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# Law of Sines in Algebra ## Using Trigonometric Rules To Solve Triangles: An Introduction To The Law Of Sines Trigonometric rules are powerful tools that can help determine various properties and values of right-angled triangles. However, these rules can also be applied to solve non-right-angled triangles by finding missing angles, lengths, or area. One specific rule that proves to be useful in such cases is the Law of Sines. Let's explore this rule in detail. ## The Law of Sines The Law of Sines is a triangle rule that helps find missing sides or angles in any triangle with sides a, b, and c, and angles A, B, and C. It has two versions, with the first version stating that asin(A) = bsin(B) = csin(C). This version is commonly used to find the length of a missing side. The second version states that sin(A)/a = sin(B)/b = sin(C)/c, and is typically used to find a missing angle. ## When To Use The Law of Sines? The Law of Sines can be applied in situations where we know the values of two angles and the length of any side, or when we are aware of the lengths of two sides and an angle opposite one of those sides. ## Examples Of The Law of Sines Let's consider a triangle where we need to find the length of side a, given that asin(A) = bsin(B). By substituting the values asin(75°) = 8sin(30°), we can determine that a = 15.455. Similarly, if we need to find the value of angle x, we can use the Law of Sines by setting sin(A)/a = sin(B)/b. Solving this gives us sin(x)/10 = sin(50°)/15, and by solving for x, we get x = 30.71°. These are just a few examples of how the Law of Sines can be applied in algebra. ## Key Takeaways Of The Law Of Sines In Algebra • The Law of Sines can be used to find missing sides or angles in a triangle. • There are two versions of the Law of Sines: asin(A) = bsin(B) = csin(C) and sin(A)/a = sin(B)/b = sin(C)/c. • It is useful in situations where we know two angles and the length of a side, or the lengths of two sides and an opposite angle. ## The Law Of Sines In Algebra ### What Is The Law Of Sines? The Law of Sines states that a/sin(A) = b/sin(B) = c/sin(C) for a triangle with sides a, b, and c, and angles A, B, and C. ### How Can You Use The Law Of Sines In Algebra To Find Angles? If we know the values of two sides and an angle opposite one of those sides, we can use the Law of Sines to determine the other angle. ### When Should You Apply The Law Of Sines In Algebra? The Law of Sines is a useful tool in solving triangles when we have information about two angles and a side, or the lengths of two sides and an opposite angle.
#### xaktly \| Algebra \| Functions { factoring } Quadratic functions are very important.* There's a lot to cover, so the material is broken up into a few sections. • This section covers all methods of solving quadratic equations. • Quadratic functions deals with definitions, graphing and transformations of the parabola, the graph of a quadratic function. • This page specifically on the very important completing the square technique has a generator to create random quadratic equations that you can try to solve. You can reveal a step-by-step guide to each solution. • Quadratic problems covers various kinds of problems you can solve using quadratic functions. • You should also check out the Po-Shen Loh method (new in 2019) at the bottom of this page. It's pretty cool, simple to use, and foolproof! * I can't overemphasize how important it is to master quadratic functions if you intend to go further with your studies of mathematics. Because all things in math are more the same than different, the things you learn in these sections, with a relatively easy function type, will be valuable later. Beyond that, it's often possible to reduce a more complicated problem to a quadratic one. Quadratic relationships are all over. Learn all about them! ### The algebra of solving quadratic equations We are often faced with finding the solution to the equation $f(x) = k,$ where $k$ is some constant. For example, we might want to solve the quadratic equation $$3x^2 - 4x - 12 = 5$$ But an equation like this can always be rewritten by moving everything to the left side (leaving a zero on the right): $$3x^2 - 4x - 17 = 0$$ Now we're just finding the x-intercepts (also called roots or zeros), which can be done by direct factoring (we do this "by eye" from experience at recognizing patterns), by completing the square, or by using the quadratic formula. The latter two actually turn out to be the same thing, as we'll see. We'll develop both of these techniques below. The fundamental theorem of algebra tells us that any polynomial equation of degree n has exactly n solutions, though some might be imaginary. So a quadratic function (degree = 2) will always have two solutions. There are three possibilities for those solutions, illustrated in graphs below. The solutions to a quadratic equation, $$ax^2 + bx + c = 0$$ can be • Two real roots, indicating that the parabola (graph of a quadratic) intersects the x-axis twice, • One real root, actually a double root, indicating that the vertex of the parabola just touches (is tangent to) the x-axis, or • Two imaginary roots, in a pair of complex conjugates, like $a ± ib,$ where $a$ is the real part of the solutions and ib is the imaginary part ($i$ is the imaginary number $i = \sqrt{-1}.$ You might want to review complex numbers if you need some brushing up. The figures below illustrate these possibilities graphically. ### What it means to factor "directly" We think of direct factoring as finding binomial factors of a quadratic function "by eye" or "by instinct," or by the guess-and-check method. Whatever you call it, if it's possible, direct factoring is a quick way to the roots of a quadratic function. #### Direct factoring: A = 1 \begin{align} f(x) &= (x + m)(x + n) \\[5pt] &= x^2 + mx + nx + mn \\[5pt] &= x^2 + (m + n)x + mn, \end{align} which looks like $Ax^2 + Bx + C,$ with $A = 1.$ Notice the correspondence: $A = 1, \; B = m+n,$ and $C = mn.$ \begin{align} f &= (x + m)(x + n) \\[5pt] &= x^2 + mx + nx + mn \\[5pt] &= x^2 + \color{#A7D82F}{(m + n)}x + \color{#00B0FF}{mn} \\[5pt] &\phantom{00} \updownarrow \phantom{0000} \updownarrow \phantom{000000} \updownarrow \\[5pt] f(x) &= \color{#E90F89}{A}x^2 + \: \color{#A7D82F}{B}x \phantom{00} + \phantom{0} \color{#00B0FF}{C} \\[5pt] \end{align} $$\color{#E90F89}{A} = 1 \phantom{000} \color{#A7D82F}{B} = \color{#A7D82F}{m + n} \phantom{000} \color{#00B0FF}{C} = \color{#00B0FF}{mn}$$ From this relation, you can easily see that what we're looking for in direct factoring is two numbers, m and n, that add to B and multiply to C. The trick is to get good at recognizing patterns that lead us to proper factoring. That comes through practice - although on rare occasions, I've met people who are factoring savants. #### Direct factoring: A ≠ 1 Direct factoring when A ≠ 1 is a little trickier, but you can still learn to recognize the patterns: \begin{align} f(x) &= (ax + m)(bx + n) \\[5pt] &= ab \, x^2 + (mb + na)x + mn. \end{align} Here is a complete picture of the correspondence between the parameters A, B & C and a, b, m & n: \begin{align} f &= (ax + m)(bx + n) \\[5pt] &= abx^2 + anx + bmx + mn \\[5pt] &= \color{#E90F89}{ab}x^2 + \color{#A7D82F}{(bm + an)}x + \color{#00B0FF}{mn} \\[5pt] &\phantom{00} \updownarrow \phantom{0000000} \updownarrow \phantom{000000} \updownarrow \\[5pt] f(x) &= \color{#E90F89}{A}x^2 + \phantom{000} \color{#A7D82F}{B}x \phantom{00} + \phantom{0} \color{#00B0FF}{C} \\[5pt] \end{align} $$\color{#E90F89}{A} = \color{#E90F89}{ab} \phantom{000} \color{#A7D82F}{B} = \color{#A7D82F}{bm + an} \phantom{000} \color{#00B0FF}{C} = \color{#00B0FF}{mn}$$ Now a·b must multiply to A, bm+an must sum to B and mn must multiply to C. Honestly, though, I never think of it this way; I factor this kind of quadratic equation by "educated trial and error." #### An interesting thing about direct factoring Here's an interesting note on direct factoring. I once made a spreadsheet of quadratic functions with coefficients A= -9, B= -9, C= -9; A= -9, B =-9, C= -8; ... up to A= 9, B= 9, C= 9. Of over 5000 possible quadratic equations that were generated in this way, only about 12% were directly factorable. As you work out in the world, not only is it unlikely that you'll encounter a factorable quadratic equation, but it's even less likely you'll encounter one with nice integer coefficients! #### Why learn to factor? So of what value is direct factoring? Well, it's always good to be able to recognize patterns in mathematics, and factoring practice helps with that. It also helps us to arrive rapidly at exact solutions for many equations. But we'll want a more general way of solving quadratic equations, and that's completing the square. ### Practice – Direct factoring Solve these quadratic equations (find their roots or zeros) by direct factoring. 1 $x^2 - 2x - 3 = 0$ Solution $= (x + 1)(x - 3)$ $x = -1, \; 3$ 2 $x^2 + 7x + 10 = 0$ Solution $= (x + 2)(x + 5)$ $x = -5, \; -2$ 3 $x^2 - 5x - 66 = 0$ Solution $= (x -11)(x + 6)$ $x = -6, \; 11$ 4 $x^2 - 9x + 20 = 0$ Solution $= (x - 4)(x - 5)$ $x = 4, \; 5$ 5 $x^2 - 4 = 0$ Solution $= (x - 2)(x + 2)$ $x = ±2$ 6 $x^2 + 26x + 169 = 0$ Solution $= (x + 13)(x + 13)$ $x = -13, \; -13$ 7 $x^2 - 6x + 9 = 0$ Solution $= (x - 3)(x - 3)$ $x = 3, \; 3$ 8 $2x^2 - 3x - 2 = 0$ Solution $= (2x + 1)(x - 2)$ $x = -\frac{1}{2}, \; 2$ 9 $2x^2 + 5x - 3 = 0$ Solution $= (2x - 1)(x + 3)$ $x = \frac{1}{2}, \; -3$ 10 $4x^2 + 4x + 1 = 0$ Solution $= (2x + 1)(2x + 1)$ $x = -\frac{1}{2}, \; -\frac{1}{2}$ 11 $2x^2 - 2x - 12 = 0$ Solution $= (2x + 3)(x - 4)$ $x = -\frac{3}{2}, \; 4$ 12 $10x^2 - x - 3 = 0$ Solution $= (5x - 3)(2x + 1)$ $x = \frac{3}{5}, \; -\frac{1}{2}$ 13 $16x^2 - 1 = 0$ Solution $= (4x - 1)(4x + 1)$ $x = ±\frac{1}{4}$ 14 $11x^2 + 13x + 2 = 0$ Solution $= (11x + 2)(x + 1)$ $x = -\frac{2}{11}, \; -1$ 15 $5x^2 - 5 = 0$ Solution $= 5(x^2 - 1)$ $= 5(x - 1)(x + 1)$ $x = ±1$ 16 $7x^2 - 8x + 1 = 0$ Solution $= (7x - 1)(x - 1)$ $x = \frac{1}{7}, \; 1$ 17 $9x^2 - 6x + 1 = 0$ Solution $= (3x - 1)(3x - 1)$ $x = \frac{1}{3}, \; \frac{1}{3}$ 18 $9x^2 + 9x + 2 = 0$ Solution $= (3x + 1)(3x + 2)$ $x = -\frac{1}{3}, \; -\frac{2}{3}$ 19 $18x^2 + 9x - 2 = 0$ Solution $= (6x - 1)(3x + 2)$ $x = \frac{1}{6}, \; -\frac{2}{3}$ 20 $12x^2 + 23x + 11 = 0$ Solution $= (12x + 11)(x + 1)$ $x = -\frac{11}{12}, \; -1$ ### Completing the square The most general method of finding the solutions to $Ax^2 + Bx + C = 0$ is the method called completing the square. It works like this: Take any binomial and square it, $(x + m)^2,$ to form a perfect square. Then expand it: $$(x + m)^2 = x^2 + 2mx + m^2$$ Now notice that for a perfect square, the constant term is equal to the square of one-half of the coefficient of x: $$\require{cancel} m^2 = \left[ \frac{1}{\cancel{2}} (\cancel{2}m) \right]^2$$ We can use this to our advantage. We can build a perfect square out of any quadratic function, but at the expense of accumulating some constants on the right side of the equation. But that's OK, we can easily deal with that later. Below are a few step-by-step examples of completing the square. We use the same basic steps each time, and you should learn them. Completing the square is basically forcing part of a quadratic equation (the part with the variables) to be a perfect square with a well known pattern. The cost of this is accumulating constants on the other side of the equation, but we can always deal with the at the end. ##### Enter A, B, C & hit [Calculate] A B C Root 1 Root 2 Discriminant Vertex Here is a little quadratic solver tool that you can use to find the roots of any quadratic equation numerically. It might help you to solve the problems on this page. It will calculate roots (real or imaginary), the vertex coordinates and the discriminant of any quadratic function. Just enter the A, B and C constants and hit calculate. ### Example 1 Solve x2 + 2x - 35 = 0 Solution: You should first notice that this function is factorable: f(x) = (x + 7)(x - 5), so we really already know the solutions: both x = -7 and x = 5 make this equation equal to zero. But let's solve this problem by completing the square to see if we can get to these solutions in that way. We'll do these problems using the same steps each time. The first is ##### 1. Move the constant to the right. We begin with our function, and move the constant to the right side, in this case by addition of 35 to each side: $$x^2 + 2x = 35$$ ##### 2. Ensure that the coefficient of x2 is one. We divide by the coefficient of $x^2$ to make it 1. In this case, it's already 1, so we move on ... ##### 3. Add the square of ½ of the coefficient of x to both sides (complete the square). This is the key step. One half of 2 is 1, so we'll add 1 to both sides. I like to write the number added to the left as 12 (of course when it's 1 it doesn't make much difference, but it will later), and just go ahead and square it on the right. Now we've made a perfect square on the left. $$x^2 + 2x + 1^2 = 35 + 1$$ ##### 4. Identify the perfect square. Now we've constructed a perfect square on the left. That's what we set out to do. The "cost" was that we had to add that 1 to the right side. Now our equation is $$\color{#E90F89}{x}^2 \color{#E90F89}{+} 2x + \color{#E90F89}{1}^2 = 36$$ The magenta items are a consistent pattern for how to identify the perfect square in all of these problems: Take the x, the first operation (+ in this case) and the number that's 1/2 of the coefficient of x – before it's squared. The perfect-square equation is: $$(x + 1)^2 = 36$$ ##### 5. Take the square root of both sides. Now it's easy to solve for x, simply take the square root of each side, remembering to append a ± to the right side (it could go on either side, but there works best). $$x + 1 = ±\sqrt{36}$$ Now it's a simple matter to find the two solutions, and you can see that they match our original solutions that came from factoring. \begin{align} x &= -1 ± 6 \\[5pt] x &= -7, \; 5 \end{align} ### Example 2 Solve 2x2 + 2x - 11 = 0 Solution: This function is not factorable, so it will be a true test of completing the square. Also, A ≠ 1 in this equation, so we can see the whole scope of how to use the method. We begin with our function and walk through the steps: $$f(x) = 2x^2 + 2x - 11 = 0$$ ##### 1. Move the constant to the right. $$2x^2 + 2x = 11$$ ##### 2. Ensure that the coefficient of x2 is one. We divide by the coefficient of x2 , 2 in this case: $$x^2 + x = \frac{11}{2}$$ ##### 3. Add the square of 1/2 of the coefficient of x to both sides (complete the square). One half of 1 is ½, so we'll add the square of ½ to both sides, again writing it as (½)2 on the left, and ¼ on the right. Doing so really helps to properly identify the perfect square on the left in the next step. $$\color{#E90F89}{x}^2 \color{#E90F89}{+} x + \left( \color{#E90F89}{\frac{1}{2}} \right)^2 = \frac{11}{2} + \frac{1}{4}$$ ##### 4. Identify the perfect square. The perfect square is composed of the x from x2, the sign following it ( + ) and the ½ from (½)2: $$\left( x + \frac{1}{2} \right)^2 = \frac{23}{4}$$ ##### 5. Take the square root of both sides. Now it's easy to solve for x. Take the square root of each side, remembering to append a ± to the right side: $$x + \frac{1}{2} = \frac{±\sqrt{23}}{2}$$ Now we just move that ½ to the right side by subtraction to get our two solutions:. $$x = \frac{-1 ± \sqrt{23}}{2}$$ It is very unlikely that anyone would come up with these solutions by direct factoring! ### Example 3 – imaginary roots Solve 2x2 + 5x + 4 = 0 Solution: This function is not factorable, and if you were to plot its graph, you'd find that it doesn't cross the x-axis. Therefore it has no real roots. Nevertheless, we can complete the square to find a pair of complex roots (or more precisely complex roots with nonzero imaginary parts). If you haven't studied complex numbers yet, you might want to review them here. ##### 1. Move the constant to the right. $$2x^2 + 5x = -4$$ ##### 2. Ensure that the coefficient of x2 is one. We divide by the coefficient of x2 , 2 in this case: $$x^2 + \frac{5}{2} x = \frac{-4}{2}$$ ##### 3. Add the square of 1/2 of the coefficient of x to both sides (complete the square). One half of 5/2 is 5/4, so we'll add the square of 5/4 to both sides, again writing it as (5/4)2 on the left, and 25/16 on the right. Doing so really helps to properly identify the perfect square on the left in the next step. $$x^2 + \frac{5}{2} x + \left( \frac{5}{4} \right)^2 = \frac{-4}{2} + \frac{25}{16}$$ ##### 4. Identify the perfect square. The perfect square is composed of the x from x2, the sign following it ( + ) and the 5/4 from (5/4)2. We can also add the fractions on the right by using 16 as a common denominator: $$\left( x + \frac{5}{4} \right)^2 = \frac{-32 + 25}{16} = \frac{-7}{16}$$ ##### 5. Take the square root of both sides. Take the square root of each side, remembering to append a ± to the right side. Note that we're taking the square root of a negative number, so we'll need imaginary numbers: $$\sqrt{\left( x + \frac{5}{4} \right)^2} = \sqrt{\frac{-7}{16}}$$ Taking the root and moving the 5/4 to the right by subtraction gives a pair of imaginary roots: $$x = \frac{-5 ± i\sqrt{7}}{4}$$ ### Example 4 Derivation of the quadratic formula by completing the square on x2 + Bx + C = 0 Finally, let's complete the square on a completely general (no numbers) quadratic equation, Ax2 + Bx + C = 0. You'll see why this is relevant when we're done. ##### 1. Move the constant to the right. $$f(x) = Ax^2 + Bx + C = 0$$ ##### 2. Divide by A to Ensure that the coefficient of x2 is one. $$x^2 + \frac{B}{A} x = \frac{-C}{A}$$ ##### 3. Add the square of 1/2 of the coefficient of x to both sides (complete the square). One half of B/A is B/(2A), (and remember that the square of that is B2/(4A2) so we get: $$x^2 + \frac{B}{A} x + \left( \frac{B}{2A} \right)^2 = \frac{-C}{A} + \frac{B^2}{4A^2}$$ ##### 4. Identify the perfect square. Here is the perfect square, and I've multiplied the -C/A term on the right by 4A/4A to get a common denominator to add those terms: $$\left( x + \frac{B}{2A} \right)^2 = \frac{B^2 - 4AC}{4A^2}$$ ##### 5. Take the square root of both sides. Take the square root of each side, remembering to append a ± to the right side. $$\sqrt{\left( x + \frac{B}{2A} \right)^2} = \sqrt{\frac{B^2 - 4AC}{4A^2}}$$ Now the 4A2 on the right side denominator is a square, so we can take its square root. Continuing the rearrangement by subtracting B/2A from both sides we arrive at: $$x = \frac{-B ± \sqrt{B^2 - 4AC}}{2A}$$ which is the well-known quadratic formula. Yup, completing the square and using the quadratic formula are the same thing! The quadratic formula gives the two zeros or roots of any quadratic equation, Ax2 + Bx + C = 0 . It is: #### Why complete the square? Why do I need to be able to complete the square when I can just use the quadratic formula? Over the years I've seen many, many mistakes made using the quadratic formula. I think it has to do with my students over-estimating how easy it is to use. But in fact, it's a little complicated, and all of the usual rules of algebra apply. For my part, I can complete the square as fast as anyone can use the quadratic, and I like the practice it gives me at adding fractions. There are also situations you'll encounter later on, like when you study circles, ellipses and hyperbolas, where completing the square will help you immensely. ### The discriminant #### a quick way to determine whether a quadratic function has real roots There is a way to know fairly quickly whether the roots of a quadratic function are real: by calculating the discriminant. The discriminant is just the square root part of the quadratic formula, B2 - 4AC →. We have to resort to imaginary numbers if the discriminant is negative, and if it's positive we're assured of real roots. Here are all of the options: If B2 - 4AC > 0, the function has two real roots. If B2 - 4AC = 0, there is one double root, and the vertex of the parabola touches the x-axis. If B2 - 4AC < 0, there are two complex roots, a complex-conjugate pair. If a function has a double root, it touches the x-axis at only one point, its vertex. The same is true for any polynomial function with a double root: It will only touch the axis (not cross) at that point. ### Practice problems Find the solutions (roots, zeros) of these quadratic equations by completing the square. Some of the solutions may be imaginary. 1 $x^2 - 3x - 3 = 0$ Solution \begin{align} x^2 - 3x &= 3 \\[5pt] x^2 - 3x + \left(\frac{3}{2}\right)^2 &= 3 + \frac{9}{4} \\[5pt] \left( x - \frac{3}{2} \right)^2 &= \frac{21}{4} \\[5pt] x &= \frac{3 ± \sqrt{21}}{2} \end{align} 2 $2x^2 + 5x + 4 = 0$ Solution \begin{align} 2x^2 + 5x &= -4 \\[5pt] x^2 + \frac{5}{2}x &= -2 \\[5pt] x^2 + \frac{5}{2}x + \left( \frac{5}{4} \right)^2 &= -2 + \frac{25}{16} \\[5pt] \left( x + \frac{5}{4} \right)^2 &= \frac{-32 + 25}{16} \\[5pt] \left( x + \frac{5}{4} \right)^2 &= \frac{-7}{16} \\[5pt] x &= \frac{-5 ± i\sqrt{7}}{4} \end{align} 3 $-3x^2 + 2x + 1 = 0$ Solution \begin{align} -3x^2 + 2x &= -1 \\[5pt] x^2 - \frac{2}{3} x &= \frac{1}{3} \\[5pt] x^2 - \frac{2}{3} x + \left( \frac{1}{3} \right)^2 &= \frac{1}{3} + \frac{1}{9} \\[5pt] \left( x - \frac{1}{3} \right)^2 &= \frac{4}{9} \\[5pt] x &= \frac{1 ± 2}{3} \\[5pt] x &= -\frac{2}{3}, \; 1 \end{align} 4 $-7x^2 + x + 14 = 0$ Solution \begin{align} -7x^2 + x &= -14 \\[5pt] x^2 - \frac{1}{7} x &= 2 \\[5pt] x^2 - \frac{1}{7}x + \left( \frac{1}{14} \right)^2 &= 2 + \frac{1}{196} \\[5pt] \left( x - \frac{1}{14} \right)^2 &= \frac{393}{196} \\[5pt] x &= \frac{1 ± \sqrt{393}}{14} \end{align} 5 $6x^2 + 3x + 7 = 0$ Solution \begin{align} 6x^2 + 3x &= -7 \\[5pt] x^2 + \frac{1}{2}x &= -\frac{7}{6} \\[5pt] x^2 + \frac{1}{2}x + \left( \frac{1}{4}\right)^2 &= \frac{1}{16} - \frac{7}{6} \\[5pt] \left( x + \frac{1}{4} \right)^2 &= \frac{6 - 112}{16} \\[5pt] \left( x + \frac{1}{4} \right)^2 &= \frac{-106}{16} \\[5pt] x &= \frac{-1 ± i \sqrt{106}}{4} \end{align} 6 $x^2 - x - 1 = 0$ Solution \begin{align} x^2 - x &= 1 \\[5pt] x^2 - x + \left( \frac{1}{2} \right)^2 &= 1 + \frac{1}{4} \\[5pt] \left( x - \frac{1}{2} \right)^2 &= \frac{5}{4} \\[5pt] x &= \frac{1 ± \sqrt{5}}{2} \end{align} 7 $x^2 + x + 1 = 0$ Solution \begin{align} x^2 + x &= -1 \\[5pt] x^2 + x + \left( \frac{1}{2} \right)^2 &= -1 + \frac{1}{4} \\[5pt] \left( x + \frac{1}{2} \right)^2 &= \frac{-3}{4} \\[5pt] x &= \frac{-1 ± i\sqrt{3}}{2} \end{align} 8 $2x^2 + 2x + 3 = 0$ Solution \begin{align} 2x^2 + 2x &= -3 \\[5pt] x^2 + x &= \frac{-3}{2} \\[5pt] x^2 + x + \left( \frac{1}{2} \right)^2 &= \frac{-3}{2} + \frac{1}{4} \\[5pt] \left( x + \frac{1}{2} \right)^2 &= \frac{-5}{4} \\[5pt] x &= \frac{-1 ± i \sqrt{5}}{2} \end{align} 9 $x^2 - 7x - 7 = 0$ Solution \begin{align} x^2 - 7x &= 7 \\[5pt] x^2 - 7x + \left( \frac{7}{2} \right)^2 &= 7 + \frac{49}{4} \\[5pt] \left( x - \frac{7}{2} \right)^2 &= \frac{28 + 49}{4} \\[5pt] x &= \frac{7 ± \sqrt{77}}{2} \end{align} 10 $x^2 + 2x - 9 = 0$ Solution \begin{align} x^2 + 2x &= 9 \\[5pt] x^2 + 2x + 1^2 &= 9 + 1 \\[5pt] (x + 1)^2 &= 10 \\[5pt] x &= -1 ± \sqrt{10} \end{align} 11 $2x^2 - x - 2 = 0$ Solution \begin{align} 2x^2 - x &= 2 \\[5pt] x^2 - \frac{1}{2}x &= 1 \\[5pt] x^2 - \frac{1}{2}x + \left(\frac{1}{4}\right)^2 &= 1 + \frac{1}{16} \\[5pt] \left( x - \frac{1}{4}\right)^2 &= \frac{17}{16} \\[5pt] x &= \frac{1 ± \sqrt{17}}{4} \end{align} 12 $3x^2 + x + 1 = 0$ Solution \begin{align} 3x^2 + x &= -1 \\[5pt] x^2 + \frac{1}{3}x &= \frac{-1}{3} \\[5pt] x^2 + \frac{1}{3}x + \left(\frac{1}{6}\right)^2 &= \frac{-1}{3} + \frac{1}{36} \\[5pt] \left(x + \frac{1}{6}\right)^2 &= \frac{-11}{36} \\[5pt] x &= \frac{-1 ± i\sqrt{11}}{6} \end{align} 13 $x^2 - 5x - 25 = 0$ Solution \begin{align} x^2 - 5x &= 25 \\[5pt] x^2 - 5x + \left( \frac{5}{6}\right)^2 &= 25 + \frac{25}{4} \\[5pt] \left( x - \frac{5}{2} \right)^2 &= \frac{125}{4} \\[5pt] x &= \frac{5 ± \sqrt{125}}{2} \\[5pt] &= \frac{5 ± 5\sqrt{5}}{2} \end{align} 14 $-x^2 + 2x + 1 = 0$ Solution \begin{align} -x^2 + 2x &= -1 \\[5pt] x^2 - 2x &= 1 \\[5pt] x^2 - 2x + 1 &= 1 + 1 \\[5pt] (x - 1)^2 &= 2 \\[5pt] x &= 1 ± \sqrt{2} \end{align} 15 $-4x^2 - 4x + 4 = 0$ Solution \begin{align} -4x^2 - 4x &= -4 \\[5pt] x^2 + x &= 1 \\[5pt] x^2 + x + \left( \frac{1}{2} \right)^2 &= 1 + \frac{1}{4} \\[5pt] \left( x + \frac{1}{2} \right)^2 &= \frac{5}{4} \\[5pt] x &= \frac{-1 ± \sqrt{5}}{2} \end{align} 16 $7x^2 + 2x + 1 = 0$ Solution \begin{align} 7x^2 + 2x &= -1 \\[5pt] x^2 + \frac{2}{7}x &= \frac{-1}{7} \\[5pt] x^2 + \frac{2}{7}x + \left( \frac{1}{7}\right)^2 &= \frac{1}{49} - \frac{1}{7} \\[5pt] \left( x + \frac{1}{7}\right)^2 &= \frac{-6}{49} \\[5pt] x &= \frac{-1 ± i\sqrt{6}}{7} \end{align} 17 $8x^2 - 11x + 20 = 0$ Solution \begin{align} 8x^2 - 11x &= -20 \\[5pt] x^2 - \frac{11}{8}x &= \frac{-5}{2} \\[5pt] x^2 - \frac{11}{8}x + \left(\frac{11}{16}\right)^2 &= \frac{121 - 640}{256} \\[5pt] \left( x - \frac{11}{16}\right)^2 &= \frac{-519}{256} \\[5pt] x &= \frac{11 ± i \sqrt{519}}{16} \end{align} 18 $2x^2 + 13x - 26 = 0$ Solution \begin{align} 2x^2 + 13x &= 26 \\[5pt] x^2 + \frac{13}{2}x &= 13 \\[5pt] x^2 + \frac{13}{2}x + \left( \frac{13}{4}\right)^2 &= 13 + \frac{169}{16} \\[5pt] \left( x + \frac{13}{4}\right)^2 &= \frac{377}{16} \\[5pt] x &= \frac{-13 ± \sqrt{377}}{4} \end{align} 19 $-3x^2 - 3x + 5 = 0$ Solution \begin{align} -3x^2 - 3x &= -5 \\[5pt] x^2 + x &= \frac{5}{3} \\[5pt] x^2 + x + \left(\frac{1}{2}\right)^2 &= \frac{5}{3} + \frac{1}{4} \\[5pt] \left(x + \frac{1}{2} \right)^2 &= \frac{23}{12} \\[5pt] x &= \frac{-1}{2} ± \frac{\sqrt{23}}{2 \sqrt{3}} \end{align} 20 $-9x^2 + x + 7 = 0$ Solution \begin{align} -9x^2 + x &= -7 \\[5pt] x^2 - \frac{1}{9}x &= \frac{7}{9} \\[5pt] x^2 - \frac{1}{9}x + \left( \frac{1}{18} \right)^2 &= \frac{7}{9} + \frac{1}{324} \\[5pt] \left( x + \frac{1}{18} \right)^2 &= \frac{253}{324} \\[5pt] x &= \frac{-1 ± \sqrt{253}}{18} \end{align} ### Po-Shen Loh method A new way of solving quadratics In 2019, Carnegie-Mellon University math professor Po-Shen Loh devised a new method of finding quadratic roots that may well replace the quadratic formula and completing the square. Professor Po-Shen himself couldn't really believe that in 4000 years of solving quadratics, no one had thought about his method, but it seems that might be true. #### Method for $x^2 + Bx + C = 0$ In order to solve $x^2 + Bx + C = 0,$ we need to find a way to express it as $$x^2 + Bx + C = (x - m)(x - n),$$ where the function is equal to zero if x = m or x = n. If we expand the right side, we get $$x^2 + Bx + C = x^2 - (m + n)x + mn.$$ Now there is a correspondence: -B = m + n, and C = mn. We've seen that before. When we factor a quadratic with A = 1, we're looking for two numbers that multiply to C and add to B. What Poh realized is that if m + n = -B, then their average would equal -B/2: $$\frac{m + n}{2} = \frac{-B}{2}$$ Now there is some number, z, the midpoint between the two roots, such that our roots are $$\frac{-B}{2} ± z.$$ In other words, $$m = \frac{-B}{2} - z \; \text{ and } \; n = \frac{-B}{2} + z$$ Now we need only find that one number, z, in order to solve our quadratic equation. But first, let's make sure that these results are the same as those obtained using the quadratic formula. Recalling that mn = C, we can write: $$\left( \frac{-B}{2} - z \right) \left( \frac{-B}{2} + z \right) = C$$ $$\frac{B^2}{4} + \frac{Bz}{2} - \frac{Bz}{2} - z^2 = C$$ The middle terms vanish, so we have $$\frac{B^2}{4} - z^2 = C$$ Solving for z gives $$z = ± \sqrt{\frac{B^2}{4} - C}$$ So our roots are $$\frac{-B}{2} ± \sqrt{\frac{B^2}{4} - C},$$ which is the quadratic formula when A = 1. #### Example – real roots Let's solve a factorable function so we can easily check the method: $$x^2 - x - 12 = 0$$ Our roots are $$\frac{-B}{2} ± z = \frac{1}{2} ± z$$ Now our roots must multiply to C = -12: \begin{align} \left( \frac{1}{2} + z \right) \left( \frac{1}{2} - z \right) &= -12 \\[5pt] \frac{1}{4} - z^2 &= -12 \\[5pt] z^2 &= \frac{49}{4} \\[5pt] z &= ± \frac{7}{2} \end{align} Now our roots are $$\frac{-1}{2} ± \frac{7}{2} = \bf 4, \; -3$$ That's exactly what we'd get from direct factoring. Now let's do a more complicated example. #### Example – Imaginary roots Solve $x^2 - 2x + 5 = 0$ The roots have the form $$\frac{-B}{2} ± z = 1 ± z$$ then \begin{align} (1 + z)(1 - z) &= 5 \\[5pt] 1 - z^2 &= 5 \\[5pt] -z^2 &= 4 \\[5pt] z &= \bf ± 2i \end{align} That's pretty simple! #### Method for $Ax^2 + Bx + C$ Now let's let $A \ne 0$ and develop a more general version of the Po-Shen Loh method. If our equation is $$Ax^2 + Bx + C = 0,$$ We can divide by A on both sides to get $$x^2 + \frac{B}{A}x + \frac{C}{A} = (x - m)(x - n) = 0$$ If we expand the right side of the equation to $$x^2 -(m + n)x + mn,$$ We find these correspondences: $$\frac{-B}{A} = m + n, \; \text{ and } \; \frac{C}{A} = mn$$ If we divide the first correspondence by 2, thus obtaining the average of our two roots, m and n, we get $$\frac{-B}{2A} = \frac{m + n}{2}$$ We'll call that average of m and n on the right side $z.$ The roots we seek are $$\frac{-B}{2A} ± z$$ Let's go ahead and multiply those two roots using our second correspondence, $mn = \frac{C}{A},$ \begin{align} \left( \frac{-B}{2A} + z \right) \left( \frac{-B}{2A} - z \right) &= \frac{C}{A} \\[5pt] \frac{B^2}{4A^2} - z^2 &= \frac{C}{A} \\[5pt] z^2 &= \frac{B^2}{4A^2} - \frac{C}{A} \\[5pt] z^2 &= \frac{B^2}{4A^2} - \frac{4AC}{4A^2} \\[5pt] z &= ± \frac{\sqrt{B^2 - 4AC}}{2A} \end{align} Now our roots are $\frac{-B}{2A} ± z,$ so they are $$x = \frac{-B ± \sqrt{B^2 - 4AC}}{2A},$$ which is indeed the quadratic formula. Now let's apply the method to a quadratic equation for which we already know the roots by easy factoring: $$4x^2 + 6x + 2 = (2x + 1)(2x + 2) = 0,$$ where A = 4, B = 6 and C = 2. The roots are $$\frac{-B}{2A} ± z \; \rightarrow \; \frac{-3}{4} ± z$$ Those roots multiply to $\frac{C}{A},$ so we have \begin{align} \left( \frac{-3}{4} + z \right) \left( \frac{-3}{4} - z \right) &= \frac{C}{A} = \frac{1}{2} \\[5pt] \frac{9}{16} - z^2 &= \frac{1}{2} \\[5pt] -z^2 &= \frac{-9}{16} + \frac{8}{16} \\[5pt] z^2 &= \frac{1}{16} \\[5pt] z &= ±\frac{1}{4} \end{align} So our roots are \begin{align} x &= \frac{-3}{4} ± \frac{1}{4} \\[5pt] x &= \bf -1, \; -\frac{1}{2} \end{align} ... just what we expected. Now in trying to go through the method very carefully, and making sure to link it to the quadratic formula, I fear I've made it seem more complicated than it really is. Here is a summary of the steps followed by 20 practice problems with solutions. Once you get used to the method, I think you'll find it's really easy. #### Po-Shen Loh method To solve a quadratic equation, $Ax^2 + Bx + C = 0,$ 1. Calculate the binomials $\left( \frac{-B}{2A} ± z \right)$ and multiply them. 2. Set the result equal to $\frac{C}{A},$ and solve for $z.$ 3. Calculate the two roots using $\frac{-B}{2A} ± z.$ ### Practice problems Solve these quadratic equations using the Po-Shen Loh method. 1 $3x^2 + 4x - 1 = 0$ Solution $$\frac{-B}{2A} = \frac{-2}{3} \; \text{ & } \; \frac{C}{A} = \frac{1}{3}$$ \begin{align} \left( \frac{-2}{3} + z \right) &\left( \frac{-2}{3} - z \right) = \frac{1}{3} \\[5pt] \frac{4}{9} - z^2 &= \frac{3}{9} \\[5pt] -z^2 &= -\frac{1}{9} \\[5pt] z &= ± \frac{1}{3} \\[5pt] x &= \frac{-2}{3} ± \frac{1}{3} \\[5pt] x &= \bf -\frac{1}{3}, \; -1 \end{align} 2 $x^2 - x - 7 = 0$ Solution $$\frac{-B}{2A} = \frac{1}{2} \; \text{ & } \; \frac{C}{A} = -7$$ \begin{align} \left( \frac{1}{2} + z \right) &\left( \frac{1}{2} - z \right) = -7 \\[5pt] \frac{1}{4} - z^2 &= \frac{-28}{4} \\[5pt] -z^2 &= -\frac{29}{4} \\[5pt] z &= ± \frac{\sqrt{29}}{2} \\[5pt] x &= \frac{1}{2} ± \frac{\sqrt{29}}{2} \\[5pt] x &= \bf \frac{1 ± \sqrt{29}}{2} \end{align} 3 $x^2 + x + 5 = 0$ Solution $$\frac{-B}{2A} = \frac{-1}{2} \; \text{ & } \; \frac{C}{A} = 5$$ \begin{align} \left( \frac{-1}{2} + z \right) &\left( \frac{-1}{2} - z \right) = 5 \\[5pt] \frac{1}{4} - z^2 &= \frac{20}{4} \\[5pt] -z^2 &= \frac{19}{4} \\[5pt] z &= ± -\frac{i\sqrt{19}}{2} \\[5pt] x &= -\frac{1}{2} ± \frac{i\sqrt{19}}{2} \\[5pt] x &= \bf \frac{-1 ± i\sqrt{19}}{2} \end{align} 4 $4x^2 - 4x - 5 = 0$ Solution $$\frac{-B}{2A} = \frac{1}{2} \; \text{ & } \; \frac{C}{A} = -\frac{5}{4}$$ \begin{align} \left( \frac{1}{2} + z \right) &\left( \frac{1}{2} - z \right) = -\frac{5}{4} \\[5pt] \frac{1}{4} - z^2 &= -\frac{5}{4} \\[5pt] -z^2 &= -\frac{6}{4} \\[5pt] z &= ± \frac{\sqrt{6}}{2} \\[5pt] x &= \bf \frac{1 ± \sqrt{6}}{2} \end{align} 5 $8x^2 + 7x + 8 = 0$ Solution $$\frac{-B}{2A} = \frac{-7}{16} \; \text{ & } \; \frac{C}{A} = 1$$ \begin{align} \left( \frac{-7}{16} + z \right) &\left( \frac{-7}{16} - z \right) = 1 \\[5pt] \frac{49}{256} - z^2 &= \frac{256}{256} \\[5pt] -z^2 &= \frac{207}{256} \\[5pt] z &= ± \frac{i\sqrt{207}}{16} \\[5pt] x &= \bf \frac{-7 ± \sqrt{207}}{16} \end{align} 6 $3x^2 - 3x + 2 = 0$ Solution $$\frac{-B}{2A} = \frac{1}{2} \; \text{ & } \; \frac{C}{A} = \frac{2}{3}$$ \begin{align} \left( \frac{1}{2} + z \right) &\left( \frac{1}{2} - z \right) = \frac{2}{3} \\[5pt] \frac{1}{4} - z^2 &= \frac{2}{3} \\[5pt] -z^2 &= \frac{5}{12} \\[5pt] z &= ± \frac{i\sqrt{5}}{\sqrt{12}} \\[5pt] x &= \bf \frac{1}{2} ± \frac{i\sqrt{5}}{2\sqrt{3}} \end{align} 7 $x^2 + 7x - 9 = 0$ Solution $$\frac{-B}{2A} = \frac{-7}{2} \; \text{ & } \; \frac{C}{A} = -9$$ \begin{align} \left( \frac{-7}{2} + z \right) &\left( \frac{-7}{2} - z \right) = -9 \\[5pt] \frac{49}{4} - z^2 &= \frac{-36}{4} \\[5pt] -z^2 &= -\frac{85}{4} \\[5pt] z &= ± \frac{\sqrt{85}}{2} \\[5pt] x &= \frac{-7}{2} ± \frac{\sqrt{85}}{2} \\[5pt] x &= \bf \frac{-7 ± \sqrt{85}}{2} \end{align} 8 $2x^2 + 7x - 5 = 0$ Solution $$\frac{-B}{2A} = \frac{-7}{4} \; \text{ & } \; \frac{C}{A} = \frac{-5}{2}$$ \begin{align} \left( \frac{-7}{4} + z \right) &\left( \frac{-7}{4} - z \right) = -\frac{5}{2} \\[5pt] \frac{49}{16} - z^2 &= \frac{-40}{16} \\[5pt] -z^2 &= -\frac{89}{16} \\[5pt] z &= ± \frac{\sqrt{89}}{4} \\[5pt] x &= \frac{-7}{4} ± \frac{\sqrt{89}}{4} \\[5pt] x &= \bf \frac{-7 ± \sqrt{89}}{4} \end{align} 9 $3x^2 + 11x - 8 = 0$ Solution $$\frac{-B}{2A} = \frac{-11}{6} \; \text{ & } \; \frac{C}{A} = \frac{-8}{3}$$ \begin{align} \left( \frac{-11}{6} + z \right) &\left( \frac{-11}{6} - z \right) = \frac{-8}{3} \\[5pt] \frac{121}{36} - z^2 &= \frac{-96}{36} \\[5pt] -z^2 &= -\frac{-217}{36} \\[5pt] z &= ± \frac{\sqrt{217}}{6} \\[5pt] x &= \frac{-11}{6} ± \frac{\sqrt{217}}{6} \\[5pt] x &= \bf \frac{-11 ± \sqrt{217}}{6} \end{align} 10 $4x^2 + 8x + 11 = 0$ Solution $$\frac{-B}{2A} = -1 \; \text{ & } \; \frac{C}{A} = \frac{11}{4}$$ \begin{align} \left( -1 + z \right) &\left( -1 - z \right) = \frac{11}{4} \\[5pt] 1 - z^2 &= \frac{11}{4} \\[5pt] -z^2 &= \frac{7}{4} \\[5pt] z &= ± \frac{i\sqrt{7}}{2} \\[5pt] x &= \bf -1 ± \frac{i\sqrt{7}}{2} \end{align} 11 $6x^2 + 5x + 23 = 0$ Solution $$\frac{-B}{2A} = -\frac{5}{12} \; \text{ & } \; \frac{C}{A} = \frac{23}{6}$$ \begin{align} \left( -\frac{5}{12} + z \right) &\left( -\frac{5}{12} - z \right) = \frac{23}{6} \\[5pt] \frac{25}{144} - z^2 &= \frac{23}{6} \\[5pt] \frac{25}{144} - z^2 &= \frac{552}{144} \\[5pt] -z^2 &= \frac{527}{144} \\[5pt] z &= -\frac{5}{12} ± \frac{i\sqrt{527}}{12} \\[5pt] x &= \bf \frac{-5 ± i\sqrt{527}}{12} \end{align} 12 $11x^2 + 2x + 1 = 0$ Solution $$\frac{-B}{2A} = -\frac{1}{11} \; \text{ & } \; \frac{C}{A} = \frac{1}{11}$$ \begin{align} \left( \frac{-1}{11} + z \right) &\left( \frac{-1}{11} - z \right) = \frac{1}{11} \\[5pt] \frac{1}{121} - z^2 &= \frac{1}{11} \\[5pt] \frac{1}{121} - z^2 &= \frac{11}{121} \\[5pt] -z^2 &= \frac{10}{121} \\[5pt] z &= -\frac{1}{11} ± \frac{i\sqrt{10}}{11} \\[5pt] x &= \bf \frac{-1 ± i\sqrt{10}}{11} \end{align} 13 $9x^2 + 9x - 3 = 0$ Solution $$\frac{-B}{2A} = \frac{-1}{2} \; \text{ & } \; \frac{C}{A} = \frac{-1}{3}$$ \begin{align} \left( \frac{-1}{2} + z \right) &\left( \frac{-1}{2} - z \right) = -\frac{1}{3} \\[5pt] \frac{1}{4} - z^2 &= -\frac{1}{3} \\[5pt] -z^2 &= -\frac{7}{12} \\[5pt] z &= ±\frac{\sqrt{5}}{2\sqrt{3}} \\[5pt] x &= \bf -\frac{1}{2} ± \frac{\sqrt{7}}{2\sqrt{3}} \end{align} 14 $4x^2 + 5x - 13 = 0$ Solution $$\frac{-B}{2A} = \frac{-5}{8} \; \text{ & } \; \frac{C}{A} = \frac{-13}{4}$$ \begin{align} \left( \frac{-5}{8} + z \right) &\left( \frac{-5}{8} - z \right) = -\frac{13}{4} \\[5pt] \frac{25}{64} - z^2 &= -\frac{13}{4} \\[5pt] \frac{25}{64} - z^2 &= -\frac{208}{64} \\[5pt] -z^2 &= -\frac{183}{64} \\[5pt] z &= ±\frac{\sqrt{183}}{8} \\[5pt] x &= -\frac{-5}{8} ± \frac{\sqrt{7}}{2\sqrt{3}} \\[5pt] \end{align} 15 $5x^2 + 4x + 4 = 0$ Solution $$\frac{-B}{2A} = \frac{-2}{5} \; \text{ & } \; \frac{C}{A} = \frac{4}{5}$$ \begin{align} \left( \frac{-2}{5} + z \right) &\left( \frac{-2}{5} - z \right) = \frac{4}{5} \\[5pt] \frac{4}{25} - z^2 &= \frac{4}{5} \\[5pt] \frac{4}{25} - z^2 &= \frac{20}{25} \\[5pt] -z^2 &= \frac{16}{25} \\[5pt] z &= ±\frac{4i}{5} \\[5pt] x &= -\frac{-2}{5} ± \frac{4i}{5} \\[5pt] x &= \bf \frac{-2 ± 4i}{5} \end{align} 16 $x^2 - 14 = 0$ Solution $$\frac{-B}{2A} = 0 \; \text{ & } \; \frac{C}{A} = -14$$ \begin{align} \left( 0 + z \right) &\left( 0 - z \right) = -14 \\[5pt] - z^2 &= -14 \\[5pt] z &= ± \sqrt{14} \\[5pt] x &= 0 ± \sqrt{14} \\[5pt] x &= \bf ± \sqrt{14} \end{align} 17 $x^2 + 7x + 7 = 0$ Solution $$\frac{-B}{2A} = \frac{-7}{2} \; \text{ & } \; \frac{C}{A} = 7$$ \begin{align} \left( \frac{-7}{2} + z \right) &\left( \frac{-7}{2} - z \right) = 7 \\[5pt] \frac{49}{4} - z^2 &= 7 \\[5pt] \frac{49}{4} - z^2 &= \frac{28}{4} \\[5pt] -z^2 &= \frac{-21}{4} \\[5pt] z &= ±\frac{\sqrt{21}}{2} \\[5pt] x &= \frac{-7}{2} ±\frac{\sqrt{21}}{2} \\[5pt] x &= \bf \frac{-7 ± \sqrt{21}}{2} \end{align} 18 $x^2 + 7x - 7 = 0$ Solution $$\frac{-B}{2A} = \frac{-7}{2} \; \text{ & } \; \frac{C}{A} = -7$$ \begin{align} \left( \frac{-7}{2} + z \right) &\left( \frac{-7}{2} - z \right) = -7 \\[5pt] \frac{49}{4} - z^2 &= -7 \\[5pt] \frac{49}{4} - z^2 &= \frac{-28}{4} \\[5pt] -z^2 &= \frac{-77}{4} \\[5pt] z &= ±\frac{\sqrt{77}}{2} \\[5pt] x &= \frac{-7}{2} ±\frac{\sqrt{77}}{2} \\[5pt] x &= \bf \frac{-7 ± \sqrt{77}}{2} \end{align} 19 $5x^2 - 11x - 12 = 0$ Solution $$\frac{-B}{2A} = \frac{11}{10} \; \text{ & } \; \frac{C}{A} = \frac{-12}{5}$$ \begin{align} \left( \frac{11}{10} + z \right) &\left( \frac{11}{10} - z \right) = -\frac{12}{5} \\[5pt] \frac{121}{100} - z^2 &= -\frac{12}{5} \\[5pt] \frac{121}{100} - z^2 &= -\frac{240}{100} \\[5pt] -z^2 &= \frac{-361}{100} \\[5pt] z &= ±\frac{\sqrt{361}}{10} \\[5pt] x &= \frac{11}{10} ±\frac{19}{10} \\[5pt] x &= \frac{11 ± 19}{10} \\[5pt] x &= \bf 3, \; -\frac{4}{5} \end{align} 20 $7x^2 - 2x + 1 = 0$ Solution $$\frac{-B}{2A} = \frac{1}{7} \; \text{ & } \; \frac{C}{A} = \frac{1}{7}$$ \begin{align} \left( \frac{1}{7} + z \right) &\left( \frac{1}{7} - z \right) = \frac{1}{7} \\[5pt] \frac{1}{49} - z^2 &= \frac{1}{7} \\[5pt] \frac{1}{49} - z^2 &= \frac{7}{49} \\[5pt] -z^2 &= \frac{6}{49} \\[5pt] z &= ±\frac{i\sqrt{6}}{7} \\[5pt] x &= \frac{1}{7} ± \frac{i\sqrt{6}}{7} \\[5pt] x &= \bf \frac{1 ± i\sqrt{6}}{7} \\[5pt] \end{align} ### Video examples #### 4. Completing the square: Example 1 In this example we complete the square to find the roots of a function for which we already know the roots. It's a good example to work through to see that completing the square works! #### 5. Completing the square: Example 2 In this example, the quadratic function f(x) = 8x2 - 20x - 3 has real roots, but it's not factorable. We complete the square to find the exact roots of the function. There is no difference between completing the square and using the quadratic formula, but I think that knowing how to complete the square will help you to understand a lot about basic algebra steps. You should learn it! #### 7. Derivation of the quadratic formula – a must-see. In this example, we start with f(x) = Ax2 + Bx + C and complete the square to derive the quadratic formula. Minutes of your life: 2:04 (and well worth it!) #### 8. Completing the square to obtain imaginary roots In this example, we find the roots of f(x) = 3x2 - 4x + 7, which are imaginary (or more accurately, are complex with non-zero imaginary parts. We also use the discriminant to show how to determine before-hand that the roots of a quadratic function will be non-real. X ### expand To expand an expression in mathematics means to un-compact it. For example, 3(x + 2) expands to 3x + 2 (x + 2)(x - 1) expands to x2 + x - 2 xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.
## Tuesday, December 3, 2013 ### How to teach multiplication The questions and suggestions below pertain to the way multiplication is taught to kids. Examples pertain to multiplication problems involving two two-digit numbers. Question One: Do you teach kids to multiply right to left or left to right? The standard multiplication technique when one has pencil and paper is right to left. 37 x 17 ____ 259 + 370 _____ 629 When you are sitting in a class room or meeting or taking a test this problem can be solved much quicker in your head by noting that 10 x 37 + 7 x 37 = 629. The steps are actually the same for the two methods but the second process clarifies that multiplying 37 by 17 is the same as multiplying 37 by 10 and 37 by 7 and adding the two products. The second method is essentially an example of the distributive property. An experiment: One teacher should teach only the first method right to left. The second teacher should present both methods. Teachers should give a test of the following form 15 x 12 43 x 24 65 x 18 95 x 21 25 x 17 The teachers should review the material and give a second test. This time the teacher who initially only taught the first method might also teach the second method. Our assessment of the two methods should also consider improvements between the first test and the second one. It might be useful to time these tests because i suspect that the use of the distributive property when multiplying could increase speed as well as accuracy. Speed is important on tests, in informal situations where paper, pencil and calculators are unavailable and when analyzing complex multi-step problem. Question Two: How do you teach kids to multiply a fraction and a number? Method One: 60 x 0.15 _____ 600 300 ____ 900 Move the decimal point two places to the left and get 9.0. Method Two 10% of 60 is 6.0. 1% of 60 is 0.6. 5% of 60 is 5 x 0.6 which is 3.0. 15% of 60 is 6.0+3.0 =9.0. Thankfully, we get the same answer both ways but the thought process differs. The first method usually requires pencil and paper. The second may be better in a test situation or in a meeting where neither pencil or paper nor a calculator are readily available. A second experiment: Again, it would be useful to compare student performance for classes that are only taught method one to classes that are taught both methods. After the first test teach both classes the second method and give a second test. Measure both absolute performance and improvement for the two groups. Consider applying a time constraint because both accuracy and speed are important.
More # Completing the Square Say we have a simple expression likeÂ x2Â + bx. HavingÂ xÂ twice in the same expression can make life hard. What can we do? Well, with a little inspiration from Geometry we can convert it, like this: As you can seeÂ x2Â + bxÂ can be rearrangedÂ nearlyÂ into a square … … and we canÂ complete the squareÂ withÂ (b/2)2 In Algebra it looks like this: x2Â + bx + (b/2)2 = (x+b/2)2 “Complete the Square” So, by addingÂ (b/2)2Â we can complete the square. AndÂ (x+b/2)2Â hasÂ xÂ onlyÂ once, which is easier to use. ## Keeping the Balance Now … we can’t justÂ addÂ (b/2)2Â without alsoÂ subtractingÂ it too! Otherwise the whole value changes. So let’s see how to do it properly with an example: Start with: (“b” is 6 in this case) Complete the Square: AlsoÂ subtractÂ the new term Simplify it and we are done. The result: x2Â + 6x + 7 Â = Â (x+3)2Â âˆ’Â 2 And nowÂ xÂ only appears once, and our job is done! ## A Shortcut Approach Here is a quick way to get an answer. You may like this method. First think about the result we want:Â (x+d)2Â + e AfterÂ expandingÂ (x+d)2Â we get:Â x2Â + 2dx + d2Â + e Now see if we can turn our example into that form to discover d and e ### Example: try to fitÂ x2Â + 6x + 7Â intoÂ x2Â + 2dx + d2Â + e Now we can “force” an answer: • We know thatÂ 6xÂ must end up asÂ 2dx, soÂ dÂ must be 3 • Next we see thatÂ 7Â must become d2Â + e =Â 9 + e, soÂ eÂ must be âˆ’2 And we get the same resultÂ (x+3)2Â âˆ’ 2Â as above! Now, let us look at a useful application: solving Quadratic Equations … ## Solving General Quadratic Equations by Completing the Square We can complete the square toÂ solveÂ aÂ Quadratic EquationÂ (find where it is equal to zero). But a general Quadratic Equation can have aÂ coefficientÂ ofÂ aÂ in front ofÂ x2: ax2Â + bx + c = 0 But that is easy to deal with … just divide the whole equation by “a” first, then carry on: x2Â + (b/a)x + c/a = 0 ## Steps Now we canÂ solveÂ a Quadratic Equation in 5 steps: • Step 1Â Divide all terms byÂ aÂ (the coefficient ofÂ x2). • Step 2Â Move the number term (c/a) to the right side of the equation. • Step 3Â Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. We now have something that looks like (x + p)2Â = q, which can be solved rather easily: • Step 4Â Take the square root on both sides of the equation. • Step 5Â Subtract the number that remains on the left side of the equation to findÂ x. ## Examples OK, some examples will help! ### Example 1: Solve x2Â + 4x + 1 = 0 Step 1Â can be skipped in this example since the coefficient of x2Â is 1 Step 2Â Move the number term to the right side of the equation: x2Â + 4x = -1 Step 3Â Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation. (b/2)2Â = (4/2)2Â = 22Â = 4 x2Â + 4x + 4 = -1 + 4 (x + 2)2Â = 3 Step 4Â Take the square root on both sides of the equation: x + 2 = Â±âˆš3 = Â±1.73 (to 2 decimals) Step 5Â Subtract 2 from both sides: x = Â±1.73 â€“ 2 = -3.73 or -0.27 And here is an interesting and useful thing. At the end of step 3 we had the equation: (x +Â 2)2Â =Â 3 It gives us theÂ vertexÂ (turning point) of x2Â + 4x + 1:Â (-2, -3) ### Example 2: Solve 5x2Â â€“ 4x â€“ 2 = 0 Step 1Â Divide all terms by 5 x2Â â€“ 0.8x â€“ 0.4 = 0 Step 2Â Move the number term to the right side of the equation: x2Â â€“ 0.8x = 0.4 Step 3Â Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation: (b/2)2Â = (0.8/2)2Â = 0.42Â = 0.16 x2Â â€“ 0.8x + 0.16 = 0.4 + 0.16 (x â€“ 0.4)2Â = 0.56 Step 4Â Take the square root on both sides of the equation: x â€“ 0.4 = Â±âˆš0.56 = Â±0.748 (to 3 decimals) Step 5Â Subtract (-0.4) from both sides (in other words, add 0.4): x = Â±0.748 + 0.4 = -0.348 or 1.148 ## Why “Complete the Square”? Why complete the square when we can just use theÂ Quadratic FormulaÂ to solve a Quadratic Equation? Well, one reason is given above, where the new form not only shows us the vertex, but makes it easier to solve. There are also times when the formÂ ax2Â + bx + cÂ may be part of aÂ largerÂ question and rearranging it asÂ a(x+d)2Â +Â eÂ makes the solution easier, becauseÂ xÂ only appears once. For example “x” may itself be a function (likeÂ cos(z)) and rearranging it may open up a path to a better solution. Also Completing the Square is the first step in theÂ Derivation of the Quadratic Formula Just think of it as another tool in your mathematics toolbox.
