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Factor a Quadratic Equation by Grouping Terms - dummies # Factor a Quadratic Equation by Grouping Terms Factoring by grouping terms is a great method to use to rewrite a quadratic equation so that you can use the multiplication property of zero and find all the solutions. The main idea behind factoring by grouping is to arrange the terms into smaller groupings that have a common factor. You go to little groupings because you can’t find a greatest common factor for all the terms; however, by taking two terms at a time, you can find something to divide them by. For example, look at the quadratic equation 2x2 + 8x – 5x – 20 = 0, which has four terms. (Although you could combine the two middle terms on the left, in this case, leave them as is for the sake of the grouping process.) The four terms in the equation don’t share a greatest common factor. You can divide the first, second, and fourth terms evenly by 2, but the third term doesn’t comply. The first three terms all have a factor of x, but the last term doesn’t. So, you group the first two terms together and take out their common factor, 2x. The last two terms have a common factor of –5. The factored form, therefore, is 2x(x + 4) – 5(x + 4) = 0. The new, factored form has two terms. Each of the terms has an (x + 4) factor, so you can divide that factor out of each term. When you divide the first term, you have 2x left. When you divide the second term, you have –5 left. Your new factored form is (x + 4)(2x – 5) = 0. Now you can set each factor equal to zero to get Keep in mind that factoring by grouping works only when you can create a new form of the quadratic equation that has fewer terms and a common factor. If the factor (x + 4) hadn’t shown up in both of the factored terms in this example, you would’ve gone in a different direction. Solving quadratic equations by grouping and factoring is even more important when the exponents in the equations get larger. For example, the equation 5x3 + x2 – 45x – 9 = 0 is a third-degree equation (the highest power on any of the variables is 3), so it has the potential for three different solutions. You can’t find a factor common to all four terms, so you group the first two terms, factor out x2, group the last two terms, and factor out –9. The factored equation is x2(5x + 1) – 9(5x + 1) = 0 The common factor of the two terms in the new equation is (5x + 1), so you divide it out of the two terms to get (5x + 1)(x2 – 9) = 0 The second factor is the difference of squares, so you can rewrite the equation as (5x + 1)(x – 3)(x + 3) = 0 The three solutions are
Courses Courses for Kids Free study material Offline Centres More Store What is ${\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{4}){\text{ + ta}}{{\text{n}}^{ - 1}}(\dfrac{3}{5})$ equal to?A. 0B. $\dfrac{\pi }{4}$ C. $\dfrac{\pi }{3}$ D. $\dfrac{\pi }{2}$ Last updated date: 13th Jul 2024 Total views: 453k Views today: 4.53k Verified 453k+ views Hint:- The inverse trigonometry formulas of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$ can be used. Given, ${\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{4}){\text{ + ta}}{{\text{n}}^{ - 1}}(\dfrac{3}{5})$ =? -(1) We know that , the inverse trigonometry formula of addition of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$ is ${\text{ta}}{{\text{n}}^{ - 1}}{\text{x + ta}}{{\text{n}}^{ - 1}}{\text{y = ta}}{{\text{n}}^{ - 1}}(\dfrac{{{\text{x + y}}}}{{1 - {\text{xy}}}})$ , xy<1 -(2) Comparing the equation (1) with the equation (2) we get, X = $\dfrac{1}{4}$ and y =$\dfrac{3}{5}$ . We need to check whether xy<1 for applying the formula of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$ $\Rightarrow {\text{xy = }}\left( {\dfrac{1}{4}} \right)\left( {\dfrac{3}{5}} \right){\text{ = }}\left( {\dfrac{3}{{20}}} \right) \\ \\$ And, $\Rightarrow \dfrac{3}{{20}} < 1 \\ \Rightarrow {\text{xy < 1}} \\$ So, the formula is applicable for a given set of x and y. Putting the value of x and y in equation (2). We get, ${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}(\dfrac{{\dfrac{1}{4}{\text{ + }}\dfrac{3}{5}}}{{1 - \left( {\dfrac{1}{4}} \right)\left( {\dfrac{3}{5}} \right)}})$ Solving right hand side , we get ${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}(\dfrac{{\dfrac{{17}}{{20}}}}{{1 - \left( {\dfrac{3}{{20}}} \right)}})$ ${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\dfrac{{17}}{{20}}}}{{\dfrac{{17}}{{20}}}}} \right)$ ${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}1$ ${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = }}\dfrac{\pi }{4}$ Hence the value of ${\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{4}){\text{ + ta}}{{\text{n}}^{ - 1}}(\dfrac{3}{5})$ is $\dfrac{\pi }{4}$. The answer is option B. Note:- The domain and the range of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$ is R and $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ respectively. And ${\text{ta}}{{\text{n}}^{ - 1}}{\text{x + ta}}{{\text{n}}^{ - 1}}{\text{y = ta}}{{\text{n}}^{ - 1}}(\dfrac{{{\text{x + y}}}}{{1 - {\text{xy}}}})$ is applicable only when xy <1.
# Calculating the Perimeter of a Rectangle Whether you are calculating the perimeter of a rectangle, or you are interested in the area of a rectangle, you will need to know how to calculate the perimeter. The perimeter is the distance of a rectangle from the edge of the rectangle. If you know the length of the side, the perimeter of the rectangle is 50 inches. ## Calculating the perimeter of a rectangle Using a perimeter calculator to calculate the perimeter of a rectangle is a useful technique. It can be done by measuring each side and using the formula below. First, determine the length of the rectangle. If you’re working off a plan, you may already have the measurements. If not, use the following formula. Secondly, determine the width of the rectangle. The width of a rectangle is congruent with the length on the opposite side. You’ll need to record the measurements in the same units as the length. If you want a longer side, the length will need to be greater than the width. The perimeter of a rectangle is a measure of the distance the rectangle covers around the outside. It is measured in linear units, such as feet and inches. The rectangle’s perimeter can be calculated with the formula: perimeter = 2(length + width). Lastly, calculate the area of a rectangle. The area of a rectangle is the sum of the squared length and width of all the squares in the rectangle. ## Finding the area of a rectangle Whether you are interested in calculating the area of a rectangle or finding the perimeter of a rectangle, you will need to know the units. These include linear units and geometric units. The area of a rectangle is equal to the length times the width. The perimeter of a rectangle is the total area covered by all four sides of the rectangle. The perimeter is a linear measure and is expressed in units. Area and perimeter of a rectangle can be calculated using several equations. In order to calculate the perimeter of a rectangle, you will need to know the length and width of one of the sides. You can also use the perimeter formula to calculate the length of the side. The perimeter of a rectangle is a linear measure of the total distance covered by all four sides of a rectangle. The perimeter is equal to the area of a rectangle and can be calculated using a formula. ## Calculating the perimeter of a rectangular patio Using a calculator, you can calculate the perimeter of a rectangle. The perimeter of a rectangle is the total length of all four sides of the rectangle. The formula for calculating the perimeter of a rectangle is P = L + W x 2. The perimeter of a rectangle can be measured in linear units such as feet, inches, and metres. To calculate the perimeter of a rectangle, you need to know the height and width of the rectangle. If you do not know these two numbers, you can use a length conversion calculator. If you are looking for a more accurate measurement of the area of a rectangle, you might need to measure it with a ruler. You can also measure the length of a rectangle and multiply it by the area. You will need to determine the best measure for your situation. This is a complex task that requires you to know what you are looking for. ## Calculating the perimeter of a rectangular garden Whether you’re measuring the perimeter of a rectangular garden or calculating the area of a rectangle, there are a few basic formulas you can use. For example, you can find the perimeter of a rectangle by adding the length of all four sides. If you’re working off a schematic or plan, you might already know the length and width of each side. Alternatively, you can use the perimeter of rectangle calculator to find the length and width of the perimeter. To calculate the perimeter of a rectangle, start by adding the length and width of all the sides. You may need to measure each side with a ruler. If you don’t have a ruler, you can use a length converter to convert from inches to meters. If you want to calculate the perimeter of a rectangular garden, you can start by adding the length and width of all four sides. You can then multiply each side by four.
# 9. Fourier Series and Fourier Transforms The Fourier transform is one of the most important mathematical tools used for analyzing functions. Given an arbitrary function $f(x)$, with a real domain ($x \in \mathbb{R}$), we can express it as a linear combination of complex waves. The coefficients of the linear combination form a complex counterpart function, $F(k)$, defined in a wave-number domain ($k \in \mathbb{R}$). It turns out that $F$ is often much easier to deal with than $f$; in particular, differential equations for $f$ can often be reduced to algebraic equations for $F$, which are much easier to solve. ## Fourier series We begin by discussing the Fourier series, which is used to analyze functions that are periodic in their inputs. A periodic function $f(x)$ is a function of a real variable $x$ that repeats itself every time $x$ changes by $a$, as shown in the figure below: The constant $a$ is called the period. Mathematically, we write the periodicity condition as $$f(x+a) = f(x), \quad\forall\;\, x\in \mathbb{R}.$$ The value of $f(x)$ can be real or complex, but $x$ should be real. You can think of $x$ as representing a spatial coordinate. (The following discussion also applies to functions of time, though with minor differences in convention that we'll discuss later.) We can also think of the periodic function as being defined over a finite segment $-a/2 \le x < a/2$, with periodic boundary conditions $f(-a/2) = f(a/2)$. In spatial terms, this is like wrapping the segment into a loop: Let's consider what it means to specify a periodic function $f(x)$. One way to specify the function is to give an explicit mathematical formula for it. Another approach might be to specify the function values in $-a/2 \le x < a/2$. Since there's an uncountably infinite number of points in this domain, we can generally only achieve an approximate specification of $f$ this way, by giving the values of $f$ at a large but finite set $x$ points. There is another interesting approach to specifying $f$. We can express it as a linear combination of simpler periodic functions, consisting of sines and cosines: $$f(x) = \sum_{n=1}^\infty \alpha_n \sin\left(\frac{2\pi n x}{a}\right) + \sum_{m=0}^\infty \beta_m \cos\left(\frac{2 \pi m x}{a}\right).$$ This is called a Fourier series. Given the set of numbers $\{\alpha_n, \beta_m\}$, which are called the Fourier coefficients, $f(x)$ can be calculated for any $x$. Note that the Fourier coefficients are real if $f(x)$ is a real function, or complex if $f(x)$ is complex. The justification for the Fourier series formula is that the sine and cosine functions in the series are, themselves, periodic with period $a$: \begin{align}\sin\left(\frac{2\pi n (x+a)}{a}\right) = \sin\left(\frac{2\pi n x}{a} + 2\pi n\right) &= \sin\left(\frac{2\pi n x}{a}\right)\\ \cos\left(\frac{2\pi m (x+a)}{a}\right) = \cos\left(\frac{2\pi m x}{a} + 2\pi m\right) &= \cos\left(\frac{2\pi m x}{a}\right).\end{align} Hence, any linear combination of them automatically satisfies the periodicity condition for $f$. (Note that in the Fourier series formula, the $n$ index does not include 0. Since the sine term with $n = 0$ vanishes for all $x$, it's redundant.) ### Square-integrable functions The Fourier series is a nice concept, but can arbitrary periodic functions always be expressed as a Fourier series? This question turns out to be surprisingly intricate, and its resolution preoccupied mathematicians for much of the 19th century. The full discussion is beyond the scope of this course. Luckily, it turns out that a certain class of periodic functions, which are commonly encountered in physical contexts, are guaranteed to always be expressible as Fourier series. These are square-integrable functions, for which $$\int_{-a/2}^{a/2} dx\; \big\|\,f(x)\,\big\|^2\;\;\text{exists and is finite}.$$ Unless otherwise stated, we will always assume that the functions we're dealing with are square-integrable. ### Complex Fourier series and inverse relations Using Euler's formula, we can re-write the Fourier series as follows: $$f(x) = \sum_{n=-\infty}^\infty e^{2\pi i n x/a}\, f_n.$$ Instead of separate sums over sines and cosines, we have a single sum over complex exponentials, which is neater. The sum includes negative integers $n$, and involves a new set of Fourier coefficients, $f_n$, which are complex numbers. (As an exercise, try working out how the old coefficients $\{\alpha_n, \beta_n\}$ are related to the new coefficients $\{f_n\}$.) If the Fourier coefficients $\{f_n\}$ are known, then $f(x)$ can be calculated using the above formula. The converse is also true: given $f(x)$, we can determine the Fourier coefficients. To see how, observe that $$\int_{-a/2}^{a/2} dx \; e^{-2\pi i m x/a}\, e^{2\pi i n x/a} = a\, \delta_{mn}\quad \mathrm{for}\;m, n \in \mathbb{Z},$$ where $\delta_{mn}$ is the Kronecker delta, defined as: $$\delta_{mn} = \left\{\begin{array}{ll}1, & \textrm{if}\; m = n\\ 0, & \mathrm{if}\;m\ne n.\end{array}\right.$$ Due to this property, the set of functions $\exp(2\pi i n x / a)$, with integer values of $n$, are said to be orthogonal functions. (We won't go into the details now, but the term "orthogonality" is used here with the same meaning as in vector algebra, where a set of vectors $\vec{v}_1, \vec{v}_2, \dots$ is said to be "orthogonal" if $\vec{v}_m \cdot \vec{v}_n = 0$ for $m\ne n$.) Hence, \begin{align}\int_{-a/2}^{\,a/2} dx\; e^{-2\pi i m x/a} \;f(x) &= \, \int_{-a/2}^{\,a/2} dx\; e^{-2\pi i m x/a} \left[\sum_{n=-\infty}^\infty e^{2\pi i n x/a}\, f_n\right] \\ &= \sum_{n=-\infty}^\infty \, \int_{-a/2}^{\,a/2} dx\; e^{-2\pi i m x/a} \, e^{2\pi i n x/a} \;f_n \\ &= \sum_{n=-\infty}^\infty \, a\, \delta_{mn} \, f_n \\ &= a \,f_m.\end{align} The procedure of multiplying by $\exp(-2\pi i m x/a)$ and integrating over $x$ acts like a sieve, filtering out all other Fourier components of $f(x)$ and keeping only the one with the matching index $m$. Thus, we arrive at a pair of relations expressing $f(x)$ in terms of its Fourier components, and vice versa: $$\left\{\;\;\begin{array}{rl}f(x) &= \displaystyle \, \sum_{n=-\infty}^\infty e^{i k_n x}\, f_n \\ f_n &= \displaystyle\,\frac{1}{a} \int_{-a/2}^{\,a/2} dx\; e^{-i k_n x}\, f(x)\end{array}\;\;\right. \quad\quad\mathrm{where}\;\; k_n \equiv \frac{2\pi n}{a}$$ The real numbers $k_n$ are called wave-numbers. They form a discrete set, with one for each Fourier component. In physics jargon, we say that the wave-numbers are "quantized" to integer multiples of $\Delta k \equiv 2\pi/a.$ ### Example: Fourier series of a square wave To get a feel for how the Fourier series behaves, let's look at a square wave: a function that takes only two values $+1$ or $-1$, jumping between the two values at periodic intervals. Within one period, the function is $$f(x) = \left\{\begin{array}{ll}-1, & -a/2 \le x < 0 \\ +1, & \quad\;\;\; 0 \le x < a/2.\end{array}\right.$$ Plugging this into the Fourier relation, and doing the straightforward integrals, gives the Fourier coefficients \begin{aligned} f_n &= -i \, \frac{\left[\sin\left(n \pi/2\right)\right]^2}{n\pi/2 } \\ &= \left\{\begin{array}{cl} -2i/n\pi ,& n \; \mathrm{odd} \\ 0,& n \; \mathrm{even}. \end{array}\right.\end{aligned} As can be seen, the Fourier coefficients become small for large $n$. We can write the Fourier series as $$f(x) \; \leftrightarrow \; \sum_{n=1,3,5,\dots} \frac{4\sin(2\pi n x / a)}{n \pi}.$$ If this infinite series is truncated to a finite number of terms, we get an approximation to $f(x)$. As shown in the figure below, the approximation becomes better and better as more terms are included. In [9]: ## Demo: compare a square wave to its Fourier series approximant %matplotlib inline from ipywidgets import interact, IntSlider, FloatSlider from numpy import linspace, sin, pi, zeros, array import matplotlib.pyplot as plt def plot_square_wave_series(N, a): col0, col1, col2 = "grey", "darkorange", "mediumblue" plt.figure(figsize=(10,5)) ## Plot several periods of the square wave plt.plot([], [], linewidth=3, color=col1, label="Square wave") for n in range(-4, 5): x = array([-0.5a, -0.5a, 0.0, 0.0, 0.5a, 0.5a]) y = array([0.0, -1.0, -1.0, 1.0, 1.0, 0.0]) plt.plot(x-na, y, linewidth=2, color=col1) ## Plot the Fourier series approximant. x = linspace(-2, 2, 600) f = zeros(len(x)) for n in range(1,N+1,2): f += (4.0/n/pi) sin(2pinx/a) plt.plot(x, f, linewidth=1, color=col2, label="Fourier series") plt.axis([x[0], x[-1], -2, 2]) plt.xlabel('x'); plt.ylabel('f(x)') plt.title('Fourier series of square wave, to order {}'.format(N)) plt.legend(loc="upper right") plt.show() interact(plot_square_wave_series, N = IntSlider(min=1, max=25, step=2, value=1, description='n_max'), a = FloatSlider(min=0.5, max=3.0, step=0.1, value=1.0, description='a')); var element = $('#95501c64-efd4-4956-845b-8c2b5b4a28db'); {"model_id": "9cd12ed17b7044d5822bab83bef1e0eb", "version_major": 2, "version_minor": 0} One amusing consequence of the above result is that we can use it to derive a series expansion for$\pi$. If we set$x = a/4$, $$f(a/4) = 1 = \frac{4}{\pi} \left[\sin(\pi/2) + \frac{1}{3}\sin(3\pi/2) + \frac{1}{5}\sin(5\pi/2) + \cdots\right],$$ and hence $$\pi = 4 \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right).$$ Let us use Python to check if the above formula really works: In [11]: from numpy import arange n = arange(10000) # n = [0, 1, 2, ... 9999] x = 4.0 sum((-1)*n / (2n + 1)) print(x) # Actually, the series isn't very useful as it converges veeeeery slowly... 3.1414926535900345 ## Fourier transforms The Fourier series applies to periodic functions defined over the interval$-a/2 \le x < a/2$. But the concept can be generalized to functions defined over the entire real line,$x \in \mathbb{R}$, if we take the limit$a \rightarrow \infty$carefully. Suppose we have a function$f$defined over the entire real line,$x \in \mathbb{R}$, such that$f(x) \rightarrow 0$for$x \rightarrow \pm\infty$. Imagine there is a family of periodic functions$\big\{f_a(x) \,\big\|\, a \in\mathbb{R}^+\big\}$, such that$f_a(x)$has periodicity$a$, and approaches$f(x)$in the limit$a\rightarrow \infty$. This is illustrated in the figure below: In mathematical terms, $$f(x) = \lim_{a \rightarrow \infty} f_a(x), \;\;\;\text{where}\;\; f_a(x+a) = f_a(x).$$ Since$f_a$is periodic, it can be expanded as a Fourier series: $$f_a(x) = \sum_{n=-\infty}^\infty e^{i k_n x}\, f_{an}, \quad\mathrm{where}\;\; k_n = n\Delta k, \;\; \Delta k = \frac{2\pi}{a}.$$ Here,$f_{an}$denotes the$n$-th complex Fourier coefficient of the function$f_a(x)$. Note that each Fourier coefficient depends implicitly on the periodicity$a$. As$a \rightarrow \infty$, the wave-number quantum$\Delta k$goes to zero, and the set of discrete$k_n$turns into a continuum. During this process, each individual Fourier coefficient$f_{an}$goes to zero, because there are more and more Fourier components in the vicinity of each$k$value, and each Fourier component contributes less. This implies that we can replace the discrete sum with an integral. To accomplish this, we first multiply the summand by a factor of$(\Delta k/2\pi) / (\Delta k/2\pi) = 1$: $$f(x) = \lim_{a\rightarrow \infty} \left[\;\,\sum_{n=-\infty}^\infty \frac{\Delta k}{2\pi} \, e^{i k_n x}\, \left(\frac{2\pi \,f_{an}}{\Delta k} \right)\;\,\right].$$ (In case you're wondering, the choice of$2\pi$factors is essentially arbitrary; we are following the usual convention.) Moreover, we define $$F(k) \equiv \lim_{a \rightarrow \infty} \left[\frac{2\pi\, f_{an}}{\Delta k}\right]_{k = k_n}.$$ In the$a \rightarrow \infty$limit, the$f_{an}$in the numerator and the$\Delta k$in the denominator both go zero, but if their ratio remains finite, we can turn the Fourier sum into the following integral: $$f(x) = \int_{-\infty}^{\infty} \frac{dk}{2\pi} \, e^{i k x}\, F(k).$$ ### The Fourier relations The function$F(k)$is called the Fourier transform of$f(x)$. Just as we have expressed$f(x)$in terms of$F(k)$, we can also express$F(k)$in terms of$f(x)$. To do this, we apply the$a \rightarrow \inftylimit to the inverse relation for the Fourier series: \begin{align}F(k_n) &= \lim_{a\rightarrow \infty} \frac{2 \pi\, f_{an}}{\Delta k} \\ &= \lim_{a\rightarrow \infty} \frac{2 \pi}{2\pi/a}\, \left(\frac{1}{a} \int_{-a/2}^{a/2} dx\; e^{-i k_n x}\right) \\ &= \int_{-\infty}^\infty dx\; e^{-i kx}\, f(x).\end{align} Hence, we arrive at a pair of equations called the Fourier relations: \left\{\;\;\begin{align}F(k) &= \;\int_{-\infty}^\infty dx\; e^{-ikx}\, f(x) &&\text{(Fourier transform)}\\ f(x) &= \int_{-\infty}^\infty \frac{dk}{2\pi}\; e^{ikx}\, F(k)&&\text{(Inverse Fourier transform).}\end{align}\;\;\right. The first equation is the Fourier transform, and the second equation is called the inverse Fourier transform. There are notable differences between the two formulas. First, there is a factor of1/2\pi$appears next to$dk$, but no such factor for$dx$; this is a matter of convention, tied to our earlier definition of$F(k)$. Second, the integral over$x$contains a factor of$e^{-ikx}$but the integral over$k$contains a factor of$e^{ikx}$. One way to remember which equation has the positive sign in the exponent is to interpret the inverse Fourier transform equation (which has the form of an integral over$k$) as the continuum limit of a sum over complex waves. In this sum,$F(k)$plays the role of the series coefficients, and by convention the complex waves have the form$\exp(ikx)$. As previously noted, all the functions we deal with are assumed to be square integrable. This includes the$f_a$functions used to define the Fourier transform. In the$a \rightarrow \infty$limit, this implies that we are dealing with functions such that $$\int_{-\infty}^{\infty} dx\; \big\|\,f(x)\,\big\|^2\;\;\text{exists and is finite}.$$ ### A simple example Consider the function $$f(x) = \left\{\begin{array}{cl}e^{-\eta x}, & x \ge 0 \\ 0, & x < 0,\end{array}\right. \qquad \eta \in \mathbb{R}^+.$$ For$x < 0$, this is an exponentially-decaying function, and for$x < 0$it is identically zero. The real parameter$\eta$is called the decay constant; for$\eta > 0$, the function$f(x)$vanishes as$x \rightarrow +\infty$and can thus be shown to be square-integrable. Larger values of$\eta$correspond to faster exponential decay. The Fourier transform can be found by directly calculating the Fourier integral: $$F(k) \;=\; \;\int_{0}^\infty dx\; e^{-i kx}\, e^{-\kappa x} \;=\; \frac{-i}{k - i \eta}.$$ It is useful to plot the squared magnitude of the Fourier transform,$\|F(k)\|^2$, against$k$. This is called the Fourier spectrum of$f(x)$. In this case, $$\big\|\,F(k)\,\big\|^2 = \frac{1}{k^2 + \eta^2}.$$ The Fourier spectrum is plotted below. It consists of a peak centered at$k = 0$, forming a curve called a Lorentzian. The width of the Lorentzian is dependent on the original function's decay constant$\eta$. For small$\eta$, i.e. weakly-decaying$f(x)$, the peak is narrow; for large$\eta$, i.e. rapidly-decaying$f(x)$, the peak is broad. In [13]: ## Demo: compare an exponential curve to its Fourier spectrum %matplotlib inline from ipywidgets import interact, FloatSlider from numpy import linspace, exp, pi, zeros import matplotlib.pyplot as plt def plot_exp_fourier(eta): ## Plotting parameters xmin, xmax = -5., 5. fmin, fmax = 0, 1.1 kmin, kmax = -5., 5. Fsqmin, Fsqmax = 0, 4 nx, nk = 50, 100 arrowdx, arrowy, bardy = 1.0, 2, 0.2 col0, col1, col2 = "black", "orangered", "mediumblue" ## Plot f(x) plt.figure(figsize=(14,5)) plt.subplot(1,2,1) ## For x < 0, f(x) = 0: plt.plot([xmin, 0, 0], [0, 0, 1], linewidth=2, color=col1) ## For x > 0, f(x) = exp(-etax): x = linspace(0, xmax, nx) f = exp(-etax) plt.plot(x, f, linewidth=2, color=col1) plt.title('Graph of f(x)') plt.xlabel('x'); plt.ylabel('f(x)') plt.xlim(xmin, xmax); plt.ylim(fmin, fmax) ## Plot Fourier spectrum using exact formula given in notes: plt.subplot(1,2,2) k = linspace(kmin, kmax, nk) F = -1j/(k-1jeta) plt.plot(k, abs(F)2, linewidth=2, color=col2) y0 = 0.5/eta2 # Half-maximum if y0 < Fsqmax: # Plot some arrow guides plt.arrow(-eta-arrowdx, y0, arrowdx, 0, color=col0, head_width=0.1, head_length=0.5, length_includes_head=True) plt.arrow(+eta+arrowdx, y0, -arrowdx, 0, color=col0, head_width=0.1, head_length=0.5, length_includes_head=True) plt.plot([-eta, -eta], [y0-bardy, y0+bardy], color=col0) plt.plot([ eta, eta], [y0-bardy, y0+bardy], color=col0) plt.text(eta + arrowdx, y0+bardy, r"$\pm\,\eta$") ## Axis labels, etc. plt.title('Fourier spectrum') plt.xlabel('k'); plt.xlim(kmin, kmax) plt.ylabel('\|F(x)\|^2'); plt.ylim(Fsqmin, Fsqmax) plt.show() interact(plot_exp_fourier, eta = FloatSlider(min=0.1, max=2.0, step=0.1, value=1.0, description='eta')); var element =$('#7adc2793-7443-4a2a-bc88-fcd4a38dd7ba'); {"model_id": "ea213cb59a324efbb0947d05568e887b", "version_major": 2, "version_minor": 0} We can quantify the width of the Lorentzian by defining the full-width at half-maximum (FWHM)—the width of the curve at half the value of its maximum. In this case, the maximum of the Lorentzian curve occurs at $k=0$ and has the value of $1/\eta^2$. The half-maximum, $1/2\eta^2$, occurs when $\delta k = \pm \eta$. Hence, the original function's decay constant, $\eta$, is directly proportional to the FWHM of the Fourier spectrum, which is $2\eta$. To wrap up this example, let's evaluate the inverse Fourier transform: $$f(x) \; = \; -i\int_{-\infty}^\infty \frac{dk}{2\pi} \; \frac{e^{i kx}}{k-i\eta}.$$ This can be solved by contour integration. The analytic continuation of the integrand has a simple pole at $k = i\eta$. For $x < 0$, the numerator $\exp(ikx)$ vanishes far from the origin in the lower half-plane, so we close the contour below. This encloses no pole, so the integral is zero. For $x > 0$, the numerator vanishes far from the origin in the upper half-plane, so we close the contour above, with a counter-clockwise arc, and the residue theorem gives $$f(x) = \left(\frac{-i}{2\pi}\right) \, \left(2\pi i\right) \, \mathrm{Res}\left[ \frac{e^{ikx}}{k-i\eta}\right]_{k=i\eta} = e^{-\eta x} \qquad(x > 0),$$ as expected. ### Aside: Numerical Fourier Transforms Fourier transforms can be obtained numerically using a discretized form called the Discrete Fourier Transform (DFT). DFTs are computed via an algorithm called the Fast Fourier Transform (FFT). We will briefly describe how to run such calculations, without going into the underlying numerical details. To convert between a Fourier transform and a DFT, we first sample $f(x)$ at $N$ equally-spaced points $[x_0, x_1, \dots, x_{N-1}]$ (note: by convention, the index starts from 0). The spacing is $\Delta x = x_{j+1} - x_j$. These gives an array of $N$ numbers, $[f_0, f_1, \dots, f_{N-1}]$, where $f_j = f(x_j)$. By definition, the DFT consists of $N$ numbers $[F_0, F_1, \dots, F_{N-1}]$ such that $$F_n = \sum_{j=0}^{N-1} \exp\left(-2\pi i \frac{nj}{N}\right)\, f_j.$$ The DFT can be computed by calling scipy.fft.fft with the $[f_0,\dots,f_{N-1}]$ array as the input. After that, there are three more things we need to do to convert $[F_0, \dots, F_{N-1}]$ into the values of the Fourier transform function $F(k)$: 1. We need the corresponding values of $k$. These can be obtained by calling scipy.fft.fftfreq with two inputs, $N$ and $\Delta x$, and multiplying the result by $2\pi$. This yields an array of $N$ numbers containing the values of $k$. 2. We must re-arrange the arrays, since the $k$ values are not in monotonic order: the positive $k$'s are in the first half of the array, and the negative $k$'s are in the second half! (This is an artifact of the conventional definition of the DFT, which we will not go into.) To fix this, we call scipy.fft.fftshift, which takes an array and returns another array arranged in the appropriate (increasing-$k$) order. This must done for both the $F$ array and the $k$ array. 3. Multiply the $F$ array by a scale factor $\Delta x$. The resulting array contains the values of $F(k)$ at the specified values of $k$, up to a constant phase factor of the form $\exp(i\theta)$, where $\theta \in \mathbb{R}$. However, we normally only care about $\|F(k)\|^2$, so the phase factor doesn't need to be determined. The code below show how to calculate numerical Fourier transforms according to the above steps, for the function discussed in the previously example. The Fourier transform of this function was previously calculated to be $F(k) = -i/(k+i\eta)$. In [3]: from numpy import linspace, exp, pi from numpy.fft import fft, fftfreq, fftshift import matplotlib.pyplot as plt eta = 0.25 x = linspace(-25, 25, 40) # List of N equally-spaced values of x. dx = x[1] - x[0] # The x-spacing f = exp(-eta x); f[x < 0] = 0. # The values of f(x) at those points. F = fft(f) # 0. Calculate the DFT of the f array. k = 2 * pi * fftfreq(len(x), dx) # 1. Get the values of k k = fftshift(k) # 2. Rearrange the k array F = fftshift(F) # 2. Rearrange the F array F = dx * F # 3. Scale the F array by dx ## Plot f vs x plt.figure(figsize=(12,4)) plt.subplot(1,2,1) plt.scatter(x, f.real, label='Sampled f-values') plt.xlabel('x'); plt.ylabel('f') plt.legend() ## Plot the Fourier spectrum, and compare it to the previously-calculated result. plt.subplot(1,2,2) k2 = linspace(-3, 3, 500) F2 = -1j / (k2 - 1jeta) plt.plot(k2, abs(F2)2, label='Exact FT') plt.plot(k, abs(F)2, 'o', label='Numerical FT') plt.xlabel('k'); plt.ylabel('\|F\|^2') plt.legend() plt.show() ## Fourier transforms for time-domain functions So far, we have been dealing with functions of a spatial coordinate $x$. Of course, mathematical relations don't care about what kind of physical variable we are dealing with, so the same equations could be applied to functions of time $t$. However, there is a important difference in convention. When dealing with functions of the time coordinate $t$, it is customary to use a different sign convention in the Fourier relations! The Fourier relations for a function of time, $f(t)$, are: \left\{\;\,\begin{align}F(\omega) &= \;\int_{-\infty}^\infty dt\; e^{i\omega t}\, f(t) \\ f(t) &= \int_{-\infty}^\infty \frac{d\omega}{2\pi}\; e^{-i\omega t}\, F(\omega).\end{align}\;\,\right. Compared to the previously-derived Fourier relations between $f(x)$ and $F(k)$, the signs of the $\pm i \omega t$ exponents are flipped. There's a good reason for this difference in sign convention: it arises from the need to describe propagating waves, which vary with both space and time. As discussed in Chapter 5, a propagating plane wave can be described by a wavefunction of the form $$f(x,t) = A e^{i(kx - \omega t)},$$ where $k$ is the wave-number and $\omega$ is the angular frequency. We write the plane wave function this way so that positive $k$ indicates forward propagation in space (i.e., in the $+x$ direction), and positive $\omega$ indicates forward propagation in time (i.e., in the $+t$ direction). This requires the $kx$ and $\omega t$ terms in the exponent to have opposite signs. Thus, when $t$ increases by some amount, a corresponding increase in $x$ leaves the exponent unchanged. As we have seen, the inverse Fourier transform relation describes how a wave-form is broken up into a superposition of elementary waves. For a wavefunction $f(x,t)$, the superposition is given in terms of plane waves: $$f(x,t) = \int_{-\infty}^\infty \frac{dk}{2\pi} \int_{-\infty}^\infty \frac{d\omega}{2\pi}\;\; e^{i(kx-\omega t)}\, F(k,\omega).$$ To be consistent with this, we need to treat space and time variables with oppositely-signed exponents: \begin{align}f(x) &= \int_{-\infty}^\infty \frac{dk}{2\pi}\; e^{ikx}\, F(k) \\ f(t) &= \int_{-\infty}^\infty \frac{d\omega}{2\pi}\; e^{-i\omega t}\, F(\omega).\end{align} ## Basic properties of the Fourier transform The Fourier transform has several important properties. These can all be derived from the definition of the Fourier transform; the proofs are left as exercises. 1. The Fourier transform is linear: if we have two functions $f(x)$ and $g(x)$, whose Fourier transforms are $F(k)$ and $G(k)$ respectively, then for any constants $a, b \in \mathbb{C}$, $$a f(x) + b g(x) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; a F(k) + b G(k).$$ 2. Performing a coordinate translation on a function causes its Fourier transform to be multiplied by a phase factor: $$f(x+b) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; e^{ikb} \, F(k).$$ As a consequence, translations leave the Fourier spectrum $\|F(k)\|^2$ unchanged. 3. If the Fourier transform of $f(x)$ is $F(k)$, then $$f^(x) \quad \overset{\mathrm{FT}}{\longrightarrow} \;\; F^(-k).$$ As a consequence, the Fourier transform of a real function must satisfy the symmetry relation $F(k) = F^(-k)$, meaning that the Fourier spectrum is symmetric about the origin in k-space: $\big\|\,F(k)\,\big\|^2 = \big\|\,F(-k)\,\big\|^2.$ 4. When you take the derivative of a function, that is equivalent to multiplying its Fourier transform by a factor of $ik$: $$\frac{d}{dx} f(x) \,\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; ik F(k).$$ For functions of time, because of the difference in sign convention discussed above, there is an extra minus sign: $$\frac{d}{dt} f(t) \;\;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; -i\omega F(\omega).$$ ## Fourier transforms of differential equations The Fourier transform is a useful tool for solving many differential equations. As an example, consider a damped harmonic oscillator subjected to an additional driving force $f(t)$. This force has an arbitrary time dependence, and is not necessarily harmonic. The equation of motion is $$\frac{d^2 x}{dt^2} + 2\gamma \frac{dx}{dt} + \omega_0^2 x(t) = \frac{f(t)}{m}.$$ To solve for $x(t)$, we first take the Fourier transform of both sides of the above equation. The result is: $$- \omega^2 X(\omega) - 2 i\gamma \omega X(\omega) + \omega_0^2 X(\omega) = \frac{F(\omega)}{m},$$ where $X(\omega)$ and $F(\omega)$ are the Fourier transforms of $x(t)$ and $f(t)$ respectively. To obtain the left-hand side of this equation, we used the properties of the Fourier transform described in the previous section, specifically linearity (1) and the Fourier transforms of derivatives (4). Note also that we are using the convention for time-domain functions. The Fourier transform has turned our ordinary differential equation into an algebraic equation which can be easily solved: $$X(\omega) = \frac{F(\omega)/m}{- \omega^2 - 2 i\gamma \omega + \omega_0^2}$$ Knowing $X(\omega)$, we can use the inverse Fourier transform to obtain $x(t)$: $$x(t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \, \frac{e^{-i\omega t}\, F(\omega)/m}{- \omega^2 - 2 i\gamma \omega + \omega_0^2}, \;\; \mathrm{where}\;\; F(\omega) = \int_{-\infty}^\infty dt\; e^{i\omega t} f(t).$$ To summarize, the solution procedure for the driven harmonic oscillator equation consists of (i) using the Fourier transform on $f(t)$ to obtain $F(\omega)$, (ii) using the above equation to find $X(\omega)$ algebraically, and (iii) performing an inverse Fourier transform to obtain $x(t)$. This is the basis for the Green's function method, a method for systematically solving differential equations that will be discussed in the next chapter. ## Common Fourier transforms To accumulate more intuition about Fourier transforms, let us examine the Fourier transforms of some interesting functions. We will just state the results; the calculations are left as exercises. ### Damped waves Previously, we saw that an exponentially decay function with decay constant $\eta \in \mathbb{R}^+$ has the following Fourier transform: $$f(x) = \left\{\begin{array}{cl}e^{-\eta x}, & x \ge 0 \\ 0, & x < 0,\end{array}\right. \;\; \overset{\mathrm{FT}}{\longrightarrow} \;\; F(k) = \frac{-i}{k-i\eta}.$$ Observe that $F(k)$ is given by a simple algebraic formula. If we "extend" the domain of $k$ to complex values, $F(k)$ corresponds to an analytic function with a simple pole in the upper half of the complex plane, at $k = i\eta$. Next, consider a decaying wave with wave-number $q \in \mathbb{R}$ and decay constant $\eta \in \mathbb{R}^+$. The Fourier transform is a function with a simple pole at $q + i \eta$: $$f(x) = \left\{\begin{array}{cl}e^{i (q + i\eta) x}, & x \ge 0 \\ 0, & x < 0.\end{array}\right. \;\; \overset{\mathrm{FT}}{\longrightarrow} \;\; F(k) = \frac{-i}{k-(q + i\eta)}.$$ In [4]: ## Demo: plot a decaying wave and its Fourier spectrum %matplotlib inline from ipywidgets import interact, FloatSlider from numpy import linspace, exp, pi, zeros import matplotlib.pyplot as plt def plot_exp_fourier_1(q, eta): ## Plot parameters: xmin, xmax = -4., 10. fmin, fmax = -1., 1. kmin, kmax = -10., 10. Fsqmin, Fsqmax = 0, 20 nx, nk = 200, 500 arrowdx, arrowy, bardy = 2, 2, 0.2 col0, col1a, col1b, col2 = "black", "orangered", "darkgreen", "mediumblue" ## Plot f(x) plt.figure(figsize=(14,5)) plt.subplot(1,2,1) ## For x < 0, f(x) = 0: plt.plot([xmin, 0, 0], [0, 0, 1], '-', linewidth=2, color=col1a, label='Re(f)') plt.plot([xmin, 0], [0, 0], '--', linewidth=2, color=col1b, label='Im(f)') ## For x > 0, f(x) = exp(i(q+ieta)x) x = linspace(0, xmax, nx) f = exp(1j * (q + 1jeta) x) plt.plot(x, f.real, '-', linewidth=2, color=col1a) # Plot the real part of f(x) plt.plot(x, f.imag, '--', linewidth=2, color=col1b) # Plot imaginary part (in same graph) plt.title('Graph of f(x)') plt.xlabel('x'); plt.xlim(xmin, xmax) plt.ylabel('f(x)'); plt.ylim(fmin, fmax) plt.legend(loc="upper right") ## Plot the Fourier spectrum (using exact formula) plt.subplot(1,2,2) k = linspace(kmin, kmax, nk) F = -1j/(k-(q+1jeta)) plt.plot(k, abs(F)2, linewidth=2, color=col2) ## Guides to the eye for eta y0 = 0.5/eta2 # Half-maximum plt.plot([q-eta, q-eta], [y0-bardy, y0+bardy], color=col0) plt.plot([q+eta, q+eta], [y0-bardy, y0+bardy], color=col0) plt.text(q + eta + arrowdx + 0.25, y0+bardy, r"$\pm\,\eta$", fontsize=22) plt.title('Fourier spectrum') plt.xlabel('k'); plt.xlim(kmin, kmax) plt.ylabel('\|F(x)\|^2'); plt.ylim(Fsqmin, Fsqmax) plt.show() interact(plot_exp_fourier_1, q = FloatSlider(min=-5.0, max=5.0, step=0.1, value=1.0, description='q'), eta = FloatSlider(min=0.2, max=1.0, step=0.1, value=0.5, description='eta')); var element = $('#307b2553-142f-439d-8411-62b87486630d'); {"model_id": "9b6f044b45a44bdaa27ee22ba3f9d97d", "version_major": 2, "version_minor": 0} On the other hand, consider a wave that grows exponentially with$x$for$x < 0$, and is zero for$x > 0$. The Fourier transform is a function with a simple pole in the lower half-plane: $$f(x) = \left\{\begin{array}{cl}0, & x \ge 0 \\ e^{i (q - i\eta) x}, & x < 0.\end{array}\right. \;\; \overset{\mathrm{FT}}{\longrightarrow} \;\; F(k) = \frac{i}{k-(q - i\eta)}.$$ From these examples, we see that oscillations and amplification/decay in$f(x)$are related to the existence of poles in the algebraic expression for$F(k)$. The real part of the pole position gives the wave-number of the oscillation, and the distance from the pole to the real axis gives the amplification or decay constant. A decaying signal produces a pole in the upper half-plane, while a signal that is increasing exponentially with$x$produces a pole in the lower half-plane. In both cases, if we plot the Fourier spectrum of$\|F(k)\|^2$versus real$k$, the result is a Lorentzian peak centered at$k = q$, with width$2\eta$. ### Gaussian wave-packets Consider a function with a decay envelope given by a Gaussian function: $$f(x) = e^{iq x} \, e^{-\gamma x^2}, \;\;\;\mathrm{where}\; q \in \mathbb{C},\; \gamma \in \mathbb{R}.$$ This is called a Gaussian wave-packet. The width of the envelope is usually characterized by the Gaussian function's standard deviation, which is where the curve reaches$e^{-1/2}$times its peak value. In this case, the standard deviation is$\Delta x = 1/\sqrt{2\gamma}$. We will show that$f(x)has the following Fourier transform: $$F(k) = \sqrt{\frac{\pi}{\gamma}} \, e^{-\frac{(k-q)^2}{4\gamma}}.$$ This prediction can be verified numerically, using the previously-discussed Discrete Fourier Transform method: In [73]: from numpy import linspace, exp, pi from numpy.fft import fft, fftfreq, fftshift import matplotlib.pyplot as plt q = 1.0 gamma = 0.01 x = linspace(-100, 100, 200) # List of N equally-spaced values of x. dx = x[1] - x[0] # The x-spacing f = exp(1jqx) exp(-gammax2) # The corresponding values of f F = dx fftshift(fft(f)) k = 2 * pi * fftshift(fftfreq(len(x), dx)) ## Plot Re(f) versus x plt.figure(figsize=(12,4)) plt.subplot(1,2,1) x2 = linspace(x[0], x[-1], 5000) f2 = exp(1jqx2) * exp(-gammax22) plt.plot(x2, f2.real) plt.scatter(x, f.real, label='Sampled x-points') plt.xlabel('x'); plt.ylabel('Re(f)') plt.legend() plt.xlim(-30, 30) ## Plot the Fourier spectrum, and compare it to the prediction plt.subplot(1,2,2) k2 = linspace(-3, 3, 500) F2 = sqrt(pi/gamma) exp(-(k2-q)2/4/gamma) plt.plot(k2, abs(F2)2, label='Exact') plt.plot(k, abs(F)*2, 'o', label='Numerical') plt.xlabel('k'); plt.ylabel('\|F\|^2') plt.legend() plt.show() To derive this result, we perform the Fourier integral as follows: \begin{align}F(k) &= \int_{-\infty}^\infty dx \, e^{-ikx}\, f(x) \\ &= \int_{-\infty}^\infty dx \, \exp\left\{-i(k-q)x -\gamma x^2\right\}.\end{align} In the integrand, the expression inside the exponential is quadratic inx. We complete the square: \begin{align}F(k) &= \int_{-\infty}^\infty dx \, \exp\left\{-\gamma\left(x + \frac{i(k-q)}{2\gamma}\right)^2 + \gamma\left(\frac{i(k-q)}{2\gamma}\right)^2\right\} \\ &= \exp\left\{ - \frac{(k-q)^2}{4\gamma}\right\}\; \int_{-\infty}^\infty dx \, \exp\left\{-\gamma\left(x + \frac{i(k-q)}{2\gamma}\right)^2\right\} \end{align}. The remaining integral is simply the Gaussian integral, with a constant shift inx$which can be eliminated by a change of variables. This yields the result stated above. The Fourier spectrum,$\|F(k)\|^2$, is a Gaussian function with standard deviation $$\Delta k = \frac{1}{\sqrt{2(1/2\gamma)}} = \sqrt{\gamma}.$$ Once again, the Fourier spectrum is peaked at a value of$k$corresponding to the wave-number of the underlying sinusoidal wave in$f(x)$, and a stronger (weaker) decay in$f(x)$leads to a broader (narrower) Fourier spectrum. These features can be observed in the plot below. In [5]: ## Demo: plot a wavepacket and its Fourier spectrum %matplotlib inline from ipywidgets import interact, FloatSlider from numpy import linspace, pi, zeros, exp, sqrt import matplotlib.pyplot as plt def plot_wavepacket_fourier(q, gamma): ## Plot parameters: xmin, xmax = -5., 5. fmin, fmax = -1., 1. kmin, kmax = -12., 12. Fsqmin, Fsqmax = 0, 15 nx, nk = 200, 500 arrowdx, arrowy, bardy = 2, 2, 0.2 col0, col1a, col1b, col2 = "grey", "orangered", "darkgreen", "mediumblue" ## Plot f(x) plt.figure(figsize=(14,5)) plt.subplot(1,2,1) x = linspace(xmin, xmax, nx) f_env = exp(-gamma x * x) f = exp(1j * q * x) * f_env plt.plot(x, f_env, '-', linewidth=1, color=col0) # Plot envelope (in same graph) plt.plot(x,-f_env, '-', linewidth=1, color=col0) plt.plot(x, f.real, '-', linewidth=2, color=col1a, label='Re(f)') # real part plt.plot(x, f.imag, '--', linewidth=2, color=col1b, label='Im(f)') # imag part plt.title('Graph of f(x)') plt.xlabel('x'); plt.xlim(xmin, xmax) plt.ylabel('f(x)'); plt.ylim(fmin, fmax) plt.legend(loc="upper right") ## Plot the Fourier spectrum (using exact formula) plt.subplot(1,2,2) k = linspace(kmin, kmax, nk) F = sqrt(pi/gamma) * exp(-0.25(k-q)2/gamma) plt.plot(k, abs(F)2, linewidth=2, color=col2) plt.title('Fourier spectrum') plt.xlabel('k'); plt.xlim(kmin, kmax) plt.ylabel('\|F(x)\|^2'); plt.ylim(Fsqmin, Fsqmax) plt.show() interact(plot_wavepacket_fourier, q = FloatSlider(min=-10.0, max=10.0, step=0.1, value=5.0, description='q'), gamma = FloatSlider(min=0.1, max=2.0, step=0.1, value=0.5, description='gamma')); var element =$('#b5f26442-0bf4-45d4-9fda-4b143cf393b4'); {"model_id": "b56c6fc0df5344f29f8cb6636f1823f2", "version_major": 2, "version_minor": 0} ## The delta function What happens when we feed the Fourier relations into one another? Plugging the Fourier transform into the inverse Fourier transform, we get \begin{align}f(x) &= \int_{-\infty}^\infty \frac{dk}{2\pi} \, e^{ikx} F(k) \\ &= \int_{-\infty}^\infty \frac{dk}{2\pi} \, e^{ikx} \int_{-\infty}^\infty dx' e^{-ikx'} f(x')\\ &= \int_{-\infty}^\infty dx' \int_{-\infty}^\infty \frac{dk}{2\pi} \, e^{ikx} e^{-ikx'} f(x')\\ &= \int_{-\infty}^\infty dx' \; \delta(x-x')\, f(x'),\end{align} In the last step, we have introduced $$\delta(x-x') = \int_{-\infty}^\infty \frac{dk}{2\pi} \, e^{ik(x-x')},$$ which is called the delta function. According to the above equations, the delta function acts as a kind of filter: when we multiply it by any function $f(x')$ and integrate over $x'$, the result is the value of that function at a particular point $x$. But here's a problem: the above integral definition of the delta function is non-convergent; in particular, the integrand does not vanish at $\pm \infty$. We can get around this by thinking of the delta function as a limiting case of a convergent integral. Specifically, let's take $$\delta(x-x') = \lim_{\gamma \rightarrow 0} \, \int_{-\infty}^\infty \frac{dk}{2\pi} \, e^{ik(x-x')} \, e^{-\gamma k^2}.$$ For $\gamma \rightarrow 0$, the "regulator" $\exp(-\gamma k^2)$ which we have inserted into the integrand goes to one, so that the integrand goes back to what we had before; on the other hand, for $\gamma > 0$ the regulator ensures that the integrand vanishes at the end-points so that the integral is well-defined. But the expression on the right is the Fourier transform for a Gaussian wave-packet, so $$\delta(x-x') = \lim_{\gamma \rightarrow 0} \; \frac{1}{\sqrt{4\pi\gamma}} \, e^{-\frac{(x-x')^2}{4\gamma}}.$$ This is a Gaussian function of width $\sqrt{2\gamma}$ and area $1$. Hence, the delta function can be regarded as the limit of a Gaussian function as its width goes to zero while keeping the area under the curve fixed at unity (which means the height of the peak goes to infinity). The most important feature of the delta function is it acts like a filter. Whenever it shows up in an integral, it picks out the value of the rest of the integrand evaluated where the delta function is centered: $$\int_{-\infty}^\infty dx \; \delta(x-x_0)\, f(x) = f(x_0).$$ Intuitively, we can understand this behavior from the above definition of the delta function as the zero-width limit of a Gaussian. When we multiply a function $f(x)$ with a narrow Gaussian centered at $x_0$, the product will approach zero almost everywhere, because the Gaussian goes to zero. The product is non-zero only in the vicinity of $x = x_0$, where the Gaussian peaks. And because the area under the delta function is unity, integrating that product over all $x$ simply gives the value of the other function at the point $x_0$. Note In physics, the delta function is commonly used to represent the density distributions of point particles. For instance, the distribution of mass within an object can be represented by a mass density function. Assuming one-dimensional space for simplicity, we define the mass density $\rho(x)$ as the mass per unit length at position $x$. By this definition,$$M = \int_{-\infty}^\infty \rho(x)\, dx$$is the total mass of the object. Now suppose the mass is distributed among $N$ point particles, which are located at distinct positions $x_1$, $x_2$, ..., $x_N$, and have masses $m_1$, $m_2$, ... $m_N$. To describe this situation, we can write the mass density function as$$\rho(x) = \sum_{j=1}^N \, m_j\, \delta(x-x_j).$$The reason for this is that if we integrate $\rho(x)$ around the vicinity of the $j$-th particle, the result is just the mass of that single particle, thanks to the features of the delta function:\begin{aligned}\lim_{\varepsilon\rightarrow 0^+}\, \int_{x_j - \varepsilon}^{x_j + \varepsilon} \rho(x) \, dx &= \sum_{i=1}^N m_i\; \Big[\lim_{\varepsilon\rightarrow 0^+}\, \int_{x_j - \varepsilon}^{x_j + \varepsilon} \delta(x-x_i) \,dx\Big]\\ &= \sum_{i=1}^N m_i\; \delta_{ij} \\ &= m_j.\end{aligned}Likewise, integrating $\rho(x)$ over all space gives the total mass $m_1 + m_2 + \cdots + m_N$. ## Multi-dimensional Fourier transforms When studying problems such as wave propagation, we often deal with Fourier transforms of several variables. This is conceptually straightforward. For a function $f(x_1, x_2, \dots, x_d)$ which depends on $d$ independent spatial coordinates $x_1, x_2, \dots x_d$, we can Fourier transform each coordinate individually: $$F(k_1, k_2, \dots, k_d) = \int_{-\infty}^\infty dx_1\; e^{-ik_1x_1}\; \int_{-\infty}^\infty dx_2\; e^{-ik_2x_2}\,\cdots\, \int_{-\infty}^\infty dx_d\; e^{-ik_d x_d}\, f(x_1,x_2, \dots,x_N)$$ Each coordinate gets Fourier-transformed into its own independent $k$ variable, so the result is also a function of $d$ independent variables. We can express the multi-dimensional Fourier transform more compactly using vector notation. If $\vec{x}$ is a $d$-dimensional coordinate vector, the Fourier-transformed coordinates can be written as $\vec{k}$, and the Fourier transform is $$F(\vec{k}) = \int d^d x \; \exp\left(-i\,\vec{k}\cdot\vec{x}\right) \, f\big(\vec{x}\big),$$ where $\int d^d x$ denotes an integral over the entire $d$-dimensional space, and $\vec{k}\cdot\vec{x}$ is the usual dot product of two vectors. The inverse Fourier transform is $$f(\vec{x}) = \int \frac{d^dk}{(2\pi)^d}\; \exp\left(i\,\vec{k}\cdot\vec{x}\right)\, F\big(\vec{k}\big).$$ The delta function can also be defined in $d$-dimensional space, as the Fourier transform of a plane wave: $$\delta^d(\vec{x}-\vec{x}') = \int \frac{d^dk}{(2\pi)^d} \, \exp\left[i\vec{k} \cdot \left(\vec{x}-\vec{x}'\right)\right].$$ Note that $\delta^d$ has the dimensions of $[x]^{-d}$. The multi-dimensional delta function has a "filtering" property similar to the one-dimensional delta function. For any $f(x_1,\dots,x_d)$, $$\int d^dx \; \delta^d(\vec{x}-\vec{x}') \, f(\vec{x}) = f(\vec{x}').$$ ## Exercises 1. Find the relationship between the coefficients $\{\alpha_n, \beta_m\}$ in the sine/cosine Fourier series and the coefficients $f_n$ in the complex exponential Fourier series:\begin{aligned}f(t) &= \sum_{n=1}^\infty \alpha_n \sin\left(\frac{2\pi n x}{a}\right) + \sum_{m=0}^\infty \beta_m \cos\left(\frac{2 \pi m x}{a}\right) \\ &= \sum_{n=-\infty}^\infty f_n \exp\left(\frac{2\pi i n x}{a}\right).\end{aligned} 2. Consider the triangular wave$$f(x) = \left\{\begin{array}{rr}- x, &-a/2 \le x < 0, \\ x, & 0 \le x < a/2\end{array}\right.$$ 1. Derive the Fourier series expansion. 2. In the incomplete Python program below, write the necessary code for plotting this Fourier series. 3. A periodic function $f(x)$ (with period $a$) is written as a complex Fourier series with coefficients $\{f_0, f_{\pm1}, f_{\pm2}, \dots\}$. Determine the relationship(s) between the Fourier coefficients under each of the following scenarios: 1. $f(x)$ is real for all $x$. 2. $f(x) = f(-x)$ for all $x$ 3. $f(x) = f(-x)^$ for all $x$. (Solution) 4. Prove the properties of the Fourier transform listed in the Basic Properties section. 5. Find the Fourier transform of $f(x) = \sin(\kappa x)/x.$ 6. Prove that if $f(x)$ is a real function, then its Fourier transform satisfies $F(k) = F(-k)^$. 7. Prove that $$\delta(ax) = \frac{1}{a}\,\delta(x),$$ where $a$ is any nonzero real number. (Solution) 8. Calculate $$\int_{-\infty}^\infty dx \int_{-\infty}^\infty dy \; x^2\, \delta\left(\sqrt{x^2+y^2}-a\right),$$ where $a$ is a real number. (Solution) In [1]: ## Here is the incomplete Python program for Problem 2: %matplotlib inline from ipywidgets import interact, IntSlider, FloatSlider from numpy import import matplotlib.pyplot as plt ## Plot a triangular wave of period 'a', as well as ## the Fourier series up to order 'N'. def plot_triangular_wave_series(N, a): np = 2 # Number of periods to plot on each side of x = 0 xmin, xmax = (-np-0.5)a, (np+0.5)a # x-axis limits of the plot ymin, ymax = 0.0, 0.6a # y-axis limits of the plot nx = 600 # Number of x-points in the Fourier series plot col1, col2 = "darkorange", "mediumblue" plt.figure(figsize=(10,5)) ## Plot the triangular wave plt.plot([], [], linewidth=3, color=col1, label="Triangular wave") for n in range(-np, np+1): x = array([-0.5a, 0, 0.5a]) y = array([0.5a, 0, 0.5a]) plt.plot(x-na, y, linewidth=2, color=col1) ## Plot the Fourier series approximant. x = linspace(xmin, xmax, nx) f = zeros(nx) ############### [[ Code for calculating f goes here ]] ############### ## plt.plot(x, f, linewidth=1, color=col2, label="Series") plt.xlabel('x') plt.ylabel('f(x)') plt.title('Fourier series of triangular wave, to order {}'.format(N)) plt.xlim(xmin, xmax) plt.ylim(ymin, ymax) plt.legend(loc="upper right") interact(plot_triangular_wave_series, N = IntSlider(min=1, max=25, step=1, value=1, description='n_max'), a = FloatSlider(min=0.5, max=4.0, step=0.1, value=1.0, description='a')); var element = \$('#12afaced-4fb1-4db5-88ae-c23eee06fcbc'); {"model_id": "7b81a6d9f57341468b270072bfe08230", "version_major": 2, "version_minor": 0}
How do you multiply (3x^2 + 2y)(7x^2 + 3y)? Jul 26, 2016 $\left(3 {x}^{2} + 2 y\right) \left(7 {x}^{2} + 3 y\right)$ $21 {x}^{4} + 9 {x}^{2} y + 14 {x}^{2} y + 6 {y}^{2}$ $21 {x}^{4} + 23 {x}^{2} y + 6 y$ Explanation: Step 1 I multiplied $3 {x}^{2}$ with $7 {x}^{2}$ Step 2 I multiplied $3 {x}^{2}$ with $3 y$. So far we have $21 {x}^{4} + 9 {x}^{2} y$ Step 3 I multiplied $2 y$ with $7 {x}^{2}$ Step 4 I multiplied $2 y$ with $3 y$. We have $14 {x}^{2} y + 6 {y}^{2}$ Step 5 In total, we have $21 {x}^{4} + 9 {x}^{2} y + 14 {x}^{2} y + 6 {y}^{2}$ Step 6 We combine like terms. You will see the only like terms are $9 {x}^{2} y + 14 {x}^{2} y$ Those are the only terms we combine. Final Step We combine everything SO our answer is $21 {x}^{4} + 23 {x}^{2} y + 6 {y}^{2}$
# Math solving website Here, we will show you how to work with Math solving website. Our website will give you answers to homework. ## The Best Math solving website Here, we will be discussing about Math solving website. In solving equations or systems of equations, substitution is often used as an effective method. Substitution involves solving for one variable in terms of the others; once a variable is isolated, the equation can be solved more easily. In general, substitution is best used when one equation in a system is much simpler than the others. However, it can also be useful in other cases where equations are not easily solved by other methods. To use substitution, one must first identify which variable will be solved for. The other variable(s) are then substituted into this equation. From there, the equation can be simplified and solved for the desired variable. Substitution can be a powerful tool in solving equations; however, it is important to ensure that all resulting equations are still consistent and have a single solution. Otherwise, the original problem may not have had a unique solution to begin with. Fractions can be a tricky concept, especially when you're dealing with fractions over fractions. But luckily, there's a relatively easy way to solve these types of problems. The key is to first convert the mixed fraction into an improper fraction. To do this, simply multiply the whole number by the denominator and add it to the numerator. For example, if you have a mixed fraction of 3 1/2, you would convert it to 7/2. Once you've done this, you can simply solve the problem as two regular fractions. So, if you're trying to solve 3 1/2 divided by 2/5, you would first convert it to 7/2 divided by 2/5. Then, you would simply divide the numerators (7 and 2) and the denominators (5 and 2) to get the answer: 7/10. With a little practice, solving fractions over fractions will become easier and more intuitive. How to solve radicals can be a tricky topic for some math students. However, with a little practice, it can be easy to understand how to solve these equations. The first step is to identify the type ofradical that is being used. There are two types of radicals, square roots and cube roots. Once the type of radical has been identified, the next step is to determine the value of the number inside the radical. This number is called the radicand. To find the value of the radicand, take the square root of the number if it is a square root radical or the cube root of the number if it is a cube root radical. The last step is to simplify the equation by cancelling out any factors that are shared by both sides of the equation. With a little practice, solving radicals can be easy! Algebra is a branch of mathematics that allows one to solve equations and systems of equations. Algebra has many applications in science and engineering and is a vital tool for solving problems. When solving algebra problems, it is important to first identify the Unknown, or the variable that one is solving for. Once the Unknown is identified, one can then use algebraic methods to solve for the Unknown. Algebraic methods include using algebraic equations and manipulating algebraic expressions. Solving algebra problems requires a strong understanding of algebraic concepts and principles. However, with practice and patience, anyone can learn how to solve algebra problems. ## We will support you with math difficulties The best app out there, but with this new update kind of lacks that goes directly into camera, you have to count the bad writers too. Just let me write the equation and save it for another use, I have to retake the photo or go to history which takes longer than just the eq. staying there. Xuxa Long It's great! It doesn't just give you the answer, but it shows step-by-step what to do! And though it's basically getting the answer with the click of a button, it still teaches the user how to do it! There's an infinitesimal number of reasons why I wouldn't use this app if I'm having trouble with a problem! ^w^ Paulina Moore Solve differential equation Solve slope intercept form calculator Math tutors around me Easy math problems with answers 3 variable solver
# How to determine if an equation is linear or not In this video lesson, we are going to learn how to determine if an equation is linear or not by giving the definition of a linear equation. • Give a rule that relates $x$ and $y$ in the following ordered pairs: a. $\{(-2,2),(-1,1),(0,0),(1,-1),(2,-2)\}$ b. $\{(1,2),(2,3),(4,5)\}$ • A linear equation in two variables is an equation that can be written in the form $Ax + By = C$ where $A,B$ and $C$ are real numbers, $A$ and $B$ not both zero. This means that for an equation to be linear, the following must be satisfied: a.Â The exponents of x and y must be both 1. b.Â The variables must not appear in the denominator c.Â No square roots, cube roots, etc. for the variables • Which of the following are linearÂ equations in two variables? 1.Â $2x + 5y = 4$ 2.Â $x^2 + 4y = 5$ 3.Â $3x - 4 = 7y$ 4.Â $x = 5$ 5. $\displaystyle{\frac{4}{5}}y - 3=2x$ 6.Â $\displaystyle{\frac{4}{2x}} + 3y=-4$ • Which of the following are linear equations in two variables? a. $y^2=y+3-4x$ b. $\displaystyle{\frac{x-4}{2}}=-6y+3$ c.Â $\displaystyle{x=\frac{1}{y}}$
Education.com Try Brainzy Try Plus # Integrals of Trigonometric Expressions Help (not rated) By — McGraw-Hill Professional Updated on Aug 31, 2011 ## Introduction to Integrals of Trigonometric Expressions Trigonometric expressions arise frequently in our work, especially as a result of substitutions. In this section we develop a few examples of trigonometric integrals. The following trigonometric identities will be particularly useful for us. I. We have The reason is that cos 2x = cos2x − sin2x = [1 − sin2x ] − sin2x = 1 − 2 sin2x . II. We have The reason is that cos 2x = cos2x − sin2x sin2x = cos2x − [1 − cos2x ] = 2 cos2x − 1. Now we can turn to some examples. #### Example 1 Calculate the integral ∫ cos2x dx . #### Solution 1 Of course we will use formula II . We write #### Example 2 Calculate the integral ∫ sin3x cos2x dx . #### Solution 2 When sines and cosines occur together, we always focus on the odd power (when one occurs). We write sin3x cos2x = sinx sin2x cos2x = sin x (1 − cos2x ) cos2x = [cos2x − cos4x ] sinx . Then ∫ sin3x cos2 dx = ∫ [cos2x − cos4x ] sin x dx . A u -substitution is suggested: We let u = cos x , du = − sin x dx . Then the integral becomes Resubstituting for the u variable, we obtain the final solution of You Try It : Calculate the integral ∫ sin23x cos53x dx . Calculate #### Solution 3 Substituting into the integrand yields Again using formula II, we find that our integral becomes Applying formula II one last time yields You Try It : Calculate the integral You Try It : Calculate the integral Integrals involving the other trigonometric functions can also be handled with suitable trigonometric identities. We illustrate the idea with some examples that are handled with the identity #### Example 4 Calculate ∫ tan3x sec3x dx . #### Solution 4 Using the same philosophy about odd exponents as we did with sines and cosines, we substitute sec 2 x − 1 for tan 2 x . The result is ∫ tan x (sec2x − l) sec3 x dx . We may regroup the terms in the integrand to obtain ∫ [sec4 x − sec2 x ] sec x tan x dx . A u -substitution suggests itself: We let u = sec x and therefore du = sec x tan x dx . Thus our integral becomes Resubstituting the value of u gives Calculate #### Solution 5 We write Letting u = tan x and du = sec2 x dx then gives the integral You Try It : Calculate the integral Further techniques in the evaluation of trigonometric integrals will be explored in the exercises. Find practice problems and solutions for these concepts at: Methods Of Integration Practice Test. 150 Characters allowed ### Related Questions #### Q: See More Questions ### Today on Education.com #### BURIED TREASURE We Dig Division! 9 Math Worksheets #### SUMMER SCIENCE Boom: An Activity Book
Dividing Numbers in Scientific Notation Here, we will learn how to divide numbers in scientific notation, which are written in the form (a x 10n) where 1 â‰¤ a < 10. The number â€˜aâ€™ is the coefficient, and â€˜bâ€™ is the power or exponent. Steps with Examples 1. Separate and divide the coefficient and exponents separately. 2. Divide the bases with the help of the division rule of exponents (am Ã· an = am – n), and thus the exponents of the denominator are subtracted from the numerator. 3. Join the result of coefficients by the new power of 10 4. If the quotient from the division of coefficients is not within the range 1 â‰¤ a < 10, convert it to scientific notation form and then multiply it by the new power of 10 With Positive Exponents Example – 1: Divide and express in scientific notation: (8 x 108) Ã· (2 x 105) 1. Separating the coefficient and the exponential part = (8 Ã· 2) Ã— (108 Ã· 105) • Dividing the coefficient and the exponential part separately = 8 Ã· 2 = 4, 108 Ã· 105 = 108-5 = 103 • The coefficient is within the range 1 â‰¤ a < 10, thus multiplying it by the new power of 10. Thus, the answer is 4 Ã— 103 Example – 2: Having Negative Exponent Divide and express in scientific notation: (2 x 104) Ã· (4 x 10-7) 1. Separating the coefficient and the exponential part = (2 Ã· 4) Ã— (104 Ã· 10-7) • Dividing the coefficient and the exponential part separately = 2 Ã· 4 = 0.5, 104 Ã· 10-8 = 104-(-7) = 1011 • The coefficient is not within the range 1 â‰¤ a < 10 and is less than 1, thus converting it to scientific notation = 0.5 = 5 Ã— 10-1 • Now, multiplying the coefficient by the new power of 10, we get = (5 Ã— 10-1) Ã— 1011 = 5 Ã— 1010 Thus, the answer is 5 Ã— 1010 Let us solve some more examples. Divide 4.2 Ã— 104 by 2.9 Ã— 102 and express your answer in scientific notation. Solution: Separating the coefficient and the exponential part = (4.2 Ã· 2.9) Ã— (104 Ã· 102) Dividing the coefficient and the exponential part separately = 4.2 Ã· 2.9 = 1.448, 104 Ã· 102 = 104-2 = 102 The coefficient is within the range 1 â‰¤ a < 10, thus multiplying it by the new power of 10. Thus, the answer is 1.448 Ã—102 Divide 3.2 Ã— 104 by 5.7 Ã— 10-2 and express your answer in scientific notation. Solution: Separating the coefficient and the exponential part = (3.2 Ã· 5.7) Ã— (104 Ã· 10-2) Dividing the coefficient and the exponential part separately = 3.2 Ã· 5.7 = 0.561, 104 Ã· 10-2 = 104-(-2) = 106 The coefficient is not within the range 1 â‰¤ a < 10 and is less than 1, thus converting it to scientific notation = 0.561 = 5.61 Ã— 10-1 Now, multiplying the coefficient by the new power of 10, we get = (5.61 Ã— 10-1) Ã— 106 = 5 Ã— 105 Thus, the answer is 5.61 Ã— 105
# Energy Transfer (When work is done) When an object is dropped from above the ground, work is done as the object is pulled to the ground. As the object is falling and work is done, the potential energy of the body is changed to kinetic energy. In principle, the quantity of potential energy stored in a body is always equal to kinetic energy produced when the body is released to do work. In other words, when energy changes, for example from potential to kinetic, there is always an accompanying work done. EXAMPLE What work is done when a mass of 6kg is raised through a vertical height of 3.5m(acceleration due to gravity is 10m/s2) Mass=6kg Acceleration (due to gravity) =10m/s2 Workdone W=Mass x acceleration x distance = 6kgx10x 3.5 =210 joules. ## POWER Power is also related to the concepts of energy and work. Power is defined as the rate of doing work, i.e. workdone divided by time. Power=Workdone/Time taken The unit of power is Watt(w) Example 1 What is the power of a child that has done work of 60 J in 20 seconds. Solution Power P=Work/Time =60/2=3 watts Example 2 Calculate the power of a pump which can lift 500kg of water through a vertical height of 12m in 0.3 minutes. Assuming g=9.8m/s2 Force=mass x acceleration(g) Force=500kg x9.8N Distance=12m Workdone =500 x9.8 x12m Time taken= 0.3minutes x60 =18 secs Workdone=mass x acceleration x distance Workdone=Force x distance Since force= mass x acceleration Workdone=500kg x 9.8N x12m =58800J Power = work done/Time taken =58800/18 Power = 3266.67 watts Convert this to kilowatts by dividing by 1000 =3266.67/1000 =3.27kilowatts EVALUATION 1.If the kinetic energy of a ball moving at a velocity of 5m/s is 60kj.What is the mass of the ball? 2.The mass of a ball is 10g and the height is 4m. Calculate the velocity of the ball just before it touches the ground. (g=10m/s) 3.If the mass of an object is 6kg, and the height is 5m. Calculate the potential energy. (assuming=10m/s). 4.If a force of 10Newton is applied by a car over a distance of 4metres, calculate the workdone by the car. 5.Calculate the work done by a ball of mass 40g which falls freely from a height 0.8m above the ground. (g-10m/s) 6.Calculate the power of a pump which can lift 500kg of water through a vertical height of 12m in 0.3minutes. Assuming g-9.8m/s2 POTENTIAL AND KINETIC ENERGY WORK, ENERGY AND POWER SKILL ACQUISITION Electrical Energy Magnetism Related Words energy transfer energy transfer stock energy transfer definition energy transfer stock price energy transfer examples energy transfer partners energy transfer lp energy transfer careers energy transfer dividend energy transfer jobs SUBSCRIBE BELOW FOR A GIVEAWAY
## How To Use Similar Triangles In Real Life #### What Are the Applications of Geometry in Real Life Proof of the Pythagorean Theorem using similar triangles This proof is based on the proportionality of the sides of two similar triangles, that is, the ratio of any corresponding sides of similar triangles is the same regardless of the size of the triangles. #### Similar Triangles Triangle Ratio Then we can use what you know about proportions and similar figures to solve the exercise. Note that, because the height-lines and the ground are (assumed to be) perpendicular, the similar triangles are also right-angled triangles. #### Similar Triangles Triangle Ratio Back here in the real world, there are three methods we can use to prove that two triangles are similar. Fortunately, they're pretty similar to a few we've already learned about. See what we did there? Aren't we a hoot? The first time we saw these methods, they looked like helpful farmhands, tending the congruent pigs and plowing the congruent fields. Now, they're a scarecrow, a tin man, and a #### 13 best Triangles in real life images on Pinterest When does it make sense to use similar triangles to measure the height and length of objects in real life? How to use similar triangles in real life #### can you follow instructions an application of congruence Similar triangles can be applied to solve real world problems. Example 28 Find the value of the height, h m, in the following diagram at which the tennis ball must be hit so that it will just pass over the net and land 6 metres away from the base of the net. #### Right Triangle Similarity Varsity Tutors arcitecture. blue prints are drawings and scale models and use similarity bc they are not the same size as the actual thing. #### when does it makes sense to use similar triangles to 16/12/2016 · This video is about I will be walking through a situation where we are solving for the hight of a tree given the length of the shadow it casts along with the shadow cast by a human standing near #### Applications of triangles Weebly There are four similarity tests for triangles. 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Yes, it takes a little longer than with any semi-automatic espresso machine, and you have to watch it to be done on the stove, and especially you have to watch the milk so you wonâ€™t burn it while steaming itâ€¦ but thatâ€™s the whole point. You slow down that much that you actually #### how to write instagram captions on different lines While itâ€™s good to play around with different lengths, remember that for longer captions, only the first 3 lines will show. 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# Concept And Tricks Of Number Series What is Number Series? Number series is a arrangement of numbers in a certain order, where some numbers are wrongly put into the series of numbers and some number is missing in that series, we need to observe and find the accurate number to the series of numbers. In competitive exams number series are given and where you need to find missing numbers. The number series are come in different types. At first you have to decided what type of series are given in papers then according with this you have to use shortcut tricks as fast as you can . Different types of Number Series There are some format of series which are given in Exams. Perfect Square Series: This Types of Series are based on square of a number which is in same order and one square number is missing in that given series. Example 1: 441, 484, 529, 576? Answer: 441 = 212, 484 = 222, 529 = 232, 576 = 242 , 625 = 252. Perfect Cube Series: This Types of Series are based on cube of a number which is in same order and one cube number is missing in that given series Example 2: 1331, 1728, 2197, ? Answer : 113 , 123 , 133 , 143 Geometric Series: This type of series are based on ascending or descending order of numbers and each successive number is obtain by multiplying or dividing the previous number with a fixed number. Example 3: 5, 45, 405, 3645,? Answer: 5 x 9 = 45, 45 x 9 = 405, 405 x 9 = 3645, 3645 x 9 = 32805. Two stage Type Series: A two tier Arithmetic series is one in which the differences of successive numbers themselves form an arithmetic series. Example 4: i. 3, 9, 18, 35, 58,—— ii. 6, 9, 17, 23,———- Mixed Series: This type of series are more than one different order are given in a series which arranged in alternatively in a single series or created according to any non-conventional rule. This mixed series Examples are describes in separately. Examples 5: 11, 24, 50, 102, 206, ? 11 x 2 = 22 +2 = 24, 24 x 2 = 48 + 2 = 50, 50 x 2 = 100 + 2 = 102, 102 x 2 = 204 + 2 = 206, 206 x 2 = 412 + 2 = 414. So the missing number is 414. June Current Affairs PDF × Thank You, Your details have been submitted we will get back to you. Join India's largest learning destination What You Will get ? • Daily Quizzes • Subject-Wise Quizzes • Current Affairs • Previous year question papers • Doubt Solving session OR Join India's largest learning destination What You Will get ? • Daily Quizzes • Subject-Wise Quizzes • Current Affairs • Previous year question papers • Doubt Solving session OR Join India's largest learning destination What You Will get ? • Daily Quizzes • Subject-Wise Quizzes • Current Affairs • Previous year question papers • Doubt Solving session Enter the email address associated with your account, and we'll email you an OTP to verify it's you. Join India's largest learning destination What You Will get ? • Daily Quizzes • Subject-Wise Quizzes • Current Affairs • Previous year question papers • Doubt Solving session Enter OTP Please enter the OTP sent to /6 Did not recive OTP? Resend in 60s Join India's largest learning destination What You Will get ? • Daily Quizzes • Subject-Wise Quizzes • Current Affairs • Previous year question papers • Doubt Solving session Join India's largest learning destination What You Will get ? • Daily Quizzes • Subject-Wise Quizzes • Current Affairs • Previous year question papers • Doubt Solving session Almost there +91 Join India's largest learning destination What You Will get ? • Daily Quizzes • Subject-Wise Quizzes • Current Affairs • Previous year question papers • Doubt Solving session Enter OTP Please enter the OTP sent to Edit Number Did not recive OTP? Resend 60 By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts Are you sure you want to skip this step? By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts Are you sure you want to skip this step?
