î² + Î²Î³ + Î³Î± = c/a Î± Î² Î³ = â€“ d/a. Therefore, putting the values of zeroes of the polynomial, Î± +Î²+Î³ = 2+1+1 = 4 = -(-4)/1 = â€“b/a Î±Î²+Î²Î³+Î³Î± = 2Ã—1+1Ã—1+1Ã—2 = 5 = 5/1= c/a Î±Î²Î³ = 2Ã—1Ã—1 = 2 = -(-2)/1 = -d/a Hence, the relationship between the zeroes and the coefficients is satisfied. 2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, â€“7, â€“14 respectively. Solution: Let us consider the cubic polynomial is ax3+bx2+cx+d, and the values of the zeroes of the polynomials are Î±, Î², Î³. As per the given question, Î±+Î²+Î³ = -b/a = 2/1 Î±Î² +Î²Î³+Î³Î± = c/a = -7/1 Î± Î²Î³ = -d/a = -14/1 Thus, from the above three expressions, we get the values of the coefficients of the polynomial. a = 1, b = -2, c = -7, d = 14 Hence, the cubic polynomial is x3-2x2-7x+14 3. If the zeroes of the polynomial x3-3x2+x+1 are a â€“ b, a, a + b, find a and b. Solution: We are given the polynomial here, p(x) = x3-3x2+x+1 And zeroes are given as a â€“ b, a, a + b Now, comparing the given polynomial with the general expression, we get; âˆ´px3+qx2+rx+s = x3-3x2+x+1 p = 1, q = -3, r = 1 and s = 1 Sum of zeroes = a â€“ b + a + a + b -q/p = 3a Putting the values q and p. -(-3)/1 = 3a a=1 Thus, the zeroes are 1-b, 1, 1+b. Now, product of zeroes = 1(1-b)(1+b) -s/p = 1-b2 -1/1 = 1-b2 b2 = 1+1 = 2 b = Â±âˆš2 Hence,1-âˆš2, 1 ,1+âˆš2 are the zeroes of x3-3x2+x+1. 4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 Â±âˆš3, find other zeroes. Solution: Since this is a polynomial equation of degree 4, there will be total of 4 roots. Let f(x) = x4-6x3-26x2+138x-35 Since 2 +âˆš3 and 2-âˆš3 are zeroes of given polynomial f(x). âˆ´ [xâˆ’(2+âˆš3)] [xâˆ’(2-âˆš3)] = 0 (xâˆ’2âˆ’âˆš3)(xâˆ’2+âˆš3) = 0 After multiplication, we get, x2-4x+1, this is a factor of a given polynomial f(x). Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x), and the remainder will be 0. So, x4-6x3-26x2+138x-35 = (x2-4x+1)(x2Â â€“2xâˆ’35) Now, on further factorizing (x2â€“2xâˆ’35) we get, x2â€“(7âˆ’5)x âˆ’35Â = x2â€“ 7x+5x+35 = 0 x(x âˆ’7)+5(xâˆ’7) = 0 (x+5)(xâˆ’7) = 0 So, its zeroes are given by: x= âˆ’5 and x = 7. Therefore, all four zeroes of the given polynomial equation are 2+âˆš3 , 2-âˆš3, âˆ’5 and 7. Q.5: If the polynomial x4 â€“ 6x3 + 16x2 â€“ 25x + 10 is divided by another polynomial x2 â€“ 2x + k, the remainder comes out to be x + a, find k and a. Solution: Letâ€™s divide x4 â€“ 6x3 + 16x2 â€“ 25x + 10 by x2 â€“ 2x + k. Given that the remainder of the polynomial division is x + a. (4k â€“ 25 + 16 â€“ 2k)x + [10 â€“ k(8 â€“ k)] = x + a (2k â€“ 9)x + (10 â€“ 8k + k2) = x + a Comparing the coefficients of the above equation, we get; 2k â€“ 9 = 1 2k = 9 + 1 = 10 k = 10/2 = 5 And 10 â€“ 8k + k2 = a 10 â€“ 8(5) + (5)2 = a [since k = 5] 10 â€“ 40 + 25 = a a = -5 Therefore, k = 5 and a = -5. NCERT Solutions for Class 10 Maths Chapter 2 â€“ Polynomials Exercise 2.4 is the fourth exercise of Chapter 2 of Class 10 Maths. Polynomials are introduced in Class 9, and this is discussed in more detail in Class 10. • This exercise is not from the examination point of view. ### Key benefits of NCERT Solutions for Class 10 Maths Chapter 2 â€“ Polynomials Exercise 2.4 • After going through the stepwise solutions given by our subject experts, you will be able to score more marks in the school as well as the board exams. • It follows NCERT guidelines which help in preparing the students accordingly. • It contains all the important questions. • These NCERT Solutions help you solve and revise all questions of Exercise 2.4.
Ascending and Descending Order of Decimals - CREST Olympiads # Ascending and Descending Order of Decimals ## Ascending and Descending Order of Decimals - Sub Topics • Ascending Order • Descending Order • Ascending and Descending Order of Decimals • ## Ascending Order Ascending numbers refer to a sequence of numbers arranged in increasing order from left to right. This means that each number in the sequence is greater than the previous one and the sequence continues to increase as you move from left to right. ## Descending Order Descending numbers refer to a sequence of numbers arranged in decreasing order, from the largest to the smallest. This means that each number in the sequence is smaller than the previous one and the sequence continues to decrease as you move from left to right. ## Ascending and Descending Order of Decimals Ascending order of decimals: It means arranging the decimal numbers from smallest to largest. Example: 2.61 < 2.75 < 2.9 < 2.93 Descending order of decimals: It means arranging the decimal numbers from largest to smallest. Example: 2.93 > 2.9 > 2.75 > 2.61 To arrange decimals in ascending/descending order, we compare the whole numbers first, then the tenths, hundredths, thousandths, and so on. Consider the following set of decimals: 0.153, 0.895, 0.721, 0.468, 0.926 To arrange these decimals in descending order by comparing the tenths and hundredths place, we start by comparing the tenths place of each decimal number. The largest tenths digit is 9, which is in the decimal 0.926. Therefore, we can put this decimal first: 0.926, ____, ____, ____, ____ Next, we compare the tenths place of the remaining decimals. The next largest tenths digit is 8, which is in the decimal 0.895. Therefore, we can put this decimal next: 0.926, 0.895, ____, ____, ____ Next, we compare the tenths place of the remaining decimals. The next largest tenths digit is 7, which is in the decimal 0.721. Therefore, we can put this decimal next: 0.926, 0.895, 0.721, ____, ____ Now we have three decimals left to compare. To compare the hundredths place, we look at the digits to the right of the tenths place. For the three remaining decimals, the hundredths digits are: • 0.153 has a hundredths digit of 5 • 0.468 has a hundredths digit of 6 We can see that the decimal with the largest hundredths digit is 0.468. Therefore, we can put this decimal next: 0.926, 0.895, 0.721, 0.468, ____ Finally, we have one decimal left to place: 0.926, 0.895, 0.721, 0.468, 0.153 Therefore, the decimals arranged in descending order are: 0.926 > 0.895 > 0.721 > 0.468 > 0.153. ## Share Your Feedback CREST Olympiads has launched this initiative to provide free reading and practice material. In order to make this content more useful, we solicit your feedback. Do share improvements at info@crestolympiads.com. Please mention the URL of the page and topic name with improvements needed. You may include screenshots, URLs of other sites, etc. which can help our Subject Experts to understand your suggestions easily. 70%
# Number Lines and the Empty Product The “Empty Product” is the product of all numbers in a group of no numbers at all. To the uninitiated, this sounds like word salad. Mostly because it is: How can you multiply no values together? How does that even make any sense? And on one level, it really doesn’t: We need the Empty Product for certain cases, such as 0!, but it’s usually defined in terms of “this is the value because that’s what it has to be.” Largely because of the development of the calculator, the logarithmic scale has lost a lot of the love it used to have. The longer I teach math, the more I wonder how this loss has gotten in the way of understanding. Today, I’m going to show why the Empty Product has the value it does by using the logarithmic scale. (If you hate logarithms, don’t worry. There are no actual logarithms being used in this article, just the scale.) First, though, I’ll provide the more formal explanation. If you want that for background, read on. If you want to ignore that, jump to “Okay, on with the show!” Here’s the somewhat formal explanation: Given two sets, A and B, we can calculate the product of all the elements of A and all the elements of B. If we have {3, 4} and {2, 5}, for instance, then 3 * 4 * 2 * 5 = 120. Likewise, if we have {3} and {4, 2, 5}, then the product is still 120, because we’re dealing with the same elements. Since we’re dealing with the same elements, the product will be the same. That’s because multiplication is associative and commutative: We can rearrange the order of the elements all we please, the product will always be the same. Hence if we have the null set and {3, 4, 2, 5}, it must be the case that the product is still 120. Since the product of the elements of {3, 4, 2, 5} is 120, the product of the elements of the null set, i.e., the Empty Product, must be a value that, when multiplied by 120, yields 120. This is a formal way of saying that the Empty Product is the number which, when multiplied by another number, yields that number. In algebraic notation, it’s the value of b which satisfies a = a * b for all a. Which is to say, 1. Another common explanation involves factorials. n! is the product of all integers from 1 to n. So, for instance, 4! = 4 * 3 * 2 * 1 = 24. Factorials are common in probability, and it is often the case that a formula will generate the need for 0!. Consider: You have five students, and want to choose three for a task. How many ways can you do so? The modern factorial was originally designed to calculate the number of ways to sort objects; we can sort five objects 5! = 120 ways. If we choose three for a task, then we fail to choose the other two, and so our formula for how many different groups of three from a larger group of five is 5!/(3!2!) = 10. For the sake of completeness, we want the formula to support any number in our subgroup from 1 to 5. If you have five students and want to choose one? That’s 5!/(1!4!) = 5. Want to choose two? 5!/(2!3!) = 10. Want to choose four? 5!/(4!1!) = 5. What happens when we want to choose all five? Logically, that’s one. There’s only one way we can choose five students from five students. Our formula gives 5!/(5!0!), which we said has to be 1. Since 5!/5! = 1, it must be the case that 1/0! = 1, so 0! = 1. But 0! is the product of all integers from 1 to 0… and there aren’t any. A challenge with stating simply that 0! is the product of all integers from 1 to 0 but not providing this “but the formula says it has to be!” explanation is that we need to prevent (-1)! = 1. A related explanation is that n!/n = (n-1)!. That is, 5!/5 = 4!, 4!/4 = 3!, and so on. So 1!/1 = 0!, which is 1, but 0!/0 is undefined, so (-1)! is also undefined. ## Okay, on with the show! So let’s consider this question from the point of view of the logarithmic scale. We should all be familiar with the number line, which is formally called the linear scale. This is what we learned in elementary school: All the numbers are spaced out evenly, and once negatives are introduced, 0 is placed in the middle. I say “in the middle” while acknowledging that, because this scale is infinitely long, it doesn’t technically have a “middle”. But, conceptually if not technically, the middle of the number line is 0, and it goes from negative infinity to positive infinity. The logarithmic scale, meanwhile, goes from 0 to infinity and has a conceptual center at 1: The linear and logarithmic scales are related in several ways: To add on the linear scale, just move that number of steps. To multiply on the logarithmic scale, just… most that number of steps. This was the basis of the slide rule, an engineering tool that was used in various forms for centuries before the development of the modern calculator. The slide rule had other functions as well, but the common scales were two logarithmic scales that could be moved relative to each other in order to multiply two values. Meanwhile, to multiply m and n on the linear scale, start at the middle (0) and take m steps of size n (or n steps of size m). To find the nth power of m on the logarithmic scale, start at the middle (1) and take n steps of size m (because the numbers aren’t evenly spread, though, taking m steps of size n gets you a different place, so m^n is not usually equal to n^m). So where are you at before you start adding anything at all? You start at 0. This is where students enter math class in kindergarten. Indeed, “adding on” is considered an indicator that math knowledge is maturing, as opposed to “adding up”. Ask a child who is first starting to add what three plus five is, and they’ll start with no fingers, then count out three, then count out another five to eight (“adding up”). Ask a slightly older child, and they will typically start at three and count up five times to eight (“adding on”). In other words, we seem to be fairly comfortable with the notion that, before we add anything, we have nothing at all. This is a large part of why, I believe, we want to say the same thing about multiplication: Before we’ve multiplied anything, we have nothing, and nothing is zero. But if we look at in terms of the scales, we reach a different conclusion. Adding means we start in the middle of the linear scale… which is 0. Multiplying means we start in the middle of the logarithmic scale… which is 1. One is where we start before we’ve multiplied anything at all. The Empty Product is the result of not multiplying anything at all. So the Empty Product is 1. The scales said so. Our method of multiplying by counting out steps on the logarithmic scale said so. Clio Corvid This site uses Akismet to reduce spam. Learn how your comment data is processed.
# Implicit Differentiation If a function is described by the equation y = f (x) where the variable y is on the left side, and the right side depends only on the independent variable x, then the function is said to be given explicitly. For example, the following functions are defined explicitly: $y = \sin x,\;\;\;y = {x^2} + 2x + 5,\;\;\;y = \ln \cos x.$ In many problems, however, the function can be defined in implicit form, that is by the equation $F\left( {x,y} \right) = 0.$ Of course, any explicit function can be written in an implicit form. So the above functions can be represented as $y - \sin x = 0,\;\;\;y - {x^2} - 2x - 5 = 0,\;\;\;y - \ln \cos x = 0.$ The inverse transformation cannot be always performed. There are often functions defined by an implicit equation that cannot be resolved with respect to the variable $$y.$$ Examples of such implicit functions are ${x^3} + {y^3} - 3{x^2}{y^5} = 0,\;\;\;\frac{{x - y}}{{\sqrt {{x^2} + {y^2}} }} - 4x{y^2} = 0,\;\;\;xy - \sin \left( {x + y} \right) = 0.$ The good news is that we do not need to convert an implicitly defined function into an explicit form to find the derivative $$y'\left( x \right).$$ If $$y$$ is defined implicitly as a function of $$x$$ by an equation $$F\left( {x,y} \right) = 0,$$ we proceed as follows: 1. Differentiate both sides of the equation with respect to $$x$$, assuming that $$y$$ is a differentiable function of $$x$$ and using the chain rule. The derivative of zero (in the right side) will also be equal to zero. Note: If the right side is different from zero, that is the implicit equation has the form $f\left( {x,y} \right) = g\left( {x,y} \right),$ then we differentiate the left and right side of the equation. 2. Solve the resulting equation for the derivative $$y'\left( x \right)$$. In the examples below find the derivative of the implicit function. ## Solved Problems ### Example 1. Find the derivative of the function given by the equation $${y^2} = 2px,$$ where $$p$$ is a parameter. Solution. This equation is the canonical equation of a parabola. Differentiating the left and right sides with respect to $$x$$, we have: ${\left( {{y^2}} \right)^\prime } = {\left( {2px} \right)^\prime },\;\; \Rightarrow 2yy' = 2p,\;\; \Rightarrow y' = \frac{p}{y},\;\;\text{where}\;\;y \ne 0.$ ### Example 2. Differentiate implicitly the function $$y\left( x \right)$$ given by the equation $$y = \cos \left( {x + y} \right).$$ Solution. Differentiate both sides with respect to $$x:$$ $\frac{{dy}}{{dx}} = \frac{d}{{dx}}\cos \left( {x + y} \right),\;\; \Rightarrow y' = - \sin \left( {x + y} \right) \cdot \left( {1 + y'} \right), \Rightarrow y' = - \sin \left( {x + y} \right) - y'\sin \left( {x + y} \right), \Rightarrow y'\left( {1 + \sin \left( {x + y} \right)} \right) = - \sin \left( {x + y} \right),$ which results in $y' = - \frac{{\sin \left( {x + y} \right)}}{{1 + \sin \left( {x + y} \right)}}.$ ### Example 3. Calculate the derivative at the point $$\left( {0,0} \right)$$ of the function given by the equation $$x = y - 2\sin y.$$ Solution. We differentiate both sides of the equation with respect to $$x$$ and solve for $$y^\prime:$$ $x^\prime = y^\prime - \left( {2\sin y} \right)^\prime,\;\; \Rightarrow 1 = y^\prime - 2\cos y \cdot y^\prime,\;\; \Rightarrow y^\prime = \frac{1}{{1 - 2\cos y}}.$ Substitute the coordinates $$\left( {0,0} \right):$$ $y^\prime\left( {0,0} \right) = \frac{1}{{1 - 2\cos 0}} = \frac{1}{{1 - 2 \cdot 1}} = - 1.$ ### Example 4. Find the equation of the tangent line to the curve $${x^4} + {y^4} = 2$$ at the point $$\left( {1,1} \right).$$ Solution. Differentiate both sides of the equation with respect to $$x:$$ $\frac{d}{{dx}}\left( {{x^4} + {y^4}} \right) = \frac{d}{{dx}}\left( 2 \right),\;\;\Rightarrow 4{x^3} + 4{y^3}y' = 0,\;\;\Rightarrow {x^3} + {y^3}y' = 0.$ Then $$y' = - {\frac{{{x^3}}}{{{y^3}}}}$$. At the point $$\left( {1,1} \right)$$ we have $$y'\left( 1 \right) = - 1.$$ Hence, the equation of the tangent line is given by $\frac{{x - 1}}{{y - 1}} = - 1\;\;\text{or}\;\;x + y = 2.$ ### Example 5. Calculate the derivative of the function $$y\left( x \right)$$ given by the equation $${x^2} + 2xy + 2{y^2} = 1$$ under condition $$y = 1.$$ Solution. We differentiate both sides of the equation implicitly with respect to $$x$$ (we consider the left side as a composite function and use the chain rule): $\frac{d}{{dx}}\left( {{x^2} + 2xy + 2{y^2}} \right) = \frac{d}{{dx}}\left( 1 \right), \Rightarrow 2x + 2\left( {y + xy'} \right) + 4yy' = 0, \Rightarrow x + y + xy' + 2yy' = 0.$ When $$y = 1,$$ the original equation becomes ${x^2} + 2x + 2 = 1,\;\; \Rightarrow {x^2} + 2x + 1 = 0,\;\; \Rightarrow {\left( {x + 1} \right)^2} = 0,\;\; \Rightarrow x = - 1.$ Substituting the values $$x = -1$$ and $$y = 1$$, we obtain: $- 1 + 1 - y' + 2y' = 0.$ It follows from here that $$y' = 0$$ at $$y = 1.$$ ### Example 6. Given the equation of a circle $${x^2} + {y^2} = {r^2}$$ of radius $$r$$ centered at the origin. Find the derivative $$y'\left( x \right).$$ Solution. Differentiate both sides of the equation with respect to $$x:$$ $\frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \frac{d}{{dx}}\left( {{r^2}} \right),\;\; \Rightarrow 2x + 2yy' = 0,\;\; \Rightarrow x + yy' = 0,\;\; \Rightarrow yy' = - x,\;\; \Rightarrow y' = - \frac{x}{y}.$ In this case, we can solve for $$y$$ directly from the equation for the upper half-circle: $$y = + \sqrt {{r^2} - {x^2}} .$$ So we get $y' = - \frac{x}{{\sqrt {{r^2} - {x^2}} }}.$ ### Example 7. ${x^2} + {y^2} - 2x - 4y = 4$ Solution. We take the derivative of each term treating $$y$$ as a function of $$x.$$ $\left( {{x^2}} \right)^\prime + \left( {{y^2}} \right)^\prime - \left( {2x} \right)^\prime - \left( {4y} \right)^\prime = 4^\prime,\;\; \Rightarrow 2x + 2yy^\prime - 2 - 4y^\prime = 0.$ Solve this equation for $$y^\prime:$$ $2yy^\prime - 4y^\prime = 2 - 2x,\;\; \Rightarrow yy^\prime - 2y^\prime = 1 - x,\;\; \Rightarrow y^\prime\left( {y - 2} \right) = 1 - x,\;\; \Rightarrow y^\prime = \frac{{1 - x}}{{y - 2}}.$ ### Example 8. ${x^3} + {y^3} = 3xy$ Solution. We differentiate the left and right sides of the equation with respect to $$x$$ considering $$y$$ as a composite function of $$x:$$ $\left( {{x^3} + {y^3}} \right)^\prime = \left( {3xy} \right)^\prime,\;\; \Rightarrow 3{x^2} + 3{y^2}y' = {\left( {3x} \right)^\prime }y + 3xy', \;\Rightarrow 3{x^2} + 3{y^2}y' = 3y + 3xy'.$ From this relation we find $$y':$$ ${y^2}y' - xy' = y - {x^2},\;\; \Rightarrow y'\left( {{y^2} - x} \right) = y - {x^2},\;\; \Rightarrow y' = \frac{{y - {x^2}}}{{{y^2} - x}}.$ This derivative exists provided ${y^2} - x \ne 0\;\;\text{or}\;\;y \ne \pm \sqrt x .$ ### Example 9. ${x^3} + 2{y^3} + y{x^2} = 3$ Solution. Differentiate both sides term-by-term with respect to $$x:$$ $\left( {{x^3}} \right)^\prime + \left( {2{y^3}} \right)^\prime + \left( {y{x^2}} \right)^\prime = 3^\prime,\;\; \Rightarrow 3{x^2} + 6{y^2}y^\prime + y^\prime{x^2} + 2yx = 0.$ Solve this equation for $$y^\prime:$$ $6{y^2}y^\prime + y^\prime{x^2} = - \left( {3{x^2} + 2yx} \right),\;\; \Rightarrow y^\prime\left( {{x^2} + 6{y^2}} \right) = - \left( {3{x^2} + 2yx} \right),\; \Rightarrow y^\prime = - \frac{{3{x^2} + 2yx}}{{{x^2} + 6{y^2}}}.$ ### Example 10. Calculate the derivative at the point $$\left( {0,0} \right)$$ of the function given by the equation ${x^5} + {y^5} - 2x + 2y = 0.$ Solution. We differentiate this equation with respect to $$x$$ and solve for $$y^\prime:$$ $\left( {{x^5}} \right)^\prime + \left( {{y^5}} \right)^\prime - \left( {2x} \right)^\prime + \left( {2y} \right)^\prime = 0^\prime,\;\; \Rightarrow 5{x^4} + 5{y^4}y^\prime - 2 + 2y^\prime = 0,\;\; \Rightarrow \left( {5{y^4} + 2} \right)y^\prime = 2 - 5{x^4},\;\; \Rightarrow y^\prime = \frac{{2 - 5{x^4}}}{{2 + 5{y^4}}}.$ Substitute the coordinates $$x = 0,$$ $$y = 0:$$ $y^\prime\left( {0,0} \right) = \frac{{2 - 5 \cdot {0^4}}}{{2 + 5 \cdot {0^4}}} = \frac{2}{2} = 1.$ See more problems on Page 2.
# How to Find LCM You will learn how to calculate the least common multiple (LCM) of two or more numbers by reading this article. But before we get into the formula or method for calculating the least common multiple of a set of numbers, let's first get a grasp on what LCM is. ## What is LCM? LCM stands for "Least Common Multiple." It is used to find the smallest positive integer that is divisible by all the numbers (from which the LCM is computed). For example, the LCM of 2 and 3 is 6, as indicated by: `LCM(2, 3) = 6` As you can see, the number 6 is divisible by both the numbers 2 and 3. LCM (Least Common Multiple) can also be called: • Lowest Common Multiple • Least Common Divisor Both expressions mean the same thing, which is to locate the smallest number such that it can be divided by the numbers that are provided (to which the LCM is to be calculated). However, there is another method we can approach to find the LCM of a given set of numbers. Let's discuss. ## Methods to find LCM To find LCM, there are multiple methods available. But here, we will only discuss two popular methods: ### Listing Multiples Method To find LCM using the listing multiples method, we have to follow the algorithm given below: • Write down the multiple of all the given numbers. • Find the smallest number that is available in every number's multiples. • That smallest number will be the LCM. Multiples of a number are numbers that are the products of a given number. For example, multiples of 3 are 3, 6, 9, 12, 15, 18, and so on. That is, when we multiply 3 by 5, we will get 15, which is the multiple of 3. Similarly, if we multiply 3 with any other number, the result we get will be the multiple of 3. For example, let's find the LCM of 10, 12, and 15. Since I already stated that, we need to write down the multiples of these numbers and the least common multiple from all the lists of common multiples. Here is the list of some multiples of all three numbers: • Multiples of 10: 10, 20, 30, 40, 50, 60, and so on. • Multiples of 12: 12, 24, 36, 48, 60, 72, and so on. • Multiples of 15: 15, 30, 45, 60, 75, 90, and so on. Since 60 is the least common multiple of all the multiples of three numbers, Therefore, the LCM of 10, 12, and 15 is 60, which can be written as follows: `LCM(10, 12, 15) = 60` This method of determining LCM is undeniably popular. It is also the easiest way to find out the LCM of two or more than two numbers. So, here is a step-by-step solution for finding the LCM of 10, 12, and 15 using the ladder method: I hope you are familiar with the above method. If you're not, then here are some main steps: • List out all the numbers. • Divide the number by least to greatest (which equals the number) until the quotient is all one. • Check if 2 divides any of the given numbers. • If it divides, then proceed and write down the quotient of the number (divided by 2) in the second line. • Otherwise, check if 3 divides any of the given numbers. • If it divides, then proceed. Otherwise, move on to the next one. • Here, the quotient of the number (if it gets divided) will be written in the next line, and the rest of the number will be written as it is. • Continue in this manner until every quotient equals 1. #### Programs on LCM of numbers in different languages « CodesCracker.com Next Topic » Liked this post? Share it!
Courses # Syllogism Notes \| Study Business Mathematics and Logical Reasoning & Statistics - CA CPT ## CA CPT: Syllogism Notes \| Study Business Mathematics and Logical Reasoning & Statistics - CA CPT The document Syllogism Notes \| Study Business Mathematics and Logical Reasoning & Statistics - CA CPT is a part of the CA CPT Course Business Mathematics and Logical Reasoning & Statistics. All you need of CA CPT at this link: CA CPT INTRODUCTION Syllogism is a ‘Greek’ word that means inference or deduction. As such inferences are based on logic, then these inferences are called logical deduction. These deductions are based on propositions (premise). Different types of questions covered in this chapter are as follows: • Two Statements and Two Conclusions ‘Syllogism’ checks basic aptitude and ability of a candidate to derive inferences from given statements using step by step methods of solving problems. Proposition Let us consider the following sentences: In all the sentences mentioned above, a relation is established between subject and predicate with the help of quantifier and copula. Now, we can define proposition as under: A proposition or premise is grammatical sentence comprising of four components. • Quantifier • Subject • Copula • Predicate Quantifier– The words ‘All’ ‘No’ and ‘Some’ are called quantifiers as they specify a quantity. Keep in mind that ‘All’ and ‘No’ are universal quantifiers because they refer to each and every object of a certain set. ‘Some’ is a particular quantifier as it refers to at least one existing object in a certain set. Subject– Subject is the part of the sentence something is said about. It is denoted by S. Copula– It is that part of a proposition that denotes the relation between subject and predicate. Predicate– It is that part of a proposition which is affirmed detail about that subject. Classification of Proposition A proposition can mainly be divided into three categories. (i) Categorical Proposition: In categorical proposition, there exists a relationship between the subject and the predicate without any condition. It means predicate is either affi rmation or denial of the subject unconditionally. Example: I. All cups are pens. II. No boy is girl. (ii) Hypothetical Proposition: In a hypothetical proposition, relationship between subject and predicate is asserted conditionally. Example: I. If it rains, he will not come. II. If he comes, I will accompany him. (iii) Disjunctive Proposition: In a disjunctive proposition, the assertion is of alteration. Example: I. Either he is sincere or he is loyal. II. Either he is educated or he is scholar. Keeping in view with the existing pattern of Syllogism in competitive examinations, we are concerned only with the categorical type of proposition. Venn Diagram Representation of Two Propositions Types of Venn diagram can be understood by the following diagram: From the above diagram, following things are very much clear: (i) Universal propositions, Either (a) completely include the subject (A-type) or (b) completely exclude the subject (E-type) (ii) Particular propositions, Either (a) partly include the subject (I-type) or (b) partly exclude the subject (O-type) Now, we can summarize the four standard types of propositions (premises) as below: Format Type All S are P (Universal Affirmative) A No S is P (Universal Negative) E Some S are P (Particular Afirmative) I Some S are not P (Particular Negative) O Venn Diagram Representation (i) A-Type (All S are P) (ii) E-Type (No S is P) (iii) E Type (No S is P) (iv) O Type (Some S are not P) Hidden Propositions The type of propositions we have discussed earlier are of standard nature but there are propositions which do not appear in standard format and yet can be classified under any of the four types. Let us now discuss the type of such propositions. I. A-Type Propositions (i) All positive propositions beginning with ‘every’ and ‘any’ are A type propositions. Example: (a) Every cat is dog ⇒ All dogs are cats (b) Each of students of class has passed ⇒ All students of class X have passed (c) Anyone can do this job ⇒ All (Women) can do this job (ii) A positive sentence with a particular person as its subject is always an A-type proposition. Example: (a) (b) (iii) A sentence with a definite exception is A type. Example: II. E-Type Propositions (i) All negative sentences beginning with ‘no one’, ‘not a single’ etc., are E-type propositions. Example: (a) Not a single student could answer the question. (b) None can cross the English channel. (ii) A negative sentence with a very definite exception is also of E-type proposition. Example: (iii) When an Introgattive sentence is used to make an assertion, this could be reduced to an E-type proposition. example: Is there any person who can scale Mount Everest? ⇒ Non can climb Mount Everest. (iv) A negative sentence with a particular person as its subject is E-type proposition. III. I-Type Propositions (i) Positive propositions beginning with words such as ‘most’, ‘a few’ ‘mostly’, ‘generally’, ‘almost’, ‘frequently’, and ‘often’ are to be reduced to the I-type propositions. Example: (a) Almost all the Vegetables have been sold. ⇒ Some vegetables have been sold. (b) Most of the students will qualify in the test. ⇒ Some of the students will qualify in the test. (c) Boys are frequently physically weak ⇒ Some boys are physically weak. (ii) Negative propositions beginning with words such as ‘few’ ‘seldom’, ‘hardly’, ‘rarely’, ‘little’ etc. are to be reduced to the I-type propositions. Example: (a) Seldom writers do not take rest. ⇒ Some writers take rest. (b) Few Teachers do not tell a lie. ⇒ Some teachers tell a lie. (c) Rarely Scientists do not get a good job ⇒ Some Scientists get a good job. (iii) A positive sentence with an exception which is not de⇒ nite, is reduced to an I-type proposition. Example: IV. O-Type Propositions (i) All negative propositions beginning with words such as ‘all’, ‘every’, ‘any’, ‘each’ etc. are to be reduced to O-type propositions. Example: (a) All Psychos are not guilty. ⇒ Some Physchos are not guilty. (b) All that glitters is not gold. ⇒ Some glittering objects are not gold. (c) Everyone is not Scientist ⇒ Some are not Scientist. (ii) Negative propositions with words as ‘most’, ‘a few’, ‘mostly, ‘generally’, ‘almost’, ‘frequently’ are to be reduced to the O-type propositions. Example: (a) Boys are usually not physically weak. ⇒ Some boys are not physically weak. (b) Priests are not frequently thiefs. ⇒ Some priests are not thiefs. (c) Almost all the questions cannot be solved. ⇒ Some questions cannot be solved. (iii) Positive propositions with starting words such as ‘few’, ‘seldom’, ‘hardly’, ‘scarcely’, ‘rarely’, ‘little’, etc., are to be reduced to the O-type propositions. Example: (a) Few boys are intelligent. ⇒ Some boys are not intelligent (b) Seldom are innocents guilty. ⇒ Some innocent are not guilty. (iv) A negative sentence with an exception, which is not definite is to be reduced to the O-type propositions. e.g. Exclusive Propositions Such propositions start with ‘only’, ‘alone’, ‘none but’, ‘none else but’ etc., and they can be reduced to either A or E or l-type. Example: Only Post-graduates are officers. (E-type) None Post-graduate is officer. (A-type) All officers are Post-graduates. (I-type) Some Post-graduates are officers Types of Inferences Inferences drawn from statements can be of two types: 1. Immediate Inference: When an inference is drawn from a single statement, then that inference is known as an immediate inference. Example: Statement: All books are pens. Conclusion: Some pens are books. In the above example, a conclusion is drawn from a single statement and does not require the second statement to be referred, hence the inference is called an immediate inference. 2. Mediate Inference: In mediate inference, conclusion is drawn from two given statements. Example: Statements: All cats are dogs. All dogs are black. Conclusion: All cats are black. In the above example, conclusion is drawn from the two statements or in other words, both the statements are required to draw the conclusion. Hence, the above conclusion is known as mediate inference. Method to Draw Immediate Inferences There are various methods to draw immediate inferences like conversion, obversion, contraposition; etc. Keeping in view the nature of questions asked in various competitive examinations, we are required to study only two methods, implications and conversion. (i) Implications (of a given proposition): Below we shall discuss the implications of all the four types of propositions. While drawing a conclusion through implication, subject remains the subject and predicate remains the predicate. A-Type: All boys are blue. From the above A-type proposition, it is very ‘clear that if all boys are blue, then some boys will definitely be blue because some is a part of all. Hence, from A-type proposition, we can draw l-type conclusion (through implication). E-Type: No cars are buses. If no’ cars are buses, it clearly means that some cars are not buses. Hence, from E-type proposition, O-type conclusion (through implication) can be drawn. I-Type: Some chairs are tables. ‘From the above l-type proposition, we cannot draw any valid conclusion (through implication). O-Type: Some A are not B. From the above O-type proposition, we can not draw any valid inference (through implication). On first look, it appears that if some A are not B, then conclusion that some A are B must be true but the possibility of this conclusion being true can be over ruled with the help of following example: Case I A = {a, b, c} and B = {d, e, f} Case II A = {a, b, c} and B ={b, c, d} The above two cases show the relationship between A and B given by O-type proposition. “Some A are not B”. Now, in case I, none of the element of set A is the element of set B. Hence, conclusion “Some A are B” cannot be valid. However, in case II, elements b and c are common to both sets A and B. Hence, here conclusion “Some A are B” is valid. But for any conclusion to be true, it should be true for all the cases. Hence, conclusion “Some A are B” is not a valid conclusion drawn from an O-type proposition. All the results derived for immediate inference through implication can be presented in the table as below: Conversion: Conversion is other way of getting immediate inferences. Unlike implication, in case of conversion, subject becomes predicate and predicate becomes subject. Let us see Conversion of A-type After conversion it becomes Clearly, A gets converted into I-type. Conversion of E-type After conversion it becomes Clearly, E gets converted into E-type. Conversion of I-type Clearly I gets converted into I- type. Conversion of O – type. O type propositions cannot be converted. Now we can make a conversion as follows: Immediate Inference Table Venn Diagram Representation of Immediate Inferences: Immediate inferences are drawn from each type of Propositions (A, E, I , O) 1) A type All S are P (i) S = { a, b, c } , P = { a, b, c, d, e} (i) S = { a, b, c} , P = { a, b, c, d, e, f} (ii) S = { a, b, c} , P = { a, b, c} The above cases show all the possibilities of two sets S and P showing the relationship by the proposition. All S are P in both cases. Some S are P.(Is true from relationship) Some P are S (Its true) Some P are not S is not valid because from it is case (i) but false from case (ii) Inference All (P are S ) is not valid because its true from case (ii) and False from case (i) 2) E type – No S Is P We can draw the inferences as (i) No P is S (ii) Some S are not P (iii) Some P are not S Any other inference drawn from E- type proposition is not valid. 3) I-type: Some S are P (i) S = { a, b, c, d} , P = {c, d, e, f} Some S are P (ii) S = { a, b, c, d} and P = { a, b} Set {a, b} is the part of S as well as Set P, hence some set S are P. (iii) S = {a, b} , P = {a, b, c, d} Set {b} is the part of the set S as well as Set P, hence some S are P. (iv) S = {a, b, c} , P = {a, b, c} The above diagrams show the relationship between S and P from I-type relationship. From the possible combinations, it’s clear that inference (Some P are S) is true. Inference (S are not P) is true from combinations (i) and (ii) but is not true from combinations (iii) and (iv). Therefore inference (Some S are not P) is not a valid inference drawn from the above proposition. Set {a, b} is part of set S and P, hence some s are P. 4) O-Type: Some S are not P (i) S = {a, b, c, d} , P = {c, d, e, f} Set {a, b} is part of the set S but not Set P Hence Some S are not P (ii) S= {a, b, c} and P = {d, e, f} Set {a, b} is the part of S but not set P Hence the above relation represented by Some S are not P . (iii) S = {a, b, c, d, e, f} , P = {e,f} Set {a, b, c} is the part of set S abut not P. Hence Proposition Some S are not P. On the basis of all possible combinations showing relationship between S and P, no valid inference can be drawn. Inference from Some P are not S) is true combination (i) and (iii) but not true for combination (iii). Hence it is invalid inference. Inference (Some P are not S) is true from combination (i) and (ii) but not true for combination (iii), hence it is also invalid. Following are the main rules for solving syllogism problems. 1) All + All = All 2) All + No = No 3) All + Some = No conclusion 4) Some + No = Some Not 5) Some + Some = No conclusion 6) No + All = Some not (Reversed) 7) No + All = Some Not (Reversed) 8) No + Some = Some Not (Reserved) 9) No + No = No conclusion 10) Some Not/ Some not reserved + Anything = No conclusion 11) If all A are B then we can say – Some B are Not A is a possibility 12) If Some B are not A then we can say – All A are B is a possibility 13) If some A are B then we can say All A are B is a possibility. All B are A is a possibility. 14) All ⇔ Some not reserved 15) Some ⇔ All 16) No conclusion = Any possibility is true Implications (In case of Conclusions from Single Statement) All ⇔ Some That means If A are B then Some B is true. Some ⇔ Some that means if Some A are B then Some B are A is true. No ⇔ No that means if No A is B then No B is true. Examples: In this type of questions two statements and two conclusions are given. Its required to check. Example 1: Statement: I. Some boys are student. II. All students are Engineers. Conclusions: I. All Engineers are students. II. Some boys are Engineers. (a) Only I follows (b) Only II follows (c) Both I and II follow (d) Neither I nor II follows. Solution (b) Statement I is an I-type proposition which distributes neither the subject nor the predicate. Statement II is an A-type propositions which distributes the subject ‘Engineers’ only. Since, the Engineers is distributed in Conclusion I without being distributed in the premises. So, Conclusion I cannot follow. In second conclusion, where it is asked that some boys are Engineers but from Statement I nit is clear that some boys are not students. These boys may not be Engineers. Example 2: Statements: I. All Lotus are flowers. II. No Lily is a Lotus. Conclusions: I. No Lily is a flower II. Some Lilies are flowers. (a) Only I follows (b) Only II follows (c) Either I or II follows (d) Neither I nor II follows Solution: (c) Here, the first premise is an A-type proposition and so the middle term ‘Lotus’ forming the subject is distributed. The second premise is an E proposition and so the middle term ‘Lotus’ forming the predicate is distributed. Since, the middle term is distributed twice, so the conclusion cannot be universal. Example 3: Statements I All A’s are C’s II All D’s are C’s Conclusion I All D’s are C’s II. Some D’s are not A’s (a) Only I follows (b) Only II follows (c) Both I and II follows (d) None follows Solution: (a) Now, taking conclusion I, it is clear that all D’s are also C’s but taking conclusion II, we cannot say that some D’s are not A’s because from Statement I it is clear that all D’s are A’s. Hence, only Conclusion I follows. Example 4: Statements: All balls are bats. All bats are stumps. The sentences are already aligned. From the above given Table, A + A = A. Hence the conclusion is of type-A whose subject is the subject of the first proposition and the predicate is the predicate of the second proposition? So the conclusion is All balls are Stumps. Example 5: Statements: All Professors are readers. All Professors are writers. This pair is not properly aligned because the subject of both the sentences is ‘Professors’. Since both the sentences are of type-A, we may convert any of them. So the aligned pair is Some readers are Professors. All Professors are writers. Here the conclusion will be of type - I because I + A = I. The conclusion is Some readers are writers. Example 6: Statements: Some Mangos are sweets. All Mangos are Fruits. The subject of both the sentences is the same. By the rule of IEA, we convert the I - type statement. So the aligned pair is, Some Sweets are Mangos. All Mangos are Fruits I+A=I. So the conclusion is Some Sweets are Fruits. Example 7: Statements: All lights are bats. No balls are lights. By changing the order of the statements itself. We can align the sentences. The aligned pair is No balls are lights. All lights are bats. E + A = O. So the conclusion is, Some bats are not balls. Example 8: Statements: Some caps are blue. No clip is blue. Here the common term is ‘blue’ which is the predicate of both the sentences. By the rule of IEA, we convert the I-type statement. After conversion, the given pair becomes, Some blue are caps. No clip is blue. Now by changing the order of the statements, we can align the sentences. So the aligned pair is, No clip is blue. Some blue are caps. The conclusion is of type O since, E + I = O*. Hence the conclusion is Some caps are not clips. Example 9: Statements: Some powders are not soaps. All soaps are detergents. The given pair is properly aligned. But no definite conclusion can be drawn from this type because it is a O+A - type combination. Complementary Pair Consider the following. Conclusions: i) Some vans are trucks. ii) Some vans are not trucks. We know that either some vans will be trucks or some vans will not be trucks. Hence either (i) or (ii) is true. Such pair of statements are called complementary pairs. So in a complementary pair, at least one of the two statements is always true. We can call a pairs a complementary pair if i) The subject and predicate of both the sentences are the same. ii) They are an I + O - type pair or an A + O type pair or an I + E - type pair. Some complementary pairs are given below. i) All birds are Pigeons. Some birds are not Pigeons. ii) Some Chairs are watches. Some Chairs are not watches. iii) Some kids are cute. No kids are cute. Note: The steps to be followed to do a syllogism problem by analytical method are mentioned below. i) Align the sentences properly. ii) Draw conclusion using the table. iii) Check for immediate inferences. iv) Check for complementary pair if steps ii and iii fail. The document Syllogism Notes \| Study Business Mathematics and Logical Reasoning & Statistics - CA CPT is a part of the CA CPT Course Business Mathematics and Logical Reasoning & Statistics. All you need of CA CPT at this link: CA CPT Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code ## Business Mathematics and Logical Reasoning & Statistics 33 videos\|101 docs\|87 tests ### Top Courses for CA CPT Track your progress, build streaks, highlight & save important lessons and more! , , , , , , , , , , , , , , , , , , , , , ;
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more. No credit card required Chapter Eighteen — Logarithmic Functions The Humongous Book of Algebra Problems 412 18.32 Solve the equation 9 x = 13 using the change of base formula and round the Express the exponential equation as a logarithmic equation and solve. 18.33 Solve the equation 7 2x + 1 = 6 using the change of base formula and round the Isolate 7 2x on the left side of the equation. Rewrite the exponential equation as a logarithmic equation. log 7 5 = 2x Calculate log 7 5 using the change of base formula. Multiply both sides of the equation by to solve for x. Logarithmic Properties Expanding, contracting, and simplifying log expressions 18.34 Expand the expression: ln (2x). Logarithmic expressions can be rewritten according to three fundamental properties. Speciﬁcally, a single logarithm can be expressed as multiple logarithms and vice versa. The ﬁrst property states that the logarithm of a product can be expressed as the sum of the logarithms of its factors: log a (xy) = log a x + log a y. Or divide both sides by 2 Which are presented one at a time in Problems 18.34, 18.36, and 18.38 Chapter Eighteen — Logarithmic Functions The Humongous Book of Algebra Problems 413 In this example, the logarithm of the product 2x may be expressed as the sum of the logarithms of its factors, 2 and x. ln (2x) = ln 2 + ln x 18.35 Demonstrate the logarithmic property presented in Problem 18.34 by verifying that ln 10 = ln 2 + ln 5. Because 2(5) = 10, the natural logarithm of the product (10) should equal the sum of the natural logarithms of the factors (2 and 5). Verify using a calculator. 18.36 Expand the expression: . A property of logarithms states that the logarithm of a quotient is equal to the difference of the logarithms of the dividend and divisor: . 18.37 Demonstrate the logarithmic property presented in Problem 18.36 by verifying that log 4 = log 12 – log 3. Because , the logarithm of the quotient (4) is equal to the difference of the logarithms of the dividend (12) and the divisor (3). Note: Problems 18.38–18.39 refer to the expression log y 3 . 18.38 Expand the logarithmic expression. A property of logarithms states that the logarithm of a quantity raised to an exponent n is equal to the product of n and the logarithm of the base: log a x n = n(log a x). log y 3 = 3(log y) There are no properties for the log of a sum. For example, ln (2 + x) ≠ (ln 2)(ln x). You also cant distribute” the letters “ln.” For example, ln (x + 2) ≠ ln x + ln 2. This is not only true for natural logs but it also works for any base, as long as all the bases in the expression match. In other words, log 10 = log 2 + log 5 and log 12 10 = log 12 2 + log 2 5. In other words, the log of a fraction equals the log of the numerator minus the log of the denominator. ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more. No credit card required
## How do you express in polar exponential form? If you have a complex number z = r(cos(θ) + i sin(θ)) written in polar form, you can use Euler’s formula to write it even more concisely in exponential form: z = re^(iθ). ### What is the exponential form of 10 12i? The expression 103 is called the exponential expression. 103 is read as “10 to the third power” or “10 cubed.” It means 10 • 10 • 10, or 1,000. 82 is read as “8 to the second power” or “8 squared.” It means 8 • 8, or 64. #### How do you write in polar form? To write complex numbers in polar form, we use the formulas x=rcosθ, y=rsinθ, and r=√x2+y2. Then, z=r(cosθ+isinθ). What is an example of exponential form? Exponential notation is an alternative method of expressing numbers. Exponential numbers take the form an, where a is multiplied by itself n times. A simple example is 8=23=2×2×2. For example, 5 ×103 is the scientific notation for the number 5000, while 3.25×102is the scientific notation for the number 325. What is an exponential signal? exponential signal is that it is an alternative representation for the real cosine signal. ## What is an example of exponential function? Exponential functions have the form f(x) = bx, where b > 0 and b ≠ 1. An example of an exponential function is the growth of bacteria. Some bacteria double every hour. If you start with 1 bacterium and it doubles every hour, you will have 2x bacteria after x hours. This can be written as f(x) = 2x. ### How do you write 1 in polar form? So the polar form of −1−i can be written as (√2,3π4) and the polar form of 1−i can be written as (√2,π4) . In the complex number a + ib, a is the real part and b is the imaginary part of the complex number. #### What is rectangular and polar form? Rectangular coordinates, or cartesian coordinates, come in the form (x,y). Polar coordinates, on the other hand, come in the form (r,θ). Instead of moving out from the origin using horizontal and vertical lines, we instead pick the angle θ, which is the direction, and then move out from the origin a certain distance r. What is the difference between standard form and exponential form? If a quantity is written as the product of a power of 10 and a number that is greater than or equal to 1 and less than 10, then the quantity is said to be expressed in standard form (or scientific notation). It is also known as exponential form. Note that we have expressed 65 as a product of 6.5 and a power of 10. How do you convert complex numbers to polar form? To write the polar form of a complex number start by finding the real (horizontal) and imaginary (vertical) components in terms of r and then find θ (the angle made with the real axis). Conversion Formula for rectangular to polar x + yi = r(cos θ + i sin θ) Example 1: convert 5 + 2i to polar form. ## How to find polar form of complex numbers? The polar form of a complex number z = a + b i is z = r ( cos θ + i sin θ) , where r = \| z \| = a 2 + b 2 , a = r cos θ and b = r sin θ , and θ = tan − 1 ( b a) for a > 0 and θ = tan − 1 ( b a) + π or θ = tan − 1 ( b a) + 180 ° for a < 0 . ### What is polar format? Polar Format. Polar format is used to view the magnitude and phase of the reflection coefficient (G) from your S11 or S22 measurement. You can use Markers to display the following: Linear magnitude (in units) or log magnitude (in dB) Phase (in degrees) The dashed circles represent reflection coefficient. #### What is the polar form of complex numbers? So we can write the polar form of a complex number as: `x + yj = r(cos θ + j\\ sin θ)`. r is the absolute value (or modulus) of the complex number. θ is the argument of the complex number. How do you express in polar exponential form? If you have a complex number z = r(cos(θ) + i sin(θ)) written in polar form, you can use Euler’s formula to write it even more concisely in exponential form: z = re^(iθ). What is the exponential form of 10 12i? The expression 103 is called the…
Вы находитесь на странице: 1из 7 # Simple Linear Regression Analysis I. Correlation Analysis Goal: measure the strength and direction of a linear association between two variables. Basic concepts Scatter diagram plot of individual pairs of observations on a two-dimensional graph; used to visualize the possible underlying linear relationship. Example: Consider the following hypothetical data: x y 1 2 3.5 4.5 2 2 4 3.5 3 3 7 6.5 3 8 3 4 6 7.9 4 5 6 7 9.4 9.3 6 6 7 7 11 10.5 12.4 11.5 7 10 8 15 8 8 11 13.7 ## The scatter plot is as follows: 16 14 12 10 8 6 4 y 2 0 0 2 4 x 6 8 10 Linear correlation coefficient () a measure of the strength of the linear relationship existing between two variables, say X and Y, that is independent of their respective scales of measurements. Some characteristics of : It can only assume values between -1 and 1. The sign describes the direction of the linear relationship between X and Y: If is positive, the line slopes upward to the right, i.e., as X increases, the value of Y also increases. If is negative, the line slopes downward to the right, and so as X increases, the value of Y decreases. If =0, then there is NO LINEAR RELATIONSHIP between X and Y. If is -1 or 1, there is perfect linear relationship between X and Y and all the points (x,y) fall on a straight line. A that is close to 1 or -1 indicates a strong linear relationship. A strong linear relationship does not necessarily imply that X causes Y or Y causes X. it is possible that a third variable may have caused the change in both X and Y, producing the observed relationship. The Pearson product moment correlation coefficient between X and Y, denoted by r, is defined as: =1 =1 =1 = 2 =1 2 =1 2 2 =1 =1 Example: Compute for r of the hypothetical data given above. Solution: x 1 2 2 2 3 3 3 3 4 4 5 6 6 6 7 7 7 8 8 8 95 y xy 3.5 3.5 4.5 9 4 8 3.5 7 7 21 6.5 19.5 8 24 6 18 7.9 31.6 7 28 9.4 47 9.3 55.8 11 66 10.5 63 12.4 86.8 11.5 80.5 10 70 15 120 11 88 13.7 109.6 171.7 956.3 x^2 y^2 1 12.25 4 20.25 4 16 4 12.25 9 49 9 42.25 9 64 9 36 16 62.41 16 49 25 88.36 36 86.49 36 121 36 110.25 49 153.76 49 132.25 49 100 64 225 64 121 64 187.69 553 1689.21 Sum ## We obtain the following values: n=20 =1 = =1 =95 =1 =171.7 956.3 2 =1 = 2 =1 = 553 1689.21 Substituting these values to the formula, we have: = 20 956.3 95(171.7) 20 553 952 (20 1689.21 (171.72 ) = 0.9511 Tests of Hypotheses for Null Hypothesis Ho =o Alternative Hypothesis Ha <o >o o = Test Statistic Critical Region (i.e., Reject Ho if) < ( = 2) > ( = 2) > /2 ( = 2) ( ) 2 1 2 Example: Consider the hypothetical data given above. Suppose that the linear correlation coefficient between X and Y in the past is 0.9. Determine if the correlation has significantly increased compared to the past. a. Ho: =0.90 b. =0.05 c. = d. = vs Ha: >0.90 ## ( ) 2 1 2 (0.9511 0.9) 18 10.95112 = 0.7019 e. Decision rule: Reject Ho if > = 2 = .05 18 = 1.734 f. Since t = 0.7019 is not greater than .05 18 = 1.734, we do not reject Ho. At 0.05 level of significance, there is a sufficient evidence to conclude that the correlation coefficient between X and Y is 0.9. NOTE: Even if two variables are highly correlated, it is not a sufficient proof of causation. One variable may cause the other or vice versa, or a third factor is involved, or a rare event may have occurred. II. ## Simple Linear Regression Analysis Goal: To evaluate the relative impact of a predictor on a particular outcome. The simple linear regression model is given by the equation = + 1 + Where - the value of the response variable for the ith element - the value of the explanatory variable for the ith element - regression coefficient that gives Y- intercept of the regression line. 1 - regression coefficient that gives the slope of the line - random error for the ith element, where are independent, normally distributed with mean 0 and variance 2 for i=1, 2, , n n number of elements Remark: The model tells us that two or more observations having the same value for X will not necessarily have the same value for Y. However, the different values of Y for a given value of X, say x i, will be generated by a normal distribution whose mean is + 1 , that is, = + 1 . This is known as the regression equation where the parameters and 1 are interpreted as follows: is the value of the mean of Y when X=0 1 is the amount of change in the mean of Y for every unit increase in the value of X. The random error It may be thought of as a representation of the effect of other factors, that is, apart from X, not explicitly stated in the model but do affect the response variable to some extent. Sources of random error: o Other response variables not explicitly stated in the model o Inherent and inevitable variation present in the response variable o Measurement errors Satisfies the following: o The error terms are independent from one another; o The error terms are normally distributed; o The error terms all have a mean of 0; and o The error terms have constant variance, 2 . Typical steps in doing a simple linear regression analysis: 1. Obtain the equation that best fits the data. 2. Evaluate the equation to determine the strength of the relationship for prediction and estimation. 3. Determine if the assumptions on the error terms are satisfied. 4. If the model fits the data adequately, use the equation for prediction and for describing the nature of the relationship between the variables. Obtaining the equation: Method of Least Squares The best-fitting line is selected as the one that minimizes the sum of squares of the deviations of the observed value of Y from its expected value. That is we want to estimate and 1 such that =1 2 is smallest, where = = + 1 Based on this criterion, the following formulas for b o , the estimate for , and b1 , the estimate for 1 , are obtained: = =1 =1 =1 2 =1 2 =1 = 1 Thus, the estimated regression equation is given by = + 1 Remarks: The estimated regression equation is appropriate only for the relevant range of X, i.e., for the values of X used in developing the regression model. If X=0 is not included in the range of the sample data, the will not have a meaningful interpretation. Example: Consider the given hypothetical example where we fit a linear model of the form = + 1 + Using the method of least squares, the following values are needed to estimate and 1 : n=20 =1 = =1 =95 =1 =171.7 956.3 2 =1 = 553 ## We get the values of bo and b1 as: 1 = 20 956.3 95 (171.7) = 1.383 20(553) 95 2 = 8.585 1.383 4.75 = 2.016 Hence, the prediction equation is given by: = 2.016 + 1.383 Interpretation: For every 1 unit increase in X, the mean of Y is estimated to increase by 1.383. Note that bo =2.016 has no meaningful interpretation since X=0 is not within the range of values used in the estimation. Mean Square Error The common variance of and Y, denoted by 2 , is given by: = 2 =1 2 2 where SSE stands for sum of squares due to error and MSE stands for mean square error. The MSE is the variance of the data, Y, about the estimated regression line, . Determining the strength of relationship between X and Y A (1-)100% Confidence Interval for 1 is (1 = 2 1 , 1 + = 2 1 ) 2 2 Where 1 = 2 =1 2 =1 ## A (1-)100% Confidence Interval for o is ( = 2 , + = 2 ) 2 2 ( 2 ) =1 2 2 =1 =1 Where = Test of Hypothesis concerning 1 Null Hypothesis Ho 1 =0 Alternative Hypothesis Ha 1 <0 1 >0 1 0 Test Statistic 1 1 Critical Region (i.e., Reject Ho if) < ( = 2) > ( = 2) > /2 ( = 2) Coefficient of Determination (R2 ) The proportion of the variability in the observed values of the response variable that can be explained by the explanatory variable through their linear relationship. The realized value of the coefficient of determination, r 2 , will be between 0 and 1. If a model has perfect predictability, then R2 =1; but if a model has no perfect predictive capability, then R2 =0. Interpretation: R2 *(100%) of the variability in the response variable, Y, can be explained by the explanatory variable, X, through the simple linear regression model. Residual (di) The difference between the observed value and predicted value of the response variable. That is, = . If indeed the variances of the error terms are constant, then the plot of the residuals versus X should tend to form a horizontal band, i.e., spread of the residuals should not increase or decrease with values of the independent variable.