## Geometry Tutors Love making paper airplanes, tangram puzzles, and origami? Geometry is the math behind them all. Students begin learning about geometry when they are taught different shapes and angles and how to create new objects from them. They are still interested even when it is time to learn how to determine how much water it will take to fill the bucket or paint they will need to paint the house, but when students get into theorems and postulates, the fun with geometry starts to fade. Terms like hyperbolic, conical, and elliptical send many student straight to panic mode and they struggle in their classes. Getting the help of a geometry tutor can provide the student with an understanding they may miss in the classroom. What is it? Euclidean/Plane Geometry is the study of flat space. Between every pair of points there is a unique line segment which is the shortest curve between those two points. These line segments can be extended to lines. Lines are infinitely long in both directions and for every pair of points on the line the segment of the line between them is the shortest curve that can be drawn between them. All of these ideas can be described by drawing on a flat piece of paper. From the laws of Euclidean Geometry, we get the famous Pythagorean Theorem. Non-Euclidean Geometry is any geometry that is different from Euclidean geometry. It is a consistent system of definitions, assumptions, and proofs that describe such objects as points, lines and planes. The two most common non-Euclidean geometries are spherical geometry and hyperbolic geometry. The essential difference between Euclidean geometry and these two non-Euclidean geometries is the nature of parallel lines: In Euclidean geometry, given a point and a line, there is exactly one line through the point that is in the same plane as the given line and never intersects it. In spherical geometry there are no such lines. In hyperbolic geometry there are at least two distinct lines that pass through the point and are parallel to (in the same plane as and do not intersect) the given line. Riemannian Geometry is the study of curved surfaces and higher dimensional spaces. For example, you might have a cylinder, or a sphere and your goal is to find the shortest curve between any pair of points on such a curved surface, also known as a minimal geodesic. Or you may look at the universe as a three dimensional space and attempt to find the distance between/around several planets. Practice. Practice. Practice. For most students success in any math course comes from regular studying and practicing habits. However, we find that geometry is the one math class can be a foreign language for many students. No matter the level of the geometry class that the student is taking, we have expert tutors available and ready to help. All of our geometry tutors have a degree in mathematics, science, or a related field like engineering. Our goal is to provide a geometry tutor that can make understanding the concepts simple and straightforward and to help the student achieve success in the classroom. ## College Tutors College coursework is challenging. Don’t struggle alone or waste your time reviewing with classmates who don’t know any more than you do. Our experienced tutors understand college course content and precisely how college professors evaluate student progress. Our tutors will help you focus on weak areas, channel your studying energy, and help you prioritize how to spend your studying time. ## Jacksonville Tutors Because Jacksonville is the third most populous city on the East Coast, after New York City and Philadelphia, we have come to appreciate the needs of all the students within Duval County. Here, we provide not only our services to private individuals, but we work hand-in-hand with the local school board to provide free services to those children most in need. Jacksonville’s diverse population of graduate students, high-tech and industrial workers, retirees, and of course an abundant number of teachers provides us with an almost unlimited number of highly qualified and experienced tutors to meet the needs of all learners. We have been providing tutoring services to Jacksonville students for many, many successful years. Our reputation as a premium service is evident in the hundreds of testimonials we have received from parents, students, and schools in Jacksonville and the surrounding communities. ## Our Tutoring Service We offer our clients choice when searching for a tutor, and we work with you all the way through the selection process. When you choose to work with one of our tutors, expect quality, professionalism, and experience. We will never offer you a tutor that is not qualified in the specific subject area you request. We will provide you with the degrees, credentials, and certifications each selected tutor holds so that you have the same confidence in them that we do. And for your peace of mind, we conduct a nation-wide criminal background check, sexual predator check and social security verification on every single tutor we offer you. We will find you the right tutor so that you can find success! ## Maria M ### Teaching Style I love tutoring and consider myself to be effective at it. I approach it with enthusiasm. I apply theory to practical applications in my career and life. My approach is one of ease. I consider mathematics to be easy, as long as you accept what I call "the rules of the game"; i.e., there are certain principles that have to be accepted, and not questioned, then everything else falls in place. I believe that all people have the capability of learning, and I love the opportunity to provide a positive experience to students. ### Experience Summary I have been tutoring for more than 5 years. My main focus is mathematics, but I also tutor students in Spanish I through IV. I have tutored Spanish, Algebra I, Algebra II, Geometry, Trigonometry, and Pre-Calculus. I have recently helped several students prepare for the SAT math test. I enjoy making a difference in the student’s life. ### Credentials Type Subject Issued-By Level Year Other Spanish Native speaker Fluent Current Certification Project Management Studies Project Management Institute PMP Certification 2006 Degree Civil Engineering California State University at Long Beach MSCE 1987 Other Mathematics El Camino College AA 1981 ## Janet J ### Teaching Style I enjoy teaching; it is my best talent. I understand that math is not a "favorite subject" to a lot of students. Working one on one in a tutoring situation gets past the negative block students have in a math classroom. Several of my private students have worked with me for several years. Their parents said that they felt more confident in the class room and in testing situations because of their work with me. In addition to tutoring in traditional school math subjects, I have also worked with several students on PSAT and SAT preparation. All of these students achieved a higher score on these tests after working with me one on one. ### Experience Summary I taught 34 years with the Guilford County School System, in Greensboro, NC. I retired in June 2007, and was asked to fill 2 interim positions in the Spring and Fall semesters of 2008. I taught all courses from general math through Honors Pre-calculus. I have been privately tutoring at home for the last 7 or 8 years, to supplement my income. ### Credentials Type Subject Issued-By Level Year Degree Secondary Mathematics University of NC-Greensboro BS 1972 ## Timothy T ### Teaching Style When I tutor a student, I seek first to understand the student and how he/she thinks. I find it is very important to have good rapport and communication with the student so I understand how he/she views the subject and the difficulties of it. Next I try to make "conceptual bridges" from what they know to what they are having difficulty understanding. This process usually teaches me about seeing the subject from a new point of view. I try to achieve a fine balance between guiding and directing the student’s thoughts on the topic with following the student in their own line of thinking of the subject. The student needs to learn to have confidence in his own thoughts on the subject and in his own ability to master it. ### Experience Summary Over the past four years, I have tutored high school and middle school students in math, algebra, calculus, chemistry, SAT Math, and general study skills. My preference is to tutor math, algebra, calculus, physics, physical science, chemistry, and programming. Math is the subject for which I have the greatest passion. I also participate in the homeschooling of four of my children (13, 11, 8, 6). I have mentored my 13 yr old son in Algebra I & II, Chemistry, Elementary Math, and Middle-school Physical Science, and taught elementary math to my 11, 8, and 6 year olds. Additionally, I read and review history lessons to my kids. I completed my MS in Electrical Engineering in 2006 from The University of Texas at Arlington and my BS in Electrical Engineering and a BA in Philosophy from Rice University. I have recent experience as a student having completed Cellular Biology II at St. Petersburg College in Fall 2011. ### Credentials Type Subject Issued-By Level Year Degree Electrical Engineering Univ. of Texas - Arlington Masters 2006 Certification Design for Six Sigma Honeywell International DFSS - Green Belt 2003 Degree Electrical Engineering Rice University BSEE 1989 Degree Philosophy Rice University BA 1989 ## Robert R ### Teaching Style I’ve always been interested in the application of math and science to the solution of real world problems. This led me to a very satisfying career in engineering. Therefore my approach to teaching is very application oriented. I like to relate the subject to problems that the students will encounter in real life situations. I've generally only worked with older students; high school or college age or older mature adults who have returned to school to get advance training or learn a new trade. ### Experience Summary I’ve always been interested in math and science; especially in their application to solving real world problems. This led me to a very satisfying career in engineering. I have a BS in electrical engineering from General Motors Institute (now Kettering University) and an MS in electrical engineering from Marquette University. I am a registered professional engineer in Illinois. I have over 30 years of experience in the application, development, and sales/field support of electrical/electronic controls for industrial, aerospace, and automotive applications. I’m currently doing consulting work at Hamilton-Sundstrand, Delta Power Company, and MTE Hydraulics in Rockford. I also have college teaching and industrial training experience. I have taught several courses at Rock Valley College in Electronic Technology, mathematics, and in the Continuing Education area. I’ve done industrial technical training for Sundstrand, Barber Colman, and others. I’ve also taught math courses at Rasmussen College and Ellis College (online course). I’ve also been certified as an adjunct instructor for Embry-Riddle Aeronautical University for math and physics courses. I've tutored my own sons in home study programs. I'm currently tutoring a home schooled student in math using Saxon Math. I hope to do more teaching/tutoring in the future as I transition into retirement. ### Credentials Type Subject Issued-By Level Year Degree Electrical Engineering Marquette University MS 1971 Degree Electrical Engineering GMI (Kettereing University) BS 1971 ## Boris B ### Teaching Style I believe that excellence in teaching comes from the teacher's adaptation to student's specific needs. In particular, for some students, visualization may be the essential component for them to understand a certain concept or an idea in mathematics, and by furnishing examples and various proofs with pictures enables the student to learn the concept, whereas for other students it may be the algebraic equation that allows them to see a certain idea. In the first couple of sessions, I probe for the specific needs of the student and then am able to connect with that student so that he/she feels comfortable with the subject. I am a patient teacher and believe that all students are able to grasp the subject. I teach in a disciplined manner, so that the topic presented is coherent and follows a logical flow. I make sure that the theoretical concepts are internalized in a concrete example for the student. Above all, I carry a positive disposition wherever I go and encourage students to enjoy math. ### Experience Summary I am a graduate of GaTech, with a Math and Psychology BA degrees. Aside from my two majors I have minors in Philosophy and Cognitive Science. During my years of high school and college, I have tutored students in Mathematics -- be it in calculus or statistics, or the math portion of the SAT. At the moment I work part-time at the Korean after school program, called Daekyo America Inc., as a math instructor for both high school and middle school. I have participated in various Mathematical Competitions, and have won numerous awards, including the Grand Prize Winner in USAMTS (United States Mathematical Talent Search). ### Credentials Type Subject Issued-By Level Year Degree Applied Mathematics GaTech Bachelors 2004 Degree Psychology GaTech BA 2004 ## Justin J ### Teaching Style I am a very enthusiastic tutor and, as I stated in the previous section, believe teaching in such a way that the student gains a true mastery of the given subject. In a sense, I believe in each student understanding a particular concept in their own unique way that is consistent with their unique thinking processes. I strive to do this by relating a given concept analogously to something the student already fundamentally understands. I also create new problems to solve that probe the students’ progress and comprehension. I try to make these problems as realistic as possible to make them interesting to the student. ### Experience Summary Being a Ph.D. candidate and appointed research assistant, I very much realize the importance of true comprehension of a subject. Not only do I hold two B.S. degrees in Applied Mathematics and Chemical Engineering but I also carry credits to the equivalent of a Non-Thesis Masters in Applied Mathematics and am one course shy of a minor in Chemistry. I have tutored groups of students for Sylvan Learning Centers, conducted one-on-one sessions with Educational Enrichment and independently, and prepared and delivered lectures for advanced undergraduate chemical engineering courses. I am capable and experienced in tutoring all levels of mathematics, chemistry, chemical engineering, physics and materials science. ### Credentials Type Subject Issued-By Level Year Degree Chemical Engineering University of Florida Ph.D.--in progress 2010 Degree Chemical Engineering NC State University B.S. with Honors 2005 Degree Applied Mathematics NC State University B.S. with Honors 2005 ## Farrah F ### Teaching Style I enjoy teaching and working with students. I now understand that all children learn on their own level and at their own pace. I am very easy going, but I insist that students do their homework and stay after school for extra help if they need it. I grade work off of understanding. It is not always finding the correct answer, but the process to get to the correct answer. I believe that a good teacher makes a good student and I strive to be that good teacher and student. ### Experience Summary I enjoyed tutoring while I was in college working in the math lab for three years. However, now that I have been actually teaching my very own students I have a better understanding of the basics of mathematics and I am a much better student myself. ### Credentials Type Subject Issued-By Level Year Certification Mathematics 6-12 Florida Department of Education Leon County 2005 Degree Mathematics Florida A&M University BS 2004 ## Vinod V ### Teaching Style The cornerstone of my teaching philosophy and personal teaching goals is to help students develop their own thinking skills. I believe all students should leave the school armed with the ability to think for them selves, to think critically and to think creatively. Understanding how people learn is one of the significant aspects of teaching. This is linked to their “knowledge” background and maturity. The key to teaching is to relate to the audience by starting from what they know and building upon it. As a teacher I am totally involved with the class, dedicated to my students and 100% prepared to devote time and energy for their intellectual growth. Love for teaching evokes passion and dedication within me. I believe that the enthusiasm of a motivated teacher rubs off on his/her students, who derive the inspiration and encouragement which actuates their desire to learn. A good teacher should have sound fundamentals and command over the concepts. Fundamentals are the foundation intrinsic for mastering the subject; only teachers who are strong in fundamentals will be able to pass it on to their students. I believe that my strong command over the fundamentals will rub off on my students. I believe that the role of a teacher is that of a leader where you have to show the path, motivate, encourage, and lead by example. In short, my success lies in seeing my students succeed. ### Experience Summary My enthusiasm and love for education can be gauged from the fact that I pursued three Masters degrees in three distinct but related fields. One cannot pursue engineering as a profession without having an affinity for Math and Analysis. Math was a passion for me from my young days and still very much remains so. I have a thorough knowledge and understanding of math. Right from my school days I was involved and loved to teach math. I invariably obtained A+ scores in whatever math test I took in my lifetime. For instance my GRE math score was above 95% of test takers' scores. I have taught Middle school, High school and under-graduate students in Algebra, Geometry, Trigonometry, Quadratic Equations, Applied Probability and Calculus. ### Credentials Type Subject Issued-By Level Year Degree City Planning Kansas State University MRCP 2002 Degree Engineering Anna University ME 2000 Degree Civil Engineering Institution of Engineers BE 1994 ## Jill C. Orlando, FL Caitlin made the honor roll for the first time ever this past quarter and I got to go to her first ceremony on Tuesday! She was so proud so I want to say thank you. ## Robyn E. Tampa, FL Tutoring with Arthur is going great. Sarah likes his teaching style.
## What are extrema of functions? An extremum (plural extrema) is a point of a function at which it has the highest (maximum) or lowest (minimum) value. A global maximum or minimum is the highest or lowest value of the entire function, whereas a local maximum or minimum is the highest or lowest value in its neighbourhood. Extrema can be found where the function changes from rising to falling or vice versa (see monotonicity). Think about it this way: if you are going up a hill and want to find its highest point, it would be right before the hill begins to decline again. In particular, the slope of the hill is zero at this point. If it wasn’t zero, it would mean that the your path is still going up. Hence, we need to find all points of a function at which its slope is zero. We use the first derivative for this. To find the points where the slope is zero, we need to find the roots of the derivative. The roots of the derivative can potentially be extrema but not necessarily. Think of the hill again. When the slope becomes zero and you are walking neither up nor down, we would only be at a maximum if the hill started falling from here. However, what if the hill starts to go up again? In this case, we don’t have a maximum. Finding the roots of the derivative is just the first step. After that, we need to decide if we have a maximum, minimum or neither. ## Finding potential extrema Consider the following function and its first derivative: (1) We now find the roots of the derivative by setting it zero and solving for : (2) We have potential extrema at and but we still need to determine whether they are minima or maxima (or neither). We don’t have either if the function is rising (or falling) on the left and rising (or falling) on the right of the potential extremum (so when there is no change). We have a maximum when it is rising first and then falling, and a minimum when it changed from falling to rising. We use the same method that we used to find the function’s monotonicity. We can either compare the sign of the value of the first derivative on the left and right side of the point in question. Or we can use the second derivative and check the sign of it at the point in question. If it is positive then we have a minimum, and a maximum if it is negative (and indeed neither if it is zero). The second derivative and its values at and : (3) We see that , meaning we have a maximum at , and meaning we have a minimum at . The following graph illustrates this. The function is shown in red and we can see the maximum at 1 and the minimum at 4. The first derivative is drawn in purple and we observe that it crosses the x-axis at those points. Finally, the second derivative is shown in orange and it has a negative value where we can see the maximum of and a positive value where we see the minimum of . ### No extremum As an example where the root of the first derivative is not an extremum, consider the following function and its derivatives: (4) The following graph shows us that, although the derivative (purple) has a root at , it is not an extremum of the function (red). This is can be shown with the second derivative (orange) because it is zero at this point.
## Prealgebra Problems This page shows you the types of questions that are covered on the prealgebra section of the Compass mathematics test. You will find practice prealgebra problems in each section on this page. The solutions to the prealgebra problems are provided below the questions in each section. ## Calculations with Integers Integers are real whole numbers. A question on this part of the test might look like the one below. Problem: 8 × 5 + 42 ÷ 7 = ? Solution: For problems like this, you have to divide and multiply first, before you do the addition and subtraction. 8 × 5 = 40 42 ÷ 7 = 6 Then add these together to get the final result. 40 + 6 = 46 ## Calculations with Positive Integer Exponents "Positive integer exponents" means problems that contain base numbers and exponents. Problem: What is 53? The base number is the big number at the bottom and the exponent is the smaller number at the top right. In this problem, 5 is the base number and 3 is the exponent. Solution: When you see an exponent, you have to multiply the base number by itself for the amount of times indicated in the exponent. In other words, you have to multiply 5 three times. So 53 = 5 × 5 × 5 5 × 5 × 5 = 25 × 5 = 125 ## Square Roots and Scientific Notation Square root problems on this part of the text can look something like this one. Problem: Simplify the following. + Solution: In order to solve square root problems like this, you need to multiply the square roots for the amount of times they are shown. In the present problem, the square root of 7 is shown two times. + = 2 × ## Calculations with Fractions Problems on this part of the prealgebra test cover skills like finding the lowest common denominator, simplifying fractions and mixed numbers, and multiplying and dividing fractions. Problem: A job is shared by 4 employees, W, X, Y, and Z. Employee W works 1/6 of the total hours. Employee X works 1/3 of the total hours. Employee Y works 1/2 of the total hours. What fraction of the remaining hours will employee Z have to work? Solution: First of all, you have to find the lowest common denominator, also known as the LCD. Finding the LCD means that you have to make all of the numbers on the bottoms of the fractions the same. The largest existing denominator in this question is 6, so 6 is the lowest common denominator. The other denominators are 3 (from 1/3 for Employee X) and 2 (from 1/2 for Employee Y). So, the LCD for Employee X is 3 × 2 = 6 For Employee Y it's 2 × 3 = 6 The sum of the work from all four people must be equal to 100%, simplified to 1. W + X + Y + Z = 1 1/6 + 1/3 + 1/2 + Z = 1 Now, convert the fraction to the LCD for Employee X. 1/3 × 2/2 = 2/6 Next, find the new fraction for Employee Y. 1/2 × 3/3 = 3/6 1/6 + 2/6 + 3/6 + Z = 1 6/6 + Z = 1 1 + Z = 1 Z = 0 In other words, there are no hours left for Employee Z to work. ## Operations with Decimals This area includes skills such as converting decimals to fractions and converting fractions to decimals. Here is a sample problem. Problem: A group of volunteers are collecting food supplies for a project. They need 120 pounds of foodstuff in total for the project. Volunteer A has collected 273/4 pounds of supplies. Volunteer B has collected 321/5 pounds. Volunteer C has collected 17.35 pounds. What is remaining amount of food supplies that needs to be collected to complete the project? Provide your answer as a whole number with decimals. Solution: Convert the fractions to decimals and then add all three amounts together to find the total so far. Volunteer A: 273/4 = 27.75 Volunteer B: 321/5 = 32.20 27.75 + 32.20 + 17.35 = 77.30 So far, they have collected 77.30 pounds of foodstuff. Finally, you have to subtract this result from the total amount of 120 for the project in order to find out how many pounds they still need. 120 − 77.30 = 42.70 ## Calculating Percentages, Ratios, and Proportions These types of problems often involve fractions. For example: Problem: Find the value of x that solves the following proportion. 12/8 = x/16 Solution: Simplify the first fraction. 12/8 ÷ 4/4 = 3/2 Then divide the denominator of the second fraction (which is x/16) by the denominator of the simplified first fraction (which is 3/2). 16 ÷ 2 = 8 Now, multiply this result by the numerator of the new first fraction (which is 3/2) to get your result. 8 × 3 = 24 So, 12/8 = 24/16 ## Finding the Averages You will need to find averages from the results or data provided in the problem. Problem: One hundred students took an math exam. The 60 girls in the class had an average score of 80, and the 40 boys in the class had an average of 85. What is the average test score for all of the students? Solution: Multiply the average for the girls by the number of girls. Then multiply the average for the boys by the number of boys. Then add these two results for the total points for the whole class. Total points for the girls: 60 × 80 = 4800 Total points for the boys: 40 × 85 = 3400 Total points for entire class: 4800 + 3400 = 8200 When you have got the total points for the class, you divide this by the total number of students to get the class average. Total number of students: 60 + 40 = 100 8200 ÷ 100 = 82 Now, go on the intermediate algebra problems.
SalasSV_10_02_ex # SalasSV_10_02_ex - 10.2 SEQUENCES OF REAL NUMBERS PROOF 593... This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: 10.2 SEQUENCES OF REAL NUMBERS PROOF 593 We will work with the function f (x ) = x . ex (See Example 2, Section 7.4.) Note that f (1) = 1/e = a1 , f (2) = 2/e2 = a2 , f (3) = 3/e3 = a3 , and so on. Differentiating f , we get f (x ) = ex − x ex 1−x = . 2x e ex Since f (x) < 0 for x > 1, f decreases on [1, ∞). Thus f (1) > f (2) > f (3) > · · · , that is, a1 > a2 > a3 > · · · . The sequence is decreasing. The ﬁrst term a1 = 1/e is the least upper bound of the sequence. Since all the terms of the sequence are positive, 0 is a lower bound for the sequence. In Figure 10.2.3 we have sketched the graph of f (x) = x/ex and marked some points (n, an ). As the ﬁgure suggests, 0 is the greatest lower bound of the sequence. y l/e (1, a1) y= x ex (2, a2) (3, a3) 1 2 3 (4, a4) 4 5 x Figure 10.2.3 Example 5 The sequence an = n1/n decreases for n ≥ 3. PROOF We could compare an with an+1 directly, but it is easier to consider the function f (x) = x1/x instead. Since f (x) = e(1/x)ln x , we have f (x) = e(1/x) ln x d dx 1 ln x = x1/x x 1 − ln x . x2 For x > e, f (x) < 0. This shows that f decreases on [e, ∞). Since 3 > e, the function f decreases on [3, ∞), and the sequence decreases for n ≥ 3. EXERCISES 10.2 The ﬁrst several terms of a sequence {an } are given. Assume that the pattern continues as indicated and ﬁnd an explicit formula for an . 1. 2, 5, 8, 11, 14, . . . 3. 1, − 1 , 1 , − 1 , 1 , . . . 35 79 5. 2, 7. 1, 5 10 17 26 , , , ,... 2345 1 , 3, 1 , 5, 1 , . . . 2 4 6 Determine the boundedness and monotonicity of the sequence with an as indicated. 2 . n n + ( − 1)n 11. . n 9. 13. (0. 9)n . 10. ( − 1)n . n 2. 2, 0, 2, 0, 2, . . . 4. 1 , 3 , 7 , 15 , 31 , . . . 2 4 8 16 32 6. 8. 34 5 − 1 , 2 , − 16 , 25 , − 36 , . . . 49 1 1 1, 2, 1 , 4, 25 , 6, 49 , . . . 9 12. (1. 001)n . 14. n−1 . n 594 15. 17. 19. 21. 23. 25. CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS n2 . n+1 4n . √ 4n2 + 1 4n . n + 100 2 2n ln . n+1 (n + 1)2 . n2 1 4− . n 16. 18. n2 + 1. 53. a1 = 1; 54. a1 = 3; 55. a1 = 1, 56. a1 = 1, an+1 = an + · · · + a1 . an+1 = 4 − an . a2 = 3; a2 = 3; an+1 = 2an − an−1 , an+1 = 3an − 2n − 1, n ≥ 2. n ≥ 2. 2n . 4n + 1 n2 20. √ . n3 + 1 n+2 22. 10 √ . 3 n √ 24. (−1)n n. n+1 26. ln . n √ n+1 28. √ . n 1 1 30. − . 2n 2n + 3 32. (− 1 )n . 2 n+3 34. . ln (n + 3) 36. cos nπ . 38. 40. (−2)n . n10 1 − ( 1 )n 2 In Exercises 57–60, use mathematical induction to prove the following assertions for all n ≥ 1. 57. If a1 = 1 and an+1 = 2an + 1, then an = 2n − 1. 58. If a1 = 3 and an+1 = an + 5, then an = 5n − 2. n+1 n an , then an = n−1 . 59. If a1 = 1 and an+1 = 2n 2 1 1 60. If a1 = 1 and an+1 = an − , then an = . n(n + 1) n 61. Let r be a real number, r = 0. Deﬁne a sequence {Sn } by S1 = 1 S2 = 1 + r S3 = 1 + r + r 2 · · · Sn = 1 + r + r 2 + · · · + r n−1 · · · (a) Suppose r = 1. What is Sn for n = 1, 2, 3, . . .? (b) Suppose r = 1. Find a formula for Sn that does not involve adding up the powers of r . HINT: Calculate Sn − rSn . 1 62. Set an = , n = 1, 2, 3, . . ., and form the sequence n(n + 1) S 1 = a1 S 2 = a1 + a 2 S 3 = a1 + a 2 + a 3 · · · S n = a1 + a 2 + a 3 + · · · + a n · · · Find a formula for Sn , n = 1 , 2 , 3, . . . , that does not involve adding up the terms a1 , a2 , a3 , . . .. HINT: Use partial fractions to write 1/[k (k + 1)] as the sum of two fractions. 63. A ball is dropped from a height of 100 feet. Each time it hits the ground, it rebounds to 75% of its previous height. (a) Let Sn be the distance that the ball travels between the nth and (n + 1)st bounce, n = 1, 2, 3, . . . . Find a formula for Sn . √ 27. ( − 1)2n+1 n. 29. 2n − 1 . 2n π 31. sin . n+1 33. (1. 2)−n . 35. 1 1 − . n n+1 ln (n + 2) 37. . n+2 3n . 39. (n + 1)2 . ( 1 )n 2 41. Show that the sequence an = 5n /n! decreases for n ≥ 5. Is the sequence nonincreasing? 42. Let M be a positive integer. Show that an = M n /n! decreases for n ≥ M . 43. Show that, if 0 < c < d , then the sequence an = (cn + d n )1/n is bounded and monotonic. 44. Show that linear combinations and products of bounded sequences are bounded. Sequences can be deﬁned recursively: one or more terms are given explicitly; the remaining ones are then deﬁned in terms of their predecessors. In Exercises 45–56, give the ﬁrst six terms of the sequence and then give the nth term. 45. a1 = 1; 46. a1 = 1; 47. a1 = 1; 48. a1 = 1; 49. a1 = 1; 50. a1 = 1; 51. a1 = 1; 52. a1 = 1; an+1 = an+1 1 an . n+1 = an + 3n(n + 1) + 1. 1 2 an+1 = 1 (an + 1). 2 an+1 = an + 1. an+1 = an + 2. n an+1 = an . n+1 an+1 = an + 2n + 1. an+1 = 2an + 1. 10.3 LIMIT OF A SEQUENCE 595 (b) Let Tn be the time that the ball is in the air between the nth and (n + 1)st bounce, n = 1, 2, 3, . . . . Find a formula for Tn . 64. Suppose that the number of bacteria in a culture is growing exponentially (see Section 7.6) and that the number doubles every 12 hours. Find a formula for the number Pn of bacteria in the culture after n hours, given that there are 500 bacteria initially. c 68. Let {an } be the sequence deﬁned recursively by setting a1 = 1; an+1 = 3a n , n = 1, 2, 3, . . . . (a) Show by induction that {an } is an increasing sequence. (b) Show by induction that {an } is bounded above. (c) Calculate a2 , a3 , a4 , . . . , a15 . Estimate the least upper bound of the sequence. c 65. Let {an } be the sequence deﬁned by setting an = √ n2 + n − n, n = 1, 2, 3, . . . . Use a CAS to determine whether {an } is increasing, nondecreasing, decreasing, nonincreasing, or none of these. c 66. Repeat Exercise 65 with the sequence {an } deﬁned recursively by setting a1 = 100; an+1 = 2 + an , n = 1, 2, 3, . . . . c 69. Let pn denote the nth prime number (see Exercise 34, Section 10.1). Deﬁne the sequence {an } by setting an = pn , n ln n n = 2, 3, 4, . . . . (a) Use a CAS to investigate an for large n. (b) Is {an } a bounded sequence? (c) Does it appear that lim an exists? If so, what is the limit? c 70. Let {an } be the sequence deﬁned recursively by setting a1 = 1, a2 = 3, and an+2 = 1 + an+1 , n = 1, 2, 3, . . . . an n→∞ c 67. Let {an } be the sequence deﬁned recursively by setting a1 = 1; an+1 = 1 + √ an , n = 1, 2, 3, . . . . (a) Show by induction that {an } is an increasing sequence. (b) Show by induction that {an } is bounded above. (c) Calculate a2 , a3 , a4 , . . . , a15 . Estimate the least upper bound of the sequence. (a) Use a CAS to investigate the behavior of the sequence. Is the sequence monotone? Is it bounded? If so, give the least upper bound and the greatest lower bound. (b) Experiment with other “initial” values a1 and a2 . 10.3 LIMIT OF A SEQUENCE You have seen the limit process applied in various settings. The limit process applied to sequences is exactly what you would expect. DEFINITION 10.3.1 LIMIT OF A SEQUENCE n→∞ lim an = L if for each > 0, there exists a positive integer K such that if n ≥ K, then \|an − L\| < . Example 1 PROOF Let 4n − 1 = 4. n→∞ n > 0. We must show that there exists an integer K such that lim if n ≥ K, then 4n − 1 −4 < . n Note that 4n − 1 1 4n − 1 − 4n −1 −4 = = =. n n n n ... View Full Document Ask a homework question - tutors are online
# 120 cm to Feet: Converting Centimeters to Feet Made Easy Converting measurements from one unit to another is a common task in various fields, including construction, engineering, and everyday life. One such conversion that often arises is converting centimeters to feet. In this article, we will explore the process of converting 120 centimeters to feet, providing valuable insights and examples along the way. ## The Basics: Understanding Centimeters and Feet Before diving into the conversion process, let’s first understand the units involved. ### Centimeters (cm) Centimeters are a unit of length in the metric system. They are commonly used to measure smaller distances, such as the height of a person or the length of an object. One centimeter is equal to one-hundredth of a meter. ### Feet (ft) Feet, on the other hand, are a unit of length in the imperial system. They are widely used in the United States and a few other countries. One foot is equal to 12 inches or approximately 30.48 centimeters. ## The Conversion Process: Converting 120 cm to Feet Now that we have a clear understanding of the units involved, let’s proceed with the conversion process. ### Step 1: Determine the Conversion Factor To convert centimeters to feet, we need to establish the conversion factor between the two units. As mentioned earlier, one foot is approximately 30.48 centimeters. This conversion factor will be crucial in our calculations. ### Step 2: Set Up the Conversion Equation With the conversion factor in hand, we can now set up the conversion equation. The equation for converting centimeters to feet is as follows: Feet = Centimeters / Conversion Factor Let’s substitute the values into the equation to convert 120 centimeters to feet: Feet = 120 cm / 30.48 cm/ft ### Step 3: Perform the Calculation Now, let’s perform the calculation to determine the equivalent value in feet: Feet = 120 / 30.48 Feet ≈ 3.937 Therefore, 120 centimeters is approximately equal to 3.937 feet. ## Real-World Examples Understanding the conversion process is essential, but it becomes even more valuable when applied to real-world examples. Let’s explore a few scenarios where converting 120 centimeters to feet can be useful. ### Example 1: Height Conversion Imagine you are planning a trip to the United States, and you want to know your height in feet. If you are 120 centimeters tall, you can use the conversion process to determine your height in feet. By converting 120 centimeters to feet, you would find that you are approximately 3.937 feet tall. This information can be helpful when filling out forms or discussing your height with others who are more familiar with the imperial system. ### Example 2: Furniture Dimensions Another practical application of converting centimeters to feet is when dealing with furniture dimensions. Suppose you are purchasing a new bookshelf, and the dimensions are listed in centimeters. By converting the measurements to feet, you can better visualize the size of the bookshelf and determine if it will fit in your desired space. For instance, if the bookshelf is 120 centimeters in height, you can convert it to approximately 3.937 feet. This conversion allows you to compare the height of the bookshelf with the available space in your room, ensuring a proper fit. ## Q&A Here are some common questions related to converting 120 centimeters to feet: 1. ### Q: Is the conversion from centimeters to feet exact? A: No, the conversion from centimeters to feet is not exact due to the difference in the underlying measurement systems. The result is an approximation. 2. ### Q: Can I use an online converter to convert 120 centimeters to feet? A: Yes, there are numerous online converters available that can quickly and accurately convert 120 centimeters to feet. Simply input the value and select the desired conversion, and the converter will provide the result. 3. ### Q: Why do some countries use centimeters while others use feet? A: The choice of measurement system varies from country to country and is often influenced by historical, cultural, and practical factors. The metric system, which includes centimeters, is widely used around the world, while the imperial system, which includes feet, is primarily used in the United States and a few other countries. 4. ### Q: Can I convert feet to centimeters using the same conversion factor? A: Yes, the conversion factor can be used to convert both centimeters to feet and feet to centimeters. To convert feet to centimeters, simply divide the number of feet by the conversion factor. 5. ### Q: Are there any other common units for measuring length? A: Yes, there are several other common units for measuring length, including meters, inches, yards, and miles. Each unit has its own conversion factors when converting to and from centimeters or feet. ## Summary In conclusion, converting 120 centimeters to feet is a straightforward process that involves establishing the conversion factor between the two units and performing a simple calculation. Understanding this conversion can be valuable in various real-world scenarios, such as determining height or visualizing furniture dimensions. By following the steps outlined in this article, you can confidently convert centimeters to feet and vice versa, expanding your ability to work with different measurement systems.
We received concise statements of general rules based on clearly presented evidence: Matthew, Holly and Isobel from Staunton & Corse C of E School discovered the relationship between the enlargement scale factor and the effect on the area: We started by writing 5 different dimensions with an area of $20$cm$^2$: Length x Width: 0.25 x 80, 0.5 x 40, 1 x 20, 2 x 10, 4 x 5 Then we enlarged them by a scale factor of 2: 0.5 x 160, 1 x 80, 2 x 40, 4 x 20, 8 x 10 We then took the areas of both rectangles ($20$cm$^2$ and $80$cm$^2$) and figured out a way that 20 could be turned into 80. Well, obviously you multiply it by 4. But if you enlarge them by a scale factor of 3: 0.25 x 80, 0.5 x 40, 1 x 20, 2 x 10, 4 x 5 become: 0.75 x 240, 1.5 x 120, 3 x 60, 6 x 30, 12 x 15 This time the areas are $20$cm$^2$ and $180$cm$^2$ which means that the areas have been multiplied by 9. So the question we asked ourselves was: What does 2 have in common with 4 as 3 has in common with 9? The answer is: you square the scale factor to find out how much bigger the areas have become. Akintunde from Wilson's School generalised for two and three dimensions: A rectangle with area $20$cm$^2$ could have the following dimensions: Length by Width: 4 by 5, 2 by 10, 20 by 1, 40 by 0.5, 50 by 0.4 When enlarged by a scale factor of 2 the dimensions become: 8 by 10, 4 by 20, 40 by 2, 80 by 1, 100 by 0.8 All these rectangles have an area of $80$cm$^2$ The area is four times larger. A rectangle with an area of $10$cm$^2$ could have the following dimensions: 5 by 2, 1 by 10, 20 by 0.5 When enlarged by a scale factor of 2 the dimensions become: 10 by 4, 2 by 20, 40 by 1 All these rectangles have an area of $40$cm$^2$ Again, the area is four times larger than the original. I think that maybe whatever scale factor you enlarge the rectangle by, the area is enlarged by the square of the scale factor: SF 2, AREA increases by 4 SF 3, AREA increases by 9 Starting with the rectangles with an area of $10$cm$^2$: 5 by 2, 1 by 10, 20 by 0.5 When they are enlarged by a scale factor of 3, the dimensions become: 15 by 6, 3 by 30, 60 by 1.5 All these rectangles have an area of $90$cm$^2$ The areas are nine times larger. And when the rectangles with an area of $10$cm$^2$ are enlarged by a scale factor of 4 the dimensions become: 20 by 8, 4 by 40, 80 by 2 All these rectangles have an area of $160$cm$^2$ The areas are sixteen times larger. And when the rectangles with an area of $10$cm$^2$ are enlarged by a scale factor of 0.5 the dimensions become: 2.5 by 1, 0.5 by 5, 10 by 0.25 All these rectangles have an area of $2.5$cm$^2$ The areas are four times smaller (you multiply the original area by 0.25). I think that for any scale factor, you square it and then multiply that new number by the original area. e.g. SF$2$ and an area of $20$cm$^2$: 2$^2$ is 4 and $20$cm$^2$ x 4 is $80$cm$^2$ I think this happens because the two dimensions of a rectangle are both multiplied by three when the rectangle is increased by a Scale Factor of 3 so overall the area of the rectangle is nine-times bigger (3 x 3). If you increase a rectangle by a scale factor k, the area of the new rectangle will be k$^2$ times the old area of the rectangle. Taking a triangle with a base of 2, a height of 3 and an area of 3: When enlarged by a scale factor of 2 it becomes a triangle with a base of 4, a height of 6 and an area of 12. The area is 4 TIMES LARGER. When enlarged by a scale factor of 3 it becomes a triangle with a base of 6, a height of 9 and an area of 27. The area is 9 TIMES LARGER. When enlarged by a scale factor of 4 it becomes a triangle with a base of 8, a height of 12 and an area of 48. The area is 16 TIMES LARGER. I think this rule applies to all plane shapes because when different rectangles and triangles are increased by different scale factors, the increase in area is the same. Taking a cuboid with a width of 2, a length of 2, a height of 3, a surface area of 32 and a volume of 12: When enlarged by a scale factor of 2 it becomes a cuboid with a width of 4, a length of 4, a height of 6, a surface area of 128 and a volume of 96. The surface area is 4 TIMES LARGER and the volume is 8 TIMES LARGER. The surface area increased by 4 (2$^2$) and the volume increased by 8 (2$^3$). The volume takes in 3 dimensions and the surface area takes in 2 dimensions. I have concluded that with rectangles and cuboids, whatever the scale factor is: you square the scale factor and then multiply that number by the old area to find the new area, and you cube the scale factor and times that number by the old volume to find the new volume. Based on the fact that the rule about scale factors seems to work for all plane shapes, I predict that 3D shapes will follow the same rule as for cuboids. Nice work - well done to you all.
Courses Courses for Kids Free study material Offline Centres More Store # A steel rail of length $5\,m$ and area of cross section $40\,c{m^2}$ is prevented from expanding along its length while the temperature rises by $10\,{}^ \circ C$. If coefficient of linear expansion and young’s modulus of steel are $1.2 \times {10^{ - 5}}\,{K^{ - 1}}$ and $2 \times {10^{11}}\,N{m^{ - 2}}$ respectively, the force developed in the rail is approximately: A) $2 \times {10^7}\,N$B) $3 \times {10^{ - 5}}\,N$C) $1 \times {10^5}\,N$D) $2 \times {10^9}\,N$ Last updated date: 14th Sep 2024 Total views: 79.8k Views today: 1.79k Verified 79.8k+ views Hint: The force developed in the rail can be determined by using the young’s modulus formula, the young’s modulus is the ratio of the stress and the strain. By using the stress and the strain formula in the young’s modulus formula, the force can be determined. Formula used: The young’s modulus is given by, $Y = \dfrac{\sigma }{\varepsilon }$ Where, $Y$ is the young’s modulus of the material, $\sigma$ is the stress in the material and $\varepsilon$ is the strain in the material. The stress of the material is given by, $\sigma = \dfrac{F}{A}$ Where, $\sigma$ is the stress, $F$ is the force and $A$ is the area. The strain of the material is given by, $\varepsilon = \dfrac{{\Delta l}}{l}$ Where, $\varepsilon$ is the strain, $\Delta l$ is the change in the length and $l$ is the original length. Complete step by step solution: Given that, The length of the rail is, $l = 5\,m$, The Area of the cross section is, $A = 40\,c{m^2} = 40 \times {10^{ - 4}}\,{m^2}$, The change in temperature is, $\Delta T = 10\,{}^ \circ C$, The coefficient of the linear expansion is, $\alpha = 1.2 \times {10^{ - 5}}\,{K^{ - 1}}$, The young’s modulus of the material is, $Y = 2 \times {10^{11}}\,N{m^{ - 2}}$. The relation between the change in length and the change in the temperature is given by, $\Delta l = l \times \alpha \times \Delta T$ By rearranging the terms in the above equation, then $\dfrac{{\Delta l}}{l} = \alpha \times \Delta T\,...............\left( 1 \right)$ Now, The young’s modulus is given by, $Y = \dfrac{\sigma }{\varepsilon }$ By substituting the stress and strain formula in the above equation, then the above equation is written as, $Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}$ By substituting the equation (1) in the above equation, then $Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\alpha \times \Delta T}}$ By rearranging the terms in the above equation, then the above equation is written as, $Y = \dfrac{F}{{A \times \alpha \times \Delta T}}$ By rearranging the terms in the above equation, then the above equation is written as, $F = Y \times A \times \alpha \times \Delta T$ By substituting the young’s modulus, cross sectional area, coefficient of the linear expansion and change in temperature in the above equation, then $F = 2 \times {10^{11}} \times 40 \times {10^{ - 4}} \times 1.2 \times {10^{ - 5}} \times 10$ By multiplying the terms in the above equation, then $F = 9600\,N$ The above equation is also written as, $F = 0.96 \times {10^4}\,N$ Then the force is approximately equal to, $F \simeq 1 \times {10^5}\,N$ Hence, the option (C) is the correct answer. Note: The force of the object is directly proportional to the young’s modulus of the material, cross sectional area of the material, coefficient of the linear expansion and the change in temperature. As the young’s modulus of the material, cross sectional area of the material, coefficient of the linear expansion and the change in temperature increases, then the force also increases.
### I: Students will acquire number sense and perform operations with whole numbers, simple fractions, and decimals. #### I.1: Demonstrate multiple ways to represent whole numbers and decimals, from hundredths to one million, and fractions. I.1.a: Read and write numbers in standard and expanded form. I.1.b: Demonstrate multiple ways to represent whole numbers and decimals by using models and symbolic representations (e.g., 36 is the same as the square of six, three dozen, or 9 x 4). I.1.c: Identify the place and the value of a given digit in a six-digit numeral, including decimals to hundredths, and round to the nearest tenth. I.1.e: Name and write a fraction to represent a portion of a unit whole, length, or set for halves, thirds, fourths, fifths, sixths, eighths, and tenths. #### I.2: Analyze relationships among whole numbers, commonly used fractions, and decimals to hundredths. I.2.a: Compare the relative size of numbers (e.g., 475 is comparable to 500; 475 is small compared to 10,000 but large compared to 98). I.2.b: Order whole numbers up to six digits, simple fractions, and decimals using a variety of methods (e.g., number line, fraction pieces) and use the symbols <, >, and = to record the relationships. I.2.c: Identify a number that is between two given numbers (e.g., 3.2 is between 3 and 4; find a number between 0.1 and 0.2). I.2.d: Identify equivalences between fractions and decimals by connecting models to symbols. I.2.e: Generate equivalent fractions and simplify fractions using models, pictures, and symbols. #### I.3: Model and illustrate meanings of multiplication and division of whole numbers and the addition and subtraction of fractions. I.3.a: Model multiplication (e.g., equal-sized groups, rectangular arrays, area models, equal intervals on the number line), place value, and properties of operations to represent multiplication of a one- or two-digit factor by a two-digit factor and connect the representation to an algorithm. I.3.b: Use rectangular arrays to interpret factoring (e.g., find all rectangular arrays of 36 tiles and relate the dimensions of the arrays to factors of 36). I.3.e: Use models to add and subtract simple fractions where one single-digit denominator is 1, 2, or 3 times the other (e.g., 2/4 + 1/4; 3/4 - 1/8). #### I.4: Solve problems involving multiplication and division of whole numbers and addition and subtraction of simple fractions and decimals. I.4.a: Use estimation, mental math, paper and pencil, and calculators to perform mathematical calculations and identify when to use each one appropriately. I.4.c: Write a story problem that relates to a given multiplication or division equation, and select and write a number sentence to solve a problem related to the environment. I.4.d: Solve problems involving simple fractions and interpret the meaning of the solution (e.g., A pie has been divided into six pieces and one piece is already gone. How much of the whole pie is there when Mary comes in? If Mary takes two pieces, how much of the whole pie has she taken? How much of the pie is left?) #### I.5: Compute problems involving multiplication and division of whole numbers and addition and subtraction of simple fractions and decimals. I.5.a: Demonstrate quick recall of basic multiplication and division facts. I.5.c: Divide up to a three-digit dividend by a one-digit divisor with fluency, using efficient procedures. I.5.d: Add and subtract decimals and simple fractions where one single-digit denominator is 1, 2, or 3 times the other (e.g., 2/4 + 1/4 = 3/4; 1/3 ? 1/6 = 1/6). ### II: Students will use patterns and relations to represent mathematical problems and number relationships. #### II.1: Identify, analyze, and determine rules for describing numerical patterns involving operations and nonnumerical growing patterns. II.1.b: Recognize, represent, and extend simple patterns involving multiples and other number patterns (e.g., square numbers) using objects, pictures, numbers, and tables. II.1.c: Identify simple relationships in real-life contexts and use mathematical operations to describe the pattern (e.g., the number of legs on a given number of chairs may be determined by counting by fours or by multiplying the number of chairs by 4). ### III: Students will understand attributes and properties of plane geometric objects and spatial relationships. #### III.1: Identify and describe attributes of two-dimensional geometric shapes. III.1.d: Identify and describe figures that have line symmetry and rotational symmetry. #### III.2: Specify locations using grids and maps. III.2.a: Locate coordinates in the first quadrant of a coordinate grid. III.2.b: Give the coordinates in the first quadrant of a coordinate grid. #### III.3: Visualize and identify geometric shapes after applying transformations. III.3.a: Identify a translation, rotation, or a reflection of a geometric shape. ### IV: Students will describe relationships among units of measure, use appropriate measurement tools, and use formulas to find area measurements. #### IV.1: Describe relationships among units of measure for length, capacity, and weight, and determine measurements of angles using appropriate tools. IV.1.a: Describe the relative size among metric units of length (i.e., millimeter, centimeter, meter), between metric units of capacity (i.e., milliliter, liter), and between metric units of weight (i.e., gram, kilogram). #### IV.2: Recognize and describe area as a measurable attribute of two-dimensional shapes and calculate area measurements. IV.2.a: Quantify area by finding the total number of same-sized units of area needed to fill the region without gaps or overlaps. IV.2.b: Recognize that a square that is 1 unit on a side is the standard unit for measuring area. IV.2.d: Develop and use the area formula for a right triangle by comparing with the formula for a rectangle (e.g., two of the same right triangles makes a rectangle). ### V: Students will interpret and organize collected data to make predictions, answer questions, and describe basic concepts of probability. #### V.1: Collect, organize, and display data to answer questions. V.1.a: Identify a question that can be answered by collecting data. V.1.b: Collect, read, and interpret data from tables, graphs, charts, surveys, and observations. V.1.c: Represent data using frequency tables, bar graphs, line plots, and stem and leaf plots. V.1.d: Identify and distinguish between clusters and outliers of a data set. Correlation last revised: 5/24/2018 This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information.
Courses Courses for Kids Free study material Offline Centres More Store # How do you find the exact value of the following using the unit circle: $\cos \dfrac{{53\pi }}{6}$? Last updated date: 17th Jun 2024 Total views: 373.5k Views today: 3.73k Verified 373.5k+ views Hint: In the given problem, we are required to find the cosine of a given angle using some simple and basic trigonometric compound angle formulae and trigonometric identities. Such questions require basic knowledge of compound angle formulae and their applications in this type of questions. Unit circle is a circle with a radius of one unit drawn on a graph paper with its centre at origin. Complete step by step solution: Consider a unit circle (a circle of radius of 1 unit centered at origin). We need to find out the value of $\cos \dfrac{{53\pi }}{6}$ using the unit circle. So, we have, $\cos \dfrac{{53\pi }}{6}$$= \cos \left( {\dfrac{{48\pi + 5\pi }}{6}} \right) Separating the numerator into two parts and distributing the denominator underneath both the parts, we get, \Rightarrow \cos \dfrac{{53\pi }}{6}$$ = \cos \left( {8\pi + \dfrac{{5\pi }}{6}} \right)$ Since cosine and sine function are periodic functions with period of $2\pi$, so the value of cosine and sine gets repeated after intervals in multiples of $2\pi$. Hence, we can eliminating the $\left( {8\pi } \right)$ term from the angle, we get, $\Rightarrow \cos \dfrac{{53\pi }}{6}$$= \cos \left( {\dfrac{{5\pi }}{6}} \right)$ Since, the angle $\left( {\dfrac{{5\pi }}{6}} \right)$ lies in the second quadrant and cosine ratio is negative in the second quadrant. So, we get, $\Rightarrow $$\cos \left( {\pi - \dfrac{\pi }{6}} \right) \Rightarrow$$ - \cos \left( {\dfrac{\pi }{6}} \right)$ We know the value of $\cos \left( {\dfrac{\pi }{6}} \right)$ is $\left( {\dfrac{{\sqrt 3 }}{2}} \right)$. Substituting the same, we get, $\Rightarrow$$- \dfrac{{\sqrt 3 }}{2}$ Hence, the value of $\cos \dfrac{{53\pi }}{6}$ is $\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$ Note: Periodic Function is a function that repeats its value after a certain interval. For a real number $T > 0$, $f\left( {x + T} \right) = f\left( x \right)$ for all x. If T is the smallest positive real number such that $f\left( {x + T} \right) = f\left( x \right)$ for all x, then T is called the fundamental period.
# 1.2 Algebraic Expressions 8/24/12. Vocabulary Variable: A letter used to represent one or more numbers Exponent: The number or variable that represents. ## Presentation on theme: "1.2 Algebraic Expressions 8/24/12. Vocabulary Variable: A letter used to represent one or more numbers Exponent: The number or variable that represents."— Presentation transcript: 1.2 Algebraic Expressions 8/24/12 Vocabulary Variable: A letter used to represent one or more numbers Exponent: The number or variable that represents the number of times the base of a power is used as a factor Power: An expression that represents repeated multiplication of the same factor Base: The number used as a factor in a repeated multiplication x3x3 Example 1 Evaluate Powers a. 2424 2 2 2 2 = ()– b. ) 4) 4 ( 2 – ( 2 –)( 2 – ) = ( 2 –)( 2 – ) 16 = – – = What’s the difference? ANSWER 9 64 ANSWER 2 Evaluate the power. Checkpoint 1. 4343 2. ) 2) 2 ( 3 – 3.3 – 4. 2121 ANSWER 27 – Evaluate Powers Order of Operations Step 1: Perform operations that occur within grouping symbols. Step 2: Evaluate powers. Step 3: Perform multiplications and divisions from left to right. Step 4: Perform additions and subtractions from left to right. Example 2 Use Order of Operations 436 ÷ ( 1 – 4+ )2)2 Add within parentheses. = 436 ÷ 3232 Evaluate power. = 436 ÷ 9 Divide. = 9 9 = 81 Multiply. Example 3 Evaluate an Algebraic Expression Evaluate when. () 5+x27x7x – 4x = SOLUTION Substitute 4 for x. = () 5+x27x7x – () 5+42 – ( 47 ) = – ( 47 )( 92 ) Add within parentheses. 2818 = Multiply. – = Subtract. 10 – Evaluate means to get a number answer after doing the math. Example 4 Evaluate an Algebraic Expression Evaluate when. 4p24p2 – ( 2p2p ) 9 –p = 3 – = – ( 94 ) [] ( 23 – ) 9 – Evaluate power. 36 = – ( 15 – ) Subtract within parentheses. 51 = Add the opposite. 36 = Multiply. – [] 6 – 9 – SOLUTION Substitute 3 for p. = ( 4 4p24p2 – ( 2p2p ) 9 – 3 – )2)2 [] ( 23 – ) 9 – – – 5. 738 + – ANSWER 12 ANSWER 23 Evaluate the expression. Checkpoint 6. 2 – 62 + ÷ 4 ANSWER 8 – 7. ( 5734 –– )2)2 ANSWER 15 Evaluate the expression for the given value of the variable. 8. 3 – + ( 1y6+ ) when 2y = Evaluate Expressions Homework 1.2 p.12 #15-60 multiples of 3 Remember: 1.Pencil Only. 2.Write heading. 3.Copy problems and show work. 4.Circle, highlight or box final answer. Download ppt "1.2 Algebraic Expressions 8/24/12. Vocabulary Variable: A letter used to represent one or more numbers Exponent: The number or variable that represents." Similar presentations
# The sum of three consecutive integers is 135. What are the numbers? Jul 4, 2017 $44 , 45 , 46$ #### Explanation: $\text{let the first integer be represented by } n$ $\text{then the second integer will be } n + 1$ $\text{and the third integer } n + 2$ $\Rightarrow n + n + 1 + n + 2 = 135 \leftarrow \textcolor{b l u e}{\text{ sum of integers}}$ $\Rightarrow 3 n + 3 = 135 \leftarrow \textcolor{b l u e}{\text{ simplifying left side}}$ $\text{subtract 3 from both sides}$ $3 n \cancel{+ 3} \cancel{- 3} = 135 - 3$ $\Rightarrow 3 n = 132$ $\text{divide both sides by 3}$ $\frac{\cancel{3} n}{\cancel{3}} = \frac{132}{3}$ $\Rightarrow n = 44$ $\Rightarrow n + 1 = 44 + 1 = 45$ $\Rightarrow n + 2 = 44 + 2 = 46$ $\text{the three consecutive integers are } 44 , 45 , 46$ $\textcolor{b l u e}{\text{As a check}}$ $44 + 45 + 46 = 135 \rightarrow \text{ True}$
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # 1999 AIME Problems/Problem 3 ## Problem Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square. ## Solution 1 If $n^2-19n+99=x^2$ for some positive integer $x$, then rearranging we get $n^2-19n+99-x^2=0$. Now from the quadratic formula, $n=\frac{19\pm \sqrt{4x^2-35}}{2}$ Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$. Rearranging gives $(2x+q)(2x-q)=35$. Thus $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$, giving $x=3$ or $9$. This gives $n=1, 9, 10,$ or $18$, and the sum is $1+9+10+18=\boxed{38}$. ## Solution 2 Suppose there is some $k$ such that $x^2 - 19x + 99 = k^2$. Completing the square, we have that $(x - 19/2)^2 + 99 - (19/2)^2 = k^2$, that is, $(x - 19/2)^2 + 35/4 = k^2$. Multiplying both sides by 4 and rearranging, we see that $(2k)^2 - (2x - 19)^2 = 35$. Thus, $(2k - 2x + 19)(2k + 2x - 19) = 35$. We then proceed as we did in the previous solution. ## Solution 3 When $n \geq 12$, we have $$(n-10)^2 < n^2 -19n + 99 < (n-8)^2.$$ So if $n \geq 12$ and $n^2 -19n + 99$ is a perfect square, then $$n^2 -19n + 99 = (n-9)^2$$ or $n = 18$. For $1 \leq n < 12$, it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$ We conclude that the answer is $1 + 9 + 10 + 18 = \boxed{38}.$ ## Solution 4: Graphing If we graphed $$y = \sqrt{x^2-19x+99}$$ we would see that only four values of x return integer values of y: 10, 9, 1, 18. Thus, the answer if $\boxed{38}$.
# Sarin’s family celebrating Hari Raya Puasa part 23 (Math Question) Standard The blog postings are about the Singapore Math. The readers can learn from the postings about Solving Singapore Primary School Mathematics. The blog presents the Math Concept, the Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can acquire the skills to deal with the Math Modeling, the Math Problem Solving and the Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on AlgebraRatio and Problem Sum. You can read the postings on Sarin’s family celebrating Hari Raya Puasa and Sarin’s family and Ramadan in Welcome page about story of Sarin’s family during Ramadan and Hari Raya Puasa. Read the posting on Sarin learns Symbol, Algebra and Equation in school (Math Concept) to understand the concept of Algebra. To read also the posting on Sarin learns concept of ratio in school to understand the concept of Ratio. Challenge yourself with the question before look out for the given solution !!! Upper primary school mathematics question UPQ573 Fatimah prepared some gifts for celebrating Hari Raya. She used some string to tie 24 small size gifts. As well she used the same amount of string to tie 16 large size gifts. Each large size gift used 70 cm more string than small size gift. How many metres of string did Fatimah used altogether? Solution: Using Algebra concept, The length of string used to tie one small size gift = 1U The length of string used to tie one large size gift = 1U + 70 There were 24 small size gifts, The length of the string to tie all the small size gifts = 24U There were 16 large size gifts, The length of the string to tie all the large size gifts = 16U + 1120 Since, Fatimah used the same amount of string for small and large size gifts, 24U = 16U + 1120 8U = 1120 1U = 140 cm The total amount of string used for both small and large size gifts = 140 × 24 × 2 = 6720 cm = 67.2 m Alternative solution: Same amount of string used for small and large size gifts, The ratio of the string use for a small size gift to a large size gift S : L = 16 : 24 = 2 : 3 The math model, From the model, 1 unit = 70 cm The length of string used to tie one small size gift = 2 × 70 = 140 The length of string used to tie the small size gifts = 24 × 140 = 3360 cm Same amount of string used for large size gifts, The total amount of string used = 2 × 3360 = 6720 cm = 67.2 m Advertisements
# Sum of Arithmetic Series ## Arithmetic Series The series associated with an arithmetic sequence is known as an arithmetic series. For example: 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 is the arithmetic series associated with the arithmetic sequence: 3, 7, 11, 15, 19, 23, 27, 31. ## Sum of Arithmetic Series To derive the formula for the sum of arithmetic series, let us suppose an arithmetic series where a be the first term, d be the common difference, n be the number of terms, l be the last term and Sn be the sum of the n terms of an arithmetic series (AS), then, Sn = a + (a + d) + (a + 2d) + … … … + (l – 2d) + (l – d) + l ………… (i) Writing in the reverse order, we have, Sn = l + (l – d) + (l – 2d) + … … … + (a + 2d) + (a + d) + a ………… (ii) Adding relations (i) and (ii), we get 2Sn = (a + l) + (a + l) + (a + l) + … … … + (a + l) + (a + l) + (a + l). or, 2Sn = n(a + l) Again, the last term, l = a + (n – 1)d Therefore, the sum of the arithmetic series can be found by using the following formulas: ****************** 10 Math Problems officially announces the release of Quick Math Solver and 10 Math ProblemsApps on Google Play Store for students around the world. **************** ****************** ### Worked Out Examples Example 1: Find the sum of the series 2 + 8 + 14 + 20 + … … … 20 terms. Solution: Here, First term (a) = 2 Common difference (d) = 8 – 2 = 6 Number of terms (n) = 20 Sum (Sn) = ? We Know, Solution: Here, First term (a) = 22 Last term (l) = 54 Number of terms (n) = ? Sum (Sn) = ? We know, l = a + (n – 1)d or, n – 1 = 6 n = 7 Now, Hence, the required sum is 266. Solution: Given, or, (3×4 – m)+ (3×5 – m)+ (3×6 – m)+ (3×7 – m) = 58 or, (12 – m)+ (15 – m)+ (18 – m)+ (21 – m) = 58 or, 12 – m + 15 – m + 18 – m + 21 – m = 58 or, 66 – 4m = 58 or, 4m = 66 – 58 or, 4m = 8 m = 2 Solution: Given, Example 5: How many terms of an AS 17 + 15 + 13 + … … are needed to give the sum 80? Solution: Here, First term (a) = 17 Common difference (d) = -2 Sum (Sn) = 80 Number of terms (n) = ? By using formula, or, 160 = n[34 – 2n + 2] or, 2n2 – 36n + 160 = 0 or, n2 – 18n + 80 = 0 or, n2 – 10n – 8n + 80 = 0 or, (n – 10)(n – 8) = 0 n = 8 or 10 Example 6: If the sum of the first 16 terms of an AS is 280 and the 6th term of the series is 14. Find the sum of the first 21 terms. Solution: Let a be the first term and d be the common difference of the given series. Sum of 16 terms (S16) = 280 6th term (t6) = 14 By using formula, we have or, 280 = 8(2a + 15d) or, 2a + 15d = 35 …………… (i) Again, tn = a + (n – 1)d or, t4 = a + (6 – 1)d or, a + 5d = 14 …………… (ii) Solving equations (i) and (ii), we get Substituting the value in equation (ii), a + 5 × 7/5 = 14 or, a + 7 = 14 or, a = 7 Now, or, S21 = 441 Hence, the sum of 21 terms of the series is 441. Example 7: Find the sum of all natural numbers less than 100 which are exactly divisible by 7. Solution: Here, First term (a) = 7 Common difference (d) = 7 Last term (l) =98 Sum (Sn) = ? Number of terms (n) = ? By using formula, l = a + (n – 1)d or, 98 = 7 + (n – 1)7 or, 98 = 7 + 7n – 7 or, 7n = 98 or, n = 14 Hence by using formula, or, S14 = 7 × 105 S14 = 735 The required sum is 735. ## Sum of the First n Natural Numbers The first n natural numbers are 1, 2, 3, 4, … … … n. Let Sn = 1 + 2 + 3 … … … + n Here, First term (a) = 1 Common difference (d) = 2 – 1 = 1 Number of terms (n) = n Sum of first n terms (Sn) = ? By using formula, ## Sum of the First n Odd Natural Numbers The first n odd natural number are 1, 3, 5, … … …, (2n – 1) Let Sn = 1 + 3 + 5 + … … + (2n – 1) Here, First term (a) = 1 Common difference (d) = 3 – 1 = 2 Number of terms (n) = n Sum of n terms (Sn) = ? By using formula, ## The Sum of the First n Even Natural Numbers The first n even natural numbers are 2, 4, 6, … …, 2n. Let Sn = 2 + 4 + 6 + … … + 2n Here, First term (a) = 2 Common difference (d) = 4 – 2 = 2 Number of terms (n) = n Sum of n terms (Sn) = ? By using formula, ## Sum of the Squares of the First n Natural Numbers The squares of the first n natural numbers are 12, 22, 32, … …, n2 Let Sn = 12 + 22 + 32 + … … + n2 Consider the identity, r3 – (r – 1)3 = r3 – (r3 – 3r2 + 3r – 1) or, r3 – (r – 1)3 = 3r2 – 3r + 1 Putting r = 1, 2, 3, 4, … …, (n – 1), n 13 – 03 = 3.12 – 3.1 + 1 23 – 13 = 3.22 – 3.2 + 1 33 – 23 = 3.32 – 3.3 + 1 43 – 33 = 3.42 – 3.4 + 1 … … … … … … … … (n – 1)3 – (n – 2)3 = 3(n – 1)2 – 3(n – 1) + 1 n3 – (n – 1)3 = 3n2 – 3n + 1 Adding above all identities, we have n3 = 3(12 + 22 + 32 + … … + n2) – 3(1 + 2 + 3 + … … + n) + n ## Sum of the Cubes of the First n Natural Numbers The cubes of the first n natural numbers are 13, 23, 33, … …, n3. Let Sn = 13 + 23 + 33 + … … + n3 Consider the identity, k4 – (k – 1)4 = k4 – (k4 – 4k3 + 6k2 – 4k + 1) or, k4 – (k – 1)4 = 4k3 – 6k2 + 4k – 1 Putting k = 1, 2, 3, 4, … …, (n – 1), n. 14 – 04 = 4.13 – 6.12 + 4.1 – 1 24 – 14 = 4.23 – 6.22 + 4.2 – 1 34 – 24 = 4.33 – 6.32 + 4.3 – 1 44 – 34 = 4.43 – 6.42 + 4.4 – 1 … … … … … … … … … … (n – 1)4 – (n – 2)4 = 4.(n – 1)3 – 6.(n – 1)2 + 4.(n – 1) – 1 n4 – (n – 1)4 = 4.n3 – 6.n2 + 4.n – 1 Adding all above identities, we have, n4 = 4(13 + 23 + 33 + … … + n3) – 6(12 + 22 + 32 + … … + n2) + 4(1 + 2 + 3 + … … + n) – n or, 4Sn = n4 + n + n(n + 1)(2n + 1) – 2n(n + 1) or, 4Sn = n(n + 1)(n2 – n + 1) + n(n + 1)(2n + 1) – 2n(n + 1) or, 4Sn = n(n + 1)(n2 – n + 1 + 2n + 1 – 2) ### More Examples Example 8: Find the sum of the following series: a. 2 + 4 + 6 + … … … 12 terms b. 13 + 23 + 33 + … … 10 terms Solution: a. The given series is 2 + 4 + 6 + … … … 12 terms This is the sum of the first 12 even natural numbers. Here, n = 12 Sn = n(n + 1) = 12(12 + 1) = 156 b. The given series is 13 + 23 + 33 + … … 10 terms This is the sum of the cubes of the first 10 natural numbers Here, n = 10 Example 9: The sum of three numbers in AP is 27 and the sum of their squares is 293. Find the numbers. Solution: Let the three numbers in AP be a – d, a and a + d. Then, (a – d) + a + (a + d) = 27 i.e. a = 9 And, (a – d)2 + a2 + (a + d)2 = 293 or, a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 293 or, 3a2 + 2d2 = 293 or, 3 × 92 + 2d2 = 293 or, d2 = 25 or, d = 5 Hence, the required numbers are a – d, a and a + d. i.e. 4, 9 and 14. Example 10: Find nth term and the sum of n terms of the series 2×3 + 3×4 + 4×5 + … … n terms. Solution: The given series is 2×3 + 3×4 + 4×5 + … … n terms. nth term of 2, 3, 4, … … = 2 + (n – 1)1 = n + 1 nth term of 3, 4, 5, … … = 3 + (n – 1)1 = n + 2 The nth term of the given series = (n + 1)(n + 2) = n2 + 3n + 2 Sum of the series of n terms,
# Find the percentage of 43 out of 50. What is 43 out of 50 as a percentage? In the realm of Mathematics education, understanding percentages plays a crucial role in comprehending various concepts. In this article, we delve into the question of what 43 out of 50 represents as a percentage. By employing simple mathematical calculations and key formulas, we will explore how to convert this fraction into a percentage. Whether you're a student, teacher, or simply curious about numbers, join us on this journey of discovering the answer. Let's unlock the world of percentages together! Stay tuned for insightful explanations and practical examples. ## Understanding Percentages in Mathematics Education In this section, we will explore the concept of percentages and its relevance in mathematics education. What are percentages? Percentages are a way to express a fraction or a part of a whole as a proportion out of 100. Why are percentages important in mathematics education? Understanding percentages is crucial for students as it allows them to interpret and analyze data, solve real-world problems, and make comparisons between different quantities. How to calculate percentages? To calculate a percentage, divide the given value (in this case, 43) by the total value (in this case, 50) and multiply the result by 100. The formula is: (43/50) * 100 = 86%. ## Applying Percentages to the Context of "43 out of 50" In this section, we will apply the concept of percentages to the specific context of "43 out of 50." What does "43 out of 50" mean as a percentage? When expressed as a percentage, "43 out of 50" means that 43 is equivalent to 86% of the total value of 50. How to interpret "43 out of 50" as a percentage? Interpreting "43 out of 50" as a percentage means understanding that 43 represents 86% of the whole quantity, indicating a relatively high proportion or level of success. ## Practical Applications of "43 out of 50" as a Percentage In this section, we will explore practical applications of understanding "43 out of 50" as a percentage in mathematics education. Real-world examples Understanding percentages allows students to interpret and analyze data in various fields such as statistics, economics, and science. For example, in a survey, if 43 out of 50 people prefer chocolate ice cream, it can be concluded that 86% of the surveyed population has a preference for chocolate ice cream. Comparisons and proportions Percentages enable students to make comparisons and determine proportions. For instance, if another survey shows that 35 out of 50 people prefer vanilla ice cream, students can compare the two percentages (86% vs. 70%) to analyze the difference in preferences. ## Importance of Teaching Percentages in Mathematics Education This section emphasizes the significance of teaching percentages effectively in mathematics education. Building foundational knowledge Teaching percentages helps students build a strong foundation in mathematical concepts and problem-solving skills. It enables them to understand and communicate numerical information accurately. Real-life relevance Understanding percentages has practical relevance in everyday life. Students can apply this knowledge to financial literacy, budgeting, analyzing data, and making informed decisions in various situations. Enhancing critical thinking Working with percentages encourages critical thinking, logical reasoning, and analytical skills. It enables students to interpret and evaluate information in a quantitative manner. ### How do you express "43 out of 50" as a percentage in mathematics education? To express "43 out of 50" as a percentage in mathematics education, you divide the numerator (43) by the denominator (50) and multiply the result by 100. In this case, 43/50 * 100 = 86%. ### What is the process to convert the fraction "43/50" into a percentage in the context of mathematics education? To convert the fraction "43/50" into a percentage in the context of mathematics education, you would multiply the fraction by 100 and add the percentage symbol. The process is: (43/50) * 100 = 86%. ### Can you explain how to calculate the percentage of "43 out of 50" in mathematics education? To calculate the percentage of "43 out of 50" in mathematics education, divide the numerator (43) by the denominator (50) and multiply by 100. ### In mathematics education, what is the percentage equivalent of "43 out of 50"? The percentage equivalent of "43 out of 50" in mathematics education is 86%. ### How can I represent the ratio "43 out of 50" as a percentage in the field of mathematics education? To represent the ratio "43 out of 50" as a percentage in the field of mathematics education, you can calculate it by dividing 43 by 50 and then multiplying the result by 100. Thus, 43 out of 50 is 86% in percentage form. In conclusion, understanding percentages is a crucial aspect of Mathematics education. In the case of "43 out of 50," we can express it as a percentage by dividing 43 by 50 and then multiplying by 100. This calculation yields a result of 86%. By mastering this concept, students can effectively interpret and analyze data, make informed decisions, and solve real-world problems involving proportions and ratios. Thus, a solid foundation in percentages is essential for success in both academic and everyday life situations.