How do you rationalize the denominator and simplify (sqrt3+7)/(8-sqrt5)? May 1, 2018 $= \frac{56 + \sqrt{15} + 7 \sqrt{5} + 8 \sqrt{3}}{59}$ Explanation: To rationalize the denominator, multiply by its reciprocal, which in this case is $8 + \sqrt{5}$. To use this operation you must apply it to both the top and the bottom: $= \frac{\sqrt{3} + 7}{8 - \sqrt{5}} \cdot \frac{8 + \sqrt{5}}{8 - \sqrt{5}}$ FOIL the top and bottom: $= \frac{8 \sqrt{3} + \sqrt{15} + 56 + 7 \sqrt{5}}{64 - 8 \sqrt{5} + 8 \sqrt{5} - 5}$ Combine all like terms: $= \frac{56 + \sqrt{15} + 7 \sqrt{5} + 8 \sqrt{3}}{59}$ This is the simplest form.
# Trigonometry can be used for two things: 1.Using 1 side and 1 angle to work out another side, or 2.Using 2 sides to work out an angle. ## Presentation on theme: "Trigonometry can be used for two things: 1.Using 1 side and 1 angle to work out another side, or 2.Using 2 sides to work out an angle."— Presentation transcript: Trigonometry can be used for two things: 1.Using 1 side and 1 angle to work out another side, or 2.Using 2 sides to work out an angle. To work out things using trigonometry we use three new buttons on the calculator labelled. SinCosTan Each of these buttons has a rule you can remember by using the word: SOH-CAH-TOA In this word each letter stands for a word. Buttons on your calculatorSides of a Triangle S = Sin A = Adjacent C = Cos O = Opposite T = Tan H = Hypotenuse The Hypotenuse is the longest side. It is the one not touching the right angle. Hypotenuse The Opposite is the side far away from the angle you are given. 60 o Opposite 30 o 57 o Opposite 33 o The Adjacent is the side next to the angle you are given. 60 o Adjacent 30 o Sin (full name sine) has the rule SOH. O H Sin θ Theta means an angle. Cos (full name cosine) has the rule CAH. A H Cos θ Tan (full name tangent) has the rule TOA. O A Tan θ 1. Label the sides of the triangle. 2. Figure out which type of side you are looking for and which sides you have. 3. Write down the formula from SOH-CAH-TOA. 4. Put in the numbers. 5. Calculate. 6. If you are trying to find an angle you need to press: Shift Sin = Shift Cos = Shift Tan = or We are looking for the Opposite. We have the Hypotenuse. We need to use SOH because it has O and H. O = Sin θ × H y = Sin 30 × 24 y = 12 Opposite 30 o Adjacent Hypotenuse 24 y We are looking for the Opposite. We have the Hypotenuse. We need to use SOH because it has O and H. O = Sin θ × H y = Sin 48 × 15 y = 11.1 Opposite 48 o Adjacent Hypotenuse 15 y We are looking for the Adjacent. We have the Hypotenuse. We need to use CAH because it has A and H. A = Cos θ × H y = Cos 28 × 20 y = 17.7 Opposite 28 o Adjacent Hypotenuse 20 y We are looking for the Opposite. We have the Adjacent. We need to use TOA because it has O and A. A = Tan θ × A y = Tan 40 × 23 y = 19.3 Opposite 40 o Adjacent Hypotenuse 23 y We are looking for the Hypotenuse. We have the Opposite. We need to use SOH because it has O and H. H = O ÷ Sin θ y = 6.0 ÷ Sin 40 y = 9.3 Opposite 40 o Adjacent Hypotenuse 6.0 y We are looking for the Adjacent. We have the Hypotenuse. We need to use CAH because it has A and H. A = Cos θ × H y = Cos 50 × 100 y = 64.3 Opposite 50 o Adjacent Hypotenuse 100 y We are looking for the Opposite. We have the Adjacent. We need to use TOA because it has O and A. O = Tan θ × A y = Tan 32 × 1.4 y = 0.9 Opposite 32 o Adjacent Hypotenuse 1.4 y We are looking for the Hypotenuse. We have the Adjacent. We need to use CAH because it has A and H. H = A ÷ Cos θ y = 32.5 ÷ Cos 40 y = 42.4 Opposite 40 o Adjacent Hypotenuse 32.5 y We are looking for the Adjacent. We have the Opposite. We need to use TOA because it has O and A. A = O ÷ Tan θ y = 3.2 ÷ Tan 40 y = 3.8 Opposite 40 o Adjacent Hypotenuse 3.2 y We are looking for an angle. We have the Adjacent and the Opposite. We need to use TOA because it has O and A. Tan θ = O ÷ A Tan θ = 4 ÷ 2.2 Tan θ = 1.82 Opposite θ Adjacent Hypotenuse 2.2 4 ShiftTan = θ = 61 o We are looking for an angle. We have the Adjacent and the Hypotenuse. We need to use CAH because it has A and H. Cos θ = A ÷ H Cos θ = 4 ÷ 7 Cos θ = 0.57 Opposite θ Adjacent Hypotenuse 7 4 Shift Cos = θ = 55 o We are looking for an angle. We have the Hypotenuse and the Opposite. We need to use SOH because it has O and H. Sin θ = O ÷ H Sin θ = 8 ÷ 10 Sin θ = 0.8 Opposite θ Adjacent Hypotenuse 10 8 Shift Sin = θ = 53 o We are looking for an angle. We have the Hypotenuse and the Adjacent. We need to use CAH because it has A and H. Cos θ = A ÷ H Cos θ = 5 ÷ 10 Cos θ = 0.5 Opposite θ Adjacent Hypotenuse 10 5 Shift Cos = θ = 60 o Download ppt "Trigonometry can be used for two things: 1.Using 1 side and 1 angle to work out another side, or 2.Using 2 sides to work out an angle." Similar presentations
# The slope of the secant line PQ for the given values of x. ### Single Variable Calculus: Concepts... 4th Edition James Stewart Publisher: Cengage Learning ISBN: 9781337687805 ### Single Variable Calculus: Concepts... 4th Edition James Stewart Publisher: Cengage Learning ISBN: 9781337687805 #### Solutions Chapter 2.1, Problem 9E (a) To determine ## To find: The slope of the secant line PQ for the given values of x. Expert Solution The slope of the secant line PQ for the following values of x is given below: x Slope mPQ 2 0 1.5 1.7321 1.4 −1.0847 1.3 −2.7433 1.2 −4.3301 1.1 −2.8173 0.5 0 0.6 −2.1651 0.7 −2.6061 0.8 −5 0.9 3.4202 ### Explanation of Solution Given: The equation of the curve is y=sin(10πx). The point P(1, 0) lies on the curve y. The Q is the point (x,sin(10πx)). Calculation: The slope of the secant lines between the points, P(1, 0) and Q(x,sin(10πx)) is mPQ=sin(10πx)0x1 (1) Obtain the slope of the secant line PQ for the value of x=2. Substitute 2 for x in sin(10πx), sin(10πx)=sin(10π2)=sin(5π)=0 Substitute Q(x,sin(10πx))=(2,0) in equation (1), mPQ=sin(10πx)0x1=0021=01=0 Thus, the slope of the secant line PQ for the value of x=2 is 0. Obtain the slope of the secant line PQ for the value of x=1.5. Substitute 1.5 for x in sin(10πx), sin(10πx)=sin(10π1.5)=sin(203π)=0.866025 Substitute Q(x,sin(10πx))=(1.5,0.866025) in equation (1), mPQ=sin(10πx)0x1=0.86602501.51=0.8660250.51.7321 Thus, the slope of the secant line PQ for the value of x=1.5 is 1.7321. Obtain the slope of the secant line PQ for the value of x=1.4. Substitute 1.4 for x in sin(10πx), sin(10πx)=sin(10π1.4)=0.43388 Substitute Q(x,sin(10πx))=(1.4,0.43388) in equation (1), mPQ=sin(10πx)0x1=0.4338801.41=0.433880.41.0847 Thus, the slope of the secant line PQ for the value of x=1.4 is 1.0847. Obtain the slope of the secant line PQ for the value of x=1.3. Substitute 1.3 for x in sin(10πx). sin(10πx)=sin(10π1.3)=0.82298 Substitute Q(x,sin(10πx))=(1.3,0.82298) in equation (1). mPQ=sin(10πx)0x1=0.8229801.31=0.822980.32.7433 Thus, the slope of the secant line PQ for the value of x=1.3 is 2.7433. Obtain the slope of the secant line PQ for the value of x=1.2. Substitute 1.2 for x in sin(10πx), sin(10πx)=sin(10π1.2)=0.86602 Substitute Q(x,sin(10πx))=(1.2,0.86602) in equation (1), mPQ=sin(10πx)0x1=0.8660201.21=0.866020.24.3301 Thus, the slope of the secant line PQ for the value of x=1.2 is 4.3301. Obtain the slope of the secant line PQ for the value of x=1.1. Substitute 1.1 for x in sin(10πx), sin(10πx)=sin(10π1.1)=0.28173 Substitute Q(x,sin(10πx))=(1.1,0.28173) in equation (1), mPQ=sin(10πx)0x1=0.2817301.11=0.281730.12.8173 Thus, the slope of the secant line PQ for the value of x=1.1 is 2.8173. Obtain the slope of the secant line PQ for the value of x=0.5. Substitute 0.5 for x in sin(10πx), sin(10πx)=sin(10π0.5)=sin(20π)=0 Substitute Q(x,sin(10πx))=(0.5,0) in equation (1), mPQ=sin(10πx)0x1=000.51=00.50 Thus, the slope of the secant line PQ for the value of x=0.5 is 0. Obtain the slope of the secant line PQ for the value of x=0.6. Substitute 0.5 for x in sin(10πx), sin(10πx)=sin(10π0.6)=0.86603 Substitute Q(x,sin(10πx))=(0.6,0.86603) in equation (1), mPQ=sin(10πx)0x1=0.8660300.61=0.866030.42.1651 Thus, the slope of the secant line PQ for the value of x=0.6 is 2.1651. Obtain the slope of the secant line PQ for the value of x=0.7. Substitute 0.7 for x in sin(10πx). sin(10πx)=sin(10π0.7)=0.78183 Substitute Q(x,sin(10πx))=(0.7,0.78183) in equation (1). mPQ=sin(10πx)0x1=0.7818300.71=0.781830.32.6061 Thus, the slope of the secant line PQ for the value of x=0.7 is 2.6061. Obtain the slope of the secant line PQ for the value of x=0.8. Substitute 0.8 for x in sin(10πx), sin(10πx)=sin(10π0.8)=1 Substitute Q(x,sin(10πx))=(0.8,1) in equation (1)., mPQ=sin(10πx)0x1=100.81=10.25 Thus, the slope of the secant line PQ for the value of x=0.8 is 5. Obtain the slope of the secant line PQ for the value of x=0.9. Substitute 0.9 for x in sin(10πx), sin(10πx)=sin(10π0.9)=0.34202 Substitute Q(x,sin(10πx))=(0.9,0.34202) in equation (1), mPQ=sin(10πx)0x1=0.3420200.91=0.342020.13.4202 Thus, the slope of the secant line PQ for the value of x=0.9 is 3.4202. Conclusion: The slope does not appear to be approaching a limit. Suppose x approaches 1, then the slope mPQ do not approach any specific value. (b) To determine Expert Solution ### Explanation of Solution The graph of the curve y=sin(10πx) is shown below in Figure 1. From Figure 1, it is observed that there seems to be the frequent oscillations of the graph. Moreover, the tangent line is so steep at the point P(1, 0). Thus, the slopes of the secant lines are not closer to the slope of the tangent line at P. Therefore, it is necessary to consider the values of x much closer to 1 for better accurate estimates of the slope. (c) To determine ### To estimate: The slope of the tangent line to the curve at P(1, 0). Expert Solution The estimated value of the slope of the tangent line to the curve at P (1, 0) is −31.4. ### Explanation of Solution The graph of the curve y=sin(10πx) between the points 0.5 and 2 is shown below in Figure 2. The secant line is close to the tangent line at P(1, 0) when x=0.999 and x=1.001. The value of the slope of the tangent line to the curve at P(1,0) is close to the average value of the slope of the secant lines closest to P. Obtain the slope of the secant line PQ for the value of x=1.001. Substitute 1.001 for x in sin(10πx). sin(10πx)=sin(10π1.001)=0.031379 Substitute Q(x,sin(10πx))=(1.001,0.031379) in equation (1). mPQ=sin(10πx)0x1=0.031379401.0011=0.03137940.00131.3794 Thus, the slope of the secant line PQ for the value of x=1.001 is 31.3794. Obtain the slope of the secant line PQ for the value of x=0.999. Substitute 0.999 for x in sin(10πx). sin(10πx)=sin(10π0.999)=0.03144219 Substitute Q(x,sin(10πx))=(0.999,0.03144219) in equation (1). mPQ=sin(10πx)0x1=0.0314421900.9991=0.031442190.00131.4422 Thus, the slope of the secant line PQ for the value of x=0.999 is 31.4422. The slope of the secant line PQ for the value of x=1.001 is 31.3794 and the slope of the secant line PQ for the value of x=0.999 is 31.4422. Take the average of two slopes of the secant lines, m31.3794+(31.4422)262.8216231.410831.4 Thus, the estimated value of the slope of the tangent line to the curve at P (1, 0) is −31.4. ### Have a homework question? Subscribe to bartleby learn! 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# GCJ – Manage Your Energy Manage Your Energy, Round 1A 2013 # GCJ – Bullseye Bullseye, Round 1A 2013 This is a largely mathematical problem: the first ring will use $(r + 1)^2 - r^2$ units of paint, the second $(r + 3)^2 - (r + 2)^2$ units, and so forth, so that the $k^{th}$ ring will require $P(k) = (r + 2k - 1)^2 - (r + 2k - 2)^2$, or $P(k) = 2r + (4k - 3)$ units of paint. I.e., each ring will take a constant amount of paint ($2r$) plus some varying amount depending on how far out it is ($4k - 3$). That second sequence is 1, 5, 9, 13, 17, 21, etc. But the question we want to answer is not how much paint each ring will take, but how many rings we can paint with a given amount of paint. One way to solve this problem would be to simply add up the paint as we paint each ring until we don’t have enough paint to go on, then report the number of the last complete ring. This solution will work for the small problem, but will fail on the large problem as it is linear in $k$, and $k$ in the large problem can be very very large. We can resolve this by finding a closed form solution for the sum of all $P(k)$ from 1 to k. We first note that there is a constant amount of paint for each ring, so we can sum up that part simply as $2rk$. The first few variable sums are 1, 6, 15, 28, 45…, which we can recognize as the sequence of hexagonal numbers with closed form solution $k(2k - 1)$. This gives us: $P_{total}(k) = 2rk + k(2k-1)$ Now that we have a closed form solution, we can try to substitute in $t$, the total amount of paint we have, for $P_{total}$, solve for $k$, and round down to obtain the maximum number of complete rings we can draw, in constant time, much better than linear! This is a quadratic, and we obtain the (positive) solution $k = \lfloor \sqrt{4r^2 - 4r + 8t + 1} - \frac{2r - 1}{4} \rfloor$ We’re done, right? Well, this solution will work fine for the small input, and in constant time. But it will fail dramatically on the large input. Why? Because we’re taking square roots of very large integers, which transforms them from integers (which are stored with perfect precision) to floating point numbers (which are stored with limited precision), and with the limits we’re given, the precision of floating points isn’t sufficient to give us the right answer this way. So we need to use a different method. Again, what we want to do is find $k$ such that $P_{total}(k) \le t$ while $P_{total}(k+1) > t$. Simply walking through increasing values of $k$ is too slow for the large input. So instead we can do a binary search. We’re guaranteed in the problem statement that $k \ge 1$. So let’s start a lower limit for our search at $k_{low} = 1$. We’ll start an upper limit at $k_{hi} = 2$. It’s probably not an upper limit yet, but we’ll make it so by iteratively doubling both our limits until it is: find the first power of 2 where $P_{total}(2^i) > t$. Then $k$ is somewhere in the interval $[k_{low}, k_{hi})$. We can find its exact value by repeatedly subdividing this interval until it is narrowed down to a single value. # GCJ – Template Code I’ve been participating in the Google Code Jam since last year, and I’ve been having a good time with it. The problems have a common format, and instead of spending a lot of time rewriting the same code to read in data from a file, write it out with the correct formatting, and so forth, I went ahead and wrote a template for new problems in Python that can handle all of that by itself. It first runs the precalculate() function to prepare anything (e.g. lookup tables and such) that I want to have before I start the main calculations. Then it reads in the input file from the command line, parses the number of cases to calculate, and uses the read_input() function to parse data for a single case. It then passes the input data to the solve_case() function, which should solve the problem, and writes the result to the output file in the correct format. Then to solve a new problem, I need only rewrite these three functions to do what I need. There’s a couple of other helpful things: read_input() has a bunch of parser functions inside it to quickly pull out specific types of data, and there’s a memoize decorator that I can use to give helper functions a cache if needed for efficiency.
Introductory Algebra for College Students (7th Edition) The solution is $(0, 2)$. The first thing we want to do is to rewrite the equations so that all the variables are on one side and the constants are on the other. For the first equation, we subtract $4y$ from both sides of the equation. For the second equation, we leave it alone because the variables are already on one side of the equation. $5x - 4y = -8$ $3x + 7y = 14$ We first need to modify the two equations so that the coefficients of one of the variables differs only in sign. With this modification, we will be able to cancel out one of the variables and solve for the other one. We see that if we multiply the first equation by $3$ and multiply the second equation by $-5$, the coefficients of the $x$ term will be $15$ and $-15$. The coefficients will now only differ in sign, so we can cancel them out when we add the two equations. Let us do the multiplication: $3(5x - 4y) = 3(-8)$ $-5(3x + 7y) = -5(14)$ Use distributive property: $3(5x) - 3(4y) = 3(-8)$ $-5(3x) - 5(7y) = -5(14)$ Let's multiply out the terms: $15x - 12y = -24$ $-15x - 35y = -70$ We can now cancel out the $x$ terms to get: $-12y = -24$ $-35y = -70$ Now we add both sides of the two equations to get: $-47y = -94$ Divide both sides of the equation by $-47$ to solve for $y$: $y = 2$ Now that we have the value for $y$, we can plug this value into one of the equations to solve for $x$. Let's use the first equation: $5x = 4(2) - 8$ Multiply first: $5x = 8 - 8$ Do the subtraction on the right-hand side of the equation: $5x = 0$ Divide both sides by $5$ to solve for $x$: $x = 0$ The solution is $(0, 2)$.
# Parallel and Transversal Lines Here we discuss how the angles formed between parallel and transversal lines. When the transversal intersects two parallel lines: • Pairs of corresponding angles are equal. • Pairs of alternate angles are equal • Interior angles on the same side of transversal are supplementary. Worked-out problems for solving parallel and transversal lines: 1. In adjoining figure l ∥ m is cut by the transversal t. If ∠1 = 70, find the measure of ∠3, ∠5, ∠6. Solution: We have ∠1 = 70° ∠1 = ∠3 (Vertically opposite angles) Therefore, ∠3 = 70° Now, ∠1 = ∠5 (Corresponding angles) Therefore, ∠5 = 70° Also, ∠3 + ∠6 = 180° (Co-interior angles) 70° + ∠6 = 180° Therefore, ∠6 = 180° - 70° = 110° 2. In the given figure AB ∥ CD, ∠BEO = 125°, ∠CFO = 40°. Find the measure of ∠EOF. Solution: Draw a line XY parallel to AB and CD passing through O such that AB ∥ XY and CD ∥ XY ∠BEO + ∠YOE = 180° (Co-interior angles) Therefore, 125° + ∠YOE = 180° Therefore, ∠YOE = 180° - 125° = 55° Also, ∠CFO = ∠YOF (Alternate angles) Given ∠CFO = 40° Therefore, ∠YOF = 40° Then ∠EOF = ∠EOY + ∠FOY = 55° + 40° = 95° 3. In the given figure AB ∥ CD ∥ EF and AE ⊥ AB. Also, ∠BAE = 90°. Find the values of ∠x, ∠y and ∠z. Solution: y + 45° = 1800 Therefore, ∠y = 180° - 45° (Co-interior angles) = 135° ∠y =∠x (Corresponding angles) Therefore, ∠x = 135° Also, 90° + ∠z + 45° = 180° Therefore, 135° + ∠z = 180° Therefore, ∠z = 180° - 135° = 45° 4. In the given figure, AB ∥ ED, ED ∥ FG, EF ∥ CD Also, ∠1 = 60°, ∠3 = 55°, then find ∠2, ∠4, ∠5. Solution: Since, EF ∥ CD cut by transversal ED Therefore, ∠3 = ∠5 we know, ∠3 = 55° Therefore, ∠5 = 55° Also, ED ∥ XY cut by transversal CD Therefore, ∠5 = ∠x we know ∠5 = 55° Therefore,∠x = 55° Also, ∠x + ∠1 + ∠y = 180° 55° + 60° + ∠y = 180° 115° + ∠y = 180° ∠y = 180° - 115° Therefore, ∠y = 65° Now, ∠y + ∠2 = 1800 (Co-interior angles) 65° + ∠2 = 180° ∠2 = 180° - 65° ∠2 = 115° Since, ED ∥ FG cut by transversal EF Therefore, ∠3 + ∠4 = 180° 55° + ∠4 = 180° Therefore, ∠4 = 180° - 55° = 125° 5. In the given figure PQ ∥ XY. Also, y : z = 4 : 5 find. Solution: Let the common ratio be a Then y = 4a and z = 5a Also, ∠z = ∠m (Alternate interior angles) Since, z = 5a Therefore, ∠m = 5a [RS ∥ XY cut by transversal t] Now, ∠m = ∠x (Corresponding angles) Since, ∠m = 5a Therefore, ∠x = 5a [PQ ∥ RS cut by transversal t] ∠x + ∠y = 180° (Co-interior angles) 5a + 4a = 1800 9a = 180° a = 180/9 a = 20 Since, y = 4a Therefore, y = 4 × 20 y = 80° z = 5a Therefore, z = 5 × 20 z = 100° x = 5a Therefore, x = 5 × 20 x = 100° Therefore, ∠x = 100°, ∠y = 80°, ∠z = 100° Lines and Angles Fundamental Geometrical Concepts Angles Classification of Angles Related Angles Some Geometric Terms and Results Complementary Angles Supplementary Angles Complementary and Supplementary Angles Linear Pair of Angles Vertically Opposite Angles Parallel Lines Transversal Line Parallel and Transversal Lines 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. ### Comparison of Three-digit Numbers \| Arrange 3-digit Numbers \|Questions Sep 13, 24 02:48 AM What are the rules for the comparison of three-digit numbers? (i) The numbers having less than three digits are always smaller than the numbers having three digits as: 2. ### Worksheet on Three-digit Numbers \| Write the Missing Numbers \| Pattern Sep 13, 24 02:23 AM Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures… 3. ### 2nd Grade Place Value \| Definition \| Explanation \| Examples \|Worksheet Sep 13, 24 01:20 AM The value of a digit in a given number depends on its place or position in the number. This value is called its place value. 4. ### Comparison of Two-digit Numbers \| Arrange 2-digit Numbers \| Examples Sep 12, 24 03:07 PM What are the rules for the comparison of two-digit numbers? We know that a two-digit number is always greater than a single digit number. But, when both the numbers are two-digit numbers
# Engineering Mathematics SEM – 1Concept Mathematics Notes for Polytechnic SEM – 1 is Designed by the ” Basics in Maths” team. Here we can learn Concepts in Basic Engineering mathematics Polytechnic Sem – I. This Material is very Useful for Basic Engineering Mathematics Polytechnic Sem – I Students. By learning These Notes, Basic Engineering Mathematics Polytechnic Sem – I Students can Write their Exam successfully and fearlessly. Engineering Mathematics SEM – 1Concept #### LOGARITHMS Logarithm: For ant two positive real numbers a, b, and a ≠ 1. If the real number x such then ax = b, then x is called logarithm of b to the base a. it is denoted by #### Standard formulae of logarithms: ##### Logarithmic Function: Let a be a positive real number and a ≠ 1. The function f: (o, ∞) → R Defined by f(x) = PARTIAL FRACTIONS Fractions: If f(x) and g(x) are two polynomials, g(x) ≠ 0, then is called rational fraction. Ex: etc. are rational fractions. ##### Proper Fraction: A rational fraction is said to be a Proper fraction if the degree of g(x) is greater than the degree of f(x). Ex: etc. are the proper fractions. Improper Fraction: A rational fraction is said to be an Improper fraction if the degree of g(x) is less than the degree of f(x). Ex: etc. are the Improper fractions. ##### Partial Fractions: Expressing rational fractions as the sum of two or more simpler fractions is called resolving a given fraction into a partial fraction. ∎ If R(x) = is proper fraction, then Case(i): – For every factor of g(x) of the form (ax + b) n, there will be a sum of n partial fractions of the form: Case(ii): – For every factor of g(x) of the form (ax2 + bx + c) n, there will be a sum of n partial fractions of the form: ∎ If R(x) = is improper fraction, then Case (i): – If degree f(x) = degree of g(x), where k is the quotient of the highest degree term of f(x) and g(x). Case (ii): – If f(x) > g(x) R(x) = ### MATRICES AND DETERMINANTS Matrix: A set of numbers arranged in the form of a rectangular array having rows and columns is called Matrix. •Matrices are generally enclosed by brackets like •Matrices are denoted by capital letters A, B, C, and so on •Elements in a matrix are real or complex numbers; real or complex real-valued functions. Oder of Matrix: A matrix having ‘m’ rows and ‘n’ columns is said to be of order m x n read as m by n. Ex: ##### Types Of Matrices Rectangular Matrix: A matrix in which the no. of rows is not equal to the no. of columns is called a rectangular matrix. Square Matrix: A matrix in which the no. of rows is equal to no. of columns is called a square matrix. Principal diagonal (diagonal) Matrix: If A = [a ij] is a square matrix of order ‘n’ the elements a11, a22, a33, ………. An n is said to constitute its principal diagonal. Trace Matrix: The sum of the elements of the principal diagonal of a square matrix A is called the trace of the matrix. It is denoted by Tr (A). Diagonal Matrix: If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix. Scalar Matrix: If each non-diagonal element of a square matrix is ‘zero’ and all diagonal elements are equal to each other, then it is called a scalar matrix. Identity Matrix or Unit Matrix: If each of the non-diagonal elements of a square matrix is ‘zero’ and all diagonal elements are equal to ‘1’, then that matrix is called unit matrix Null Matrix or Zero Matrix: If each element of a matrix is zero, then it is called a null matrix. Row matrix & column Matrix: A matrix with only one row s called a row matrix and a matrix with only one column is called a column matrix. ###### Triangular matrices: A square matrix A = [aij] is said to be upper triangular if aij = 0 ∀ i > j A square matrix A = [aij] is said to be lower triangular matrix aij = 0 ∀ i < j ###### Equality of matrices: matrices A and B are said to be equal if A and B are of the same order and the corresponding elements of A and B are equal. If A and B are two matrices of the same order, then the matrix obtained by adding the corresponding elements of A and B is called the sum of A and B. It is denoted by A + B. Subtraction of matrices: If A and B are two matrices of the same order, then the matrix obtained by subtracting the corresponding elements of A and B is called the difference from A to B. ##### Product of Matrices: Let A = [aik]mxn and B = [bkj]nxp be two matrices, then the matrix C = [cij]mxp where Note: Matrix multiplication of two matrices is possible when no. of columns of the first matrix is equal to no. of rows of the second matrix. A m x n . Bp x q = AB mx q; n = p Transpose of Matrix: If A = [aij] is an m x n matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of A. It is denoted by AI or AT. Note: (i) (AI)I = A (ii) (k AI) = k . AI (iii) (A + B )T = AT + BT (iv) (AB)T = BTAT Symmetric Matrix: A square matrix A is said to be symmetric if AT =A If A is a symmetric matrix, then A + AT is symmetric. Skew-Symmetric Matrix: A square matrix A is said to be skew-symmetric if AT = -A If A is a skew-symmetric matrix, then A – AT is skew-symmetric. Minor of an element: Consider a square matrix the minor element in this matrix is defined as the determinant of the 2×2 matrix obtained after deleting the rows and the columns in which the element is present. Ex: – minor of a3 is = b1c2 – b2c1 Minor of b2 is = a1c3 – a3c1 Cofactor of an element: The cofactor of an element in i th row and j th column of A3×3 matrix is defined as its minor multiplied by (- 1) i+j. ###### Properties of determinants: If each element of a row (column) f a square matrix is zero, then the determinant of that matrix is zero. Ex: – A = ⇒ det(A) = 0 If A is a square matrix of order 3 and k is scalar then. If two rows (columns) of a square matrix are identical (same), then Det. Of that matrix is zero. Ex: – A = ⇒ det(A) = 0 If each element in a row (column) of a square matrix is the sum of two numbers then its determinant can be expressed as the sum of the determinants. Ex: – If each element of a square matrix are polynomials in x and its determinant is zero when x = a, then (x-a) is a factor of that matrix. For any square matrix A Det(A) = Det (AI). Det (AB) = Det(A). Det(B). For any positive integer n Det (An) = (DetA)n. ###### Singular and non-singular matrices: A Square matrix is said to be singular if its determinant is is zero, otherwise it is said to be non-singular matrix. Ex: – A = det(A) = 4 – 4 = 0 ∴ A is singular matrix B = Det(B) = 4 + 4 = 8≠ 0 ∴ B is non-singular Adjoint of a matrix: The transpose of the matrix formed by replacing the elements of a square matrix A with the corresponding cofactors is called the adjoint of A. Let A = and cofactor matrix of A = Invertible matrix: Let A be a square matrix, we say that A is invertible if there exists a matrix B such that AB =BA = I, where I is a unit matrix of the same order as A and B. 1. (A– I)– I = A 2. (AI)– I = (A-I)I 3. (AB)-I = B-I A-I 4. A-I = #### Compound Angles The algebraic sum of two or more angles is called a ‘compound angle’. Thus, angles A + B, A – B, A + B + C etc., are Compound Angles For any two real numbers A and B sin (A + B) = sin A cos B + cos A Cos B sin (A − B) = sin A cos B − cos A Cos B cos (A + B) = cos A cos B − sin A sin B cos (A − B) = cos A cos B + sin A sin B tan (A + B) = tan (A − B) = cot (A + B) = cot (A − B) = tan ( + A) = tan ( − A) = cot (+ A) = cot (− A) = sin (A + B + C) = ∑sin A cos B cos C − sin A sin B sin C cos (A + B + C) = cos A cos B cos C− ∑cos A sin B sin C tan (A + B + C) = ⋇ cot (A + B + C) = ⋇ sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A ⋇ cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A ##### Multiple and Sub Multiple Angles If A is an angle, then its integral multiples 2A, 3A, 4A, … are called ‘multiple angles ‘of A and the multiple of A by fraction like are called ‘submultiple angles. ∎ If is not an add multiple of #### PROPERTIES OF TRIANGLES In ∆ABC, Lengths AB = c; BC = a; AC =b The area of the triangle is denoted by ∆. Perimeter of the triangle = 2s = a + b + c A = ∠CAB; B = ∠ABC; C = ∠BCA. Sine rule: In ∆ABC, ⟹ a = 2R sin A; b = 2R sin B; c = 2R sin C Where R is the circumradius and a, b, c, are lengths of the sides of ∆ABC. Cosine rule: In ∆ABC, ###### Projection rule: In ∆ABC, a = b cos C + c cos B b = a cos C + c cos A c = a cos B + b cos A ###### Tangent rule (Napier’s analogy): Area of the triangle: In ∆ABC, a, b, and c are sides S = and area of the triangle ### HYPERBOLIC FUNCTIONS The function f: R→R defined by f(x) = ∀ x ∈ R is called the ‘hyperbolic sin’ function. It is denoted by Sinh x. ∴Sinh x = Similarly, ###### Identities: cosh2x – sinh2 x = 1 cosh2x = 1 + sinh2 x sinh2 x = cosh2 x – 1 sech2 x = 1 – tanh2 x tanh2 x = 1 – sesh2 x cosech2 x = coth2 x – 1 coth2 x = 1 + coth2 x ###### Addition formulas of hyperbolic functions: Sinh (x + y) = Sinh x Cosh y + Cosh x Sinh y Sinh (x − y) = Sinh x Cosh y − Cosh x Sinh y Cosh (x + y) = Cosh x Cosh y + Sinh x Sinh y Cosh (x − y) = Cosh x Cosh y − Sinh x Sinh y tanh (x + y) = tanh (x − y) = coth (x + y) = sinh 2x = 2 sinh x cosh 2x = cosh 2x = cosh2x + sinh2 x = 2 cosh2x – 1 = 1 + 2 sinh2x = tanh 2x = ###### Inverse hyperbolic functions: Sinh−1x = ∀ x ∈ R Cosh−1x = ∀ x ∈ (1, ∞) Tanh−1x = ∀ < 1 #### COMPLEX NUMBERS The equation x2 + 1 = 0 has no roots in real number system. ∴ scientists imagined a number ‘i’ such that i2 = − 1. #### Complex number: if x, y are any two real numbers then the general form of the complex number is z = x + i y; where x real part and y is the imaginary part. 3 + 4i, 2 – 5i, – 3 + 2i are the examples for Complex numbers. • z = x +i y can be written as (x, y). • If z1 = x1 + i y1, z2 = x2 + i y2, then • z1 + z2 = (x1 + x2, y1 + y2) = (x1 + x2) + i (y1 + y2) • z1 − z2 = (x1 − x2, y1 − y2) = (x1 − x2) + i (y1 − y2) • z1∙ z2 = (x1 x2 −y1 y2, x1y2 + x2y1) = (x1x2 −y1 y2) + i (x1y2 +x2 y1) • z1/ z2 = (x1x2 + y1 y2/x22 +y22, x2 y1 – x1y2/ x22 +y22) = (x1x2 + y1 y2/x22 +y22) + i (x2 y1 – x1y2/ x22 +y22) ###### Multiplicative inverse of complex number: The multiplicative inverse of the complex number z is 1/z. z = x + i y then 1/z = x – i y/ x2 + y2 Conjugate complex numbers: The complex numbers x + i y, x – i y are called conjugate complex numbers. Conjugate of z is denoted by The sum and product of two conjugate complex numbers are real. If z1, z2 are two complex numbers then ###### Modulus and amplitude of complex numbers: Modulus: – If z = x + i y, then the non-negative real number is called modulus of z and it is denoted by or ‘r’. Amplitude: – The complex number z = x + i y represented by the point P (x, y) on the XOY plane. ∠XOP = θ is called amplitude of z or argument of z. x = r cos θ, y = r sin θ x2 + y2 = r2 cos2θ + r2 sin2θ = r2 (cos2θ + sin2θ) = r2(1) ⇒ x2 + y2 = r2 ⇒ r = and = r. • Arg (z) = tan−1(y/x) • Arg (z1.z2) = Arg (z1) + Arg (z2) + nπ for some n ∈ { −1, 0, 1} • Arg(z1/z2) = Arg (z1) − Arg (z2) + nπ for some n ∈ { −1, 0, 1} Note: ∎ e = cos θ + i sin θ ∎ e−iθ = cos θ − i sin θ ##### De- Moiver’s theorem For any integer n and real number θ, (cos θ + i sin θ) n = cos n θ + i sin n θ. cos α + i sin α can be written as cis α cis α.cis β= cis (α + β) 1/cisα = cis(-α) cisα/cisβ = cis (α – β) (cos θ + i sin θ) -n = cos n θ – i sin n θ (cos θ + i sin θ) (cos θ – i sin θ) = cos2θ – i2 sin2θ = cos2θ + sin2θ = 1. cos θ + i sin θ = 1/ cos θ – i sin θ and cos θ – i sin θ = 1/ cos θ + i sin θ (cos θ – i sin θ) n = (1/ (cos θ –+i sin θ)) n = (cos θ + i sin θ)-n = cos n θ – i sin n θ nth root of a complex number: let n be a positive integer and z0 ≠ 0 be a given complex number. Any complex number z satisfying z n = z0 is called an nth root of z0. It is denoted by z01/n or let z = r (cos θ + i sin θ) ≠ 0 and n be a positive integer. For k∈ {0, 1, 2, 3…, (n – 1)} let Then a0, a1, a2, …, an-1 are all n distinct nth roots of z and any nth root of z is coincided with one of them. nth root of unity: Let n be a positive integer greater than 1 and Note: • The sum of the nth roots of unity is zero. • The product of nth roots of unity is (– 1) n – 1. • The nth roots of unity 1, ω, ω2, …, ωn-1 are in geometric progression with common ratio ω. ##### Cube root of unity: x3 – 1 = 0 ⇒ x3 = 1 x =11/3 cube roots of unity are: 1 ω = , ω2 = ω2 +ω + 1 = 0 and ω3 = 1 ### TRANSFORMATIONS For A, B∈ R ⋇ sin (A + B) + sin (A – B) = 2sin A cos B ⋇ sin (A + B) −sin (A – B) = 2cos A sin B ⋇ cos (A + B) + cos (A – B) = 2 cos A cos B ⋇ cos (A + B) − cos (A – B) = − 2sin A sin B For any two real numbers C and D ⋇ sin C + sin D = 2sin cos ⋇ sin C −sin D= 2cos sin ⋇ cos C + cos D = 2 cos cos ⋇ cos C − cos D = − 2sin sin If A + B + C = π or 1800, then ⋇ sin (A + B) = sin C; sin (B + C) = sin A; sin (A + C) = sin B ⋇ cos (A + B) = − cos C; cos (B + C) = −cos A; cos (A + C) = − cos B If A + B + C = 900 or then ⋇ sin = cos; sin = cos ; sin = cos ⋇ cos = sin; cos = sin; cos = sin If then ⋇ sin (A + B) = cos C; sin (B + C) = cos A; sin (A + C) = cos B ⋇ cos (A + B) = sin C; cos (B + C) = sin A; cos (A + C) = sin B ### INVERSE TRIGONOMETRIC RATIOS If A, B are two sets and f: A→ B is a bijection, then f-1 is existing and f-1: B → A is an inverse function. ### Solutions of Simultaneous Equations #### Matrix Inversion Method: Let a system of simultaneous equations be a1 x + b1 y + c1z = d1 a2 x + b2 y + c2z = d2 a3 x + b3 y + c3z = d3 The matrix form of the above equations is Therefore, the matrix equation is AX = B If Det A ≠ 0, A-1 is exists X = A-1 B By using above Condition, we get the values of x, y and z This Method is called as Matrix Inversion Method #### Cramer’s Method: Let system of simultaneous equations be a1 x + b1 y + c1z = d1 a2 x + b2 y + c2z = d2 a3 x + b3 y + c3z = d3 1 is obtained by replacing the coefficients of x (1st column elements of ∆) by constant values 2 is obtained by replacing the coefficients of y (2nd column elements of ∆) by constant values 3 is obtained by replacing the coefficients of z (3rd column elements of ∆) by constant values Now This method is called Cramer’s Method #### Gauss-Jordan Method: Let a system of simultaneous equations be a1 x + b1 y + c1z = d1 a2 x + b2 y + c2z = d2 a3 x + b3 y + c3z = d3 Augmented matrix: The coefficient matrix (A) augmented with the constant column matrix (B) is called the augmented matrix. It is denoted by [AD]. This Matrix is reduced to the standard form ofby using row operations 1. Interchanging any two rows 2. Multiplying the elements of any two elements by a constant. 3. Adding to the elements of one row with the corresponding elements of another row multiplied by a constant. ∴ The solution of a given system of simultaneous equations is x = α, y = β, and z = γ. ##### Procedure to get the standard form: 1. Take the coefficient of x as the unity as a first equation. 2. If 1 is there in the first-row first column, then make the remaining two elements in the first column zero. 3. After that, if one element in R2 or R3 is 1, then make the remaining two elements in that column C2 or C3 as zeroes. 4. If any row contains two elements as zeros and only non-zero divide that row elements with the non-zero element to get unity and make the remaining two elements in that column as zeros.
Survey Thank you for your participation! * Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project Document related concepts no text concepts found Transcript ```Factoring by GCF Remember factor Trees?? Create a factor tree for the following two numbers. 60 24 6 * 10 4 * 6 23 25 22 * 2*3 Are there any numbers common to each?? Sure! When we break the numbers down using factor trees, we can see the common factors. In the example to the left, 60 = 2·2·3·5 24 = 2·2·2·3 The common parts are 2, 2, and 3. If we multiply these, we get 2·2·3 = 12. So, 12 is the Greatest Common Factor (GCF). Now, think through the same process using the following sets of numbers: 60 60 =2·2·3·5 75 75 = 3·5·5 45 Break each number down into factors…. 45 = 3·3·5 Now, pick the factors that are in all three… 3·5 = 15 The Greatest Common Factor is 15. We can do the same with monomials!! Find the Greatest Common Factor: 24x 3 y2 = 2·2·2·3·x·x·x·y·y 56x 2 y4 Break each into factors…. = 2·2·2·7·x·x·y·y·y·y Now pick the factors that are in both… 2·2·2·x·x·y·y The GCF = 8x 2 y2 Summarize: What do you notice about the powers of the variables in the GCF of the two monomials above?? The GCF variables have the lowest power for the variable from the monomials! (What is the lowest power of the x for the monomials? x 2 What is the lowest power of the y for the monomials? y 2 These are the powers of the variables in the GCF) When you pull the Greatest Common Factor out of all terms of a polynomial and place the left over parts inside of a parenthesis, it is called Factoring by GCF: 24x 3 y2 + 56x 2 y4 = 8x 2 y2 ( 3x + 7y 2 ) This is called reverse distributive or factoring by GCF Here is another example to factor 60 x3 y 2 z 4  24 xy3 z 3 First find the GCF for 60 x 3 y 2 z 4 and 24 xy3 z 3 ………….. 12 x1 y 2 z 3 is the GCF!!! 60 x 3 y 2 z 4  24 xy3 z 3 factors as 12 xy2 z 3 (5x 2 z  2 y) That means: More examples: 3x + 6xy – 9 x 2 Factors into 3x(1 + 2y - 3 x )  m 2  5m  5 Factors into -( m 2  5m  5 ) ……. Try to leave highest power term + 9x + 6y Factors into 3(3x + 2y) Problems to try: Factor by GCF. 1. 40 x 5  24 x 3 y 2.  18x 3 y 2  45x 2 y  9 xy 3. 32 x 5  48x 3 y  80 xy 2 4. 64 x 4  16 x 3 5.  x12 8x 5  12 xy 6. 36 x 5 y 3 z  18x 4 yz  42 xyz  24 xz
# Real Life Applications Of Right Angle Triangle Geometry provides us with an elementary introduction to shapes around us. This branch of maths is useful in mastering shape reasoning. If you ever wonder how math is going to be used in real life, geometry offers the answer most readily. We know that circles, triangles, and rectangles are the 2D figures that we see around most commonly. Another 2D figure, which is a ramification of the triangle is a right triangle. It is one of the most stable structures and has its utility in the fields of carpentry, architecture, etc. ## How right triangles are used in real-life? What exactly is a right triangle? If you visualize elaborately, a right triangle is two line segments placed at 90o angle to each other and their free ends are joined by a diagonally oriented segment. This is a shape that we may locate quite effortlessly around us. Staircases when observed from the side may give you an impression of several right angles juxtaposed to each other. Apart from this common example locatable in homes, listed here are the other real-life uses of the right triangles. • Shadow cast by a tall object: A tree forms right angle to the land. Its top and the open end of the shadow form the hypotenuse allowing you to visualize the right triangle structure. • A climbing ladder leaning to the wall: A slanting ladder to the wall forms right triangle shape where hypotenuse corresponds to the ladder. • Angles for placing a slab: Wall mounted shelves are placed on right triangular holding points. • A person standing at a tower and locating a ball on the land : The gaze of a person standing on the tower forms the hypotenuse of the right triangle shape whose two vertical and horizontal arms are height of the tower and distance between the tower and ball. ## Solving right triangle and its application in various professions Right triangles fit various spheres of our lives like a glove. These geometrical figures have an entire branch of math dedicated to it. We all know it by the name of ‘trigonometry’. Various relationships between the two from the three sides of the right triangle (trigonometric functions) help solve real-life problems various professionals face[1]. Let’s take a look at those. ### 1. Carpentry Carpenters need to ascertain that the walls are perfectly straight and corners are squares to fit in the elements as a part of the interior designing process. They may apply the Pythagorean theorem that relates the sides of the right triangle to find the required measurements. ### 2. Finding height or length In architecture, the experts solve right triangle problems to find the required height or length of any structure. This calculation is needed to assure compliance with the building plan. In navigation, finding the distance of the ship from the shore, or between two ships or other objects can help in finding the probable duration of the tour. It may also help find the safe speed and the suitable maneuvering angles to keep the vessel protected from hits or accidents. ### 4. Surveying Surveyor uses the trigonometry functions or solves right triangle based problems to estimate the height of a building. Using a total station theodolite (TST), the surveyor finds the exact angle of elevation from a chosen distance. ### 5. Astronomy Several astronomical calculations like estimating the radius of the Earth or the movement of certain objects in a particular period can be found using the principles and functions of trigonometry. The study of distances between the two astral bodies becomes possible to do by treating the distance as one arm of the right triangle and applying trigonometry principles. ### 6. Astronauts and space scientists Astronauts, during their expedition to space, need to find the characteristics of weather, soil, air pressure, etc. of the places where they cannot visit physically. They use tools like robotic arms for such purposes. With the help of right triangles-based calculations, they can find the correct angle and orientation for maneuvering the robotic arm or other similar tools. ## Teaching right triangles the activity way All the examples mentioned above indicate the possibility of teaching right triangles by using activities. High school students have to hone their trigonometry basic skills. Just solving equations and mugging up values of cos, tan and sine may not produce the required outcomes. Better learning is achieved when they get to apply the concepts. Activities based on right triangles help learn by doing. A few activities that involve solving right triangles’ variables are: ### 1. Estimating heights of objects using Lego Robot kit Lego Robot Kit act as a portable counterpart of the things students may see around. They apply trigonometry principles to estimate the height of various objects given in the kit. On scaling up the measurements received, they can apply the learning to find the heights of buildings, towers, poles, etc. ### 2. Estimating boundaries of a plot Prepare the model of a landscape with dummy hills, buildings, empty land, etc. Assign a few values to the objects’ distance from each other and contours of the land. By making hypothetical right triangles from the top of the dummy hill, building, etc., students can learn about how to find values of various variables like height, speed, distance, etc. ### 3. Parallax activity to guess the distance between stars and sun Parallax means the difference in the position of an object when viewed from different points of view. This displacement can create an angle of view, and two arms that resemble a triangle. By using calculators and trigonometric functions’ tables, they can get the idea of the distance between astronomical objects. Moving forth, the distance between two landmarks can also be estimated. ## Conclusion Studying right triangles can help students understand how the geometric and trigonometric principles can be applied to solve real-life problems. It is this practical knowledge attained that helps blur the line between bookish knowledge and applied science. So, next time you are done mugging up the definitions and functions’ values, start making hypothetical models and apply various combinations of values to strengthen your learning. It brings fun into the learning hour and makes the study of mathematical concepts more interesting. References: 1. Libretexts. (2021, January 2). 1.3: Applications and Solving Right Triangles. Mathematics LibreTexts. https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Elementary_Trigonometry_(Corral)/01%3A_Right_Triangle_Trigonometry_Angles/1.03%3A_Applications_and_Solving_Right_Triangles
## College Algebra (10th Edition) $a_n = 3 \cdot (\frac{1}{3})^{n-1}$ RECALL: (1) The $n^{th}$ term $a_n$ of a geometric sequence is given by the formula $a_n=a_1 \cdot r^{n-1}$ where $a_1$ = first term $r$ = common ratio (2) To find the next term in a geometric sequence, the common ratio $r$ is multiplied by the current term. Part (2) above implies that if the value of $a_3$ is known, the common ratio must be multiplied by $a_3$ three times to find the value of $a_6$. Thus, $a_6 = a_3 \cdot r \cdot r \cdot r \\a_6 = a_3 \cdot r^3$ Substitute the given values of $a_3$ and $a_6$ into the equation above to obtain: $a_6 = a_3 \cdot r^3 \\\frac{1}{81} = \frac{1}{3} \cdot r^3$ Multiply by $3$ on both sides to obtain: $3 \cdot \frac{1}{81} = \frac{1}{3} \cdot r^3 \cdot 3 \\\frac{1}{27} = r^3 \\(\frac{1}{3})^3=r^3$ Take the cube root of both sides to obtain: $\sqrt[3]{(\frac{1}{3})^3} = \sqrt[3]{r^3} \\\frac{1}{3} = r$ Note that: $a_3=a_1 \cdot r \cdot r \\a_3 = a_1 \cdot r^2$ Substitute the known values of $a_3$ and $r$ into the equation above to obtain: $a_3 = a_1 \cdot r^2 \\\frac{1}{3} = a_1 \cdot (\frac{1}{3})^2 \\\frac{1}{3} = a_1 \cdot \frac{1}{9}$ Multiply by $9$ on both sides of the equation to obtain: $9(\frac{1}{3}) = a_1 \cdot \frac{1}{9} \cdot 9 \\3 = a_1$ Thus, the $n^{th}$ term of the geometric sequence is: $a_n=a_1 \cdot r^{n-1} \\a_n = 3 \cdot (\frac{1}{3})^{n-1}$
• Written By Akash_Anand # Maths Formulas for Class 9 Maths Formulas for Class 9: Maths is one of the most important subjects for Class 9 students, and a clear understanding of the Maths Formulas for Class 9 is essential. Students must have outstanding Maths skills to contribute to great inventions in the field of Engineering, Science and Technology. It takes a lot of practice, a deep comprehension of topics, and memorising formulas to do well in the CBSE Class 9 Maths exam. The Maths Formulas for Class 9 will help students easily score good marks in maths exams. Students who find difficulties in solving maths problems should go through the topic-wise Class 9 Maths formulas provided by Embibe. From this article, students can download the NCERT Class 9 Maths Formulas PDF, which will help them study better. ## Chapter-wise Maths Formulas for Class 9 Mathematical formulas aren’t just for closing your eyes and learning. Conceptual clarity is crucial, as is understanding all the formulas of Math, implementing them, and analysing them. Before getting into the list of the formulae, let us check out the important chapters of Class 9 Maths for which formulas are needed: • Numbers • Polynomials • Coordinate Geometry • Algebra • Triangles • Areas of Parallelograms and Triangles • Circles • Heron’s Formula • Surface Areas and Volumes • Statistics • Probability ### Topic-wise Maths Formulas for Class 9 Here are some of the essential formulae for Class 9 polynomial identities and all Class 9 identities in Mathematics. Any number that can be written in the form of p ⁄ q where p and q are integers and q ≠ 0 are rational numbers. Irrational numbers cannot be written in the p ⁄ q form. • There is a unique real number that can be represented on a number line. • If r is one such rational number and s is an irrational number, then (r + s), (r – s), (r × s) and (r ⁄ s) are irrational. • For positive real numbers, the corresponding identities hold together: 1. $$\sqrt{ab}$$ = $$\sqrt{a} × \sqrt{b}$$ 2. $$\sqrt{\tfrac{a}{b}}$$ = $$\frac{\sqrt{a}}{\sqrt{b}}$$ 3. $$(\sqrt{a}+\sqrt{b})\times(\sqrt{a}-\sqrt{b})=a-b$$ 4. $$(a+\sqrt{b})\times(a-\sqrt{b})=a^2-b$$ 5. $$(\sqrt{a}+\sqrt{b})^2=a^2+2\sqrt{ab}+b$$ • If you want to rationalize the denominator of 1 ⁄ √ (a + b), then we have to multiply it by √(a – b) ⁄ √(a – b), where a and b are both integers. • Suppose a is a real number (greater than 0) and p and q are the rational numbers. 1. ap x b= (ab)p+q 2. (ap)q = apq 3. ap / a= (a)p-q 4. ap / bp = (ab)p #### Maths Formulas for Class 9 – Polynomial A polynomial p(x) denoted for one variable ‘x’ comprises an algebraic expression in the form: p(x) = anxn + an-1xn-1 + ….. + a2x2 + a1x + a0 ; where a0, a1, a2, …. an are constants where an ≠ 0 • Any real number; let’s say ‘a’ is considered to be the zero of a polynomial ‘p(x)’ if p(a) = 0. In this case, a is said to be the mysqladmin of the equation p(x) = 0. • Every one variable linear polynomial will contain a unique zero, a real number which is a zero of the zero polynomial and a non-zero constant polynomial which does not have any zeros. • Remainder Theorem: If p(x) has the degree greater than or equal to 1 and p(x) when divided by the linear polynomial x – a will give the remainder as p(a). • Factor Theorem: x – a will be the factor of the polynomial p(x), whenever p(a) = 0. The vice-versa also holds true every time. #### Maths Formulas for Class 9 – Coordinate Geometry Whenever you have to locate an object on a plane, you need two divide the plane into two perpendicular lines, thereby, making it a Cartesian Plane. • The horizontal line is known as the x-axis, and the vertical line is called the y-axis. • The coordinates of a point are in the form of (+, +) in the first quadrant, (–, +) in the second quadrant, (–, –) in the third quadrant and (+, –) in the fourth quadrant; where + and – denotes the positive and the negative real number respectively. • The coordinates of the origin are (0, 0) and thereby it gets up to move in the positive and negative numbers. #### Maths Formulas for Class 9 – Algebraic Identities Once the students have a hold over all Algebraic identities class 9, they will be able to solve all the Algebra related problems in their exams. Given below are Algebraic identities for class 9 which are considered very important Maths formulas for Class 9: • (a + b)2 = a2 + 2ab + b2 • (a – b)2 = a2 – 2ab + b2 • (a + b) (a – b) = a2 -b2 • (x + a) (x + b) = x2 + (a + b) x + ab • (x + a) (x – b) = x2 + (a – b) x – ab • (x – a) (x + b) = x2 + (b – a) x – ab • (x – a) (x – b) = x2 – (a + b) x + ab • (a + b)3 = a3 + b3 + 3ab (a + b) • (a – b)3 = a3 – b3 – 3ab (a – b) • (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz • (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2xz • (x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz • (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz • x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz -xz) • x+ y2 = $$\frac{1}{2}$$ [(x + y)2 + (x – y)2] • (x + a) (x + b) (x + c) = x+ (a + b + c)x2 + (ab + bc + ca)x + abc • x3 + y3 = (x + y) (x– xy + y2) • x3 – y3 = (x – y) (x+ xy + y2) • x2 + y2 + z2 – xy – yz – zx = $$\frac{1}{2}$$ [(x – y)2 + (y – z)2 + (z – x)2] #### Maths Formulas for Class 9 – Triangles A triangle is a closed geometrical figure formed by three sides and three angles. • Two figures are congruent if they have the same shape and same size. • If the two triangles ABC and DEF are congruent under the correspondence that A ↔ D, B ↔ E and C ↔ F, then symbolically, these can be expressed as ∆ ABC ≅ ∆ DEF. #### Right Angled Triangle: Pythagoras Theorem Suppose ∆ ABC is a right-angled triangle with AB as the perpendicular, BC as the base and AC as the hypotenuse; then Pythagoras Theorem will be expressed as: (Hypotenuse)2 = (Perpendicular)2 + (Base)2 i.e. (AC)2 = (AB)2 + (BC)2 #### Maths Formulas for Class 9 – Areas of Parallelograms and Triangles A parallelogram is a type of quadrilateral that contains parallel opposite sides. • Area of parallelogram = Base × Height • Area of Triangle = $$\frac{1}{2}$$ × Base × Height #### Maths Formulas for Class 9 – Circle A circle is a closed geometrical figure. All points on the boundary of a circle are equidistant from a fixed point inside the circle (called the centre). • Area of a circle (of radius r) = π × r2 • The diameter of the circle, d = 2 × r • Circumference of the circle = 2 × π × r • Sector angle of the circle, θ = (180 × l ) / (π × r ) • Area of the sector = (θ/2) × r2; where θ is the angle between the two radii • Area of the circular ring = π × (R2 – r2); where R – radius of the outer circle and r – radius of the inner circle #### Maths Formulas for Class 9 – Heron’s Formula Heron’s Formula is used to calculate the area of a triangle whose all three sides are known. Let us suppose the length of the three sides is a, b and c. • Step 1 – Calculate the semi-perimeter, $$s=\frac{a+b+c}{2}$$ • Step 2 – Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$ #### Maths Formulas for Class 9 – Surface Areas & Volumes In this section, students will find Class 9 surface area and volume formulas at one place. Below, LSA stands for Lateral/Curved Surface Area and TSA stands for Total Surface Area. #### Maths Formulas for Class 9 – Statistics Certain facts or figures which can be collected or transformed into some useful purpose are known as data. These data can be graphically represented to increase readability for people. Three measures of formulae to interpret the ungrouped data: #### Maths Formulas for Class 9 – Probability Probability is the possibility of any event likely to happen. The probability of any event can only be from 0 to 1 with 0 being no chances and 1 being the possibility of that event happening. $$Probability=\frac{Number\: of\: favourable\: outcomes}{Total\: Number\: of\: outcomes}$$ ### FAQ on Maths Formulas for Class 9 Q.1: Where can I practice the Class 9 Maths questions? Ans: You can practice for Class 9 Maths questions at Embibe. Embibe offers you topic-wise questions which are available for free. Q.2: Is the NCERT Maths book enough for the Class 9 exam? Ans: Yes, for Class 9 NCERT Maths book is enough for students preparing for exams. Make sure you understand all the concepts and solve the questions diligently. Note that regular practice is a must. Q.3: How can I learn these Maths formulas for Class 9? Ans: Mathematics is a subject of logic. Therefore, it should be interpreted in the same way. You can learn these formulae by understanding them logically. Then, you can try solving the questions by implementing these formulae. Q.4: Are the Class 9 Maths formulas based on the CBSE curriculum? Ans: We have compiled Class 9 Maths formulas in this article so that students can understand them. These formulae are based on CBSE, ICSE, and other respective boards. Q.5: How can you learn all the polynomial Class 9 formulas for Maths? Ans: You can try to remember everything you are trying to learn in the form of a story. Sequencing will help you to memorise the formulas in a particular order. Also, make sure to understand the derivations of the formulas rather than rote learning. This way, you will be able to remember all formulas of Maths Class 9 for a long time. Free Unlimited 3D Learning & Practice With Solutions For Every Question in 9th CBSE
# How many factors does 75 have? ## How many factors does 75 have? The factors of 75 are 1, 3, 5, 15, 25 and 75. ### What are the pairs factors? Factors are often given as pairs of numbers, which multiply together to give the original number. These are called factor pairs. A square number will have one factor pair consisting of one factor multiplied by itself. This factor is called the square root of the given number. What are the multiples of 75? The first five multiples of 75 are 75, 150, 225, 300, and 375. What are the factors of 72 in pairs? Here we can see all the Positive pair factors of 72. • x 72= 72. therefore, 1 and 72 are a factor pair of 72. • x 36= 72. therefore, 2 and 36 are a factor pair of 72. • x 24= 72. therefore, 3 and 24 are a factor pair of 72. • x 18= 72. ## What is the greatest factor of 75? We also know that 75=1×75,3×25, or 5×15 . All of these numbers are prime numbers so they are all factors of 75 . Thus we see that the factors of 75 are 1 , 3 , 5 , 15 , 25 and 75 . ### What is the factor of 3 and 75? For 3 and 75 those factors look like this: Factors for 3: 1 and 3. Factors for 75: 1, 3, 5, 15, 25, and 75. What are factor pairs of 74? Factors of 74 in Pairs • 1 × 74 = 74. • 2 × 37 = 74. How many pairs are there in 26? The pair factors of 26 are (1, 26), (2, 13). ## What is the LCM of 75? To find the LCM of 75 and 100 using prime factorization, we will find the prime factors, (75 = 3 × 5 × 5) and (100 = 2 × 2 × 5 × 5). LCM of 75 and 100 is the product of prime factors raised to their respective highest exponent among the numbers 75 and 100. ⇒ LCM of 75, 100 = 22 × 31 × 52 = 300. ### What are the factors of 76? The factors of 76 are 1, 2, 4, 19, 38, and 76. What is a factor of 74? The factors of 74 are 1, 2, 37 and 74. What are the greatest common factors of 75? The factors of 75 are 1, 3, 5, 15, 25 and 75. The factors of 20 are 1, 2, 4, 5, 10 and 20. So, the Greatest Common Factor for these numbers is 5 because it divides all them without a remainder. Read more about Common Factors below. ## What is the greatest common factor of 75 and 125? The first step to find the gcf of 75 and 125 is to list the factors of each number. The factors of 75 are 1, 3, 5, 15, 25 and 75. The factors of 125 are 1, 5, 25 and 125. So, the Greatest Common Factor for these numbers is 25 because it divides all them without a remainder. ### Which are factor pair has a product of 75? Factors of 75 that add up to 24 = 1 + 3 + 5 + 15 . Factor of 75 in pairs 1 x 75, 3 x 25, 5 x 15, 15 x 5, 25 x 3, 75 x 1. 1 and 75 are a factor pair of 75 since 1 x 75= 75 . 3 and 25 are a factor pair of 75 since 3 x 25= 75 . 5 and 15 are a factor pair of 75 since 5 x 15= 75 . 15 and 5 are a factor pair of 75 since 15 x 5= 75 What are the positive factors of 75? Positive Integer factors of 75 = 3, 5, 15, 75 divided by 3, 5, 5, gives no remainder. They are integers and prime numbers of 75, they are also called composite number.