# Problem Solving With Proportions We are trying to get our unknown number, x, on the left side of the equation, all by itself. When we talk about the speed of a car or an airplane we measure it in miles per hour. A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours. A ratio can be written in three different ways and all are read as "the ratio of x to y" $$x\: to\: y$$ $$x:y$$ $$\frac$$ A proportion on the other hand is an equation that says that two ratios are equivalent. will help you set up and solve proportions that represent everyday, real-life situations involving integers and fractions. Each math worksheet is accompanied by an answer key, is printable, and can be customized to fit your needs. worksheets will help students meet Common Core Standards for Expressions & Equations as well as Ratios & Proportional Relationships. The same mathematics applies when we wish to enlarge. Depicting something in the scale of 2:1 all measurements then become twice as large as in reality. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .and .are unblocked. $$\frac=\frac\: \: or\: \: \frac=\frac$$ If we write the unknown number in the nominator then we can solve this as any other equation $$\frac=\frac$$ Multiply both sides with 100 $$\, \frac=\, \frac$$ $$x=\frac$$ $$x=10$$ If the unknown number is in the denominator we can use another method that involves the cross product. The cross product is the product of the numerator of one of the ratios and the denominator of the second ratio. ## Comments Problem Solving With Proportions • ###### Solving Proportions - Texas Instruments Proportions are tools for solving a variety of problems with two variables involved in a proportional relationship. Students need to examine situations carefully to determine if they describe a proportional relationship.… • ###### Solving Proportions by Cross Multiplication. An interactive math lesson about solving proportions by cross multiplication. Cross multiplication to solve proportions. A proportion is an equation which states that.… • ###### Math Mammoth Ratios, Proportions & Problem Solving Math Mammoth Ratios, Proportions & Problem Solving is a worktext that concentrates, first of all, on two important concepts ratios and proportions, and then on problem solving. The book is best suited for grade levels 6 and 7.… • ###### Solving Proportions - Grade 7 - Practice with Math Games Earn up to 5 stars for each level The more questions you answer correctly, the more stars you'll unlock!… • ###### Solving Proportions - Nearpod Solving Proportions. Students will solve proportions using multiplication and division, including cross-multiplication, write proportions using a variable, and apply proportions as a problem solving tool. Math. Middle School.… • ###### Solving Proportions Examples - Shmoop Solve the following proportion for x using multiplication. To get rid of those denominators, we multiply both sides of the equation by 515. The 5's cancel on the left, and the 15's cancel on the right.… • ###### Solving Proportions Worksheets - Solving Proportions Worksheets Learn how to solve proportions with this set of worksheets that are specially designed for students of grades 6 and 7. A series of multi-level worksheets require students to solve proportions using the cross product method and the answers so derived will be in the form of whole numbers, fractions or decimals.… • ###### Solving Proportions Worksheets Math Worksheets These solving proportions worksheets will help students meet Common Core Standards for Expressions & Equations as well as Ratios & Proportional Relationships. I would recommend these exercise for 6th grade, 7th grade, and 8th grade math students.…
MOTION - Applications of the Derivative - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC ## Master AP Calculus AB & BC Part II. AP CALCULUS AB & BC REVIEW CHAPTER 6. Applications of the Derivative MOTION Calculus has its long, threatening talons in just about every aspect of day-to-day life; luckily, most of us are blissfully ignorant of it and unaware of it stalking us, waiting until we go to sleep, and then messing with our stuff—like putting CDs in the wrong cases and breaking all the points off your pencils on test day. But not even calculus can hide its influence in the topic of motion. Because a derivative describes a rate of change, we have already seen its influence many times and alluded to this very moment: describing how a derivative affects a position equation. NOTE. A position equation describes an object’s motion by giving its position at any time. A position equation is typically denoted as s(t) or x(t); for any time t, its output is the object’s position relative to something else. For example, output may represent how far a projectile is off the ground or how far away a particle is from the origin. The derivative of position, s'(t) or v(t), gives the velocity of the object. In other words, v(t) tells how fast the object is moving and in what direction. For example, if we are discussing a ball thrown into the air and v(3 seconds) = —4 ft/sec, when time equals 3, the ball is traveling at a rate of 4 ft/sec downward. The derivative of velocity, v\t) or a(t), gives the acceleration of the object. This ties in directly to the section you just completed. If an object has positive acceleration, then the position equation (two derivatives “above” a(t)) must be concave up, and the velocity equation (one derivative “above” a(t)) must be increasing. The most common motion questions on the AP test focus on the motion of a particle on a line, usually horizontal (although the direction of the line doesn’t matter). For example, consider a particle moving along the x-axis whose position at any time t is given by s(t) = t3 — 10t2 + 25t — 1, t > 0. The graph of the position equation looks like but the particle itself never leaves the v-axis. Let’s look at this problem in depth to better understand a typical particle motion problem. Example 6: If the position (in feet) of a particle moving horizontally along the x-axis is given by the equation s(t) = t3 — 10t2 + 25t, t > 0 seconds, answer the following: (a) Evaluate s(1), s(4), and s(5), and interpret your results. By simple substitution, s(1) = 16, s(4) = 4, and s(5) = 0. In other words, when 1 second has elapsed, the particle is 16 feet to the right of the origin, but 3 seconds later at t = 5, the particle is back to the origin. (b) At what time(s) is the particle temporarily not moving, and why? The particle will be temporarily stopped when its velocity equals 0—this makes a lot of sense, doesn’t it? Since the derivative of position is velocity, take the derivative and set it equal to 0: v(t) = s'(t) = 3t2 - 20t + 25 = 0 ALERT! Same students don’t believe me when I say that it is necessary to briefly stop when changing direction, and in demonstrating their point, many of these students end up with neck injuries. Now, factor the quadratic equation to complete the solution: Therefore, at these two moments, the particle is stopped because it is in the process of changing direction. (Remember, part (a) showed you that it changed direction between t = 1 and t = 4.) (c) On what interval of time is the particle moving backward? The particle moves backward when it has negative velocity. Therefore, we will draw a velocity (first derivative) wiggle graph. We already know the critical numbers from part (b), so all that remains is to choose some test points from among the intervals. Because v(.5) is positive, v(3) is negative, and v(6) is positive (of course you wouldn’t have to pick the same test points), you get the following wiggle graph: Therefore, the particle is moving backward on (5/3,5). This makes sense if you consider the graph of s(t). Remember, this position graph tells how far the particle is away from the origin. On the interval (5/3,5), the particle’s distance from the origin is decreasing, indicating backward movement. It should be no surprise that the velocity is negative then, since velocity is the derivative of that graph, and derivatives have a nasty habit of describing the direction of things. (d) How far does the particle travel in its first 4 seconds of motion? You may be tempted to answer 4 feet, since s(4) = 4; however, that is what’s called the displacement of the particle. The displacement is the net change in position. Because s(0) = 0 and s(4) = 4, no matter what happened in between, the particle ended up a total of 4 units from where it started. However, the problem doesn’t ask for displacement—it asks for total distance traveled. We need to measure how far it swung out to the right of the origin when it changed direction at t = 5/3 and then how far back toward the origin it came. We already know s(0) = 0, but it is essential to know that s(5/3) ≈ 18.518518518, because it tells us that the particle traveled 18.518518518 feet in the first seconds. At this point, the particle changes direction and ends up 4 feet from the origin. In the return trip, then, it traveled 18.518518518 — 4 = 14.518518518 feet. The total distance it traveled was 18.518518518 + 14.518518518 ≈ 33.037 feet. NOTE. Air resistance has been neglected torso long in theoretical mathematics that it is rumored to have joined a 12-step program. The other type of motion problem the AP test enjoys inflicting upon you is the dreaded trajectory problem. Did you know that anything thrown, kicked, fired, or otherwise similarly propelled follows a predetermined position equation on the earth? It’s true, neglecting air resistance of course. The generic projectile position equation is where g is the gravitational acceleration constant (32 ft/sec2 in the English system and 9.8 m/sec2 in the metric), v0 is the object’s initial velocity, and h0 is the object’s initial height. It is probably a good idea to memorize this equation in case you ever need it, although the questions typically asked for this sort of problem are extremely similar to those asked in Example 6. EXERCISE 5 Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book. YOU MAY USE A GRAPHING CALCULATOR FOR BOTH OF THESE PROBLEMS. 1. A very neurotic particle moves up and down the y-axis according to the position equation y = (t2 — 6t + 8) ∙ sin t, t > 0, where position is in centimeters and time is in seconds. Knowing this, answer the following questions: (a) When is the particle moving down on the interval [0,5]? (b) At what values of t is the particle moving at a rate equal to the average rate of change for the particle on the interval [0,5]? (c) At what time(s) is the particle exactly 2 cm away from the origin on the interval [0,5]? (d) What is the acceleration of the particle the first time it comes to rest? 2. The practice of shooting bullets into the air—for whatever purpose—is extremely dangerous. Assuming that a hunting rifle discharges a bullet with an initial velocity of 3,000 ft/sec from a height of 6 feet, answer the following questions (neglecting wind resistance): (a) How high will the bullet travel at its peak? (b) How long will it take the bullet to hit the ground? (c) At what speed will the bullet be traveling when it slams into the ground, assuming that it hits nothing in its path? (d) What vertical distance does the bullet travel in the first 100 seconds? 1. (a) The particle is moving down when its position equation is decreasing—when the velocity is negative. You should make a first derivative wiggle graph, so begin by finding the critical numbers of the derivative: v(t) = (t2 — 6t + 8)(cos t) + (2t — 6)(sin t) = 0 It’s best to solve this using your graphing calculator. The solutions are t = .738, 2.499, and 3.613. The wiggle graph is Therefore, the particle is moving down on (.738,2.499) (3.613,5). (b) The average rate of change of the particle will be cm/sec. To determine when the particle travels this speed, set the velocity equal to this value and solve with your calculator. This is actually the Mean Value Theorem in disguise; we know at least one t will satisfy the requirements in the question, but it turns out that the instantaneous rate of change equals the average rate of change three times, when t = .813, 2.335, and 3.770 sec. (c) The particle will be two cm away from the origin when its position is 2 (two cm above) or —2 (2 cm below). So, you need to solve both the equations (t2 — 6t + 8)sin t = 2 and (t2 — 6t + 8)sin t = —2 with your calculator. The solutions are t = .333, 1.234, and 4.732 sec. (d) The particle first comes to rest at its first critical number, t = .73821769. To find acceleration, you need to differentiate the velocity and substitute in the critical number. Rather than doing this by hand, why not use the graphing calculator and the nDeriv function? You would type the following on your TI-83: nDeriv((x2 — 6x + 8)(cos (x)) + (2x — 6)(sin (x)),x,.73821769). The resulting acceleration is —8.116 cm/sec2. 2. (a) You will need to apply the projectile position equation. The initial height and velocity are stated by the problem, and since the question uses English system units (feet), you should use g = 32 ft/sec2 as the acceleration due to gravity. You put your left foot in, you take your left foot out, you put your left foot in, shake it all about, and the position equation is s(t) = —16t2 + 3,000t + 6 The bullet will reach its peak at the maximum of the position equation (since it is an upside-down parabola, there will be only one extrema point). To find the t value at which the peak occurs, find the derivative and set it equal to 0 (since the bullet will have a velocity of 0 at its highest point before it begins to fall toward the ground): v(t) = s'(t) = -32t + 3,000 = 0 t = 93.75 seconds This, however, is not the answer. The height the bullet reaches at this point is the solution: 140,631 feet, or 26.635 miles. (b) The bullet will hit the ground when s(t) = 0—literally, when the bullet is 0 feet off of the ground. So, set the position equation equal to 0, and solve (if you use the calculator, you’ll have to ZoomOut a few times before the graph appears—these are big numbers). The bullet will remain in the air 187.502 seconds, or 3.125 minutes. (c) \|s'(187.502)\| = 3,000.064 ft/sec (since speed is the absolute value of velocity, the answer is not negative). The bullet will hit at a speed slightly greater than that at which it was fired. Therefore, being hit by a bullet that was fired into the air will have the same impact as being hit by a bullet at point-blank range. (d) You already know that the bullet travels 140,631 feet in the first 93.75 seconds. Because s(100) = 140,006, the bullet falls 140,631 — 140,006 = 625 feet between t = 93.75 and t = 100. Therefore, the bullet travels a total vertical distance of 140,631 feet up + 625 feet down = 141,256 feet.
# UNIT 9 OPERATIONS WITH DECIMALS Save this PDF as: Size: px Start display at page: Download "UNIT 9 OPERATIONS WITH DECIMALS" ## Transcription 1 UNIT 9 OPERATIONS WITH DECIMALS INTRODUCTION In this Unit, we will use our understanding of operations, decimals, and place value to perform operations with decimals. The table below shows the learning objectives that are the achievement goal for this unit. Read through them carefully now to gain initial exposure to the terms and concept names for the lesson. Refer back to the list at the end of the lesson to see if you can perform each objective. Learning Objective Media Examples You Try Add decimals in the tenths and hundreds place using decimal grids 1 2 Add decimals using a place value chart 3 5 Use an algorithm to add decimals 4 5 Subtract decimals in the tenths and hundreds place using decimal grids 6 7 Subtract decimals using a place value chart 8 10 Use an algorithm to subtract decimals 9 10 Add and subtract signed decimals Multiply a whole number times a decimal using decimal grids Multiply two decimals using a decimal grid Multiply decimal using place value Divide decimals using a decimal grid Divide decimals using place value Multiply decimals by powers of ten Divide decimals by powers of Perform decimal operations on a calculator Solve application problems with decimals 2 UNIT 9 MEDIA LESSON SECTION 9.1: ADDING DECIMALS USING THE AREA MODEL In this section, we will learn to visualize the addition of decimals using the area model with the 10 by 10 grid. Problem 1 MEDIA EXAMPLE Adding Decimals in the Tenths and Hundredths Place Use the decimal grids to shade the addends of the addition problem. Then combine your addends in a new grid to find the sum. (Note: We call the numbers we are adding in an addition problem the addends. We call the simplified result the sum.) a) Sum: b) Sum: c) Sum: d) Sum: 2 3 Problem 2 YOU TRY - Adding Decimals Using the Area Model Unit 9 Media Lesson Use the decimal grids to shade the decimal portions of the addends of the addition problem. Then combine your addends in a new grid to find the sum. a) Sum: SECTION 9.2: ADDING DECIMALS USING PLACE VALUE In the last section, we were actually using place value to add decimals by grouping according to the place value of the decimals. In this section, we will streamline this process, by adding using a place value chart and then learning how to add without the place value chart. Problem 3 MEDIA EXAMPLE Adding Decimals Using a Place Value Chart Place the numbers in the place value chart and then use the chart as an aid to add the numbers Sum: Problem 4 MEDIA EXAMPLE Adding Decimals Using Place Value Add the decimals without a place value chart by aligning the decimals points and adding 4 Problem 5 You Try Adding Decimals Using Place Value Unit 9 Media Lesson In the first problem, add the decimals using the place value chart. In the second problem, align the decimal points to add. a) b) SECTION 9.3: SUBTRACTING DECIMALS USING THE AREA MODEL In this section, we will learn to visualize the subtraction of decimals using the 10 by 10 grid. Problem 6 MEDIA EXAMPLE Subtracting Decimals in the Tenths and Hundredths Place Use the decimal grids to shade the given decimals in the subtraction problem. Then find the difference by taking away the second quantity from the first quantity. a) Difference: b) Difference: 4 5 c) Difference: d) Difference: Problem 7 YOU TRY - Subtracting Decimals Using the Area Model Use the decimal grids to shade the given decimals in the subtraction problem. Then find the difference by taking away the second quantity from the first quantity Difference: 5 6 SECTION 9.4: SUBTRACTING DECIMALS USING PLACE VALUE In the last section, we were actually using place value to subtract decimals by grouping according to the place value of the decimals. In this section, we will streamline this process, by subtracting using a place value chart and then learning how to subtract without the place value chart. Problem 8 MEDIA EXAMPLE Subtracting Decimals Using a Place Value Chart Place the numbers in the place value chart and then use the chart as an aid to subtract the numbers Difference: Problem 9 MEDIA EXAMPLE Subtracting Decimals Using Place Value Subtract the decimals without a place value chart by aligning the decimals points and subtracting Problem 10 You Try Subtracting Decimals Using Place Value In the first problem, subtract the decimals using the place value chart. In the second problem, align the decimal points to subtract. a) b) 7 SECTION 9.5: ADDING AND SUBTRACTING SIGNED DECIMALS In this section, we will add and subtract signed decimals. The same rules that apply to these processes on integers can be extended to decimals. These procedures are summarized below. A. When adding two or more numbers, all with the same sign, 1. Add the absolute values of the numbers 2. Keep the common sign of the numbers B. When adding two numbers with different signs. 1. Find the absolute value of the numbers 2. Subtract the smaller absolute value from the larger absolute value 3. Keep the original sign of the number with the larger absolute value. C. When subtracting two decimals, we can use following fact. Fact: Subtracting a decimal from a number is the same as adding the decimal s opposite to the number. 1. If given a subtraction problem, rewrite it as an addition problem. 2. Use the rules for addition to add the signed numbers as summarized above. Problem 11 MEDIA EXAMPLE Adding and Subtracting Signed Decimals Use the rules for signed numbers to add or subtract the decimals. a) ( 0.27) b) ( 7.24) c) 4.2 ( 3.8) Problem 12 You Try Adding and Subtracting Signed Decimals Use the rules for signed numbers to add or subtract the decimals. a) ( 0.14) b) c) 5.2 ( 2.7) 7 8 SECTION 9.6: MULTIPLYING DECIMALS USING THE AREA MODEL In this section, we will learn to visualize the multiplication of decimals using the area model with the 10 by 10 grid. Problem 13 MEDIA EXAMPLE Multiplying a Whole Number Times a Decimal Rewrite the multiplication statements using copies of language and word names. Then represent the decimal problems using the decimal grids. a) 3 4 Copies Language: Picture: Product: b) Copies Language: Product: c) Copies Language: Product: d) Describe the pattern that you see in a through c. 8 9 Problem 14 MEDIA EXAMPLE Multiplying Two Decimals Unit 9 Media Lesson Rewrite the multiplication statements using copies of language and word names. Then represent the decimal problems using the decimal grids. a) b) Copies Language: Copies Language: Product: Product: c) Describe the pattern that you see. Problem 15 You Try Multiplying Two Decimals Rewrite the multiplication statements using copies of language and word names. Then represent the decimal problems using the decimal grids. a) Copies Language: Product: b) Copies Language: Product: 9 10 SECTION 9.7: MULTIPLYING DECIMALS USING PLACE VALUE In this section, we will multiply decimals by using the patterns we saw in Section 4.1. In particular, we will use the strategy below. To multiply two decimals: 1. Multiply the two numbers as if they were whole numbers (disregard the decimals for now). 2. Determine the total number of digits that were to the right of the decimal points in your two original factors and add them. 3. Take your product from step one. Starting from the right, count as many place values as you found in step 2 and place the decimal point in this spot. Problem 16 MEDIA EXAMPLE Multiplying Decimals Using Place Value Multiply the decimals. a) = b) = c) = a) = e) = f) = g) = e) = f) = Problem 17 You Try Multiplying Decimals Using Place Value Multiply the decimals. a) = b) = c) = 10 11 SECTION 9.8: DIVIDING DECIMALS USING THE AREA MODEL Problem 18 MEDIA EXAMPLE Dividing Decimals using the Area Model Rewrite the division statements using copies of language and word names. Then represent the decimal problems using the decimal grids. a) 12 3 Copies Language: Picture: Quotient b) Copies Language: Quotient: c) Copies Language: Quotient: 11 12 Problem 19 You Try Dividing Decimals Using the Area Model Unit 9 Media Lesson Rewrite the division statements using copies of language and word names. Then represent the decimal problems using the decimal grids Copies Language: Quotient: SECTION 9.9: DIVIDING DECIMALS USING PLACE VALUE In this section, we will look at quotients that are not whole numbers. We will use the patterns developed to create a general method for dividing numbers involving decimals. Problem 20 Divide the decimals. MEDIA EXAMPLE Dividing Decimals Using Place Value a) 24 8 = b) = c) = d) = e) = f) = Problem 21 Divide the decimals. You Try Dividing Decimals Using Place Value a) 56 8 = b) = c) = 12 13 SECTION 9.10: MULTIPLYING AND DIVIDING DECIMALS BY POWERS OF 10 In this section, we will investigate patterns when multiplying or dividing by powers of ten. Some examples of powers of ten are 10 1 = 10, 10 2 = 100, and 10 3 = Problem 22 MEDIA EXAMPLE Multiplying by Powers of Ten Multiply the numbers by the given powers of 10 by moving the decimal point the appropriate number of places. a) = b) = c) = d) = e) = f) = Problem 23 MEDIA EXAMPLE Dividing by Powers of Ten Divide the numbers by the given powers of 10 on your calculator then look for patterns to make a general strategy. a) = b) = c) = d) = e) = f) = g) Look for patterns in the examples above and complete the statement below. To divide a decimal number by a power of 10, you move the decimal place Problem 24 YOU TRY - Multiplying and Dividing by Powers of Ten Multiply the numbers by the given powers of 10 by moving the decimal point the appropriate number of places. a) = b) = c) = d) = e) = f) = 13 14 SECTION 9.11: DECIMAL OPERATIONS ON THE CALCULATOR When performing the mathematical operations of addition, subtraction, multiplication, and division using decimals, our calculator is a great support tool. Once the given numbers are combined, rounding often comes into play when presenting the final result. Problem 25 MEDIA EXAMPLE Decimal Operations on the Calculator Use your calculator to compute each of the following. Round as indicated. a) Multiply then round the result to the nearest tenth. b) Divide then round the result to the nearest thousandth. c) Evaluate(0.1) 2. Write your result first in decimal form. Then, convert to a simplified fraction. d) Combine the numbers below. Round your final result to the nearest whole number Problem 26 YOU TRY - Decimal Operations on the Calculator Use your calculator to combine the numbers below. Round your final result to the nearest hundredth. When computing, try to enter the entire expression all at once. (6.41) 15 SECTION 9.12: APPLICATIONS WITH DECIMALS Problem 27 MEDIA EXAMPLE Applications with Decimals In preparation for mailing a package, you place the item on your digital scale and obtain the following readings: 6.51 ounces, 6.52 ounces, and 6.60 ounces. What is the average of these weights? Round to the nearest hundredth of an ounce. GIVEN: GOAL: MATH WORK: CHECK: FINAL ANSWER AS A COMPLETE SENTENCE: Problem 28 YOU TRY - Applications with Decimals Rally went to Target with \$40 in his wallet. He bought items that totaled \$1.45, \$2.15, \$7.34, and \$ If the tax comes to \$2.26, how much of his \$40 would he have left over? Round to the nearest cent (hundredths place). 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All the problems are the same type, so that you can More information ### Math and FUNDRAISING. Ex. 73, p. 111 1.3 0. 7 Standards Preparation Connect 2.7 KEY VOCABULARY leading digit compatible numbers For an interactive example of multiplying decimals go to classzone.com. Multiplying and Dividing Decimals Gr. 5 NS 2.1 More information ### Addition. They use numbered number lines to add, by counting on in ones. Children are encouraged to start with the larger number and count on. Year 1 add with numbers up to 20 Children are encouraged to develop a mental picture of the number system in their heads to use for calculation. 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Eligible Content Grade 5 Math PA Core Standards - CC.2: PA Core: Mathematics Eligible Content Standard Area - CC.2.1: Numbers and Operations Standard - CC.2.1.5.B.1: Apply place value to show an understanding of operations More information ### The Partial-Quotients Division Algorithm, Part 1 The Partial-Quotients Division Algorithm, Part 1 Objectives To introduce and provide practice with a low-stress division algorithm for 1-digit divisors. www.everydaymathonline.com epresentations etoolkit More information ### Maths Level Targets. This booklet outlines the maths targets for each sub-level in maths from Level 1 to Level 5. Maths Level Targets This booklet outlines the maths targets for each sub-level in maths from Level 1 to Level 5. Expected National Curriculum levels for the end of each year group are: Year 1 Year 2 Year More information ### Year 1. Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on. Year 1 Add with numbers up to 20 Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on. +1 +1 +1 Children should: Have access to a wide More information ### numerical place value additional topics rounding off numbers power of numbers negative numbers addition with materials fundamentals Math Scope & Sequence fundamentals number sense and numeration of the decimal system Count to 10 by units Associate number to numeral (1-10) KN 1 KN 1 KN 2 KN 2 Identify odd and even numbers/numerals and More information ### Math Review. Numbers. Place Value. Rounding Whole Numbers. Place value thousands hundreds tens ones Math Review Knowing basic math concepts and knowing when to apply them are essential skills. You should know how to add, subtract, multiply, divide, calculate percentages, and manipulate fractions. This More information ### Maths Assessment Year 4: Fractions Name: Maths Assessment Year : Fractions 1. Recognise and show, using diagrams, families of common equivalent fractions. 2. Count up and down in hundredths. 3. Solve problems involving increasingly harder More information ### ARITHMETIC. Overview. Testing Tips ARITHMETIC Overview The Arithmetic section of ACCUPLACER contains 17 multiple choice questions that measure your ability to complete basic arithmetic operations and to solve problems that test fundamental More information ### 3. ROUNDING OFF DECIMAL NUMBERS TO THE NEAREST TENTH 3. ROUNDING OFF DECIMAL NUMBERS TO THE NEAREST TENTH Material: Decimal Board and materials Small gold circle Prepared problems Paper and pencil Presentation: 1. Form a number on the board with discs: 0.68. More information ### Solving Equations With Fractional Coefficients Solving Equations With Fractional Coefficients Some equations include a variable with a fractional coefficient. Solve this kind of equation by multiplying both sides of the equation by the reciprocal of More information ### Consultant: Lynn T. Havens. Director of Project CRISS Kalispell, Montana Teacher Annotated Edition Study Notebook Consultant: Lynn T. Havens SM Director of Project CRISS Kalispell, Montana i_sn_c1fmtwe_893629.indd i 3/16/09 9:17:03 PM Copyright by The McGraw-Hill Companies, More information ### Change Number Stories Objective To guide children as they use change diagrams to help solve change number stories. Number Stories Objective To guide children as they use change diagrams to help solve change number stories. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game More information ### FSCJ PERT. Florida State College at Jacksonville. assessment. and Certification Centers FSCJ Florida State College at Jacksonville Assessment and Certification Centers PERT Postsecondary Education Readiness Test Study Guide for Mathematics Note: Pages through are a basic review. Pages forward More information ### Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Solve word problems that call for addition of three whole numbers More information ### Decimals are absolutely amazing We have only 10 symbols, yet can represent any number, large or small We use zero (0) as a place holder to allow us Decimals 1 Decimals are absolutely amazing We have only 10 symbols, yet can represent any number, large or small We use zero (0) as a place holder to allow us to do this 2 Some Older Number Systems 3 Can More information ### Rational Number Project Rational Number Project Fraction Operations and Initial Decimal Ideas Lesson : Overview Students estimate sums and differences using mental images of the 0 x 0 grid. Students develop strategies for adding More information ### 1 ENGAGE. 2 TEACH and TALK GO. Round to the Nearest Ten or Hundred Lesson 1.2 c Round to the Nearest Ten or Hundred Common Core Standard CC.3.NBT.1 Use place value understanding to round whole numbers to the nearest 10 or 100. Lesson Objective Round 2- and 3-digit numbers More information ### Grade 4 - Module 5: Fraction Equivalence, Ordering, and Operations Grade 4 - Module 5: Fraction Equivalence, Ordering, and Operations Benchmark (standard or reference point by which something is measured) Common denominator (when two or more fractions have the same denominator) More information ### Maths Targets Year 1 Addition and Subtraction Measures. N / A in year 1. Number and place value Maths Targets Year 1 Addition and Subtraction Count to and across 100, forwards and backwards beginning with 0 or 1 or from any given number. Count, read and write numbers to 100 More information ### Grades 4 and 5 Math. TEKS and TAKS Daily Distributive Practice Grades 4 and 5 Math TEKS and TAKS Daily Distributive Practice 90 days of cumulative TEKS/TAKS practice tests Nine-question tests designed to meet 3 levels of achievement in a single classroom I'm the largest More information ### Calculation strategies for subtraction Calculation strategies for subtraction 1 Year 1 Subtract numbers up to 20. Children are encouraged to develop a mental picture of the number system in their heads to use for calculation. They develop ways More information ### MACMILLAN/McGRAW-HILL. MATH CONNECTS and IMPACT MATHEMATICS WASHINGTON STATE MATHEMATICS STANDARDS. ESSENTIAL ACADEMIC LEARNING REQUIREMENTS (EALRs) MACMILLAN/McGRAW-HILL MATH CONNECTS and IMPACT MATHEMATICS TO WASHINGTON STATE MATHEMATICS STANDARDS ESSENTIAL ACADEMIC LEARNING REQUIREMENTS (EALRs) And GRADE LEVEL EXPECTATIONS (GLEs) / Edition, Copyright More information
# Parallel Lines Lines are the greatest thing ever to be created. JUST KIDDING. No, I'm serious, l think they are pretty cool. Guess what? There are different kinds of lines. There are food lines, waiting lines, hair lines, finish lines, and straight lines. In graphs and equations there are two different types of lines that can be noticed. The two types of lines are perpendicular and parallel. On this page, it is all about parallel lines. Parallel lines are the same distance apart, and they will never meet or touch. Here are some examples of parallel lines: Horizontal Parallel Lines Vertical Parallel Lines (Train tracks) Parallel Lines on a graph Y=mx+b is the slope-intercept form. X and Y are the coordinates, m is the slope, and b is the y-intercept Parallel lines always have to have the same slope in equations in order to never touch and be parallel. You can look at the coefficient of the x variable to see if they have the same slope. The y-intercept, (the b variable) does not affect whether or not the lines will be parallel. It is all about the slope. Examples- Ex. 1) Are these two line parallel to each other Y=2X-1 Y=2X+3 To see if these two equations are parallel, you have to find the slope of both equations. The slope is the coefficient of X so in both equations, the slope would be 2. Since, both equations have the same slope, the two lines are parallel. Ex. 2) What line is parallel to the line of Y=5x+12 and crosses through the point of (4,2)? -Since the eqaution of y=5x+12 is already in slope-intercept form, you don't need to change anything. -Since the line also crosses through (4, 2), they are your x and y variables. -Also, you know that if two lines are parallel, they have to have the same slope. This slope in the first line is 5, so the new line's slope must be 5 too. -Now, you have the x and y variables, and the slope, so all you have to do is solve for b. -Plug in all the numbers you found in the slope-intercept form of y=mx+b and solve. -So the new equation would be 2=5(4) +b 2=20+b -18=b -Now that you have found b, write the equation with only the slope and the y-intercept for the line. Y=5x-18. Yay! Graphing Graphing parallel lines are pretty easy to do, and they are easy to check too. Let's graph the parallel lines of Y=(1/2)X+2 and Y=(1/2)X+1 -To graph parallel lines, first make sure that the equations are in the slope-intercept form and they are indeed parallel lines, with the same slope. Y=(1/2)X+2 √ Y=(1/2)X+1√ -Once you have checked everything, we can finally get to the fun stuff -Graph the lines just like you would graph a regular line. - But in case you forgot how to graph a line, that's what we're here for! -First, find the y intercept, if there is one. The y-intercept is the b variable. Y=(1/2)X+2 So the y intercepts are (0,2) and (0,1) Y=(1/2)X+1 -Then start plugging in some numbers for X. Lets plug in the number 7 because everyone loves that number. -Substitute 7 for X in both equations so they would become Y=(1/2)(7)+2 Y=(1/2)(7)+1 -Then solve the equations for y Y=(1/2)(7)+2 =5.5 Y=(1/2)(7)+1 =4.5 Now you have a point for each equation. the points are (7, 5.5) and (7, 4.5) -Plot the coordinate points. -Now, you can either connect the dots of the point and its corresponding y-intercepts, or you can plot more points to determine the line. -But you don't need to plot more points because parallel lines are always straight. The resulting graph should look like this... Parallel lines, equations and graphs are all pretty easy to understand. Basically, just remember that in order to have parallel lines you must have: -same slope -lines NEVER intersect That is it. Now you know all about parallel lines and how they are created through different equations, slopes, and graphs. And don't forget to credit us when you ace your tests! (And don't mention us if you fail.)
# What is a pentagon? Learn all about this 5-sided shape, its properties, examples, and real-life application. Author Taylor Hartley Expert Reviewer Published: July 2024 # What is a pentagon? Learn all about this 5-sided shape, its properties, examples, and real-life application. Author Taylor Hartley Expert Reviewer Published: July 2024 # What is a pentagon? Learn all about this 5-sided shape, its properties, examples, and real-life application. Author Taylor Hartley Expert Reviewer Published: July 2024 Key takeaways • Pentagons are 5-sided polygons – “Penta” means “five,” so it’s easy to remember how many sides you’re dealing with when you see the word “pentagon.” • Finding the perimeter of a regular pentagon is easy – Since regular pentagons have equal sides, just multiply the length of the sides by 5 to find the answer. • Figuring out the area of irregular pentagons is a bit tricky – While there’s a straightforward formula for the area of a regular pentagon (read on to find out more!), irregular pentagons require knowledge of other area formulas, like how to calculate the area of a triangle! Today, we’ll be discussing our five-sided friend, the pentagon! From a common stop sign to the tiles on your bathroom floor, pentagons are a common shape you’ll run into in your everyday life. It might be nice, then, to understand how they work and the kinds of formulas we can use to calculate their perimeter and their area. Ready to dive in? Let’s get going. ## What is a pentagon? A pentagon is a five-sided polygon with five interior angles that add up to 540°. It’s another one of our classic 2-D shapes, meaning it exists on a flat plane just like your average triangle, square or circle ### What does a pentagon look like? A pentagon resembles a house with a pointed roof! There are five sides in total with five angles at each of the shape’s vertices (the points where lines join to form angles). ### Parts of a pentagon There are a few key parts you’ll need to know in order to identify the perimeter or area of a pentagon. These include: • Side – These are the five line segments that form the shape of a pentagon. • Vertex – The vertices are the points at which each of the five line segments meet. Each pentagon has five vertices. Treat a vertex the same way you would treat a corner. • Diagonal – A diagonal is a line that connects vertices that are non-adjacent (not right next to each other). When these two corners are connected, the line segment that connects them is known as a diagonal. There are five diagonals as well. • Interior angle – An interior angle includes any of the angles that form at the point where the sides meet (the vertex). There are five interior angles to match with the five vertices and five sides, and the sum of these interior angles is 540°. • Exterior angle – There are also five exterior angles, each of which is formed outside of the pentagon’s shape. For every interior angle, there is a corresponding exterior angle. • Apothem – A line drawn from the center of any polygon to the midpoint of one of the sides. ### Real-life examples of pentagons There are a few key parts you’ll need to know in order to identify the perimeter or area of a pentagon. These include: Did you know? The headquarters for the U.S Department of Defense is actually named The Pentagon, and it’s shaped just like one too! Unlock unlimited maths questions Put your skills to the test with fun exercises + maths games that are proven to boost ability! ## Practise Geometry with DoodleMaths! DoodleLearning is an award-winning app that’s filled with thousands of questions and games aligned. tothe national curriculum! Designed by teachers, it creates each child a unique work programme tailored to their needs, doubling their progression with just 10 minutes of use a day. Try it for free! ## Types of pentagons There are a few different categories of pentagons we’ll encounter including regular pentagons, irregular pentagons, convex pentagons, and concave pentagons. You can identify the kind of pentagon you’re dealing with by identifying the types of angles the pentagon has, what the sums of these angles are, and how the pentagon itself is shaped. Follow along as we examine the four different categories of our five-sided friends. ### Regular pentagons Regular pentagons have five equal sides and five equal interior angles. If each of the five sides have an equal length, and each of the five angles have an equal measurement, the pentagon is considered regular. Just remember: regular means equal. ### Irregular pentagons Irregular pentagons are defined as pentagons with unequal values for both side length and angle measurements. Because of this, irregular pentagons aren’t symmetrical and will look more elongated or flattened. ### Convex pentagons Convex pentagons are unique because all of their vertices point outward. This means that none of the sides of a convex pentagon point inward. Each of them point outward to create an outward facing vertex. ### Concave pentagons Concave pentagons are the opposite of convex, because they include at least one vertex that points inward. Think of the meeting of each side as the formation of an arrow. If the point of the arrow (the vertex) is facing inward, the shape is concave. If it faces outward, the shape is convex! Select a year group ## Perimeter of a pentagon Finding the perimeter of a pentagon is the same process as finding the perimeter of any other kind of shape. All you need to do is find the sum of each of the five sides, and you’ll have calculated the perimeter of the pentagon. This process is the same regardless of whether the pentagon is regular, irregular, convex or concave. However, finding the perimeter of a regular pentagon is a whole lot easier than finding the perimeter of an irregular pentagon: all you have to do is multiply the side length by 5! ## Area of a pentagon Finding the area of a pentagon is a bit more complicated than finding its perimeter, as you’ve probably already expected. The method you use to calculate the pentagon’s area depends on whether you’re dealing with a regular or irregular pentagon. Let’s walk through the steps of how you’d calculate the area of both. ### Area of a regular pentagon Finding the area of a regular pentagon is easy, so long as you have the apothem and the measure of one of the sides! Later, you may be asked to find the apothem of a pentagon. For now, though, let’s assume that you are given the apothem when you’re asked to find the area. 1. There’s an easy formula that will help you find the area of a regular pentagon: 2. So, first we need to find the perimeter. This should be easy! A regular pentagon has 5 equal sides, so we’ll just need to multiply the length of the sides by 5 to find our perimeter. 3. Then, we’ll take the apothem and the perimeter and plug in the values to find our area! So let’s say we have a regular pentagon with a side length 5 cm and an apothem of 3.44 cm. 1. We will first find the perimeter of the regular pentagon: 5 x 5 = 25 2. Then, we will use our formula: 1/2 x Perimeter x Apothem sq units = 1/2 x 25 x 3.44 cm² 3. 25 x 3.44 cm² = 86 cm² x 1/2 = 43 cm² Remember, this formula is for finding the area of a regular pentagon. The method for finding the area of an irregular pentagon is something you won’t see until much later in your math career! ## Let’s practise together! 1. A regular pentagon has a side length of 8 cm. What is its perimeter? We’ve got this one for sure! But let’s go through it just in case. 1. Remember, a regular pentagon has equal sides. 2. That means that we can do one of two things: we can multiply 8 by 5 to get the perimeter, or we can add 8 together 5 times. 3. Let’s multiply: 8 x 5 = 40 cm. 4. The perimeter is 40 cm! ### 2. A regular pentagon has a side length of 6 cm and an apothem of 4.18 cm. Find its area. Let’s walk through it together, step by step. 1. We know we can find the perimeter of a regular pentagon by multiplying 6 by 5. That gives us a perimeter of 30. 2. We have an apothem of 4.18 cm. So, then we plug those values into our equation. 1/2 x Perimeter x Apothem sq units 1. When we do that, we get: 1/2 x 30 x 4.18 cm = 62.7 square centimeters ### 3. The apothem of a regular pentagon is 6.96 cm and the length of each side is 10 cm. What is the area of the pentagon? Easy peasy, right? Let’s walk through it! 1. Remember, the formula for finding the area of a regular pentagon is: Area = 1/2 x Perimeter x Apothem 2. Let’s find the perimeter. For a regular pentagon, the perimeter (the sum of the lengths of the sides) is simply 5 times the length of one side, because all sides are equal. So, since each side is 10 cm, the perimeter is 5 x 10 cm = 50 cm. 3. Apply the formula for the area. We know the apothem is 6.96 cm, and we’ve found the perimeter to be 50 cm. Now, we substitute these values into the formula: 1. So, the area of the pentagon is 174 cm². Ready to give it a go? Now that we’ve practiced together, it’s time to try things out on your own! Try the next few practice problems to test your knowledge of pentagons. Don’t be afraid if you get a little lost – just go back through this guide to help you! You’ve got this! ## Practice problems Click to reveal the answer. The answer is false. The answer is 45 cm The answer is 250.8 cm². ## FAQs about pentagon shapes We understand that diving into new information can sometimes be overwhelming, and questions often arise. That’s why we’ve meticulously crafted these FAQs, based on real questions from students and parents. We’ve got you covered! A pentagon shape is a geometric figure with five straight sides and five angles. Not always! Only a “regular” pentagon has five equal sides. An “irregular” pentagon has sides of different lengths. The total degrees of all angles inside any pentagon, whether regular or irregular, is 540 degrees. A five-sided pentagon is unique because of its five edges. In contrast, triangles have three sides, rectangles have four, and hexagons feature six sides. Parents, sign up for a DoodleMaths subscription and see your child become a maths wizard! ## Parent Guide The answer is false How did we get here? Only regular pentagons have five equal sides. The answer is 30 cm How did we get here? 1. This is an irregular pentagon, which means we need to add up the measure of all sides of the shape. 2. So we add 4 cm + 5 cm + 6 cm + 7 cm + 8 cm = 30 cm. The answer is 45 cm. How did we get here? 1. We’re dealing with a regular pentagon, which means all sides are equal. 2. That means we can multiply the length of the side by 5 to get the perimeter. 3. 9 × 5 = 45 cm. The answer is 250.8 cm². How did we get here? 1. Since this is a regular pentagon, we can use the formula 1/2 x Perimeter x Apothem to find the area. 2. First, let’s find the perimeter. 12 x 5 = 60 cm. 3. 1/2 x 60 cm x 8.36 cm² = 1/2 x 501.6 cm² = 250.8 cm². Lesson credits Taylor Hartley Taylor Hartley is an author and an English teacher. When she's not writing, you can find her on the rowing machine or lost in a good novel. Jill Padfield has 7 years of experience teaching mathematics, ranging from Algebra 1 to Calculus. She is currently working as a Business Analyst, working to improve services for Veterans while earning a masters degree in business administration. Taylor Hartley Taylor Hartley is an author and an English teacher. When she's not writing, you can find her on the rowing machine or lost in a good novel. Jill Padfield has 7 years of experience teaching mathematics, ranging from Algebra 1 to Calculus. She is currently working as a Business Analyst, working to improve services for Veterans while earning a masters degree in business administration. # Are you a parent, teacher or student? 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# Can someone explain this question? ##### 2 Answers Mar 9, 2017 You need to derive your function with respect to $t$ and then solve the resulting equation when you set the derivative equal to zero: #### Explanation: Ok, I can be wrong but your function can be derived as: $f ' \left(t\right) = {e}^{2 {t}^{2}} 3 \cdot {10}^{3 t - 5} \ln \left(10\right) + {10}^{3 t - 5} \cdot 4 t {e}^{2 {t}^{2}}$ rearranging: $f ' \left(t\right) = {e}^{2 {t}^{2}} \cdot {10}^{3 t - 5} \left[3 \ln \left(10\right) + 4 t\right]$ Let us set this equal to zero: ${e}^{2 {t}^{2}} \cdot {10}^{3 t - 5} \left[3 \ln \left(10\right) + 4 t\right] = 0$ when: $t = - \frac{3 \ln \left(10\right)}{4}$ that makes the square bracket equal to zero.... I think.... Mar 9, 2017 This is what I got. #### Explanation: Given expression is $f \left(t\right) = {10}^{3 t - 5} \times {e}^{2 {t}^{2}}$ Using product rule $f ' \left(t\right) = {10}^{3 t - 5} \times \left({e}^{2 {t}^{2}} \times 4 t\right) + {e}^{2 {t}^{2}} \times \ln 10 \times {10}^{3 t - 5} \times 3$ $\implies f ' \left(t\right) = \left(4 t + 3 \ln 10\right) \times {10}^{3 t - 5} \times {e}^{2 {t}^{2}}$ To meet the given condition $\left(4 t + 3 \ln 10\right) \times {10}^{3 t - 5} \times {e}^{2 {t}^{2}} = 0$ .....(1) $\implies f \left(t\right) \left[4 t + 3 \ln 10\right] = 0$ either $f \left(t\right) = 0$ or$\left[4 t + 3 \ln 10\right] = 0$ $\implies t = - \frac{3 \ln 10}{4}$ ....(2) To find roots of $f \left(t\right) = 0$, ${10}^{3 t - 5} \times {e}^{2 {t}^{2}} = 0$ either ${10}^{3 t - 5} = 0$ .......(3) or ${e}^{2 {t}^{2}} = 0$ ......(4) From both (3) and (4) we get $t = - \infty$. This solution is purely theoretical as such this equation has no solution in real or complex numbers. As such only possible solution is as given in equation (2)
Associated Topics \|\| Dr. Math Home \|\| Search Dr. Math ### Dividing Fractions ``` Date: 03/16/99 at 12:22:32 From: Alex Subject: Division of Fractions I know how to divide fractions. I want to know WHY you cannot divide ``` ``` Date: 03/16/99 at 21:12:40 From: Doctor Peterson Subject: Re: Division of Fractions I am not quite sure what you mean. Of course you can divide fractions, but I suspect you mean "why don't you divide fractions directly, but instead have to do it by multiplying?" There are several ways to divide fractions, some more direct than others. The method of multiplying the reciprocal is usually the easiest. Suppose we want to divide 4/9 by 2/3. We COULD do it just by dividing the numerators and denominators: 4 2 4 / 2 2 --- / --- = ----- = --- 9 3 9 / 3 3 You can see that it works. (I am using "/" where you would normally use the "bar and two dots" division symbol.) Why do we usually not do this? Because I chose that problem carefully so the divisions would work out. More typically you would have trouble: 3 5 3 / 5 3/5 --- / --- = ----- = --- 4 6 4 / 6 2/3 You have not gained anything, since you still have one fraction divided by another. We have to fix this up; we can do this by multiplying numerator and denominator by the least common denominator (LCD) of the two fractions: 3 5 3/5 3/5 * 15 9 --- / --- = --- = -------- = -- 4 6 2/3 2/3 * 15 10 Or, we could first convert both fractions to use the same denominator: 3 5 9 10 9 / 10 9 --- / --- = -- / -- = ------- = -- 4 6 12 12 12 / 12 10 Since I made the denominators the same, the "denominator" of the quotient works out to 1. But there is really no division in that method, is there? It is still multiplication. A variation on this method is: 3 9 --- -- 4 12 9 ----- = ---- = -- 5 10 10 --- -- 6 12 You just multiply numerator and denominator by 12 in the last step to eliminate the fractions. Do you see what I did there? Let us write it out more carefully and you will see where "multiplying by the reciprocal" is hidden: 3 36 --- --- 4 46 36 33 9 ----- = ----- = --- = --- = -- 5 54 54 52 10 --- --- 6 64 Here I have ignored LEAST common denominators and just used the simplest common denominator, the product of the denominators 4 and 6. The product is simply the product of 3/4 and 6/5. To answer your question simply, we divide fractions by multiplying, not because we cannot divide, but because multiplication is easier than division, and because division by itself does not always produce whole numbers. All these methods just make you multiply the numerator of the dividend by the denominator of the divisor, and vice versa, with some division to simplify the fraction; the difference is only in how much you have to think about what you are doing. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ ``` Associated Topics: Elementary Division Elementary Fractions Middle School Division Middle School Fractions Search the Dr. Math Library: Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words Submit your own question to Dr. Math Math Forum Home \|\| Math Library \|\| Quick Reference \|\| Math Forum Search
--> # Differentiation of Exponential and Logarithmic Functions The exponential function derivative is not easily solved directly with the definition of the derived function. However, it can be solved using the inverse function derivative, namely the logarithmic function. So, in this article the derivative of the logarithmic function will take precedence before obtaining the derivative function derivative. ### Derivatives of Logarithmic Functions For example, a function f is stated by the following formula, f(x) = alog x, x > 0 then the derivative of the function can be obtained by the following steps,. Note that for h → 0, then u → 0. if it is known that then the form above can be simplified to become, So, the function f(x) = alog x, x > 0 is differentiable and the derivative is expressed by a formula. Especially for the logarithmic function with a base number e expressed by f(x) = ln x, x > 0, the derivative formula is as follows. In general, the derivative of function f(x) = alog h(x) is The function derivative f (x) = ln g (x) is ### Derivative Function f(x) = ex To obtain a function formula in the form of f(x) = ex, we can use the derivative formula of the logarithmic function. However, we can also use the inverse function f(x) = ln x first. Suppose y = ex then applies x = ln y. By using Leibniz notation in derivatives, the following form is obtained, So, the derivative of the exponential function f(x) = ex is expressed by the formula: f’(x) = ex Whereas, for exponential functions with numbers and past bases, namely f(x) = ax in the same way as above we can find the formula as follows. f’(x) = ax nlog a or f’(x) = ax ln a In general, the derivative of the function f(x) = eh(x) is f’(x)’=h’(x) . eh(x) the derivative of function f(x) = ag(x) is f’(x) = ag(x) . g’(x) ln a Example 1. Find the function derivative f(x) = 3log (2x – 4). Example 2. Find the function derivative f(x) = ln (2x3 + 7). Example 3. Find the function derivative f(x) = log 3x. Example 4. Find the function derivative f(x) = e2x +3