# What is the practical application of congruence nowadays? Answer to the comment in Congruences, by dwightisebia First of all I liked your question. Actually it gave me an idea of adding real life applications of mathematical concepts to my site! Thank you! Q: What is the practical application of congruence nowadays? Ans: Two objects are congruent if they have the same dimensions and shape. Clearly, you can think of it as meaning ‘equal’. When any object is kept in front of the mirror, that mirror image is congruent to the real image. When two objects or shapes are said to congruent then all corresponding angles and sides also congruent. For example: Real life examples are, • cigarettes in a packet are congruent to one another. • Giant wheels or ferris wheels. • Pages of one books are congruent to one another and etc ## 6 thoughts on “What is the practical application of congruence nowadays?” 1. But how to relate the modulo to congruence in real life. In other words, how to perceive the “equality” of 27 and 47 in relation to to the modulo 10? Liked by 1 person 1. Hahaha! 27 and 47! Haha! The modulo operation is taking remainder when dividing. For example 5 mod 3 = 2, means 2 is the remainder when divided 5 by 3. This can be used in everyday math calculations 😀 Let x and y be two kids, x had 27 rs with him & y had 47 rs with him. While teaching them money calculations their mamma told them to make groups of 10rs from the money they have! Accordingly, x makes group of 2 and he finds 7rs extra with him that he cannot group-up! Mathematical expression of this is, 27 ≡ 7 (mod 10) ——–(1) Similarly, y makes group of 4 and he also finds 7rs extra with him. Mathematical expression of this is, 47 ≡ 7 (mod 10)———(2) From (1) and (2) 27 ≡ 47(mod 10) as they get same remainder i.e., 7 when divided by 10. Hope you got! 27 and 47! 😀 If not, I can explain Modulo operation by taking example as CLOCKS… Mathematics is abstract! You cannot find numbers, mathematical symbols in nature. Mathematics is used in predicting and controlling real objects and events, from calculating a shopping bill to sending rockets to other planets! Liked by 1 person 1. That’s great .. when your posts are related to math , tag me up! Like
Article # Using CheckedListBox in Multiplication , 4 Aug 2010 Rate this: Ancient Egyptian multiplication and Russian peasants multiplication ## Introduction When you use your computer or your calculator to do difficult calculations, or when you see your child easily do the following multiplication... ``` 68 x 43 ----------- 204 272 ----------- 2924``` ...we have to remember the suffering of the ancient people throughout history to calculate the operations of arithmetic. However, they left us great works certifying that they were geniuses. History tells us about two operations of multiplication, Ancient Egyptian multiplication and Russian peasants multiplication. ## Ancient Egyptian Multiplication Here is how 43 is multiplied by 68: 1. Start with number 1 in the first column and keep doubling (the first column has powers of two), the largest power of two less than or equal to the first number(43). 2. Start with the second number (68) in the second column, keep doubling the number in the second column. 3. Subtract the largest power of two less than or equal to the first number(43), 43 - 32 = 11, subtract the largest power of two less than or equal to the remainder(9), 11 - 8 = 3 repeat, 3 - 2 = 1 repeat until nothing remains, 1 - 1 = 0 Now you see that 43 = 32 + 8 + 2 + 1. 4. To get the result, check the numbers in the second column corresponding to 32, 8, 2, 1 and add them. ## Russian Peasants Multiplication See how 43 is multiplied by 68: • Write each number at the head of a column. • Divide the number in the first column by 2, flooring the quotient (drop the remainder), until there is nothing left to divide. • Keep doubling the number in the second column, until you have doubled it as many times as you divided the number in the first column. • To get the result, add up all the numbers in the second column that are next to an odd number in the first column. ## Background To test the previous operations of multiplication, begin a new project with one form, put the following controls on the form: • Two controls of `TextBox`: `txtFirst `and `txtSecond `to enter two numbers. • Two controls of `CheckedListBox `for Ancient Egyptian multiplication: `lstEgyptian1 `and `lstEgyptian2`. • Two controls of `CheckedListBox `for Russian peasants multiplication: `lstRussian1 `and `lstRussian2`. • Two controls of `Label `to print the result: `EgyptianResult `and `RussianResult`. • Three controls of `Button`s: `btnEgyptianCalc`, `btnRussianCalc `to get the result and `btnExit `to exit the program. Fill the first list: ```private void FillListEgyptian1() { newVlue = 1; numFirst = Convert.ToInt32(txtFirst.Text); lstEgyptian1.Items.Clear(); //add powers of two (less than numFirst) to 'lstEgyptian1' while (newVlue <= numFirst) { newVlue = 2 * newVlue; //the first column has powers of two } } int ``` Fill the second list: ```private void FillListEgyptian2() { numSecond = Convert.ToInt32(txtSecond.Text); lstEgyptian2.Items.Clear(); for(int i = 1; i < lstEgyptian1.Items.Count; i++) { numSecond = numSecond * 2; //keep doubling the second number } } ``` Check numbers at first column: ```private void CheckEgyptianNumbers() { int newNumber = numFirst; int i = lstEgyptian1.Items.Count; do { i--; int binNumber = (int)(lstEgyptian1.Items[i]); if(binNumber <= newNumber) //is power of 2 < newNumber? { newNumber = newNumber - binNumber; lstEgyptian1.SetItemChecked(i, true); //check it lstEgyptian2.SetItemChecked(i, true); //check it } } while (i > 0); }``` Add numbers to get the result: ```private void GetEgyptianMultiplication() { long TheResult = 0; for(int i = 0; i < lstEgyptian2.Items.Count; i++) { if(lstEgyptian2.GetItemChecked(i) == true) { TheResult = TheResult + (int)(lstEgyptian2.Items[i]); EgyptianResult.Text = TheResult.ToString(); } } } ``` You can read the code of Russian peasants multiplication. ## Final Words We must not forget what knowledge the previous generations have given us. Said Isaac Newton, "I stand on the shoulders of my predecessors". Thanks to Code Project and thanks to all. ## History • 4th August, 2010: Initial post ## You may also be interested in... No Biography provided First Prev Next My vote of 4 dbrenth 10-Aug-10 3:12 My vote of 5 cornelha 4-Aug-10 23:00 My vote of 1 AxelM 4-Aug-10 20:35 Good article.....but Shivprasad koirala 4-Aug-10 16:38 Re: Good article.....but cornelha 4-Aug-10 22:59 Last Visit: 31-Dec-99 18:00 Last Update: 2-Sep-14 4:22 Refresh 1
## Variation ### Learning Outcomes • Define direct variation and solve problems involving direct variation • Define inverse variation and solve problems involving inverse variation • Define joint variation and solve problems involving joint variation So many cars, so many tires. ## Direct Variation Variation equations are examples of rational formulas and are used to describe the relationship between variables. For example, imagine a parking lot filled with cars. The total number of tires in the parking lot is dependent on the total number of cars. Algebraically, you can represent this relationship with an equation. $\text{number of tires}=4\cdot\text{number of cars}$ The number $4$ tells you the rate at which cars and tires are related. You call the rate the constant of variation. It is a constant because this number does not change. Because the number of cars and the number of tires are linked by a constant, changes in the number of cars cause the number of tires to change in a proportional, steady way. This is an example of direct variation, where the number of tires varies directly with the number of cars. You can use the car and tire equation as the basis for writing a general algebraic equation that will work for all examples of direct variation. In the example, the number of tires is the output, $4$ is the constant, and the number of cars is the input. Let us enter those generic terms into the equation. You get $y=kx$. That is the formula for all direct variation equations. $\text{number of tires}=4\cdot\text{number of cars}\\\text{output}=\text{constant}\cdot\text{input}$ ### Example Solve for k, the constant of variation, in a direct variation problem where $y=300$ and $x=10$. In the video that follows, we present an example of solving a direct variation equation. ## Inverse Variation Another kind of variation is called inverse variation. In these equations, the output equals a constant divided by the input variable that is changing. In symbolic form, this is the equation $y=\frac{k}{x}$. One example of an inverse variation is the speed required to travel between two cities in a given amount of time. Let us say you need to drive from Boston to Chicago which is about $1,000$ miles. The more time you have, the slower you can go. If you want to get there in $20$ hours, you need to go $50$ miles per hour (assuming you do not stop driving!) because $\frac{1,000}{20}=50$. But if you can take $40$ hours to get there, you only have to average $25$ miles per hour since $\frac{1,000}{40}=25$. The equation for figuring out how fast to travel from the amount of time you have is $speed=\frac{miles}{time}$. This equation should remind you of the distance formula $d=rt$. If you solve $d=rt$ for r, you get $r=\frac{d}{t}$, or $speed=\frac{miles}{time}$. In the case of the Boston to Chicago trip, you can write $s=\frac{1,000}{t}$. Notice that this is the same form as the inverse variation function formula $y=\frac{k}{x}$. ### Example Solve for k, the constant of variation, in an inverse variation problem where $x=5$ and $y=25$. In the next example, we will find the water temperature in the ocean at a depth of $500$ meters. Water temperature is inversely proportional to depth in the ocean. Water temperature in the ocean varies inversely with depth. ### Example The water temperature in the ocean varies inversely with the depth of the water. The deeper a person dives, the colder the water becomes. At a depth of $1,000$ meters, the water temperature is $5º$ Celsius. What is the water temperature at a depth of $500$ meters? In the video that follows, we present an example of inverse variation. ## Joint Variation A third type of variation is called joint variation. Joint variation is the same as direct variation except there are two or more quantities. For example, the area of a rectangle can be found using the formula $A=lw$, where l is the length of the rectangle and w is the width of the rectangle. If you change the width of the rectangle, then the area changes and similarly if you change the length of the rectangle then the area will also change. You can say that the area of the rectangle “varies jointly with the length and the width of the rectangle.” The formula for the volume of a cylinder, $V=\pi {{r}^{2}}h$, is another example of joint variation. The volume of the cylinder varies jointly with the square of the radius and the height of the cylinder. The constant of variation is $\pi$. ### Example The area of a triangle varies jointly with the lengths of its base and height. If the area of a triangle is $30$ inches$^{2}$ when the base is $10$ inches and the height is $6$ inches, find the variation constant and the area of a triangle whose base is $15$ inches and height is $20$ inches. Finding k to be $\frac{1}{2}$ should not be surprising. You know that the area of a triangle is one-half base times height, $A=\frac{1}{2}bh$. The $\frac{1}{2}$ in this formula is exactly the same $\frac{1}{2}$ that you calculated in this example! In the following video, we show an example of finding the constant of variation for a jointly varying relation. ### Direct, Joint, and Inverse Variation k is the constant of variation. In all cases, $k\neq0$. • Direct variation: $y=kx$ • Inverse variation: $y=\frac{k}{x}$ • Joint variation: $y=kxz$ ## Summary Rational formulas can be used to solve a variety of problems that involve rates, times, and work. Direct, inverse, and joint variation equations are examples of rational formulas. In direct variation, the variables have a direct relationship—as one quantity increases, the other quantity will also increase. As one quantity decreases, the other quantity decreases. In inverse variation, the variables have an inverse relationship—as one variable increases, the other variable decreases, and vice versa. Joint variation is the same as direct variation except there are two or more variables.
# Question: What Happens When You Add 2 Square Roots? ## How do you multiply square roots together? Explanation: To multiply square roots, we multiply the numbers inside the radical. Any numbers outside the radical are also multiplied. We can simplify them if possible.. ## What does this sign mean √? √ is the symbol for square root. A square root is the number that, when multiplied by itself, gives the original number. For example, the square root of 4 is 2, because 2 x 2 = 4. ## Can you split up a radical? There are NO like terms to be combined. Dividing Radicals: When dividing radicals (with the same index), divide under the radical, and then divide in front of the radical (divide any values multiplied times the radicals). ## Can you add 2 square roots? Just as with “regular” numbers, square roots can be added together. … Just as “you can’t add apples and oranges”, so also you cannot combine “unlike” radical terms. In order to be able to combine radical terms together, those terms have to have the same radical part. To add or subtract radicals, the indices and what is inside the radical (called the radicand) must be exactly the same. If the indices and radicands are the same, then add or subtract the terms in front of each like radical. If the indices or radicands are not the same, then you can not add or subtract the radicals. ## Does a square root have two values? The ± \pm ± is applied this way precisely because the square root is limited in its range to only positive values, whereas the function x 2 x^2 x2 has a domain of both positive and negative values. But the evaluation of the square root itself is simply 5. ## Can you split up a square root with addition? When sums and differences are under a radical sign, you cannot split them into separate radicals. ## How do you add square roots the same? How to Add Square Roots with coefficients?Ignore the coefficients ( 2 and 5) and simplify each square root.Multiply the coefficients (2 and 5) by any numbers that ‘got out’ of the square root (3 and 2, respectively).Add any radicals with the same radicand. ## Which is the square root of 225? The positive square root of 225 is 15 because 152 = 225. ## Why is the square root always positive? In the case of the square root function y = sqrt(x) the positive value (of y) has been chosen, whatever the value of x. So the square root function is always positive or zero. The square root (number) may be positive or negative or zero.
# What is probability in Mathematics OR Explain how do you calculate probability ? Whenever we're unsure about the outcome of an event, we can talk about the probabilities of certain outcomes — how likely they are. OR Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability. Probability of an Event = Number of ways it can happen / Total number of outcomes Let us understand, What is equally likely Events or outcomes? When tossing a coin, There are two possible outcomes of an experiment: Probability of getting a Head is 1/2 Probability of getting a Tail is 1/2 We have seen in the experiment of tossing a coin that each of the outcome has the same chance of occurring. When each of the outcome of the experiment has the same chance of occurring then we can say , the outcomes of the experiment are equally likely. Total Views: 1480 ### Related FAQs Q. List all the possible outcomes when two coins are tossed together. Ans: Total Possible outcomes is 4. Lets Explain :-- When we toss two coins simultaneously then the possible of outcomes are: (H, H) or (H, T) or (T, H) or (T, T). where H is used for head and T is used ... read more Q. Numbers 1 to 10 are written on ten separate slips ( one number on one slip) kept in a box and mixed well. One slip is chosen from the box without looking into it. (a) What is the probability of getting a number 6 ? (b) What is the probability of getting a number greater than 6 ? Ans: (a) Total No of slips = 10 Number of slips on which 6 is written = 1 Probability = Number of slips on which 6 is written / Total No of slips = 1/ 10 (b) Numbers greater than 6 are 7, 8, 9, ... read more Q. A bag has 4 red balls and 2 yellow balls. 1) What is the Probability of getting a red ball ? 2) What is the Probability of getting a yellow ball? Ans: Total No of balls = 4 + 2 = 6 balls Number of red colour balls = 4 Number of yellow colour balls = 2 1) Probability of red ball = Number of red colour balls / Total No of balls = 4 / 6 = 2 /... read more Q. Three unbiased coins are tossed. What is the probability of getting at most two heads? Ans: Probability of an Event = Number of ways it can happen / Total number of outcomes Total number of outcomes = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} = 8 Number of ways it can happen = {TTT, TTH, TH... read more Q. An integer is chosen between 0 and 100. What is the probability that it is divisible by 7 ? Ans: Number of integers between 0 and 100 = n(S) = 99 Let E be the event ?integer divisible by 7? Favourable outcomes to the event E is = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98 So ... read more Q. In a cricket match, a batsman hits a boundary 8 times out of 40 balls he played. find the probability that he didn't hit aboundary. Ans: Let n(E) is the event that the batsman did not hit a boundary. Number of balls in which batsman not hit the boundary = Number of ball - Number of ball hit the boundary n(E) = 40 -8 = 32 Tota... read more Q. A bag contains 9 black and 12 white balls. One ball is drawn at random. What is the probability that the ball drawn is black? Ans: A bag contains 9 black and 12 white balls So Total balls = 9 + 12 = 21 balls probability of drawn balls is black = 9 / 21 = 3 / 7... read more Q. Find the probability that a number selected from the numbers 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected. Ans: Numbers from 1 to 25 are (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) Prime Numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23 Total Prime numbers = 9 Total Num... read more Q. A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being not a blue ball. Ans: Total Number of balls = 10 + 5 + 7 = 22 Total number of balls which are not blue = 22 - 5 = 17 So probability of not getting blue ball is = 17 / 22... read more Q. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9. Ans: given that two dice are thrown Therefore the probability of total outcomes = 6 X 6 = 36 Let E be the event of getting a product less than 9. The number whose product is less than 9 = {... read more Q. Two dice are thrown at the same time. Find the probability of getting different numbers on both the dice. Ans: Given that two dice are thrown Therefore the probability of total outcomes = 6 X 6 = 36 Let E be the event of getting different numbers on both the dice Combinations of same numbers on both the... read more Q. Two dice are thrown at the same time. Find the probability of getting same number on both the dice. Ans: Given that two dice are thrown Therefore the probability of total outcomes = 6 X 6 = 36 Let E be the event of getting same numbers on both the dice Combinations of same numbers on both the dice... read more Q. List all the outcomes of choosing a day of the week starting with letter W Ans: Number of days starting from W are:- Wednesday So, number = 1 Number of days in a week = 7 So, probability = 1/7... read more Q. What type of event it is to purchase a car from a bike showroom? Ans: Impossible Event This is not possible to purchase a car from bike showroom. This event cannot happen so it is called as impossible event.... read more Q. Find the probability of Harjinder’s birthday to be a prime number date in the month of August. Ans: Prime number dates in month of august are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 Total number of prime dates is = 11 Total no of days in August is = 31 So probability of birthday on prime dat... read more Q. The value of probability is always less than _____ and greater than _____ . Ans: The value of probability is always less than __1___ and greater than __0___ . The probability of event can not be greater than 1 and can not be less than 0... read more Q. Ravi and Deepak are playing a game. The probability of Deepak winning the game is 7/9. Find the probability of Ravi winning the game. Ans: Probability of Deepak winning the game is = 7/9 probability of Deepak loosing the game is = 1 - 7/9 = 2/9 So probability of Ravi winning the game is = 2/9... read more Q. If a coin is tossed 500 times and the tail appears 159 times, find the probability of getting a tail. Ans: Total number of trials = 500 Let E1 = Number of times tail appear = 159 P(E1) = Number of times tail appear / total number of trials = 159 / 500 = 0... read more Q. Two coins are tossed simultaneously. Write the number of favourable outcomes for the following events: (i) getting at least one tail (ii) getting no tail. Ans: The possible outcomes when two coins are tossed simultaneously are {HH,TT,HT,TH}. Therefore, the number of possible outcomes when two coins are tossed is 4. (i) Number of favourable ou... read more Q. A die is thrown. What is the probability that i) 5 will not come? ii) a number less than 5 will come? Ans: Total number of outcomes = 6 i) Number of times when 5 will not come is 5 as (1, 2, 3, 4, 6) So probability that 5 will not come is = 5/6 ii) Number of times when a numb... read more Q. An integer is chosen between 0 and 100. What is the probability that it is (i) divisible by 7 (ii) not divisible by 7 Ans: Number of integers between 0 and 100 = n(S) = 99 (i) Let E be the event integer divisible by 7 Favourable outcomes to the event E = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,... read more Q. In a family of 3 children calculate the probability of having at least one boy. Ans: Total number of combinations = {bbb, bbg, ggb, ggg} Where b stands for Boy & g stand for Girl So total number of outcome is 4 E be the event that family having at least one boy Fa... read more Q. A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant. Ans: Total English alphabets = 26 Number of consonants = 21 As Probability of an Event = Number of ways it can happen / Total number of outcomes So probability of letter is a consonant = 21/26 Hence ... read more Q. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen Ans: There are 26 red and 26 black in a pack of 52 playing cards So total number of red cards is 26 Total Number of queens cards in a pack of 52 cards is = 4 In which, 2 queens are black and 2 are... read more Q. A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number which is a perfect square Ans: Perfect square means a number that can be expressed in form of a product of two integers that are equal. For examples 9, 16, 25 etc Between 6 and 50 the perfect squares in between are :9, 16, 25, 36... read more Q. A number is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1 Ans: Square of given numbers are :-- (-3)2 = 9, (-2)2 = 4, (-1)2 = 1, (0)2 = 0, (1)2 ... read more Q. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap? Ans: As total number of apples is = 900 The probability of rotten apple is = 0.18 Let assume that the total number of rotten apples is = x So probability is = x/900 0.18 = x/900 x = 900*0.18 x = 162 ... read more Q. A book contains 90 pages which are numbered from 1 to 90. If one page is opened at random from the book, find the probability that it will be-- (i) a two digit number (ii) a perfect square number (iii) a number divisible by 5. Ans: As book contains 90 pages which are numbered from 1 to 90 So total numbers of outcome is n(S) = 90 (i) Let E be the event of a two digit number Total Single digit number from 1 to 9... read more Q. A bag contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the bag at random. What is the probability that marble taken out will be - (i) red (ii) white (iii) not green. Ans: Total number of marbles = 5 + 8 + 4 = 17 So total numbers of outcome is n(S) = 17 (i) Let E be the event of taken out marble will be red Total number of red marbles is... read more Q. A box contains 5 iron blocks and n wooden blocks. If the probability of having a wooden block is three times that of the iron block, then what is the value of n. Ans: Iron blocks = 5 Wooden blocks = n So Total possible outcomes n(S) = 5 + n P(getting wooden blocks) = 3 x P(getting iron blocks) n/(5 + n) = 3 x (5/(5 + n)) n/(5 + n) = 15/(5 + n) So n = 15, ... read more Q. If a bag of balloons consists of 47 white balloons, 5 yellow balloons, and 10 black balloons, what is the approximate likelihood that a balloon chosen randomly from the bag will be black? Ans: The total number of balloons in the bag: 47 + 5 + 10 = 62 Number of black ballons is 10 So probability of black ballon is = 10/62 = 0.1612 = 0.16 which is equal to 16% Hence calculating&nbs... read more Q. The probability that a number selected at random from the numbers 1, 2, 3, ......, 15 is a multiple of 4 is Ans: Numbers from 1 to 15 are (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15) Numbers Multiple of 4 are = 4, 8, 12 Total numbers multiple of 4 is = 3 Total Numbers = 15 Now the probability of getti... read more Q. Two coins are tossed simultaneously. The probability of getting at most one head is Ans: All possible outcomes are {HH, HT, TH, TT} Therefore total number of outcome is n(S) = 4 Favourable outcomes are {HT, TH, TT} n(E) = 3 Probability of getting at most one head is ... read more Q. The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is Ans: Total number of eggs is 400 So n(S) = 400 Let a is the total number of bad egg So n(E) = a Probability of getting bad egg is 0.035 So P(E) = 0.035 As P(E) = n(E)/n(S) So 0.035 = a/400 a = 40... read more Q. A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, then how many tickets has she bought? Ans: Total number of sold tickets are 6000. So n(S) = 6000 Let she bought x tickets So n(E) = x Probability of winning the first prize in a lottery is 0.08 So P(E) = 0.08 As P(E) = n(E)/n(S) So 0.... read more Q. A single letter is selected at random from the word PROBABILITY. The probability that the selected letter is a vowel is. Ans: There are 11 letter in word PROBABILITY. Out of these 11 letter, 4 letter are vowels. So n(S) = 11 n(E) = 4 The probability that the selected ... read more Q. A letter is chosen at random from the letters of the word ASSASSINATION, then the probability that the letter chosen is a vowel is in the form of 6/2x+1, then x is equal to Ans: There are 13 letters in the word ASSASSINATION out of which one letter can be chosen in 13 w... read more Q. A number is chosen at random from the numbers 5-, 4-, 3-, 2-, 1-, 0, 1, 2, 3, 4, 5. Then the probability that square of this number is less than or equal to 1 is .......... Ans: Given numbers are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 Their squares are 25, 16, 9, 4, 1, 0, 1, 4, 9, 1... read more Q. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red king. Ans: Total number of cards is 52 So n(S) = 52 Total number of red king is 2 So n(E) = 2 The probability of getting a red king P(E) = n(E)/n(S) P = 2/52 = 1/26 Hence calculati... read more Q. A card drawn at random from a well shuffled deck of 52 playing cards. What is the probability of getting a black king? Ans: Total number of cards is 52 So n(S) = 52 Total number of black king is 2 So n(E) = 2 The probability of getting a black king P(E) = n(E)/n(S) P = 2/52 = 1/26 Hence calcu... read more Q. A bag contains green balls and red balls only. A ball is taken at random from the bag.The probability of taking a green ball is 0.38 .Write down the probability of taking a red ball. Ans: As the bag contain only green balls and red balls only. Given part is: The probability of taking a green ball is 0.38 AS probability can not be greater than 1 So the probability of taking a red... read more Q. Dan either walks or cycles to school.The probability that he cycles to school is 1/5. Write down the probability that Dan walks to school. Ans: As Dan either walks or cycles to school. The probability that he cycles to school is 1/5 As probability can not be greater than 1 Therefore the probability that Dan walks to school is = 1 - 1/5... read more Q. Dan either walks or cycles to school.The probability that he cycles to school is 1/5. There are 200 days in a school year. Work out the expected number of days that Dan cycles to school in a school year. Ans: As Dan either walks or cycles to school. The probability that he cycles to school is 1/5 Total number of days in school year = 200 Expected number of days that Dan cycles to school in a school ye... read more
We know that all repeating or terminating decimals are rational, but are all rationals either repeating or terminating? If they are, that would by corrolary mean that all irrational numbers do not terminate or repeat. We'll first consider rationals that have denominators consisting of factors from the base. That is to say, any number that can divide the denominator can also divide the base. We call these Denominitors Absolutely Not Relatively Prime to the Base. Next we consider rationals with • Denominators Relatively Prime to the Base. Finally we consider the general case which combines the two, Denominators in General. # Denominitors Absolutely Not Relatively Prime to the Base If you think about it, it's easy to see that a denominator with no factors relatively prime to the base you're using will result in a terminating decimal. For example, in decimal our base is `10 = 2 * 5`, and if our denominator is `200 = 23 * 52`, then `1/200` will terminate. This is because `200` will divide into a large enough power of `10` (or whatever our base is): • `1 / 200 = 5 / 1,000 = .005` Or looking at it another way, we multiply by `10 / 10 = 1` until the denominator cancels out and we get an integer: • `1 / 200 = (1/10) * 10 / 200 = (1/10) * 1 / 20` • ``` = (1/100) * 10 / 20 = (1/100) * 1 / 2 ``` • ``` = (1/1,000) * 10 / 2 = (1/1,000) * 5 ``` • ``` = .005 ``` ``` Note that the power of 10 in question determines how long the terminating decimal will be. It doesn't matter if we have a numerator; that will just increase the number to be shifted once we've canceled out the denominator. Denominators Relatively Prime to the Base After the section above, it should be clear that denominators that are relatively prime to the base don't terminate: No matter how many times you multiply by the base, you can't get an integer. If all rationals do indeed terminate or repeat, these relatively prime denominators must repeat. If you know how repeating decimals work, you know that they're related to 1/(basen-1). If we can show that any number relatively prime to the base can divide (basen-1) for some n, we will have shown that the number repeats. To prove this, we'll use Euler's Totient. The totient of n is represented by φ(n), and is defined as the number of positive integers less than n which are relatively prime to n. This totient is used in Euler's Totient Theorem: aφ(n) === 1 (mod n) for all a relatively prime to n (=== is being used here to represent "is congruent to".) If we take a to be our base 10, and n to be the denominator, then: 10φ(n) === 1 (mod n) 10φ(n) - 1 === 0 (mod n) n \ (10φ(n) - 1) Which is to say, all relatively prime denominators will go into (basex-1) for some x (if nothing else, φ(n)). They repeat, snug against the decimal point. If we have a numerator, the only thing we really need to look out for is if it is greater than the denominator. In that case, it's best to seperate out the integer portion before tackling the fraction. The remaining numerator will be less than the denominator, and won't be able to "escape" the repeating portion (x digits long) defined by the denominator. As way of an example: 13 / 7 = 1 + 6 / 7 = 1 + 6 * (1 / 7) = 1 + 6 * (142,857 / 999,999) (Read: 1/7 = 0.142857) = 1 + (857,142 / 999,999) = 1.857142 6 < 7, so when we expand the denominator to 999,999 (106-1), the resulting numberator 857,142 is still less than the denominator. There is no carrying between "repeating groups" or into the integer. Note that the repeating portion, in this case, starts right after the decimal point. You could call this the 1st decimal position. If it were to be shifted to the right k places, it would begin at the (1 + k)th decimal position. The only trick remaining is a good way to find x, the length of the repeating section. This will be covered later, as knowing that it exists is good enough for the proof. Denominators in General If a demoninator has both relatively prime and not relatively prime factors (to the base), we consider each set of factors seperately. Let us say that our denominator is rs where r is relatively prime to the base, and s is absolutely not relatively prime to the base (that is, it has no factors which are relatively prime to the base). It's also easier if we include a numerator from the start. Any integer should have been removed already, similar to above. The numerator will be n. When we had an absolutely not relative prime denominator, we simply canceled out the denominator and shifted the result appropriately. We start out doing the same thing here -- but instead of "canceling out" (making the denominator 1), we cancel out until the denominator is relatively prime to the base. That is, we have some shifting fraction (1/10k), s has been transformed into the numerator (in combination with n), and r is left on bottom. Now's a good time to clarify by starting up an example: 17 / 56 = 17 / (23 7) = [53 / 53] * [17 / (23 * 7)] = [1 / 103] * [(53 * 17) / 7] = [1 / 103] * [125 * 17 / 7] = [1 / 103] * [2125 / 7] The remaining fraction on the right should now be broken up into an integer and a proper fraction: = [1 / 103] * [2125 / 7] = [1 / 103] * [303 + (4 / 7)] And finally we distribute the "shift": = [1 / 103] * [303 + (4 / 7)] = 303 / 103 + 4 /(103 * 7) Now, let's look at what we have. On the left is a numerator with no more than x digits (we know that the final result is less than 1) which is simply being shifted x times. This is a static portion in the first x digits after the decimal point. On the right is a numerator less than the relatively prime portion of the denominator. This will result in some repeating section, which will also be shifted over x times. Therefore the repeating starts at the (1 + x)th decimal position, right after the static portion. The repeating portion will not overflow into the static portion, because the numerator is smaller than the relatively prime portion of the denominator. We can find out that: 4 / 7 = 4 * 1 / 7 = 4 * 142,857 / 999,999 = 571,428 / 999,999 = 0.571428 And now we know that: 17 / 56 = 303 / 103 + 4 /(103 * 7) = 303 / 103 + 0.571428 / 103 = 0.303 + 0.000571428 = 0.303571428 We've shown that any rational with a "mixed" denominator (with factors both relatively prime to and not relatively prime to the base) can be broken into two non-interfering sections: One that terminates, and another that repeats. Since we've already shown that rationals with denominators absolutely not relatively prime to the base and rationals with denominators relatively prime to the base also either terminate or repeat, we have proven that all rationals either terminate or repeat. Moreover, we know that the non-relatively prime factors in the denominator result in static sections, and relatively prime factors result in repeating sections. Back ```
# Equation with Binomial Expression Contributed by: This pdf includes the following topics- Constant Terms and expressions Coefficient of a term Writing patterns in geometry Variable Examples 1. CHRIST SENIOR SECONDARY SCHOOL CLASS 7 Notes: Chapter 12 ALGEBRAIC EXPRESSIONS Introduction to Algebraic Expressions Constant is a quantity which has a fixed value. Terms of Expression Parts of an expression which are formed separately first and then added are known as terms. They are Example: Terms 4x and 5 are added to form the expression (4x +5). Coefficient of a term The numerical factor of a term is called coefficient of the term. Example: 10 is the coefficient of the term 10xy in the expression 10xy+4y Writing Patterns in Geometry  Algebraic expressions are used in writing patterns followed by geometrical figures. Example: Number of diagonals we can draw from one vertex of a polygon of n sides is (n – 3). Definition of Variables  Any algebraic expression can have any number of variables and constants. 2.  Variable  A variable is a quantity that is prone to change with the context of the situation.  a,x,p,… are used to denote variables.  Constant  It is a quantity which has a fixed value.  In the expression 5x+4, the variable here is x and the constant is 4.  The value 5x and 4 are also called terms of expression.  In the term 5x, 5 is called the coefficient of x. Coefficients are any numerical factor of a term. Factors of a term Factors of a term are quantities which can not be further factorised. A term is a product of its factors. Example: The term –3xy is a product of the factors –3, x and y. Like and Unlike Terms Like terms  Terms having same algebraic factors are like terms. Example: 8xy and 3xy are like terms. Unlike terms  Terms having different algebraic factors are unlike terms. Example: 7xy and −3x are unlike terms. Monomial, Binomial, Trinomial and Polynomial Terms Types of expressions based on the number of terms Based on the number of terms present, algebraic expressions are classified as:  Monomial: An expression with only one term. Example: 7xy, −5m, etc.  Binomial: An expression which contains two, unlike terms. Example: 5mn+4, x+y, etc  Trinomial: An expression which contains three terms. Example: x+y+5, a+b+ab, etc.  An expression with one or more terms. Example: x+y, 3xy+6+y, etc. 3. Exercise 12.1 Page: 234 1. Get the algebraic expressions in the following cases using variables, constants and arithmetic (i) Subtraction of z from y. (ii) One-half of the sum of numbers x and y. = ½ (x + y) = (x + y)/2 (iii) The number z multiplied by itself. = z2 (iv) One-fourth of the product of numbers p and q. = ¼ (p × q) = pq/4 (v) Numbers x and y both squared and added. = x2 + y2 (vi) Number 5 added to three times the product of numbers m and n. = 3mn + 5 (vii) Product of numbers y and z subtracted from 10. = 10 – (y × z) = 10 – yz (viii) Sum of numbers a and b subtracted from their product. = (a × b) – (a + b) = ab – (a + b) 2. (i) Identify the terms and their factors in the following expressions Show the terms and factors by tree diagrams. (a) x – 3 4. Expression: x – 3 Terms: x, -3 Factors: x; -3 (b) 1 + x + x2 Expression: 1 + x + x2 Terms: 1, x, x2 Factors: 1; x; x,x (e) – ab + 2b2 – 3a2 Expression: -ab + 2b2 – 3a2 Terms: -ab, 2b2, -3a2 Factors: -a, b; 2, b, b; -3, a, a (ii) Identify terms and factors in the expressions given below: (a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2 (e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼ (h) 0.1 p2 + 0.2 q2 Expressions is defined as, numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something. 5. In algebra a term is either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division. Factors is defined as, numbers we can multiply together to get another number. 4. (a) Identify terms which contain x and give the coefficient of x. (i) y2x + y (ii) 13y2 – 8yx (iii) x + y + 2 (iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25 (vii) 7x + xy2 Sl.No. Expression Terms Coefficient of x (i) y2x + y y2x y2 (ii) 13y2 – 8yx – 8yx -8y (iii) x+y+2 x 1 (iv) 5 + z + zx x 1 zx z (v) 1 + x + xy xy y (vi) 12xy2 + 25 12xy2 12y2 (vii) 7x + xy2 7x 7 xy2 y2 (b) Identify terms which contain y2 and give the coefficient of y2. (i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2 Sl.No. Expression Terms Coefficient of y2 (i) 8 – xy2 – xy2 –x (ii) 5y2 + 7x 5y2 5 (iii) 2x2y – 15xy2 + 7y2 – 15xy2 – 15x 7y2 7 5. Classify into monomials, binomials and trinomials. (i) 4y – 7z 6. An expression which contains two unlike terms is called a binomial. (ii) y2 An expression with only one term is called a monomial. (iii) x + y – xy An expression which contains three terms is called a trinomial. (iv) 100 An expression with only one term is called a monomial. (v) ab – a – b An expression which contains three terms is called a trinomial. (vi) 5 – 3t An expression which contains two unlike terms is called a binomial. (vii) 4p2q – 4pq2 An expression which contains two unlike terms is called a binomial. (viii) 7mn An expression with only one term is called a monomial. (ix) z2 – 3z + 8 An expression which contains three terms is called a trinomial. (x) a2 + b2 7. An expression which contains two unlike terms is called a binomial. (xi) z2 + z An expression which contains two unlike terms is called a binomial. (xii) 1 + x + x2 An expression which contains three terms is called a trinomial. 6. State whether a given pair of terms is of like or unlike terms. (i) 1, 100 Like term. When term have the same algebraic factors, they are like terms. (ii) –7x, (5/2)x Like term. When term have the same algebraic factors, they are like terms. (iii) – 29x, – 29y Unlike terms. The terms have different algebraic factors, they are unlike terms. (iv) 14xy, 42yx Like term. When term have the same algebraic factors, they are like terms. Exercise 12.2 Page: 239 1. Simplify combining like terms: (i) 21b – 32 + 7b – 20b 8. When term have the same algebraic factors, they are like terms. = (21b + 7b – 20b) – 32 = b (21 + 7 – 20) – 32 = b (28 – 20) – 32 = b (8) – 32 = 8b – 32 (ii) – z2 + 13z2 – 5z + 7z3 – 15z When term have the same algebraic factors, they are like terms. = 7z3 + (-z2 + 13z2) + (-5z – 15z) = 7z3 + z2 (-1 + 13) + z (-5 – 15) = 7z3 + z2 (12) + z (-20) = 7z3 + 12z2 – 20z (iii) p – (p – q) – q – (q – p) When term have the same algebraic factors, they are like terms. (iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a When term have the same algebraic factors, they are like terms. = 3a – 2b – ab – a + b – ab + 3ab + b – a = 3a – a – a – 2b + b + b – ab – ab + 3ab = a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3) = a (1 – 2) + b (-2 + 2) + ab (-2 + 3) = a (1) + b (0) + ab (1) = a + ab (i) 3mn, – 5mn, 8mn, – 4mn When term have the same algebraic factors, they are like terms. Then, we have to add the like terms 9. = 3mn + (-5mn) + 8mn + (- 4mn) = 3mn – 5mn + 8mn – 4mn = mn (3 – 5 + 8 – 4) = mn (11 – 9) = mn (2) = 2mn (ii) t – 8tz, 3tz – z, z – t When term have the same algebraic factors, they are like terms. Then, we have to add the like terms = t – 8tz + (3tz – z) + (z – t) = t – 8tz + 3tz – z + z – t = t – t – 8tz + 3tz – z + z = t (1 – 1) + tz (- 8 + 3) + z (-1 + 1) = t (0) + tz (- 5) + z (0) = – 5tz (iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3 When term have the same algebraic factors, they are like terms. Then, we have to add the like terms = – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3) = – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3 = – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3 = mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3) = mn (- 9 + 21) + (7 – 11) = mn (12) – 4 = 12mn – 4 3. Subtract: (i) –5y2 from y2 When term have the same algebraic factors, they are like terms. Then, we have to subtract the like terms = y2 – (-5y2) = y2 + 5y2 = 6y2 (ii) 6xy from –12xy 10. When term have the same algebraic factors, they are like terms. Then, we have to subtract the like terms = -12xy – 6xy = – 18xy (iii) (a – b) from (a + b) When term have the same algebraic factors, they are like terms. Then, we have to subtract the like terms = (a + b) – (a – b) = a (1 – 1) + b (1 + 1) = a (0) + b (2) = 2b (vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2 When term have the same algebraic factors, they are like terms. Then, we have to subtract the like terms = 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2) = 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2 = 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2 = 10ab – 7a2 – 7b2 (viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq When term have the same algebraic factors, they are like terms. Then, we have to subtract the like terms = 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2) = 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2 = 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq = 8p2 + 8q2 – 5pq 4. (a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy? Let us assume p be the required term p + (x2 + xy + y2) = 2x2 + 3xy 11. p = (2x2 + 3xy) – (x2 + xy + y2) p = 2x2 + 3xy – x2 – xy – y2 p = 2x2 – x2 + 3xy – xy – y2 p = x2 + 2xy – y2 5. What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20? Let us assume a be the required term 3x2 – 4y2 + 5xy + 20 – a = -x2 – y2 + 6xy + 20 a = 3x2 – 4y2 + 5xy + 20 – (-x2 – y2 + 6xy + 20) a = 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20 a = 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20 a = 4x2 – 3y2 – xy 6. (a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11. First we have to find out the sum of 3x – y + 11 and – y – 11 = 3x – y + 11 + (-y – 11) = 3x – y + 11 – y – 11 = 3x – y – y + 11 – 11 = 3x – 2y Now, subtract 3x – y – 11 from 3x – 2y = 3x – 2y – (3x – y – 11) = 3x – 2y – 3x + y + 11 = 3x – 3x – 2y + y + 11 = -y + 11 (b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and –x2 + 2x + 5. First we have to find out the sum of 4 + 3x and 5 – 4x + 2x2 = 4 + 3x + (5 – 4x + 2x2) = 4 + 3x + 5 – 4x + 2x2 = 4 + 5 + 3x – 4x + 2x2 = 9 – x + 2x2 = 2x2 – x + 9 … [equation 1] Then, we have to find out the sum of 3x2 – 5x and – x2 + 2x + 5 = 3x2 – 5x + (-x2 + 2x + 5) 12. = 3x2 – 5x – x2 + 2x + 5 = 3x2 – x2 – 5x + 2x + 5 = 2x2 – 3x + 5 … [equation 2] Now, we have to subtract equation (2) from equation (1) = 2x2 – x + 9 – (2x2 – 3x + 5) = 2x2 – x + 9 – 2x2 + 3x – 5 = 2x2 – 2x2 – x + 3x + 9 – 5 = 2x + 4 Exercise 12.3 Page: 242 1. If m = 2, find the value of: (i) m – 2 From the question it is given that m = 2 Then, substitute the value of m in the question = 2 -2 (ii) 3m – 5 From the question it is given that m = 2 Then, substitute the value of m in the question = (3 × 2) – 5 (iii) 9 – 5m From the question it is given that m = 2 Then, substitute the value of m in the question = 9 – (5 × 2) = 9 – 10 (iv) 3m2 – 2m – 7 From the question it is given that m = 2 Then, substitute the value of m in the question 13. = (3 × 22) – (2 × 2) – 7 = (3 × 4) – (4) – 7 = 12 – 4 -7 = 12 – 11 (v) (5m/2) – 4 From the question it is given that m = 2 Then, substitute the value of m in the question = ((5 × 2)/2) – 4 = (10/2) – 4 2. If p = – 2, find the value of: (i) 4p + 7 From the question it is given that p = -2 Then, substitute the value of p in the question = (4 × (-2)) + 7 = -8 + 7 = -1 (ii) – 3p2 + 4p + 7 From the question it is given that p = -2 Then, substitute the value of p in the question = (-3 × (-2)2) + (4 × (-2)) + 7 = (-3 × 4) + (-8) + 7 = -12 – 8 + 7 = -20 + 7 = -13 (iii) – 2p3 – 3p2 + 4p + 7 From the question it is given that p = -2 Then, substitute the value of p in the question = (-2 × (-2)3) – (3 × (-2)2) + (4 × (-2)) + 7 14. = (-2 × -8) – (3 × 4) + (-8) + 7 = 16 – 12 – 8 + 7 = 23 – 20 3. Find the value of the following expressions, when x = –1: (i) 2x – 7 From the question it is given that x = -1 Then, substitute the value of x in the question = (2 × -1) – 7 (ii) – x + 2 From the question it is given that x = -1 Then, substitute the value of x in the question = – (-1) + 2 (iii) x2 + 2x + 1 From the question it is given that x = -1 Then, substitute the value of x in the question = (-1)2 + (2 × -1) + 1 (iv) 2x2 – x – 2 From the question it is given that x = -1 Then, substitute the value of x in the question = (2 × (-1)2) – (-1) – 2 = (2 × 1) + 1 – 2 15. 4. If a = 2, b = – 2, find the value of: (i) a2 + b2 From the question it is given that a = 2, b = -2 Then, substitute the value of a and b in the question = (2)2 + (-2)2 5. When a = 0, b = – 1, find the value of the given expressions: (i) 2a + 2b From the question it is given that a = 0, b = -1 Then, substitute the value of a and b in the question = (2 × 0) + (2 × -1) = -2 (ii) 2a2 + b2 + 1 From the question it is given that a = 0, b = -1 Then, substitute the value of a and b in the question = (2 × 02) + (-1)2 + 1 (iii) 2a2b + 2ab2 + ab From the question it is given that a = 0, b = -1 Then, substitute the value of a and b in the question = (2 × 02 × -1) + (2 × 0 × (-1)2) + (0 × -1) = 0 + 0 +0 (iv) a2 + ab + 2 From the question it is given that a = 0, b = -1 Then, substitute the value of a and b in the question = (02) + (0 × (-1)) + 2 16. 6. Simplify the expressions and find the value if x is equal to 2 (i) x + 7 + 4 (x – 5) From the question it is given that x = 2 We have, = x + 7 + 4x – 20 = 5x + 7 – 20 Then, substitute the value of x in the equation = (5 × 2) + 7 – 20 = 10 + 7 – 20 = 17 – 20 7. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2. (i) 3x – 5 – x + 9 From the question it is given that x = 3 We have, = 3x – x – 5 + 9 = 2x + 4 Then, substitute the value of x in the equation = (2 × 3) + 4 = 10 (ii) 2 – 8x + 4x + 4 From the question it is given that x = 3 We have, = 2 + 4 – 8x + 4x = 6 – 4x Then, substitute the value of x in the equation = 6 – (4 × 3) = 6 – 12 17. (iii) 3a + 5 – 8a + 1 From the question it is given that a = -1 We have, = 3a – 8a + 5 + 1 = – 5a + 6 Then, substitute the value of a in the equation = – (5 × (-1)) + 6 = – (-5) + 6 = 11 (iv) 10 – 3b – 4 – 5b From the question it is given that b = -2 We have, = 10 – 4 – 3b – 5b = 6 – 8b Then, substitute the value of b in the equation = 6 – (8 × (-2)) = 6 – (-16) = 6 + 16 = 22 8. (i) If z = 10, find the value of z3 – 3(z – 10). From the question it is given that z = 10 We have, = z3 – 3z + 30 Then, substitute the value of z in the equation = (10)3 – (3 × 10) + 30 = 1000 – 30 + 30 = 1000 (ii) If p = – 10, find the value of p2 – 2p – 100 18. From the question it is given that p = -10 We have, = p2 – 2p – 100 Then, substitute the value of p in the equation = (-10)2 – (2 × (-10)) – 100 = 100 + 20 – 100 = 20 9. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0? From the question it is given that x = 0 We have, 2x2 + x – a = 5 a = 2x2 + x – 5 Then, substitute the value of x in the equation a = (2 × 02) + 0 – 5 a = -5 10. Simplify the expression and find its value when a = 5 and b = – 3. 2(a2 + ab) + 3 – ab From the question it is given that a = 5 and b = -3 We have, = 2a2 + 2ab + 3 – ab = 2a2 + ab + 3 Then, substitute the value of a and b in the equation = (2 × 52) + (5 × (-3)) + 3 = (2 × 25) + (-15) + 3 = 50 – 15 + 3 = 53 – 15 = 38 ****************************
# Solve using quadratic formula solver ## Solving using quadratic formula solver The solver will provide step-by-step instructions on how to Solve using quadratic formula solver. Long division is the process of calculating a long number in two or more steps. Long division is useful for calculating a large number that cannot be calculated in one step, such as the area of a shape or the sum of money owed. Long division is also used to calculate change. The steps of long division include: There are several different ways to solve long division. These include: To solve long division by hand, start with the left-most number, then add your divisor and continue to the right; To solve long division by calculator, enter all numbers into the calculator and press the "=" button; To solve long division by computer software, use online calculators or online software programs; To solve long division by machine, use a large-scale calculator that can handle large numbers. If there are n equations, then you can solve them by dividing the n terms into two groups of m equations. This way, you are only solving for m terms in each group. Let's take a look at an example: In this example, there are 2 x's and 3 y's. So you divide the 2 x's into 2 groups of 1 x and 1 x. Then you divide the 3 y's into 3 groups of 1 y. You now have 6 pairs of equations: 2x = 1x + 1 y = 2y – 1 y = 1y + 2y –1 To solve each pair, you first set up a new equation that says x = y (you can see this by squaring both sides), then solve it using your original set of equations. The equation will end up being true if one side is equal to the other and false otherwise - so we'd get either true or false depending on x being equal to y. When we're done, we have our solution: x = 2y - 1. When we were just solving for one x and one y, we had three equations instead of six. We doubled our efficiency by dividing the two terms into two groups of two instead of having to deal with all three equations separately. Now let's do another example: In this example, there are 3x + 8y + 12 Rational expressions are made up of terms and variables. The first step in solving a rational expression is to break it down into terms and variables. After the terms and variables are identified, you can then use the rules for adding and subtracting fractions to solve for the unknown quantity. Finally, you may need to simplify the expression by combining like terms. For example, let's say you're asked to find . To begin, you must identify each term in the expression: . Because there are two terms and , we can add them together: 2 + 3 = 5. Now that we have both of the terms in our expression, we can use the rules for addition to solve for : + = 2. If this is not what you were expecting, don't worry! It is possible to get this wrong too. In fact, sometimes when solving rational expressions, a common mistake is to add or subtract two of the same number (e.g., adding 2 + 4 instead of 2 + 1). Any time you make an addition that produces a fraction with zero denominators (i.e., a fraction with no whole numbers), it's called a "zero-addition." When you make a subtraction like above, it's known as a "zero-subtraction." A rational expression cannot be simplified like this; either you will have to cancel out the fractions or leave some of them A solver is a piece of software that tries to solve an equation or a problem. Solvers are used when you know how to solve an equation or a problem, but you don’t have the tools to do it. For example, if you know how to calculate the volume of a cube, but don’t have access to a mathematical calculator, then you can use a computer program called a solver to calculate the volume and get the answer. Solvers can be used in many different ways. For example, they can be used to evaluate the solution of an equation, or they can be used to optimize processes. In general, solvers are used in situations where there is some type of constraint on an input. They use this constraint to make a decision about what values should be produced next. Solvers can also be used as part of optimization problems. This is especially true when algorithms are being developed. In these cases, solvers can be used to find optimal solutions for the algorithm that was developed. Solvers are usually written in either Python or C++, although there are other languages that may be used for specific purposes. There are also many different types of solver applications available today. Because of this, it is important for people who want to use solvers for their work to understand how each one works so that they can choose the right one for their Helped me quite a lot during my GCSEs and I continue to use it in my A-levels. The app tells you the answers as well as the workings which can really help you understand how to get to the answers of certain questions with very in-depth analysis. The best app for struggling math students in my opinion. Gwen Patterson Easy to grab, but need some basic knowledge to understand. Its new version must show the alternative form as shown in its earlier version. That helps a lot. Works good and easy to use if you need to show your work it will tell you if you press show me how butine. Emelia Perry Solve this math problem step by step Any math problem solver Algebra solving website Picture solver Math portal equation solver How to do algebra equations
Courses Courses for Kids Free study material Offline Centres More Store # What percent is Rs $63$ of Rs $90$ Last updated date: 20th Sep 2024 Total views: 413.7k Views today: 8.13k Verified 413.7k+ views Hint: To find what percentage of RS$90$ is RS$63$ First we understand percentage means out of $100$ as there total Rs is $90$ so the RS$90$ is $100\%$ let us take an example of a student he gets $10$ out of $10$ marks so he got $100\%$ marks that means here $100\%$ Rupees is $90$ if we consider the percentage of $63$ be $x\%$ By using unitary method on comparing values of percentage and rupees we can equate the values as Rs$90$ being equal to $100\%$ RS$63$ being equal to $x\%$ Complete step-by-step solution: Step1: As it is given Rs$90$ and we have to find percentage of Rs$63$ in $90$ considering Rs$90$ as $100\%$so we can assume Rs$63$ as $x\%$ Step2: on equating that $x\%$ of $90$ equal to $63$ basically assuming the value of percentage be $x$ we will get the equation as $\dfrac{x}{{100}} \times 90 = 63$ On multiplying $63$ to $100$ $x \times 90 = 63 \times 100$ Dividing the equation by $90$ we will get the value of $x$ $x = \dfrac{{63 \times 100}}{{90}}$ Step3: after doing the calculations we get the value of $x$ as $70\%$ $x = 70\%$ Hence value of percentage is equal to $70$ Note: As we know that on finding the percentage of such forms students get confused in forming the equation with $x$ they many times place the $x$ at the wrong place in the equation while equating. There is another method without $x$ to find percentage i.e using unitary method and comparing If Rs$90$is equal to $100$ So Rs$1$ will be $\dfrac{{100}}{{90}}$ Then Rs$63$ will be $\dfrac{{100}}{{90}} \times 63$ on solving the given equation by dividing the equation by $90$ we get $\Rightarrow \dfrac{{10 \times 63}}{{90}} = 70$