# What Are Exponent Rules? I will give you some apples and ask you to make 3 groups containing 2. Now if you multiply it, The answer you got is 8. You can write the same using numbers like 2 2 2 = 8. Now I will write it as 23, where 2 is the number of apples in each group and 3 is the total number of times the groups are multiplied. Here 2 is called the base and 3 as an exponent. You can represent this form generally as an. There are certain rules made to solve the expressions involving exponents easily. These are called exponent rules or Laws of exponents. Let us learn more about exponent rules now. With the help of many examples, we will discuss six of the most important laws of exponents in this article. ## Laws or Properties of Exponents: These are some of the laws or properties of exponents that are needed to be followed while solving the exponents. 1. Product Rule: As per this rule for real numbers, when bases are the same while doing the multiplication the exponents are added. a) am an = am+n, Eg: 23 22 = 25. 2. b) am a-n = am-n, Eg: 23 2-2 = 2. 3. Quotient Rule: As per this rule for real numbers, when bases are the same while doing the division the exponents are subtracted. a) am / an = am-n, when m > n. Eg: 23 / 22 = 2. 4. b) am / an = 1/an-m, when n > m. Eg: 23 / 25 = 1/22. 5. Power Rule: As per this rule, when the exponent of an exponent is given for the same base then the exponents are multiplied. (am)n = am n. Eg: (23)2 = 26. 6. Zero Law of Exponents: As per this rule, when the exponent of a number is zero. Then irrespective of the base value the value of that number is 1. I,e a0 = 1. This law is not applicable for 0 as a base. I.e 00 = undefined. 7. Power of the Product Rule: As per this rule, If there is a product of two different bases with the same power then it can be written as (a b)n = an bn. Here both a and b are non-zero numbers and n is an integer. Ex: (2 3)2 = 22 32. 8. Power of the Quotient Rule: As per this rule, If there is a division of two different bases with the same power then it can be written as (a b)n = an bn. Here both a and b is a non-zero numbers and n is an integer. Ex: (2 3)2 = 22 32 = 4/9. 9. Negative Law of Exponents: As per the law, the reciprocal of an exponent makes the exponent positive. I.e a-m = 1/ am. Eg: 2-6 = 1/26. 10. Root law of Exponents: According to this law, = x(1/n) and = x(m/n). Eg: = 3(1/3) and = 5(2/3).. Also Read: 12 Characteristics of Economics Students Who Are Successful ## Solved Examples: Example 1: Solution: = = 2, = = 2 = ⅖ Example 2: 23 – )0 (-3)3 Solution: 23 = 8, )0 = 1 and (-3)3 = – 27 23 – )0 (-3)3= 8 – 1 (-27) = 8 + 27 = 35. This is all about the rules or properties of exponents. For more curious information and solved problems about exponents log on to the Cuemath website. Exponent laws are also known as properties of exponents. Exponentiation simplifies multiplication and division operations and makes them easier to solve. Share via Send this to a friend
# Construct a 2 × 3 matrix Question: Construct a $2 \times 3$ matrix whose elements $a_{i j}$ are given by : (i) $a_{i j}=i . j$ (ii) $a_{i j}=2 j-j$ (iii) $a_{i j}=i+j$ (iv) $\mathrm{a}_{i j}=\frac{(i+j)^{2}}{2}$ Solution: (i) Here, $a_{i j}=i \cdot j, 1 \leq i \leq 2$ and $1 \leq j \leq 3$ $a_{11}=1 \times 1=1, a_{12}=1 \times 2=2, a_{13}=1 \times 3=3$ $a_{21}=2 \times 1=2, a_{22}=2 \times 2=4$ and $a_{23}=2 \times 3=6$ Required matrix $=A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6\end{array}\right]$ $(i i)$ Here, $a_{i j}=2 i-j$ $a_{11}=2(1)-1=2-1=1, a_{12}=2(1)-2=2-2=0, a_{13}=2(1)-3=2-3=-1$ $a_{21}=2(2)-1=4-1=3, a_{22}=2(2)-2=4-2=2$ and $a_{23}=2(2)-3=4-3=1$ Required matrix $=A=\left[\begin{array}{ccc}1 & 0 & -1 \\ 3 & 2 & 1\end{array}\right]$ $(i v)$ Here, $a_{i j}=\frac{(i+j)^{2}}{2}$ $a_{11}=\frac{(1+1)^{2}}{2}=\frac{(2)^{2}}{2}=\frac{4}{2}=2, a_{12}=\frac{(1+2)^{2}}{2}=\frac{(3)^{2}}{2}=\frac{9}{2}, a_{13}=\frac{(1+3)^{2}}{2}=\frac{(4)^{2}}{2}=\frac{16}{2}=8$ $a_{21}=\frac{(2+1)^{2}}{2}=\frac{(3)^{2}}{2}=\frac{9}{2}, a_{22}=\frac{(2+2)^{2}}{2}=\frac{(4)^{2}}{2}=\frac{16}{2}=8$ and $a_{23}=\frac{(2+3)^{2}}{2}=\frac{(5)^{2}}{2}=\frac{25}{2}$ Required matrix $=A=\left[\begin{array}{ccc}2 & \frac{9}{2} & 8 \\ \frac{9}{2} & 8 & \frac{25}{2}\end{array}\right]$
Add and Subtract Fractions With Unlike Denominators ```In order to add and subtract fractions with unlike denominators, you have to convert them into fractions with like denominators and corresponding numerators. The steps for adding and subtracting fractions are very similar until the very end, when you have to either add or subtract the numerators of the fractions. If you want to know how to add and subtract fractions with unlike denominators, just follow these steps. ``` Steps Finding a Common Denominator 1. Place the fractions side by side. Write the fractions you are using side by side. Keep the numerators (top numbers) level with each other on top, and the denominators (bottom numbers) on the line beneath them. Let's use the fractions 9 / 11 and 2 / 4 as our example. 2. Understand equivalent fractions. If you multiply the numerator and denominator in a fraction by the same number, you end up with an equivalent fraction, exactly equal to the first. For example, if you take 2 / 4, and multiply each number by 2, you get 4 / 8, which is an equal ("equivalent") fraction to 2/4. You can test this yourself by drawing the fractions: • Draw a circle, divide it into four equally sized pieces, then color in two of them (2 / 4). • Draw a new circle, divide it into 8 equal pieces, then color in four of them (4 / 8). • Compare the colored areas in the two circles, representing 2/4 and 4/8. These two areas have an equal size. 3. Multiply the two denominators to find a common denominator. Before we can add or subtract the fractions, we need to write them so they have the same denominator (a "common denominator") that is divisible by both numbers. The quickest way to find this is to multiply the two denominators together. Once you've written the answer down, you can skip ahead to the section on Add-and-Subtract-Fractions-With-Unlike-Denominators, or try the step below to find a different common denominator that may be easier to use. • For example, we started with the fractions 9 / 11 and 2 / 4. 11 and 4 are the denominators. • Multiply the two denominators together: 11 x 4 = 44. 4. Find a smaller common denominator instead (optional). The method above is quick, but you can instead find the "least common denominator," meaning the smallest possible answer. To do this, write down the multiples for each of the original denominators. Circle the smallest number that appears on both lists. Here's a new example, which we could use if we were solving "5/6 + 2/9": • The denominators are 6 and 9, so we want to "count by sixes" and "count by nines" to find the multiples: • Multiples of 6: 6, 12, 18, 24 • Multiples of 9: 9, 18, 27, 36 • Since 18 is in both tables, it can be used as a common denominator. Finishing the Problem 1. Change the first fraction to use the common denominator. In our first example, using 9/11 and 2/4, we decided to use 44 as the common denominator. But remember, we can't just change the denominator without multiplying the numerator by the same amount as well. Here's how we turn it into an equivalent fraction: • We know 11 x 4 = 44 (this is how we found the number 44 to begin with, but you can solve 44 ÷ 11 if you forgot). • Multiply both sides of the fraction by the same number to get the result: • (9 x 4 ) / (11 x 4) = 36/44 2. Do the same for the second fraction. Here's the second fraction in our example, 2 / 4, transformed into an equivalent fraction using 44 as the denominator: • 4 x 11 = 44 • (2 x 11) / (4 x 11) = 22/44. 3. Add or subtract the numerators of the fractions to get the answer. Once both fractions use the same denominator, you can add or subtract the numerators to get the answer: • Addition: 36 / 44 + 22 / 44 = (36 + 22) / 44 = 58/44 • Or subtraction: 36 / 44 - 22/44 = (36 - 22) / 44 = 14 / 44 4. Convert improper fractions into a mixed number. If the numerator ends up larger than the denominator, you have a fraction larger than 1 (an "improper fraction). You can make these into a mixed number, which is easier to read, by dividing the numerator by the denominator, and keeping the remainder as a fraction. For example, using the fraction 58 / 44, we get 58 ÷ 44 = 1, with remainder 14 left over. This means our final mixed number is 1 and 14/44. • If you're not sure how to divide the numbers, you can keep subtracting the bottom number from the top, writing down how many times you've subtracted. For example, convert 317 / 100 like this: • 317 - 100 = 217 (subtracted 1 time). 217 - 100 = 117 (subtracted 2 times). 117 - 100 = 17 (3 times). We can't subtract any more, so the answer is 3 and 17/100. 5. Simplify the fraction. Simplifying a fraction means writing it in its smallest equivalent form, to make it easier to use. Do this by dividing the numerator and denominator by the same number. If you can find a way to simplify the answer even further, keep doing it until you can't find another. For example, to simplify 14/44: • The numbers 14 and 44 are both divisible by 2, so let's use that. • (14 ÷ 2 ) / (44 ÷ 2) = 7 / 22 • There are no numbers that divide evenly into both 7 and 22, so this is our final, simplified answer. Example Problems • Try to solve these problems on your own. When you think you have the answer, Select-Phrases-of-Text-on-an-iOS-Device the invisible text after the equal sign to read the answer and check your work. The problems in each section get harder as you move down the list. The last ones are tricky, so don't expect to get every one on the first try: • 1 / 2 + 3 / 8 = 7 / 8 • 2 / 5 + 1 / 3 = 11 / 15 • 3 / 4 + 4 / 8 = 1 and 1/4 • 10 / 3 + 3 / 9 = 3 and 2/3 • 5 / 6 + 8 / 5 = 2 and 13/30 • 2 / 17 + 4 / 5 = 78 / 85 Practice subtraction problems: • 2 / 3 - 5 / 9 = 1 / 9 • 15 / 20 - 3 / 5 = 3 / 20 • 7 / 8 - 7 / 9 = 7 / 72 • 3 / 5 - 4 / 7 = 1 / 35 • 7 / 12 - 3 / 8 = 5 / 24 • 16 / 5 - 1 / 4 = 2 and 19/20 Tips • The Least Common Denominator is shortened as "LCD." "Least" means "smallest" and "common" means "shared", so in more modern language you can think of it as the "smallest shared denominator" you can use for both fractions.
# Module Focus: Grade 5 Math Module 5 The Grade 5 presentation focused on the material covered in Module 5: Addition and Multiplication with Volume and Area. Participants practiced lots of hands-on activities in order to experience the progression students experience with developing their concept of volume and area. Students first experience calculating volume by building figures and counting unit cubes. Students construct open boxes and calculate volumes by “filling” in the box. They then experience volume pictorially through the use of dot paper and constructing cubes. There is much discussion in the module about composing and decomposing right rectangular prisms using layers, which helps with students’ conceptual knowledge of what volume actually means. There is no mention of a volume formula in Topic A. Topic B is where the multiplication formula is introduced with the concept of layers. Students also explore the connection between volume in cm and liquid volume in mL. We were able to see the liquid volume increase by 1 mL after the dropping of a cubic centimeter – very cool! Application problems were presented at this point, such as: • A small fish tank is filled to the top with water. If the tank measures 15 cm x 10 cm x 10 cm, what is the volume of the water in the tank? Express answer in Liters. What if after a week, water evaporates so that the water level in the tank is 9 cm high? What effect does that have on the volume of the water? How many Liters? • This is an interesting problem in that students can just take off the “layer” from the original water level, or they can re-calculate with a new height of 9 cm. • A shed in the shape of a right rectangular prism measures 6 ft. long by 5 ft. wide by 8 ft. high. The owner realizes that he needs 480 cubic feet of storage. Will he achieve this goal if he doubles each dimension? If he wants to keep the height the same, what could the other dimensions be for him to get the volume that he wants? • This problem lends itself to discussing what happens to volume when you double one dimension, two dimensions, or all three. The “create a sculpture” activity in lessons 8-9 is an opportunity for students to express their creativity, while at the same time apply the concepts and formula of volume to design a sculpture within a given set of parameters. The activity is graded with a rubric used by the students. Participants discussed the value of having students use a rubric. Peer review always holds students more accountable, but the peer review also ties into the Mathematical Practice of critiquing the reasoning of others. Topic C shifts the focus from volume to calculating the area of rectangles with fractional side lengths. Once again, this demonstrated an excellent transition from concrete, pictorial to abstract. Students tile a rectangular region using patty paper, then draw the image on white paper (area model), and then use prior knowledge of area (partial products) and the multiplication of fractions to calculate the area. Participants practiced this transition using mystery rectangles. The topic ends with application problems that ask students to decide which process is more efficient and whether they should deal with improper fractions or convert to mixed numbers. We want them to say “It depends.” Topic D uses the cutting apart of trapezoids and parallelograms in order to take a look at the properties that exist for each, leading the student towards success in being able to create a hierarchy of quadrilaterals that go from most general to specific. Excellent visual activities were done here with parallelograms constructed by the group so that participants had a wide range of parallelograms. Activities showed the angle relationships that exist within these shapes (consecutive angles supplementary, all four angles add up to 360). Participants looked at diagonals for parallelograms and great questioning techniques were modeled in regards to answering the question, “will the diagonals always be bisected, or are they ever the same?” Angle measurement was recommended as a fluency activity. An excellent end to an excellent week here at NTI.
## SQUARE ROOT TRICKS FOR FAST CALCULATION square root is the important chapter in maths. 2-3 question are always based on this topic in every exams. While in banking 8 – 10 question are based on this square chapter or on the bases of square root tricks we going to solve within seconds. we provide tricks how to calculate SQUAREfast within seconds if you practice, practice and practice I think you will be able to solve square to 30 to 130 within few seconds and you can save time is exams. squares • number with last digit 5. • from 30 to 50. • from 50 to 80. • from 80 to 100. • from 100 to 130.FIRST OF ALL YOU CAN LEARN SQUARE BETWEEN 1 TO 30 IN MIND NUMBER SQUARE NUMBER SQUARE 1 1 16 256 2 4 17 289 3 9 18 324 4 16 19 361 5 25 20 400 6 36 21 441 7 49 22 484 8 64 23 529 9 81 24 576 10 100 25 625 11 121 26 676 12 144 27 729 13 169 28 784 14 196 29 841 15 225 30 900 FIRST CASE:- NUMBER WITH LAST DIGIT 5 SECOND CASE:- SQUARE BETWEEN 30 TO 50 THIRD CASE:- SQUARE BETWEEN 50 TO 80 FOURTH CASE:- SQUARE BETWEEN 80 TO 100 FIFTH CASE:- SQUARE BETWEEN 100 TO 130 MOCK TEST Q1. Find the square of 95? a. 9125 b. 9025 c. 9225 d. 9175 Q2. Find the square of 42? a. 1763 b. 1764 c. 1765 d. 1766 Q3. Find the square of 79? a. 6233 b. 6255 c. 6241 d. 6251 Q4. Find the square of 121? a. 15252 b. 12625 c. 14625 d. 14641 Q5. find the square of 98? a. 9604 b. 9609 c. 9603 d. 9600
# 6.12: Expression Evaluation with Mixed Numbers Difficulty Level: At Grade Created by: CK-12 Estimated11 minsto complete % Progress Practice Expression Evaluation with Mixed Numbers MEMORY METER This indicates how strong in your memory this concept is Progress Estimated11 minsto complete % Estimated11 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Have you ever had to combine pieces of something to make a whole? Travis is doing exactly that. He has three pieces of pipe that has to be clamped together. It will be connected by a professional, but Travis needs to combine the pieces of pipe and figure out the total length that he has. The first piece of pipe measures \begin{align}5 \frac{1}{3}\end{align} feet. The second piece of pipe measures \begin{align}6 \frac{1}{2}\end{align} feet. The third piece of pipe measures \begin{align}2 \frac{1}{3}\end{align} feet. If Travis is going to combine these together, then he has to add mixed numbers. This Concept will teach you how to evaluate numerical expressions involving mixed numbers. Then we will return to this original problem once again. ### Guidance Sometimes, we can have numerical expressions that have both addition and subtraction in them. When this happens, we need to add or subtract the mixed numbers in order from left to right. \begin{align}4\frac{1}{6}+3\frac{4}{6}-1\frac{4}{6}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Here is a problem with two operations in it. These operations are addition and subtraction. All of these fractions have the same common denominator, so we can begin right away. We start by performing the first operation. To do this, we are going to add the first two mixed numbers. \begin{align}4\frac{1}{6}+3\frac{4}{6}=7\frac{5}{6}\end{align} Now we can perform the final operation, subtraction. We are going to take the sum of the first two mixed numbers and subtract the final mixed number from this sum. \begin{align}7\frac{5}{6}-1\frac{4}{6}=6\frac{1}{6}\end{align} Our final answer is \begin{align}6\frac{1}{6}\end{align}. What about when the fractions do not have a common denominator? When this happens, you must rename as necessary to be sure that all of the mixed numbers have one common denominator before performing any operations. After this is done, then you can add/subtract the mixed numbers in order from left to right. \begin{align}2\frac{4}{6}+1\frac{1}{6}-1\frac{1}{2}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} The fraction parts of these mixed numbers do not have a common denominator. We must change this before performing any operations. The lowest common denominator between 6, 6 and 2 is 6. Two of the fractions are already named in sixths. We must rename the last one in sixths. \begin{align}1\frac{1}{2}=1\frac{3}{6}\end{align} Next we can rewrite the problem. \begin{align}2\frac{4}{6}+1\frac{1}{6}-1\frac{3}{6}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Add the first two mixed numbers. \begin{align}2\frac{4}{6}+1\frac{1}{6}=3\frac{5}{6}\end{align} Now we can take that sum and subtract the last mixed number. \begin{align}3\frac{5}{6}-1\frac{3}{6}=2\frac{2}{6}\end{align} Don’t forget to simplify. \begin{align}2\frac{2}{6}=2\frac{1}{3}\end{align} Now it's time for you to try a few on your own. Be sure your answer is in simplest form. #### Example A \begin{align}6\frac{4}{8}+2\frac{2}{8}-1\frac{1}{8}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Solution: \begin{align}7 \frac{5}{8}\end{align} #### Example B \begin{align}4\frac{3}{9}+2\frac{1}{3}-1\frac{2}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Solution: \begin{align}5 \frac{4}{9}\end{align} #### Example C \begin{align}2\frac{1}{3}+ 5\frac{1}{3}-6\frac{1}{4}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Solution: \begin{align}1 \frac{5}{12}\end{align} Now back to Travis and the pipe. Here is the original problem once again. Travis is doing exactly that. He has three pieces of pipe that has to be clamped together. It will be connected by a professional, but Travis needs to combine the pieces of pipe and figure out the total length that he has. The first piece of pipe measures \begin{align}5 \frac{1}{3}\end{align} feet. The second piece of pipe measures \begin{align}6 \frac{1}{2}\end{align} feet. The third piece of pipe measures \begin{align}2 \frac{1}{3}\end{align} feet. If Travis is going to combine these together, then he has to add mixed numbers. To solve this, we can begin by writing an expression that shows all three mixed numbers being added together. \begin{align}5\frac{1}{3}+ 6\frac{1}{2} + 2\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Now we can convert all of the mixed numbers to improper fractions. \begin{align} \frac{16}{3} + \frac{13}{2} + \frac{7}{3}\end{align} Next, we rename each fraction using the lowest common denominator. The LCD of 3 and 2 is 6. \begin{align} \frac{32}{6} + \frac{39}{6} + \frac{14}{6}\end{align} \begin{align} \frac{85}{6} = 14 \frac{1}{6}\end{align} feet. ### Vocabulary Here are the vocabulary words in this Concept. Mixed Number a number that has a whole number and a fraction. Numerical Expression a number expression that has more than one operation in it. Operation ### Guided Practice Here is one for you to try on your own. \begin{align}2\frac{1}{8}+ 3\frac{1}{4}-2\frac{1}{2}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} To start, we need to convert all of the mixed numbers to improper fractions. \begin{align} \frac{17}{8} + \frac{13}{4} - \frac{5}{2}\end{align} Now we rename each fraction using the lowest common denominator. The LCD of 8, 4 and 2 is 8. \begin{align} \frac{17}{8} + \frac{26}{8} - \frac{20}{8}\end{align} Now we can combine and simplify. \begin{align} \frac{23}{8} = 2 \frac{7}{8}\end{align} ### Video Review Here are videos for review. ### Practice Directions: Evaluate each numerical expression. Be sure your answer is in simplest form. 1. \begin{align}2\frac{1}{3}+4\frac{1}{3}-1\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 2. \begin{align}6\frac{2}{5}+6\frac{2}{5}-1\frac{1}{5}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 3. \begin{align}7\frac{3}{9}+8\frac{1}{9}-1\frac{2}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 4. \begin{align}8\frac{3}{10}+2\frac{5}{10}-6\frac{4}{10}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 5. \begin{align}6\frac{1}{5}+2\frac{3}{5}-1\frac{1}{5}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 6. \begin{align}9\frac{4}{9}+2\frac{4}{9}-3\frac{5}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 7. \begin{align}6\frac{9}{12}+3\frac{2}{12}-8\frac{4}{12}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 8. \begin{align}7\frac{8}{9}-1\frac{1}{9}+1\frac{3}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 9. \begin{align}6\frac{4}{8}+3\frac{4}{8}-6\frac{6}{8}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 10. \begin{align}14\frac{2}{3}-2\frac{1}{3}+1\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 11. \begin{align}12\frac{6}{9}+12\frac{8}{9}-10\frac{7}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 12. \begin{align}9\frac{1}{7}+12\frac{3}{7}+1\frac{2}{7}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 13. \begin{align}14\frac{3}{4}+2\frac{1}{4}-1\frac{3}{4}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 14. \begin{align}18\frac{6}{15}+2\frac{3}{15}-4\frac{2}{15}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 15. \begin{align}12\frac{1}{9}+2\frac{1}{3}-1\frac{1}{6}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Mixed Number A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$. Numerical expression A numerical expression is a group of numbers and operations used to represent a quantity. operation Operations are actions performed on variables, constants, or expressions. Common operations are addition, subtraction, multiplication, and division. Operations Operations are actions performed on variables, constants, or expressions. Common operations are addition, subtraction, multiplication, and division. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects: ## Concept Nodes: Date Created: Oct 29, 2012 Sep 04, 2016 Image Detail Sizes: Medium \| Original MAT.ARI.316.4.L.1
NCERT Solutions (Ex - 4.4, 4.5) - Practical Geometry, Maths, Class 8 Notes - Class 8 Class 8: NCERT Solutions (Ex - 4.4, 4.5) - Practical Geometry, Maths, Class 8 Notes - Class 8 The document NCERT Solutions (Ex - 4.4, 4.5) - Practical Geometry, Maths, Class 8 Notes - Class 8 is a part of Class 8 category. All you need of Class 8 at this link: Class 8 Exercise 4.4 Question 1: Construct the following quadrilaterals: DE = 4 cm, EA = 5 cm, AR = 4.5 cm, ∠E = 60o ,∠A = 90o TR = 3.5 cm, RU = 3 cm, UE = 4 cm, ∠ R= 75°, ∠ U = 120° (i) Given: DE = 4cm, EA = 5cm, AR = 4.5 cm, ∠E = 60% ,∠A = 90o To construct: A quadrilateral DEAR. Steps of construction: (a) Draw a line segment DE = 4 cm. (b) At point E, construct an angle of 60o. (c) Taking radius 5 cm, draw an arc from point E which intersects at A. (d) Construct , draw an arc of radius 4.5 cm with centre A which intersect at R. (e) Join RD. It is the required quadrilateral DEAR. (ii) Given: TR = 3.5 cm, RU = 3 cm, UE = 4 cm, ∠R = 75°, ∠U = 120o Steps of construction: 1. Draw a line segment TR = 3.5 cm. 2. Construct an angle 75o at R and draw an arc of radius 3 cm with R as centre, which intersects at U. 3. Construct an angle of 120° at U and produce the side UE. 4. Draw an arc of radius 4 cm with U as centre. 5. Join UE and TE. It is the required quadrilateral TRUE. Exercise 4.5 Question 1: Draw the following: The square READ with RE = 5.1 cm. Given: RE = 5.1 cm. Steps of construction: (i) Draw RE = 5.1 cm. (ii) At point E, construct an angle of 90o and draw an arc of radius 5.1 cm, which intersects at point A. (iii) At point R, draw an arc of radius 5.1 cm at point A, draw another arc of radius 5.1 cm which intersects the first arc at point D. It is the required square READ, Question 2: Draw the following: A rhombus whose diagonals are 5.2 cm and 6.4 cm. Given: Diagonals of a rhombus AC = 5.2 cm and BD = 6.4 cm. To construct: A rhombus ABCD. Steps of construction: (a) Draw AC = 5.2 cm and draw perpendicular bisectors on AC. (b) Since, diagonals bisect at mid-point O, therefore get half of 6.4 cm, i.e., 3.2 cm. (c) Draw two arcs on both sides of AC of radius 3.2 cm from intersection point O, which intersects at B and D. (d) Join AB, BC, CD and DA. It is required rhombus ABCD. Question 3: Draw the following: A rectangle with adjacent sides of length 5 cm and 4 cm. Given: MN = 5 cm and MP = 4 cm. To construct: A rectangle MNOP Steps of construction: (a) Draw a segment MN = 5 cm. (b) At points M and N, draw perpendiculars of lengths 4 cm and produce them. (c) Taking centres M and N, draw two arcs of 4 cm each, which intersect P and Q respectively. (d) Join side PO. It is required rectangle MNOP. Question 4: Draw the following: A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Given: OK = 5.5 cm and KA = 4.2 cm. To construct: A parallelogram OKAY. Steps of construction: (a) Draw a line segment OK = 5.5 cm. (b) Draw an angle of 90 at K and draw an arc of radius KA = 4.2 cm, which intersects at point A. (c) Draw another arc of radius AY = 5.5 cm and at point O, draw another arc of radius 4.2 cm which intersect at Y. (d) Join AY and OY. It is the required parallelogram OKAY. The document NCERT Solutions (Ex - 4.4, 4.5) - Practical Geometry, Maths, Class 8 Notes - Class 8 is a part of Class 8 category. All you need of Class 8 at this link: Class 8 Use Code STAYHOME200 and get INR 200 additional OFF Top Courses for Class 8 Track your progress, build streaks, highlight & save important lessons and more! , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ;
# 9.2: Inferences for Two Population Means- Large, Independent Samples $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ ##### Learning Objectives • To understand the logical framework for estimating the difference between the means of two distinct populations and performing tests of hypotheses concerning those means. • To learn how to construct a confidence interval for the difference in the means of two distinct populations using large, independent samples. • To learn how to perform a test of hypotheses concerning the difference between the means of two distinct populations using large, independent samples. Suppose we wish to compare the means of two distinct populations. Figure $$\PageIndex{1}$$ illustrates the conceptual framework of our investigation in this and the next section. Each population has a mean and a standard deviation. We arbitrarily label one population as Population $$1$$ and the other as Population $$2$$, and subscript the parameters with the numbers $$1$$ and $$2$$ to tell them apart. We draw a random sample from Population $$1$$ and label the sample statistics it yields with the subscript $$1$$. Without reference to the first sample we draw a sample from Population $$2$$ and label its sample statistics with the subscript $$2$$. ##### Definition: Independence Samples from two distinct populations are independent if each one is drawn without reference to the other, and has no connection with the other. Our goal is to use the information in the samples to estimate the difference $$\mu _1-\mu _2$$ in the means of the two populations and to make statistically valid inferences about it. ## Confidence Intervals Since the mean $$x-1$$ of the sample drawn from Population $$1$$ is a good estimator of $$\mu _1$$ and the mean $$x-2$$ of the sample drawn from Population $$2$$ is a good estimator of $$\mu _2$$, a reasonable point estimate of the difference $$\mu _1-\mu _2$$ is $$\bar{x_1}-\bar{x_2}$$. In order to widen this point estimate into a confidence interval, we first suppose that both samples are large, that is, that both $$n_1\geq 30$$ and $$n_2\geq 30$$. If so, then the following formula for a confidence interval for $$\mu _1-\mu _2$$ is valid. The symbols $$s_{1}^{2}$$ and $$s_{2}^{2}$$ denote the squares of $$s_1$$ and $$s_2$$. (In the relatively rare case that both population standard deviations $$\sigma _1$$ and $$\sigma _2$$ are known they would be used instead of the sample standard deviations.) ## $$100(1-\alpha )\%$$ Confidence Interval for the Difference Between Two Population Means: Large, Independent Samples The samples must be independent, and each sample must be large: ##### Example $$\PageIndex{1}$$ To compare customer satisfaction levels of two competing cable television companies, $$174$$ customers of Company $$1$$ and $$355$$ customers of Company $$2$$ were randomly selected and were asked to rate their cable companies on a five-point scale, with $$1$$ being least satisfied and $$5$$ most satisfied. The survey results are summarized in the following table: Company 1 Company 2 $$n_1=174$$ $$n_2=355$$ $$x-1=3.51$$ $$x-2=3.24$$ $$s_1=0.51$$ $$s_2=0.52$$ Construct a point estimate and a 99% confidence interval for $$\mu _1-\mu _2$$, the difference in average satisfaction levels of customers of the two companies as measured on this five-point scale. ###### Solution The point estimate of $$\mu _1-\mu _2$$ is $\bar{x_1}-\bar{x_2}=3.51-3.24=0.27 \nonumber$ In words, we estimate that the average customer satisfaction level for Company $$1$$ is $$0.27$$ points higher on this five-point scale than it is for Company $$2$$. To apply the formula for the confidence interval, proceed exactly as was done in Chapter 7. The $$99\%$$ confidence level means that $$\alpha =1-0.99=0.01$$ so that $$z_{\alpha /2}=z_{0.005}$$. From Figure 7.1.6 "Critical Values of " we read directly that $$z_{0.005}=2.576$$. Thus $(\bar{x_1}-\bar{x_2})\pm z_{\alpha /2}\sqrt{\frac{s_{1}^{2}}{n_1}+\frac{s_{2}^{2}}{n_2}}=0.27\pm 2.576\sqrt{\frac{0.51^{2}}{174}+\frac{0.52^{2}}{355}}=0.27\pm 0.12 \nonumber$ We are $$99\%$$ confident that the difference in the population means lies in the interval $$[0.15,0.39]$$, in the sense that in repeated sampling $$99\%$$ of all intervals constructed from the sample data in this manner will contain $$\mu _1-\mu _2$$. In the context of the problem we say we are $$99\%$$ confident that the average level of customer satisfaction for Company $$1$$ is between $$0.15$$ and $$0.39$$ points higher, on this five-point scale, than that for Company $$2$$. ## Hypothesis Testing Hypotheses concerning the relative sizes of the means of two populations are tested using the same critical value and $$p$$-value procedures that were used in the case of a single population. All that is needed is to know how to express the null and alternative hypotheses and to know the formula for the standardized test statistic and the distribution that it follows. The null and alternative hypotheses will always be expressed in terms of the difference of the two population means. Thus the null hypothesis will always be written $H_0: \mu _1-\mu _2=D_0 \nonumber$ where $$D_0$$ is a number that is deduced from the statement of the situation. As was the case with a single population the alternative hypothesis can take one of the three forms, with the same terminology: Form of Ha Terminology $$H_a: \mu _1-\mu _2<D_0$$ Left-tailed $$H_a: \mu _1-\mu _2>D_0$$ Right-tailed $$H_a: \mu _1-\mu _2\neq D_0$$ Two-tailed As long as the samples are independent and both are large the following formula for the standardized test statistic is valid, and it has the standard normal distribution. (In the relatively rare case that both population standard deviations $$\sigma _1$$ and $$\sigma _2$$ are known they would be used instead of the sample standard deviations.) ## Standardized Test Statistic for Hypothesis Tests Concerning the Difference Between Two Population Means: Large, Independent Samples $Z=\frac{(\bar{x_1}-\bar{x_2})-D_0}{\sqrt{\frac{s_{1}^{2}}{n_1}+\frac{s_{2}^{2}}{n_2}}} \nonumber$ The test statistic has the standard normal distribution. The samples must be independent, and each sample must be large: $$n_1\geq 30$$ and $$n_2\geq 30$$. ##### Example $$\PageIndex{2}$$ Refer to Example $$\PageIndex{1}$$ concerning the mean satisfaction levels of customers of two competing cable television companies. Test at the $$1\%$$ level of significance whether the data provide sufficient evidence to conclude that Company $$1$$ has a higher mean satisfaction rating than does Company $$2$$. Use the critical value approach. Solution: • Step 1. If the mean satisfaction levels $$\mu _1$$ and $$\mu _2$$ are the same then $$\mu _1=\mu _2$$, but we always express the null hypothesis in terms of the difference between $$\mu _1$$ and $$\mu _2$$, hence $$H_0$$ is $$\mu _1-\mu _2=0$$. To say that the mean customer satisfaction for Company $$1$$ is higher than that for Company $$2$$ means that $$\mu _1>\mu _2$$, which in terms of their difference is $$\mu _1-\mu _2>0$$. The test is therefore $H_0: \mu _1-\mu _2=0 \nonumber$ $vs. \nonumber$ $H_a: \mu _1-\mu _2>0\; \; @\; \; \alpha =0.01 \nonumber$ • Step 2. Since the samples are independent and both are large the test statistic is $Z=\frac{(\bar{x_1}-\bar{x_2})-D_0}{\sqrt{\frac{s_{1}^{2}}{n_1}+\frac{s_{2}^{2}}{n_2}}} \nonumber$ • Step 3. Inserting the data into the formula for the test statistic gives $Z=\frac{(\bar{x_1}-\bar{x_2})-D_0}{\sqrt{\frac{s_{1}^{2}}{n_1}+\frac{s_{2}^{2}}{n_2}}}=\frac{(3.51-3.24)-0}{\sqrt{\frac{0.51^{2}}{174}+\frac{0.52^{2}}{355}}}=5.684 \nonumber$ • Step 4. Since the symbol in $$H_a$$ is “$$>$$” this is a right-tailed test, so there is a single critical value, $$z_\alpha =z_{0.01}$$, which from the last line in Figure 7.1.6 "Critical Values of " we read off as $$2.326$$. The rejection region is $$[2.326,\infty )$$. Figure $$\PageIndex{2}$$: Rejection Region and Test Statistic for Example $$\PageIndex{2}$$ • Step 5. As shown in Figure $$\PageIndex{2}$$ the test statistic falls in the rejection region. The decision is to reject $$H_0$$. In the context of the problem our conclusion is: The data provide sufficient evidence, at the $$1\%$$ level of significance, to conclude that the mean customer satisfaction for Company $$1$$ is higher than that for Company $$2$$. ##### Example $$\PageIndex{3}$$ Perform the test of Example $$\PageIndex{2}$$ using the $$p$$-value approach. Solution: The first three steps are identical to those in Example $$\PageIndex{2}$$ • Step 4. The observed significance or $$p$$-value of the test is the area of the right tail of the standard normal distribution that is cut off by the test statistic $$Z=5.684$$. The number $$5.684$$ is too large to appear in Figure 7.1.5, which means that the area of the left tail that it cuts off is $$1.0000$$ to four decimal places. The area that we seek, the area of the right tail, is therefore $$1-1.0000=0.0000$$ to four decimal places. See Figure $$\PageIndex{3}$$. That is, $$p$$-value=$$0.0000$$ to four decimal places. (The actual value is approximately $$0.000000007$$.) • Step 5. Since $$0.0000<0.01$$, $$p-value <\alpha$$ so the decision is to reject the null hypothesis: The data provide sufficient evidence, at the $$1\%$$ level of significance, to conclude that the mean customer satisfaction for Company $$1$$ is higher than that for Company $$2$$. ##### Key Takeaway • A point estimate for the difference in two population means is simply the difference in the corresponding sample means. • In the context of estimating or testing hypotheses concerning two population means, “large” samples means that both samples are large. • A confidence interval for the difference in two population means is computed using a formula in the same fashion as was done for a single population mean. • The same five-step procedure used to test hypotheses concerning a single population mean is used to test hypotheses concerning the difference between two population means. The only difference is in the formula for the standardized test statistic. 9.2: Inferences for Two Population Means- Large, Independent Samples is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.
## Deploy your apps to a supercloud in a few clicks This Engineering Education program is supported by Section. Instantly deploy your GitHub apps, Docker containers or K8s namespaces to a supercloud. # Solving Linear Equations Using Matlab ##### September 15, 2021 In mathematics, equations in the form `Ax=b` are linear algebra equations. In such equations, `A` is a matrix while `x` and `b` are column vectors. A matrix is a two-dimensional arrangement of numbers. Such equations are common in engineering and scientific disciplines. Thus, understanding the setup of these equations and finding solutions to the problems is an essential skill. Matlab gives a powerful and reliable way to find solutions to these problems. But, we will also realize that the solutions provided are not always what they appear to be. In this article, we will learn how to solve these problems using Matlab. It is a matrix laboratory, hence the best environment for solving matrix problems. ### Prerequisites To follow along with this tutorial, you’ll need: • MATLAB installed. • Proper understanding of MATLAB basics. • Basic understanding of matrix. ### Solving simultaneous equations We are going to look at how to solve simultaneous equations using Matlab. Simultaneous equations are finite sets of equations for which common solutions are set. What we mean is that they are conditions that define the relationship between two unknowns through an equal number of equations. We are going to use the matrix method. A matrix is a two-dimensional arrangement of numbers. For example, Defining the matrix is by dimensions `m x n`, where `m` is the number of rows while `n` is the number of columns. A row is a horizontal arrangement, while a column is the vertical arrangement of the numbers. If we have a 3x2 matrix, then what that means is that it has 3 rows and 2 columns. These matrices are used compactly to work with linear equations. The different forms of the matrix are; • Row vector are matrices with a single row. • Column vector are matrices with a single column. • Square matrices are matrices where `m=n`, that is, the number of columns equals the number of rows. The basic form of linear equations is; Where `A_ij` are the elements of the `MxN` matrix, `X_j` are the elements of Nx1 matrix column vectors, and `b_i` are the elements of the `Mx1` row vector. For example, given a simultaneous equation shown below; The simplification of this equation is; For you to solve these simultaneous equations using the matrix method, `m` of the second matrix must equalize `n` of the first matrix after the simplification done above. It is because solving simultaneous equations using Matlab involves the multiplication of the matrix. These equations are simultaneous because one set of `x_i` must satisfy all the equations of `M`. Assume that you have the value of `A` and `x` to find `b`, then the equation is easy to solve. You apply the matrix multiplication method. The big problem is finding `x` given `A` and `b`; focusing on such problems, we will see how to handle them. ### Matlab’s solution The basic operations that you use to solve these equations in Matlab depend on the variable provided. When; • `A` and `x` are provided, the solution is `b = Ax`. The `n` of `A` must equal `m` of ` x` for this operation to work. • `A` and `b` is provided, the solution is `A/b`. Here, `m` of `A` must equal to `m` of `b`. Example: Below is the first matrix, `A`. Below is the second matrix `b`. Now that you have `A` and `b`, we are supposed to find `x`. When you have `A` and `b`, we use `x =A\b`. So to get this done, execute the following command in the command window. ``````A = [4 5; 3 -2]; b = [6; 14]; x = A\b `````` We have assigned our matrices to variables `A` and `b` then gave the formulae used to solve the problem. In Matlab, the separation of rows is by semicolons. When you execute the command, the result is displayed as shown below; ``````x = 3.5652 -1.6522 `````` Matlab provides a solution to linear algebra. But, at some point, it doesn’t produce the solution, or the solutions provided are less trustworthy. So, Matlab will give the user a warning at some point, but this happens in rare cases. Now, the fault here may not be due to poor syntax or Matlab’s fault, but it could be due to the user’s failure to understand linear algebra. But don’t worry. I am here to show you the causes of such problems and the meaning of the output given by Matlab. The problem we solved before had two unknowns, `x` and `y`, and since there were two unknowns, the output had a length of two(column vector). There are various types of simultaneous equations: • Inconsistent equations. • Undetermined equations. • Overdetermined equations. ### Inconsistent equations Inconsistent equations are equations in which `m=n`(number of rows equals the number of columns), but the solution does not exist. In these types of equations, the left sides are equal, but right sides are not e.g What we mean is that for the left side we have `4x+y` for both equation 1 and 2 while for the right side we have 6 and 14 for equation 1 and 2 respectively. It shows that the equations are inconsistent. Solving simultaneous is finding the point at which two equations meet when plotted. When you plot these two equations, the lines are parallel. Let’s make a plot to visualize what we are talking about in the above statement. Since we are plotting `x` against `y`, we will first make `y` the subject for each equation then plot. As shown below: To plot, we give the `x` variables first ``````x = [-10:10]; % x ranges from -10 to 10 y1 = (6 -4x)/5; %y variables y2 = (14 -4x)/5; plot(x, y1, x, y2) %plotting the two equations legend('y1', 'y2') %Adding legends to the plot `````` The two lines are parallel. It means that the two lines intersect at infinity. So when you try to get the solution to this problem using Matlab, the output given for the unknowns is `inf`, which means infinity. ``````x = -Inf Inf `````` ### Undetermined equations These are equations in which `m>n`. It means that the provided information is insufficient to give a solution to the problem example; 4x + 5y =6 Mathematically, the solution is `y = (6-4x)/5`. It means that the `x` value can range from `-inf` to `inf` as long as it works with the provided `y`. If Matlab is used to solve such equations, it will give only one value and the other set to 0. ``````A = [4 5]; b = 6; x = A\b `````` The output will be; ``````x = 0 1.2000 `````` Inadequate information not only occurs in cases where `m=n`, but it can also occur for problems in which `m>n`. It could be due to the redundancy of other equations. e.g In the above equation, the information is insufficient since equations (i) and (ii) can be generated from equation (i). If you take equation(i) and multiply by 2, we get equation (ii). If you also take equation(i) and divide by -2, we get equation(iii). ### Overdetermined equations These kind of equations mainly occur when `m>n`. Here, the information provided by the equations is too much. The equation `Ax=b` cannot be satisfied simultaneously by any value of the vector `x`. When you solve the equation using Matlab, it will give an output, but it does not satisfy the matrix rule `ax=b`. To understand this, we start by plotting the equations below; For the first equation, `y1 = -(4x - 6)/5`, plot `y2` and `y1` against `x`. ``````x = [-10;10]; y1 = (-4x + 6)/5; y2 = (3x -14)/2; plot(x, y1, x, y2); grid on xlabel('x'); ylabel('y'); legend('y1', 'y2') `````` We will use the `intersect function` to find the point at which the lines intersects. The intersect function uses the gradients and the constants as the inputs as shown below; ``````function [x0, y0] = intersectPoints(m1,m2,b1,b2) %m is the gradient while b is the constant. x0 = (b2-b1)/(m1-m2); %find the x point y0 = m1x0+b1; end `````` To call this function in the command window; ``````intersectPoints(-4/5, 3/2, 6/5, -14/2) `````` The output will be; ``````ans = 3.5652 `````` This function gives a single output corresponding to the `x` value for the point of intersection. To find `y`, you can replace `x` in the equation and get your `y`. The diagram shows that the solution is similar to the one we found before(3.5652,-1.6522). It means that the solution is correct. Now assume that we have the equation below; Let’s plot the line of the three equations. To do that, we add the code below; ``````y3 = 7x - 25; hold on plot(x, y1, x,y2, x, y3) `````` When you execute the above commands, we get a 3rd line(the green line) which is for the third equation. Looking at the plots, you may think these three lines intersect at a common point, but this is not the case. To see this, zoom out the plots by clicking on the magnifying lens icon having `+` inside it. After clicking it, move to the point at which the lines seems to intersect a few times, and the result will be; This shows that there is no perfect solution to these problems. Let’s try solving it using matlab and plot this output to see the position that matlab gives as the output. This can be done by the code below; ``````A = [4 5; 3 -2; 7 -1]; b = [6; 14;25]; x = A\b; hold on plot(x(1),x(2),'k') plot(x(1),x(2),'r') `````` As you can see, Matlab is trying to locate a point close to all three lines, and this is an approximation. You can find the error to the solution by; ``````error = A*x -b `````` Matlab locates this closest point by finding the square root of the sum of squares of elements of the input vector. Locating this point is by use of the `norm` function. ``````norm(error) `````` The output is `0.7941` and is known as `optimum solution in the least square sense`. It is the point at which the `norm(error)` is as small as possible. If we try other points, we get `norm(error)` to be higher than `0.7941`. ### Conclusion Solving simultaneous equations in Matlab depends on the type of problem that you are handling. As a user, you should know the type of equation or problem you are trying to solve. Besides, it will help you know the output you expect from Matlab. What you must know is that Matlab will provide a solution to all these problems. Also, knowing the type of equation helps you avoid errors in linear equations. That’s it. Happy Coding. Peer Review Contributions by: Espira Marvin
# Solving $x\|x-1\|<3$? Seems pretty straight-forward but I'm a bit confused with the modulus and inequality. So I know that if $\|x-1\|>3$ because the entire equation on the left is within the modulus we can write $x-1>3$ or $x-1<-3$ I also know that if $y<0$ $y\|x-1\|>3$ Can be written as $\|x-1\|<3/y$ But what do we do with $x\|x-1\|<3$ Can we break it down further without trying to input values? • What is the range of your solutions? Do you include real numbers, integers or something else? Commented Jul 20, 2018 at 7:15 • Lets assume all integers Commented Jul 20, 2018 at 7:17 • There's actually a smarter way to do this problem – which is to separate it into $x < 0$, $0 < x < 1$, and $1, 2, 3 \cdots$. Commented Jul 20, 2018 at 7:30 • @TobyMak I don't think I understand the smarter way that you are talking about. Do you mean putting values? Commented Jul 20, 2018 at 7:35 • If $x < 0$, $x$ is negative but $\|x-1\|$ is positive, so $x\|x-1\|$ is negative and $<0$. If $0<x<1$, $x < 1$ and $\|x-1\| ≤ 1$, so $x\|x-1\|$ is at most $1$. Now you can check the cases $1, 2, 3$ separately. Commented Jul 20, 2018 at 7:37 Let's do it like this. For $x>1$ the equation is equivalent to $x(x-1)<3$ this gives $x^2-x-3<0$ Roots of this equation are $$\frac{1+\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}$$ So this gives the range $x\in(1, \frac{1+\sqrt{13}}{2})$ For $x<1$ the equation is $-x(x-1)<3$ or $x^2-x+3>0$ this always satisfied for any values of $x$. Hence $x\in(-\infty, \frac{1+\sqrt{13}}{2})$, taking $x$ to be integers would make this $x\le2$. • The inequality trivially satisfied for non-positive $x$. – A.Γ. Commented Jul 20, 2018 at 7:25 • @A.Γ. How about when $0 < x < 1$? (I just realised when that happens, $x\|x-1\| < 1 \cdot 1 < 3$.) Commented Jul 20, 2018 at 7:27 • @PiyushDivyanakar $x=3$ doesn't work as well. Commented Jul 20, 2018 at 7:34 • I have corrected. Commented Jul 20, 2018 at 7:36 • Nitpicking: you're neglecting the case $x=1$. Commented Jul 20, 2018 at 13:44 This method doesn't generalize, but it may be instructive. The inequality surely holds for $x\le0$, so we can concentrate on $x>0$. Dividing both sides by $x$, we obtain $$\|x-1\|<\frac{3}{x}$$ that's equivalent to $$-\frac{3}{x}<x-1<\frac{3}{x}$$ that in turn becomes \begin{cases} x^2-x+3 > 0 \\[4px] x^2-x-3 < 0 \end{cases} The top one is true for every $x$; the bottom one is satisfied for $$0<x<\frac{1+\sqrt{13}}{2}$$ (recall we're restricting to $x>0$). Putting back the interval $x\le0$ we discussed before, we can conclude the solution set is $$x<\frac{1+\sqrt{13}}{2}$$ Consider the cases where $x > 1$ and where $x < 1$. If $x > 1$, we have: $$x(x-1) < 3$$ $$\Rightarrow x^2-x-3 < 0$$ With $x=2$, we have $2^2-2-3 = -1 < 0$, and with $x = 3$ we have $3^2 - 3 - 3 = 3 > 0$, so $x ≤ 2$. If $x < 1$, we have: $$x(x+1) < 3$$ $$\Rightarrow x^2+x-3 < 0$$ Can you continue? Consider the graph of $y=x(x-1)$: $\hspace{2cm}$ If the absolute values are applied to get $y=x\|x-1\|$, the graph becomes: $\hspace{2cm}$ The point $A$ has coordinates $x(x-1)=3 \Rightarrow x=\frac{1+\sqrt{13}}{2}$ and $y=3$. Hence, the solution of the inequality $x\|x-1\|<3$ is: $$x\in \left(-\infty,\frac{1+\sqrt{13}}{2}\right).$$
# CLASS-8RELATION & MAPPING - FUNCTION & RELATION Function & Relation - We have already discussed earlier about the relation & mapping but now we would like to discussed about the function & relation. There are some example are given below This is to be remembered that – 1) All functions are relations but all relations are not functions. The domain of relation (1) = {1, 2, 3, 4, 5} = set A and the domain of relation (5) = {1, 2, 3, 4, 5} = set A 2) The domain of a mapping or function from set A to set B = set A The range of relation (1) = {a, b, c, d, e} = set B, while the range of relation (5) = {a, b, c, d, e} set B Next, consider the relation “is half of” from set A = {3, 5, 7, 9} to set B = {6, 10, 14, 18}, illustrated by the arrow diagram alongside. It is clear that, 2 R 4, 5 R 10, 7 R 14, 9 R 18, the relation is a function because each element of A is associated with only one member of B. If we denote this relation by ‘f ’ then we can represent the function ‘f’ from set A to set B as f : A → B. We can write f(3) = 6, similarly f(5) = 10, f(7) = 14, and f(9) = 18. In other words the images of 3, 5, 7 & 9 are 6, 10, 14, & 18 respectively. We can represent f in the roster form as – f = {(3, 6), (5, 10), (7, 14), (9, 81)} where 3, 5, 7, 9 belongs to set A and 6, 10, 14, 18 belongs to set B or, f = {(x, y) : x Є A, y Є B and y = 2x} if ‘x’ is an element of A then the image of ‘x’ is 2x, i.e., f(x) = 2x so, f = {(x, f(x)) such that x Є A} = {(x, f(x) : x Є A} we can also write, f : A → B such that f(x) = 2x, x Є A this is the equation form of the functions, The set A is called the domain of the function ‘f ’, the set B is called the co-domain of the function ‘f’ and the set of all the images of the elements of A is called the range of the function ‘f’. Next, consider the relation “is double of” from set A = {49, 64, 81, 121} to set B = {7, 8, 9, 11}, illustrated by the arrow diagram alongside. It is clear that, 49 R 7, 64 R 8, 81 R 9, 121 R 11, the relation is a function because each element of A is associated with only one member of B. If we denote this relation by ‘f ’ then we can represent the function ‘f’ from set A to set B as f : A → B. We can write f(49) = 7, similarly f(64) = 8, f(81) = 9, and f(121) = 11. In other words the images of 49, 64, 81 & 121 are 7, 8, 9, & 11 respectively. We can represent f in the roster form as – f = {(49, 7), (64, 8), (81, 9), (121, 11)} where 49, 64, 81, 121 belongs to set A and 7, 8, 9, 11 belongs to set B or, f = {(x, y) : x Є A, y Є B and 2x = y} if ‘x’ is an element of A then the image of ‘x’ is 2x, i.e., f(2x) = x so, f = {(x, f(2x)) such that x Є A} = {(x, f(2x) : 2x Є A} we can also write, f : A → B such that f(2x) = x, 2x Є A this is the equation form of the functions, The set A is called the domain of the function ‘f ’, the set B is called the co-domain of the function ‘f’ and the set of all the images of the elements of A is called the range of the function ‘f’.
Question Construct 5 equivalent equations for the equation – 3 + 3x = – 3 Describe which value you multiplied by for each equivalent equation 1. thuhuong The five equivalent equations for the given equation are as follows: 1) -6 + 6x = -6; multiplied by 2 2) -1 + x = -1; divided by 3 3) 1 + 3x = 1; added by 4 4) -8 + 3x = -8; subtracted by 5 5) 1 – x = 1; taking -3 as common ### What are equivalent equations? The equations that have the same solution are called equivalent equations. ### Calculation: The given equation is – 3 + 3x = – 3 1) Multiplying with 2: 2(- 3 + 3x) = 2 × – 3 ⇒ -6 + 6x = -6 2) Dividing with 3: – 3/3 + 3x/3 = – 3/3 ⇒ -1 + 3x = -1 3) Adding 4 on both sides: – 3 + 3x + 4 = – 3 + 4 ⇒ 1 + 3x = 1 4) Subtracting 5 from both sides: – 3 + 3x -5 = – 3 -5 ⇒ -8 + 3x = -8 5) Taking -3 common: – 3 + 3x = – 3 ⇒ -3(1 – x) = -3 ⇒ 1 – x = 1 Therefore, the required 5 equivalent equations are framed.
# Math in Focus Grade 8 Chapter 6 Lesson 6.3 Answer Key Understanding Linear and Nonlinear Functions Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 6 Lesson 6.3 Understanding Linear and Nonlinear Functions to score better marks in the exam. ## Math in Focus Grade 8 Course 3 A Chapter 6 Lesson 6.3 Answer Key Understanding Linear and Nonlinear Functions ### Math in Focus Grade 8 Chapter 6 Lesson 6.3 Guided Practice Answer Key Tell whether each table of values represents a linear or nonlinear function. Explain. Question 1. Because the rate of change for the function is , the table represents a function. Because the rate of change for the function is constant, the table represents a linear function. Question 2. Because the rate of change for the function is , the table represents a function. Because the rate of change for the function is non-constant, the table represents a non-linear function. Tell whether each graph represents a linear function. If so, find the rate of change. Question 3. Because the graph is a , it represents a function. The line passes through (, ) and (, ). Rate of change = $$\frac{?-?}{?-?}$$ = So, the rate of change of the graph is . Because the graph is a straight line, it represents a linear function. The line passes through (0, 3) and (6, 8). Rate of change = $$\frac{8-3}{6-0}$$ = 5/6 So, the rate of change of the graph is constant. Question 4. Because the graph is a , it represents a function. Because the graph is a curve, it represents a non-linear function. Describe the function. Sketch a graph for the function. Question 5. A cruise ship traveling at a constant speed consumes 4,000 gallons of gasoline per hour. When fully filled, it has a fuel capacity of 330,000 gallons. The amount of gasoline consumed, y gallons, is a function of the total traveling time, x hours. a) Give the least possible input value and the corresponding output value. Tell whether the function is linear or nonlinear. Then tell whether the function is increasing or decreasing. Explain. At the beginning of the trip, the cruise ship has 330,000 gallons of gasoline. So, the least possible input value is 0 and the corresponding output value is 330,000. For every hour of traveling time, the cruise ship consumes 4,000 gallons of gasoline. So, the rate of change of the function is constant. As the total traveling time increases, the amount of gasoline left decreases. Thus the function is a linear and decreasing function. b) Sketch a graph for the function. Hands-On Activity Materials: number cards (from -5 to 5) SKETCH LINEAR FUNCTIONS Work in pairs. STEP 1: Shuffle the cards and place them face down on the table. STEP 2: Each player draws two cards. Use your cards to write an equation in slope-intercept form. Use one of the cards for the slope and one for the y-intercept of the equation. For example: Slope-intercept form: y = -2x + 3 STEP 3: Graph the equation you wrote. STEP 4: Copy and complete the table. For each equation that you and your partner write, record the slope, the y-intercept, and whether the function is increasing or decreasing. STEP 5: Repeat STEP 2 to STEP 4 until a player has written two equations that represent increasing functions and two equations that represent decreasing functions. If no player has done this, reshuffle the cards and repeat STEP 2 to STEP 4. The player who reaches this goal first wins the game. Math Journal What can you conclude about the slope of the graph of an increasing function and the slope of the graph of a decreasing function? Explain. ### Math in Focus Course 3A Practice 6.3 Answer Key Tell whether each table of values represents a linear or nonlinear function. Question 1. Question 2. Linear Function Question 3. Non-linear function Question 4. Tell whether each graph represents a linear function. If so, find the rate of change. Question 5. Yes, 1/2 Question 6. Tell whether each function is linear or nonlinear. Then tell whether the function is increasing or decreasing. Question 7. Non-linear and increasing function Question 8. Answer: straight line and linear function Question 9. Question 10. Answer: Straight line and linear function Describe the function. Sketch a graph for the function. Question 11. A machine at a factory pours juice into bottles at a constant rate of 6 liters per minute. The total amount of juice poured, y liters is a function of the number of minutes that the juice is poured, x. a) Give the least possible input value and the corresponding output value. Tell whether the function is linear or nonlinear. Then tell whether the function is increasing or decreasing. Explain. If the time taken for the machine to pour juice into bottles is 0 min, the amount of juice poured will be 0L. So, the least possible input value is 0 and the corresponding output value is 0. Because the machine pours juice at 6L/min, the rate of change of the function is constant. As the time taken for the machine to pour juice into bottles increases, the amount of juice poured also increases. Hence the function is a linear and increasing function. b) Sketch a graph for the function. Question 12. Aidan was 100 miles from Town P. He traveled to Town P by car at a constant speed. The distance from Town P, y miles, is a function of the traveling time, x hours. a) Give the least possible input value and the corresponding output value. Tell whether the function is linear or nonlinear. Then tell whether the function is increasing or decreasing. Explain. When Aidan [eaves to Town P, the corresponding distance from his position to Town P is 100, therefore the least possible input value is 0 and the corresponding output is 100. Because each hour he is driving by the same speed, the rate of change of the function is constant. As time passes (the input increases), the distance to Town P (the output) decreases, so the function is decreasing. Therefore the function is linear and decreasing. b) Sketch a graph for the function. The distance to Town P starts at (0, 100) and falls from left to right The slope of the line, which is constant, is the distance traveLed each hour (the speed): Solve. Use graph paper. Question 13. The table shows the number of students, y, as a function of the number of teachers, x. a) Tell whether the function is linear or nonlinear. Then tell whether the function is increasing or decreasing. Explain. The function has a constant rate of change, 25. Thus it is a linear and increasing function. As the number of students increases, the number of teachers also increases. b) Graph the table of values and draw a line through the points. Use 1 unit on the horizontal axis to represent 1 teacher for the x interval, and 1 unit on the vertical axis to represent 25 students for the y interval from 100 to 250. Do the coordinates of every point on the line make sense for the function? Explain. No, because the number of teachers and students must be whole numbers. Question 14. A cyclist starts riding from home to another town. His cycling speed, y miles per hour, is a function of the amount of time he takes to cycle, x hours. a) Tell whether the function is linear or nonlinear. Then tell whether the function is increasing or decreasing. Explain. Non-linear function and decreasing b) Graph the table of values and draw a curve through the points. Use 1 unit on the horizontal axis to represent 1 hour for the x interval, and 1 unit on the vertical axis to represent 1 mile per hour for the y interval. Do the coordinates of every point on the curve make sense for the function? Explain.