## Maryland Standards for Eighth Grade Math Collecting and describing dataCollecting and describing data refers to the different ways to gather data and the different ways to arrange data whether it is in a table, graph, or pie chart. Data can be collected by either taking a sample of a population or by conducting a survey. Describing data looks at data after it has been organized and makes conclusions about the data. Read more...iWorksheets: 3Study Guides: 1 Linear relationshipsLinear relationships refer to two quantities that are related with a linear equation. Since a linear equation is a line, a linear relationship refers to two quantities on a line and their relationship to one another. This relationship can be direct or inverse. If y varies directly as x, it means if y is doubled, then x is doubled. The formula for a direct variation is y = kx, where k is the constant of variation. Read more...iWorksheets: 3Study Guides: 1 Solving equations and inequalitiesAlgebraic equations are mathematical equations that contain a letter or variable which represents a number. To solve an algebraic equation, inverse operations are used. Algebraic inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to, ≥; less than, <; and less than or equal to, ≤. When multiplying or dividing by a negative number occurs, the inequality sign is reversed from the original inequality sign in order for the inequality to be correct. Read more...iWorksheets: 3Study Guides: 1 Solving linear equationsWhen graphed, a linear equation is a straight line. Although the standard equation for a line is y = mx + b, where m is the slope and b is the y-intercept, linear equations often have both of the variables on the same side of the equal sign. Linear equations can be solved for one variable when the other variable is given. Read more...iWorksheets: 5Study Guides: 1 FunctionsFreeA function is a rule that is performed on a number, called an input, to produce a result called an output. The rule consists of one or more mathematical operations that are performed on the input. An example of a function is y = 2x + 3, where x is the input and y is the output. The operations of multiplication and addition are performed on the input, x, to produce the output, y. By substituting a number for x, an output can be determined. Read more...iWorksheets: 5Study Guides: 1 Patterns in geometryPatterns in geometry refer to shapes and their measures. Shapes can be congruent to one another. Shapes can also be manipulated to form similar shapes. The types of transformations are reflection, rotation, dilation and translation. With a reflection, a figure is reflected, or flipped, in a line so that the new figure is a mirror image on the other side of the line. A rotation rotates, or turns, a shape to make a new figure. A dilation shrinks or enlarges a figure. A translation shifts a figure to a new position. Read more...iWorksheets: 3Study Guides: 1 Plane figuresPlane figures refer to points, lines, angles, and planes in the coordinate plane. Lines can be parallel or perpendicular. Angles can be categorized as acute, obtuse or right. Angles can also be complementary or supplementary depending on how many degrees they add up to. Plane figures can also refer to shapes in the coordinate plane. Triangles, quadrilaterals and other polygons can be shown in the coordinate plane. Read more...iWorksheets: 4Study Guides: 1 Rational numbers and operationsA rational number is a number that can be made into a fraction. Decimals that repeat or terminate are rational because they can be changed into fractions. A square root of a number is a number that when multiplied by itself will result in the original number. The square root of 4 is 2 because 2 · 2 = 4. Read more...iWorksheets: 3Study Guides: 1 ### MD.1.0. Knowledge of Algebra, Patterns, and Functions: Students will algebraically represent, model, analyze, or solve mathematical or real-world problems involving patterns or functional relationships. #### 1.A.1. Patterns and Functions: Identify, describe, extend, and create patterns, functions and sequences. ##### 1.A.1.a. Determine the recursive relationship of arithmetic sequences represented in words, in a table or in a graph (Assessment limit: Provide the nth term no more than 10 terms beyond the last given term using common differences no more than 10 with integers (-100 to 5000)). SequencesA sequence is an ordered list of numbers. Sequences are the result of a pattern or rule. A pattern or rule can be every other number or some formula such as y = 2x + 3. When a pattern or rule is given, a sequence can be found. When a sequence is given, the pattern or rule can be found. Read more...iWorksheets :5Study Guides :1 ##### 1.A.1.b. Determine the recursive relationship of geometric sequences represented in words, in a table, or in a graph (Assessment limit: Provide the nth term no more than 5 terms beyond the last given term using the recursive relationship of geometric sequences with whole numbers and a common ratio of no more than 5:1 (0 - 10,000)). SequencesA sequence is an ordered list of numbers. Sequences are the result of a pattern or rule. A pattern or rule can be every other number or some formula such as y = 2x + 3. When a pattern or rule is given, a sequence can be found. When a sequence is given, the pattern or rule can be found. Read more...iWorksheets :5Study Guides :1 #### 1.B.1. Expressions, Equations, and Inequalities: Write, simplify, and evaluate expressions. ##### 1.B.1.a. Write an algebraic expression to represent unknown quantities (Assessment limit: Use one unknown and no more than 3 operations and rational numbers (-1000 to 1000)). Introduction to AlgebraAlgebra is the practice of using expressions with letters or variables that represent numbers. Words can be changed into a mathematical expression by using the words, plus, exceeds, diminished, less, times, the product, divided, the quotient and many more. Algebra uses variables to represent a value that is not yet known. Read more...iWorksheets :4Study Guides :1 Equations and inequalitiesAn equation is mathematical statement that shows that two expressions are equal to each other. The expressions used in an equation can contain variables or numbers. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Inequalities are also solved by using inverse operations. Read more...iWorksheets :3Study Guides :1 ##### 1.B.1.c. Evaluate numeric expressions using the order of operations (Assessment limit: Use no more than 5 operations including exponents of no more than 3 and 2 sets of parentheses, brackets, a division bar, or absolute value with rational numbers (-100 to 100). Using IntegersIntegers are negative numbers, zero and positive numbers. To compare integers, a number line can be used. On a number line, negative integers are on the left side of zero with the larger a negative number, the farther to the left it is. Positive integers are on the right side of zero on the number line. If a number is to the left of another number it is said to be less than that number. In the coordinate plane, the x-axis is a horizontal line with negative numbers, zero and positive numbers. Read more...iWorksheets :4Study Guides :1 ##### 1.B.1.d. Simplify algebraic expressions by combining like terms (Assessment limit: Use no more than 3 variables with integers (-50 to 50), or proper fractions with denominators as factors of 20 (-20 to 20)). Introduction to AlgebraAlgebra is the practice of using expressions with letters or variables that represent numbers. Words can be changed into a mathematical expression by using the words, plus, exceeds, diminished, less, times, the product, divided, the quotient and many more. Algebra uses variables to represent a value that is not yet known. Read more...iWorksheets :4Study Guides :1 Equations and inequalitiesAn equation is mathematical statement that shows that two expressions are equal to each other. The expressions used in an equation can contain variables or numbers. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Inequalities are also solved by using inverse operations. Read more...iWorksheets :3Study Guides :1 Polynomials and ExponentsFreeA polynomial is an expression which is in the form of ax<sup>n</sup>, where a is any real number and n is a whole number. If a polynomial has only one term, it is called a monomial. If it has two terms, it is a binomial and if it has three terms, it is a trinomial. The standard form of a polynomial is when the powers of the variables are decreasing from left to right. Read more...iWorksheets :6Study Guides :1 ##### 1.B.1.e. Describe a real-world situation represented by an algebraic expression. Mathematical processesMathematical processes refer to the skills and strategies needed in order to solve mathematical problems. If one strategy does not help to find the solution to a problem, using another strategy may help to solve it. Problem solving skills refer to the math techniques that must be used to solve a problem. If a problem were to determine the perimeter of a square, a needed skill would be the knowledge of what perimeter means and the ability to add the numbers. Read more...iWorksheets :3Study Guides :1 #### 1.B.2. Expressions, Equations, and Inequalities: Identify, write, solve, and apply equations and inequalities. ##### 1.B.2.a. Write equations or inequalities to represent relationships (Assessment limit Use a variable, the appropriate relational symbols (>, is greater than or equal to, <, is less than or equal to, =), and no more than 3 operational symbols (+, -, x, /) on either side and rational numbers (-1000 to 1000)). Equations and InequalitiesAlgebraic equations are mathematical equations that contain a letter or variable, which represents a number. To solve an algebraic equation, inverse operations are used. The inverse operation of addition is subtraction and the inverse operation of subtraction is addition. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Read more...iWorksheets :6Study Guides :1 Algebraic InequalitiesFreeAlgebraic inequalities are mathematical equations that compare two quantities using these criteria: greater than, less than, less than or equal to, greater than or equal to. The only rule of inequalities that must be remembered is that when a variable is multiplied or divided by a negative number the sign is reversed. Read more...iWorksheets :3Study Guides :1 Equations and inequalitiesAn equation is mathematical statement that shows that two expressions are equal to each other. The expressions used in an equation can contain variables or numbers. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Inequalities are also solved by using inverse operations. Read more...iWorksheets :3Study Guides :1 ##### 1.B.2.d. Identify or graph solutions of inequalities on a number line (Assessment limit: Use one variable once with a positive whole number coefficient and integers (-100 to 100)). Equations and InequalitiesAlgebraic equations are mathematical equations that contain a letter or variable, which represents a number. To solve an algebraic equation, inverse operations are used. The inverse operation of addition is subtraction and the inverse operation of subtraction is addition. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Read more...iWorksheets :6Study Guides :1 Algebraic InequalitiesFreeAlgebraic inequalities are mathematical equations that compare two quantities using these criteria: greater than, less than, less than or equal to, greater than or equal to. The only rule of inequalities that must be remembered is that when a variable is multiplied or divided by a negative number the sign is reversed. Read more...iWorksheets :3Study Guides :1 Equations and inequalitiesAn equation is mathematical statement that shows that two expressions are equal to each other. The expressions used in an equation can contain variables or numbers. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Inequalities are also solved by using inverse operations. Read more...iWorksheets :3Study Guides :1 ##### 1.B.2.f. Apply given formulas to a problem-solving situation (Assessment limit: Use no more than four variables and up to three operations with rational numbers (-500 to 500)). Measurement, Perimeter, and CircumferenceThere are two systems used to measure objects, the U.S. Customary system and the metric system. The U.S. Customary system measures length in inches, feet, yards and miles. The metric system is a base ten system and measures length in kilometers, meters, and millimeters. Perimeter is the measurement of the distance around a figure. It is measured in units and can be measured by inches, feet, blocks, meters, centimeters or millimeters. To get the perimeter of any figure, simply add up the measures of the sides of the figure. Read more...iWorksheets :3Study Guides :1 ##### 1.B.2.g. Write equations and inequalities that describe real-world problems. Equations and InequalitiesAlgebraic equations are mathematical equations that contain a letter or variable, which represents a number. To solve an algebraic equation, inverse operations are used. The inverse operation of addition is subtraction and the inverse operation of subtraction is addition. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Read more...iWorksheets :6Study Guides :1 Algebraic InequalitiesFreeAlgebraic inequalities are mathematical equations that compare two quantities using these criteria: greater than, less than, less than or equal to, greater than or equal to. The only rule of inequalities that must be remembered is that when a variable is multiplied or divided by a negative number the sign is reversed. Read more...iWorksheets :3Study Guides :1 Equations and inequalitiesAn equation is mathematical statement that shows that two expressions are equal to each other. The expressions used in an equation can contain variables or numbers. Inequalities are mathematical equations that compare two quantities using greater than, >; greater than or equal to ≥; less than, <; and less than or equal to, ≤. Inequalities are also solved by using inverse operations. Read more...iWorksheets :3Study Guides :1 #### 1.C.1. Numeric and Graphic Representations of Relationships: Locate points on a number line and in a coordinate plane. ##### 1.C.1.a. Graph linear equations in a coordinate plane (Assessment limit: Use two unknowns having integer coefficients (-9 to 9) and integer constants (-20 to 20)). Introduction to FunctionsA function is a rule that is performed on a number, called an input, to produce a result called an output. The rule consists of one or more mathematical operations that are performed on the input. An example of a function is y = 2x + 3, where x is the input and y is the output. The operations of multiplication and addition are performed on the input, x, to produce the output, y. By substituting a number for x, an output can be determined. Read more...iWorksheets :7Study Guides :1 Linear equationsLinear equations are equations that have two variables and when graphed are a straight line. Linear equation can be graphed based on their slope and y-intercept. The standard equation for a line is y = mx + b, where m is the slope and b is the y-intercept. Slope can be found with the formula m = (y2 - y1)/(x2 - x1), which represents the change in y over the change in x. Read more...iWorksheets :6Study Guides :1 #### 1.C.2. Numeric and Graphic Representations of Relationships: Analyze linear relationships. ##### 1.C.2.a. Determine the slope of a graph in a linear relationship (Assessment limit: Use an equation with integer coefficients (-9 to 9) and integer constants (-20 to 20) and a given graph of the relationship). Introduction to FunctionsA function is a rule that is performed on a number, called an input, to produce a result called an output. The rule consists of one or more mathematical operations that are performed on the input. An example of a function is y = 2x + 3, where x is the input and y is the output. The operations of multiplication and addition are performed on the input, x, to produce the output, y. By substituting a number for x, an output can be determined. Read more...iWorksheets :7Study Guides :1 Linear equationsLinear equations are equations that have two variables and when graphed are a straight line. Linear equation can be graphed based on their slope and y-intercept. The standard equation for a line is y = mx + b, where m is the slope and b is the y-intercept. Slope can be found with the formula m = (y2 - y1)/(x2 - x1), which represents the change in y over the change in x. Read more...iWorksheets :6Study Guides :1 ##### 1.C.2.b. Determine the slope of a linear relationship represented numerically or algebraically. Introduction to FunctionsA function is a rule that is performed on a number, called an input, to produce a result called an output. The rule consists of one or more mathematical operations that are performed on the input. An example of a function is y = 2x + 3, where x is the input and y is the output. The operations of multiplication and addition are performed on the input, x, to produce the output, y. By substituting a number for x, an output can be determined. Read more...iWorksheets :7Study Guides :1 Linear equationsLinear equations are equations that have two variables and when graphed are a straight line. Linear equation can be graphed based on their slope and y-intercept. The standard equation for a line is y = mx + b, where m is the slope and b is the y-intercept. Slope can be found with the formula m = (y2 - y1)/(x2 - x1), which represents the change in y over the change in x. Read more...iWorksheets :6Study Guides :1 ### MD.2.0. Knowledge Geometry: Students will apply the properties of one-, two-, or three-dimensional geometric figures to describe, reason, or solve problems about shape, size, position, or motion of objects. #### 2.A.1. Properties of Plane Geometric Figures: Analyze the properties of plane geometric figures. ##### 2.A.1.b. Identify and describe the relationship among the parts of a right triangle (Assessment limit: Use the hypotenuse or the legs of right triangles). The Pythagorean TheoremPythagorean Theorem is a fundamental relation in Euclidean geometry. It states the sum of the squares of the legs of a right triangle equals the square of the length of the hypotenuse. Determine the distance between two points using the Pythagorean Theorem. Read more...iWorksheets :10Study Guides :2 #### 2.A.2. Properties of Plane Geometric Figures: Analyze geometric relationships. ##### 2.A.2.b. Apply right angle concepts to solve real-world problems (Assessment limit: Use the Pythagorean Theorem). The Pythagorean TheoremPythagorean Theorem is a fundamental relation in Euclidean geometry. It states the sum of the squares of the legs of a right triangle equals the square of the length of the hypotenuse. Determine the distance between two points using the Pythagorean Theorem. Read more...iWorksheets :10Study Guides :2 ##### 2.A.2.c. Determine whether three given side lengths form a right triangle. The Pythagorean TheoremPythagorean Theorem is a fundamental relation in Euclidean geometry. It states the sum of the squares of the legs of a right triangle equals the square of the length of the hypotenuse. Determine the distance between two points using the Pythagorean Theorem. Read more...iWorksheets :10Study Guides :2 #### 2.D.1. Congruence and Similarity: Apply the properties of similar polygons. ##### 2.D.1.a. Determine similar parts of polygons (Assessment limit: Use the length of corresponding sides or the measure of corresponding angles and rational numbers with no more than 2 decimal places (0 - 1000)). Geometric ProportionsGeometric proportions compare two similar polygons. Similar polygons have equal corresponding angles and corresponding sides that are in proportion. A proportion equation can be used to prove two figures to be similar. If two figures are similar, the proportion equation can be used to find a missing side of one of the figures. Read more...iWorksheets :4Study Guides :1 Ratios, proportions and percentsNumerical proportions compare two numbers. A proportion is usually in the form of a:b or a/b. There are 4 parts to a proportion and it can be solved when 3 of the 4 parts are known. Proportions can be solved using the Cross Product Property, which states that the cross products of a proportion are equal. Read more...iWorksheets :4Study Guides :1 Similarity and scaleSimilarity refers to similar figures and the ability to compare them using proportions. Similar figures have equal corresponding angles and corresponding sides that are in proportion. A proportion equation can be used to prove two figures to be similar. If two figures are similar, the proportion equation can be used to find a missing side of one of the figures. Read more...iWorksheets :7Study Guides :1 ### MD.3.0. Knowledge of Measurement: Students will identify attributes, units, or systems of measurements or apply a variety of techniques, formulas, tools, or technology for determining measurements. #### 3.C.1. Applications in Measurement: Estimate and apply measurement formulas. ##### 3.C.1.a. Estimate and determine the circumference or area of a circle (Assessment limit: Include circles using rational numbers with no more than 2 decimal places (0 - 10,000)). Measurement, Perimeter, and CircumferenceThere are two systems used to measure objects, the U.S. Customary system and the metric system. The U.S. Customary system measures length in inches, feet, yards and miles. The metric system is a base ten system and measures length in kilometers, meters, and millimeters. Perimeter is the measurement of the distance around a figure. It is measured in units and can be measured by inches, feet, blocks, meters, centimeters or millimeters. To get the perimeter of any figure, simply add up the measures of the sides of the figure. Read more...iWorksheets :3Study Guides :1 Exploring Area and Surface AreaArea is the amount of surface a shape covers. Area is measured in square units, whether the units are inches, feet, meters or centimeters. The area formula for a triangle is: A = 1/2 · b · h, where b is the base and h is the height. The area formula for a circle is: A = π · r², where π is usually 3.14 and r is the radius of the circle. The area formula for a parallelogram is: A = b · h, where b is the base and h is the height. Read more...iWorksheets :4Study Guides :1 Perimeter and areaWhat Is Perimeter and Area? Perimeter is the measurement of the distance around a figure. It is measured in units and can be measured by inches, feet, blocks, meters, centimeters or millimeters. To find the perimeter of any figure, simply add up the measures of the sides of the figure. Area is the amount of surface a shape covers. Area is measured in square units, whether the units are inches, feet, meters or centimeters. The area formula for a parallelogram is: A = b · h, where b is the base and h is the height. Read more...iWorksheets :4Study Guides :1 ##### 3.C.1.c. Estimate and determine the volume of a cylinder (Assessment limit: Use cylinders, the given the formula, and whole number dimensions (0 - 10,000)). Finding VolumeVolume measures the amount a solid figure can hold. Volume is measured in terms of cubed units and can be measured in inches, feet, meters, centimeters, and millimeters. The formula for the volume of a rectangular prism is V = l · w · h, where l is the length, w is the width, and h is the height. Read more...iWorksheets :4Study Guides :1 Three dimensional geometry/MeasurementThree-dimensional geometry/measurement refers to three-dimensional (3D) shapes and the measurement of their shapes concerning volume and surface area. The figures of prisms, cylinders, pyramids, cones and spheres are all 3D figures. Volume measures the amount a solid figure can hold. Volume is measured in terms of units³ and can be measured in inches, feet, meters, centimeters, and millimeters. Read more...iWorksheets :11Study Guides :1 ##### 3.C.1.d. Determine the volume of cones, pyramids, and spheres. Finding VolumeVolume measures the amount a solid figure can hold. Volume is measured in terms of cubed units and can be measured in inches, feet, meters, centimeters, and millimeters. The formula for the volume of a rectangular prism is V = l · w · h, where l is the length, w is the width, and h is the height. Read more...iWorksheets :4Study Guides :1 Three dimensional geometry/MeasurementThree-dimensional geometry/measurement refers to three-dimensional (3D) shapes and the measurement of their shapes concerning volume and surface area. The figures of prisms, cylinders, pyramids, cones and spheres are all 3D figures. Volume measures the amount a solid figure can hold. Volume is measured in terms of units³ and can be measured in inches, feet, meters, centimeters, and millimeters. Read more...iWorksheets :11Study Guides :1 ##### 3.C.1.e. Determine the surface area of cylinders, prisms, and pyramids. Exploring Area and Surface AreaArea is the amount of surface a shape covers. Area is measured in square units, whether the units are inches, feet, meters or centimeters. The area formula for a triangle is: A = 1/2 · b · h, where b is the base and h is the height. The area formula for a circle is: A = π · r², where π is usually 3.14 and r is the radius of the circle. The area formula for a parallelogram is: A = b · h, where b is the base and h is the height. Read more...iWorksheets :4Study Guides :1 Three dimensional geometry/MeasurementThree-dimensional geometry/measurement refers to three-dimensional (3D) shapes and the measurement of their shapes concerning volume and surface area. The figures of prisms, cylinders, pyramids, cones and spheres are all 3D figures. Volume measures the amount a solid figure can hold. Volume is measured in terms of units³ and can be measured in inches, feet, meters, centimeters, and millimeters. Read more...iWorksheets :11Study Guides :1 #### 3.C.2. Applications in Measurement: Analyze measurement relationships. ##### 3.C.2.a. Use proportional reasoning to solve measurement problems (Assessment limit: Use proportions, scale drawings with scales as whole numbers, or rates using whole numbers or decimals (0 - 1000)). Numerical ProportionsNumerical proportions compare two numbers. The numbers can have the same units such as a ratio or the numbers can have different units such as rates. A proportion is usually in the form of a:b or a/b. Ratios are used to compare objects, wins and losses, sides of a figure to its area and many more. Rates are used to compare miles per hour, words per minute, and many others. A unit rate is when the denominator of a proportion is one. Read more...iWorksheets :4Study Guides :1 Geometric ProportionsGeometric proportions compare two similar polygons. Similar polygons have equal corresponding angles and corresponding sides that are in proportion. A proportion equation can be used to prove two figures to be similar. If two figures are similar, the proportion equation can be used to find a missing side of one of the figures. Read more...iWorksheets :4Study Guides :1 Ratios, proportions and percentsNumerical proportions compare two numbers. A proportion is usually in the form of a:b or a/b. There are 4 parts to a proportion and it can be solved when 3 of the 4 parts are known. Proportions can be solved using the Cross Product Property, which states that the cross products of a proportion are equal. Read more...iWorksheets :4Study Guides :1 Similarity and scaleSimilarity refers to similar figures and the ability to compare them using proportions. Similar figures have equal corresponding angles and corresponding sides that are in proportion. A proportion equation can be used to prove two figures to be similar. If two figures are similar, the proportion equation can be used to find a missing side of one of the figures. Read more...iWorksheets :7Study Guides :1 ### MD.4.0. Knowledge of Statistics: Students will collect, organize, display, analyze, or interpret data to make decisions or predictions. #### 4.A.1. Data Displays: Organize and display data. ##### 4.A.1.b. Organize and display data to make box-and-whisker plots (Assessment limit: Use no more than 12 pieces of data and whole numbers (0 - 1000)). Using graphs to analyze dataThere are different types of graphs and ways that data can be analyzed using the graphs. Graphs are based on the coordinate plane. Data are the points on the plane. If collecting data about the ages of people living on one street, the data is all the ages. The data can then be organized into groups, and evaluated. Mean, mode and median are different ways to evaluate data. Read more...iWorksheets :7Study Guides :1 ##### 4.A.1.c. Organize and display data to make a scatter plot (Assessment limit: Use no more than 10 points and whole numbers (0 - 1000)). Analyzing, Graphing and Displaying DataThere are many types of graphs such as, bar graphs, histograms and line graphs. A bar graph compares data in categories and uses bars, either vertical or horizontal. A histogram is similar to a bar graph, but with histograms the bars touch each other where with bar graphs the bars do not touch each other. A line graph is useful for graphing how data changes over time. With a line graph, data is plotted as points and lines are drawn to connect the points to show how the data changes. Read more...iWorksheets :6Study Guides :1 Using graphs to analyze dataThere are different types of graphs and ways that data can be analyzed using the graphs. Graphs are based on the coordinate plane. Data are the points on the plane. If collecting data about the ages of people living on one street, the data is all the ages. The data can then be organized into groups, and evaluated. Mean, mode and median are different ways to evaluate data. Read more...iWorksheets :7Study Guides :1 Displaying dataDisplaying data refers to the many ways that data can be displayed whether it is on a bar graph, line graph, circle graph, pictograph, line plot, scatter plot or another way. Certain data is better displayed with different graphs as opposed to other graphs. E.g. if data representing the cost of a movie over the past 5 years were to be displayed, a line graph would be best. A circle graph would not be appropriate to use because a circle graph represents data that can add up to one or 100%. Read more...iWorksheets :4Study Guides :1 #### 4.B.1. Data Analysis: Analyze data. ##### 4.B.1.a. Interpret tables (Assessment limit: Use no more than 5 categories having no more than 2 quantities per category and whole numbers or decimals with no more than 2 decimal places (0 - 100)). Organizing DataThe data can be organized into groups, and evaluated. Mean, mode, median and range are different ways to evaluate data. The mean is the average of the data. The mode refers to the number that occurs the most often in the data. The median is the middle number when the data is arranged in order from lowest to highest. The range is the difference in numbers when the lowest number is subtracted from the highest number. Data can be organized into a table, such as a frequency table. Read more...iWorksheets :3Study Guides :1 Analyzing, Graphing and Displaying DataThere are many types of graphs such as, bar graphs, histograms and line graphs. A bar graph compares data in categories and uses bars, either vertical or horizontal. A histogram is similar to a bar graph, but with histograms the bars touch each other where with bar graphs the bars do not touch each other. A line graph is useful for graphing how data changes over time. With a line graph, data is plotted as points and lines are drawn to connect the points to show how the data changes. Read more...iWorksheets :6Study Guides :1 Using graphs to analyze dataThere are different types of graphs and ways that data can be analyzed using the graphs. Graphs are based on the coordinate plane. Data are the points on the plane. If collecting data about the ages of people living on one street, the data is all the ages. The data can then be organized into groups, and evaluated. Mean, mode and median are different ways to evaluate data. Read more...iWorksheets :7Study Guides :1 ##### 4.B.1.b. Interpret box-and-whisker plots (Assessment limit: Use minimum, first (lower) quartile, median (middle quartile, third (upper) quartile, or maximum and whole numbers (0 - 100)). Using graphs to analyze dataThere are different types of graphs and ways that data can be analyzed using the graphs. Graphs are based on the coordinate plane. Data are the points on the plane. If collecting data about the ages of people living on one street, the data is all the ages. The data can then be organized into groups, and evaluated. Mean, mode and median are different ways to evaluate data. Read more...iWorksheets :7Study Guides :1 ##### 4.B.1.c. Interpret scatter plots (Assessment limit: Use no more than 10 points using whole numbers or decimals with no more than 2 decimal places (0 - 100)). Analyzing, Graphing and Displaying DataThere are many types of graphs such as, bar graphs, histograms and line graphs. A bar graph compares data in categories and uses bars, either vertical or horizontal. A histogram is similar to a bar graph, but with histograms the bars touch each other where with bar graphs the bars do not touch each other. A line graph is useful for graphing how data changes over time. With a line graph, data is plotted as points and lines are drawn to connect the points to show how the data changes. Read more...iWorksheets :6Study Guides :1 Using graphs to analyze dataThere are different types of graphs and ways that data can be analyzed using the graphs. Graphs are based on the coordinate plane. Data are the points on the plane. If collecting data about the ages of people living on one street, the data is all the ages. The data can then be organized into groups, and evaluated. Mean, mode and median are different ways to evaluate data. Read more...iWorksheets :7Study Guides :1 Displaying dataDisplaying data refers to the many ways that data can be displayed whether it is on a bar graph, line graph, circle graph, pictograph, line plot, scatter plot or another way. Certain data is better displayed with different graphs as opposed to other graphs. E.g. if data representing the cost of a movie over the past 5 years were to be displayed, a line graph would be best. A circle graph would not be appropriate to use because a circle graph represents data that can add up to one or 100%. Read more...iWorksheets :4Study Guides :1 ##### 4.B.1.d. Interpret circle graphs (Assessment limit: Use no more than 8 categories (0 - 1000)). Plane Figures: Lines and AnglesPlane figures in regards to lines and angles refer to the coordinate plane and the various lines and angles within the coordinate plane. Lines in a coordinate plane can be parallel or perpendicular. Angles in a coordinate plane can be acute, obtuse, right or straight. Adjacent angles are two angles that have a common vertex and a common side but do not overlap. Read more...iWorksheets :3Study Guides :1 Plane Figures: Closed Figure RelationshipsPlane figures in regards to closed figure relationships refer to the coordinate plane and congruent figures, circles, circle graphs, transformations and symmetry. Congruent figures have the same size and shape. Transformations are made up of translations, rotations and reflections. A translation of a figure keeps the size and shape of a figure, but moves it to a different location. A rotation turns a figure about a point on the figure. A reflection of a figure produces a mirror image of the figure when it is reflected in a given line. Read more...iWorksheets :3Study Guides :1 Using graphs to analyze dataThere are different types of graphs and ways that data can be analyzed using the graphs. Graphs are based on the coordinate plane. Data are the points on the plane. If collecting data about the ages of people living on one street, the data is all the ages. The data can then be organized into groups, and evaluated. Mean, mode and median are different ways to evaluate data. Read more...iWorksheets :7Study Guides :1 ##### 4.B.1.e. Analyze multiple box-and-whisker plots using the same scale. Using graphs to analyze dataThere are different types of graphs and ways that data can be analyzed using the graphs. Graphs are based on the coordinate plane. Data are the points on the plane. If collecting data about the ages of people living on one street, the data is all the ages. The data can then be organized into groups, and evaluated. Mean, mode and median are different ways to evaluate data. Read more...iWorksheets :7Study Guides :1 ### MD.5.0. Knowledge of Probability: Students will use experimental methods or theoretical reasoning to determine probabilities to make predictions or solve problems about events whose outcomes involve random variation. #### 5.A.1. Sample Space: Identify a sample space. ##### 5.A.1.a. Describe the difference between independent and dependent events. Using ProbabilityProbability is the possibility that a certain event will occur. Probability is the chance of an event occurring divided by the total number of possible outcomes. Probability is based on whether events are dependent or independent of each other. An independent event refers to the outcome of one event not affecting the outcome of another event. A dependent event is when the outcome of one event does affect the outcome of the other event. Probability word problems. Read more...iWorksheets :3Study Guides :1 Theoretical probability and countingProbability word problems worksheets. Theoretical probability is the probability that a certain outcome will occur based on all the possible outcomes. Sometimes, the number of ways that an event can happen depends on the order. A permutation is an arrangement of objects in which order matters. A combination is a set of objects in which order does not matter. Probability is also based on whether events are dependent or independent of each other. Read more...iWorksheets :3Study Guides :1 #### 5.B.1. Theoretical Probability: Determine the probability of an event comprised of no more than 2 independent events. ##### 5.B.1.a. Express the probability of an event as a fraction, a decimal, or a percent (Assessment limit: Use a sample space of 36 to 60 outcomes). Introduction to ProbabilityProbability is the possibility that a certain event will occur. An event that is certain to occur has a probability of 1. An event that cannot occur has a probability of 0. Therefore, the probability of an event occurring is always between 0 and 1. Probability word problems worksheets. Read more...iWorksheets :4Study Guides :1 Using ProbabilityProbability is the possibility that a certain event will occur. Probability is the chance of an event occurring divided by the total number of possible outcomes. Probability is based on whether events are dependent or independent of each other. An independent event refers to the outcome of one event not affecting the outcome of another event. A dependent event is when the outcome of one event does affect the outcome of the other event. Probability word problems. Read more...iWorksheets :3Study Guides :1 Ratios, proportions and percentsNumerical proportions compare two numbers. A proportion is usually in the form of a:b or a/b. There are 4 parts to a proportion and it can be solved when 3 of the 4 parts are known. Proportions can be solved using the Cross Product Property, which states that the cross products of a proportion are equal. Read more...iWorksheets :4Study Guides :1 Experimental ProbabilityFreeExperimental probability is the probability that a certain outcome will occur based on an experiment being performed multiple times. Probability word problems worksheets. Read more...iWorksheets :3Study Guides :1 #### 5.B.2. Theoretical Probability Determine the probability of a second event that is dependent on a first event of equally likely outcomes. ##### 5.B.2.a. Express the probability as a fraction, a decimal, or a percent (Assessment limit: Use a sample space of no more than 60 outcomes). Introduction to ProbabilityProbability is the possibility that a certain event will occur. An event that is certain to occur has a probability of 1. An event that cannot occur has a probability of 0. Therefore, the probability of an event occurring is always between 0 and 1. Probability word problems worksheets. Read more...iWorksheets :4Study Guides :1 Using ProbabilityProbability is the possibility that a certain event will occur. Probability is the chance of an event occurring divided by the total number of possible outcomes. Probability is based on whether events are dependent or independent of each other. An independent event refers to the outcome of one event not affecting the outcome of another event. A dependent event is when the outcome of one event does affect the outcome of the other event. Probability word problems. Read more...iWorksheets :3Study Guides :1 Ratios, proportions and percentsNumerical proportions compare two numbers. A proportion is usually in the form of a:b or a/b. There are 4 parts to a proportion and it can be solved when 3 of the 4 parts are known. Proportions can be solved using the Cross Product Property, which states that the cross products of a proportion are equal. Read more...iWorksheets :4Study Guides :1 Experimental ProbabilityFreeExperimental probability is the probability that a certain outcome will occur based on an experiment being performed multiple times. Probability word problems worksheets. Read more...iWorksheets :3Study Guides :1 #### 5.C.1. Experimental Probability: Analyze the results of a survey or simulation. ##### 5.C.1.a. Make predictions and express the probability of the results as a fraction, a decimal with no more than 2 decimal places, or a percent (Assessment limit: Use 20 to 500 results). Introduction to ProbabilityProbability is the possibility that a certain event will occur. An event that is certain to occur has a probability of 1. An event that cannot occur has a probability of 0. Therefore, the probability of an event occurring is always between 0 and 1. Probability word problems worksheets. Read more...iWorksheets :4Study Guides :1 Using ProbabilityProbability is the possibility that a certain event will occur. Probability is the chance of an event occurring divided by the total number of possible outcomes. Probability is based on whether events are dependent or independent of each other. An independent event refers to the outcome of one event not affecting the outcome of another event. A dependent event is when the outcome of one event does affect the outcome of the other event. Probability word problems. Read more...iWorksheets :3Study Guides :1 Ratios, proportions and percentsNumerical proportions compare two numbers. A proportion is usually in the form of a:b or a/b. There are 4 parts to a proportion and it can be solved when 3 of the 4 parts are known. Proportions can be solved using the Cross Product Property, which states that the cross products of a proportion are equal. Read more...iWorksheets :4Study Guides :1 Experimental ProbabilityFreeExperimental probability is the probability that a certain outcome will occur based on an experiment being performed multiple times. Probability word problems worksheets. Read more...iWorksheets :3Study Guides :1 #### 5.C.2. Experimental Probability: Conduct a probability experiment. Introduction to ProbabilityProbability is the possibility that a certain event will occur. An event that is certain to occur has a probability of 1. An event that cannot occur has a probability of 0. Therefore, the probability of an event occurring is always between 0 and 1. Probability word problems worksheets. Read more...iWorksheets :4Study Guides :1 Experimental ProbabilityFreeExperimental probability is the probability that a certain outcome will occur based on an experiment being performed multiple times. Probability word problems worksheets. Read more...iWorksheets :3Study Guides :1 #### 5.C.3. Experimental Probability: Compare outcomes of theoretical probability with the results of experimental probability. Introduction to ProbabilityProbability is the possibility that a certain event will occur. An event that is certain to occur has a probability of 1. An event that cannot occur has a probability of 0. Therefore, the probability of an event occurring is always between 0 and 1. Probability word problems worksheets. Read more...iWorksheets :4Study Guides :1 #### 5.C.4. Experimental Probability: Describe the difference between theoretical and experimental probability. Introduction to ProbabilityProbability is the possibility that a certain event will occur. An event that is certain to occur has a probability of 1. An event that cannot occur has a probability of 0. Therefore, the probability of an event occurring is always between 0 and 1. Probability word problems worksheets. Read more...iWorksheets :4Study Guides :1 Experimental ProbabilityFreeExperimental probability is the probability that a certain outcome will occur based on an experiment being performed multiple times. Probability word problems worksheets. Read more...iWorksheets :3Study Guides :1 Theoretical probability and countingProbability word problems worksheets. Theoretical probability is the probability that a certain outcome will occur based on all the possible outcomes. Sometimes, the number of ways that an event can happen depends on the order. A permutation is an arrangement of objects in which order matters. A combination is a set of objects in which order does not matter. Probability is also based on whether events are dependent or independent of each other. Read more...iWorksheets :3Study Guides :1 ### MD.6.0. Knowledge of Number Relationships and Computation/Arithmetic: Students will describe, represent, or apply numbers or their relationships or will estimate or compute using mental strategies, paper/pencil, or technology. #### 6.A.1. Knowledge of Number and Place Value: Apply knowledge of rational numbers and place value. ##### 6.A.1.a. Read, write, and represent rational numbers (Assessment limit: Use exponential notation or scientific notation (-10,000 to 1,000,000,000)). Exponents, Factors and FractionsFreeIn a mathematical expression where the same number is multiplied many times, it is often useful to write the number as a base with an exponent. Exponents are also used to evaluate numbers. Any number to a zero exponent is 1 and any number to a negative exponent is a number less than 1. Exponents are used in scientific notation to make very large or very small numbers easier to write. Read more...iWorksheets :8Study Guides :1 Polynomials and ExponentsFreeA polynomial is an expression which is in the form of ax<sup>n</sup>, where a is any real number and n is a whole number. If a polynomial has only one term, it is called a monomial. If it has two terms, it is a binomial and if it has three terms, it is a trinomial. The standard form of a polynomial is when the powers of the variables are decreasing from left to right. Read more...iWorksheets :6Study Guides :1 ##### 6.A.1.b. Compare, order, and describe rational numbers with and without relational symbols (<, >, =) (Assessment limit: Use no more than 4 integers (-100 to 100) or positive rational numbers (0-100) using equivalent forms or absolute value). Rational and Irrational NumbersA rational number is a number that can be made into a fraction. Decimals that repeat or terminate are rational because they can be changed into fractions. An irrational number is a number that cannot be made into a fraction. Decimals that do not repeat or end are irrational numbers. Pi is an irrational number. Read more...iWorksheets :3Study Guides :1 Exponents, Factors and FractionsFreeIn a mathematical expression where the same number is multiplied many times, it is often useful to write the number as a base with an exponent. Exponents are also used to evaluate numbers. Any number to a zero exponent is 1 and any number to a negative exponent is a number less than 1. Exponents are used in scientific notation to make very large or very small numbers easier to write. Read more...iWorksheets :8Study Guides :1 #### 6.C.1. Number Computation: Analyze number relations and compute. ##### 6.C.1.a. Add, subtract, multiply and divide integers (Assessment limit: Use one operation (-1000 to 1000)). Using IntegersIntegers are negative numbers, zero and positive numbers. To compare integers, a number line can be used. On a number line, negative integers are on the left side of zero with the larger a negative number, the farther to the left it is. Positive integers are on the right side of zero on the number line. If a number is to the left of another number it is said to be less than that number. In the coordinate plane, the x-axis is a horizontal line with negative numbers, zero and positive numbers. Read more...iWorksheets :4Study Guides :1 Integer operationsInteger operations are the mathematical operations that involve integers. Integers are negative numbers, zero and positive numbers. Adding and subtracting integers are useful in everyday life because there are many situations that involved negative numbers such as calculating sea level or temperatures. Equations with integers are solved using inverse operations. Addition and subtraction are inverse operations, and multiplication and division are inverse operations of each other. Read more...iWorksheets :4Study Guides :1 ##### 6.C.1.b. Calculate powers of integers and square roots of perfect square whole numbers (Assessment limit: Use powers with bases no more than 12 and exponents no more than 3, or square roots of perfect squares no more than 144). Rational and Irrational NumbersA rational number is a number that can be made into a fraction. Decimals that repeat or terminate are rational because they can be changed into fractions. An irrational number is a number that cannot be made into a fraction. Decimals that do not repeat or end are irrational numbers. Pi is an irrational number. Read more...iWorksheets :3Study Guides :1 Exponents, Factors and FractionsFreeIn a mathematical expression where the same number is multiplied many times, it is often useful to write the number as a base with an exponent. Exponents are also used to evaluate numbers. Any number to a zero exponent is 1 and any number to a negative exponent is a number less than 1. Exponents are used in scientific notation to make very large or very small numbers easier to write. Read more...iWorksheets :8Study Guides :1 The Pythagorean TheoremPythagorean Theorem is a fundamental relation in Euclidean geometry. It states the sum of the squares of the legs of a right triangle equals the square of the length of the hypotenuse. Determine the distance between two points using the Pythagorean Theorem. Read more...iWorksheets :10Study Guides :2 Real numbersReal numbers are the set of rational and irrational numbers. The set of rational numbers includes integers, whole numbers, and natural numbers. A rational number is a number that can be made into a fraction. Decimals that repeat or terminate are rational because they can be changed into fractions. An irrational number is a number that cannot be made into a fraction. Decimals that do not repeat or end are irrational numbers. Read more...iWorksheets :4Study Guides :1 ##### 6.C.1.c. Identify and use the laws of exponents to simplify expressions (Assessment limit: Use the rules of power times power or power divided by power with the same integer as a base (-20 to 20) and exponents (0-10)). Polynomials and ExponentsFreeA polynomial is an expression which is in the form of ax<sup>n</sup>, where a is any real number and n is a whole number. If a polynomial has only one term, it is called a monomial. If it has two terms, it is a binomial and if it has three terms, it is a trinomial. The standard form of a polynomial is when the powers of the variables are decreasing from left to right. Read more...iWorksheets :6Study Guides :1 ##### 6.C.1.d. Use properties of addition and multiplication to simplify expressions (Assessment limit: Use the commutative property of addition or multiplication, associative property of addition or multiplication, additive inverse property, the distributive property, or the identity property for one or zero with integers (-100 to 100). Using IntegersIntegers are negative numbers, zero and positive numbers. To compare integers, a number line can be used. On a number line, negative integers are on the left side of zero with the larger a negative number, the farther to the left it is. Positive integers are on the right side of zero on the number line. If a number is to the left of another number it is said to be less than that number. In the coordinate plane, the x-axis is a horizontal line with negative numbers, zero and positive numbers. Read more...iWorksheets :4Study Guides :1 #### 6.C.3. Number Computation: Analyze ratios, proportions, and percents. ##### 6.C.3.a. Determine unit rates (Assessment limit: Use positive rational numbers (0 - 100)). Numerical ProportionsNumerical proportions compare two numbers. The numbers can have the same units such as a ratio or the numbers can have different units such as rates. A proportion is usually in the form of a:b or a/b. Ratios are used to compare objects, wins and losses, sides of a figure to its area and many more. Rates are used to compare miles per hour, words per minute, and many others. A unit rate is when the denominator of a proportion is one. Read more...iWorksheets :4Study Guides :1 Ratios, proportions and percentsNumerical proportions compare two numbers. A proportion is usually in the form of a:b or a/b. There are 4 parts to a proportion and it can be solved when 3 of the 4 parts are known. Proportions can be solved using the Cross Product Property, which states that the cross products of a proportion are equal. Read more...iWorksheets :4Study Guides :1 ##### 6.C.3.b. Determine or use percents, rates of increase and decrease, discount, commission, sales tax, and simple interest in the context of a problem (Assessment limit: Use positive rational numbers (0 - 10,000)). Introduction to PercentWhat Is Percent? A percent is a term that describes a decimal in terms of one hundred. Percent means per hundred. Percents, fractions and decimals all can equal each other, as in the case of 10%, 0.1 and 1/10. Percents can be greater than 100% or smaller than 1%. A markup from the cost of making an item to the actual sales price is usually greater than 100%. A salesperson's commission might be 1/2% depending on the item sold. Read more...iWorksheets :4Study Guides :1 Applying PercentsApplying percents is a term that refers to the different ways that percents can be used. The percent of change refers to the percent an amount either increases or decreases based on the previous amounts or numbers. Applying percents also means to calculate simple interest using the interest equation, I = P · r · t, where P is the principal; r is the rate and t is the time. Read more...iWorksheets :3Study Guides :1 Applications of percentPercent increase or decrease can be found by using the formula: percent of change = actual change/original amount. The change is either an increase, if the amounts went up or a decrease if the amounts went down. If a number changes from 33 to 89, the percent of increase would be: Percent of increase = (89 -33) ÷ 33 = 56 ÷ 33 ≈ 1.6969 ≈ 170% Read more...iWorksheets :4Study Guides :1 ##### 6.C.3.c. Solve problems using proportional reasoning (Assessment limit: Use positive rational numbers (0 - 1000)). Numerical ProportionsNumerical proportions compare two numbers. The numbers can have the same units such as a ratio or the numbers can have different units such as rates. A proportion is usually in the form of a:b or a/b. Ratios are used to compare objects, wins and losses, sides of a figure to its area and many more. Rates are used to compare miles per hour, words per minute, and many others. A unit rate is when the denominator of a proportion is one. Read more...iWorksheets :4Study Guides :1 Ratios, proportions and percentsNumerical proportions compare two numbers. A proportion is usually in the form of a:b or a/b. There are 4 parts to a proportion and it can be solved when 3 of the 4 parts are known. Proportions can be solved using the Cross Product Property, which states that the cross products of a proportion are equal. Read more...iWorksheets :4Study Guides :1 Similarity and scaleSimilarity refers to similar figures and the ability to compare them using proportions. Similar figures have equal corresponding angles and corresponding sides that are in proportion. A proportion equation can be used to prove two figures to be similar. If two figures are similar, the proportion equation can be used to find a missing side of one of the figures. Read more...iWorksheets :7Study Guides :1 Numbers and percentsNumbers and percents refer to the relationship between fractions, decimals, and percents. A percent is a term that describes a decimal in terms of one hundred. Percent means per hundred. Percents, fractions and decimals all can equal each other, as in the case of 10%, 0.1 and 1/10. Fractions and decimals can easily be changed into percent. There are three cases of percent. Read more...iWorksheets :3Study Guides :1 ### MD.7.0. Processes of Mathematics: Students demonstrate the processes of mathematics by making connections and applying reasoning to solve problems and to communicate their findings. #### 7.A.1. Problem Solving: Apply a variety of concepts, processes, and skills to solve problems ##### 7.A.1.c. Make a plan to solve a problem Applying PercentsApplying percents is a term that refers to the different ways that percents can be used. The percent of change refers to the percent an amount either increases or decreases based on the previous amounts or numbers. Applying percents also means to calculate simple interest using the interest equation, I = P · r · t, where P is the principal; r is the rate and t is the time. Read more...iWorksheets :3Study Guides :1 ##### 7.A.1.d. Apply a strategy, i.e., draw a picture, guess and check, finding a pattern, writing an equation Applying PercentsApplying percents is a term that refers to the different ways that percents can be used. The percent of change refers to the percent an amount either increases or decreases based on the previous amounts or numbers. Applying percents also means to calculate simple interest using the interest equation, I = P · r · t, where P is the principal; r is the rate and t is the time. Read more...iWorksheets :3Study Guides :1 Algebraic EquationsWhat are algebraic equations? Algebraic equations are mathematical quations that contain a letter or variable, which represents a number. When algebraic equations are written in words, the words must be changed into the appropriate numbers and variable in order to solve. Read more...iWorksheets :5Study Guides :1 ##### 7.A.1.e. Select a strategy, i.e., draw a picture, guess and check, finding a pattern, writing an equation Applying PercentsApplying percents is a term that refers to the different ways that percents can be used. The percent of change refers to the percent an amount either increases or decreases based on the previous amounts or numbers. Applying percents also means to calculate simple interest using the interest equation, I = P · r · t, where P is the principal; r is the rate and t is the time. Read more...iWorksheets :3Study Guides :1 Algebraic EquationsWhat are algebraic equations? Algebraic equations are mathematical quations that contain a letter or variable, which represents a number. When algebraic equations are written in words, the words must be changed into the appropriate numbers and variable in order to solve. Read more...iWorksheets :5Study Guides :1 ##### 7.A.1.f. Identify alternative ways to solve a problem Applying PercentsApplying percents is a term that refers to the different ways that percents can be used. The percent of change refers to the percent an amount either increases or decreases based on the previous amounts or numbers. Applying percents also means to calculate simple interest using the interest equation, I = P · r · t, where P is the principal; r is the rate and t is the time. Read more...iWorksheets :3Study Guides :1 #### 7.B.1. Reasoning: Justify ideas or solutions with mathematical concepts or proofs ##### 7.B.1.a. Use inductive or deductive reasoning Mathematical processesMathematical processes refer to the skills and strategies needed in order to solve mathematical problems. If one strategy does not help to find the solution to a problem, using another strategy may help to solve it. Problem solving skills refer to the math techniques that must be used to solve a problem. If a problem were to determine the perimeter of a square, a needed skill would be the knowledge of what perimeter means and the ability to add the numbers. Read more...iWorksheets :3Study Guides :1 ##### 7.B.1.b. Make or test generalizations Mathematical processesMathematical processes refer to the skills and strategies needed in order to solve mathematical problems. If one strategy does not help to find the solution to a problem, using another strategy may help to solve it. Problem solving skills refer to the math techniques that must be used to solve a problem. If a problem were to determine the perimeter of a square, a needed skill would be the knowledge of what perimeter means and the ability to add the numbers. Read more...iWorksheets :3Study Guides :1 ##### 7.B.1.d. Use methods of proof, i.e., direct, indirect, paragraph, or contradiction Mathematical processesMathematical processes refer to the skills and strategies needed in order to solve mathematical problems. If one strategy does not help to find the solution to a problem, using another strategy may help to solve it. Problem solving skills refer to the math techniques that must be used to solve a problem. If a problem were to determine the perimeter of a square, a needed skill would be the knowledge of what perimeter means and the ability to add the numbers. Read more...iWorksheets :3Study Guides :1 #### 7.D.1. Connections: Relate or apply mathematics within the discipline, to other disciplines, and to life ##### 7.D.1.c. Identify mathematical concepts in relationship to life Mathematical processesMathematical processes refer to the skills and strategies needed in order to solve mathematical problems. If one strategy does not help to find the solution to a problem, using another strategy may help to solve it. Problem solving skills refer to the math techniques that must be used to solve a problem. If a problem were to determine the perimeter of a square, a needed skill would be the knowledge of what perimeter means and the ability to add the numbers. Read more...iWorksheets :3Study Guides :1
Search 74,167 tutors # Areas of Circles and Sectors In order to work on the final section of our study of areas, we must first learn about a shape that we have not discussed at all in the past. In fact, this shape is not a polygon at all, which means that it doesn't have vertices or sides. The shape we are going to focus on in this section is called a circle. Let's begin this lesson by examining the definition of a circle. Definition: A circle is a plane curve that is equidistant from a given fixed point, called the center. Every point on the curve is a distance of x units away from center C. There are countless properties we can use in order to describe circles, but we will just focus on two of them: radius and diameter. Although we do not directly use diameter to find the area of a circle, understanding how it compares to the radius can help us figure out areas of circles. Let's look at the definitions of radius and diameter, as well as the illustration below to see how they relate. Definition: A radius is a line segment that joins the center of a circle with any point on the circumference of the circle. Definition: A diameter is a straight line segment that passes through the center of a circle. Notice that the radius is only extended from the center of the circle to the outside edge of the circle, whereas the diameter goes all the way through from one side to the other. Because the definition of a circle describes the locus of points that are equidistant from the center, we know that all of the radians of a circle are equal. So, we can essentially just break down the diameter at the center of the circle to create two radians. Therefore, we know that where d is the length of the diameter of a circle and r is the length of the radius. Because all of the radians of a circle are equal, we know that two of them make up the diameter of a circle. Now that we've discussed the important parts of a circle, we can learn how to measure the areas of circles. Because circles don't have sharp edges, their areas in square units will almost never come out to an even number, so we will just round areas to the hundredths place. Let's learn how to use the area formula for circles. ## Areas of Circles The area of a circle is equal to the product of pi (?), and the radius of the circle squared. We describe this mathematically as where A is the area in square units, ? is a mathematical constant (approximately equal to 3.14), and r is the radius. Recall that we were able to put the diameter of a circle in terms of its radius. Because every problem will not give us the radius of a circle, we might need to use our knowledge of their diameters to help us figure out their areas. In other words, if we are given the diameter of a circle, we know that half of the diameter is equal to the radius, which we can plug into our area formula. Let's work on some exercises now. ### Exercise 1 Find the area of circle p. Let's analyze the information we've been given in order figure out what we must do to find the area of circle p. We are given that the diameter is 18 inches, and we know that the diameter of a circle is twice and long as its radius, so we all we have to do in order to find the radius is take half of the diameter. We see that the radius of our circle is 9 inches. Remember, that ? is not a variable; it is a mathematical constant. Also, we will not worry about much precision when it comes to the value of ? . We can just define ? as 3.14 since our final answer will be rounded to the hundredths place. We are ready to solve for the area of circle p. So, the area of circle p is approximately 254.34 square inches. Now, let's look at another example that will require a bit more work. ### Exercise 2 Find the value of x if the area of circle g is approximately 1661 square feet. In this example, we have been given the area of circle g, so we will have to work backwards in order to find the radius. Let's fill out the area formula, substituting in for the variables we know. While ? is not equal to 3.14, we will use 3.14 as an approximate value to help us solve for x. We have In order to get rid of the square, we need to take the square root of both sides: We know that the radius of circle g is approximately 23 feet, but we haven't solved for the variable x. We just need to subtract 7 from both sides of the equation, and we get So, the value of x is approximately 16. Let's learn about sectors of circles now. ## Areas of Sectors Sometimes, we will not want to find the areas of full circles and instead need to find smaller sections of a circle. In these cases, we will need a way to calculate these parts of circles called sectors. Let's study the definition of sectors and see what they look like before we introduce the area formula. Definition: A circular sector is the portion of a circle enclosed by two radii and an arc of the circle. Notice that the arc of a circle is just the part of the circumference enclosed by the endpoints of both radians. Working with the sectors of circles can be quite simple if we know how to apply the area formula for circles. If we know that the circle is split up into a certain amount of congruent areas, we can just put the corresponding factor into our area formula. For instance, if we have a circle that is split up into four equal sections, and we want to find the area of one of those sections, our area formula would be Because one-fourth of the circle is shaded, we just multiply the area formula of circle c by a factor of ¼ to find the area of the circle sector. In other cases, we may be given the measure of the angle at the radius of a circle, called the central angle. For those exercises, we can apply the area of sectors formula, which is where A represents area, x represents the degree measure of the central angle, and r is the radius. This formula essentially does the same as what we've done in the previous example because it just converts the degree measure of the interior angle into an equivalent fraction. Circles have degree measures of 360°. Therefore, when we divide a given measure by 360°, we are just taking the fraction of the circle we desire and multiplying it by our regular area formula. Let's take a look at one final example to make sure we understand how to apply the area of sectors formula. ### Exercise 3 Find the area of the shaded sector below. Because circle t has not been split into even sections for us, we cannot just multiply the area formula of circles by a fraction. Rather, we need to use the degree measure of the central angle and plug it into the area of sectors formula. Remember, we need to also use the fact that the radius is 14 meters long in order to solve for the area of the sector. Let's do this now. So, the area of the shaded sector is approximately 230.79 square meters. ## Thinking Further Now, let's go back and analyze the relationship between the degree measure of the central angle and the portion of the circle that was shaded. The first factor of the area formula for sectors eventually simplified to ? because The fact that this fraction simplifies to ? means that the sector's area is three-eighths the area of the entire circle. If we split up circle t into eight congruent pieces, we see that a 135° central angle creates a sector with three-eighths the area of the entire circle. Now, we know how to measure smaller sections of a circle and can compare these sections to the area of the circle as a whole. While we have yet to discuss circles in depth, working with sectors of circles can help us learn how to relate the degrees and radians of circles, which are significant components of subjects like precalculus and calculus. Sign up for free to access more geometry resources like . WyzAnt Resources features blogs, videos, lessons, and more about geometry and over 250 other subjects. Stop struggling and start learning today with thousands of free resources!
• Save Effective Problem On Ages Tricks For Bank And SSC Exams Last year 2.4K Views Mathematics is a complex subject for those students are preparing for SSC and Banking exams. Problem ages is an important topic of this subject. In this topic, we get two and more person ages and ratios or relation between their age. To solve these questions, we have a very short time in the exam. In this blog, I am given some questions to practice and along with a short trick to solve them effectively with speed. By using these tricks you can save your important time during the exams. As well as these tricks, you can visit problems on ages with solutions to know how to solve these type of questions. Problem On Ages Tricks with Solution for Competitive Exams: Example 1: Present age of Sameer and Anand are in the ratio of 5: 4 respectively. Three years hence, the ratio of their ages will become 11: 9 respectively. What is Anand’s present age in years? (A) 24 (B) 27 (C) 40 (D) None of these Solution: In this question, we have to get Anand present age. The ratio of Sameer and Anand is 5: 4. Short trick: We have to divide all the options by the ratio of Anand = 4. Option (A) and (C) values 24 and 27 are fully divided by ratio of Anand. Next, Statement is Three years hence, the ratio of their ages will become 11: 9. So, we have to add 3 year in Option (A) and Option (C). Then the option (A) and option (C) Values are = 24+3 year = 27 years, = 40+3 years = 43 years, Next, Option (A) and (C) are divided of ratio of Anand = 9 Option (A) is fully divided by ratio of Anand 9. Example 2: Present age of A and B are in the ratio of 5: 6 respectively. Seven years hence this ratio will become 6: 7 respectively. What is A’s present age in years? (A) 35 (B) 42 (C) 49 (D) None of these Solution. In this question, we have to get A present age. The ratio of A and B is 5: 6. Short trick We have to divide all the options by the ratio of A = 5. Option (A) value 35 is fully divided by ratio of A. Next, Statement is Seven years hence; the ratio of their ages will become 6: 7. So, we have to added 7 year in Option and Option (A). Then the option (A) Value is = 35+7 year = 42 years, Next, Option (A) is divided of ratio of A = 6 Option (A) is fully divided by ratio of Anand 6. Example 3: The age of two persons differ by 16 years. If 6 years ago, the elder one be 3 times as old as the younger one, find their present ages. Solution Let the age of the younger person be X years. Then, age of the elder person = (x+16) years. ⸫ 3(X -16) = (X+16-6) ↔ 3X – 18 = X+10 ↔2X = 28 ↔ X=14. Hence, their present age are 14 years and 30 years. Example 4: The present age of a father is 3 years more than three times the age of his son. Three years hence, father’s age will be 10 years more than twice the age of the son. Find the present age of the father. Solution ⸫ Let the son’s present age be x years. Then, Father’s present age = (3X+3) years. ⸫ (3x+3+3) = 2(x+3)+10 ↔ 3X+6) = (3×10+3) years = 33 years. Problem On Ages Practice Question: Q.1. Madhav is younger than Mohan by 4 years. If their ages are in the respectively ratio of 7 : 9, how old is Madhav? (A) 16.5 years (B) 18 years (C) 19.5 years (D) 24.5 years Ans . D Q.2. The ratio between the present age of Ram and Shyam is 6 : 7. If Shyam is 4 years old than Ram, what will be the ratio of the ages of Ram and Shyam after 4 years? (A) 7 : 8 (B) 9 : 7 (C) 7 : 6 (D) 8 : 5 Ans . A Q.3. Six years ago, the ratio of the ages Salman and Sharukh Was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sharukh’s age at present ? (A) 16 years (B) 19 years (C) 20 years (D) 14 years. Ans . A Q.4. The total of the ages of Mayank, Lucky and Naman is 93 years. Ten years ago, the ratio of their age was 2 : 3 : 4. What is the present age of Naman? (A) 30 years (B) 35 Years (C) 34 Years (D) 38 Years Ans . D Q.5. Hansraj is 40 Year old and Ronnie is 60 years old. How many years ago was the ratio of their ages 3 : 5 ? (A) 5 Years (B) 10 Years (C) 25 Years (D) 35 Years Ans . B If you want to practice more questions related to Problem on Ages then you can go to the link provided below. In this post I have made every effort to explain the Problem on Ages in a simple way, even if you are having problems in any manner or question, you can Ask me by writing in the comment box below.