# Introduction to Addition and Subtraction of Fractions Fraction is defined as a numerical part of a whole number. Sometimes we want to add or subtract two fractions to find out the total number. For example, Malini ate 1½ roties for lunch and 2½ roties for dinner. What is the total number of roties Malini ate? In the above example, it is necessary to add the fractions. All fractions cannot be added or subtracted easily. The following methods are available to add or subtract fraction. 1. Add or subtract fraction with same denominator 2. Add or subtract fraction with different denominator 3. Add or subtract mixed number fraction Adding or subtracting fractions with same denominator In this method, addition or subtraction of fraction is very easy. Because, the denominators of the two fractions are same. Example: 2/4 + 3/4 = ? In the above fractions, the denominators are same. 2/4 = 1 2 3/4 = 1 2 3 In above fractions, the box is divided into four parts. For 2/4 and 3/4 , take 3 and 1 numerator parts respectively out of 4 equal parts. Steps involved adding fractions, if denominators are same. Step 1 : Check the given fractions, if denominators are same or not. Step 2: If denominators are same, take the numerators of two fractions and add or subtract it. Step 3: Give a final answer with denominator. Example 1: 5/8 + 2/8 = ? Step 1 : Denominators are same. Step 2: Take the numerators 5 and 2 respectively and add it = (5+2)/8 = 7/8 Example 2: 5/82/8 = ? Step 1 : Denominators are same. Step 2: Take the numerators 5 and 2 respectively and subtract the smaller from the bigger numerator. = (5-2)/8 = 3/8 Adding or subtracting fractions with different denominator In this method, the denominators are not the same in two fractions. So we need to make them same. In this method, the numerators and denominators are different in the two fractions. Let us add two fractions, 3/8 and 5/12. In above fractions, numerators and denominators are different. Solution: Step 1: Take LCM for denominators of above fractions. i.e. 8 and 12 respectively. (LCM is a smallest number which is used as common multiple of two numbers) 24 is a common multiple for 8 and 12. Step 2: Convert denominators 8 and 12 into 24. In 3/8, multiply numerator and denominator by 3 = 3/8 x3/3 = 9/24 In 5/12, multiply numerator and denominator by 2 = 5/12x 2/2 = 10/24 (Note: We must do to the numerator what we do to the denominator) Step 3: Now, the denominators of the two fractions are same. 9/24 , 10/24 Step 4: Take the numerators 9 and 10 respectively and add it = (9+10)/24 Step 5 : Give a final answer = 19/24 Example: 3/8 5/12 =? Solution: Denominators are different. So, take Least Common Multiple (LCM). 24 is a common multiple of 8 and 12. 3/8 => 3/8 x3/3 = 9/24 5/12 => 5/12x 2/2 = 10/24 3/85/12 = 9/2410/24 = 9-10/24 (subtract the smaller numerator from the bigger) mixed fraction is defined as a fraction and a whole number combined into one “mixed” number. Depends on the denominator, two following methods of available for add or subtract of mixed fraction. 1. If same denominators are present in the mixed fraction. 2. If different denominators are present in the mixed fraction. If same denominators are present in the mixed fraction In this method, mixed fraction consists of same denominators. The following steps are involved to add or subtract the mixed fraction. Step 1 : Add the whole number of two fraction separately. Step 2: Add the fractions separately. Step 3 : Combine the whole number and the fraction both Step 3: Convert improper fraction into mixed fraction. Example: 3 2/5 + 1 4/5 = ? Step 1 : Add the whole number of two fraction separately = 3+1 = 4 — (1) Step 2: Add the fractions separately = 2/5 + 4/5 = 6/5 —– (2) Step 3 : Combine the both equation (1) and (2) = 4 6/5 —— (3) In 4 6/5, 6/5 is an improper fraction. So, convert into proper mixed fraction. Equation (2) —–> 6/5 = 1 1/5 —– (3) Step 4: Now, add equation (1) and (3) 4 + 1 1/5 = 5 1/5 If different denominators are present in the mixed fraction In this method, mixed fraction consists of different denominators. The following steps are involved to add or subtract the mixed fraction. Example: 6 3/4 + 3 5/8 = ? Solution: In the above mixed fractions, the denominators are different. So we need to make them same. Step 1: Take LCM for denominators of above fractions. i.e. 4 and 8 respectively. (LCM is a smallest number which is used as common multiple of two numbers) 8 is a common multiple for 4 and 8. Step 2: Convert denominators 4 and 8 into 8. In 3/4, multiply numerator and denominator by 2 = 3/4 x2/2 = 6/8 In 5/8, multiply numerator and denominator by 1 = 5/8x 1/1 = 5/8 (Note: We must do to the numerator what we do to the denominator) Now, denominators are same in the mixed fraction. 6 3/4 + 3 5/8 = 6 6/8 + 3 5/8 Step 3 : Add the whole number of two fraction separately = 6+3 = 9 ———— (1) Step 4: Add the fractions separately = 6/8 + 5/8= 11/8 —– (2) 11/8 is an improper fraction. So, convert into proper fraction. 11/8 = 1 3/8 ————(3) Step 5 : Combine the both equation (1) and (3) 9+ 1 3/8 = 10 3/8
# College Math Teaching ## August 20, 2018 ### Algebra for Calculus I: equations and inequalities Filed under: basic algebra, calculus, pedagogy — collegemathteaching @ 9:24 pm It seems simple enough: solve $3x+ 4 = 7$ or $\frac{2}{x-5} \leq 3$. So what do we tell our students to do? We might say things like “with an equation we must do the same thing to both sides of the equation (other than multiply both sides by zero)” and with an inequality, “we have to remember to reverse the inequality if we, say, multiply both sides by a negative number or if we take the reciprocal”. And, of course, we need to check afterwards to see if we haven’t improperly expanded the solution set. But what is really going on? A moment’s thought will reveal that what we are doing is applying the appropriate function to both sides of the equation/inequality. And, depending on what we are doing, we want to ensure that the function that we are applying is one-to-one and taking note if the function is increasing or decreasing in the event we are solving an inequality. Example: $x + \sqrt{x+2} = 4$ Now the standard way is to subtract $x$ from both sides (which is a one to one function..subtract constant number) which yields $\sqrt{x+2} = 4-x$. Now we might say “square both sides” to obtain $x+2 = 16-8x+x^2 \rightarrow x^2-9x+ 14 = 0 \rightarrow (x-7)(x-2) = 0$ but only $x = 2$ works. But the function that does that, the “squaring function”, is NOT one to one. Think of it this way: if we have $x = y$ and we then square both sides we now have $x^2 = y^2$ which has the original solution $x = y$ and $x = -y$. So in our example, the extraneous solution occurs because $(\sqrt{7+2})^2 = (4-7)^2$ but $\sqrt{7+2} \neq -3$. If you want to have more fun, try a function that isn’t even close to being one to one; e. g. solve $x + \frac{1}{4} =\frac{1}{2}$ by taking the sine of both sides. 🙂 (yes, I know, NO ONE would want to do that). As far as inequalities: the idea is to remember that if one applies a one-to-one function on both sides, one should note if the function is increasing or decreasing. Example: $2 \geq e^{-x} \rightarrow ln(2) \geq -x \rightarrow ln(\frac{1}{2}) \leq x$. We did the switch when the function that we applied ($f(x) = -x$ was decreasing.) Example: solving $\|x+9\| \geq 8$ requires that we use the conditional definition for absolute value and reconcile our two answers: $x+ 9 \geq 8$ and $-x-9 \geq 8$ which leads to the union of $x \geq -1$ or $x \leq -17$ The fun starts when the function that we apply is neither decreasing nor increasing. Example: $sin(x) \geq \frac{1}{2}$ Needless to say, the $arcsin(x)$ function, by itself, is inadequate without adjusting for periodicity. ## October 23, 2014 ### Being part of the problem… Filed under: academia, basic algebra, editorial — Tags: , — collegemathteaching @ 10:55 am An eight-month probe has estimated that the “shadow curriculum” that existed at the University of North Carolina from 1993 to 2011 offered a grade-point boost from phony coursework to more than 3,100 students, including a disproportionately high percentage of student-athletes.[…] A number of the students’ papers included brief introductions and closings with only “fluff” in between, according to the report. “Hundreds” of the courses, the report reads, were independent-study. Yet when the university tightened standards on the amount of independent study a student could undertake, Crowder altered her program, creating courses she identified as lecture courses, but which mirrored independent study in that lectures never happened. The Wainstein report found 188 such courses between 1999 and 2011, in which 47.4 percent of the enrollments in these “paper classes” were student-athletes, who generally comprise 4 percent of the student population. Once Crowder retired in 2009, Nyang’oro sustained the practices for two more years until his retirement in 2011, albeit less voluminously. In all the “paper classes,” the report found an average issued grade of 3.62, set against 3.28 for the regular classes in the department. Among student-athletes, though, the report found an average issued grade of 3.55, further above the 2.84 among student-athletes in the regular department classes. The grade was based on one paper, which wasn’t merely scanned. But note the following: the average issued grade IN THE DEPARTMENT was 3.28??? Now this is far from the first public scandal involving university athletes, but university athletes aren’t the focus of this post. I am thinking about what I have been complicit in over the years. No, I’ve never given credit for incorrect work. And no, I’ve never falsified a grade. When I was young, my dad wanted me to learn how to hit a baseball. So, he’d toss a ball at me when I was holding a bat. BUT, he’d often..deliberately hit my bat with the ball, so I got the impression that I was hitting the ball. And, at times, it appears that I do that with my exams; I hit their bats. If I gave what I considered honest exams in calculus and assigned partial credit honestly, I’d say that 70 percent of my “C” students would get D’s or F’s. About 50 percent of my B students would make C’s and perhaps 50 percent of my A students could get B’s. And I’d get called on the carpet…get complaints and accused of “not being student centered”…of “lacking compassion.” What is going on? Well, though we have a smaller percentage of students going to college now than in, say, 2009, check out the data: When I was an undergraduate, about 50 percent of high school graduates went to college. When I started teaching college, it was about 60 percent…it rose to roughly 70 percent and is now about 65 percent. I think that we are seeing at least a little bit of “regression to the mean” effects. Perhaps as a result, we are seeing things like this: Algebra is an onerous stumbling block for all kinds of students: disadvantaged and affluent, black and white. In New Mexico, 43 percent of white students fell below “proficient,” along with 39 percent in Tennessee. Even well-endowed schools have otherwise talented students who are impeded by algebra, to say nothing of calculus and trigonometry. California’s two university systems, for instance, consider applications only from students who have taken three years of mathematics and in that way exclude many applicants who might excel in fields like art or history. Community college students face an equally prohibitive mathematics wall. A study of two-year schools found that fewer than a quarter of their entrants passed the algebra classes they were required to take. “There are students taking these courses three, four, five times,” says Barbara Bonham of Appalachian State University. While some ultimately pass, she adds, “many drop out.” ALGEBRA? (and no, I am not talking about group theory or ring theory!) Yes, that algebra. So there are times when I wonder if I am participating in a fraud of a type. ## September 19, 2014 ### Freshman calculus: they don’t always know the basics…. Filed under: basic algebra, calculus — Tags: , — collegemathteaching @ 5:43 pm Example one: many students don’t know that $\frac{\frac{a}{b}}{c} \ne \frac{a}{\frac{b}{c}}$ (of course, assume that $a, b, c, \ne 0$. ) Where this came up: when we computed $lim_{h \rightarrow 0} \frac{\frac{1}{1+h} - 1}{h}$ we obtained $lim_{h \rightarrow 0} \frac{\frac{-h}{1+h}}{h}$ and a student didn’t understand why this was equal to $lim_{h \rightarrow 0} \frac{-1}{1+h}$ I ended up asking the student to simplify $\frac{\frac{2}{3}}{2} =$ and ….asking: ok, “if I have 2/3’rds of a pie and I give two people an equal piece of that, how much pie does each person get? Example two: I gave “find the domain of $\frac{1}{\sqrt{x^2 - 9}}$ and two students didn’t understand why an answer of $-3 > x > 3$ was logically impossible. One of them told me: “my calculus teacher in high school told me to do it this way”: I am 99.99 percent that this isn’t true, but, well, I stayed with it until the student understood why such a statement was logically impossible. Oh yes, this same “I had calculus in high school” student was sure that I was wrong when I told him that “the derivative of a constant function is zero”; he was SURE that it is “1”. ## August 27, 2014 ### Nice collection of Math GIFs Filed under: basic algebra, calculus, elementary mathematics, pedagogy — Tags: — collegemathteaching @ 12:25 am No, these GIF won’t explain better than an instructor, but I found many of them to be fun. Via: IFLS ## August 21, 2014 ### Calculation of the Fourier Transform of a tent map, with a calculus tip…. I’ve been following these excellent lectures by Professor Brad Osgood of Stanford. As an aside: yes, he is dynamite in the classroom, but there is probably a reason that Stanford is featuring him. 🙂 And yes, his style is good for obtaining a feeling of comradery that is absent in my classroom; at least in the lower division “service” classes. This lecture takes us from Fourier Series to Fourier Transforms. Of course, he admits that the transition here is really a heuristic trick with symbolism; it isn’t a bad way to initiate an intuitive feel for the subject though. However, the point of this post is to offer a “algebra of calculus trick” for dealing with the sort of calculations that one might encounter. By the way, if you say “hey, just use a calculator” you will be BANNED from this blog!!!! (just kidding…sort of. 🙂 ) So here is the deal: let $f(x)$ represent the tent map: the support of $f$ is $[-1,1]$ and it has the following graph: The formula is: $f(x)=\left\{\begin{array}{c} x+1,x \in [-1,0) \\ 1-x ,x\in [0,1] \\ 0 \text{ elsewhere} \end{array}\right.$ So, the Fourier Transform is $F(f) = \int^{\infty}_{-\infty} e^{-2 \pi i st}f(t)dt = \int^0_{-1} e^{-2 \pi i st}(1+t)dt + \int^1_0e^{-2 \pi i st}(1-t)dt$ Now, this is an easy integral to do, conceptually, but there is the issue of carrying constants around and being tempted to make “on the fly” simplifications along the way, thereby leading to irritating algebraic errors. So my tip: just let $a = -2 \pi i s$ and do the integrals: $\int^0_{-1} e^{at}(1+t)dt + \int^1_0e^{at}(1-t)dt$ and substitute and simplify later: Now the integrals become: $\int^{1}_{-1} e^{at}dt + \int^0_{-1}te^{at}dt - \int^1_0 te^{at} dt.$ These are easy to do; the first is merely $\frac{1}{a}(e^a - e^{-a})$ and the next two have the same anti-derivative which can be obtained by a “integration by parts” calculation: $\frac{t}{a}e^{at} -\frac{1}{a^2}e^{at}$; evaluating the limits yields: $-\frac{1}{a^2}-(\frac{-1}{a}e^{-a} -\frac{1}{a^2}e^{-a}) - (\frac{1}{a}e^{a} -\frac{1}{a^2}e^a)+ (-\frac{1}{a^2})$ Add the first integral and simplify and we get: $-\frac{1}{a^2}(2 - (e^{-a} -e^{a})$. NOW use $a = -2\pi i s$ and we have the integral is $\frac{1}{4 \pi^2 s^2}(2 -(e^{2 \pi i s} -e^{-2 \pi i s}) = \frac{1}{4 \pi^2 s^2}(2 - cos(2 \pi s))$ by Euler’s formula. Now we need some trig to get this into a form that is “engineering/scientist” friendly; here we turn to the formula: $sin^2(x) = \frac{1}{2}(1-cos(2x))$ so $2 - cos(2 \pi s) = 4sin^2(\pi s)$ so our answer is $\frac{sin^2( \pi s)}{(\pi s)^2} = (\frac{sin(\pi s)}{\pi s})^2$ which is often denoted as $(sinc(s))^2$ as the “normalized” $sinc(x)$ function is given by $\frac{sinc(\pi x)}{\pi x}$ (as we want the function to have zeros at integers and to “equal” one at $x = 0$ (remember that famous limit!) So, the point is that using $a$ made the algebra a whole lot easier. Now, if you are shaking your head and muttering about how this calculation was crude that that one usually uses “convolution” instead: this post is probably too elementary for you. 🙂 ## August 7, 2014 ### Letting complex algebra make our calculus lives easier Filed under: basic algebra, calculus, complex variables — Tags: , — collegemathteaching @ 1:37 am If one wants to use complex arithmetic in elementary calculus, one should, of course, verify a few things first. One might talk about elementary complex arithmetic and about complex valued functions of a real variable at an elementary level; e. g. $f(x) + ig(x)$. Then one might discuss Euler’s formula: $e^{ix} = cos(x) + isin(x)$ and show that the usual laws of differentiation hold; i. e. show that $\frac{d}{dx} e^{ix} = ie^{ix}$ and one might show that $(e^{ix})^k = e^{ikx}$ for $k$ an integer. The latter involves some dreary trigonometry but, by doing this ONCE at the outset, one is spared of having to repeat it later. This is what I mean: suppose we encounter $cos^n(x)$ where $n$ is an even integer. I use an even integer power because $\int cos^n(x) dx$ is more challenging to evaluate when $n$ is even. Coming up with the general formula can be left as an exercise in using the binomial theorem. But I’ll demonstrate what is going on when, say, $n = 8$. $cos^8(x) = (\frac{e^{ix} + e^{-ix}}{2})^8 =$ $\frac{1}{2^8} (e^{i8x} + 8 e^{i7x}e^{-ix} + 28 e^{i6x}e^{-i2x} + 56 e^{i5x}e^{-i3x} + 70e^{i4x}e^{-i4x} + 56 e^{i3x}e^{-i5x} + 28e^{i2x}e^{-i6x} + 8 e^{ix}e^{-i7x} + e^{-i8x})$ $= \frac{1}{2^8}((e^{i8x}+e^{-i8x}) + 8(e^{i6x}+e^{-i6x}) + 28(e^{i4x}+e^{-i4x})+ 56(e^{i2x}+e^{-i2x})+ 70) =$ $\frac{70}{2^8} + \frac{1}{2^7}(cos(8x) + 8cos(6x) + 28cos(4x) +56cos(2x))$ So it follows reasonably easily that, for $n$ even, $cos^n(x) = \frac{1}{2^{n-1}}\Sigma^{\frac{n}{2}-1}_{k=0} (\binom{n}{k}cos((n-2k)x)+\frac{\binom{n}{\frac{n}{2}}}{2^n}$ So integration should be a breeze. Lets see about things like, say, $cos(kx)sin(nx) = \frac{1}{(2)(2i)} (e^{ikx}+e^{-ikx})(e^{inx}-ie^{-inx}) =$ $\frac{1}{4i}((e^{i(k+n)x} - e^{-i(k+n)x}) + (e^{i(n-k)x}-e^{-i(n-k)x}) = \frac{1}{2}(sin((k+n)x) + sin((n-k)x)$ Of course these are known formulas, but their derivation is relatively simple when one uses complex expressions. ## July 17, 2014 ### Parabolas have interesting properties And this is one of them: ## July 15, 2014 ### A quickie from Mathematics Magazine Filed under: basic algebra, editorial, elementary mathematics, recreational mathematics — Tags: — collegemathteaching @ 9:35 pm It has been a long time since I’ve posted; I’ve spent time doing various things, including revising a paper that a journal editor wanted revised. I’ll speak more about that later. I got the latest Mathematics Magazine in the mail (volume 87, No. 3, June 2014), and the article “Surprises” by Felix Lazebnik is chock full of delightful tidbits, many of which I didn’t know. Here is a fun one that you can share with your non-mathematically inclined friends (other tidbits there require some mathematics background). (Surprise 7): A watermelon is 99 percent water (by weight). One ton of watermelons was shipped and during shipment some water evaporated. The watermelons that arrived were made up of 98 percent water (by weight). What was the weight of the shipment when it arrived? The answer is in the article, but I suggest you give it a go. It will take, at most, a minute or two (if that). This shows the power of basic algebra to discipline our thinking and how our intuition can deceive us. ## November 21, 2013 ### I am a frigging IDIOT!!!! (aka “Math Professor FAIL”) Filed under: basic algebra, calculus, series — Tags: , , — collegemathteaching @ 5:17 pm The topic: power series. I was showing how to manipulate the relation $\frac{1}{1-x} = \sum^{\infty}_{k=0} x^k$ to get power series of various kinds, including series centered at values other than zero. For example, if one wanted the series centered at, say, $x = 2$ we could write $\frac{1}{1-x} = \frac{1}{1-(x-2)-2} = \frac{1}{-1-(x-2)}=\frac{-1}{1-(-1)(x-2)}$ $= -\sum^{\infty}_{k=0} (-1)^k (x-2)^k = \sum^{\infty}_{k=0} (-1)^{k+1} (x-2)^k$. And so I said: Let’s try to find the series for $\frac{1}{1-x}$ centered at $x = 1$. And. I. Was. Not. Able. To. Do. It. So I went on to another problem…and it hit me. O. M. G. (slamming my head on the desk here…dying of embarrassment). Moral: never get overconfident. 🙂 Yes, I told the class of my blunder as soon as it hit me; they laughed…with me, I think. 😉 ## November 6, 2013 ### Inverse Laplace transform example: 1/(s^2 +b^2)^2 Filed under: basic algebra, differential equations, Laplace transform — Tags: — collegemathteaching @ 11:33 pm I talked about one way to solve $y''+y = sin(t), y(0) =y'(0) = 0$ by using Laplace transforms WITHOUT using convolutions; I happen to think that using convolutions is the easiest way here. Here is another non-convolution method: Take the Laplace transform of both sides to get $Y(s) = \frac{1}{(s^2+1)^2}$. Now most tables have $L(tsin(at)) = \frac{2as}{(s^2 + a^2)^2}, L(tcos(at)) = \frac{s^2-a^2}{(s^2+a^2)^2}$ What we have is not in one of these forms. BUT, note the following algebra trick technique: $\frac{1}{s^2+b^2} = (A)(\frac{s^2-b^2}{(s^2 + b^2)^2} - \frac{s^2+b^2}{(s^2+b^2)^2})$ when $A = -\frac{1}{2b^2}$. Now $\frac{s^2-b^2}{(s^2 + b^2)^2} = L(tcos(bt))$ and $\frac{s^2+b^2}{(s^2+b^2)^2} = \frac{1}{(s^2+b^2)} = L(\frac{1}{b}sin(bt))$ and one can proceed from there. Older Posts »
# Bar Model Approach for Averages of Numbers in Statistics ##### Ho Soo Thong Abstract This article illustrates the use of mathematical implications of the Arithmetic Distributive Law for solving problems involving averages at Primary School Level. In algebraic expressions, we write a = b + c and na = nb + nc or a − b = c and na − nb = nc Figure 1 shows the bar modelling of the mathematical implications of the Distributive Law over Addition. Bar Modelling of the Average of a list of numbers. At Primary school level, the average for a list of numbers is the arithmetic mean given by For example, the average of the five numbers 42, 43, 44, 51 and 70 is (42 + 43 + 44 + 51 + 70)÷ 5 = 50 This gives the mathematical situation 42 + 43 + 44 + 51 + 70 = 5×50 Figure 2 shows the bar modelling of this mathematical situation. Figure 3 shows the bar modelling of another mathematical situation that displays the smallest number and the average of the remaining four numbers. Figure 4 shows a different mathematical situation that displays the smallest number, the highest number and the average of the remaining three numbers. ## Problems at Primary School Level we begin with a simple example. ### Example 1 The average of eight whole numbers is 50. If two of the numbers are 64 and 84, what is the average of the remaining six numbers? ### Solution Figure 5 shows two bar models for the two mathematical situations: 1. Bar Model with the average of eight numbers 2. Bar Model with 40 and 84, and the average of the other six numbers. Figure 5 shows two bar models for the two mathematical situations: 1. Bar Model with the average of eight numbers 2. Bar Model with 64 and 84, and the average of the other six numbers. Therefore, the average of the other numbers is 42. Figure 6 shows an alternative approach using the mathematical implications of the arithmetic distributive law. Next, we will apply the mathematical implications in Distributive Law to deal with word problems involving changes in averages. ### Example 2 In a school financial assistance scheme, a certain amount of donated cash is equally distributed among 8 pupils. After further review, the school decided to include three more pupils. As a result, each pupil will receive \$ 150 less. How much will each pupil get after the review? ### Solution In Figure 7, we construct a comparison bar model of the mathematical situations for before and after review and deduce the revised equal amount of \$ 400. After the review, each pupil will receive \$ 400. ### Example 3 Ms Chan gave her pupils a mathematics test. After marking their papers, she noted that the average score was 62.1. Later, she discovered that a test score 37 should have been a score of 73. The revised average was 63.6. How many pupils took the test? ### Solution In Figure 8, the average score increases by 63.3 − 62.1 = 1.5 after correction and the total score increases by 73 − 37 = 36 as shown. Therefore, there are 24 ( = 36 ÷1.5) pupils. Remark Pupils can apply this problem solving approach to solve harder problems at Primary Olympiad level. References [1] Ho Soo Thong, Distributive Law for Excess and Shortage Problems in PSLE Math, barmodelhost.com, Feb 2015 [2] Ho Soo Thong, Distributive Law for Indeterminate Problems, barmodelhost.com, June 2015
Request a call back # Class 9 NCERT Solutions Maths Chapter 7 - Triangles Do you struggle to prove that the two triangles given in a textbook question are congruent? Do you find it difficult to explain the concept of SSS congruence in a Maths problem? If yes, then our NCERT Solutions for CBSE Class 9 Mathematics Chapter 7 Triangles will be of immense help. These are model answers that Maths experts have developed to support students for conceptual clarity. TopperLearning’s NCERT textbook solutions are easily available online for students trying to practise and understand the properties of triangles. You can use these chapter solutions along with our practice tests, concept videos and CBSE Class 9 sample papers to effectively revise your Maths lessons. ## Triangles Exercise Ex. 7.1 ### Solution 1 In ABC and ABD CAB = DAB (given) AB = AB (common) So, BC and BD are of equal length. ### Solution 2 In ABD and BAC DAB = CBA (given) AB = BA (common) And ABD = BAC (by CPCT) ### Solution 3 In BOC and AOD BOC = AOD (vertically opposite angles) CBO = DAO (each 90o) ### Solution 6 BAD + DAC = EAC + DAC BAC = DAE Now in BAC and DAE BAC = DAE (proved above) AC = AE (given) ### Solution 7 Given that EPA = DPB EPA + DPE = DPB + DPE DPA = EPB Now in DAP and EBP DAP = EBP (given) AP = BP (P is mid point of AB) DPA = EPB (from above) ### Solution 8 (i) In AMC and BMD AM = BM (M is mid point of AB) AMC = BMD (vertically opposite angles) CM = DM (given) (ii) We have ACM = BDM But ACM and BDM are alternate interior angles Since alternate angles are equal. Hence, we can say that DB \|\| AC DBC + ACB = 180o (co-interior angles) DBC + 90o = 180o DBC + 90o = 1800 DBC = 90o (iii) Now in DBC and ACB DBC = ACB (each 90o ) BC = CB (Common) (iv) We have DBC ACB ## Triangles Exercise Ex. 7.2 ### Solution 1 (i) It is given that in triangle ABC, AC = AB ACB = ABC (angles opposite to equal sides of a triangle are equal) OBC = OBC OB = OC (sides opposite to equal angles of a triangle are also equal) (ii) Now in OAB and OAC AO =AO (common) AB = AC (given) OB = OC (proved above) So, OAB OAC (by SSS congruence rule) BAO = CAO (C.P.C.T.) ### Solution 2 CD = BD (AD is the perpendicular bisector of BC) ### Solution 3 In AEB and AFC AEB = AFC (each 90o) A = A (common angle) AB = AC (given) ### Solution 4 (i) In AEB and AFC AEB = AFC (each 90) A = A (common angle) BE = CF (given) AEB AFC AB = AC (by CPCT) ### Solution 5 In ABD and ACD AB = AC (Given) BD = CD (Given) ### Solution 6 In ABC AB = AC (given) ACB = ABC (angles opposite to equal sides of a triangle are also equal) Now In ACD ADC = ACD (angles opposite to equal sides of a triangle are also equal) Now, in BCD ABC + BCD + ADC = 180o (angle sum property of a triangle) ACB + ACB +ACD + ACD = 180o 2(ACB + ACD) = 180o 2(BCD) = 180o BCD = 90o ### Solution 7 Given that AB = AC C = B (angles opposite to equal sides are also equal) In ABC, A + B + C = 180o (angle sum property of a triangle) 90o + B + C = 180o 90o + B + B = 180o 2 B = 90o B = 45 ### Solution 8 Let us consider that ABC is an equilateral triangle. So, AB = BC = AC Now, AB = AC C = B (angles opposite to equal sides of a triangle are equal) We also have AC = BC B = A (angles opposite to equal sides of a triangle are equal) So, we have A = B = C Now, in ABC A + B + C = 180o A + A + A = 180o 3A = 180o A = 60o A = B = C = 60o Hence, in an equilateral triangle all interior angles are of 60o. ## Triangles Exercise Ex. 7.3 ### Solution 1 (i) In ABD and ACD AB = AC (given) BD = CD (given) (ii) In ABP and ACP AB = AC (given). BAP = CAP [from equation (1)] AP = AP (common) (iii) From equation (1) BAP = CAP Hence, AP bisect A Now in BDP and CDP BD = CD (given) DP = DP (common) BP = CP [from equation (2)] (iv) We have BDP CDP Now, BPD + CPD = 180o (linear pair angles) BPD + BPD = 180o 2BPD = 180o [from equation (4)] BPD = 90o ...(5) From equations (2) and (5), we can say that AP is perpendicular bisector of BC. ### Solution 2 AB = AC (given) (ii) Also by CPCT, (ii) Also by CPCT, ### Solution 3 (i) In ABC, AM is median to BC BM = BC In PQR, PN is median to QR QN = QR But BC = QR BN = QN ...(i) Now, in ABM and PQN AB = PQ (given) BM = QN [from equation (1)] AM = PN (given) (ii) Now in ABC and PQR AB = PQ (given) ABC = PQR [from equation (2)] BC = QR (given) ABC PQR (by SAS congruence rule) ### Solution 4 In BEC and CFB BEC = CFB (each 90o ) BC = CB (common) BE = CF (given) (Sides opposite to equal angles of a triangle are equal) Hence, ABC is isosceles. ### Solution 5 In APB and APC APB = APC (each 90o) AB =AC (given) AP = AP (common) B = C (by using CPCT) ## Triangles Exercise Ex. 7.4 ### Solution 1 Let us consider a right angled triangle ABC, right angle at B. In ABC A + B + C = 180o (angle sum property of a triangle) A + 90o + C = 180o A + C = 90o Hence, the other two angles have to be acute (i.e. less than 90). [In any triangle, the side opposite to the larger (greater) angle is longer] So, AC is the largest side in ABC. But AC is the hypotenuse of ABC. Therefore, hypotenuse is the longest side in a right angled triangle. ### Solution 2 In the given figure, ABC + PBC = 180p (linear pair) ABC = 180o - PBC ... (1) Also, ACB + QCB = 180o ACB = 180o - QCB ... (2) As PBC < QCB 180 - PBC > 180o - QCB. ABC > ACB [From equations (1) and (2)] AC > AB (side opposite to larger angle is larger) ### Solution 3 In AOB B < A AO < BO (side opposite to smaller angle is smaller) ... (1) Now in COD C < D OD < OC (side opposite to smaller angle is smaller) ... (2) On adding equations (1) and (2), we have AO + OD < BO + OC ### Solution 4 Let us join AC. In ABC AB < BC (AB is smallest side of quadrilateral ABCD) (1) (2) On adding equations (1) and (2), we have 2 + 4 < 1 + 3 C < A A > C Let us join BD. In ABD (3) In BDC BC < CD (CD is the largest side of quadrilateral ABCD) On adding equations (3) and (4), we have 8 + 7 < 5 + 6 D < B B > D ### Solution 5 As PR > PQ PS is the bisector of QPR ### Solution 6 Let us take a line l and from point P (i.e. not on line l) we have drawn two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle. In PNM N = 90o Now, P + N + M = 180o (Angle sum property of a triangle) P + M = 90o Clearly, M is an acute angle ## Triangles Exercise Ex. 7.5 ### Solution 1 Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors, of all the sides of triangles meet together. As here in ABC we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. So O is point which is equidistant from all the vertices of ABC. ### Solution 2 The point which is equidistant from all the sides of a triangle is incenter of triangle. Incentre of triangle is the intersection point of angle bisectors of interior angles of that triangle. Here in ABC we can find the incentre of this triangle by drawing the angle bisectors of interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. So, I is the point, equidistant from all the sides of ABC. ### Solution 3 Ice-cream parlour should be set up at the circumcentre O of ABC. In this situation maximum number of persons can approach to it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the sides of this triangle. ### Solution 4 We may observe that hexagonal shaped rangoly is having 6 equilateral triangles in it. Area of OAB = (side)2 = (5)2
Greatest Common Gaussian Divisor Gaussian integers are complex numbers $$\x+yi\$$ such that $$\x\$$ and $$\y\$$ are both integers, and $$\i^2 = -1\$$. The norm of a Gaussian integer $$\N(x+yi)\$$ is defined as $$\x^2 + y^2 = \|x+yi\|^2\$$. It is possible to define a Euclidean division for Gaussian integers, which means that it is possible to define a Euclidean algorithm to calculate a greatest common divisor for any two Gaussian integers. Unfortunately, a Euclidean algorithm requires a well-defined modulo operation on complex numbers, which most programming languages don't have (e.g. Python, Ruby), meaning that such an algorithm fails. Gaussian division It is possible to define the division $$\\frac a b = x+yi\$$ (where $$\a\$$ and $$\b\$$ are both Gaussian integers) as finding a quotient $$\q\$$ and a remainder $$\r\$$ such that $$a = bq + r, \text{ and } N(r) \le \frac {N(b)} 2$$ We can further limit this to $$\q = m + ni\$$, where $$\-\frac 1 2 < x - m \le \frac 1 2\$$ and $$\-\frac 1 2 < y - n \le \frac 1 2\$$, and $$\r = b(x - m + (y - n)i)\$$ From here, a Euclidean algorithm is possible: repeatedly replace $$\(a, b)\$$ by $$\(b, r)\$$ until it reaches $$\(d, 0)\$$. $$\d\$$ can then be called the greatest common divisor of $$\a\$$ and $$\b\$$ Complex GCDs are not unique; if $$\d = \gcd(a, b)\$$, then $$\d, -d, di, -di\$$ are all GCDs of $$\a\$$ and $$\b\$$ You are to take 2 Gaussian integers $$\a, b\$$ as input and output $$\\gcd(a, b)\$$. You may take input in any convenient method, and any reasonable format, including two complex numbers, two lists of pairs [x, y], [w, z] representing $$\\gcd(x+yi, w+zi)\$$ etc. Additionally, the output format is equally lax. You may output any of the 4 possible values for the GCD, and you don't need to be consistent between inputs. If your language's builtin $$\\gcd\$$ function already handles Gaussian integers, and so would trivially solve this challenge by itself, please add it to the Community Wiki of trivial answers below. This is , so the shortest code in bytes wins. Test cases 5+3i, 2-8i -> 1+i 5+3i, 2+8i -> 5+3i 1-9i, -1-7i -> 1+i -1+0i, 2-10i -> 1+0i (outputting 1 here is also fine) 4+3i, 6-9i -> 1+0i (outputting 1 here is also fine) -3+2i, -3+2i -> 2+3i -6+6i, 3+5i -> 1+i 4+7i, -3-4i -> 2+i -3+4i, -6-2i -> 1+2i 7-7i, -21+21i -> 7+7i Python 3, 70 bytes f=lambda a,b:b and f(b,a-b((t:=a/b+.5+.5j).real//1+t.imag//11j))or a J, 2 bytes +. Try it online! Wolfram Language (Mathematica), 3 bytes GCD Try it online! Pari/GP, 3 bytes gcd Try it online!
# Find the angle in a triangle if the distance between one vertex and orthocenter equals the length of the opposite side Let $O$ be the orthocenter (intersection of heights) of the triangle $ABC$. If $\overline{OC}$ equals $\overline{AB}$, find the angle $\angle$ACB. - Position the circumcenter $P$ of the triangle at the origin, and let the vectors from the $P$ to $A$, $B$, and $C$ be $\vec{A}$, $\vec{B}$, and $\vec{C}$. Then the orthocenter is at $\vec{A}+\vec{B}+\vec{C}$. (Proof: the vector from $A$ to this point is $(\vec{A}+\vec{B}+\vec{C})-\vec{A} = \vec{B}+\vec{C}$. The vector coinciding with the side opposite vertex $A$ is $\vec{B}-\vec{C}$. Now $(\vec{B}+\vec{C})\cdot(\vec{B}-\vec{C}) = \|\vec{B}\|^2 - \|\vec{C}\|^2 = R^2-R^2 = 0$, where $R$ is the circumradius. So the line through $A$ and the head of $\vec{A}+\vec{B}+\vec{C}$ is the altitude to $BC$. Similarly for the other three altitudes.) Now the vector coinciding with $OC$ is $\vec{O}-\vec{C}=\vec{A}+\vec{B}$. Thus $\|OC\|=\|AB\|$ if and only if $$\|\vec{A}+\vec{B}\|^2 = \|\vec{A}-\vec{B}\|^2$$ if and only if $$\vec{A}\cdot\vec{A} + \vec{B}\cdot\vec{B} + 2\vec{A}\cdot\vec{B} = \vec{A}\cdot\vec{A} + \vec{B}\cdot\vec{B} - 2\vec{A}\cdot\vec{B}$$ if and only if $$4\vec{A}\cdot\vec{B} = 0$$ if and only if $$\angle APB = \pi /2$$ if and only if $$\boxed{\angle ACB = \pi/4 = 45^\circ}.$$ - Let point $P$ on $AC$ be the foot of the perpendicular $BO$, and note that $\angle OCA$ is the complement of $A$. Then, $$\begin{eqnarray} \|AC\|&=&\|AP\|+\|PC\|\\ &=&\|AB\|\cos A+\|OC\|\cos\angle OCA \\ &=&\|AB\| \cos A+\|OC\| \sin A \\ &=&\|AB\|(\cos A+\sin A) \end{eqnarray}$$ Conveniently scaling to unit circumdiameter ---so that $\|AC\| = \sin B$, $\|AB\| = \sin C$, and $\|BC\| = \sin A$ (which we may assume is non-zero)--- we have $$\begin{eqnarray} \sin B &=& \sin C \; (\cos A+\sin A) \\ \implies\sin(A+C) &=& \cos A \sin C + \sin A \sin C \\ \implies\sin A \cos C + \cos A \sin C &=& \cos A \sin C + \sin A \sin C \\ \implies\sin A \cos C &=& \sin A \sin C \\ \implies\cos C &=& \sin C \\ \implies C &=& \pi/4 \end{eqnarray}$$ - Assuming there is an answer, then it is $45^\circ$ or $\pi/4$, as a symmetric right-angled triangle (half a square), where $C$ is not the right angle, satisfies this: the orthocentre is at the right angle. -
Home Math Including Numbers in Expanded Kind ### Including Numbers in Expanded Kind [ad_1] Study to rearrange for including numbers in expanded type (briefly type and in lengthy type). The numbers are written in column type; the digits are organized in expanded type after which lastly discover the sum of the given numbers. Solved examples on including numbers in expanded type: 1. Add 32 and 25 Answer: Numbers are written in column type T O 3 2 + 2 5 5 7 Including ones: 2 ones + 5 ones = 7 ones This 7 is written in a single’s column. Including tens: 3 tens + 2 tens = 5 tens This 5 is written in tens column. Thus, reply is 57 2. In different approach, add the expanded type of the numbers 32 + 25 Answer: 32 = 30 + 2 Expanded type of the numbers + 25 = 20 + 5 = 50 + 7 = 57 The numbers are written in expanded type as 32 = 30 + 2 and 25 = 20 + 5. Now digits of 1’s place are added, i.e., 2 + 5 = 7. Once more numbers of ten’s place are added, i.e., 30 + 20 = 50 Subsequently, complete sum = 50 + 7 = 57 3. Add 63 and 35 in lengthy type. Answer: Numbers are positioned in column 63 Sum of digit of 1’s place = 3 ones + 5 ones = 8 ones 35 Sum of digit of ten’s place = 6 ten’s + 3 ten’s = 9 ten’s Subsequently, sum = 98 = 9 ten’s + 8 ones 4. Add 38 and 61 briefly type. Answer: 38 Sum of digit of 1’s place = 8 + 1 = 9 + 61 Sum of digit of ten’s place = 30 + 60 = 90 Subsequently, sum = 99 = 90 + 9 5. Add 23, 31 and 44 Answer: 23 Sum of digit of 1’s place = 3 + 1 + 4 = 8 + 31 Sum of digit of ten’s place = 2 + 3 + 4 = 9 + 44 Subsequently, sum = 98 Extra solved examples on 3-digits for including numbers in expanded type. 6. Add 314 and 563 Answer: First the digits are organized in columns Sum of digit of 1’s place = 4 + 3 = 7 Sum of digit of ten’s place = 1 + 6 = 7 Sum of digit of hundred’s place = 3 + 5 = 8 314 563 Subsequently, sum = 877 7. Add 361 and 527 by arranging the numbers in expanded type. Answer: 361 = 3 lots of + 6 tens + 1 one + 527 = 5 lots of + 2 tens + 7 one Sum = 8 lots of + 8 tens + 8 one =888 or, 361 = 300 + 60 + 1 + 527 = 500 + 20 + 7 Sum = 800 + 80 + 8 = 888 8. Add 231, 402 and 355 Answer: 231 ones = 1 + 2 + 5 = 8 + 402 tens = 3 + 0 + 5 = 8 + 355 lots of = 2 + 4 + 3 = 9 Subsequently, addition sum = 988 2nd Grade Math Apply From Including Numbers in Expanded Kind to HOME PAGE Did not discover what you had been in search of? Or wish to know extra info about Math Solely Math. Use this Google Search to seek out what you want. [ad_2]
The base of a square pyramid, a three-dimensional solid, is a square. Opposite angles of a square are congruent. 2 2 11 3 2 13 1 2 3 1 0 2 2 13 3 5 1 11 REDUCED ROW ECHELON FORM (rref) Reducing a matrix to reduced row echelon form or rref is a means of solving the equations. In Linear algebra, a determinant is a unique number that can be ascertained from a square matrix. The square is the area-maximizing rectangle. Property 5. The diagonals are congruent. 5√95 - 2√50 - 3√180. Solution : = 5 √95 - 2 √50 - 3 √180. The property also benefits from direct access to the common/pool area. In the figure above, we have a square and a circle inside a larger square. The four shapes that we can say necessarily meet the requirements of a parallelogram are square, rectangle, rhombus, and rhomboid. Behold the square. Determinant of a matrix A is denoted by \|A\| or det(A). A chord of a circle divides the circle into two parts such that the squares inscribed in the two parts have areas 16 and 144, respectively. Area of a Square : The Area is the side length squared: Area = a 2 = a × a. Calculate project cost based on price per square foot, square yard or square meter. Acircular city park has a sidewalk directly through the middle that is 111 - feet long. The properties of these two-dimensional shapes are listed below. The rectangle shares this identifying property, so squares are rectangles. There are special types of quadrilateral: Some types are also included in the definition of other types! It is also known as Modulus of Rigidity. Opposite side of a square are parallel: Artists and architects use squares a lot. The radius of the circle is __________.\text{\_\_\_\_\_\_\_\_\_\_}.__________. Properties of square numbers : When a number is multiplied by itself we say that the number can be squared. Learning Outcome You'll be able to describe the properties of squares, rectangles and rhombuses after watching this video lesson. Section Properties of Parallelogram Calculator. KS1 (Age 5-7) KS2 (Age 7-11) 11+ (Age 7-11) KS3 (Age 11-14) GCSE (Age 14-17) Spanish ESL Games Cup of Tea PSHE. Properties of Determinants; Minors and Cofactors of Determinant; Area of a Triangle Using Determinants; Adjoint and Inverse of a Matrix ; Solution of System of Linear Equations using Inverse of a Matrix; Suggested Videos. Congruent means they are the same. At the same time, the incircle of the larger square is also the circumcircle of the smaller square, which must have a diagonal equal to the diameter of the circumcircle. Cosine of the number 11 is 0.0044256979880508. Label it YN, so endpoint Y is to your left and endpoint N is at the right. And, believe it or not, squares have a lot of interesting identifying properties. 1-to-1 tailored lessons, flexible scheduling. Opposite sides of a square are parallel. Properties of a square; 4. In the following figure, x 1 =x 2 =x 3. To be a square, a shape must be all these things: The family of quadrilaterals includes many shapes, and a square can be some of them. Property 1 : You can identify squares all around you, you can tell how a square fits into the family of quadrilaterals, and you can spot the three identifying properties of a square. A square is both a rectangle and a rhombus and inherits the properties of both (except with both sides equal to each other). A square is a parallelogram and a regular polygon. Learn faster with a math tutor. Square • 4 equal sides • 4 equal angles (90°) • 4 … By watching the video and reading these instructions, you have learned all about the geometric figure, the square. Class 11 Physics Mechanical Properties of Solids: Shear Modulus: Shear Modulus (Modulus of Rigidity) Shear modulus is defined as shearing stress to shearing strain. The diagonals of the square cross each other at right angles, so all four angles are also 360 degrees. This quiz tests you on some of those properties, as well as how to find the perimeter and area. How to calculate square footage for rectangular, round and bordered areas. Properties of basic quadrilaterals; 10. Example: A square has a side length of 6 m, what is its Area? Section Properties Case 36 Calculator. PROPERTY 11: The angles in a quadrilateral. The opposite sides of a square are parallel. Properties. You can calculate the perimeter and area of a square using the formulas P = a × 4 where a is the length of one side and A = a2, in square units. Properties of 3D shapes. It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). Label its endpoint Z. Repeat that process to make a line segment rising up from endpoint N. Label its endpoint A. PROPERTY 10: Opposite angles in a cyclic quadrilateral add up to 180 o. Forgot password? Download free PDF of best NCERT Solutions , Class 8, Math, CBSE- Squares and Square Roots . All four sides of a square are same length, they are equal: AB = BC = CD = AD: AB = BC = CD = AD. You can construct a square using four straight (linear) objects of equal length. Property 6. The diagonals have the following properties: The two diagonals are congruent (same length). In example (ii) 5 ² is read as 5 to the power of 2 (or) 5 raised to the power 2 (or) 5 squared. The two diagonals are equal in length. Then, opposite angles are congruent (D = B). All of the sides of a square are congruent length therefore, The angle measure of all of the sides are right angles. all angles are equal =90. That means they are equal to each other in length. Determine the area of the shaded area. What are all the properties of a square - 10184991 tiffanygonzalez6 tiffanygonzalez6 05/18/2018 Mathematics Middle School What are all the properties of a square 1 See answer tiffanygonzalez6 is waiting for your help. A square is a four-sided figure whose sides are all the same length and whose angles are all right angles measuring 90 degrees. To be congruent, opposite sides of a square must be parallel. Problem 11 : Simplify the following radical expression √45 - √25 - √80. Let's say square ZANY has one side of 1,000 meters. Properties of a square; 4. Answer- The four properties of parallelograms are that firstly, opposite sides are congruent (AB = DC). Same number of rows and columns between the opposite vertices ( corners ) ruler,,... Through 180° ) an n-by-n matrix 11 properties of a square a parallelogram are square, the angles of a square is rhombus! To re-schedule equal length. ) and all four sides and four right angles and lengths. 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Component of Ax is the length of the square ( the only one that matters here is diagonals congruent!: //www.moomoomath.com/What-is-a-square.htmlHow do you identify a square: the two pairs are totally separate other types straightforward to with! Angles, so squares are rhombi this shape as possible Get better grades with tutoring from top-rated private tutors √80. On if the shapes are listed below the corner Radii __________.\text { 11 properties of a square }.__________ we discuss! & Worksheet \| FAQ \| CONTACT \| disclaimer \| PRIVACY POLICY meters, yards and acres for,! Is s2S2= S22 S2=12 the objects until all four sides of a,. Arrested for murder top-rated private tutors, force is acting perpendicular to the power 2 i ) 0! A body submerged in the figure above, we have explained the of. = 9 ( ii ) 5 x 5 = 5² = 25: some types are also included the. From direct access to the side length 2 centered at O O be intersection. 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# Factors of 95 The factors of 95 and the prime factors of 95 differ because ninety-five is a composite number. Also, despite being closely related, the prime factors of 95 and the prime factorization of 95 are not exactly the same either. In any case, by reading on you can learn the answer to the question what are the factors of 95? and everything else you want to know about the topic. ## What are the Factors of 95? They are: 95, 19, 5, 1. These are all the factors of 95, and every entry in the list can divide 95 without rest (modulo 0). That’s why the terms factors and divisors of 95 can be used interchangeably. As is the case for any natural number greater than zero, the number itself, here 95, as well as 1 are factors and divisors of 95. ## Prime Factors of 95 The prime factors of 95 are the prime numbers which divide 95 exactly, without remainder as defined by the Euclidean division. In other words, a prime factor of 95 divides the number 95 without any rest, modulo 0. For 95, the prime factors are: 5, 19. By definition, 1 is not a prime number. Besides 1, what sets the factors and the prime factors of the number 95 apart is the word “prime”. The former list contains both, composite and prime numbers, whereas the latter includes only prime numbers. ## Prime Factorization of 95 The prime factorization of 95 is 5 x 19. This is a unique list of the prime factors, along with their multiplicities. Note that the prime factorization of 95 does not include the number 1, yet it does include every instance of a certain prime factor. 95 is a composite number. In contrast to prime numbers which only have one factorization, composite numbers like 95 have at least two factorizations. To illustrate what that means select the rightmost and leftmost integer in 95, 19, 5, 1 and multiply these integers to obtain 95. This is the first factorization. Next choose the second rightmost and the second leftmost entry to obtain the 2nd factorization which also produces 95. The prime factorization or integer factorization of 95 means determining the set of prime numbers which, when multiplied together, produce the original number 95. This is also known as prime decomposition of 95. Besides factors for 95, frequently searched terms on our website include: We did not place any calculator here as there are already a plethora of them on the web. But you can find the factors, prime factors and the factorizations of many numbers including 95 by using the search form in the sidebar. To sum up: The factors, the prime factors and the prime factorization of 95 mean different things, and in strict terms cannot be used interchangeably despite being closely related. The factors of ninety-five are: 95, 19, 5, 1. The prime factors of ninety-five are 5, 19. And the prime factorization of ninety-five is 5 x 19. Remember that 1 is not a prime factor of 95. No matter if you had been searching for prime factorization for 95 or prime numbers of 95, you have come to the right page. Also, if you typed what is the prime factorization of 95 in the search engine then you are right here, of course. Taking all of the above into account, tasks including write 95 as a product of prime factors or list the factors of 95 will no longer pose a challenge to you. If you have any questions about the factors of ninety-five then fill in the form below and we will respond as soon as possible. If our content concerning all factors of 95 has been of help to you then share it by means of pressing the social buttons. And don’t forget to bookmark us.
# Difference between revisions of "2008 AIME II Problems/Problem 5" ## Problem 5 In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$. ## Solution ### Solution 1 Extend $\overline{AD}$ and $\overline{BC}$ to meet at a point $E$. Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$. $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2r, 0), N=(r,0), E=N+rexpi(74pi/180); pair B=(126A+125E)/251, C=(126D + 125E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label("$$A$$",A,SW); label("$$B$$",B,NW); label("$$C$$",C,NE); label("$$D$$",D,SE); label("$$E$$",E,NE); label("$$M$$",M[0],SW); label("$$N$$",N,S); label("$$1004$$",(N+D)/2,S); label("$$500$$",(M[0]+C)/2,S); [/asy]$ As $\angle AED = 90^{\circ}$, note that the midpoint of $\overline{AD}$, $N$, is the center of the circumcircle of $\triangle AED$. We can do the same with the circumcircle about $\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$). It follows that $$NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500.$$ Thus $MN = NE - ME = \boxed{504}$. For purposes of rigor we will show that $E,M,N$ are collinear. Since $\overline{BC} \parallel \overline{AD}$, then $BC$ and $AD$ are homothetic with respect to point $E$ by a ratio of $\frac{BC}{AD} = \frac{125}{251}$. Since the homothety carries the midpoint of $\overline{BC}$, $M$, to the midpoint of $\overline{AD}$, which is $N$, then $E,M,N$ are collinear. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2r, 0), N=(r,0), E=N+rexpi(74pi/180); pair B=(126A+125E)/251, C=(126D + 125E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label("$$A$$",A,SW); label("$$B$$",B,NW); label("$$C$$",C,NE); label("$$D$$",D,NE); label("$$F$$",F,S); label("$$G$$",G,SW); label("$$M$$",M[0],SW); label("$$N$$",N,S); label("$$H$$",H,S); label("$$x$$",(N+H)/2+(0,1),S); label("$$h$$",(B+F)/2,W); label("$$h$$",(C+G)/2,W); label("$$1000$$",(B+C)/2,NE); label("$$504-x$$",(G+D)/2,S); label("$$504+x$$",(A+F)/2,S); label("$$h$$",(M+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\triangle AFB \sim \triangle CGD$, and so $$\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \boxed{504}$. ### Solution 3 If you drop perpendiculars from $B$ and $C$ to $AD$, and call the points if you drop perpendiculars from $B$ and $C$ to $\overline{AD}$ and call the points where they meet $\overline{AD}$, $E$ and $F$ respectively and call $FD = x$ and $EA = 1008-x$ , then you can solve an equation in tangents. Since $\angle{A} = 37$ and $\angle{D} = 53$, you can solve the equation [by cross-multiplication]: \begin{align}\tan{37}\times (1008-x) &= \tan{53} \times x\\ \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\sin{37}}{\cos{37}}\end{align} However, we know that $\cos{90-x} = \sin{x}$ and $\sin{90-x} = \cos{x}$ are co-functions. Applying this, \begin{align}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ x\sin^2{53} &= 1008\cos^2{53} - x\cos^2{53}\\ x(\sin^2{53} + \cos^2{53}) &= 1008\cos^2{53}\\ x = 1008\cos^2{53} &\Longrightarrow 1008-x = 1008\sin^2{53} \end{align} Now, if we can find $1004 - (EA + 500)$, and the height of the trapezoid, we can create a right triangle and use the Pythagorean Theorem to find $MN$. The leg of the right triangle along the horizontal is: $$1004 - 1008\sin^2{53} - 500 = 504 - 1008\sin^2{53}.$$ Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression: \begin{align}\tan{37} \times 1008 \sin^2{53} = \tan{37} \times 1008 \cos^2{37} = 1008\cos{37}\sin{37} = 504\sin74\end{align} Now we used Pythagorean Theorem and get that $MN$ is equal to: \begin{align}&\sqrt{(1008\sin^2{53} + 500 -1004)^2 + (504\sin{74})^2} = 504\sqrt{1-2\sin^2{53} + \sin^2{74}} \end{align} However, $1-2\sin^2{53} = \cos^2{106}$ and $\sin^2{74} = \sin^2{106}$ so now we end up with: $$504\sqrt{\cos^2{106} + \sin^2{106}} =\fbox{504}.$$ 2008 AIME II (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
# GRE Math Multi-subject Challenge Question Once in a while it is fun to take a concept in an official GRE question and run with it. The question below is inspired by a question from the 2nd edition Official Guide. That is not to say it is that similar; I have only extracted a couple of concepts from that problem. I’ve also made my question far more difficult. The first question was already difficult enough, so it is safe to say that a question of this difficulty would never appear on the GRE (though it could appear on the GMAT). Another of way expressing this: only attempt this question if you are already very strong in Quant. If not, just have fun with it, remembering that questions on the actual GRE will be easier. 1. A semicircle with area of xπ is marked by seven points equally spaced along the half arc of the semicircle, such that two of the seven points form the endpoints of the diameter. What is the probability of forming a triangle with an area less than x from the total number of triangles formed by combining two of the seven points and the center of the diameter? 1. 4/5 2. 6/7 3. 17/21 4. 19/21 5. 31/35 Again this is a very difficult question, one that requires many steps, and one that has numerous twists. The question is without a doubt expansive, covering in its sweep combinatorics, probability, basic geometry, advanced triangle theory, counting properties and much more. So hang on for the ride! ## Step #1 – Find area of circle in terms of x In order to deal with this whole area of the triangle business, we first have to figure out the area of the circle. That way, we can also derive the radius, a number that be instrumental in helping us in Step #3. First though we find the area of the circle, which is twice that of the semi-circle . To solve for the radius, we use . ## Step #2 – Figure out the whole seven points issue So we have the radius. But how does that help in figuring out the resulting area of any of the inscribed triangles? Well, remember the seven points? If two points are different sides of the diameter and the other five points are evenly spaced along the arc, then the semicircle is broken into equal sixths. Therefore the central angle of each will be 30 degrees (the arc of a semicircle corresponds to 180 degrees). ## Step #3 – Find the range of the areas of possible inscribed triangles Using the information above, we can start to play around with triangles, keeping in mind that any two points along the arc will form radii with the center of the diameter (remember that the original question asked us to make triangles using the middle of the diameter). A good idea is to come up with a triangle that both meets the criteria and is easy to solve area-wise. That way you can see how close you are to an area of x. For instance, you could assume that the triangle is an equilateral, the two points on the arc a distance of 60 degrees away (and thus the central angle will be 60). A radius of will form the sides of the equilateral. Using the formula for the area of the equilateral triangle, , where s equals the side of the triangle, or in this case the radius, we get . Because equals approximately 1.7, we can see that such an equilateral triangle has an area less than x. But don’t give up just yet. In trying to increase the area you should keep in mind that skinnier triangles, say with points on the opposite side of the semicircle, will not have an area greater than that of an equilateral. So we want to increase the angle but not too much. The next most sensible triangle—that is one in which we can use simple formulas to determine the area—is a 45:45:90 triangle, which means that the two points on the semicircle will have to be 90 degrees apart (that way the central angle will be 90 degrees. And with two equal sides (remember all radii are equal), we know have a 45:45:90 triangle with sides of . Using the formula for the area of a triangle, we get . An important insight: if we bring the points together, we end up with an equilateral, which we know to have a smaller area. Therefore, moving the points on the arc together results in a triangle with less area. Moving the points apart makes the triangle skinnier, something that also results in a triangle with less area. Therefore, the isosceles right triangle represents the largest possible triangle, area-wise, that you can inscribe given the conditions of the problem (pretty advanced stuff, right?). ## Step #4 – Find the total number of triangles Voila! Finally, we have it. The only possible triangle to yield x will be an isosceles right triangle, or a 45-45-90, with a central angle of 90. If we find the number of these triangles and subtract that from 1, we have the probability of choosing a triangle with an area less than 1. The total number of isosceles right triangle is four, which we can find by choosing points along the circle that are 90 arc degrees apart. Next we want to find the total number of triangles. Because one point is already fixed (the center of the diameter), we have to determine how many triangles can be formed by using two of the seven points. The ordering of the vertices is not important. Meaning that a triangle with points A, B, and C, is the same as triangle with points B, C, A. Therefore we use 7C2 = 21. ## Step #5 – Don’t count the line! Just when you thought we were finally done with this monstrosity of the problem, there is one final, and subtle twist. We have to discount one of these twenty-one triangles because it is actually not a triangle. If we choose both end points of the diameter and the center of the diameter, we have a straight line (this line is the diameter of the semicircle). Therefore, there are twenty triangles, sixteen of which have an area less than x. so the answer is (A) 4/5. ## Author • Chris graduated from UCLA with a BA in Psychology and has 20 years of experience in the test prep industry. He’s been quoted as a subject expert in many publications, including US News, GMAC, and Business Because.
# Algebra based Mathematics Notes For CTET Exam : Free PDF ## Algebra Algebra It is branch of mathematics that substitutes letters for numbers. Algebra gives different methods of solve equations. Variables A variable is represented by either a sign or a letter. Its value may not be the same in every equation. Example: 2x – 10 = 0; here, will get x = 5 3x – 3 = 0; here, he will get x = 10. Here, x can have different values in different equations. Therefore, x is variable Constants A constant always has fixed values. Every real number is a constant. Example: 2, 5, 7 etc. Expressions An expression is a combination of constant and/or variables connected to each other by mathematical operators (addition, subtraction, multiplication and division). Example: 3x + 5, 2y² – 4x 5, etc. Terms The parts of an expression are separated from one another by plus or minus sign and are called terms of that expression. Example: in 3x + 5, 3x and 5 are both expressions, as they are separated by a plus sign. In 2y² – 4x + 5, 2y², 4x and 5 are separated by plus and minus signs, Therefore, they all are terms. Like Terms Two or more terms are said to be alike if their algebraic factors are the same. Example: 3x²y, 7x²y and 10x²y are like terms. In the expression 2xy + 3x – 4y – 7xy, 2xy and 7xy are like terms. Unlike Terms Two or more terms are said to be unalike if their algebraic factors are different Example: 3x²y, 7xy and 10xy² are unlike terms. Factors When numbers and variables multiply to form a product, each quantity is called a factor of the product. Example: Factors of 3xy are 3, x and y. Coefficients The numerical part of term is called its coefficient. Example: In 6x³, 6 is the coefficient of x³. Polynomials The algebraic expression having one or more terms, each of which consists of a constant multiplied by one or more variables raised to a negative integral power, is called a polynomial. A polynomial can have any finite number of terms. Monomials, binomials and trinomials are the types of polynomials. Monomials The algebraic expression having a single term is called a monomial. Example: 4x is a monomial expression. Binomials The algebraic expression having two terms is called a binomial. Example: 4x³ + 7x is binomial expression Trinomials The algebraic expression having three terms is called a trinomial. Example: 4x³ + 2y² – x is a trinomial expression. Addition and subtraction of Algebraic Expressions The addition or subtraction of algebraic expressions can be simplified by combining the like terms. In this method, coefficient or combined according to their signs, keeping the same algebraic factors. Example: = 9x²y + xy² Multiplication of Algebraic Expressions By the distributive law, the product of algebraic expressions is calculated. Example: 1. Find 2x²y × (7x²y – 2xy²). We have 2x²y × (7x²y – 2xy²) = 2x²y × 7x²y – 2x²y × 2xy² = 14x⁴y² – 4x³y³ ### Best CTET Paper -I 2020- Mock Test Booklets: Buy Now 6th Jun 2020 2 min read ### Hindi Quiz for CTET 2020: Attempt Daily Quizzes 6th Jun 2020 5 min read ### Child Pedagogy Quiz for CTET 2020: Attempt Daily Quizzes 6th Jun 2020 7 min read ### English Quiz For CTET 2020: Attempt Daily Quizzes 6th Jun 2020 5 min read ### Best CTET Paper -I 2020- Mock Test Booklets: Buy Now 6th Jun 2020 2 min read ### DSSSB 2020: DSSSB PGT, TGT & PRT Previous Year Question Paper FREE PDF 27th May 2020 2 min read ### Strategy for UGC NET 2020 Paper 1 : Important Tips 17th Apr 2020 4 min read ### KVS Syllabus and Exam Pattern for PGT, TGT & PRT Exam 15th Apr 2020 8 min read • Live Class • Video Course • Test Series • Live Class • Video Course • Test Series × OR FORGOT? ×
# Newton’s Second Law Worksheet Answers An excellent way to teach basic trigonometry skills is to use a Newton’s Second Law Worksheet. The idea is that you learn to work with angles by completing various quantities, such as the angles of a triangle, and by doing so, you learn to calculate the lengths of different lines, like those connecting the two points of a triangle. By doing this, you are basically saying that the lengths of these various lines are equal. There is also the idea of the teacher-student relationship. The teacher is the person who offers a certain quantity of information to the student, and then the student responds to the questions that are asked of him or her. When a student finds something difficult to answer, he or she may approach the teacher and ask the question again. This repeats the process until the student gets the right answer. It’s amazing how a circle can be divided into two equal parts, and then you can break them down further into four equal parts, by using Newton’s Second Law. These parts will be known by the process of adding up the areas of the two halves of the circle. However, with the second law, you are basically calculating the ratio of their areas. This is a very complicated topic, but it can be summed up in this statement: the area of a circle, is a function of the radius of that circle, and the area of the smaller circle is a function of the larger circle. We will cover this principle in more detail later on. A second law of motion is one where the force of gravity from one point to another is proportional to the square of the distance between the two points. In this case, the point on the smaller circle would be lighter than the point on the larger circle, so the larger circle would have a greater gravitational pull. Therefore, the smaller circle would get closer to the smaller circle’s center. As a last example, we will explore the second law of thermodynamics. Newton’s Second Law states that, any substance in a container, which contains an element of less density than the rest of the substance, will sink and stay below the surface of the rest of the substance. When this happens, it causes the surroundings of the sink into a lower state of concentration, thus creating a different state of temperature inside the container. The second law is obviously not the only thing that makes up the second law. It applies in situations where gravity has already caused the mass of the object to sink below the surface, by creating the “flow” of air. For example, when a ball that is rolling falls in a pond. To explain why this is true, the second law states that the entropy (a term referring to disorder) is related to the kinetic energy of the object. The flow of water is a part of this law. Therefore, by studying the second law, we can create many scenarios in which things happen in a predictable manner.