Measures of Variability David M. Lane Prerequisites Percentiles, Distributions, Measures of Central Tendency Learning Objectives 1. Determine the relative variability of two distributions 2. Compute the range 3. Compute the inter-quartile range 4. Compute the variance in the population 5. Estimate the variance from a sample 6. Compute the standard deviation from the variance What is Variability? Variability refers to how "spread out" a group of scores is. To see what we mean by spread out, consider graphs in Figure 1. These graphs represent the scores on two quizzes. The mean score for each quiz is 7.0. Despite the equality of means, you can see that the distributions are quite different. Specifically, the scores on Quiz 1 are more densely packed and those on Quiz 2 are more spread out. The differences among students were much greater on Quiz 2 than on Quiz 1. Quiz 1 Quiz 2 Figure 1. Bar charts of two quizzes. The terms variability, spread, and dispersion are synonyms, and refer to how spread out a distribution is. Just as in the section on central tendency where we discussed measures of the center of a distribution of scores, in this chapter we will discuss measures of the variability of a distribution. There are four frequently used measures of variability: the range, interquartile range, variance, and standard deviation. In the next few paragraphs, we will look at each of these four measures of variability in more detail. Range The range is the simplest measure of variability to calculate, and one you have probably encountered many times in your life. The range is simply the highest score minus the lowest score. Let’s take a few examples. What is the range of the following group of numbers: 10, 2, 5, 6, 7, 3, 4? Well, the highest number is 10, and the lowest number is 2, so 10 - 2 = 8. The range is 8. Let’s take another example. Here’s a dataset with 10 numbers: 99, 45, 23, 67, 45, 91, 82, 78, 62, 51. What is the range? The highest number is 99 and the lowest number is 23, so 99 - 23 equals 76; the range is 76. Now consider the two quizzes shown in Figure 1. On Quiz 1, the lowest score is 5 and the highest score is 9. Therefore, the range is 4. The range on Quiz 2 was larger: the lowest score was 4 and the highest score was 10. Therefore the range is 6. Interquartile Range The interquartile range (IQR) is the range of the middle 50% of the scores in a distribution. It is computed as follows: IQR = 75th percentile - 25th percentile For Quiz 1, the 75th percentile is 8 and the 25th percentile is 6. The interquartile range is therefore 2. For Quiz 2, which has greater spread, the 75th percentile is 9, the 25th percentile is 5, and the interquartile range is 4. Recall that in the discussion of box plots, the 75th percentile was called the upper hinge and the 25th percentile was called the lower hinge. Using this terminology, the interquartile range is referred to as the H-spread. A related measure of variability is called the semi-interquartile range. The semi-interquartile range is defined simply as the interquartile range divided by 2. If a distribution is symmetric, the median plus or minus the semi-interquartile range contains half the scores in the distribution. Variance Variability can also be defined in terms of how close the scores in the distribution are to the middle of the distribution. Using the mean as the measure of the middle of the distribution, the variance is defined as the average squared difference of the scores from the mean. The data from Quiz 1 are shown in Table 1. The mean score is 7.0. Therefore, the column "Deviation from Mean" contains the score minus 7. The column "Squared Deviation" is simply the previous column squared. Table 1. Calculation of Variance for Quiz 1 scores. Scores Deviation from Mean Squared Deviation 9 2 4 9 2 4 9 2 4 8 1 1 8 1 1 8 1 1 8 1 1 7 0 0 7 0 0 7 0 0 7 0 0 7 0 0 6 -1 1 6 -1 1 6 -1 1 6 -1 1 6 -1 1 6 -1 1 5 -2 4 5 -2 4 Means 7 0 1.5 One thing that is important to notice is that the mean deviation from the mean is 0. This will always be the case. The mean of the squared deviations is 1.5. Therefore, the variance is 1.5. Analogous calculations with Quiz 2 show that its variance is 6.7. The formula for the variance is: where σ2 is the variance, μ is the mean, and N is the number of numbers. For Quiz 1, μ = 7 and N = 20. If the variance in a sample is used to estimate the variance in a population, then the previous formula underestimates the variance and the following formula should be used: where s2 is the estimate of the variance and M is the sample mean. Note that M is the mean of a sample taken from a population with a mean of μ. Since, in practice, the variance is usually computed in a sample, this formula is most often used. The simulation "estimating variance" illustrates the bias in the formula with N in the denominator. Let's take a concrete example. Assume the scores 1, 2, 4, and 5 were sampled from a larger population. To estimate the variance in the population you would compute s2 as follows: M = (1 + 2 + 4 + 5)/4 = 12/4 = 3. s2 = [(1-3)2 + (2-3)2 + (4-3)2 + (5-3)2]/(4-1) = (4 + 1 + 1 + 4)/3 = 10/3 = 3.333 There are alternate formulas that can be easier to use if you are doing your calculations with a hand calculator. You should note that these formulas are subject to rounding error if your values are very large and/or you have an extremely large number of observations. and For this example, Standard Deviation The standard deviation is simply the square root of the variance. This makes the standard deviations of the two quiz distributions 1.257 and 2.203. The standard deviation is an especially useful measure of variability when the distribution is normal or approximately normal (see Chapter on Normal Distributions) because the proportion of the distribution within a given number of standard deviations from the mean can be calculated. For example, 68% of the distribution is within one standard deviation of the mean and approximately 95% of the distribution is within two standard deviations of the mean. Therefore, if you had a normal distribution with a mean of 50 and a standard deviation of 10, then 68% of the distribution would be between 50 - 10 = 40 and 50 +10 =60. Similarly, about 95% of the distribution would be between 50 - 2 x 10 = 30 and 50 + 2 x 10 = 70. The symbol for the population standard deviation is σ; the symbol for an estimate computed in a sample is s. Figure 2 shows two normal distributions. The red distribution has a mean of 40 and a standard deviation of 5; the blue distribution has a mean of 60 and a standard deviation of 10. For the red distribution, 68% of the distribution is between 35 and 45; for the blue distribution, 68% is between 50 and 70. Figure 2. Normal distributions with standard deviations of 5 and 10. q1=c(9,9,9,8,8,8,8,7,7,7,7,7,6,6,6,6,6,6,5,5) IQR(q1, type = 6) [1] 2 x=c(1,2,4,5) var(x) [1] 3.333333 sd(q1) [1] 1.256562 q2=c(10,10,9,9,9,8,8,8,7,7,7,6,6,6,5,5,4,4,3,3) sd(q2) [1] 2.202869
# How to Find the Surface Area of a 3D Shape Using Its Net ## Introduction The surface area of a three-dimensional shape is the total area of all its faces. To find the surface area of a 3D shape, we can use a method called net. A net is a two-dimensional representation of a 3D shape that shows all its faces when unfolded. By using the net of a 3D shape, we can easily calculate the area of each face and then add them up to get the surface area. ## What is a Net? A net is a pattern made when the surface of a 3D shape is laid out flat. It shows each face of the shape in two dimensions. For example, the net of a cube is shown below: A net can help us visualize how a 3D shape is formed by folding along its edges. A net can also help us measure the dimensions of each face of the shape. For example, in the net of the cube above, we can see that each face is a square with side length 4 cm. A 3D shape can have more than one possible net. For example, the net of a rectangular prism can be arranged in different ways, as shown below: ## How to Find the Surface Area Using a Net? To find the surface area of a 3D shape using its net, we need to follow these steps: • Identify the shape and its net. • Find the area of each face using the appropriate formula for its shape. • Add up the areas of all the faces to get the surface area. Let’s see an example of how to apply this method. ### Example: Find the surface area of a triangular prism using its net. Solution: • The shape is a triangular prism and its net is shown above. • The net has five faces: two triangular bases and three rectangular lateral faces. • To find the area of each face, we need to use the formula for the area of a triangle and the area of a rectangle. • The area of a triangle is given by �=��12�ℎA=frac12bh , where �b is the base and ℎh is the height. • The area of a rectangle is given by �=��A=lw , where �l is the length and �w is the width. • The area of each face is calculated as follows: • The surface area is the sum of the areas of all the faces: ��=15+15+48+40+40=��158��2textSurfaceArea=15+15+48+40+40=boxed158textcm2 ## Conclusion In this article, we learned how to find the surface area of a three-dimensional shape using its net. A net is a two-dimensional representation of a three-dimensional shape that shows all its faces when unfolded. By using the net of a shape, we can easily calculate the area of each face and then add them up to get the surface area. This method can be applied to any polyhedron, such as cubes, prisms, pyramids, etc. However, for shapes with curved surfaces, such as cylinders, cones, spheres, etc., we need to use different formulas to find their surface areas.
Request a call back # Class 9 FRANK Solutions Maths Chapter 5 - Factorisation Achieve high marks with the support of Frank Solutions for ICSE Class 9 Mathematics Chapter 5 Factorisation. Prepared by TopperLearning’s experts, the solutions will help you understand how to factorise mathematical expressions by removing the common factors. Also, learn the method of using the difference of two squares to factorise the given data. Explore ICSE Class 9 Maths Frank textbook solutions any time online on our online education portal. Additionally, with the help of our concept videos and online practice tests designed by subject experts, you can get ahead in your Maths learning. ## Factorisation Exercise Ex. 5.1 ### Solution 2(a) 15xy - 9x - 25y + 15 = (15xy - 9x) - (25y + 15) = 3x(5y - 3) - 5(5y - 3) = (5y - 3)(3x - 5) ### Solution 2(b) 15x2 + 7y - 3x - 35xy = 15x2 - 3x - 35xy + 7y = (15x2 - 3x) - (35xy - 7y) = 3x(5x - 1) - 7y(5x - 1) = (5x - 1)(3x - 7y) ### Solution 2(c) 9 + 3xy + x2y + 3x = 9 + 3xy + 3x + x2y = (9 + 3xy) + (3x + x2y) = 3(3 + xy) + y(3 + xy) = (3 + xy)(3 + x) ### Solution 2(d) 8(2a + b)2 - 8a - 4b = 8(2a + b)2 - (8a + 4b) = 8(2a + b)2 - 4(2a + b) = 4(2a + b)[2(2a + b) - 1] = 4(2a + b)[4a + 2b - 1] ### Solution 2(e) x(x - 4) - x + 4 = x(x - 4) - 1(x - 4) = (x - 4)(x - 1) ### Solution 2(f) 2m3 - 5n2 - 5m2n + 2mn = 2m3 + 2mn - 5m2n - 5n2 = (2m3 + 2mn) - (5m2n + 5n2) = 2m(m2 + n) - 5n(m2 + n) = (m2 + n)(2m - 5n) ### Solution 2(g) 4abx2 + 49aby2 + 14xy(a2 + b2) = 4abx2 + 49aby2 + 14a2xy + 14b2xy = (4abx2 + 14a2xy) + (14b2xy + 49aby2) = 2ax(2bx + 7ay) + 7by(2bx + 7ay) = (2bx + 7ay)(2ax + 7by) ### Solution 2(h) 9x3 + 6x2y2 - 4y3 - 6xy = 9x3 + 6x2y2 - 6xy - 4y3 = (9x3 + 6x2y2) - (6xy + 4y3) = 3x2(3x + 2y2) - 2y(3x + 2y2) = (3x + 2y2)(3x2 - 2y) ### Solution 2(i) 3ax2 - 5bx2 + 9az2 + 6ay2 - 10by2 - 15bz2 = 3ax2 + 6ay2 + 9az2 - 5bx2 - 10by2 - 15bz2 = (3ax2 + 6ay2 + 9az2) - (5bx2 + 10by2 + 15bz2) = 3a(x2 + 2y2 + 3z2) - 5b(x2 + 2y2 + 3z2) = (x2 + 2y2 + 3z2)(3a - 5b) ### Solution 2(j) 8x3 - 24x2y + 54xy2 - 162y3 = (8x3 - 24x2y) + (54xy2 - 162y3) = 8x2(x - 3y) + 54y2(x - 3y) = (x - 3y)(8x2 + 54y2) ### Solution 2(k) 2a + b + 3c - d + (2a + b)3 + (2a + b)2(3c - d) = (2a + b + 3c - d) + [(2a + b)3 + (2a + b)2(3c - d)] = 1(2a + b + 3c - d) + (2a + b)2(2a + b + 3c - d) = (2a + b + 3c - d)[1 + (2a + b)2] ### Solution 2(l) xy(a2 + 1) + a(x2 + y2) = a2xy + xy + ax2 + ay2 = (a2xy + ax2) + (ay2 + xy) = ax(ay + x) + y(ay + x) = (ay + x)(ax + y) ### Solution 2(m) p2x2 + (px2 + 1)x + p = p2x2 + px3 + x + p = (p2x2 + px3) + (p + x) = px2(p + x) + 1(p + x) = (p + x)(px2 + 1) ### Solution 2(n) x2 - (p + q)x + pq = x2 - px - qx + pq = (x2 - px) - (qx + pq) = x(x - p) - q(x - p) = (x - p)(x - q) ### Solution 2(p) x + y + m(x + y) = (x + y) + m(x + y) = (x + y)(1 + m) ### Solution 2(r) 2p(a2 - 2b2) - 14p + (a2 - 2b2)2 - 7(a2 - 2b2) = 2p(a2 - 2b2) + (a2 - 2b2)2 - 14p - 7(a2 - 2b2) = [2p(a2 - 2b2) + (a2 - 2b2)2] - [14p + 7(a2 - 2b2)] = (a2 - 2b2)(2p + a2 - 2b2) - 7(2p + a2 - 2b2) = (2p + a2 - 2b2)(a2 - 2b2 - 7) ## Factorisation Exercise Ex. 5.2 ### Solution 1(a) x2 + 6x + 8 = x2 + 4x + 2x + 8 = x(x + 4) + 2(x + 4) = (x + 4)(x + 2) ### Solution 1(b) x2 - 11x + 24 = x2 - 8x - 3x + 24 = x(x - 8) - 3(x - 8) = (x - 8)(x - 3) ### Solution 1(c) x2 + 5x - 6 = x2 + 6x - x - 6 = x(x + 6) - 1(x + 6) = (x + 6)(x - 1) ### Solution 1(d) p2 - 12p - 64 = p2 - 16p + 4p - 64 = p(p - 16) + 4(p - 16) = (p - 16)(p + 4) ### Solution 1(e) y2 - 2y - 24 = y2 - 6y + 4y - 24 = y(y - 6) + 4(y - 6) =(y - 6)(y + 4) ### Solution 1(f) 3x2 + 19x - 14 = 3x2 + 21x - 2x - 14 = 3x(x + 7) - 2(x + 7) = (x + 7)(3x - 2) ### Solution 1(g) 15a2 - 14a - 16 = 15a2 - 24a + 10a - 16 = 3a(5a - 8) + 2(5a - 8) = (5a - 8)(3a + 2) ### Solution 1(h) 12 + x - 6x2 = 12 + 9x - 8x - 6x2 = 3(4 + 3x) - 2x(4 + 3x) = (4 + 3x)(3 - 2x) ### Solution 1(i) 7x2 + 40x - 12 = 7x2 + 42x - 2x - 12 = 7x(x + 6) - 2(x + 6) = (x + 6)(7x - 2) ### Solution 2(a) 5x2 - 17xy + 6y2 = 5x2 - 15xy - 2xy + 6y2 = 5x(x - 3y) - 2y(x - 3y) = (x - 3y)(5x - 2y) ### Solution 2(b) 9x2 - 22xy + 8y2 = 9x2 - 18xy - 4xy + 8y2 = 9x(x - 2y) - 4y(x - 2y) = (x - 2y)(9x - 4y) ### Solution 2(c) 2x3 + 5x2y - 12xy2 = 2x3 + 8x2y - 3x2y - 12xy2 = 2x2(x + 4y) - 3xy(x + 4y) = (x + 4y)(2x2 - 3xy) = (x + 4y)x(2x - 3y) = x(x + 4y)(2x - 3y) ### Solution 2(d) x2y2 + 15xy - 16 = x2y2 + 16xy - xy - 16 = xy(xy + 16) - 1(xy + 16) = (xy + 16)(xy - 1) ### Solution 2(e) (2p + q)2 - 10p - 5q - 6 = (2p + q)2 - (10p - 5q) - 6 = (2p + q)2 - 5(2p + q) - 6 = (2p + q)2 - 6(2p + q) + (2p + q) - 6 = (2p + q)(2p + q - 6) + 1(2p + q - 6) = (2p + q - 6)(2p + q + 1) ### Solution 2(f) y2 + 3y + 2 + by + 2b = y2 + y + 2y + 2 + by + 2b = y2 + y + by + 2y + 2 + 2b = y(y + 1 + b) + 2(y + 1 + b) = (y + 1 + b)(y + 2) ### Solution 2(g) x3y3 - 8x2y2 + 15xy = x3y3 - 3x2y2 - 5x2y2 + 15xy = x2y2(xy - 3) - 5xy(xy - 3) = (xy - 3)(x2y2 - 5xy) = (xy - 3)xy(xy - 5) = xy(xy - 3)(xy - 5) ### Solution 3(a) 5(3x + y)2 + 6(3x + y) - 8 = 5(3x + y)2 + 10(3x + y) - 4(3x + y) - 8 = 5(3x + y)(3x + y + 2) - 4(3x + y + 2) = (3x + y + 2)[5(3x + y) - 4] ### Solution 3(b) 5 - 4(a - b) - 12(a - b)2 = 5 - 10(a - b) + 6(a - b) - 12(a - b)2 = 5[1 - 2(a - b)] + 6(a - b)[1 - 2(a - b)] = [5 + 6(a - b)][1 - 2(a - b)] = (5 + 6a - 6b)(1 - 2a + 2b) ### Solution 3(c) (3a - 2b)2 + 3(3a - 2b) - 10 = (3a - 2b)2 + 5(3a - 2b) - 2(3a - 2b) - 10 = (3a - 2b)(3a - 2b + 5) - 2(3a - 2b +5) = (3a - 2b + 5)((3a - 2b - 2) ### Solution 3(d) (a2 - 2a)2 - 23(a2 - 2a) + 120 = (a2 - 2a)2 - 15(a2 - 2a) - 8(a2 - 2a) + 120 = (a2 - 2a)(a2 - 2a - 15) - 8(a2 - 2a - 15) = (a2 - 2a - 15)(a2 - 2a - 8) = (a2 - 5a + 3a - 15)(a2 - 4a + 2a - 8) = [a(a - 5) + 3(a - 5)][a(a - 4) + 2(a - 4)] = [(a - 5)(a + 3)][(a - 4)(a + 2)] = (a - 5)(a + 3)(a - 4)(a + 2) = (a + 2)(a + 3)(a - 4)(a - 5) ### Solution 3(e) (x + 4)2 - 5xy - 20y - 6y2 = (x + 4)2 - 5y(x + 4) - 6y2 = (x + 4)2 - 6y(x + 4) + y(x + 4) - 6y2 = (x + 4)(x + 4 - 6y) + y(x + 4 - 6y) = (x + 4 - 6y)(x + 4 + y) = (x - 6y + 4)(x + y + 4) ### Solution 3(f) 7(x - 2)2 - 13(x - 2) - 2 = 7(x - 2)2 - 14(x - 2) + (x - 2) - 2 = 7(x - 2)(x - 2 - 2) + 1(x - 2 - 2) = 7(x - 2)(x - 4) + 1(x - 4) = (x - 4)[7(x - 2) + 1] = (x - 4)(7x - 14 + 1) = (x - 4)(7x - 13) ### Solution 3(g) 12 - (y + y2)(8 - y - y2) = 12 - a(8 - a) [Taking y + y2 = a] = 12 - 8a + a2 = 12 - 6a - 2a + a2 = 6(2 - a) - a(2 - a) = (2 - a)(6 - a) = [2 - (y + y2)][6 - (y + y2)] = (2 - y - y2)(6 - y - y2) = (2 - 2y + y - y2)(6 - 3y + 2y - y2) = [2(1 - y) + y(1 - y)][3(2 - y) + y(2 - y)] = [(1 - y)(2 + y)][(2 - y)(3 + y)] = (1 - y)(2 + y)(2 - y)(3 + y) = (y - 1)(y + 2)(y - 2)(y + 3) ### Solution 3(h) (p2 + p)2 - 8(p2 + p) + 12 = (p2 + p)2 - 6(p2 + p) - 2(p2 + p) + 12 = (p2 + p)(p2 + p - 6) - 2(p2 + p - 6) = (p2 + p - 6)(p2 + p - 2) = (p2 + 3p - 2p - 6)(p2 + 2p - p - 2) = [p(p + 3) - 2(p + 3)][p(p + 2) - 1(p + 2)] = [(p + 3)(p - 2)][(p + 2)(p - 1)] = (p + 3)(p - 2)(p + 2)(p - 1) ### Solution 4(a) (y2 - 3y)(y2 - 3y + 7) + 10 = a(a + 7) + 10 [taking (y2 - 3y) = a] = a2 + 7a + 10 = a2 + 5a + 2a + 10 = a(a + 5) + 2(a + 5) = (a + 5)(a + 2) = (y2 - 3y + 5)(y2 - 3y + 2) = (y2 - 3y + 5)(y2 - 2y - y + 2) = (y2 - 3y + 5)[y(y - 2) - 1(y - 2)] = (y2 - 3y + 5)[(y - 2)(y - 1)] = (y - 1)(y - 2)(y2 - 3y + 5) ### Solution 4(b) (t2 - t)(4t2 - 4t - 5) - 6 = (t2 - t)[4(t2 - t) - 5] - 6 = a[4a - 5] - 6 [Taking (t2 - t) = a] = 4a2 - 5a - 6 = 4a2 - 8a + 3a - 6 = 4a(a - 2) + 3(a - 2) = (a - 2)(4a + 3) = (t2 - t - 2)[4(t2 - t) + 3] = (t2 - 2t + t - 2)(4t2 - 4t + 3) = [t(t - 2) + 1(t - 2)](4t2 - 4t + 3) = [(t - 2)(t + 1)](4t2 - 4t + 3) = (t + 1)(t - 2)(4t2 - 4t + 3) ### Solution 4(c) 12(2x - 3y)2 - 1(2x - 3y) - 1 = 12a2 - a - 1 [Taking (2x - 3y) = a] = 12a2 - 4a + 3a - 1 = 4a(3a - 1) + 1(3a - 1) = (3a - 1)(4a + 1) = [3(2x - 3y) - 1][4(2x - 3y) + 1] = (6x - 9y - 1)(8x - 12y + 1) ### Solution 4(d) 6 - 5x + 5y + (x - y)2 = 6 - 5(x - y) + (x - y)2 = 6 - 3(x - y) - 2(x - y) + (x - y)2 = 3[2 - (x - y)] - (x - y)[2 - (x - y)] = 3(2 - x + y) - (x - y)(2 - x + y) = (2 - x + y)(3 - x + y) ### Solution 4(f) P4 + 23p2q2 + 90q4 = p4 + 18p2q2 + 5p2q2 + 90q4 = p2(p2 + 18q2) + 5q2(p2 + 18q2) = (p2 + 18q2)(p2 + 5q2) ### Solution 4(g) 2a3 + 5a2b - 12ab2 = 2a3 + 8a2b - 3a2b - 12ab2 = 2a2(a + 4b) - 3ab(a + 4b) = (a + 4b)(2a2 - 3ab) = (a + 4b)a(2a - 3b) = a(a + 4b)(2a - 3b) ## Factorisation Exercise Ex. 5.3 x2 - 16 = x2 - 42 = (x - 4)(x + 4) ### Solution 1(b) 64x2 - 121y2 = (8x)2 - (11y)2 = (8x - 11y)(8x + 11y) ### Solution 1(c) 441 - 81y2 = (21)2 - (9y)2 = (21 - 9y)(21 + 9y) = 3(7 - 3y)3(7 + 3y) = 9(7 - 3y)(7 + 3y) ### Solution 1(d) x6 - 196 = (x3)2 - (14)2 = (x3 - 14)(x3 + 14) ### Solution 1(e) 625 - b2 = (25)2 - (b)2 = (25 - b)(25 + b) ### Solution 1(g) 8xy2 - 18x3 = 2x(4y2 - 9x2) = 2x[(2y)2 - (3x)2] = 2x[(2y - 3x)(2y + 3x)] = 2x(2y - 3x)(2y + 3x) ### Solution 1(h) 16a4 - 81b4 = (4a2)2 - (9b2)2 = (4a2 - 9b2)(4a2 + 9b2) = [(2a)2 - (3b)2](4a2 + 9b2) = [(2a - 3b)(2a + 3b)](4a2 + 9b2) = (2a - 3b)(2a + 3b)(4a2 + 9b2) ### Solution 1(i) a(a - 1) - b(b - 1) = a2 - a - b2 + b = a2 - b2 - a + b = (a2 - b2) - (a - b) = (a - b)(a + b) - (a - b) = (a - b)(a + b - 1) ### Solution 1(j) (x + y)2 - 1 = (x + y)2 - (1)2 = (x + y + 1)(x + y - 1) ### Solution 1(k) x2 + y2 - z2 - 2xy = x2 + y2 - 2xy - z2 = (x2 + y2 - 2xy) - z2 = (x - y)2 - (z)2 = (x - y - z)(x - y + z) ### Solution 1(l) (x - 2y)2 - z2 = (x - 2y)2 - (z)2 = (x - 2y - z)(x - 2y + z) ### Solution 2(a) 9(a - b)2 - (a + b)2 = [3(a - b)]2 - (a + b)2 = [3(a - b) - (a + b)][3(a - b) + (a + b)] = (3a - 3b - a - b)(3a - 3b + a + b) = (2a - 4b)(4a - 2b) = 2(a - 2b)2(2a - b) = 4(a - 2b)(2a - b) ### Solution 2(b) 25(x - y)2 - 49(c - d)2 = [5(x - y)]2 - [7(c - d)]2 = [5(x - y) - 7(c - d)][5(x - y) + 7(c - d)] = (5x - 5y - 7c + 7d)(5x - 5y + 7c - 7d) ### Solution 2(c) (2a - b)2 - 9(3c - d)2 = (2a - b)2 - [3(3c - d)]2 = [(2a - b) - 3(3c - d)][(2a - b) + 3(3c - d)] = (2a - b - 9c + 3d)(2a - b + 9c - 3d) ### Solution 2(d) b2 - 2bc + c2 - a2 = (b2 - 2bc + c2) - a2 = (b - c)2 - (a)2 = (b - c - a)(b - c + a) ### Solution 2(f) (x2 + y2 - z2)2 - 4x2y2 = (x2 + y2 - z2)2 - (2xy)2 = (x2 + y2 - z2 - 2xy)(x2 + y2 - z2 + 2xy) = [(x2 + y2 - 2xy) - z2][(x2 + y2 + 2xy) - z2] = [(x - y)2 - z2][(x + y)2 - z2] = [(x - y - z)(x - y + z)][(x + y - z)(x + y + z)] = (x - y - z)(x - y + z)(x + y - z)(x + y + z) ### Solution 2(g) a2 + b2 - c2 - d2 + 2ab - 2cd = (a2 + b2 + 2ab) - (c2 + d2 + 2cd) = (a + b)2 - (c + d)2 = (a + b + c + d)(a + b - c - d) ### Solution 2(h) 4xy - x2 - 4y2 + z2 = z2 - x2 - 4y2 + 4xy = z2 - (x2 + 4y2 - 4xy) = z2 - (x - 2y)2 = [z - (x - 2y)][z + (x - 2y)] = (z - x + 2y)(z + x - 2y) ### Solution 2(i) 4x2 - 12ax - y2 - z2 - 2yz + 9a2 = (4x2 - 12ax + 9a2) - (y2 + z2 + 2yz) = (2x - 3a)2 - (y + z)2 = [(2x - 3a) + (y + z)][(2x - 3a) - (y + z)] = (2x - 3a + y + z)(2x - 3a - y - z) ### Solution 2(j) (x + y)3 - x - y = (x + y)(x + y)2 - (x + y) = (x + y)[(x + y)2 - 1] = (x + y)[(x + y + 1)(x + y - 1)] = (x + y)(x + y + 1)(x + y - 1) ### Solution 2(k) y4 + y2 + 1 = y4 + 2y2 + 1 - y2 = (y2 + 1)2 - y2 = (y2 + 1 + y)(y2 + 1 - y) ### Solution 2(l) (a2 - b2)(c2 - d2) - 4abcd = a2c2 - a2d2 - b2c2 + b2d2 - 4abcd = a2c2 + b2d2 - 2abcd - a2d2 - b2c2 - 2abcd = (a2c2 + b2d2 - 2abcd) - (a2d2 + b2c2 + 2abcd) = (ac - bd)2 - (ad + bc)2 = [(ac - bd) + (ad + bc)][(ac - bd) - (ad + bc)] = (ac - bd + ad + bc)(ac - bd - ad - bc)
# Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving. ## Presentation on theme: "Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving."— Presentation transcript: Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.2 Introduction to Problem Solving Copyright © 2013, 2009, 2005 Pearson Education, Inc. Objectives Solving a Formula for a Variable Steps for Solving a Problem Percentages Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example A parallelogram has an area of 35 cm 2 and a base of 7 cm. a. Write a formula to find the height h of the parallelogram with known area A and base b. b. Use the formula to find h. Solution a. We must solve A = bh for h. b. Substitute 35 for A and 7 for b into this formula. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example The formula is used to convert degrees Fahrenheit to degrees Celsius. Use this formula to convert 23°F to an equivalent Celsius temperature. Solution = 5°C Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example Solve the equation for y and write a formula for a function f defined by y = f(x). 5(x – 2y) = 8x Solution 5x +10y = 8x 5(x – 2y) = 8x 5x + 5x 10y = 8x + 5x 10y = 3x Thus Copyright © 2013, 2009, 2005 Pearson Education, Inc. Step 1: Read the problem carefully to be sure that you understand it. (You may need to read the problem more than once.) Assign a variable to what you are being asked to find. If necessary, write other quantities in terms of this variable. Step 2: Write an equation that relates the quantities described in the problem. You may need to sketch a diagram, make a table, or refer to known formulas. STEPS FOR SOLVING A PROBLEM Copyright © 2013, 2009, 2005 Pearson Education, Inc. Step 3: Solve the equation and determine the solution. Step 4: Look back and check your answer. Does it seem reasonable? Did you find the required information? STEPS FOR SOLVING A PROBLEM Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example The sum of three consecutive integers is 126. Find the three numbers. Solution Step 1: Assign a variable to an unknown quantity. n: smallest of the three integers n + 1: next integer n + 2: largest integer Step 2: Write an equation that relates these unknown quantities. n + (n + 1) + (n + 2) = 126 Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) Step 3: Solve the equation in Step 2. n + (n + 1) + (n + 2) = 126 (n + n + n) + (1 + 2) = 126 3n + 3 = 126 3n = 123 n = 41 So the numbers are 41, 42, and 43. Step 4: Check your answer. The sum of these integers is 41 + 42 + 43 = 126. The answer checks. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example The perimeter of a basketball court is 288 ft. The width is 44 ft less than the length. Find the dimensions of the court. Solution Step 1: Assign a variable to an unknown quantity. x: width of the court x + 44: length of the court Step 2: Write an equation that relates these unknown quantities. x + (x + 44) + x + (x + 44) = 288 Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) The perimeter of a basketball court is 288 ft. The width is 44 ft less than the length. Find the dimensions of the court. Step 3: Solve x + (x + 44) + x + (x + 44) = 288 4x + 88 = 288 4x = 200 x = 50 The width is 50 ft and the length is 50 + 44 = 94 ft. Step 4: Check your answer. The perimeter is 50 + 94 + 50 + 94 = 288 feet. It checks. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example Two trucks leave Perrysburg at the same time and travel in opposite directions. After 4 hours, they are 320 miles apart. If one truck is traveling 20 miles per hour faster than the other truck, find the speeds of the two trucks. Solution x: speed of the slower truck x + 20: speed of the faster truck RateTimeDistance Slower Truckx44x4x Faster Truckx + 2044 (x + 20) Total320 4x + 4(x + 20) = 320 Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) Solve. The speed of the slower truck is 30 mph and the speed of the faster truck is 20 miles per hour faster, or 50 mph. 4x + 4(x + 20) = 320 4x + 4x + 80 = 320 8x + 80 = 320 8x = 240 x = 30 Check: Distance traveled by the slower truck 30 × 4 = 120 Distance traveled by the faster truck 50 × 4 = 200 120 + 200 = 320 miles Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example Convert each percentage to fraction and decimal notation. a. 47%b. 9.8%c. 0.9% Solution Fraction NotationDecimal Notation a. b. c. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example A car salesman sells a total of 85 cars in the first and second quarter of the year. In the second quarter, he had an increase of 240% over the previous quarter. How many cars did the salesman sell in the first quarter? Solution Step 1: Assign a variable. x: the amount sold in the first quarter. Step 2: Write an equation. x + 2.4x = 85 Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) Step 3: Solve the equation in Step 2. x + 2.4x = 85 3.4x = 85 x = 25 In the first quarter the salesman sold 25 cars. Step 4: Check your answer. An increase of 240% of 25 is 2.4 × 25 = 60. Thus the amount of cars sold in the second quarter would be 25 + 60 = 85. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used? Solution Let x = milliliters of 40% Let x + 100 = milliliters of 36% ConcentrationSolution Amount (milliliters) Pure alcohol 0.2810028 0.40x0.4x 0.36x + 1000.36x + 36 Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used? Step 2: Write an equation. 0.28(100) + 0.4x = 0.36(x + 100) ConcentrationSolution Amount (milliliters) Pure alcohol 0.2810028 0.40x0.4x 0.36x + 1000.36x + 36 Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) Step 3: Solve the equation in Step 2. 0.28(100) + 0.4x = 0.36(x + 100) 28(100) + 40x = 36(x + 100) 2800 + 40x = 36x + 3600 2800 + 4x = 3600 4x = 800 x = 200 200 mL of 40% alcohol solution was added to the 100 mL of the 28% solution. Copyright © 2013, 2009, 2005 Pearson Education, Inc. Example (cont) Step 4: Check your answer. If 200 mL of 40% solution are added to the 100 mL of 28% solution, there will be 300 mL of solution. 200(0.4) + 100(0.28) = 80 + 28 = 108 of pure alcohol. The concentration is or 36%.
# Class 11 Maths NCERT Solutions for Chapter 4 Principle of Mathematical Induction Exercise 4.1 ### Principle of Mathematical Induction Exercise Solutions 1. Prove the following by using the principle of mathematical induction for all n ∈ N: 1 + 3 + 1 + 3 + 32 + ….. + 3n-1 = (3n – 1)/2 Solution Let the given statement be P(n), i.e., P(n): 1 + 3 + 32 + …. + 3n-1 = (3n – 1)/2 For n = 1 we have P(1): 1 = (31 – 1)/2 = (3-1)/2 = 2/2 = 1, which is true. Let P(k) be true for some positive integer k, i.e., 1 + 3 + 32 + …. + 3k-1 = (3k – 1)/2 ...(i) We shall now prove that P(k + 1) is true. consider 1 + 3 + 32 + ….+ 3k-1 + (3(k+1)-1 = (1 + 3 + 32 + .... + 3k - 1) + 3k Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 2. Prove the following by using the principle of mathematical induction for all n ∈ N: 13 + 23 + 33+ ….+ n3 = [n(n+1)/2]2 Solution Let the given statement be P(n) , i.e., P(n) : 13 + 23 + 33 + …. + n3 = [n(n+1)/2]2 For n = 1, we have Let P(k) be true for some positive integer k, i.e., 13 + 23 + 33 + ….+ k3 = [k(k+1)/2]2 ...(i) We shall now prove that P(k + 1) is true. Consider 13 + 23 + 33 + ....+ k3 + (k + 1)3 = (13 + 23 + 33 + .... + k3 ) + (k + 1)3 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural number i.e., n. 3. Prove that following by using the principle of mathematical induction for all n ∈ N : 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + .... + 1/(1 + 2 + 3 + ...n) = 2n/(n + 1) Solution Let the given statement be P(n), i.e., P(n): For n = 1, we have P(1) : 1 = 2.1/(1 + 1) = 2/2 = 1 which is true. Let P(k) be true for some positive integer k, i.e., We shall now prove that P(k + 1) is true. Consider Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 4. Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) = [n(n + 1)(n + 2)(n +3)]/4 Solution Let the given statement be P(n), i.e., P(n) : 1.2.3 + 2.3.4 + .... + n(n + 1)(n+2) = [n(n+1)(n+2)(n+3)]/4 For n = 1 , we have P(1) : 1.2.3 = 6 = [1(1 + 1)(1 + 2)(1 + 3)]/4 = (1.2.3.4)/4 = 6, which is true. Let P(k) be true for some positive integer k, i.e., 1.2.3 + 2.3.4 + ...+k(k + 1)(k + 2) = [k(k+1)(k + 2)(k + 3)]/4 ...(i) We shall now prove that P(k + 1) is true. Consider, 1.2.3 + 2.3.4 + ... +k(k + 1)(k + 2)+ (k+1)(k + 2)(k + 3) ={1.2.3 + 2.3.4 + ....+k(k+1)(k + 2)} + (k + 1)(k + 2)(k + 3) Thus, P(k+1) is true whenever P(k) is true . Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 5. Prove the following by using the principle of mathematical induction for all n ∈ N: 1.3 + 2.32 + 3.33 + .... + n.3n =[(2n - 1)3n+1 + 3]/4 Solution Let the given statement be P(n), i.e., P(n) : 1.3 + 2.32 + 3.33 + .... + n.3n =[(2n - 1)3n+1 + 3]/4 For n = 1, we have Let P(k) be true for some positive integer k, i.e., 1.3 + 2.32 + 3.33 + .... + k.3k[(2k-1)3k+1 + 3]/4 ...(i) We shall now prove that P(k + 1) is true. Consider, 1.3 + 2.32 + 3.33 + .... + k3k = [(2k -1)3k+1 + 3]/4 ...(i) We shall now prove that P(k + 1)is true. Consider, 1.3 + 2.32 + 3.33 + ...+k3k + (k +1)3k+1 = (1.3 + 2.32 + 3.33 + .... + k.3k ) + (k + 1)3k+1 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 6. Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.3+ 3.4 + ....+ n.(n + 1) = [{n(n+1)(n+2)}/3] Solution Let the given statement be P(n), i.e., P(n) : 1.2 + 2.3+ 3.4 + .....+ n.(n + 1) = [{n(n+1)(n+2)}/3] For n = 1, we have P(1) : 1.2 = 2 = [1(1 +1)(1 + 2)]/3 = 1.2.3/3 = 2 , which is true. Let P(k) be true for some positive integer k, i.e., 1.2 + 2.3 + 3.4 + .... + k.(k + 1) = [{k(k+1)(k+2)}/3] ...(i) We shall now prove that P(k + 1) is true. Consider 1.2 + 2.3 + 3.4 + ....+ k.(k + 1) + (k + 1).(k + 2) = [1.2 + 2.3 + 3.4 + ..... + k.(k + 1)] + (k + 1).(k + 2) Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 7. Prove the following by using the principle of mathematical induction for all n ∈ N: 1.3 + 3.5 + 5.7 + .... + (2n - 1)(2n + 1) = [n(4n2 + 6n - 1)]/3 Solution Let the given statement be P(n) , i.e., P(n): 1.3+ 3.5 + 5.7 + ....+(2n - 1)(2n + 1) = [n(4n2 + 6n - 1)]/3 For n = 1, we have Let P(k) be true for some positive integer k, i.e., 1.3 + 3.5 + 5.7 + ....+(2k - 1)(2k + 1) = [k(4k2 + 6k - 1)]/3 ...(i) We shall now prove that P(k + 1) is true. Consider (1.3 + 3.5 + 5.7 + .... +(2k - 1)(2k + 1) + {2(k + 1) - 1}{2(k+1) +1} Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 8. Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2 . Solution Let the given statement be P(n), i.e., P(n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2 For n = 1, we have P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2, which is true. Let P(k) be true for some positive integer k, i.e., 1.2 + 2.22 + 3.22 + … + k.2k = (k – 1) 2k + 1 + 2 …(i) We shall now prove that P(k + 1) is true. Consider {1.2 + 2.22 + 3.23 + ..... +k.2k } + {k + 1}. 2k+1 = (k - 1) 2k+1 + 2 + (k + 1)2k+1 = 2k+1 {(k - 1) + (k + 1)} + 2 = 2k+1 . 2k + 2 = k.2(k+1) + 1 + 2 = {(k + 1) - 1}2(k+1)+1 + 2 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 9. Prove the following by using the principle of mathematical induction for all n ∈ N : 1/2 + 1/4 + 1/8 + .... + 1/2n = 1 - 1/2n Solution Let the given statement be P(n), i.e., P(n) : 1/2 + 1/4 + 1/8 + .... + 1/2n = 1 - 1/2n For n = 1, we have P(1) : 1/2 = 1 - 1/21 = 1/2 , which is true. Let P(k) be true for some positive integer k, i.e., 1/2 + 1/4 + 1/8 + ....+ 1/2k = 1 - 1/2k ...(i) We shall now prove that P(k + 1) is true. Consider, Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 10. Prove the following by using the principle of mathematical induction for all n ∈ N: 1/2.5 + 1/5.8 + 1/8.11 + .... + 1/[(3n - 1)(3n + 2)] = n/(6n + 4) Solution Let the given statement be P(n), i.e., P(n) : 1/2.5 + 1/5.8 + 1/8.11 + .... + 1/[(3n - 1)(3n + 2)] = n/(6n + 4) For n = 1, we have P(1) = 1/2.5 = 1/10 = 1/(6.1 + 4) = 1/10, which is true. Let P(k) be true for some positive integer k, i.e., 1/2.5 + 1/5.8 + 1/8.11 + .... + 1/[(3k - 1)(3k+ 2)] = k/(6k + 4) ...(i) We shall now prove that P(k + 1) is true. Consider, Thus, P(k + 1)is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 11. Prove the following by using the principle of mathematical induction for all n ∈ N: 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + ....+ 1/[n(n+1)(n+2)] = n(n+3)/[4(n+1)(n+2)]. Solution Let the given statement be P(n), i.e, For n = 1, we have Let P(k) be true for some positive integer k, i.e., We shall now prove that P(k + 1) is true. Consider, Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 12. Prove the following by using the principle of mathematical induction for all n ∈ N: a + ar + ar2 + .....+ arn-1 = a(rn - 1)/(r - 1) Solution Let the given statement be P(n), i.e., P(n): a + ar + ar2 + .....+ arn-1 = a(rn - 1)/(r - 1) For n = 1, we have P(1): a = a(r1 - 1)/(r - 1) = a , which is true. Let P(k) be true for some positive integer k, i.e., a + ar + ar2 + ....+ ark-1 = a(rk - 1)/(r - 1) ...(i) We shall now prove that P(k + 1) is true. Consider {a + ar + ar2 + .....+ ark-1 } + ar(k+1)-1 = a(rk - 1)/(r - 1) + ark [Using (i)] 13. Prove the following by using the principle of mathematical induction for all n ∈ N: (1+3/1)(1+ 5/4)(1+7/9)...[1 + (2n+1)/n2 ) = (n+ 1)2 Solution Let the given statement be P(n), i.e., For n = 1, we have P(1): (1 + 3/1) = 4 = (1 + 1)2 = 22 = 4, which is true. Let P(k) be true for some positive integer k, i.e., We shall now prove that p(k + 1) is true. Consider, Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 14. Prove the following by using the principle of mathematical induction for all n ∈ N (1 + 1/1)(1 + 1/2)(1+ 1/3).....(1 + 1/n) = (n + 1) Solution Let the given statement be P(n), i.e., P(n) : (1 + 1/1)(1 + 1/2)(1+ 1/3).....(1 + 1/n) = (n + 1) For n = 1, we have P(1): (1 + 1/1) = 2 = (1 + 1) , which is true. Let P(k) be true for some positive integer k, i.e., We shall now prove that P(k + 1) is true. Consider, Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 15. Prove the following by using the principle of mathematical induction for all n ∈ N 12 + 32 + 52 + ... + (2n - 1)2 = [n(2n - 1)(2n+1)]/3 Solution Let the given statement be P(n), i.e., P(n) = 12 + 32 + 52 + ... + (2n - 1)2 = [n(2n - 1)(2n+1)]/3 For n = 1, we have P(1) = 12 = 1 = [1(2.1 - 1)(2.1 + 1)]/3 = 1.1.3/3 = 1, which is true. Let P(k) be true for some positive integer k, i.e., P(k) = 12 + 32 + 52 + ...+(2k - 1)2 = [k(2k-1)(2k+1)]/3 ...(1) We shall now prove that P(k+1) is true. Consider, Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 16. Prove the following by using the principle of mathematical induction for all n ∈ N 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/[(3n-2)(3n+1)] = n/(3n+1) Solution Let the given statement be P(n), i.e., P(n) : 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/[(3n-2)(3n+1)] = n/(3n+1) For n = 1, we have P(1) = 1/1.4 = 1/(3.1 + 1) = 1/4 = 1/1.4 , which is true. Let P(k) be true for some positive integer k, i.e., We shall now prove that P(k+1) is true. Consider, Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 17. Prove the following by using the principle of mathematical induction for all n ∈ N 1/3.5 + 1/5.7 + 1/7.9 + ....+ 1/[(2n+1)(2n+3)] = n/3(2n+3) Solution Let the given statement be P(n), i.e., For n = 1, we have Let P(k) be true for some positive integer k, i.e., We shall now prove that P(k + 1) is true. Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 18. Prove the following by using the principle of mathematical induction for all n ∈ N: 1+ 2 + 3 + ... + n <(1/8)(2n + 1)2. Solution Let P(k) be true for some positive integer k, i.e., 1+2+....+ k < (1/8)(2n + 1)2 Adding (k + 1) on both the sides of the inequality , we have, Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 19. Prove the following by using the principle of mathematical induction for all n ∈ Nn (n + 1) (n + 5) is a multiple of 3. Solution Let the given statement be P(n), i.e., P(n): n (n + 1) (n + 5), which is a multiple of 3. It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3. Let P(k) be true for some positive integer k, i.e., k (k + 1) (k + 5) is a multiple of 3. ∴ k (k + 1) (k + 5) = 3m, where m ∈ N …(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider (k + 1){(k+1) + 1}{(k+1) + 5} = (k+1)(k+2){(k+5) + 1} = (k + 1)(k + 2)(k+5)+(k+1)(k+2) = {k(k+1)(k+5)+2(k+1)(k+5)}+ (k+1)(k+2) = 3m + (k + 1){2(k+5)+(k+2)} = 3m + (k+1){2k+10+k+2} = 3m+ (k+1)(3k+12) = 3m+ 3(k+1)(k+4) = 3[m + (k+1)(k+4)] = 3 × q, where q = {m+(k+ 1)(k+4)} is some natural number Therefore, (k+1)[(k+1)+1][(k+1) + 5] is a multiple of 3. thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 20. Prove the following by using the principle of mathematical induction for all n ∈ N: 102n–1 + 1 is divisible by 11. Solution Let the given statement be P(n), i.e., P(n): 102n–1 + 1 is divisible by 11. It can be observed that P(n) is true for n = 1 since P(1) = 102.1–1 + 1 = 11, which is divisible by 11. Let P(k) be true for some positive integer k, i.e., 102k–1 + 1 is divisible by 11. ∴102k–1 + 1 = 11m, where m ∈ N …(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider 102(k+1)-1 + 1 = 102k+2-1 + 1 = 102k+1 + 1 = 102 (102k-1 + 1 - 1) + 1 = 102 (102k-1 + 1) - 102 + 1 = 102 .11 m - 100 + 1 [Using (1)] = 100 × 11m - 99 = 11(100m - 9) = 11r, where r = (100m - 9) is some natural number Therefore, 102(k+1)-1 + 1 is divisible by 11. Thus, p(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 21. Prove the following by using the principle of mathematical induction for all n ∈ Nx2n – y2n is divisible by x y. Solution Let the given statement be P(n), i.e., P(n): x2n – y2n is divisible by x + y. It can be observed that P(n) is true for n = 1. This is so because x× 1 – y× 1 = x2 – y2 = (x + y) (x – y) is divisible by (x + y). Let P(k) be true for some positive integer k, i.e., x2k – y2k is divisible by x + y. ∴x2k – y2k = m (x + y), where m ∈ N …(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 22. Prove the following by using the principle of mathematical induction for all n ∈ N: 32n+2 – 8n– 9 is divisible by 8. Solution Let the given statement be P(n), i.e., P(n): 32n+2 – 8n – 9 is divisible by 8. It can be observed that P(n) is true for n = 1 since 3× 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8. Let P(k) be true for some positive integer k, i.e., 32k+2 – 8k – 9 is divisible by 8. ∴32k+2 – 8k – 9 = 8m; where m ∈ N …(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider 32(k+1)+2 - 8(k+1) - 9 = 32k+2.32 -8k-8-9 = 32 (32k+2 -8k - 9 + 8k + 9) - 8k - 17 = 32 (32k+2 - 8k - 9) + 32 (8k + 9) - 8k - 17 = 9.8m + 9(8k + 9) - 8k - 17 = 9.8m + 72k + 81 - 8k - 17 = 9.8m + 64k + 64 = 8(9m + 8k + 8) = 8r, where r = (9m + 8k + 8) is a natural number Therefore, 32(k+1)+2 - 8(k+1) - 9 is divisible by 8. Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle mathematical induction, statement P(n) is true for all natural number i.e., n. 23. Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27. Solution Let the given statement be P(n), i.e., P(n):41n – 14nis a multiple of 27. It can be observed that P(n) is true for n = 1 since 411 - 141 = 27, which is a multiple of 27. Let P(k) be true for some positive integer k, i.e., 41k – 14kis a multiple of 27 ∴41k – 14k = 27m, where m ∈ N …(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider 41k+1 - 14k+1 = 41k . 41- 14k .14 = 41(41k - 14k + 14k ) - 14k .14 = 41(41k - 14k ) + 41.14k - 14k .14 = 41.27m + 14k (41-14) = 41.27m + 27.14k = 27(41m - 14k ) = 27× r, where r = (41m - 14k ) is a natural number Therefore, 41k+1 - 14k+1 is a multiple of 27. Thus, P(k+1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n. 24. Prove the following by using the principle of mathematical induction for all n ∈ N (2+7) < (n + 3)2 Solution Let the given statement be P(n), i.e., P(n): (2+7) < (n + 3)2 It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true. Let P(k) be true for some positive integer k, i.e., (2k + 7) < (k + 3)2 …(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider [2(k+1)+7] = (2k + 7) + 2 ∴ [2(k+1)+7] = (2k+7)+2 <(k+3)2 + 2 [using (1)] ⇒ 2(k+1) + 7 < k2 + 6k + 9 + 2 ⇒ 2(k+1) + 7 < k2 + 6k + 11 Now, k2 + 6k + 11 < k2 + 8k + 16 ∴ 2(k+1) + 7 < (k + 4)2 2(k + 1) + 7 < {(k+1) + 3}2 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural number i.e., n.