# Math Riddles with Answers: Are you Genius? Find All Missing Numbers Math Riddles with Answers: Can you find the missing numbers in these tricky math puzzles? Only 1% of people with a high IQ can get the answer right. Check out these math riddles for geniuses! Math Riddles with Answers: Are you Genius? Find All Missing Numbers Math Riddles are so challenging, but that makes them worthwhile to solve. Math riddles are logical problems that require strong analytical abilities, high IQ, knowledge of math concepts, and good calculation skills. Math riddles have also been known to make Mathematics enjoyable for kids, students, and even adults to solve. Riddles, puzzles, and brain teasers can develop strategic thinking, logical reasoning, and problem-solving skills. Can you solve all the math puzzles? You have 20 seconds for each riddle! ## Math Riddle #5 Also Read: Math Riddles with Answers: Only the 1% Smartest Can Find All Missing Numbers Math Riddle #1 You have to divide 30 by half, not by 2. So, the equation will be = 30 x 2 + 12 = 72 Math Riddle #2 1 + 9 = 10 1 x 9 = 9 Math Riddle #3 1 + 2 + 3 = 6 1 x 2 x 3 = 6 Math Riddle #4 444 + 44 + 4 + 4 + 4 = 500 Math Riddle #5 Column #1 Column #2 Column #3 Row #1 32 24 18 Row #2 16 14 12 Row #3 4 ?? 3 Row #4 128 112 72 To solve this puzzle, we first understand that: For Column #1: row #4 ÷ row #2 = quotient and quotient x row #3 = row #1 128 ÷ 16 = 8 and 8 x 4 = 32 For Column 3: row #4 ÷ row #2 = quotient and quotient x row #3 = row #1 72 ÷ 12 = 6 and 6 x 3 = 18 Similarly, for Column 2, we will find the missing number with the same equation: row #4 ÷ row #2 = quotient and quotient x row #3 = row #1 112 ÷ 14 = 8 and 8 x ?? = 24 (so we know that to get the product as in row #1, we need to multiply the quotient by 3 which will equal 24. ## Tell us in comments: Did you solve all these tricky math riddles? Also Read: Math Riddles: Only High IQ Genius Can Solve These Also Read: Tricky IQ Riddles: Only 1 in 5 people with high IQ can solve these Also Read: What Am I? Riddles: Only Smartest 1% Can Solve These Tricky Puzzles Also Read: World’s Toughest Riddle: Only High IQ Genius Can Solve This Math Riddle
# What is the unit vector that is normal to the plane containing ( i +k) and (i+2j+2k) ? Jun 10, 2017 $\vec{n} = \frac{2}{3} i + \frac{1}{3} j - \frac{2}{3} k$ #### Explanation: The vector we're looking for is $\vec{n} = a \vec{i} + b \vec{j} + c \vec{k}$ where $\vec{n} \cdot \left(i + k\right) = 0$ AND $\vec{n} \cdot \left(i + 2 j + 2 k\right) = 0$, since $\vec{n}$ is perpendicular to both of those vectors. Using this fact, we can make a system of equations: $\vec{n} \cdot \left(i + 0 j + k\right) = 0$ $\left(a i + b j + c k\right) \left(i + 0 j + k\right) = 0$ $a + c = 0$ $\vec{n} \cdot \left(i + 2 j + 2 k\right) = 0$ $\left(a i + b j + c k\right) \cdot \left(i + 2 j + 2 k\right) = 0$ $a + 2 b + 2 c = 0$ Now we have $a + c = 0$ and $a + 2 b + 2 c = 0$, so we can say that: $a + c = a + 2 b + 2 c$ $0 = 2 b + c$ $\therefore a + c = 2 b + c$ $a = 2 b$ $\frac{a}{2} = b$ Now we know that $b = \frac{a}{2}$ and $c = - a$. Therefore, our vector is: $a i + \frac{a}{2} j - a k$ Finally, we need to make this a unit vector, meaning we need to divide each coefficient of the vector by its magnitude. The magnitude is: $\| \vec{n} \| = \sqrt{{a}^{2} + {\left(\frac{a}{2}\right)}^{2} + {\left(- a\right)}^{2}}$ $\| \vec{n} \| = \sqrt{\frac{9}{4} {a}^{2}}$ $\| \vec{n} \| = \frac{3}{2} a$ So our unit vector is: $\vec{n} = \frac{a}{\frac{3}{2} a} i + \frac{\frac{a}{2}}{\frac{3}{2} a} j + \frac{- a}{\frac{3}{2} a} k$ $\vec{n} = \frac{2}{3} i + \frac{1}{3} j - \frac{2}{3} k$
# FRACTIONS Fractions Any unit can be divided into any numbers of equal parts, one or more of this parts is called fraction of that unit. e.g. one-forth (1/4), one-third (1/3), three-seventh (3/7) etc. The lower part indicates the number of equal parts into which the unit is divided, is called denominator. The upper part, which indicates the number of parts taken from the fraction is called the numerator. The numerator and the denominator of a fraction are called its terms. • A fraction is unity, when its numerator and denominator are equal. • A fraction is equal to zero if its numerator is zero. • The denominator of a fraction can never be zero. • The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number.e.g. 2/3 = 2/6 = 8/12 = (2/4)/(3/4) • When there is no common factor between numerator and denominator it is called in its lowest terms.e.g. 15/25 = 3/5 • When a fraction is reduced to its lowest term, its numerator and denominator are prime to each other. • When the numerator and denominator are divided by its HCF, fraction reduces to its lowest term. Proper fraction: A fraction in which numerator is less than the denominator. e.g. 1/4, 3/4, 11/12 etc. Improper Fraction: A fraction in which numerator is equal to or more than the denominator. e.g. 5/4, 7/4, 13/12 etc. Like fraction: Fractions in which denominators are same is called like fractions. e.g. 1/12, 5/12, 7/12, 13/12 etc. Unlike fraction: Fractions in which denominators are not same is called, unlike fractions. e.g. 1/12, 5/7, 7/9 13/11 etc. Compound Fraction: Fraction of a fraction is called a compound fraction. e.g. 1/2 of 3/4 is a compound fraction. Complex Fractions: Fractions in which numerator or denominator or both are fractions, are called complex fractions. Continued fraction: Fraction that contain additional fraction is called continued fraction. e.g. Rule: To simplify a continued fraction, begin from the bottom and move upwards. Decimal Fractions: Fractions in which denominators are 10 or multiples of 10 is called, decimal fractions. e.g. 1/10, 3/100, 2221/10000 etc. Recurring Decimal: If in a decimal fraction a digit or a set of digits is repeated continuously, then such a number is called a recurring decimal. It is expressed by putting a dot or bar over the digits. e.g. Pure recurring decimal: A decimal fraction in which all the figures after the decimal point is repeated is called a pure recurring decimal. Mixed recurring decimal: A decimal fraction in which only some of the figures after the decimal point is repeated is called a mixed recurring decimal. Conversion of recurring decimal into proper fraction: CASE I: Pure recurring decimal Write the repeated digit only once in the numerator and put as many nines as in the denominator as the number of repeating figures. e.g. CASE II: Mixed recurring decimal In the numerator, take the difference between the number formed by all the digits after the decimal point and that formed by the digits which are not repeated. In the denominator, take the number formed as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. e.g. Questions Level-I 1. Evaluate : (2.39)2 – (1.61)2 2.39 – 1.61 A. 2 B. 4 C. 6 D. 8 2. What decimal of an hour is a second ? A. 0.0025 B. 0.0256 C. 0.00027 D. 0.000126 3. The value of (0.96)3 – (0.1)3 is: (0.96)2 + 0.096 + (0.1)2 A. 0.86 B. 0.95 C. 0.97 D. 1.06 4. The value of 0.1 x 0.1 x 0.1 + 0.02 x 0.02 x 0.02 is: 0.2 x 0.2 x 0.2 + 0.04 x 0.04 x 0.04 A. 0.0125 B. 0.125 C. 0.25 D. 0.5 5. If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ? A. 0.172 B. 1.72 C. 17.2 D. 172 6. When 0.232323….. is converted into a fraction, then the result is: A. 1 5 B. 2 9 C. 23 99 D. 23 100 7. .009 = .01 ? A. 0.0009 B. 0.09 C. 0.9 D. 9 8. The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to: A. 0.02 B. 0.2 C. 0.04 D. 0.4 9. (0.1667)(0.8333)(0.3333) is approximately equal to: (0.2222)(0.6667)(0.1250) A. 2 B. 2.40 C. 2.43 D. 2.50 10. 3889 + 12.952 – ? = 3854.002 A. 47.095 B. 47.752 C. 47.932 D. 47.95 11. Level-II 0.04 x 0.0162 is equal to: A. 6.48 x 10-3 B. 6.48 x 10-4 C. 6.48 x 10-5 D. 6.48 x 10-6 12. 4.2 x 4.2 – 1.9 x 1.9 is equal to: 2.3 x 6.1 A. 0.5 B. 1 C. 20 D. 22 13. If 144 = 14.4 , then the value of x is: 0.144 x A. 0.0144 B. 1.44 C. 14.4 D. 144 14. The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ? A. 2010 B. 2011 C. 2012 D. 2013 15. Which of the following are in descending order of their value ? A. 1 , 2 , 3 , 4 , 5 , 6 3 5 7 5 6 7 B. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 7 6 7 C. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 6 7 7 D. 6 , 5 , 4 , 3 , 2 , 1 7 6 5 7 5 3 16. Which of the following fractions is greater than 3 and less than 5 ? 4 6 A. 1 2 B. 2 3 C. 4 5 D. 9 10 17. The rational number for recurring decimal 0.125125…. is: A. 63 487 B. 119 993 C. 125 999 D. None of these 18. 617 + 6.017 + 0.617 + 6.0017 = ? A. 6.2963 B. 62.965 C. 629.6357 D. None of these 19. The value of 489.1375 x 0.0483 x 1.956 is closest to: 0.0873 x 92.581 x 99.749 A. 0.006 B. 0.06 C. 0.6 D. 6 20. 0.002 x 0.5 = ? A. 0.0001 B. 0.001 C. 0.01 D. 0.1 Level-I Explanation: Given Expression = a2 – b2 = (a + b)(a – b) = (a + b) = (2.39 + 1.61) = 4. a – b (a – b) Explanation: Required decimal = 1 = 1 = .00027 60 x 60 3600 Explanation: Given expression = (0.96)3 – (0.1)3 (0.96)2 + (0.96 x 0.1) + (0.1)2 = a3 – b3 a2 + ab + b2 = (a – b) = (0.96 – 0.1) = 0.86 Explanation: Given expression = (0.1)3 + (0.02)3 = 1 = 0.125 23 [(0.1)3 + (0.02)3] 8 Explanation: 29.94 = 299.4 1.45 14.5 = 2994 x 1 [ Here, Substitute 172 in the place of 2994/14.5 ] 14.5 10 = 172 10 = 17.2 Explanation: 0.232323… = 0.23 = 23 99 Explanation: Let 0.009 = .01; Then x = .009 = .9 = .9 x 0.01 1 Explanation: Given expression = (11.98)2 + (0.02)2 + 11.98 x x. For the given expression to be a perfect square, we must have 11.98 x x = 2 x 11.98 x 0.02 or x = 0.04 Explanation: Given expression = (0.3333) x (0.1667)(0.8333) (0.2222) (0.6667)(0.1250) = 3333 x 1 x 5 6 6 2222 2 x 125 3 1000 = 3 x 1 x 5 x 3 x 8 2 6 6 2 = 5 2 = 2.50 Explanation: Let 3889 + 12.952 – x = 3854.002. Then x = (3889 + 12.952) – 3854.002 = 3901.952 – 3854.002 = 47.95. Level-II Explanation: 4 x 162 = 648. Sum of decimal places = 6. So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10-4 Explanation: Given Expression = (a2 – b2) = (a2 – b2) = 1. (a + b)(a – b) (a2 – b2) Explanation: 144 = 14.4 0.144 x 144 x 1000 = 14.4 144 x x = 14.4 = 0.0144 1000 Explanation: Suppose commodity X will cost 40 paise more than Y after z years. Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40 0.25z = 0.40 + 2.10 z = 2.5 = 250 = 10. 0.25 25 X will cost 40 paise more than Y 10 years after 2001 i.e., 2011. Explanation: 3 = 0.75, 5 = 0.833, 1 = 0.5, 2 = 0.66, 4 = 0.8, 9 = 0.9. 4 6 2 3 5 10 Clearly, 0.8 lies between 0.75 and 0.833. 4 lies between 3 and 5 . 5 4 6 Explanation: 0.125125… = 0.125 = 125 999 Explanation: 617.00 6.017 0.617 + 6.0017 ——– 629.6357 ——— Explanation: 489.1375 x 0.0483 x 1.956 489 x 0.05 x 2 0.0873 x 92.581 x 99.749 0.09 x 93 x 100 = 489 9 x 93 x 10 = 163 x 1 279 10 = 0.58 10 = 0.058 0.06.
Limits Practice Limits of Functions. As gets closer and closer and closer, but never quite reaches its goal, how does your function behave? Expect to see and learn how to solve questions like this one: (Be careful, most people who try this problem don’t get it right and/or get it right, but without understanding the full picture.) Limits are used to describe the most extreme cases of a function’s behavior. They also allow you to use the “smoothness” of a function as a tool for understanding how a function behaves around strange singularities. Being able to solve limit problems requires a combination of both knowing the essential strategies for simplifying functions and also understanding the rules for when limits do and don’t exist. In other cases, such as the there is no real, numerical limit as is not a real number, however, conceptually, we say that since, in this case, the left-hand and right-hand limits both equal. Exercises - Continuity. Given the following function, $$f(x) = \left\{ \begin{array}{ccc} x^2-3x & \textrm{ if } & x \lt -1\\\\ 2x-3 & \textrm{ if } & -1 \le x \lt 1\\\\ 2 & \textrm{ if } & x = 1\\\\ \displaystyle{\frac{1}{x-2}} & \textrm{ if } & x \ge 1\\\\ \end{array} \right.$$ Evaluate the following expressions. When a limit below fails to exist, cite the reason why this is the case. The Oxford Math Center. Consider the graph of the function $\displaystyle{f(x)=\frac{1}{x}}$ shown below. Clearly, looking at the graph, the height of this function appears to get closer and closer to a height of zero, as we look at $x$ values farther and farther to the right. That is to say, it appears that $f(x)$ "goes to" $0$ as $x$ "goes to" $+\infty$. The Oxford Math Center. The Limit Laws Epsilon-delta (ε, δ) limit definition. Types of discontinuities. Limits. Calculus means “a method of calculation or reasoning.” When one computes the sales tax on a purchase, one employs a simple calculus. When one finds the area of a polygonal shape by breaking it up into a set of triangles, one is using another calculus. Proving a theorem in geometry employs yet another calculus. Despite the wonderful advances in mathematics that had taken place into the first half of the 17th century, mathematicians and scientists were keenly aware of what they could not do. (This is true even today.) Area seems innocuous enough; areas of circles, rectangles, parallelograms, etc., are standard topics of study for students today just as they were then. Rates of change were also important. It turns out that these two concepts were related. The foundation of “the calculus” is the limit. Chapter Summary In this chapter we: Why? Later, we will want to add up an infinite list of numbers. These are just two quick examples of why we are interested in limits. One-sided limits from graphs. One-sided limits from graphs (practice) One-sided limit. The limit as x decreases in value approaching a (x approaches a "from the right" or "from above") can be denoted: or The limit as x increases in value approaching a (x approaches a "from the left" or "from below") can be denoted: In probability theory it is common to use the short notation: for the left limit and. Calculus I - Limits. Show Mobile NoticeShow All NotesHide All Notes You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. The topic that we will be examining in this chapter is that of Limits. This is the first of three major topics that we will be covering in this course. Here is a list of topics that are in this chapter. Tangent Lines and Rates of Change –In this section we will introduce two problems that we will see time and again in this course : Rate of Change of a function and Tangent Lines to functions.
# Calculus : Stationary Point Find the value of $\displaystyle x$ between $\displaystyle 0$ and $\displaystyle \frac{\pi}{2}$ for which the curve $\displaystyle y={{e}^{{\sqrt{3}x}}}\cos x$ has a stationary point. Determine whether it is a maximum or minimum point. Solution $\displaystyle y={{e}^{{\sqrt{3}x}}}\cos x$ $\displaystyle \frac{{dy}}{{dx}}=-\sin x\cdot {{e}^{{\sqrt{3}x}}}+\sqrt{3}\cdot {{e}^{{\sqrt{3}x}}}\cos x$ $\displaystyle \ \ \ \ \ ={{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)$ $\displaystyle \frac{{dy}}{{dx}}=0$, when $\displaystyle {{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)=0$ Since $\displaystyle {{e}^{{\sqrt{3}x}}}>0$ for every $\displaystyle x\in R$, $\displaystyle \sqrt{3}\cos x-\sin x=0$ $\displaystyle \therefore \sqrt{3}\cos x=\sin x$ $\displaystyle \therefore \tan x=\sqrt{3}\Rightarrow x=\frac{\pi }{3}$ $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}={{e}^{{\sqrt{3}x}}}(-\sqrt{3}\sin x-\cos x)+\sqrt{3}\cdot {{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)$ $\displaystyle \ \ \ \ \ \ ={{e}^{{\sqrt{3}x}}}(-\sqrt{3}\sin x-\cos x+3\cos x-\sqrt{3}\sin x)$ $\displaystyle \ \ \ \ \ \ =2{{e}^{{\sqrt{3}x}}}(\cos x-\sqrt{3}\sin x)$ $\displaystyle {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right\|}_{{x=\frac{\pi }{3}}}}=2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}(\cos \frac{\pi }{3}-\sqrt{3}\sin \frac{\pi }{3})$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}\left( {\frac{1}{2}-\frac{3}{2}} \right)$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =-2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}$ $\displaystyle \therefore {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right\|}_{{x=\frac{\pi }{3}}}}$ $\displaystyle <$ $\displaystyle 0$ Therefore the stationary point is a maximum turning point.
# Difference between revisions of "1990 AHSME Problems/Problem 19" ## Problem For how many integers $N$ between $1$ and $1990$ is the improper fraction $\frac{N^2+7}{N+4}$ $\underline{not}$ in lowest terms? $\text{(A) } 0\quad \text{(B) } 86\quad \text{(C) } 90\quad \text{(D) } 104\quad \text{(E) } 105$ ## Solution What we want to know is for how many $n$ is $$\gcd(n^2+7, n+4) > 1.$$ We start by setting $$n+4 \equiv 0 \mod m$$ for some arbitrary $m$. This shows that $m$ evenly divides $n+4$. Next we want to see under which conditions $m$ also divides $n^2 + 7$. We know from the previous statement that $$n \equiv -4 \mod m$$ and thus $$n^2 \equiv (-4)^2 \equiv 16 \mod m.$$ Next we simply add $7$ to get $$n^2 + 7 \equiv 23 \mod m.$$ However, we also want $$n^2 + 7 \equiv 0 \mod m$$ which leads to $$n^2 + 7\equiv 23 \equiv 0 \mod m$$ from the previous statement. Since from that statement $23$ divides $m$ evenly, $m$ must be of the form $23x$, for some arbitrary integer $x$. After this, we can set $$n+4=23x$$ and $$n=23x-4.$$ Finally, we must find the largest $x$ such that $$23x-4<1990.$$ This is a simple linear inequality for which the answer is $x=86$, or $\fbox{B}$. Solution by hjklhjklhjkl2008. 1990 AHSME (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions
# 1st PUC Maths Question Bank Chapter 15 Statistics Students can Download Maths Chapter 15 Statistics Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. ## Karnataka 1st PUC Maths Question Bank Chapter 15 Statistics Question 1. Find the mean deviation about the mean for the following data (i) 6, 7, 10, 12, 13, 4, 8, 12 (ii) 4, 7, 8, 9, 10, 12,13,17 (iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 Question 2. Find the mean deviation about the mean for the data. 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5. $$N=20, \bar{x}=10, \Sigma\left\|x_{i}-\bar{x}\right\|=124, \mathrm{M} \cdot \mathrm{D} \cdot(\bar{x})=6 \cdot 2$$ Question 3. Find the mean deviation about the median for the data: (i) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 (ii) 36, 72, 46, 60, 45, 53, 46, 51, 49, 42 (iii) 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21 (i) Arrange the data in ascending order as Question 4. Find the mean deviation about the mean for the following data. Question 5. Find the mean deviation about the median for the data, Question 6. Find the mean deviation about the mean for the following data. (i) Income per day Number of persons 0 – 100 4 100 – 200 8 200 – 300 9 300 – 400 10 400 – 500 7 500 – 600 5 600 – 700 4 700 – 800 3 (ii) Height (cms) Number of boys 95 – 105 9 105 – 115 13 115 – 125 26 125 – 135 30 135 – 145 12 145 – 155 10 (iii) Marks obtained Number of students 0 – 10 12 10 – 20 18 20 – 30 27 30 – 40 20 40 – 50 17 50 – 60 6 (iv) Marks obtained Number of students 10 – 20 2 20 – 30 3 30 – 40 8 40 – 50 14 50 – 60 8 60 – 70 3 70 – 80 2 Method (II) Method (I) Method (ii) Here a = 130, h = 10 (iii) Method (I) Question 7. Find the mean deviation about the median for the data: (i) Class Frequency 0 – 10 6 10 – 20 7 20 – 30 15 30 – 40 16 40 – 50 4 50 – 60 2 (ii) Marks Number of Girls 0 – 10 6 10 – 20 8 20 – 30 14 30 – 40 16 40 – 50 4 50 – 60 2 (ii) Question 8. Calculate the mean deviation about median age for the age distribution of 100 persons given below. Age Number 16 – 20 5 21 – 25 6 26 – 30 12 31 – 35 14 36 – 40 26 41- 45 12 46 – 50 16 51 – 55 9 First, modify the classes to make the data. continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. Question 9. Find Mean and variance for each data: (i) 6, 8, 10, 12, 14, 16, 18, 20, 22, 24. (ii) 6, 7, 10, 12, 13, 4, 8, 12 (iii) First n natural numbers. (iv) First 10 multiples of 3. Question 10. Find mean and variance. (i) xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3 (ii) xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3 (i) (ii) Question 11. Find the mean and standard deviation using short-cut method. xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5 Question 12. Find the mean and variance (i) Class Frequency 0 – 30 2 30 – 60 3 60 – 90 5 90 – 120 10 120 – 150 3 150 – 180 5 180 – 210 2 (ii) Class Frequency 0 – 10 5 10 – 20 8 20 – 30 15 30 – 40 16 40 – 50 6 Question 13. Find the mean, variance and standard deviation using short-cut method. Height (cms) No of Children 70 – 75 3 75 – 80 4 80 – 85 7 85 – 90 7 90 – 95 15 95 – 100 9 100 – 105 6 105 – 110 6 110 – 115 3 Question 14. The diameters of circles (in mm ) drawn in a design are given below calculate S.D and mean diameter. Diameter Circles 70 – 75 15 75 – 80 17 80 – 85 21 85 – 90 22 90 – 95 25 Question 15. From the data given below state which group is more variable, A or B? Marks Group A Group B 10 – 20 9 10 20 – 30 17 20 30 – 40 32 30 40 – 50 33 25 50 – 60 40 143 60 – 70 10 15 70 – 80 9 7 We know that, the group having greater C.V.(co-efficient of variation) is said w be more variable than other. So, we compute mean and S.D. for groups A and B. First we compute mean and S.D. for A. Clearly, B having greater C. V. than A. Therefore B is more variable. Note: For two frequency distributions with equal means, the series with greater S.D. is more variable than the other. Question 16. From the prices of shares X and Y below, find out which is more stable in value. Here, number of items for both = 10. First we compute mean and S.D. for x, taking a = 56, N = 10. Question 17. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results. A B No. of wage earners 586 648 Mean of monthly wages (Rs ) 5253 5253 Variance of the distribution of wages 100 121 (i) Which firm A or B pays larger amount as monthly wages? (ii) Which firm A or B shows greater variability in individual wages? (i) Amount of monthly wages paid by firm A. = 586 x mean wages = 586 x 5253 = Rs.3078258 And amount of monthly wages paid by form B = 648 x mean wages = 648 x 5253 = Rs. 3403944. Clearly, firm B pays more wages. Alternatively, The variance of wages in A is 100. ∴ S.D. of distribution of wage in A = 10 Also the variance of distribution of wages in firm B is 121. ∴ S.D. of distribution of wages in B = 11. since the average monthly wages in both firms is same i.e., Rs. 5253, therefore the firm with greater S.D. will have more variability. ‍∴ Firm B pays more wages. (ii) Firm B with greater S.D. shows greater variability in individual wages. Question 18. Two plants A and B of a factory show following results about the number of workers and the wages paid to them. A B No – of workers 5000 6000 Average monthly wages 2500 2500 Variance distribution of wages 81 100 In which plant A or B is there greater variability in individual wages? B. (Try yourself) (Ex. 13 Text book) Question 19. The following is the record of goals scored by team A in a football session: For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent? First, compute mean number of goals and S.D. for team A. Given : Mean number of goals for the team B is 2 and S.D. is 1.25 Since S.D. For A is less than S.D for B. ∴ A is more consistent. Question 20. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below: Which is more varying, the length or weight? Question 21. Co-efficient of variation of two distributions are 60 and 70, and the standard deviations are 21 and 16, respectively. What are their arithmetic means. Let $$\bar{x}_{1} \text { and } \bar{x}_{2}$$ be arithmetic means of first and second distribution respectively. Then Question 22. The following values are calculated in respect of heights and weight of the students of a section of class XI. Height Weight Mean 162 – 6 cm 52-36 kg Variance 127 – 69 cm2 23. 1361kg2 Can we say that the weight show greater variation than the height. Clearly C.V. (weights) is greater than C.V. (heights). Weights show more variability than heights. Miscellaneous Examples Question 1. The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13. Find the remaining two observations. Let the remaining two observation be x and y. Question 2. The mean of 5 observations is 4.4 and their variance is 8.24. If three observations are 1, 2 and 6, find the remaining two observations. Let the remaining two observations be x and ∴ Five observations are 1, 2, 6, x, y. Question 3. The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations. 6 and 8 (Try yourself) Here, x + y = 14, x2 + y2 = 100 Question 4. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation. Let six observations be Question 5. Given that $$\bar{x}$$is the mean and σ2 is the variance of n observations x1 , x2,……….. ,xn . Prove that the mean and variance of the observations ax1,ax2,………………,axn are $$a \bar{x}$$ and a2a1 respectively (a ≠ 0) Question 6. If each of the observation x1,x2,………. ,xn is increased by ‘a’, where a is positive or negative number, show that the variance remains unchanged. Question 7. The mean and standard deviation 20 observations are found to be 10 and 2 respectively. On rechecking, it was fond that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases : (i) If wrong item is omitted, (ii) If it is replaced by 12. Question 8. The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation? Given: N = 100 Incorrect mean = 40 = $$a \bar{x}$$ Incorrect S.D. = 5.1 We have, $$\bar{x}=\frac{1}{N} \Sigma x_{i}$$ ⇒$$\Sigma x_{i}=N \bar{x}=100 \times 40=4000$$ ∴ Incorrect sum of observations = 4000. ∴ Correct sum of observations = 4000 – 50 + 40 = 3990. Question 9. The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find and standard deviation if the incorrect observations are omitted. Given N = 100,New N = 97 Question 10. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below. Subject Mathematics Physics Chemistry Mean 42 32 40 – 9 S D 12 15 20 Which of three subjects shows the highest variability in marks and which shows the lowest?
# Difference between revisions of "2017 AIME I Problems/Problem 13" ## Problem 13 For every $m \geq 2$, let $Q(m)$ be the least positive integer with the following property: For every $n \geq Q(m)$, there is always a perfect cube $k^3$ in the range $n < k^3 \leq m \cdot n$. Find the remainder when $$\sum_{m = 2}^{2017} Q(m)$$is divided by 1000. ## Solution 1 Lemma 1: The ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases. Lemma 2: If the range $(n,mn]$ includes $y$ cubes, $(p,mp]$ will always contain at least $y-1$ cubes for all $p$ in $[n,+\infty)$. If $m=14$, the range $(1,14]$ includes one cube. The range $(2,28]$ includes 2 cubes, which fulfills the Lemma. Since $n=1$ also included a cube, we can assume that $Q(m)=1$ for all $m>14$. Two groups of 1000 are included in the sum modulo 1000. They do not count since $Q(m)=1$ for all of them, therefore $$\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{17} Q(m) \mod 1000$$ Now that we know this we will find the smallest $n$ that causes $(n,mn]$ to contain two cubes and work backwards (recursion) until there is no cube in $(n,mn]$. For $m=2$ there are two cubes in $(n,2n]$ for $n=63$. There are no cubes in $(31,62]$ but there is one in $(32,64]$. Therefore $Q(2)=32$. For $m=3$ there are two cubes in $(n,3n]$ for $n=22$. There are no cubes in $(8,24]$ but there is one in $(9,27]$. Therefore $Q(3)=9$. For $m$ in $\{4,5,6,7\}$ there are two cubes in $(n,4n]$ for $n=7$. There are no cubes in $(1,4]$ but there is one in $(2,8]$. Therefore $Q(4)=2$, and the same for $Q(5)$, $Q(6)$, and $Q(7)$ for a sum of $8$. For all other $m$ there is one cube in $(1,8]$, $(2,16]$, $(3,24]$, and there are two in $(4,32]$. Therefore, since there are 10 values of $m$ in the sum, this part sums to $10$. When the partial sums are added, we get $\boxed{059}\hspace{2 mm}QED\hspace{2 mm} \blacksquare$ This solution is brought to you by a1b2 ## Solution 2 We claim that $Q(m) = 1$ when $m \ge 8$. When $m \ge 8$, for every $n \ge Q(m) = 1$, we need to prove there exists an integer $k$, such that $n < k^3 \le mn$. That because $\sqrt[3]{mn} - \sqrt[3]{n} \ge 2\sqrt[3]{n} - \sqrt[3]{n} = \sqrt[3]{n} \ge 1$, so k exists between $\sqrt[3]{mn}$ and $\sqrt[3]{n}$ $\sqrt[3]{n} < k \le \sqrt[3]{mn}$. We can then hand evaluate $Q(m)$ for $m = 2,3,4,5,6,7$, and get $Q(2) = 32$, $Q(3) = 9$, and all the others equal 2. There are a total of 2010 integers from 8 to 2017. $$\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{7} Q(m) + 2010 \equiv 32+9+2+2+2+2+10 = \boxed{059} \mod 1000$$ -AlexLikeMath 2017 AIME I (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
# RS Aggarwal Solutions Class 10 Ex 19B ## RS Aggarwal Class 10 Ex 19B Chapter 19 All these RS Aggarwal class 10 solutions Chapter 19 Exercise 19.2 : Volume and Surface Areas of Solids are solved by Byju's top ranked professors as per CBSE guidelines. QUESTION-1: The dimension of a metallic cuboid are 100cm x 80cm x 64cm. It is melted and recast into a cube. Find the surface area of the cube. Solution: Let the edge of the cube be a. As, Volume of cube = volume of cuboid $\Rightarrow a^{3}=100\times 80\times 64$ $\Rightarrow a^{3}=512000$ $\Rightarrow a=\sqrt[3]{512000}$ => a = 80cm Now, the surface area of the cube = $6a^{2}$ = 6 x 80 x 80 = 38400 $cm^{2}$ So, the surface area of the cube is 38400$cm^{2}$. QUESTION-2: A cone of height 20cm and radius of base 5cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere. Solution: We have, The radius of the cone, r = 5cm and the height of the cone, h = 20cm Let the radius of the sphere be R. As, Volume of sphere = Volume of cone $\Rightarrow \frac{4}{3}\pi R^{3}=\frac{1}{3}\pi r^{2}h$ $\Rightarrow R^{3}=\frac{\pi r^{2}h\times 3}{3\times 4\pi }$ $\Rightarrow R^{3}=\frac{r^{2}h}{4}$ $\Rightarrow R^{3}=\frac{5\times 5\times 20}{4}$ $\Rightarrow R^{3}=125$ $\Rightarrow R=\sqrt[3]{125}=5cm$ => Diameter of the sphere = 2R = 2 x 5 = 10cm So, the diameter of the sphere is 10cm QUESTION-3: Metallic spheres of radii 6cm, 8cm and 10cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere. Solution: We have, The radii $r_{1}=6cm,r_{2}=8cm\:and\:r_{3}=10cm$ Let the radius of the resulting sphere be R. As, Volume of resulting sphere = Volume of three metallic spheres $\Rightarrow \frac{4}{3}R^{3}=\frac{4}{3}\pi r_{1}^{3}+\frac{4}{3}\pi r_{2}^{3}+\frac{4}{3}\pi r_{3}^{3}$ $\Rightarrow \frac{4}{3}R^{3}=\frac{4}{3}\pi \left ( r_{1} ^{3}+r_{2}^{3}+r_{3}^{3}\right )$ $\Rightarrow R^{3}=r_{1}^{3}+r_{2}^{3}+r_{3}^{3}$ $\Rightarrow R^{3}=6^{3}+8^{3}+10^{3}$ $\Rightarrow R^{3}=216+512+1000$ $\Rightarrow R^{3}=1728$ $\Rightarrow R=\sqrt[3]{1728}$ => R = 12cm So, the radius of the resulting sphere is 12cm. QUESTION-4: A solid metal cone with the radius of base 12cm and height 24cm is melted to form solid spherical balls of diameter 6cm each. Find the number of balls thus formed. Solution: Height of the cone = 24cm Volume = $\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 12\times 12\times 24=48\times 24\times \pi cm^{3}$ Radius of each ball = 3cm Volume of each ball = $\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \times 3\times 3\times 3=36\pi cm^{3}$ Total number of balls formed by melting the cone = $\frac{Volume\:of\:cone}{Volume\:of\:a\:ball}=\frac{48\times 24\pi }{36\pi }=32$ QUESTION-5: The radii of internal and external surfaces of a hollow spherical shell are 3cm and 5cm respectively. It is melted and recast into a solid cylinder of diameter 14cm. Find the height of the cylinder. Solution: We have, The internal base radius of spherical shell, $r_{1}$ = 3cm, The external base radius of spherical shell, $r_{2}$ = 5cm and The base radius of solid cylinder, r = 14/2 = 7cm Let the height of the cylinder be h. As, Volume of solid cylinder = Volume of spherical shell $\Rightarrow \pi r^{2}h=\frac{4}{3}\pi r_{2}^{3}-\frac{4}{3}\pi r_{1}^{3}$ $\Rightarrow \pi r^{2}h=\frac{4}{3}\pi \left ( r_{2}^{3}- r_{1}^{3} \right )$ $\Rightarrow r^{2}h=\frac{4}{3} \left ( r_{2}^{3}- r_{1}^{3} \right )$ $\Rightarrow 49\times h=\frac{4}{3}\left ( 125-27 \right )$ $\Rightarrow h=\frac{4}{3}\times \frac{98}{49}$ $Therefore, h=\frac{8}{3}$ So, the height of the cylinder is $\frac{8}{3}$ QUESTION-6: The internal and external diameters of a hollow hemispherical shell are 6cm and 10cm respectively. It is melted and recast into a solid cone of base diameter 14cm. Find the height of the cone so formed. Solution: Volume of material in the shell = $\frac{2}{3} \pi \times [5^{3} – 3^{3}] cm^{2}$ = $\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times h cm^{3} = \frac{154h}{3} cm^{3}$ $\Rightarrow \frac{154h}{3} = \frac{616}{3}$ $Rightarrow h = \frac{616}{154} = 4 cm$ Hence, height of the cone = 4 cm. QUESTION-7: A copper rod of diameter 2cm and length 10cm is drawn into a wire of uniform thickness and length 10m. Find the thickness of the wire. Solution: We have, The radius of the copper rod, R = 2/2 = 1cm, The height of the copper rod, H = 10cm and The height of the wire , h = 10m = 1000cm Let the radius of the wire be r. As, Volume of the wire = Volume of the rod $\Rightarrow \pi r^{2}h=\pi R^{2}H$ $\Rightarrow r^{2}h= R^{2}H$ $\Rightarrow r^{2}\times 1000=1\times 10$ $\Rightarrow r^{2}=\frac{10}{1000}$ $\Rightarrow r^{2}=\frac{1}{100}$ $\Rightarrow r=\sqrt{\frac{1}{100}}$ $\Rightarrow r=\frac{1}{10}=0.1cm$ => The diameter of the wire = 2r = 2 x 0.1 = 0.2cm Therefore, The thickness of the wire = 0.2cm So, the thickness of the wire is 0.2cm or 2mm. QUESTION-8: A hemispherical bowl of internal diameter 30cm contains some liquid. This liquid is to be filled into cylindrical-shaped bottles each of diameter 5cm and height 6cm. Find the number of bottles necessary to empty the bowl. Solution: Inner diameter of the bowl = 30cm Inner radius of the bowl = 15 cm Volume of liquid in it = $\frac{2}{3} \pi \times r^{3} = \frac{2}{3} \pi \times 15^{3} cm^{3}$ Radius of each cylinder bottle = 2.5 cm and its height = 6 cm. Volume of each cylindrical bottle = $\pi r^{2} h = \pi \times (\frac{5}{2})^{2} \times 6 cm^{2}$ $= \frac{25}{4} \times 6 \pi = \frac{75 \pi }{2} cm^{3}$ Required number of bottles = $\frac{Volume \, of \, liquid}{Volume \, of \, each \, cylindrical \, bottle}$ $\frac{\frac{2}{3} \times \pi \times 15 \times 15 \times 15}{\frac{75}{2} \times \pi} = 60$ Hence, bottles required = 60. QUESTION-9: A solid metallic sphere of diameter 21cm is melted and recast into a number of smaller cones, each of diameter 3.5cm and height 3cm. Find the number of cones so formed. Solution: Radius of the sphere = $\frac{21}{2} cm$ Volume of the sphere = $\frac{4}{3} \pi \times r^{3} = \frac{4}{3} \times (\frac{21}{2})^{3} cm^{3}$ Radius of cone = (7/4) cm and height 3 cm Volume of cone = $\frac{1}{3} \pi r^{2} h = \frac{1}{3} \times \pi \times (\frac{7}{4})^{2} \times 3 cm^{3}$ Let the number of cones formed be n, then, $n \times \frac{1}{3} \pi \times (\frac{7}{4})^{2} \times 3 = \frac{4}{3} \pi (\frac{21}{2})^{3}$ $n = \frac{4}{3} \pi \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} \times \frac{3}{\pi} \times \frac{4}{7} \times \frac{4}{7} \times \frac{1}{3}$ n = 504 Hence, number of cones formed = 504 QUESTION-10: A spherical cannon ball 28cm in diameter is melted and recast into a right circular conical mould, base of which is 35cm in diameter. Find the height of the cone. Solution: Diameter of cannon ball = 28cm So, Radius of cannon ball = 14 cm Volume of cannon ball = $\frac{4}{3} \pi r^{3} = \frac{4}{3} \pi \times 14^{3}$ Radius of the cone = $\frac{35}{2} cm$ Let the height of cone be h cm. Volume of cone = $\frac{1}{3} \pi (\frac{35}{2})^{2} \times h$ cm3 Hence, $\frac{4}{3} \pi \times 14^{3} = \frac{1}{3} \pi \times (\frac{35}{2})^{2} \times h$ $h = \frac{4}{3} \pi \times 14 \times 14 \times 14 \times \frac{3}{\pi} \times \frac{2}{35} \times \frac{2}{35}$ = 35.84 cm Hence, the height of the cone = 35.84 cm. QUESTION-11: A spherical ball of lead 3cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5cm and 2cm. Find the radius of the third ball. Solution: Let the radius of the third ball be r cm. Then, Volume of third ball = Volume of spherical ball – volume of 2 small balls Volume of the third ball = $\frac{4}{3} \pi \times 3^{3} – \frac{4}{3} \pi (\frac{3}{2})^{2} – \frac{4}{3} \pi ??\times 2^{3}$ $36 \pi – \frac{9 \pi }{2} – \frac{32 \pi}{3} cm^{3} ??= \frac{125 \pi }{6} cm^{3}$ i.e $\frac{4}{3} \pi r^{3} = \frac{125 \pi }{6}$ i.e $r^{3} = \frac{125 \pi \times 3}{6 \times 4 \times \pi } = \frac{125}{8}$ i.e $r = \frac{5}{2} cm = 2.5 cm$ Hence, the radius of the third ball is 2.5cm. QUESTION-12: A spherical shell of lead, whose external and internal diameters are respectively 24cm and 18cm, is melted and recast into a right circular cylinder 37cm high. Find the diameter of the base of the cylinder. Solution: External diameter of shell = 24cm and internal diameter of shell = 18cm So, External radius of shell = 12 cm and internal radius = 9 cm Volume of lead in the shell = $\frac{4}{3} \pi [12^{3} – 9^{3}] cm^{3}$ Let the radius of the cylinder be r cm Its height = 37 cm Volume of cylinder = $\pi r^{2}h=\pi r^{2}\left ( 37 \right )$ $\frac{4}{3} \pi [12^{3} – 9^{3}] = \pi r^{2} \times 37$ $\frac{4}{3} \pi \times 999 = \pi r^{2} \times 37$ $r^{2} = \frac{4}{3} \times \pi \times 999 \times \frac{1}{37 \pi } = 36 cm^{2}$ $r = \sqrt{36} = 6 cm$ Hence, diameter of the base of the cylinder = 12 cm QUESTION-13: A hemisphere of lead of radius 9cm is cast into a right circular cone of height 72cm. Find the radius of the base of the cone. Solution: Volume of hemisphere of radius 9 cm = $\frac{2}{3} \times \pi \times 9 \times 9 \times 9 cm^{3}$ Volume of circular cone (height = 72 cm) $\frac{1}{3} \times \pi \times r^{2} \times 72 cm$ Volume of cone = volume of hemisphere $\frac{1}{3} \times \pi \times r^{2} \times 72 = \frac{2}{3} \pi \times 9 \times 9 \times 9$ $r^{2} = \frac{2 \pi }{3} \times 9 \times 9 \times 9 \times \frac{1}{24 \pi } = 20.25$ $r = \sqrt{20.25} = 4.5 cm$ Hence, radius of the base of the cone = 4.5 cm QUESTION-14: A spherical ball of diameter 21cm is melted and recast into cubes, each of side 1cm. Find the number of cubes so formed. Solution: Diameter of sphere = 21 cm Hence, radius of the sphere = $\frac{21}{2}$ Volume of sphere = $\frac{4}{3} \pi \times r^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$ Volume of cube = a3 = 13 Let number of cubes formed be n Volume of sphere = n(volume of cube) $\frac{4}{3} \times \frac{22}{7} \times \frac{441}{4} \times \frac{21}{2} = n \times 1$ = $441 \times 11 = n$ 4851 = n Hence, number of cubes is 4851. Solution: Volume of sphere (when r = 1 cm) = $\frac{4}{3} \pi r^{3} = \frac{4}{3} \times 1 \times 1 \times 1 \times \pi cm^{3}$ Volume of sphere (when r = 8 cm) = $\frac{4}{3} \pi r^{3} = \frac{4}{3} \times 8 \times 8 \times 8 \times \pi cm^{3}$ Let the number of balls = n $n \times \frac{4}{3} \times 1 \times 1 \times 1 \times \pi = \frac{4}{3} \times 8 \times 8 \times 8 \times \pi$ $n = \frac{4 \times 8 \times 8 \times 8 \times 3}{3 \times 4} = 512$ QUESTION-16: A solid sphere of radius 3cm is melted and then cast into small spherical balls, each of diameter 0.6cm. Find the number of small balls so obtained. Solutions: Radius of sphere = 3 cm Volume of sphere = $\frac{4}{3} \pi r^{3} = \frac{4}{3} \times \pi \times 3 \times 3 \times 3 cm^{3} = 36 \pi cm^{3}$ Radius of small sphere = $\frac{0.6}{2} cm$ = 0.3 cm Volume of small sphere = $\frac{4}{3} \times \pi \times 0.3 \times 0.3 \times 0.3 cm^{3} cm^{3}$ = $\frac{4}{3} \pi \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10} cm^{3}$ Let the number of small balls be n $n \times \frac{4}{3} \pi \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10} = \frac{4}{3} \times \pi \times 3 \times 3 \times 3$ n = 1000 Hence, the number of small balls = 1000. QUESTION-17: The diameter of a sphere is 42cm. It is melted and drawn into a cylindrical wire of diameter 2.8cm. Find the length of the wire. Solution: Diameter of sphere = 42 cm Radius of sphere = 21 cm Volume of sphere = $\frac{4}{3} \pi r^{3} = \frac{4}{3} \times \pi \times 21 \times 21 \times 21 cm^{3}$ Diameter of cylindrical wire = 2.8 cm Radius of cylindrical wire = 1.4 cm Volume of cylindrical wire = $\pi r^{2} h = \pi \times 1.4 \times 1.4 \times h cm^{3} = 1.96 \pi h cm^{3}$ Volume of cylindrical wire = volume of sphere $1.96 \pi h = \frac{4}{3} \times \pi \times 21 \times 21 \times 21$ $h = \frac{4}{3} \times \pi \times 21 \times 21 \times 21 \times \frac{1}{1.96} \times \frac{1}{\pi}$ cm h = 6300 $h (\frac{6300}{100}) m = 63 m$ Hence, length of the wire = 63 m. QUESTION-18: The diameter of a sphere is 18cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108m, find its diameter. Solution: Diameter of sphere = 18 cm Radius of copper sphere = $\frac{3600}{100} m = 36 m$ Volume of sphere =??$\frac{4}{3} \pi r^{3} = \frac{4}{3} \times \pi \times 9 \times 9 \times 9 cm^{3} = 972 \pi cm^{3}$ Length of wire = 108 m = 10800 cm Let the radius of wire be r cm = $\pi r^{2} l = \pi r^{2} \times 10800$ But the volume of wire = volume of sphere $\Rightarrow \pi r^{2} \times 10800 = 972 \pi ??$ $r^{2} = \frac{972 \pi }{10800 \pi } = 0.09 cm^{2}$ $r = \sqrt{0.09} cm = 0.3$ Hence, the diameter = 2r = 0.6 cm. #### Practise This Question u – object distance v – image distance f – focal length
### Factors, Fractions, and Exponents ```Warm-Up: EOC Prep What is RS? What is m<ABC? Bisectors in Triangles 5.2 Today’s Goals By the end of class today, YOU should be able to… 1. Define and use the properties of perpendicular bisectors and angle bisectors to solve for unknowns. 2. Locate places equidistant from two given points on a map. Review… We learned in chapter 4 that ≅ ΔCBD. Therefore, we can conclude that CA ≅ CB, that CA = CB, or simply that C is equidistant from points A and B. Perpendicular Bisector Theorem  If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. Converse of the Perpendicular Bisector Theorem  If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. Ex.1: Using the Perpendicular Bisector Theorem If CD is the perpendicular bisector of both XY and ST, and CY = 16. Find the length of TY. Ex.1: Solution CS = CT CY – CT = TY We know from the Perpendicular Bisector Theorem that CS is equivalent to CT Subtract to find the value of TY You Try… If CD is the perpendicular bisector of both XY and ST, and CY = 16. Find the length of CX. Angle Bisector Theorem  If a point is on the bisector of an angle, then the point is equidistant from the sides of the angle. Converse of the Angle Bisector Theorem  If a point in the interior of an angle is equidistant from the sides of the angle, then the point is on the angle bisector. Ex.2: Using the Angle Bisector Theorem Find the value of x, then find FD and FB: Ex.2: Solution From the diagram we see that F is on the bisector of ACE. Therefore, FB = FD. FB = FD 5x = 2x + 24 Substitute 3x = 24 Subtract 2x x=8 Divide by 3 FB = 5x = 5(8) = 40 Substitute FD = 40 Substitute You Try… Does point A lie on an angle bisector of <TXR? Explain. Angle bisector constructions Construct an angle ABC Place the compass point on the angle vertex with the compass set to any convenient width Draw an arc that falls across both legs of the angle *The compass can then be adjusted at this point if desired From where an arc crosses a leg, make an arc in the angle's interior, then without changing the compass width, repeat for the other leg Draw a straight line from B to point D, where the arcs cross Done. The line just drawn bisects the angle ABC Practice A mysterious map has come into your possession. The map shows the Sea Islands off the coast of Georgia. But that’s not all! The map also contains three clues that tell where a treasure is supposedly buried!    The 1st clue: Draw a line from Baxley to Savannah. From Savannah, draw a southwesterly line that forms a 60° angle with the first line. The treasure is on an island that lies along the second line. On which islands could the treasure be buried? The 2nd clue: The treasure is on an island 22 miles from Everett. Construct a figure that contains all the points 22 miles from Everett. According to the first two clues, on which islands could the treasure be buried? Explain. The 3rd clue: The perpendicular bisector of the line segment between Baxley and Jacksonville passes through the island. The treasure is buried by the lighthouse on that island. On which island is the treasure buried? Explain. Homework  Page 251 #s 8-11, 16, 18  Page 252 # 30  The assignment can also be found at: • http://www.pearsonsuccessnet.com/snpap p/iText/products/0-13-037878-X/Ch05/0502/PH_Geom_ch05-02_Ex.pdf ```
Question The equation of the plane containing the lines 2x - 5y + z = 3, x + 4z = 5 and parallel to the plane x + 3y + 6z = 1, is A 2x+6y+12z=13 B x+3y+6z=7 C x+3y+6z=7 D 2x+6y+12z=13 Solution The correct option is D $$x + 3y + 6z = 7$$Let equation of plane containing the lines $$2x - 5y + z = 3$$ and $$x+y+4z=5$$ be $$\left( {2x - 5y + z - 3} \right) + \lambda \left( {x + y + 4z - 5} \right) = 0$$or, $$\left( {2 + \lambda } \right)x + \left( {\lambda - 5} \right)y + \left( {4\lambda + 1} \right)z - 3 - 5\lambda = 0........\left( i \right)$$This planes parallel to the plane $$x+3y+6z=1$$$$\therefore \frac{{2 + \lambda }}{1} = \frac{{\lambda - 5}}{3} = \frac{{4\lambda + 1}}{6}$$Considering first two qualities, we get$$\begin{array}{l} \Rightarrow 6+3\lambda =\lambda -5 \\ \Rightarrow 2\lambda =-11 \\ \Rightarrow \lambda =\frac { { -11 } }{ 2 } \end{array}$$Considering the last two qualities ,we get$$\begin{array}{l} \Rightarrow 6\lambda -30=3+12\lambda \\ \Rightarrow -6\lambda =33 \\ \Rightarrow \lambda =\frac { { -11 } }{ 2 } \end{array}$$So, the equation of required plane is $$\left( {2 - \frac{{11}}{2}} \right)x + \left( {\frac{{ - 11}}{2} - 5} \right)y + \left( {\frac{{ - 44}}{2} + 1} \right)z - 3 + 5 \times \frac{{11}}{2} = 0$$$$\begin{array}{l} \Rightarrow \frac { { -7 } }{ 2 } x-\frac { { 21 } }{ 2 } y-\frac { { 42 } }{ 2 } z+\frac { { 49 } }{ 2 } =0 \\ \Rightarrow x+3y+6z-7=0 \\ \Rightarrow x+3y+6z=7 \end{array}$$Hence,Option $$C$$ is correct answerMaths Suggest Corrections 0 Similar questions View More People also searched for View More
```OpenStax-CNX module: m38348 1 Rational Numbers - Grade 10 [CAPS] * Free High School Science Texts Project Based on Rational Numbers by Free High School Science Texts Project Mark Horner Heather Williams This work is produced by OpenStax-CNX and licensed under the 1 Introduction As described in Review of past work, a number is a way of representing quantity. The numbers that will be used in high school are all real numbers, but there are many dierent ways of writing any single real number. This chapter describes rational numbers. Khan Academy video on Integers and Rational Numbers Figure 1 * Version 1.4: Jun 13, 2011 9:25 pm -0500 http://cnx.org/content/m31331/1.5/ http://cnx.org/content/m38348/1.4/ OpenStax-CNX module: m38348 2 2 The Big Picture of Numbers Figure 2 The term whole number does not have a consistent denition. Various authors use it in many dierent ways. We use the following denitions: • natural numbers are (1, 2, 3, ...) • whole numbers are (0, 1, 2, 3, ...) • integers are (... -3, -2, -1, 0, 1, 2, 3, ....) 3 Denition The following numbers are all rational numbers. 10 , 1 −1 , −3 21 , 7 10 , 20 −3 6 (2) You can see that all denominators and all numerators are integers. Denition 2: Rational Number A rational number is any number which can be written as: a b (2) where a and b are integers and b 6= 0. a as −a Note that because we can write −b b (in other words, one can always nd an equivalent rational expression where b > 0) mathematicians typically dene rational numbers not as both a and b being integers, but rather that a is an integer and b is a natural number. This avoids having to worry about zero in the denominator. tip: Only fractions which have a numerator and a denominator (that is not 0) that are integers are rational numbers. This means that all integers are rational numbers, because they can be written with a denominator of 1. Therefore √ 2 , 7 π 20 (2) are not examples of rational numbers, because in each case, either the numerator or the denominator is not an integer. A number may not be written as an integer divided by another integer, but may still be a rational number. This is because the results may be expressed as an integer divided by an integer. The rule is, if a number can be written as a fraction of integers, it is rational even if it can also be written in another way as well. Here are two examples that might not look like rational numbers at rst glance but are because there are equivalent forms that are expressed as an integer divided by another integer: −1, 33 133 = , −3 300 http://cnx.org/content/m38348/1.4/ −3 −300 −100 = = 6, 39 639 213 (2) OpenStax-CNX module: m38348 3 3.1 Rational Numbers 1. If a is an integer, b is an integer and c is irrational, which of the following are rational numbers? a. 56 b. a 3 c. b 2 d. 1c 2. If a1 is a rational number, which of the following are valid values for a? a. 1 b. −10 √ c. 2 d. 2, 1 4 Forms of Rational Numbers All integers and fractions with integer numerators and denominators are rational numbers. There are two more forms of rational numbers. 4.1 Investigation : Decimal Numbers You can write the rational number 1. 2. 3. 4. 5. 1 2 as the decimal number 0,5. Write the following numbers as decimals: 1 4 1 10 2 5 1 100 2 3 Do the numbers after the decimal comma end or do they continue? If they continue, is there a repeating pattern to the numbers? You can write a rational number as a decimal number. Two types of decimal numbers can be written as rational numbers: 4 1. decimal numbers that end or terminate, for example the fraction 10 can be written as 0,4. 2. decimal numbers that have a repeating pattern of numbers, for example the fraction 31 can be written as 0, 3̇. The dot represents recurring 3's i.e., 0, 333... = 0, 3̇. For example, the rational number 56 can be written in decimal notation as 0, 83̇ and similarly, the decimal number 0,25 can be written as a rational number as 14 . You can use a bar over the repeated numbers to indicate that the decimal is a repeating decimal. tip: 1 http://www.fhsst.org/l35 2 http://www.fhsst.org/l3N http://cnx.org/content/m38348/1.4/ OpenStax-CNX module: m38348 4 5 Converting Terminating Decimals into Rational Numbers A decimal number has an integer part and a fractional part. For example 10, 589 has an integer part of 10 and a fractional part of 0, 589 because 10 + 0, 589 = 10, 589. The fractional part can be written as a rational number, i.e. with a numerator and a denominator that are integers. Each digit after the decimal point is a fraction with a denominator in increasing powers of ten. For example: • • 1 10 1 100 is 0, 1 is 0, 01 This means that: 10, 589 = = = 10 + 8 100 589 10 1000 10589 1000 5 10 + + 9 1000 (2) 5.1 Fractions 1. Write the following as fractions: a. 0, 1 b. 0, 12 c. 0, 58 d. 0, 2589 6 Converting Repeating Decimals into Rational Numbers When the decimal is a repeating decimal, a bit more work is needed to write the fractional part of the decimal number as a fraction. We will explain by means of an example. If we wish to write 0, 3̇ in the form ab (where a and b are integers) then we would proceed as follows x = 0, 33333... 10x = 3, 33333... 9x = 3 3 9 x = = multiply by 10 on both sides (subtracting the second equation from the rst equation) (2) 1 3 And another example would be to write 5, 4̇3̇2̇ as a rational fraction. x = 1000x = 5, 432432432... 5432, 432432432... multiply by 1000 on both sides (2) 999x = x = 5427 5427 999 3 http://www.fhsst.org/l3R http://cnx.org/content/m38348/1.4/ = 201 37 (subtracting the second equation from the rst equation) OpenStax-CNX module: m38348 5 For the rst example, the decimal was multiplied by 10 and for the second example, the decimal was multiplied by 1000. This is because for the rst example there was only one digit (i.e. 3) recurring, while for the second example there were three digits (i.e. 432) recurring. In general, if you have one digit recurring, then multiply by 10. If you have two digits recurring, then multiply by 100. If you have three digits recurring, then multiply by 1000. Can you spot the pattern yet? The number of zeros is the same as the number of recurring digits. √ Not all decimal numbers can be written as rational numbers. Why? Irrational decimal numbers like 2 = 1, 4142135... cannot be written with an integer numerator and denominator, because they do not have a pattern of recurring digits. However, when possible, you should try to use rational numbers or fractions 6.1 Repeated Decimal Notation 1. Write the following using the repeated decimal notation: a. 0, 11111111... b. 0, 1212121212... c. 0, 123123123123... d. 0, 11414541454145... 2. Write the following in decimal form, using the repeated decimal notation: a. 23 3 b. 1 11 5 c. 4 6 d. 2 19 3. Write the following decimals in fractional form: a. 0, 6333̇ b. 5, 313131 c. 0, 999999̇ 7 Summary Real numbers can be either rational or irrational. A rational number is any number which can be written as The following are rational numbers: a. b. c. d. a b where a and b are integers and b 6= 0 Fractions with both denominator and numerator as integers. Integers. Decimal numbers that end. Decimal numbers that repeat. 4 http://www.fhsst.org/l3U 5 http://www.fhsst.org/l3n 6 http://www.fhsst.org/l3Q http://cnx.org/content/m38348/1.4/ OpenStax-CNX module: m38348 8 End of Chapter Exercises 1. If a is an integer, b is an integer and c is irrational, which of the following are rational numbers? a. 56 b. a3 c. 2b d. 1c 2. Write each decimal as a simple fraction: a. 0, 5 b. 0, 12 c. 0, 6 d. 1, 59 e. 12, 277̇
Q: # How do I find the ratio between two numbers? A: Ratios give the relation between two quantities. For example, if two quantities A and B have a ratio of 1:3, it means that for every quantity of A, B has three times as much. Ratios can be represented in various formats, such as 1:3, 1/3 or "1 to 3." If two numbers are known, here are some simple steps to find the ratio between them. Know More 1. ### Determine if the two numbers have a common factor Ratios are usually the simplest representation of two quantities. The numbers that form the ratio should therefore be reduced to their simplest terms by dividing the numbers by their greatest common factor, if they have one. For example, for the ratio of 20 red balloons to 15 blue balloons, the numbers would be divided by 5. The number 5 is their greatest common factor, and it reduces the quantities to their equivalents of 4 red balloons and 3 blue balloons. 2. ### Arrange your numbers in the desired ratio format Choose the format in which the ratio is to be represented. In the example of red and blue balloons stated above, there are various forms of representation. For example, it can be stated that "the ratio of red to blue balloons is 4:3," "the ratio is 4/3" or "the ratio of red to blue balloons is 4 to 3." 3. ### Manipulate the ratio (if necessary) This step is optional. Ratios can be expanded to calculate the desired number of an object in a given scenario. In the balloon example, assume a birthday party is being planned so the ratio of red to blue balloons is 4:3 in all rooms and 70 balloons are to be used. Now knowing the required ratio, it can be calculated that 40 balloons need to be red and 30 need to be blue to maintain the ratio. Sources: ## Related Questions • A: Since a ratio is just a comparison between two quantities, figure out ratios by dividing the initial quantity by the quantity that follows. For example, if there are 12 males in a classroom and 15 females, the ratio of males to females is obtained by dividing 12 by 15. Filed Under: • A: The ratio 7 to 12 is written as 7/12 as a fraction in its simplest form. It is not possible to simplify this fraction any further because 7 and 12 have no common factors. Filed Under: • A: A rational expression in mathematics is an expression that is the ratio of two polynomials. Most often, it is expressed as a fraction, such as (x^2 + 5)/(x - 2).
• #### Class 10 Maths Study Material An Educational platform for Preparation and Practice Class 10. Kidsfront provide unique pattern of learning Maths with free online comprehensive study material in the form of QUESTION & ANSWER for each Chapter of Maths for Class 10. This study material help Class 10, Maths students in learning every aspect of Constructions. Students can understand Constructions concept easily and consolidate their learning by doing Online Practice Tests on Maths,Constructions chapter repeatedly till they excel in Class 10, Constructions. Free ONLINE PRACTICE TESTS on Class 10, Constructions comprise of Hundreds of Questions on Constructions, prepared by the highly professionals team. Every repeat test of Constructions will have new set of questions and help students to prepare themselves for exams by doing unlimited Online Test exercise on Constructions. Attempt ONLINE TEST on Class 10,Maths,Constructions in Academics section after completing this Constructions Question Answer Exercise. Unique pattern • Topic wise:Constructions preparation in the form of QUESTION & ANSWER. • Evaluate preparation by doing ONLINE TEST of Class 10, Maths,Constructions. • Review performance in PRACTICE TEST and do further learning on weak areas. • Attempt repeat ONLINE TESTS of Maths Constructions till you excel. • Evaluate your progress by doing ONLINE MOCK TEST of Class 10, Maths, All TOPICS. ##### The scale factor means:. a) The ratio of angles of the triangle to be constructed with the corresponding angle of the given triangle b) The ratio of sides of the triangle to be constructed with the corresponding sides of the given triangle c) Both a and b d) None of these a) 0 b) 90 c) 180 d) 360 ##### In drawing triangle ABC, it is given that AB = 3 cm, BC = 2 cm and AC = 6 cm. It is not possible to draw the triangle as: a) AB b) AC>AB+BC c) AB d) AB>BC a) 40 b) 50 c) 60 d) 70 ##### The exterior angle of a triangle is equal to a) Sum of all the angles in a triangle b) Sum of vertically opposite angles c) Sum of opposite interior angles in a triangle d) None of these a) 70 b) 105 c) 135 d) 145 a) 2 cm b) 6 cm c) 8 cm d) 12 cm ##### Identify the incorrect statement. a) All circles are similar b) All squares are similar c) All equilateral triangles are similar d) All similar figures are congruent a) 13 b) 14 c) 15 d) 12 a) S.360 b) 90 c) 180 d) 270 #### Other Topics script type="text/javascript">
# 180 Days of Math for Fourth Grade Day 16 Answers Key By accessing our 180 Days of Math for Fourth Grade Answers Key Day 16 regularly, students can get better problem-solving skills. ## 180 Days of Math for Fourth Grade Answers Key Day 16 Directions Solve each problem. Question 1. 32 – 7 = ___ 32 – 7 = 25 Explanation: Perform subtraction operation on above two given numbers. Subtract 7 from 32 the difference is 25. Question 2. Calculate one-tenth of 40. ____________ 1/10 Ã— 40 = 1 Ã— 4 = 4 So, one-tenth of 40 is 4. Explanation: One-tenth of number can be written in 1/10. So, multiply 1/10 with 40 the product is 4. Question 3. How many groups of 7 are in 14? ____________ Perform division operation. Divide 14 by 7 the result is 2. 14/7 = 2 There are 2 sevensÂ in 14. So, 2 groups of 7 are in 14. Question 4. Explanation: Perform division operation on above two given numbers. Divide 36 by 9 the quotient is 4. Question 5. Write 563 in expanded notation. ______________ The expanded notation of 563 is 500 + 60 + 3. Explanation: Expanded notation of the number is separated into individual place values. The given number 563 in expanded notation form is 500 + 60 + 3. Question 6. Fill in the missing number. 49, _____, 29, 19, 9 The given series is 49, _____, 29, 19, 9 The difference between 9 and 19 is 10. The difference between 19 and 29 is 10. So, add 10 to the number 29 in the series. 29 + 10 = 39 The missing number in the given series is 39. Question 7. You can put either a ball or a brick into a bucket of water. Circle the one that would raise the water level higher. Explanation: In the above image we can observe a bucket of water, ball and a brick. When we put a brick into a bucket of water the water level is higher compared to ball. So, the brick should be circled. Question 8. How many days are in January? ______________ There are 31 days in January. Question 9. Dollars Earned in May Audrey $15 Dameon$23 Jason $12 Lauren$18 How much money did Lauren earn?