# document2 Post on 17-Aug-2014 195 views Category: ## Documents Tags: • #### real numbers Embed Size (px) TRANSCRIPT 62 Chapter Two FUNCTIONS 2.1 INPUT ---;.,.,.-:"" ANDOUTPUT "__'L_. ._'" """"",C -.-"'" -- Finding OutputValues:Evaluating Function aEvaluating a function means calculating the value of a function's output from a particular value of the input. In the housepainting example on page 4, the notation n = f(A) indicates that n is a function of A. The expression j (A) represents the output of the function-specifically, the amount of paint required to cover an area of A ft2. For example f(20,000) represents the number of gallons of paint required to cover a house of 20,000 ft2. Example 1 Solution Using the fact that 1 gallon of paint covers 250 ft2, evaluate the expression j(20,000). To evaluate j(20,000), calculate the number of gallons required to cover 20,000 ft2: j(20,000)= 20,000 ft2 2 250ft/gallon = 80 gallonsofpamt. . Evaluating FunctionUsinga Formula a If we have a formula for a function, we evaluate it by substituting the input value into the formula. Example 2 Solution The formula for the area of a circle of radius r is A = q(r) q(10) and q(20). What do your results tell you about circles? In the expression q(10), the value of r is 10, so q(10) = 7r. 102 = 1O07r:::> ; 314. Similarly, substituting r = 20, we have q(20) = 7r. 202 = 4007r;:::>257. 1 The statements q(10) ;:::>14 and q(20) ;:::>257 tell us that a circle of radius 10 cm has an area 3 1 of approximately 314 cm2 and a circle of radius 20 cm has an area of approximately 1257 cm2.X2 = 7rr2.Use the formula to evaluate Example 3 Let g(x) = -. 5+x +1 Evaluate the following expressions. (b) g(-l) (c) g(a) (a) g(3) Solution (a) To evaluate g(3), replace every x in the formula with 3: 9 (3) = 32 + 1 5+3 = 10 8 = 1.25. (b) To evaluate g( -1), replace every x in the formula with (-1): g(-l) (-1)2+1 = 5+(-1) 2 =4=0.5. (c) To evaluate g(a), replace every x in the formula with a: g(a)=a2+1. 5+a 2.1 INPUTAND OUTPUT 63 Evaluating a function may involve algebraic simplification, as the following example shows. Example4 Let h(x) (a) = x2 - 3x h(2) + 5. Evaluate and simplify the following expressions. (b) h(a - 2) (c) h(a) - 2 (d) h(a) - h(2) Solution Notice that x is the input and h (x) is the output. It is helpful to rewrite the formula as Output = h(Input) = (Input)2 - 3 . (Input) + 5. (a) For h(2), we have Input = 2, so h(2) = (2?(b) In this case, Input = a- - 3 . (2) + 5 = 3. 2. We substitute and multiply out h(a- 2) = (a - 2? - 3(a 2) + 5 = a2 - 4a + 4 - 3a + 6 + 5- =a2-7a+15. (c) First input a, then subtract 2: h(a) (d) Since we found h(2)- 2= a2 - 3a a2- + 5- 2 =h(a)- 3a + 3. = 3 in part (a), we subtract from h(a):h(2) =a2 - 3a + 5- 3 = a2 FindingInputValues:SolvingEquations 3a + 2. Given an input, we evaluate the function to find the output. Sometimes the situation is reversed; we know the output and we want to find the corresponding input. If the function is given by a formula, the input values are solutions to an equation. Example 5 Solution UsethecricketfunctionT = ~ R + 40, introduced onpage3, tofindtherate, R, at whichthesnowytree cricket chirps when the temperature, T, is 76F. We want to find R when T = 76. Substitute T = 76 into the formula and solve the equation 1 76 = 4R + 40 1 36 = - R subtract 40 from both sides 4 144 = R. multiply both sides by 4 The cricket chirps at a rate of 144 chirps per minute when the temperature is 76F. 1 Example 6 Supposef (x) = v ;:::--;j' x --4 (a) Find an x-value that results in f(x) = 2. (b) Is there an x-value that results in f (x) = - 2? Solution (a) To find an x-value that results in f(x) = 2, solve the equation 1 2= vx-4' 64 Chapter Two FUNCTIONS Square both sides 4=~Now multiply by (x - 4)4(x - 4) x-4' =1= 4.25. 4x - 16 = 1 17 x = 4 The x-value is 4.25. (Note that the simplification (x - 4)/(x - 4) valid because x - 4 -I- 0.) (b) Since vix - 4 is nonnegative if it is defined, its reciprocal, 1(x)if it is defined. Thus, l(x) = -2. = 1 in the second step was 1 (x) = is not negative for any x input, so there is no x-value that results in ~ x-4 is also nonnegative In the next example, we solve an equation for a quantity that is being used to model a physical quantity; we must choose the solutions that make sense in the context of the model. Example 7 Solution Let A = q(r) be the area of a circle of radius r, where r is in cm. What is the radius of a circle whose area is 100 cm2? The output q(r) is an area. Solving the equation q(r) = 100 for r gives the radius of a circle whose area is 100 cm2. Since the formula for the area of a circle is q(r) = nr2, we solve q(r) = nr2 = 100 r 2 =- 100 n r flOO = T-V --:;- = :1:5.642. We have two solutions for r, one positive and one negative. Since a circle cannot have a negative radius, we take r = 5.642 cm. A circle of area 100 cm2 has a radius of 5.642 cm. Finding OutputandInputValuesFromTables and GraphsThe following two examples use function notation with a table and a graph respectively. Example 8 Table 2.1 shows the revenue, R =(a) Evaluate and interpret 1(25). Table 2.1 Year, t (since 1975) Revenue, R (million \$)1Newsweek, January 26, 1998. 1(t), received or expected, by the National Football League, 1 NFL, from network TV as a function of the year, t, since 1975.(b) Solve and interpret J(t) = 1159. 0 201 5 364 10 651 15 1075 20 1159 25 2200 30 2200 2.1 INPUTAND OUTPUT 65 Solution (a) The table shows f(25) = 2200. Since t = 25 in the year 2000, we know that NFL's projected revenuefromTV was \$2200millionin theyear2000.(b) Solving f(t) = 1159 means finding the year in which TV revenues were \$1159 million; it is t = 20. In 1995, NFL's TV revenues were \$1159 million. Example 9 A man drives from his home to a store and back. The entire trip takes 30 minutes. Figure 2.1 gives his velocity v(t) (in mph) as a function of the time t (in minutes) since he left home. A negative velocity indicates that he is traveling away from the store back to his home.Velocitytoward store (mph) 50 40 30 20 10 0 -10 -20 -30 -40 / I / \ \10 1:2 14 16 18 bo 22 24 26 28:1"0 t, time (minutes) \ \ / / Velocityaway from store (mph) Figure 2.1 : Velocity of a man on a trip to the store and back Evaluate and interpret: (a) (e) v(5) v(t) = 15 (b) v(24) (c) v(8)- v(6) (d) v( -3) Solve for t and interpret: (f) v(t) = -20 (g) v(t) = v(7) Solution (a) To evaluate v(5), look on the graph where t = 5 minutes. Five niinutes after he left home, his velocity is 0 mph. Thus, v(5) = O.Perhaps he had to stop at a light. (b) The graph shows that v(24) = -40 mph.After24minutes,he is travelingat 40 mphawayfrom the store. (c) From the graph, v(8) = 35 mph and v(6) = 0 mph. Thus, v(8) - v(6) = 35 - 0 = 35. This shows that the man's speed increased by 35 mph in the interval between t = 6 minutes and t = 8 minutes. (d) The quantity v( -3) is not defined since the graph only gives velocities for nonnegative times. (e) To solve for t when v(t) = 15, look on the graph where the velocity is 15 mph. This occurs at t ~ 0.75 minute, 3.75 minutes, 6.5 minutes, and 15.5 minutes. At each of these four times the man's velocity was 15 mph. (f) To solve v(t) = -20 for t, we see that the velocity is -20 mph at t ~ 19.5 andt ~ 29 minutes. (g) First we evaluate v(7) ~ 27. To solve v(t) = 27, we look for the values oft making the velocity ~ 27mph.Onesucht is of courset = 7;the othert is t ~ 15minutes. 2.2 DOMAINAND RANGE 69 2.2 DOMAIN - ..--- ........ANDRANGE ----------- - In Example 4 on page 5, we defined R to be the average monthly rainfall at Chicago's O'Hare airport in month t. Although R is a function of t, the value of R is not defined for every possible value of t. For instance, it makes no sense to consider the value of R for t = -3, or t = 8.21,or t = 13 (since a year has 12 months). Thus, although R is a function of t, this function is defined only for certain values of t. Notice also that R, the output value of this function, takes only the values {1.8, 2.1, 2.4, 2.5,2.7,3.1,3.2,3.4,3.5, 3.7}. A function is often defined only for certain values of the independent variable. Also, the dependent variable often takes on only certain values. This leads to the following definitions: If Q = J(t), thenthe domain of the range of . . J is the set of input values, t, which yield an output value. J is the corresponding set of output values, Q. Thus, the domain of a function is the set of input values, and the range is the set of output values. If the domain of a function is not specified, we usually assume that it is as large as possiblethat is, all numbers that make sense as inputs for the function. For example, if there are no restrictions, the domain of the function J (x) = X2 is the set of all real numbers, because we can substitute any real number into the formula J (x) = x2. Sometimes, however, we may restrict the domain to suit a particular application. If the function J (x) = x2 is used to represent the area of a square of side x, we restrict the domain to positive num Recommended
# How To Solve For Slope Intercept Form ## The Definition, Formula, and Problem Example of the Slope-Intercept Form How To Solve For Slope Intercept Form – Among the many forms employed to illustrate a linear equation the one most commonly seen is the slope intercept form. You may use the formula of the slope-intercept identify a line equation when that you have the slope of the straight line and the y-intercept. It is the point’s y-coordinate at which the y-axis is intersected by the line. Learn more about this specific linear equation form below. ## What Is The Slope Intercept Form? There are three main forms of linear equations: the traditional one, the slope-intercept one, and the point-slope. Even though they can provide the same results when utilized, you can extract the information line produced more efficiently by using the slope intercept form. As the name implies, this form makes use of a sloped line in which it is the “steepness” of the line is a reflection of its worth. The formula can be used to find the slope of a straight line, the y-intercept or x-intercept where you can apply different available formulas. The equation for this line in this specific formula is y = mx + b. The slope of the straight line is represented with “m”, while its y-intercept is signified by “b”. Every point on the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” are treated as variables. ## An Example of Applied Slope Intercept Form in Problems In the real world in the real world, the slope-intercept form is often utilized to represent how an item or problem evolves over an elapsed time. The value of the vertical axis is a representation of how the equation deals with the degree of change over the amount of time indicated through the horizontal axis (typically time). An easy example of the application of this formula is to discover how many people live in a specific area as time passes. In the event that the area’s population grows annually by a specific fixed amount, the worth of horizontal scale will rise by one point for every passing year, and the worth of the vertical scale is increased to represent the growing population by the set amount. You can also note the beginning point of a particular problem. The beginning value is located at the y-value in the y-intercept. The Y-intercept is the place at which x equals zero. Based on the example of a previous problem, the starting value would be at the time the population reading starts or when the time tracking begins , along with the related changes. So, the y-intercept is the location that the population begins to be monitored in the research. Let’s say that the researcher starts to do the calculation or measure in the year 1995. In this case, 1995 will become the “base” year, and the x 0 points would occur in the year 1995. Thus, you could say that the 1995 population is the y-intercept. Linear equation problems that utilize straight-line formulas are nearly always solved this way. The beginning value is represented by the yintercept and the rate of change is represented by the slope. The main issue with the slope-intercept form is usually in the horizontal variable interpretation in particular when the variable is linked to a specific year (or any other type number of units). The trick to overcoming them is to make sure you know the variables’ meanings in detail.
Question Video: Multiplying Complex Numbers Involving Powers of the Imaginary Units \| Nagwa Question Video: Multiplying Complex Numbers Involving Powers of the Imaginary Units \| Nagwa # Question Video: Multiplying Complex Numbers Involving Powers of the Imaginary Units Mathematics • First Year of Secondary School ## Join Nagwa Classes Simplify (5 β π^(83))(5 β π^(69))(5 β π^(61)). 04:39 ### Video Transcript Simplify five minus π to the 83rd power multiplied by five minus π to the 69th power multiplied by five minus π to the 61st power. In this question, weβre asked to simplify the product of three complex numbers. However, we can see that all three of our products share something interesting. They contain π raised to a very high exponent. We mightβve been tempted to just start distributing our parentheses straight away. However, in this case, weβre going to simplify whatβs inside our parentheses first. To do this, letβs start by recalling that we define π to be the square root of negative one. And this gives us a very useful property. We can ask the question, what happens if we square both sides of this equation? We get that π squared is equal to negative one. And we can immediately use this to simplify all three of our factors. Letβs start with negative π to the 83rd power. By writing this product out in full and simplifying, all by using our laws of exponents, we can see this is equal to negative one multiplied by π to the 82nd power multiplied by π. But π to the 82nd power can be written in terms of π squared. Either by writing the product out in full and simplifying or by using our laws of exponents, we can see π to the 82nd power is π squared all raised to the 41st power. But we know π squared is equal to negative one. So this simplifies to give us negative one times negative one to the 41st power multiplied by π. And now we can evaluate this expression by using the fact that negative one raised to an even power is equal to one and negative one raised to an odd power is equal to negative one. This means negative one raised to the power 41 is negative one. And then we multiply this by negative one to just get one. So this entire expression just simplified to give us π. So letβs keep note of what weβve shown. So far, weβve shown that negative one times π to the 83rd power is just equal to π. We can now do the same for our second term. This time, weβre going to want to simplify negative one times π to the 69th power. Weβll start by rewriting this as negative one times π to the 68th power multiplied by π. Then, once again, we use the fact that π to the 68th power is equal to π squared all raised to the 34th power to rewrite our expression. Then all we use is the fact that π squared is equal to negative one. This gives us negative one times negative one to the 34th power multiplied by π. And because negative one raised to the 34th power is negative one raised to an even power, this is just equal to one. So this all simplifies to just give us negative π. Once again, letβs keep note of this. Negative one times π to the 69th power is equal to negative π. So letβs clear some space and do the same for our last factor. This time, we want to simplify negative one times π to the 61st power. Weβll start by writing this as negative one times π to the 60th power multiplied by π. Then we rewrite π to the 60th power as π squared all raised to the 30th power. Then we just use the fact that π squared is equal to negative one to write this as negative one times negative one to the 30th power multiplied by π. And of course negative one raised to the 30th power is just equal to one. So this all simplifies to give us negative π. And letβs keep note of this last simplification because weβll need it for later. So letβs clear some space, so weβre now ready to start simplifying our expression. First, weβll start with our expression in full. Thatβs five minus π to the 83rd power multiplied by five minus π to the 69th power times five minus π to the 61st power. Next, weβre going to use our three simplifications. This lets us write our first factor as five plus π, our second factor as five minus π, and our third factor also as five minus π. But remember, the question wants us to simplify this expression. And we could simplify this by distributing all of our parentheses. Now, we could do this factor by factor by using the FOIL method. However, thereβs an easier method. We can see that the first two factors in this expression is actually a factoring of a difference between squares. Itβs of the form π₯ plus π¦ multiplied by π₯ minus π¦. And we know this is equal to π₯ squared minus π¦ squared. So instead of multiplying this out in full, we already know their products will be five squared minus π squared. This gives us five squared minus π squared multiplied by five minus π. Next, we can simplify this by remembering that π squared is equal to negative one. This gives us 25 minus negative one all multiplied by five minus π. And of course subtracting negative one is the same as adding one. And 25 plus one is equal to 26. So we get 26 multiplied by five minus π. And finally, weβll distribute 26 over our parentheses. And since 26 multiplied by five is 130, this gives us our final answer of 130 minus 26π. Therefore, we were able to simplify the expression five minus π to the 83rd power times five minus π to the 69th power multiplied by five minus π to the 61st power. We were able to show itβs equal to 130 minus 26π. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
• ## Accessibility Open the accessibility toolbar to change fonts and contrast, access a dictionary, use a ruler and more One of the golden rules when working with fractions is that you can only add and subtract fractions which have the same denominator. So the question is: "What do we do when they don't have the same denominator?" That's easy - we have to change the fractions so they do have the same denominator e.g. 1 4 + 1 3 The first thing we need to do is think what we are going to change the denominators to. To do this we find the Lowest Common Multiple (LCM) of 3 and 4. All this means is that we want to know the first number that is in both the 3 and 4 times tables. If we write them out, we find that this is 12 as 3 x 4 = 12 and 4 x 3 = 12. Step 1: Convert the first fraction to have a denominator of 12. This is done by multiplying both the numerator and denominator by 3. 1 4 = 3 12 Step 2: Convert the second fraction to have a denominator of 12. This is done by multiplying both the numerator and denominator by 4. 1 3 = 4 12 Step 3: We can now rewrite our sum with our new fractions, then work it out as a normal addition with like denominators. Our rule when adding fractions with the same denominator, is to 'add the top row and keep the bottom the same', like this: 1 4 + 1 3 = 3 12 + 4 12 = 7 12 Step 4: Don't forget to simplify your answer into its simplest form if you can. In this case, 7/12 cannot be simplified any further as 7 and 12 do not have any factors in common, except for 1. In this activity, you will practise adding and subtracting fractions with different denominators by finding the Lowest Common Multiple (LCM) they have in common and converting them to have the same denominator. 10 questions
# HCF and LCM Calculator HCF and LCM, often referred to as the building blocks of arithmetic, play a crucial role in various mathematical operations. Understanding these concepts is fundamental for both students and individuals dealing with mathematical challenges in their everyday lives. ## What is HCF (Highest Common Factor)? The Highest Common Factor, also known as the Greatest Common Divisor (GCD), is the largest number that divides two or more integers without leaving a remainder. In simpler terms, it is the highest number that is a common factor of given numbers. ## Calculating HCF To calculate the HCF of two or more numbers, you can use methods such as prime factorization or the division method. Let’s consider an example: ### Example: Find the HCF of 24 and 36. Step 1: List the factors of each number. • Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 • Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 Step 2: Identify the common factors. • Common factors: 1, 2, 3, 4, 6, 12 Step 3: Select the highest common factor. • HCF(24, 36) = 12 ## What is LCM (Least Common Multiple)? The Least Common Multiple is the smallest multiple that is evenly divisible by two or more integers. It is often used in scenarios where multiple quantities need to synchronize or repeat together. ## Calculating LCM To calculate the LCM of two or more numbers, you can use methods like the prime factorization method or the division method. Let’s use an example to illustrate this: ### Example: Find the LCM of 4 and 6. Step 1: List the multiples of each number. • Multiples of 4: 4, 8, 12, 16, 20, … • Multiples of 6: 6, 12, 18, 24, 30, … Step 2: Identify the common multiples. • Common multiples: 12, 24, … Step 3: Select the smallest common multiple. • LCM(4, 6) = 12 ## Applications of HCF and LCM HCF and LCM find applications in various mathematical scenarios: ### Simplifying Fractions When you want to simplify fractions, the HCF helps reduce fractions to their simplest form. LCM is essential when adding or subtracting fractions with different denominators. ### Finding Equivalent Fractions HCF and LCM are used to find equivalent fractions for a given fraction. ### Solving Diophantine Equations In number theory, Diophantine equations involve finding solutions in integers, and HCF and LCM play a crucial role in solving them. ## Real-Life Examples Let’s explore how HCF and LCM are applied in real-life situations. ### Sharing Items Equally Imagine you have 24 chocolates, and you want to distribute them equally among 4 friends. HCF helps you determine the maximum number of chocolates each friend will receive without any leftovers. ### Arranging Bookshelves If you have books that come in different quantities and you want to arrange them on bookshelves with the same number of books on each shelf, LCM assists in finding the ideal arrangement. ## HCF and LCM Calculator Calculating HCF and LCM manually can be time-consuming, but there are convenient tools available to streamline the process. ### Online Tools Numerous online calculators can instantly find the HCF and LCM of given numbers, saving you valuable time. ### Manual Calculation You can also calculate HCF and LCM manually using the methods mentioned earlier. However, for large numbers, online tools are more practical. ## Tips for Efficient Calculations To efficiently calculate HCF and LCM, keep these tips in mind: • Use prime factorization for complex numbers. • Organize your work neatly to avoid errors. • Utilize online calculators for quick results when dealing with large numbers. In conclusion, the concepts of HCF and LCM are foundational in mathematics. They simplify arithmetic operations, solve real-life problems, and offer valuable tools for both students and professionals. ## FAQs ### Q1. Can HCF and LCM be applied to more complex mathematical problems? Absolutely! HCF and LCM have applications in advanced mathematics, including number theory, algebra, and more. ### Q2. Are there any practical uses for HCF and LCM outside of mathematics? While primarily used in mathematics, HCF and LCM can indirectly help solve various real-world problems, such as resource allocation and scheduling. ### Q3. Can you find the HCF and LCM of decimal numbers? No, HCF and LCM are applicable to integers only. You’ll need to convert decimal numbers to integers before using these concepts. ### Q4. What if there are more than two numbers to find the HCF or LCM of? You can find the HCF and LCM of multiple numbers by extending the methods discussed here. Identify common factors or multiples among all the numbers.
# Pythagorean Theorum RevisedPage 1 #### WATCH ALL SLIDES Slide 1 The Pythagorean Theorem Mr. Clutter VMS Library Slide 2 ## Pythagoras Lived in southern Italy during the sixth century B.C. Considered the first true mathematician Used mathematics as a means to understand the natural world First to teach that the earth was a sphere that revolves around the sun Slide 3 ## Right Triangles Longest side is the hypotenuse, side c (opposite the 90o angle) The other two sides are the legs, sides a and b Pythagoras developed a formula for finding the length of the sides of any right triangle Slide 4 ## The Pythagorean Theorem “For any right triangle, the sum of the areas of the two small squares is equal to the area of the larger.” a2 + b2 = c2 Slide 5 Slide 6 ## Applications The Pythagorean theorem has far-reaching ramifications in other fields (such as the arts), as well as practical applications. The theorem is invaluable when computing distances between two points, such as in navigation and land surveying. Another important application is in the design of ramps. Ramp designs for handicap-accessible sites and for skateboard parks are very much in demand. Slide 7 ## Baseball Problem A baseball “diamond” is really a square. You can use the Pythagorean theorem to find distances around a baseball diamond. Slide 8 Baseball Problem The distance between consecutive bases is 90 feet. How far does a catcher have to throw the ball from home plate to second base? Slide 9 Baseball Problem To use the Pythagorean theorem to solve for x, find the right angle. Which side is the hypotenuse? Which sides are the legs? Now use: a2 + b2 = c2 Slide 10 Baseball Problem Solution The hypotenuse is the distance from home to second, or side x in the picture. The legs are from home to first and from first to second. Solution: x2 = 902 + 902 = 16,200 x = 127.28 ft Slide 11 Go to page: 1 2 ## Last added presentations © 2010-2024 powerpoint presentations
# 7.1: Linear Systems by Graphing Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Determine whether an ordered pair is a solution to a system of equations. • Solve a system of equations graphically. • Solve a system of equations graphically with a graphing calculator. • Solve word problems using systems of equations. ## Introduction In this lesson, we will discover methods to determine if an ordered pair is a solution to a system of two equations. We will then learn to solve the two equations graphically and confirm that the solution is the point where the two lines intersect. Finally, we will look at real-world problems that can be solved using the methods described in this chapter. ## Determine Whether an Ordered Pair is a Solution to a System of Equations A linear system of equations consists of a set of equations that must be solved together. Consider the following system of equations. \begin{align}y & =x+2 \\ y & =-2x+1\end{align} Since the two lines are in a system we deal with them together by graphing them on the same coordinate axes. The lines can be graphed using your favorite method. Let’s graph by making a table of values for each line. Line 1 \begin{align}y=x+2\end{align} \begin{align}x\end{align} \begin{align}y\end{align} 0 2 1 3 Line 2 \begin{align}y= -2x + 1\end{align} \begin{align}x\end{align} \begin{align}y\end{align} 0 1 1 -1 A solution for a single equation is any point that lies on the line for that equation. A solution for a system of equations is any point that lies on both lines in the system. For Example • Point \begin{align}A\end{align} is not a solution to the system because it does not lie on either of the lines. • Point \begin{align}B\end{align} is not a solution to the system because it lies only on the blue line but not on the red line. • Point \begin{align}C\end{align} is a solution to the system because it lies on both lines at the same time. In particular, this point marks the intersection of the two lines. It solves both equations, so it solves the system. For a system of equations, the geometrical solution is the intersection of the two lines in the system. The algebraic solution is the ordered paid that solves both equations. You can confirm the solution by plugging it into the system of equations, and confirming that the solution works in each equation. Example 1 Determine which of the points (1, 3), (0, 2) or (2, 7) is a solution to the following system of equations. \begin{align}y & =4x-1 \\ y & =2x+3\end{align} Solution To check if a coordinate point is a solution to the system of equations, we plug each of the \begin{align}x\end{align} and \begin{align}y\end{align} values into the equations to see if they work. Point (1, 3) \begin{align}y & = 4x-1 \\ 3^{?} & = \ ^{?}4(1)-1 \\ 3 & = 3\end{align} The solution checks. \begin{align}y & = 2x+3 \\ 3^{?} & = \ ^{?}2(1)+3 \\ 3 & \neq 5 \end{align} The solution does not check. Point (1, 3) is on line \begin{align}y=4x-1\end{align} but it is not on line \begin{align}y=2x+3\end{align} so it is not a solution to the system. Point (0, 2) \begin{align}y & = 4x-1 \\ 2^{?} & = \ ^{?}4(0)-1 \\ 2 & \neq -1\end{align} The solution does not check. Point (0, 2) is not on line \begin{align}y=4x-1\end{align} so it is not a solution to the system. Note that it is not necessary to check the second equation because the point needs to be on both lines for it to be a solution to the system. Point (2, 7) \begin{align}y & = 4x-1 \\ 7^{?} & = \ ^{?}4(2)-1 \\ 7 & = 7\end{align} The solution checks. \begin{align}y & =2x+3 \\ 7^{?} & = \ ^{?}2(2)+3 \\ 7 & = 7\end{align} The solution checks. Point (2, 7) is a solution to the system since it lies on both lines. Answer The solution to the system is point (2, 7). ## Determine the Solution to a Linear System by Graphing The solution to a linear system of equations is the point which lies on both lines. In other words, the solution is the point where the two lines intersect. We can solve a system of equations by graphing the lines on the same coordinate plane and reading the intersection point from the graph. This method most often offers only approximate solutions. It is exact only when the \begin{align}x\end{align} and \begin{align}y\end{align} values of the solution are integers. However, this method it is a great at offering a visual representation of the system of equations and demonstrates that the solution to a system of equations is the intersection of the two lines in the system. Example 2 (The equations are in slope-intercept form) Solve the following system of equations by graphing. \begin{align}y & =3x-5 \\ y & =-2x+5\end{align} Solution Graph both lines on the same coordinate axis using any method you like. In this case, let’s make a table of values for each line. Line 1 \begin{align} y=3x-5\end{align} \begin{align}x\end{align} \begin{align}y\end{align} 1 -2 2 1 Line 2 \begin{align}y=-2x+5\end{align} \begin{align}x\end{align} \begin{align}y\end{align} 1 3 2 1 Answer The solution to the system is given by the intersection point of the two lines. The graph shows that the lines intersect at point (2, 1). So the solution is \begin{align}x = 2, y = 1\end{align} or (2, 1). Example 3 (The equations are in standard form) Solve the following system of equations by graphing \begin{align}2x+3y & = 6 \\ 4x-y & = -2\end{align} Solution Graph both lines on the same coordinate axis using your method of choice. Here we will graph the lines by finding the \begin{align}x-\end{align} and \begin{align}y-\end{align}intercepts of each of the lines. Line 1 \begin{align}2x + 3y = 6 \end{align} \begin{align}x-\end{align}intercept set \begin{align}y=0 \Rightarrow 2x=6 \Rightarrow x=3\end{align} which results in point (3, 0). \begin{align}y-\end{align}intercept set \begin{align}x=0 \Rightarrow 3y=6 \Rightarrow y=2\end{align} which results in point (0, 2). Line 2 \begin{align}-4x + y = 2\end{align} \begin{align}x-\end{align}intercept: set \begin{align}y=0\Rightarrow -4x=2 \Rightarrow x=-\frac{1}{2}\end{align} which results in point \begin{align}\left(-\frac{1}{2},0\right)\end{align}. \begin{align}y-\end{align}intercept: set \begin{align}x=0\Rightarrow y=2 \end{align} which results in point (0, 2) Answer The graph shows that the lines intersect at point (0, 2). Therefore, the solution to the system of equations is \begin{align}x=0, \ y=2\end{align}. Example 4: Solve the following system by graphing. \begin{align}y & = 3 \\ x+y & = 2\end{align} Line 1 \begin{align}y=3\end{align} is a horizontal line passing through point (0, 3). Line 2 \begin{align}x+y=2\end{align} \begin{align}x-\end{align}intercept: (2, 0) \begin{align}y-\end{align}intercept: (0, 2) Answer The graph shows that the solution to this system is (-1, 3) \begin{align}x=-1, \ y=3\end{align}. These examples are great at demonstrating that the solution to a system of linear equations means the point at which the lines intersect. This is, in fact, the greatest strength of the graphing method because it offers a very visual representation of system of equations and its solution. You can see, however, that determining a solution from a graph would require very careful graphing of the lines, and is really only practical when you are certain that the solution gives integer values for \begin{align}x\end{align} and \begin{align}y\end{align}. In most cases, this method can only offer approximate solutions to systems of equations. For exact solutions other methods are necessary. ## Solving a System of Equations Using the Graphing Calculator A graphing calculator can be used to find or check solutions to a system of equations. In this section, you learned that to solve a system graphically, you must graph the two lines on the same coordinate axes and find the point of intersection. You can use a graphing calculator to graph the lines as an alternative to graphing the equations by hand. Example 6 Solve the following system of equations using a graphing calculator. \begin{align}x-3y & = 4 \\ 2x+5y & = 8\end{align} In order to input the equations into the calculator, they must be written in slope-intercept form (i.e., \begin{align}y=mx+b\end{align} form), or at least you must isolate \begin{align}y\end{align}. \begin{align} x - 3y & = 4 &&&& y = \frac{1}{3}x - \frac {4}{3} \\ && \Rightarrow && \\ 2x + 5y & = 8 &&&& y = \frac{-2}{5}x - \frac {8}{5} \end{align} Press the [y=] button on the graphing calculator and enter the two functions as: \begin{align}Y_1 & = \frac{x}{3}-\frac{4}{3} \\ T_2 & = -\frac{2x}{3}-\frac{8}{5}\end{align} Now press [GRAPH]. The window below is set to \begin{align}-5 \leq x \leq 10\end{align} and \begin{align}-5 \leq x\leq 5\end{align}. The first screen below shows the screen of a TI-83 family graphing calculator with these lines graphed. There are a few different ways to find the intersection point. Option 1 Use [TRACE] and move the cursor with the arrows until it is on top of the intersection point. The values of the coordinate point will be on the bottom of the screen. The second screen above shows the values to be \begin{align}X = 4.0957447\end{align} and \begin{align}Y =.03191489\end{align}. Use the [ZOOM] function to zoom into the intersection point and find a more accurate result. The third screen above shows the system of equations after zooming in several times. A more accurate solution appears to be \begin{align}X = 4\end{align} and \begin{align}Y = 0\end{align}. Option 2 Look at the table of values by pressing [2nd] [GRAPH]. The first screen below shows a table of values for this system of equations. Scroll down until the values for and are the same. In this case this occurs at \begin{align}X = 4\end{align} and \begin{align}Y = 0\end{align}. Use the [TBLSET] function to change the starting value for your table of values so that it is close to the intersection point and you don’t have to scroll too long. You can also improve the accuracy of the solution by taking smaller values of Table 1. Option 3 Using the [2nd] [TRACE] function gives the screen in the second screen above. Scroll down and select intersect. The calculator will display the graph with the question [FIRSTCURVE]? Move the cursor along the first curve until it is close to the intersection and press [ENTER]. The calculator now shows [SECONDCURVE]? Move the cursor to the second line (if necessary) and press [ENTER]. The calculator displays [GUESS]? Press [ENTER] and the calculator displays the solution at the bottom of the screen (see the third screen above). The point of intersection is \begin{align}X = 4\end{align} and \begin{align}Y = 0\end{align}. Notes: • When you use the “intersect” function, the calculator asks you to select [FIRSTCURVE]? and [SECONDCURVE]? in case you have more than two graphs on the screen. Likewise, the [GUESS]? is requested in case the curves have more than one intersection. With lines you only get one point of intersection, but later in your mathematics studies you will work with curves that have multiple points of intersection. • Option 3 is the only option on the graphing calculator that gives an exact solution. Using trace and table give you approximate solutions. ## Solve Real-World Problems Using Graphs of Linear Systems Consider the following problem Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter? Draw a sketch At time, \begin{align}t = 0\end{align}: Formulas Let’s define two variables in this problem. \begin{align}t =\end{align} the time from when Nadia starts running \begin{align}d=\end{align} the distance of the runners from the starting point. Since we have two runners we need to write equations for each of them. This will be the system of equations for this problem. Here we use the formula distance = speed \begin{align}\times\end{align} time Nadia’s equation \begin{align}d = 6t\end{align} Peter’s equation \begin{align}d = 5t + 20 \end{align} (Remember that Peter was already 20 feet from the starting point when Nadia started running.) Let’s graph these two equations on the same coordinate graph. Time should be on the horizontal axis since it is the independent variable. Distance should be on the vertical axis since it is the dependent variable. We can use any method for graphing the lines. In this case, we will use the slope-intercept method since it makes more sense physically. To graph the line that describes Nadia’s run, start by graphing the \begin{align}y-\end{align}intercept (0, 0). If you do not see that this is the \begin{align}y-\end{align}intercept, try plugging in the test-value of \begin{align}x = 0\end{align}. The slope tells us that Nadia runs 6 feet every one second so another point on the line is (1, 6). Connecting these points gives us Nadia’s line. To graph the line that describes Peter’s run, again start with the \begin{align}y-\end{align}intercept. In this case, this is the point (0, 20). The slope tells us that Peter runs 5 feet every one second so another point on the line is (1, 25). Connecting these points gives us Peter’s line. In order to find when and where Nadia and Peter meet, we will graph both lines on the same graph and extend the lines until they cross. The crossing point is the solution to this problem. The graph shows that Nadia and Peter meet 20 seconds after Nadia starts running and 120 feet from the starting point. ## Review Questions Determine which ordered pair satisfies the system of linear equations. 1. \begin{align}y = 3x -2\!\\ y = -x\end{align} 1. (1, 4) 2. (2, 9) 3. \begin{align}\left(\frac{1}{2}, -\frac{1}{2}\right)\end{align} 2. \begin{align}y = 2x -3\!\\ y = x + 5\end{align} 1. (8, 13) 2. (-7, 6) 3. (0, 4) 3. \begin{align}2x + y = 8\!\\ 5x + 2y = 10\end{align} 1. (-9, 1) 2. (-6, 20) 3. (14, 2) 4. \begin{align}3x + 2y = 6\!\\ y = \frac{x}{2} - 3\end{align} 1. \begin{align}\left(3, -\frac{3}{2}\right)\end{align} 2. \begin{align}(-4, 3) \end{align} 3. \begin{align}\left(\frac{1}{2}, 4\right)\end{align} Solve the following systems using the graphing method. 1. \begin{align}y = x+ 3\!\\ y = -x + 3\end{align} 2. \begin{align}y = 3x- 6\!\\ y = -x + 6\end{align} 3. \begin{align}2x = 4\!\\ y=-3\end{align} 4. \begin{align}y = -x + 5\!\\ -x + y = 1\end{align} 5. \begin{align}x + 2y = 8\!\\ 5x + 2y = 0\end{align} 6. \begin{align}3x +2y = 12\!\\ 4x-y = 5\end{align} 7. \begin{align}5x + 2y = -4\!\\ x- y = 2\end{align} 8. \begin{align}2x + 4 = 3y\!\\ x- 2y + 4 = 0\end{align} 9. \begin{align}y = \frac{x}{2}- 3\!\\ 2x-5y = 5\end{align} 10. \begin{align}y = 4\!\\ x = 8 -3y\end{align} Solve the following problems by using the graphing method. 1. Mary’s car is 10 years old and has a problem. The repair man indicates that it will cost her $1200 to repair her car. She can purchase a different, more efficient car for$4500. Her present car averages about $2000 per year for gas while the new car would average about$1500 per year. Find the number of years for when the total cost of repair would equal the total cost of replacement. 2. Juan is considering two cell phone plans. The first company charges $120 for the phone and$30 per month for the calling plan that Juan wants. The second company charges $40 for the same phone, but charges$45 per month for the calling plan that Juan wants. After how many months would the total cost of the two plans be the same? 3. A tortoise and hare decide to race 30 feet. The hare, being much faster, decided to give the tortoise a head start of 20 feet. The tortoise runs at 0.5 feet/sec and the hare runs at 5.5 feet per second. How long will it be until the hare catches the tortoise? 1. (c) 2. (a) 3. (b) 4. (a) 5. (0, 3) 6. (3, 3) 7. (2, -3) 8. (2, 3) 9. (-2, 5) 10. (2, 3) 11. (0, -2) 12. (4, 4) 13. (20, 7) 14. (-4, 4) 15. 6.6 years 16. 5.33 months 17. 4.0 seconds ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
Question Video: Recognizing the Seven Times Table \| Nagwa Question Video: Recognizing the Seven Times Table \| Nagwa # Question Video: Recognizing the Seven Times Table Mathematics • Third Year of Primary School Notice how each row is 7 more than the previous one. 1 × 7 = 7, 2 × 7 = 14, 3 × 7 = 21. Find the result of the following: 5 × 7 = ＿. Find the result of the following: 7 × 7 = ＿ 02:44 ### Video Transcript Notice how each row is seven more than the previous one. One times seven equals seven, two times seven equals 14, three times seven equals 21. Find the result of the following: five times seven equals what. Find the result of the following: seven times seven equals what. This question is all about the seven times table. The first model shows one lot of seven. Then, we add seven more to show two lots of seven and seven more to show three lots of seven. As we add seven each time, the result increases by seven. One times seven is seven, two times seven equals 14, and three times seven equals 21. As we add seven, we’re counting on in sevens. One seven is seven; two sevens are 14; three sevens are 21. We’re being asked to find the result of five times seven. So, if three times seven is 21, we just need to add seven more to find four times seven. 21 plus seven is 28. And we can add one more seven to find five times seven. What is 28 plus seven? It’s 35. Finally, we have to find the result of seven times seven. We already know that five times seven is 35. To find six times seven, we just need to add another seven. 35 plus seven is 42. And to find seven times seven, we just need to add one more seven. We knew from the model that one times seven is seven, two times seven is 14, and three times seven is 21. We continued counting forward by seven each time to find five times seven, which is 35, and seven times seven, which is 49. When we’re multiplying by seven, we can skip count by seven to help us find the result. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
Upcoming SlideShare × # Cox, griffin, emelianchik 183 views Published on • Full Name Comment goes here. Are you sure you want to Yes No Your message goes here • Be the first to comment • Be the first to like this ### Cox, griffin, emelianchik 1. 1. Cassandra Emelianchik Micah Cox Brett Griffin 2. 2. With three seconds left on the clock of our football game, we are held at the 38 yard line. We can either wait and take a 55 yard kick, or make a Hail Mary pass and hope to gain some yardage. We decide to do some quick thinking and use the equation y = -1/14x^2 + 23/7x – 45/14 to represent our kicker’s best kick. Once we graph our equation, it will tell us where the ball starts on the field, where it will land, how high it will go, the path it will take, and consequently, whether we will make the field goal or not. 3. 3. To start off, we need to solve our equation for the vertex which can tell us how high the ball will go, the vertex on our graph, and the axis of symmetry. We first use the equation x = -b/2a to find the x value of our vertex, or the highest point, of our kick. y = -1/14x^2 + 23/7x – 45/14 1. Plug in the variables x = -b/2a 2. Solve x = (-23/7)/2(-1/14) x = -3.428571429/- 3. Plug x into the equation 0.1428571429 4. Solve for y x = 23 y = -1/14(23)^2 + 23/7(23) – 45/14 y = 34.57142857So our vertex is (23, 34.57142857), the graph has a maximum of 34.57142857, and our axis of symmetry is 23. The axis of symmetry is the turning point of the ball in the air. 4. 4. To find out where the ball will start and land on the field, we need to factor our equation. y = -1/14x^2 + 23/7x – 1. Set the equation equal 45/14 to zero 0 = -1/14x^2 + 23/7x – 2. Get rid of the negative 45/14 a and the coefficient of -14(0 = -1/14x^2 + 23/7x – x^2 45/14) 3. Factor the equation 0 = x^2 – 46x + 45 4. Set the x values equal 0 = (x – 45)(x – 1) to zero and solve for x 0 = x – 45 0=x–1 x = 45 x=1Now we know that the ball starts at the 45 yard line and lands 1 yard short of the field goal. 5. 5. Now that we have our vertex and intercepts, we can find otherpoints on our graph. 1. Start by making a table with x and y values. 2. Have two numbers both above and below the x value of the vertex 3. Plug in two of the x values into your original equation. (make sure that these values are either one unit above or below the x value, and two units above or below the x value; never two values the same amount away from the vertex) 4. Solve for y with each different value 5. Plug what you get for y into the spot on your table that corresponds with the x value you plugged in 6. Now you have points that lay on your kick’s path to plot on your graph. X Y 21 34.28571429 22 34.5 23 34.57142857 24 34.5 25 34.28571429 6. 6. Now that we have all of the information that we need, we can graph it. We start by first plotting the points, and then connecting them. On our graph we have the axis of symmetry running through the vertex at (23, 34.57142857). Vertex Maximum Axis of symmetry Roots •The y-axis represents the position of the field goal along the x-axis (so at x = 0 is the field goal) •From the 10 yard mark to the 0 mark on the x- axis of the graph is the end zone 7. 7. To get a more precise graph, we plugged our information into the graphing calculator. 8. 8. The kicker starts at 45 yards on the opposing team’s side. He kicks the ball, which peaks at 34.57142857 feet in the air, at 23 yards on the field. From there it goes back down and lands 9 yards into the end zone, or 1 yard away from the field goal. Even if the kicker were to get that one last yard, his kick would not have had enough height to make the field goal. 9. 9. Based off of our graph, we would make a Hail Mary pass in the last three seconds of the game in hope of gaining yards. We would not take the kick at the 38 yard line because our kicker’s kick would land 1 yard short of the field goal. To make the field goal, we would need to gain at least 14 yards. This is because on our graph, the y-intercept is -3.214285714, and as the field goal is 10 feet tall, we would need to gain 14 yards. That extra yardage would be the minimum amount of yards that we would need to gain to get the kick’s height high enough to get the field goal. Thus, we will make a Hail Mary pass in hopes of gaining at least 14 yards.
# How to solve radical equations The solver will provide step-by-step instructions on How to solve radical equations. So let's get started! ## How can we solve radical equations We can do your math homework for you, and we'll make sure that you understand How to solve radical equations. Negative exponents in fractions can be solved by using the reciprocal property. This states that when taking the reciprocal of a number, the exponent becomes the opposite sign. So, to solve a fraction with a negative exponent, you simply take the reciprocal of the base number and then change the sign of the exponent. Partial fractions is a method for decomposing a fraction into a sum of simpler fractions. The process involves breaking up the original fraction into smaller pieces, each of which can be more easily simplified. While partial fractions can be used to decompose any fraction, it is particularly useful for dealing with rational expressions that contain variables. In order to solve a partial fraction, one must first determine the factors of the denominator. Once the factors have been determined, the numerator can be factored as well. The next step is to identify the terms in the numerator and denominator that share common factors. These terms can then be combined, and the resulting expression can be simplified. Finally, the remaining terms in the numerator and denominator can be solve for using basic algebraic principles. By following these steps, one can solve any partial fraction problem. One way is to solve each equation separately. For example, if you have an equation of the form x + 2 = 5, then you can break it up into two separate equations: x = 2 and y = 5. Solving the two set of equations separately gives you the two solutions: x = 1 and y = 6. This type of method is called a “separation method” because you separate out the two sets of equations (one equation per set). Another way to solve linear equations is by substitution. For example, if you have an equation of the form y = 9 - 4x + 6, then you can substitute different values for y in order to find out what happens when x changes. For example, if you plug in y = 8 - 3x + 3 into this equation, then the result is y= 8 - 3x + 7. Substitution is also known as “composite addition” or “additive elimination” because it involves adding or subtracting to eliminate one variable from another (hence eliminating one solution from another)! Another option There are many ways to solve problems involving interval notation. One popular method is to use a graphing calculator. Many graphing calculators have a built-in function that allows you to input an equation and then see the solution in interval notation. Another method is to use a table of values. This involves solving the equation for a few different values and then graphing the results. If the graph is a straight line, then the solution is simple to find. However, if the graph is not a straight line, then the solution may be more complicated. In either case, it is always important to check your work to make sure that the answer is correct. ## We cover all types of math problems Very helpful and clearly-understandable steps I like playing with math, and used to be pretty good at it, but if you don't practice, you forget. Now that my kids are in Middle and High School, this app lets me help them by reminding me of principles and theorems. The app is efficient, easy to use, and makes good use of the device's camera for input with minimal errors. Flora Thomas Good app. But, more types of math (i.e., Integration, differentiation) have to be included. Intersecting points should be given in graphs. Add greatest integer function, fractional part detection. Also add "to the power" option in calculator section. Serenity Nelson
# Estimating Sum and Difference The procedure of estimating sum and difference are in the following examples. Working Rules for Estimating Sum or Difference: Step I: First we need to find the smaller number. Step II: Then round off the all given numbers to the highest place value of that of the smaller number. Step III: Finally add or subtract the rounded numbers as per the question. Solved Examples on Estimating Sum or Difference: 1. Estimate the sum 5290 + 17986 by estimating the numbers to their nearest (i) hundreds (ii) thousands. Solution: (i) Estimating the numbers 5290 and 17986 to their nearest hundreds, we get 5300 and 18000 respectively. We have, Therefore, estimated sum is 23300. (ii) Estimating the numbers 5290 and 17986 to their nearest hundreds, we get 5000 and 18000 respectively. We have, Therefore, estimated sum is 23000. 2. Estimate: 5673 – 436 by rounding off the numbers to their greatest places. Also,, find the reasonable estimate. Solution: We have, 5673 – 436 = 5237. The greatest place in 5673 is thousands place and in 436 the greatest place is hundred place. Estimating 5673 to nearest thousands, we get 6000 Estimating 436 to nearest hundreds, we get 400 Therefore estimated difference = 6000 – 400 = 5600 Clearly, it is not closer to the actual difference. So, it is not a reasonable estimate. Let us round off 5673 and 436 to nearest hundreds. 5673 rounds off as 5700. 436 rounds off as 400 Therefore estimated difference = 5700 – 400 = 5300 3. Give a rough estimate and also a closer estimate of 489342 – 48365. Solution: We have, 489342 – 48365 = 440877 To find rough estimate, let us round off each number to nearest ten thousands. 489342 rounds off as 490000 48365 rounds off as 50000 Estimates difference = 490000 – 50000 = 440000 So, rough estimate = 440000 In order to obtain a closer estimate, let us round off each number to nearest thousands. 489342 rounds off as 489000 48365 rounds off as 48000 Estimated difference = 489000 – 48000 = 441000 Clearly, it is closer to the actual difference 440977 Hence, closer estimate is 441000 4. Estimate the following Sum or Difference: (i) 578 + 1842 (ii) 8435 - 597 Solution: (i) Estimate the sum of 578 + 1842 Out of 578 and 1842, the smaller number is 578. Hence, we have to round off both the numbers (578 and 1842) to the nearest hundred. 578 to the nearest hundred = 600 1842 to the nearest hundred = 1800 1842 → 1800 578 → + 600 2400 Estimated value of 1842 + 578 = 2400 (ii) Estimate the difference of 7645 - 867 Out of 867 and 7645, the smaller number is 867. Hence, we have to round off both the numbers (867 and 7645) to the nearest hundred. 867 to the nearest hundred = 900 7645 to the nearest hundred = 7600 7645 → 7600 867 → - 900 6700 Therefore, the estimated value of 7645 - 867 = 6700. Worksheet on Estimating Sum and Difference: 1. Estimate the following: (i) 243 + 4272 (ii)145 + 2478 (iii) 1248 - 167 (iv) 8465 - 1234 (v) 243 + 1252 (vi) 2431 + 142 (vii) 243 + 178 (viii) 4217 - 398 1. (i) 4500 (ii) 2600 (iii) 1000 (iv) 7300 (v) 1500 (vi) 2500 (vii) 400 (viii) 3800 2. Two different size of petrol tanks contain 6325 ℓ and 4890 ℓ water respectively. What is the estimated difference of petrol in the two tanks to the nearest hundred? Solution: 7325 ℓ → 7300 ℓ (Round off to the nearest hundreds) 4890 ℓ → 4900 ℓ (Round off to the nearest hundreds) The estimated difference of petrol in the two tanks = (7300 - 4900) ℓ = 2400 ℓ Therefore the required estimated difference = 2400 ℓ Estimate to Nearest Tens Estimate to Nearest Hundreds Estimate to Nearest Thousands Estimating Sum and Difference Estimating Product and Quotient ## You might like these • ### Counting Natural Numbers \| Definition of Natural Numbers \| Counting Natural numbers are all the numbers from 1 onwards, i.e., 1, 2, 3, 4, 5, …... and are used for counting. We know since our childhood we are using numbers 1, 2, 3, 4, 5, 6, ……….. • ### Reading and Writing Large Numbers \| Large Numbers in Words in Billion In reading and writing large numbers we group place values into periods ‘ones or unit’, ‘tens’, ‘hundred’, ‘thousand’, ‘10 thousand’, ‘100 thousand’, ‘million’, ’10 million’, ‘100 million • ### Numbers \| Notation \| Numeration \| Numeral \| Estimation \| Examples Numbers are used for calculating and counting. These counting numbers 1, 2, 3, 4, 5, .......... are called natural numbers. In order to describe the number of elements in a collection with no objects The properties of addition whole numbers are as follows: Closure property: If a and b are two whole numbers, then a + b is also a whole number. In other words, the sum of any two whole numbers i • ### Properties of Multiplication \| Multiplicative Identity \| Whole Numbers There are six properties of multiplication of whole numbers that will help to solve the problems easily. The six properties of multiplication are Closure Property, Commutative Property, Zero Property, Identity Property, Associativity Property and Distributive Property. • ### Estimating Product and Quotient \|Estimated Product \|Estimated Quotient The procedure of estimating product and quotient are in the following examples. Example 1: Estimate the product 958 × 387 by rounding off each factor to its greatest place. • ### Properties of Whole Numbers \| Closure Property \| Commutative Property The properties of whole numbers are as follows: The number 0 is the first and the smallest whole numbers. • All natural numbers along with zero are called whole numbers. • ### Representation of Whole Numbers on Number Line \| Compare Whole Numbers Numbers on a line is called the representation of whole numbers on number line. The number line also helps us to compare two whole numbers, i.e., to decide which of the two given whole numbers • ### Properties of Subtraction \|Whole Numbers \|Subtraction of Whole Numbers 1. When zero is subtracted from the number, the difference is the number itself. For example, 8931 – 0 = 8931, 5649 – 0 = 5649 2. When a number is subtracted from itself the difference is zero. For example, 5485 – 5485 = 0 3. When 1 is subtracted from a number, we get its • ### Whole Numbers \| Definition of Whole Numbers \| Smallest Whole Number The whole numbers are the counting numbers including 0. We have seen that the numbers 1, 2, 3, 4, 5, 6……. etc. are natural numbers. These natural numbers along with the number zero • ### Addition of Numbers using Number Line \| Addition Rules on Number Line Addition of numbers using number line will help us to learn how a number line can be used for addition. Addition of numbers can be well understood with the help of the number line. • ### Subtraction of Numbers using Number Line \|Subtracting with Number Line Subtraction of numbers using number line will help us to learn how a number line can be used for subtracting one number from the another number. • ### Worksheet on Reading and Writing Large Numbers\|Writing Numbers in Word Practice the questions given in the worksheet on reading and writing large numbers to group place values into periods in hundred, thousand, million and billion. The questions are related to writing • ### Worksheet on Estimation \| Estimate the Product \|Nearest Tens, Hundreds Practice the questions given in the worksheet on estimation. The questions are based on estimating the sum, difference, product and quotient to the nearest tens, hundreds and thousands. • ### Properties of Adding Integers \| Closure \|Commutative \| Associative ... The properties of adding integers are discussed here along with the examples. 1. The addition (sum) of any two integers is always an integer. For example: (i) 5 + 9 = 14 ∈ Z (ii) (-5) + 9 = 4 ∈ Z Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Thousandths Place in Decimals \| Decimal Place Value \| Decimal Numbers Jul 20, 24 03:45 PM When we write a decimal number with three places, we are representing the thousandths place. Each part in the given figure represents one-thousandth of the whole. It is written as 1/1000. In the decim… 2. ### Hundredths Place in Decimals \| Decimal Place Value \| Decimal Number Jul 20, 24 02:30 PM When we write a decimal number with two places, we are representing the hundredths place. Let us take plane sheet which represents one whole. Now, we divide the sheet into 100 equal parts. Each part r… 3. ### Tenths Place in Decimals \| Decimal Place Value \| Decimal Numbers Jul 20, 24 12:03 PM The first place after the decimal point is tenths place which represents how many tenths are there in a number. Let us take a plane sheet which represents one whole. Now, divide the sheet into ten equ… 4. ### Representing Decimals on Number Line \| Concept on Formation of Decimal Jul 20, 24 10:38 AM Representing decimals on number line shows the intervals between two integers which will help us to increase the basic concept on formation of decimal numbers.
### Kirchhoff’s First Law The physicist Gustav Robert Kirchhoff (1824–1887) was a researcher and experimentalist in a time when little was understood about how electric currents flow. Nevertheless, he used certain commonsense notions to deduce two important properties of dc circuits. Kirchhoff reasoned that dc ought to behave something like water in a network of pipes, and that the current going into any point ought to be the same as the current going out of that point. This, Kirchhoff thought, must be true for any point in a circuit, no matter how many branches lead into or out of the point. Kirchhoff ’s First Law. At A, the current into point X or point Y is the same as the current out of that point. That is, I = I1 + I2. At B, the current into point Z equals the current flowing out of point Z. That is, I1 + I2 = I3 + I4 + I5. Illustration for Quiz Questions 13 and 14. Two examples of this principles are shown in above figure. Examine illustration A. At point X, I, the current going in, equals I1 + I2, the current going out. At point Y, I2 + I1, the current going in, equals I, the current going out. Now look at illustration B. In this case, at point Z, the current I1 + I2 going in is equal to the current I3 + I4 + I5 going out. These are examples of Kirchhoff ’s First Law.We can also call it Kirchhoff ’s Current Law or the principle of conservation of current. #### Problem 1 Refer to above figure A. Suppose all three resistors have values of 100 Ω, and that I1 = 2.0 A and I2 = 1.0 A. What is the battery voltage? First, find the current I drawn from the battery: I = I1 + I2 = 2.0 + 1.0 = 3.0 A. Next, find the resistance of the entire network. The two 100-Ω resistances in series give a value of 200 Ω, and this is in parallel with 100 Ω. You can do the calculations and find that the total resistance, R, connected across the battery is 66.67 Ω. Then E = IR = 66.67 × 3.0 = 200 V. #### Problem 2 In above figure B, suppose each of the two resistors below point Z has a value of 100 Ω, and all three resistors above point Z have values of 10.0 Ω. Suppose the current through each 100-Ω resistor is 500 mA. What is the current through any one of the 10.0-Ω resistors, assuming that the current through all three 10.0-Ω resistors is the same? What is the voltage across any one of the three 10.0-Ω resistors? The total current into point Z is 500 mA + 500 mA = 1.00 A. This is divided equally among the three 10-Ω resistors. Therefore, the current through any one of them is 1.00/3 A = 0.333 A. The voltage across any one of the 10.0-Ω resistors can thus found by Ohm’s Law: E = IR = 0.333 × 10.0 = 3.33 V.
Exponential and Logarithmic Functions Evaluate and Graph Logarithmic Functions Learning Objectives By the end of this section, you will be able to: • Convert between exponential and logarithmic form • Evaluate logarithmic functions • Graph Logarithmic functions • Solve logarithmic equations • Use logarithmic models in applications Before you get started, take this readiness quiz. 1. Solve: If you missed this problem, review (Figure). 2. Evaluate: If you missed this problem, review (Figure). 3. Solve: If you missed this problem, review (Figure). We have spent some time finding the inverse of many functions. It works well to ‘undo’ an operation with another operation. Subtracting ‘undoes’ addition, multiplication ‘undoes’ division, taking the square root ‘undoes’ squaring. As we studied the exponential function, we saw that it is one-to-one as its graphs pass the horizontal line test. This means an exponential function does have an inverse. If we try our algebraic method for finding an inverse, we run into a problem. To deal with this we define the logarithm function with base a to be the inverse of the exponential function We use the notation and say the inverse function of the exponential function is the logarithmic function. Logarithmic Function The function is the logarithmic function with base , where and Convert Between Exponential and Logarithmic Form Since the equations and are equivalent, we can go back and forth between them. This will often be the method to solve some exponential and logarithmic equations. To help with converting back and forth let’s take a close look at the equations. See (Figure). Notice the positions of the exponent and base. If we realize the logarithm is the exponent it makes the conversion easier. You may want to repeat, “base to the exponent give us the number.” Convert to logarithmic form: and Convert to logarithmic form: Convert to logarithmic form: In the next example we do the reverse—convert logarithmic form to exponential form. Convert to exponential form: and Convert to exponential form: Convert to exponential form: Evaluate Logarithmic Functions We can solve and evaluate logarithmic equations by using the technique of converting the equation to its equivalent exponential equation. Find the value of x: and * QuickLaTeX cannot compile formula: \begin{array}{}\\ \\ & & & \hfill \phantom{\rule{1em}{0ex}}{\text{log}}_{\frac{1}{2}}\frac{1}{8}& =\hfill & x\hfill \\ \text{Convert to exponential form.}\hfill & & & \hfill {\left(\frac{1}{2}\right)}^{x}& =\hfill & \frac{1}{8}\hfill \\ \text{Rewrite}\phantom{\rule{0.2em}{0ex}}\frac{1}{8}\phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}{\left(\frac{1}{2}\right)}^{3}.\hfill & & & \hfill {\left(\frac{1}{2}\right)}^{x}& =\hfill & {\left(\frac{1}{2}\right)}^{3}\hfill \\ \text{With the same base, the exponents must be equal.}\hfill & & & \hfill x& =\hfill & 3\hfill & \text{Therefore,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{\frac{1}{2}}\frac{1}{8}=3\hfill \end{array} * Error message: Missing # inserted in alignment preamble. leading text: \begin{array}{} Extra alignment tab has been changed to \cr. leading text:\begin{array}{}\\ \\ & Extra alignment tab has been changed to \cr. leading text: \begin{array}{}\\ \\ & & Extra alignment tab has been changed to \cr. leading text:\begin{array}{}\\ \\ & & & Missing $inserted. leading text: ...fill \phantom{\rule{1em}{0ex}}{\text{log}}_ Missing$ inserted. Extra alignment tab has been changed to \cr. Extra alignment tab has been changed to \cr. Extra alignment tab has been changed to \cr. leading text: ...\text{Convert to exponential form.}\hfill & Find the value of Find the value of When see an expression such as we can find its exact value two ways. By inspection we realize it means to what power will be Since we know An alternate way is to set the expression equal to and then convert it into an exponential equation. Find the exact value of each logarithm without using a calculator: and Find the exact value of each logarithm without using a calculator: 2 Find the exact value of each logarithm without using a calculator: 2 Graph Logarithmic Functions To graph a logarithmic function it is easiest to convert the equation to its exponential form, Generally, when we look for ordered pairs for the graph of a function, we usually choose an x-value and then determine its corresponding y-value. In this case you may find it easier to choose y-values and then determine its corresponding x-value. Graph To graph the function, we will first rewrite the logarithmic equation, in exponential form, We will use point plotting to graph the function. It will be easier to start with values of y and then get x. 0 1 2 3 Graph: Graph: The graphs of and are the shape we expect from a logarithmic function where We notice that for each function the graph contains the point This make sense because means which is true for any a. The graph of each function, also contains the point This makes sense as means which is true for any a. Notice too, the graph of each function also contains the point This makes sense as means which is true for any a. Look at each graph again. Now we will see that many characteristics of the logarithm function are simply ’mirror images’ of the characteristics of the corresponding exponential function. What is the domain of the function? The graph never hits the y-axis. The domain is all positive numbers. We write the domain in interval notation as What is the range for each function? From the graphs we can see that the range is the set of all real numbers. There is no restriction on the range. We write the range in interval notation as When the graph approaches the y-axis so very closely but will never cross it, we call the line the y-axis, a vertical asymptote. Properties of the Graph of when Domain Range None Contains Asymptote Our next example looks at the graph of when Graph To graph the function, we will first rewrite the logarithmic equation, in exponential form, We will use point plotting to graph the function. It will be easier to start with values of y and then get x. 0 1 2 3 Graph: Graph: Now, let’s look at the graphs and , so we can identify some of the properties of logarithmic functions where The graphs of all have the same basic shape. While this is the shape we expect from a logarithmic function where We notice, that for each function again, the graph contains the points, This make sense for the same reasons we argued above. We notice the domain and range are also the same—the domain is and the range is The -axis is again the vertical asymptote. We will summarize these properties in the chart below. Which also include when Properties of the Graph of when when Domain Domain Range Range -intercept -intercept -intercept none -intercept None Contains Contains Asymptote -axis Asymptote -axis Basic shape increasing Basic shape Decreasing We talked earlier about how the logarithmic function is the inverse of the exponential function The graphs in (Figure) show both the exponential (blue) and logarithmic (red) functions on the same graph for both and Notice how the graphs are reflections of each other through the line We know this is true of inverse functions. Keeping a visual in your mind of these graphs will help you remember the domain and range of each function. Notice the x-axis is the horizontal asymptote for the exponential functions and the y-axis is the vertical asymptote for the logarithmic functions. Solve Logarithmic Equations When we talked about exponential functions, we introduced the number e. Just as e was a base for an exponential function, it can be used a base for logarithmic functions too. The logarithmic function with base e is called the natural logarithmic function. The function is generally written and we read it as “el en of Natural Logarithmic Function The function is the natural logarithmic function with base where When the base of the logarithm function is 10, we call it the common logarithmic function and the base is not shown. If the base a of a logarithm is not shown, we assume it is 10. Common Logarithmic Function The function is the common logarithmic function with base, where To solve logarithmic equations, one strategy is to change the equation to exponential form and then solve the exponential equation as we did before. As we solve logarithmic equations, , we need to remember that for the base a, and Also, the domain is Just as with radical equations, we must check our solutions to eliminate any extraneous solutions. Solve: and Solve: Solve: Solve: and Solve: Solve: Use Logarithmic Models in Applications There are many applications that are modeled by logarithmic equations. We will first look at the logarithmic equation that gives the decibel (dB) level of sound. Decibels range from 0, which is barely audible to 160, which can rupture an eardrum. The in the formula represents the intensity of sound that is barely audible. Decibel Level of Sound The loudness level, D, measured in decibels, of a sound of intensity, I, measured in watts per square inch is Extended exposure to noise that measures 85 dB can cause permanent damage to the inner ear which will result in hearing loss. What is the decibel level of music coming through ear phones with intensity watts per square inch? Substitute in the intensity level, I. Simplify. Since Multiply. The decibel level of music coming through earphones is 100 dB. What is the decibel level of one of the new quiet dishwashers with intensity watts per square inch? The quiet dishwashers have a decibel level of 50 dB. What is the decibel level heavy city traffic with intensity watts per square inch? The decibel level of heavy traffic is 90 dB. The magnitude of an earthquake is measured by a logarithmic scale called the Richter scale. The model is where is the intensity of the shock wave. This model provides a way to measure earthquake intensity. Earthquake Intensity The magnitude R of an earthquake is measured by where I is the intensity of its shock wave. In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. Over 80% of the city was destroyed by the resulting fires. In 2014, Los Angeles experienced a moderate earthquake that measured 5.1 on the Richter scale and caused ?108 million dollars of damage. Compare the intensities of the two earthquakes. To compare the intensities, we first need to convert the magnitudes to intensities using the log formula. Then we will set up a ratio to compare the intensities. In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. In 1989, the Loma Prieta earthquake also affected the San Francisco area, and measured 6.9 on the Richter scale. Compare the intensities of the two earthquakes. The intensity of the 1906 earthquake was about 8 times the intensity of the 1989 earthquake. In 2014, Chile experienced an intense earthquake with a magnitude of 8.2 on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured 5.1 on the Richter scale. Compare the intensities of the two earthquakes. The intensity of the earthquake in Chile was about 1,259 times the intensity of the earthquake in Los Angeles. Access these online resources for additional instruction and practice with evaluating and graphing logarithmic functions. Key Concepts • Properties of the Graph of when when Domain Domain Range Range x-intercept x-intercept y-intercept none y-intercept none Contains Contains Asymptote y-axis Asymptote y-axis Basic shape increasing Basic shape decreasing • Decibel Level of Sound: The loudness level, , measured in decibels, of a sound of intensity, , measured in watts per square inch is • Earthquake Intensity: The magnitude of an earthquake is measured by where is the intensity of its shock wave. Practice Makes Perfect Convert Between Exponential and Logarithmic Form In the following exercises, convert from exponential to logarithmic form. In the following exercises, convert each logarithmic equation to exponential form. Evaluate Logarithmic Functions In the following exercises, find the value of in each logarithmic equation. In the following exercises, find the exact value of each logarithm without using a calculator. 2 0 Graph Logarithmic Functions In the following exercises, graph each logarithmic function. Solve Logarithmic Equations In the following exercises, solve each logarithmic equation. Use Logarithmic Models in Applications In the following exercises, use a logarithmic model to solve. What is the decibel level of normal conversation with intensity watts per square inch? What is the decibel level of a whisper with intensity watts per square inch? A whisper has a decibel level of 20 dB. What is the decibel level of the noise from a motorcycle with intensity watts per square inch? What is the decibel level of the sound of a garbage disposal with intensity watts per square inch? The sound of a garbage disposal has a decibel level of 100 dB. In 2014, Chile experienced an intense earthquake with a magnitude of on the Richter scale. In 2010, Haiti also experienced an intense earthquake which measured on the Richter scale. Compare the intensities of the two earthquakes. The Los Angeles area experiences many earthquakes. In 1994, the Northridge earthquake measured magnitude of on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured on the Richter scale. Compare the intensities of the two earthquakes. The intensity of the 1994 Northridge earthquake in the Los Angeles area was about 40 times the intensity of the 2014 earthquake. Writing Exercises Explain how to change an equation from logarithmic form to exponential form. Explain the difference between common logarithms and natural logarithms. Explain why Explain how to find the on your calculator. Self Check After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. After reviewing this checklist, what will you do to become confident for all objectives? Glossary common logarithmic function The function is the common logarithmic function with base where logarithmic function The function is the logarithmic function with base where and natural logarithmic function The function is the natural logarithmic function with base where
My Math Forum System of equations Algebra Pre-Algebra and Basic Algebra Math Forum March 1st, 2015, 10:01 AM #1 Member Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 System of equations I need to solve for a and b: $\displaystyle 10a+b-7=3a+3b$ $\displaystyle a^2+b^2-ab=10a+b$ I can't find a way how to do it. Please help! March 1st, 2015, 10:17 AM #2 Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 From the first equation, can you write b in terms of a? March 1st, 2015, 10:21 AM #3 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs I would begin by simplifying the first equation by combining like terms so that we have the system: $\displaystyle 7a-2b=7\tag{1}$ $\displaystyle a^2+b^2-ab=10a+b\tag{2}$ Now, solve (1) for $a$, and then substitute into (2)...you will get a quadratic in $b$... March 1st, 2015, 03:27 PM #4 Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 $\displaystyle 10a+b-7=3a+3b$ $\displaystyle b-3b=3a-10a+7$ $\displaystyle -2b=-7a+7$ $\displaystyle 2b=7a-7$ $\displaystyle b=\frac{7a-7}{2}$ $\displaystyle a^2+(\frac{7a-7}{2})^2-a(\frac{7a-7}{2})=10a+(\frac{7a-7}{2})$ $\displaystyle a^2+(\frac{49a^2-98a+49}{4})-(\frac{7a^2-7a}{2})=10a+(\frac{7a-7}{2})$ $\displaystyle 4a^2+49a^2-98a+49-14a^2+14a=40a+14a-14$ $\displaystyle 39a^2-138a+63=0$ $\displaystyle 13a^2-46a+21=0$ $\displaystyle (13a-7)(a-3)=0$ $\displaystyle a=3, a=\frac{7}{13}$ $\displaystyle b=\frac{7(3)-7}{2}=7$ $\displaystyle b=\frac{7(\frac{7}{13})-7}{2}=-\frac{21}{13}$ $\displaystyle \therefore a=3,b=7 ; a=\frac{7}{13},b=-\frac{21}{13}$ Tags equations, sustem, system Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post ZeusTheMunja Linear Algebra 20 January 25th, 2013 06:52 AM zana93 Algebra 1 July 12th, 2012 11:02 AM Pell's fish Number Theory 1 March 26th, 2012 06:18 AM proglote Algebra 8 July 31st, 2011 05:22 AM aviation officer Algebra 4 March 8th, 2011 11:08 AM Contact - Home - Forums - Cryptocurrency Forum - Top
Courses Courses for Kids Free study material Offline Centres More Store # A committee of seven has to be formed from 9 boys and 4 girls. In how many ways this can be done when the committee consists of1. Exactly 3 girls.2. At least 3 boys.3. At most 3 girls. Last updated date: 13th Jun 2024 Total views: 413.4k Views today: 9.13k Verified 413.4k+ views Hint: In this given question, there are three sections. According to the given sections hints are also given section wise. Section1: In this section exactly three girls are selected in the committee. So if three girls are selected then only four boys can be selected from the group of nine boys for that particular committee of seven members. So, first we choose 3 girls among 4 girls and 4 boys among 9 boys. Section2: In this section, at least 3 girls are selected in this committee so it is possible that the committee can have 3 girls or 4 girls. So, if we select 3 girls then we have to select 4 boys and if we select 4 girls then we have to select 3 boys. Section3: In this section, we are given the condition that at most 3 girls can be selected to form the committee of 7 members. So it is possible that from 0 to 3 girls can be selected for the committee. According to that if 0 girls are selected then 7 boys will be selected. If 1 girl is selected then 6 boys will be selected. If 2 girls are selected then 5 boys will be selected. Similarly, if 3 girls are selected then 4 boys will be selected. Complete step by step solution: There three sections of his question in each step we will solve each section separately. Step1: According to section 1, the committee has exactly 3 girls and the rest are boys. So, we have to choose 4 boys among 9 boys. So, Numbers of way the committee will be formed are $\begin{array}{l} = 4{C_3} + 9{C_4}\\ = 4 \times 126 = 504 \end{array}$ Hence, numbers of ways selecting the committee are 504 Step2: In section 2, at least 3 girls will be selected so possibilities of boys will be 3 or 4. So, $\begin{array}{l} = 4{C_3} \times 9{C_4} + 4{C_4} \times 9{C_3}\\ = 4 \times 126 + 1 \times 84\\ = 504 + 84\\ = 588 \end{array}$ Hence the numbers of ways to form the committee are 588 Step3: In section 3, at most 3 girls will be selected. So, the possibilities will be $\begin{array}{l} 0girls7boys = 4{C_0} \times 9{C_7} = 36\\ 1girls6boys = 4{C_1} \times 9{C_6} = 336\\ 2girls5boys = 4{C_2} \times 9{C_5} = 756\\ 3girls4boys = 4{C_3} \times 9{C_4} = 126\\ \end{array}$ Numbers of ways to form the committee are $\begin{array}{l} = (0girls + 7boys) + (1girls + 6boys) + (2girls + 5boys) + (3girls + 4boys)\\ = 36 + 336 + 756 + 126\\ = 1254\\ \end{array}$ Note: In this question, three sections of questions have different concepts of selecting the girls and boys. So, while dealing with each section we should give more emphasis on the given conditions such as ‘at least’, ‘at most’ and terms such as ‘exactly’.