# If a ball is dropped on planet Krypton from a height of 20 " ft" hits the ground in 2" sec", at what velocity and how long will it take to hit the ground from the top of a 200 " ft"-tall building? Jun 22, 2017 $6.3246$ seconds #### Explanation: Let accelaration due to gravity on Krypton be ${g}_{k}$. Hence, when an object at a state of rest is dropped and it reaches ground in $2$ seconds, then we have $S = \frac{1}{2} {g}_{k} {t}^{2}$ or $20 = \frac{1}{2} {g}_{k} \times {2}^{2}$ or $20 = 2 {g}_{k}$ and hence ${g}_{k} = 10$ $\frac{f t}{{\sec}^{2}}$ Let $t$ be the time taken by the object to hit the ground from the top of a $200$ $f t$ tall building. Then $200 = \frac{1}{2} \times {g}_{k} \times {t}^{2}$ or $200 = \frac{1}{2} \times 10 \times {t}^{2}$ or $5 {t}^{2} = 200$ or $t = \sqrt{\frac{200}{5}} = \sqrt{40} = 2 \sqrt{10} = 2 \times 3.1623 = 6.3246$ seconds Jun 22, 2017 $t \approx 6.32 \text{ s}$ $v \approx 63.2 \text{ft"/"s}$ #### Explanation: On Krypton, the velocity as a function of time is $v \left(t\right) = a t$ The position equation is the anti-derivative of the velocity, so $x \left(t\right) = \int \left(a t\right) \mathrm{dt}$ $x \left(t\right) = \frac{1}{2} a {t}^{2} + C$ Let $t = 0$ when the ball is first dropped, at height $x \left(0\right) = 20$ feet. $\frac{1}{2} a \left(0\right) + C = 20$ $C = 20$ So, the position equation becomes $x \left(t\right) = \frac{1}{2} a {t}^{2} + 20$ When the ball hits the ground, the position will be $x \left(2\right) = 0$ feet $\frac{1}{2} a {\left(2\right)}^{2} + 20 = 0$ $2 a = - 20$ $a = - 10 {\text{ft"/"s}}^{2}$ You were asked for the time and velocity when the ball hits the ground from $200$ feet. $x \left(t\right) = \frac{1}{2} \left(- 10\right) {t}^{2} + 200$ $\frac{1}{2} \left(- 10\right) {t}^{2} + 200 = 0$ $- 5 {t}^{2} = - 200$ ${t}^{2} = \sqrt{40}$ $t \approx 6.32 \text{ sec}$ The velocity when $t = 6.32$ seconds is $v \left(6.32\right) = - 10 \left(6.32\right) = 63.2 \text{ft"/"s}$
# Chapter 12 - Exponents and Logarithms This chapter discusses exponents and logarithms. It contains the following sections: • section 12.1 - In this section we complete the discussion on the properties of exponents that was begun in section 3.3 on integer exponents. • section 12.2 - In this section we introduce logarithms, state their properties and the change of base formula. • section 12.3 - In this section we talk about exponential functions and their various forms. • section 12.4 - In this section we talk about logarithmic functions and applications such as decibels and the Richter scale. • section 12.5 - In this section we talk about logarithmic and semi-logarithmic graphs. • section 12.6 - In this section we solve exponential equations of various types. • section 12.7 - In this section we solve logarithmic equations. ## 12.1 - Exponents Section 3.3 discussed integer exponents. This section continues from there and explains non-integer exponents. We saw that if n is a natural number (i.e. 1, 2, 3, …) then the exponential b n is defined to mean multiplying b by itself n times, like this: The number b is called the base, n is called the exponent, and we say that we are raising b to the n th power. We also saw that any base raised to a negative integer power is the reciprocal of the same base raised to the corresponding positive power: and that any base raised to the 0th power equals 1: b 0 = 1. In that section we also saw that exponentials have these three properties: Multiplication property Division property Exponentiation property Let's assume that these properties can be generalized to exponents that are not necessarily integers. We will find that we can also give meaning to rational and real exponents and even complex exponents. The meaning of b 1/n : Let m = 1/n in the exponentiation property. This gives: On the other hand taking the nth power of the nth root of b gives the same result: Putting this together we find that: In other words, b 1/n is the nth root of b. The most important case is: In other words, b 1/2 is the square root of b. Note: If we are working over the real numbers and the base b is negative then n must be an odd integer; otherwise we can't take the nth root of b. If we are working over the complex numbers then there is no problem and no restriction on n. Click here for more information. For example the following exponential gives a real value: but this exponential gives a complex value: The meaning of b m/n : Using the exponentiation property we can write b m/n two ways: But in the previous section we saw that b 1/n is the nth root of b. Thus this equation says that b m/n may be thought of as the nth root of the mth power of b or as the mth power of the nth root of b. For example: A note on negative bases: If we are working over the real numbers and the base b is negative then n must be an odd integer; otherwise we can't take the nth root of b. If we are working over the complex numbers then there is no problem and no restriction on n. Click here for more information. The meaning of b r where r is any real number: We have made sense of exponents that are positive or negative or fractions but what about exponents that are real numbers? We will explain this case using an example. We will show that 10 1.2 equals 16 (approximately). To do this we make a graph of the function y = 10 x. Here is the table of values and the graph with a smooth curve interpolated through the points. We see that the interpolated curve goes through the point ( x = 1.2, y = 16 ). It is in this sense that we can say that 10 1.2 = 16. It is easy to see that this use of interpolation doesn't depend on the base being 10; any other base would produce a similar result. Note: If we are working over the real numbers then the base b must be a positive number. If we are working over the complex numbers then the base can have any value, with one exception (namely the base can't be zero if the real part of the exponent is negative or zero, because this causes division by zero; see below.) Click here for more information. The meaning of 0 x : This is an interesting situation because depending on the exponent x, there is zero either in the numerator or the denominator, so the expression can be either 0 or undefined. And 0 0 cannot be computed; it is defined to equal 1. In summary, if the exponent x is a real number then there are the following cases: If the exponent x is a complex number then there are the following cases: ## 12.2 - Logarithms Before reading this section you may want to review the previous section on exponents, since logarithms are based on them. Introduction: It is a fact that every positive number, y, can be expressed as 10 raised to some power, x. We write this relationship in equation form, like this: y = 10 x For example it is obvious that the number 1000 can be expressed as 10 3, because the exponent 3 means to multiply 10 by itself 3 times and 10·10·10=1000. It is not so obvious that the number 16 can be expressed as 10 1.2. Clearly this cannot mean to multiply 10 by itself 1.2 times! What does it mean and how can we calculate that this is the correct power of 10? We start by giving it a name: logarithm. We say that 1.2 is the logarithm of 16. It is the subject of this chapter. Let’s start by making a graph of the equation y = 10 x. To make this graph we make a table of a few obvious values of y = 10 x as shown below, left. Then we plot the values in the graph (they are the red dots) and draw a smooth curve through them. Then we observe that the curve goes through y = 16 and x = 1.2 (the black dot). This is what we mean when we say that 16 = 10 1.2. We next define a function called the logarithm function that takes a number like 16 as input, calculates that it can be expressed as 10 1.2 (click here to see exactly how this is done by your calculator), and returns the exponent 1.2 as its output value: Here is the formal definition of the logarithm function. Definition: The logarithm is the function that takes any positive number x as input and returns the exponent to which the base 10 must be raised to obtain x. It is denoted log(x). Example 1: Evaluate log ( 10 5.7 ). In this example the argument of the log function (i.e. the quantity in brackets) is already expressed as 10 raised to an exponent, so the log function simply returns the exponent. log ( 10 5.7 ) = 5.7 Example 2: Evaluate log ( 1000 ). The argument is a number that is easily expressed as 10 raised to an exponent. We do this and the log function then returns the exponent. log ( 1000 ) = log ( 10 3 ) = 3 Example 3: Evaluate log ( 16 ). The argument is a number which we don't know how to express as 10 raised to an exponent (unless we remember the above discussion which said that 16 = 10 1.2 ). Therefore we use a calculator or the Algebra Coach to evaluate it. log ( 16 ) = 1.2 Example 4: log ( x + 4 ). The argument is an expression. Until we can evaluate that expression we have no choice but to leave this logarithm as is. Note that the number 16 can be expressed in exponential form using various bases, so various types of logarithms can be defined. Each type of logarithm is still denoted ‘log’ but we now include the base, written as a subscript, as part of the notation. Here are some examples: 16 = 2 4 → log 2 (16) = 4 16 = 4 2 → log 4 (16) = 2 16 = 16 1 → log 16 (16) = 1 Guided by these examples, we now give the following, more general definition of a logarithm in any base: Definition: The logarithm to base b is the function that takes any positive number x as input and returns the exponent to which the base b must be raised to obtain x. We denote it as logb(x). Notes: • If the subscript after the word ‘log’ is omitted then the base of the logarithm is understood to be 10. Thus log10(x) and log(x) are the same thing. • Suppose that x and y are related by the equation x = b y. This says that ‘x is b raised to the exponent y’. This is called the exponential form of this relationship. But this same relationship can also be written as log b(x) = y, This says that ‘the logarithm of x to base b is y’. This is called the logarithmic form of this relationship. These two equations are equivalent. They are like saying ‘Mary is John’s mother’ and ‘John is Mary’s son’. We say that we are ‘taking logs’ when converting from exponential to logarithmic form, and ‘taking antilogs’ when converting from logarithmic to exponential form. • If we substitute the exponential form x = b y into its own logarithmic form log b(x) = y we get the identity log b( b y ) = y, and if we substitute the logarithmic form log b(x) = y into its own exponential form b y = x we get the identity These identities show how the inverse operations of logging and antilogging undo each other. Example: For our previous example 16 = 10 1.2 the above two identities read log10(10 1.2 ) = 1.2, and Example: Evaluate each of the following logarithms without using a calculator: Solution: The key is to express the argument of the log function (i.e. the quantity in brackets) in exponential form with the base chosen to match the base of the log. Then we use the fact that taking logs and exponentiation are inverse operations: In the above section on exponents we stated these three properties of exponents: Multiplication property Division property Exponentiation property If we rewrite them in logarithmic form then they become the properties of logarithms. To do this, make these substitutions on the left side of each of the three properties: b m = x and b n = y Note for later reference that these substitutions expressed in logarithmic form are: m = log b(x) and n = log b(y). With the substitutions the three properties of exponents read: Now take logs of these three equations (i.e. write them in logarithmic form): Now substitute log b(x) for m and log b(y) for n on the right side of each property (but only for m in the third one). The result is three properties of logarithms. The properties of logarithms are: Property 1: the logarithm of a product: Property 2: the logarithm of a quotient: Property 3: the logarithm of an exponential: These properties are very useful for simplifying a logarithm or for combining several logarithms into one logarithm. Another useful property can be gotten by letting x = 1 in property 2 or by letting m = −1 in property 3: Property 4: the logarithm of a reciprocal: Examples: For each of the following expressions, use the properties of logarithms (or exponents) to combine the logarithms into a single logarithm: Solutions: • Step 1: get rid of the coefficient 3 by using property 3 to make it an exponent of 3; Step 2: combine the sum of logs using property 1: • Step 1: get rid of the coefficients 2 and 5 by using property 3; Step 2: combine the difference of logs using property 2: • Step 1: combine the sum of the first two logs using property 1; Step 2: combine the difference of the remaining logs using property 2; • This could be handled exactly the same way as the previous problem but a shortcut is to notice that all positive logs go into the numerator and all negative logs go into the denominator. • Step 1: get rid of the coefficients 3 and 6 by using property 3; Step 2: combine the sum of logs using property 1; Step 3: write a 3 b 6 as (a b 2 ) 3. Click here to see why you can; Step 4; use property 3 of logarithms again, but this time in reverse: A shortcut is to notice that a common factor of 3 can be factored out of the two terms at the very beginning: ### Common logarithms and natural logarithms Suppose that we wish to express an arbitrary positive number y in exponential form y = b x. The base b that we use could be any positive number whatsoever except 0 or 1. This is because 0x can only equal 0 and 1x can only equal 1 for any value of x. Also, if we try a negative b then we run into trouble with b1/2 since this is the square root of a negative number. Of all the remaining possibilities for the base b there are two special values: • base b = 10 This would seem to be the obvious choice since our number system is based on 10 (probably due to the fact that humans have 10 fingers with which they first learned to count!) Logarithms to base 10 are called common logarithms. For convenience we omit the subscript 10 when using common logs. Thus log(x) is understood to mean log10(x). All scientific calculators are programmed to compute logarithms to base 10 (on most calculators you use the log button) and antilogs to base 10 (use the 10 x button.) • base b = e ≈ 2.71828… This may not seem like a natural choice for the base but nevertheless logarithms to base e are called natural logarithms. The importance of base e (which is the symbol for an irrational number whose value is approximately equal to 2.718) results from the fact that the exponential growth function y = e x, with that particular base, is the only function whose slope equals its own height everywhere. This makes it important in the study of any quantity whose rate of growth is proportional to its present value. (An example is the balance in an interest bearing bank account.) Click here for more information on the function y = e x. For convenience we will use the abbreviation ln(x) instead of the longer form log e(x) to represent the natural logarithm function. (LN stands for Log Natural.) All scientific calculators are programmed to compute logarithms to base e (on most calculators use the ln button) and antilogs to base e (use the e x button.) Here is a comparison table for common logarithms and natural logarithms: Common logarithms Natural logarithms the base 10 e ≈ 2.718 an example in exponential form y = 10 3 = 1000 y = e 3 ≈ 20.08 the same example in logarithmic form log(y) = 3 ln(y) = 3 ### The Change of Base Formulas Logarithms to base 10 and logarithms to base e are proportional to each other (just like miles and kilometers). Logarithms in one base can be changed to logarithms in any other base using a ‘change of base formula’ which basically just uses a proportionality constant. There is also a change of base formula that can be applied to a relationship expressed in exponential form. Logarithmic form of the Change of Base Formula: Exponential form of the Change of Base Formula: These formulae are used to convert from an inconvenient base c to a convenient base b (usually 10 or e). The left side is usually converted to the right side. The quantity log b(c) is the conversion factor. Proof: To prove the second formula notice that it is just the identity discussed previously, namely , with x replaced by c. To prove the first formula, start with the second formula in the form: Example: Compute log 3 (8). Solution: Use the logarithmic form of the change of base formula: Example: Express the function y = 7 x using the base e instead of the base 7. Solution: Use the exponential form of the change of base formula on the number 7. Remember that ln(7) means the same thing as log e(7). 7 = e ln (7) = e 1.946 Substitute this expression in for 7 in y = 7 x. y = (7) x = (e 1.946 ) x = e 1.946 x ## 12.3 - Exponential Functions Click here to review the definition of a function. Click here to see how exponential functions compare with other types of functions in the gallery of functions. Exponential functions are closely related to geometric sequences. A geometric sequence is a list of numbers in which each number is obtained by multiplying the previous number by a fixed factor m. An example is the sequence {1, 3, 9, 27, 81, …}. If we label the numbers in the sequence as {y0, y1, y2, …} then their values are given by the formula yn = y0 · m n. A geometric sequence is completely described by giving its starting value y0 and the multiplication factor m. For the above example y0 = 1 and m = 3. Another example of a geometric sequence is the sequence {40, 20, 10, 5, 2.5, …}. For this sequence y0 = 40 and m = 0.5. An exponential function is obtained from a geometric sequence by replacing the counting integer n by the real variable x. The graph below shows the exponential functions corresponding to these two geometric sequences. Thus we define an exponential function to be any function of the form y = y0 · m x. It gets its name from the fact that the variable x is in the exponent. The “starting valuey0 may be any real constant but the base m must be a positive real constant to avoid taking roots of negative numbers. The exponential function y = y0 · m x has these two properties: • When x = 0 then y = y0. • When x is increased by 1 then y is multiplied by a factor of m. This is true for any real value of x, not just integer values of x. To prove this suppose that y has some value ya when x has some value xa. That is, Now increase x from xa to xa + 1. We get We see that y is now m times its previous value of ya. If the multiplication factor m > 1 then we say that y grows exponentially, and if m < 1 then we say that y decays exponentially. ### The Graph of the Exponential Function We have seen graphs of exponential functions before: • In the section on real exponents we saw a saw a graph of y = 10 x. • In the gallery of basic function types we saw five different exponential functions, some growing, some decaying. The graph below shows two more examples. The first example is the exponential growth function y = 1 (3)x. For it y = 1 when x = 0 and y grows by a factor of 3 when x increases by 1. The second example is the exponential decay function y = 40 (0.5)x. For it y = 40 when x = 0 and y is multiplied by a factor of 0.5 when x is increased by 1. (Equivalently we could say that y decays by a factor of 2 when x is increased by 1.) Notice that the two curves have the same general shape but are reversed left to right and that neither ever touches the x axis. Notice also that if we interchange the x and y axes then the graph of an exponential function turns into the graph of a logarithmic function. The reason for this is that if we take the exponential function y = b x, then interchange x and y to get x = b y, and then solve for y we get y = log b(x). ### The Special Propery of y = e x We have seen graphs of various exponential functions y = b x with various bases b. Notice that for all of them, as we go higher and higher up the curves, they get steeper and steeper. Of all the possible bases there is one particular base, namely e, that causes the curve to have the property that the slope exactly equals the height at every point along the curve. This property defines the number e and makes the curve y = e x an important standard in calculus where we study the slopes of different functions. Let’s prove this important property. ### Constructing the Function y = e x In this section we will construct a function whose slope equals its own height everywhere and then show that it is indeed the function y = e x. We begin in the figure to the right with a single straight-line segment whose left endpoint is at (x = 0, y = 1). Since the height there is 1 we demand that the slope equal 1. The problem of course is that this line segment rises to the right but the slope is not increasing as it should if we wish the slope to equal the height everywhere along the curve. So we improve on this by dividing the region 0 < x < 1 into two equal intervals as shown here. At x = 0.5 we change the slope to 1.5 to reflect the fact that the height there is 1.5. Now the slope equals the height at two points, namely the left endpoints of both line segments. Now we improve even more by dividing the region 0 < x < 1 into n equal intervals as shown here. We can find the height of the right endpoint of each line segment by the following method: In this formula yR is the height at the right endpoint and yL is the height at the left endpoint of the line segment. Now we demand that the slope of each line segment equals the height of its left endpoint. Substitute this in. Now solve for yR. Knowing yL we can use this formula to find the yR. We can iterate this formula across all n line segments, working from left to right. The heights of the right endpoints of all n line segments work out to be: These heights are shown in the third figure. Now we let n → ∞ (i.e. we let the number of intervals become infinite). The curve will become smooth and the slope will equal the height everywhere. We claim that the resulting curve is the function y = e x. To prove this let’s first find the value of y at x = 1. The third figure says that this value is (1 + 1 / n) n. In the following table we have used a calculator to find the value of the expression (1 + 1 / n) n for various values of n: n (1 + 1 / n) n 1 2 2 2.25 10 2.59374 1000 2.71692 1,000,000 2.71828 The values are approaching a definite limit as n becomes large. We define the number e to be the value of the expression (1 + 1 / n) n in the limit as n → ∞. (The number is named e in honor of Leonard Euler who first discovered it.) We express this using this notation: Now let’s find the value of y at arbitrary x. We must begin at x = 0 and go right a total of n · x intervals or steps. From the figure above it is clear that the value of y after n x intervals is y = (1 + 1 / n) n x. By the rules of exponents this can be written in the limit as n → ∞. Thus the function whose slope equals its own height everywhere and which goes through the point (x = 0 , y = 1) is y = e x as claimed. Summary: The function y = e x is the function whose slope equals its own height everywhere and which goes through the point (x = 0 , y = 1). Generalization: The function y = y0 · e bx is the function whose slope is b times its own height everywhere and which goes through the point (x = 0 , y = y 0 ). (b may be positive or negative.) This generalization can be understood as follows: • The function y = y0 · e x is just the function y = e x but stretched vertically by the factor y0. Thus both its slope and height are everywhere multiplied by the factor y0. Its slope still equals its own height everywhere but it goes through the point (x = 0 , y = y0) . • The function y = y0 · e bx is just the function y = y0 · e x but squeezed horizontally by the factor b. Thus its slope is everywhere multiplied by factor b but its height is not changed. ### Alternative Forms for Exponential Growth and Decay Let us now replace the independent variable x in the exponential function by the variable t and let it represent time. Now the exponential function looks like this: y = y0 · m t. We already know that y has the value y0 when t = 0 and that y is multiplied by a factor m when time t is increased by 1. If m > 1 then y grows with time and if m < 1 then y decays with time. We call the growth or decay exponential because t is in the exponent. We have seen how we can change the base by using the change of base formula. In this section we want to show how using various bases or putting the exponent in various forms can reveal various features of the exponential growth or decay. Form 1: Base Greater than 1. First, we can always choose the base m to be greater than 1. The reason for doing this is that then y = y0 · mt, with positive exponent makes it obvious that we have exponential growth, and y = y0 · m− t, with negative exponent makes it obvious that we have exponential decay. The growth or decay is by the factor m each time t increases by 1. For example the function y = (¼) t describes exponential decay but the base is smaller than 1. But using the rules of exponents we can rewrite it as y = (4−1) t, or as y = 4 − t. Now the base is bigger than 1 and the exponent is negative. Notice that these two forms show that the statements “y is multiplied by a factor of 1/4 ” and “y decays by a factor of 4 ” are equivalent. Form 2: Growth or Decay by Given Factor in Given Time. Next we want to give t units, say seconds. To do this we can write an exponential growth in the form y = y0 · b t / T, where b again is a base > 1, and T is a positive constant with the same units of time as t. This form is useful because it makes it plain that y grows by a factor b in a time of T seconds as we can see from this table of values: t y0 · b t / T 0 y0 · b 0 = y0 T y0 · b T / T = y0 · b 2 T y0 · b 2 T / T = y0 · b 2 Note: We can always change b but we must change T accordingly. For example exponential growth by a factor of 10 every 1 second is equivalent to growth by a factor of 100 every 2 seconds. Similarly, any exponential decay can be written as y = y0 · b − t / T, making it plain that y decays by a factor b in a time of T seconds. Form 3: The Time Constant Form. This is a special case of Form 2. If b = e then the constant T is called the time constant and is denoted by the greek letter τ (tau). The exponential growth formula now reads y = y0 · et / τ, making it plain that y grows by a factor of e or approximately 2.7 every τ seconds. The decay formula reads y = y0 · e − t / τ, making it plain that y decays by a factor of e or approximately 2.7 every τ seconds (or decays to e − 1 ≈ 37% of its former value every τ seconds). This can be seen from the table of values: t y0 · e − t / τ 0 y0 · e 0 = y0 τ y0 · e − τ / τ = y0 · e − 1 ≈ 37% of y0 5 τ y0 · e − 5 τ / τ = y0 · e − 5 ≈ 0.7% of y0 Electrical engineers prefer this form because τ is easy to measure and to calculate. Summary: The time constant τ is the period of time that it takes for an exponentially decaying quantity to decay to a fraction e − 1 of its initial value, or to about 37% of its initial value. After a period of 5 time constants it has decayed to less than 1% of its initial value and for many engineering purposes equals zero. Form 4: The Rate Form. Any exponential growth can be written in the form y = y0 · er t. Comparing this with the time constant form we see that r = 1 / τ. Assuming τ has units of seconds, then r has units of 1/seconds. r is called the instantaneous growth rate. Bankers and people interested in rates of growth prefer this form. The equation y = y0 · er t represents a quantity y whose initial value is y0 and whose rate of growth at any instant equals r times its value at that instant. Similarly any exponential decay can be written in the form y = y0 · e − r t, This equation represents a quantity y whose initial value is y0 and whose rate of decay at any time equals r times its value at that time. Example: Describe the function y = \$50 · e 0.20 t in words and sketch its graph. Assume that t is measured in years. Solution: This is an exponential growth function expressed in rate form. Its value is \$50 at time 0 and it grows at a rate of 20% per year. In the sketch below we have shown that the slope of the curve is 0.20 times the height at three different points along the curve. Example: Describe the function y = 100 · e − t / 1.5 and sketch its graph. Solution: This is an exponential decay function expressed in time constant form. Its value at time 0 is 100 and it decays to 37% of its former value in any 1.5 second interval. In the sketch we have shown this for two different intervals. Example: Take the exponential function y = 12 · (1/3) t and put it into the decay by a given factor in a given time” form which will show at a glance how long it takes a) to decay by a factor of 9, and b) to decay by a factor of 100. Solution: a) We wish to change the base to 9 so the formula reads Then the value of T will be the time required to decay by the factor of 9. Notice that 1/3 = 9 −1/2. Substituting this into the original exponential function gives This form clearly shows that y decays by a factor of 9 every 2 seconds. b) We wish to change the base to 100 so the formula reads No simple short-cut with exponents is possible in this case. Instead we can equate the given original form for y with the desired form. This gives We must solve this equation for T. To do this divide by 12 and take log 10 on both sides Use property 3 of logarithms to bring down to exponent: Now divide both sides by t and solve for T. We get T = 4.19 seconds. Substituting this into the desired form gives This form clearly shows that y decays by a factor of 100 every 4.19 seconds. Example: Consider the function y = 1000 · (1/4) t which is graphed below: Using the exponential form of the change of base formula and some simple algebra it is possible to rewrite this function in the following equivalent forms: Discuss the merits of each of the forms. Solution: All of the forms have a base greater than 1 so the negative exponent indicates exponential decay (as opposed to growth). All of the forms have y with an initial value of 1000. • Form (a) shows that y decays by a factor of 4 each time t increases by 1 second. • Form (b) shows that y decays by a factor of 2 each time t increases by 0.5 sec. • Form (c) shows that y decays at an instantaneous rate of 138.6% per second. (The dashed triangle in the figure shows that at the initial value of 1000 this implies an initial slope of −1386.) • Form (d) shows that y decays by a factor of e (or decays to 1/e ≈ 37% of of its former value) each time t increases by 0.721 sec. In other words y decays with a time constant of 0.721 sec. Note that these statements hold anywhere along the exponential curve. It is also interesting to note that if the decay were linear (followed a straight line) then y would hit zero at t = τ. Example: Suppose that we put \$1500 into a bank account which receives interest at a rate of 8% per year and which is compounded continuously. (Compounded continuously is just another way of saying grows exponentially.) Let y denote the amount of money in the account at any time t. Then y can be expressed in the rate form where y0 = \$1500 is called the principal and r = 0.08/yr is the interest rate. The units of y will also be dollars and t must be given in years so that the units of r and t cancel. Questions: (a) What amount will be in the account at the end of 15 months? (b) After how many years will there be \$ 4000 in the account? Solution: (a) Substituting t = 15 months = 1.25 yrs into the equation and evaluating gives (b) Substituting y = \$4000 into the equation and solving for t gives t = 12.26 yrs. (a) After 15 months the account holds \$ 1657.75. (b) The account holds \$ 4000 after 12.26 yrs. Example: Make a sketch of the exponential decay function y = 45 · e − t / 20 (a) Draw a smooth decay curve from upper-left to lower-right. Label the axes but donâ€™t put any numbers on the axes yet. (b) Draw a bracket (or just imagine it) to indicate where y has 100% (or all) of its initial value. Then draw brackets Â˝ and ÂĽ as high where y has 50% and 25% of its initial value. About half-way between 25% and 50% lies 37%. Remember that e−1 is about 37%. (c) Go across at 37% until you hit the curve then go down. This value on the t axis is the time constant. Our function has a time constant of 20 so put that on the t axis. (d) Accuracy check: a straight line with the initial slope should hit the same spot on the t axis. (e) Finish up the graph by putting more tick marks and values on both axes. ### Analyzing Exponential Growth and Decay Just as 2 points determine a straight line, so 2 points determine an exponential function. To derive its equation follow these steps: 1. choose the desired form of the equation, 2. substitute in the 2 points, and 3. solve the resulting 2 equations for the 2 unknowns, one being the initial value and the other being the growth or decay rate or time constant. Example: The electric current, i, flowing in a certain electric circuit decays exponentially with time, t, as shown. Two points on the curve are given. Find an exponential equation of the time constant form i = i0 · e − t / τ to describe the current. Solution: We will explain two methods of solving this problem. Method 1: Substitute the values of i and t at the 2 given points into the equation. This yields a system of 2 equations in the 2 unknowns. The unknowns are i0 and τ: We can eliminate i0 by dividing these equations Take the natural logarithm of both sides and solve for τ ln(3.222) = 2.85 / τ τ = 2.436 Now back-substitute this value of τ into, say, the first of the two equations to get i0: 8.7 = i0 · e − 1.25 / 2.436 i0 = 14.5 Thus the equation is i = 14.5 · e − t / 2.44. Method 2: This method uses the fact that an exponential function decays by a given factor in a given time anywhere along the curve. Thus for the purposes of finding the time constant τ we may use the value 8.7 as i0 and the time difference 4.1 − 1.25 = 2.85 sec. as the time taken for the function to decay to the value 2.7. Substituting these numbers into the equation yields 2.7 = 8.7 · e − 2.85 / τ This equation can be solved for τ τ = 2.436 Now get the actual i0 by substituting τ and one of the points on the curve, say (t = 1.25, i = 8.7), into the equation i = i0 · e − t / τ to get: 8.7 = i0 · e − 1.25 / 2.436 i0 = 14.5 Thus again we find the equation to be i = 14.5 · e − t / 2.44. ### Exponential Decay Toward a Limiting Value The figure to the right shows four functions whose differences from the limiting value y = 5 decay exponentially at various rates. The functions are: (a) y = 5 − 5 e − t / 2 (b) y = 5 + 3 e − t / 2 (c) y = 5 + 3 e − t / 6 (d) y = 5 − 7 e − t / 6 These functions could describe the temperature of hot drinks cooling and cold ones warming toward room temperature. (c) and (d) are in better insulated containers so they take longer to heat or cool. These functions are all of the form y = y + a e − t / τ, There are 3 parameters: y , a and τ. If we write transpose y to the left-hand-side and write the equation as yy = a e − t / τ, then we see that the right side is the familiar exponential decay in time constant form, and that the difference of y from y equals a when t = 0 and that this difference decays with time constant τ. When t = ∞ then y = y. Because there are 3 parameters, we must be given the value of y at 3 different times to fix them. We look at two examples. In this first example the limiting value is given so we only need 2 more points. Example: The voltage in a certain electric circuit decays exponentially toward the limiting value v = 8.2. Two points on the curve are given. The curve may be described by an equation of the form v = v + a e − t / τ, Calculate the values of a and τ. Solution: Similar to the previous example, there are two methods of solving this problem. Method 1: Substitute v = 8.2 and the values of v and t at the 2 given points into the equation. This yields a system of 2 equations in the 2 unknowns a and τ: In the previous example we divided one equation by the other in order to eliminate one of the variables. Here we must first move the 8.2 to the left-hand-side: Now we can divide the equations. We get 0.3226 = e −1.8 / τ Solving gives τ = 1.591 and back-substituting gives a = −13.18. Method 2: In this method use the fact that the difference of v from v decays exponentially to find the time constant τ first. At the first point this difference is 6.2, and 1.8 seconds later the difference is 2.0. Substituting these values into the time constant form of the exponential decay formula, y = y0 e − t / τ, gives 2.0 = 6.2 e − 1.8 / τ Solving for τ gives τ = 1.591 Now find the value of a by substituting τ, v and the value of t and v at either one of the given points into the equation v = v + a e − t / τ. This again gives a = −13.18. In this second example we are given 3 points separated by equal time intervals. If the time intervals were unequal the resulting system of equations could only be solved by computer. Example: The curve to the right is described by the equation v = v + a e − t / τ, Calculate the values of v , a and τ accurate to 3 significant figures. Solution: Move v to the left-hand-side as we did in the previous example. Then substitute in the values of v and t at the 3 given points. This yields the following system of 3 non-linear equations in the 3 unknowns v , a and τ: We can eliminate a by dividing Eq.(3) by (2), and (2) by (1), to get: Now we can eliminate τ by equating Eqs.(4) and (5). (Note that they are equal only because the 3 points were separated by equal time intervals.) We get After cross-multiplying we can solve to get v =116.45 volts. Substituting this back into Eq.(4) or (5) we get τ = 8.151 secs and substituting this back into Eq.(1), (2) or (3) we get a = 152.9 volts, so the final equation to 3 sig. figs. is v = 116 − 153 e − t / 8.15. ## 12.4 - Logarithmic Functions Logarithmic functions are often used to describe quantities that vary over immense ranges. Large ranges of numbers occur in many settings: the amplification abilities of different electronic amplifiers, the sensitivity range of the human eye or the human ear, or the range of energies released by earthquakes. Logarithmic scales such as the decibel scale and the Richter scale are designed to describe large ranges of numbers. Any logarithmic function can be expressed in the form: y = a ln (x) + b, where x and y are variables and a and b are constants. An example is: y = 2.79 ln (x) + 5.80. This function can be expressed in many equivalent forms using the change of base formula. For example: y = 2.79 ln (x) + 5.80 = 6.43 log (x) + 5.80 = 5 log 6 (8 x). #### The Graph of the Logarithmic Function Here are graphs of the functions y = log (x) and y = ln (x). Notice that immense variations in x correspond to small changes in y (at least in the region x > 1). Notice also that log(0) = −∞, and that logarithmic functions exist only to the right of the y axis. Thus the logarithm of a negative number does not exist. The graph of any logarithmic function has the same shape as these and will differ only in a vertical shift and/or a vertical stretch. #### The Electromagnetic Spectrum The electromagnetic spectrum is mentioned here because it is a nice example of a logarithmic scale. Electromagnetic (EM) waves can be produced and detected in a variety or spectrum of wavelengths. The table below shows the EM spectrum and the names given to EM radiation of various wavelengths. The spectrum is actually infinite - we have simply shown the range from 10 −15m to 10 6m that has been intensively exploited by mankind so far. The scale used to present the wavelengths is called a logarithmic scale. In this type of scale each factor of 10 in wavelength is allocated one line of the table. That is, the range from 10 1 to 10 2 is allocated the same amount of space as the range from 10 2 to 10 3, namely 1 line. By contrast in a linear scale the range from 10 2 to 10 3 (a range of 900) would be allocated 10 times as much space as the range from 10 1 to 10 2 (a range of 90). wavelength(in meters) name of region ofEM spectrum 10 −15 10 −14 10 −13 gamma ray 10 −12 10 −11 10 −10 x-ray 10 −9 10 −8 10 −7 ultraviolet 10 −6 visible region 10 −5 infrared 10 −4 10 −3 10 −2 microwave 10 −1 10 0 short radio waves 10 1 FM radio 10 2 AM radio 10 3 10 4 10 5 long radio waves 10 6 Here is a brief history of the electromagnetic spectrum and the significance of some of its regions. In 1820 Oerstad and Ampere discovered that any wire carrying an electric current was surrounded by a magnetic field. A practical application of this effect was the electromagnet. In 1831 Michael Faraday discovered that a changing magnetic field near a wire could induce a voltage in the wire. A practical application of this effect was the electric generator. In 1865 James Clerk Maxwell was able to combine the mathematical equations describing these two effects and to theoretically predict the existence of waves of electricity and magnetism, which we now call electromagnetic (EM) radiation. His equations also predicted the velocity of these waves. You can imagine his excitement when he discovered that their speed was exactly the speed of light! Twenty-two years later Hertz first produced and detected electromagnetic waves, and ten years after that Marconi invented the radio. Our eyes are sensitive only to a very small part of the EM spectrum called the visible region. Different wavelengths in this region are perceived by us as the different colors of the rainbow. Visible light is caused by transitions of the outermost electrons of atoms from one orbit to another. We feel as heat on our skin (rather than see) the infrared region of the EM spectrum. And our skin becomes tanned by ultraviolet radiation. EM radiation from the microwave region causes the water molecule to vibrate. This effect can be used to cook food in a microwave oven. (In case you wondered, the metal screen on the oven door has a fine enough mesh to short circuit the electric component of the microwaves so they can't escape from the oven but is coarse enough to let the shorter wavelength visible light through so that we can see what's cooking inside.) X-rays are emitted by the inner electrons of atoms as they fall from a higher orbit to a lower one. Gamma rays are caused by the rearrangement of particles inside an unstable atomic nucleus. The energy of radiation depends on two things: the intensity (brightness) of the radiation, and the wavelength; short wavelengths being more energetic than long wavelengths. Thus gamma rays are the most energetic and dangerous form of radiation. Astronomy is one area of science that uses every part of the EM spectrum. Radio telescopes as big as several kilometers across gather the long-wavelength radiation emitted by molecules in interstellar space. X-ray telescopes (launched into earth orbit to avoid absorption of x-rays by the earth's atmosphere) observe short-wavelength radiation emitted by black holes, neutron stars and other exotic objects. #### The Decibel Power Scale An ideal amplifier multiplies the height of an input signal by some factor and outputs it without distortion. If the height of the signal represents power then the amplification factor is the ratio Pout /Pin where Pout is the output power and Pin is the input power, both measured in Watts at the same instant in time. For lasers the ratio can be up to 10 12. Attenuators, of which microphones are an example, are devices for which the ratio Pout /Pin is much less than 1. To compress these large variations we define the decibel scale: N is called the gain of the amplifier or attenuator. It is measured in units called decibels (abbreviated db). Notice that the gain is positive for amplifiers, negative for attenuators and zero if Pout /Pin = 1. Example: What is the output power Pout of an amplifier with a gain of 40 db if Pin = 50 mW? Solution: Substitute the given values into the gain formula and then solve for Pout : Amplifiers in Cascade: To achieve very large amplification factors amplifiers are cascaded, which means that the output of one amplifier becomes the input of the next amplifier: The amplification factor of the whole system is the product of the amplification factors of the components. The whole system in the figure has an amplification factor of 6. Equivalently, the gain of the system equals the sum of the gains of its components. To prove this statement let Pin be the power going into amplifier # 1, Pout be the power coming out of amplifier # 2 and Pint be the intermediate power (coming out of Amplifier # 1 and going into # 2). The gain of the whole system is: Now put a factor of Pint /Pint inside the brackets: Now use property 1 of logarithms: The first term in the final expression is just the gain of amplifier # 2 and the second term is the gain of amplifier # 1. This proves that the gain of a system equals the sum of the gains of its components. Example: Find the output power of the system shown: Solution: Adding the three gains gives the gain of the whole system, N = 45 db. Therefore: 45 = 10 log 10 (Pout / 0.05W) Solving for Pout gives: Pout = 0.05 W · 10 4.5 = 1581 W Sometimes the gain formula is used to compare the signal power to the noise power. For example a signal-to-noise ratio of 1000:1 may be quoted as +30 db. #### The Decibel Loudness Scale Fluctuations in air pressure cause our eardrums to vibrate and we interpret these vibrations as sound. The loudness L of a sound (measured in decibels) is defined as: where P is the pressure fluctuation of the sound and Po is the pressure fluctuation of a sound at the threshold of human hearing, which is 20 μPa. Example: How many times larger are the pressure fluctuations of average traffic noise (with a loudness of 85 db) than average livingroom noise (with a loudness of 40 db)? Solution: Substituting each noise into the loudness formula, we have: We can eliminate Po and compare Ptraffic and Proom directly by subtracting these equations and using property 2 of logarithms: Solving for the ratio Ptraffic /Proom gives: The acoustic loudness scale can also be defined in terms of the intensity I of the sound (the energy/unit area/unit time falling on the eardrum): The threshold intensity for human hearing is Io = 10 −12 Watts / m2. A factor of 10 appears in this formula (rather than 20) since the energy of a wave is proportional to the square of the pressure fluctuation. #### The Richter Earthquake Scale The Richter number R describing the strength of an earthquake is defined as: where E is the energy released by the earthquake and Eo is a reference arbitrarily set at the limit of sensitivity of Richter's original seismic measuring apparatus. #### The pH Acidity Scale The pH number of an acid or base is defined as: where [ H + ] is the concentration in moles/liter of the H + ion responsible for acidity. A solution with pH = 7 is neutral, pH < 7 is an acid and pH > 7 is a base. #### The Astronomical Brightness Scale The magnitude M of a star describes its brightness as it appears to us on earth and is defined as: where I is the intensity of the light from the star. I1 is the intensity of a first magnitude star (the 20 brightest stars in the sky are approximately magnitude 1). The most powerful telescopes can detect stars as faint as magnitude +24. ## 12.5 - Logarithmic Graphs In section 6.1 we learned about the cartesian plane, rectangular coordinates, graphs and how to identify points on graphs. Before reading this section you may want to review that section. Henceforth we will call the graphs described in section 6.1 linear graphs. In this section we will learn how to plot quantities on a new type of graph called a logarithmic graph. We will see how this can make it easier to determine the functional relationship between certain quantities. We will motivate logarithmic graphs by giving two examples. t i ln(i) 0.00 5.00 1.64 2.00 2.57 0.94 4.00 1.32 0.28 6.00 0.68 -0.39 8.00 0.35 -1.05 10.00 0.18 -1.71 Example 1: Let t represent time and let i represent the electric current flowing in some electric circuit. The data shown in the table was collected for t and i. (For later use, the natural logarithm of the entries in column 2 was taken and entered in column 3.) Find an equation relating i and t. Solution: First we try plotting i versus t. But this only produces a curve as shown to the right and from this curve it is difficult to find the equation. Next we try plotting ln (i) versus t. This does produce a straight line as shown to the right and we know how to find the equation of a straight line. To find the equation of this line we will pretend for a moment that the axes are labelled y and x. Then this line must have the equation y = m x + b, where m and b are yet to be determined. Eventually we’ll put back the proper label ln (i) for y and t for x. To determine m and b we will use the elimination method discussed in section 6.2. Let’s use these two points which lie on the line: (x = 2.00 , y = 0.94) and (x = 8.00, y = −1.05). Substituting the first point into the equation of the straight line, y = m x + b, gives the equation 0.94 = 2 m + b. Substituting the second point into the equation of the straight line, y = m x + b, gives the equation −1.05 = 8 m + b. This pair of equations, 0.94 = 2 m + b −1.05 = 8 m + b is a system of two equations for the two unknowns m and b. We can solve them by elimination. Subtracting the first equation from the second equation eliminates b and gives: −1.99 = 6 m m = −0.33 Back-substituting this value of m into, say the first equation, 0.94 = 2 m + b, then gives 0.94 = (2)(−0.33) + b which gives b = 1.6. So the equation of the straight line is y = −0.33 x + 1.6. Now replace back y by ln(i) and x by t: ln (i) = −0.33 t + 1.6. Antilogging this equation gives the equation i = e −0.33 t + 1.6, or simplifying, i = 5 e −0.33 t. The key lesson learned from this example is that the graph of any exponential function y = a e b x is a straight line when we plot ln (y) versus x. Example 2: The curve to the right could be the graph of y = x 2, y = x 3 or y = x 4. From a plot of y versus x it is difficult to tell which it is. But if we plot ln (y) versus ln (x) then these three functions become straight lines. These functions are called power functions because the variable x is raised to some power. The key lesson learned from this example is that the graph of any power function y = a x b is a straight line when we plot ln (y) versus ln (x). ### Semilog And log-log graph paper The figure above shows a logarithmic scale of the kind used on logarithmic graph paper. It is not a linear scale because larger numbers are allotted less space than smaller numbers. The lower part of the picture shows the logarithmic scale in more detail. The numbers along the axis are located where their logarithms would be placed on linear graph paper. This means that we can plot x itself on logarithmic graph paper rather than plot log (x) on linear graph paper. We avoid the job of taking logarithms on the calculator. Notice that moving to the left along a logarithmic axis only makes the numbers smaller and smaller. Zero is located an infinite distance to the left and negative numbers do not exist. The range from one power of 10 to the next is called a cycle. Two types of logarithmic graphs are useful: the semi-log graph, which has a logarithmic vertical scale and a linear horizontal scale, as shown below. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. ### The equations of straight lines on logarithmic graph paper One purpose of logarithmic graph paper is simply to put wide ranges of data on one graph. Another purpose is to quickly check if a function follows an exponential law or a power law. We saw one example showing that plotting ln (y) versus ln (x) caused power law functions of the form y = a x b to become straight lines. And we saw another example showing that plotting ln (y) versus x caused an exponential function of the form y = a e b x to become a straight line. From these two examples we conclude the following: A straight line on a semilog graph of y versus x represents an exponential function of the form y = a e b x. A straight line on a log-log graph of y versus x represents a power law function of the form y = a x b. To find the constants a and b, we can substitute two widely-spaced points which lie on the line into the appropriate equation. This gives two equations for the two unknowns a and b which can be solved by elimination. Example: Find the equation of the straight line in the graph to the right. Solution: A straight line on a log-log graph of Q versus T represents the power law function Q = a T b. We must find the values of a and b. To do this we will use a variation of the method described in section 6.2. Take the two points shown in the graph and substitute them, one at a time, into the equation Q = a T b. This gives two equations in the two unknown constants a and b: Here is the variation: first take natural logarithms of both equations and then subtract the two equations. This eliminates ln (a). Then solve for b: Note that b is the power of the power function and is often called the “slope of the log-log graph” and this equation is often used as a shortcut to compute it. Back-substituting b into either of the previous equations gives ln (a) = 0.7120, and anti-logging gives a = 2.04. The constant a is often called the “intercept” although it occurs at T = 1, not at T = 0. (Remember that 0 does not exist in a logarithmic scale.) So to 3 sig. figs. the equation describing the line is Q = 2.04 t 0.558. Example: Find the equation of the straight line in the graph to the right. Solution: A straight line on a semi-log graph of P versus t represents the exponential function P = a e b t. We must find the values of a and b. To do this we will use a variation of the method described in section 6.2. Take the two points shown in the graph and substitute them, one at a time, into the equation P = a e b t. This gives two equations in the two unknown constants a and b: Here is the variation: first take natural logarithms of both equations and then subtract the two equations. This eliminates ln (a). Then solve for b: Note that b is the rate of the exponential function and is often called the “slope of the semilog graph” and this equation is often used as a shortcut to compute it. Back-substituting b into either of the previous equations gives ln (a) = 4.377, and anti-logging gives a = 79.6. Note that the constant a is the “y intercept”. So to 3 sig. figs. the equation describing the line is P = 79.6 e −0.461 t. ## 12.6 - Exponential Equations Before reading this section you may find it helpful to review the following topics: An exponential equation is one in which the variable to be solved for (call it x) is in an exponent. For example 3 x = 5 is an exponential equation while x 3 = 5 is not. Only certain types of exponential equations can be solved using algebra. We will study the following types. (Hover over the links to see examples.) ### Exponential equations in which the unknown occurs just once To solve this type of equation, follow these steps: • Identify the exponential. • Invert the operations that were applied to the exponential in the reverse order in which they were applied. The result is that the exponential stands alone on one side of the equation, which now has the form f = a, where the exponent f contains the unknown x. • If the base of the exponential is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent. This puts the equation into one of these forms: f · ln (b) = ln (a) or f · log (b) = log (a). • In either case exponents are no longer involved. Finish solving for the unknown, x, by using the basic steps for solving equations. • Check the solution. Example: Solve the exponential equation 2 · 5 x + 3 = 21 for x. Solution: This equation contains a single exponential, 5 x. (Notice that the number 2 is not part of it!) To isolate the exponential, subtract 3 and then divide by 2 on both sides of the equation. The result is that the exponential is isolated on the left-hand-side of the equation: 5 x = 9. Take logarithms to base 10 of both sides: log(5 x) = log(9) Use property 3 of logarithms to bring down the exponent: x · log(5) = log(9) Solve for x: ### Exponential equations of the form a · b x = c · d x To solve this type of equation, follow these steps: • Make sure that the equation is of precisely the form a · b x = c · d x. There can only be two terms and one must be on each side of the equation. • If the base of either exponential is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent on both sides of the equation. This puts the equation into one of these forms: ln (a) + x · ln (b) = ln (c) + x · ln (d) or log (a) + x · log (b) = log (c) + x · log (d). • In either case this is now a linear equation in x. It is solved by collecting x terms on the left-hand-side and factoring out x and collecting constant terms on the right: x (log (b) − log (d)) = log (c) − log (a) and then isolating x: • Check the solution. Example: Solve the exponential equation 5 · e 1.7 x = 2 · 4 2.9 x for x. Solution: Because one of the exponentials has base e, take natural logarithms of both sides of the equation: ln (5 · e 1.7 x ) = ln (2 · 4 2.9 x ). On both sides of the equation, use property 1 of logarithms to split up the logarithm of the product ln (5) + ln (e 1.7 x ) = ln (2) + ln (4 2.9 x ). On the right-hand-side, use property 3 of logarithms to bring down the exponent. On the left-hand-side you could do the same, but instead you can just use the fact that the natural log and the exponential function cancel. This gives: ln (5) + 1.7 x = ln (2) + 2.9 x ln (4) Simplify: 1.609 + 1.7 x = 0.6931 + 4.02 x This is a linear equation in x. Solve it by collecting x terms on the left-hand-side and constant terms on the right, and then isolating x. The solution is x = 0.3949. Check this solution by substituting it into the original equation. This gives 9.784 = 9.784 so it checks out. ### Exponential equations in which the same exponential occurs several times To solve this type of equation, follow these steps: • Identify the exponentials in the equation and make sure that they are all identical. Suppose that they are all equal to f, where the exponent f is an expression containing the unknown, x. • Use an alias. Replace each occurrence of f with, say Q. This may turn the equation into a linear or fractional or some other non-exponential type of equation in the variable Q. Solve this equation for Q using the usual techniques. Then substitute back f for Q. At this point the equation has the form f = a, where the exponent f contains the unknown x. • If the base is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent. This puts the equation either into one of these two forms: f · ln (b) = ln (a) or f · log (b) = log (a). • In either case exponents are no longer involved. Finish solving for the unknown x by using the usual techniques. • Check the solution. Example: Solve this exponential equation for t: Solution: This equation contains two identical exponentials, ek t. We replace both occurrences of ek t with Q: This is now a fractional equation in Q. In order to clear the denominators, multiply both sides by 1 − Q: a (1 − Q) = b (1 + Q). This is now a linear equation in Q. In order to collect like terms, distribute on both sides: aa Q = b + b Q. Collect the constant terms on the left-hand-side. Then collect the terms containing Q on the right-hand-side and factor out Q: ab = Q (a + b). Solve for Q: Now substitute back ek t for Q. The result is a single exponential that is isolated: We are finally at the point where we can take natural logarithms. Doing so gives: The unknown, t is no longer in the exponent. Now divide through by −k to solve to t: We could leave the answer in this form or we could get fancy and use the fact that to get rid of a − sign and write the answer as: ### Exponential equations containing exponentials b x, b 2 x, b 3 x, … Suppose that an exponential equation contains the exponentials b x, b 2 x, b 3 x, etc., where b could be any base. This type of equation can be solved by a slight extension of the alias method used above. It depends on the following observation: If b x is replaced by the alias, Q, then b 2 x = (b x ) 2 = Q 2, b 3 x = (b x ) 3 = Q 3, b 4 x = (b x ) 4 = Q 4, etc. Thus all the exponentials can be replaced with positive integer powers of Q. Even exponentials like b 2 x + 1 can be expressed in terms of an integer power of Q since b 2 x + 1 = b 1 · b 2 x = b Q 2 Here are the steps to solve this type of equation: • Identify the exponentials in the equation and make sure that naming one of them Q makes all the others become positive integer powers of Q. (This may well be the hardest step!) • Use the alias. Let b x be Q and express all the other exponentials in terms of Q as well. This hopefully turns the equation into a quadratic or polynomial or some other non-exponential type of equation in the variable Q. Solve this equation for Q using the techniques for that type of equation. Then substitute back b x for Q. At this point the equation (or equations) have the form b x = a. • If the base is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent. This puts the equation either into one of these two forms: x · ln (b) = ln (a) or x · log (b) = log (a). • In either case exponents are no longer involved. Finish solving for the unknown x by using the usual techniques. • Check the solution. Example: Solve the exponential equation e − 2 t + e − t + 1 − 1 = 0. Solution: This is a quadratic equation in e − t as can be easily seen if we use the properties of exponents to rewrite it as: (e − t ) 2 + e 1 · e − t − 1 = 0. Now replace e − t with the alias Q. The result is: Q 2 + e Q − 1 = 0. (Recall that e is the number approximately equal to 2.718.) Solve this quadratic equation for x using the quadratic formula with a = 1, b = e and c = −1: Now substitute back e − t for Q: Since we can’t take the logarithm of a negative number the second solution is extraneous. That leaves the only solution: t = −ln (0.3282) = 1.11 Note that not all exponential equations can be solved using algebra. For example consider the seemingly simple equation x = 10x. We cannot get the x out of the exponent without putting the other x into a logarithm. This equation can only be solved approximately using a computer. ## 12.7 - Logarithmic Equations Before reading this section you may want to review the following topics: A logarithmic equation is one where the variable to be solved for (call it x) is in the argument of a logarithm function. For example log(x) = 5 is a logarithmic equation while log(5) = x is not. To solve a logarithmic equation follow these steps: • If the equation contains several logarithms then you must first use the properties of logarithms to combine them into a single logarithm. • Isolate the logarithm. This means put the equation into the form log bf ) = a so that the log function is alone on one side of the equation. (The expression f contains the unknown x.) • Antilog both sides, thus putting the equation into the exponential form b a = f. • The unknown x is no longer inside a logarithm. Now you can finish solving for x by using the basic procedures for solving equations. • Check the solution. Example: Solve the logarithmic equation 2 log 3 (x − 1) = 4 for x. Solution: There is only a single logarithm so the first step is to isolate the logarithm. To do this divide both sides by 2: log 3 (x − 1) = 2 The next step is to antilog both sides: x − 1 = 3 2 The equation is no longer logarithmic and we can finish solving for x by simply adding 1 to both sides: x = 10 Example: Solve the logarithmic equation log (3 x + 1) − 2 log (x) = 1 for x: Solution: The first step is to combine the logarithms using the properties of logarithms. First use property 3, then property 2: log(3 x + 1) − log (x) = 1 The next step is to antilog both sides. Note that the base of the logarithm is understood to be 10. The equation is no longer logarithmic - it is fractional, so we can proceed to solve for x using techniques for fractional equations. Clear the denominator by multiplying through by x and then move all terms to the left side: 10 x 2 − 3 x − 1 = 0 The result is a quadratic equation in standard form. The left-hand-side can be factored: (5 x + 1) (2 x − 1) = 0. We can replace this equation by two new equations, each of which results from setting one of the factors equal to zero. Solving them yields the solutions: Now we must check the solutions. Substituting x = 1/2 into the original equation and simplifying yields the equation 1 = 1, so it checks out. But substituting x = −1/5 into the original equation means that we must evaluate the logarithm of a negative number and this cannot be done over the real numbers. Thus this solution is extraneous; which leaves us with the only solution, x = 1/2. Note that not all logarithmic equations can be solved using algebra. For example consider the seemingly simple equation x = log (x). We cannot get the x out of the logarithm without putting the other x into an exponential. This equation can only be solved approximately using a computer.
# GCSE Science/Parallel and series circuits You will have already studied series and parallel circuits before, so should be familiar with what a series and parallel circuit is and their basic properties.But in this module we will be looking at them in a little more detail. We will apply Ohm's law to see how we can work out the resistance of a whole circuit that is made up of a large number of components. Before we begin You might want to try some revision questions. Follow the link then come back here when you are finished. ```GCSE Science/Parallel and series circuits revision questions. ``` ## Resistors in Series Circuits As we know, 'in a series circuit the current in all parts of the circuit is the same and the current has a one way system. The current depends on the applied voltage and the number of and nature of other components in the circuit. Consider two resistors in a series circuit with a battery. As you might expect the total resistance in this circuit is higher than the resistance of each resistor, because the battery has to "push" the charge through both resistors one after the other. So the total potential difference of the supply is "shared" between the two resistors. Think about what is happening to the current as it flows around the circuit. There are no branches, nowhere for the electric current to escape to, so obviously the same current must flow through both resistors. Let's call this current I. The total resistance for the whole circuit is very simple. It's just R1 plus R2. ### Formula for resistance of two resistors in series RTotal = R1 + R2 The total resistance in a series circuit is the sum of the resistances of all the components. Using this formula we can calculate the voltage across each resistor. But it could depend on how many resistors there are. ### Example: Calculating voltages in a series circuit Question: Suppose R1 = 1Ω and R2= 4Ω. If the battery supplies 2.5 Volts what is the voltage across each resistor. First use the total resistance of the circuit to work out how much current is flowing through the circuit. Step 1: Work out the total resistance • identify formula: RTotal = R1 + R2 • insert numbers and units RTotal = 1Ω + 4Ω = 5Ω Step 2: Use Ohms's law to calculate the current in the circuit. Decide which resistance to use. • identify formula(using the triangle): I = V / R (Ohms's law) • decide which resistance to use: Use RTotal; so formula becomes I = V / RTotal • insert numbers and units I amps = 2.5V / 5Ω = 0.5A Now we know how much current is flowing through both resistors, we can work out the voltage across each resistor. Step 3: Use Ohm's law to calculate the voltage across each resistor: • identify formula(using the triangle): V = I x R • insert numbers for resistor 1: V1 volts = 0.5A x 1Ω = 0.5 Volts • insert numbers for resistor 2: V2 volts = 0.5A x 4Ω = 2.0 Volts ### Practice questions • Q1)Let R1 = 2Ω, R2= 3Ω, and V=5V what is the voltage across R1 and R2 ? • Q2)Let R1 = 3Ω, R2= 3Ω, and V=6V what is the voltage across R1 and R2 ? • Q3)Let R1 = 2Ω, V=5V and I= 1A what is the value of resistor R2 ? ## Resistors in Parallel Circuits Parallel circuits are a bit more complicated than series circuits, because they contain a branch - the electric current will take more than one route. Look at the diagram. At point X the current splits into 2 paths, and flows through both resistor R1 and resistor R2. It may not split in equal amounts. When several(two or more) components are connected in parallel branches, the voltage (potential difference) across each parallel branch is the same. And this is the same as the voltage across the battery. The current through each component is the same as if it were the only component present. So the total current flowing through the battery is the sum of the currents flowing through each branch. Here is the formula for the currents flowing through a parallel circuit. IMain = I1 + I2 Because the voltage is the same for all branches, the branch with the lowest resistance has the highest current flowing through it. Because there are more paths for the charge to flow along, the total resistance is less than either of the two paths on their own. And therefore (with the same battery) the current is bigger. To find out the resistance of the whole circuit , we can't just add together the resistors as we did in the series circuit, we have to apply Ohm's law to each branch of the circuit. Imagine we replaced the resistors with bulbs. You should now be able to answer the following questions from your previous knowledge. • Q4)You disconnect one of the bulbs, does the other stay lit? • Q5)You reconnect the disconnected bulb, does the other dim or brighten? • Q6)Would it be reasonable to say "each branch of the circuit behaves as if the other branch weren't there" ? If you answered the above questions correctly you should find this next section easy! ### Example: Calculating the resistance of several resistors in parallel It's best to break the process down into a number of simple steps. Some books may give you a formula to use, but you shouldn't use any formula without understanding where it comes from (otherwise you are likely to remember it incorrectly or apply it inappropriately). By using a simple step by step method instead you can get a feel for why it works, and you will be far less likely to make a mistake in the exam. Step1 considering each branch on its own, as if the other branch didn't exist use Ohm's law to work out the current flowing through each branch. Step 2 Add the currents together to find out the total amount of current flowing through the whole circuit. Step 3 Apply Ohm's law again to work out the total resistance of the circuit. ### Example Suppose V=2V R1 = 1Ω and R2= 1Ω. Applying Ohm's law to the branch containing R1 gives I1=V/R1 =2/1 =2A Applying Ohm's law to the branch containing R2 gives I2=V/R2 =2/1 =2A Total current = I1+I2 = 4A Applying Ohm's law again, to the whole circuit. RTotal =V/Itotal = 2/4 =0.5Ω Notice that the resistance of the whole circuit is lower than the resistance of either branch! ### Practice questions • Q7) Calculate the resistance of the above pair of resistors if they had been in series rather than parallel. • Q8) A parallel circuit has two resistors, one is 2Ω the other is 3Ω. The voltage is 6V. What is the total resistance of the circuit? • Q9) Three resistors are in parallel (three separate branches) their resistances are 2Ω, 2Ω and 3Ω. The voltage is 6V. What is the resistance of the circuit? Summary Series Circuits The current through each component is the same. To find the total resistance just add the individual ones up. The total voltage is the sum of the voltages across the individual components. Parallel Circuits Each branch acts as if the other branches were not there. The Voltage across each branch is the same. Ohm's law can be applied to each branch separately. The total current, is just the sum of the currents through each branch.
# nth Term of an Arithmetic Sequence – Examples and Practice The nth term of an arithmetic sequence is found using the value of the common difference, the position of the term, and the value of the first term. We subtract 1 from the term position and multiply by the common difference. Then, we add the value of the first term to find the nth term. Here, we will solve some examples of the nth term of arithmetic sequences. In addition, you can practice your knowledge of this topic with practice problems. ##### ALGEBRA Relevante para Solving some examples of arithmetic sequences. See examples ##### ALGEBRA Relevant for Solving some examples of arithmetic sequences. See examples ## 10 Examples of the nth term of arithmetic sequences with answers If you need a quick revision of the formulas of the nth term of arithmetic sequences, you can explore this article. ### EXAMPLE 1 Find the value of the 6th term of an arithmetic sequence that has an initial value of 2 and a common difference of 3. In this example, we know the values of the initial value and the common difference directly. Also, we know we have to find the 6th term. Then, • $latex a=2$ • $latex d=3$ • $latex n=6$ Using these values in the formula for the nth term of an arithmetic sequence, we have: $latex a_{n}=a+(n-1)d$ $latex a_{6}=2+(6-1)3$ $latex a_{6}=2+(5)3$ $latex a_{6}=2+15$ $latex a_{6}=17$ ### EXAMPLE 2 What is the value of 8th term of an arithmetic sequence in which the first term is 4 and the common difference is 5? Similar to the previous example, here we also know the values of the first term and the common difference. Then, we have: • $latex a=4$ • $latex d=5$ • $latex n=8$ By applying these values in the formula of the nth term of an arithmetic sequence, we have: $latex a_{n}=a+(n-1)d$ $latex a_{8}=4+(8-1)5$ $latex a_{8}=4+(7)5$ $latex a_{8}=4+35$ $latex a_{8}=39$ ### EXAMPLE 3 If the first term of an arithmetic sequence is 10 and its common difference is -2, find the value of the 8th term. Using the given information, we can identify the following values: • $latex a=10$ • $latex d=-2$ • $latex n=8$ In this case, we have a negative common difference. However, we can use the same formula, since it also applies in these cases: $latex a_{n}=a+(n-1)d$ $latex a_{8}=10+(8-1)(-2)$ $latex a_{8}=10+(7)(-2)$ $latex a_{8}=10-14$ $latex a_{8}=-4$ ### EXAMPLE 4 The first four terms of an arithmetic sequence are 3, 6, 9, 12. What is the value of the 11th term of the sequence? In this example, we know the value of the first term directly, but not the common difference. To find the common difference, we can subtract any term by its previous term. For example, we have $latex 12-9=3$. Then, • $latex a=3$ • $latex d=3$ • $latex n=11$ Using these values with the formula for the nth term, we have: $latex a_{n}=a+(n-1)d$ $latex a_{11}=3+(11-1)3$ $latex a_{11}=3+(10)3$ $latex a_{11}=3+30$ $latex a_{11}=33$ ### EXAMPLE 5 The first four terms of an arithmetic sequence are 10, 6, 2, -2. Find the 8th term of the sequence. We start by finding the value of the common difference. Therefore, we have $latex -2-2=-4$. Then, we have the following: • $latex a=10$ • $latex d=-4$ • $latex n=8$ Applying the formula for the nth term with these values, we have: $latex a_{n}=a+(n-1)d$ $latex a_{8}=10+(8-1)(-4)$ $latex a_{8}=10+(7)(-4)$ $latex a_{8}=10-28$ $latex a_{8}=-18$ ### EXAMPLE 6 An arithmetic sequence starts with terms 3, 9, 15, … Find the 9th term of the sequence. Finding the common difference, we have $latex 15-9=6$. Therefore, we have the following values: • $latex a=3$ • $latex d=6$ • $latex n=9$ Using these values in the formula for the nth term, we have: $latex a_{n}=a+(n-1)d$ $latex a_{9}=3+(9-1)6$ $latex a_{9}=3+(8)6$ $latex a_{9}=3+48$ $latex a_{9}=51$ ### EXAMPLE 7 Find the 12th term of the arithmetic sequence that starts with the terms 12, 7, 2, … The common difference value is $latex 2-7=-5$. Thus, we observe the following values: • $latex a=12$ • $latex d=-5$ • $latex n=12$ Using these values with the formula for the nth term, we have: $latex a_{n}=a+(n-1)d$ $latex a_{12}=12+(12-1)(-5)$ $latex a_{12}=12+(11)(-5)$ $latex a_{12}=12-55$ $latex a_{12}=-43$ ### EXAMPLE 8 An arithmetic sequence begins with the terms -13, -9, -5, … What is the value of 7th term? The common difference of the given sequence is $latex -5-(-9)=4$. Then, we have: • $latex a=-13$ • $latex d=4$ • $latex n=7$ Applying the formula for the nth term with these values, we have: $latex a_{n}=a+(n-1)d$ $latex a_{7}=-13+(7-1)4$ $latex a_{7}=-13+(6)4$ $latex a_{7}=-13+24$ $latex a_{7}=11$ ### EXAMPLE 9 An arithmetic sequence has the first four terms: 5, 11, 17, 23. Find the 30th term. The common difference of the given sequence is $latex 23-17=6$. Therefore, we have the following information: • $latex a=5$ • $latex d=6$ • $latex n=30$ Using the formula for the nth term with these values, we have: $latex a_{n}=a+(n-1)d$ $latex a_{30}=5+(30-1)6$ $latex a_{30}=5+(29)6$ $latex a_{30}=5+174$ $latex a_{30}=179$ ### EXAMPLE 10 Find the 26th term of the arithmetic sequence that starts with the numbers 20, 17, 14, … The common difference of the given sequence is $latex 14-17=-3$. Then, we have the following values: • $latex a=20$ • $latex d=-3$ • $latex n=26$ Using the formula with the given values, we have: $latex a_{n}=a+(n-1)d$ $latex a_{26}=20+(26-1)(-3)$ $latex a_{26}=20+(25)(-3)$ $latex a_{26}=20-75$ $latex a_{26}=-55$ ## nth term of arithmetic sequences – Practice problems nth term of arithmetic sequences quiz You have completed the quiz! #### What is the 16th term of an arithmetic sequence that begins with the terms 19, 12, 5, …? Write the answer in the input box. $latex a_{16}=$ ## See also Interested in learning more about arithmetic sequences? You can take a look at these pages: LEARN MORE
# Integration Scheme of work: Year 12 A-Level: Pure 1: Integration #### Prerequisite Knowledge • From GCSE • Calculate with roots, and with integer and fractional indices • Simplify and manipulate algebraic expressions • Simplifying expressions involving sums, products and powers, including the laws of indices • Calculate exactly with surds • Simplify surd expressions involving squares and rationalise denominators • From Pure Year 1 • Find the gradient of curves from first principals and understand this as differentiation. • Differentiate, for rational values of n, and related constant multiples, sums and differences #### Success Criteria • Understand integration as the reverse of differentiation. • Use the gradient function and a point on the curve to find the equation of a curve. • Understand that indefinite integration involves an arbitrary constant. • Evaluate definite integrals. • Use definite integration to find the area enclosed within a curve and x-axis. • Understand that area below the x-axis is negative. #### Key Concepts • Integration as the reverse of differentiation If dy/dx = xn, then y=\frac{1}{n+1} x^{n+1}+c, n \neq-1 • Indefinite Integration \int k f(x) d x=h \int f(x) dx \int(f(x)+g(x)) d x=\int f(x) d x+\int g(x) d x • Definite Integration of the form xn \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{n}=\frac{b^{n+1}-a^{n+1}}{n+1}, n \neq 1 • Area Under a Curve The value of the definite integral represents the area enclosed within the x-axis, the function and the two limits. When a graph goes below the x-axis, the corresponding values of y are negative, so the area becomes negative. To find the area where some parts of the curve are above the axis and others are below it, you need to separate the integrals so that the negative and positive values do not cancel each other out. #### Common Misconceptions • Some students attempt to answer exam questions using the integral function on their calculator, which results in zero marks awarded. • Poor notation, namely dropping the dx part of the integral, can result in lost marks. • When asked to integrate questions of the form below, some students attempt to integrate the numerator and denominator separately. \int\left(\frac{2 x^{3}-5}{x^{2}}\right) d x • Some students forget to include the ‘c’ term when evaluating an indefinite integral. • Integrals must be fully simplified, as marks are lost for not leaving fractions in their simplest form. Some students also include the integral sign as part of their final answer. • While most students know to separate integrals when the curve goes from above the axis to below it, some add to the two areas, not knowing the area below the axis is negative. ## Integration Resources ### Mr Mathematics Blog #### Planes of Symmetry in 3D Shapes Planes of Symmetry in 3D Shapes for Key Stage 3/GCSE students. Use isometric paper for hands-on learning and enhanced understanding. #### GCSE Trigonometry Skills & SOH CAH TOA Techniques Master GCSE Math: Get key SOH-CAH-TOA tips, solve triangles accurately, and tackle area tasks. Ideal for students targeting grades 4-5. #### Regions in the Complex Plane Explore Regions in the Complex Plane with A-Level Further Maths: inequalities, Argand diagrams, and geometric interpretations.
# Difference between revisions of "Problem Solver's Toolbox" The goal of this page is to collect simple problem solving strategies and tools. We hope that students interested in the Wisconsin Math Talent Search would find the described ideas useful. Our hope is that this page and the discussed topics can be used as a starting point for future exploration. ## General ideas There is no universal recipe for math problems that would work every time, that's what makes math fun! There are however a number of general strategies that could be useful in most cases, here is a short list of them. (Many of these ideas were popularized by the Hungarian born Mathematician George Pólya in his book How to Solve It.) • Make sure that you understand the problem. • If possible, draw a figure. • Can you connect the problem to a problem you have solved before? • If you have to show something for all numbers (or a large number) then try to check the statement for small values first. • Can you solve the problem in a special case first? Can you solve a modified version of the problem first? • Is there some symmetry in the problem that you can exploit? • Is it possible to work backward? • Is it possible to generalize the problem? (Sometimes the generalized is easier to prove.) ## Modular arithmetic When we have divide two integers, they don't always divide evenly, and there is a quotient and a remainder. For example when we divide 10 by 3 we get a remainder of 1. It turns out that these remainders behave very well under addition, subtraction, and multiplication. We say two numbers are the same "modulo $m$" if they have the same remainder when divided by $m$. If $a$ and $x$ are the same modulo $m$, and $b$ and $y$ are the same modulo $m$, then $a+b$ and $x+y$ are the same modulo $m$, and similarly for subtraction and multiplication. This often makes calculation much simpler. For example, see 2016-17 Set #2 problem 3. See Art of Problem Solving's introduction to modular arithmetic for more information. ## Mathematical induction Suppose that you want to prove a statement for all positive integers, for example that for each positive integer $n$ the following is true: $1\cdot 2+2\cdot 3+3\cdot 4+\cdots+n\cdot (n-1)=\frac{n(n+1)(n+2)}{3}.\qquad\qquad()$ Mathematical induction provides a tool for doing this. You need to show the following two things: 1. (Base case) The statement is true for $n=1$. 2. (Induction step) If the statement is true for $n$ then it must be true for $n+1$ as well. If we can show both of these parts, then it follows that the statement is true for all positive integer $n$. Why? The first part (the base case) shows that the statement is true for $n=1$. But then by the second part (the induction step) the statement must be true for $n=2$ as well. Using the second part again and again we see that the statement is true for $n=3, 4, 5, \cdots$ and repeating this sufficiently times we can prove that the statement is true for any fixed value of $n$. You can visualize proof by induction as the infinite chain $\text{the statement for 1}\to \text{the statement for 2} \to \cdots \to \text{the statement for }n \to \cdots$ The base case shows that the first statement (the first link in the chain) is true, while the induction step shows that if a statement in the chain is true then the statement to its right is also true. • Try to use induction to show the identity $()$ above for all positive integer $n$.
# Question: What is the probability of getting a sum of 8 when you roll 2 dice? Contents ## What is the probability that the sum is 8 when throwing a dice given that the first die shows a 3? The total number of ways to roll an 8 with 3 dice is therefore 21, and the probability of rolling an 8 is 21/216, which is less than 5/36. heads out of 20 is (20 10 ) /220 ≈ 17.6%. The first scenario is more likely. ## What is the probability of rolling a sum of 8 on a standard pair of six sided dice? Two (6-sided) dice roll probability table Roll a… Probability 6 5/36 (13.889%) 7 6/36 (16.667%) 8 5/36 (13.889%) 9 4/36 (11.111%) ## When two dice are rolled what is the probability of getting a sum of 7? For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6. ## What is the probability of not getting a sum of 8 if a pair of dice is rolled? The probability of any number occurring is 1 in 36 or 1 / 36. Then the probability an 8 will not occur is: 1 – 5 / 36 or 31 / 36. ## What is the probability of getting sum of 8? Probabilities for the two dice Total Number of combinations Probability 8 5 13.89% 9 4 11.11% 10 3 8.33% 11 2 5.56% ## What is the probability formula? In general, the probability is the ratio of the number of favorable outcomes to the total outcomes in that sample space. It is expressed as, Probability of an event P(E) = (Number of favorable outcomes) ÷ (Sample space). ## What is the probability of getting a sum of 8 when rolling a pair of dice Brainly? Therefore theprobability that we get the sum as8 when two dice are thrown is 5/36. ## What is the probability of rolling 2 dice and getting a sum of 5? The probability of rolling a pair of dice whose numbers add to 5 is 4/36 = 1/9. ## How do you find the probability of a sum of dice? If you want to know how likely it is to get a certain total score from rolling two or more dice, it’s best to fall back on the simple rule: Probability = Number of desired outcomes ÷ Number of possible outcomes. ## How do you find the sum of probability? The sum rule is given by P(A + B) = P(A) + P(B). Explain that A and B are each events that could occur, but cannot occur at the same time.