# FRACTIONAL EXPONENTS - SOLVED EXAMPLES AND EXERCISES ON RATIONAL EXPONENTS Please study the Basics of Fractional Exponents, if you have not already done so. It is a prerequisite here. There, we provided the explanation for Rational Exponents. We applied the same 7 Laws and the 2 Rules in solving problems for Rational Exponents. We provided a few solved examples and problems for practice with answers. Here we provide many more Solved Studying the worked out problems will help remember and apply the 7 Laws and the 2 Rules for Rational Exponents. Practice makes one perfect. This is especially true for remembering Algebra Formulas (Math Formulas). So, take the exercises seriously and practice solving the problems. Here is a collection of proven tips, tools and techniques to turn you into a super-achiever - even if you've never thought of yourself as a "gifted" student. and remember large chunks of information with the least amount of effort. If you apply what you read from the above collection, you can achieve best grades without giving up your fun, such as TV, surfing the net, playing video games or going out with friends! Know more about the Speed Study System. ## Set of Solved Examples : Fractional Exponents ### Solved Example 1 of Fractional Exponents If ax = by = cz and ba = cb, show that yx = 2z⁄(x + z) Solution to Example 1 of Fractional Exponents: Let ax = by = cz = k a = k1⁄x; b = k1⁄y; c = k1⁄z; By data, ba = cb ⇒ (k1⁄y)⁄(k1⁄x) = (k1⁄z)⁄(k1⁄y) We know aman = am - n Applying this here, we get (k1⁄y - 1⁄x) = (k1⁄z - 1⁄y) Since the bases are same, the exponents have to be equal. ∴ (1⁄y - 1⁄x) = (1⁄z - 1⁄y) ⇒ 1⁄y + 1⁄y = 1⁄z + 1⁄x ⇒ 2⁄y = (x + z)⁄xz Multiplying both sides with x⁄2, we get xy = (x + z)⁄2zyx = 2z ⁄(x + z) (Proved.) Target="_Blank"> ### Solved Example 2 of Fractional Exponents If y = 31⁄3 + 1⁄(31⁄3), show that 3y3 - 9y = 10. Solution to Example 2 of Fractional Exponents: Let a = 31⁄3. Then a3 = 3........(i) and b = 1⁄(31⁄3). Then b3 = 1⁄3 .......(ii) Also, ab = 1.......(iii) and a + b = y........(iv). y = 31⁄3 + 1⁄(31⁄3) = a + b y3 = (a + b)3 = a3 + b3 + 3ab(a + b) Using (i), (ii), (iii) and (iv) here, we get y3 = 3 + 1⁄3 + 3(1)(y) Multiplying both sides with 3, we get 3y3 = 9 + 1 + 9y ⇒ 3y3 - 9y = 10. (Proved.) ### Solved Example 3 of Fractional Exponents Solve 2x1⁄3 + 2x-1⁄3 = 5. Solution to Example 3 of Fractional Exponents: Let a = x1⁄3. Then x-1⁄3 = 1⁄a. The given equation becomes 2a + 2⁄a = 5. Multiplying both sides with a, we get 2a2 + 2 = 5a. ⇒ 2a2 - 5a + 2 = 0. We know The solution of the quadratic equation ax2 + bx + c = 0 is x = {-b ± √( b2 - 4ac) }⁄2a Applying this here, we get a = [-(-5) ± √{(-5)2 - 4(2)(2)}]⁄[2(2)] = [5 ± √{ 25 - 16 }]⁄[4] = [5 ± √{ 9 }]⁄[4] = [5 ± 3]⁄[4] = (5 + 3)⁄4 or (5 - 3)⁄4 = 8⁄4 or 2⁄4 = 2 or 1⁄2 . a = x1⁄3a3 = x; x = 23 or (1⁄2)3 = 8 or 1⁄8. Ans. ## Progressive Learning of Math : Fractional Exponents Recently, I have found a series of math curricula (Both Hard Copy and Digital Copy) developed by a Lady Teacher who taught everyone from Pre-K students to doctoral students and who is a Ph.D. in Mathematics Education. This series is very different and advantageous over many of the traditional books available. These give students tools that other books do not. Other books just give practice. These teach students “tricks” and new ways to think. These build a student’s new knowledge of concepts from their existing knowledge. These provide many pages of practice that gradually increases in difficulty and provide constant review. These also provide teachers and parents with lessons on how to work with the child on the concepts. The series is low to reasonably priced and include Elementary Math curriculum and Algebra Curriculum. ## Exercise on Fractional Exponents 1. Show that 1⁄(1 + xa - b + xa - c) + 1⁄(1 + xb - c + xb - a) + 1⁄(1 + xc - a + xc - b) = 1. 2. If ax = by = cz = dw and ab = cd, show that: 1⁄x + 1⁄y = 1⁄w + 1⁄z . 3. If a1⁄3 + b1⁄3 + c1⁄3 = 0, show that: (a + b + c)3 = 27abc. For Answers, see at the bottom of the page. ### Answers to Exercise on Fractional Exponents 1. (i) 1. (ii) 1. 2. (i) 1⁄2. (ii) -8. ## Research-based personalized Math Help tutoring program : Fractional Exponents Here is a resource for Solid Foundation in Math Fundamentals from Middle thru High School. You can check your self by the ### FREE TRIAL. Are you spending lot of money for math tutors to your child and still not satisfied with his/her grades ? Do you feel that more time from the tutor and more personalized Math Help to identify and fix the problems faced by your child will help ? 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# Long division Long division is a method of dividing two numbers, using repeated multiplications and subtractions in a tableau.[1] Because it is easy to do, it is usually taught in schools. There are other methods which are faster, or easier to program with a computer, but they are more difficult to understand and perform manually. If we have a large or complicated division problem, we can use long division to break it down into a series of easier calculations. Long division can also be done on polynomials as well.[2] ## Overview As with most division problems we have three numbers: the dividend, our first number; the divisor, the second number we divide it by; and the quotient, which is the result.[1][3] Long division is a kind of algorithm, which means it helps us to find the solution to a problem by following a set of clearly-defined steps: 1. First, we need to split our first number (dividend) into a separate number for each digit. If our dividend is 123, then we would split this into 1, 2 and 3. 2. Next, we need to divide each of these digits by our second number (divisor). If our divisor is 8, then we would do 1 / 8, 2 / 8, followed by 3 / 8. • If the division has a remainder, then the remainder is carried to the next step. • If the division is less than 0 (for example. when trying to divide a small number by a large number), then the dividend, instead of the remainder, is carried to the next step. 3. Once all of the numbers have been processed, every result (quotient) would then be combined into a single number again. For example. if the quotients are 7, 8 and 2, then the final result would be 782. • Any leftover remainders make up the decimal part of the answer. ## Examples ### Basic case Let's assume we want to divide 780 by 4. To do so with long division, we need to split 780 into digits -- 7 and 8 and 0 -- and then divide each one by 4, carrying any remainders to the next step. 7 / 4 = 1 R 3 -- since we have a remainder of 3, we have to carry this down. 38 / 4 = 9 R 2 -- we carry our remainder of 2 down. 20 / 4 = 5 R 0 -- we have reached the end. By using long division, we have found that 780 / 4 = 195. ### Complex case Let's perform a similar calculation where our numbers do not divide easily: 468 / 12 4 / 12 = 0 R 0 -- we cannot perform this division as 4<12, so we have to carry our dividend like a remainder. 46 / 12 = 3 R 10 -- we carry our remainder of 10 down. 108 / 12 = 9 R 0 -- we have reached the end. By the same process we have found that 468 / 12 = 39. Let's follow the same process for numbers that give an answer with a decimal because they don't have common factors: 123 / 8 1 / 8 = 0 R 0 -- we cannot perform this division as 1<8, so we have to carry our dividend like a remainder. 12 / 8 = 1 R 4 -- we carry our remainder of 4 down. 43 / 8 = 5 R 3 -- we have reached the end with a remainder, which we have to add to our final answer. 123 / 8 = 15 R 3, which equals 15.375 (15 3/8). ## Related pages • Chunking, a different type of long division done in the UK. • Divisor, a number which evenly divides another number • Short division, a faster version of long division done with smaller numbers. • Synthetic division, an alternate algorithm for polynomial long division ## References 1. "The Definitive Higher Math Guide to Long Division and Its Variants — for Integers". Math Vault. 2019-02-24. Retrieved 2020-08-26. 2. Weisstein, Eric W. "Long Division". mathworld.wolfram.com. Retrieved 2020-08-26. 3. "Long Division". www.mathsisfun.com. Retrieved 2020-08-26.
# Proof by induction using summation [duplicate] I'm trying to figure out how to solve this equation by induction and I really don't know where to begin. I have seen some YouTube tutorials, but can't understand how I can go from $$k(k+1)$$ to $$n+1$$ in the equation. The task is: Use induction to show that: $$\sum_{k=1}^{n} {1 \over k(k+1)} = {n \over n+1}$$ Can someone help me solve this equation? Or give me some tips for where to start? I would really appreciate it. • Induction aside, familiar with telescoping? Dec 14, 2019 at 22:05 To prove this you would first check the base case $$n = 1$$. This is just a fairly straightforward calculation to do by hand. Then, you assume the formula works for $$n$$. This is your "inductive hypothesis". So we have $$\begin{equation} \sum_{k = 1}^n \frac 1{k(k + 1)} = \frac n{n + 1}. \end{equation}$$ Now we can add $$\frac 1{(n + 1)(n + 2)}$$ to both sides: \begin{align} \sum_{k = 1}^{n + 1} \frac 1{k(k + 1)} &= \frac n{n + 1} + \frac 1{(n + 1)(n + 2)} \\ &= \frac{n(n + 2) + 1}{(n + 1)(n + 2)} \\ &= \frac{(n + 1)^2}{(n + 1)(n + 2)} \\ &= \frac{(n + 1)}{(n + 1) + 1} \end{align} But this is exactly the same formula again, just with $$n$$ replaced by $$n + 1$$. So if the formula works for $$n$$, it also works for $$n + 1$$. And then, by induction, we are done. The hint is $$\sum_{k=1}^{n+1}\frac{1}{k(k+1)} =\biggl(\,\sum_{k=1}^n\frac{1}{k(k+1)}\biggr)+\frac{1}{(n+1)(n+2)}$$ which should equal $$\frac{n+1}{n+2}$$ Define the hypothesis $$H_n$$ as $$H_n : \sum_{k = 1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$ For $$n = 1$$, we have LHS = $$\sum \limits_{k = 1}^1 \frac{1}{k(k+1)} = \frac{1}{2}$$ and RHS = $$\frac{1}{1 + 1} = \frac{1}{2}$$ so $$H_1$$ is true. Now suppose $$H_m$$ is true for some integer $$m \geq 1$$, that is, we have $$\sum \limits_{k = 1}^m\frac{1}{k(k+1)} = \frac{m}{m+1}$$ as a given. Then, consider $$\sum_{k = 1}^{m+1} \frac{1}{k(k+1)}$$ $$= \frac{1}{ (m+1)(m + 2)} + \sum_{k=1}^m \frac{1}{k(k+1)}$$ By induction hypothesis, we have: $$= \frac{1}{ (m+1)(m + 2)} + \frac{m}{m+1}$$ $$= \frac{1 + m(m+2)}{(m+1)(m+2)}$$ $$= \frac{(m+1)^2}{(m+1)(m+2)}$$ $$= \frac{m+1}{(m + 1) + 1}$$ Therefore, $$\sum_{k=1}^{m+1} \frac{1}{k(k+1)} = \frac{m+1}{(m + 1) + 1}$$ So $$H_m \implies H_{m+1}$$ Since we had $$H_1$$ true, by induction, $$H_n$$ is true for all integers $$n \geq 1$$
## What is the perimeter? The perimeter indicates the distance we will walk if we start from a certain point, complete a full lap, and return exactly to the starting point. For example, if we are asked what the perimeter of the waist is, we will take a tape measure and measure the perimeter from a certain point until completing a full lap and returning to the same point from which we started the measurement. It works exactly the same way in mathematics. The perimeter of any shape is the distance from a specific point back to it after having completely surrounded it. If this is our figure: Its perimeter will be the distance we cover if we travel along its line from a certain point, and return to it after making a full lap. Imagine that you are surrounding the figure: ## Examples with solutions for Perimeter ### Exercise #1 Look at the rectangle below. Side AB is 2 cm long and side BC has a length of 7 cm. What is the perimeter of the rectangle? ### Step-by-Step Solution Since in a rectangle every pair of opposite sides are equal to each other, we can claim that: $AB=CD=2$ $AD=BC=7$ Now we can add all the sides together and find the perimeter: $2+7+2+7=4+14=18$ 18 cm ### Exercise #2 Look at the following rectangle: Find its perimeter. ### Step-by-Step Solution Since in a rectangle all pairs of opposite sides are equal: $AD=BC=5$ $AB=CD=9$ Now we calculate the perimeter of the rectangle by adding the sides: $5+5+9+9=10+18=28$ 28 ### Exercise #3 Look at the rectangle below. Side DC has a length of 1.5 cm and side AD has a length of 9.5 cm. What is the perimeter of the rectangle? ### Step-by-Step Solution Since in a rectangle every pair of opposite sides are equal to each other, we can claim that: $AD=BC=9.5$ $AB=CD=1.5$ Now we can add all the sides together and find the perimeter: $1.5+9.5+1.5+9.5=19+3=22$ 22 cm ### Exercise #4 Calculate the perimeter of the given parallelogram: ### Step-by-Step Solution As is true for a parallelogram every pair of opposite sides are equal: $AB=CD=6,AC=BD=4$ The perimeter of the parallelogram is equal to the sum of all sides together: $4+4+6+6=8+12=20$ 20 ### Exercise #5 Calculate the perimeter of the given parallelogram. ### Step-by-Step Solution As is true for a parallelogram each pair of opposite sides are equal and parallel, Therefore it is possible to argue that: $AC=BD=7$ $AB=CD=10$ Now we can calculate the perimeter of the parallelogram by adding together all of its sides: $10+10+7+7=20+14=34$ 34 ### Exercise #6 Look at the rectangle below: Calculate its perimeter. ### Step-by-Step Solution Since in a rectangle every pair of opposite sides are equal to each other, we can claim that: $AB=CD=10$ $BC=AD=7$ Now let's add all the sides together to find the perimeter of the rectangle: $10+7+10+7=20+14=34$ 34 ### Exercise #7 Look at the trapezoid in the diagram. What is its perimeter? ### Step-by-Step Solution To calculate the perimeter, we'll add up all the sides of the trapezoid: 7+10+7+12 = 36 And that's the solution! 36 ### Exercise #8 Calculate the perimeter of the parallelogram ABCD. CD is parallel to AB. ### Step-by-Step Solution Let's recall the properties of parallelograms, where pairs of opposite sides are parallel and equal. Therefore, AB is parallel to CD Therefore, BC is parallel to AD From this, we can conclude that AB=CD=7 Now we can calculate the perimeter by adding all the sides together: $7+7+12+12=14+24=38$ 38 ### Exercise #9 What is the perimeter of the trapezoid in the figure? ### Step-by-Step Solution To find the perimeter we will add all the sides: $4+5+9+6=9+9+6=18+6=24$ 24 ### Exercise #10 Look at the circle in the figure: Its radius is equal to 4. What is its circumference? ### Step-by-Step Solution The formula for the circumference is equal to: $2\pi r$ ### Exercise #11 Look at the circle in the figure. What is its circumference if its radius is equal to 6? ### Step-by-Step Solution Formula of the circumference: $P=2\pi r$ We insert the given data into the formula: $P=2\times6\times\pi$ $P=12\pi$ $12\pi$ ### Exercise #12 O is the center of the circle in the figure below. What is its circumference? ### Step-by-Step Solution We use the formula:$P=2\pi r$ We replace the data in the formula:$P=2\times8\pi$ $P=16\pi$ $16\pi$ cm ### Exercise #13 Is it possible that the circumference of a circle is 8 meters and its diameter is 4 meters? ### Step-by-Step Solution To calculate, we will use the formula: $\frac{P}{2r}=\pi$ Pi is the ratio between the circumference of the circle and the diameter of the circle. The diameter is equal to 2 radii. Let's substitute the given data into the formula: $\frac{8}{4}=\pi$ $2\ne\pi$ Therefore, this situation is not possible. Impossible ### Exercise #14 Look at the isosceles triangle below: What is its perimeter? ### Step-by-Step Solution Since we are referring to an isosceles triangle, the two legs are equal to each other. In the drawing, they give us the base which is equal to 4 and one side is equal to 6, therefore the other side is also equal to 6. The perimeter of the triangle is equal to the sum of the sides and therefore: $6+6+4=12+4=16$ 16 ### Exercise #15 Look at the trapezoid in the figure. The long base is 1.5 times longer than the short base. Find the perimeter of the trapezoid. ### Step-by-Step Solution First, we calculate the long base from the existing data: Multiply the short base by 1.5: $5\times1.5=7.5$ Now we will add up all the sides to find the perimeter: $2+5+3+7.5=7+3+7.5=10+7.5=17.5$ 17.5
# How do you differentiate arctan(8^x)? Sep 28, 2017 $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{8}^{x} \left(\ln 8\right)}{{8}^{2 x} + 1}$ #### Explanation: This kind of problem is usually done with implicit differentiation. We begin by writing the function in terms of $x$ and $y$ as such: $y = \arctan \left({8}^{x}\right)$ Next, we use the definition of $\arctan$ to rewrite the inverse in terms of the original $\tan$: $\tan y = {8}^{x} \text{ } \left[A\right]$ We can now proceed with the implicit differentiation with respect to $x$: $\frac{d}{\mathrm{dx}} \left(\tan y\right) = \frac{d}{\mathrm{dx}} \left({8}^{x}\right)$ ${\sec}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {8}^{x} \left(\ln 8\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{8}^{x} \left(\ln 8\right)}{{\sec}^{2} y}$ $\frac{\mathrm{dy}}{\mathrm{dx}} = {8}^{x} \left(\ln 8\right) \cdot \left({\cos}^{2} y\right) \text{ } \left[B\right]$ In line [B] we note that the definition of $\sec y$ is $\frac{1}{\cos y}$ and make the suitable substitution. The issue with this is the derivative still contains a term of $y$, which we must remove. The clever trick here is to go back to line [A] in the solution and recognize that we can use the trigonometric definition of tangent to define a right triangle using this information: Note how the picture illustrates the fact that $\tan y = {8}^{x} = \left(\text{opposite")/("adjacent}\right)$ from trigonometry. Using trigonometry, we can now evaluate what the value of $\cos y$ should be, and thus also ${\cos}^{2} y$ which needs to be replaced in our form of the derivative. Note that $c$ (the hypotenuse) can be derived from the Pythagorean Formula: ${c}^{2} = {a}^{2} + {b}^{2}$ ${c}^{2} = {\left({8}^{x}\right)}^{2} + \left({1}^{2}\right)$ ${c}^{2} = {8}^{2 x} + 1$ Thus: $\cos y = \left(\text{adjacent")/("hypotenuse}\right) = \frac{1}{c}$ ${\cos}^{2} y = {\left(\frac{1}{c}\right)}^{2} = \frac{1}{c} ^ 2 = \frac{1}{{8}^{2 x} + 1}$ Finally: $\frac{\mathrm{dy}}{\mathrm{dx}} = {8}^{x} \left(\ln 8\right) \cdot \left({\cos}^{2} y\right) = {8}^{x} \left(\ln 8\right) \cdot \left(\frac{1}{{8}^{2 x} + 1}\right)$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{8}^{x} \left(\ln 8\right)}{{8}^{2 x} + 1}$
Here we’ll look at some more probability calculations before moving on to permutations, combinations and the binomial theorem. More Probability… If you recall from the last part, we used set notation to describe a general way of calculating simple probability: $$P(A)={{\|A\|} \over {\|S\|}}.$$ Where $$A$$ is a set of events and $$S$$ is a set of all possible outcomes. Here’s a sample space set describing two dice: $S=\{1,2,3,4,5,6\}^2$ …for all $$\|S\|=6^2=36$$ possible outcomes. This set contains repeated values and ordering, so let’s directly enumerate the whole set: \begin{array} ((1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5),(5,1),(1,6),(6,1) \\ (2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2),(2,6),(6,2) \\ (3,3),(3,4),(4,3),(3,5),(5,3),(3,6),(6,3) \\ (4,4),(4,5),(5,4),(4,6),(6,4) \\ (5,5),(5,6),(6,5) \\ (6,6) \end{array} Each tuple represents a probable roll of a pair of dice. Normally we roll a pair of dice at the same time so that we don’t care what the order is – a five and a three are read the same as a three and a five, for example. However, if we roll each dice individually you can see that the order counts – I may roll a one first followed by a six, or the other way around. Notice how the doubles occur only once in the above table but the different rolls occur twice – (3,6) and (6,3), for example. This leads to a counter-intuitive probability problem. You’d naturally think that any pair of numbers (including the doubles) would have an equal chance, but this is not so as can be seen here: $A=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$ …which is an event set for all the possible double throws. If we calculate the probability: $P(A) = {{\|A\|} \over {\|S\|}} = {6 \over 36} \approx 0.17$ …a 0.17 or 17% chance of rolling any double, and as there are 30 possible outcomes for rolling two different numbers: $P(A) = {{\|A\|} \over {\|S\|}} = {30 \over 36} \approx 0.83$ …or an 83% chance of rolling anything other than a double. Finally, you can see that you are twice as likely to roll a differing pair of dice than a double: $A=\{(6,6)\} {\;\;\;\;} B=\{(5,3),(3,5)\}$ $P(A) = {1 \over 36} \approx 0.03 {\;\;\;\;} P(B) = {2 \over 36} = {1 \over 18} \approx 0.06$ …a one in thirty-six chance of rolling a particular double verses a one in eighteen chance for rolling two different numbers. This explains why doubles are so highly prized in dice games. The Sum Of Probabilities If set $$S$$ contains all of the possible outcomes from some random event, like tossing a coin, then if we sum up all of the probabilities of those outcomes we get the number 1 or 100%. We can see this easily with a coin sample set: $$S=\{\text{Heads,Tails}\}$$. I think that we can all work out that there is a 50% chance that the outcome is either heads or tails. The sum of both these probabilities is, of course, 100%. Mathematically, we can write it like this: $\sum _{x \in S} p(x) = 1$ …where the function $$p(x)$$ returns a probability value between 0 and 1, inclusive $$(p(x)=[0,1])$$, for any event $$x$$ inside of set $$S$$. The sigma $$\sum$$ symbol simply means to add all of the probabilities for all possible events together: $p(\text{Heads}) + p(\text{Tails}) = 0.5 + 0.5 = 1$ The event set is a subset of the sample space set. If $$A$$ is the event set then $$A \subseteq S$$. This, logically, means that the events inside of set $$A$$ can be any of the subsets of the power set of $$S$$, $$\wp(S)$$. Including the empty-set, $$\emptyset$$. In that case, we haven’t flipped a coin so we have no chance of winning or losing! 🙂 All of the possible event sets (subsets) $$A$$ from a heads/tails set $$S$$ give us these probabilities for $$p(x)$$: $p(\emptyset)=0 {\;\;} p(\text{Heads})=0.5 {\;\;} p(\text{Tails})=0.5 {\;\;} p(\{\text{Heads, Tails}\})=1$ …and therefore: $P(A) = \sum _{x \in A} p(x)$ In the next lesson we’ll look at the factorial function, its interaction with the prime numbers and ways to solve it for positive real numbers.
# Percentage Error It is hard for one to take a measurement and get an exact result. Percentage error is a little degree in percent at which a known length is measured above/below its actual value. The error that might be made during the measurement could be either positive or negative, in any ways just know that error still remains error provided that it has a value order than the actual result. For instance if the actual length of a room is 3.0m. It is said that three students on their industrial Training measured it differently and they got: 2.85m, 3.01m and 2.91m respectively. What are their errors? Error = (Wrong value – Actual value) or (Actual value – Wrong value) The first student’s error = 3.0 – 2.85 = 0.15m Second student’s error = 3.01 – 3.0 = 0.01m Third student’s error = 3.0 – 2.91 = 0.09m Note that the ones we just did above are the errors. For one to calculate the percentage error, this formula will be used: Percentage error (PE) = Absolute error (AE) divided by Actual value (AV) and you multiply by 100. (PE) = (AE/AV) x 100 Now, the percentage error of the example above are: For the first student: AE = 0.15m. AV = 3.0m PE = ? PE = AE/AV x 100 = 0.15/3 x 100 = 15/3 = 5% For the second student: AE = 0.01m. AV = 3.0m PE = ? PE = AE/AV x 100 = 0.01/3 x 100 = 1/3% = 0.33% For the third student: AE = 0.09m. AV = 3.0m PE = ? PE = AE/AV x 100 = 0.09/3 x 100 = 9/3 = 3% ## QUESTIONS AND SOLUTIONS: ### Q1: A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error. Solution: Let the certain number be y. He was meant to subtract 15 from it and mistakenly added 25. y + 25 = 145 y = 120 = the certain number Actual value = 120 – 15 = 105 Absolute Error = 145 – 105 = 40 PE = 40/105 x 100 = 38.1% ### Q2: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1 decimal place? AE = 6.65 – 6.35 = 0.3 AV = 6.35 PE = ? PE = AE/AV x 100 = 0.3/6.35 x 100 = 4.7% (1 dp) ### Q3: If the age of a man 64 years is written as 71 years, calculate the percentage error to 3 significant figures. Solution: AE = 71 – 64 = 7 years AV = 64 PE = ? PE = AE/AV x 100 = 7/64 x 100 = 10.9% (3 SF) ### Q4: Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m. Solution: AE = 1.26 – 1.24 = 0.02m AV = 1.24 PE = ? PE = AE/AV x 100 = 0.02/1.24 x 100 = 1.6% ### Q5: A man underestimated his expenses by 6.5% but actually spent #400.00. What was his estimate? Solution: Let his estimate be D AV = #400 AE = (400 – D) PE = 6.5% PE = [(400 – D)/400] x 100 = 6.5 Solve for D. PE = (400 – D) = 26 D = 400 – 26 = #374 ### Q6: An error of 4% was made in finding the length of rope that was actually 25m. By how many metres was the measurement wrong? Solution: Let the wrong value be T AV = 25m AE = (T – 25)m PE = 4% PE = [(T – 25)/25] x 100 = 4 Solve for T. 100(T – 25) = 4 x 25 100(T – 25) = 100 T – 25 = 1 T = 1 + 25 = 26m Therefore the measurement is ±1m wrong. Students by now will be wondering why my absolute error (AE) is = (T – 25)m and not (25 – T)m. Note that in which ever way you place it, it is still the same provided that the question comes in that nature. Now, let’s use the reverse (25 – T)m as our AE. PE = [(25 – T)/25] x 100 = 4 100(25 – T) = 25 x 4 100(25 – T) = 100 Solve for T 25 – T = 1 -T = 1 – 25 -T = -24 = 24m Actual length = 25m So the measurement is ±1m wrong. ### Q7: What is the percentage error in an area of a lawn that actually measures 750m² but found to be 690m²? Solution: AV = 750m² AE = 750 – 690 = 60m² PE = ? PE = 60/750 x 100 = 8% ### Q8: The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope and by how many metres was the measurement wrongto 1 decimal place. Solution: Let the actual length be x. PE = 6% AE = (x – 35)m [(x – 35)/x] X 100 = 6 Solve for x. 100(x – 35) = 6x 100x – 3500 = 6x 94x = 3500 x = 37.2m So, the measurement is 37.2 – 35 = ±2.2% wrong Solution: AE = 15cm AV = 165cm PE = ? PE = AE/AV x 100 = (15/165) x 100 = 9.1% ### Q10: The length and breath of a rectangle was mistakenly measured as 40m and 35m instead of 42.5m and 34.2m respectively. Find the percentage error in:(a) the area(b) the perimeter Solution: (a) the area Actual area = 40 x 35 = 1400m² Wrong area = 42.5 x 34.2 = 1453.5m² AV = 1400 AE = 1453.5 – 1400 = 53.5m PE = (53.5/1400) x 100 = 3.8% (b) the perimeter Perimeter of the actual rectangular = 40 + 35 = 75m Wrong Perimeter = 42.5 + 34.2 = 76.7m AE = 76.7 – 75 = 1.7m Therefore: PE = (1.7/75) x 100 = 2.3% ### Q11: The length of a file 25cm was measured as 27.5cm. Calculate the percentage error. Solution: Actual value = 25cm Wrong value = 27.5cm Absolute Error =2.5cm Percentage error = ? PE = (2.5/25) x 100 = 250/25 = 10% Scroll to Top
# How to write average rate of change The slope of a line tells us how something changes over time. The Average Rate of Change function describes the average rate at which one quanity is changing with respect to something else changing. John would like to find out how much money he saved per month for the year. This means that the average of all the slopes of lines tangent to the graph of f x between the points 0, —2 and 3, f 3 is For example, to calculate the average rate of change between the points: A line passing through the points 2, 5 and -3, 1 has a slope of Since this is a positive number, the line will appear to slope upwards to the right when graphed. John may want to analyze his finances a little more and figure out about how much he was saving per month. Take a look at how this can be solved. Find the average annual rate of change in dollars per year in the value of the house. When create a process or series of steps to do a certain task we are often creating a function. A man has driven a car 50 miles in one hour. This is called the rate of change per month. We use the two points 1, 50 and 4, If we find the slope we can find the rate of change over that period. The exact slope at one point defies our basic formula for slope since we need to know TWO points, and this will be approached differently. What is his average velocity speed over that next 3 hours? This is most likely the initial year or the year the house was built. The easiest way is to use point-slope form which is. The average velocity is the average rate of change of this distance with respect to time. I am given information about the year in which Linda purchased a house and the amount that the house is worth. Therefore, our two ordered pairs are 0, and 12, We use t and d to represent time in hours and distance in miles. So, we need another method! Connecting Slope to Real Life Why do we need to find the slope of a line in real life? In fact, if you take any two distinct points on a curve, x1,y1 and x2,y2the slope of the line connecting the points will be the average rate of change from x1 to x2 Example 1: In most real life problems, your units will not be the same on the x and y axis. Find the slope of the line going through the curve as x changes from 3 to 0.The calculator will find the average rate of change of the given function on the given interval, with steps shown. Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. You might have noticed that the Average Rate of Change function looks a lot like the formula for the slope of a line. In fact, if you take any two distinct points on a curve, (x 1,y 1) and (x 2,y 2), the slope of the line connecting the points will be the average rate of change from x 1 to x 2. The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in the table above did not change by the same amount each year, so the rate of change was not constant. The slope is the rate of change from one month to the next. Take a look at how this can be solved. InLinda purchased a house for \$, Inthe house was worth \$, Find the average annual rate of change in dollars per year in the value of the house. I can write them as an ordered pair. Special Note: If time is. Introductory Calculus: Average Rate of Change, Equations of Lines AVERAGE RATE OF CHANGE AND SLOPES OF SECANT LINES: The average rate of change of a function f (x) over an interval between two points (a, f (a)) and (b, f (b)) is the slope of the secant line connecting the two points. You divide the change in concentration by the time interval. > Consider a reaction "aA + bB → cC + dD" You measure the rate by determining the concentration of a component at various times. Please contact [email protected] with any questions. How do you calculate the average rate of a reaction? Chemistry Chemical Kinetics Rate Law. 1. How to write average rate of change Rated 0/5 based on 8 review
The fastest action by step guide for calculating what is 30 percent that 80 We already have our very first value 30 and also the 2nd value 80. Let"s i think the unknown worth is Y i m sorry answer we will find out. You are watching: 30 is what percent of 80 As we have actually all the required values us need, currently we have the right to put castle in a straightforward mathematical formula as below: STEP 1Y = 30/100 STEP 2Y = 30/100 × 80 STEP 3Y = 30 ÷ 100 × 80 STEP 4Y = 24 Finally, we have discovered the value of Y which is 24 and also that is ours answer. If you want to use a calculator to know what is 30 percent that 80, simply get in 30 ÷ 100 × 80 and also you will get your answer which is 24 Here is a calculator come solve percent calculations such together what is 30% the 80. You deserve to solve this type of calculation through your values by start them right into the calculator"s fields, and also click "Calculate" to obtain the an outcome and explanation. Calculate Question: in ~ a high school 30 percent of seniors walk on a mission trip. There to be 80 seniors. How many seniors go on the trip? Answer: 24 seniors go on the trip. See more: Convert 83 Cm Is How Many Inches Is 83 Cm? Convert 83 Centimeters To Inches Question: 30 percent the the kids in kindergarten prefer Thomas the Train. If there are 80 youngsters in kindergarten, how plenty of of them favor Thomas? Answer: 24 youngsters like cutting board the Train. Another step by action method Step 1: Let"s solve the equation because that Y by an initial rewriting it as: 100% / 80 = 30% / Y Step 2: fall the percentage marks to leveling your calculations: 100 / 80 = 30 / Y Step 3: main point both political parties by Y to move it top top the left next of the equation: Y ( 100 / 80 ) = 30 Step 4: To isolate Y, multiply both political parties by 80 / 100, we will have: Y = 30 ( 80 / 100 ) Step 5: computing the best side, we get: Y = 24 This leaves us v our final answer: 30% that 80 is 24 30 percent that 80 is 24 30.01 percent of 80 is 24.008 30.02 percent of 80 is 24.016 30.03 percent of 80 is 24.024 30.04 percent of 80 is 24.032 30.05 percent of 80 is 24.04 30.06 percent of 80 is 24.048 30.07 percent of 80 is 24.056 30.08 percent the 80 is 24.064 30.09 percent of 80 is 24.072 30.1 percent of 80 is 24.08 30.11 percent that 80 is 24.088 30.12 percent that 80 is 24.096 30.13 percent the 80 is 24.104 30.14 percent that 80 is 24.112 30.15 percent that 80 is 24.12 30.16 percent of 80 is 24.128 30.17 percent the 80 is 24.136 30.18 percent that 80 is 24.144 30.19 percent of 80 is 24.152 30.2 percent the 80 is 24.16 30.21 percent the 80 is 24.168 30.22 percent that 80 is 24.176 30.23 percent the 80 is 24.184 30.24 percent that 80 is 24.192 30.25 percent of 80 is 24.2 30.26 percent the 80 is 24.208 30.27 percent of 80 is 24.216 30.28 percent the 80 is 24.224 30.29 percent the 80 is 24.232 30.3 percent the 80 is 24.24 30.31 percent that 80 is 24.248 30.32 percent of 80 is 24.256 30.33 percent the 80 is 24.264 30.34 percent that 80 is 24.272 30.35 percent the 80 is 24.28 30.36 percent that 80 is 24.288 30.37 percent the 80 is 24.296 30.38 percent the 80 is 24.304 30.39 percent of 80 is 24.312 30.4 percent that 80 is 24.32 30.41 percent that 80 is 24.328 30.42 percent the 80 is 24.336 30.43 percent of 80 is 24.344 30.44 percent that 80 is 24.352 30.45 percent the 80 is 24.36 30.46 percent of 80 is 24.368 30.47 percent that 80 is 24.376 30.48 percent that 80 is 24.384 30.49 percent the 80 is 24.392 30.5 percent of 80 is 24.4 30.51 percent that 80 is 24.408 30.52 percent that 80 is 24.416 30.53 percent of 80 is 24.424 30.54 percent that 80 is 24.432 30.55 percent that 80 is 24.44 30.56 percent of 80 is 24.448 30.57 percent of 80 is 24.456 30.58 percent of 80 is 24.464 30.59 percent the 80 is 24.472 30.6 percent that 80 is 24.48 30.61 percent of 80 is 24.488 30.62 percent the 80 is 24.496 30.63 percent of 80 is 24.504 30.64 percent that 80 is 24.512 30.65 percent of 80 is 24.52 30.66 percent of 80 is 24.528 30.67 percent of 80 is 24.536 30.68 percent of 80 is 24.544 30.69 percent of 80 is 24.552 30.7 percent that 80 is 24.56 30.71 percent of 80 is 24.568 30.72 percent the 80 is 24.576 30.73 percent the 80 is 24.584 30.74 percent of 80 is 24.592 30.75 percent of 80 is 24.6 30.76 percent of 80 is 24.608 30.77 percent the 80 is 24.616 30.78 percent that 80 is 24.624 30.79 percent the 80 is 24.632 30.8 percent that 80 is 24.64 30.81 percent the 80 is 24.648 30.82 percent of 80 is 24.656 30.83 percent that 80 is 24.664 30.84 percent the 80 is 24.672 30.85 percent of 80 is 24.68 30.86 percent of 80 is 24.688 30.87 percent that 80 is 24.696 30.88 percent that 80 is 24.704 30.89 percent that 80 is 24.712 30.9 percent the 80 is 24.72 30.91 percent that 80 is 24.728 30.92 percent of 80 is 24.736 30.93 percent of 80 is 24.744 30.94 percent that 80 is 24.752 30.95 percent the 80 is 24.76 30.96 percent that 80 is 24.768 30.97 percent of 80 is 24.776 30.98 percent that 80 is 24.784 30.99 percent of 80 is 24.792
1.2: Order of Real Numbers $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ Order of Real Numbers Suppose that you and three friends were playing a game where you each drew a number from a hat and the person with the highest number won. Let's say that you drew the number 32, while your friends drew the numbers 3–√, 1.7, and π3, respectively. Could you figure out who won the game? Ordering Real Numbers Let's use the chart to categorize the following numbers: 1. 0 Zero is a whole number, an integer, a rational number, and a real number. 1. –1 –1 is an integer, a rational number, and a real number. 1. π/3 π/3 is an irrational number and a real number. 1. √36/9 √36/9=6/9=2/3. This is a rational number and a real number. Graphing and Ordering Real Numbers Every real number can be positioned between two integers. Many times you will need to organize real numbers to determine the least value, greatest value, or both. This is usually done on a number line. Let's plot the following rational numbers on a number line: 1. 2/3 2/3=0.667, which is between 0 and 1. [Figure 2] 1. −3/7 −3/7 is between –1 and 0. [Figure 3] 1. 57/16 57/16=3.5625 Now, let's compare π/15 and √3/√75: First we simplify in order to better compare: √3/√75=√3/(5√3)=1/5. Now we rewrite π/15 to compare it to 1/5: π/15=π/(3×5)=π/3×1/5. Since π>3, π/3>1 so π/3×1/5>1/5. Therefore, π/15>√3/√75. Examples Example 1 $$\PageIndex{1}$$ Earlier, you were asked to determine which number out of 3/2, √3, 1.7, and π/3 is the highest. Solution Notice that 3/2 = 1.5. Also, since π≈3.14, π/3≈1. Using a calculator, √3≈1.73. Thus the order of the numbers is π/3 < 3/2 < 1.7 < √3 and √3 is the largest number. Example 2 $$\PageIndex{1}$$ For the numbers: √12, (1.5)⋅(√3), 3/2, 2√5/√20, classify each number. Solution We need to simplify the numbers in order to classify them: √12 = √((4x3)/2) = (2√3)/2 = √3. This is an irrational number. An irrational numbers is a type of real number. (1.5)⋅(√3). This number cannot be simplified, but since it is a multiple of an irrational number, it is also irrational. In other words, we cannot get rid of the irrational part, and so we cannot write it as a rational number. It is also a real number. 3/2. Since this number is in the form of a proper fraction, it is also a rational number and real number. (2√5)/√20 = (2√5)/(√4x5) = (2√5)/(2√5) = 1. This number can be simplified to an integer. All integers can be expressed as rational numbers and are a special kind of real number. Example 3 $$\PageIndex{1}$$ For the numbers: √(12)/2, (1.5)⋅(√3), 3/2, (2√5)/√20, order the four numbers. Solution The four numbers are ordered as follows: 1<3/2<√3<(1.5)⋅√3. 1<3/2 since the numerator is larger then the denominator and 3/2=1.5. 3/2<√3 since we can see on our calculators that √3≈1.7 √3<(1.5)⋅√3 since multiplying by 1.5 makes any number larger. Review Classify the following numbers. Include all the categories that apply to the number. 1. √0.25 2. √1.35 3. √20 4. √25 5. √100 6. Place the following numbers in numerical order from lowest to highest. 6–√261501.5−−−√1613 7. Find the value of each marked point; Mixed Review 1. Simplify (9/4)÷6. 2. The area of a triangle is given by the formula A=b(h)/2, where b= base of the triangle and h= height of the triangle. Determine the area of a triangle with base =3 feet and height =7 feet. 3. Reduce the fraction 144/6. 4. Construct a table for the following situation: Tracey jumps 60 times per minute. Let the minutes be {0,1,2,3,4,5,6}. What is the range of this function? PLIX: Play, Learn, Interact, eXplore: Number Lines: Freezing Cold Comparison Real World Application: It's Elementary! Practice: Order Real Numbers Video: Graphing Real Numbers on a Number Line - Example 1 This page titled 1.2: Order of Real Numbers is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
# Lesson video In progress... Hi, I'm Miss Davies. In this lesson, we're going to be adding and subtracting decimals. when adding and subtracting decimals, it's just like adding and subtracting integers. We can work these two questions out in the same way. Note this, with the right hand question that has the decimal points in it, all of the decimal points are aligned with each other. When we are adding either decimals or integers, we start from the right hand side. Seven add five is 12. We can write this as a two in the ones column and then exchange the one for one ten. We can do the same with the decimal question. Seven hundreds add five hundreds is the same as 12 hundreds. These can be written as two hundreds and one tenth. Three add one in the left hand question is the same as three tens add to one ten. Then we need to add the other one ten. So we have five tens in total. In the right hand decimal question, we've got three tenths, add one tenth, add one tenth, which is five tenths. We've now got zero hundreds, add seven hundreds, which is 700. On the right time question, Zero ones add seven ones, which is seven ones. Finally, for the integer question, 2000 add 4,000 is 6,000. And for the decimal, two tens add four tens is six tens. You can see that our answers have the same digits in the same order. The right hand question simply have a decimal point in it. Subtracting decimals is just like subtracting integers. We look at these two questions, all of the digits are the same. The right hand question, this has decimal points in it. Notice that all of these decimal points are in line with each other. When we are subtracting, whether it is integers or decimals, we start from the right hand column. In our integer question, this is seven ones subtract four ones, which is three ones. In our decimal question, We've got seven hundredths subtract four hundredths, which is three hundredths. Our next column in the integer question is three tens subtract one ten, which is two tens. In our decimal question, we have three tenths subtract one tenth, which is two tenths. Our next column in the integer question, we have zero hundreds subtract two hundreds. In order to complete this, to give a positive result, we need to exchange our two thousands for a one thousand and 10 hundreds. We now have 10 hundreds subtract two hundreds, which is eight hundreds. On our decimal question, we're going to do the same. We're going to exchange 20 for 10 and 10 ones. The 10 ones subtract two ones is eight ones. Our final calculation is 1000 subtract 1000, which is zero. And in our decimal question, it is one ten subtract one ten, which is zero. You can see that the answers are made from the same digits in the same order. But our decimal question has a decimal point. Here's some questions for you to try. Pause the video to complete your task and resume once you're finished. Here are the answers. Make sure that your decimal points are in line vertically, this ensures that all of the place values are in the correct place. Here's some questions for you to try, pause the video to complete your task and resume once you're finished. Here are the answers. Make sure that you've aligned your decimal point vertically. Also remember to include any exchanges that you have done. When working with decimals with a different number of decimal places, it is important to line up the decimal point and the place values. If we are calculating 45. 67 subtract 1. 63, this is not the correct way to set it out. You can see that our decimal points aren't in line and the place values also aren't in line. This is the correct way to set this out. You can see that our three decimal points are all aligned with each other. The place values are also aligned. Seven hundredths subtract three hundredths is four hundredths. Six tenths subtract six tenths is zero tents. Five ones subtract one one is four ones. Then we have four tens subtract nothing, which is four tens. We can check that this answer is correct by doing the inverse. We can calculate 44. 63. If we've done our question correctly, we should get an answer of 45. 67. Let's start from the right column. Four hundredths add three hundredths, is seven hundredths. Zero tenths add six tenths is six tenths. Four ones add one one is five ones. Four tens we're not adding anything, so that leaves us with four tens. This means that our working out is correct. Here's some questions for you to try, pause the video to complete your task and resume once you're finished. Here are the answers. Here is for the difference between the numbers in each column, in the tens column, five subtract seven is not equal to two. The correct answer is 0. 83. Here's some questions for you to try, pause the video to complete your task and resume once you're finished. Here are the answers. For part A you need add together 1. 69 and 2. 49, this gives an answer \$4. 18. Impart B, you need to add together 1. 39, 2. 49 and 1. 89. This gives an answer of 5. 77. That's all for this lesson. Thanks for watching.
Instruction 1 In the first case known to such data in the triangle, as the angle and length of sides that form this angle. The side opposite the known angle, you need to find the cosine theorem, according to which you want the length of the known sides squared up and folded, then subtract from the obtained sum the product of these sides multiplied by two and the cosine of known angle. The formula for this calculation looks like the following: h = √(e2+f2 – 2efcosA) where: e and f are the lengths of the known sides; h – the unknown leg (or side); A – the angle formed by the known sides. 2 In the second case, when we know two angles and the side between them of the triangle to use the theorem of sines. According to this theorem, if you divide the sine of the angle to the length of the opposite leg, we get a ratio equal to any other in this triangle. Also, if you do not know the right side, you can easily find knowing the fact that the sum of the angles of a triangle is equal to hundred eighty degrees. This statement can be represented in a formula as follows: SinD/d = a sinF/f = sinE/e, where: D, F, E – values of the opposite angles; d, f, e are the legs, the opposite corresponding angles. 3 In the third case, known only to the corners of this triangle, so it is impossible to know the length of all sides of this triangle. But you can find the attitude of these parties and the method of selection is to find a similar triangle. The aspect ratio of this triangle is found using the formulation of the system of three equations with three unknowns. Here is the formula for compiling: d/sinD f/sinF e/sinE, where: d, f, e are the unknown sides of the triangle; D, F, E – the angle opposite the unknown sides. 4 This equation is solved as follows: d/sinD = f/sinF = e/sinE (dsinFsinE-f sinD* sinE-e* sinD* sinF)/ sinD* sinE* sinF.
# Formula For Data Sufficiency ## Formula for Data Sufficiency and Definitions Important Formula for Data Sufficiency is given here on this page. In data sufficiency, questions are framed to examine students reasoning, logical and decision-making skills. For this, you must read the questions properly and based on the information given in the options you have to decide whether the information is sufficient to answer or not. ## Understanding Solution Selections: For solving data sufficiency questions you need to first understand the question and based on that, selection should be made. The questions will be asked with certain information with two statements labeled I and II. You need to analyze whether the information given in the label I and II is adequate to come to the solution or not. You should use all the data given in the statements along with your mathematics skill and well-known facts (such as sun rises in the east or the meaning of niece), the selection is to be made from below given options: The answer can be obtained from Statement I alone but statement II alone is not sufficient to answer the question asked; A. Data in statement 1 is sufficient alone to determine the answer. B. Data in statement 2 is sufficient alone to determine the answer. C. Data in either of the statements is sufficient to determine the answer. D. Data provided in both the statements together are not sufficient to determine the answer. E. Data in both statements are required to determine the answer. Note: In data sufficiency problems, you can mark as “the data given in the statements is satisfactory” only when there is outcome or solution is in the numerical value. ### Question 1. Who is the father of A. Statement I: x and y are brothers. Statement II: y’s wife is the sister of A’s wife. A. Data in statement 1 is sufficient alone to determine the answer. B. Data in statement 2 is sufficient alone to determine the answer. C. Data in either of the statements is sufficient to determine the answer. D. Data provided in both the statements together are not sufficient to determine the answer. E. Data in both statements are required to determine the answer. Explanation: The statement I gives the relation between X and Y. So statement I alone cannot determine the answer. Through statement II we can determine that Y is the brother-in-law of A. Thus using both the statements you cannot determine the father of A. hence option D is correct. ### Question 2. In which year was Dhoni born? Statement : At present Dhoni is 21 years younger to his father. Dhoni’s brother, who was born in 2001, is 28 years younger to his father. 1. 1 alone is sufficient while 2 alone is not sufficient 2. 2 alone is sufficient while 1 alone is not sufficient. 3. Either I or II is sufficient 4. Neither I nor II is sufficient 5. Both I and II are sufficient Answer: Both 1 and 2 are sufficient Explanation – From both I and II, we find that Dhoni is (28 – 21) = 7 years older than his brother, who was born in 2001. So, Dhoni was born in 1996. ### Question 3. What will be the total weight of 80 books, each of the same weight ? Statements: One-fourth of the weight of each book is 5 kg. The total weight of three books is 20 kilograms more than the total weight of two books. 1. I alone is sufficient while II alone is not sufficient. 2. II alone is sufficient while I alone is not sufficient 3. Either I or II is sufficient 4. Neither I nor II is sufficient 5. Both I and II are sufficient Answer: Either 1 or 2 is sufficient Explanation From I, we conclude that the weight of each book = (4×5) kg = 20 kg. So, total weight of 10 books = (20 x 10) kg = 200 kg. From II, we conclude that: Weight of each book = (weight of 3 book) – (weight of 2 book) = 20 kg. So, total weight of 10 books = (20 x 10) kg = 200 kg. ### Question 4. How many children does Mahi have ? Statements: Harsha is the only daughter of Y who is the wife of Mahi. Karan and Jivan are brothers of Mahi. 1. I alone is sufficient while II alone is not sufficient 2. II alone is sufficient while I alone is not sufficient 3. Either I or II is sufficient 4. Neither I nor II is sufficient 5. Both I and II are sufficient Explanation From I, we conclude that Harsha is the only daughter of Mahi. But this does not indicate that Mahi has no son. The information given in II is immaterial. ### Question 5. How much was the total sale of the shop ? Statements: The shop sold 2500 units of masks each costing Rs. 15. This company has no other product line. 1. I alone is sufficient while II alone is not sufficient 2. II alone is sufficient while I alone is not sufficient 3. Either I or II is sufficient 4. Neither I nor II is sufficient 5. Both I and II are sufficient Answer: Either 1 or 2 is sufficient Explanation – From I, total sale of 8000 units of masks = Rs. (2500 x 15) = Rs. 37500. From II, we know that the company deals only in masks This implies that sale of masks is the total sale of the shop, which is Rs. 37500. ### Related Banners Get PrepInsta Prime & get Access to all 200+ courses offered by PrepInsta in One Subscription ## Get over 200+ course One Subscription Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others
# Surface area of 3-dimensional shapes This chapter talks about line symmetry, rotational symmetry and transformations, and surface area of 3-dimensional shapes. We will learn how to find the lines of symmetry in 2-dimensional shapes, and whether a 2-dimensional shape has rotational symmetry. Also, we will learn how symmetry can help us find surface area of 3-dimensional shapes. Symmetry can be found all around us. The fact that we are learning symmetry now is because our ancestors were so amazed by the symmetry in nature that a stream of study was developed! A line of symmetry is a line that divides an object into two and produces two mirror images of each other. The lines of symmetry can be in any directions: horizontal, vertical or diagonal. An object can have more than one line of symmetry. The number of lines of symmetry varies in different polygons. In the second part of the lesson, we will then explore rotational symmetry in 2-dimensional shapes. To say a figure has rotational symmetry means that when this figure turns about its centre of rotation, it will still look the same after a certain amount of rotation. Different shapes have different order of rotation and angle of rotation. Order of rotation refers to the number of times that the figure looks the same in one complete turn; while the angle of rotation is the smallest angle that the figure needs to turn so that it still looks the same. Like the line of symmetry, the order and angle of rotation can be different in different polygons too. For example, a regular hexagon has an order of rotation of 6 and its angle of rotation is 60°. The last part of this lesson focuses on how to find the surface area of 3-dimensional shapes. Surface area is the total area of all faces of a shape. Thanks to our ancient mathematicians, we have formulas to help us find surface areas of many regular objects. ### Surface area of 3-dimensional shapes In this section, we will learn how to calculate the surface area of 3D objects. We will also look further into the subject ? What would happen to the surface area if the shapes are cut into pieces? How about a piece is cut out of the shape?