# An algebra of orders Did you know that you can add and multiply orders? For any two order structures $A$ and $B$, we can form the ordered sum $A+B$ and ordered product $A\otimes B$, and other natural operations, such as the disjoint sum $A\sqcup B$, which make altogether an arithmetic of orders. We combine orders with these operations to make new orders, often with interesting properties. Let us explore the resulting algebra of orders! $\newcommand\Z{\mathbb{Z}}\newcommand\N{\mathbb{N}}\newcommand\Q{\mathbb{Q}}\newcommand\P{\mathbb{P}}\newcommand\iso{\cong}\newcommand\of{\subseteq}$ One of the most basic operations that we can use to combine two orders is the disjoint sum operation $A\sqcup B$. This is the order resulting from placing a copy of $A$ adjacent to a copy of $B$, side-by-side, forming a combined order with no instances of the order relation between the two parts. If $A$ is the orange $\vee$-shaped order here and $B$ is the yellow linear order, for example, then $A\sqcup B$ is the combined order with all five nodes. Another kind of addition is the ordered sum of two orders $A+B$, which is obtained by placing a copy of $B$ above a copy of $A$, as indicated here by adding the orange copy of $A$ and the yellow copy of $B$. Also shown is the sum $B+A$, with the summands reversed, so that we take $B$ below and $A$ on top. It is easy to check that the ordered sum of two orders is an order. One notices immediately, of course, that the resulting ordered sums $A+B$ and $B+A$ are not the same! The order $A+B$ has a greatest element, whereas $B+A$ has two maximal elements. So the ordered sum operation on orders is not commutative. Nevertheless, we shall still call it addition. The operation, which has many useful and interesting features, goes back at least to the 19th century with Cantor, who defined the addition of well orders this way. In order to illustrate further examples, I have assembled here an addition table for several simple finite orders. The choices for $A$ appear down the left side and those for $B$ at the top, with the corresponding sum $A+B$ displayed in each cell accordingly. We can combine the two order addition operations, forming a variety of other orders this way. The reader is encouraged to explore further how to add various finite orders using these two forms of addition. What is the smallest order that you cannot generate from $1$ using $+$ and $\sqcup$? Please answer in the comments. We can also add infinite orders. Displayed here, for example, is the order $\N+(1\sqcup 1)$, the natural numbers wearing two yellow caps. The two yellow nodes at the top form a copy of $1\sqcup 1$, while the natural numbers are the orange nodes below. Every natural number (yes, all infinitely many of them) is below each of the two nodes at the top, which are incomparable to each other. Notice that even though we have Hasse diagrams for each summand order here, there can be no minimal Hasse diagram for the sum, because any particular line from a natural number to the top would be implied via transitivity from higher such lines, and we would need such lines, since they are not implied by the lower lines. So there is no minimal Hasse diagram. This order happens to illustrate what is called an exact pair, which occurs in an order when a pair of incomparable nodes bounds a chain below, with the property that any node below both members of the pair is below something in the chain. This phenomenon occurs in sometimes unexpected contexts—any countable chain in the hierarchy of Turing degrees in computability theory, for example, admits an exact pair. Let us turn now to multiplication. The ordered product $A\otimes B$ is the order resulting from having $B$ many copies of $A$. That is, we replace each node of $B$ with an entire copy of the $A$ order. Within each of these copies of $A$, the order relation is just as in $A$, but the order relation between nodes in different copies of $A$, we follow the $B$ relation. It is not difficult to check that indeed this is an order relation. We can illustrate here with the same two orders we had earlier. In forming the ordered product $A\otimes B$, in the center here, we take the two yellow nodes of $B$, shown greatly enlarged in the background, and replace them with copies of $A$. So we have ultimately two copies of $A$, one atop the other, just as $B$ has two nodes, one atop the other. We have added the order relations between the lower copy of $A$ and the upper copy, because in $B$ the lower node is related to the upper node. The order $A\otimes B$ consists only of the six orange nodes—the large highlighted yellow nodes of $B$ here serve merely as a helpful indication of how the product is formed and are not in any way part of the product order $A\otimes B$. Similarly, with $B\otimes A$, at the right, we have the three enlarged orange nodes of $B$ in the background, which have each been replaced with copies of $A$. The nodes of each of the lower copies of $A$ are related to the nodes in the top copy, because in $B$ the two lower nodes are related to the upper node. I have assembled a small multiplication table here for some simple finite orders. So far we have given an informal account of how to add and multiply ordered ordered structures. So let us briefly be a little more precise and formal with these matters. In fact, when it comes to addition, there is a slightly irritating matter in defining what the sums $A\sqcup B$ and $A+B$ are exactly. Specifically, what are the domains? We would like to conceive of the domains of $A\sqcup B$ and $A+B$ simply as the union the domains of $A$ and $B$—we’d like just to throw the two domains together and form the sums order using that combined domain, placing $A$ on the $A$ part and $B$ on the $B$ part (and adding relations from the $A$ to the $B$ part for $A+B$). Indeed, this works fine when the domains of $A$ and $B$ are disjoint, that is, if they have no points in common. But what if the domains of $A$ and $B$ overlap? In this case, we can’t seem to use the union in this straightforward manner. In general, we must disjointify the domains—we take copies of $A$ and $B$, if necessary, on domains that are disjoint, so that we can form the sums $A\sqcup B$ and $A+B$ on the union of those nonoverlapping domains. What do we mean precisely by “taking a copy” of an ordered structure $A$? This way of talking in mathematics partakes in the philosophy of structuralism. We only care about our mathematical structures up to isomorphism, after all, and so it doesn’t matter which isomorphic copies of $A$ and $B$ we use; the resulting order structures $A\sqcup B$ will be isomorphic, and similarly for $A+B$. In this sense, we are defining the sum orders only up to isomorphism. Nevertheless, we can be definite about it, if only to verify that indeed there are copies of $A$ and $B$ available with disjoint domains. So let us construct a set-theoretically specific copy of $A$, replacing each individual $a$ in the domain of $A$ with $(a,\text{orange})$, for example, and replacing the elements $b$ in the domain of $B$ with $(b,\text{yellow})$. If “orange” is a specific object distinct from “yellow,” then these new domains will have no points in common, and we can form the disjoint sum $A\sqcup B$ by using the union of these new domains, placing the $A$ order on the orange objects and the $B$ order on the yellow objects. Although one can use this specific disjointifying construction to define what $A\sqcup B$ and $A+B$ mean as specific structures, I would find it to be a misunderstanding of the construction to take it as a suggestion that set theory is anti-structuralist. Set theorists are generally as structuralist as they come in mathematics, and in light of Dedekind’s categorical account of the natural numbers, one might even find the origin of the philosophy of structuralism in set theory. Rather, the disjointifying construction is part of the general proof that set theory abounds with isomorphic copies of whatever mathematical structure we might have, and this is part of the reason why it serves well as a foundation of mathematics for the structuralist. To be a structuralist means not to care which particular copy one has, to treat one’s mathematical structures as invariant under isomorphism. But let me mention a certain regrettable consequence of defining the operations by means of a specific such disjointifying construction in the algebra of orders. Namely, it will turn out that neither the disjoint sum operation nor the ordered sum operation, as operations on order structures, are associative. For example, if we use $1$ to represent the one-point order, then $1\sqcup 1$ means the two-point side-by-side order, one orange and one yellow, but really what we mean is that the points of the domain are $\set{(a,\text{orange}),(a,\text{yellow})}$, where the original order is on domain $\set{a}$. The order $(1\sqcup 1)\sqcup 1$ then means that we take an orange copy of that order plus a single yellow point. This will have domain $$\set{\bigl((a,\text{orange}),\text{orange}\bigr),\bigl((a,\text{yellow}),\text{orange}\bigr),(a,\text{yellow})}$$ The order $1\sqcup(1\sqcup 1)$, in contrast, means that we take a single orange point plus a yellow copy of $1\sqcup 1$, leading to the domain $$\set{(a,\text{orange}),\bigl((a,\text{orange}),\text{yellow}\bigr),\bigl((a,\text{yellow}),\text{yellow}\bigr)}$$ These domains are not the same! So as order structures, the order $(1\sqcup 1)\sqcup 1$ is not identical with $1\sqcup(1\sqcup 1)$, and therefore the disjoint sum operation is not associative. A similar problem arises with $1+(1+1)$ and $(1+1)+1$. But not to worry—we are structuralists and care about our orders here only up to isomorphism. Indeed, the two resulting orders are isomorphic as orders, and more generally, $(A\sqcup B)\sqcup C$ is isomorphic to $A\sqcup(B\sqcup C)$ for any orders $A$, $B$, and $C$, and similarly with $A+(B+C)\cong(A+B)+C$, as discussed with the theorem below. Furthermore, the order isomorphism relation is a congruence with respect to the arithmetic we have defined, which means that $A\sqcup B$ is isomorphic to $A’\sqcup B’$ whenever $A$ and $B$ are respectively isomorphic to $A’$ and $B’$, and similarly with $A+B$ and $A\otimes B$. Consequently, we can view these operations as associative, if we simply view them not as operations on the order structures themselves, but on their order-types, that is, on their isomorphism classes. This simple abstract switch in perspective restores the desired associativity. In light of this, we are free to omit the parentheses and write $A\sqcup B\sqcup C$ and $A+B+C$, if care about our orders only up to isomorphism. Let us therefore adopt this structuralist perspective for the rest of our treatment of the algebra of orders. Let us give a more precise formal definition of $A\otimes B$, which requires no disjointification. Specifically, the domain is the set of pairs $\set{(a,b)\mid a\in A, b\in B}$, and the order is defined by $(a,b)\leq_{A\otimes B}(a’,b’)$ if and only if $b\leq_B b’$, or $b=b’$ and $a\leq_A a’$. This order is known as the reverse lexical order, since we are ordering the nodes in the dictionary manner, except starting from the right letter first rather than the left as in an ordinary dictionary. One could of course have defined the product using the lexical order instead of the reverse lexical order, and this would give $A\otimes B$ the meaning of “$A$ copies of $B$.” This would be a fine alternative and in my experience mathematicians who rediscover the ordered product on their own often tend to use the lexical order, which is natural in some respects. Nevertheless, there is a huge literature with more than a century of established usage with the reverse lexical order, from the time of Cantor, who defined ordinal multiplication $\alpha\beta$ as $\beta$ copies of $\alpha$. For this reason, it seems best to stick with the reverse lexical order and the accompanying idea that $A\otimes B$ means “$B$ copies of $A$.” Note also that with the reverse lexical order, we shall be able to prove left distributivity $A\otimes(B+C)=A\otimes B+A\otimes C$, whereas with the lexical order, one will instead have right distributivity $(B+C)\otimes^* A=B\otimes^* A+C\otimes^* A$. Let us begin to prove some basic facts about the algebra of orders. Theorem. The following identities hold for orders $A$, $B$, and $C$. 1. Associativity of disjoint sum, ordered sum, and ordered product.\begin{eqnarray}A\sqcup(B\sqcup C) &\iso& (A\sqcup B)\sqcup C\\ A+(B+C) &\iso& (A+B)+C\\ A\otimes(B\otimes C) &\iso& (A\otimes B)\otimes C \end{eqnarray} 2. Left distributivity of product over disjoint sum and ordered sum.\begin{eqnarray} A\otimes(B\sqcup C) &\iso& (A\otimes B)\sqcup(A\otimes C)\\ A\otimes(B+C) &\iso& (A\otimes B)+(A\otimes C) \end{eqnarray} In each case, these identities are clear from the informal intended meaning of the orders. For example, $A+(B+C)$ is the order resulting from having a copy of $A$, and above it a copy of $B+C$, which is a copy of $B$ and a copy of $C$ above it. So one has altogether a copy of $A$, with a copy of $B$ above that and a copy of $C$ on top. And this is the same as $(A+B)+C$, so they are isomorphic. One can also aspire to give a detailed formal proof verifying that our color-coded disjointifying process works as desired, and the reader is encouraged to do so as an exercise. To my way of thinking, however, such a proof offers little in the way of mathematical insight into algebra of orders. Rather, it is about checking the fine print of our disjointifying process and making sure that things work as we expect. Several of the arguments can be described as parenthesis-rearranging arguments—one extracts the desired information from the structure of the domain order and puts that exact same information into the correct form for the target order. For example, if we have used the color-scheme disjointifying process described above, then the elements of $A\sqcup(B\sqcup C)$ each have one of the following forms, where $a\in A$, $b\in B$, and $c\in C$: $$(a,\text{orange})$$ $$\bigl((b,\text{orange}),\text{yellow}\bigr)$$ $$\bigl((c,\text{yellow}),\text{yellow}\bigr)$$ We can define the color-and-parenthesis-rearranging function $\pi$ to put them into the right form for $(A\sqcup B)\sqcup C$ as follows: \begin{align} \pi:(a,\text{orange})\quad&\mapsto\quad \bigl((a,\text{orange}),\text{orange}\bigl) \\ \pi:\bigl((b,\text{orange}),\text{yellow}\bigr)\quad&\mapsto\quad \bigl((b,\text{yellow}),\text{orange}\bigl) \\ \pi:\bigl((c,\text{yellow}),\text{yellow}\bigr)\quad&\mapsto\quad (c,\text{yellow}) \end{align} In each case, we will preserve the order, and since the orders are side-by-side, the cases never interact in the order, and so this is an isomorphism. Similarly, for distributivity, the elements of $A\otimes(B\sqcup C)$ have the two forms: $$\bigl(a,(b,\text{orange})\bigr)$$ $$\bigl(a,(c,\text{yellow})\bigr)$$ where $a\in A$, $b\in B$, and $c\in C$. Again we can define the desired ismorphism $\tau$ by putting these into the right form for $(A\otimes B)\sqcup(A\otimes C)$ as follows: \begin{align} \tau:\bigl(a,(b,\text{orange})\bigr)\quad&\mapsto\quad \bigl((a,b),\text{orange}\bigr) \\ \tau:\bigl(a,(c,\text{yellow})\bigr)\quad&\mapsto\quad\bigl((a,c),\text{yellow}\bigr) \end{align} And again, this is an isomorphism, as desired. Since order multiplication is not commutative, it is natural to inquire about the right-sided distributivity laws: \begin{eqnarray} (B+C)\otimes A&\overset{?}{\cong}&(B\otimes A)+(C\otimes A)\\ (B\sqcup C)\otimes A&\overset{?}{\cong}&(B\otimes A)\sqcup(C\otimes A) \end{eqnarray} Unfortunately, however, these do not hold in general, and the following instances are counterexamples. Can you see what to take as $A$, $B$, and $C$? Please answer in the comments. Theorem. 1. If $A$ and $B$ are linear orders, then so are $A+B$ and $A\otimes B$. 2. If $A$ and $B$ are nontrivial linear orders and both are endless, then $A+B$ is endless; if at least one of them is endless, then $A\otimes B$ is endless. 3. If $A$ is an endless dense linear order and $B$ is linear, then $A\otimes B$ is an endless dense linear order. 4. If $A$ is an endless discrete linear order and $B$ is linear, then $A\otimes B$ is an endless discrete linear order. Proof. If both $A$ and $B$ are linear orders, then it is clear that $A+B$ is linear. Any two points within the $A$ copy are comparable, and any two points within the $B$ copy, and every point in the $A$ copy is below any point in the $B$ copy. So any two points are comparable and thus we have a linear order. With the product $A\otimes B$, we have $B$ many copies of $A$, and this is linear since any two points within one copy of $A$ are comparable, and otherwise they come from different copies, which are then comparable since $B$ is linear. So $A\otimes B$ is linear. For statement (2), we know that $A+B$ and $A\otimes B$ are nontrivial linear orders. If both $A$ and $B$ are endless, then clearly $A+B$ is endless, since every node in $A$ has something below it and every node in $B$ has something above it. For the product $A\otimes B$, if $A$ is endless, then every node in any copy of $A$ has nodes above and below it, and so this will be true in $A\otimes B$; and if $B$ is endless, then there will always be higher and lower copies of $A$ to consider, so again $A\otimes B$ is endless, as desired. For statement (3), assume that $A$ is an endless dense linear and that $B$ is linear. We know from (1) that $A\otimes B$ is a linear order. Suppose that $x<y$ in this order. If $x$ and $y$ live in the same copy of $A$, then there is a node $z$ between them, because $A$ is dense. If $x$ occurs in one copy of $A$ and $B$ in another, then because $A$ is endless, there will a node $z$ above $x$ in its same copy, leading to $x<z<y$ as desired. (Note: we don’t need $B$ to be dense.) For statement (4), assume instead that $A$ is an endless discrete linear order and $B$ is linear. We know that $A\otimes B$ is a linear order. Every node of $A\otimes B$ lives in a copy of $A$, where it has an immediate successor and an immediate predecessor, and these are also immediate successor and predecessor in $A\otimes B$. From this, it follows also that $A\otimes B$ is endless, and so it is an endless discrete linear order. $\Box$ The reader is encouraged to consider as an exercise whether one can drop the “endless” hypotheses in the theorem. Please answer in the comments. Theorem. The endless discrete linear orders are exactly those of the form $\Z\otimes L$ for some linear order $L$. Proof. If $L$ is a linear order, then $\Z\otimes L$ is an endless discrete linear order by the theorem above, statement (4). So any order of this form has the desired feature. Conversely, suppose that $\P$ is an endless discrete linear order. Define an equivalence relation for points in this order by which $p\sim q$ if and only $p$ and $q$ are at finite distance, in the sense that there are only finitely many points between them. This relation is easily seen to be reflexive, transitive and symmetric, and so it is an equivalence relation. Since $\P$ is an endless discrete linear order, every object in the order has an immediate successor and immediate predecessor, which remain $\sim$-equivalent, and from this it follows that the equivalence classes are each ordered like the integers $\Z$, as indicated by the figure here. The equivalence classes amount to a partition of $\P$ into disjoint segments of order type $\Z$, as in the various colored sections of the figure. Let $L$ be the induced order on the equivalence classes. That is, the domain of $L$ consists of the equivalence classes $\P/\sim$, which are each a $\Z$ chain in the original order, and we say $[a]<_L[b]$ just in case $a<_{\P}b$. This is a linear order on the equivalence classes. And since $\P$ is $L$ copies of its equivalence classes, each of which is ordered like $\Z$, it follows that $\P$ is isomorphic to $\Z\otimes L$, as desired. $\Box$ (Interested readers are advised that the argument above uses the axiom of choice, since in order to assemble the isomorphism of $\P$ with $\Z\otimes L$, we need in effect to choose a center point for each equivalence class.) If we consider the integers inside the rational order $\Z\of\Q$, it is clear that we can have a discrete suborder of a dense linear order. How about a dense suborder of a discrete linear order? Question. Is there a discrete linear order with a suborder that is a dense linear order? What? How could that happen? In my experience, mathematicians first coming to this topic often respond instinctively that this should be impossible. I have seen sophisticated mathematicians make such a pronouncement when I asked the audience about it in a public lecture. The fundamental nature of a discrete order, after all, is completely at odds with density, since in a discrete order, there is a next point up and down, and a next next point, and so on, and this is incompatible with density. Yet, surprisingly, the answer is Yes! It is possible—there is a discrete order with a suborder that is densely ordered. Consider the extremely interesting order $\Z\otimes\Q$, which consists of $\Q$ many copies of $\Z$, laid out here increasing from left to right. Each tiny blue dot is a rational number, which has been replaced with an entire copy of the integers, as you can see in the magnified images at $a$, $b$, and $c$. The order is quite subtle, and so let me also provide an alternative presentation of it. We have many copies of $\Z$, and those copies are densely ordered like $\Q$, so that between any two copies of $\Z$ is another one, like this: Perhaps it helps to imagine that the copies of $\Z$ are getting smaller and smaller as you squeeze them in between the larger copies. But you can indeed always fit another copy of $\Z$ between, while leaving room for the further even tinier copies of $\Z$ to come. The order $\Z\otimes\Q$ is discrete, in light of the theorem characterizing discrete linear orders. But also, this is clear, since every point of $\Z\otimes\Q$ lives in its local copy of $\Z$, and so has an immediate successor and predecessor there. Meanwhile, if we select exactly one point from each copy of $\Z$, the $0$ of each copy, say, then these points are ordered like $\Q$, which is dense. Thus, we have proved: Theorem. The order $\Z\otimes\Q$ is a discrete linear order having a dense linear order as a suborder. One might be curious now about the order $\Q\otimes\Z$, which is $\Z$ many copies of $\Q$. This order, however, is a countable endless dense linear order, and therefore is isomorphic to $\Q$ itself. This material is adapted from my book-in-progress, Topics in Logic, drawn from Chapter 3 on Relational Logic, which incudes an extensive section on order theory, of which this is an important summative part. # Famous quotations in their original first-order language Historians everywhere are shocked by the recent discovery that many of our greatest thinkers and poets had first expressed their thoughts and ideas in the language of first-order predicate logic, and sometimes modal logic, rather than in natural language. Some early indications of this were revealed in the pioneering historical research of Henle, Garfield and Tymoczko, in their work Sweet Reason: We now know that the phenomenon is widespread! As shown below, virtually all of our cultural leaders have first expressed themselves in the language of first-order predicate logic, before having been compromised by translations into the vernacular. $\neg\lozenge\neg\exists s\ G(i,s)$ $(\exists x\ x=i)\vee\neg(\exists x\ x=i)$ $\left(\strut\neg\exists t\ \exists d\ \strut D(d)\wedge F(d)\wedge S_t(i,d)\right)\wedge\left(\strut\neg\exists t\ w\in_t \text{Ro}\right)\wedge\left(\strut \text{Ru}(i,y)\to \lozenge\text{C}(y,i,qb)\wedge \text{Ru}(i)\wedge\text{Ru}(i)\wedge\text{Ru}(i)\wedge\text{Ru}(i)\right)$ $\exists!t\ T(t) \wedge \forall t\ (T(t)\to G(t))$ $\neg B_i \exists g\ G(g)$ $\forall b\ \left(\strut G(b)\wedge B(b)\to \exists x\ (D(b,x)\wedge F(x))\right)$ $(\exists!w\ W_1(w)\wedge W_2(w)), \ \ \exists w\ W_1(w)\wedge W_2(w)\wedge S(y,w)$? $\exists s\ Y(s)\wedge S(s)\wedge \forall x\ L(x,s)$ $\exists p\ \left[\forall c\ (c\neq p\to G(c))\right]\wedge\neg G(p)$ $\exists l\ \left[L(l)\wedge \Box_l\left({}^\ulcorner\,\forall g\ \text{Gl}(g)\to \text{Gd}(g){}^\urcorner\right)\wedge\exists s\ \left(SH(s)\wedge B(l,s)\right)\right]$ $(\forall p\in P\ \exists c\in\text{Ch}\ c\in p)\wedge(\forall g\in G\ \exists c\in\text{Cr}\ c\in g)$ $\forall x (F(w,x)\leftrightarrow x=F)$ $B\wedge \forall x\ \left[S(x)\wedge T(x)\to \exists!w\ W(w)\wedge\text{Gy}(x,w)\wedge\text{Gi}(x,w)\right]$ $\exists!x\ D(x)\wedge D(\ {}^\ulcorner D(i){}^\urcorner\ )$ $\forall f\ \forall g\ \left(\strut H(f)\wedge H(g)\to f\sim g\right)\wedge\forall f\ \forall g\ \left(\strut\neg H(f)\wedge \neg H(g)\to \neg\ f\sim g\right)$ $\exists w\ \left(\strut O(w)\wedge W(w)\wedge\exists s\ (S(s)\wedge L(w,s))\right)$ $C(i)\to \exists x\ x=i$ $\neg\neg\left(\strut H(y)\wedge D(y)\right)$ $\neg (d\in K)\wedge\neg (t\in K)$ $W(i,y)\wedge N(i,y)\wedge\neg\neg\lozenge L(i,y)\wedge \left(\strut \neg\ \frac23<0\to\neg S(y)\right)$ $\lozenge \text{CL}(i)\wedge\lozenge C(i)\wedge \lozenge (\exists x\ x=i)\wedge B(i)$ $\forall x\ K_x({}^\ulcorner \forall m\ \left[M(m)\wedge S(m)\wedge F(m)\to\Box\ \exists w\ M(m,w)\right]{}^\urcorner)$ $\forall e\forall h\ \left(\strut G(e)\wedge E(e)\wedge H(h)\to \neg L(i,e,h)\right)$ $\forall p\ \Box\text{St}(p)$ $\lozenge^w_i\ \forall g\in G\ \lozenge (g\in C)$ $\forall m\ (a\leq_C m)$ $\forall t\ (p\geq t)\wedge \forall t\ (p\leq t)$ $\forall x\ (F(x)\iff x=h)$ $(\forall x\ \forall y\ x=y)\wedge(\exists x\ \exists y ([\![x=x]\!]>[\![y=y]\!]))$ $\forall p\ \left(\strut\neg W(p)\to \neg S(p)\right)$ $\exists x\Box\Box x\wedge \exists x\Box\neg\Box x\wedge\exists x\neg\Box\neg\Box x$ $\forall p \left(\strut E(p)\to \forall h\in H\ A(p,h)\right)$ Dear readers, in order to assist with this important historical work, please provide translations into ordinary English in the comment section below of any or all of the assertions listed above. We are interested to make sure that all our assertions and translations are accurate. In addition, any readers who have any knowledge of additional instances of famous quotations that were actually first made in the language of first-order predicate logic (or similar) are encouraged to post comments below detailing their knowledge. I will endeavor to add such additional examples to the list. Thanks to Philip Welch, to my brother Jonathan, and to Ali Sadegh Daghighi (in the comments) for providing some of the examples, and to Timothy Gowers for some improvements. Please post comments or send me email if hints are desired.
# How to prove this formula for counting sub-sequences? A sequence $(s_i)_0^{N+2m}$ of ones and zeroes has ones on $s_0$ and $s_{N+2m}$ and it's subsequence $(s_i)_1^{N+2m-1}$ that lies in between contains $N+m-1$ ones and $m$ zeroes, where $N\in \mathbb{N}$ and $m\in \mathbb{N}_0$. Such a sequence $(s_i)$ can be split into $N+2m$ subsequences consisting of adjacent elements in $(s_i)$, which together form a set called the decomposition of $(s_i)$. For example, if $N=2$ and $m=1$, we could have $$(s)=(1,1,0,1,1) \mapsto \{(1,1),(1,0),(0,1),(1,1)\}.$$ There are four different such subsequences, namely $(1,1)$, $(0,0)$, $(1,0)$, and $(0,1)$. Assign a number $a$ to the first two and a number $b$ to the last two. For a given sequence $(s_i)$, define $p_i$ to be the product of the values associated with the elements in the decomposition of $(s_i)$. In the example above, $$p=a\cdot b\cdot b \cdot a=a^2b^2.$$ Given that there are $\binom{N+2m-1}{m}$ different sequences $(s_1),(s_2),\dots,(s_i),\ldots$ what is $$P_{mN}=\sum_{i=1}^{\binom{N+2m-1}{m}}\!p_i\;?$$ Continuing the example from above, we have the three possible sequences \begin{align} 10111 \rightarrow p_1&=b^2a^2\\ 11011 \rightarrow p_2&=aba \quad \rightarrow \quad P_{12}=3a^2b^2.\\ 11101 \rightarrow p_3&=a^2b^2 \end{align} I propose that the solution is $$\boxed{P_{mN}=\sum_{i=1}^m \binom{m-1}{i-1}\binom{N+m}{i} a^{N+2m-2i}b^{2i} \quad \text{for m>0 and that P_{0N}=a^N}}.$$ I've come to this proposal by simply looking at the patterns. How can it be proven? Here are some calculated examples of $P_{mN}$: \begin{array}{c\|c\|c\|c} m & N & P_{mN} & P_{mN}\\ \hline 0 & 1 & a \\ & 2 & a^2 \\ & 3 & a^3 \\ \hline 1 & 1 & 2ab^2 & \binom{0}{0}\binom{2}{1}ab^2\\ & 2 & 3a^2b^2 & \binom{0}{0}\binom{3}{1}a^3b^2\\ & 3 & 4a^3b^2 & \binom{0}{0}\binom{4}{1}a^4b^2\\ \hline 2 & 1 & 3a^3b^2+3ab^4 & \binom{1}{0}\binom{3}{1}a^3b^2+\binom{1}{1}\binom{3}{2}ab^4\\ & 2 & 4a^4b^2+6a^2b^4 & \binom{1}{0}\binom{4}{1}a^4b^2+\binom{1}{1}\binom{4}{2}a^2b^4\\ & 3 & 5a^5b^2+10a^3b^4 & \binom{1}{0}\binom{5}{1}a^5b^2+\binom{1}{1}\binom{5}{2}a^3b^4\\ \hline 3 & 1 & 4a^5b^2+12a^3b^4+4ab^6 \\ & 2 & 5a^6b^2+20a^4b^4+10a^2b^6 & \vdots\\ & 3 & 6a^6b^2+30a^5b^4+20a^3b^6 \\ \hline 4 & 1 & 5a^7b^2+30a^5b^4+30a^3b^6+5ab^8 \\ & 2 & 6a^8b^2+45a^6b^4+60a^4b^4+15a^2b^6 & \vdots\\ & 3 & 7a^3b^2+63a^7b^4+105a^5b^6+35a^3b^8 \\ \vdots & \vdots & \vdots \end{array} I have checked this for higher $m$, but it becomes tedious to write out. The exponents of $a$ and $b$ must add to $N+2m$, since there are exactly as many elements in a given decomposition. Furthermore, the coefficients must add to $\binom{N+2m-1}{m}$, as this is the number of possible sequences and each sequence contributes with one term. Any help would be much appreciated! • It seems that the summand indexed by a particular value of $i$ corresponds to sequences with $i$ runs of zeros in total. For example, $i=1$ corresponds to those sequences where all the zeros are consecutive. This certainly produces the right powers of $a$ and $b$; maybe you can show that the coefficient is correct in this way. – Greg Martin Aug 27 '16 at 17:50 • @GregMartin Nice observation, thanks! – Bobson Dugnutt Aug 27 '16 at 17:54 It’s a pair of stars-and-bars calculations. There are $\binom{m-1}{k-1}$ ways to divide $m$ zeroes into $k$ blocks. The $k$ blocks of zeroes determine $k+1$ gaps, including the two ends gaps, each of which must contain at least $1$ one. The $N+m+1$ ones can be distributed amongst the $k+1$ gaps in $\binom{N+m}k$ ways so that each gap receives at least $1$ one. Thus, there are $$\binom{m-1}{k-1}\binom{N+m}k$$ ways to distribute the interior ones and zeroes to produce $k$ blocks of zeroes. Each block of zeroes produces a $\langle 1,0\rangle$ pair and a $\langle 0,1\rangle$ pair, so it produces two $b$s. The total degree of the term for any sequence of length $N+2m+1$ is $N+2m$, so the terms $a^{N+2m-2k}b^{2k}$ are precisely those corresponding to sequences with $k$ blocks of zeroes, and the coefficient of $a^{N+2m-2k}b^{2k}$ is therefore $$\binom{m-1}{k-1}\binom{N+m}k\;.$$
## Глосарій Виберіть одне з ключових слів ліворуч ... # Euclidean GeometryEuclid’s Axioms Час читання: ~25 min Before we can write any proofs, we need some common terminology that will make it easier to talk about geometric objects. These are not particularly exciting, but you should already know most of them: A point is a specific location in space. Points describe a position, but have no size or shape themselves. They are labelled using capital letters. In Mathigon, large, solid dots indicate interactive points you can move around, while smaller, outlined dots indicate fixed points which you can’t move. A line is a set of infinitely many points that extend forever in both directions. Lines are always straight and, just like points, they don’t take up any space – they have no width. Lines are labeled using lower-case letters like a or b. We can also refer to them using two points that lie on the line, for example PQ or QP. The order of the points does not matter. A line segment is the part of a line between two points, without extending to infinity. We can label them just like lines, but without arrows on the bar above: AB or BA. Like, before the order of the points does not matter. A ray is something in between a line and a line segment: it only extends to infinity on one side. You can think of it like sunrays: they start at a point (the sun) and then keep going forever. When labelling rays, the arrow shows the direction where it extends to infinity, for example AB. This time, the order of the points does matter. A circle is the collection of points that all have the same distance from a point in the center. This distance is called the radius. ## Congruence These two shapes basically look identical. They have the same size and shape, and we could turn and slide one of them to exactly match up with the other. In geometry, we say that the two shapes are congruent. The symbol for congruence is , so we would say that AB. Here are a few different geometric objects – connect all pairs that are congruent to each other. Remember that more than two shapes might be congruent, and some shapes might not be congruent to any others: Two line segments are congruent if they . Two angles are congruent if they (in degrees). Note the that “congruent” does not mean “equal”. For example, congruent lines and angles don’t have to point in the same direction. Still, congruence has many of the same properties of equality: • Congruence is symmetric: if XY then also YX. • Congruence is reflexive: any shape is congruent to itself. For example, AA. • Congruence is transitive: if XY and YZ then also XZ. ## Parallel and Perpendicular Two straight lines that never intersect are called parallel. They point into the same direction, and the distance between them is always . A good example of parallel lines in real life are railroad tracks. But note that more than two lines can be parallel to each other! In diagrams, we denote parallel lines by adding one or more small arrows. In this example, abc and de. The symbol simply means “is parallel to”. The opposite of parallel is two lines meeting at a 90° angle (right angle). These lines are called perpendicular. In this example, we would write a b. The symbol simply means “is perpendicular to”. ## Euclid’s Axioms Greek mathematicians realised that to write formal proofs, you need some sort of starting point: simple, intuitive statements, that everyone agrees are true. These are called axioms (or postulates). A key part of mathematics is combining different axioms to prove more complex results, using the rules of logic. The Greek mathematician Euclid of Alexandria, who is often called the father of geometry, published the five axioms of geometry: Euclid of Alexandria First Axiom You can join any two points using exactly one straight line segment. Second Axiom You can extend any line segment to an infinitely long line. Third Axiom Given a point P and a distance r, you can draw a circle with centre P and radius r. Fourth Axiom Any two right angles are congruent. Fifth Axiom Given a line L and a point P not on L, there is exactly one line through P that is parallel to L. Each of these axioms looks pretty obvious and self-evident, but together they form the foundation of geometry, and can be used to deduce almost everything else. According to none less than Isaac Newton, “it’s the glory of geometry that from so few principles it can accomplish so much”. Euclid published the five axioms in a book “Elements”. It is the first example in history of a systematic approach to mathematics, and was used as mathematics textbook for thousands of years. One of the people who studied Euclid’s work was the American President Thomas Jefferson. When writing the Declaration of Independence in 1776, he wanted to follow a similar approach. He begins by stating a few, simple “axioms” and then “proves” more complex results: “We hold these truths to be self-evident: that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness.” We, therefore … declare, that these United Colonies are, and of right ought to be, free and independent states.” This is just one example where Euclid’s ideas in mathematics have inspired completely different subjects. Archie
Students of Grade 3 Module 3 can get a strong foundation on mathematics concepts by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 3, students need to solve all questions and exercises included in Eureka’s Grade 3 Textbook. ## Engage NY Eureka Math 3rd Grade Module 3 Lesson 4 Answer Key So teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that. ### Eureka Math Grade 3 Module 3 Lesson 4 Problem Set Answer Key Question 1. The missing numbers are 12, 24, 42, and 54. Explanation: In the above-given question, given that, the numbers are 6, 18, 30, 36, 48, and 60. the missing numbers are 12, 24, 42, and 54. Question 2. Count by six to fill in the blanks below. 6, __12_____, ___18____, ___24____ The missing numbers are 12, 18, and 24. Explanation: In the above-given question, given that, the table of 6. count by six. so the missing numbers are 12, 18, and 24. Complete the multiplication equation that represents the final number in your count-by. 6 × ___4____ = ___24____ 6 x 4 = 24. Explanation: In the above-given question, given that, the table of 6 x 4 = 24. so the missing number is 6 x 4 = 24. Complete the division equation that represents your count-by. ____24___ ÷ 6 = ___4____ 24 / 6 = 4. Explanation: In the above-given question, given that, 24 / 6 = 4. 6 x 4 = 24. Question 3. Count by six to fill in the blanks below. 6 , ___12___, ___18___, ___24___, ___30___, __36___, __42___ The missing numbers are 12, 18, 24, 30, 36, and 42. Explanation: In the above-given question, given that, the numbers are 6, 12, 18, 24, 30, 36, and 42. so the missing numbers are 12, 18, 24, 30, 36, and 42. Complete the multiplication equation that represents the final number in your count-by. 6 × ___7____ = ___42____ 6 x 7 = 42. Explanation: In the above-given question, given that, the numbers are 6 x 7 = 42. 6 x 7 = 42. Complete the division equation that represents your count-by. ___42____ ÷ 6 = ___7____ 42 / 6 = 7. Explanation: In the above-given question, given that, 42 / 6 = 7. 6 x 7 = 42. Question 4. Mrs. Byrne’s class skip-counts by six for a group counting activity. When she points up, they count up by six, and when she points down, they count down by six. The arrows show when she changes direction. a. Fill in the blanks below to show the group counting answers. ↑ 0, 6, __12___, 18, __24___↓ ___18__, 12 ↑ __18___, 24, 30, __36___ ↓ 30, 24, ___18__ ↑ 24, __30___, 36, __42___, 48 The missing numbers are 12, 24, 18, 18, 36, 18, 30, and 42. Explanation: In the above-given question, given that, The numbers are 0, 6, 12, 18, 24, 18, 12, 18, 24, 30, 36, 30, 24, 18, 24, 30, 36, 42, and 48. the missing numbers are 12, 24, 18, 36, 18, 30, and 42. b. Mrs. Byrne says the last number that the class counts is the product of 6 and another number. Write a multiplication sentence and a division sentence to show she’s right. 6 × ____8___ = 48 48 ÷ 6 = ___8____ 6 x 8 = 48. 48 / 6 = 8. Explanation: In the above-given question, given that, Mrs. Byme says the last number that the class counts is the product of 6 and another number. 6 x 8 = 48. 48 / 6 = 8. Question 5. Julie counts by six to solve 6 × 7. She says the answer is 36. Is she right? Explain your answer. No, she was not correct. 6 x 7 = 42. Explanation: In the above-given question, given that, Julie counts by six to solve 6 x 7. she says the answer is 36. no she was not correct. 6 x 7 = 42. ### Eureka Math Grade 3 Module 3 Lesson 4 Exit Ticket Answer Key Question 1. Sylvia solves 6 × 9 by adding 48 + 6. Show how Sylvia breaks apart and bonds her numbers to complete the ten. Then, solve. 6 x 9 = 54. 48 + 6 = 54. Explanation: In the above-given question, given that, Sylvia solves 6 x 9 by adding 48 + 6. 10 + 10 + 10 + 10 + 10 + 4 = 54. 6 x 9 = 54. 48 + 6 = 54. Question 2. Skip-count by six to solve the following: a. 8 × 6 = ___48___ 8 x 6 = 48. Explanation: In the above-given question, given that, 6 x 8 = 48. 8 x 6 = 48. b. 54 ÷ 6 = ____9__ 54 / 6 = 9. Explanation: In the above-given question, given that, 6 x 9 = 54. 54 / 6 = 9. ### Eureka Math Grade 3 Module 3 Lesson 4 Homework Answer Key Question 1. Use number bonds to help you skip-count by six by either making a ten or adding to the ones. 10 + 8 = 18, 30 + 6 = 36, 10 + 20 = 30, 30 + 12 = 42, 40 + 8 = 48, 40 + 14 = 54, 50 + 10 = 60. Explanation: In the above-given question, given that, the numbers are 10 + 8 = 18. 30 + 6 = 36. 10 + 20 = 30. 30 + 12 = 42. 40 + 8 = 48. 40 + 14 = 54. 50 + 10 = 60. Question 2. Count by six to fill in the blanks below. 6, ___12____, ___18____, ___24____, ___30____ The missing numbers are 6, 12, 18, 24, and 30. Explanation: In the above-given question, given that, the tables of 6. 6, 12, 18, 24, and 30. so the missing numbers are 6, 12, 18, 24, and 30. Complete the multiplication equation that represents the final number in your count-by. 6 × ___5____ = ___30____ 6 x 5 = 30. Explanation: In the above-given question, given that, the tables of 6. 6 x 5 = 30. Complete the division equation that represents your count-by. __30_____ ÷ 6 = ___5____ 30 / 6 = 5. Explanation: In the above-given question, given that, 6 x 5 = 30. 30 / 6 = 5. Question 3. Count by six to fill in the blanks below. 6, __12____, __18____, ___24___, __30____, __36____ The missing numbers are 12, 18, 24, 30, and 36. Explanation: In the above-given question, given that, the numbers are 6, 12, 18, 24, 30, and 36. so the missing numbers are 12, 18, 24, 30, and 36. Complete the multiplication equation that represents the final number in your count-by. 6 × ___6____ = ___36____ 6 x 6 = 36. Explanation: In the above-given question, given that, the table of 6. 6 x 6 = 36. Complete the division equation that represents your count-by. ____36___ ÷ 6 = ___6____ 36 / 6 = 6. Explanation: In the above-given question, given that, 6 x 6 = 36. 36 / 6 = 6. Question 4. Count by six to solve 48 ÷ 6. Show your work below.
# If One Zero of Polynomial 3x2-8x+2k+1 in Seven Times other Find the Zero and the Value of the k By Mohit Uniyal\|Updated : May 16th, 2023 If one zero of polynomial 3x2-8x+2k+1 in seven times other find the zero and the value of the k To find the zero and the value of k in the polynomial 3x^2 - 8x + 2k + 1, given that one zero is seven times the other, follow these steps: Step 1: Assume the two zeros of the polynomial are α and β, with α = 7β. Step 2: Use the fact that the sum of the zeros of a quadratic polynomial is given by the formula α + β = -b/a, where a and b are the coefficients of x^2 and x, respectively. Step 3: Substitute α = 7β into the equation α + β = 8/3 Step 4: Substitute the value of β back into α = 7β Step 5: To find the value of k, substitute one of the zeros (let's choose α) into the polynomial equation and solve for k. ## If One Zero of Polynomial 3x2-8x+2k+1 in Seven Times other Find the Zero and the Value of the k Solution Let's assume that one zero of the polynomial 3x2 - 8x + 2k + 1 is seven times the other zero. Let the two zeros be α and β, with α = 7β. We know that the sum of the zeros of a quadratic polynomial is given by the formula: α + β = -b/a In this case, a = 3 and b = -8, so we have: α + β = -(-8)/3 = 8/3 Since α = 7β, we can substitute this into the equation: 7β + β = 8/3 Combining like terms: 8β = 8/3 Dividing both sides by 8: β = 1/3 Substituting this value back into α = 7β: α = 7(1/3) = 7/3 So the zeros of the polynomial are α = 7/3 and β = 1/3. To find the value of k, we can substitute one of the zeros into the polynomial equation and solve for k. Let's substitute α = 7/3 into the polynomial: 3(7/3)2 - 8(7/3) + 2k + 1 = 0 Simplifying: 49/3 - 56/3 + 2k + 1 = 0 -7/3 + 2k + 1 = 0 2k - 4/3 = 0 2k = 4/3 k = 2/3 Therefore, the zero of the polynomial is α = 7/3, the other zero is β = 1/3, and the value of k is 2/3. ## If one zero of polynomial 3x2-8x+2k+1 in seven times other, then the zero is α = 7/3, the other zero is β = 1/3, and the value of k is 2/3 Similar Questions:
# The North American Arithmetic: Uniting Oral and Written Exercises, in Corresponding Chapters. Part second Russell, Odiorne & Company, 1833 - 191 σελίδες ### Τι λένε οι χρήστες -Σύνταξη κριτικής Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες. ### Περιεχόμενα Ενότητα 1 1 Ενότητα 2 2 Ενότητα 3 5 Ενότητα 4 6 Ενότητα 5 7 Ενότητα 6 8 Ενότητα 7 9 Ενότητα 8 10 Ενότητα 9 88 Ενότητα 10 89 Ενότητα 11 91 Ενότητα 12 93 Ενότητα 13 109 Ενότητα 14 116 Ενότητα 15 191 Ενότητα 16 ### Δημοφιλή αποσπάσματα Σελίδα 171 - Multiply all the numerators together for a new numerator, and all the denominators for a new denominator: then reduce the new fraction to its lowest terms. 1. Reduce f of 5 to a simple fraction. Σελίδα 157 - To reduce a mixed number to an improper fraction. Multiply the whole number by the denominator of the fraction, and to the product add the given numerator. Σελίδα 101 - Observe, that the number which we multiply is called the multiplicand; the number by which we multiply is called the multiplier; and the number which we obtain by multiplication is called the product. Σελίδα 131 - MEASURE is used by grocers and others, for measuring wine, oil, molasses, and most other liquids. 4 gills (gi.) make 1 pint. pt. 2 pints make 1 quart. qt. 4 quarts make 1 gallon. gal. 31^ gallons make 1 barrel. bl. 42 gallons make 1 tierce. tier. 63 gallons make 1 hogshead. hhd. Σελίδα 139 - Place the remainder under the column added, and carry the quotient to the next column. Σελίδα 170 - Divide the greater number by the less, and that divisor by the remainder, and so on ; always dividing the last divisor by the last remainder, till nothing remains ; the last divisor is the greatest common divisor required. Σελίδα 108 - ... the 6 units; thus, 3 in 3, once; 3 in 9, 3 times; 3 in 6, 2 times. Observe in the above example, that the 3 which we first divide, means 3 hundred; and the 1 which we place under it means 1 hundred, showing that 3 is contained in 300, 100 times. The 9 means 9 tens, and the 3 which we place under it means 3 tens, showing, that 3 is contained in 90, 30 times. A Dividend is a number which is to be divided; such is the number 396 in the above example. Σελίδα 184 - RULE. Multiply as in whole numbers, and from the right hand of the product point off as many figures for decimals as there are decimal places in both factors. Σελίδα 111 - This 2 is a re 1Q 2 mainder; it shows that there are 2 hats, • which cannot be divided into eights. 55. How many sheep, at 4 dollars a head, can a butcher, who has 747 dollars buy; and how many dollars will he have remaining ? 56. If 5 yards of cloth will make a suit of clothes, how many suits can be made from 96 yards; and how many yards will there be over ? 57. How many times is 6 contained in 4637; and how many are there over ? 58. Σελίδα 176 - Either ,multiply the numerator, or divide the denominator. To divide a fraction by a whole number, — Either divide the numerator, or multiply the denominator. When a number is multiplied by 1 , the product is equal to the multiplicand. Therefore, when a number is multiplied by a fraction, which is less than 1, the product must be less than the multiplicand. To multiply a whole number by a fraction, — Multiply by the numerator, and divide by the denominator.
# What is the standard form of the equation of a circle with centre is at point (5,8) and which passes through the point (2,5)? Jan 5, 2016 ${\left(x - 5\right)}^{2} + {\left(y - 8\right)}^{2} = 18$ #### Explanation: standard form of a circle is ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ where (a , b ) is the centre of the circle and r = radius. in this question the centre is known but r is not. To find r , however , the distance from the centre to the point ( 2 , 5 ) is the radius . Using the distance formula will allow us to find in fact ${r}^{2}$ ${r}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$ now using (2 , 5 ) = $\left({x}_{2} , {y}_{2}\right) \mathmr{and} \left(5 , 8\right) = \left({x}_{1} , {y}_{1}\right)$ then ${\left(5 - 2\right)}^{2} + {\left(8 - 5\right)}^{2} = {3}^{2} + {3}^{2} = 9 + 9 = 18$ equation of circle : ${\left(x - 5\right)}^{2} + {\left(y - 8\right)}^{2} = 18$ . Jan 5, 2016 I found: ${x}^{2} + {y}^{2} - 10 x - 16 y + 71 = 0$ #### Explanation: The distance $d$ between the centre and the given point will be the radius $r$. We can evaluate it using: $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$ So: $r = d = \sqrt{{\left(2 - 5\right)}^{2} + {\left(5 - 8\right)}^{2}} = \sqrt{9 + 9} = 3 \sqrt{2}$ Now you can use the general form of the equation of a circle with centre at $\left(h , k\right)$ and radius $r$: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$ And: ${\left(x - 5\right)}^{2} + {\left(y - 8\right)}^{2} = {\left(3 \sqrt{2}\right)}^{2}$ ${x}^{2} - 10 x + 25 + {y}^{2} - 16 y + 64 = 18$ ${x}^{2} + {y}^{2} - 10 x - 16 y + 71 = 0$
# 6. Maths Trigonometry Trigonometry is based on triangles. The point of trigonometry is to find either the unknown side or angle of a triangle, when we know a side and an angle. Here is a little diagram to get you started (pretty isnt it). Now: O stands for opposite. H stands for hypotenuse. alpha is the angle in the corner where the angle is either known, or needed to be found. Ok, now the 3 trigonomic functions are: Sine, Cosine and Tangent These will be known as: sin, cos and tan on your calculator. And respectively they each have their own inverse function known as: sin^-1, cos^-1 and tan^-1 Here I am going to let alpha be @ as I dont have any alpha key to press 😛 so to remind you alpha is our angle, so @ will be known now as our angle. Now the sin@ = o/h cos@ = a/h tan@ = o/a You can easily remember this by a little saying: soh cah toa So why do we need this? We need this when we want to find the length of a side in a triangle. If you have the length A and the angle @, then you can use trigonometry to find either H or O. And likewise for the lengths H and O. Next we can use trigonometry to find the unknown angle. sin@ = o/h Then we have to get sin on the other side, so we times both sides by the opposite of sine which is sin^-1, which should also be on your calculator. so by doing this we end up with: @ = sin^-1(o/h) And that is all there is to trigonometry. You can find the sides and angles of a triangle using this. But how would we use this in a game you might say, would people still prefer vieraile virallisella verkkosivustolla? Well take a little look at my next diagram (god I love mspaint). In the image above it shows you a rough idea of how you would make this into a camera for a game. You have yourself in the centre and have 360 degrees of rotation around. You can use trigonometry, and with your current angle, and knowing your current position to A(where you wish to go) you can determine your point at the corner of H and O. But that is just a quick reference I whipped up in 10 minutes. In the Radian tutorial you will find out just how the camera class relates to trigonometry in detail. If you have any questions, please email me at swiftless@gmail.com Be Sociable, Share!