# Unit 3: Modeling using Equations Minds On. Unit 3: Modeling using Equations Solving Polynomial Equations (2) Learning Goal I can solve equations. ## Presentation on theme: "Unit 3: Modeling using Equations Minds On. Unit 3: Modeling using Equations Solving Polynomial Equations (2) Learning Goal I can solve equations."— Presentation transcript: Unit 3: Modeling using Equations Minds On Unit 3: Modeling using Equations Solving Polynomial Equations (2) Learning Goal I can solve equations Success Criteria Expand using distributive property to eliminate fractions or to eliminate brackets Collect like terms Isolate the variable by adding or subtracting terms, so that the unknown variable is isolated to one side of the equation Multiply or divide both sides of the equation so that the coefficient of the variable is 1 Unit 3: Modeling using Equations Solving Polynomial Equations (2) Example 1: 5(2 + x) – 2(3x – 6) = 0 Unit 3: Modeling using Equations Solving Polynomial Equations (2) Step One: Expand using distributive property to eliminate fractions or to eliminate brackets Unit 3: Modeling using Equations Solving Polynomial Equations (2) 5(2 + x) – 2(3x – 6) = 0 Step Two: Collect like terms Unit 3: Modeling using Equations Solving Polynomial Equations (2) 10 + 5x – 6x + 12 = 0 Step Three: Isolate by adding or subtracting terms Unit 3: Modeling using Equations Solving Polynomial Equations (2) 22 – x = 0 Step Four: Multiply or divide both sides of the equation so that the coefficient of the variable is 1 Unit 3: Modeling using Equations Solving Polynomial Equations (2) – x = -22 Unit 3: Modeling using Equations Solving Polynomial Equations (2) Step One: Multiply each term by the lowest common denominator to eliminate fractions Unit 3: Modeling using Equations Solving Polynomial Equations (2) Step Two: Collect like terms Unit 3: Modeling using Equations Solving Polynomial Equations (2) Step Three: Isolate the variable by adding or subtracting terms Unit 3: Modeling using Equations Solving Polynomial Equations (2) Step Four: Multiply or divide both sides of the equation so that the coefficient of the variable is 1 Unit 3: Modeling using Equations Solving Polynomial Equations (2) Step Five: Perform a check to see if your answer is correct Unit 3: Modeling using Equations Solving Polynomial Equations (2) Practice  Read Page 312 “Key Ideas” and Example #1  Do Page 313 Q# 3aeim, 4adg, 5acg, 6bd, 7aceg,  Do proper “CHECKS” for questions with the “*” Unit 3: Modeling using Equations Solving Polynomial Equations (2) Download ppt "Unit 3: Modeling using Equations Minds On. Unit 3: Modeling using Equations Solving Polynomial Equations (2) Learning Goal I can solve equations." Similar presentations
How do you graph f(x)=x/(x-2) using holes, vertical and horizontal asymptotes, x and y intercepts? Jul 26, 2018 Below Explanation: $f \left(x\right) = \frac{x}{x - 2}$ $f \left(x\right) = \frac{\left(x - 2\right) + 2}{x - 2}$ $f \left(x\right) = 1 + \frac{2}{x - 2}$ For vertical asymptotes, $x - 2 \ne 0$ since the denominator cannot equal to 0 as the function will be undefined at that point. To find at what point the function is undefined, we can go $x - 2 = 0$ so $x = 2$ is our vertical asymptote. For horizontal asymptotes, we let $x \to \infty$. As $x \to \infty$, $\frac{2}{x - 2} \to 0$. Hence, $f \left(\infty\right) = 1 + 0 = 1$. Therefore, there is an horizontal asymptote at $y = 1$ For intercepts, When $y = 0$, $x = 0$ When $x = 0$, $y = 0$ Plotting your intercepts and drawing in your asymptotes (remember that the asymptotes influence the endpoints of your graph ONLY and nothing else) graph{x/(x-2) [-10, 10, -5, 5]}
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of "2008 AMC 12A Problems/Problem 7" The following problem is from both the 2008 AMC 12A #7 and 2008 AMC 10A #11, so both problems redirect to this page. ## Problem While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing towards the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking? $\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10$ ## Solution 1 It will take $\frac{1}{4}$ of an hour or $15$ minutes to get to shore. Since only $30$ gallons of water can enter the boat, only $\frac{30}{15}=2$ net gallons can enter the boat per minute. Since $10$ gallons of water enter the boat each minute, LeRoy must bail $10-2=8$ gallons per minute $\Rightarrow\mathrm{(D)}$. ## Solution 2 We set up the following equation, where $x$ is the answer: $\frac{30}{10-x} = 15 \Rightarrow\mathrm{x=8}$, so the answer is $(D)$. ## Solution 3 Let $r$ be the slowest rate LeRoy can bail out water. It will take them $15$ minutes to get to shore. As stated in solution 1, the number of gallons that enter the boat can be at most $2$ for the boat to stay afloat. Water enters at a rate of $10$ gallons per minute. Therefore: $2 \geq 10-r$ $r \geq 8$ The minimum possible value for $r$ is $8 \mathrm{(D)}$ ## See Also 2008 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions 2008 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS
Survey * Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project Document related concepts Elementary mathematics wikipedia, lookup Moiré pattern wikipedia, lookup Patterns in nature wikipedia, lookup Transcript ```13 SEQUENCES W LO 1.1 Section title S E R C I P Neptune was the first planet to be found by mathematical prediction. It was found by looking at the number patterns of the other planets in the Solar System, and its position was correctly predicted to within a degree. Two scientists were eventually jointly credited with the discovery, one British and one French, but it has since been shown that the Brits took a bit too much of the credit! Objectives Before you start In this chapter you will: continue number patterns using the four rules of number continue patterns using pictures and give the rule for continuing the pattern use number machines to produce a number pattern and write down the rule complete a table of values using a number machine and write down the rule use the first difference to find the nth term and use the nth term to find any number in a sequence identify whether or not a number is in a sequence. You need to know: how to spot how a number pattern continues that the numbers in number patterns can get bigger and get smaller that multiples are members of a multiplication table, e.g. 3, 6, 9, 12 are multiples of 3. 241 Chapter 13 Sequences 13.1 Sequences Objectives Why do this? You can continue number patterns by adding or subtracting, or multiplying or dividing by a number. You can give the term to term rules for continuing number patterns. You can continue patterns using pictures and give the rule for continuing the pattern. You may use sequences when learning a dance routine or a note sequence when playing a musical instrument. Even numbers form a pattern Odd numbers also form a pattern 2, 4, 6, 8, 10, 12, ... They go up in twos. 1, 3, 5, 7, 9, 11, ... These also go up in twos. 1. Write down all the even numbers up to 20. 2. Write down all the odd numbers up to 20. 3. Check that you have written all the numbers from 1 to 20. Key Points A sequence is a pattern of numbers or shapes that follows a rule. Number patterns can be continued by adding, subtracting, multiplying and dividing. Patterns with pictures can be continued by finding the rule for continuing the pattern. The numbers in a number pattern are called terms. The term to term rules for continuing number patterns can be given. (This means you can say how you find a term from the one before it.) Example 1 a 14  4  18 18  4  22 a Write down the next two numbers in this number pattern. 2 6 10 14 b What is the rule you use to find the next number in the number pattern? c Find the 10th number in this pattern. 2 6 �4 10 �4 14 �4 18 �4 b To get the next number you add 4 each time. c The 10th number in the pattern is 38. 2 6 10 14 18 22 26 30 34 38 Carry on the number pattern until you get to the 10th number in the pattern. 242 terms term to term rules sequence 13.1 Sequences Questions in this chapter are targeted at the grades indicated. Exercise 13A 1 Find the two missing numbers in these number patterns. For each pattern, write down the term to term rule. a 3, 6, 9, __ , __ , 18, 21 b 3, 7, 11, __ , __ , 23, 27 c 5, 10, 15, 20, __ , __ , 35, 40 d 2, 7, 12, 17, __ , __ , 32, 37 e 1, 4, 7, 10, __ , __ , 19, 22 f 5, 7, 9, 11, __ , __ , 17, 19 g 3, 8, 13, 18, __ , __ , 33, 38 h 4, 7, 10, 13, __ , __ , 22, 25 i 2, 6, 10, 14, __ , __ , 26, 30 j 10, 20, 30, __ , __ , 60, 70 2 a Write down the next two numbers in these sequences. Examiner’s Tip i 1, 5, 9, 13, 17, … ii 2, 5, 8, 11, 14, … iii 3, 7, 11, 15, 19, … iv 4, 8, 12, 16, 20, … … means that the sequence v 5, 8, 11, 14, 17, … vi 5, 11, 17, 23, … carries on. vii 2, 6, 10, 14, 18, … viii 1, 7, 13, 19, 25, … ix 3, 11, 19, 27, 35, … x 5, 9, 13, 17, 21, … b Write down the rule you used to find the missing numbers in each sequence. 3 Find the 10th number of each of the number patterns in questions 1 and 2. 4 Jenny saves £2 each week in her piggy bank. Here is the pattern of how her money grows. Week 1 2 3 Money in piggy bank 2 4 6 F E 4 5 a Copy and complete the table. b Jenny is saving for a present for her Mum’s birthday that costs £20. How many weeks will this take? AO3 Continuing patterns by subtracting Key Point Number patterns can be continued by subtracting the same number from each term. Example 2 a Write down the next two numbers in this number pattern. 60 54 48 42 36 b What is the rule you use to find the next number in the number pattern? c Find the 8th number in this pattern. a 36  6  30 30  6  24 60 54 �6 48 �6 42 �6 36 �6 To get to the next number you take away 6. Take away 6 from 36 to get 30 then take away 6 from 30 to get 24. 243 Chapter 13 Sequences b To get the next number you subtract 6 each time. c The 8th number in the pattern is 18. 60 54 48 42 36 30 24 18 Carry on the number pattern until you get to the 8th number in the pattern. Exercise 13B F 1 Find the two missing numbers in these number patterns. Write down the rule for each number pattern. a 20, 18, 16, 14, __ , __ , 8 b 17, 15, 13, 11, __ , __ , 5 c 55, 50, 45, 40, __ , __ , 25 d 42, 37, 32, 27, __ , __ , 12 e 22, 19, 16, 13, __ , __ , 4 f 19, 17, 15, 13, __ , __ , 7 g 45, 38, 31, 24, __ , __ , 3 h 25, 22, 19, 16, __ , __ , 7 i 29, 25, 21, 17, __ , __ , 5 j 80, 70, 60, __ , __ , 30 2 a Write down the next two numbers in these sequences. i 41, 37, 33, 29, … ii 27, 24, 21, 18, … iii 59, 55, 51, 47, … iv 34, 31, 28, 25, … v 30, 27, 24, 21, … vi 61, 55, 49, 43, … vii 22, 20, 18, 16, … viii 51, 46, 41, 36, … ix 64, 57, 50, 43, … x 8, 6, 4, 2, 0, 2, … b Write down the rule you used to find the missing numbers in each sequence. E 3 Find the 10th number of each of the number patterns in questions 1 and 2. 4 Abdul’s mother gives him £20 each week to buy his lunch. His lunch costs him £3 each day. Here is the pattern of how he spends his money. Day M Tu Money left at end of day 17 14 W Th F a Copy and complete the table. AO3 244 b How much money will Abdul have left at the end of the week? 13.1 Sequences Continuing number patterns by multiplying Example 3 a Write down the next two numbers in this number pattern. 1 3 9 27 81 b What is the rule you use to find the next number in the number pattern? c Find the 8th number in this pattern. a 81  3  243 1 243  3  729 3 �3 9 �3 27 �3 81 �3 To get to the next number you multiply by 3. Multiply 81 by 3 to get 243 then multiply 243 by 3 to get 729. b To get the next number you multiply by 3 each time. c The 8th number in the pattern is 2187. 1 3 9 27 81 243 729 2187 Carry on the number pattern until you get to the 8th number in the pattern. Exercise 13C 1 Find the missing numbers in these number patterns. For each pattern, write down the rule. a 1, 2, 4, 8, __ , __ , 64 b 1, 4, 16, 64, __ , 1024 c 1, 5, __ , 125, __ , 3125 d 1, 10, 100, __ , __ , 100 000 e 3, 6, 12, 24, __ , __ , 192 f 2, 6, 18, __ , __ , 486 g 2, 8, 32, __ , __ , 2048 h 2, 20, 200, 2000, __ , __ , 2 000 000 i 2, 10, 50, __ , __ , 6250 j 3, 15, 75, __ , 1875 2 a Write down the next two numbers in these sequences. i 2, 4, 8, 16, … ii 3, 9, 27, 81, … iii 4, 16, 64, 256, … iv 5, 25, 125, 625, … v 5, 10, 20, 40, … vi 4, 12, 36, 108, … vii 10, 30, 90, 270, … viii 5, 50, 500, 5000, … ix 10, 20, 40, 80, … x 6, 36, 216, 1296, … b Write down the rule you used to find the missing number in each sequence. 3 Find the 10th number of each of the number patterns in questions 1 and 2. 4 The number of rabbits in a particular colony doubled every month for 10 months. The table shows the beginning of the pattern. Month 1 2 3 Number of rabbits 2 4 8 4 F E 5 a Copy and complete the table. b How many rabbits were in the colony in month 10? AO2 245 Chapter 13 Sequences Continuing number patterns by dividing Example 4 a Write down the next two numbers in this number pattern. 729 243 81 27 b What is the rule you use to find the next number in the number pattern? c Find the 8th number in this pattern. a 27  3  9 933 81 27 �3 9 �3 3 �3 1 �3 To get to the next number you divide by 3. Divide 27 by 3 to get 9 then divide 9 by 3 to get 3. b To get the next number you divide by 3 each time. 1 729 243 81 27 9 3 1 __ 3 Carry on the number pattern until you get to the 8th number in the pattern. 1 1  3  __ 3 1 c The 8th number in the pattern is 1  3  __ 3 Exercise 13D F E 1 Find the missing numbers in these number patterns. Write down the rule for each number pattern. a 64, 32, 16, 8, __ , __ , 1 b 1024, 256, 64, __ , 4 c 3125, 625, 125, __ , __ , 1 d 100 000, 10 000, 1000, __ , __ , 1 e 192, 96, 48, 24, __ , __ , 3 f 486, 162, 54, 18, __ , 2 g 1024, 512, 256, 128, __ , __ , 16 h 300 000, 30 000, 3000, __ , __ , 3 i 6250, 1250, 250, __ , __ , 2 j 2000, 200, 20, __ , __ , 0.02 2 a Write down the next two numbers in these sequences. i 64, 32, 16, 8, … ii 243, 81, 27, 9, … iii 128, 64, 32, 16, … iv 625, 125, 25, 5, … v 80, 40, 20, 10, … vi 972, 324, 108, 36, … vii 2430, 810, 270, 90, … viii 50 000, 5000, 500, 50, … ix 160, 80, 40, 20, … x 1296, 216, 36, 6, … b Write down the rule you used to find the missing number in each sequence. 3 Find the 8th number of each of the number patterns in questions 1 and 2. 4 The number of radioactive atoms in a radioactive isotope halves every 10 years. The table shows the beginning of the pattern. Years Number of atoms AO3 246 0 10 20 2560 1280 640 30 40 a Copy and complete the table. b How many radioactive atoms were in the isotope in year 100? 13.1 Sequences Continuing patterns in pictures Example 5 a Copy and complete the table for the number of matches used to make each member of the pattern. Pattern number 1 2 3 Number of matches used 4 7 10 4 5 6 7 b Write down the rule to get the next number in the pattern. c How many matches are there in pattern number 10? a Pattern number Number of matches used 1 2 4 7 3 4 5 6 7 10 13 16 19 22 Count the number of matches in each pattern and write down the number of matches used. 4 7 �3 10 �3 10  3  13 16  3  19 �3 �3 13  3  16 19  3  22 b Add 3 to the previous number. c Pattern number 10 has 31 matches. Continue the patterns. 22 25 28 31 Exercise 13E 1 For these patterns: i draw the next two patterns ii write down the rule in words to find the next pattern iii use your rule to find the 10th term. E a b c d e 247 Chapter 13 Sequences E 2 a Write down the number of matches in each of these patterns. Pattern 1 Pattern 2 Pattern 3 b Draw the next two patterns. c Write down the rule in words to continue the pattern. d Use your rule to find the number of matches needed for pattern number 10. 3 Repeat question 2 with the hexagon shape shown below. 13.2 Using input and output machines to investigate number patterns Objectives Why do this? You can use number machines to produce a number pattern and write down the rule (term number to term). You can complete a table of values using a number machine and write down the rule (term number to term). You can find missing values in a table of values and use the term number to term rule. When baking, you take your ingredients, mix them together and bake them in the oven, and you end up with a cake. Input Action This process applies to anything from baking a cake to making a motor car. 1. Put the following numbers into this number machine and write down the answers. a 5 b 3 c 11 d 17 2. Draw a number machine for the process 6 and use it to find the answer when the following numbers are put into it. a 10 b 6 c 2 248 Output �3 13.2 Using input and output machines to investigate number patterns Key Points In this number pattern 3 Term 1 7 Term 2 11 Term 3 15 Term 4 19 23 … 3, 7, 11, 15, 19, 23, … are the terms. The term number tells you the position of each term in the pattern. In the sequence 3, 7, 11, 15, 19, … term 1 is 3, term 2 is 7, etc. You can use number machines to produce a number pattern and write down the rule (term number to term rule). You can complete a table of values using a number machine and write down the rule (term number to term). You can find missing values in a table of values and use the term number to term rule. Sometimes you can put two number machines together to make a sequence. One-stage input and output machines Example 6 This number machine has been used to produce the terms of a pattern. �5 a Complete the term numbers and terms in this table of values for the number machine. Term number Term 1 5 2 3 4 b What is the rule for working out the term from the term number? c Write down the rule for finding the next term from the term before it. a Term number 1 Term 5 Term number 1 Term 5 2 10 2 10 3 15 3 15 4 20 4 20 5 5 5 The rule for the number machine is multiply the term number by 5 so the terms will be 5, 10, 15, 20. b Multiply the term number by 5. To get to the term from the term number you multiply by 5. To get to the next term from the term before it you have to add 5 since the pattern is 5, 10, 15, 20, … table of values one-stage input and output machines 249 Chapter 13 Sequences Exercise 13F For each of these questions: a copy and complete the table of values for the number machine b write down the rule for finding the term from the term number c write down the rule for finding the next term from the term before it. F �3 Term number 1 2 3 4 Term 3 6 �7 Term number 1 2 3 4 Term 7 14 �4 Term number 1 2 3 4 Term 4 8 �2 Term number 1 2 3 4 Term 2 4 �8 Term number 1 2 3 4 Term 8 16 �10 Term number 1 2 3 4 Term 10 20 �12 Term number 1 2 3 4 Term 12 24 �50 Term number 1 2 3 4 Term 50 100 1 2 3 4 5 6 7 8 250 13.2 Using input and output machines to investigate number patterns Two-stage input and output machines Example 7 Term number �3 Term �1 This number machine has been used to produce the terms of a pattern. Examiner’s Tip Use the rule from the number machine on the term number to get to the term. You feed the result of the first machine into the second machine. a Complete the terms in a table of values for the number machine. Term number Term 1 4 2 3 4 b What is the rule for working out the term from the term number? c Write down the rule for finding the next term from the term before it. a Term number Term Term number Term Working 1 4 1 4 1314 2 7 2 7 2317 3 10 3 10 3  3  1  10 4 13 4 13 4  3  1  13 b Multiply by 3 and add 1. 3 3 3 To get to the term from the term number you 3 and 1. To get to the next term from the term before it you have to add 3 since the pattern is 4, 7, 10, 13, … 251 Chapter 13 Sequences Exercise 13G For each of these questions: a copy and complete the table of values for the number machine b write down the rule for finding the term from the term number c write down the rule for finding the next term from the term before it. E 1 �3 �2 Term number 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 Term 2 �2 �1 Term number Term 3 �3 �1 Term number Term 4 �4 �3 Term number Term 5 �5 �2 Term number Term 6 �3 �4 Term number Term 7 �3 �5 Term number Term 8 �5 �1 Term number Term 9 �4 �3 Term number Term 10 �5 �4 Term number Term 252 13.2 Using input and output machines to investigate number patterns Example 8 4 Complete this table of values for the number pattern with term number to term rule, ‘Multiply by 4 and subtract 2’. 2 2 Term number 1 6 2 6 2426 3 3 10 3  4  2  10 4 4 14 4  4  2  14 5 5 18 5  4  2  18 ↓ 30 8  4  2  30 ↓ 38 10  4  2  38 Term number 1 Term 2 ↓ Term Working 2 1422 ↓ ↓ 8 8 ↓ ↓ 38 10 ↓ You can find these terms by using the rule 4 then 2. Examiner’s Tip Don’t forget Bidmas: you do the  before the . You met Bidmas in Chapter 8. Exercise 13H Copy and complete these tables of values. 1 3 Term number 1 2 3 4 5 ↓ 10 ↓ 1 Term 4 ↓ ↓ 34 2 2 Term number 1 2 3 4 5 ↓ 10 ↓ 1 Term 1 ↓ ↓ 25 3 5 Term number 1 2 3 4 5 ↓ 10 ↓ 3 E Term 8 ↓ ↓ 78 253 Chapter 13 Sequences E 4 4 Term number 1 2 3 4 5 ↓ 10 ↓ 3 Term 1 ↓ ↓ 45 5 10 Term number 1 2 3 4 5 ↓ 10 ↓ 1 Term 11 ↓ ↓ 151 7 a Find the 10th number in this number pattern. 3, 7, 11, 15, … b What is the term number for the term that is 47? 8 a Find the 10th number in this number pattern. 4, 9, 14, 19, … b What is the term number for the term that is 69? 9 a Find the 10th number in this number pattern. 8, 11, 14, 17, … b What is the term number for the term that is 50? 6 5 Term number 1 2 3 4 5 ↓ 10 ↓ 3 Term 2 ↓ ↓ 67 13.3 Finding the nth term of a number pattern Objective Why do this? You can use the first difference to find the nth term of a number pattern and use the nth term to find any number in a number pattern or sequence. This may be useful when your teacher is dividing the class into groups, so that you can work out which group you are going to be in, or make sure you will be in a group with your friends. 1. Write down the difference between each term in these number patterns. a 5, 10, 15, 20, 25, 30, … b 40, 35, 30, 25, 20, … c 4, 7, 10, 13, 16, … d 7, 11, 15, 19, 21, … e 50, 47, 44, 41, 37, … 2. Find the 10th term in each of the number patterns in question 1. Key Point The first difference can be used to find the nth term of a number pattern and then the nth term can be used to find any number in a sequence. 254 first difference 13.3 Finding the nth term of a number pattern Example 9 a Term number 1 Here is a number pattern 4, 7, 10, 13, 16, … a Find the nth term in this pattern. b Find the 20th term in this number pattern. Term Difference 4 2 3 4 5 7 10 13 16 n 3n  1 b The 20th term is 61. 3 3 3 3 Step 1 Put the number pattern into a table of values. Step 2 Find the difference between the terms in the number pattern. In this case it is 3. Step 3 Multiply each term number by the difference to get a new pattern. 3, 6, 9, 12, 15 … Step 4 Compare your new pattern with the original one and see what number you need to add or subtract to/from each term to get the original number pattern. In this case it is 1. The nth term is 3n  1. You replace the n by 20 in the nth term to find the 20th term. It is 3  20  1  61 Exercise 13I 1 For questions 1, 2 and 3 in Exercise 10H, find the nth term of each of the number patterns. 2 Write each pattern in a table and use the table to find the nth term of these number patterns. Use your nth term to find the 20th term in each of these number patterns. a 1, 3, 5, 7, 9, 11, … b 3, 5, 7, 9, 11, 13, … c 2, 5, 8, 11, 14, 17, … d 5, 8, 11, 14, 17, 20, … e 1, 5, 9, 13, 17, 21, … f 2, 6, 10, 14, 18, 22, … g 2, 7, 12, 17, 22, 27, … h 4, 9, 14, 19, 24, 29, … i 8, 13, 18, 23, 28, … j 5, 7, 9, 11, 13, … k 40, 35, 30, 25, 20, … l 38, 36, 34, 32, 30, … m 35, 32, 29, 26, 23, … n 20, 18, 16, 14, 12, … o 19, 17, 15, 13, 11, … p 190, 180, 160, 150, … C Examiner’s Tip To find the nth term of a sequence that gets smaller you subtract a multiple of n from a fixed number. e.g. 15  2n is the nth term of 13, 11, 9, 7, … 255 Chapter 13 Sequences C 3 Here is a pattern made from sticks. Pattern number 1 Pattern number 2 Pattern number 3 a Draw pattern number 4. b Copy and complete this table of values for the number of sticks used to make the patterns. Pattern number 1 2 Number of sticks 6 10 3 4 5 6 c Write, in terms of n, the number of sticks needed for pattern number n. d How many sticks would be needed for pattern number 20? 13.4 Deciding whether or not a number is in a number pattern Objective Why do this? You can use number patterns or use the nth term to identify whether a number is in the pattern. This is useful when you want to work out what will happen in the future, for example, you could work out whether next year will be a leap year as this happens every four years. 1. Write each of these patterns in a table and use the table to find the nth term. Use your nth term to find the 20th term in each pattern. a 4, 7, 10, 13, … b 3, 8, 13, 18, … c 13, 15, 17, 19, … Key Points Number patterns or the nth term can be used to identify whether a number is in the pattern. Sometimes you will be asked how you know if a number is part of a sequence. You would then have to explain why the number is in the sequence or, even, why it is not in the sequence. 256 Chapter review Example 10 Here is a number pattern. 3 8 3 18 23 a Explain why 423 is in the pattern. b Explain why 325 is not in the pattern. a 423 is in the number pattern. Every odd term ends in 3 and goes up 3, 13, 23, etc, so 423 will be a member as it ends in a 3. b 325 is not in the number pattern. 325 ends in a 5 and every member of the pattern ends in either a 3 or an 8 so 325 cannot be in the pattern. There are other ways of answering questions like these. For example, you could identify the nth term The nth term is 5n  2 if 5n  2  423 5n  425 so n  85 so 423 is the 85th term. The nth term is 5n  2 so if 325 is in the pattern 5n  2  325 5n  327 so n  65.4 If 325 is in the pattern n must be a whole number. 65.4 is not a whole number so 325 is not in the pattern. Exercise 13J For each of these number patterns, explain whether each of the numbers in brackets are members of the number pattern or not. 1 1, 3, 5, 7, 9, 11, … (21, 34) 2 3, 5, 7, 9, 11, 13, … (63, 86) 3 2, 5, 8, 11, 14, 17, … (50, 66) 4 5, 8, 11, 14, 17, 20, … (50, 62) 5 1, 5, 9, 13, 17, 21, … (101, 150) 6 2, 6, 10, 14, 18, 22, … (101, 98) 7 2, 7, 12, 17, 22, 27, … (97, 120) 8 4, 9, 14, 19, 24, 29, … (168, 169) 9 40, 35, 30, 25, 20, … (85, 4) 10 38, 36, 34, 32, 30, … (71, 82) 11 3, 7, 11, 15, 19, 21, … (46, 79) 12 5, 11, 17, 23, 29, … (119, 72) AO3 C Chapter review A sequence is a number or shape pattern which follows a rule. Number patterns can be continued by adding, subtracting, multiplying and dividing. The term to term rules for continuing number patterns can be given. Patterns using pictures can be continued by finding the rule for continuing the pattern. You can use number machines to produce a number pattern and write down the rule (term number to term rule) You can complete a table of values using a number machine and write down the rule (term number to term). You can find missing values in a table of values and use the term number to term rule. The first difference can be used to find the nth term of a number pattern and then the nth term can be used to find any number in a sequence. Number patterns or the nth term can be used to identify whether a number is in the pattern. 257 Chapter 13 Sequences Review exercise F 1 Here are some patterns made of squares. Pattern number 1 Pattern number 2 Pattern number 3 The diagram below shows part of Pattern number 4. a Copy and complete Pattern number 4 Pattern number 4 AO2 D AO3 b Find the number of squares used for Pattern number 10 2 Here are the first 4 terms in a number sequence. 124 122 120 118 a Write down the next term in this number sequence. b Write down the 7th term in this number sequence. c Can 9 be a term in this number sequence? You must give a reason for your answer. AO3 AO3 May 2009 3 The nth term of a sequence is n2  4. Alex says ‘The nth term of the sequence is always a prime number when n is an odd number’. 4 Here are the first 5 terms of a sequence. 1 1 2 3 5 The rule for the sequence is ‘The first two terms are 1 and 1. To get the next term add the two previous terms’. a Find the 6th term and the 7th term. b Find the 10th term. c Explain why after the first two terms the other terms of the sequence are alternately even and odd. The rule for another sequence is ‘The first two terms are 2 and 2. To get the next term multiply the two previous terms’. d Find an expression for the 10th term of this sequence. You do not have to wok out the expression. C AO3 5 Here are the first four terms of an arithmetic sequence. 5 8 11 14 Is 140 a term in the sequence? You must five a reason for your answer AO3 258 6 The first term of a sequence is x. To get the next term, multiply the previous term by 2 and add 1. The third term of the sequence is 21. Find the value of x. ``` Related documents
# How do you simplify 15\div 5( 8- 6+ 3) \times 5 ? Dec 9, 2016 3 #### Explanation: $15 \div 5 \left(8 - 6 + 3\right) \times 5$ $15 \div 5 \left(5\right) \times 5$ $15 \div 5 \left(5\right) \times 5$ $15 \div 25 \times 5$ $\frac{x}{a} = x \cdot \left(\frac{1}{a}\right)$ $15 \cdot \frac{1}{25} \cdot 5$ $\frac{15}{25} \cdot 5$ $\frac{75}{25}$ = 3 May 30, 2017 It depends... #### Explanation: Given: $15 \div 5 \left(8 - 6 + 3\right) \times 5$ I think we are all agreed that the content of the parentheses should be evaluated first. Subtraction and addition are given the same priority, so the expression in parentheses is to be evaluated from left to right: $8 - 6 + 3 = 2 + 3 = 5$ Now we have: $15 \div 5 \left(5\right) \times 5$ This is where it gets interesting. Here are three possibilities, in no particular order: $\textcolor{w h i t e}{}$ Possible interpretation 1 - "Historical" Historically the obelus $\div$ was used to express a division of everything on the left by everything on the right. In our example, that means that we have: $15 \div 5 \left(5\right) \times 5 = \frac{15}{5 \left(5\right) \times 5} = \frac{15}{5 \times 5 \times 5} = \frac{15}{125} = \frac{3}{25}$ $\textcolor{w h i t e}{}$ Possible interpretation 2 - "Pure PEMDAS" PEMDAS does not distinguish multiplication by juxtaposition from any other kind of multiplication. So in full we can write: $15 \div 5 \left(5\right) \times 5 = 15 \div 5 \times 5 \times 5$ This is then evaluated from left to right (multiplication and division having the same priority). So we get: $15 \div 5 \times 5 \times 5 = 3 \times 5 \times 5 = 15 \times 5 = 75$ $\textcolor{w h i t e}{}$ Possible interpretation 3 - "Skewed PEMDAS" This common practice gives higher priority to multiplication by juxtaposition. This is sometimes "justified" by people who claim that the "Parentheses first" includes any multiplier outside the parentheses. Such a justification seems spurious to me, but the visual proximity does suggest a higher priority. Following this interpretation, we get: $15 \div 5 \left(5\right) \times 5 = 15 \div 25 \times 5 = \frac{3}{5} \times 5 = 3$ $\textcolor{w h i t e}{}$ Which is right? They are all "right" or "wrong" or neither. The fact is that the given expression is ambiguous. Operator precedence rules are intended to clarify communication by providing agreed rules of interpretation. They do not work well all of the time - especially if the particular rule set is not shared between the writer and reader. It is best to use extra parentheses to make the meaning clear. Sep 3, 2017 In addition to the other answers, here is an additional comment and warning about just applying $P E D M A S \mathmr{and} B O D M A S$ without considering the meaning of an expression. Consider the terms in the parentheses: $\left(8 - 6 + 3\right)$ Following the [order of operations] (https://socratic.org/prealgebra/arithmetic-and-completing-problems/order-of-operations)at this point states that ADDITION is to be done first. However, many students will apply this incorrectly to do: (8" "-color(blue)(6+3)) = 8-color(blue)(9)" " and then subtract to get $\text{ } - 1$ The negative sign applies ONLY to the $6$ and not the sum of $6 + 3$ If this was intended it would be written as $\text{ } \left(8 - \left(6 + 3\right)\right)$ A safer approach is to place the additions at the beginning and the subtracts at the end. $\left(8 - 6 + 3\right)$ $= \left(8 + 3 - 6\right) = \left(11 - 6\right) = 5$ My approach for the calculation would now be $15 \div \textcolor{red}{5 \left(5\right)} \times 5$ $= 15 \div \textcolor{red}{25} \times 5$ $= \frac{15}{25} \times \frac{5}{1}$ $= 3$ However, this is one interpretation, as discussed by George C. Sep 8, 2017 15-:5(8-6+3)xx5=color(blue)(75 #### Explanation: Simplify: $15 \div 5 \left(8 - 6 + 3\right) \times 5$ It is important to follow the order of operations: Parentheses/Brackets, Exponents, Multiplication and Division from left to right, Addition and Subtraction from left to right. An acronym like PEMDAS or BODMAS can be helpful in learning to memorize the order of operations. Simplify the parentheses. $15 \div 5 \left(\textcolor{red}{8} - \textcolor{red}{6} + 3\right) \times 5$ Simplify. $15 \div 5 \left(\textcolor{red}{2} + 3\right) \times 5$ Simplify the parentheses. $15 \div 5 \left(5\right) \times 5$ Multiplication and division are equal, so they are performed from left to right. First do the division, and next multiply by $5 \times 5$. $\frac{15}{5} \times 5 \times 5$ Simplify. $3 \times 5 \times 5$ Simplify. $75$
## If $2 {p}^{2} - 3 p - 4 = 0$ and $2 {q}^{2} - 3 q - 4 = 0$ (p does not equal to 0), find the value of $2 p + 2 q$ and ${p}^{2} {q}^{2}$ May 18, 2018 I tried this...the procedure should be ok...BUT check my maths anyway. #### Explanation: Have a look: May 18, 2018 $\left(\frac{3}{2}\right) \cdot 2 = 3$ and ${\left(- \frac{4}{2}\right)}^{2} = 4$ thus, $2 p + 2 q = 3$ and ${p}^{2} {q}^{2} = 4$ #### Explanation: Quick way: You may use Vieta's Formulas First notice that p and q have the exact same equation and thus will have the same solution, $p + q = - \frac{b}{a}$, $p q = \frac{c}{a}$ proof: $a \left(x - {r}_{1} \setminus\right) \left(x - {r}_{2}\right) = a {x}^{2} + b x + c$ $a {x}^{2} - a \left({r}_{1} + {r}_{2}\right) x + a \left({r}_{1}\right) \left({r}_{2}\right) = a {x}^{2} + b x + c$ Thus ${r}_{1} + {r}_{2} = - \frac{b}{a} \mathmr{and} \left({r}_{1}\right) \left({r}_{2}\right) = \frac{c}{a}$ $p + q = - \frac{3}{2} , p q = \frac{4}{2} = 2$ Long way: solve for $2 {p}^{2} - 3 p - 4 = 0$ $p = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$ Sub in a =2, b = -3 and c = -4 p = \frac{3 \pm \sqrt{9 - 4(2)(-4}}{2(2)} $p = \setminus \frac{3 \setminus \pm \setminus \sqrt{9 + 32}}{4}$ $p = \setminus \frac{3 \setminus \pm \setminus \sqrt{41}}{4}$ $p = \setminus \frac{3 + \setminus \sqrt{41}}{4}$, $p = \setminus \frac{3 - \setminus \sqrt{41}}{4}$ q has the exact same equation and is thus have the same solution: $q = \setminus \frac{3 + \setminus \sqrt{41}}{4}$, $q = \setminus \frac{3 - \setminus \sqrt{41}}{4}$ $p + q = \setminus \frac{3 + \setminus \sqrt{41} + 3 - \setminus \sqrt{41}}{4} = \setminus \frac{6}{4} = \frac{3}{2}$ $p q = \setminus \frac{- 32}{16} = - 2$ $2 \left(p + q\right) = 3 \mathmr{and} {p}^{2} {q}^{2} = 4$
#### Need Help? Get in touch with us # Add Fractions With Like Denominators Sep 19, 2022 ## Key Concepts • Common Denominator • Numerator • Standard form or simplest form • Mixed fractions ## Introduction: ### What is a common denominator? The denominator of a fraction is the number on the bottom. When we say that fractions have a common denominator, it means they have the same number on the bottom. ### What is a denominator? The denominator of a fraction is the number of equal parts that the whole has been split into. For example, When two or more fractions have a common denominator, that means the wholes have been divided into the same number of equal pieces, and each piece is the same size. For example, Here, the whole has been divided into 8 equal parts, and each part is the same size. Fractions with the same denominator are called like fractions. For example, 1/8, 2/8 , 3/8 , 4/8……… are like fractions. Addition of fractions with Like denominators: In order to add fractions, the fractions must have a common denominator. That means, we need the pieces of each fraction to be the same size to combine them together. For example, if we need to add 3/5 and 1/5 together we need to check the denominator. These two fractions have the same denominator. That means the equal parts that the whole has been split into are the same size. Since the pieces are all the same size, we can add these two fractions together. Let us check another example, If we have to add 1/9, 7/9 and 4/9 together, we can see that the denominators are the same. So, we can add these fractions together. 1/9+7/9+4/9= 1+7+4/9= 12/9 If you check the answer, the numerator is greater than the denominator. Hence, 12/9 à Improper fractions Which can be converted to a mixed fraction. We can show this calculation visually as below An easy way to add fractions with like denominators is that you can combine them together by adding the numerators together (the top numbers). The denominator will always stay the same because the size of the equal pieces does not change when you combine the two fractions together • Keep the denominator the same Example: Solution: They have the same denominator, so they can be combined together. Add the numerators (1 + 6 = 7). Keep the denominator the same. Remember: The denominator does not change because the sizes of the pieces stay the same. You’re just counting the total number of pieces between the two fractions. 6/10+ 1/10 = 6+1/10= 7/10 Visually Example: The table shows the distance Shannon ran over a week. How far did Shannon run on Wednesday and Friday together? Solution: Distance ran on Wednesday = 62 miles Distance ran on Friday = 42 miles Total distance = 6/2+4/2= 6+4/2= 102 = 5 miles ## Exercise: 1. Find 3/12+4/12 using a model. 2. Robin bought a bamboo plant that was 4/5 feet high. After a month, it has grown another 7/5 feet. What was the total height of the plant after a month? 3. On Monday, Tiffany spent 2 ⁄8 hours studying. On Tuesday, she spent another 1 ⁄8 hours studying. What is the combined length of time she spent studying? 4. Emma likes chocolate. One day she bought a chocolate and ate 5/8 of it in the morning and 2/8 in the evening. What part of the chocolate did she eat? 5. Sophia completed 2/5 of her homework before going out for play. She did 1/5 of her homework after the play. How much homework did she complete altogether? 6. Mary read 2/9 of her book in the morning and 5/9 in the evening. What fraction of the book did she read? 7. If Sarah’s friend Abby took 2/10 of Sarah’s cupcakes, and her friend Allison took 1/10 of the cupcakes, how much of the cupcakes did Abby and Allison take altogether? 8. You go out for a long walk. You walk 3/7 mile and then sit down to take a rest. Then you walk another 3/7 of a mile. How far did you walk altogether? 9. Use the bar models to add 1. Dora has 3/8 liters of juice. She gave 9/8 liters of juice to Doreen. How many liters of juice does both of them have together? ### What have we learned: • Fractions with same denominator • Like fractions • Addition of fractions with like denominators #### Addition and Multiplication Using Counters & Bar-Diagrams Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […] #### Dilation: Definitions, Characteristics, and Similarities Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […] #### How to Write and Interpret Numerical Expressions? Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
Great news! We will be upgrading our calculator and lesson pages over the next few months. If you notice any issues, you can submit a contact form by clicking here. ##### Related Content Thank you! On behalf of our dedicated team, we thank you for your continued support. It's fulfilling to see so many people using Voovers to find solutions to their problems. Thanks again and we look forward to continue helping you along your journey! Nikkolas and Alex Founders and Owners of Voovers ax2 + bx + c = 0 a = b = c = Help ? ## Lesson on the Quadratic Formula ### What is the Quadratic Formula? The quadratic formula is a formula that is used to solve a quadratic equation of the standard form ax2 + bx + c = 0. The quadratic formula is given as: $$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$$ Where x denotes the solution(s) to the quadratic equation, and a, b, and c are the coefficients of the quadratic equation. The part inside the square root (b2 – 4ac) is called the discriminant and it tells us how many roots the quadratic equation has. ### How to Solve a Quadratic To solve a quadratic equation, we must make sure it is in standard form. As shown earlier, standard form for a quadratic is ax2 + bx + c = 0. Once in standard form, we plug the equation’s coefficients into the quadratic formula. If the discriminant is a positive number, continue solving until the two real solutions are obtained. Sometimes the discriminant will not be positive which results in us either taking the square root of zero or a negative number. If the discriminant is zero, continue solving the quadratic formula as usual until the final, single solution is obtained. That final number is the solution, but we call it a repeated real solution. This means the graph of the quadratic touches the x-axis twice in the same spot. If the discriminant is a negative number, taking the square root of it results in an imaginary number. Therefore, the quadratic equation has two imaginary solutions and no real solutions. ## How the Calculator Works The quadratic formula solver on this page is written in JavaScript (JS) and is powered by a JS native computer algebra system (CAS). The calculations all occur within your device’s internet browser JS engine, which allows for near-instantaneous answers and no waiting on the page to communicate with a server or refresh. When the calculate button is pressed, your coefficients are fed to the CAS which then uses quadratic solving steps to symbolically calculate the solution(s). Being symbolic in nature, the CAS treats the quadratic like a human would with paper and pencil and maintains near-perfect accuracy. When the CAS is done solving, the answer is converted to LaTeX (a math rendering language) and displayed in the answer area. If an error occurs or the inputs do not allow for a solution to be calculated, an error notice will instead be displayed. Scroll to Top
# Subtraction in Columns Practice for Grade 3 Students Teaching subtraction in columns to third-grade students can be made simple and easy to understand. Here's a step-by-step guide on how to teach column subtraction. Let's subtract 416 from 559. ## Step 2: start from the right column (ones place) Subtract the digit in the bottom number (subtrahend) from the digit in the top number (minuend). ## Step 3: move to the tens place Now, subtract the digit in the tens place of the subtrahend from the digit in the tens place of the minuend. ## Step 4: finally, move to the hundreds place Again, subtract the digit in the hundreds place of the subtrahend from the digit in the hundreds place of the minuend. Add the difference (143) to the subtrahend (416) to verify that the sum is equal to the minuend (559). ## Important points to remember Borrowing/Regrouping: Sometimes, you'll need to borrow from the next higher place value if the digit on top is smaller than the digit being subtracted from it. Ensure the kids understand this concept. Place Value: Emphasize the significance of the place value in determining the value to be subtracted. Practice: Provide plenty of practice problems to reinforce the concept. Make sure to use various numbers and conduct exercises to solidify their understanding. Visualization aids like base-10 blocks, drawings, or even online interactive tools can be used to make the concept more tangible and engaging for the students. We have a number of different worksheets for students to practice subtraction in column at the grade 3 level. ## Subtracting 3-digit numbers in columns In these worksheets students subtract up to 3-digit numbers in columns. ## Subtracting 4-digit numbers in columns In these worksheets students practice subtracting numbers up to 9,999. ## Regrouping across two zeros These more difficult worksheets involve borrowing across zeros to solve the equations. ## Regrouping across three zeros These are the most difficult of our subtraction in columns worksheets where students subtract across three zeros. Become a Member This content is available to members only.
Paul's Online Notes Home / Calculus II / Parametric Equations and Polar Coordinates / Polar Coordinates 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 3-6 : Polar Coordinates 1. For the point with polar coordinates $$\displaystyle \left( {2,\frac{\pi }{7}} \right)$$ determine three different sets of coordinates for the same point all of which have angles different from $$\displaystyle \frac{\pi }{7}$$ and are in the range $$- 2\pi \le \theta \le 2\pi$$. Solution 2. The polar coordinates of a point are $$\left( { - 5,0.23} \right)$$. Determine the Cartesian coordinates for the point. Solution 3. The Cartesian coordinate of a point are $$\left( {2, - 6} \right)$$. Determine a set of polar coordinates for the point. Solution 4. The Cartesian coordinate of a point are $$\left( { - 8,1} \right)$$. Determine a set of polar coordinates for the point. Solution For problems 5 and 6 convert the given equation into an equation in terms of polar coordinates. 1. $$\displaystyle \frac{{4x}}{{3{x^2} + 3{y^2}}} = 6 - xy$$ Solution 2. $$\displaystyle {x^2} = \frac{{4x}}{y} - 3{y^2} + 2$$ Solution For problems 7 and 8 convert the given equation into an equation in terms of Cartesian coordinates. 1. $$6{r^3}\sin \theta = 4 - cos\theta$$ Solution 2. $$\displaystyle \frac{2}{r} = \sin \theta - \sec \theta$$ Solution For problems 9 – 16 sketch the graph of the given polar equation. 1. $$\displaystyle \cos \theta = \frac{6}{r}$$ Solution 2. $$\displaystyle \theta = - \frac{\pi }{3}$$ Solution 3. $$r = - 14\cos \theta$$ Solution 4. $$r = 7$$ Solution 5. $$r = 9\sin \theta$$ Solution 6. $$r = 8 + 8\cos \theta$$ Solution 7. $$r = 5 - 2\sin \theta$$ Solution 8. $$r = 4 - 9\sin \theta$$ Solution
St Nicholas Church of EnglandPrimary Academy Search Translate # Maths ## Number - addition and subtraction By the end of year 2 the children should be able to: • solve problems with addition and subtraction: • using concrete objects and pictorial representations, including those involving numbers, quantities and measures • applying their increasing knowledge of mental and written methods • recall and use addition and subtraction facts to 20 fluently, and derive and use related facts up to 100 • add and subtract numbers using concrete objects, pictorial representations, and mentally, including: • a two-digit number and 1s • a two-digit number and 10s • 2 two-digit numbers • show that addition of 2 numbers can be done in any order (commutative) and subtraction of 1 number from another cannot • recognise and use the inverse relationship between addition and subtraction and use this to check calculations and solve missing number problems This year the children will extend their understanding of the language of addition and subtraction to include sum and difference. Children practise addition and subtraction to 20 to become increasingly fluent in deriving facts such as using 3 + 7 = 10; 10 − 7 = 3 and 7 = 10 − 3 to calculate 30 + 70 = 100; 100 − 70 = 30 and 70 = 100 − 30. They will check their calculations, including by adding to check subtraction and adding numbers in a different order to check addition (for example, 5 + 2 + 1 = 1 + 5 + 2 = 1 + 2 + 5). This establishes commutativity and associativity of addition. When ready the children may start to record addition and subtraction in columns which supports place value and prepares them for formal written methods with larger numbers in Year 3. ## Place Value A good understanding of place value (the value of each digit in a number) is vital in primary-school maths. Here is how your child will be taught about units, tens, hundreds and thousands with number lines, arrow cards and more, as well as outlining how place value is used to help children visualise calculations. ## What is place value? Place value is the value of each digit in a number. It means understanding that 582 is made up of 500, 80 and 2, rather than 5, 8 and 2. ## How are children taught to understand place value in KS1? In school two maths aids are used to help make place value clear to children. Deines blocks are blocks in which cubes represent units / ones, rods of ten cubes represent tens, flats of 100 cubes represent hundreds and blocks of 1000 cubes represent thousands: In Key Stage 1, a child might be given some ten and units (ones) Deines blocks and asked to make a number such as 43. They would need to select 4 tens rods and 3 ones blocks. This makes it very clear to them that a two-digit number it made up of tens and ones. It also helps them to practise counting in tens. Arrow cards look like this: A child might be asked to make the number 34 using arrow cards. They would need to take the 30 and the 4 and put them together so that the arrows were lined up. This again helps to make clear that a two-digit number is made up of tens and ones. It is absolutely vital that children understand place value before they can go onto adding and subtracting two-digit numbers. ### Counting In Year 2 we continue on from the learning from Year 1 and count forward and backwards in ones from any number within the range of 100. Please practise daily counting with your children! Sometimes children have difficulties counting backwards and bridging the tens numbers (eg. 32,31,30,29 or 58 59 60 61 ) We then develop more fluency with our counting in different amounts such as in 2's,3's,5's and 10's which lead on to learning about the times tables and division facts of this numbers. We learn how to read and write our numbers in numerals and in the written form. We develop our reasoning skills and our problem solving skills throughout each lesson in order for us to help us understand more about the mathematical concepts we are learning about. We will learn about odd and even numbers and investigate the difference between them and the patterns that occur. Top
Wednesday, April 8, 2015 Section 15-3: The Inscribed Angle Theorem (Day 142) Today we finally reach the last chapter of the U of Chicago text. Chapter 15 of the U of Chicago text is on Further Work With Circles, and this chapter belongs here at the end because it's the only chapter that appears on the PARCC End-of-Year (EOY) but not the Performance-Based Assessment (PBA). Yet of the three circle lessons that are tested on the EOY, Section 15-3 is the only one that actually appears in Chapter 15. The other two lessons are Section 13-5 on tangents to circles and yesterday's Section 11-3 on equations of circles. Therefore today's worksheet contains the only lesson that I'll include from Chapter 15. Section 15-3 of the U of Chicago text is on the Inscribed Angle Theorem. I admit that I often have trouble remembering all of the circle theorems myself, but this one is the most important: Inscribed Angle Theorem: In a circle, the measure of an inscribed angle is one-half the measure of its intercepted arc. The text divides the proof into three cases -- depending on whether the center of the circle is inside, outside, or on the inscribed angle. The easiest case occurs when the center is on the angle, and this case is used to prove the other two cases. The exact same thing occurred when we proved the Triangle Area Formula back in Chapter 8. Then, the three cases were whether the altitude was inside, outside, or aside of the triangle -- and the case when the altitude was aside of the triangle (i.e., when the triangle was a right triangle) was used to prove the other two. I will reproduce the paragraph proof of the Inscribed Angle Theorem from the U of Chicago: Given: Angle ABC inscribed in Circle O Prove: Angle ABC = 1/2 * Arc AC Proof: Case I: The auxiliary segment OA is required. Since Triangle AOB is isosceles [both OA and OB are radii of the circle -- dw], Angle B = Angle A. Call this measure x. By the Exterior Angle Theorem, Angle AOC = 2x. Because the measure of an arc equals the measure of its central angle, Arc AC = 2x = 2 * Angle B. Solving for Angle B, Angle B = 1/2 * Arc AC. QED Case I. Notice that the trick here was that between the central angle (whose measure equals that of the arc) and the inscribed angle is an isosceles triangle. We saw the same thing happen in yesterday's proof of the Angle Bisector Theorem -- the angle bisector of a triangle is a side-splitter of a larger triangle, and cutting out the smaller triangle from the larger leaves an isosceles triangle behind. Let's move onto Case II. Well, the U of Chicago almost gives us a two-column proof here, so why don't we complete it into a full two-column proof. For Case II, O is in the interior of Angle ABC. Statements Reasons 1. O interior ABC 1. Given 2. Draw ray BO inside ABC 2. Definition of interior of angle 3. Angle ABC = Angle ABD + Angle DBC 3. Angle Addition Postulate 4. Angle ABC = 1/2 * Arc AD + 1/2 * Arc DC 4. Case I and Substitution 5. Angle ABC = 1/2(Arc AD + Arc DC) 5. Distributive Property 6. Angle ABC = 1/2 * Arc AC 6. Arc Addition Property and Substitution The proof of Case III isn't fully given, but it's hinted that we use subtraction rather than addition as we did in Case II. Once again, I bring up the Triangle Area Proof -- the case of the obtuse triangle involved subtracting the areas of two right triangles, whereas in the case where that same angle were acute, we'd be adding the areas of two right triangles. The text mentions a simple corollary of the Inscribed Angle Theorem: Theorem: An angle inscribed in a semicircle is a right angle. The text motivates the study of inscribed angles by considering camera angles and lenses. According to the text, a normal camera lens has a picture angle of 46 degrees, a wide-camera lens has an angle of 118 degrees, and a telephoto lens has an angle of 18. I briefly mention this on my worksheet. But a full consideration of camera angles doesn't occur until the next section of the text, Section 15-4 -- but we're only really doing Section 15-3.
Courses # RD Sharma Solutions (Part - 2) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes \| EduRev ## Class 7: RD Sharma Solutions (Part - 2) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes \| EduRev The document RD Sharma Solutions (Part - 2) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes \| EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics. All you need of Class 7 at this link: Class 7 #### QUESTION 10: A table top is 9 dm 5 cm long 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre? We have, Length of the table top = 9 dm 5 cm = (9 x 10 + 5) cm = 95 cm [ Since 1 dm = 10 cm] Breadth of the table top = 6 dm 5 cm = (6 x 10 + 5) cm = 65 cm ∴ Area of the table top = Length x Breadth = (95 cm x 65 cm) = 6175 cm2 Rate of polishing per square centimetre = 20 paise = Rs. 0.20 Total cost = Rs. (6175 x 0.20) = Rs. 1235 #### Question 11: A room is 9.68 m long and 6.2 m wide. Its floor is to be covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs 2.50 per tile. We have, Length of the floor of the room = 9.68 m Breadth of the floor of the room = 6.2 m Area of the floor = 9.68 m x 6.2 m = 60.016 m2 Length of the tile = 22 cm Breadth of the tile = 10 cm Area of one tile = 22 cm x 10 cm = 220 cm2 = 0.022 m2 [Since 1 m2 = 10000 cm2] Thus, Number of tiles = Cost of one tile = Rs. 2.50 Total cost = Number of tiles x Cost of one tile = Rs. (2728 x 2.50) = Rs. 6820 #### Question 12: One side of a square field is 179 m. Find the cost of raising a lown on the field at the rate of Rs 1.50 per square metre. We have, Side of the square field = 179 m Area of the field = (Side)2 = (179 m)2 = 32041 m2 Rate of raising a lawn on the field per square metre = Rs. 1.50 Thus, Total cost of raising a lawn on the field = Rs.(32041 x 1.50) = Rs. 48061.50 #### Question 13: A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec? We have, Length of the rectangular field = 290 m Breadth of the rectangular field = 210 m Perimeter of the rectangular field = 2(Length + Breadth) = 2(290 + 210) = 1000 m Distance covered by the girl = 2 x Perimeter of the rectangular field = 2 x 1000 = 2000 m The girl walks at the rate of 1.5 m/sec. or, Rate = 1.5 x 60 m/min = 90 m/min Thus, Required time to cover a distance of 2000 m = Hence, the girl will take min to go two times around the field. #### Question 14: A corridor of a school is 8 m long and 6 m wide. It is to be covered with convas sheets. If the available canvas sheets have the size 2 m × 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs 8 per sheet. We have, Length of the corridor = 8 m Breadth of the corridor = 6 m Area of the corridor of a school = Length x Breadth = (8 m x 6 m) = 48 m2 Length of the canvas sheet = 2 m Breadth of the canvas sheet = 1 m Area of one canvas sheet = Length x Breadth = (2 m x 1 m) = 2 m2 Thus, Number of canvas sheets = Cost of one canvas sheet = Rs. 8 ∴ Total cost of the canvas sheets = Rs. (24 x 8) = Rs. 192 #### Question 15: The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metres per second. We have, Length of a playground = 62 m 60 cm = 62.6 m [ Since 10 cm = 0.1 m] Breadth of a playground = 25 m 40 cm = 25.4 m Area of a playground = Length x Breadth= 62.6 m x 25.4 m = 1590.04 m2 Rate of turfing = Rs. 2.50/m2 ∴ Total cost of turfing = Rs. (1590.04 x 2.50) = Rs. 3975.10 Again, Perimeter of a rectangular field = 2(Length + Breadth) = 2(62.6 + 25.4) = 176 m Distance covered by the man in 3 rounds of a field = 3 x Perimeter of a rectangular field = 3 x 176 m = 528 m The man walks at the rate of 2 m/sec. or, Rate = 2 x 60 m/min = 120 m/min Thus, Required time to cover a distance of 528 m = = 4 minutes 24 seconds [ since 0.1 minutes = 6 seconds] The document RD Sharma Solutions (Part - 2) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes \| EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics. All you need of Class 7 at this link: Class 7 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code All Tests, Videos & Notes of Class 7: Class 7 ## RD Sharma Solutions for Class 7 Mathematics 97 docs ### Top Courses for Class 7 Track your progress, build streaks, highlight & save important lessons and more! , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ;
# What is the middle number? Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle. ## Which number is the middle number? The median of a set of numbers is the middle number in the set (after the numbers have been arranged from least to greatest) -- or, if there are an even number of data, the median is the average of the middle two numbers. ## What is the middle of 7? The middle of 7 is 4: How would you explain this to a child? : r/math. ## Is median The middle number? The median is the middle number in an ordered data set. The mean is the sum of all values divided by the total number of values. ## What's the middle of two numbers? The midpoint between two numbers is the number exactly in the middle of the two numbers. Calculating the midpoint is the same thing as calculating the average of two numbers. Therefore, you can calculate the midpoint between any two numbers by adding them together and dividing by two. ## What is the middle of 15? We see that the number in the middle is 7.5, which is exactly 7.5 units from 0 and exactly 7.5 units from 15. Therefore, half of 15 is 7.5. ## What is the middle of 1 and 9? “Ask adults from the industrialized world what number is halfway between 1 and 9, and most will say 5. But pose the same question to small children, or people living in some traditional societies, and they're likely to answer 3. ## What is the median of 4 and 7? The mean of these middle values is (4 + 7) / 2 = 5.5 , so the median is 5.5. ## How do I find the median? The median is calculated by arranging the scores in numerical order, dividing the total number of scores by two, then rounding that number up if using an odd number of scores to get the position of the median or, if using an even number of scores, by averaging the number in that position and the next position. ## What is the middle number of a data set? from least to greatest or greatest to least; the median is the data value in the middle; if there is an even number of data values in the set, the median is the mean of the two middle values. ## Why is 7 important in the Bible? Seven was symbolic in ancient near eastern and Israelite culture and literature. It communicated a sense of “fullness” or “completeness” (שבע “seven” is spelled with the same consonants as the word שבע “complete/full”). This makes sense of the pervasive appearance of “seven” patterns in the Bible. ## What's the middle of 7 and 3? Example: what is the central value for 3 and 7? Answer: Half-way between, which is 5. ## Does 7 have a line in the middle? Most people in Continental Europe, and some in Britain and Ireland as well as Latin America, write 7 with a line in the middle ("7"), sometimes with the top line crooked. ## Is 0 the middle of a number line? Zero is the middle point of a number line. All (natural numbers) positive numbers occupy the right side of the zero whereas negative numbers occupy the left side of zero on the number line. As we move on to the left side value of a number decreases. ## How do you find the middle of 4 numbers? If we have an odd number of terms then the middle term is the median. If we have an even number of terms then we have to add two middle terms and then divide the sum by 2. The mean we get is the median. ## Is the mode the highest number? Place all numbers in a given set in order; this can be from lowest to highest or highest to lowest, and then count how many times each number appears in the set. The one that appears the most is the mode. ## What is the median of 10? Hence, the required median of the first ten natural numbers = 5 . 5 . ## What is the median of 5 and 7? Hence, the median of 5 and 7 is 6 . ## What is the median of 1 to 50? Hence, the median of first 50 whole numbers is 24.5. ## What is the median in 1 to 10? Here we have to find the median of the first 10 natural numbers. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Where the number of terms is in even. Therefore, the median of the first 10 natural numbers is 5.5. ## What is the median of 23 and 23? Since there are an even number of values, the median will be the average of the two middle numbers, in this case, 23 and 23, the mean of which is 23. ## What is the median of 25? The median is the middle number, for an even set of numbers there will be two middle numbers, to find the median in an even set of numbers we can average the two middle numbers. So since 25 and 25 are both middle numbers we can average them to get the median, which is 25 . ## What's in the middle of 2 and 5? The class midpoint of 2-5 is 3.5. ## What is in the middle of 28 and 42? 18 is the HCF of 28 and 42. The HCF of 28 and 42 is 14. ## What is in the middle of 0 and 1? This is a good definition for intervals, and the middle point of (0,1) would be 1/2 with this definition.
## What are the five properties of exponents? Understanding the Five Exponent Properties • Product of Powers. • Power to a Power. • Quotient of Powers. • Power of a Product. • Power of a Quotient. ## What are the 7 properties of exponents? Make sure you go over each exponent rule thoroughly in class, as each one plays an important role in solving exponent based equations. • Product of powers rule. … • Quotient of powers rule. … • Power of a power rule. … • Power of a product rule. … • Power of a quotient rule. … • Zero power rule. … • Negative exponent rule. ## What are the three properties of exponents? Check out this video and this video. • Product of powers. This property states that when multiplying two powers with the same base, we add the exponents. … • Quotient of powers. This property states that when dividing two powers with the same base, we subtract the exponents. … • Power of a power property. ## What are the properties of exponents and power? When you raise a quotient to a power you raise both the numerator and the denominator to the power. When you raise a number to a zero power you’ll always get 1. Negative exponents are the reciprocals of the positive exponents. The same properties of exponents apply for both positive and negative exponents. ## What are the 10 rules of exponents? Laws Of Exponents • Powers with Same Base. • Quotient with Same Base. • Power of a Power. • Product to a Power. • Quotient to a Power. • Zero Power Rule. • Negative Exponent Rule. • Fractional Exponent Rule. ## What are the rules of exponents? The Power Rule for Exponents: (am)n = am*n. To raise a number with an exponent to a power, multiply the exponent times the power. Negative Exponent Rule: xn = 1/xn. Invert the base to change a negative exponent into a positive. ## What are the properties of exponents explain with example? An exponent (also called power or degree) tells us how many times the base will be multiplied by itself. For example ‘, the exponent is 5 and the base is . This means that the variable will be multiplied by itself 5 times. You can also think of this as to the fifth power. ## What is an exponent simple definition? Definition of exponent 1 : a symbol written above and to the right of a mathematical expression to indicate the operation of raising to a power. 2a : one that expounds or interprets. ## What is the exponent for 7? Exponent 7 Number Table Find the exponent 7 of…The exponent 7 7 7=823543 8 7=2097152 9 7=4782969 10 7=10000000 24 sept 2020 ## What does property of exponents mean? An exponent (also called power or degree) tells us how many times the base will be multiplied by itself. For example ‘, the exponent is 5 and the base is . This means that the variable will be multiplied by itself 5 times. You can also think of this as to the fifth power. ## What properties are in math? The properties of math are the rules governing the relationship and interaction of numbers with each other. There are four basic properties: commutative, associative, distributive, and identity. ## What is an exponent simple definition? Definition of exponent 1 : a symbol written above and to the right of a mathematical expression to indicate the operation of raising to a power. 2a : one that expounds or interprets. ## What are exponents called? Exponents are also called Powers or Indices. ## What is an exponent and example? Exponent is defined as the method of expressing large numbers in terms of powers. That means, exponent refers to how many times a number multiplied by itself. For example, 6 is multiplied by itself 4 times, i.e. 6 × 6 × 6 × 6. This can be written as 64. Here, 4 is the exponent and 6 is the base. ## What is another name for exponent in math? Other names for exponent are index or power. ## How do you represent exponents? The caret (^) is used as the exponentiation operator. Note: The exponent operator should not be confused with the base-10 exponent symbol. An uppercase letter “E”, or lowercase letter “e” can be used as a base-10 exponent (scientific notation) symbol in a numeric literal. ## What is the difference between exponents and powers? Power denotes the repeated multiplication of a factor and the number which is raised to that base factor is the exponent. This is the main difference between power and exponent. For example, 32 is the power where 3 is the base and 2 is the exponent. ## Who is the father of exponents? Nicolas Chuquet used a form of exponential notation in the 15th century, which was later used by Henricus Grammateus and Michael Stifel in the 16th century. The word exponent was coined in 1544 by Michael Stifel. ## What is the function of exponential? An exponential function is a Mathematical function in the form f (x) = ax, where “x” is a variable and “a” is a constant which is called the base of the function and it should be greater than 0. The most commonly used exponential function base is the transcendental number e, which is approximately equal to 2.71828.