### By: Ishani Sharma • Introduction to Parabolas • Key Features of Quadratic Relations • Types of Equations: Vertex form • Types of Equations: Standard form • Types of Equations: Factored form • Ways to Represent a Quadratic Relation • Investigating Vertex Form • Graphing in Vertex Form • Finding an Equation in Vertex Form Using a Graph 4.3 Transformations of Quadratics in Vertex Form • Transformations of Parabolas • Writing Equations from Given Transformations • Graphing Transformations of Quadratic Relations • Mapping Notation 4.4 Graphing Quadratics in Vertex Form • Graph using Step Pattern • Graph using Mapping Notation • Word problems for quadratics in vertex form • Finding x and y intercepts in vertex form Mini Test #1 5.1 Multiplying Binomials 5.2 Special Products 5.3 Common Factoring 5.4 Factoring Simple Trinomials 5.5 Factoring Complex Trinomials 6.2 Solving Quadratics by Factoring (finding the zeros) 6.3 Graphing Quadratics in Factored Form • Graph using x intercepts • Finding the Vertex • Word Problems for Quadratics in Factored Form Mini Test #2 6.1 Maximum and Minimum Values (Completing the Square) • How to go from Standard Form to Vertex Form • Finding the x intercepts when factoring is not possible 6.5 Word Problems in Standard Form Mini Test #3 Reflection • Math Homework Help ## Introduction to Parabolas • Parabolas can open up/down • The zero is when the parabola crosses the x-axis • Zeros can also be called x-intercepts or roots • The axis of symmetry divides the parabola in half equally • The vertex of a parabola is where the parabola and axis of symmetry meet • The vertex is where the parabola is at its maximum or minimum value • The optimal value is the value of the y co-ordinate of the vertex • The y-intercept is where the graph crosses the y-axis ## 1) Table of Values A table of values is used to find the first and second differences, When the first differences are the same it is a linear relation and when the second differences are the same it is a quadratics reaction. In order to find the first differences you must subtract the terms in the y-value from each other, and to find the second differences you must subtract the first difference terms from each other. ## 2) Graphs When graphing or looking at graphs the line unlike a linear relation must be curved. ## Investigating Vertex Form The Vertex Form y= a(x-h)^2 + k gives us the the value of the axis of symmetry and optimal value. The value of "h" gives us the axis of symmetry. The value of "k" gives us the optimal value, and together "h" and "k" gives us the vertex of a parabola. The vertex form also gives away, the vertical stretch or compression by a factor of "a". ## Graphing in Vertex Form INSERT VIDEO HERE (DONE) ## Finding an Equation in Vertex Form Using a Graph INSERT VIDEO HERE (DONE) ## Transformations of Quadratics in Vertex Form Consider a parabola: y = 3(x-1)^2+2 y = a(x-h)^2+k Translations: The parabola will open up , it is vertically compressed by a factor of 3. It is horizontally translated to the right by 1 unit and vertically translated up by 2 units. Vertex Form: • The "a" determines weather it is vertically compressed or stretched (narrow or wide). If the "a" value is greater than 1, the parabola will be vertically stretched and if the value of "a" is less than 1 the parabola will be vertically compressed. • The "h" value will gives us the horizontal translation of right or left. • The "k" value will give us the vertical translation of up or down. • The sign in front of the "a" value will determine if the parabola open ups or down, If the sign is a positive it will open up, and if the sign is a negative, it will open down. • Remember that the "h" value will change signs when brought out from the bracket, if it is a negative in the bracket it will become a positive outside the bracket, therefore will move to the right, and vice-versa. ## Writing Equations from Given Transformations If given the transformation of a parabola, for example: The graph of y = x^2 is opening downward, stretched vertically by a factor of 3, translated horizontally to the left by 4 units and translated vertically down by 5 units. The equation would become: y = -3(x+4)^2 - 5 ## Mapping Notation Mapping notation is used to accurately graph any quadratic relation. In general, to go from the graph of y = x^2 to y = a(x-h)^2 + k you can use the mapping notation: (x,y) = x + h, ay + 1 Example: a) y =5(x-4)^2 + 2 will become x + 4, 5y + 2 b) y =(x+7)^2 will become x - 7, y c) y =(x-3)^2 + 5 will become x + 3, y +5 Untitled ## Word Problems for Quadratics in Vertex Form November 30, 2015 ## Finding x and y intercepts in Vertex Form November 30, 2015 ## 5.1 Multiplying Binomials Simplifying polynomials includes both collecting like terms and use of the distributive property. It is important to be sure to use BED MAS at all times. Exponent rules must be used for both multiplication and division when simplifying. When multiplying binomials there are two rules you must follow 1. Distributive Property 2. Collect like terms Example 1: (x+3)(x+4) 1. We will use the distributive property to multiply the first two terms 2. Then we will multiply the two inner terms with the outer terms and add the outcome, in this case the outer term is 1 so you would simply add 3 and 4 3. Then we will multiply the two inner terms with each other x^2 + 4x + 3x + 12 1. Lastly we will add the like terms together x^2 + 7x + 12 Reminder - Multiply coefficient + add power 2x^2 (2x) 4x^3 Example 2: 2x(3x-4) + (x+3) - (2x-7) (6x^2-8x) + (x+3) - (2x-7) (6x^2-7x+3) - (2x-7) 6x^2 - 7x - 2x + 3 + 7 6x^2 - 9x + 10 ## 5.2 Special Products The formula for factoring a special product is a^2 + 2ab + b^2 = (a+b)^2 and (a-b)^2 Example: (3y+7x) = a^2 + 2ab + b^2 (3y)^2 + 2(3y)(7x) + (7x)^2 9y^2 + 42xy^2 + 49x^2 ## 5.3 Common Factoring Factoring is the opposite of expanding, the three methods of factoring are: 1. Finding the GCF 2. Common Factoring 3. Grouping 1. When factoring by GCF, you simply find the greatest common factor of the terms , for example the terms 5c + 10d both have 5 in common, so you divide both terms by 5. You always keep the common factor outside of the bracket and place everything else inside of the bracket, SO 5(c+2d). 2. Common Factoring is when the two binomials are exactly the same, for example 5x(3x+2) + 4(3x+2), since the two binomials are the same you will put them as one bracket and place the remaining into another bracket so, (3x+2)(5x+4). 3. When grouping factors you group a set of two like terms into brackets • ax + ay + 2x + 2y • (ax+ay) + (2x+2y) Then you will remove the like term to the outside of the bracket • a(x+y) + 2(x+y) Now since you can see that there is a common binomial you follow the rules for Common factoring • (a+2)(x+y) ## 5.4 Factoring Simple Trinomials The equation used for factoring simple trinomials is ax^2 + bx + c Where x is a variable and "a", "b", "c" are constants. You can factor a quadratic in standard form to get factored form: x^2 + bx + c = (x+r)(x-s) STD FORM Factored Form Where "r+s" is" b" and " rs" is "c" Step 1: Find the product and sum • find the two numbers whose product is "c" • find the two numbers whose sum is "b" Example 1: x^2 + 6x + 5 1. Find the two numbers whose product when multiplied is 5 (1)(5) = 5 2. Find the two numbers when added together equals to 6 5+1 = 6 3. The factored form will be (x+5)(x+6) Look at the signs of "b" and "c" to make it easier to figure out the sign of the two numbers • if "b" and "c" are positive, both "r" and "s" • if "b" is negative and "c" is positive , both "r" and "s" are negative • if "c" is negative , only ONE of "r" or "s" is negative • if both "c" and "b" are negative, only ONE of "r" or "s" is negative Example 2: n^2 + 6n + 8 (4)(2) = 8 4 + 2 = 6 (n+4)(n+2) ## 5.5 Factoring Complex Trinomials Remember we are trying to break down the middle term so that it equates to 4 terms at which we can proceed to factor by grouping. 1. Always look for a common factor 2. To factor ax^2 + bx + c, find the two integers whose product is "ac" and whose sum is "b", similar to simple trinomial factoring Example 1: 3x^2 + 8x + 4 (3)(4) = 12 (2)(6) = 12 2 + 6 = 8 Break up the middle term (3x+6x) + (2x+4) Factor by Grouping 3x(x+2) + 2(x+2) Collect the binomials (3x+2)(x+2) You follow the same steps when factoring a trinomial with two variables. ## 6.2 Solving Quadratics in Factored Form When solving quadratics in factored form or better known as finding the zeros you must follow the following steps: 1. Make one side equal to zero 2. Set each bracket equal to zero 3. Solve for "x" Example 1: x^2 + 9x + 14 = 0 x^2 + 7x + 2x + 14 = 0 x(x+7) + 2(x+7) = 0 (x+2)(x+7) = 0 x+2=0 x+7= 0 x= -2 x= -7 (-2,0) (-7,0) ## 6.3 Graphing Quadratics in Factored Form Using the example from above, you can than proceed find the axis of symmetry by adding the two x intercepts together and dividing by two (-2)+(-7) / 2 = -4 The -4 is the "x" value The AOS can used to find the optimal value (y intercept) y = (x+7)(x+1) y = (-4+7)(-4+1) y = (3)(-3) y = -9 This is the "y" value Now since you have the "x" and "y" values you can put them together to make the vertex, so the vertex in this case would be (-4,-9). At the end of this you have 4 points: 1. (0,7) 2. (-7,0) 3. (-1,0) 4. (-4,-9) You can use the points above to plot your graph ## 6.1 Maximum and Minimum Values Since not all quadratic functions are written in Vertex Form, when functions are written in standard form you can covert them into vertex form using a process called completing the square. Completing the Square Steps: 1. Group the x terms 2. Divide the coefficient of the middle term by 2, square it, then add and subtract that number inside the brackets 3. remove the subtracted term from the brackets 4. factor the brackets as a perfect square trinomial Example 1: y = x^2 - 6x + 4 y = (x^2 -6x) + 4 y = (x^2 - 6x + 9 - 9) + 4 y= (x^2 - 6x + 9) - 9 + 4 y = (x-3)^2 -5 Vertex = (3,-5) AOS = 3 Max or Min value = min -5 Values that "x" may take = all real numbers Values that "y" may take = y greater than -5 When there is a number in front of the "a" value you must first divide the middle term by that number and place then proceed with the steps. You must also multiply the square outside of the brackets. When there is a negative in front of the "a" value you must multiply the middle term by the negative, if it is already a negative it will become a positive and if it is a positive it will become a negative. You must also multiply the square outside of the bracket by the negative. The quadratic can be useful when a quadratic function cannot be factored. ## The discriminant All quadratic equations of the from ax^2 + bx + c can be solved using the quadratic equation. But you cannot figure out how many solutions a quadratic equation has. In order to figure that out you must find the discriminant. The discriminant is the value under the square root in the quadratic formula. The formula to find the discriminant is D = b^2 - 4ac When the equation gives you a positive number or a number greater than 0, there will be 2 solutions (x intercepts). When the equation gives you a negative number or a number less than 0, there will no solutions (x intercepts). When the equation gives a number equal to 0, there will be 1 solution (x intercepts). ## 6.5 Word Problems in Standard Form November 30, 2015 ## Reflection This unit was quite interesting for me. Although many students question why we must learn things that seem useless at the time, learning about parabolas has been a real eye opener. Before this unit i was living ignorant to the vast majority of things in the world around me, i just recently noticed that the M in the McDonalds sign uses two parabolas or that the curves in your windows are parabolas, now many can say that's useless information but its amazing how everything in the world is connected, you cannot learn one subject and not expect there to be bits of another tied into it. For example in math, communication counts for a part of your overall mark and communication is basically English. So similar to that math is related to the world, obviously in a bigger picture and when you learn something like quadratics you should always except to encounter it at some point in your life and i am glad that i have the opportunity to get an education and learn about such information. As for the unit itself, i must say it was a struggle and required long nights of studying and preparation but in the end i am satisfied with how much effort i put in and i have learnt a very important life lesson from the hard work i put into this unit, which is that nothing in life comes easy, however perseverance will lead can lead you to victory. Although i did not achieve the nest marks i believe i tried and never gave up which for me means that i am winner.
# Properties of Multiplication ### Text-only Preview Properties of Multiplication Properties of Multiplication An easier insight into how problems are factored can be shown through the four properties of multiplication. The name commutative holds definition in the prefix of the word, commute, meaning to move from place to place.The Commutative Property shows how two numbers in a multiplication number will result in the same answer despite the order, 32=6 and 23=6. Multiplication and addition both use the Commutative Property, for example: a+b=b+a and abc=bca.An addition or multiplication problem may cite the Commutative Property, which wil simply mean that it is okay to move numbers around. The Associative Property is named for associating or grouping and is also used in addition and multiplication.The Associative Property states the irrelevance of number order within a group of numbers that are multiplied or summed. In the example, a(cb)=c(ba) and a+(c+b)=b+(a+c), the Associative Property is demonstrated. Only the order of numbers would be need to be regrouped if a problem asked for rearrangement through the Associative Property. Know More About :- Independent Variable Examples Math.Tutorvista.com Page No. :- 1/4 Any number multiplied by one remains the same number, as stated by the Multiplicative Identity Property.15=5 as well as 110=10as defined by this property. The most complex of the four properties, the Distributive Property, is used to factor out a set of numbers.The term distributive comes from the properties use of distribution of multiplication This property can be show through an example, such as: a(b+c)=ab+ac, and also as ax+b=a(x+b). This property can also be used for multiplication over subtraction as displayed in the example: 5x-15=5(x-3), in that the 5 is distributed to factor into parentheses just as it is factored out of the parentheses. Flexibility within the Distributive Property's definition is shown through the above example as subtraction is used; subtraction can also be the addition of a negative number. The examples and definitions of these four properties show how and why math rules are used.Math skills are sharpened and ready for more complex quantitative problems once these essentials of multiplication are understood. Multiplication Methods Below are the multiplication methods: Method 1: Multiplying integers of 345 x 6 Solution: Math.Tutorvista.com Page No. :- 2/4 Here, 5 is at unit place, so we do 5 x 1 = 5 x 6 = 30 4 is at 10's place, so we do 4 x 10 = 40 x 6 = 240 3 is at 100's place, so we do 3 x 100 = 300 x 6 = 1800 The product result = 2070 Method 2: Multiplying integers of 345 x 6 Solution: 345 x 6 -------------- 3 0 2 4 0 1 8 0 0 -------------- 2 0 7 0 -------------- Math.Tutorvista.com Page No. :- 4/4 ThankYouForWatching Presentation • ﾿
In this episode of TuesdACT, we are taking a look at the principles of multiplying and adding even and odd numbers using an example from The Real ACT Prep Guide that lots of students struggle with. Real ACT Prep Guide Test 4 Question 48 on page 599, we are coming for you! Check out the video or read below for the explanation and a little lesson on even and odd integers. If is a positive integer, which of the following expressions must be an odd integer? F. G. H. J. K. Alright, let’s pause and go over principles of adding, subtracting, and multiplying odd and even numbers. Here’s how it shakes out. Even + Even = Even Even + Odd = Odd Odd + Odd = Even Even x Even = Even Even x Odd = Even Odd x Odd = Odd You don’t need to memorize the rules; if you forget, just go through the examples with easy numbers like 2 and 3 and see what happens. It will be the same for all even and odd numbers. So, for example: Even + Even = Even 2 + 2 = 4 Even + Odd = Odd 2 + 3 = 5 Odd + Odd = Even 3 + 3 = 6 Even x Even = Even 2 x 2 = 4 Even x Odd = Even 2 x 3 = 6 Odd x Odd = Odd 3 x 3 = 9 Question 48 is also a great example of how you can find important clues in the answer choices on the ACT. Notice that all of our answer choices have the number 3 in them, so all of our answer choices are doing something with an odd number. Now let’s test with values for . Don’t forget that we don’t know whether is even or odd, so we need to test for both cases. 3 is raised to a power of . If is even (2), this gives us 3 x 3 (ODD x ODD = ODD). If is odd (3), this gives us 3 x 3 x 3 (ODD x ODD x ODD), which is also always odd. Since an odd number multiplied by an odd number is always an odd number, and because in this case we are just multiply 3 (an odd number) by itself a number of times, it will always be odd. So F is our answer, but let’s check the rest. is raised to a power of 3. So if is odd, then we have 3 x 3 x 3 (ODD x ODD x ODD) and we know that ODD x ODD = ODD. But if is even then we have 2 x 2 x 2, and EVEN x EVEN = EVEN. So is not necessarily odd. . So if is even (2), we have 3 x 2 (ODD x EVEN = EVEN); if is odd, we have 3 x 3 (ODD x ODD = ODD). So is not necessarily odd. . This is an interesting one, because if we pick a value for that does not divide evenly by 3, then it is actually neither even nor odd. By rule, the terms even or odd apply only to integers not to fractions.
## Finding the Slope of a Line There might be a question on the GED math test that gives you two point coordinates and/or a line on a coordinate plane and asks you to provide the slope of the line. The formula for finding the slope is provided for you in the formula sheet: This might look more complicated than it really is. All you have to do is count the up and down difference between the two points, and the left and right difference between the two points. Then divide that first number you got by that second number. If the slope is going up from left to right then the slope is positive, if down, negative. Let’s look at an example. What is the slope of the line on this graph? First count how many spaces are between the two points vertically (up and down), then the spaces horizontally (left and right). So the change in y is 4 and the change in x is 3. Now you divide the y change by the x change. So the slope is one and a third. We know that this should not be a negative number since the line is going up from left to right rather than down. If it were going down then the slope would be If there is a totally flat line: then the slope is zero. If there is a line that goes straight up and down: then there is no slope. ## Finding the Slope by Counting Depending on your learning style, it might be easier and save you time to find the slope of a line just by counting and remembering that slope is “rise over run”. This is demonstrated in the video below. Share and Enjoy: 1. Patti says: Your graph above is great, but you have incorrectly identified the point in the 3rd quadrant. It should be (-1, -2) instead of (-2, -1). That’s an easy mistake to make and it would be great if you corrected it! Thank you! ðŸ™‚ • Ziyi says: You’re right, it should be (x-axis,y-axis), and thanks for catching that and pointing it out to me! It should be fixed now. • nekke says: I thank God I found this website, I’ve struggled with math for a while, so glad to have found it. 2. K says: But when you divided 3 by 4, why did you get 1 1/3? I just got 1. • Ziyi says: I divided 4 by 3. One way to think about it is how many threes can go into a 4. One and a third threes can go into a four. 3. jennifer says: the only book compared to ur teaching is the actual ged book and its great but after awhile u already memorized the answers this one help out quite a bit is there a way for to send a example of the ged or can u do that? i just have to pass my math too 4. Yvonne says: Thank you for sharing this information for GED. I have passed all of my test except the math.After studing your formats, I beleive I’ll be able to past when I take my next test. 5. Jacqueline says: I would like to say that the plane grids were my weakness beside fractions, i can’t believe how detailed and easy to follow these are. I truly appreciate this learning experience. I just have my math to get, i failed twice and hopefully the third is a charm Thanks again ðŸ™‚ 6. olivia says: All I need to take is my math want to know what web site I should go to so that I can practice n my math a little more before I take my tset and hopefully pass it 7. liz says: I could never get this part in the ged workbook I studied. This makes it super simple to understand. 8. this web site is helpful . 9. Colleen says: Wow thank you so much….my son read your page and was able to refresh his head and do his weekend homework! 10. moises says: what if the the change in y were 3 units and x 4? what would you have gotten if you divided 3/4?
# Which pair of numbers has an LCM of 16 • 3 and 16 2 and 4 4 and 8 4 and 16 In this question, we have to find the pair of numbers for which the LCM is 16.LCM stands for Least Common Multiple, defined as the smallest multiple common number between the required numbers for which LCM is to be determined. It is the smallest positive number that is divisible by all given numbers. LCM can be determined between 2 or more than 2 numbers.LCM can be found by three methods: 1. LCM by using prime factorization 2. LCM by using repeated division 3. LCM by using multiple Here, we will find the LCM by using the method of multiples i.e. finding the common multiplies between the 2 given numbers and then selecting the smallest among them as the LCM for that pair. The LCM for each pair is calculated as followsThe LCM of $3$ and $16$ will be:$3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, …$$16 = 16, 32, 48, …$Common Multiple is $48$. As it is the smallest common multiple, hence:$LCM = 48$The LCM of $2$ and $4$ will be:$2 = 2, 4, 6, 12, …$$4 = 4, 8, 12, …$Common Multiples are $4,8, …$. As the smallest common multiple is $4$, hence$LCM = 4$The LCM of $4$ and $8$ will be:$4 = 4, 8, 12, 16, 20, 24, …$$8 = 8, 16, 24, …$Common Multiples are $8,16, …$. As the smallest common multiple is $8$, hence$LCM = 8$The LCM of $4$ and $16$ will be:$4 = 4, 8, 12, 16, 20, 24, 28, 32, …$$16 = 16, 32, …$Common Multiples are $16, 32, …$. As the smallest common multiple is $16$, hence$LCM = 16$ ## Numerical Results: So the required pair of numbers for which the LCM is $16$ is $4$ and $16$ ## Example: Find out which of the following pairs has the LCM of $24$.$a)$ $3$ and $8$$b) 2 and 12$$c)$ $6$ and $4$$d)$ $4$ and $12$ ### Solution: The LCM of $3$ and $8$ will be:$3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, …$$8 = 8, 16, 24, 32, 40, 48, …$$LCM = 24$The LCM of $2$ and $12$ will be:$2 = 2 ,4, 6, …$$12 = 12, 24, 36, 48, 60, 72, …$$LCM = 12$The LCM of $4$ and $6$ will be:$4 = 4, 8, 12, 16, 20, …$$6 = 6, 12, 18, 24, …$$LCM = 12$The LCM of $4$ and $12$ will be:$4 = 4, 8, 12, 16, 20, …$$12 = 12, 24, 36, 48, 60, 72, …$$LCM = 12$So the required pair is $3$ and $8$.Image/Mathematical drawings are created in Geogebra.
# A right cicular cylindrical can is to be constructed to have a volume of 57.749 cubic inches (one quart). The sides of the can are to be formed by rolling and welding a strip of metal, which may be purchased in rolls with width equal to the desired height of the can. The material for the sides costs 20 cents per square foot. The welding cost is 1.1 cents per inch. Top and bottom of the can are circles cut from hexagons to minimize the waste. the width of the hexagon is the diameter of the can plus 0.4 inches. the extra inches is crimpled over the sides to form the seal. Crimping costs are 1.6 cents per inch and the material for the ends sells for 30 cents per square foot. The metal for the ends of the cans may be purchased in a rolls which allows for exactly 4 hexagons and exactly 3 hexagons in alternate strips. Find the dimensions of the most economical can which can be constructed to meet these specifications. Write equations for the amount of both rolls of material needed to produce seven cans, as a function of the diameter of the can. The height can be expressed in terms of the required volume, V, and the diameter, D. V = (1/4) pi D^2 h h = 4 V/[pi D^2] Then compute the total cost of the material and processing for seven cans, as a function of diameter only, with V as a constant. Differentiate Cost vs(D) and set the derivative = 0. Then solve for the optimum diameter. The reason you should do this for seven cans is that you get seven lids at a time, with minimum waste. Remember than seven cans require a total of 14 tops and bottoms ## To find the dimensions of the most economical can, we will first need to write equations for the amount of material needed to produce seven cans, as a function of the diameter of the can. Let's use the following variables: - V: Required volume of one can (57.749 cubic inches) - D: Diameter of the can - h: Height of the can Firstly, we can express the height, h, in terms of the required volume, V, and the diameter, D: h = (4V) / (πD^2) Secondly, we need to consider the material for the sides of the can. The sides are formed by rolling and welding a strip of metal, which has a width equal to the desired height of the can. The cost of this material is 20 cents per square foot. We can calculate the amount of material required for the sides as follows: - Length of one side strip = circumference of the can = πD - Width of one side strip = height of the can = h - Area of one side strip = Length * Width = πD * h Since we need to produce seven cans, the total amount of material needed for the sides is 7 times the area of one strip: Material for sides = 7 * (πD * h) Next, let's consider the material for the top and bottom of the can. The top and bottom are circles cut from hexagons, and the width of the hexagon is the diameter of the can plus 0.4 inches. The extra inches are crimped over the sides to form the seal. Crimping costs are 1.6 cents per inch and the material for the ends sells for 30 cents per square foot. We can calculate the amount of material required for the ends as follows: - Length of the hexagon side = circumference of the can = πD - Width of the hexagon = diameter of the can plus 0.4 inches = D + 0.4 - Area of one hexagon = Length * Width = πD * (D + 0.4) - Since we get 4 hexagons for every 7 cans, the total amount of material needed for the ends is 4/7 times the area of one hexagon: Material for ends = (4/7) * (πD * (D + 0.4)) Now, let's compute the total cost of the material and processing for seven cans, as a function of diameter only, with V as a constant. Total cost = (Material cost for sides + Material cost for ends) + (Welding cost + Crimping cost) Since the costs are given in different units (cents per square foot and cents per inch), we also need to convert the dimensions to the appropriate units before calculating the costs. Finally, we can differentiate the total cost with respect to D (the diameter) and set the derivative equal to zero. By solving this equation, we can find the optimum diameter that minimizes the total cost and therefore gives us the most economical can.
1 / 6 # § 9.2 § 9.2. Subtracting Signed Numbers. The opposite of 4 is – 4. – 5. – 4. – 3. – 2. – 1. 0. 1. 2. 3. 4. 5. The sum of a number and its opposite is zero. 10 – 4 = 6. 10 + ( –4) = 6. Subtracting is the same as adding the opposite. 15 – 8 = 7. 15 + ( –8) = 7. 12 – 2 = 10. ## § 9.2 E N D ### Presentation Transcript 1. § 9.2 Subtracting Signed Numbers 2. The opposite of 4 is – 4. – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 The sum of a number and its opposite is zero. 10 – 4 = 6 10 + (–4) = 6 Subtracting is the same as adding the opposite. 15 – 8 = 7 15 + (–8) = 7 12 – 2 = 10 12 + (–2) = 10 Opposite Numbers The opposite of a positive number is a negative number with the same absolute value. 4 + (– 4) = 0 3. The opposite of 14 is –14. Change the subtraction to addition. Perform the addition of the two negative numbers. Subtraction of Signed Numbers Subtraction of Signed Numbers To subtract signed numbers, add the opposite of the second number to the first number. Example: Subtract. –6 – 14 –6 + (–14) –20 4. The opposite of –13 is 13. Perform the addition. Change the subtraction to addition. Subtraction of Signed Numbers Example: Subtract. –21 – (–13) –21 + (13) –8 Remember when subtracting two signed numbers: • The first number does not change. • The subtraction sign is changed to addition. • Write the opposite of the second number. • Find the result of the addition problem. 5. Subtraction of Signed Numbers Example: Subtract. Subtraction changed to addition. The LCD is 55. Multiply the fractions. Add the fractions. 6. 5628 feet Difference in altitude 5628 feet – (–350 feet ) Sea level 350 feet Applied Problems with Signed Numbers Example: Find the difference in altitude between a mountain 5628 feet high and a gorge 350 feet below sea level. = 5978 feet 5628 – (–350) = 5628 + 350 = 5978 More Related
# What is 1/643 as a decimal? ## Solution and how to convert 1 / 643 into a decimal 1 / 643 = 0.002 Fraction conversions explained: • 1 divided by 643 • Numerator: 1 • Denominator: 643 • Decimal: 0.002 • Percentage: 0.002% 1/643 or 0.002 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Both are used to handle numbers less than one or between whole numbers, known as integers. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Once we've decided the best way to represent the number, we can dive into how to convert 1/643 into 0.002 1 / 643 as a percentage 1 / 643 as a fraction 1 / 643 as a decimal 0.002% - Convert percentages 1 / 643 1 / 643 = 0.002 ## 1/643 is 1 divided by 643 The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 1 is being divided into 643. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators and Denominators. This creates an equation. To solve the equation, we must divide the numerator (1) by the denominator (643). Here's how you set your equation: ### Numerator: 1 • Numerators are the portion of total parts, showed at the top of the fraction. With a value of 1, you will have less complexity to the equation; however, it may not make converting any easier. The bad news is that it's an odd number which makes it harder to covert in your head. Ultimately, having a small value may not make your fraction easier to convert. Now let's explore X, the denominator. ### Denominator: 643 • Denominators represent the total parts, located at the bottom of the fraction. 643 is one of the largest two-digit numbers to deal with. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. So grab a pen and pencil. Let's convert 1/643 by hand. ## How to convert 1/643 to 0.002 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 643 \enclose{longdiv}{ 1 }$$ Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 643 \enclose{longdiv}{ 1.0 }$$ Uh oh. 643 cannot be divided into 1. So that means we must add a decimal point and extend our equation with a zero. This doesn't add any issues to our denominator but now we can divide 643 into 10. ### Step 3: Solve for how many whole groups you can divide 643 into 10 $$\require{enclose} 00.0 \\ 643 \enclose{longdiv}{ 1.0 }$$ How many whole groups of 643 can you pull from 10? 0 Multiply by the left of our equation (643) to get the first number in our solution. ### Step 4: Subtract the remainder $$\require{enclose} 00.0 \\ 643 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$ If there is no remainder, youâ€™re done! If there is a remainder, extend 643 again and pull down the zero ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value. ### Why should you convert between fractions, decimals, and percentages? Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Here are examples of when we should use each. ### When you should convert 1/643 into a decimal Dining - We don't give a tip of 1/643 of the bill (technically we do, but that sounds weird doesn't it?). We give a 0% tip or 0.002 of the entire bill. ### When to convert 0.002 to 1/643 as a fraction Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'. ### Practice Decimal Conversion with your Classroom • If 1/643 = 0.002 what would it be as a percentage? • What is 1 + 1/643 in decimal form? • What is 1 - 1/643 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.002 + 1/2? ### Convert more fractions to decimals From 1 Numerator From 643 Denominator What is 1/644 as a decimal? What is 2/643 as a decimal? What is 1/645 as a decimal? What is 3/643 as a decimal? What is 1/646 as a decimal? What is 4/643 as a decimal? What is 1/647 as a decimal? What is 5/643 as a decimal? What is 1/648 as a decimal? What is 6/643 as a decimal? What is 1/649 as a decimal? What is 7/643 as a decimal? What is 1/650 as a decimal? What is 8/643 as a decimal? What is 1/651 as a decimal? What is 9/643 as a decimal? What is 1/652 as a decimal? What is 10/643 as a decimal? What is 1/653 as a decimal? What is 11/643 as a decimal? What is 1/654 as a decimal? What is 12/643 as a decimal? What is 1/655 as a decimal? What is 13/643 as a decimal? What is 1/656 as a decimal? What is 14/643 as a decimal? What is 1/657 as a decimal? What is 15/643 as a decimal? What is 1/658 as a decimal? What is 16/643 as a decimal? What is 1/659 as a decimal? What is 17/643 as a decimal? What is 1/660 as a decimal? What is 18/643 as a decimal? What is 1/661 as a decimal? What is 19/643 as a decimal? What is 1/662 as a decimal? What is 20/643 as a decimal? What is 1/663 as a decimal? What is 21/643 as a decimal? ### Convert similar fractions to percentages From 1 Numerator From 643 Denominator 2/643 as a percentage 1/644 as a percentage 3/643 as a percentage 1/645 as a percentage 4/643 as a percentage 1/646 as a percentage 5/643 as a percentage 1/647 as a percentage 6/643 as a percentage 1/648 as a percentage 7/643 as a percentage 1/649 as a percentage 8/643 as a percentage 1/650 as a percentage 9/643 as a percentage 1/651 as a percentage 10/643 as a percentage 1/652 as a percentage 11/643 as a percentage 1/653 as a percentage
# How do you simplify 5m^0? May 25, 2016 5 #### Explanation: $\textcolor{g r e e n}{\text{Note that you are 'not allowed' to divide by 0}}$ $\textcolor{g r e e n}{\text{So "x/x=0/0 !=1" It is 'undefined' (not allowed)}}$ Consider $\frac{4}{4} = 1$ now compare that to $\frac{x}{x} = 1$ Considering the same thing but by using powers $\textcolor{b r o w n}{\text{If you have "1/x" you can write this as } {x}^{- 1}}$ '..................................................... Now think about $x \times x = {x}^{2}$ But $x$ can be written as ${x}^{1}$ So $x \times x = {x}^{2} \to {x}^{1} \times {x}^{1} = {x}^{2}$ $\textcolor{b r o w n}{\text{So } {x}^{1} \times {x}^{1} = {x}^{1 + 1} = {x}^{2}}$ '............................................................. $\textcolor{b l u e}{\text{Putting it all together}}$ So if we have $\frac{x}{x}$ this can be written as $x \times \frac{1}{x}$ $= {x}^{1} \times {x}^{- 1}$ $= {x}^{1 - 1}$ $= {x}^{0} = 1$ '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Answering your question}}$ Thus ${m}^{0} = 1$ so $5 {m}^{0} = 5 \times 1 = 5$
## A Multiplication Based Logic Puzzle ### 635 Some Multiplication Facts to Know Backwards and Forwards Something amazing happens in the multiplication table when an even number is multiplied by 6. Something just as amazing (but backwards) happens in the multiplication table when a multiple of 3 is multiplied by 7. Here is a graphic that may be helpful for students memorizing some of those 6 and 7 multiplication facts. It is meant to be read from left to right and contains some cool coincidences: Knowing those 6 and 7 multiplication facts will help any student know the multiplication table backwards and forwards! While it isn’t necessary to memorize the following number facts, some of the patterns in the graphic above continue with them: • 10 + 5 = 15, and 15 x 7 = 105 • 12 + 6 = 18, and 18 x 7 = 126 • 14 + 7 = 21, and 21 x 7 = 147 • 16 + 8 = 24, and 24 x 7 = 168 • 18 + 9 = 27, and 27 x 7 = 189 ——————————————————— What makes 635 an interesting number? 635 is the sum of the nine prime numbers from 53 to 89. It is also the sum of the thirteen prime numbers from 23 to 73. 635 can be written as the sum of 5 consecutive numbers: 125 + 126 + 127 + 128 + 129 = 635. 635 is a part of exactly 4 Pythagorean triples. Which factors of 635 are greatest common factors for the non-primitive triples? • 635 is the hypotenuse of the Pythagorean triple 381-508-635 • 635 is the short leg in the triple 635-1524-1650 • 635 is the short leg in the triple 635-40320-40325 • 635 is the short leg in the primitive triple 635-201612-201613 ———————————————————————– • 635 is a composite number. • Prime factorization: 635 = 5 x 127 • The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 635 has exactly 4 factors. • Factors of 635: 1, 5, 127, 635 • Factor pairs: 635 = 1 x 635 or 5 x 127 • 635 has no square factors that allow its square root to be simplified. √635 ≈ 25.199206. ### 199 and My List of Multiplication Rhymes 199 is a prime number. Factor pairs: 199 = 1 x 199 Factors of 199: 1, 199 Prime factorization: none How do we know that 199 is a prime number? The square root of 199 is an irrational number approximately equal to 14.11. If 199 were not a prime number, then it would be divisible by at least one prime number less than or equal to 14.11. Since 199 cannot be divided evenly by 2, 3, 5, 7, 11, or 13, we know that 199 is a prime number. ———————————————————————————– I recently wrote a post that asked “Are Multiplication Rhymes Able to Help Kids Learn the Multiplication Table?” Here is a list of rhymes that I’ve put together. Most of the rhymes have rhyming clues to help students remember both numbers being multiplied as well as their product. I wrote the rhymes that are highlighted in yellow. Those highlighted in peach, purple, or blue are from other sources, but I did alter the rhymes for 24, 28, and 81. A black and white pdf of the rhymes is also available here. ### 194 and Are Multiplication Rhymes Able to Help Kids Learn the Multiplication Table? 194 is a composite number. Factor pairs: 194 = 1 x 194 or 2 x 97. Factors of 194: 1, 2, 97, 194. Prime factorization: 194 = 2 x 97. Are multiplication rhymes able to help kids learn the multiplication table? The obvious answer is, of course, YES! Duh! However, sometimes this obvious answer may not be correct. I recently read a researched article on the effectiveness of certain teaching practices in helping first grade children learn mathematics. The points made in the article would likely apply to third and fourth grade students learning multiplication as well. The article states that alternative techniques (music, movement, m&m’s, other manipulatives, etc.) do reinforce concepts for students who already understand what is being taught, but students who struggle actually make no gains when unproven techniques are used. The researcher hypothesized that when students struggle learning mathematical concepts “alternative techniques tend to demand more, cognitively, from [these] students.” Demanding more cognitively can backfire: Requiring a student to memorize a rhyme and the math fact could put that student’s brain on overload with even less facts learned. I have now modified my opinion of multiplication rhymes. If rhyme-and-rhythm is helpful, use it, if not, don’t. What is good for some of the class, may not be good for all of the class. If a student doesn’t make progress using the rhymes, it may be better to stick with more direct instruction to learn the math facts. Additionally, each rhyme should be examined individually and its own pros and cons considered. Here are two of my FAVORITE sources of multiplication rhymes: American Academy’s MULTIPLICATION-RHYMES.pdf now requires registration to view, but it has rhymes printed in pretty colors and contains a few of my favorite multiplication rhymes. Kids would probably like looking at this pdf multiple times. Multiplication Rhymes 1 is an entertaining You-tube video made by Mrs. Rice that children would probably enjoy watching over and over again, and it also has a few very memorable rhymes. Mrs. Rice shares 4 x 4 = 16 rhyme. Although these two sources are great, I think some of the rhymes are too similar: • From the pdf we have “8 and 4 were sad and blue, 8 x 4 = 32” while the video recites “6 and 7 are sad and blue, they make number 42.” If students learned either of those rhymes, they may have difficulty remembering the product or which two numbers were sad and blue several weeks later. • The video also gave two other rhymes that were too similar to each other: “9 and 8 what do they do? They go play (tag) with 72.” and “9 and 9 are having fun. They play tag with 81.” “Tag” is in parentheses because it wasn’t on the note card that was shown, but it was spoken on the video. Again weeks later, groups or individuals may not remember which numbers were playing tag or what the product was when they were multiplied together. I compiled my own List of Multiplication Rhymes which focus on the multiplication facts highlighted in the multiplication table below. I included some rhymes from the above pdf and the video and added some from the song Five Pennies Make a Nickel. All of those rhymes are in italics. The rhymes that I wrote myself are in regular print. If I modified an existing rhyme, the modifications are in regular print while the rest of the rhyme is in italics. I also ordered the rhymes in a way that should make finding the rhyme for any particular fact much easier. I hope there will be many rhymes on this list that you enjoy as well, and that they get used to help many students learn the very important multiplication facts highlighted in the table below.
# Product to Sum Trigonometry Identities Calculator To use the product to sum identities calculator, enter the values of angles U & V, and hit calculate Give Us Feedback ## Product to Sum Identities Calculator Product to sum identities calculator finds the result of two angles using the formulas of product to sum conversions. ## What are Trigonometric Identities? Trigonometric identities are equations that relate various trigonometric functions to each other. These identities are derived from the geometric properties of triangles and the unit circle. They play a crucial role in simplifying trigonometric expressions and solving trigonometric equations. One such set of identities is the product to sum trigonometry identities. ## What is Product to Sum Trigonometry Identities? Product to sum trigonometry identities are formulas that express the product of two trigonometric functions as the sum or difference of two trigonometric functions. These identities are helpful when we need to simplify trigonometric expressions involving products. By using these identities, we can rewrite a product of trigonometric functions as a sum or difference of trigonometric functions, which is often easier to work with. ## Product to Sum Identities for Sine and Cosine: The product to sum identities for sine and cosine are as follows: • sin(u) * sin(v) = (1/2) [cos (u - v) – cos (u + v)] • cos(u) * cos(v) = (1/2) [cos (u - v) + cos (u + v)] • sin(u) * cos(v) = (1/2) [sin (u - v) + sin (u + v)] • cos(u) * sin(v) = (1/2) [sin (u + v) - sin (u - v)] ## Examples: Applying Product to Sum Trigonometry Identities Example 1: Simplify the expression: sin(3x) * sin(2x) Solution: Using Identity 1: sin(u) * sin(v) = (1/2) [cos (u - v) – cos (u + v)] We have: sin(3x) * sin(2x) = (1/2) [cos (3x - 2x) – cos (3x + 2x)] sin(3x) * sin(2x) = (1/2) [cos(x) - cos(5x)] Example 2: Simplify the expression: cos(4x) * cos(3x) Solution: Using Identity 2: cos(u) * cos(v) = (1/2) [cos (u - v) + cos (u + v)] We have: cos(4x) * cos(3x) = (1/2) [cos (4x - 3x) + cos (4x + 3x)] cos(4x) * cos(3x) = (1/2) [cos (x) + cos (7x)] Example 3: Simplify the expression: sin(2x) * cos(2x) Solution: Using Identity 3: sin(u) * cos(v) = (1/2) [sin (u - v) + sin (u + v)] We have: sin(2x) * cos(2x) = (1/2) [sin (2x - 2x) + sin (2x + 2x)] sin(2x) * cos(2x) = (1/2) [sin (0) + sin(4x)] sin(2x) * cos(2x) = (1/2) [0 + sin(4x)] sin(2x) * cos(2x) = (1/2) sin(4x)
2021-12-20 Using algebra, find the solution to the system of inequalities below $y\le 5-3x$ $y+19\ge {x}^{2}-8x$ Jonathan Burroughs Step 1 Given, $y\le 5-3x,y+19\ge {x}^{2}-8x$ Step 2 $y+19-19\ge {x}^{2}-8x-19$ [Subtract by 19] $y\ge {x}^{2}-8x-19$ ${x}^{2}-8x-19\le y$ So, ${x}^{2}-8x-19\le y\le 5-3x$ ${x}^{2}-8x-19-5+3x\le 5-3x-5+3x$ [Subtract 5 and add 3x] ${x}^{2}-8x+3x-24\le 0$ $x\left(x-8\right)+3\left(x-8\right)\le 0$ $\left(x+3\right)\left(x-8\right)\le 0$ Step 3 The region is possible when So, $-3\le x\le 8$. In interval notation, $x\in \left[-3,8\right]$ Step 4 Now, $\left(-3\right)\left(-3\right)\ge -3x\ge \left(-3\right)8$ [Multiply by -3] $9\ge -3x\ge -24$ $-24\le -3x\le 9$ $5-24\le 5-3x\le 5+9$ [Add 5] $-19\le y\le 14$ In interval notation, $y\in \left[-19,14\right]$ Step 5 Hence, the solution of system of inequalities is $x\in \left[-3,8\right],y\in \left[-19,14\right]$. Becky Harrison Step 1 Given: $y\le 5-3x$ (1) $y+19\ge x-8x$ (2) Step 2 Simplify the equation (2) $y+19\ge x-8x$ $y+19\ge -7x$ $y\ge -7x-19$ $y\ge -\left(7x+19\right)$ $-y\le 7x+19$ (3) Step 3 $0\le 4x+24$ $-24\le 4x$ $4x\ge -24$ $x\ge -6$ Step 4 Put the value of x in equation (3) $-y\le 7\left(-6\right)+19$ $-y\le -23$ $y\ge 23$ Step 5 Answer: $\left\{\left(x,y\right)\in {R}^{2}\mid x\ge -6,y\ge 23\right\}$ nick1337 Step 1 Given the system of inequalities $y\le 5-3x$ $y+19\ge {x}^{2}-8x$ Simplify the inequality $y+19\ge {x}^{2}-8x$ as, $y+19\ge {x}^{2}-8x$ $y\ge {x}^{2}-8x-19$ ${x}^{2}-8x-19\le 5-3x$ ${x}^{2}-8x-19+3x\le 5-3x+3x$ ${x}^{2}-5x-19\le 5$ ${x}^{2}-5x-24\le 0$ $\left(x+3\right)\left(x-8\right)\le 0$ $-3\le x\le 8$ Step 2 Simplify the inequality $y\le 5-3x$ as, $y\le 5-3x$ $5-3x\ge y$ $-3x\ge y-5$ $x\le \frac{-y+5}{3}$ Thus, $-3\le x\le 8$ Since $-3\le x\le 8$obtain the inequality for y as, Thus,$-19\le y\le 14$ Therefore, the solution is Do you have a similar question?
# How to Solve Matrices A matrix is a very useful way of representing numbers in a block format,[1] which you can then use to solve a system of linear equations. If you only have two variables, you will probably use a different method. See Solve a System of Two Linear Equations and Solve Systems of Equations for examples of these other methods. But when you have three or more variables, a matrix is ideal. By using repeated combinations of multiplication and addition, you can systematically reach a solution. Part 1 Part 1 of 4: ### Setting up the Matrix for Solving 1. 1 Verify that you have sufficient data. In order to get a unique solution for each variable in a linear system using a matrix, you need to have as many equations as the number of variables that you are trying to solve. For example, with variables x, y and z, you would need three equations. If you have four variables, you need four equations. • If you have fewer equations than the number of variables, you will be able to learn some limiting information about the variables (such as x = 3y and y = 2z), but you cannot get a precise solution. For this article, we will be working toward getting a unique solution only. 2. 2 Write your equations in standard form. Before you can transfer information from the equations into matrix form, first write each equation in standard form. The standard form for a linear equation is Ax+By+Cz=D, where the capital letters are the coefficients (numbers), and the last number - in this example, D - is on the right side of the equals sign.[2] • If you have more variables, you will just continue the line as long as necessary. For example, if you are trying to solve a system with six variables, your standard form would look like Au+Bv+Cw+Dx+Ey+Fz =G. For this article, we will focus on systems with only three variables. Solving a larger system is exactly the same, but just takes more time and more steps. • Note that in standard form, the operations between the terms is always addition. If your equation has subtraction instead of addition, you will need to work with this later my making your coefficient negative. If it helps you remember, you can rewrite the equation and make the operation addition and the coefficient negative. For example, you can rewrite the equation 3x-2y+4z=1 as 3x+(-2y)+4z=1. 3. 3 Transfer the numbers from the system of equations into a matrix. A matrix is a group of numbers, arranged in a block-looking format, that we will work with to solve the system.[3] It actually carries the same data as the equations themselves, but in a simpler format. To create the matrix from your equations in standard form, just copy the coefficients and result of each equation into a single row, and stack those rows one on top of each other. • For example, suppose you have a system that consists of the three equations 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. The top row of your matrix will contain the numbers 3,1,-1,9, since these are the coefficients and solution of the first equation. Note that any variable that has no coefficient showing is assumed to have a coefficient of 1. The second row of the matrix will be 2,-2,1,-3, and the third row will be 1,1,1,7. • Be sure to align the x-coefficients in the first column, the y-coefficients in the second, the z-coefficients in the third, and the solution terms in the fourth. When you finish working with the matrix, these columns will be important in writing your solution. 4. 4 Draw a large square bracket around your full matrix. By convention, a matrix is designated with a pair of square brackets, [ ], around the entire block of numbers. The brackets do not factor into the solution in any way, but they do illustrate that you are working with matrices. A matrix can consist of any number of rows and columns. As we work through this article, we will use brackets around terms in a row to help join them.[4] 5. 5 Use common symbolism. In working with matrices, it is common convention to refer to the rows by the abbreviation R and the columns with the abbreviation C. You can use numbers together with these letters to indicate a specific row or column. For example, to indicate Row 1 of a matrix, you can write R1. Row 2 would be R2.[5] • You can indicate any specific position in a matrix by using a combination of R and C. For example, to pinpoint the term in the second row, third column, you could call it R2C3. Part 2 Part 2 of 4: ### Learning the Operations for Solving a System with a Matrix 1. 1 Recognize the form of the solution matrix. Before you begin doing any work to solve your system of equations, you should recognize what you will be trying to do with the matrix. Right now, you have a matrix that looks like this: • 3 1 -1 9 • 2 -2 1 -3 • 1 1 1 7 • You will be working with some basic operations to create the “solution matrix.” The solution matrix will look like this:[6] • 1 0 0 x • 0 1 0 y • 0 0 1 z • Notice that the matrix consists of 1’s in a diagonal line with 0’s in all other spaces, except the fourth column. The numbers in the fourth column will be your solution for the variables x, y and z. 2. 2 Use scalar multiplication. The first tool at your disposal for solving a system using a matrix is scalar multiplication. This is simply a term that means you will be multiplying the items in a row of the matrix by a constant number (not a variable). When you use scalar multiplication, you must remember to multiply every term of the entire row by whatever number you select. If you forget and only multiply the first term, you will ruin the entire solution. You are not required, however, to multiply the entire matrix at the same time. You are only working on one row at a time with scalar multiplication.[7] • It is common to use fractions in scalar multiplication, because you often want to create that diagonal row of 1s. Get used to working with fractions. It will also be easier, for most steps in solving the matrix, to be able to write your fractions in improper form, and then convert them back to mixed numbers for the final solution. Therefore, the number 1 2/3 is easier to work with if you write it as 5/3. • For example, the first row (R1) of our sample problem begins with the terms [3,1,-1,9]. The solution matrix should contain a 1 in the first position of the first row. In order to “change” our 3 into a 1, we can multiply the entire row by 1/3. Doing this will create the new R1 of [1,1/3,-1/3,3]. • Be careful to keep any negative signs where they belong. 3. 3 Use row-addition or row-subtraction. The second tool you can use is to add or subtract any two rows of the matrix. In order to create the 0 terms in your solution matrix, you will need to add or subtract numbers that get you to 0. For example, if R1 of a matrix is [1,4,3,2] and R2 is [1,3,5,8], you can subtract the first row from the second row and create the new row of [0,-1,2,6], because 1-1=0 (first column), 3-4=-1 (second column), 5-3=2 (third column), and 8-2=6 (fourth column). When you perform a row-addition or row-subtraction, rewrite your new result in place of the row you started with. In this case, we would take out row 2 and insert the new row [0,-1,2,6].[8] • You can use some shorthand and indicate this operation as R2-R1=[0,-1,2,6]. • Recognize that adding and subtracting are merely opposite forms of the same operation. You can either think of adding two numbers or subtracting the opposite. For example, if you begin with the simple equation 3-3=0, you could consider this instead as an addition problem of 3+(-3)=0. The result is the same. This seems basic, but it is sometimes easier to think of a problem in one form or the other. Just keep track of your negative signs. 4. 4 Combine row-addition and scalar multiplication in a single step. You cannot expect the terms always to match so you can use simple addition or subtraction to create 0s in your matrix. More often, you will need to add (or subtract) a multiple of another row. To do this, you perform the scalar multiplication first, then add that result to the target row that you are trying to change.[9] • Suppose you have a Row 1 of [1,1,2,6] and a Row 2 of [2,3,1,1]. You want to create a 0 term in the first column of R2. That is, you want to change the 2 into a 0. To do this, you need to subtract a 2. You can get a 2 by first multiplying Row 1 by the scalar multiplication 2, and then subtract the first row from the second row. In shorthand, you can think of this as R2-2R1. First multiply R1 by 2 to get [2,2,4,12]. Then subtract this from R2 to get [(2-2),(3-2),(1-4),(1-12)]. Simplify this and your new R2 will be [0,1,-3,-11]. 5. 5 Copy down rows that are unchanged as you work. As you work with the matrix, you will be changing a single row at a time, either through scalar multiplication, row-addition or row-subtraction, or a combination step. When you change the one row, make sure to copy the other rows of your matrix in their original form. • A common mistake occurs when conducting a combined multiplication and addition step in one move. Suppose, for example, you need to subtract double R1 from R2. When you multiply R1 by 2 to do this step, remember that you are not changing R1 in the matrix. You are only doing the multiplication to change R2. Copy R1 first in its original form, then make the change to R2. 6. 6 Work from top to bottom first. To solve your system, you will work in a very organized pattern, essentially “solving” one term of the matrix at a time. The order for a three-variable matrix will begin as follows: • 1. Create a 1 in the first row, first column (R1C1). • 2. Create a 0 in the second row, first column (R2C1). • 3. Create a 1 in the second row, second column (R2C2). • 4. Create a 0 in the third row, first column (R3C1). • 5. Create a 0 in the third row, second column (R3C2). • 6. Create 1 in the third row, third column (R3C3). 7. 7 Work back up from bottom to top. At this point, if you have done the steps correctly, you are halfway to the solution. You should have the diagonal line of 1’s, with 0’s beneath them. The numbers in the fourth column are really irrelevant at this point. Now you will work your way back up to the top as follows: • Create a 0 in the second row, third column (R2C3). • Create a 0 in first row, third column (R1C3). • Create a 0 in the first row, second column (R1C2). 8. 8 Check that you have created the solution matrix. If your work is correct, you will have created the solution matrix with 1’s in a diagonal line of R1C1, R2C2, R3C3 and 0’s in the other positions of the first three columns. The numbers in the fourth column are the solutions to your linear system.[10] Part 3 Part 3 of 4: ### Putting the Steps Together to Solve the System 1. 1 Begin with a sample system of linear equations.[11] To practice these steps, begin with the sample we used before: 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. When you write this into a matrix, you will have R1= [3,1,-1,9], R2=[2,-2,1,-3], and R3=[1,1,1,7]. 2. 2 Create a 1 in the first position R1C1. Notice that R1 currently begins with a 3. You need to change it into a 1. You can do this by scalar multiplication, by multiplying all four terms of R1 by 1/3. In shorthand, you can note this as R11/3. This will give a new result for R1 as R1=[1,1/3,-1/3,3]. Copy R2 and R2, unchanged, as R2=[2,-2,1,-3] and R3=[1,1,1,7]. • Notice that multiplication and division are merely inverse functions of each other. We can say we are multiplying by 1/3 or dividing by 3, and the result is the same. 3. 3 Create a 0 in the second row, first column (R2C1). Currently, R2=[2,-2,1,-3]. To move closer to the solution matrix, you need to change the first term from a 2 to a 0. You can do this by subtracting twice the value of R1, since R1 begins with a 1. In shorthand, the operation is R2-2R1. Remember, you are not changing R1, but just working with it. So first, copy R1 as R1=[1,1/3,-1/3,3]. Then, when you double each term of R1, you will get 2R1=[2,2/3,-2/3,6]. Finally, subtract this result from the original R2 to get your new R2. Working through term by term, this subtraction is (2-2), (-2-2/3), (1-(-2/3)), (-3-6). These simplify to give the new R2=[0,-8/3,5/3,-9]. Notice that first term is 0, which was your objective. • Copy down the unaffected row 3 as R3=[1,1,1,7]. • Be very careful with subtracting negative numbers, to make sure you keep the signs correct. • For now, leave the fractions in their improper forms. This will make later steps of the solution easier. You can simplify fractions in the final step of the problem. 4. 4 Create a 1 in the second row, second column (R2C2). To continue forming the diagonal line of 1’s, you need to transform the second term -8/3 into 1. Do this by multiplying the entire row by the reciprocal of that number, which is -3/8. Symbolically, this step is R2(-3/8). The resulting second row is R2=[0,1,-5/8,27/8]. • Notice that as the left half of the row starts looking like the solution with the 0 and 1, the right half may start looking ugly, with improper fractions. Just carry them along for now. • Remember to continue copying the unaffected rows, so R1=[1,1/3,-1/3,3] and R3=[1,1,1,7]. 5. 5 Create a 0 in the third row, first column (R3C1). Your focus now moves to the third row, R3=[1,1,1,7]. To create a 0 in the first position, you will need to subtract a 1 from the 1 that is in that position currently. If you look up, there is a 1 in the first position of R1. Therefore, you simply need to subtract R3-R1 to get the result you need. Working term by term, this will be (1-1), (1-1/3), (1-(-1/3)), (7-3). These four mini-problems simplify to give the new R3=[0,2/3,4/3,4]. • Continue to copy along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8]. Remember that you only change one row at a time. 6. 6 Create a 0 in the third row, second column (R3C2). This value is currently 2/3, but it needs to be transformed into a 0. At first glance, it looks like you might be able to subtract double the R1 values, since the corresponding column of R1 contains a 1/3. However, if you double all the values of R1 and subtract them, you will affect the 0 in the first column of R3, which you do not want to do. This would be taking a step backward in your solution. So you need to work with some combination of R2. If you subtract 2/3 of R2, you will create a 0 in the second column, without affecting the first column. In shorthand notation, this is R3- 2/3R2. The individual terms become (0-0), (2/3-2/3), (4/3-(-5/32/3)), (4-27/82/3). Simplifying gives the result R3=[0,0,42/24,42/24]. 7. 7 Create a 1 in the third row, third column (R3C3). This is a simple step of multiplying by the reciprocal of the number that is there. The current value is 42/24, so you can multiply by 24/42 to create the desired value of 1. Notice that the first two terms are 0’s, so any multiplication will remain 0. The new value of R3=[0,0,1,1]. • Notice that the fractions, which appeared quite complicated in the previous step, have already begun to resolve themselves. • Continue to carry along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8]. • Notice that at this point, you have the diagonal of 1’s for your solution matrix. You just need to transform three more items of the matrix into 0’s to find your solution. 8. 8 Create a 0 in the second row, third column. R2 currently is [0,1,-5/8,27/8], with a value of -5/8 in the third column. You need to transform it to a 0. This means conduction some operation involving R3 that will consist of adding 5/8. Because the corresponding third column of R3 is a 1, you need to multiply all of R3 by 5/8 and add the result to R2. In shorthand, this is R2+5/8R3. Working term by term, this is R2=(0+0), (1+0), (-5/8+5/8), (27/8+5/8). These simplify to R2=[0,1,0,4]. • Copy along R1=[1,1/3,-1/3,3] and R3=[0,0,1,1]. 9. 9 Create a 0 in the first row, third column (R1C3). The first row is currently R1=[1,1/3,-1/3,3]. You need to transform the -1/3 in the third column to a 0, by using some combination of R3. You do not want to use R2, because the 1 in the second column of R2 would affect R1 in the wrong way. So, you will multiply R31/3 and then add the result to R1. The notation for this is R1+1/3R3. Working it out term by term results in R1=(1+0), (1/3+0), (-1/3+1/3), (3+1/3). These simplify to give a new R1=[1,1/3,0,10/3]. • Copy the unchanged R2=[0,1,0,4] and R3=[0,0,1,1]. 10. 10 Create a 0 in the first row, second column (R1C2). If everything has been done properly, this should be your final step. You need to transform the 1/3 in the second column to a 0. You can get this by multiplying R21/3 and subtracting. In shorthand, this is R1-1/3R2. The result is R1=(1-0), (1/3-1/3), (0-0), (10/3-4/3). Simplifying gives the result of R1=[1,0,0,2]. 11. 11 Look for the solution matrix. At this point, if all has gone well, you should have the three rows R1=[1,0,0,2], R2=[0,1,0,4] and R3=[0,0,1,1]. Notice, if you write this in the block matrix form with the rows on top of each other, you will have the diagonal 1’s, with 0’s everywhere else, and your solutions in the fourth column. The solution matrix should look like this: • 1 0 0 2 • 0 1 0 4 • 0 0 1 1 12. 12 Make sense of your solution. When you translated your linear equations into a matrix, you put the x-coefficients in the first column, the y-coefficients in the second column, the z-coefficients in the third column. There, to rewrite your matrix back into equation form, these three lines of the matrix really mean the three equations 1x+0y+0z=2, 0x+1y+0z=4, and 0x+0y+1z=1. Since we can drop the 0-terms and don’t need to write the 1 coefficients, these three equations simplify to give you the solution, x=2, y=4 and z=1. This is the solution to your system of linear equations.[12] Part 4 Part 4 of 4: 1. 1 Replace the solution values into each variable in each equation. It is always a good idea to check that your solution actually is correct. You do this by testing your results in the original equations.[13] • Recall that the original equations for this problem were 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. When you replace the variables with their solved values, you get 32+4-1=9, 22-24+1=-3, and 2+4+1=7. 2. 2 Simplify each equation. Perform the operations in each equation according to the basic rules of operations. The first equation simplifies to 6+4-1=9, or 9=9. The second equation simplifies as 4-8+1=-3, or -3=-3. The final equation is simply 7=7. • Because each equation simplifies to a true mathematical statement, your solutions are correct. If any of them did not resolve correctly, you would have to go back through your work and look for any errors. Some common mistakes occur in dropping negative signs along the way or confusing the multiplication and addition of fractions. 3. 3 Write out your final solutions. For this given problem, the final solution is x=2, y=4 and z=1.[14] ## Expert Q&A 200 characters left ## Tips • If your system of equations is very complicated, with many variables, you may be able to use a graphing calculator instead of doing the work by hand. For information on this, see Use a Graphing Calculator to Solve a System of Equations. Thanks Submit a Tip All tip submissions are carefully reviewed before being published Thanks for submitting a tip for review! ## Video.By using this service, some information may be shared with YouTube. Co-authored by: wikiHow Staff Writer This article was co-authored by wikiHow Staff. Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 154,968 times. Co-authors: 9 Updated: October 7, 2022 Views: 154,968 Categories: Linear Algebra Article SummaryX By properly setting up a matrix, you can use them to solve a system of linear equations. Start by writing out your equations and then transfer the numbers from them into a matrix by copying the coefficients and results into a single row. Stack the rows one on top of each other to form a block-looking format. Add a large square bracket around your full matrix and use the abbreviation “R” for the rows and “C” for the columns. This allows you to refer to a specific position in the matrix with a combination of R and C, such as R4C1. To solve the matrix, you can use different operations. For instance, you could use row-addition or row-subtraction, which allows you to add or subtract any two rows of the matrix. To learn about other ways to create a solution matrix, keep reading!