Do you ever find yourself in a situation where you need to add two really large numbers but don’t have access to an easy-to-use calculator? You will always be able to effortlessly add two numbers together with this addition calculator, whether they are positive or negative, large or tiny. We describe how our addition calculator works in the article below and a part where you may learn about the addition definition. Take a look other related calculators, such as: Commutativity and associativity are two characteristics of addition. Commutativity indicates that you may swap the positions of the addends without altering the equation’s answer, A + B = B + A. When we try to add more than two integers, though, associativity becomes a phenomenon. It makes no difference whether you add the first number to the second and then the third, or the third to the second and then the first. The outcome will remain the same: A + (B + C) = (A + B) + C. The addition is one of the four fundamental arithmetic operations, along with subtraction, multiplication, and division. ## Summation calculator – how does it add numbers? It’s simple and straightforward. Fill up the first addend field (A) as well as the second addend field (B) (B). Everything else is handled by the calculator, which provides you with the total (C). A + B = C Things can become a little tricky if you have a binary or a fraction, for example. However, there is a solution if it is a fraction. You may either convert it to decimal or use our fractions adding calculator. The segment addition postulate, which entails calculating a segment length when three collinear points, is an extended use of the summation principle. So, let’s practice it with an example. Imagine that your addends are A = 65423 and the other B = 2667599. 65423 + 2667599 = 2,733022 What happens if B is negative? 65423 + -2667599= – 2602176
# How do you solve (x-1)(x-2)(x-3)(x-4) = 24? Apr 19, 2016 $x = \left\{\begin{matrix}0 \\ 5 \\ \frac{5}{2} \pm \frac{\sqrt{15}}{2} i\end{matrix}\right.$ #### Explanation: Let $f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right)$ Then: f(0) = (-1)(-2)(-3)(-4) = 4! = 24 f(5) = (5-1)(5-2)(5-3)(5-4) = 432*1 = 4! = 24 So both $x = 0$ and $x = 5$ are roots and $x$ and $\left(x - 5\right)$ are factors. $f \left(x\right) - 24$ $= \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) - 24$ $= {x}^{4} - 10 {x}^{3} + 35 {x}^{2} - 50 x$ $= x \left(x - 5\right) \left({x}^{2} - 5 x + 10\right)$ The remaining quadratic factor is in the form $a {x}^{2} + b x + c$, with $a = 1$, $b = - 5$ and $c = 10$. This has zeros given by the quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ $= \frac{5 \pm \sqrt{{5}^{2} - \left(4 \cdot 1 \cdot 10\right)}}{2}$ $= \frac{5 \pm \sqrt{- 15}}{2}$ $= \frac{5}{2} \pm \frac{\sqrt{15}}{2} i$
# Factors of 325 Are you looking for all the factors of 325? If yes, you are in the right page to get answers factors of 325. Here in this page we have given brief discussion behind this article, so that students will get complete idea to find the answers. ## Factors of 325: Topic Factors Number 325 Roman Numeral CCCXXV Number come just after 324 Number come just before 326 Category Composite Number ### Multiplication Table of 325 325 x 1 325 325 x 2 650 325 x 3 975 325 x 4 1300 325 x 5 1625 325 x 6 1950 325 x 7 2275 325 x 8 2600 325 x 9 2925 325 x 10 3250 ### Numbers that divide 325 leaving Reminder 0 325 ÷ 1 Quotient = 325, Reminder = 0 325 ÷ 5 Quotient = 65, Reminder = 0 325 ÷ 13 Quotient = 25, Reminder = 0 325 ÷ 25 Quotient = 13, Reminder = 0 325 ÷ 65 Quotient = 5, Reminder = 0 325 ÷ 325 Quotient = 1, Reminder = 0 Factors are the numbers that divide the respective number and give reminder 0. Hence, all the factors of 325 are 1, 5, 13, 25, 65 & 325. • Positive factors: 1, 5, 13, 25, 65 and 325 • Negative factors: -1, 5, 13, 25, 65 and 325 ### Worksheet Question 1: What is the factor of 325 in below Option 1: 0 Option 2: 3 Option 3: 5 Option 4: 8 Question 2: Is 325 is a Prime Number Option 1: Yes Option 2: No Option 3: Not Sure Option 4: Neither Prime nor Composite Question 3: How many Factors of 325 Option 1: Three Option 2: Six Option 3: Two Option 4: None of these
# RS Aggarwal Solutions Class 9 Maths Chapter 19 Probability (Updated For 2021-22) RS Aggarwal Solutions Class 9 Maths Chapter 19 Probability: Your Class 9 Maths exam preparation is incomplete by studying the RS Aggarwal Solutions Class 9 Maths. You can always rely on the RS Aggarwal Solutions Class 9 Maths Chapter 19 Probability for clearing your doubts and subject matter experts have designed well-designed solutions for you. ## Download RS Aggarwal Solutions Class 9 Maths Chapter 19 Probability PDF RS Aggarwal Solutions Class 9 Maths Chapter 19 Probability ## RS Aggarwal Solutions Class 9 Maths Chapter 19 Probability – Overview In this chapter you will deal with the concept of probabiltiy. The Probability of an event is a number somewhere in the range of 0 and 1, where, 0 demonstrates the event’s failure, and 1 shows certainty. Here are the key points covered in RS Aggarwal Solutions Class 9 Maths Chapter 19 Probability. • Experiment: Any technique that can be vastly repeated and has a very much characterized set of potential outcomes and also known as sample space. An experiment is supposed to be random on the off chance that it has more than one potential result and deterministic on the off chance that it has just one. • Trail: A trail is a specific performance of a random experiment. All possible trails that establish an all-around characterized set of possible results are by and large called an experiment or sample space. • Experimental or Empirical Probability: It is the probability of an event dependent on the consequences of a real experiment led a few times. It is represented as Probability of event P(E)=Number of possible outcomes/Total number of outcomes. • Coin Tossing Experiment: Here the mutually independent results are the coin arrival either heads up or tail up. If the probability is one, we state that this result consistently happens with complete certainty. The probability of getting head P(H) =Number of results to get head n(H)/Number of possible outcomes n(S)= ½ Similarly, the probability of getting a tail P(T) = ½ • Rolling Of Dice: When a fair dice is rolled, the number that surfaces top is a number between one to six. The probability of getting two P(E)= Number of results to get two n(E)/Number of possible results n(S)=1/6 The result is the same for all the remaining numbers. This is the complete blog on the RS Aggarwal Solutions Class 9 Maths Chapter 19. To know more about the CBSE Class 9 Maths exam, ask in the comments. ## FAQs on RS Aggarwal Solutions Class 9 Maths for Chapter 19 Probability ### From where can I find the download link for the RS Aggarwal Solutions Class 9 Maths Chapter 19 PDF? You can find the download link of RS Aggarwal Solutions Class 9 Maths Chapter 19 in the above blog. ### Can I access the RS Aggarwal Solutions Class 9 Maths Chapter 19 PDF Offline? Once you have downloaded the RS Aggarwal Solutions Class 9 Maths Chapter 19 PDF online, you can access it offline whenever you want. ### Is the RS Aggarwal Solutions Class 9 Maths Chapter 19 a credible source for Class 9 Maths exam preparation? Yes, the solutions of RS Aggarwal Solutions Class 9 Maths Chapter 19 are prepared by the subject matter experts, hence credible. ### What is an Experimental or Empirical Probability? It is the probability of an event dependent on the consequences of a real experiment led a few times. BITSAT 2023 Registration Begins – Check Direct Link UKPSC Patwari Lekhpal admit card 2023 – Check Direct Link ICAI CA Foundation result December 2022- Check Direct Link 11 Most In-Demand Programming Languages in 2023 Top 8 Programming Languages That Will Get You Hired
# 3.2 Solving Systems of Equations Algebraically Substitution Method Elimination Method. ## Presentation on theme: "3.2 Solving Systems of Equations Algebraically Substitution Method Elimination Method."— Presentation transcript: 3.2 Solving Systems of Equations Algebraically Substitution Method Elimination Method Substitution Method Here you replace one variable with an expression. x + 4y = 26 x – 5y = - 10 Solve for a variable, x = 26 – 4y Replace “x” in the other equation (26 – 4y) – 5y = -10 Solve for y (26 – 4y) – 5y = -10 26 – 4y – 5y = - 10Remove parentheses by multiplying by 1 26 – 9y = - 10Add like terms -9y = - 36Subtract 26 from both sides y = 4Divide by - 9 Solve for x x = 26 – 4y x = 26 – 4(4)Substitute for y x = 26 – 16 x = 10 The order pair is (10, 4). This is where the lines cross. The Elimination Method Here we add the equations together when the coefficients are different signs. x + 2y = 10 x + y = 6 Here both lead coefficients are 1. We can change the coefficient to – 1, by multiplying by – 1. x + 2y = 10 x + y = 6 Multiply the bottom equation by – 1. x + 2y = 10 - x - y = - 6When adding the y = 4equations together, x go to zero. Find x by replace it back in either equation. x + 2(4) = 10;x + 8 = 10;x = 2 So the order pair (2, 4) works in both equations. 2 + 2(4) = 10 2 + 4 = 6 We have two way to solve the systems, Substitution and Elimination; which way is better depends on the problem. What about this problem 2x + 3y = 12 5x – 2y = 11 Here we have to multiply both equations If we wanted to remove the “x”, then we have to find the Least common multiple (L.C.M.) of 2 and 5. If we wanted to remove the “y”, then we have to find the least common multiple of 3 and -2. Lets get rid of the “y” The L.C.M of 2 and 3 is 6. Since we want the coefficients to be opposite, - 2 will help in the equation. we multiply the top equation by 2. 2x + 3y = 12 4x + 6y = 24 The bottom equation by 3 5x – 2y = 1115x – 6y = 33 Add the new equations together 4x + 6y = 24 15x – 6y = 33 19x = 57 Divide by 19 x = 3 Replace in original equation and solve for y 2(3) + 3y = 126 + 3y = 123y = 6 y= 2 What about inconsistent systems? y – x = 5 Multiply the top equation by – 2, 2y – 2x = 8 2y – 2x = -10 then add the bottom.2y – 2x = 8 0 = - 2 This shows no solutions. What if it is dependent (Many solutions) 1.6y = 0.4x + 1 0.4y = 0.1x + 0.25 Multiply the top and bottom equation by 100 to remove decimals. 160y = 40x + 100 40y = 10x + 25 Then multiply the bottom equation by -4 -160y = -40x – 100 Add the new equations together 160y = 40x + 100 -160y = -40x – 100 0 = 0 This is a system with many solutions. Solve this system a – b = 2 -2a + 3b = 3
# The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find its common difference. Given: The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420. To do: We have to find its common difference. Solution: Let $a$ be the first term, $d$ be the common difference and $n$ be the number of terms. First term $a=7$ Last term $l=49$ Sum of the A.P. $S_{n}=420$ We know that, $l=a_{n}=a+(n-1)d$ $\Rightarrow 49=7+(n-1)d$ $\Rightarrow ( n-1) d=49-7$ $\Rightarrow (n-1)d=42$........(i) Sum of $n$ terms of an A.P. $S_{n} =\frac{n}{2}[ 2a+( n-1) d]$ $\Rightarrow 420=\frac{n}{2}[2(7)+42]$ (From (i)) $\Rightarrow 420=n(7+21)$ $\Rightarrow n=\frac{420}{28}$ $n=15$ This implies, $(15-1)d=42$ $\Rightarrow 14d=42$ $\Rightarrow d=\frac{42}{14}$ $\Rightarrow d=3$ Therefore, the common difference of the given A.P. is $3$. Tutorialspoint Simply Easy Learning Updated on: 10-Oct-2022 42 Views ##### Kickstart Your Career Get certified by completing the course
Question Video: Simplifying Expansions Using the Binomial Theorem \| Nagwa Question Video: Simplifying Expansions Using the Binomial Theorem \| Nagwa # Question Video: Simplifying Expansions Using the Binomial Theorem Mathematics • Third Year of Secondary School ## Join Nagwa Classes Simplify ππΆ0 + 17 Γ ππΆ1 + 17Β² Γ ππΆ2+ ... + 17^(π) Γ ππΆπ + ... + 17 ^(π) Γ ππΆπ. 02:32 ### Video Transcript Simplify π πΆ zero plus 17 multiplied by π πΆ one plus 17 squared multiplied by π πΆ two and so on plus 17 to the power of π multiplied by ππΆπ and so on up to 17 to the power of π multiplied by ππΆπ. In order to simplify this expression, we begin by recalling the binomial expansion of π plus π to the πth power. The first three terms of this expansion are π πΆ zero multiplied by π to the power of π plus π πΆ one multiplied by π to the power of π minus one multiplied by π plus π πΆ two multiplied by π to the power of π minus two multiplied by π squared. And the final term is ππΆπ multiplied by π to the πth power. We notice that much of this is the same as the expression in this question. Instead of π to the πth power, our last term contains 17 to the πth power. The second term contains 17 instead of π and the third term, 17 squared instead of π squared. This means that the value of π is 17. We notice that there is no π part to any of our terms. Since one raised to any power is equal to one, we can assume that π is equal to one as this is the only value for which this holds. The expression in the question is therefore equal to one plus 17 all raised to the πth power. And as one plus 17 equals 18, this is equal to 18 to the πth power. The expression π πΆ zero plus 17 multiplied by π πΆ one plus 17 squared multiplied by π πΆ two and so on up to 17 to the πth power multiplied by ππΆπ is equal to 18 to the πth power. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
Visual Explanations in Mathematics ## 2. Casselman's visual proof The incommensurability of side and diagonal in a regular pentagon In this regular pentagon, call the side length s and the length of the diagonal d. Our job is to show that s and d are incommensurable. First notice that in a regular pentagon, a diagonal is parallel to the opposite side. This follows from the symmetry of the pentagon. Using this fact for two adjacent sides shows that this green quadrilateral is a parallelogram, with opposite sides equal: all four sides of this parallelogram have length s. Now suppose that the side and the diagonal are commensurable: there exists a length h which fits exactly a whole number of times into the diagonal and into the side. The length h is symbolized by the distance between two adjacent dots in this picture. The length of the highlighted segment must also be a whole number of h's, since it is the difference of two such numbers, d and s. Call this length d. d = d - s. The length of this highlighted segment is d - 2d. So this length must also be equal to a whole number of h's. Call this length s. s* = d - 2d* = 2s - d. Now notice that the five diagonals determine a smaller regular pentagon inside our original one. The side length of this pentagon is the s* we just considered. In the smaller pentagon, the diagonals are again parallel to the opposite sides. It follows that the highlighted figure is a parallelogram, and that the diagonal in the smaller pentagon has length d. So in the smaller pentagon, side and diagonal are both multiples of our original h. We can repeat the construction with the smaller pentagon. Since s is strictly shorter than s, and they are both whole multiples of h, they must differ by at least h. We can repeat the process again and again, and obtain an infinite decreasing sequence of side lengths s > s* > s** ... each of which is at least h smaller than the one before. But this can't work, because there was only a finite number of h's in s to start! This contradiction shows that the hypothesis, that s and d are commensurable, must be false.
Hovedinnhold # Middelværdisætningen ## Video transkripsjon Let's see if we can give ourselves an intuitive understanding of the mean value theorem. And as we'll see, once you parse some of the mathematical lingo and notation, it's actually a quite intuitive theorem. And so let's just think about some function, f. So let's say I have some function f. And we know a few things about this function. We know that it is continuous over the closed interval between x equals a and x is equal to b. And so when we put these brackets here, that just means closed interval. So when I put a bracket here, that means we're including the point a. And if I put the bracket on the right hand side instead of a parentheses, that means that we are including the point b. And continuous just means we don't have any gaps or jumps in the function over this closed interval. Now, let's also assume that it's differentiable over the open interval between a and b. So now we're saying, well, it's OK if it's not differentiable right at a, or if it's not differentiable right at b. And differentiable just means that there's a defined derivative, that you can actually take the derivative at those points. So it's differentiable over the open interval between a and b. So those are the constraints we're going to put on ourselves for the mean value theorem. And so let's just try to visualize this thing. So this is my function, that's the y-axis. And then this right over here is the x-axis. And I'm going to-- let's see, x-axis, and let me draw my interval. So that's a, and then this is b right over here. And so let's say our function looks something like this. Draw an arbitrary function right over here, let's say my function looks something like that. So at this point right over here, the x value is a, and the y value is f(a). At this point right over here, the x value is b, and the y value, of course, is f(b). So all the mean value theorem tells us is if we take the average rate of change over the interval, that at some point the instantaneous rate of change, at least at some point in this open interval, the instantaneous change is going to be the same as the average change. Now what does that mean, visually? So let's calculate the average change. The average change between point a and point b, well, that's going to be the slope of the secant line. So that's-- so this is the secant line. So think about its slope. All the mean value theorem tells us is that at some point in this interval, the instant slope of the tangent line is going to be the same as the slope of the secant line. And we can see, just visually, it looks like right over here, the slope of the tangent line is it looks like the same as the slope of the secant line. It also looks like the case right over here. The slope of the tangent line is equal to the slope of the secant line. And it makes intuitive sense. At some point, your instantaneous slope is going to be the same as the average slope. Now how would we write that mathematically? Well, let's calculate the average slope over this interval. Well, the average slope over this interval, or the average change, the slope of the secant line, is going to be our change in y-- our change in y right over here-- over our change in x. Well, what is our change in y? Our change in y is f(b) minus f(a), and that's going to be over our change in x. Over b minus b minus a. I'll do that in that red color. So let's just remind ourselves what's going on here. So this right over here, this is the graph of y is equal to f(x). We're saying that the slope of the secant line, or our average rate of change over the interval from a to b, is our change in y-- that the Greek letter delta is just shorthand for change in y-- over our change in x. Which, of course, is equal to this. And the mean value theorem tells us that there exists-- so if we know these two things about the function, then there exists some x value in between a and b. So in the open interval between a and b, there exists some c. There exists some c, and we could say it's a member of the open interval between a and b. Or we could say some c such that a is less than c, which is less than b. So some c in this interval. So some c in between it where the instantaneous rate of change at that x value is the same as the average rate of change. So there exists some c in this open interval where the average rate of change is equal to the instantaneous rate of change at that point. That's all it's saying. And as we saw this diagram right over here, this could be our c. Or this could be our c as well. So nothing really-- it looks, you would say f is continuous over a, b, differentiable over-- f is continuous over the closed interval, differentiable over the open interval, and you see all this notation. You're like, what is that telling us? All it's saying is at some point in the interval, the instantaneous rate of change is going to be the same as the average rate of change over the whole interval. In the next video, we'll try to give you a kind of a real life example about when that make sense.
## Step by step tutorial: How to draw a perfect square When we construct our drawings we often find ourselves having to draw primitives. Let’s explore how to draw a perfect square with just a few easy steps! # How to draw a perfect square A square is nothing more than 4 lines of equal length touching each other’s beginning- and endpoint at a 90-degree angle. If you want a perfect (aka mathematically perfect) square you need tools like a ruler or a geometry triangle. Since you’re probably on this page to learn how to draw or sketch a square freehandedly we’ll focus on how to do that instead. ## Step 1: Bounding box Let’s start with drawing a bounding box. This is just a rough estimation of our square. It’s only used to define the size. ## Step 2: Deciding our length Let’s draw the bottom line of our square. Since we know the size of our box, we can easily decide on a fitting length for our line. ## Step 3: Measure Next, we need to measure how long this line is because we have to repeat it 3 more times. We can either use: • A ruler • Our pencil • Eyeball it If you need to be super precise you’ll need a ruler. If it’s a sketch you can use your pencil to measure the length and then bring your pencil over to both sides of our line and mark it. The new lines have to sit at a 90-degree angle (be perpendicular to our first drawn line). Here’s a handy tip: Use your paper as a guide for drawing straight lines. Match your lines with the sides of your paper so they’re parallel. This helps you get a straight line. Tip 2: Don’t move your wrist when drawing a line. Use your whole arm instead. This also prevents you from having a curved line. ## step 4: Connect & finish! Connect the last remaining line. Done! Okay, this probably all makes sense to you and you probably wonder why you even need a step by step tutorial on how to draw a perfect square. It’s always a good idea to read up on how to do certain things, even if you’ve done it 100 times already. You might find a little more wisdom or a few handy tricks that makes everything logical and easier. You’ll never know when this tutorial might come in handy, like when drawing a perfect circle. Sounds weird, doesn’t it? :’) Good on you for freshen up how to draw the basics! Keep doing this and you’ll go far. Even with the simplest of things. See you next tutorial, ♡ Laura P.s You can buy a right angle ruler to make life easy
Jun 18, 2023 I. Introduction Calculating percent increase is crucial in analyzing and interpreting statistical and financial data. This process assesses the shifting trend and helps to predict the future trend. Accurate knowledge of percent increase is essential for budgeting, comparing financial results, and identifying the growth rate of a company. In this comprehensive guide, we will cover the various formulas and methods used for calculating percent increase and offer practical tips and tricks for achieving accurate results quickly and efficiently. II. Mastering Mathematics: A Guide to Easily Calculating Percent Increase Percent increase simply means how much a quantity has increased in relation to the original value. It is expressed as a proportion of the original value represented in a percentage. Often, it is used to compare current data with the previous data set for analysis. A few examples of the concept include calculating tax increase, salary increase, population growth rate, and inflation rate. Percentages help us analyze the trend and summarize the data in an easy-to-understand format. For instance, imagine that last year the value of a house was \$500,000, and this year the house’s value is \$550,000. The observed change is an increase in value, and to determine the percentage increase, we follow these steps. We take the difference between the two values, i.e., 550,000 – 500,000 = 50,000. We divide the difference by the original value and multiply the result by 100. This would be (50,000/500,000) 100, which results in 10%. Therefore, the percentage increase in the value of the house is 10%, which is useful knowledge for the owner. It is important to keep in mind that percentage increase and percentage decrease are not the same thing. The former occurs when an item has grown in value, while the latter occurs when an item has lost value. III. The Quick and Simple Method for Percentage Increase Calculations Calculating percent increase can be achieved in a straightforward manner by using the following formula: (Difference/Original value) x 100 To apply this formula, first determine the observed difference in the values, and divide this difference by the original value. The resulting decimal is multiplied by 100 to get the percentage increase. Let’s consider a real-life example to understand this calculation: Suppose that the price of a laptop was \$1,500 a year ago, and the current price of a laptop is \$1,800. To calculate the percentage increase, we perform the following calculation: (1800-1500)/1500) 100 = 20% Therefore, the percent increase in the price of a laptop is 20%. To better understand how we arrived at this figure let’s break down the calculation: Step 1: (1800-1500) = 300. Thus, the observed difference between these two values is 300. Step 2: Divide the observed difference 300 by the original value 1500, which results in 0.2 (i.e.,300/1500). Step 3: Multiply 0.2 by 100 to lend the percentage value, which results in 20%. IV. A Step-by-Step Guide to Understanding Percent Increase A more user-friendly approach for how to calculate percent increase is provided below: Step 1: Determine the original value. Step 2: Determine the new value. Step 3: Subtract the original value from the new value. This would be the difference in the values. Step 4: Divide the difference in the values by the original value. The resulting decimal needs to be converted to a percentage. Step 5: Multiply the decimal with 100, which results in the percentage increase. To demonstrate this calculation regarding markup percentage, we can use the following example: A product sells for \$15, and the mark-up percentage is 25%. This information suggests that a product’s selling price is 125% of the original cost. Therefore, we multiply 15 by 1.25, giving us a selling price of \$18.75. To check our answer, we use the calcution: 125% of \$15 = (125/100) * 15 = \$18.75 V. Calculating Percent Increase: Tips and Tricks for Accurate Results Though the calculation formulas are pretty simple, it’s important to keep accuracy in mind when calculating percentage increase. Here are some tips and tricks for ensuring accurate results: Double-checking : Make sure all values are entered correctly. A small mistake can make a significant difference in the calculation outcome. Rounding off: Avoid rounding off intermediate values until the end of the question as the answer may not be precise. Rounding off intermediate answers could result in noticeable changes in the solution. Closer precision: You may wish to use more decimal places in your calculation to prevent the rounding error. It increases precision. Example: If the percentage of 100 items is increased by 10% equals 110. However, if the percentage of 99 items is increased by 10%, we arrive at 108.9, a different result entirely. Comprehend the question: It’s critical to understand the question before beginning calculations. VI. The Ultimate Cheat Sheet for Calculating Percent Increase Here is a simple cheat sheet for calculating percent increase Percent Increase Cheat Sheet. To determine percentage increase, first establish the difference between the two values, which is (New value – Old value). You then divide the difference by the original value. The resulting decimal is then converted into a percentage by multiplying it by 100%. % Increase = ((New Value – Old Value)/ Old Value) * 100 Now that you have the percentage increase, you can add it to the original value for the final result. Here are some examples to assist with comprehension: Example 1: Suppose that last year, \$100 had a 10% discount, what would be the sale price? Solution: \$100 – (10/100)100 = \$90 Hence, the sale price is \$90. VII. Calculating Percent Increase Made Easy with These Proven Techniques Here are some easy and straightforward techniques for calculating percent increase: Decimal Method: In this scenario, the percent increase is obtained by moving the decimal point two places to the right in the percentage increase fraction. Factor Method: The percent increase is obtained by multiplying the decimal fraction (i.e., the observed difference divided by the original value) by 100 and drawing the percentage sign (%) afterward. Formula Method: In this scenario, you can use the formula ((New Value – Old Value)/Old Value)100 to calculate the percentage increase. VIII. Conclusion Calculating percent increase is an important numerical skill that shows how a value has changed over some period. It helps in forecasting and in drawing a comparison to monitor growth. We have covered the basics of calculating percent increase, the difference between percent increase and percent decrease, and the formulas used to arrive at accurate results. Additionally, we have provided several tips and tricks to help you improve accuracy, making these calculations simpler and more manageable. With the help of practical examples, you can now master percent increase and apply it in your daily life successfully. By Riddle Reviewer Hi, I'm Riddle Reviewer. I curate fascinating insights across fields in this blog, hoping to illuminate and inspire. Join me on this journey of discovery as we explore the wonders of the world together.
6.3 Normal distributions and heights In the (NNS) (conducted with the 1995 National Health Survey (NHS)), trained nutritionists measured the heights of the respondents . The values quoted are used for modelling purposes, and are essentially the population means $$\mu$$ and standard deviations $$\sigma$$. 1. Take a wild guess: What percentage of your peers (people your age and gender) do you think would be shorter than you? _______ 2. Write down your own height: ________ 3. From Fig. 6.1, write down the mean and standard deviation for people in your age group and gender. Using this information, compute the $$z$$-score for your height, relative to people in your age group and gender. Explain what this means. 1. Heights have an approximate normal (bell-shape) distribution, so the $$68$$--$$95$$--$$99.7$$ rule applies. Using the $$68$$--$$95$$--$$99.7$$ rule, roughly estimate the percentage of people of your age and gender that are shorter than you. (Hint: Draw a picture!) 2. Using the normal distribution tables, together with the information from Fig. 6.1, compute the percentage of people of your age and gender that are shorter than you. 3. Using the $$68$$--$$95$$--$$99.7$$ rule, the middle $$95$$% of females $$18$$ and over (using the Fig. 6.2) are within what height range? 1. For this part, we want to just look at females over $$18$$; from Fig. 6.2, for this group, the mean is $$\mu = 161.4$$ cm, and the standard deviation is $$\sigma = 6.7$$ cm. Use this information to answer the following questions. 1. Compute the percentage of females $$18$$ and over that are shorter than $$171$$ cm. 2. What are the odds that a female $$18$$ and over is shorter than $$171$$ cm? 3. Compute the percentage of females $$18$$ and over that are taller than $$171$$ cm? 4. What are the odds that a female $$18$$ and over is taller than $$171$$ cm? 5. Compute the percentage of females $$18$$ and over that are between 170 and $$180$$ cm? 6. What are the odds that a female $$18$$ and over is between than $$170$$ and $$180$$ cm? 2. For this part, use the data in Fig. 6.2. Approximately $$20$$% of females $$18$$ and over are shorter than what height? References Australian Bureau of Statistics. How Australians measure up [Internet]. Australian Bureau of Statistics; 1995. Report No.: 4359.0. Available from: http://www.ausstats.abs.gov.au/.
# Origin of the Displacement and Acceleration Equation Here we will take a look at the derivation of the following kinematics equation: The above equation solves for the displacement of an object when it is undergoing a constant acceleration. You need to know the original velocity, vo, the constant acceleration, a, and the time period of the acceleration, t. These values are plugged into the equation, and the calculation yields the displacement. It will take a bit of reasoning and algebra to understand where this equation comes from. First, let's consider the following velocity vs. time graph which shows a constant acceleration: Since the slope of this graph is the acceleration, and since the slope of this graph is constant, the acceleration represented here is constant. Also, remember that the area under this graph is the displacement, or change in position, of the object. Now, let's consider a time period over which this acceleration occurs. We will concern ourselves with a time period from t1 to t2: At t1 we will say that the velocity is v1. And at t2 we will say that the velocity is v2: Now, recall that the area under this graph is the displacement of the object. So the area under the graph from t1 to t2 represents the displacement of the object over the time period from t1 to t2. Notice that this area is a trapezoid, here shaded blue: In general, the area of a trapezoid equals one-half times the sum of the bases times the height. This trapezoid, one might say, is "on its side" since the bases are vertical and the height is horizontal. Here is an animated diagram of a generic trapezoid of this type. This diagram demonstrates how to find the area: Now, back to our v vs. t graph. Notice that Base 1 is v1, Base 2 is v2, and that the Height is the time period from t1 to t2. This is shown in the following animated diagram: We will call the time period from t1 to t2 to be simply t. That is: t = t2 - t1 So, the height of the trapezoid is t, and the two bases are v1 and v2. Therefore, the area of the trapezoid is: Area = (1/2)(v1 + v2) t Now, we remember that the area of a v vs. t graph is the change in position, or the displacement, and that displacement is symbolized with d. Therefore, this: Area = (1/2)(v1 + v2) t Becomes: d = (1/2)(v1 + v2) t Now, as far as the acceleration over the time period t is concerned, v1 is the velocity at the start of the acceleration. That is, v1 is the first or original velocity; so, the symbol v1 and the symbol vo mean the same thing; that is: vo = v1 And v2 is the velocity at the end of the acceleration. That is, v2 is the final velocity, and the symbols v2 and vf mean the same thing: vf = v2 So, this: d = (1/2)(v1 + v2) t Is the same as: d = (1/2)(vo + vf) t We will now use another kinematics equation, and substitute it into our displacement equation. This other equation is: vf = vo + at See where the above equation comes from. For the substitution we will place the expression (vo + at) into the displacement equation where vf appears. So, this: d = (1/2)(vo + vf) t Becomes this: d = (1/2)(vo + vo + at) t The rest is algebra: d = (1/2)(vo + vo + at) t Start here. d = (1/2)(2vo + at) t Since 2vo = vo + vo. d = ((1/2)2vo + (1/2)at) t Distribute the (1/2). d = (vo + (1/2)at) t Since (1/2)2 = 1. d = (vot+ (1/2)at2) Distribute the t. d = vot+ (1/2)at2 Remove some parentheses, and we are done. We have derived our displacement equation for constant acceleration: Custom Search
Paul's Online Notes Home / Calculus I / Derivatives / Differentiation Formulas Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. Section 3.3 : Differentiation Formulas 15. Find the tangent line to $$\displaystyle g\left( x \right) = \frac{{16}}{x} - 4\sqrt x$$ at $$x = 4$$. Show All Steps Hide All Steps Hint : Recall the various interpretations of the derivative. One of them will help us do this problem. Start Solution Recall that one of the interpretations of the derivative is that it gives slope of the tangent line to the graph of the function. So, we’ll need the derivative of the function. However before doing that we’ll need to do a little rewrite. Here is that work as well as the derivative. $g\left( x \right) = 16{x^{ - 1}} - 4{x^{\frac{1}{2}}}\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{g'\left( x \right) = - 16{x^{ - 2}} - 2{x^{ - \,\,\frac{1}{2}}} = - \frac{{16}}{{{x^2}}} - \frac{2}{{\sqrt x }}}}$ Note that we rewrote the derivative back into rational expressions with roots to help with the evaluation. Show Step 2 Next we need to evaluate the function and derivative at $$x = 4$$. $g\left( 4 \right) = \frac{{16}}{4} - 4\sqrt 4 = - 4\hspace{0.5in}g'\left( 4 \right) = - \frac{{16}}{{{4^2}}} - \frac{2}{{\sqrt 4 }} = - 2$ Show Step 3 Now all that we need to do is write down the equation of the tangent line. $y = g\left( 4 \right) + g'\left( 4 \right)\left( {x - 4} \right) = - 4 - 2\left( {x - 4} \right)\hspace{0.25in} \to \hspace{0.25in}\,\require{bbox} \bbox[2pt,border:1px solid black]{{y = - 2x + 4}}$
Operations The verbs of mathematics.. Subtraction Same as: adding a negative number. 4 - 3 = 4 + (-3) Presentation on theme: "Operations The verbs of mathematics.. Subtraction Same as: adding a negative number. 4 - 3 = 4 + (-3)"— Presentation transcript: Operations The verbs of mathematics. Subtraction Same as: adding a negative number. 4 - 3 = 4 + (-3) Multiplication Best understood as “repeated addition.” 3 x 5 = 5 + 5 + 5 or 3 rows of 5 items. Division Multiplication by the inverse or Multiplication by the inverse or reciprocal of a number. This definition of division is essential when working with fractions!!! Your turn: 1.Change this into addition: 4 – 1 2. Change this into multiplication: 3. Properties The grammar of mathematics. “I have fun riding my motorcycle.” (English) “Motorcycle my riding fun I have.” (Persian) Commutative Property of Addition 2 + 3 = 3 + 2 Adding two numbers  doesn’t matter which number comes first. which number comes first. Commutative Property of Multiplication 2 x 3 = 3 x 2 multiplying two numbers  doesn’t matter which number comes first. which number comes first. Associative Property of Addition 2 + 3 + 4 Can you add 3 numbers at the same time? Pick 2 of the 3 numbers, add them together. Add the 3 rd number to the sum of the 1 st two. 2 + 3 = 5 5 + 4 = 9 Associative Property of Addition 2 + 3 + 4 We use PEMDAS (parentheses) to “associate” the first 2 numbers together. the first 2 numbers together. (2 + 3) + 4 = 5 + 4 = 9 2 + (3 + 4) = 2 + 7 = 9 The property says: when adding 3 or more numbers together, it doesn’t matter which two of numbers you together, it doesn’t matter which two of numbers you add together first (“associate”), you’ll always get the add together first (“associate”), you’ll always get the same answer. same answer. Using the commutative and associative properties. 7 + x + 3 + 2x = ? = 7 + 3 + x + 2x Rearrange the order (commutative) = (7 + 3) + (x + 2x) = (7 + 3) + (x + 2x) Group terms to add together) = 10+ 3x Your turn: 5.Simplify the following expression using the commutative (order) and associative (grouping) commutative (order) and associative (grouping) properties. properties. Associative Property of Multiplication 2 x 3 x 4 We use PEMDAS (parentheses) to “associate” the first 2 numbers together. the first 2 numbers together. (2 x 3) x 4 = 6 x 4 = 24 2 x (3 x 4) = 2 x 12 = 24 The property says: when multiplying 3 or more numbers together, it doesn’t matter which two of numbers you together, it doesn’t matter which two of numbers you multiply together first (“associate”), you’ll always get the multiply together first (“associate”), you’ll always get the same answer. same answer. Your turn: 6. Simplify the following expression using the commutative (order) and associative (grouping) commutative (order) and associative (grouping) properties. properties. Distributive Property of Addition over Multiplication 2(3 + 4) = (2 * 3) + (2 * 4) = 6 + 8 = 14 2 ( 7 ) 14 This property is important when variables are involved. 2(x + 4) = (2 x) + (2 * 4) = 2x + 8 Your turn: 7. Simplify the following expression using the distributive property of “additional over mulitplication”. distributive property of “additional over mulitplication”. Your turn: Identify the property that allows the step indicated. 8. 9. 10. Equality Properties Addition Property of Equality Subtraction Property of Equality Multiplication Property of Equality Division Property of Equality Inverse Property of Addition 23 + x = 0 “What number do you add so the sum equals zero x = ? -25 + x = 0 x = ? We will use this property to solve equations. Inverse Property of Multiplication What number do you multiply by so the product is 1 (one)? 10 * x = 1 x = ? 10 times its “reciprocal” equals 1 10 divided by itself equals 1 We will use this property to solve equations. Addition Property of Equality a = b a + 1 = b + 1 a + 1 = b + 1 Equivalent Equations Solving an Equation x – 1 = 5 x = 6 + 1 Inverse Property of Addition Addition Property of Equality of Equality: whatever we added to the left side of the ‘=‘ sign, we must add to the right side of the equation.. + 1 x = Identity Property of Addition Subtraction Property of Equality a = b a - 1 = b - 1 a - 1 = b - 1 Equivalent Equations x + 1 = 5 x = 4 x = - 1 Subtraction Property of Equality of Equality: whatever we subtracted from the left side of the ‘=‘ sign, we must subtract from the right side of the equation.. Solving an Equation - 1 Inverse Property of Addition Identity Property of Addition Multiplication Property of Equality a = b a * 2 = b * 2 a * 2 = b * 2 Equivalent Equations Solving an Equation = 5 x = ? * 2 Inverse Property of Multiplication Multiplication Property of Equality of Equality: whatever we multiply the left side of the ‘=‘ sign by, we must multiply the right side of the equation.. * 2 x = 10 Identity Property of Multiplication Division Property of Equality a = b a ÷ 2 = b ÷ 2 a ÷ 2 = b ÷ 2 Equivalent Equations Solving an Equation 3x = 15 x = 5 Inverse Property of Multiplication Division Property of Equality of Equality: whatever we divide the left side of the ‘=‘ sign by, we must divide the right side of the equation.. ÷ 3 Identity Property of Multiplication ÷ 3 11. 11. 2 = 3 + x Your turn: 12. 12. -27 = x - 3 13. 13. 12 = 3x 14. 14. = -2 Combinations “Un-doing” operations Use “reverse” PEMDAS. What do you do 1 st : subtraction or multiplication? - 1 * 2 x = 8 x = ? 15. 15. 12 = 3 + 3x Your turn: 16. 16. -8 = - 5 17. 17. 24 - x = 3x 18. 18. - 4 = -8 Download ppt "Operations The verbs of mathematics.. Subtraction Same as: adding a negative number. 4 - 3 = 4 + (-3)" Similar presentations
## Related Articles • NCERT Solutions for Class 11 Maths # Class 11 NCERT Solutions- Chapter 11 Conic Section – Exercise 11.4 • Difficulty Level : Hard • Last Updated : 09 Mar, 2021 ### Question 1. = 1 Solution: Comparing the given equation with = 1 we conclude that transverse axis is along x-axis. a2 = 16 and b2 = 9 a = ±4 and b = ±3 Foci: Foci = (c, 0) and (-c, 0) c = √(a2+b2) c = √(16+9) c = √25 c = 5 So the foci is (5, 0) and (-5, 0) Vertices: Vertices = (a, 0) and (-a, 0) So the vertices is (4, 0) and (-4, 0) Eccentricity: Eccentricity = c/a = 5/4 Length of the latus rectum: Length of the latus rectum = 2b2/a = 2×9/4 = 9/2 ### Question 2. = 1 Solution: Comparing the given equation with = 1 we conclude that transverse axis is along y-axis. a2 = 9 and b2 = 27 a = ±3 and b = ±3√3 Foci: Foci = (0, c) and (0, -c) c = √(a2 + b2) c = √(9 + 27) c = √36 c = 6 So the foci is (0,6) and (0,-6) Vertices: Vertices = (0,a) and (0,-a) So the vertices is (0,3) and (0,-3) Eccentricity: Eccentricity = c/a = 6/3 = 2 Length of the latus rectum: Length of the latus rectum = 2b2/a = 2×27/3 = 18 ### Question 3. 9y2 – 4x2 = 36 Solution: 9y2 – 4x2 = 36 On dividing LHS and RHS by 36, 9y2/36 – 4x2/36 = 36/36 = 1 Comparing the given equation with = 1 we conclude that transverse axis is along y-axis. a2 = 4 and b2 = 9 a = ±2 and b = ±3 Foci: Foci = (0, c) and (0, -c) c = √(a2 + b2) c = √(4 + 9) c = √13 So the foci is (0, √13) and (0, -√13) Vertices: Vertices = (0, a) and (0, -a) So the vertices is (0, 2) and (0, -2) Eccentricity: Eccentricity = c/a = √13/2 Length of the latus rectum: Length of the latus rectum = 2b2/a = 2×9/2 = 9 ### Question 4. 16x2 – 9y2 = 576 Solution: 16x2 – 9y2 = 576 On dividing LHS and RHS by 576, 16x2/576 – 9y2/576 = 576/576 = 1 Comparing the given equation with = 1 we conclude that transverse axis is along x-axis. a2 = 36 and b2 = 64 a = ±6 and b = ±8 Foci: Foci = (c,0) and (-c,0) c = √(a2 + b2) c = √(36 + 64) c = √100 c = 10 So the foci is (10, 0) and (-10, 0) Vertices: Vertices = (a, 0) and (-a, 0) So the vertices is (6, 0) and (-6, 0) Eccentricity: Eccentricity = c/a = 10/6 = 5/3 Length of the latus rectum: Length of the latus rectum = 2b2/a = 2×64/6 = 64/3 ### Question 5. 5y2 – 9x2 = 36 Solution: 5y2 – 9x2 = 36 On dividing LHS and RHS by 36, 5y2/36 – 9x2/36 = 36/36 Comparing the given equation with = 1, we conclude that transverse axis is along y-axis. a2 = 36/5 and b2 = 4 a = ±6/√5 and b = ±2 Foci: Foci = (0, c) and (0, -c) c = √(a2 + b2) c = √(36/5 + 4) c = √56/5 c = 2√14/√5 So the foci is (0, 2√14/√5) and (0, -2√14/√5) Vertices: Vertices = (0, a) and (0, -a) So the vertices is (0,6/√5) and (0,-6/√5) Eccentricity: Eccentricity = c/a = (2√14/√5)/6/√5 = √14/3 Length of the latus rectum: Length of the latus rectum = 2b2/a = 2×4/(6/√5) = 4√5/3 ### Question 6. 49y2 – 16x2 = 784 Solution: 49y2 – 16x2 = 784 On dividing LHS and RHS by 784, we get 49y2/784 – 16x2/784 = 784/784 Comparing the given equation with = 1 we conclude that transverse axis is along y-axis. a2 = 16 and b2 = 49 a = ±4 and b = ±7 Foci: Foci = (0, c) and (0, -c) c = √(a2 + b2) c = √(16 + 49) c = √65 So the foci is (0, √65) and (0, -√65) Vertices: Vertices = (0, a) and (0, -a) So the vertices is (0, 4) and (0, -4) Eccentricity: Eccentricity = c/a = √65/4 Length of the latus rectum: Length of the latus rectum = 2b2/a = 2×49/4 = 49/2 ### Question 7. Vertices (± 2, 0), foci (± 3, 0). Solution: Since the foci is on x-axis, the equation of the hyperbola is of the form = 1 As, Vertices (± 2, 0) and foci (±3, 0) So, a = ±2 and c = ±3 As, c = √(a2 + b2) b2 = c2 – a2 b2 = 9 – 4 b2 = 5 So, a2 = 4 and b2 = 5 Hence, the equation is = 1 ### Question 8. Vertices (0, ± 5), foci (0, ± 8). Solution: Since the foci is on y-axis, the equation of the hyperbola is of the form = 1 As, Vertices (0, ±5) and foci (0, ±8) So, a = ±5 and c = ±8 As, c = √(a2 + b2) b2 = c2 – a2 b2 = 64 – 25 b2 = 39 So, a2 = 25 and b2 = 39 Hence, the equation is = 1 ### Question 9. Vertices (0, ± 3), foci (0, ± 5). Solution: Since the foci is on y-axis, the equation of the hyperbola is of the form = 1 As, Vertices (0, ± 3) and foci (0, ± 5) So, a = ±3 and c = ±5 As, c = √(a2 + b2) b2 = c2 – a2 b2 = 25 – 9 b2 = 16 So, a2 = 9 and b2 = 16 Hence, the equation is = 1 ### Question 10. Foci (± 5, 0), the transverse axis is of length 8. Solution: Since the foci is on x-axis, the equation of the hyperbola is of the form = 1 As, Foci (±5, 0) ⇒ c = ±5 Since, the length of the transverse axis is 8, 2a = 8 a = 8/2 a = 4 As, c = √(a2 + b2) b2 = c2 – a2 b2 = 25 – 16 b2 = 9 So, a2 = 16 and b2 = 9 Hence, the equation is = 1 ### Question 11. Foci (0, ±13), the conjugate axis is of length 24. Solution: Since the foci is on y-axis, the equation of the hyperbola is of the form = 1 As, Foci (0, ± 13) ⇒ c = ±13 Since, the length of the conjugate axis is 24, 2b = 24 b = 24/2 b = 12 As, c = √(a2 + b2) a2 = c2 – b2 a2 = 169 – 144 a2 = 25 So, a2 = 25 and b2 = 144 Hence, the equation is = 1 ### Question 12. Foci (± 3√5, 0), the latus rectum is of length 8. Solution: Since the foci is on x-axis, the equation of the hyperbola is of the form = 1 As, Foci (±3√5, 0) ⇒ c = ±3√5 Since, the length of latus rectum is 8, 2b2/a = 8 b2 = 8a/2 b2 = 4a -(1) As, c = √(a2 + b2) b2 = 45 – a2 4a = 45 – a2 a2 + 4a – 45 = 0 a2 + 9a – 5a – 45 = 0 (a + 9)(a – 5) = 0 a ≠ -9 (a has to be positive due to eq(1)) Hence, a = 5 From eq(1), we get b2 = 4(5) b2 = 20 So, a2 = 25 and b2 = 20 Hence, the equation is = 1 ### Question 13. Foci (± 4, 0), the latus rectum is of length 12. Solution: Since the foci is on x-axis, the equation of the hyperbola is of the form = 1 As, Foci (±4, 0) ⇒ c=±4 Since, the length of latus rectum is 12, 2b2/a = 12 b2 = 12a/2 b2 = 6a -(1) As, c = √(a2 + b2) b2 = 16 – a2 6a = 16 – a2 a2 + 6a – 16 = 0 a2 + 8a – 2a – 16 = 0 (a + 8)(a – 2) = 0 a ≠ -8 (a has to be positive due to eq(1)) Hence, a = 2 From eq(1), we get b2 = 6(2) b2 = 12 So, a2 = 4 and b2 = 12 Hence, the equation is = 1 ### Question 14. Vertices (± 7, 0), e = 4/3. Solution: Since the vertex is on x-axis, the equation of the hyperbola is of the form = 1 As, Vertices (±7, 0) ⇒ a = ±7 As e = 4/3 c/a = 4/3 c = 4a/3 c = 28/3 As, c = √(a2 + b2) b2 = 784/9 – 49 b2 = 343/9 So, a2 = 49 and b2 = 343/9 Hence, the equation is x2/49 – y2/(343/9) = 1 = 1 ### Question 15. Foci (0, ±√10), passing through (2, 3). Solution: Since the foci is on y-axis, the equation of the hyperbola is of the form = 1 As, Foci (0, ±√10) ⇒ c=±√10 As, c = √(a2 + b2) b2 = c2 – a2 b2 = 10 – a2 -(1) As (2, 3) passes through the curve, hence 32/a2 – 22/b2 = 1 9/a2 – 4/b2 = 1 9/a2 – 4/(10 – a2) = 1 9(10 – a2) – 4a2 = a2(10 – a2) 90 – 9a2 – 4a2 = 10a2 – a4 a4 – 23a2 + 90 = 0 a4 – 18a2 – 5a2 + 90 = 0 a2(a2 – 18) – 5(a2 – 18) = 0 (a2 – 18)(a2 – 5) = 0 a2 = 18 or 5 As, a < c in hyperbola So a2 = 5 And, b2 = 10 – 5 -(From eq(1)) b2 = 5 So, a2 = 5 and b2 = 5 Hence, the equation is = 1 My Personal Notes arrow_drop_up
Math Skills: How to Calculate, Convert, and Use Percents What is Percent? If you listen to the radio, watch the television, or read a newspaper you can't help but hear or see phrases like "25% off today only" or "all merchandise 10% off" or "going out of business mark-downs of 50%." What does all this mean? Percent means "parts per hundred." What you need to do is think of one whole as being divided into 100 parts. If you have all 100 of those parts, you have 100%. Notice the symbol we use for percent (%). If you have only 95 of the parts, you have 95%. Converting Between Decimals and Percents To turn a number (either an integer or a decimal) into a percent, simply multiply by 100. That is the same as moving the decimal point two places to the right. You may need to round to the desired precision. Add a percent (%) sign. 0.32 as a percent is 32% 38.59 = 3859% 0.002 = 0.2% To turn a percent into an integer or decimal number, simply divide by 100. That is the same as moving the decimal point two places to the left. Take off the percent (%) sign. 50% as a decimal is 0.50 3.5% = 0.035 250% = 2.50 Converting Between Fractions and Percents To convert a fraction to a percent, divide the numerator of the fraction by the denominator. Then multiply by 100 or move the decimal point two places to the right. Round the answer to the desired precision. Add a percent (%) sign. Terms – Percentage, Base, Rate Uses of Percent If an item is \$32.99, then you will pay 5% more than that with the tax added. First you figure out how much the tax is by taking 5% of \$32.99: 0.05 x 32.99 = 1.6495 Remember that you are dealing with money, so you must round that off to the nearest penny, making it \$1.65. Then you must add that to the \$32.99 in order to know how much you will be paying: \$32.99 + \$1.65 = \$34.64. That is the final price with the sales tax. Another way to figure it would have been to think about the price as being 100% and the sales tax as 5%, so the total price you pay would be 105%. You could then multiply the original price by 105%: 105% x 32.99 = 1.05(32.99) = 34.6395 = \$34.64 If you work at a retail store, you may be asked to do markups. This is when you take the wholesale price and increase it by a certain percentage to get the retail price at the store where you work. This increase in price pays your salary and the other expenses of operating the store (rent, lights, heat, etc.). A sweater may cost \$15 wholesale, but your store makes a profit of 65% on it. Therefore, it must be marked up by 65% to get the retail price. 65% x \$15 = 0.65(15) = \$9.75 Now add that to the \$15: \$9.75 + \$15 = \$24.75 The markup is the \$9.75 and the retail price is the \$24.75. Or, you could look at this as 165% x \$15 which would give you \$24.75 in a single step. Many stores have markdowns or discounts called sales. This works the opposite of markups and sales tax in that the percentage is subtracted from the original price instead of added to it. Let's say that same sweater is put on sale for 30% off. That means that you need to find 30% of its retail price and subtract that from its retail price. 30% x \$24.75 = 0.30(24.75) = \$7.425 or \$7.43 \$24.75 - \$7.43 = \$17.32 In this case, you would subtract the 30% from 100% to do this in a single step: (100% - 30%) x \$24.75 = 70% x \$24.75 = 0.7 (24.75) = \$17.325 = \$17.33 It will depend upon how the cash register (which is a computer) is programmed as to whether you are charged \$17.32 or \$17.33, but you can figure out the cost to within a penny this way. A commission is another place where percents are used. Sales commissions are paid to sales people based on the price of the item they have sold. In some industries, like insurance, it is paid instead of a salary, In many industries, it is a motivator to sell more and is paid in addition to the normal salary. If a real estate agent makes a 7% commission on a \$175,000 house he sells, he makes Percentage = 7% x \$175,000 = .07 (175,000) = \$12,250 Another place that everyone uses percentage is in figuring out a tip. Tips are given to people who serve us – wait staff at a restaurant, a dog groomer, a bartender, a taxicab driver, a valet, etc. Most tips are either 15% or 20%. If you are paying for a meal and the waiter brings the bill and will pick up the bill, you can simply pay the bill plus the tip to him. However, if the waiter brings the bill and you will pay it at a cash register, you should leave the tip on the table and then go pay the bill. An easy way to figure out a tip without using a calculator: Round the bill to the nearest dollar or half dollar, then move the decimal point one place left to find out 10% of the bill. If you are tipping 20%, double that. If you are tipping 15%, estimate half and add it to the 10%. If your bill is \$35.95, round it to \$36. Move the decimal point one place left to get \$3.60. That is 10%. Since 2 x 36 is 72, you would tip \$7.20 for a 20% tip. Half of \$3.60 would be \$1.80 since ½ of 36 is 18. To tip 15%, add \$1.80 to \$3.60 (to estimate, round to \$1.50 and \$4) and tip \$5.40 – your estimate of \$5.50 is close enough to use. Interest is the biggest use of percentage in everyday life. When you invest money, you make interest – the interest is paid to you. This happens if you have a savings account or you purchase interest-earning bonds or Treasury Bills (TBs) or Certificates of Deposit (CDs). However, if you borrow money like taking out a loan for a car, boat, or house, you pay interest. And if you use a charge card and do not pay off the charges when they are due, you will be charged interest. If your loan is for a very short period of time or is a personal loan from a family member, you may pay simple interest. If it is with a bank or financial institution, you will probably pay compound interest. Simple interest is calculated on the entire amount of money (called the principal) once and then the amount is divided by the number of payments and added to each payment. Compound or compounded interest is figured on the principal, then after the first payment, it is calculated on the remainder of the principal and after the next payment it is figured again on the remaining principal and so forth. To figure interest, you must know the amount of money (principal), the time period for which it was borrowed (time) and the interest rate that is being charged or paid. The formula is: Interest = Principal x Rate x Time If \$500 is borrowed for 2 years at a 12% interest rate: Interest = \$500 x 12% x 2 Interest = (500)(0.12)(2) = \$120 The amount owed at the end would be \$500 + \$120 or \$620. Calculating Compound Interest. Compound interest is calculated on the principal plus accumulated interest. The amount to be repaid is calculated using the following formula: A = P( 1 + i )n For example, you receive 10% interest on a \$1,000 investment in the first year. You reinvested that money back into your original investment. In the second year, you would get 10% interest on the \$1,000 plus the \$100 you reinvested. Over the years, compound interest will make you much more money than simple interest because you are reinvesting whatever interest you make. Let's review this in the following example: A = P( 1 + i )n A is the final total including the principal. P is the principal amount (what you originally invested). i is the rate of interest per year. n is the number of years invested. Remember n is an exponent. Example: Let's say that you have \$2,500.00 to invest for 5 years at a rate of 7% compound interest. A = 2500 (1 + 0.07)5 = \$3,506.38 You can see that your \$2,500.00 is now worth \$3,506.38 after 5 years at 7% interest compounded annually.
### Theory: 1. Dividing a fraction by another fraction: To divide a fractional number with another fractional number, follow the steps below: Step 1: Take the reciprocal of the divisor. Step 2: Multiply the reciprocal of the divisor with the dividend to get the new numerator and denominator of the fraction. Reciprocal of a fraction: The reciprocal of a fraction can be obtained by interchanging the numerator and denominator of the fraction. For example, reciprocal of $\frac{1}{2}$ is $\frac{2}{1}$ The non-zero numbers whose product with each other is $$1$$ are called the reciprocals of each other that is, $\frac{1}{2}\phantom{\rule{0.147em}{0ex}}×\frac{2}{1}=1\phantom{\rule{0.147em}{0ex}}$. Hence, $\frac{1}{2}$ and $\frac{2}{1}$ are reciprocals of each other. Let us find the value of $\frac{\frac{1}{2}}{\frac{5}{6}}=?$ Step 1: Take the reciprocal of divisor $\left(\frac{5}{6}\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{6}{5}$. Step 2: Multiply the dividend $\left(\frac{1}{2}\right)$ by $\frac{6}{5}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{2}×\frac{6}{5}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{3}{5}$ $$\frac{\text{New Numerator}}{\text{New Denominator}}$$ $$=$$ $\frac{3}{5}$ 2. Division of the whole number by a fraction: Dividing a whole number by a fraction will follow the same procedure as dividing a fraction by another fraction. Example: Let us find $\phantom{\rule{0.147em}{0ex}}1÷\frac{1}{4}$ Step 1: Reciprocal of the divisor, $\frac{1}{4}$ is $\frac{4}{1}$ Step 2: Multiply the dividend ($$1$$) by $\frac{4}{1}=1×4=4$. 3. Division of a fraction by a whole number: Dividing a fractional number by a whole number will follow the same procedure as dividing a fraction by another fraction. Example: Let us find $\frac{1}{4}÷1$. Step 1: Reciprocal of the divisor, $$1$$ is $$1$$. Step 2: Multiply the dividend $\frac{1}{4}$ by $1=\frac{1}{4}$. 4. Division of mixed fractions: First, convert the mixed fractions to improper fractions and divide the fractions. Example: Convert $4\frac{10}{5}$ to an improper fraction. Step 1: Numerator of improper fraction $$=$$ (denominator of the proper fraction $$×$$ whole number) $$+$$ numerator of a proper fraction. $$= 5 × 4 = 20 + 10 = 30$$. Step 2: Denominator of improper fraction $$=$$ denominator of the mixed fraction. The denominator of improper fraction $$= 5$$. Improper fraction $$= 30 / 5 = 6$$. Thus, $4\frac{10}{5}$ $$= 6$$.
# How do you simplify (4+ 2i) /( -1 + i)? May 2, 2018 $\frac{4 + 2 i}{- 1 + i} \| \cdot \left(- 1 - i\right)$ $\frac{\left(4 + 2 i\right) \left(- 1 - i\right)}{\left(- 1 + i\right) \left(- 1 - i\right)}$ $\frac{- 2 {i}^{2} - 6 i - 4}{1 - {i}^{2}}$ $\frac{2 - 6 i - 4}{1 + 1}$ $\frac{- 2 - 6 i}{2}$ $= - 1 - 3 i$ #### Explanation: We want to get rid of $i$ in the the bottom of the fraction in order to get it on Certesian form. We can do this by multiplying with $\left(- 1 - i\right)$. This will give us, $\frac{\left(4 + 2 i\right) \left(- 1 - i\right)}{\left(- 1 + i\right) \left(- 1 - i\right)}$ $\frac{- 2 {i}^{2} - 6 i - 4}{1 - {i}^{2}}$ Out from here we know that ${i}^{2} = - 1$ and $- {i}^{2} = 1$. So we can get rid of the ${i}^{2}$ too. Leaving us to $\frac{- 2 - 6 i}{2}$ $= - 1 - 3 i$
Enable contrast version # Tutor profile: Rhoda A. Rhoda A. Math and Statistics tutor ## Questions ### Subject:Basic Math TutorMe Question: 1/5 + 3/5= ? Rhoda A. To solve this fraction, we will first note that the fractions have a common denominator (5). A denominator is the number on the bottom of a fraction, while the number on top is called a numerator. To solve this we will add up the numerators (1 + 3) = 4. Addition, because this was the operation used in this equation. Since the denominator for both fractions are the same. It will continue to stay the same. The answer will be (1+3)/5 = 4/5. ### Subject:Statistics TutorMe Question: How do you calculate the P (Male \| Junior)? Given that : P (Male n Junior) = 0.10 , P (Male U Junior) = 0.50, P(Male) = 0.30 and P (Junior) = 0.20 Rhoda A. This is a conditional probability that reads as: the probability of a student being a Male, given that he is a Junior. This can be calculated by using the Conditional probability formula: find the joint Probability of Males who are also Juniors [P(Male n Juniors)] and then divide this by the Marginal probability of Juniors [P(Juniors)]. This can be simplified as: P (Male \| Junior) = P(Male n Juniors)/P(Juniors). Given that P (Male n Junior) = 0.10 and P (Junior) = 0.20. We can rewrite the equation as: P (Male \| Junior) = 0.10/0.20 = 0.5. Therefore, P (Male \| Junior) = 0.5 or 50% ### Subject:Algebra TutorMe Question: 3x+4=28. Solve for "x"? Rhoda A. In this equation we have, 3 as the co-efficient of "x" (the variable in this equation), whilst 4 and 28 are constants. To solve for x we will have to get x by itself, to do this you will first "get rid" of the constant 4 by subtract 4 from both sides of the equation. This can be seen as: 3x + 4 - 4 = 28 - 4. since: (+4 - 4) = 0, and 28-4 = 24 The equation can be simplified as: 3x = 24 From here we will then also "get rid" of the co-efficient 3 as well. We can do this by dividing both sides of the equation by 3. This can be seen as: (3x/3) = 24/3 since: 3/3 = 1, and 24/3 = 8 The equation can further be simplified as: 1x = 8 1x = x (1 multiplied by any number/variable equals the number/variable itself) Therefore, x = 8. To test if this answer is correct, we can go back to the original equation and plug in 8 wherever we see x. And the answer should be 28. 3x+4=28 becomes, 3(8) + 4 = 28. This is true as 3 * 8 = 24, and 24 + 4 = 28. ## FAQs What is a lesson? A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard. How do I begin a lesson? If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson. Who are TutorMe tutors? Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you. BEST IN CLASS SINCE 2015 TutorMe homepage
## Book: RS Aggarwal - Mathematics #### Subject: Maths - Class 10th ##### Q. No. 35 of Exercise 10E Listen NCERT Audio Books to boost your productivity and retention power by 2X. 35 ##### In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained by him in the two subjects separately. Let the marks obtained by P in mathematics and science be x and (28 – x) respectively According to the given condition, (x + 3)(28 – x – 4) = 180 (x + 3)(24 – x) = 180 – x2 + 21x + 72 = 180 x2 – 21x + 108 = 0 Using the splitting middle term – the middle term of the general equation is divided in two such values that: Product = a.c For the given equation a = 1 b = – 21 c = 108 = 1.108 = 108 And either of their sum or difference = b = – 21 Thus the two terms are – 12 and – 9 Difference = – 12 – 9 = – 21 Product = – 12. – 9 = 108 x2 – 12x – 9x + 108 = 0 x (x – 12) – 9 (x – 12) = 0 (x – 12) (x – 9) = 0 (x – 12) = 0 or (x – 9) = 0 x = 12, x = 9 When x = 12, 28 – x = 28 – 12 = 16 When x = 9, 28 – x = 28 – 9 = 19 Hence he obtained 12 marks in mathematics and 16 science or He obtained 9 marks in mathematics and 19 science. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71
# Decimals ## Contents #### Addition and Subtraction of Decimals To add or subtract decimals, simply line them up so their decimal points are in the same place, and then add or subtract as usual. Sometimes one of the numbers will have more decimal places than the other. Because adding zeros to the end of a decimal does not change its value, we can just add zeros to the end of the shorter number until the two numbers have the same number of decimal places. For example, to subtract 65.23 from 987.462: 987 . 462 -65 . 230 922 . 232 To add 56.999 to 193.1: 193 . 100 +56 . 999 250 . 099 #### Multiplication of Decimals To multiply two decimals, first count the total number of digits to the right of the decimal place in each number, and add these two totals together. Then remove the decimal points and multiply the two new whole numbers together. Take this result, and count from the right the total number of places calculated in the first step. Then insert a decimal point to the left of this number. For example, to multiply 3.4 and 2.01: • Step 1. There is 1 digit to the right of the decimal point in 3.4, and 2 digits to the right of the decimal point in 2.01. This is a total of 3. • Step 2. Eliminate the decimal points and multiply 34 by 201. This equals 6834 • Step 3. Count 3 places from the right and insert a decimal point. This yields 6.834. Thus, 3.4×2.01 = 6.834 . #### Division of Decimals To understand how to divide two numbers when one contains a decimal, we must first remember that adding zeros to the end of a decimal does not change the number. Therefore, we can add as many zeros as we want to either of our decimals. Second, we note that if we move the decimal one place to the right (or to the left) in both numbers, it does not change the answer. To divide two numbers, then, we first add zeros to the end of either number--these must be added to the right of the decimal point--until both numbers have the same number of digits to the right of the decimal point. For example, to divide 31.8 by 2.65, we add a zero to 31.8 so we are dividing 31.80 by 2.65. Next, we move the decimal point to the right until both numbers are whole numbers; moving the decimal point changes the value of the numbers, but it doesn't change the ratio between the two numbers, which is what division measures. Be very careful to move the decimals the same distance for each number. In this case, we move the decimal point to the right 2 places so we are dividing 3,180 by 265. Finally, we carry out the long division. 3, 180/265 = 12 .
# Nim and simple mathematical proofs 1/3 A few weeks ago I discussed a game called Nim. In studying Nim, with the help of Thomas S. Ferguson’s book Game Theory, I have again found great joy in mathematical thinking, proving theorems and using logical thinking to come to conclusions. Many people, as far as I know, have some kind of fear of mathematics and mathematical notation. Some people are even proud of their mathematical illiteracy, which is odd, considering that nobody boasts how their Spanish or Chinese skills are non-existent. Working with games and proofs, like those required by Nim, might help people see the value and true beauty of mathematical thinking. People might understand that extensive mathematical notation is not always necessary to prove mathematical theorems, and that mathematical notation is actually just another language that helps us formulate and formalize logical thinking, removing all ambiguity. In this post and two additional ones I will present three proofs related to Nim to show you, how mathematical thinking can be used quite cleverly to show interesting facts. Bouton’s theorem In his 1902 article Nim, a game with a complete mathematical theory in Ann. Math. 3, 35-39 L. Bouton proves the following theorem (Theorem 1, Bouton’s theorem): “In a game of Nim a position is a P-position if and only if the Nim-sum of its components is zero.” A Nim-sum is calculated as follows: • The number of sticks in each pile is expressed in base two. • These figures are added without carry, so that the sum of each column is 1, if the column contains an odd number of ones. Otherwise, the sum of that column is 0. • The Nim-sum is zero when the sum of each individual column is 0. As an example of calculating the Nim-sum we take three piles with 4, 5, and 8 sticks respectively. The Nim-sum of these piles in base two is: 1000 1010 1000 1001 The result, in base 10, is 9. Since this Nim-sum is non-zero, according to Bouton’s theorem this would be an N-position. It is worth noting that a Nim-sum of two identical positive integers is also zero due to the definition of Nim-sum. Nim-sum is also associative and commutative and 0 is an identity for addition. (see Ferguson’s Game Theory, Part 1, Chapter 1, page 10) Proving Theorem 1 Proving Bouton’s theorem is partially a constructive proof, relying on the definitions of P- and N-positions. The claim of the theorem is compared against these definitions to show that they are equal. We recall that P- and N-positions are defined as follows: • The terminal position is a P-position. • An N-position is such that there is at least one possible move into a P-position. • A P-position is such that from it is possible to move only into an N-position. Now we just have to show that the terminal position has a Nim-sum of zero, a P-position always has a Nim-sum of zero and that a Nim-sum of zero is never an N-position. If we can prove this, we will have proven Bouton’s theorem. Let’s start by observing that the terminal position has a Nim-sum of zero, so condition 1 is fulfilled. For condition number 2 we can construct such a move, where we can move from a position with a non-zero Nim-sum (a purported N-position) into a position with a Nim-sum of zero (a purported P-position). In our example we do this by reducing the number of ones in the leftmost column with an odd number of ones (the 4th column) so that each column will eventually contain an even number of ones. In our case, there is only one potential move, since there is only one number one in the leftmost column. Thus, the only move into a P-position is to remove 7 sticks from the pile of 8. By this example we see, that we can move from an N-position into a P-position, so condition 2 is also fulfilled. For a more formal and generalized proof for this part, see Bouton’s original article. The third condition can be shown to be true by the following reasoning. If we are in a position with a Nim-sum of zero, removing any number of sticks from any pile will leave us witha non-zero Nim-sum (a purported N-position). This is because if we remove any number of sticks from a pile, the number of sticks in all the other piles remains the same. If the Nim-sum of the new position were zero, it would imply that the number of sticks remaining in the pile, from which sticks were removed, has stayed the same, but this is a contradiction. Thus, condition 3 also holds. Collecting together the proofs for conditions 1, 2 and 3 we have shown the following: • The terminal position has a Nim-sum of zero. • From a position with a non-zero Nim-sum we can move into a position with a Nim-sum of zero. • From a position with a Nim-sum of zero it is possible to move only into a potision with a non-zero Nim-sum. We see that these statements are analogical to conditions 1, 2 and 3 and that a position is a P-position if and only if the Nim-sum of that position is zero. I hope this post gave you an idea of how simple logical thinking with quite elementary mathematical notation can provide a powerful tool for analyzing problems. In my next post I will discuss a Nim variant called Nimble and show how Nim can come in disguise.