# Basic math glossary-D Basic math glossary-D define words beginning with the letter D Data: Information that we collect Decimal places: The positions to the right of the decimal point Decimal number: All numbers in the base 10 number system that have one or more numbers in the decimal places. Decimal point: In a decimal number, it is a period that is used to separate the whole number from the numbers in the decimal places Degree: A measure of angles. It is equal to 1/360 of a circle The amount a container or a unit will hold when full Denominator: In a fraction, It is the number below the fraction bar Diameter: The distance across a circle through the center Difference: The answer to a subtraction problem. Digit: Any of the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Dimensions: Length, width,or height of the size of an geometric figure Discount: A reduction made from the regular price Discount rate: The percent that the price is reduced Dividend: The number being divided Divisibility: Able to be divided evenly Divisible: Able to be divided without a remainder Division: The process of dividing two numbers to find how many times one number is contained into another number Divisor: The number by which you are dividing ## Recent Articles 1. ### Writing an Algebraic Expression Sep 25, 17 09:22 AM Writing an algebraic expression when a phrase is given is the goal of this lesson
Question Video: Integration of Rational Functions by Partial Fractions \| Nagwa Question Video: Integration of Rational Functions by Partial Fractions \| Nagwa Question Video: Integration of Rational Functions by Partial Fractions Use partial fractions to evaluate β«(π β΄ + 81)/(π (π Β² + 9)Β²) dπ . 09:11 Video Transcript Use partial fractions to evaluate the indefinite integral of π to the fourth power plus 81 over π times π squared plus nine squared with respect to π . Letβs recall what we actually mean when we write a rational expression in partial fraction form. Essentially, weβre going to reverse the process for adding fractions to write π to the fourth power plus 81 over π times π squared plus nine all squared as the sum of some simpler rational functions. Now, there are several steps that we need to take to do so. The first thing weβre going to check is that the order of the expression on our denominator is greater than that of our numerator. The order or the degree of the polynomial on our denominator is five, since distributing the parentheses would give us the highest exponent of π as five, whereas the order of our numerator is four. Since the order of the polynomial on the denominator is higher than that of the numerator, we donβt need to perform any polynomial long division. And we can move on to our second step. Now, our second step is to fully factor the expression on the denominator. But this is done for us. And so weβre going to begin the next step. The first thing we do is we look for any nonrepeated factors in our denominator. The repeated factor here is π squared plus nine because itβs being squared. π , however, is a nonrepeated factor, and in fact itβs a linear factor. And so our first rational function is π΄, which is some constant, over π . We then look at our repeated factor. Since itβs being squared, weβre going to have two rational functions in increasing exponents of π squared plus nine. So their respective denominators are π squared plus nine and π squared plus nine squared. Remember though, these denominators are not linear; theyβre quadratic. And so their numerators have to reflect this. We write these as π΅π plus πΆ plus π·π plus πΈ, where π΅, πΆ, π·, and πΈ are constants. Our next job is to work out the value of π΄, π΅, πΆ, π·, and πΈ. And so what we do now is we add these three rational expressions. And we do so by creating a common denominator. That common denominator will match the denominator of our original rational function. And so we see that, for our first fraction, weβll need to multiply its denominator by π squared plus nine squared to achieve this. Of course, to create an equivalent fraction, we need to do the same to the numerator, giving us π΄ times π squared plus nine squared over π times π squared plus nine squared. For our second fraction, weβll need to multiply the numerator and denominator by π times π squared plus nine. This gives us a numerator of π΅π plus πΆ times π times π squared plus nine. Finally, weβll multiply our third fraction by π only. And this gives us a numerator of π times π·π plus πΈ. Now, since the denominators of our three fractions are equal, we simply add their numerators as shown. Now, of course, this is still equal to our earlier fraction. So we see that since the denominator of these two fractions is equal, for the fractions themselves to be equal, the numerators must also be equal. That is, π to the fourth power plus 81 must be equal to π΄ times π squared plus nine squared plus π΅π plus πΆ times π times π squared plus nine plus π times π·π plus πΈ. Now, to find the values of our constants, we have a couple of options. One method is to choose values of π . And weβre allowed to do this because, actually, we could use an identity symbol here. This expression is true for all values of π . The other method is to distribute our parentheses on the right-hand side and equate coefficients. Now, there are very few values of π that will help us to eliminate any terms. So weβre going to use the second method and equate coefficients. Distributing the parentheses, π squared plus nine squared gives us π to the fourth power plus 18π squared plus 81. Of course, this is all being multiplied by π΄. And so we can write this as π΄π to the fourth power plus 18π΄π squared plus 81π΄. Then when we distribute π times π squared plus nine, we get π cubed plus nine π . We can then distribute again, and we get π΅π to the fourth power plus nine π΅π squared plus πΆπ cubed plus nine πΆπ . Letβs distribute our final set of parentheses. When we do, we get π·π squared plus πΈπ . Weβre now going to equate coefficients of π . Letβs begin by comparing the coefficients of π to the fourth power on both sides. On the left-hand side, we simply have π to the fourth power. So its coefficient is one. On our right-hand side, we have π΄π to the fourth power plus π΅π to the fourth power. So our coefficient is π΄ plus π΅. And we formed an equation; that is, one equals π΄ plus π΅. Weβll now compare coefficients of π cubed. On the left, thatβs zero. And on the right, we only have πΆ. And so weβve established that πΆ is equal to zero. Weβll now look at coefficients of π squared. Once again, on the left, thatβs zero. And on the right, we have 18π΄, we have nine π΅, and we have π·. Next, itβs π to the power of one or simply π . On the left-hand side, thatβs zero. And on the right, itβs nine πΆ plus πΈ. But of course, remember, we said πΆ is equal to zero. So this becomes zero equals nine times zero plus πΈ or zero equals zero plus πΈ, meaning that πΈ itself must also be zero. Now, weβll consider coefficients of π to the power of zero. Remember, π to the power of zero is one. So weβre really looking at constant terms. On the left-hand side, itβs 81. And on the right-hand side, we have 81π΄. So, 81 equals 81π΄. And dividing through by 81 gives us π΄ equals one. We know the value of πΆ, πΈ, and π΄. So letβs go back and work out π΅ and π·. We saw that π΄ was equal to one. So our first equation becomes one equals one plus π΅, meaning π΅ must be equal to zero. Then letting π΄ be equal to one and π΅ be equal to zero in our third equation, and we get zero equals 18 plus zero plus π·, meaning π· is equal to negative 18. Weβve now done enough to express our original function in partial fraction form. By replacing π΄, π΅, πΆ, π·, and πΈ with their respective values, we find π to the fourth power plus 81 over π times π squared plus nine squared can be written as one over π minus 18π over π squared plus nine squared. Now, of course, weβre not finished. We were looking to integrate this. So weβre going to integrate one over π minus 18π over π squared plus nine squared with respect to π . Letβs clear some space. When integrating the sum or difference of two functions, we can find the sum or difference of the integrals of their respective functions. This essentially means we can integrate term by term. And this is great because the integral of one over π with respect to π is quite straightforward. Itβs the natural log of the absolute value of π plus some constant of integration π΄. But what about our second integral? Well, the key here is spotting that the derivative of part of the denominator is some multiple of our numerator. That tells us we can use integration by substitution to evaluate this integral. We let our new variable π’ be equal to the inner part of the composite function on the denominator. π’ is π squared plus nine. We then differentiate this with respect to π . So dπ’ by dπ is two π . Now, of course, this isnβt a fraction. But in integration by substitution, we treat it a little like one. And we can write this equivalently as dπ’ equals two π dπ . Notice the numerator of our fraction is 18π . And if we multiply through dπ’ equals two π dπ by nine, we get the expression nine dπ’ on the left and 18π dπ on the right, meaning we can now replace 18π dπ with nine dπ’. And of course we can replace π squared plus nine with π’. And so weβre now going to integrate nine over π’ squared with respect to π’. One over π’ squared is the same as π’ to the power of negative two. So we can write our integrand as nine π’ to the power of negative two. When we integrate an expression of this form, we add one to the exponent and divide by that new value. So thatβs nine π’ to the power of negative one divided by negative one, which is negative nine over π’. Remember, we have that constant of integration π΅ since weβre performing an indefinite integral. Now, we were actually integrating with respect to π . So letβs go back and replace π’ with our substitution, giving us negative nine over π squared plus nine plus our constant of integration π΅. Meaning that when we integrate π to the fourth power plus 81 over π times π squared plus nine squared with respect to π , we get the natural log of the absolute value of π plus π΄ minus negative nine over π squared plus nine plus π΅, which can be simplified further by distributing the parentheses and collecting together our constants to form one new constant πΎ. And we have the natural log of the absolute value of π plus nine over π squared plus nine plus πΎ. 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
# 3.3. Algorithm Analysis Examples¶ ## 3.3.1. An Anagram Detection Example¶ A good example problem for showing algorithms with different orders of magnitude is the classic anagram detection problem for strings sometimes called the anagram detection problem. One string is an anagram of another if the second is simply a rearrangement of the first. For example, "heart" and "earth" are anagrams. The strings "python" and "typhon" are anagrams as well. For the sake of simplicity, we will assume that the two strings in question are of equal length and that they are made up of symbols from the set of 26 lowercase alphabetic characters. Our goal is to write a Boolean function that will take two strings and return whether they are anagrams. ### 3.3.1.1. Solution 1: Checking Off¶ Our first solution to the anagram problem will check the lengths of the strings and then to see that each character in the first string actually occurs in the second. If it is possible to “checkoff” each character, then the two strings must be anagrams. The first step in the process will be to convert the second string to a local second string for checking off. Each character from the first string can be checked against the characters in the local second string and if found, checked off by replacement. ActiveCode 1 shows this function. To analyze this algorithm, we need to note that each of the n characters in s1 will cause an iteration through up to n characters in the array from s2. Each of the n positions in the array will be visited once to match a character from s1. The number of visits then becomes the sum of the integers from 1 to n. We stated earlier that this can be written as: $\begin{split}\sum_{i=1}^{n} i &= \frac {n(n+1)}{2} \\ &= \frac {1}{2}n^{2} + \frac {1}{2}n\end{split}$ As $$n$$ gets large, the $$n^{2}$$ term will dominate the $$n$$ term and the $$\frac {1}{2}$$ can be ignored. Therefore, this solution is $$O(n^{2})$$. ### 3.3.1.2. Solution 2: Sort and Compare¶ Another solution to the anagram problem will make use of the fact that even though s1 and s2 are different, they are anagrams only if they consist of exactly the same characters. So, if we begin by sorting each string alphabetically, from a to z, we will end up with the same string if the original two strings are anagrams. ActiveCode 2 shows this solution. At first glance you may be tempted to think that this algorithm is $$O(n)$$, since there are three consecutive simple iterations: the first two to convert strings to char arrays and the last to compare the n characters after the sorting process. Sorting is typically either $$O(n^{2})$$ or $$O(n\log n)$$, so the sorting operations dominate the iteration. In the end, this algorithm will have the same order of magnitude as that of the sorting process. ### 3.3.1.3. Solution 3: Brute Force¶ A brute force technique for solving a problem typically tries to exhaust all possibilities. For the anagram detection problem, we can simply generate an array of all possible strings using the characters from s1 and then see if s2 occurs. However, there is a difficulty with this approach. When generating all possible strings from s1, there are n possible first characters, $$n-1$$ possible characters for the second position, $$n-2$$ for the third, and so on. The total number of candidate strings is $$n(n-1)(n-2)...321$$, which is $$n!$$. Although some of the strings may be duplicates, the program cannot know this ahead of time and so it will still generate $$n!$$ different strings. It turns out that $$n!$$ grows even faster than $$2^{n}$$ as n gets large. In fact, if s1 were 20 characters long, there would be $$20!=2,432,902,008,176,640,000$$ possible candidate strings. If we processed one possibility every second, it would take us 77,146,816,596 years to go through the entire array. This is probably not going to be a good solution. ### 3.3.1.4. Solution 4: Count and Compare¶ Our final solution to the anagram problem takes advantage of the fact that any two anagrams will have the same number of a’s, the same number of b’s, the same number of c’s, and so on. In order to decide whether two strings are anagrams, we will first count the number of times each character occurs. Since there are 26 possible characters, we can use an array of 26 counters, one for each possible character. Each time we see a particular character, we will increment the counter at that position. In the end, if the two arrays of counters are identical, the strings must be anagrams. ActiveCode 3 shows this solution. Again, the solution has a number of iterations. However, unlike the first solution, none of them are nested. The first two iterations used to count the characters are both based on n. The third iteration, comparing the two arrays of counts, always takes 26 steps since there are 26 possible characters in the strings. Adding it all up gives us $$T(n)=2n+26$$ steps. That is $$O(n)$$. We have found a linear order of magnitude algorithm for solving this problem. Before leaving this example, we need to say something about space requirements. Although the last solution was able to run in linear time, it could only do so by using additional storage to keep the two arrays of character counts. In other words, this algorithm sacrificed space in order to gain time. This is a common occurrence. On many occasions you will need to make decisions between time and space trade-offs. In this case, the amount of extra space is not significant. However, if the underlying alphabet had millions of characters, there would be more concern. As a computer scientist, when given a choice of algorithms, it will be up to you to determine the best use of computing resources given a particular problem. ## 3.3.2. The Traveling Salesperson Problem¶ Let’s consider a famous problem in computer science for a bit and let’s return to the brute force method. Imagine that a salesperson needs to travel to a set of places and find the shortest path to do so. The Traveling Salesperson problem (TSP) has numerous direct applications in a number of fields, including transportation and logistics. The example of arranging school bus routes to pick up the children in a school district is of important historical significance since it provided motivation for Merrill Flood to do pioneering of TSP research in the 1940s. More current applications involve the scheduling of service calls or the delivery of packages or meals. Although transportation applications are clearly a natural setting for TSP, there are applications in other areas such as the scheduling of a machine to drill holes in a circuit board. If the time it takes to move the head of the drill is a significant portion of the overall manufacturing process, then the TSP is important in reducing costs. To be concrete, let’s imagine that a salesperson needs to travel to each country in the European Union and find the shortest path to do so. At the time of this writing, there were 28 European countries are members of the EU. Without the UK, there will be 27. Applying the Brute Force solution to the Traveling Salesperson problem is a really terrible idea. an algorithm is said to scale well or be scalable if it is suitably efficient and practical when applied to an input with a large n, and brute force does not scale at all well because n = 28 is quite a small number. As we know a brute force technique for solving a problem typically tries to exhaust all possibilities for our salesperson, that means trying every set of routes and checking the path distance. Like the anagram detection example, the total number of paths that a salesperson could try is $$n(n-1)(n-2)...321$$, which is $$n!$$ because they have to choose a first city from the $$n$$ choices, then there are only $$n-1$$ choices for the next city. Then $$n-2$$ choices for the third etc. That is $$O(n!)$$. Using brute force to solve this problem requires we check the 27 factorial ways for the salesperson to consider traveling to each country. This means that there are 10,888,869,450,418,352,160,768,000,000 possible paths for the salesperson to check to travel when using the brute force solution. Some of the fastest readily available processors currently are around 5GHz, where 1GHz represents 1 billion cycles per second. If you could do two computations in a cycle, then a computation would take 2/5,000,000,000 seconds which equals 0.0000000004 seconds. We can call this computation rate. Note that processors typically take more than one cycle to complete an instruction, but for the last decade or so most processors have been multicore… So, this is a rough estimate. Next you should take, computation rate times number of paths to give the amount of time in seconds, then divide that by the number of seconds in a year $$((2460)60)*365$$. It would take 345,283,785,211,134,961,972.6 years for the Brute Force Solution to find the shortest path for our salesperson. If you have 6 or 12 cores, you can just multiply by three or six, but that is clearly not going to help us a lot. What if we use a super computer? Summit is the fastest supercomputer in the world and can deliver as much as 200 petaflops at peak. This is equivalent to 200 quadrillion floating-point operations per second. Since a quadrillion is a million billion, Summit is 40 million times faster than the fastest regular processors. However, 8,632,094,630,278.3 years is clearly still far too many to wait. And this was just to visit the 28 European Union Countries. How long would it take to find the shortest path to visit all of the 48 states in the continental United States? Using brute force on problems with numbers even as small as 28 is clearly unworkable. And, unfortunately for our salesperson, there is no known tractable solution for finding the best route for TSP, so solutions are used that are not best but are good enough. These are called heuristics. The moral of the story here is that algorithms matter and algorithm analysis can help you decide not to choose a particular algorithm to use or not to use. Self Check
Home \| \| Statistics 12th Std \| Method of Least Squares # Method of Least Squares Method of least squares can be used to determine the line of best fit in such cases. It determines the line of best fit for given observed data by minimizing the sum of the squares of the vertical deviations from each data point to the line. METHOD OF LEAST SQUARES In most of the cases, the data points do not fall on a straight line (not highly correlated), thus leading to a possibility of depicting the relationship between the two variables using several different lines. Selection of each line may lead to a situation where the line will be closer to some points and farther from other points. We cannot decide which line can provide best fit to the data. Method of least squares can be used to determine the line of best fit in such cases. It determines the line of best fit for given observed data by minimizing the sum of the squares of the vertical deviations from each data point to the line. ## 1. Method of Least Squares To obtain the estimates of the coefficients ‘a’ and ‘b’, the least squares method minimizes the sum of squares of residuals. The residual for the ith data point ei is defined as the difference between the observed value of the response variable, yi, and the estimate of the response variable, ŷi, and is identified as the error associated with the data. i.e., ei = yiŷi , i =1 ,2, ..., n. The method of least squares helps us to find the values of unknowns ‘a’ and ‘b’ in such a way that the following two conditions are satisfied: Sum of the residuals is zero. That is Sum of the squares of the residuals E ( a , b ) = is the least ## 2. Fitting of Simple Linear Regression Equation The method of least squares can be applied to determine the estimates of a’ and ‘b’ in the simple linear regression equation using the given data (x1,y1), (x2,y2), ..., (xn,yn) by minimizing Here, yˆi = a + bx i is the expected (estimated) value of the response variable for given xi. It is obvious that if the expected value (y^ i) is close to the observed value (yi), the residual will be small. Since the magnitude of the residual is determined by the values of a’ and ‘b’, estimates of these coefficients are obtained by minimizing the sum of the squared residuals, E(a,b). Differentiation of E(a,b) with respect to ‘a’ and ‘b’ and equating them to zero constitute a set of two equations as described below: These equations are popularly known as normal equations. Solving these equations for ‘a’ and ‘b’ yield the estimates ˆa and ˆb. It may be seen that in the estimate of ‘ b’, the numerator and denominator are respectively the sample covariance between X and Y, and the sample variance of X. Hence, the estimate of ‘b’ may be expressed as Further, it may be noted that for notational convenience the denominator of bˆ above is mentioned as variance of nX. But, the definition of sample variance remains valid as defined in Chapter I, that is, From Chapter 4, the above estimate can be expressed using, rXY , Pearson’s coefficient of the simple correlation between X and Y, as ## Important Considerations in the Use of Regression Equation: 1. Regression equation exhibits only the relationship between the respective two variables. Cause and effect study shall not be carried out using regression analysis. 2. The regression equation is fitted to the given values of the independent variable. Hence, the fitted equation can be used for prediction purpose corresponding to the values of the regressor within its range. Interpolation of values of the response variable may be done corresponding to the values of the regressor from its range only. The results obtained from extrapolation work could not be interpreted. ### Example 5.1 Construct the simple linear regression equation of Y on X if ### Solution: The simple linear regression equation of Y on X to be fitted for given data is of the form ˆY = a + bx ……..(1) The values of ‘a’ and ‘b’ have to be estimated from the sample data solving the following normal equations. Substituting the given sample information in (2) and (3), the above equations can be expressed as 7 a + 113 b = 182 (4) 113 a + 1983 b = 3186 (5) (4) × 113 791 a + 12769 b = 20566 (5) × 7 791 a + 13881 b = 22302 Substituting this in (4) it follows that, 7 a + 113 × 1.56 = 182 7 a + 176.28 = 182 7 a = 182 – 176.28 = 5.72 Hence, a = 0.82 ### Example 5.2 Number of man-hours and the corresponding productivity (in units) are furnished below. Fit a simple linear regression equation ˆY = a + bx applying the method of least squares. ### Solution: The simple linear regression equation to be fitted for the given data is ˆYˆ = a + bx Here, the estimates of a and b can be calculated using their least squares estimates From the given data, the following calculations are made with n=9 Substituting the column totals in the respective places in the of the estimates aˆ and bˆ , their values can be calculated as follows: Thus, bˆ = 1.48 . Now aˆ can be calculated using bˆ as aˆ = 121/9 – (1.48× 62.1/9) = 13.40 – 10.21 Hence, aˆ = 3.19 Therefore, the required simple linear regression equation fitted to the given data is ˆYˆ = 3.19 +1.48x It should be noted that the value of Y can be estimated using the above fitted equation for the values of x in its range i.e., 3.6 to 10.7. In the estimated simple linear regression equation of Y on X ˆYˆ = aˆ + ˆbx we can substitute the estimate aˆ = − bˆ . Then, the regression equation will become as It shows that the simple linear regression equation of Y on X has the slope bˆ and the corresponding straight line passes through the point of averages ( , ). The above representation of straight line is popularly known in the field of Coordinate Geometry as ‘Slope-Point form’. The above form can be applied in fitting the regression equation for given regression coefficient bˆ and the averages and . As mentioned in Section 5.3, there may be two simple linear regression equations for each X and Y. Since the regression coefficients of these regression equations are different, it is essential to distinguish the coefficients with different symbols. The regression coefficient of the simple linear regression equation of Y on X may be denoted as bYX and the regression coefficient of the simple linear regression equation of X on Y may be denoted as bXY. Using the same argument for fitting the regression equation of Y on X, we have the simple linear regression equation of X on Y with best fit as The slope-point form of this equation is Also, the relationship between the Karl Pearson’s coefficient of correlation and the regression coefficient are Tags : Example Solved Problems \| Regression Analysis , 12th Statistics : Chapter 5 : Regression Analysis Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail 12th Statistics : Chapter 5 : Regression Analysis : Method of Least Squares \| Example Solved Problems \| Regression Analysis
# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 3375 Exercise Given the differentiable function $$z(x,y)=e^yf(ye^{\frac{x^2}{2y^2}})$$ Prove the equation $$(x^2-y^2)\cdot z'_x+xy\cdot z'_y=xyz$$ Proof When we have a function and in its parentheses there is a complex expression instead of a simple variable, we will define a new variable like this $$t=ye^{\frac{x^2}{2y^2}}$$ We get the function $$z(x,y)=e^yf(t)$$ Or simply $$z(x,y)=e^yf$$ We will use the chain rule to calculate the partial derivatives of z. $$z'_x=e^y\cdot f'_t\cdot t'_x$$ $$z'_y=e^y\cdot f + e^y\cdot f'_t\cdot t'_y$$ Similarly, we will calculate the partial derivatives of t. $$t'_x=ye^{\frac{x^2}{2y^2}}\cdot \frac{2x}{2y^2}=$$ $$=ye^{\frac{x^2}{2y^2}}\cdot \frac{x}{y^2}=$$ $$=e^{\frac{x^2}{2y^2}}\cdot \frac{x}{y}$$ $$t'_y=1\cdot e^{\frac{x^2}{2y^2}}+y\cdot e^{\frac{x^2}{2y^2}}\cdot \frac{-x^2}{4y^4}\cdot 4y=$$ $$t'_y=e^{\frac{x^2}{2y^2}}(1-y\cdot \frac{x^2}{y^3})=$$ $$t'_y=e^{\frac{x^2}{2y^2}}(1-\frac{x^2}{y^2})$$ We put the results in the partial derivatives of z and get $$z'_x=e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}\cdot \frac{x}{y}$$ $$z'_y=e^y\cdot f + e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}(1-\frac{x^2}{y^2})$$ We will put the partial derivatives in the left side of the equation we need to prove. $$(x^2-y^2)\cdot z'_x+xy\cdot z'_y=$$ $$=(x^2-y^2)\cdot e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}\cdot \frac{x}{y}+xy\cdot (e^y\cdot f + e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}(1-\frac{x^2}{y^2})=$$ $$=\frac{x^3}{y}e^ye^{\frac{x^2}{2y^2}}f'_t-xye^ye^{\frac{x^2}{2y^2}}f'_t+xye^yf +xye^yf'_t e^{\frac{x^2}{2y^2}}-xye^yf'_t\frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}=$$ $$=\frac{x^3}{y}e^ye^{\frac{x^2}{2y^2}}f'_t+xye^yf +-xye^yf'_t\frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}=$$ $$=\frac{x^3}{y}e^ye^{\frac{x^2}{2y^2}}f'_t+xye^yf +-e^yf'_t\frac{x^3}{y}e^{\frac{x^2}{2y^2}}=$$ $$=xye^yf=$$ We put the equation $$z=e^yf$$ And get $$=xyz$$ Hence, we got $$(x^2-y^2)\cdot z'_x+xy\cdot z'_y=xyz$$ As required. Have a question? Found a mistake? – Write a comment below! Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! Share with Friends
# Set Theory Contents ## Summary • The modern study of set theory was initiated by Georg Cantor and Richard Dedekind in the 1870s. • A set is a collection of distinct objects, called elements of the set. • Set’s are denoted by upper case letters and they are enclosed in curly brackets {….}. • When we write $x\quad \in \quad A$, this means x belongs to the set A. • An empty set is denoted by ∅. • Subset contains fewer or all elements of only its parent set and it is denoted by the symbol ⊆. If B is a subset of A, we write BA. • Some important set operations are: Union: A ⋃ B Intersection: A ⋂ B Complement: Complement of set A is written as ${ A }^{ c }$ The modern study of set theory was initiated by Georg Cantor and Richard Dedekind in the 1870s. Set theory includes a diverse collection of topics, ranging from the structure of the real number line to the study of the consistency of large cardinals. Set Theory is fundamental to modern mathematics. It is the alphabet of mathematics. #### What is a set? A set is a collection of distinct objects, called elements of the set. When we know all members of the set, it is known as definite set. Set’s are denoted by upper case letters (variables) and the elements of the set are denoted by lower case letters. These are enclosed in curly brackets. We can say two sets are equal if they have the same elements. The symbol $\in$ means ”is an element of”. I.e when we write $x\quad \in \quad A$, this means x belongs to the set A. The set that contains no elements is the empty set, and is denoted by ∅. Some examples of sets are listed below: • {1, 2, 3, 4} • {cat, dog, bird, fish} • {orange, yellow, blue, green, white} • {50, 55, 86, 88, 89, 90} Some of the sets that have special notations for different types are: N = Set of Natural Numbers Q = Set of Rational Numbers R = Set of Real Numbers P = Set of Prime Numbers Z = Set of Integers E = Set of Even Integers O = Set of Odd Integers #### What is a Subset? Subset contains fewer or all elements of only its parent set. For example if it is said that B is a subset of A, then this means all of the elements of B are also in A. Symbol ⊆ is used to denote a subset. If B is a subset of A, we write B ⊆ A. i.e A = {5, 10, 15, 20, 25} B = {10 ,15} We can see that B is a subset of A written as B ⊆ A. #### Venn Diagrams A very common method to represent operations of sets are through venn diagrams. This way we can easily visualise the relation between different sets. For example we can represent set A={5, 10, 15, 20, 25} through a venn diagram in the following way. Note that set A (the circle) is a subset of the Universal set (the rectangle). Some important set operations are: i) Union The union of two sets contains all the elements contained in either set (or both sets). The union is notated A ⋃ B. For example if we have two sets A = {2, 4, 6, 8} and B = (2, 3, 4, 5, 6}, then A ⋃ B ={2, 3, 4, 5, 6, 8} note we list 2, 4, 6 only once. This can be shown on a venn diagram in the following way: ii) Intersection The intersection of two sets contains only the elements that are in both sets. The intersection is notated A ⋂ B. For example if we have two sets A = {2 , 4, 6, 8} and B = (2, 3, 4, 5, 6}, then A ⋂ B = {2, 4, 6}. This can be shown on a venn diagram in the following way: iii) Complement The complement of a set A contains everything that is not in the set A. The complement is notated A’, or ${ A }^{ c }$. A complement is relative to the universal set, so ${ A }^{ c }$ contains all the elements in the universal set that are not in A.
# The Mode of Ungrouped Data The Mode This is the final type of average, and the easiest one to work out… so long as you remember how! The mode is the number or value that appears most frequently in the data set. How to work out the Mode: Find the most common piece of data (number or letter) and this is your model NOTE: You can have no modes or more than one mode, and you must write them all down! For discrete numerical data, the mode is the most frequently occurring value in the data set. For continuous numerical data, we cannot talk about a mode in this way because no two data Values will be exactly equal. Instead we talk about a modal class, which is the class or group that occurs most frequently. Definition: Mode The mode of a data set is the value that occurs most often in the set. The mode can also be described as the most frequent or most common value in the data set. To calculate the mode, we simply count the number of times that each value appears in the data set and then find the value that appears most often. A data set can have more than one mode if there is more than one value with the highest count. For example, both 2 and 3 are modes in the data set {1; 2; 2; 3; 3}. If all points in a data set occur with equal frequency, it is equally accurate to describe the data set as having many modes or no mode. Worked example: Finding the mode. QUESTION Find the mode of the data set {2; 2; 3; 4; 4; 4; 6; 6; 7; 8; 8; 10; 10}. SOLUTION Step 1: Count the number of times that each value appears in the data set Step 2: Find the value that appears most often From the table above we can see that 4 is the only value that appears 3 times. All the other values appear less than 3 times. Therefore the mode of the data set is 4. One problem with using the mode as a measure of central tendency is that we can usually not compute the mode of a continuous data set. Since continuous values can lie anywhere on the real line, any particular value will almost never repeat. This means that the frequency of each value in the data set will be 1 and that there will be no mode. We will look at one way of addressing this problem in the section on grouping data. Q1. In a Mathematics class, 23 learners completed a test out of 25 marks. Here is a list of their results: Find the mode of this data. solution: The mode of the results of the test is 13 (13 appears 4 times). Q2. Find the mode of the following data: (i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6 (ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8 solution: (i) Mode =7 since 7 occurs 4 times. (ii) Mode =11 since it occurs 4 times Q3. The following table shows the frequency distribution of heights of 50 boys: Find the mode of heights. solution: Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18. Q4. Find the median and mode for the set of numbers: 2, 2, 3, 5, 5, 5, 6, 8 and 9 solution: Median =(9+1)/2=5th term which is 5. Mode =5 because it occurs maximum number of times. Let’s read the post ‘Examples of The Median for Ungrouped Data‘. Q5. Calculate the mode of the following data set: {10; 10; 10; 18; 7; 10; 3; 10; 7; 10; 7} solution: We first sort the data set: {3; 7; 7; 7; 10; 10; 10; 10; 10; 10; 18}. The mode is the value that occurs most often in the data set. Therefore the mode is: 10. Q6. The learners in Ndeme’s class have the following ages: {5; 6; 7; 5; 4; 6; 6; 6; 7; 4} Find the mode of their ages. solution: We first sort the data set: {4; 4; 5; 5; 6; 6; 6; 6; 7; 7}. The mode is the value that occurs most often in the data set. Therefore the mode is: 6. Q7. Calculate the mode of the following data set: {6; 10; 6; 6; 13; 12; 12; 7; 13; 6} solution: We first sort the data set: {6; 6; 6; 6; 7; 10; 12; 12; 13; 13}. The mode is the value that occurs most often in the data set. Therefore the mode is: 6
How to Use Algebra in Arithmetic USING ALGEBRA IN ARITHMETIC It’s interesting that algebra can be useful in arithmetic. Since arithmetic is a prerequsite to learning algebra, most algebra courses don’t focus on how to apply algebra to do arithmetic. However, you can apply your algebra skills to arithmetic problems. Following is an example. Suppose that you would like to multiply 58 times 58, but don’t have a calculator (or cell phone) handy. One way to do this, which is often taught in elementary school, is basically to multiply 8 times 58, then multiply 50 times 58, and add these together. Algebraically, what your elementary school teacher taught you is that 58 × 58 = (50 + 8) × 58 = 50 × 58 + 8 × 58. You may not have realized it at the time, but your teacher was applying the distributive property of mathematics. When you solve the problem this way, first you multiply 8 × 58 to get 464. Then you multiply 50 × 58 to get 2900. Finally, you add these together to get 3364. Now let’s see how algebra can supply an alternative solution. I’d prefer to work with nice round numbers. Hey, 60 is a round number close to 58. If I could work with a round number, like 60, and a small number, like 2, that would make the arithmetic much simpler to do without a calculator. So let’s write 58 as 60 minus 2. Then the problem is 58 × 58 = (60 – 2) × (60 – 2). Use the f.o.i.l. method from algebra. Recall that the f.o.i.l. method stands for “first,” “outside,” “inside,” “last.” One example of the f.o.i.l. method from algebra is (x + y)(x + y) = x² + xy + xy +y² = x² + 2xy +y². Let’s apply this to the arithmetic problem. We can think of (60 – 2) as (x + y), where x = 60 and y = –2. 58 × 58 = (60 – 2) × (60 – 2) = 60 × 60 + 2 (60)(–2) + (–2) × (–2). Look what we did. We wrote the original problem, 58 × 58, in terms of three simple multiplication problems: 60 × 60, 2 × 60 × (–2), and (–2) × (–2). 58 × 58 = 3600 – 240 + 4 = 3364. Note that (–2) × (–2) = +4 because the two minus signs cancel out. CHRIS MCMULLEN, PH.D. Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks Newest release: • 300+ Mathematical Pattern Puzzles Improve Your Math Fluency. Build fluency in: • arithmetic • long division • fractions • algebra • trigonometry • graphing Mathematical Pattern Puzzle: Fill in the Missing Numbers NUMBER PATTERN PUZZLE Here is an exercise in pattern recognition. It’s not a linear pattern. This is an array, so there is a slight geometric element to the pattern. See if you can figure out the missing numbers in the above puzzle. Study the four arrays. See if you can recognize the pattern. Once you identify the pattern, apply it to the fifth array. Spoiler alert. If you scroll down too far… You will run into the answer. So stop scrolling down… If you would like more time to solve the puzzle. Ready or not. Here it comes. First the answer: Now the solution. Begin with the top left number. Double the top left number. That makes the top right number. 5 doubled = 10. Now multiply the top two numbers. That makes the bottom left number. 5 times 10 = 50. Now add the bottom left number to the top right number. That makes the bottom right number. 50 plus 10 equals 60. CHRIS MCMULLEN, PH.D. Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks Newest releases: • 300+ Mathematical Pattern Puzzles • Basic Linear Graphing Skills Practice Workbook • Systems of Equations: Simultaneous, Substitution, Cramer’s Rule Improve Your Math Fluency. Build fluency in: • arithmetic • long division • fractions • algebra • trigonometry • graphing Five Math Puzzles (pattern recognition): Can You Solve Them? MATH PUZZLES Here is a math puzzle challenge. Hint: Each of the 5 patterns below has something in common. Directions: See if you can figure out which numbers go in the blanks. • 1, 2, 4, 6, 10, 12, 16, 18, 22, _, _ • 4, 6, 10, 14, 22, 26, 34, 38, _, _ • 3, 7, 13, 19, 29, 37, _, _ • 4, 9, 25, 49, 121, 169, 289, _, _ • 5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, _, _ If you need help, you can find hints below. But don’t scroll too far or you’ll run into the answers and explanations. PUZZLE HINT Each pattern above has something in common. They all involve prime numbers. A prime number is only evenly divisible by two integers: 1 and itself. For example, 7 is a prime number because the only integers that can multiply together to make 7 are 1 and 7. In contrast, 6 isn’t a prime number because 2 x 3 = 6 (in addition to 1 x 6). Here are the first several prime numbers. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 Each of the puzzles above relates to these prime numbers. When you’re ready, you can find answers and explanations below. PUZZLE ANSWERS Here are the answers and explanations to the math puzzles: • 28, 30. Explanation: Subtract 1 from each prime number: 2 – 1 = 1, 3 – 1 = 2, 5 – 1 = 4, 7 – 1 = 6, 11 – 1 = 10, etc. • 46, 58. Explanation: Double each prime number: 2 x 2 = 4, 3 x 2 = 6, 5 x 2 = 10, 7 x 2 = 14, 11 x 2 = 22, etc. • 43, 53. Explanation: Every other prime number: 3 (skip 5) 7 (skip 11) 13 (skip 17) 19 (skip 23) 29 etc. • 361, 529. Explanation: Square each prime number: 2² = 4, 3² = 9, 5² = 25, 7² = 49, 11² = 121, etc. • 78, 84. Explanation: Add consecutive prime numbers together: 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34, etc. WANT MORE MATH PUZZLES? One way is to follow my blog. I will post occasional math puzzles in the future. Another way is to check out my newest book, 300+ Mathematical Pattern Puzzles. It starts out easy and the level of challenge grows progressively so that puzzlers of all abilities can find many puzzles to enjoy. A wide variety of topics are covered, including: • visual patterns • arithmetic • repeating patterns • Roman numerals • Fibonacci sequence • prime numbers • arrays • analogies • and much more The cover was designed by Melissa Stevens at www.theillustratedauthor.net. CHRIS MCMULLEN, PH.D. Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks Newest releases: • 300+ Mathematical Pattern Puzzles • Basic Linear Graphing Skills Practice Workbook • Systems of Equations: Simultaneous, Substitution, Cramer’s Rule Improve Your Math Fluency. Build fluency in: • arithmetic • long division • fractions • algebra • trigonometry • graphing Tips for Finding the Slope of a Straight Line HOW TO FIND THE SLOPE OF A STRAIGHT LINE Slope is a measure that indicates how steep or shallow a straight line is: • A line with greater slope is steeper. • A line with less slope is shallower. • A horizontal line has zero slope. Slope can be positive or negative: • A line with positive slope slants upward. • A line with negative slope slants downward. • A line with zero slope is horizontal. The slope of a straight line equals rise over run. • The rise between two points is vertical. It’s the change in y. • The run between two points is horizontal. It’s the change in x. TIPS FOR FINDING SLOPE From the graph of a straight line, determine the slope as follows: • Mark two points on the line. • Read the x- and y-coordinates of the two points, (x1, y1) and (x2, y2). • Subtract y2 – y1 to get the rise. • Subtract x2 – x1 to get the run. • Divide the rise by the run. Here are a few tips: • When choosing the two points, try to find points where both x and y are easy to read without interpolating. This isn’t always possible: In that case, at least one coordinate should be easy to read without interpolating. • Choose two points far apart. This reduces the relative error in interpolating. • Make sure that both points lie on the straight line. Don’t choose a point that’s close to the line, but doesn’t lie on it. EXAMPLE OF HOW TO DETERMINE SLOPE Example: Find the slope of the straight line in the graph below. Solution: First, choose two points on the line. Ideally, these points should be far apart and easy to read. In this case, it’s easy to read both the x- and y-coordinates for the leftmost and rightmost points shown in the graph. So let’s choose those. • The leftmost point has coordinates (0, 3). • The rightmost point has coordinates (10, 8). Subtract the y-values to determine the rise: y2 – y1 = 8 – 3 = 5 Subtract the x-values to determine the run: x2 – x1 = 10 – 0 = 10 (In coordinate graphing, recall that x is horizontal and y is vertical.) Divide the rise by the run to find the slope: The slope of the line is 0.5. Check: You can check your answer as follows. Look at the graph. Starting from (0, 3), the next point that’s easy to read is (2, 1). From (0, 3) to (2, 1), the line goes one unit up (vertically) and 2 units over (horizontally). The ratio of the rise to the run is 1 to 2. Divide the rise (of 1) by the run (of 2). The slope is 0.5. ♦ CHRIS MCMULLEN, PH.D. Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks Newest release: • Basic Linear Graphing Skills Practice Workbook Related books: • Trigonometry Essentials Practice Workbook with Answers • Learn or Review Trigonometry Essential Skills • Algebra Essentials Practice Workbook with Answers • Systems of Equations: Simultaneous, Substitution, Cramer’s Rule Memory Tip for Sine, Cosine, and Tangent of Special Angles (Trigonometry) TRIGONOMETRY MEMORY TIP There is a simple way to remember the sine, cosine, and tangent of special trigonometry angles. The special trig angles are 0º, 30º, 45º, 60º, and 90º. What makes these angles special? The 30º-60º-90º triangle is one-half of an equilateral triangle, while the 45º-45º-90º triangle is one-half of a square. In both cases, the trig functions (sine, cosine, and tangent) can be expressed as simple ratios. Here is the trick for quickly working out the sine, cosine, and tangent of 0º, 30º, 45º, 60º, and 90º. STEP 1: Special angles. Write the special angles in order. STEP 2: Integers. Write the integers 0 thru 4 in order. STEP 3: Squareroots. Squareroot each number. STEP 4: Find the sine of theta. Divide each number by 2. These are the sine of 0º, 30º, 45º, 60º, and 90º. It’s that simple. Here’s a recap: • Write the numbers 0 thru 4. • Squareroot each number. • Divide each number by 2. STEP 5: Find the cosine of theta. Just write the previous numbers in reverse order. Why does this work? Because the sine of theta equals the cosine of the complement of theta: sin(θ)=cos(90º–θ). What’s opposite to theta is adjacent to its complement. STEP 6: Find the tangent of theta. Divide sine theta by cosine theta. TRIG CHART This chart shows all of the steps together. • Write the special angles. • Write the integers 0 thru 4. • Squareroot each number. • Divide each number by 2. This gives you sine of theta. • Write the numbers in reverse order. This gives you cosine of theta. • Divide the previous two rows (sine over cosine). This gives you tangent theta. CHRIS MCMULLEN, PH.D. Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks • Trigonometry Essentials Practice Workbook with Answers • Learn or Review Trigonometry Essential Skills • Trigonometry Flash Cards (for Kindle) • Algebra Essentials Practice Workbook with Answers • Systems of Equations: Simultaneous, Substitution, Cramer’s Rule • Other volumes cover fractions, long division, arithmetic, and more • Also look for books on the fourth dimension, astronomy, conceptual chemistry, and more Multiplication Facts 0-10 Memory Tips MULTIPLICATION FACTS 0-12 When students learn multiplication, it starts out easy. Too easy, as 0 times anything equals 0 (e.g. 0 x 8 = 0, while anything times 1 equals itself (e.g. 4 x 1 = 4). But it doesn’t stay easy. Most students struggle to remember multiplication facts where one of the numbers is 6, 7, 8, or 9. However, there are some tips to aid in memorization. For one, mirror images don’t matter. The word for this is commutative. This means 2 x 8 = 8 x 2, for example. Both equal 16. So if you know 6 x 8 = 48, you also know that 8 x 6 = 48. You don’t need to memorize both. One of the tricky multiplication facts is 7 x 8. But it’s easy if you remember the trick. Remember 5678. These are 5 thru 8 in order. 56 = 7 x 8. Piece of cake, huh? The 9’s are easy. The answer is one decade less than multiplying by 10, then make the two digits add up to 9. For example, 9 x 7 is in the 60’s (because 10 x 7 = 70, one decade less is 60). It’s 63 since 6 + 3 = 9. Another example is 5 x 9. It’s in the 40’s (one decade less than 5 x 10 = 50). It’s 45 since 4 + 5 = 9. To do the trick, after you figure out the decade (by multiplying by 10 and then subtracting 10), subtract the tens digit from 9 to get the units digit. For example, consider 9 x 6. Multiply 10 x 6 to get 60, and subtract 10 to make 50 (one decade less). Now subtract 5 (the tens digit of 50) from 9 to get 4. Therefore, 9 x 6 = 54. Once you get the hang of it, this makes remembering the 9’s easy. Try out all the 9’s to get some practice. If you’re good at doubling numbers quickly, try writing 6 as 2 x 3. Then 6 x 7 = 2 x 3 x 7. If you know 3 x 7 = 21, double 21 to get 42. Similarly, for 6 x 4 = 2 x 3 x 4, start with 3 x 4 = 12 and double 12 to make 24. You can use the doubling trick for the 8’s, too. Just double the number 3 times. For example, consider 5 x 8. Double 5 three times: 10, 20, 40. So 5 x 8 = 40. Try 8 x 6. Double 6 three times: 12, 24, 48. Therefore, 8 x 6 = 48. It works with the 4’s, also. Just double twice. With 4 x 9, double 9 twice: 18, 36. So 4 x 9 = 36. That leaves 7 x 7 = 49. You should know 7 x 9 from the 9’s trick. You can make 7 x 6 and 7 x 8 from the doubling tricks. (The latter you can also know from the 5678 trick.) 5 and under are easier. So to complete the 7’s, you really just need to memorize 7 x 7 = 49. This covers the 6 thru 9’s, which tend to be the trickier multiplication facts. The 10’s and 11’s are easy. For the 10’s, just add a zero, as in 8 x 10 = 80 or 6 x 10 = 60. For the 11’s, with 1 thru 9 just double the digits, like 3 x 11 =33 or 11 x 8 = 88. Get 11 x 10 = 110 from the 10’s trick (add a zero). Then you just need to memorize that 11 x 11 = 121 to complete the 11’s. You can get the 12’s by doubling the 6’s. For example, knowing that 6 x 5 = 30, you can find that 12 x 5 = 60 by doubling 6 x 5. Do you know any other tips for remembering multiplication facts 0-12? If so, please share them in the comments. CHRIS MCMULLEN, PH.D. Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks • arithmetic facts • multi-digit arithmetic • long division with remainders • fractions, decimals, and percents • algebra and trigonometry Uses Of Anagrams In The Past Here is a fascinating history of anagrams. Traveling Problem Here is a neat little math puzzle from the Mathemagical Site (which lives up to its great name). Fibonacci Sequence & a Cool Pattern Image from ShutterStock. FIBONACCI SEQUENCE The Fibonacci sequence adds consecutive terms: • 1 • 1 • 1 + 1 = 2 • 2 + 1 = 3 • 3 + 2 = 5 • 5 + 3 = 8 • 8 + 5 = 13 • 13 + 8 = 21 • 21 + 13 = 34 • 34 + 21 = 55 • 55 + 34 = 89 Since the last two terms were 55 and 89, we would add these together to get 89 + 55 = 144. Then you would add 144 and 89 to make 233, and so on. I saw a cool pattern involving the Fibonacci sequence recently at the Mathemagical Site: Fibonacci Triples via Mathemagical Site This involves Fibonacci triples. A Fibonacci triple consists of three consecutive numbers from the Fibonacci sequence, such as: • 1, 1, 2 • 1, 2, 3 • 2, 3, 5 • 3, 5, 8 • 5, 8, 13 • 8, 13, 21 • 13, 21, 34 As shown on the Mathemagical Site, the square of the middle number is always one less or one more than the product of the first and third numbers: Here are a few examples: • (2, 3, 5): 3 x 3 = 2 x 5 – 1 • (3, 5, 8): 5 x 5 = 3 x 8 + 1 • (5, 8, 13): 8 x 8 = 5 x 13 – 1 Curious about this, I’ve been working through the algebra, and finally came up with an algebraic proof, which follows. My proof is algebraic and not necessarily obvious, but since I worked it out, I thought I would share it here. 🙂 We begin with two facts about the Fibonacci sequence: These are two different ways of saying that if you add two consecutive numbers from the Fibonacci sequence, you get the next number in the sequence. Now solve for xn in each sequence: Multiply these together: Foil this out: Recall that Plug this into the first term on the right-hand side of the previous equation: Distribute: Two of these terms cancel (the remaining terms are rearranged): Believe it or not, this basically concludes the proof. The remainder is basically interpreting this result. This is a recursion relation that relates the square of the n-th term to the square of the previous term (xn-1 times itself). Following is the Fibonacci sequence, labeling values of n: • n = 1 is 1. • n = 2 is 1. • n = 3 is 2. • n = 4 is 3. • n = 5 is 5. • n = 6 is 8. • n = 7 is 13. • n = 8 is 21. Let’s plug in n = 3 and see what happens: If instead you plug in n = 4, you get: Now just plug in these two expressions (x3x3 – x4x2 and x4x4 – x5x3) into the previous recursion relation and you can prove that all of the Fibonacci triples satisfy one of the following relations: That is, if x3x3 – x4x2 = 1 and x4x4 – x5x3 = -1, the previous recursion relation gives similar expressions for x5x5 – x6x4, x6x6 – x7x5, and so on. CHRIS MCMULLEN, PH.D. Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks • Algebra Essentials Practice Workbook with Answers • Trigonometry Essentials Practice Workbook with Answers • Learn or Review Trigonometry: Essential Skills A Challenging Dissection Challenge What a cool visual puzzle from Mathemagical Site. Better think out of the box for this one.
# Buffon's Needle Buffon's Needle The Buffon's Needle problem is a mathematical method of approximating the value of pi $(\pi = 3.1415...)$involving repeatedly dropping needles on a sheet of lined paper and observing how often the needle intersects a line. # Basic Description The method was first used to approximate π by Georges-Louis Leclerc, the Comte de Buffon, in 1777. Buffon apparently first tried throwing bread sticks over his shoulder and counting how often they intersected a line on his tile floor. He was intrigued by the relationship he observed between the probability of an intersection and π. Subsequent mathematicians have used the method with needles instead of bread sticks, or with computer simulations. In the case where the distance between the lines is equal the length of the needle, an approximation of π can be calculated using the equation $\pi \approx {2\mbox{number of drops} \over \mbox{number of hits}}$ # A More Mathematical Explanation #### Will the Needle Intersect a Line? [[Image:willtheneedlecros [...] #### Will the Needle Intersect a Line? To prove that the Buffon's Needle experiment will give an approximation of π, we can consider which positions of the needle will cause an intersection. The variable θ is the acute angle between 0 and $\tfrac {\pi}{2}$ made by the needle and an imaginary line parallel to the ones on the paper. Finally, d is the distance between the center of the needle and the nearest line. Since the needle drops are random, there is no reason why the needle should be more likely to intersect one line than another. As a result, we can simplify our proof by focusing on a particular strip of the paper bounded by two lines. We can extend line segments from the center and tip of the needle to meet at a right angle. A needle will cut a line if the green arrow, d, is shorter than the leg opposite θ. More precisely, it will intersect when $d \leq \left( \frac{1}{2} \right) \sin(\theta). \$ See case 1, where the needle falls at a relatively small angle with respect to the lines. Visually we can see that a needle lying at such a small angle will only intersect if the center falls close to one of the parallel lines. In case 2, the needle intersects even though the center of the needle is far from both lines because the angle is so large. #### The Probability of an Intersection To approximate π using Buffon's method, we need to know the probability of an intersection. If we graph the outcomes of θ along the X axis and d along the Y, we have the sample space for the trials. In the diagram below, the sample space is contained by the dashed lines. The sample space is useful in this type of simulation because it gives a visual representation of all the possible ways the needle can fall. Each point on the graph represents some combination of an angle and distance that a needle might occupy. We divide the area that contains combinations that will cause an intersection by the total possible positions to calculate the probability. There will be an intersection if $d \leq \left ( \frac{1}{2} \right ) \sin(\theta) \$, which is represented by the blue region. The area under this curve represents all the combinations of distances and angles that will cause the needle to intersect a line. The area under the blue curve, which is equal to $\tfrac {2}{\pi}$ in this case, can found by evaluating the integral $\int_0^{\frac {\pi}{2}} \frac{1}{2} \sin(\theta) d\theta$ Then, the area of the sample space can be found by multiplying the length of the rectangle by the height. $\frac {1}{2} \frac {\pi}{2} = \frac {\pi}{4}$ The probability of a hit can be calculated by taking the number of total ways an intersection can occur over the total number possible outcomes (the number of trials). For needle drops, the probability is proportional to the ratio of the two areas in this case because each possible value of θ and d is equally probable. The probability of an intersection is $P_{hit} = \cfrac{ \frac{1}{2} }{\frac{\pi}{4}} = \frac {2}{\pi} = .6366197...$ #### Using Random Samples to Approximate Pi The original goal of the Buffon's needle method, approximating π, can be achieved by using probability to solve for π. If a large number of trials is conducted, the proportion of times a needle intersects a line will be close to the probability of an intersection. That is, the number of line hits divided by the number of drops will equal approximately the probability of hitting the line. $\frac {\mbox{number of drops}}{\mbox{number of hits}} \approx P_{hit} = \frac {2}{\pi}$ So $\frac {\mbox{number of drops}}{\mbox{number of hits}} \approx \frac {2}{\pi}$ Therefore, we can solve for π: $\pi \approx \frac {2 * {\mbox{number of drops}}}{\mbox{number of hits}}$ # Why It's Interesting #### Monte Carlo Methods The Buffon's needle problem was the first recorded use of a Monte Carlo method. These methods employ repeated random sampling to approximate a probability, instead of computing the probability directly. Monte Carlo calculations are especially useful when the nature of the problem makes a direct calculation impossible or unfeasible, and they have become more common as the introduction of computers makes randomization less laborious. π is a mathematical constant whose value is the ratio of any circle's circumference to its diameter in Euclidean space. π is an irrational number, which means that its value cannot be expressed exactly as a fraction a/b, where a and b are integers. Since π is irrational, mathematicians have been challenged with trying to determine increasingly accurate approximations. In the past 50 years especially, improvements in computer capability allow mathematicians to determine more decimal places. Nonetheless, better methods of approximation are still desired. A recent study conducted the Buffon's Needle experiment to approximate π using computer software. The researchers administered 30 trials for each number of drops, and averaged their estimates for π. They noted the improvement in accuracy as more trials were conducted. These results show that Buffon's Needle approximation is a relatively tedious process because the estimates converge towards π rather slowly compared to other computer generated techniques. Nonetheless, the intriguing relationship between the probability of a needle's intersection and the value of π has attracted mathematicians to study the Buffon's Needle method since its introduction in the 18th century. Buffon's approach has since been used in studies ranging from the behavior of ants to the Manhattan Project of developing atomic bombs in WWII. The problem remains relevant as variations can make the its accessible to a wide range of mathematical abilities. The method can be found everywhere from a simple demonstration in a middle school classroom to an exam for graduate students. #### Generalization of the problem The Buffon’s needle problem has been generalized so that the probability of an intersection can be calculated for a needle of any length and paper with any spacing. It has been proven that for a needle shorter than the distance between the lines, the probability of a intersection is $\frac {2l}{\pid}$. This equation makes sense when we consider the normal case, where l =1 and d =1, so these variables disappear and the probability is $\frac {2}{\pi}$. The generalization of the problem is useful because it allows us to examine the relationship between length of the needle, distance between the lines, and probability of an intersection. The variable for length is in the numerator, so a longer needle will have a greater probability of an intersection. The variable for distance is in the denominator, so greater space between lines will decrease the probability. To see how a longer needle will affect probability, follow this link: http://whistleralley.com/java/buffon_graph.htm #### Needles in Nature Applications of the Buffon's Needle method are even found naturally in nature. The Centre for Mathematical Biology at the University of Bath found uses of the Buffon's Needle algorithm in a recent study of ant colonies. The researchers found that an ant can estimate the size of an anthill by visiting the hill twice and noting how often it recrosses its first path. Ants generally nest in groups of about 50 or 100, and the size of their nest preference is determined by the size of the colony. When a nest is destroyed, the colony must find a suitable replacement, so they send out scouts to find new potential homes. In the study, scout ants were provided with "nest cavities of different sizes, shapes, and configurations in order to examine preferences" [2]. From their observations, researchers were able to draw the conclusion that scout ants must have a method of measuring areas. A scout initially begins exploration of a nest by walking around the site to leave tracks. Then, the ant will return later and walk a new path that repeatedly intersects the first tracks. The first track will be laced with a chemical that causes the ant to note each time it crosses the original path. The researchers believe that these scout ants can calculate an estimate for the nest's area using the number of intersections between its two visits. "In effect, an ant scout applies a variant of Buffon's needle theorem: The estimated area of a flat surface is inversely proportional to the number of intersections between the set of lines randomly scattered across the surface." [7] This idea can be related back to the generalization of the problem by imagining if the parallel lines were much further apart. A larger distance between the two lines would mean a much smaller probability of intersection. We can see in case 3 that when the distance between the lines is greater than the length of the needle, even very large angle won’t necessarily cause an intersection. The ants can measure the size of their hill using a related and fairly intuitive method: If they are constantly intersecting their first path, the area must be small. If they rarely reintersects the first track, the area of the hill must be much larger so there is plenty of space for a non-intersecting second path. This natural method of random motion in nature allows the ants to gauge the size of their potential new hill regardless of its shape. Scout ants are even able to asses the area of a hill in complete darkness. The animals show that algorithms can be used to make decisions where an array of restrictions may prevent other methods from being effective. The Buffon's needle problem began the study of geometric probability. The study of geometric probability allowed mathematicians to calculate probability of events without counting each possible outcome which paved the way for study of complex events like ants searching for new nests. # References [4] The Number Pi. Eymard, Lafon, and Wilson. [5] Monte Carlo Methods Volume I: Basics. Kalos and Whitlock. [6] Heart of Mathematics. Burger and Starbird
13. Equations with absolute values Earlier we learned \|±a\|=a \|a\|, also called the "absolute value" of a, returns the value of a after removing its sign. So, how do we now solve an equation of the form \$\|x+3\|=5\$? One way would be to square both sides, since squaring also effectively removes a negative sign. (squares of negative and positive numbers are both positive) This gives: \$\|x+3\|^2=5^2\$ or \$x^2+6x+9=25\$ or \$x^2+6x-16=0\$ \$x^2+6x-16=(x+8)(x-2)=0\$, from which we get \$x=-8\$ or \$x=2\$ Substituting, we see \$\|-8+3\|=\|-5\|=5\$ and \$\|2+3\|=\|5\|=5\$ e.g. solve: \$\|x+3\|<5\$ Squaring both sides, \$\|x+3\|^2<5^2\$ or \$x^2+6x+9<25\$ or \$x^2+6x-16<0\$ This gives, \$x^2+6x-16=(x+8)(x-2)<0\$, which means only one of \$(x+8)\$ or \$(x-2)\$ is less than 0 (for their product to be negative) if \$(x+8)<0\$, then \$x<-8\$, and \$x-2\$ is also negative, which wouldn't work. so \$(x-2)<0\$, in which case \$x<2\$ and \$(x+8)>0\$ or \$x>-8\$, which gives us a range of \$-8<x<2\$ as the desired solution.
# How do you simplify (-2x^-5) x^7? Jul 1, 2016 $= - 2 {x}^{2}$ #### Explanation: the above $\left(- 2 {x}^{- 5}\right) {x}^{7}$ can be written like this : $= - 2 \cdot {x}^{- 5} \cdot {x}^{7}$] {so here we have two identical variables and they are being multiplied so we will add there exponents like this} $= 2 \cdot {x}^{- 5 + 7}$ $= - 2 \cdot {x}^{2}$ $= - 2 {x}^{2}$ {remember when same variables with different exponents are multiplies there exponents are added and when they are divided there exponents are subtracted}
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 # NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 are the part of NCERT Solutions for Class 11 Maths. In this post, you will find the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5. ## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 #### Ex 1.5 Class 11 Maths Question 1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = (2, 4, 6, 8} and C = {3, 4, 5, 6}. Find (i) A’ (ii) B’ (iii) (A C)’ (iv) (A B)’ (v) (A’)’ (vi) (B – C)’ Solution. Here U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6} (i) A’= U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4} = {5, 6, 7, 8, 9} (ii) B’ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9} (iii) A C = {1, 2, 3, 4} {3, 4, 5, 6} = (1, 2, 3, 4, 5, 6} (A C)’ = U – (A C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6} = {7, 8, 9} (iv) A B = {1, 2, 3, 4} {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8} (A B)’ = U – (A B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8} = {5, 7, 9} (v) We know that A’ = {5, 6, 7, 8, 9} (A’)’ = U – A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9} = {1, 2, 3, 4} (vi) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8} (B – C)’ = U – (B – C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8} = {1, 3, 4, 5, 6, 7, 9}. #### Ex 1.5 Class 11 Maths Question 2. If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets: (i) A = {a, b, c} (ii) B = {d, e, f, g} (iii) C = {a, c, e, g} (iv) D = {f, g, h, a} Solution. (i) A’ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c} = {d, e, f, g, h} (ii) B’ = U – B = {a, b, c, d, e, f, g, h} – {d, e, f, g} = {a, b, c, h} (iii) C’ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g} = {b, d, f, h} (iv) D’ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a} = {b, c, d, e} #### Ex 1.5 Class 11 Maths Question 3. Taking the set of natural numbers as the universal set, write down the complements of the following sets: (i) {x: x is an even natural number} (ii) {x: x is an odd natural number} (iii) {x: x is a positive multiple of 3} (iv) {x: x is a prime number} (v) {x: x is a natural number divisible by 3 and 5} (vi) {x: x is a perfect square} (vii) {x: x is a perfect cube} (viii) {x: x + 5 = 8} (ix) (x: 2x + 5 = 9) (x) {x: x ≥ 7} (xi) {x: x W and 2x + 1 > 10} Solution. #### Ex 1.5 Class 11 Maths Question 4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that (i) (A B)’ = A’ ∩ B’ (ii) (A ∩ B)’ = A’ B’ Solution. #### Ex 1.5 Class 11 Maths Question 5. Draw appropriate Venn diagram for each of the following: (i) (A B)’ (ii) A’ ∩ B’ (iii) (A ∩ B)’ (iv) A’ B’ Solution. #### Ex 1.5 Class 11 Maths Question 6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’? Solution. Here U = {x: x is a triangle} A = {x: x is a triangle and has at least one angle different from 60°} A’ = U – A = {x: x is a triangle} – {x: x is a triangle and has at least one angle different from 60°} = {x: x is a triangle and has all angles equal to 60°} = {x: x is an equilateral triangle} #### Ex 1.5 Class 11 Maths Question 7. Fill in the blanks to make each of the following a true statement: (i) A A’ = ……. (ii) Ï†’ ∩ A = ……. (iii) A ∩ A’ = ……. (iv) U’ ∩ A = ……. Solution. (i) A A’= U (ii) Ï†’ ∩ A = U ∩ A = A (iii) A ∩ A’ = Ï† (iv) U’ ∩ A = Ï† ∩ A = Ï†
# Natural Numbers are Comparable ## Theorem Let $\N$ be the natural numbers, defined as the minimal infinite successor set $\omega$. Let $m, n \in \N$. Then exactly one of the following is the case: $(1): \quad m \in n$ $(2): \quad m = n$ $(3): \quad n \in m$ That is, two natural numbers are always comparable by the ordering $\le$ where: $m \le n \iff \begin{cases} m = n & \text{or}\\ m \in n & \end{cases}$ ## Proof Proof by induction: For each $n \in \omega$, let $S \left({n}\right)$ be the set of all $m \in \omega$ which are comparable with $n$. Let $S$ be the set of all $n \in \N$ for which $S \left({n}\right) = \omega$. Thus the proof is equivalent to demonstrating that $S = \omega$. ### Basis for the Induction First consider $S \left({0}\right)$. Clearly $0 \in S \left({0}\right)$ as $0 = 0$. Let $m \in S \left({0}\right)$. Then $m \notin 0$ as $0 = \varnothing$. So either: $(1): \quad 0 = m$, in which case $0 \in m^+$ by definition of successor set or: $(2): \quad 0 \in m$, in which case, because $m \in m^+$ by definition of successor set, again $0 \in m^+$. In all cases, $m \in S \left({0}\right) \implies m^+ \in S \left({0}\right)$. So $S \left({0}\right) = \omega$ by induction. This is the basis for the induction. ### Induction Hypothesis The induction hypothesis is that: $S \left({k}\right) = \omega$ for some $k \in \omega$. It is now necessary to show that it follows that: $S \left({k^+}\right) = \omega$ ### Induction Step This is the induction step: Consider the set $S \left({k^+}\right)$, given that $S \left({k}\right) = \omega$. From the basis for the induction, we have that $k^+ \in S \left({0}\right)$. That is, $k^+$ is comparable with $0$. So $0$ is comparable with $k^+$ and so $0 \in S \left({k^+}\right)$. Suppose $m \in S \left({k^+}\right)$. Then either: $(1): \quad k^+ \in m$, in which case $k^+ \in m^+$ or: $(2): \quad k^+ = m$, in which case the same applies: $k^+ \in m^+$ or: $(3): \quad m \in k^+$. In case $(3)$, either: $(3 a): \quad m = k$, in which case $m^+ = k^+$ or: $(3 b): \quad m \in k$. Case $(3 b)$ is treated as follows. We have that $m^+ \in S \left({k}\right)$ by the induction hypothesis. Therefore, either: $(4 a): \quad k \in m^+$ or: $(4 b): \quad k = m^+$ or: $(4 c): \quad m^+ \in k$. $(4 a)$ is not compatible with $m \in k$, because either: $k \in m$ or $k = m$ and so $k \subseteq m$ which contradicts Finite Ordinal is not Subset of one of its Elements. Both $(4 b)$ and $(4 c)$ imply that $m^+ \in n^+$. Hence $S \left({k^+}\right) = \omega$. It follows by the Principle of Mathematical Induction that $S = \omega$. $\Box$ It then follows from Finite Ordinal is not Subset of one of its Elements that it is not possible for more than one of: $(1): \quad m \in n$ $(2): \quad m = n$ $(3): \quad n \in m$ to be true at the same time. $\blacksquare$
# Slope-Intercept Form of Straight Lines • Last Updated : 21 Dec, 2020 Straight lines can be viewed as a point extended indefinitely in any two opposite directions. A straight line is one that has no curves and covers an infinite number of points. ### Properties of a straight line • Infinite number of straight lines pass through a point • Infinite numbers of planes contain a straight line (when viewed in 3-Dimensions) • A straight line has infinite number of points • Line segment joining any two points on a straight line lies on the straight line ### Basic Terminologies • Point: A point is a dimensionless geometrical figure which has a location (with respect to some reference e.g: points in a coordinate system). • Plane: A flat surface of the infinite area. It is 2-Dimensional. • Ray: A point when extended indefinitely in any one direction in a plane forms a ray. • Line segment: The shortest curve joining two points is called a line segment. Straight lines are studied in 2-D and 3-D. We are going to discuss straight lines in 2-D and for this purpose, we use the coordinate system. ### Coordinate axes Two perpendicular lines which divide the plane (of our consideration) into 4 regions called quadrants. Let AB be a straight line in this plane. Let the angle between line AB and the X-axis be w°. Then line AB can be: • Parallel to X-axis, w = 0° • Parallel to Y-axis, w = 90° • At an angle w° with the X-axis (see Fig. below), 0° < w° < 180° ### Slope of a Straight Line It is defined as the tangent of the angle (here tan(w)) between the line and the X-axis as shown below: slope of line AB = tan(w) = (0 – t) / (s – 0) = t/s i.e. if (x1, y1) and (x2, y2) are two points on line AB then, slope of line AB = tan(w) = (y2 – y1) / (x2 – x1) = (y1 – y2) / (x1 – x2) = distance between the y coordinate of both points / distance between the x coordinate of both points ### X-intercept The point of intersection of a line and the x-axis is called the x-intercept. Example: Here, s is the x-intercept of line AB ### Y-intercept The point of intersection of a line and the y-axis is called the y-intercept. Example: Here, t is the y-intercept of line AB ### Equations Of Straight Lines Straight lines can be represented in various forms: • Slope-intercept form • Point-slope form • Two-point form, etc. We are going to discuss (i) slope-intercept from keeping the above figure in mind. Equation of the straight line in slope-intercept form is: y = slope * x + c y = tan(w) * x + c y = mx + c Here m = tan(w) = slope of line AB and c is the y-intercept. ### Sample Problems on Straight Lines Now, we can easily do the following tasks: Problem 1: Write the equation of a straight line whose slope is p and y-intercept is g. Solution: y = p * x + g Problem 2: Write the equation of a straight line whose slope is 1/2 and y-intercept is 11. Solution: y = (1/2) * x + 11 Problem 3: Write the equation of a straight line that passes through the origin and has slope 1/6. Solution: Since the line passes through the origin, Y-intercept = 0 y = (1/6) * x + 0 y = (1/6)x, which is the required equation. Problem 4: Find the slope and y-intercept of the line represented by the equation y = tx + 4 Solution: The equation of the given line is in slope-intercept form. Thus, slope of the given line is t and its y-intercept is 4. Problem 5: Find the slope and y-intercept of the line represented by the equation y + 4 = x Solution: The given equation is not in slope-intercept form. Therefore, we convert it to that form. y = x – 4 y = 1 * x – 4 Thus, slope of the given line is 1 and its y-intercept is -4. Problem 6: Find the slope and y-intercept of the line represented by the equation x = y/q Solution: The given equation is not in slope-intercept form. Therefore, we convert it to that form. y = qx y = qx + 0 Thus, slope of the given line is q and its y-intercept is 0. Problem 7: Find the slope and y-intercept of the line represented by the equation x = d Solution: The given equation is not in slope-intercept form. Moreover, the variable y is not present in the equation i.e. the equation of this line is independent of the value of variable y. Equations like this represent lines which are parallel to y-axis. For lines that are parallel to y-axis w = 90°, thus the slope of these lines is not defined because tan(90°) is not defined. Since, the line is parallel to y-axis, it never intersects the y-axis and hence does not have y-intercept. Problem 8: Find the slope and y-intercept of the line represented by the equation y = h Solution: The given equation is not in slope-intercept form. Moreover, the variable x is not present in the equation i.e. the equation of this line is independent of the value of variable x. Equations like this represent lines which are parallel to x-axis. Given equation can be rewritten as: y = 0x + h, which is the slope intercept form of the given line. Thus, slope of given line is 0 and its y-intercept is h. ### Sample Word Problems Problem 1: The base fare for riding a taxi is Rs 30.00 for the first 2 km. After, this the rider is charged Rs 4 per km. Calculate the fare for traveling the distance of 14 km by forming a linear equation. Solution: Base fare = Rs 30, this means that Rs 30 is the minimum fare. Therefore, the fare for the distances from 1m to 2000m (2km) is Rs 30. Fare for distances after 2 km is Rs. 4/km. Let the total distance traveled be d km and total cost be Rs c. Charge up to the first 2 km charge is Rs 30.00. Charge for the rest of the (d-2) km charge is Rs 4 * (d – 2). Therefore, total fare is given by the equation, c = 4 * (d – 2) + 30 c = 4 * d – 8 + 30 c = 4 * d + 22, (which is of the slope intercept form) Given, d = 14km, Therefore, total cost, c = 4 * 14 + 22 = 56 + 22 = 78 Thus, the fare for travelling the distance of 14km is Rs 78. Problem 2: The base fare for riding a taxi is Rs 30.00 for the first 2 km. After, this the rider is charged Rs 4 per 600m. Calculate the fare for traveling the distance of 14 km by forming a linear equation. Solution: Base fare = Rs 30, this means that Rs 30 is the minimum fare. Therefore, the fare for the distances 1m to 2000m (2km) is Rs 30. Fare for distances after 2 km is Rs4 per 600m. Let the total distance traveled be d km and total cost be Rs c. Up to the first 2 km charge is Rs 30.00. Remaining distance = (d – 2)km Charge for 600 m = Rs 4 Charge for 1 m = Rs (4/600) ———(unitary method) Charge for 1000m = Rs. (4 / 600) * 1000 = Rs (40/6) Therefore, the charge per km after the first 2 km is Rs (40/6) per km. Charge for the rest of the (d – 2) km is Rs (40/6) * (d – 2). Therefore, total fare is given by the equation, c = (40/6) * (d – 2) + 30 c = (40/6) * d – 20/6 + 30 c = (20/3) * d + (-20 + 180) / 6 c = (20/3) * d + (160/6) c = (20/3) * d + (80/3), (which is of the slope-intercept form) Given, d = 14km, Therefore, total cost, c = (20/3) * 14 + (80/3) = (280/3) + (80/3) = 360/3 = 120 (approx.) Thus, the total fare for traveling the distance of 14km is Rs 120 (approximately). Problem 3: Plane number 13 owned by Earth Airlines is undergoing repair. The engineer is informed that the airlines could suffer a loss of Rs 1300 per 12 minutes until it is repaired. The engineer was told that the airlines would not suffer a loss greater than Rs5000 and that the extra amount would be deducted from his salary. Calculate • The maximum time in minutes within which the engineer has to repair the airplane. • The total time taken if the total loss is Rs 6000 Solution: Loss in 12 min = Rs. 1300 Loss in 1 min = Rs. (1300/12) = Rs. (325/3) Therefore, loss per minute is Rs. (325/3). (i) Let the maximum time be t minutes. Therefore, t * (325/3) = 5000 or t = (5000 * 3)/325 or t = 46.15 minutes = 46 minutes (approx.) Thus, the engineer has to repair the airplane within 46 minutes. (ii) Total loss = Rs 6000 (Given) Let the time taken for this loss be k minutes. Time required for loss of Rs(325/3) = 1 min (unitary method) Time required for loss of Rs 1 = 1 * (3/325)min Therefore, time required for loss of Rs 6000 = 6000 * (3/325) minutes = 55.384 minutes Problem 4: X and y are two 2-digit numbers which satisfy the following conditions: • y > x • x + y = m, where m is the 2-digit number obtained by reversing the digits of y • y – x = x – 1 • Tens digit of y – Units digit of y = -2 Find the two numbers. Solution: y – x = x – 1, ——-condition(iii) can be rewritten as y = x + x – 1 y = 2x – 1 ———-(1) Let the tens and units digits of the number y be a and b respectively. This means that y = 10a + b and m = 10b + a From condition(ii), x + y = m x + 10a + b = 10b + a 9a – 9b + x = 0 9(a – b) + x = 0 9 * (-2) + x = 0, ——-condition (iv) -18 + x = 0 Therefore, x = 18 Substituting the value of x in equation (1), y = 2 * 18 – 1 y = 36 – 1 Therefore, y = 35 Problem 5: The sum of the scores of two friends in a subject is 100. The difference of their scores in that same subject is 10. Find their scores. Solution: Let the scores be x and y such that x > y. Given, x + y = 100 —-(1) and x – y = 10 —-(2) Adding equations (1) and (2), x + y + x – y = 100 + 10 2x + 0y = 110 2x = 110 x = 110/2 = 55 Substituting value of x in equation (2) 55 – y = 10 55 – 10 = y y = 45 Therefore, their scores are 45 and 55. My Personal Notes arrow_drop_up
# Warm-Up Solve the following Inequalities: 1. 2. 3. ## Presentation on theme: "Warm-Up Solve the following Inequalities: 1. 2. 3."— Presentation transcript: Warm-Up Solve the following Inequalities: 1. 2. 3. Warm-Up Answers 1. 2. 3. Lesson 6.2 Graphing Inequalities In Slope-Intercept Form It could be greater than (>) or less than (<) or (>) or (<). Graphing linear inequalities with 2 variables The graph of the solutions to a single inequality is called a ____________ because it includes all the points in the coordinate plane that fall on one side of the __________ line. half-plane boundary To Graph an Inequality: You must follow these 5 steps: 1. Put in y = mx + b form. 2. Find m and b. 3. Decide if the line is solid or dotted. dotted line solid line 4.Graph using your m and b 5.Shade above or below the line using: Test Point Method!  (0, 0) is the easiest! But we cannot always test this point so we have to find another point that is clearly above or below the line. If the point you plug in makes a TRUE statement, shade where that point is included. If the point you plug in makes a FALSE statement, shade away from that point. Example 1: 1. Find m and b. m =-2/3 b = (0, 2) Dotted line 2. Is the line dotted or solid? 3. Next, let’s graph so can pick a test point to try! Use (0, 0) as a test point, and plug into the formula Is this a true statement? NO! Which means we shade AWAY from that point… Example 2: 1. m = ¼ 2. b = (0, 3) 3.Solid Line 4.Test Point (0, 0) Which is true so shade where that point exists! Example 3: y < – 3x + 1 Try this one on your own! m = b = (0, ) Dotted or Solid Line? Test Point? Lesson 6.3 Solving and Graphing Inequalities in Standard Form You have two options for graphing in standard form. Solve for y by getting into slope intercept form. Plug in your zeros to get x and y intercepts (assuming equal sign). Then use a test point on either side for shading. Ex. 4 2x + y < 4 Easily Solve for y -2x Graph the y intercept (0,4) Now, graph two more points using the slope (m=-2) Draw a dotted line through the points (It is dotted because there is not an = sign under the inequality) Pick a test point to see which side of the line to shade. Let’s use (4, 5) Plug the ordered pair into the original equation 5<-2(4) + 4. Is this true? Only shade the true side. Since this false we will shade the other side of the line Ex. 5 -2x -4 Pick a test point to see which side of the line to shade. Let’s use (0, 0) Plug the ordered pair into the original equation. Is this true? Only shade the true side. You try! Ex. 6 Ex. 7 Summary/Reflection How will you remember when to use a solid or dotted line? and How will you remember where to shade on a linear inequality? Classwork Worksheet 6.2-6.3 Homework Worksheet 6.2-6.3 Download ppt "Warm-Up Solve the following Inequalities: 1. 2. 3." Similar presentations
# Critical point solver We'll provide some tips to help you select the best Critical point solver for your needs. We can solve math word problems. ## The Best Critical point solver There is Critical point solver that can make the technique much easier. The best math with steps is when you can practice a new skill over and over again. This helps you to remember what’s going on and how to do it correctly. It also gives you the chance to make mistakes so that you can learn from them. When you’re working with steps, it can be very easy to get lost or confused. The best thing to do is take your time and make sure that you understand what you’re doing before moving on. Once you’ve got the hang of it, then you can start making bigger leaps forward in your development. When it comes to math, there are lots of different ways to practice. You could try out some different apps, such as a Sudoku app or a word search game. You could also go through a basic book like “Fifty Simple Things You Can Do To Improve Your Math Skills.” Or you could even try something more hands-on like building a tower out of blocks or designing your own LEGO model. Solve for x right triangles by using a Pythagorean formula. This calculator is useful for determining the length of a side of a right triangle, known as the hypotenuse. The Pythagorean relationship between sides x and y is: The ratio or proportion between sides x and y is given by: Substituting this into the above equation gives: or in other words: This can be simplified further as shown below: Therefore, solving for "x" right triangles involves applying this formula to any right triangle with lengths equal to 1, 2, and 3. If the hypotenuse (AB) is known then "x" can be determined from the equation. For example, if AB = 9 then "x" = 9. On the other hand, if AB = 16 then "x" = 16. For example, if AB = 12 then "x" = 12.