Continuous Functions: Properties & Definition Instructor: Kimberlee Davison Kim has a Ph.D. in Education and has taught math courses at four colleges, in addition to teaching math to K-12 students in a variety of settings. In this lesson, you will learn what a continuous function is and how to identify one by tracing your finger or pencil on a graph or by using three properties of continuous functions. Definition A continuous function is a function, or equation, which is smooth when graphed over the entire domain (set of x-values) that you care about. If you were to run your finger along the graph, you would never need to lift your finger. There are no interruptions or gaps in the graph. Continuous Functions and the Real World Imagine that you live next door to the local football stadium and you want to set up a stand to sell snow cones as people arrive for the game. You are trying to figure out where to buy your ice. Green Grocery Store sells ice by the 10 pound bag. Each bag costs you \$2. On the other hand, Red Ice Delivery Service can bring you any amount of ice you want. It costs you \$0.25 per pound. If you buy ice from Green, you will have some waste if you need 15 or 25 pounds because you can only buy ten pound bags, but the cost is a little lower per pound. With Red, you pay a little more per pound but you can get exactly how much you want. If you look at the graphs of the Red and the Green ice costs, you will see a couple big differences. One difference is that the red graph is a straight line with no breaks in it. The green graph looks like a staircase. There are breaks in the graph at each ten pounds of ice needed. Using Green, if you need 40 pounds of ice, or just under, you can get away with 4 bags of ice - \$8 in cost. But as soon as you need a little more than 40, you have to buy one more bag and your cost jumps by \$2. Because the green function has breaks, or jumps, in it, it is not continuous. You could not trace a pencil across the entire length of the green graph without having to pick up your pencil. The red function, however, could be drawn without lifting your pencil. It is continuous. There are no sudden jumps in your costs at any point. Graphs and Continuity If you look at the four small graphs (Graphs A-D), you will see that only one is continuous (Graph A) over the intervals (x-values) shown. The other three all have breaks in them. You could not draw the entire length of the graph without lifting your pencil. Continuous Functions and Calculus You can also use calculus to determine whether a function is continuous. Property 1. There is a connection between continuous functions and limits, a topic studied in calculus. If a function is continuous at some point, then the limit at that point is the same whether you approach the point from the right or left. In other words, if you move your pencil toward that point from either side, you end up at the same place. In Graph D above, you end up 'higher' (larger y-value) if you move your pencil toward the '2' along the blue line from the left rather than along the red line from the right. If a is the x-value you are interested in, then mathematically you would write this as: It's a bit like saying that, whether you approach Rome from the North or from the South on a road labeled 'To Rome,' you still are approaching the same place. To unlock this lesson you must be a Study.com Member. Register to view this lesson Are you a student or a teacher?
# Three Special Types of Parallelograms Save Parallelograms are a specific type of quadrilateral – which is a four-sided shape – but what distinguishes parallelograms from other quadrilaterals is that both pairs of opposite sides of a parallelogram are parallel. Additionally, some parallelograms are special -- rhombuses, rectangles and squares -- because these shapes have additional properties that distinguish them from other parallelograms. ## Properties of a Parallelogram • Parallelograms are quadrilaterals that have two sets of parallel sides and two sets of congruent sides. A parallelogram’s opposite angles are congruent; its consecutive angles are supplementary; its diagonals bisect each other and its diagonals form two congruent triangles. So, in a hypothetical parallelogram ABCD, moving clockwise, starting from point A at the top left of the parallelogram, you see that side AB is parallel to side DC and side BC is parallel to side AD. The parallelogram's opposite angles are congruent to each other and its consecutive angles are supplementary to each other. The parallelogram's diagonals AC and BD bisect each other and its diagonals form two congruent triangles. ## Properties of a Rectangle • A rectangle is a quadrilateral that has four right angles -- but unlike a square -- a rectangle's four sides are not all the same length. A rectangle has two sets of parallel sides, with two sides the same length and the other two sides equal to each other, but not to the first set of equal sides. A rectangle is also a parallelogram, so that it contains all the properties of a parallelogram and also includes additional properties. These additional properties are that its four angles are right angles and that its diagonals are congruent to each other. In a hypothetical rectangle ABCD, moving clockwise, starting from point A on the top left, you see that the rectangle's four angles are all right angles and that its two diagonals are congruent, with diagonal AC congruent to diagonal BD. ## Properties of a Rhombus • A rhombus is a quadrilateral that has four congruent sides and includes all the properties of a parallelogram. A rhombus has additional properties, which are that its consecutive sides are congruent; its diagonals bisect pairs of opposite angles; and its diagonals are perpendicular to each other. In a hypothetical rhombus ABCD, moving clockwise, starting from point A on the top left, you see that side AB is congruent to side BC and side CD is congruent to side DA. You can also see that the rhombus' diagonals bisect pairs of opposite angles and that diagonal AC is perpendicular to diagonal DB. ## Properties of a Square • A square is a quadrilateral and a parallelogram that has four congruent sides and four congruent angles. The definition of a square also combines the definitions of both a rectangle and a rhombus, so that all the properties that apply to a rectangle and a rhombus also apply to a square. A square has four 90-degree angles, four equal sides, equal diagonal lengths, perpendicular diagonals and bisected opposite angles. In a hypothetical square, ABCD, moving clockwise, starting from point A at the top left, you see that side AB = side BC; side BC = side CD; side CD = side DA and therefore, side DA = side AB. Diagonal AC is congruent to BD. ## References • Photo Credit Zhenikeyev/iStock/Getty Images Promoted By Zergnet ## Related Searches Check It Out ### Can You Take Advantage Of Student Loan Forgiveness? M Is DIY in your DNA? Become part of our maker community.
How To Multiply Fractions # How To Multiply Fractions ## How To Multiply Fractions - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. How To Multiply Fractions Introducing: • factor • product • reciprocal • inverse • identity • inverse 2. Multiply Fractions 1 The parts of this multiplication example are the first factor3/8 , and a second factor 3. There are 3 rows with 3/8 in each row. 3. Multiply Fractions 2 Multiplication is a form of addition. This picture shows that 3/8 is added 3 times. The product can be found by addition of like amounts: 3/8 + 3/8 + 3/8 = 9/8 4. Multiply Fractions 3 To calculate the product, first writing both factors in fraction form. Then multiply the numerators 3 and 3 for 9 in the product numerator and the denominators 8 and 1 for 8 in the product denominator. 5. Multiply Fractions 4 The product9/8 can be written in fraction form 1 1/8 6. Multiply Fractions 5 It is easy to tell the product 4 4/5 from this picture. Notice the 4 complete circles and the 2/5 + 2/5 circles for a product of 4 4/5. 7. Multiply Fractions 6 This picture shows 3 square units. Two 1/2 squares are selected. This gives a first factor of 2 and a second factor of 1/2. Together 1/2 and 1/2 squares give a sum of 1 unit. 8. Multiply Fractions 7 The factors 2 and 1/2 are special because their product is 1. Factors that have a product of 1 are called reciprocals. Another name for reciprocal is multiplicative inverse, or inverse for short. 9. Multiply Fractions 8 To find the reciprocal of a fraction, replace the denominator with the numerator and the numerator with the denominator. The reciprocal or inverse of 2/1 is 1/2. 10. Multiply Fractions 9 The factors 1 1/4 and 4/5 are also reciprocals. As you can see, multiplying 5/4 by 4/5 gives a product of 1. If you are asked to invert or write the reciprocal of 5/4 you will write 4/5 . 11. Multiply Fractions 10 Both factors are greater than 1. Notice the product is greater than 4 x 1 but less than 5 x 2 by rounding down and rounding up both factors. 12. Multiply Fractions 11 You can tell by the picture that there are 4 whole units, five 1/2 units, and one 1/4 units. The sum of the units is 4 + 5/2 + 1/4 = 6 3/4 . 13. Multiply Fractions 12 The second factor has been decreased to 1. The product has been decreased to 4 1/2 . 14. Multiply Fractions 13 When 1 is used as a factor, the product is equal to the other factor. One is called the identity for multiplication. 15. Multiply Fractions 14 The second factor has been decreased to 1/2. Notice the product has been decreased to 2 1/4. When one of the factors is smaller than 1, the product is smaller than the other factor. 16. Multiply Fractions 15 Here both factors are less than 1. The product1/3 is smaller than either factor.
dr.carroll@optonline.net ### How do I draw a simple random sample? There are free random number generators on the internet that you can use. Two are: www.random.org and www.randomizer.org If you want to draw your own sample, there is a method you must abide by in order for your sample to be representative of the population. The first step is to identify all of the members in your population. You must be able to list them in what is called a sampling frame. The frame should have the names without order to them and non-overlapping (no duplicates). Alphabetizing the list by surname is a way to insure a random order in the sampling frame. Second, you must give each name an identification number. Start with “1” and continue. Third, you must decide what the size will be for your sample. You can use the table suggested in the Krejcie (1970) article, or whatever feels right for you to believe in the results you obtain. As a rule of thumb, use as large a sample size as possible. Fourth, you need to get a Table of Random Numbers. Many are located at the end of statistical or mathematical textbooks. The Table of Random Numbers consists of rows and columns of numbers arranged at random so that they can be used as any point by reading in any direction left or right, up or down. We are now ready to draw our random sample. Here is a simple example for you. Sampling Frame ID Name 1. Alex 2. Allen 3. Lea 4. Joey 5. Auggie 6. Mary 7. Ellen 8. Emily 9. Felice 10. Juan 11. Julia 12. Ruthie 13. Annie 14. Mark 15. Miguel 16. Maura 17. Olivia 18. Rachael 19. Rebecca 20. Seth Fifth, we know that our largest ID number has two digits (20). So we are going to need a two-digit column in the Table of Random Numbers. We start anyplace in the Table of Random numbers. We have decided in advance whether we would use two digits going up, down, left or right. So we begin. Sixth, Number #6 is the first ID number that is within our band of between 1 and 20. This is the first member of our random sample and it is Mary. We continue 82, 56, 96, 66, 46 until we and come up with the next is ID #13 Annie. Our three last members of the sample are ID #8 (Emily), ID #5 (Auggie) and ID #4 (Joey). We have five randomly selected members: Mary, Annie, Emily, Auggie and Joey.
## Saturday, May 09, 2015 ### The problem about the garden area The problem about the garden area Why such a strange image to task about the area of the garden? Because the problem itself, to put it mildly, very strange. Here's how it sounds: The garden occupies 80 hectares. Apple trees occupy 5/8 of the area, and 31% of cherry trees. How many hectares of area under apple trees larger than the area under the cherry trees? Let's first solve this problem for those students who want to write off the stupid decision, and only then talk about the strangeness of this problem. The first action we define the area that is occupied by apple trees in the garden. To do this, the total area of the garden should be multiplied by the fractional expression of the area under apple trees. 80 * 5/8 = 50 ha The second area of action is determined that the garden occupied cherry trees. Take the common garden area, multiplied by the number of cherry trees per cent and 100 per cent share. Interest on interest reduced and as a result we get an area of hectares. 80 * 31%: 100% = 24.8 ha The area under apple trees we really get more than the area under the cherry trees. Takes away from smaller and larger area of the results. 50 - 24.8 = 25.2 ha Answer: The area under apple trees on 25.2 hectares more than the area under the cherry trees. Without checking any decision can be considered incorrect. How to check the solution to this problem? It is necessary to put together the area under apple trees and the area under the cherry trees. This result should be compared with the total area of the garden. If the sum is greater than the total area, so we decided not to challenge properly. If the sum is equal to or less than the total area, so our decision is correct. 50 + 24.8 = 74.8 hectares of less than 80 hectares In most curious students immediately arises a natural question: what else is growing in this delicious daze, about what we were afraid to tell? It was a children's nursery problem solution. Now the conversation for adults. This is the task of the textbook, which approved the Ministry of Education as an educational tool. The condition of this problem and at the same time used the percentage of fractional parts of a whole. Normal literate people'd never allow it. They used a ratio or percentage. Only an idiot is able to dump everything into one pile. The author of this task is illiterate idiot who either do not understand what he was doing, or for approval of a textbook command is ready to do anything. Stati, the quality of textbooks is very well characterizes the quality of all education. Our education is built on the principle of "one fools compose tutorials, others say they are fools". For example, I'll write down the number on the same principle that is used by the author about the problem area of the garden. I simultaneously use two forms of records: number and letter. That's what I got: 2 thousand three hundred 45 As you can see, only idiots can do so. Competent people will write down this number as follows: 2345 or two thousand three hundred forty-five Why am I so vehemently opposed to such problems? Children - they are like a sponge that absorbs everything. If it is written in the book, then you can do so. As a result, we get the next portion of the idiots who write stupid textbooks that stupid leadership approves a flock of idiots. Just because they are so used to be taught. What would you say there is not mathematics, but knowledge of mathematics lies not in the ability to accurately repeat all the things the teacher taught. Knowledge of mathematics - is the ability to competently and simply express their thoughts in the language of mathematics.
Select Page 1. Find the compound amount and the interest earned for the following: Amount: $980 Rate: 8%, compounded quarterly Time: 5 years 2. Find the interest earned if$2750 is deposited on June 12 and withdrawn on August 30. Assume 3.5% interest compounded daily. 3. Find the present value for the following: Amount needed: $20,000 Time: 10 years Rate: 5%, compounded semiannually 4. What is the effect of inflation on spendable income? s. Application Problem: Describe a practical example where you do, or could, use the concepts of compound interest found in this chapter in your life. ## Sample Answer ## Sample Answer Title: Unraveling the World of Compound Interest: Calculations and Real-Life Applications Compound Amount and Interest Earned:Amount:$980 Rate: 8% Compounded Quarterly Time: 5 years To calculate the compound amount and interest earned, we can use the formula for compound interest: [ A = P(1 + \frac{r}{n})^{nt} ] Where: A = Amount after t years P = Principal amount ($980) r = Annual interest rate (8% or 0.08) n = Number of times interest is compounded per year (quarterly, so 4) t = Time in years (5) Plugging in the values: [ A = 980(1 + \frac{0.08}{4})^{45} ] [ A = 980(1 + 0.02)^{20} ] [ A ≈ 980(1.02)^{20} ] [ A ≈ 980 1.48594673 ] [ A ≈$1,456.03 ] To find the interest earned: Interest = Compound Amount – Principal Interest = $1,456.03 –$980 Interest ≈ $476.03 Interest Earned on Deposit and Withdrawal:Deposit Amount:$2750 Interest Rate: 3.5% Compounded Daily Time: June 12 to August 30 To calculate the interest earned, we can use the formula for compound interest with the given dates in mind. Present Value Calculation:Amount Needed: $20,000 Rate: 5% Compounded Semiannually Time: 10 years To find the present value, we need to determine how much money needs to be invested now to reach$20,000 in 10 years with a semiannual compounding interest rate of 5%. Effect of Inflation on Spendable Income: Inflation erodes the purchasing power of money over time, meaning that the same amount of money will buy fewer goods and services as time goes on. This decrease in purchasing power reduces the spendable income of individuals, as they need more money to maintain the same standard of living. Real-Life Application of Compound Interest: One practical example where compound interest concepts come into play is in long-term savings or investments, such as retirement funds or college savings plans. By understanding how compound interest works, individuals can make informed decisions about where to invest their money to maximize growth over time. For instance, contributing regularly to a retirement account and taking advantage of compounding can help build a substantial nest egg for the future.
# Boats and Streams Questions Answers • #### 1. A man can row upstream 10 kmph and downstream 20 kmph. Find the man rate in still water and rate of the stream. 1. 0,5 2. 5,5 3. 15,5 4. 10,5 Explanation: If a is rate downstream and b is rate upstream Rate in still water = 1/2(a+b) Rate of current = 1/2(a-b) => Rate in still water = 1/2(20+10) = 15 kmph => Rate of current = 1/2(20-10) = 5 kmph • #### 2. A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the current. 1. 1 km/hr 2. 2 km/hr 3. 3 km/hr 4. 4 km/hr Explanation: First of all, we know that speed of current = 1/2(speed downstream - speed upstream) [important] So we need to calculate speed downstream and speed upstream first. Speed = Distance / Time [important] \begin{aligned} \text {Speed upstream =}\\ (\frac{15}{3\frac{3}{4}}) km/hr \\ = 15 \times \frac{4}{15} = 4 km/hr \\ \text{Speed Downstream = } (\frac{5}{2\frac{1}{2}}) km/hr \\ = 5 \times \frac{2}{5} = 2 km/hr \\ \text {So speed of current = } \frac{1}{2}(4-2) \\ = 1 km/hr \end{aligned} • #### 3. In one hour, a boat goes 11km along the stream and 5 km against it. Find the speed of the boat in still water 1. 6 2. 7 3. 8 4. 9 Explanation: We know we can calculate it by 1/2(a+b) => 1/2(11+5) = 1/2(16) = 8 km/hr • #### 4. A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is 1. 2 km/hr 2. 3 km/hr 3. 4 km/hr 4. 5 km/hr Explanation: Let the speed of the stream be x km/hr. Then, Speed downstream = (15 + x) km/hr, Speed upstream = (15 - x) km/hr So we know from question that it took 4(1/2)hrs to travel back to same point. So, \begin{aligned} \frac{30}{15+x} - \frac{30}{15-x} = 4\frac{1}{2} \\ => \frac{900}{225 - x^2} = \frac{9}{2} \\ => 9x^2 = 225 \\ => x = 5 km/hr \end{aligned} • #### 5. If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the stream is 1. 5 km/hr 2. 4 km/hr 3. 2 km/hr 4. 1 km/hr Explanation: Rate upstream = (15/3) kmph Rate downstream (21/3) kmph = 7 kmph. Speed of stream (1/2)(7 - 5)kmph = 1 kmph • #### 6. A man can row \begin{aligned} 9\frac{1}{3} \end{aligned} kmph in still water and finds that it takes him thrice as much time to row up than as to row, down the same distance in the river. The speed of the current is. 1. \begin{aligned} 3\frac{2}{3}kmph \end{aligned} 2. \begin{aligned} 4\frac{2}{3}kmph \end{aligned} 3. \begin{aligned} 5\frac{2}{3}kmph \end{aligned} 4. \begin{aligned} 6\frac{2}{3}kmph \end{aligned} Explanation: Friends first we should analyse quickly that what we need to calculate and what values we require to get it. So here we need to get speed of current, for that we will need speed downstream and speed upstream, because we know Speed of current = 1/2(a-b) [important] Let the speed upstream = x kmph Then speed downstream is = 3x kmph [as per question] \begin{aligned} \text{speed in still water = } \frac{1}{2}(a+b) \\ => \frac{1}{2}(3x+x) \\ => 2x \\ \text{ as per question we know, }\\ 2x = 9\frac{1}{3} \\ => 2x = \frac{28}{3} => x = \frac{14}{3} \\ \end{aligned} So, Speed upstream = 14/3 km/hr, Speed downstream 14 km/hr. Speed of the current \begin{aligned} =\frac{1}{2}[14 - \frac{14}{3}]\\ = \frac{14}{3} = 4 \frac{2}{3} kmph \end{aligned} • #### 7. A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water is 1. 4 kmph 2. 5 kmph 3. 6 kmph 4. 7 kmph Explanation: Rate upstream = (750/675) = 10/9 m/sec Rate downstream (750/450) m/sec = 5/3 m/sec Rate in still water = (1/2)[(10/9) + (5/3)] m/sec. = 25/18 m/sec = (25/18)(18/5) kmph = 5 kmph • #### Akshay Kheral 6 years ago Indeed Helpful Questions for all govt competitive exam. #### mastguru 6 years ago replied Thank you Akshay for feedback ! • #### Akshay Prabhu 7 years ago nice questions..pls upload some questions for non verbal reasoning.. • #### rakesh 7 years ago not good............ • #### Dinesh 7 years ago Good Tutorial #### mastguru 7 years ago replied Thank you Dinesh.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Congruent Angles and Angle Bisectors ## Bisectors split the angle into two equal halves. 0% Progress Practice Congruent Angles and Angle Bisectors Progress 0% Congruent Angles and Angle Bisectors What if you knew that an angle was split exactly in half? How could you use this information to help you solve problems? After completing this Concept, you'll be able to bisect an angle and solve problems related to angle bisectors. ### Watch This Watch the first part of this video. ### Guidance When two rays have the same endpoint, an angle is created. Here, BA\begin{align}\overrightarrow{BA}\end{align} and \begin{align}\overrightarrow{BC}\end{align} meet to form an angle. An angle is labeled with an “\begin{align}\angle\end{align}” symbol in front of the three letters used to label it. This angle can be labeled \begin{align}\angle ABC\end{align} or \begin{align}\angle CBA\end{align}. Always put the vertex (the common endpoint of the two rays) in the middle of the three points. It doesn’t matter which side point is written first. An angle bisector is a ray that divides an angle into two congruent angles, each having a measure exactly half of the original angle. Every angle has exactly one angle bisector. \begin{align}\overline{BD}\end{align} is the angle bisector of \begin{align}\angle ABC\end{align} Label equal angles with angle markings, as shown below. ##### Investigation: Constructing an Angle Bisector 1. Draw an angle on your paper. Make sure one side is horizontal. 2. Place the pointer on the vertex. Draw an arc that intersects both sides. 3. Move the pointer to the arc intersection with the horizontal side. Make a second arc mark on the interior of the angle. Repeat on the other side. Make sure they intersect. 4. Connect the arc intersections from #3 with the vertex of the angle. To see an animation of this construction, view http://www.mathsisfun.com/geometry/construct-anglebisect.html. #### Example A How many angles are in the picture below? Label each one two different ways. There are three angles with vertex \begin{align}U\end{align}. It might be easier to see them all if we separate them out. So, the three angles can be labeled, \begin{align}\angle XUY\end{align} or \begin{align} \angle YUX, \ \angle YUZ\end{align} or \begin{align} \angle ZUY\end{align}, and \begin{align}\angle XUZ\end{align} or \begin{align} \angle ZUX\end{align}. #### Example B What is the measure of each angle? From the picture, we see that the angles are congruent, so the given measures are equal. To find the measure of \begin{align}\angle ABC\end{align}, plug in \begin{align}x = 8^\circ\end{align} to \begin{align}(5x + 7)^\circ\end{align}. Because \begin{align}m \angle ABC = m \angle XYZ, \ m \angle XYZ = 47^\circ\end{align} too. #### Example C Is \begin{align}\overline{OP}\end{align} the angle bisector of \begin{align} \angle SOT\end{align}? If \begin{align}m \angle ROT = 165^\circ\end{align}, what is \begin{align}m \angle SOP\end{align} and \begin{align}m \angle POT\end{align}? Yes, \begin{align}\overline{OP}\end{align} is the angle bisector of \begin{align}\angle SOT\end{align} according to the markings in the picture. If \begin{align}m \angle ROT = 165^\circ\end{align} and \begin{align}m \angle ROS = 57^\circ\end{align}, then \begin{align}m \angle SOT = 165^\circ - 57^\circ = 108^\circ\end{align}. The \begin{align}m \angle SOP\end{align} and \begin{align}m \angle POT\end{align} are each half of \begin{align}108^\circ\end{align} or \begin{align}54^\circ\end{align}. Watch this video for help with the Examples above. ### Vocabulary When two geometric figures have the same shape and size then they are congruent. An angle bisector is a ray that divides an angle into two congruent angles, each having a measure exactly half of the original angle. ### Guided Practice For exercises 1 and 2, copy the figure below and label it with the following information: 1. \begin{align}\angle A \cong \angle C\end{align} 2. \begin{align}\angle B \cong \angle D\end{align} 3. Use algebra to determine the value of d: 1. You should have corresponding markings on \begin{align}\angle A \end{align} and \begin{align}\angle C\end{align}. 2. You should have corresponding markings on \begin{align}\angle B \end{align} and \begin{align}\angle D\end{align} (that look different from the markings you made in #1). 3. The square marking means it is a \begin{align}90^\circ\end{align} angle, so the two angles are congruent. Set up an equation and solve: ### Practice For 1-4, use the following picture to answer the questions. 1. What is the angle bisector of \begin{align}\angle TPR\end{align}? 2. What is \begin{align}m\angle QPR\end{align}? 3. What is \begin{align}m\angle TPS\end{align}? 4. What is \begin{align}m\angle QPV\end{align}? For 5-6, use algebra to determine the value of variable in each problem. For 7-10, decide if the statement is true or false. 1. Every angle has exactly one angle bisector. 2. Any marking on an angle means that the angle is \begin{align}90^\circ\end{align}. 3. An angle bisector divides an angle into three congruent angles. 4. Congruent angles have the same measure. In Exercises 11-15, use the following information: \begin{align}Q\end{align} is in the interior of \begin{align}\angle ROS\end{align}. \begin{align}S\end{align} is in the interior of \begin{align}\angle QOP\end{align}. \begin{align}P\end{align} is in the interior of \begin{align}\angle SOT\end{align}. \begin{align}S\end{align} is in the interior of \begin{align}\angle ROT\end{align} and \begin{align}m \angle ROT = 160^\circ, \ m \angle SOT = 100^\circ,\end{align} and \begin{align}m \angle ROQ = m \angle QOS = m \angle POT\end{align}. 1. Make a sketch. 2. Find \begin{align}m \angle QOP\end{align} 3. Find \begin{align}m \angle QOT\end{align} 4. Find \begin{align}m \angle ROQ\end{align} 5. Find \begin{align}m \angle SOP\end{align} ### Vocabulary Language: English angle bisector angle bisector An angle bisector is a ray that splits an angle into two congruent, smaller angles. Congruent Congruent Congruent figures are identical in size, shape and measure.
# The Numbers Magician: Teaching Children Math Skills, Part 1 ## http://parentedge.in/the-numbers-magician-teaching-children-math-skills-part-1/ Math is one subject we cannot distance ourselves from – whether we make a career out of it or not, Math is an essential part of life. Math is more than just mastery over basic number functions. It is not all about + – x and divide. It is about understanding how numbers ‘fit’ into our day to day lives and the environment around us. Does an early schooler notice car number plates and their function? Does she recognise the changes that an added family member brings into the home? Does she notice the change in weight before and after eating? Or how one apple relates to a bean, or a pencil in terms of volume and weight? I wasn’t fortunate enough to get that understanding of math until the time I learnt a new approach to Math. This article is inspired by the teaching at my daughter’s school – Great Hearts Archway. I thank them for introducing these concepts to me. They follow Singapore Math as their module for teaching Mathematics. In Singapore math, they focus on laying a strong foundation of Math concepts. The chart you see below is called a ‘Ten’s Frame’. In a Ten’s frame, there are 2 rows of 5 blocks and rows are filled with varying number of black dots. One block has one big black dot. Children are taught to understand how many dots are there without counting on their fingers. E.g. in Fig. 1 kids are taught to see that all 5 blocks on top are filled; which makes 5 + one block at the bottom which makes 1. Hence, 6 blocks out of 10 are filled and 4 are empty. They are also taught to approach this problem in multiple ways. One could see it as 6 filled out of ten or 4 less than 10 or as times 5, where each row is seen as a multiple of 5. In Fig. 2, there are 3 dots filled out of 5 at the top. Remember that always, the blocks on the top are filled FIRST before filling blocks in the bottom. However, kids are challenged to think in different ways to fill 3 in blocks of ten. So the teacher will typically discuss how the same 3 dots could be placed in any of 10 blocks below. In Figure 3, one is filled and 9 are empty. In Fig. 4, I have shown a 20s frame. After the ten’s frame concepts are founded, one can move on to the 20s frame. I have noticed how quickly kids are able to pick this concept. I observed 4 kids and it took them less than 5 secs to say 16! They did not have to wait to count on their fingers. Here they are quick to see 5 + 5 + 5 + 1 GAME The Ten’s Frame War game is a game my daughter’s teacher plays in school. It is an excellent way to repeat and integrate that concept while also challenging them to be quick to evaluate without counting on one’s fingers. Split kids in pairs. Make an equal number of ten’s frame cards and give each child a pile. Have them open the top card and call out their number quickly without counting. The one who has the bigger number gets both the cards. You can decide who wins – the one with the most or the one with the least number of cards. MATH FOR BABIES AND TODDLERS For pre-schoolers, there is a different way to introduce this math concept. Make cardboard or hard paper squares approximately palm sized, each with different number of dots, beginning from 1 to 50. Start out with holding card dots of 1 to 10 in your hand in front of your child at his or her eye level. Face the child and show him one card at a time, quickly moving on to the next one. With each dot, say the number aloud. E.g. – This dot below is number 1. Hence, just say ‘1’ while showing it and so on. After 15 days of repetition, increase the number from 10 to 15 or up to 20 based on the child’s attention. Repeating this activity with infants starting 4 months, can strengthen math concepts early and set the stage for greater math understanding. Remember that for children of any age, do not do math and reading activities when they are tired, hungry, sleepy or frustrated. It beats the purpose and is energy wasted because the brain is too busy to fix on something else and cannot attend to the learning. Another way to teach Math is through ‘Hands-On Learning’. When teaching 2 + 2, have concrete objects to show them. 2 spoons + 2 spoons makes a lot more sense to any brain than just the numbers 2 + 2. You can further have 4 toys on one part of the sofa or on a chair on inside a hoola hoop. Then show the kid what it looks like when 1 is removed and then another 1. Then put the 2 removed toys on another part of the sofa, chair or into another hoola hoop. Now the child knows concretely what 4 looks like and how it can be divided into 2 groups of 2. You can also teach measuring skills from 2 years onwards. One can use one’s palm stretch from the thumb to the pinky finger to measure a table or a pencil. One can use a pencil to measure the length of the scale, the hand, the chair. Then one can weigh objects on both palms or on a weighing scale. How many erasers does it take to measure up to an apple? Do all apples measure the same? And so on. Measuring tapes and scales for real measuring are fun too. Kids love it! Measure their dresses, their favorite toys and even measure them while standing, sitting or sleeping. We shall continue our math concepts in the following month’s article… https://mystyrimz.wordpress.com/2015/10/31/the-numbers-magician-part-2-teaching-children-maths-skills-through-board-games/
# Blog Archives ## Fractions A fraction is any ratio between whole numbers. In mathematical terms, it is called a rational number. An irrational number is any number such as π which cannot be represented as the ratio of two whole numbers. 🙂 🙂 In today’s blog, I will go over a single lemma regarding fractions: Lemma: for any given rational number let’s say a/b, there exists an integer let’s say c such that: absolute (a/b – c) ≤ (1/2). (1) To prove this, we need only consider the case where abs(a) is greater abs(b). [If abs(a) ≤ abs(b), then the conclusion follows from Corollary 2.1, here] (2) From the division algorithm (see Theorem 1, here), we know that a = bq + r where r ≥ 0 and less than abs(b). [For example, if a=-3, b=-2, then q=2, r=1 where r is greater than b but less than abs(b).] (3) Let a’=abs(a), b’=abs(b) (4) We know that a’ – b’q is less than b’ (since a’ – b’q = r and r is less than b’) (5) So, it follows that: a’/b’ – q is less than 1. (6) Now, if both a,b are positive or a,b are negative, it follows that a/b = a’/b’ and abs(a/b – q) is less than 1. (7) If a,b are of different sign, than -a/b = a’/b’ and -a/b – q = -(a/b + q) so that abs(a/b + q) is less than 1. (8) The conclusion follows from Lemma 2, here. QED Corollary: if a/b is a rational number, then there exists an integer c such that: absolute(a/b – c/2) ≤ (1/4). (1) From the lemma above, we know that for 2(a/b), there exists a number c such that: abs(2(a/b) – c) ≤ (1/2). (2) Dividing both sides by 2, gives us: abs(a/b – c/2) ≤ (1/4) QED Now, it turns out that any number with a repeating decimal can be represented as a rational number. (1) Let’s assume that we have a decimal such as 5.234523452345… We can represent this decimal as a repeating decimal such as 5.2345. (2) Now, we know if we multiply the number by 104 we get: 52345.2345 (3) So, subtracting (2) by (1) gives us: 104 – 1 = 52345 – 5 = 52340. (4) So, the rational form of this repeating decimal is: 52340/9999. Now, let’s look at the proof that demonstrates this: Lemma: Any number with a repeating decimal is rational. (1) Any number with a repeating decimal can be represented with the following form: d1…dm.a1…an NOTE: If a number has a nonrepeating portion, then we multiply this number by the number of nonrepeating digits, to get a number of the above form. Later, we divide our result by this same number. (2) We can get an integer result by subtracting 10nthe number by the original number which after canceling for the repeating decimal gives us: 10n (d1…dm.a1…an…) – (d1…dm.a1…an…) = (d1..dma1..an) – (d1..dm). (3) Now, our rational number is equal to the value in step #2 divided by 10n – 1. QED THANK YOU !
# Class 10 Maths notes for Coordinate geometry In this page we will explain the topics for the chapter 7 of Coordinate geometry Class 10 Maths.We have given quality Coordinate geometry Class 10 Notes to explain various things so that students can benefits from it and learn maths in a fun and easy manner, Hope you like them and do not forget to like , social share and comment at the end of the page. Table of Content ## Introduction • We require two perpendicular axes to locate a point in the plane. One of them is horizontal and other is Vertical.The plane is called Cartesian plane and axis are called the coordinates axis • The horizontal axis is called x-axis and Vertical axis is called Y-axis • The point of intersection of axis is called origin. • The distance of a point from y axis is called x –coordinate or abscissa and the distance of the point from x –axis is called y – coordinate or Ordinate • The x-coordinate and y –coordinate of the point in the plane is written as (x, y) for point and is called the coordinates of the point • Above all in details can be read in Class IX Maths Coordinate Geometry notes ## Distance formula Distance between the points AB is given by Distance of Point A from Origin (1) Find the distance of point (3,4) and ( 4,3) from Origin? Solution Distance from Origin is given by $D=\sqrt {x^2 + y^2}$ for (3,4), $D=\sqrt {3^2 + 4^2} =5$ for (4,3), $D=\sqrt {4^2 + 3^2} =5$ Practice Questions (1) Arrange these points in ascending order of distance from origin (a) (1,2) (b) (5,5) (c) (-1, 3) (d) (1/2, 1) (e) (.5 ,.5) (f) (-1,1) (2) Find the distance between the points (a) (1,2) and (3,4) (b) (5,6) and ( 1,1) (c) (9,3) and ( 3,1) ## Section Formula A point P(x,y) which divide the line segment AB in the ratio m1 and m1 is given by The mid point P is given by ## Area of Triangle ABC Area of triangle ABC of coordinates A(x1,y1) , B(x2,y2) and C(x3,y3) is given by For point A,B and C to be collinear, The value of A should be zero ## How to Solve the line segment bisection ,trisection and four-section problem's 1. You will be given coordinates of the two point A , B 2. If the problem is to find bisection, then you can simply found the mid point using 3. If the problem is to find trisection(three equal parts of the line ).Let us assume the point are P and Q, then AP=PQ=QB Now P divides the line AB into 1:2 part While Q divides the line AB into 2:1 part So we can use section formula to get the coordinate of point P and Q 4. If the problem is to find four equal parts .Let us assume the point are P ,Q And R such that AP=PQ=QR=RB Method 1: Now P divides the line AB into 1:3 part Q divides the line AB into 1:1 part R divides the line AB into 3:1 part So we can use section formula to get the coordinate of point P ,Q and R Method 2: We can find using Mid point also as Q divides A and B in 1:1 and P divides A and Q in 1:1 and R divides Q and B in 1:1 ## How to Prove three points are collinear 1) We need to assume that if they are not collinear,they should be able to form triangle. We will calculate the area of the triangle,if it comes zero, that no triangle can be found and they are collinear ## Solved Examples Question 1 The points (4, 5), (7, 6) and (6, 3) are collinear. True or false Solution False. Since the area of the triangle formed by the points is 4 sq. units, the points are not collinear Question 2 If the mid-point of the line segment joining the points A (3, 4) and B (k, 6) is P (x, y) and $x + y - 10 = 0$, find the value of k. Solution Mid-point of the line segment joining A (3, 4) and B (k, 6) = $\frac {(3+k)}{2}$, 5 Now mid -point is P(x,y) $x=\frac {(3+k)}{2}$ or 3+k=2x y=5 Since $x + y - 10 = 0$, we have $\frac {3+k}{2} +5 - 10 = 0$ 3 + k = 10 or k=7 Question 3 ABCD is a parallelogram with vertices A (x1, y1), B (x2, y2) and C (x3, y3). Find the coordinates of the fourth vertex D in terms of x1,x2,x3,y1,y2,y3 Solution Let the coordinates of D be (x, y). We know that diagonals of a parallelogram bisect each other. Therefore, mid-point of AC = mid-point of BD $[ \frac{(x_1 + x_3)}{2} ,\frac{(y_1 + y_3)}{2}]=[ \frac{(x + x_2)}{2} ,\frac {(y + y_2)}{2}]$ i.e $x_1 + x_3 = x_2 + x$ and $y_1 + y_3 = y_2 + y$ $x= x_1 + x_3 - x_2$ and $y=y_1+ y_3 -y_2$ So coordinates are $(x_1 + x_3 - x_2, y_1+ y_3 - y_2)$ Question 4< The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is Solution $A=\frac {1}{2} [a(c+a -a-b) + b(a+b-b-c) + c( b+c -c-a)]$ $A=\frac {1}{2} [ac-bc + ba-bc + bc-ac)]$ A=0 ### Quiz Time Question 1 Point on y axis and x axis has coordinates: ? A) (a,0) & (0,b) B) (0,a) & (0,b) C) (a,b) & (a,b) D) (0,a) & (b,0) Question 2 The perimeter of a PQR triangle with vertices P(0, 4), Q(0, 0) and R(3, 0) is: ? A) 6 B) 7 C) 11 D) 12 Question 3 Which point is at minimum distance from origin A) (2, -3) B) (2,2) C) (6,8) D) (6,6) Question 4 The end points of diameter of circle are (0, 0) & (24, 7). The radius of the circle is: A) 12.5 B) 15.5 C) 12 D) 10 Question 5 The mid point of point A(4,5) and B(-8,7) lies in quadrant? A) I B) II C)III D) IV Question 6 The point on the x-axis which is equidistant from (-2,5) and (2, -3) is? A)(0,3) B) (0,0) C)(5,0) D) (-2,0) Crossword Puzzle Across 1. the distance of point(3,4) from origin 3. x-coordinate is called 5. the y -axis is a .....line 6. The (-6,-2) lies in .....quadrant 7. the x-axis is a ......line 9. The point of intersection of axis 10. The figure formed by joining the point (0,0), (2,0),(2,2) and (0,2) Down 2. the coordinates (3,0) (-3,0) and $(0,3 \sqrt {3})$ formed an ...... triangle 4. The three point are said to be ....if the area of triangle formed by them is zero 8. y-coordinate is called (1)five (2)equilateral (3)abscissa (4)collinear (5)vertical (6)fourth (7)horizontal (8)Ordinate (9) origin (10) square ## Summary Here is Coordinate geometry Class 10 Maths Notes Summary • The distance between two points is given by $D= \sqrt {(x_2 -x_1)^2 + (y_2 -y_1)^2}$ • The distance of a point from origin is given by $D= \sqrt {x^2 + y^2}$ • The coordinates of the point O(x, y) which divides the line segment joining the points P(x1, y1) and Q(x2, y2) internally in the ratio m : n is given by $x=\frac {mx_2 + nx_1}{m+n}$ and $y=\frac {my_2 + ny_1}{m+n}$ • The area of the triangle is given by $A= \|\frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y-1) + x_3(y_1 -y_2)]\|$ Go back to Class 10 Main Page using below links ### Practice Question Question 1 What is $1 - \sqrt {3}$ ? A) Non terminating repeating B) Non terminating non repeating C) Terminating D) None of the above Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is? A) 19.4 cm3 B) 12 cm3 C) 78.6 cm3 D) 58.2 cm3 Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ? A) 2 ,21,11 B) 1,10,19 C) -1 ,8,17 D) 2 ,11,20 Synthetic Fibres and Plastics Class 8 Practice questions Class 8 science chapter 5 extra questions and Answers Mass Calculator 3 Fraction calculator Garbage in Garbage out Extra Questions7
Saturday, November 26, 2022, 8:55 PM Site: Dearborn Educational Curriculum Course: Dearborn Educational Curriculum (DEC) Glossary: Mi Math Standards 3 #### 3.MD.C.7c Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning. 03 #### 3.MD.C.7d Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems. 03 #### 3.MD.D Grade 3 » Measurement & Data » Geometric measurement: recognize perimeter. #### 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters. 03 #### 3.NBT.A Grade 3 » Number & Operations in Base Ten » Use place value understanding and properties of operations to perform multi-digit arithmetic.¹ #### 3.NBT.A.1 Use place value understanding to round whole numbers to the nearest 10 or 100. 03 #### 3.NBT.A.2 Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction. 03 #### 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 in the range 10-90 (e.g., 9 × 80, 5 × 60) using strategies based on place value and properties of operations. 03 #### 3.NF.A Grade 3 » Number & Operations—Fractions¹ » Develop understanding of fractions as numbers. #### 3.NF.A.1 Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
# Diameter of a Sphere – Formulas and Examples The diameter of a sphere is equal to the line segment connecting two ends of the sphere and passing through the center. Diameter is an important measure of the sphere since, similar to the radius, we can also use the diameter to calculate the volume and surface area of the sphere. This means that it is also possible to find the length of the diameter if we know the volume or the surface area of the sphere. Here, we will learn about three different methods that we can use to find the length of the diameter. In addition, we will use these methods to solve some problems. ##### GEOMETRY Relevant for Learning to calculate the diameter of a sphere. See methods ##### GEOMETRY Relevant for Learning to calculate the diameter of a sphere. See methods ## Diameter of a sphere using the radius The radius of a sphere is the line segment that connects the center of the sphere to any point on its surface. The radius is probably the most important dimension of a sphere. This means that in most cases, we will know the length of the radius. We can calculate the length of the diameter starting from the radius simply by multiplying the radius by 2. Therefore, if we have the length of the radius, we use the following expression to get the length of the diameter: $latex d=2r$ #### EXAMPLE 1 If a sphere has a radius of 5 m, what is its diameter? Solution: We have the length $latex r=5$. Therefore, we use the formula with this value: $latex d=2r$ $latex d=2(5)$ $latex d=10$ The diameter is 10 m. #### EXAMPLE 2 What is the diameter of a sphere that has a radius of 11 m? Solution: We use the value $latex r=11$ in the given formula: $latex d=2r$ $latex d=2(11)$ $latex d=22$ The diameter measures 22 m. ## Diameter of a sphere using the volume The diameter of a sphere can be calculated if we have its volume. Recall that the following is the formula for the volume of a sphere: $latex V=\frac{4}{3}\pi {{r}^3}$ Writing this in terms of the diameter, we have: $latex V=\frac{1}{6}\pi {{d}^3}$ Therefore, we use this formula and solve for d. #### EXAMPLE 1 If a sphere has a volume of 100 m³, what is its diameter? Solution: We use the value $latex V=100$ in the volume formula and solve for d: $latex V=\frac{1}{6}\pi {{d}^3}$ $latex 100=\frac{1}{6}\pi {{d}^3}$ $latex {{d}^3}=\frac{6(100)}{\pi}$ $latex {{d}^3}=\frac{600}{\pi}$ $latex {{d}^3}=191$ $latex d \approx 5.76$ The diameter measures 5.76 m. #### EXAMPLE 2 What is the diameter of a sphere that has a volume of 240 m³? Solution: We substitute the value $latex V=240$ in the volume formula and solve for d: $latex V=\frac{1}{6}\pi {{d}^3}$ $latex 240=\frac{1}{6}\pi {{d}^3}$ $latex {{d}^3}=\frac{6(240)}{\pi}$ $latex {{d}^3}=\frac{1440}{\pi}$ $latex {{d}^3}=458.4$ $latex d \approx 7.71$ The diameter measures 7.71 m. ## Diameter of a sphere using the surface area We can also use the surface area to calculate the diameter of a sphere. For this, we use the formula for surface area and solve for the diameter. Recall that the formula for the surface area of a sphere is: $latex A_{s}=4\pi {{r}^2}$ Writing in terms of the diameter, we have: $latex A_{s}=\pi {{d}^2}$ #### EXAMPLE 1 What is the diameter of a sphere that has a surface area of 100 m²? Solution: We use the value $latex A_{s}=100$ in the formula for surface area and solve for d: $latex A_{s}=\pi {{d}^2}$ $latex 100=\pi {{d}^2}$ $latex {{d}^2}=\frac{100}{\pi}$ $latex {{d}^2}=31.83$ $latex d\approx 5.64$ The diameter of the sphere measures 5.64 m. #### EXAMPLE 2 If a sphere has a surface area of 240 m², what is its diameter? Solution: We have the value $latex A_{s}=240$, so we use it in the formula for surface area and solve for d: $latex A_{s}=\pi {{d}^2}$ $latex 240=\pi {{d}^2}$ $latex {{d}^2}=\frac{240}{\pi}$ $latex {{d}^2}=76.4$ $latex d\approx 8.74$ The diameter of the sphere measures 8.74 m.
# Solving Ratios Word Problems You can answer even the most complex word problems, provided you understand the mathematical concepts addressed.While the degree of difficulty may change, the way to solve word problems involves a planned approach that requires identifying the problem, gathering the relevant information, creating the equation, solving and checking your work.Double-check your calculations along the way to prevent any mistakes. You can answer even the most complex word problems, provided you understand the mathematical concepts addressed.While the degree of difficulty may change, the way to solve word problems involves a planned approach that requires identifying the problem, gathering the relevant information, creating the equation, solving and checking your work.Double-check your calculations along the way to prevent any mistakes. Tags: Cie Igcse English CourseworkAnthem Essays IndividualitySites To Solve Math ProblemsDissertation Chair Typically SelectedDefense Of Thesis ProposalInvestment Group Business PlanTaxidermy Business Plan In certain circumstances, the comparison by the process of division makes good sense when compared to performing their difference. Hence, the comparison of two quantities by the process of division method is called as ‘ Ratio and proportions are said to be faces of the same coin. Suppose we have two quantities or two numbers or two entities and we have to find the ratio of these two, then the formula for ratio is defined as; a : b ⇒ a/b where, a and b could be any two . Now, let us assume that, in proportion, the two ratios are a:b & c:d. In the example, you know by adding up all the numbers for the sisters that you have a maximum of 23 socks. Since the problem mentions that the little sister lost two pairs, the final answer must be less than 23. Subtract the two missing pairs for a final equation of (8 6 9) - 2 = n, where n is the number of pairs of socks the sisters have left. Using the equation, solve the problem by plugging in the values and solving for the unknown variable. Question 2: Are the two ratios and in proportion? Solution: = 8/10= 0.8 and = 7/10= 0.7 Since both the ratios are not equal, they are not in proportion. Use a letter for the unknown variable, and create an algebraic equation that represents the problem. In the example, take the total number of pairs of socks Suzy owns -- eight plus six. ## Comments Solving Ratios Word Problems • ###### Ratio word solving problem? Yahoo Answers Ratio word solving problem? During a sale the price of a dishwasher is reduced from \$640 to 560\$. A Fridge costing \$896 is reduced in the same ratio as the dishwasher. What does the fridge sell for? Please answer and explain, it would mean a lot!… • ###### Ratio Word Problems Math Salamanders Ratio Word Problems. Here you will find a range of problem solving worksheets about ratio. The sheets involve using and applying knowledge to ratios to solve problems. The sheets have been put in order of difficulty, with the easiest first. Each problem sheet comes complete with an answer sheet.… • ###### How do you solve ratio's word problems The rules for solving word problems are read the problem, decide what you need to do, solve the problem, and check your the ratio and make that the denominator of the proportion and cross multiply. A proportion will help you solve problems like the one below.… • ###### Ratios Word Problems - Formulae & Examples Ratios Word Problems. Ratio is a mathematical concept that we use most frequently in day to day life. If you are out to buy groceries and you find that one pack of oats costs say6 A continued ratio is a ratio of three or more quantities. It can be written as \$abcde\$ Word Problems. Back to Top.… • ###### Solving Word Problems with Ratios and Proportions Want to know something exciting? I taught ratios and proportions in Algebra 1 without once mentioning cross multiplying. Two kids brought it up as an option, but I quickly told them that there was a way to do these problems without cross multiplication.… • ###### Word Problems Calculator Math Celebrity This word problem lesson solves for a quantity of two coins totaling a certain value with a certain amount more or less of one coin than another. Solves a ratio word problem using a given ratio of 2 items in proportion to a whole number. Features Calculator Practice Problem GeneratorTags.…
Courses # Independent Events and Their Important Properties JEE Notes \| EduRev ## JEE : Independent Events and Their Important Properties JEE Notes \| EduRev The document Independent Events and Their Important Properties JEE Notes \| EduRev is a part of the JEE Course Mathematics (Maths) Class 12. All you need of JEE at this link: JEE Independent Events Two events A & B are said to be independent if occurrence or non occurrence of one does not effect the probability of the occurrence or non occurence of other. (a) If the occurrence of one event affects the probability of the occurrence of the other event then the events are said to be dependent or Contingent. For two independent events A and B . P(B). Often this is taken as the definition of independent events. (b) Three events A, B & C are independent if & only if all the following conditions hold ; i.e., they must be pairwise as well as mutually independent. Similarly for n events A1, A2, A3, ........... An to be independent, the number of these conditions is equal to nC2 + nC3 + ............. + nCn = 2n - n - 1. Note : Independent events are not in general mutually exclusive & vice versa. Mutually exclusiveness can be used when the events are taken from the same experiment & independence can be used when the events are taken from different experiments. Ex.1 The probability that an anti aircraft gun can hit an enemy plane at the first, second and third shot are 0.6, 0.7 and 0.1 respectively. The probability that the gun hits the plane is Sol. Let the events of hitting the enemy plane at the first, second and third shot are respectively A, B and C. Then as given P(A) = 0.6, P(B) = 0.7, P(C) = 0.1 Since events A, B, C are independent, so â‡’ = 1 - (1 - 0.6) (1 - 0.7) (1 - 0.1)) = 1 - (0.4)(0.3)(0.9) = 1 - 0.108 = 0.892 Ex.2 If two events A and B are such that , P(B) = 0.4 and then equals Sol. Probability of Three Events For any three events A, B and C we have (a) P(A or B or C) = P(A) + P(B) + P(C) â€“ P(Aâˆ©B) â€“ P(B âˆ© C) â€“ P(C âˆ© A) + P(A âˆ© B âˆ© C) (b) P (at least two of A, B, C occur) = P(B âˆ© C) + P (C âˆ© A) + P(A âˆ© B) â€“ 2P (A âˆ© B âˆ© C) (c) P (exactly two of A, B, C occur) = P(B âˆ© C) + P (C âˆ© A) + P(A âˆ© B) â€“ 3P (A âˆ© B âˆ© C) (d) P (exactly one of A, B, C occurs) = P(A) + P(B) + P(C) â€“ 2P (B âˆ© C) â€“ 2P (C âˆ© A) â€“ 2P (A âˆ© B) + 3P (A âˆ© B âˆ© C) Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity! ## Mathematics (Maths) Class 12 209 videos\|222 docs\|124 tests , , , , , , , , , , , , , , , , , , , , , ;
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 # NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 ## NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1. ### Ex 3.1 Class 8 Maths Question 1. Given here are some figures. Classify each of them on the basis of the following: (a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon Solution: (a) Simple curve: (1), (2), (5), (6) and (7) are simple curves. (b) Simple closed curve: (1), (2), (5), (6) and (7) are simple curves. (c) Polygon: (1) and (2) are polygons. (d) Convex polygon: (2) is convex polygon. (e) Concave polygon: (1) is concave polygon. ### Ex 3.1 Class 8 Maths Question 2. How many diagonals does each of the following have? (b) A regular hexagon (c) A triangle Solution: (a) In the following figure, ABCD is a convex quadrilateral which has two diagonals AC and BD. (b) In the following figure, ABCDEF is a regular hexagon which has nine diagonals AE, AD, AC, BF, BE, BD, CF, CE and DF. (c) In the following figure, ABC is a triangle which has no diagonal. ### Ex 3.1 Class 8 Maths Question 3. What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and verify) Solution: In the following figure, we have a quadrilateral ABCD. Join the diagonal AC, which divides the quadrilateral into two triangles ABC and ACD. In ∆ABC, 3 + 4 + 6 = 180° …(i) (angle sum property) In ∆ACD, 1 + 2 + 5 = 180° …(ii) (angle sum property) Adding equations (i) and (ii), we get 1 + 3 + 2 + 4 + 5 + 6 = 180° + 180° A + C + D + B = 360° Hence, the sum of the measures of the angles of a convex quadrilateral is 360°. Let us draw a non-convex quadrilateral. Join the diagonal BD and prove as above in triangles BCD and ABD. Thus, this property also holds true for a non-convex quadrilateral. ### Ex 3.1 Class 8 Maths Question 4. Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that). What can you say about the angle sum of a convex polygon with number of sides? (a) 7 (b) 8 (c) 10 (d) n Solution: From the given table, we can conclude that the sum of the angles of a convex polygon with side ‘n’ = (n – 2) × 180° (a) If the number of sides = 7 Then the angle sum = (7 – 2) × 180° = 5 × 180° = 900° (b) If the number of sides = 8 Then the angle sum = (8 – 2) × 180° = 6 × 180° = 1080° (c) If the number of sides = 10 Then the angle sum = (10 – 2) × 180° = 8 × 180° = 1440° (d) If the number of sides = n Then the angle sum = (n – 2) × 180° ### Ex 3.1 Class 8 Maths Question 5. What is a regular polygon? State the name of a regular polygon of (i) 3 sides (ii) 4 sides (iii) 6 sides Solution: A polygon, whose all sides and all angles are equal, is called a regular polygon. (i) A polygon with 3 sides is called an equilateral triangle. (ii) A polygon with 4 sides is called a square. (iii) A polygon with 6 sides is called a regular hexagon. ### Ex 3.1 Class 8 Maths Question 6. Find the angle measure x in the following figures: Solution: (a) Sum of the angles of a quadrilateral = 360° 50° + 130° + 120° + x = 360° 300° + x = 360° x = 360° – 300° x = 60° (b) Sum of the angles of a quadrilateral = 360° x + 70° + 60° + 90° = 360° [ 180° – 90° = 90°] x + 220° = 360° x = 360° – 220° x = 140° (c) Sum of the angles of a pentagon = 540° 30° + x + 110° + 120° + x = 540° [ 180° – 70° = 110°; 180° – 60° = 120°] 2x + 260° = 540° 2x = 540° – 260° 2x = 280° x = 140° (d) Sum of the angles of a regular pentagon = 540° x + x + x + x + x = 540° [All the angles of a regular pentagon are equal] 5x = 540° x = 108° ### Ex 3.1 Class 8 Maths Question 7. (a) Find x + y + z (b) Find x + y + z + w Solution: (a) a + 30° + 90° = 180° [Angle sum property of a triangle] a + 120° = 180° a = 180° – 120° = 60° Now, y = 180° – a [Linear pair] y = 180° – 60° y = 120° and, z + 30° = 180° [Linear pair] z = 180° – 30° = 150° also, x + 90° = 180° [Linear pair] x = 180° – 90° = 90° Thus, x + y + z = 90° + 120° + 150° = 360° (b) r + 120° + 80° + 60° = 360° [Angle sum property of a quadrilateral] r + 260° = 360° r = 360° – 260° = 100° Now, x + 120° = 180° (Linear pair) x = 180° – 120° = 60° and y + 80° = 180° (Linear pair) y = 180° – 80° = 100° again, z + 60° = 180° (Linear pair) z = 180° – 60° = 120° also, w = 180° – r = 180° – 100° = 80° (Linear pair) Therefore, x + y + z + w = 60° + 100° + 120° + 80° = 360°. You can also like these: NCERT Solutions for Maths Class 9 NCERT Solutions for Maths Class 10 NCERT Solutions for Maths Class 11 NCERT Solutions for Maths Class 12
Standard and vertex form of the equation of parabola and how it relates to a parabola's graph. # Equation of a Parabola Standard Form and Vertex Form Equations The equation of a parabola can be expressed in either standard or vertex form as shown in the picture below. # Standard Form Equation The standard form of a parabola's equation is generally expressed: $$y = ax^2 + bx + c$$ The role of 'a' The role of 'a' The larger the $\|a\|$ is (when $\|a\|$ is greater than 1), the more the graphs narrows. Case I : When $\|a\| > 1$ Case II : When $\|a\| < 1$ The larger the $\|a\|$ is (when $\|a\|$ is greater than 1), the more the graphs narrows. The axis of symmetry The axis of symmetry is the line $x = -\frac{b}{2a}$ Picture of Standard form equation Axis of Symmetry from Standard Form # Vertex Form The vertex form of a parabola's equation is generally expressed as: $y = a(x-h)^2 +k$ • (h,k) is the vertex as you can see in the picture below • If a is positive then the parabola opens upwards like a regular "U". • If a is negative, then the graph opens downwards like an upside down "U". • And, just like standard form, the larger the $\|a\|$, the more narrow the parabola's graph gets. The role of 'a' Case I : When $\|a\| > 1$ The larger the $\|a\|$ is, the more the graphs narrows. Case II : When $\|a\| < 1$ The larger the $\|a\|$ is, the more narrow the parabola is. Or, another way to think of it, the closer that the value of $a$ gets to zero, the wider the parabola becomes. # Vertex Form Practice Problems ##### Problem 1 The parabola's vertex is the point (1,1). #### Identifying the vertex in vertex form ##### Problem 4.1 The vertex is the point (-3,4) ##### Problem 4.2 (3,4) is the vertex. ##### Problem 4.3 vertex is (2, –3) #### Part II ##### Problem 5.1 The vertex is (3,4) and it opens upwards since a is positive( it is 2), it opens upwards. ##### Problem 5.2 Vertex = (-3, 4), and it opens upwards since a is positive. ##### Problem 5.3 Vertex = (9, 5) and since a is negative (it is -22), it opens downwards.