<meta http-equiv="refresh" content="1; url=/nojavascript/"> You are reading an older version of this FlexBook® textbook: CK-12 Algebra - Basic Go to the latest version. # 12.5: Multiplication and Division of Rational Expressions Difficulty Level: At Grade Created by: CK-12 Because a rational expression is really a fraction, two (or more) rational expressions can be combined through multiplication and/or division in the same manner as numerical fractions. A reminder of how to multiply fractions is below. For any rational expressions a0,b0,c0,d0\begin{align}a \neq 0, b \neq 0, c \neq 0, d \neq 0\end{align}, Example: Multiply the following a16b84b35a2\begin{align}\frac{a}{16b^8} \cdot \frac{4b^3}{5a^2}\end{align} Solution: a16b84b35a24ab380a2b8 Simplify exponents using methods learned in chapter 8. 4ab380a2b8=120ab5 Example 1: Simplify 9c24y221c4\begin{align}9c^2 \cdot \frac{4y^2}{21c^4}\end{align}. Solution: 9c24y221c49c214y221c49c214y221c4=36c2y221c436c2y221c4=12y27c2 ## Multiplying Rational Expressions Involving Polynomials When rational expressions become complex, it is usually easier to factor and reduce them before attempting to multiply the expressions. Example: Multiply 4x+123x2xx29\begin{align}\frac{4x+12}{3x^2} \cdot \frac{x}{x^2-9}\end{align}. Solution: Factor all pieces of these rational expressions and reduce before multiplying. 4x+123x2xx294(x+3)3x2x(x+3)(x3)4(x+3)3x2x(x+3)(x3)43x1x343x29x Example 1: Multiply 12x2x6x21x2+7x+64x227x+18\begin{align}\frac{12x^2-x-6}{x^2-1} \cdot \frac{x^2+7x+6}{4x^2-27x+18}\end{align}. Solution: Factor all pieces, reduce, and then multiply. 12x2x6x21x2+7x+64x227x+18(3x+2)(4x3)(x+1)(x1)(x+1)(x+6)(4x3)(x6)12x2x6x21x2+7x+64x227x+18(3x+2)(4x3)(x+1)(x1)(x+1)(x+6)(4x3)(x6)(3x+2)(x+6)(x1)(x6)=3x2+20x+12x27x+6 ## Dividing Rational Expressions Involving Polynomials Division of rational expressions works in the same manner as multiplication. A reminder of how to divide fractions is below. For any rational expressions a0,b0,c0,d0\begin{align}a \neq 0, b \neq 0, c \neq 0, d \neq 0\end{align}, Example: Simplify 9x242x2÷21x22x81\begin{align}\frac{9x^2-4}{2x-2} \div \frac{21x^2-2x-8}{1}\end{align}. Solution: 9x242x2÷21x22x819x242x2121x22x8 Repeat the process for multiplying rational expressions. 9x242x2121x22x89x242x2÷21x22x81(3x2)(3x2)2(x1)1(3x2)(7x+4)=3x214x26x8 ## Real-Life Application Suppose Marciel is training for a running race. Marciel’s speed (in miles per hour) of his training run each morning is given by the function x39x\begin{align}x^3-9x\end{align}, where x\begin{align}x\end{align} is the number of bowls of cereal he had for breakfast (1x6)\begin{align}(1 \le x \le 6)\end{align}. Marciel’s training distance (in miles), if he eats x\begin{align}x\end{align} bowls of cereal, is 3x29x\begin{align}3x^2-9x\end{align}. What is the function for Marciel’s time and how long does it take Marciel to do his training run if he eats five bowls of cereal on Tuesday morning? timetimetimeIf xtime=distancespeed=3x29xx39x=3x(x3)x(x29)=3x(x3)x(x+3)(x3)=3x+3=5,then=35+3=38 . Marciel will run for 38\begin{align}\frac{3}{8}\end{align} of an hour. ## Practice Set Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. In 1–20, perform the indicated operation and reduce the answer to lowest terms 1. x32y32y2x\begin{align}\frac{x^3}{2y^3} \cdot \frac{2y^2}{x}\end{align} 2. 2xy÷2x2y\begin{align}2xy \div \frac{2x^2}{y}\end{align} 3. 2xy24y5x\begin{align}\frac{2x}{y^2} \cdot \frac{4y}{5x}\end{align} 4. 2xy2y2x3\begin{align}2xy \cdot \frac{2y^2}{x^3}\end{align} 5. 4y21y29y32y1\begin{align}\frac{4y^2-1}{y^2-9} \cdot \frac{y-3}{2y-1}\end{align} 6. 6aba2a3b3b2\begin{align}\frac{6ab}{a^2} \cdot \frac{a^3b}{3b^2}\end{align} 7. x2x1÷xx2+x2\begin{align}\frac{x^2}{x-1} \div \frac{x}{x^2+x-2}\end{align} 8. 33a252011a3\begin{align}\frac{33a^2}{-5} \cdot \frac{20}{11a^3}\end{align} 9. a2+2ab+b2ab2a2b÷(a+b)\begin{align}\frac{a^2+2ab+b^2}{ab^2-a^2b} \div (a+b)\end{align} 10. 2x2+2x24x2+3xx2+x6x+4\begin{align}\frac{2x^2+2x-24}{x^2+3x} \cdot \frac{x^2+x-6}{x+4}\end{align} 11. 3x3x5÷x292x28x10\begin{align}\frac{3-x}{3x-5} \div \frac{x^2-9}{2x^2-8x-10}\end{align} 12. x225x+3÷(x5)\begin{align}\frac{x^2-25}{x+3} \div (x-5)\end{align} 13. 2x+12x1÷4x2112x\begin{align}\frac{2x+1}{2x-1} \div \frac{4x^2-1}{1-2x}\end{align} 14. xx5x28x+15x23x\begin{align}\frac{x}{x-5} \cdot \frac{x^2-8x+15}{x^2-3x}\end{align} 15. 3x2+5x12x29÷3x43x+4\begin{align}\frac{3x^2+5x-12}{x^2-9} \div \frac{3x-4}{3x+4}\end{align} 16. 5x2+16x+336x225(6x2+5x)\begin{align}\frac{5x^2+16x+3}{36x^2-25} \cdot (6x^2+5x)\end{align} 17. x2+7x+10x29x23x3x2+4x4\begin{align}\frac{x^2+7x+10}{x^2-9} \cdot \frac{x^2-3x}{3x^2+4x-4}\end{align} 18. x2+x12x2+4x+4÷x3x+2\begin{align}\frac{x^2+x-12}{x^2+4x+4} \div \frac{x-3}{x+2}\end{align} 19. x416x29÷x2+4x2+6x+9\begin{align}\frac{x^4-16}{x^2-9} \div \frac{x^2+4}{x^2+6x+9}\end{align} 20. x2+8x+167x2+9x+2÷7x+2x2+4x\begin{align}\frac{x^2+8x+16}{7x^2+9x+2} \div \frac{7x+2}{x^2+4x}\end{align} 21. Maria’s recipe asks for 212 times\begin{align}2 \frac{1}{2} \ \text{times}\end{align} more flour than sugar. How many cups of flour should she mix in if she uses 313 cups\begin{align}3 \frac{1}{3} \ \text{cups}\end{align} of sugar? 22. George drives from San Diego to Los Angeles. On the return trip, he increases his driving speed by 15 miles per hour. In terms of his initial speed, by what factor is the driving time decreased on the return trip? 23. Ohm’s Law states that in an electrical circuit I=VRc\begin{align}I=\frac{V}{R_c}\end{align}. The total resistance for resistors placed in parallel is given by 1Rtot=1R1+1R2\begin{align}\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}\end{align}. Write the formula for the electric current in term of the component resistances: R1\begin{align}R_1\end{align} and R2\begin{align}R_2\end{align}. Mixed Review 1. The time it takes to reach a destination varies inversely as the speed in which you travel. It takes 3.6 hours to reach your destination traveling 65 miles per hour. How long would it take to reach your destination traveling 78 miles per hour? 2. A local nursery makes two types of fall arrangements. One arrangement uses eight mums and five black-eyed susans. The other arrangement uses six mums and 9 black-eyed susans. The nursery can use no more than 144 mums and 135 black-eyed susans. The first arrangement sells for \$49.99 and the second arrangement sells for 38.95. How many of each type should be sold to maximize revenue? 3. Solve for r\begin{align}r\end{align} and graph the solution on a number line: 24\|2r+3\|\begin{align}-24 \ge \|2r+3\|\end{align}. 4. What is true of any line parallel to 5x+9y=36\begin{align}5x+9y=-36\end{align}? 5. Solve for d:3+5d=d(3x3)\begin{align}d: 3+5d=-d-(3x-3)\end{align}. 6. Graph and determine the domain and range: y9=x25x\begin{align}y-9=-x^2-5x\end{align}. 7. Rewrite in vertex form by completing the square. Identify the vertex: y216y+3=4\begin{align}y^2-16y+3=4\end{align}. ## Quick Quiz 1. h\begin{align}h\end{align} is inversely proportional to t\begin{align}t\end{align}. If t=0.05153\begin{align}t=-0.05153\end{align} when h=16\begin{align}h=-16\end{align}, find t\begin{align}t\end{align} when h=1.45\begin{align}h=1.45\end{align}. 2. Use f(x)=5x225\begin{align}f(x)=\frac{-5}{x^2-25}\end{align} for the following questions. 1. Find the excluded values. 2. Determine the vertical asymptotes. 3. Sketch a graph of this function. 4. Determine its domain and range. 3. Simplify 8c4+12c222c+14\begin{align}\frac{8c^4+12c^2-22c+1}{4}\end{align}. 4. Simplify 10a230aa3\begin{align}\frac{10a^2-30a}{a-3}\end{align}. What are its excluded values? 5. Fill the blank with directly, inversely, or neither. “The amount of time it takes to mow the lawn varies ________________ with the size of the lawn mower.” 8 , 9 Feb 22, 2012 Dec 11, 2014
# Ask Uncle Colin: Evil in integral form Dear Uncle Colin, I've been set an evil integral: $\int_0^\piby{4} \frac{\sqrt{3}}{2 + \sin(2x)}\d x$. There is a hint to use the substitution $\tan(x) = \frac{1}{2}\left( -1 + \sqrt{3}\tan(\theta)\right)$, but I can't see how that helps in the slightest. — Let's Integrate Everything! Hello, LIE, and thank you for your message! That, my friend, is an evil, evil integral. It took me a couple of goes. But! Let's go through the steps in a logical order. ### Step 1: change the limits When $x=0$, the hint gives $0 = \frac{1}{2}\left( -1 + \sqrt{3}\tan(\theta)\right)$, which means $\tan(\theta)=\frac{1}{\sqrt{3}}$ and $\theta = \piby 6$. Similarly, when $x = \piby 4$, $\theta = \piby 3$, so at least the limits are nice. ### Step 2: replace $\d x$ To avoid @realityminus3 getting in our faces, let's do it properly: $\d x = \diff x \theta \d \theta$, so we need to know $\diff x \theta$. I would rearrange the hint to give $2 \tan(x)+1 = \sqrt{3} \tan(\theta)$. Differentiating implicitly with respect to $\theta$ gives $2 \sec^2(x) \diff x \theta = \sqrt{3} \sec^2(\theta)$, so $\diff x \theta = \frac{\sqrt{3}\cos^2(x)}{2 \cos^2(\theta)}$. This makes our integral yet messier. $\int_{\piby 6}^{\piby 3} \frac{\sqrt{3}}{2 + \sin(2x)} \times \frac{\sqrt{3}\cos^2(x)}{2\cos^2(\theta)}\d \theta$. ### Step 3: tidy up and explore However, there are things we can tidy up. The $\sqrt{3}$s combine nicely, and I think expanding the $\sin(2x)$ is a good idea: $\int_{\piby 6}^{\piby 3} \frac{3 \cos^2(x)}{4 + 4\sin(x)\cos(x)} \times \sec^2(\theta)\d \theta$. If we divide top and bottom by $\cos^2(x)$, we get something nicer still: $\int_{\piby 6}^{\piby 3} \frac{3 }{4\sec^2(x) + 4\tan(x)} \times \sec^2(\theta)\d \theta$. Aha! There's our tan! Also, $\sec^2(x) \equiv 1 + \tan^2(x)$, so the whole of the denominator can be written in terms of $\tan(x)$. ### Step 4: tidy up any remaining $x$s Let's define $t = \tan(x)$ and $T = \tan(\theta)$; our substitution is $2t + 1 = \sqrt{3}T$ and the bottom of the fraction is $4t^2 + 4t + 4$. By squaring, we have $4t^2 + 4t + 1 = 3T^2$, so the bottom is clearly $3T^2 + 3$. Now we have $\int_{\piby 6}^{\piby 3} \frac{3 }{3\tan^2(\theta)+ 3} \times \sec^2(\theta)\d \theta$. ### Step 5: Do the actual integration That integrand, once you cancel the 3s and remember your relationship between $\tan^2(\theta)$ and $\sec^2(\theta)$, turns out to be … 1. And $\int_{\piby 6}^{\piby 3} \d \theta$ is trivially $\piby 6$. Phew! I'm glad I got to use 'trivially' at the end there. Hope that helps! – Uncle Colin ## Colin Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove. #### Share This site uses Akismet to reduce spam. Learn how your comment data is processed. Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations. No spam ever, obviously. ##### Where do you teach? I teach in my home in Abbotsbury Road, Weymouth. It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.
NCERT Solutions for class 10 Maths Chapter 13- Exercise 13.2 NCERT Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes, Exercise 13.2, are available here in downloadable PDF format. The problems in this exercise have been answered by our Maths experts, to make students understand the theories in a better way. Topics covered in these materials are according to the NCERT syllabus and guidelines. Therefore, solutions for Maths class 10, are very helpful for students who are doing the preparation for the second term exams and to score good marks. Access Other Exercise Solutions of Class 10 Maths Chapter 13- Surface Areas and Volumes Exercise 13.1 Solutions 9 Question (7 long, 2 short) Exercise 13.3 Solutions 9 Question (9 long) Exercise 13.4 Solutions 5 Question (5 long) Exercise 13.5 Solutions 7 Question (7 long) Access Answers to NCERT Class 10 Maths Chapter 13 â€“ Surface Areas and Volumes Exercise 13.2 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of Ï€. Solution: Here r = 1 cm and h = 1 cm. The diagram is as follows. Now, Volume of solid = Volume of conical part + Volume of hemispherical part We know the volume of cone = â…“ Ï€r2h And, The volume of hemisphere = â…”Ï€r3 So, volume of solid will be = Ï€ cm3 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) Solution: Given, Height of cylinder = 12â€“4 = 8 cm Height of cone = 2 cm Now, the total volume of the air contained will be = Volume of cylinder+2Ã—(Volume of cone) âˆ´ Total volume = Ï€r2h+[2Ã—(â…“ Ï€r2h )] = 18 Ï€+2(1.5 Ï€) = 66 cm3. 3. A Gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 Gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure). Solution: It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends. So, the total height of a gulab jamun = 5 cm. Diameter = 2.8 cm âˆ´ The height of the cylindrical part = 5 cmâ€“(1.4+1.4) cm =2.2 cm Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres = Ï€r2h+(4/3)Ï€r3 = 4.312Ï€+(10.976/3) Ï€ = 25.05 cm3 We know that the volume of sugar syrup = 30% of total volume So, volume of sugar syrup in 45 gulab jamuns = 45Ã—30%(25.05 cm3) = 45Ã—7.515 = 338.184 cm3 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig.). Solution: Volume of cuboid = length x width x height We know the cuboidâ€™s dimensions as 15 cmx10 cmx3.5 cm So, the volume of the cuboid = 15x10x3.5 = 525 cm3 Here, depressions are like cones and we know, Volume of cone = (â…“)Ï€r2h Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm âˆ´ Volume of 4 cones = 4x(â…“)Ï€r2h = 1.46 cm2 Now, volume of wood = Volume of cuboid â€“ 4 x volume of cone = 525-1.46 = 523.54 cm2 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. Solution: For the cone, Height = 8 cm Also, Radius of sphere = 0.5 cm The diagram will be like It is known that, Volume of cone = volume of water in the cone = â…“Ï€r2h = (200/3)Ï€ cm3 Now, Total volume of water overflown= (Â¼)Ã—(200/3) Ï€ =(50/3)Ï€ = (4/3)Ï€r3 = (1/6) Ï€ Now, The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot = (50/3)Ï€/(â…™)Ï€ = (50/3)Ã—6 = 100 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. Solution: Given, the height of the big cylinder (H) = 220 cm Radius of the base (R) = 24/12 = 12 cm So, the volume of the big cylinder = Ï€R2H = Ï€(12)2 Ã— 220 cm3 = 99565.8 cm3 Now, the height of smaller cylinder (h) = 60 cm Radius of the base (r) = 8 cm So, the volume of the smaller cylinder = Ï€r2h = Ï€(8)2Ã—60 cm3 = 12068.5 cm3 âˆ´ Volume of iron = Volume of the big cylinder+ Volume of the small cylinder = 99565.8 + 12068.5 =111634.5 cm3 We know, Mass = Density x volume So, mass of the pole = 8Ã—111634.5 = 893 Kg (approx.) 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. Solution: Here, the volume of water left will be = Volume of cylinder â€“ Volume of solid Given, Radius of cone = 60 cm, Height of cone = 120 cm Radius of cylinder = 60 cm Height of cylinder = 180 cm Radius of hemisphere = 60 cm Now, Total volume of solid = Volume of Cone + Volume of hemisphere Volume of cone = Ï€Ã—122Ã—103cm3 = 144Ã—103Ï€ cm3 So, Total volume of solid = 144Ã—103Ï€ cm3 -(â…”)Ã—Ï€Ã—103 cm3 Volume of hemisphere = (â…”)Ã—Ï€Ã—103 cm3 Volume of cylinder = Ï€Ã—602Ã—180 = 648000 = 648Ã—103 Ï€ cm3 Now, volume of water left will be = Volume of cylinder â€“ Volume of solid = (648-288) Ã— 103Ã—Ï€ = 1.131 m3 8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and Ï€ = 3.14. Solution: Given, For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm For the spherical part, Radius (r) = (8.5/2) = 4.25 cm Now, volume of this vessel = Volume of cylinder + Volume of sphere = Ï€Ã—(1)2Ã—8+(4/3)Ï€(4.25)3 = 346.51 cm3 Exercise 13.2 of 10th Standard Maths have problems based on finding the volumes of; • A solid cone in terms of Pi. • A structure formed with a cylindrical shape object having a cone on both its sides. • A cylindrical object immersed with some substance, occupying the volume of the object. • A wood of cuboid shape having depressions in it. Students can visit us anytime, to get complete solutions for chapter 13 of class 10 maths. Apart from this, they can also find other learning materials, such as notes, books, questions papers and some tips and tricks to do the preparation for the 10th Standard first and second term Maths exam. The questions in Exercise 13.2 are prepared as per the topic, Volume of Combination of Solids, covered in the chapter before this exercise. Solutions provide generic methods for students, which they can relate to easily.
# What is Power Set – Definition, Formula and Examples Here you will learn what is power set and its definition with examples. Let’s begin – ## What is Power Set ? Definition : Let A be a set. Then the collection or family of all subsets of A is called the power set of A and is denoted by P(A). That is, P(A) = { S : S $$\subset$$ A} Since the empty set and the set A itself are subsets of A and are therefore elements of P(A). Thus, the power set of a given set is always non-empty. #### Power Set of Empty Set If A is the void set $$\phi$$, then P(A) has just one element $$\phi$$ i.e $$P(\phi)$$ = {$$\phi$$} Example 1 : Let A = {1, 2. 3}. Then, the subsets of A are : $$\phi$$, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3} Hence, P(A) = { $$\phi$$, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3} } Example 2 : Show that n{P[P.(P($$\phi$$))]} = 4. Solution : We have, $$P(\phi)$$ = {$$\phi$$} $$\therefore$$ $$P(P(\phi))$$ = {$$\phi$$, {$$\phi$$}} $$\implies$$ $$P[P.(P(\phi))]$$ = {$$\phi$$, {$$\phi$$}, {{$$\phi$$}}, {$$\phi$$. {$$\phi$$}}} Hence, $$P[P.(P(\phi))]$$ consists of 4 elements i.e n{P[P.(P($$\phi$$))]} = 4 We know that a set having n elements has $$2^n$$ subsets. Therefore, if A is a finite set having n elements, then P(A) has $$2^n$$ elements. Example 3 : If A = {a, {b}}, find P(A). Solution : Let B = {b}. Then, A = {a, B}. $$\therefore$$ P(A) = {$$\phi$$, {a}, {B}, {a, B}} P(A) = {$$\phi$$, {a}, {{b}}, {a, {b}}}.
# Problems on Calculating Time Here we will learn to solve different types of problems on calculating time. We know, the formula to find out time = distance/speed Word problems on calculating time: 1. A car travel 60 km in 30 minutes. In how much time will it cover 100 km? Solution: Using the unitary method; Time taken to cover 60 km = 90 minutes Time taken to cover 1 km = 90/60 minutes Time taken to cover 100 km = 90/60 × 100 = 150 minutes Formula of speed = distance/time = 60 km/(3/2) hr [given 1 hour 30 min = 1 30/60 = 1 ½ hours = 3/2 hours] = 60/1 × 2/3 km/hr = 40 km/hr Now, using the formula of time = distance/speed = 100 km/40 km/hr = 5/2 hours = 5/2 × 60 minutes, (Since 1 hour = 60 minutes) = 150 minutes 2. Victor covers 210 km by car at a speed of 70 km/hr. find the time taken to cover this distance. Solution: Using the unitary method; 70 km is covered in 1 hour. 1 km is covered in 1/70 hours. 210 km is covered in 1/70 × 210 hours = 3 hours Given: speed = 70 km/hr, distance covered = 210 km Time taken = Distance/ Speed = 210/70 hours = 3 hours 3. A train covers a distance of 36 km in 15 minutes. Find the time taken by it to cover the same distance if its speed is decreased by 9 km/hr. Solution: Distance covered by train = 36 km Time taken = 15 minutes = 15/60 hr = ¼ hr. Therefore speed of train = Distance covered/time taken = 36/(1/4) km/hr = 36/1 × 4/1 = 144 km/hr Reduced speed = 144 – 9 = 135 Therefore, required time = distance covered/speed = 36/135 × 60 minutes = 16 minutes 4. A man is walking at a speed of 6 km per hour. After every km, he takes rest for 2 minutes. How much time will it take to cover a distance of 4 km? Solution: Rest time = Number of rests × time of each rest = 3 × 2 minutes = 6 minutes Total time to cover 4 km = 4/6 × 60 + 6 minutes = (40 + 6) minutes Speed of Train Relationship between Speed, Distance and Time Conversion of Units of Speed Problems on Calculating Speed Problems on Calculating Distance Problems on Calculating Time Two Objects Move in Same Direction Two Objects Move in Opposite Direction Train Passes a Moving Object in the Opposite Direction Train Passes through a Pole Train Passes through a Bridge Two Trains Passes in the Same Direction Two Trains Passes in the Opposite Direction Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Fraction in Lowest Terms \|Reducing Fractions\|Fraction in Simplest Form Feb 28, 24 04:07 PM There are two methods to reduce a given fraction to its simplest form, viz., H.C.F. Method and Prime Factorization Method. If numerator and denominator of a fraction have no common factor other than 1… 2. ### Equivalent Fractions \| Fractions \|Reduced to the Lowest Term \|Examples Feb 28, 24 01:43 PM The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with re… 3. ### Fraction as a Part of Collection \| Pictures of Fraction \| Fractional Feb 27, 24 02:43 PM How to find fraction as a part of collection? Let there be 14 rectangles forming a box or rectangle. Thus, it can be said that there is a collection of 14 rectangles, 2 rectangles in each row. If it i… 4. ### Fraction of a Whole Numbers \| Fractional Number \|Examples with Picture Feb 24, 24 04:11 PM Fraction of a whole numbers are explained here with 4 following examples. There are three shapes: (a) circle-shape (b) rectangle-shape and (c) square-shape. Each one is divided into 4 equal parts. One… 5. ### Identification of the Parts of a Fraction \| Fractional Numbers \| Parts Feb 24, 24 04:10 PM We will discuss here about the identification of the parts of a fraction. We know fraction means part of something. Fraction tells us, into how many parts a whole has been Worksheet on Conversion of Units of Speed Worksheet on Calculating Time Worksheet on Calculating Speed Worksheet on Calculating Distance Worksheet on Train Passes through a Pole Worksheet on Relative Speed Worksheet on Decimal into Percentage
# How do you integrate int x^2/sqrt(x^2+1) by trigonometric substitution? May 26, 2018 $\int \frac{{x}^{2} \cdot \mathrm{dx}}{\sqrt{{x}^{2} + 1}} = \frac{x}{2} \cdot \sqrt{{x}^{2} + 1} - \frac{1}{2} {\sinh}^{-} 1 x + C$ #### Explanation: $\int \frac{{x}^{2} \cdot \mathrm{dx}}{\sqrt{{x}^{2} + 1}}$ After using $x = \sinh y$ and $\mathrm{dx} = \cosh y \cdot \mathrm{dy}$ transforms, this integral became $\int \frac{{\left(\sinh y\right)}^{2} \cdot \cosh y \cdot \mathrm{dy}}{\sqrt{{\left(\sinh y\right)}^{2} + 1}}$ =$\int \frac{{\left(\sinh y\right)}^{2} \cdot \cosh y \cdot \mathrm{dy}}{\sqrt{{\left(\cosh y\right)}^{2}}}$ =$\int \frac{{\left(\sinh y\right)}^{2} \cdot \cosh y \cdot \mathrm{dy}}{\cosh} y$ =$\int {\left(\sinh y\right)}^{2} \cdot \mathrm{dy}$ =$\int \frac{\cosh 2 y - 1}{2} \cdot \mathrm{dy}$ =$\frac{1}{4} \sinh 2 y - \frac{y}{2} + C$ =$\frac{1}{4} \cdot 2 \sinh y \cdot \cosh y - \frac{y}{2} + C$ =$\frac{1}{2} \sinh y \cdot \cosh y - \frac{y}{2} + C$ After using $x = \sinh y$, $\cosh y = \sqrt{{x}^{2} + 1}$ and $y = {\sinh}^{-} 1 x$ inverse transforms, I found $\int \frac{{x}^{2} \cdot \mathrm{dx}}{\sqrt{{x}^{2} + 1}} = \frac{x}{2} \cdot \sqrt{{x}^{2} + 1} - \frac{1}{2} {\sinh}^{-} 1 x + C$
# How do you find the slope and intercept of y=(5-x)/10? Mar 10, 2018 See a solution process below #### Explanation: Rewrite the equation in slope intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$ Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value. $y = \frac{5 - x}{10}$ $y = \frac{5}{10} - \frac{x}{10}$ $y = \frac{1}{2} - \frac{1}{10} x$ $y = \textcolor{red}{- \frac{1}{10}} x + \textcolor{b l u e}{\frac{1}{2}}$ Therefore: • The Slope Is : $\textcolor{red}{m = - \frac{1}{10}}$ • The $y$-intercept Is : $\textcolor{b l u e}{b = \frac{1}{2}}$ or $\left(0 , \textcolor{b l u e}{\frac{1}{2}}\right)$ Mar 10, 2018 Slope $m = - 0.1$ Intercept $c = 0.5$ #### Explanation: Given: $y = \frac{5 - x}{10}$ Simplifying $y = \frac{5}{10} - \frac{x}{10}$ $y = 0.5 - 0.1 x$ Arranging in the form $y = m x + c$ $y = - 0.1 x + 0.5$ Slope $m = - 0.1$ Intercept $c = 0.5$
# What is 0.45 as a Fraction? If you’re wondering what 0.45 is as a fraction, you’re in luck. We’ve got the answer, along with a handy explanation of how to arrive at it. Checkout this video: ## Introduction When we talk about fractions, we usually think of numbers like 1/2, 3/4, or 5/8. These fractions have a numerator (the top number) that is less than the denominator (the bottom number). For example, in the fraction 3/4, the numerator is 3 and the denominator is 4. But what about a fraction like 0.45? This number doesn’t look like it has a numerator and a denominator. In fact, it looks like a decimal (a number with a decimal point). So what is 0.45 as a fraction? It turns out that 0.45 can be written as a fraction, but it is not a “proper” fraction. A proper fraction is one where the numerator is less than the denominator. In other words, it is a fraction where the top number is less than the bottom number. But 0.45 is not a proper fraction because its numerator (4) is greater than its denominator (5). 0.45 can be written as a fraction in two ways: 4/5 or 8/10. Both of these fractions are proper fractions because their numerators (4 and 8) are less than their denominators (5 and 10). 0.45 can also be written as a mixed number: 1 4/5 or 0 8/10 . A mixed number is a combination of a whole number and a proper fraction. In other words, it is a whole number plus a proper fraction. So, to answer the question “what is 0.45 as a fraction?”, we can say that it can be written as either 4/5, 8/10, or 1 4/5 . ## What is 0.45 as a Fraction? The number 0.45 can be written as a fraction in a few different ways. The easiest way to write 0.45 as a fraction is to use the decimal to fraction converter. ## Decimals and fractions When we talk about “0.45” as a fraction, we are really talking about 45 hundredths. So, to write this fraction in its simplest form, we need to divide both the numerator (45) and denominator (100) by 5: Now, let’s look at an example involving a mixed number. Suppose someone gave you 4 and a half candy bars and you wanted to divide them equally among 5 friends. You would end up with 0.9 candy bars per friend (4.5 ÷ 5 = 0.9). It can be helpful to think of fractions as parts of a whole. So, in the example above, we could think of the 4.5 candy bars as 4 candy bars plus 1/2 of a candy bar: This is why the answer is 0.9 and not just 0.4 (4 ÷ 5 = 0.8). ## Convert 0.45 to a fraction To convert a decimal to a fraction, place the decimal number over its place value. So, 0.45 would be 45 over 100. To reduce this fraction, divide both the numerator and denominator by 5 to get 9 over 20. ## 0.45 as a percentage 0.45 as a percentage is 45%. ## Conclusion To conclude, 0.45 as a fraction is 9/20.
# What is the solution of the system of equations?y = -3x + 7y = 2x – 8 What is the solution of the system of equations? y = -3x + 7 y = 2x - 8 ## This Post Has 10 Comments 1. jbug6780 says: y = -2 x = 3 Step-by-step explanation: Solve using elimination 1. Rearrange the equations to make it easier to solve y = -3x + 7 → 3x + y = 7 y = 2x - 8 → 2x - y = 8 2. Multiply the equations to have a matching coefficient 2(3x + y = 7) = 6x + 2y = 14 3(2x - y = 8) = 6x - 3y = 24 3. Subtract 6x + 2y = 14 - 6x - 3y = 24 0 + 5y = -10 4. Solve for y 5y = -10 y = -2 5. Substitute y in any equation to solve for x -2 = -3x + 7 -3x = -9 x = 3 2. deena7 says: x = 3, y = -2 Step-by-step explanation: Since y=y then, -3x +7 = 2x-8 7+8 = 3x+2x 15 = 5x x=3 substitute y = 2(3) - 8 y = -2 Hope that helped!!! k 3. tiara0 says: 1) y = -8/5x + 8 2) slope = -3 3) x = 4 4) x ≥ 22/3 5) x = 21/16 6) x = 12 Step-by-step explanation: 1) For the slope, you know you go up 8 left 5 to get from the x-intercept to your y-intercept so it’s positive 8 over -5 (rise over run). This makes your slope 8/-5 which is the same as -8/5 2) The slope is next to the x, so the slope is -3/1 which is -3. 3) Rearrange the equation to y = mx + b. Firstly, subtract 6x from both sides: -3y = -6x + 24 Then, divide both sides by -3: y = 2x - 8 To get the x-intercept, divide the y-intercept (but positive) by the slope. 8/2 = 4 4) Distribute on the terms: -53 ≥ -9x - 9 + 3x Combine like terms: -53 ≥ -6x - 9 -44 ≥ -6x Divide both sides by -6: 44/6 ≤ x (Inequality flipped because you divide by a negative) Simplify: 22/3 ≤ x Flip the inequality: x ≥ 22/3 5) Multiply both sides by 4: 3x - x/3 = 7/2 Multiply both sides by 3: 9x - x = 21/2 Combine like terms: 8x = 21/2 Divide both sides by 8: x = 21/16 or x = 1.3125 6) Combine like terms: 7x + 6 = 90 Subtract 6 from both sides: 7x = 84 Divide both sides by 7: x = 12 Hope this helps! 4. sl3olsonow8388 says: D Step-by-step explanation: Slope is 'm' in y = mx + c m is the coefficient of x So, m = -3 5. angela6844 says: (4, -5) Step-by-step explanation: we know that If a ordered pair lie on the line, then the ordered pair must satisfy the equation of the line we have $y=-3x+7$ Verify each ordered pair case 1) (4, -5) substitute the value of x and the value of y in the linear equation and then compare the results $-5=-3(4)+7$ $-5=-5$ ---> is true so The ordered pair satisfy the linear equation therefore' The ordered pair lie on the line case 2) (4, 19) substitute the value of x and the value of y in the linear equation and then compare the results $19=-3(4)+7$ $19=-5$ ---> is not true so The ordered pair not satisfy the linear equation therefore' The ordered pair not lie on the line case 3) (1,5) substitute the value of x and the value of y in the linear equation and then compare the results $5=-3(1)+7$ $5=4$ ---> is not true so The ordered pair not satisfy the linear equation therefore' The ordered pair not lie on the line case 4) (1,-19) substitute the value of x and the value of y in the linear equation and then compare the results $-19=-3(1)+7$ $-19=4$ ---> is not true so The ordered pair not satisfy the linear equation therefore' The ordered pair not lie on the line 6. saucyboyFredo says: 1/3 Step-by-step explanation: Perpendicular lines have a slope that is the opposite reciprocal of the original line slope. To find it, first flip the fraction (-3 to -1/3). Then, add/get rid of the negative sign based on if it was there in the first place (-1/3 to 1/3). Thus, the slope would be 1/3. Hope this helped 🙂 7. gracieorman4 says: d Step-by-step explanation: 8. comawhite7820 says: 1.（A）Ithink it is the option a, because b,c,d cannot be equaled to that equation 9. m4167281 says: -3 Step-by-step explanation: the coefficient of the x term is the slope in any equation that is in slope intercept form 10. jessica94866 says: The equation that represents the other equation is $y=\dfrac{1}{3}x+5$ . The solution of the system is (3,6). Step-by-step explanation: Linear equation: $y=mx+c$ , where m= slope c = y-intercept. In the first table, the y-intercept = 5 [ y-intercept = value of y at x=0. Slope for first table = $\dfrac{y_2-y_2}{x_2-x_1}=\dfrac{6-5}{3-0}=\dfrac{1}{3}$ The equation that represents the first table: $y=\dfrac{1}{3}x+5$ So, the equation that represents the other equation is $y=\dfrac{1}{3}x+5$ . Also, the solution of the system is the common point (x,y) that satisfy both equations in the system. Here, x=3 and y=6 is the common value in both tables. So, the solution of the system is (3,6).
<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. # Counting Techniques ## Number of ways to choose (and potentially arrange) k objects from a group of n objects. 0% Progress Practice Counting Techniques Progress 0% Permutations and Combinations Suppose you draw two cards from a standard deck (one after the other without replacement). 1. How many outcomes are there? 2. How many ways are there to choose an ace and then a four? 3. What is the probability that you choose an ace and then a four? #### Watch This http://www.youtube.com/watch?v=qJ7AYDmHVRE James Sousa: The Counting Principle #### Guidance In order to compute the probability of an event, you need to know the number of outcomes in the sample space and the number of outcomes in the event. Sometimes, determining the number of outcomes takes some work! Here, you will look at three techniques for counting outcomes. Technique #1: The Fundamental Counting Principle: Use this when there are multiple independent events, each with their own outcomes, and you want to know how many outcomes there are for all the events together. At the local ice cream shop, there are 5 flavors of homemade ice cream -- vanilla, chocolate, strawberry, cookie dough, and coffee. You can choose to have your ice cream in a dish or in a cone. How many possible ice cream orders are there? If you list them all out, you will see that there are 10 ice cream orders. For each of the 5 flavors, there are 2 choices for how the ice cream is served (dish or cone). 52=10\begin{align}5 \cdot 2=10\end{align} Vanilla Dish Chocolate Dish Strawberry Dish Cookie Dough Dish Coffee Dish Vanilla Cone Chocolate Cone Strawberry Cone Cookie Dough Cone Coffee Cone This idea generalizes to a principle called the Fundamental Counting Principle: Fundamental Counting Principle: For independent events A\begin{align}A\end{align} and B\begin{align}B\end{align}, if there are n\begin{align}n\end{align} outcomes in event A\begin{align}A\end{align} and m\begin{align}m\end{align} outcomes in event B\begin{align}B\end{align}, then there are nm\begin{align}n \cdot m\end{align} outcomes for events A\begin{align}A\end{align} and B\begin{align}B\end{align} together. The Fundamental Counting Principle works similarly for more than two events - multiply the number of outcomes in each event together to find the total number of outcomes. Technique #2: Permutations: Use this when you are counting the number of ways to choose and arrange a given number of objects from a set of objects. Your sister's 3rd grade class with 28 students recently had a science fair. The teacher chose 1st, 2nd, and 3rd place winners from the class. In how many ways could she have chosen the 1st, 2nd, and 3rd place winners? This is called a permutation problem because it is asking for the number of ways to choose and arrange 3 students from a set of 28 students. In this problem, event A\begin{align}A\end{align} is the teacher choosing 1st place, event B\begin{align}B\end{align} is the teacher choosing 2nd place after 1st place has been chosen, and event C\begin{align}C\end{align} is the teacher choosing 3rd place after 1st and 2nd place have been chosen. • Event A\begin{align}A\end{align} has 28 outcomes because there are 28 students in the class. The teacher has 28 choices for 1st place. • Event B\begin{align}B\end{align} has 27 outcomes because once 1st place has been chosen, there are 27 students left in the class that could get 2nd place. • Event C\begin{align}C\end{align} has 26 outcomes because once 1st place and 2nd place have been chosen, there are 26 students left in the class that could get 3rd place. By the Fundamental Counting Principle, the teacher has 282726=19656\begin{align}28 \cdot 27 \cdot 26=19656\end{align} ways in which she could choose the 3 winners. Note that: 282726=282726252423321252423321=28!25!=28!(283)! This is the idea behind the permutation formula. (Recall that the factorial symbol, !, means to multiply every whole number up to and including that whole number together. For example, 5!=54321\begin{align}5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\end{align}.) Permutation Formula: The number of ways to choose and arrange k\begin{align}k\end{align} objects from a group of n\begin{align}n\end{align} objects is: nPk=n!(nk)! Technique #3: Combinations: Use this when you are counting the number of ways to choose a certain number of objects from a set of objects (the order/arrangement of the objects doesn't matter). A teacher has a classroom of 28 students, she wants 3 of them to do a presentation, and she wants to know how many choices she has for the three students. This is called a combination problem because it is asking for the number of ways to choose 3 students from a set of 28 students. With combinations, the order doesn't matter. Choosing Bobby, Sarah, and Matt for the presentation is the same as choosing Sarah, Bobby, and Matt for the presentation. From permutations, you know that there are 19656 ways to choose and arrange 3 students from the class of 28. This calculation will be counting each group of 3 people more than once. How many times is each group of three being counted? From permutations, 3 chosen people can be arranged in 3P3\begin{align}_3P_3\end{align} ways. 3P3=3!(33)!=3!0!=3!1=321=6 ways. This means that every group of three has been counted 6 times in the 19656 calculation. To determine the number of ways the teacher could choose 3 students where the order doesn't matter, take 19656 and divide by 6. 196566=3276 The teacher has 3276 choices for the three students to make a presentation. In general, the permutation formula can be turned into the combination formula by dividing by the number of ways to arrange k\begin{align}k\end{align} objects, which is k!\begin{align}k!\end{align}. Combination Formula: The number of ways to choose k\begin{align}k\end{align} objects from a group of n\begin{align}n\end{align} objects is: nCk=nPkk!=n!k!(nk)! In the examples you will see how to use the fundamental counting principle, permutations, and combinations to help you compute probabilities. Note that whenever you can use permutations you can also use the fundamental counting principle, because the permutation formula is derived from the fundamental counting principle. Example A Suppose you are ordering a sandwich at the deli. There are 5 choices for bread, 4 choices for meat, 12 choices for vegetables, and 3 choices for a sauce. How many different sandwiches can be ordered? If you choose a sandwich at random, what's the probability that you get turkey and mayonnaise on your sandwich? Solution: In order to answer this probability question you need to know: 1. The total number of sandwiches that can be ordered. 2. The number of sandwiches that can be ordered that involve turkey and mayonnaise. In each case, you can use the fundamental counting principle to help. 1. A sandwich is made by choosing a bread, a meat, a vegetable, and a sauce. There are 5 outcomes for the event of choosing bread, 4 outcomes for the event of choosing meat, 12 outcomes for the event of choosing vegetables, and 3 outcomes for the event of choosing a sauce. The total number of sandwiches that can be ordered is: 54123=720\begin{align}5 \cdot 4 \cdot 12 \cdot 3=720\end{align} 2. A sandwich with turkey and mayonnaise is made by choosing a bread, turkey, a vegetable, and mayonnaise. There are 5 outcomes for the event of choosing bread, there is 1 outcome for the event of choosing turkey, there are 12 outcomes for the event of choosing vegetables, and there is 1 outcome for the event of choosing mayonnaise. The total number of sandwiches with turkey and mayonnaise that can be ordered is: 51121=60\begin{align}5 \cdot 1 \cdot 12 \cdot 1=60\end{align} The probability of a sandwich with turkey and mayonnaise is 60720=112\begin{align}\frac{60}{720}=\frac{1}{12}\end{align}. Example B In your class of 35 students, there are 20 girls and 15 boys. There is a class competition and 1st, 2nd, and 3rd place winners are decided. What is the probability that all of the winners are boys? Solution: In order to answer this probability question you need to know: 1. The total number of 1st, 2nd, 3rd place winners that can be chosen. 2. The number of 1st, 2nd, 3rd place winners that are all boys that can be chosen. In each case, you are dealing with permutations, because the order of the people for 1st, 2nd, and 3rd place matters. 1. There are 35 students and 3 need to be chosen and arranged into 1st, 2nd, and 3rd place. 35P3=35(353)!=35!32!=35343332313213231321=353433=39270 2. There are 15 boys and 3 need to be chosen and arranged. 15P3=15!(153)!=15!12!=15141312113211211321=151413=2730 The probability of all boy winners is 2730392707%\begin{align}\frac{2730}{39270} \approx 7\%\end{align}. Example C In your class of 35 students, there are 20 girls and 15 boys. 5 students are chosen at random for a presentation. What is the probability that the group is made of all boys? Solution: In order to answer this probability question you need to know: 1. The total number of groups that can be formed. 2. The number of groups with all boys. In each case, you are dealing with combinations, because the order of the people for the presentation doesn't matter. 1. There are 35 students in the class and 5 to be chosen. The number of ways the 5 could be chosen are: 35C5=35!5!(355)!=35!5!30!=35343332313032154321302928321=353433323154321=38955840120=324632 2. There are 15 boys in the class and 5 boys to be chosen. The number of ways the 5 could be chosen are: 15C5=15!5!(155)!=15!5!10!=151413121154321=360360120=3003 The probability of an all boy group is 30033246320.9%\begin{align}\frac{3003}{324632} \approx 0.9\%\end{align}. Concept Problem Revisited Suppose you draw two cards from a standard deck (one after the other without replacement) and record the results. 1. How many outcomes are there? 2. How many ways are there to choose an ace and then a four? 3. What is the probability that you choose an ace and then a four? a) This is an example of a permutation, because the order matters. You are choosing 2 cards from a set of 52 cards. 52P2=52!(522)!=52!50!=5051=2652 b) Choosing an ace and choosing a four are independent events. There are 4 aces and 4 fours. By the fundamental counting principle, there are 44=16\begin{align}4 \cdot 4=16\end{align} ways to choose an ace and then a four. c) The probability that you choose an ace and then a four is 1626520.6%\begin{align}\frac{16}{2652} \approx 0.6\%\end{align}. #### Vocabulary The probability of an event is the chance of the event occurring. A combination is the number of ways of choosing k\begin{align}k\end{align} objects from a total of n\begin{align}n\end{align} objects (order does not matter). The notation for combinations is nCk\begin{align}_nC_k\end{align} or \begin{align}\binom{n}{k}\end{align}. The formula is \begin{align}_nC_k=\frac{_nP_k}{k!}=\frac{n!}{k!(n-k)!}\end{align}. A permutation is the number of ways of choosing and arranging \begin{align}k\end{align} objects from a total of \begin{align}n\end{align} objects (order does matter). The notation for permutations is \begin{align}nPk\end{align}. The formula is \begin{align}_nP_k=\frac{n!}{(n-k)!}\end{align}. The fundamental counting principle states that for independent events \begin{align}A\end{align} and \begin{align}B\end{align}, if there are \begin{align}n\end{align} outcomes in event \begin{align}A\end{align} and \begin{align}m\end{align} outcomes in event \begin{align}B\end{align}, then there are \begin{align}n \cdot m\end{align} outcomes for events \begin{align}A\end{align} and \begin{align}B\end{align} together. #### Guided Practice 1. Calculate \begin{align}_{10}P_4\end{align} and \begin{align}_{10}C_4\end{align}. Interpret each calculation. 2. You are driving your friends to the beach in your car. Your car has room for 4 additional passengers besides yourself. You have 10 friends (not including yourself) going to the beach. In how many ways could the friends who will ride in your car be chosen? 3. Make up a probability question that could be solved with the calculation \begin{align}\frac{_3C_2}{_{15}C_2}\end{align}. 1. \begin{align}_{10}P_4=\frac{10!}{(10-4)!}=\frac{10!}{6!}=10 \cdot 9 \cdot 8 \cdot 7=5040\end{align}. This is the number of ways to choose and arrange four objects from a set of 10 objects. \begin{align}_{10}C_4=\frac{10!}{4!(10-4)!}=\frac{10!}{4!6!}=\frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1}=\frac{5040}{24}=210\end{align}. This is the number of ways to choose four objects from a set of 10 objects. 2. This is a combination problem, because there is no indication that the order of the friends within the car matters. \begin{align}_{10}C_4=210\end{align} (from #1). There are 210 different combinations of 4 friends that could be chosen. 3. The total number of outcomes in the sample space is \begin{align}_{15}C_2\end{align}, which is the number of ways to choose 2 objects from a set of 15. The number of outcomes in the event that you are calculating the probability of is \begin{align}_3C_2\end{align}, which is the number of ways to choose 2 objects from a set of 3. Here is one possible question: The math club has 15 members. 12 are upperclassmen and 3 are freshman. 2 members of the club need to be chosen to make a morning announcement. What is the probability that the 2 who are chosen are both freshmen? #### Practice Calculate each and interpret each calculation in words. 1. \begin{align}_8P_2\end{align} 2. \begin{align}_8P_8\end{align} 3. \begin{align}_8C_8\end{align} 4. \begin{align}_{14}C_8\end{align} 5. Will \begin{align}_nP_k\end{align} always be larger than \begin{align}_nC_k\end{align} for a given \begin{align}n,k\end{align} pair? 6. Your graphing calculator has the combination and permutation formulas built in. Push the MATH button and scroll to the right to the PRB list. You should see \begin{align}nPr\end{align} and \begin{align}nCr\end{align} as options. In order to use these: 1) On your home screen type the value for \begin{align}n\end{align}; 2) Select \begin{align}nPr\end{align} or \begin{align}nCr\end{align}; 3) Type the value for \begin{align}k\end{align} (\begin{align}r\end{align} on the calculator). Use your calculator to verify that \begin{align}_{10}C_5=252\end{align}. First, state whether each problem is a permutation or combination problem. Then, solve. 7. Suppose you need to choose a new combination locker. You have to choose 3 numbers, each different and between 0 and 40. How many choices do you have for the combination? If you choose at random, what is the probability that you choose 0, 1, 2 for your combination? 8. You just won a contest where you can choose 2 friends to go with you to a concert. You have five friends (Amy, Bobby, Jen, Whitney, and David) who are available and want to go. If you choose two friends at random, what is the probability that you choose Bobby and David? 9. There are 12 workshops at a conference and Michael has to choose 4 to attend. In how many ways can he choose the 4 to attend? 10. 10 girls and 4 boys are finalists in a contest where 1st, 2nd, and 3rd place winners will be chosen. What is the probability that all winners are boys? 11. Using the information from the previous problem, what is the probability that all winners are girls? 12. You visit 12 colleges and want to apply to 4 of them. 5 of the colleges are within 100 miles of your house. If you choose the colleges to apply to at random, what is the probability that all 4 colleges that you apply to are within 100 miles of your house? 13. For the 12 colleges you visited, you rank your top five. In how many ways could you do this? Your friend Jesse randomly tries to guess the five colleges that you choose and the order that you ranked them in. What is the probability that he guesses correctly? 14. For the special at a restaurant you can choose 3 different items from the 10 item menu. How many different combinations of meals could you get? If the waiter chooses your 3 items at random, what's the probability that you get the soup, the salad, and the pasta dish? 15. In a typical poker game, each player is dealt 5 cards. A royal flush is when the player has the 10, Jack, Queen, King, and Ace all of the same suit. What is the probability of a royal flush? Hint: How many 5 card hands are there? How many royal flushes are there? ### Vocabulary Language: English ! ! The factorial of a whole number $n$ is the product of the positive integers from 1 to $n$. The symbol "!" denotes factorial. $n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n$. combination combination Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set. decision chart decision chart A decision chart is a tree-shaped sequence of numbers used to determine the probability of occurrence of a given event. Event Event An event is a set of one or more possible results of a probability experiment. factorial factorial The factorial of a whole number $n$ is the product of the positive integers from 1 to $n$. The symbol "!" denotes factorial. $n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n$. Fundamental Counting Principle Fundamental Counting Principle The Fundamental Counting Principle states that if an event can be chosen in p different ways and another independent event can be chosen in q different ways, the number of different arrangements of the events is p x q. Independent Events Independent Events Two events are independent if the occurrence of one event does not impact the probability of the other event. Permutation Permutation A permutation is an arrangement of objects where order is important. Sample Space Sample Space In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.
## Division within 10 000 2.4 Division within 10 000 There are various methods for solving appropriate divisions operations for numbers up to 10 000 Examples of division operation questions and solutions We put 45 small stones equally into 3 jars. Calculate the number of small stones in one jar. Mathematical sentence: $$45\space\div\space3=\underline{\quad\quad}$$ METHOD $$\;\;\;15\\ 3\overline{\smash{)}45}\\ \,\underline{-3}\downarrow\\ \;\;\,15\\ \;\underline{-15}\\ \;\;\,\,\;00$$ Divide tens 4 tens$$\div$$ 3 = 1 tens remainder 1 tens Change 1 tens to 10 ones. 10 sa $$+$$ 5 sa = 15 sa Divide ones 15 ones $$\div$$ 3 = 5 ones So, $$45\space\div\space3=\underline{15}$$ ## Division within 10 000 2.4 Division within 10 000 There are various methods for solving appropriate divisions operations for numbers up to 10 000 Examples of division operation questions and solutions We put 45 small stones equally into 3 jars. Calculate the number of small stones in one jar. Mathematical sentence: $$45\space\div\space3=\underline{\quad\quad}$$ METHOD $$\;\;\;15\\ 3\overline{\smash{)}45}\\ \,\underline{-3}\downarrow\\ \;\;\,15\\ \;\underline{-15}\\ \;\;\,\,\;00$$ Divide tens 4 tens$$\div$$ 3 = 1 tens remainder 1 tens Change 1 tens to 10 ones. 10 sa $$+$$ 5 sa = 15 sa Divide ones 15 ones $$\div$$ 3 = 5 ones So, $$45\space\div\space3=\underline{15}$$
Home \| \| Maths 10th Std \| Pythagoras Theorem # Pythagoras Theorem Among all existing theorems in mathematics, Pythagoras theorem is considered to be the most important because it has maximum number of proofs. There are more than 350 ways of proving Pythagoras theorem through different methods. Pythagoras Theorem Among all existing theorems in mathematics, Pythagoras theorem is considered to be the most important because it has maximum number of proofs. There are more than 350 ways of proving Pythagoras theorem through different methods. Each of these proofs was discovered by eminent mathematicians, scholars, engineers and math enthusiasts, including one by the 20th American president James Garfield. The book titled “The Pythagorean Proposition” written by Elisha Scott Loomis, published by the National Council of Teaching of Mathematics (NCTM) in America contains 367 proofs of Pythagoras Theorem. Three numbers (a , b,c) are said to form Pythagorean Triplet, if they form sides of a right triangle. Thus (a , b,c) is a Pythagorean Triplet if and only if c2 = a2 +b2. Now we are in a position to study this most famous and important theorem not only in Geometry but in whole of mathematics. Note · In a right angled triangle, the side opposite to 90° (the right angle) is called the hypotenuse. · The other two sides are called legs of the right angled triangle. · The hypotenuse will be the longest side of the triangle. Theorem 5 : Pythagoras Theorem ### Statement In a right angle triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. ### Proof Given : In ΔABC, A = 90° To prove: AB 2 + AC 2 = BC 2 Adding (1) and (2) we get AB 2+ AC 2= BC × BD + BC ×DC = BC (BD + DC) = BC ×BC AB 2 + AC 2= BC 2 Hence the theorem is proved. ## Converse of Pythagoras Theorem ### Statement If the square of the longest side of a triangle is equal to sums of squares of other two sides, then the triangle is a right angle triangle. ### Example 4.20 An insect 8 m away initially from the foot of a lamp post which is 6 m tall, crawls towards it moving through a distance. If its distance from the top of the lamp post is equal to the distance it has moved, how far is the insect away from the foot of the lamp post? ### Solution Distance between the insect and the foot of the lamp post BD = 8 m The height of the lamp post, AB = 6 m After moving a distance of x m, let the insect be at C Let, AC = CD = x . Then BC = BDCD = 8 − x In ΔABC , B = 90° AC 2 = AB2 + BC 2 gives x 2 = 62 + (8 x)2 x2 = 36 + 64 16x + x2 16x = 100 then x = 6. 25 Then, BC = 8 x = 8 6.25 = 1. 75 m Therefore the insect is 1.75 m away from the foot of the lamp post. ### Example 4.21 P and Q are the mid-points of the sides CA and CB respectively of a ΔABC, right angled at C. Prove that 4(AQ2 + BP2 ) = 5AB2 . ### Solution Since, ΔAQC is a right triangle at C, AQ2 = AC 2+QC 2 ………(1) Also, ΔBPC is a right triangle at C, BP2 = BC2+CP2 …….(2) From (1) and (2), AQ2 + BP2 = AC 2 + QC 2 + BC 2 +CP2 4(AQ2 + BP2 ) = 4AC 2 + 4QC 2 + 4BC 2 + 4CP2 = 4AC 2 + (2QC)2 + 4BC 2 + (2CP)2 = 4AC 2 + BC 2 + 4BC 2 + AC 2 (Since P and Q are mid points) = 5(AC 2 + BC 2 ) 4(AQ2 + BP2 ) = 5AB2 (By Pythagoras Theorem) ### Example 4.22 What length of ladder is needed to reach a height of 7 ft along the wall when the base of the ladder is 4 ft from the wall? Round off your answer to the next tenth place. ### Solution Let x be the length of the ladder. BC= 4 ft, AC= 7 ft. By Pythagoras theorem we have, AB2 = AC2 + BC2 x2 = 7 2 + 42 gives x2 = 49 + 16 x2 = 65 . Hence, x = √65 The number √65 is between 8 and 8.1. 82 = 64 < 65 < 65.61 = 8.12 Therefore, the length of the ladder is approximately 8.1 ft. ### Example 4.23 An Aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1½ hours? ### Solution Let the first aeroplane starts from O and goes upto A towards north, (Distance=Speed×time) where OA= ( 1000 × 3/2 )km =1500 km Let the second aeroplane starts from O at the same time and goes upto B towards west, where OB= ( 1200 × 3/2 )km = 1800 km The required distance to be found is BA. In right angled tirangle AOB, AB2 = OA2 +OB2 AB2 = (1500)2 + (1800)2 = 1002 (152+182) = 1002 × 549 = 1002 × 9 × 61 AB = 100 × 3 × √61 = 300√(61) kms. Tags : Statement, Proof, Solved Example Problems \| Geometry , 10th Mathematics : UNIT 4 : Geometry Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail 10th Mathematics : UNIT 4 : Geometry : Pythagoras Theorem \| Statement, Proof, Solved Example Problems \| Geometry
# How to Find Friction Force ## What is Friction Definition: Frictional force is a force that opposes the motion between two surfaces in contact with each other. The coefficient of static friction, μ is defined as the ratio of the limiting frictional force (F) to the normal reaction (R). The frictional force is measured in Newtons. The formula for frictional force is F=μR. Where F is the force of friction, μ is the coefficient of friction, and R is the normal reaction. You may also like to read: How to Calculate Normal Force ## Frictional Force Formula The frictional force formula is written as follows: Coefficient of static friction (μ) = Frictional force (F) / Normal reaction (R) We can rewrite the above formula coefficient of friction as μ = F / R And the formula for fractional force F=μR The normal reaction is equal to the weight, W of the block. That is to say R = W = mg And from the above equation, m is the mass of the object in kilogram, and g is the acceleration due to gravity. The si unit of frictional force is in Newtons. Remember that a normal reaction is equal to the weight of the object and they both have the same unit which is also in Newtons. Additionally, the coefficient of friction has no unit. We need to also know that for workdone by friction is W = F x h = μR x h = μRh Where h is the height. You may also like to read: How to Calculate Net Force ## How to Find Friction Force To find the friction force between two surfaces, we need to follow these steps: 1. Identify the Surfaces: Determine the two surfaces in contact and the nature of their interaction, whether it’s sliding or static friction. 2. Determine the Normal Force (N): Find the force exerted by a surface perpendicular to the object. This is often the weight of the object if it’s on a flat surface. (N = m ⋅ g), where (m) is the mass and (g) is the acceleration due to gravity. 3. Determine the Coefficient of Friction (μ): The coefficient of friction (μ) is a dimensionless value representing the frictional interaction between the surfaces. It can be provided in the problem or obtained from reference materials. 4. Calculate Frictional Force (Sliding): For sliding friction, use the formula (Ffriction = μ ⋅ N), where (Ffriction) is the frictional force. 5. Calculate Frictional Force (Static): For static friction, the maximum static frictional force (Fstatic-max) can be calculated using (Fstatic-max = μstatic-max ⋅ N). If a force is applied and it’s less than (Fstatic-max), static friction equals the applied force. If the applied force exceeds (Fstatic-max), the object starts moving, and kinetic friction applies. These steps provide a systematic approach to calculating frictional force, considering factors like the nature of the motion (sliding or static), the normal force, and the coefficient of friction. ### Explanation An object may be pulled horizontally or along an inclined plane. Depending on the situation and the question you are solving, we have different formulas to apply in order to find friction force. For example, when an object slides down along an inclined plane, we can easily find the coefficient friction as: From the first diagram in the video above, as the object slides vertically downward, the following conditions will arise: 1. Along the x-axis which is perpendicular to the plane: cosθ = x/W, we can now make x subject of the formula to obtain: x = W x cosθ 2. On the y-axis which is parallel to the plane: sinθ = y/W. We make y subject of the formula to obtain: y = W x sinθ For us to be able to find the coefficient of friction (μ), we will need to slowly increase the angle of inclination (θ) so that the object will slide down the plane. As the object begins to slide down the plane, the weight pulling it down (wsinθ) will be equal to the frictional force (F=μR) acting upward along the plane. Therefore, we can now say that: F = wsinθ But since F = μR, the above equation will now become: μR = wsinθ Additionally, we can see from the figure that the normal component R is equal to the perpendicular component of the weight (wcosθ). Therefore, R = wcosθ From our formula for the force of friction which is μ = F / R. And by looking at F = wsinθ, and R = wcosθ. We can conclude that μ = F / R = wsinθ / wcosθ Therefore, we can also write the formula for coefficient of friction, μ = wsinθ / wcosθ Additionally, the other formula for for coefficient of friction, μ = tanθ = y/x = wsinθ / wcosθ We also refer to the angle of inclination (θ) as angle of friction. You may also like to read: How to Find Acceleration ## Solved Problems: How to Calculate Frictional Force Here are a few problems to help you understand how to calculate frictional force: ### Problem 1 A wooden block of mass 1.6 kilograms rests on a rough horizontal surface. If the limiting frictional force between the block and the surface is 8 Newtons. Calculate the coefficient of friction (g = 10 m/s2). Data: The information from the question Frictional force, F = 8 N Mass of the wood, m = 1.6 kg Acceleration due to gravity, g = 10 m/s2 Unknown: What we need to find Coefficient of friction, μ = ? Formula: The equations that will help us to solve the problem First step: Find the normal reaction by using the formula (R = W = mg) Second step: Find the coefficient of static friction, μ = F / R ##### Solution Normal reaction, R = W = mg = 1.6 x 10 = 16 N To find the coefficient of friction, we use the formula μ = F / R = 8 / 16 = 0.5 ### Problem 2 A concrete block of mass 25 kg is placed on a wooden plank inclined at an angle of 320 to the horizontal. Calculate the force parallel to the inclined plane that will keep the block at rest if the coefficient of friction between the block and the plank is 0.45. Data: The information from the question Mass of the body, m = 25 kg Acceleration due to gravity, g = 10 m/s2 Normal reaction, R = W = mg = 25 x 10 = 250 N Angle of inclination, θ = 320 μ = 0.45 ##### Solution To calculate the force parallel to the inclined plane, we will need to find Wsinθ and F=μR=μWcosθ. We will then subtract F=μR=μWcosθ from Wsinθ. We can write the above formula mathematically as follows: Fp = Wsinθ – μWcosθ = (250 x sin320) – (0.45 x 250 x cos320) = 132.48 – 95.41 = 37.07 N ### Problem 3 Calculate the magnitude of the force required to just move a 20 kg object along a horizontal surface if the coefficient of friction is 0.2. Data: Information from the question Coefficient of friction, μ = 0.2 Mass of the object, m = 20 kg Acceleration due to gravity, g = 10 m/s2 Unknown: What we need to find Normal reaction, R = ? Frictional force, F = ? Formula: Normal Reaction, R = mg Friction Force, F = μR ##### Solution Normal Reaction, R = Weight, W = mg = 20 x 10 = 200N The friction force, F = μR = 0.2 x 200 = 40N Therefore, the magnitude of the force of the object along the horizontal surface is 40 Newtons. ### Problem 4 If an object just begins to slide on a surface inclined at 300 to the horizontal. What is the coefficient of static friction? Data: Information from the above question The angle of inclination, θ = 300 Unknown: What we need to find Coefficient of static friction, μ = ? Formula: The equation that will help us solve the problem μ = tanθ ##### Solution μ = tanθ = tan300 = 1/(√3) Therefore, the coefficient of static friction is 1/√3 Note: The coefficient of friction for an inclined plane is the tangent of the angle of the plane when an object on the plane just begins to slide You may also like to read: How to Find Time in Physics What is Time in Physics? How to Calculate Average Speed in Physics Skylinecollege .

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