Part 4: Division 4.2: Word problems for dividing integers Notes Word problems for the division problem, A ÷ B, can either use the “how-many-groups” interpretation of division or the “how-many-units-in-one-group” interpretation. How many groups? A ÷ B represents, “How many groups are there if A units are divided into groups of B units?” For example, • Arithmetic problem: Solve 12 ÷ 4. • Word problem: A pizza cut into 12 slices is to be divided among friends such that each friend gets 4 slices. How many friends can share the pizza? • 12 ÷ 4 = 3, so 3 friends can share the pizza. • Here, each friend is a “group,” the pizza slices are the units, and we’re finding the number of friends: 12 slices ÷ 4 slices per fried = 3 friends. • Equivalently, 3 friends times 4 slices per friend equals 12 slices in total. How many units in one group? A ÷ B represents, “How many units are in one group if A units are divided into B groups?” For example, • Arithmetic problem: Solve 12 ÷ 4. • Word problem: A pizza cut into 12 slices is to be divided among 4 friends. How many slices do they each get? • 12 ÷ 4 = 3, so each friend gets 3 slices. • Again, each friend is a “group” and the pizza slices are the units, but now we’re finding the number of slices per friend: 12 slices ÷ 4 friends = 3 slices per friend. • Equivalently, 3 slices per friend times 4 friends equals 12 slices in total. The video below works through some examples of word problems for dividing integers. Practice Exercises Do the following exercises to practice matching integer division problems and word problems.
## With Ease No more spending hours on rote memorization and repetitive worksheets. Smartick’s personalized math training method is designed to boost your child’s skills in a fraction of the time. ## With Ease Welcome, young learners, to the fascinating world of triangles! Triangles are amazing shapes that can be found all around us. From the pyramids of Egypt to the roofs of houses, triangles are everywhere. No credit card required No credit card required # Learn About Triangles with Smartick What is a Triangle? Basic Properties of Triangles What are the Different Types of Triangles? Classification on the basis of Angles What is a Triangle? A triangle is a polygon with three sides and three angles. It is one of the simplest and most common shapes in geometry. The sides of a triangle are line segments, and the angles are the spaces between these line segments. Triangles are named based on their side lengths and angle measurements. Basic Properties of Triangles Basic Properties of Triangles: • Three Sides: A triangle always has three sides, which are line segments joining three non-collinear points. • Three Angles: A triangle also has three angles formed by the intersection of its sides. • Sum of Angles: The sum of all the angles in a triangle is always 180 degrees. • Interior and Exterior: The region enclosed by the triangle is called the interior, and the space outside the triangle is called the exterior. • Vertices: The corners of a triangle are called vertices (singular: vertex). What are the Different Types of Triangles? Let’s explore some common types: • Equilateral Triangle: An equilateral triangle has three equal sides and three equal angles of 60 degrees each. • Isosceles Triangle: An isosceles triangle has two equal sides and two equal angles. • Scalene Triangle: A scalene triangle has three unequal sides and three unequal angles. Classification on the basis of Angles Right Triangle: A right triangle has one angle measuring 90 degrees, forming a right angle. Acute Triangle: An acute triangle has three angles measuring less than 90 degrees. Obtuse Triangle: An obtuse triangle has one angle measuring more than 90 degrees. Q1: Can a triangle have two right angles? A1: No, a triangle cannot have two right angles. The sum of the angles in a triangle is 180 degrees, and if two angles are right angles (90 degrees each), the third angle would be 0 degrees, which is not possible. Q2: What is the longest side in a right triangle called? A2: The longest side in a right triangle is called the hypotenuse. Q3: Are all equilateral triangles also isosceles triangles? A3: Yes, every equilateral triangle is also an isosceles triangle because it has two equal sides. Q4: How many acute angles does an acute triangle have? 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Pythagoras Theorem # Pythagoras Theorem ## Pythagoras Theorem In a right-angled triangle, the side opposite to the right angle is called the hypotenuse. It is the longest side of the right-angled triangle. In the following figure, the side AB, which is opposite to C (right angle), is the hypotenuse. We generally use the small letter of an angle in a triangle to denote the side opposite the angle. In the above figure, a, b, and c denote the sides opposite A, B, and C respectively. Pythagoras theorem relates the lengths of the three sides of a right-angled triangle. According to Pythagoras theorem, In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.” Pythagoras theorem states that, In ABC, if C = 90°, then AB2 = BC2 + AC2 or c2 = a2 + b2 Example 1: In ABC, C = 90°, AC = 8 cm, and BC = 6 cm. Find the length of AB. Solution: C = 90° (given) c2 = a2 + b2 (Pythagoras Theorem) = (6)2 + (8)2 = 36 + 64 = 100 c = 100 = 10 The length of AB is 10 cm. Example 2: In PQR, P = 90°, PQ = 24 cm, and QR = 25 cm. Find the length of PR. Solution: P = 90° (given) QR2 = PQ2 + PR2 (Pythagoras Theorem) 252 = 242 + PR2 PR2 = 252 – 242 = 625 – 576 = 49 PR = 49 = 7 cm The length of PR is 7 cm. ## The Converse of Pythagoras Theorem In a triangle, if the square of the longest side is equal to the sum of the squares of the other two sides, then the angle opposite the longest side is a right angle. For any triangle with sides a, b, and c, if c2 = a2 + b2 , then the angle between a and b measures 90° and the triangle is a right-angled triangle. This is called the converse of Pythagoras theorem. Example 3: Determine whether the following triangle is a right-angled triangle. Solution: QR2 = 302 = 900 PR2 + PQ2 = 202 + 212 = 841 QR2 PR2 + PQ2 Hence, PQR is not a right-angled triangle. Example 4: The sides of a triangle are 10 cm, 24 cm and 26 cm. Determine whether the triangle is a right-angled triangle or not. Solution: Here, 102 + 242 = 100 + 576 = 676 Again, 262 = 676 102 + 242 262 In the given triangle, (Base)2 + (Perpendicular)2 (Hypotenuse)2 Hence, the triangle is a right-angled triangle. Please do not enter any spam link in the comment box.
## Solve Lesson 8 Page 62 SBT Math 10 – Kite> Topic Find real numbers a and b such that each of the following pairs of vectors are equal: a) $$\overrightarrow m = (2a + 3;b – 1)$$ and $$\overrightarrow n = (1; – 2)$$ b) $$\overrightarrow u = (3a – 2;5)$$and $$\overrightarrow v = (5;2b + 1)$$ c) $$\overrightarrow x = (2a + b;2b)$$ and $$\overrightarrow y = (3 + 2b;b – 3a)$$ Solution method – See details $$\overrightarrow a = ({x_1};{y_1})$$ and $$\overrightarrow b = ({x_2};{y_2})$$ are equal if and only if $$\left\{ \begin{ array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.$$ Detailed explanation a) $$\overrightarrow m = (2a + 3;b – 1)$$ and $$\overrightarrow n = (1; – 2)$$ $$\overrightarrow m = \overrightarrow n \Leftrightarrow \left\{ \begin{array}{l}2a + 3 = 1\\b – 1 = – 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = – 1\\b = – 1\end{array} \right.$$ b) $$\overrightarrow u = (3a – 2;5)$$and $$\overrightarrow v = (5;2b + 1)$$ $$\overrightarrow u = \overrightarrow v \Leftrightarrow \left\{ \begin{array}{l}3a – 2 = 5\\5 = 2b + 1\end{array} \right. \Leftrightarrow \left\{ \ begin{array}{l}a = \frac{7}{3}\\b = 2\end{array} \right.$$ c) $$\overrightarrow x = (2a + b;2b)$$ and $$\overrightarrow y = (3 + 2b;b – 3a)$$ $$\overrightarrow x = \overrightarrow y \Leftrightarrow \left\{ \begin{array}{l}2a + b = 3 + 2b\\2b = b – 3a\end{array} \right. \Leftrightarrow \left\ { \begin{array}{l}2a – b = 3\\3a + b = 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a = \frac{3} {5}\\b = – \frac{9}{5}\end{array} \right.$$
1 / 34 Welcome: Please have your math supplies and homework ready. Sec 1.1.3; page 16. Function Notation Domain and Range Intersections of Graphs. Try the following on your calculator and record what your calculator displays. Who is this?. Function Notation. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript ### Sec 1.1.3; page 16 Function Notation Domain and Range Intersections of Graphs ### Function Notation Find function values using f(x) notation or graphs. What is function notation? Function notation is another way to write “y=“ The notation looks like this: f(x) f(x) is read “f of x” Solving Problems with Function Notation f(x) = 2x + 5, find f(5) f(5) = 2(5) + 5 Plug in 5 in place of x f(5) = 10 + 5 Simplify f(5) = 15 Simplify This means that when x = 5, y = 15 (5,15) Try this problem f(x) = -3x + 5, find f(-2) f( -2 ) = -3( -2 ) + 5 f( -2 ) = 6 + 5 f( -2 ) = 11 Other forms of function notation: Function notation is useful if you are studying more than one function. You may use other letters to describe the function and the variable. The following are some examples: expressed as: g(x), h(x), or s(t) They are all used in place of “y=“ . Two Functions at One Time f(x) = 6x + 3 g(r) = 2r – 4 Find the following: • f(4) 2. g(-1) 3. g(3) 4. f(0) See this problem worked See this problem worked See this problem worked See this problem worked For each function determine if there are any restrictions on the values that can be assigned to x. I will demonstrate with k h- 7 j(x) = Domain The domain of a function is the complete set of possible values of the independent variable in the function. • The domain of a function is the set of all possible x-values which will make the function "work" and will output real y-values. • When finding the domain, remember: • The denominator of a fraction cannot be zero. • The values under a square root sign ) must be zero or positive. Range The range of a function is the complete set of all possible resulting values of the dependent variable (y ) of a function, after we have substituted the domain values. • The rangeof a function is the possible y values of a function that result when we substitute all the possible x-values into the function. When finding the range, remember: • Substitute different x-values into the expression for y to see what is happening. (Is y always positive? Always negative? Or maybe not equal to certain values?) • Make sure you look for minimum and maximum values of y. BESTEST HINT EVER: • A picture is worth a thousand words. Graph the function! Examples: • Graphical Examples • Not Graphical Examples First on your calculator and then on your graph paperDraw a complete graph of y=(x+1)(x-3) State the domain and range of the function. ### II. Intersection of Graphs Systems of Equations. First on your calculator and then on your graph paperDraw a complete graph functions below, on the same grid. • and x-9 Find the solution to system of equations • and x-9 How does graphing two functions and finding the solution to a system of equations related? System of Equations Word Problem • Kristin spent \$262 on shirts. Dress shirts cost \$56 each and Dri-Fit shirts cost \$30 each. If she bought a total of 7 shirts then how many of each kind did she buy? • What Are our two equations? • 56D+30F=262 and D+F=7 Solution to Word Problem F=7-D ; We can rearrange one of the equations to solve for one variable in terms of the other 56D+30(7-D)=262 ; we can then substitute this into the other equation, there is only one variable to solve for now! 56D+210-30D=262 ; simplify 26D=52 ; simplify D=2 ; solve for D (2)+F=7 ; original equation with our value for D substituted in F=5; solve for F This means Kristen bought 5 Dri-Fit shirts and 2 Dress shirts Graph of Problem What is the graph telling us? Before you start on your homework… Make sure you know how to do the following: Solve for y: x= 3y-4
Easiest Proof Pythagorean Theorem with LEGO Pythagorean Theorem is a widely used theorem in math and geometry with many applications. It shows the relationship of the 3 sides of a right triangle. It is used so much, that it is hard to not know it. However, many students just know it as it is. They never know how to prove it. How do you know Pythagorean Theorem formula is true? Today we are going to show you, with the toy you have at home, LEGO. Created by my son, this is the easiest proof of Pythagorean Theorem, so easy that a 3rd grader will be able to do it. Sometimes kids have better ideas, and this is one of them. To prove Pythagorean Theorem following the strict mathematics method, some one needs understand many advanced math concepts first, such as square root. For elementary students who have no knowledge of these concepts, it is hard to understand the mathematic proof of the theorem. However, many schools instroduce Pythagorean Theorem at elementary school level. To avoiding confusing kids, it goes without a proof. It is just introduce as it is. As kids see it again in middle and high school, they are not shown the proof, because they “already know it well”. This leads to the void in students’ math path through high school that it is never be proved. With this method of Pythagorean Theorem proof with LEGO, kids don’t need many advanced knowledge. All they need know is the area of a square, which is the product of the two sides. Since the two sides are the same length for the square shape, it is square of the sides. If kids don’t have the concept of square yet, you can just stop at the concept of the product of the two sides. Pythagorean Theorem Definition To understand why the proof works, let’s first look at the Pythagorean Theorem: a² + b² = c² As show in the equation, Pythagorean Theorem is about the relationship among the 3 sides of a right triangle. If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, the side opposite to the right angle, then the sum of the squares of the lengths of the legs (a² + b²) or (a x a + b x b) is equal to the square of the length of the hypotenuse c² or (c x c). With this, we can see that a² is the area of a square A with all sides length being a, and b² is the area of a square B with the side length as b, and c² is the area of a square C of the side length c. To prove the Pythagorean Theorem, we just need prove the sum of the areas of square A and square B is equal to the area of square C. Now, with LEGO, we can make 3 squares that the length of their sides satisfy a² + b² = c². How do we find these squares? Find Pathagorean Triples Luckily, there are so many research on Pythagorean Theorem, a quick search on Google can give us many of these square triples. To save your time, I have listed out all 16 primitive Pythagorean Triples under 100. Yes, there are just 16 primitive triples under 100. You can download the list, together with a LEGO proof template at the bottom of this post. I will explain about the template shortly. Primitive means the 3 numbers don’t share common factors greater than 1. An example is (3, 4, 5). You may ask, how about (6, 8, 10)? Or (9, 12, 15)? They are also Pythagorean triples. But they are not primitive triples, since they are multiple of (3, 4, 5). Now you know, besides the primitive triples, there are many more Pythagorean triples. For example, based on the 2nd triple on our list (5, 12, 13), you know (10, 24, 26) is also a Pythagorean Triple. Move on to the proof. Prove Pythagorean Theorem with LEGO These are the steps to prove Pythagorean Theorem with LEGO: • Step 1. Print out the Pythagorean Triples list. You can find the download at the bottom of this post. • Step 2. Pick one set of triple, and build 3 squares with LEGO. Each square has a side length of one of the numbers in the triple. If you picked the triple (3, 4, 5), then build 3 squares, one with a side of 3, one with a side of 4, one 5. • Step 3. Pick enough LEGO pieces of 1×1 to cover both the smaller squares • Step 4. Lay out the 3 squares in the way shown as the template shows, having the sides of the squares forming a triangle with one side from each square. Make sure the ends of the sides touch each other. You can find one of many possible templates from the download of the Pythagorean Triples list at the bottom of this post • Step 5. Point it out to kids, the angle formed betweeen the shorter sides of the triangle is a right angle. • Step 6. Move the 1×1 LEGO pieces from the two smaller squares to the biggest square. • What do you find? You see the big square is completely covered by the 1×1 LEGO pieces, and no LEGO 1×1 pieces is left on the two smaller squares. What is the conclusion of this activity? The conclusion is the sum of the areas of the two smaller squares equals the area of the biggest square, ie., a² + b² = c² Now, you just proved the Pythagorean Theorem with LEGO! Looking for more LEGO math activities? Try LEGO Math to Learn COordinate Plane and LEGO Movie Math for Areas and Perimeters. .
# Factors Of 34 – With Division and Prime Factorization ## Factors Of 34 An integer factor of 34 is a number that divides the original number evenly, resulting in a whole number as a quotient. The number 34 has more than two factors since it is a composite number. The factors and multiples of 34 are different from each other. In comparison, factors of a number divide the original number, whereas multiples are divisible by the original. In this case, 34 is both a factor and multiple. Let’s go on to learn how to find the prime factors and pair factors of 34. ## How to Find Factors of 34? Numbers that divide 34 are called factors. You will also get the original number when finding the factors, we must divide the original number by natural numbers until the quotient is 1. Our first step is to divide 34 by the smallest natural number, 1. • 34 ÷ 1 = 34 • 34 ÷ 2 = 17 • 34 ÷ 17 = 2 • 34 ÷ 34 = 1 Accordingly, the factors of 34: 1, 2, 17, and 34. ## What are the factors of -34? Factors of -34: -1, -2, -17, and -34. Multiplying two numbers to get the product -34 gives us the following: -34 x 1 = -34 34 x -1 = -34 -17 x 2 = -34 17 x -2 = -34 Therefore, we have two possible pairs of factors of -34 as (-34 x 1), (34 x -1), (-17 x 2), and (17 x -2). ## Pair Factors of 34 Using the product of the two numbers, we can find the pair factors of 34. Positive Pair Factors • 1 × 34 = 34 • 2 × 17 = 34 Consequently, the positive pair factors are (1, 34) and (2, 17). Negative Pair Factors When you multiply 34 by its negative pair factors, you will also get the original number. Thus, • -1 × -34 = 34 • -2 × -17 = 34 As a result, the negative pair factors are (-1, -34) and (-2, -17). ## Prime Factorization of 34 A prime factor is a number that has only two factors. Typical examples include 2, 3, 5, 7, 11, etc. As we are looking for prime factors for 34, we need to determine whether that prime number is completely divisible by 34. So, we’ll start by dividing 34 by the smallest prime factor, 2. Step 1: Divide 34 by the smallest prime number. 34/2 = 17 Step 2: Since 17 is a prime number, it can only be divided by 17. 17/17 = 1 Step 3: No more division can be made. Therefore, 2 and 7 will be considered the prime factors of 34. Prime factorization of 34 = 2 x 17 ## Solved Examples Emma is giving out chocolates at her birthday party in her class. How many students are present if there are 34 chocolates and each gets two? Number of chocolates = 34 Number of chocolates each student gets = 2 Therefore, number of students = 34/2 = 17 Calculate the sum of all the factors of 34. Factors of 34 are 1, 2, 17, and 34. Sum = 1+2+17+34 = 54 As a result, 54 is the required sum. What are 33 and 34’s common factors? They are both composite numbers. Accordingly, the factors are 33 → 1, 3, 11, and 33 34 → 1, 2, 17, and 34 As a result, 33 and 34 have the same common factor of 1. ## FAQs ### 1. There are how many factors of 34? 34 has four factors. These are 1, 2, 17, and 34. ### 2. Is the number 34 a prime or a composite number? Because it has more than two factors, 34 is not a prime number but a composite number. ### 3. What are the two factors that make 34? Multiplying 2 by 17 gives us 34. As a result, our original number comprises two such factors, 2 and 17. ### 4. Which is the largest prime factor of 34? 34 has a prime factor of 17 as its largest prime. ### 5. What are the multiples of 34? These are the first ten multiples of 34: 34, 68, 102, 136, 170, 204, 238, 272, 306 and 340 ### 6. How many prime factors does 34 have? 34 has a prime factorization of 2 x 17. Hence, the prime factors of 34 are 2 and 17.
Anda di halaman 1dari 9 # Concerning “Solving Mathematical Problems: ## A Personal Perspective” by Terence Tao Tom Verhoeff June 2007 Introduction Terence Tao, 2006 Fields medal winner, wrote a delightful book [6] on prob- lem solving in (elementary) mathematics. It includes an excellent selection of 26 problems with fully documented solutions, That is, Terence not only provides an answer and a proof, but explains in detail the reasoning that led him to the answer. The book also includes a set of exercises (without solutions) to practice the illustrated techniques. ## Problems for which full solutions are presented The following problems —mostly taken from well-known sources— are treated in depth. ## Problem 1.1 (p. 1) A triangle has its lengths in an arithmetic progression, with difference d. The area of the triangle is t. Find the lengths and angles of the triangle. Problem 2.1 (p. 11) Show that among any 18 consecutive 3-digit numbers there is at least one which is divisible by the sum of its digits. [7, p. 7] Problem 2.2 (p. 14) Is there a power of 2 such that its digits could be re- arranged and made into another power of 2? (No zeroes are allowed in the leading digit: e.g. 0032 is not allowed.) [7, p. 37] Problem 2.3 (p. 19) Find all integers n such that the equation 1/a + 1/b = n/(a + b) is satisfied for some non-zero integer values of a and b (with a + b 6= 0). [2, p. 15] 1 Problem 2.4 (p. 20) Find all solutions of 2n + 7 = x2 where n and x are integers. [7, p. 7] Problem 2.5 (p. 23) Prove that for any nonnegative integer n, the number 1n + 2n + 3n + 4n is divisible by 5 if and only if n is not divisible by 4. [4, p. 74] Problem 2.6 (p. 24) (*) Let k, n be natural numbers with k odd. Prove that the sum 1k + 2k + · · · + nk is divisible by 1 + 2 + · · · + n. [5, p. 14] Problem 2.7 (p. 27) Let p be a prime number greater than 3. Show that the numerator of the (reduced) fraction 1/1 + 1/2 + 1/3 + · · · + 1/(p?1) is divisible by p2 . For example, when p is 5, the fraction is 1/1 + 1/2 + 1/3 + 1/4 = 25/12, and the numerator is obviously divisible by 52 . [5, p. 17] Problem 3.1 (p. 36) () Suppose f is a function mapping the positive inte- gers to the positive integers, such that f satisfies f (n+1) > f (f (n)) for all positive integers n. Show that f (n) = n for all positive integers n. [3, p. 19] Problem 3.2 (p. 38) Suppose f is a function on the positive integers which takes integer values with the following properties: (a) f (2) = 2 (b) f (mn) = f (m)f (n) for all positive integers m and n (c) f (m) > f (n) if m > n. Find f (1983) (with reasons, of course). [2, p. 7] Problem 3.3 (p. 43) Let a, b, c be real numbers such that 1 1 1 1 + + = a b c a+b+c with all denominators non-zero. Prove that 1 1 1 1 5 + 5+ = a b c5 (a + b + c)5 [2, p. 13] Problem 3.4 (p. 45) (*) Prove that any polynomial of the form f (x) = (x − a0 )2 (x − a1 )2 · · · (x − an )2 + 1 where a0 , a1 , . . . , an are all integers, cannot be factorized into two non-trivial polynomials, each with integer coefficients. 2 Problem 4.1 (p. 50) ABC is a triangle that is inscribed in a circle. The angle bisectors of A, B, C meet the circle at D, E, F , respectively. Show that AD is perpendicular to EF . [2, p. 12] Problem 4.2 (p. 52) In triangle BAC the bisector of the angle at B meets AC at D; the angle bisector of C meets AB at E. These bisectors meet at O. Suppose that \|OD\| = \|OE\|. Prove that either 6 BAC = 60◦ or that BAC is isosceles (or both). [7, p. 8, Q1] Problem 4.3 (p. 55) () Let ABF E be a rectangle and D be the intersec- tion of the diagonals AF and BE. A straight line through E meets the extended line AB at G and the extended line F B at C so that \|DC\| = \|DG\|. Show that \|AB\|/\|F C\| = \|F C\|/\|GA\| = \|GA\|/\|AE\|. [2, p. 13] Problem 4.4 (p. 58) Given three parallel liines, construct (with straight- edge and compass) an equilateral triangle with each parallel line con- taining one of the vertices of the triangle. Problem 4.5 (p. 62) A square is divided into five rectangles as shown be- low. The four outer rectangles R1 , R2 , R3 , R4 all have the same area. Prove that the inner rectangle R0 is a square. [7, p. 10, Q4] R2 R1 R0 R3 R4 Problem 4.6 (p. 66) Let ABCD be a square, and let k be the circle with centre B passing through A, and let lbe the semicircle inside the square with diameter AB. Let E be a point on l and let the extension of B meet circle k at F . Prove that 6 DAF = 6 EAF . [1, Q1] 3 Problem 5.1 (p. 69) A regular polygon with n vertices is inscribed in a circle of radius 1. Let L be the set of all possible distinct lengths of all line segments joining the vertices of the polygon. What is the sum of the squares of the elements of L? [2, p. 14] Problem 5.2 (p. 74) () A rectangle is partitioned into several smaller rect- angles. Each of the smaller rectangles has at least one side of integer length. Prove that the big rectangle has at least one side of integer length. ## Problem 5.3 (p. 77) On a plane we have a finite collection of points, no three of which are collinear. Some points are joined to others by line segments, but each point has at most on line segment attached to it. Now we perform the following procedure: We take two intersecting line segments, say AB and CD, and remove them an replace them with AC and BD. Is it possible to perform this procedure indefinitely? [7, p. 8] Problem 5.4 (p. 79) In the centre of a square swimming pool is a boy, while his teacher (who cannot swim) is at one corner of the pool. The teacher can run three times faster than the boy can swim, but the boy can run faster than the teacher can. Can the boy escape from the teacher? (Assume both persons are infinitely manoeuvrable.) [7, p. 34, Q2] Problem 6.1 (p. 83) Suppose on a certain island there are 13 grey, 15 brown, and 17 crimson chameleons. If two chameleons of different colour meet, they both change to the third colour (e.g. a brown and crimson pair would both change to grey). This is the only time they change colour. Is it possible for all chameleons to eventually be the same colour? [7, p. 25, Q5] Problem 6.2 (p. 86) () Alice, Betty, and Carol took the same series of examinations. For each examination there was one mark of x, one mark of y, and one mark of z, where x, y, z are distinct positive integers. After all the examinations, Alice had a total score of 20, Betty a total score of 10, and Carol a total score of 9. If Betty was placed first in Algebra, who was placed second in Geometry? Problem 6.3 (p. 90) Two people play a game with a bar of chocolate mad of 60 pieces, in a 6 × 10 rectangle. The first person breaks off a part of the chocolate bar along the grooves dividing the pieces, and discards (eats) the part he broke off. The the second breaks off a part of the remaining part and discards her part. The game continues until one piece is left. The winner is the one who leaves the other with the single 4 piece (i.e. is the last to move). Which person has a perfect winning strategy? [7, p. 16, Q3] Problem 6.4 (p. 95) Two brothers sold a herd of sheep. Each sheep sold for as many rubles as the number of sheep originally in the herd. The money was then divided in the following manner. First the older borther took 10 rubles, the the younger brother took 10 rubles, then the older brother took another 10 rubbles, and so on. At the end of the division the younger brother, whose turn it was, found that there were fewer than 10 roubles left, so he took what remained. To make the division fair, the older brother gave the younger his penknife, which was worth an integer number of roubles. How much was the penknife worth? [5, p. 9] ## Errata and remarks p. 2, l. 18 “Problem 1.1 question”: delete “question” ## p. 5, l. 7 “gung-ho”: ‘Gung-ho is a phrase borrowed from Chinese language, frequently used in Chinese as an adjective meaning enthusiastic.’ (from Wikipedia) ## p. 9, l. –9 “which are exactly the same”: change “are” to “have” (?) p. 16, l. –5 “the digit sum of 217 is a mere 14”: change “217” to “217 ## p. 25, l. 9 “n”: change to “n” p. 27, Problem 2.7 The given p > 3 is used rather late. On p. 30, l. –9, “p is an odd prime” appears, and on p. 33, l. –14, “p is a prime greater than 3” appears. p. 31, l. 9 “we are now reduced”: change to “we have now reduced it to” ## p. 33, l. 5 “Thus”: why is this inference valid? 5 p. 34, l. 3 “trick”: I don’t like that terminology ## p. 35, l. 4 (of quote) “that”: change to “than” p. 39, l. 17 “f (1) = 1”: follows immediately from (b) and (a): f (2) = f (2 · 1) = f (2)f (1) and f (2) = 2 6= 0. ## p. 42, l. –2 “one has”: change to “these have” p. 44, l. –14 and –3 “5ab”: I don’t understand this; what is the signifi- cance/role/purpose of 5ab? ## p. 44 From the solution it follows that Problem 3.3 can be generalized, by replacing the exponent 5 by any odd n > 0. The relevance of 5 was already doubtful from the beginning. Note that we have a + b = 0 implies 1/a + 1/b = 0, and hence also 1/an + 1/bn = 0 for odd n > 0. ## p. 46, l. –7 Why is it not possible that p(x) = 1 for all x? p. 47, Exercise 3.7 The problem statement is wrong. One of many coun- terexamples: (x − 0)(x + 2) + 1 = (x + 1)(x + 1). The hint suggest that if f (x) = p(x)q(x) then p(x) ≡ q(x), which is not what was asked. ## p. 49, Figure It might have helped the unexperienced reader to label 6 P AO = 6 OP A = α and 6 OP B = 6 P BO = β, and stating that α+(α+β)+β = ## p. 53, Figure 6 ADO labeled (γ + β)/2: change to γ + β/2 6 AEO labeled (β + γ)/2: change to β + γ/2 p. 56, l. 2 below (15) “the third ratio”: note that given \|DC\| = \|DG\| has not yet been used. ## p. 57, Proof “P QT is similar to P T R” may not be obvious. This follows from ‘same chords subtend same angles’ (a picture would help) ## p. 58, l. –1 “as best we can”: insert “as” after “best” (?) 6 p. 61, l. 1 “so long as they are not at 60◦ angles”: this is not correct; in that case there is only one solution, not two. ## p. 63, l. 3 “square to be equal”: change to “rectangle to be a square” p. 64, l. 4 “a rectangle with a fixed area”: this concerns the outer rectangles (?) ## p. 77, l. –8 “assertion is plausible”: it is not clear what is meant; there is no assertion in the problem statement, only a question; presumably, the assertion that ‘the procedure cannot be performed indefinitely’ is meant p. 79, l. 4 “triangle inequality”: the uninitated reader may need some help −→ applying it (translate e.g. AB over vector AC to yield A0 B 0 with A0 = C; now consider 4DB 0 C; a picture would help) ## p. 83, Problem 6.1 Concerning solution: observe that pairwise differences in the number of chameleons modulo 3 are invariant. ## p. 92, l. 22 “that it is”: delete “is” 7 p. 95, l. 2, 4, 5, 7, 9 “ruble”, “rouble”: inconsistent spelling (both are correct) p. 95, Problem 6.4 Not clear whether sheep are sold one-by-one or all at once; From the solution it follows that ‘all at once’ is intended. ## p. 95, l. 22 “windfall”: ‘unexpected or sudden gain or advantage’ (accord- ing to Webster) p. 96, l. –1 “is restricted to between 1 and 9”: insert “lie” after “to” ## References I cannot confirm the reference to Taylor 1989. References [1] AMOC (Australian Mathematical Olympiad Committee) Correspondence Programme, 1986–1987, Set One. ## [2] Australian Mathematics Competition. Mathematical Olympiads: The 1987 Australian Scene. Canberra College of Advanced Education, 1987 ## [3] Samual L. Greitzer. International Mathematical Olympiads, 1955-1977. Mathematical Association of America, 1978. ## [4] József Kürshák; G. Hajós, G. Neukomm, L. Surányi (Eds.). Hungarian Problem Book I, based on the Eötvös competitions 1894–1905. Mathemat- ical Association of America, 1963/1967. ## [5] D. O. Shklarsky, N. N. Chentzov, I. M. Yaglom, (I. Sussman, ed.). The USSR Olympiad Problem Book: Selected Problems and Theorems in Ele- mentary Mathematics, Freeman, 1962. ## [6] Terence Tao. Solving Mathematical Problems: A Personal Perspective. Oxford University Press, 2006. ## [7] Peter J. Taylor. International Mathematics Tournament of the Towns 1984–1989: Questions and Solutions. Australian Mathematics Trust Pub- lishing, 1989. (A 2nd corrected edition is available.) ## [8] Stan Wagon. “Fourteen Proofs of a Result about Tiling a Rectangle”, American Mathematical Monthly, 94(14):601–617 (Aug-Sep 1987). 8 [9] I. M. Yaglom. Geometric Transformations I. Mathematical Association of America, 1962.
# Find a Missing Side Using the Tan Ratio In this worksheet, students will use the Tan ratio to find the missing side of a triangle. Key stage: KS 4 Year: GCSE GCSE Subjects: Maths GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR, Curriculum topic: Geometry and Measures, Mensuration Curriculum subtopic: Properties and Constructions Mensuration and Calculation Triangle Mensuration Difficulty level: #### Worksheet Overview Great, this is what we have been waiting for. Time to top up the tan! Okay so maybe not that sort of "tan". We can use the tan ratio to help us find a missing side length in a right angled triangle. We are going to look at Tan here and what the "TOA" means. T is for Tan (which will be given as an angle) O is the opposite side The line in the middle means divide. To use this triangle we cover up what side it is we want to find and we are left with a formula to follow. If we want to find the opposite side, cover up the O. The formula we are left with is Tan(angle) x Adjacent. O = T x A If we want to find the adjacent, cover up the A. The formula we are left with is opposite divided by Tan(angle) A = O ÷ T Let's look at some examples. IMPORTANT NOTE: Make sure your calculator is set to degrees otherwise this formula will not work correctly. (You should see a D at the top of the calculator screen if it is in the correct mode). Example 1 1. Label the lengths of the triangle: A, O, and H 2. Work out the side you want to find. In this case, x is the opposite length, so we want O 3. The formula triangle shows that O = A x T 4. Substitute the angle and the length into the formula. This gives O = 7 x tan47 5. In your calculator type in 7 x tan 47 Example 2 1. Label the lengths of the triangle: A, O, and H 2. Work out the side you want to find. In this case, x is the adjacent length, so we want A 3. The formula triangle shows that A = O ÷ T 4. Substitute the angle and the length into the formula. This gives A = 15 ÷ tan 32 5. In your calculator type in 15 ÷ tan 32
8.01SC \| Fall 2016 \| Undergraduate # Classical Mechanics Week 1: Kinematics ## PS.1.5 Worked Example: Pedestrian and Bike at Intersection « Previous \| Next » You are in a car standing by a traffic light and at time $$\displaystyle t=0$$ the light turns green. You start to accelerate during the first $$\displaystyle t_1$$ seconds so that the acceleration of your car is given by: $$\displaystyle a_1(t) = \left\{ \begin{array}{ll} b_1 & \quad 0 \leq t \leq t_1 \\ 0 & \quad t_1 < t \leq t_2 \end{array} \right.$$ where $$\displaystyle b_1$$ is a positive constant. At the instant the light turns green a cyclist passes through the intersection moving with a speed $$\displaystyle v_0$$ in the same direction as your car is moving. At that instant, the cyclist starts to brake with a constant acceleration of magnitude $$\displaystyle b_2$$. At time $$\displaystyle t = t_2$$ the cyclist stops at the same location where you are. The goal of the problem is to calculate the value of $$\displaystyle b_2$$ in terms of the given variables, $$\displaystyle b_1$$, $$\displaystyle v_0$$, $$\displaystyle t_1$$ and $$\displaystyle t_2$$. First: Describe the motion of the car. Given the car’s acceleration $$\displaystyle a_1(t)$$, find its velocity and its position as a function of time: (Part b) Calculate $$\displaystyle x_1(t)$$ , the car’s position as a funcion of time. Express your answer in terms of $$\displaystyle t$$, $$\displaystyle t_1$$, and $$\displaystyle b_1$$ as needed. Second: Describe the motion of the bicycle. Given the bicycle’s acceleration $$\displaystyle a_2(t)$$, find its velocity and its position: (Part c) Calculate $$\displaystyle v_2(t)$$ , the bicycle’s velocity as a funcion of time. Express your answer in terms of $$\displaystyle t$$, $$\displaystyle b_2$$, $$\displaystyle v_0$$ as needed. (Part d) Calculate $$\displaystyle x_2(t)$$ , the bicycle’s position as a funcion of time. Express your answer in terms of $$\displaystyle t$$,$$\displaystyle b_2$$, and $$\displaystyle v_0$$ as needed. Third: Find the value of $$\displaystyle b_2$$. We know that the bicycle stops at $$\displaystyle t=t_2$$, this condition is expressed as: $$\displaystyle v_2(t_2) = 0$$ (eq. 1) We also know that the bicycle and the car are at the same location when the bicycle stops. This condition implies: $$\displaystyle x_2(t_2) = x_1(t_2)$$ (eq. 2) Use (eq.1) and (eq. 2) to obtain the value of $$\displaystyle b_2$$. Express your answer in terms of $$\displaystyle t_1$$, $$\displaystyle b_1$$, and $$\displaystyle v_0$$. Do not use $$\displaystyle t_2$$ in your answer. « Previous \| Next » Fall 2016 Lecture Videos Problem Sets Online Textbook
# RCTEXSCC - Editorial Author: Temirlan Baibolov # PREREQUISITES: Probability, Euler’s Formula # PROBLEM: Given a M \times N matrix of random integers between 1 and K, find the expected number of strongly connected components if we have a directed edges between every two adjacent cells iff the value in the first cell is greater than or equal to the value in the second cell. # Quick Explanation We notice that every strongly connected component will have all of its vertices/cells equal to the same value. Then we can just solve the problem of finding the number of connected components if the graph was undirected and we had edges iff two adjacent cells had the same value. This has a simple solution because we can use Euler’s formula for planar graphs. # Explanation We’ll first show why our problem is equivalent to counting the number of connected components in the undirected graph formed by edges between cells with the same value. Let’s prove this by contradiction: Assume there are two adjacent cells (x_1, y_1) and (x_2, y_2) that are in the same strongly connected component and A_{x_1, y_1} < A_{x_2, y_2}. Then by the definition of strong connectivity, there should exist a path from (x_1, y_1) to (x_2, y_2). However, all edges “decrease” the value (i.e. they go from a larger value to a lower one). This means that all cells reachable from (x_1, y_1) will have a value of at most A_{x_1, y_1}. Therefore, (x_2, y_2) is not reachable from (x_1, y_1), because A_{x_1, y_1} < A_{x_2, y_2}. This contradicts with our assumption of the cells being in the same strongly connected component\$. From the above proof, it follows that two cells can be in the same strongly connected component iff their values are the same (then we have a bidirectional edge). This by itself is equivalent to counting the number of connected components in an undirected graph. To solve this new problem (the undirected version), we first notice that the graph is planar by construction. This is useful, because of one of the most popular theorems for planar graphs - Euler’s formula. It states that v - e + f = 2, where v is the number of vertices in the graph, e is the number of edges and f is the number of faces. This formula works for any connected planar graph, but it can be easily generalised for graphs with more than one connected component (which is our case). Let k be the number of connected components. Then the generalisation states that v - e + f = 1 + k. Let’s get back to our problem. We are interested in \mathop{\mathbb{E}}[k] = \mathop{\mathbb{E}}[v - e + f - 1]. Using linearity of expectation we get \mathop{\mathbb{E}}[k] = \mathop{\mathbb{E}}[v] - \mathop{\mathbb{E}}[e] + \mathop{\mathbb{E}}[f] - \mathop{\mathbb{E}}[1]. Now we’re only left with calculating the different terms of the previous equation: • \mathop{\mathbb{E}}[v] is the expected number of vertices. This is clearly always constants, so \mathop{\mathbb{E}}[v] = N * M. • \mathop{\mathbb{E}}[e] is the expected number of edges. We’ll use the contribution idea (with linearity of expectation) to calculate it. Consider an edge between two different cells. It’ll exist iff the values in the two cells are the same. Therefore, the probability of it existing is \frac{1}{K}, as we can think of this as if we’ve fixed the first value and then finding the probability of the second one being the same. Now we can simply use linearity of expectation to get that \mathop{\mathbb{E}}[e] = \frac{N * (M - 1) + (N - 1) * M}{K}, because we have N * (M - 1) + (N - 1) * M edges and the contribution of each one is equal to \frac{1}{K}. • \mathop{\mathbb{E}}[f] is the expected number of faces. The main observation here is that we can only have a face that looks like a 2 \times 2 square with the same value. Also we shouldn’t forget that we always have 1 additional face on the “outside”. Clearly, that M = 1, we have \mathop{\mathbb{E}}[f] = 1, because the only face is the “outside” one. Now let’s consider the M = 2 case. Then we can again use linearity of expectation. The number of 2 \times 2 squares is N - 1 and the probability of each of them forming a face is \frac{1}{K^3} (using the same idea of fixing one of the values). Then \mathop{\mathbb{E}}[f] = \frac{N - 1}{K^3} + 1. • \mathop{\mathbb{E}}[1] = 1. This one is quite obvious. Now we are pretty much done. We just calculate the 4 above-mentioned values in constant* time and then fine the required expectation for the number of connected components. (we have to find the inverse of K, so the complexity actually logarithmic). The given constraints might look a bit weird, because they allow linear solutions. This is because initially the author’s solution was using FFT to calculate the number of configurations with 1, 2, ... , N M connected components. This didn’t look very natural, so the problem was converted to finding the expectation for the number of connected components (where the intended solution was again FFT). Then while testing the problem, I solved it with the Euler’s formula, but we decided to still let the FFT/some other linear solutions pass and so we didn’t increase the constraints to 10^9. Also, there is a way to solve the M = 3 case (which wasn’t included in the problem) again using the Euler’s formula solution, as then we only need to additionally count the 3 \times L faces which always look like a border with a “hole” inside. An open question would be whether it’s possible to count the number of components for larger M's. If you have some ideas, I’m open to discuss! Tester's solution #include <bits/stdc++.h> #define endl '\n' #define SZ(x) ((int)x.size()) #define ALL(V) V.begin(), V.end() #define L_B lower_bound #define U_B upper_bound #define pb push_back using namespace std; template<class T, class T1> int chkmin(T &x, const T1 &y) { return x > y ? x = y, 1 : 0; } template<class T, class T1> int chkmax(T &x, const T1 &y) { return x < y ? x = y, 1 : 0; } const int MAXN = (1 << 20); const int mod = 998244353; int pw(int x, int p) { int r = 1; while(p) { if(p & 1) r = r * 1ll * x % mod; x = x * 1ll * x % mod; p >>= 1; } return r; } int n, m, k; cin >> m >> n >> k; } void solve_1() { int ik_c = (1 - pw(k, mod - 2) + mod) % mod; int ans = (1ll + ik_c * 1ll * (n - 1)) % mod; cout << ans << endl; } void fix(int &x) { if(x >= mod) x -= mod; if(x < 0) x += mod; } void solve_2() { int ik = pw(k, mod - 2); // We'll use Euler's theorem: // v - e + f = 1 + k // E[k] = 2 * n - 1 - E[e] + E[f] // // E[e] = ik * (3 * n - 1) int answer = 2 * n - 1; answer -= ik * 1ll * (3ll * n - 2) % mod; fix(answer); int Ef = ik * 1ll * ik % mod; Ef = Ef * 1ll * ik % mod; Ef = Ef * 1ll * (n - 1) % mod; Ef += 1; // main face } void solve() { if(m == 1) solve_1(); else solve_2(); } int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); solve(); return 0; } # VIDEO EDITORIAL: 8 Likes \frac{p}{q}=\frac{k+(n-1)(k-1)}{k} for m=1 \frac{p}{q}=\frac{k^2(2k-1)+(n-1)(2k^3-3k^2+1)}{k^3} for m=2 5 Likes I wrote an NK^2 dp getting inspiration from @carre 's NRWP subtask 2 where K^2 states for each i can be easily converted to only 2 states giving us O(N) solution- implementation. 4 Likes I solved the first subtask with an MN2 DP solution and then ran the code for all N ranging from 1 to ~200 and printed the outputs locally. I observed a pattern: ans(N) for N>3 can be expressed as ans(2) + (ans(3)-ans(2))(N-2) So for the bigger subtasks with M=2, I simply calculated answer for N=3 and N=2 and then used this formula for the given N. Not the best way to solve the problem because it allowed me to get AC without really understanding the underlying theory about Euler’s formula (link to my implementation). But learnt something from reading the editorial. Nice question and neat editorial! 3 Likes Very simple and innovative. During contest I tried to come up with NxM dp without any luck. Thank you for editorial! Can you please explain it a little bit more ? I’d love to but in a few hours …I’ll write in detail. What is linearity of expectation? Okay, Will be waiting for it It’s great that I could have inspired you, sadly I didn’t inspire myself and wrote a really ugly solution 3 Likes I got some hint from this paper. I managed to derive a formula using the methods stated in it. Find the increase in the expected number of connected components after adding one column. Cool Problem You can also solve this problem using elementary probability theory. The M = 1 case is pretty easy: Let random variable I_i = 1 if A[i] \neq A[i+1] and 0 otherwise. Filling up the one dimensional array A from 1 to N, you can notice that every time a new component is created if and only if A[i] \neq A[i+1]. So the number of components is 1 + \sum_{i=1}^{N-1} I_i. Since the variables inside the sum are independent, the sum itself is a binomial distribution with parameters N-1 and 1 - 1/K. Formally, \sum_{i=1}^{N-1} I_i \sim B(N-1, 1 - 1/K). See Binomial distribution on Wikipedia. We know that the expectation value of the binomial distribution is the product of the two parameters so the expected number of components are: \displaystyle 1 + (N-1) \cdot \frac{K-1}{K}. Notice that it is in the form of a constant plus (N-1) times a value that only depends on K: c + (N-1) \cdot f(K). Update: The discussions below this point are only valid for M=2, which has been pointed out in the comment below. I apologize for my mistake. This is also true for higher values of M although the calculation needs a bit more work than using Euler’s theorem. The observation is that, while filling the matrix column by column, the increase in the number of components only depends on the values of the last column and the newly added column. That is, it only depends on M and K. For a fixed M, the dependence is only on K. So for M=2 we can determine f(K) by studying the configurations of 4 random variables. The constant c in the above expression is the expectation for one column which we have already calculated above. So, I think it is possible to solve this for higher values of M but it will get complicated. 2 Likes I don’t think this can be generalised for larger M's. This is mainly because of the “The observation is that, while filling the matrix column by column”. That’s indeed true for the M = 2 case, but for larger values we can actually have something like this: 1111 0001 0111 Consider the first 2 columns. We have no way of figuring out that the cells (1, 1) and (3, 2) will be in the same connected component based only on two adjacent rows. 2 Likes Yes, you are right. I incorrectly generalized the M=2 case to higher. So, it is a good brain teaser to think about the M \geq 3 case. Can anyone help me understanding the time complexity accepted for various subtasks . How do we come to know what should be the time complexity to get a particular subtask accepted. Can anyone explain with current question as an example , it would be great help for a newbie like me. Hi @ashu_3916 take a look at @betlista’s post in this thread: how many approx loops are allowed in 1 sec time limit. With that in mind, lets take a look at what time complexities would be accepted for various subtasks of this problem: Subtask #1: since N and K are very small (both less than 5), solutions with really bad time complexity like exponential will also be accepted. Subtask #2: Here N can be as large as 105 and K as large as 108. So O(N) would work, O(K) probably won’t work and O(NK) or anything bigger will definitely not work. Subtask #3: Here N and K can be at most 500, so complexities like O(NK2), O(N2K) or lower will work because these will amount to ~107 operations but higher than that like O(NK3) won’t work. Subtask #4: Constraints on N and K are the same as the second subtask but with M=2. So again, O(N) would work but O(NK) or bigger will not work. So with the given constraints, the worst time complexity that could give you an AC for all subtasks and a 100 score for this problem is linear in N i.e. O(N). Hope that helps! 1 Like Here it is! 4 Likes Thanks 1 Like Thanks. It’s a really well written editorial. 1 Like I think I also a wrote a dp solution like you but looks really ugly as I manually added all possible cases. Solution: 41189552 \| CodeChef
# Prealgebra for Two-Year Colleges/Appendix (procedures)/Reducing fractions Equivalent fractions represent the same number, although they are written differently. For example, ${\displaystyle {\frac {2}{4}}}$ and ${\displaystyle {\frac {1}{2}}}$ represent the same amount. The latter fraction has a smaller denominator, so we say the fraction has been reduced. You can reduce a fraction by dividing the numerator and denominator by the same number. For example, ${\displaystyle {\frac {75}{100}}={\frac {75{\color {Red}\div 5}}{100{\color {Red}\div 5}}}={\frac {15}{20}}}$ and ${\displaystyle {\frac {15}{20}}={\frac {15{\color {Red}\div 5}}{20{\color {Red}\div 5}}}={\frac {3}{4}}}$. When we ask you to simplify a fraction or reduce a fraction such as 75/100, we want you to find the equivalent fraction with the smallest possible denominator. In this case, the fully reduced form is 3/4. The only number that we could divided into both 3 and 4 would be 1, but that would not reduce the fraction any further. ${\displaystyle {\frac {3}{4}}={\frac {3{\color {Red}\div 1}}{4{\color {Red}\div 1}}}={\frac {3}{4}}}$. You are also allowed to multiply the numerator and denominator by the same number. We usually call this finding equivalent fractions.
## The Second Derivative and the Shape of a Function f(x) Read this section to learn how the second derivative is used to determine the shape of functions. Work through practice problems 1-9. ### f'' and Extreme Values The concavity of a function can also help us determine whether a critical point is a maximum or minimum or neither. For example, if a point is at the bottom of a concave up function (Fig. 7), then the point is a minimum. Fig. 7 The Second Derivative Test for Extremes: (a) If $\mathrm{f}^{\prime}(\mathrm{c})=0$ and $\mathrm{f}^{\prime \prime}(\mathrm{c}) < 0$ then $\mathrm{f}$ is concave down and has a local maximum at $\mathrm{x}=\mathrm{c}$. (b) If $\mathrm{f}^{\prime}(\mathrm{c})=0$ and $\mathrm{f}^{\prime \prime}(\mathrm{c}) > 0$ then $\mathrm{f}$ is concave up and has a local minimum at $\mathrm{x}=\mathrm{c}$. (c) If $f^{\prime}(c)=0$ and $f^{\prime \prime}(c)=0$ then $f$ may have a local maximum, a minimum or neither at $x=c$. Proof: (a) Assume that $\mathrm{f}^{\prime}(\mathrm{c})=0$. If $\mathrm{f}^{\prime \prime}(\mathrm{c}) < 0$ then $\mathrm{f}$ is concave down at $\mathrm{x}=\mathrm{c}$ so the graph of $\mathrm{f}$ will be below the tangent line $L(x)$ for values of $x$ near $c$. The tangent line, however, is given by $L(x)=f(c)+f^{\prime}(c)(x-c)=f(c)+0(x-c)=f(c)$, so if $x$ is close to $c$ then $f(x) < L(x)=f(c)$ and $f$ has a local maximum at $\mathrm{x}=\mathrm{c}$. The proof of (b) for a local minimum of $\mathrm{f}$ is similar. (c) If $\mathrm{f}^{\prime}(\mathrm{c})=0$ and $\mathrm{f}^{\prime \prime}(\mathrm{c})=0$, then we cannot immediately conclude anything about local maximums or minimums of $f$ at $x=c$. The functions $f(x)=x^{4}, g(x)=-x^{4}$, and $h(x)=x^{3}$ all have their first and second derivatives equal to zero at $x=0$, but $f$ has a local minimum at $0, g$ has a local maximum at $0$, and $\mathrm{h}$ has neither a local maximum nor a local minimum at $\mathrm{x}=0$. The Second Derivative Test for Extremes is very useful when $\mathrm{f}^{\prime \prime}$ is easy to calculate and evaluate. Sometimes, however, the First Derivative Test or simply a graph of the function is an easier way to determine if we have a local maximum or a local minimum – it depends on the function and on which tools you have available to help you. Practice 2: $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-15 \mathrm{x}^{2}+24 \mathrm{x}-7$ has critical numbers $\mathrm{x}=1$ and $4$. Use the Second Derivative Test for Extremes to determine whether $\mathrm{f}(1)$ and $\mathrm{f}(4)$ are maximums or minimums or neither.
# Square root of an integer ## Introduction The square root is not only one of the central operations in mathematics (used almost as often as addition, multiplication, or division), but it is also at the core of countless everyday gadgets and cool technology we use daily, like the radio and GPS systems. The square root of a number $$x$$, denoted with the $$\sqrt{x}$$ symbol, is formally defined to be a number $$y$$ such that $$y^2 = y\times y=x$$. For example: $$\sqrt{4} = 2$$ and $$\sqrt{1253} \approx 35.3977$$. The square root is defined for every positive real number but in this lesson, we will derive an algorithm for calculating the square root for integers. As for almost every coding interview problem, there are several possible solutions and approaches we can take to tackle this problem. In this lesson, we will learn how to write a simple and yet sub-optimal solution that runs in $$O(\sqrt{n})$$ time, as well as a much faster and elegant logarithmic time solution. ## Problem statement Write a function that calculates the integral part of the square root of an integer $$n$$ i.e. $$\lfloor \sqrt{n}\rfloor$$. You cannot use any library functions. Given $$n=9$$ the function returns $$3$$: $$\lceil\sqrt{9 \rceil}=3$$ Given $$n=11$$ the function returns $$3$$: $$\lceil\sqrt{11 \rceil}\approx\lceil3.316624 \rceil=3$$ Given $$n=18$$ the function returns $$4$$: $$\lceil\sqrt{11 \rceil}\approx\lceil4.242640 \rceil=4$$ ## Clarification Questions What is the maximum value the parameter $$n$$ can take? The greatest input is guaranteed to be smaller than $$2^{32}$$. Is $$n$$ guaranteed to be always positive? Yes, there is no need to check for invalid input. ## Discussion A brute-force solution is quickly derivable from the definition of square root given above ($$\sqrt{x} = y$$ where $$y^2 = x$$.) and the interviewer is very likely expecting to see it mentioned or appearing on the whiteboard within the first few minutes of the interview. ### Brute-Force We know that if $$y = \sqrt{x}$$ then $$y^2 = x$$. Moreover, $$y$$ is an integer only when $$x$$ is a perfect square[^1]. If $$x$$ is not a perfect square, then $$y$$ is a real number and the following holds true: $$\lfloor{y}^2 \rfloor \leq x$$ and $$\lceil{y} \rceil^2 > x$$. For instance, $$\sqrt{5} \approx 2.2360$$ and $$2^2=4 \leq 5$$ and $$3^2=9 > 5$$. We can use this last property to blindly loop through all the integers $$k=0,1,2,\ldots$$ until the following is true: $$k^2\leq n$$ and $$(k+1)^2 > n$$. A solution is guaranteed to be found because eventually, $$k$$ will be equal to $$\lfloor y \rfloor$$. Moreover, it is clear that no more than $$\sqrt{n}$$ numbers will be tested, which proves that the time complexity of this approach is $$O(\sqrt{n})$$. Listing [list:square_root_brute_force] shows a C++ implementation of this idea. Listing 1: $O(\sqrt{n})$ solution to the problem of finding the square root of an integer. int square_root_brute_force(const int n) { long i = 0; while ((i * i) <= n) i++; // i at this point is the smallest element s.t. i*i > n return i - 1; } It is worth noticing that the variable $$i$$ has a type that is larger in size than an . This is necessary in order to prevent overflows during the calculation of $$i^2$$ (see the highlighted line). One of the constraints of the problem is that the largest input can be $$n=2^{32}-1$$; The square of that number does not fit in a $$4$$ bytes . ### Logarithmic Solution Binary search can be effectively used to solve this problem: in order to show that, we are going to look at the problem from a slightly different angle. Let $F(k)=\begin{cases} 0 & \text{: } k^2 \leq n \\ 1 & \text{: } k^2 > n \end{cases} \label{eq:square_root_piecewice}$ be a piece-wise function that partition the search space $$[0\ldots n]$$ into two parts, as shown in Table 2.1: 1. the numbers less or equal than $$\sqrt{n}$$ 2. the numbers strictly greater or equal than $$\sqrt{n}$$ Partition of the search space according to the function in Eq. [eq:square_root_piecewice] $$0$$ $$1$$ $$2$$ $$\lfloor \sqrt{n \rfloor}$$ $$\lfloor \sqrt{n \rfloor}+1$$ $$n$$ $$0$$ $$0$$ $$1$$ $$1$$ $$1$$ Clearly, the answer we are looking for is the greatest value $$k$$ s.t. $$F(k) = 0$$. Notice that every number in the left part of the search space, $$0 \leq l \leq \lfloor n \rfloor$$ has $$F(l) = 0$$, while the elements in the right side,$$\lfloor n \rfloor +1 \leq r \leq n$$, have $$F(r) = 1$$. Because the function $$F(k)$$ splits the search space into two parts, we can use binary search to find the end of the first partition (this is true in general, and if we ever recognize a problem that presents these characteristics, we can apply binary search to it). We can do that because if we pick an integer from in $$[0,n]$$, say $$k$$, and $$F(k) = 1$$ we know that $$k$$ is not the solution and crucially, also that all the values greater than $$k$$ are not good candidates because they all belong to the right partition. On the other hand, if $$F(k) = 0$$, we know that $$k$$ might be the solution but also that, all the elements smaller than $$k$$ are not good candidates as $$k$$ is already a better answer than any of those numbers would be. The idea above is implemented in Listing [list:square_root_binary_search]. The algorithm works by maintaining an interval (defined by the variables and ): inside of it lies the solution, which is initially set to be the entire search space $$[0,n]$$. It iteratively shrinks this range by testing the middle element of $$[l,r]$$ (value hold by ), and this can lead to one of the following three scenarios: 1. $$^2 = n$$: is the solution and also that $$n$$ is a perfect square. 2. $$^2 > n$$: is not the solution and we can also exclude all numbers $$k \geq$$ from the search (by setting ). 3. $$^2 < n$$: is the best guess we have found so far (it might be the solution). We can, however, exclude every number $$k <$$ (by assigning ) as when squared, they would also be smaller than $$^2$$. Pay attention to the way the midpoint between $$l$$ and $$r$$ is calculated. It is common to see it calculated by using the following formula: $$(l+r)/2$$. However, this can lead to overflow problems when $$l+r$$ does not fit in an . Finally, the time and space complexity of this algorithm is $$O(log(n))$$ and $$O(1)$$, respectively. A good improvement w.r.t. to the complexity of the brute-force solution.

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