# How do you find the slope and intercept of y=-x-7? Aug 3, 2018 Slope: -1 x-intercept: (-7, 0) y-intercept: (0, -7) #### Explanation: $y = - x - 7$ This equation is in slope-intercept form: Based on the image, we know that the slope is the value multiplied by $x$, so the slope is $- 1$. We know the $y$-intercept is $b$, or $- 7$, so it is at $\left(0 , - 7\right)$. To find the $x$-intercept, plug in $0$ for $y$ and solve for $x$: $0 = - x - 7$ $7 = - x$ $x = - 7$ The $x$-intercept is at $\left(- 7 , 0\right)$.' Hope this helps! Aug 4, 2018 Slope $- 1$, $y$-int $- 7$, $x$-int $- 7$ #### Explanation: The good thing is that this equation is in slope-intercept form $y = m x + b$, with slope $m$ and a $y$-intercept of $b$. By pattern matching, we see that our slope is $- 1$, and our $y$-intercept is $- 7$. We can find our $x$-intercept by setting $y$ equal to zero. We get $- x - 7 = 0 \implies - x = 7 \implies x = - 7$ Therefore, our slope is $- 1$, our $y$-intercept is $- 7$, and our $x$-intercept is $- 7$. Hope this helps!
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Systems of Linear Equations in Three Variables Solve multiple equations by evaluating them in pairs 0% Progress Practice Systems of Linear Equations in Three Variables Progress 0% Solving Linear Systems in Three Variables Guided Practice 1. Is the point, (-3, 2, 1), a solution to the system: x+y+z4x+5y+z3x+2y4z=0=1?=8 2. Solve the following system using linear combinations: 5x3y+zx+6y4z8xy+5z=1=17=12 3. Solve the following system using linear combinations: 2x+yzx2y+z6x+3y3z=3=5=6 1. Check to see if the point satisfies all three equations. Equation 1: (3)+(2)+(1)=3+2+1=0\begin{align}(-3)+(2)+(1)=-3+2+1=0 \end{align} Equation 2: 4(3)+5(2)+(1)=12+10+1=1\begin{align}4(-3)+5(2)+(1)=-12+10+1=-1 \end{align} Equation 3: 3(3)+2(2)4(1)=9+44=98\begin{align}3(-3)+2(2)-4(1)=-9+4-4=-9 \neq -8 \end{align} Since the third equation is not satisfied by the point, the point is not a solution to the system. 2. Combine the first and second equations to eliminate z\begin{align}z\end{align} . Then combine the first and third equations to eliminate z\begin{align}z\end{align} . 4(5x3y+z=1)20x12y+4z=4x+6y4z=17 x+6y4z=17 21x6y=21 Result from equations 1 and 2: 21x6y=21\begin{align}21x-6y=-21\end{align} 5(5x3y+z=1) 25x+15y5z=58xy+5z=12 8xy+5z=1217x+14y=17 Result from equations 1 and 3: 17x+14y=17\begin{align}-17x+14y=17\end{align} Now we have reduced our system to two equations in two variables. We can eliminate y\begin{align}y\end{align} most easily next and solve for x\begin{align}x\end{align} . 7(21x6y=21) 147x42y=1473(17x+14y=17) 51x+42y=51 96x=96 x=1 Now find y\begin{align}y\end{align} : 21(1)6y216y6yy=21=21=0=0 Finally, we can go back to one of the original three equations and use our y\begin{align}y\end{align} and z\begin{align}z\end{align} values to find x\begin{align}x\end{align} . 5(1)3(0)+z5+zz=1=1=4 Therefore the solution is (-1, 0, 4). Don’t forget to check your answer by substituting the point into each equation. Equation 1: 5(1)3(0)+(4)=5+4=1\begin{align}5(-1)-3(0)+(4)=-5+4=-1 \end{align} Equation 2: (1)+6(0)4(4)=116=17\begin{align}(-1)+6(0)-4(4)=-1-16=-17 \end{align} Equation 3: 8(1)(0)+5(4)=8+20=12\begin{align}8(-1)-(0)+5(4)=-8+20=12 \end{align} 3. Combine equations 1 and two to eliminate z\begin{align}z\end{align} . Then combine equations 2 and 3 to eliminate z\begin{align}z\end{align} . 2x+yz=3x2y+z=5 3xy=8 Result from equations 1 and 2: 3xy=8\begin{align}3x-y=8\end{align} 3(x2y+z=5)3x6y+3z=156x+3y3z=66x+3y3z=69x3y=21 Result from equations 2 and 3: 9x3y=21\begin{align}9x-3y=21\end{align} Now we have reduced our system to two equations in two variables. Now we can combine these two equations and attempt to eliminate another variable. 3(3xy=8) 9x+3y=24 9x3y=219x2y=21 0=3 Since the result is a false equation, there is no solution to the system. Explore More 1. Is the point, (2, -3, 5), the solution to the system: 2x+5yz5xy3z3x+2y+4z=16=2=20 1. Is the point, (-1, 3, 8), the solution to the system: 8x+10yz11x+4y3z2x+3y+z=14=23=10 1. Is the point, (0, 3, 5), the solution to the system: 5x3y+2z7x+2yzx+4y3z=1=1=3 1. Is the point, (1, -1, 1), the solution to the system: x2y+2z6x+y4z4x3y+z=5=1=8 Solve the following systems in three variables using linear combinations. 1. . 3x2y+z4x+y3z9x2y+2z=0=9=20 1. . 11x+15y+5z3x+4y+z7x+13y+3z=1=2=3 1. . 2x+y+7z3x2yz4xy+3z=5=1=5 1. . x+3y4z2x+5y3zx3y+z=3=3=3 1. . 3x+2y5z3x+2y+5z6x+4y10z=8=8=16 1. . x+2yz2x+4y+z3xy+8z=1=10=6 1. . x+y+z2xyz4x+y+z=3=6=0 1. . 4x+y+3z8x+2y+6z3x3yz=8=15=5 1. . 2x+3yzx2y+3zx+y2z=1=4=3 1. . x3y+4zx+2y5z2x+5y3z=14=13=5 1. . x+y+zx+yz2x+2y+z=3=3=6 Vocabulary Language: English Consistent Consistent A system of equations is consistent if it has at least one solution. eliminating a variable eliminating a variable Eliminating the variable means making the coefficient of the variable equal to zero thereby removing that variable from that particular system and reducing the number of unknown quantities. Inconsistent Inconsistent A system of equations is inconsistent if it has no solutions. linearly independent linearly independent Three equations are linearly independent if each equation cannot be produced by a linear combination of the other two. Systems that are linearly independent will have just one solution. scaling a row scaling a row Scaling a row means multiplying every coefficient in the row by any number you choose (besides zero). This can be helpful for getting coefficients to match so that they can be eliminated. system of equations system of equations A system of equations is a set of two or more equations.
# CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 2 POLYNOMIALS EX – 2.4 ## Polynomials Question 1. Determine which of the following polynomials has (x +1) a factor: (i) x+x2 + 1 (ii) x4 + x3+x2+x+1 (iii) x4 + 3x3 + 3x2 + x + 1 (iv) x3 -x2 -(2 $\sqrt { 2 }$ )x + $\sqrt { 2 }$ Solution: The zero of x +1 is -1. (i) Let p(x) = x3 + x2 + x +1 Then, p(-1) = (-1)3 + (-1)2 + (-1) +1 = -1+1-1+1 => p(-1)= 0 So, by the Factor theorem (x +1) is a factor of x3 + x2 + x+1. (ii) Let p(x)= x4 + x3 + x2 + x +1 Then, p(-1) = (-1)4 + (-1)3 + (-1)2 + (-1) +1 =1-1+1-1+1 => p(-1) = 1 So, by the.Factor theorem (x +1) is not a factor of x4 + x3 + x2 + x +1 (iii) Let p(x) = x4 + 3x3 + 3x2 + x +1 Then, p(-l) = (-1)4 + 3(-1)3 + 3(-1)2 + (-1) +1 =1-3+3-1+1 => P(-1) = 1 So, by the Factor theorem (x +1) is not a factor of x4 + 3x3 + 3x2 + x +1 (iv) Let p(x) = x3-x2-(2 + $\sqrt { 2 }$ )x + $\sqrt { 2 }$ Then, p(-1) = (-1)3 – (-1)2 – (2 + $\sqrt { 2 }$ )(-1) + $\sqrt { 2 }$ = —1—1+2 + $\sqrt { 2 }$ + $\sqrt { 2 }$ = 2$\sqrt { 2 }$ So, by the Factor theorem (x + 1) is not a factor ofx3 -x2 -(2 + $\sqrt { 2 }$ x + $\sqrt { 2 }$ Question 2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases : (i) p(x) – 2x3 + x2 -2x-1, g(x) = x + 1 (ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x3– 4x2 + x + 6, g(x) = x – 3 Solution: (i) The zero of g(x) = x +1 is x = -1. Then, p(-1)= 2(-1)3 +(-1)2 -2(-1)-1 [∵ p(x) = 2x3 + x2 -2x-1] = -2 +1+2-1 ⇒ p(-1) = 0 Hence, g(x) is a factor of p(x). (ii) The zero of g(x) = x + 2 is -2. Then, p(-2) = (-2)3 + 3(-2)2 + 3(-2) +1 [∵ p(x) = x3 + 3x2 + 3x +1] = -8+12-6 +1 =>p(-2)=-1 Hence, g(x) is not a factor of p(x). (iii) The zero of g(x) = x – 3 is 3. Then,p(3) = 33 – 4(3)2 +3 + 6 [∵ p(x) = x3 – 4x2 + x + 6] = 27-36 +3 + 6 =» p(3) = 0 Hence, g(x) is a factor of p(x). Question 3. Find the value of k, if x -1 is a factor of p(x) in each of the following cases: (i) p(x) = x2+x + k . (ii) p(x) = 2x2 + kx + V2= (iii) p(x) = kx2 -42x + 1 (iv) p(x) = kx2 -3x + k Solution: Question 4. Factorise: (i) 12x2 – 7x+1 (ii) 2x2+7x+3 (iii) 6x2+5x-6 (iv) 3x2-x-4 Solution: Question 5. Factorise: (i) x3-2x2-x+2 (ii) x3-3x2-9x-5 (iii) x3-13x2+32x+20 (iv) 2y3+y2-2y-1 Solution: (i) Let p(x) = x3-2x2-x+2, constant term of P(x) is 2. Factors of 2 are ± 1 and ± 2. Now, p(1)=13-2(1)2-1+2 =1-2-1+2 By trial we find that p(l) = 0, so (x -1) is a factor of p(x). So, x3-2x2-x+2= x3-x2-x2 + x-2x+2 = x2(x-1)-x(x-1)-2(x-1) = (x—1)(x2 -x-2) = (x -1)(x2 -2x + x -2) = (x -1) [x(x -2) + l(x -2)] = (x-1)(x-2)(x+1) (ii) Let p(x)= x3 -3x2 -9x-5 By trial, we find that p(5) = (5)3 – 3(5)2 – 9(5) – 5 = 125 – 75 – 45 – 5 = 0 So, (x – 5) is a factor of p(x). So, x3-3x2-9x-5=x3-5x2+2x2 -10x + x – 5 = x2 (x – 5) + 2x(x – 5) + 1(x – 5) = (x – 5) (x2+2x+1) = (x – 5) (x2 + x + x +1) = (x + 5) [x(x +1) + 1(x +1)] = (x-5)(x +1) (x +1) = (x – 5)(x +1)2 (iii) Let p(x)= x3 +13x2 +32x +20 By trial, we find that p(-1) = (-1)3 +13(-1)2 + 32(-1) + 20 = -1 +13 -32 + 20 = -33 + 33 = 0 So (x +1) is a factor of p(x). So, x3 + 13x2 + 32x + 20 = x3 + x2 + 12x2 + 12x + 20x + 20 = x2 (x +1) + 12x(x +1) + 20(x +1) = (x+1)(x2 + 12x + 20) = (x+1)(x2 + 10x + 2x + 20) = (x +1)[x(x +10) + 2(x +10)] = (x +l)(x +10)(x + 2) (iv) Let p(y) = 2y3 + y2 – 2y -1 By trial we find that p(l) = 2(1)3 + (1)2 -2(1) -1, =2+1-2-1=0 So (y -1) is a factor of p(y). So, 2y3 + y2 -2y-1 = 2y3 -2y2 +3y2 -3y + y-1 = 2y2(y-1) + 3y(y-1) + 1(y-1) = (y – 1)(2y2 +3y + 1) = (y – 1)(2y2 + 2y + y + 1) = (y -1) [2y(y +1) + l(y +1) = (y-1)(y+ l)(2y+ 1)
# Matrix multiplication Matrix multiplication can either refer to multiplying a matrix by a scalar, or multiplying a matrix by another matrix. If necessary, refer to the matrix notation page for a review of the notation used to describe the sizes and entries of matrices. ## Matrix-Scalar multiplication The first kind of matrix multiplication is the multiplication of a matrix by a scalar, which will be referred to as matrix-scalar multiplication. A scalar is a number that makes things larger, smaller, or even negative (think of a negative scalar as "pointing backwards" or "flipping" something in the opposite direction). Given an m × n matrix, A, and a scalar, c, the matrix, cA, or Ac, is the m × n matrix whose i,jth entry is c times the i,jth entry of A. In other words, for all i=1, ..., m and j = 1, ..., n. For example, ### Properties of Matrix-Scalar multiplication • Commutative: cA = Ac • Associative: (cd)A = c(dA) • Distributive over matrix addition: c(A + B) = cA + cB • Distributive over scalar addition: (c + d)A = cA + dA ## Matrix-Matrix multiplication Multiplying two matrices involves the use of an algebraic operation called the dot product. A vector can be seen as a 1 × matrix (row vector) or an n × 1 matrix (column vector). To use the dot product, the vectors must be of equal length, meaning that they have the same number of entries. Given two matrices, A and B, if and the dot product of A and B is: Note that in the equation above, the "·" represents the dot product, not multiplication. To multiply matrices, the dot product of the corresponding rows and columns of the matrices being multiplied are computed to determine the entries of the resulting matrix; this is described in detail below. It follows that, in order to multiply two matrices, A and B, the number of columns of A must equal the number of rows of B. For a product matrix, AB, to be defined in general, A must be m × n, and B must be n × p for some positive integers m, n, and p. If A is m × n, and B is n × p, the product matrix, AB, is the m × p matrix such that the i,jth entry of AB, denoted (AB)ij, is the dot product of the ith row vector of A with the jth column vector of B Therefore, (AB)ij is given by the formula: = = = for all i = 1,2 ..., m and j = 1,2, ..., p (i runs through the rows of A, of which there are m, and j runs through the columns of B, of which there are p). The row and column highlighted in blue in the matrices below represent the ith row and jth column of the matrices being multiplied: Thus, the (AB)ij entry is the sum of the products of the entries of row i of matrix A and column j of matrix B: Example Multiply A and B if A is the following 2 × 3 matrix and B is the following 3 × 2 matrix: = = = = corresponds to: = = = corresponds to: = = = = corresponds to: and = = = = corresponds to: Assembling these entries into the matrix product AB yields then AB is not defined since A has 3 columns and B has 2 rows, and 3 ≠ 2. ### Properties of matrix multiplication • Commutative with scalars (i.e. matrix-scalar multiplication above): If A is m × n, B is n × p, and c is a scalar, cAB = AcB = ABc. Note: matrix-matrix multiplication is not commutative. For example, matrix A × matrix B does not necessarily equal matrix B × matrix A and more typically does not. • Associative: If A is m × n, B is n × p, C is p × q, then (AB)C = A(BC)
Euler equations characteristic polynomial $$\left[r(r-1) + a r + b\right] x^r = 0.$$ $$\tag{2} r(r-1) + a r + b = 0.$$ (2) 這個方程就是 Euler equations 的特徵多項式. 理論上如果 (2) 的根是 $r_1$, 那 $y = x^{r_1}$ 就是方程式的解. A. 兩相異實根 $$\tag{3} y(x) = c_1 x^{r_1} + c_2 x^{r_2},$$ 其中 $c_1$, $c_2$ 為常數. 備註: $$x^r = e^{r\ln x}.$$ 這樣即使 $r$ 是無理數, 我們也可以輕易地計算 $x^r$. B. 兩複數根 \begin{align} x^{\lambda + i\mu} &= e^{(\lambda + i\mu)\ln x} \\ &= e^{\lambda \ln x}\left[\cos(\mu\ln x) + i\sin(\mu\ln x)\right] \\ &= x^r\cos(\mu\ln x) + ix^r\sin(\mu\ln x). \end{align} $$\tag{4} y(x) = c_1 x^{\lambda}\cos(\mu\ln x) + c_2 x^{\lambda}\sin(\mu\ln x),$$ 其中 $c_1$, $c_2$ 為常數. C. 一重根 $$\tag{5} x^2 y’’ + (1 - 2r_1) xy’ + r^2_1 y = 0.$$ $$\frac{d}{dx} y_1 = r_1 x^{r_1-1} = \frac{r_1}{x} x^{r_1} = \frac{r_1}{x} y_1.$$ $$\tag{6} \left(x\frac{d}{dx} - r_1\right) y = 0.$$ Observation 1 $$\tag{7} \left(x\frac{d}{dx} - r_1\right)\left(x\frac{d}{dx} - r_1\right) y = 0.$$ Observation 2 $$\tag{8} \left(x\frac{d}{dx} - r_1\right) y_2 = cy_1,$$ 其中 $c$ 是個常數. 那我們就可以得到 $$\left(x\frac{d}{dx} - r_1\right)\left(x\frac{d}{dx} - r_1\right) y_2 = \left(x\frac{d}{dx} - r_1\right) cy_1 = 0.$$ Observation 3 $$\tag{9} xu’ = cu.$$ $$\tag{10} y(x) = c_1 x^{r_1} + c_2 x^{r_1}\ln(x),$$ 其中 $c_1$, $c_2$ 為常數. Te-Sheng Lin (林得勝) Associate Professor The focus of my research is concerned with the development of analytical and computational tools, and further to communicate with scientists from other disciplines to solve engineering problems in practice.
# 2016 AMC 12A Problems/Problem 11 ## Problem Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents? $\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$ ## Solution Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act. Let $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act. From the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$. Adding these equations together, we get $2(a + b + c) + ab + bc + ac = 136$. Since there are a total of $100$ students, $a + b + c + ab + bc + ac = 100$. Subtracting these equations, we get $a + b + c = 36$. Our answer is $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}$ ## Solution 2 An easier way to solve the problem: Since $42$ students cannot sing, there are $100-42=58$ students who can. Similarly $65$ students cannot dance, there are $100-65=35$ students who can. And $29$ students cannot act, there are $100-29=71$ students who can. Therefore, there are $58+35+71=164$ students in all ignoring the overlaps between $2$ of $3$ talent categories. There are no students who have all $3$ talents, nor those who have none $(0)$, so only $1$ or $2$ talents are viable. Thus, there are $164-100=\boxed{\textbf{(E) }64}$ students who have $2$ of $3$ talents.
# Less Than vs. Equal To: Improve Math Vocabulary In mathematics, the concepts of less than vs. equal to are foundational for understanding numerical relationships and inequalities. Understanding the exact meaning and proper usage of these symbols is crucial because even a small misinterpretation can completely change the meaning of a mathematical expression. Contents ## The Difference between Less than and Equal to ### Key Takeaways • Less than (<): This symbol indicates that one number is smaller than another. For example, if we say 3 < 5, we mean that 3 is less than 5. • Equal to (=): The equality symbol shows that two values are the same. So when we express that 4 = 4, it’s clear that both sides of the equation represent the identical value. ### Less than vs. Equal to: the Definition What Does Less than Mean? “Less than” is a term we use when one value is smaller than another. In mathematics, we use the symbol < to denote this relationship. For instance, if we say 3 < 5, we’re asserting that 3 is less than 5. Here’s another example: if there are 2 apples and 6 oranges, we express that with 2 < 6, indicating fewer apples than oranges. What Does Equal to Mean? “Equal to” indicates that two values are exactly the same. We signify this with the symbol =. For example, if we have two groups of marbles and each contains exactly 4 marbles, we write this as 4 = 4. Similarly, if a recipe calls for 2 cups of sugar and you measure out precisely 2 cups, you have 2 cups equal to the required amount, represented by 2 = 2. ### Tips to Remember the Difference 1. L Method: Remember that “less than” starts with an “L”; similarly, the symbol “<” resembles an “L”. Hence, “<” means “less than”. 2. Equal Sign Recall: Notice that the equals sign ” = ” consists of two parallel lines. Since the lines are the same length, they symbolize the concept of things being the same, or equal. ## Less than vs. Equal to: Examples ### Usage and Examples Here are some real-life scenarios to demonstrate: 1. Age Restrictions: If a ride requires you to be under 12 years of age, we write that as age < 12. 2. Budget Limits: If our budget for a project is up to \$500, we denote this as budget ≤ \$500. 3. Temperature Thresholds: Saying that water freezes at 0 degrees Celsius or lower is written as temperature ≤ 0°C. Comparison Type Symbol Example Interpretation Less than < 3 < 4 3 is less than 4 Less than or equal to 7 ≤ 7 7 is equal to 7 (hence, ≤ 7) ### Example Sentences Using Less than • Our team completed less than 20 projects this year, marking a decrease from last year’s total. • I have less than \$50 in my wallet, which isn’t enough for the concert ticket. • We need to ensure that the temperature stays less than 25°C to preserve the samples. • The distance between our office and the client’s location is less than 10 kilometers. • It took us less than two hours to organize the warehouse, which was quicker than expected. ### Example Sentences Using Equal to • The amount of coffee in each cup is equal to 300 milliliters for consistency in serving. • Our quarterly earnings this year were surprisingly equal to last quarter’s, despite market fluctuations. • Each participant in the race receives a time slot equal to 15 minutes to complete their practice run. • The total number of votes cast for both candidates was unexpectedly equal to one another. • For our recipe, the ratio of flour to sugar is equal to one, providing the perfect balance for our cookies. ## Related Confused Math Terms ### Less than vs. Fewer than Less than” is typically used for quantities that can’t be counted individually, known as uncountable nouns. On the other hand, “fewer than” is correct when referring to countable nouns. Example sentence: • We used less than a gallon of paint to cover the wall. • We had fewer than twenty participants in the survey. ### Equal to vs. Equivalent to Equal to” conveys that two quantities are the same in number, amount, or degree, often used in mathematics or when comparing measurable attributes. “Equivalent to” implies that two entities are comparable in value or function, even if they aren’t the exact same thing. • The number of apples in this basket is equal to the number in that basket. • Our online subscription service is equivalent to accessing a large library. Related:
# How do you find the equation of a tangent line at a given point? ## How do you find the equation of a tangent line at a given point? 1) Find the first derivative of f(x). 2) Plug x value of the indicated point into f ‘(x) to find the slope at x. 3) Plug x value into f(x) to find the y coordinate of the tangent point. 4) Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line. ### How do you find the equation of the tangent to a parabola? Equation of tangent in slope (m) form In (6) replacing t by 1/m we have y = mx + which is equation of tangent in terms of slope and the point of contact is . Thus if line y = mx + c touches parabola y2 = 4ax we must have c = a/m (comparing equation with y = mx + a/m). READ: What should teachers do when students make mistakes? What is the tangent line equation? The equation of the tangent line can be found using the formula y – y1 = m (x – x1), where m is the slope and (x1, y1) is the coordinate points of the line. What is a tangent of a parabola? A secant of a parabola is a line, or line segment, that joins two distinct points on the parabola. A tangent is a line that touches the parabola at exactly one point. ## How do you find the tangent line using implicit differentiation? Take the derivative of the given function. Evaluate the derivative at the given point to find the slope of the tangent line. Plug the slope of the tangent line and the given point into the point-slope formula for the equation of a line, ( y − y 1 ) = m ( x − x 1 ) (y-y_1)=m(x-x_1) (y−y1)=m(x−x1), then simplify. ### Is the equation of a tangent line the derivative? The derivative of a function gives us the slope of the line tangent to the function at any point on the graph. This can be used to find the equation of that tangent line. READ: Can you get AIDS from a piece of glass? How do you find the equation for the tangent line? 1) Find the first derivative of f(x). 2) Plug x value of the indicated point into f ‘(x) to find the slope at x. 3) Plug x value into f(x) to find the y coordinate of the tangent point. 4) Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line. What is the slope of the tangent line at x = 2? The equation for the slope of the tangent line to f(x) = x2 is f ‘(x), the derivative of f(x). Using the power rule yields the following: Therefore, at x = 2, the slope of the tangent line is f ‘(2). ## How to find the gradient of the tangent of a point? Use the rules of differentiation: To determine the gradient of the tangent at the point \\ (\\left (1;3ight)\\), we substitute the \\ (x\\)-value into the equation for the derivative. Substitute the gradient of the tangent and the coordinates of the given point into the gradient-point form of the straight line equation. READ: Is chairman appropriate for a woman? ### How do you find the normal of a tangent curve? Substitute the gradient of the tangent and the coordinates of the given point into an appropriate form of the straight line equation. Make \\ (y\\) the subject of the formula. The normal to a curve is the line perpendicular to the tangent to the curve at a given point.
### 2-digit Square A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number? ### Consecutive Squares The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false? ### Plus Minus Can you explain the surprising results Jo found when she calculated the difference between square numbers? # Hollow Squares ##### Age 14 to 16 Challenge Level: Kunal from Ecole Internationale de Manosque in France used Charlie's method to find the numbers of dots in the first hollow squares: Pablo from King's College Alicante in Spain and Adithya from Hymers College in the UK both used Charlie's method and some algebra. This is Pablo's working: Since they are two squares, one inside the other, we can express the total number of soldiers as a difference of squares. In this case: $20^2 - 8^2$ and $16^2 -10^2$ A simpler version of the difference of squares is $a^2-b^2 = (a+b)(a-b)$ So: $20^2 - 8^2 = (20+8)(20-8) = 28\times12 = 336$ $16^2 - 10^2 = (16+10)(16-10) = 26\times6 = 156$ Persassy from Annie Gale School in Canada sent in a method for finding hollow squares using $960$ soldiers: 1) The side lengths (large-small) must equal an even number 2) The difference between the large and small square areas must be the number ($960$) Solving (where I started) e.g., the square root of $960$ is $30.98...$ $31\times31=961$ $961-960=1,$ and $961-1=960$ (even number, can have equal "walls" around) $1$ is also a perfect square Then, I would move to the next number for the outside length (in this case $32,$ and would continue with the multiplying ($32\times32=1024$)) REACHING A NUMBER THAT CAN'T BE MADE INTO A HOLLOW SQUARE: e.g., $33$ $33\times33=1089$ $1089-960= 129$ $\sqrt{129}=11.357...$ because it is not a whole number, it isn't an applicable hollow square formation. Adam from Nower Hill High School in the UK also used a detailed trial approach. Click here to see Adam's work. Jonathan from Chairo Christian School in Australia and 10Ma1 at Woodrush High in England also found some solutions for 960 soldiers. This is 10Ma1's work and observations. For 960 soldiers we found 5 solutions for a symmetrical hollow square. These pairs were: 31 and 1 32 and 8 34 and 14 38 and 22 46 and 34 We then looked for patterns and noticed all but 1 pair had a difference in multiples of 4. We then decided that all multiples of 4 would work for the value of $n$ (the total number of soldiers). Tony and Laurenc from Greenacre Public School in Australia, Agathiyan from Hymers College in the UK, Parth, Nicolas and Zawad from Perth Modern School in Australia, Weijie from BSKL in Malaysia and Sumayyah, Anshika, Shyam, Hafsah and Suhasini from Nower Hill High School in the UK all used Alison's method to find hollow squares using $960$ soldiers. Anskih'a work is shown below. At the bottom, when listing the possibilities for the hollow squares, Anshika has forgotten to put $^2$ outside the brackets, but otherwise the work is correct. Pablo used a different method to get to $240$, and Pablo's method explains 10Ma1's assertion that mutliples of $4$ can be divided into symmertrical hollow squares. This is Pablo's diagram and working: If we divide the square up like in the picture, we can say that: \begin{align}(a+b)^2-(b-a)^2 =& 960\\ [(a+b)+(b-a)][(a+b)-(b-a)] =& 960\\ [(a+b)+(b-a)] = b + b + a - a = 2b\\ [(a+b)-(b-a)] = b - b + a + a = 2a\\ (2b)(2a) =& 960\\ 4ab = &960\\ ab = &240\end{align} $$4ab = [\text{Number of Soldiers}]$$From this we can see that only battalions with sizes that are multiples of $4$ can be arranged. It also generalises this problem, so all the possible hollow square formations can be obtained! Adithya from Hymers College in the UK and Emer, Emily, Kayaan, Kanusha and Shyam from Nower Hill High School in the UK continued to use Charlie's method to find hollow squares using $960$ soldiers. Adithya said: Using $x$ and $y$ as integer values of the hollow square lengths, wherein $x>y$, $x^2-y^2=960$. When factorised this gives $(x+y)(x-y)=960$. As a result, this means that $x+y$ is equal to a certain factor of $960$ and $x-y$ is the corresponding factor of $960$ (e.g. if $x+y=48$ then $x-y$ must equal $20$ to satisfy the equation). The factor pairs of $960$ are as follows: $1$ and $960,$ $2$ and $480,$ $3$ and $320,$ $4$ and $240,$ $5$ and $192,$ $6$ and $160,$ $8$ and $120,$ $10$ and $96,$ $12$ and $80,$ $15$ and $64,$ $16$ and $60,$ $20$ and $48,$ $24$ and $40,$ $30$ and $32.$ This means that there are $14$ factor pairs of $960$ without reversing the order of the numbers in the factor pairs. Emer noticed that the midpoint of the factor pair - so the midpont of $x+y$ and $x-y$ - is $x$, ie the side length of the larger square, so The pair $5, 192$ did not work because the midpoint was not an integer, so this factor pair can not be turned into a hollow square. So I looked at the factor pairs of $960$ again and eliminated any factor pairs that would not create a midpoint that is an integer (so any where one number was even and one was odd). This is Emily's method for finding hollow squares for differently sized battalions: 1) Find all the factor pairs 2) Get rid of any where the pair is an odd and an even number 3) For each remaining factor pair, firstly find the difference between the two numbers 4) Next, divide this answer by 2 - the answer is the $y$ value of the equation $x^2-y^2=\text{total number of soldiers}$ 5) Subtract the $y$ value from the larger factor to get the $x$ value In order to find the number of possibilities you simply count the number of remaining factor pairs after step 2. Shyam pointed out that All batallions consisting of a prime number, or odd number or a number from which you can factorise a single 2 only cannot be written or constructed as a symmetrical hollow square. Agathiyan explained how to use Alison's method to make hollow squares out of battalions of $n$ soldiers, where $n$ is a multiple of $4$: Since we need 4 rectangles take $n$ soldiers and split into to four, then using the four groups, form them into as many differently sized rectangles as possible, but not squares (because if it was a square, then the shape wouldn’t form a hollow square, but just a square, as four squares would form that shape) and all groups should be in the same shaped rectangle. Then arranging them all in the same way Alison did, it should be that there will be every single possible combination for a symmetrical hollow square. Shyam and Agathiyan both used Charlie's method to find out more about hollow squares that are not necessarily symmetrical. Shyam said: Integers with only one 2 able to be factored out of it is not possible to make any hollow square because: when the number is written as $(x+y)(x-y)$, the one 2 must either be a factor of $(x+y)$ or of $(x-y)$, so only one of them is even. But this does not work as any 2 pairs of numbers when added and subtracted from each other have to create either 2 odd numbers or 2 even numbers, meaning the given circumstance is impossible. However, every other number that can't make a symmetrical hollow square (all odd numbers) can make a non-symmetrical square as: $(x+y)(x-y)=x^2-y^2$ and even for primes, you can make $x-y=1$, so $x+y=the \text{ }prime$, so $x$ will be $\frac{prime}{2}+0.5$ and $y$ will be $\frac{prime}2-0.5$. However when $x-y=1$, it will only be a right angle (L shape), not a proper hollow square as for it to be a proper hollow square $x-y\ge2$, $1$ for each border so no primes can make a symmetrical hollow square. Meanwhile other odd numbers can work if $x-y$ is one of their factors and $x+y$ is the other factor in the factor pair. Agathiyan considered moving the smaller square to different positions inside the larger square, since this is possible when the squares do not have to be symmetrical. This is Agathian's work, where $a$ represents the side length of the larger square and $b$ represents the side length of the smaller square. Letting $a-b=f_1$ and $a+b=f_2$, $$f_2-f_1=2b\\ b=\frac{f_2-f_1}2\\ a-b=f_1\\ a=f_1+\frac{f_2-f_1}2$$ I knew that there must be at least one soldier on each of the square's sides, thus the width of the square in which the inner square can move into is: $a-2$ (As there needs to be one soldier at each edge, thus the width is reduced by $2$) $$f_1+\frac{f_2-f_1}2-2=w$$ And since both the range and the inner square are both squares, then the range in which the square can move laterally squared is equal to the range the square can move in 2 dimensions since area of a square is equal to its length squared. So to find the range the square can travel laterally, I tried a few widths and derived the following equation for the range of lateral travel: $w-b+1=r$, where $w$ is the 'width' the square can occupy, as defined above, $b$ is still the side length of the smaller square, and $r$ is the range of lateral travel, or number of possible positions when moving horizontally. Or: $$f_1+\frac{f_2-f_1}2-2-\frac{f_2-f_2}2+1=r\\f_1-1=r$$ Therefore the total 2 dimensional area the inner square can be placed is: $$(f_1-1)^2$$ Now that I had finally found the number of possible combinations relative to the square sizes, I formed an equation. $f_{x_1},f_{x_2}$ are pairs of factors with even difference of a number and $n$ is the number of pairs of factors with even difference the number has. Since for each pair of factors with an even difference [there are $f_1-1)^2$ possible positions for the 'hole'] then the number of possible arrangements any number has is: $$(f_{1_1}-1)^2+(f_{2_1}-1)^2+...+(f_{n_1}-1)^2$$
# Tyra Banks Under The Microscope (01/16/2020) How will Tyra Banks do on 01/16/2020 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is not scientifically verified – do not take this too seriously. I will first work out the destiny number for Tyra Banks, and then something similar to the life path number, which we will calculate for today (01/16/2020). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology practitioners. PATH NUMBER FOR 01/16/2020: We will analyze the month (01), the day (16) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 01 and add the digits together: 0 + 1 = 1 (super simple). Then do the day: from 16 we do 1 + 6 = 7. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 1 + 7 + 4 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the path number for 01/16/2020. DESTINY NUMBER FOR Tyra Banks: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for Tyra Banks we have the letters T (2), y (7), r (9), a (1), B (2), a (1), n (5), k (2) and s (1). Adding all of that up (yes, this can get tedious) gives 30. This still isn’t a single-digit number, so we will add its digits together again: 3 + 0 = 3. Now we have a single-digit number: 3 is the destiny number for Tyra Banks. CONCLUSION: The difference between the path number for today (3) and destiny number for Tyra Banks (3) is 0. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is not scientifically verified. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind. ### Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. #### Latest posts by Abigale Lormen (see all) Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
# wikiHow to Understand Calculus Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of mathematics, and underpins many of the equations that describe physics and mechanics. You will probably need a college level class to understand calculus well, but this article can get you started and help you watch for the important concepts as well as technical insights. ### Part 1 Reviewing the Basics of Calculus 1. 1 Know that calculus is the study of how things are changing. Calculus is a branch of mathematics that looks at numbers and lines, usually from the real world, and maps out how they are changing. While this might not seem useful at first, calculus is one of the most widely used branches of mathematics in the world. Imagine having the tools to examine how quickly your business is growing at any time, or plotting the course of a spaceship and how fast it is burning fuel. Calculus is an important tool in engineering, economics, statistics, chemistry, and physics, and has helped create many real-world inventions and discoveries. 2. 2 Remember that functions are relationships between two numbers, and are used to map real-world relationships. Functions are rules for how numbers relate to one another, and mathematicians use them to make graphs. In a function, every input has exactly one output. For example, in ${\displaystyle y=2x+4,}$ every value of ${\displaystyle x}$ gives you a new value of ${\displaystyle y.}$ If ${\displaystyle x=2,}$ then ${\displaystyle y=8.}$ If ${\displaystyle x=10,}$ then ${\displaystyle y=24.}$[1] All calculus studies functions to see how they change, using functions to map real-world relationships. • Functions are often written as ${\displaystyle f(x)=x+3.}$ This means that the function ${\displaystyle f(x)}$ always adds 3 to the number you input for ${\displaystyle x.}$ If you want to input 2, write ${\displaystyle f(2)=(2)+3,}$ or ${\displaystyle f(2)=5.}$ • Functions can map complex motions too. NASA, for example, has a function that describes how fast a rocket will go based on how much fuel it burns, the wind resistance, and the weight of the rocket itself. 3. 3 Think about the concept of infinity. Infinity is when you repeat a process over and over again. It is not a specific place (you can’t go to infinity), but rather the behavior of a number or equation if it is done forever. This is important to study change: you might want to know how fast your car is moving at any given time, but does that mean how fast you were at that current second? Millisecond? Nanosecond? You could find infinitely smaller amounts of time to be extra precise, and that is where calculus comes in. 4. 4 Understand the concept of limits. A limit tells you what happens when something is near infinity. Take the number 1 and divide it by 2. Then keep dividing it by 2 again and again. 1 would become 1/2, then 1/4, 1/8, 1/16, 1/32, and so on. Each time, the number gets smaller and smaller, getting “closer” to zero. But where would it end? How many times do you have to divide by 1 by 2 to get zero? In calculus, instead of answering this question, you set a limit. In this case, the limit is 0. • Limits are easiest to see on a graph – are the points that a graph almost touches, for example, but never does? • Limits can be a number, infinity, or not even exist. For example, if you add 1 + 2 + 2 + 2 + 2 + ... forever, your final number would be infinitely large. The limit would be infinity. 5. 5 Review essential math concepts from algebra, trigonometry, and pre-calculus. Calculus builds on many of the forms of math you’ve been learning for a long time. Knowing these subjects completely will make it much easier to learn and understand calculus. Some topics to refresh include: • Algebra. Understand different processes and be able to solve equations and systems of equations for multiple variables. Understand the basic concepts of sets. Know how to graph equations. • Geometry. Geometry is the study of shapes. Understand the basic concepts of triangles, squares, and circles and how to calculate things like area and perimeter. Understand angles, lines, and coordinate systems • Trigonometry. Trigonometry is branch of maths which deals with properties of circles and right triangles. Know how to use trigonometric identities, graphs, functions, and inverse trigonometric functions. 6. 6 Purchase a graphing calculator. Calculus is not easy to understand without seeing what you are doing. Graphing calculators take functions and display them visually for you, allowing you to better comprehend the equations you are writing and manipulating. Oftentimes, you can see limits on the screen and calculate derivatives and functions automatically. • Many smartphones and tablets now offer cheap but effective graphing apps if you do not want to buy a full calculator. ### Part 2 Understanding Derivatives 1. 1 Know that calculus is used to study “instantaneous change.” Knowing why something is changing at an exact moment is the heart of calculus. For example, calculus tells you not only the speed of your car, but how much that speed is changing at any given moment. This is one of the simplest uses of calculus, but it is incredibly important. Imagine how useful that knowledge would be for the speed of a spaceship trying to get to the moon! [2] • Finding instantaneous change is called differentiation. Differential calculus is the first of two major branches of calculus. 2. 2 Use derivatives to understand how things change instantaneously. A "derivative" is a fancy sounding word that inspires anxiety. The concept itself, however, isn't that hard to grasp -- it just means "how fast is something changing." The most common derivatives in everyday life relate to speed. You likely don’t call it the “derivative of speed,” however – you call it "acceleration." • Acceleration is a derivative – it tells you how fast something is speeding up or slowing down, or how the speed is changing. 3. 3 Know that the rate of change is the slope between two points. This is one of the key findings of calculus. The rate of change between two points is equal to the slope of the line connecting them. Think of a basic line, such as the equation ${\displaystyle y=3x.}$ The slope of the line is 3, meaning that for every new value of ${\displaystyle x,}$ ${\displaystyle y}$ changes by 3. The slope is the same thing as the rate of change: a slope of three means that the line is changing by 3 for every change in ${\displaystyle x.}$ When ${\displaystyle x=2,y=6;}$ when ${\displaystyle x=3,y=9.}$ 4. 4 Know that you can find the slope of curved lines. Finding the slope of a straight line is relatively straightforward: how much does ${\displaystyle y}$ change for each value of ${\displaystyle x?}$ Yet complex equations with curves, like ${\displaystyle y=x^{2}}$ are much harder to find. However, you can still find the rate of change between any two points – simply draw a line between them and calculate the slope. • For example, in ${\displaystyle y=x^{2},}$ you can take any two points and get the slope. Take ${\displaystyle (1,1)}$ and ${\displaystyle (2,4).}$ The slope between them would equal ${\displaystyle {\frac {4-1}{2-1}}={\frac {4}{2}}=2.}$ This means that the rate of change between ${\displaystyle x=1}$ and ${\displaystyle x=2}$ is 2. 5. 5 Make your points closer together for a more accurate rate of change. The closer your two points, the more accurate your answer. Say you want to know how much your car accelerates right when you step on the gas. You don’t want to measure the change in speed between your house and the grocery store, you want to measure the change in speed the second after you hit the gas. The closer your measurement is to that split-second moment, the more accurate your reading will be. • For example, scientists study how quickly some species are going extinct to try to save them. However, more animals often die in the winter than the summer, so studying the rate of change across the entire year is not as useful – they would find the rate of change between closer points, like from July 1st to August 1st. 6. 6 Use infinitely small lines to find the “instantaneous rate of change,” or the derivative. This is where calculus often becomes confusing, but this is actually the result of two simple facts. First, you know that the slope of a line equals how quickly it is changing. Second, you know that closer the points of your line are, the more accurate the reading will be. But how can you find the rate of change at one point if slope is the relationship of two points? The answer: you pick two points infinitely close to one another. • Think of the example where you keep dividing 1 by 2 over and over again, getting 1/2, 1/4, 1/8, etc. Eventually you get so close to zero, the answer is "practically zero." Here, your points get so close together, they are "practically instantaneous." This is the nature of derivatives. 7. 7 Learn how to take a variety of derivatives. There are a lot of different techniques to find a derivative depending on the equation, but most of them make sense if you remember the basic principles of derivatives outlined above. All derivatives are is a way to find the slope of your "infinitely small" line. Now that your know the theory of derivatives, a large part of the work is finding the answers. 8. 8 Find derivative equations to predict the rate of change at any point. Using derivatives to find the rate of change at one point is helpful, but the beauty of calculus is that it allows you to create a new model for every function. The derivative of ${\displaystyle y=x^{2},}$ for example, is ${\displaystyle y^{\prime }=2x.}$ This means that you can find the derivative for every point on the graph ${\displaystyle y=x^{2}}$ simply by plugging it into the derivative. At the point ${\displaystyle (2,4),}$ where ${\displaystyle x=2,}$ the derivative is 4, since ${\displaystyle y^{\prime }=2(2).}$ • There are different notations for derivatives. In the previous step, derivatives were labeled with a prime symbol – for the derivative of ${\displaystyle y,}$ you would write ${\displaystyle y^{\prime }.}$ This is called Lagrange's notation. • There is also another popular way of writing derivatives. Instead of using a prime symbol, you write ${\displaystyle {\frac {\mathrm {d} }{{\mathrm {d} }x}}.}$ Remember that the function ${\displaystyle y=x^{2}}$ depends on the variable ${\displaystyle x.}$ Then, we write the derivative as ${\displaystyle {\frac {{\mathrm {d} }y}{{\mathrm {d} }x}}}$ – the derivative of ${\displaystyle y}$ with respect to ${\displaystyle x.}$ This is called Leibniz's notation. 9. 9 Remember real-life examples of derivatives if you are still struggling to understand. The easiest example is based on speed, which offers a lot of different derivatives that we see every day. Remember, a derivative is a measure of how fast something is changing. Think of a basic experiment. You are rolling a marble on a table, and you measure both how far it moves each time and how fast it moves. Now imagine that the rolling marble is tracing a line on a graph – you use derivatives to measure the instantaneous changes at any point on that line. • How fast does the marble change location? What is the rate of change, or derivative, of the marble’s movement? This derivative is what we call “speed.” • Roll the marble down an incline and see how fast in gains speed. What is the rate of change, or derivative, of the marble’s speed? This derivative is what we call “acceleration.” • Roll the marble along an up and down track like a roller coaster. How fast is the marble gaining speed down the hills, and how fast is it losing speed going up hills? How fast is the marble moving exactly halfway up the first hill? This would be the instantaneous rate of change, or derivative, of that marble at its one specific point. ### Part 3 Understanding Integrals 1. 1 Know that you use calculus to find complex areas and volumes. Calculus allows you to measure complex shapes that are normally too difficult. Think, for example, about trying to find out how much water is in a long, oddly shaped lake – it would be impossible to measure each gallon of water separately or use a ruler to measure the shape of the lake. Calculus allows you to study how the edges of the lake change, and use that information to learn how much water is inside. • Making geographic models and studying volume is using integration. Integration is the second major branch of calculus. 2. 2 Know that integration finds the area underneath a graph. Integration is used to measure the space underneath any line, which allows you to find the area of odd or irregular shapes. Take the equation ${\displaystyle y=4-x^{2},}$ which looks like an upside-down “U.” You might want to find out how much space is underneath the U, and you can use integration to find it. While this may seem useless, think of the uses in manufacturing – you can make a function that looks like a new part and use integration to find out the area of that part, helping you order the right amount of material. 3. 3 Know that you have to select an area to integrate. You cannot just integrate an entire function. For example, ${\displaystyle y=x}$ is a diagonal line that goes on forever, and you cannot integrate the whole thing because it would never end. When integrating functions, you need to choose an area, such as all points between ${\displaystyle x=2}$ and ${\displaystyle x=5.}$ 4. 4 Remember how to find the area of a rectangle. Imagine you have a flat line above a graph, like ${\displaystyle y=4.}$ To find the area underneath it, you would be finding the area of a rectangle between ${\displaystyle y=0}$ and ${\displaystyle y=4.}$ This is easy to measure, but it will never work for curvy lines that cannot be turned into rectangles easily. 5. 5 Know that integration adds up many small rectangles to find area. If you zoom in very close to a curve, it looks flat. This happens every day – you cannot see the curve of the earth because we are so close to its surface. Integration makes an infinite number of little rectangles under a curve that are so small they are basically flat, which allows you to measure them. Add all of these together to get the area under a curve. • Imagine you are adding together a lot of little slices under the graph, and the width of each slice is ‘’almost’’ zero. 6. 6 Know how to correctly read and right integrals. Integrals come with 4 parts. A typical integral looks like this: ${\displaystyle \int f(x){\mathrm {d} }x}$ • The first symbol, ${\displaystyle \int ,}$ is the symbol for integration (it is actually an elongated S). • The second part, ${\displaystyle f(x),}$ is your function. When it is inside the integral, it is called the integrand. • Finally, the ${\displaystyle {\mathrm {d} }x}$ at the end tells you what variable you are integrating with respect to. Because the function ${\displaystyle f(x)}$ depends on ${\displaystyle x,}$ that is what you should integrate with respect to. • Remember, the variable you are integrating is not always going to be ${\displaystyle x,}$ so be careful what you write down. 7. 7 Learn how to find integrals. Integration comes in many forms, and you will need to learn a lot of different formulas to integrate every function. However, they all follow the principles outlined above: integration sums up an infinite number of things. 8. 8 Know that integration reverses differentiation, and vice versa. This is an ironclad rule of calculus that is so important, it has its own name: the Fundamental Theorem of Calculus. Since integration and differentiation are so closely related, a combination of the two of them can be used to find rate of change, acceleration, speed, location, movement, etc. no matter what information you have. • For example, remember that the derivative of speed is acceleration, so you can use speed to find acceleration. But if you only know the acceleration of something (like objects falling due to gravity), you can integrate it to find the speed! 9. 9 Know that integration can also find the volume of 3D objects. Spinning a flat shape around is a way to create 3D solids. Imagine spinning a coin on the table in front of you – notice how it appears to form a sphere as it spins. You can use this concept to find volume in a process known as “volume by rotation.” • This lets you find the volume of any solid in the world, as long as you have a function that mirrors it. For example, you can make a function that traces the bottom of a lake, and then use that to find the volume of the lake, or how much water it holds. ## Community Q&A 200 characters left ## Sources and Citations • Calculus Made Easy by Silvanus P. Thompson and Martin Gardner ## Article Info Categories: Calculus In other languages: Español: entender cálculo, Italiano: Comprendere il Calcolo Infinitesimale, Português: Entender Cálculo, Русский: понять вычисления, Deutsch: Differenzial und Integralrechnung, Français: comprendre le calcul infinitésimal, Bahasa Indonesia: Memahami Kalkulus Thanks to all authors for creating a page that has been read 129,651 times.
# AP Statistics Curriculum 2007 StudentsT (Difference between revisions) Revision as of 04:39, 4 February 2008 (view source)IvoDinov (Talk \| contribs)← Older edit Revision as of 04:42, 4 February 2008 (view source)IvoDinov (Talk \| contribs) (→Activities)Newer edit → Line 65: Line 65: * Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types what is the probability that exactly 6 subjects have type O blood type in the sample? * Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types what is the probability that exactly 6 subjects have type O blood type in the sample? - Approach I (exact!) : $P(X=6)=$? Where $X\sim B(12, 0.44)$ Approach I (exact!) : $P(X=6)=$? Where $X\sim B(12, 0.44)$ : $P(X=6)={12\choose 6}p^6(1-p)^{6}$, with ${12\choose 6}=\frac{12!}{6!(6)!}=0.2068$, using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator]. : $P(X=6)={12\choose 6}p^6(1-p)^{6}$, with ${12\choose 6}=\frac{12!}{6!(6)!}=0.2068$, using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator]. - Approach II (Approximate): $X \sim B(n=12, p=0.44)$ Approach II (Approximate): $X \sim B(n=12, p=0.44)$ - : $X (approx.) \sim N [m = n p = 5.28; (np(1-p))1/2=1.7]$ $P(X=6) \approx P(Z_1\leq Z \leq Z_2)$, where $Z = (X – 5.28)/1.7$ and $X_1=5.5$, $X_2=6.5$. So, $P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.$ + : $X (approx.) \sim N [m = n p = 5.28; (np(1-p))1/2=1.7]$. $P(X=6) \approx P(Z_1\leq Z \leq Z_2)$, where $Z = (X – 5.28)/1.7$ and $X_1=5.5$, $X_2=6.5$. So, $P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.$ - + Approach III (Approximate): $X \sim B(n=12, p=0.44)$ The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$ Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where $p_1=0.5 – 1/24$ and $p_2=0.5+1/24$. Standardize $Z = (p-0.44)/0.1433$ to get: Approach III (Approximate): $X \sim B(n=12, p=0.44)$ The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$ Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where $p_1=0.5 – 1/24$ and $p_2=0.5+1/24$. Standardize $Z = (p-0.44)/0.1433$ to get: $P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$ $P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$ ## General Advance-Placement (AP) Statistics Curriculum - Student's T Distribution Very frequently in practice we do now know the population variance and therefore need to estimate it using the sample-variance. This requires us to introduce the T-distribution, which is a one-parameter distribution connecting $Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}$. ### Student's T Distribution The Student's t-distribution arises in the problem of estimating the mean of a normally distributed population when the sample size is small and the population variance is unknown. It is the basis of the popular Student's t-tests for the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means. Suppose X1, ..., Xn are independent random variables that are Normally distributed with expected value μ and variance σ2. Let $\overline{X}_n = {X_1+X_2+\cdots+X_n \over n}$ be the sample mean, and ${S_n}^2=\frac{1}{n-1}\sum_{i=1}^n\left(X_i-\overline{X}_n\right)^2$ be the sample variance. We already discussed the following statistic: $Z=\frac{\overline{X}_n-\mu}{\sigma/\sqrt{n}}$ is normally distributed with mean 0 and variance 1, since the sample mean $\scriptstyle \overline{X}_n$ is normally distributed with mean μ and standard deviation $\scriptstyle\sigma/\sqrt{n}$. Gosset studied a related quantity under the pseudonym Student), $T=\frac{\overline{X}_n-\mu}{S_n / \sqrt{n}},$ which differs from Z in that the (unknown) population standard deviation $\scriptstyle \sigma$ is replaced by the sample standard deviation Sn. Technically, $\scriptstyle(n-1)S_n^2/\sigma^2$ has a Chi-square distribution $\scriptstyle\chi_{n-1}^2$ distribution. Gosset's work showed that T has a specific probability density function, which approaches Normal(0,1) as the degree of freedom (df=sample-size -1) increases. ### Example Suppose a researcher wants to examine CD4 counts for HIV(+) patients seen at his clinic. She randomly selects a sample of n = 25 HIV(+) patients and measures their CD4 levels (cells/uL). Suppose she obtains the following results and we are interested in calculating a 95% (α = 0.025) confidence interval for μ: Variable N N* Mean SE of Mean StDev Minimum Q1 Median Q3 Maximum CD4 25 0 321.4 14.8 73.8 208.0 261.5 325.0 394.0 449.0 What do we know from the background information? $\overline{y}= 321.4$ s = 73.8 SE = 14.8 n = 25 $CI(\alpha)=CI(0.025): \overline{y} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}.$ $321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}$ $321.4 \pm 2.064\times 14.8$ [290.85,351.95] #### CI Interpretation Still, does this CI (290.85, 351.95) mean anything to us? Consider the following information: The U.S. Government classification of AIDS has three official categories of CD4 counts – asymptomatic = greater than or equal to 500 cells/uL • AIDS related complex (ARC) = 200-499 cells/uL • AIDS = less than 200 cells/uL • Now how can we interpret our CI? ### SOCR CI Experiments The SOCR Confidence Interval Experiment provides empirical evidence that the definition and the construction protocol for Confidence intervals are consistent. ### Activities • A biologist obtained body weights of male reindeer from a herd during the seasonal round-up. He measured the weight of a random sample of 102 reindeer in the herd, and found the sample mean and standard deviation to be 54.78 kg and 8.83 kg, respectively. Suppose these data come from a normal distribution. Calculate a 99% confidence interval. • Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types what is the probability that exactly 6 subjects have type O blood type in the sample? • Approach I (exact!) : P(X = 6) = ? Where $X\sim B(12, 0.44)$ $P(X=6)={12\choose 6}p^6(1-p)^{6}$, with ${12\choose 6}=\frac{12!}{6!(6)!}=0.2068$, using SOCR Binomial interactive GUI or calculator. • Approach II (Approximate): $X \sim B(n=12, p=0.44)$ $X (approx.) \sim N [m = n p = 5.28; (np(1-p))1/2=1.7]$. $P(X=6) \approx P(Z_1\leq Z \leq Z_2)$, where Failed to parse (lexing error): Z = (X – 5.28)/1.7 and X1 = 5.5, X2 = 6.5. So, $P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.$ • Approach III (Approximate): $X \sim B(n=12, p=0.44)$ The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$ Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where Failed to parse (lexing error): p_1=0.5 – 1/24 and p2 = 0.5 + 1 / 24. Standardize Z = (p − 0.44) / 0.1433 to get: $P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$
# Cuisenaire Fractions ##### Math Lesson Plan Objectives Students will be introduced to fractions by comparing ratios between different Cuisenaire rods. Materials Procedure 1. Let students experiment with the Cuisenaire rods for a few minutes. 2. Ask students to predict how many orange rods it will take to make a line across a table or bulletin board. 3. Let students predict and then measure. 4. Next ask students to try to find something in the room that is a certain number of orange rods in length, i.e. ten orange rods long. 5. After something has been found, have students work in pairs to decide how many yellow rods it would take without measuring it. 6. Have them write down their answer and explain how they got it. 7. Discuss the responses. Students should realize that a yellow rod is half the length of the orange rod. Measure to verify the outcomes that students get. 8. Discuss the relationship between the two different rods with the class. Introduce the words twice, half, whole, and fraction. Talk about the yellow rod being a fraction of the orange one, being part of the whole, and being half of the whole. Introduce the symbol ½. 9. Have your kids experiment with the Cuisenaire rods, looking for relationships among them. 10. Let students express their discoveries to the class. Have them do so using fractions and the appropriate numerical representations (1/2, 1/4, 1/3, etc.). 11. Have students work in pairs on the worksheet provided which will give students experience in beginning fractions. Closure Review the worksheet when students are finished. Let students explain their answers. Discuss the reasoning for the answers. Have them reflect on this new concept of fractions. Do they like it? Is it hard/easy? When are fractions used in the real world? Evaluation Did students grasp the concept of the ratios between the Cuisenaire rods? Were they able to figure out the length of something based on the yellow rod without having to actually measure it? Notice which students understood easily and those that had difficulties. Were students able to grasp the concept of a whole when you introduced fractions? Were they able to successfully complete the worksheet? Notice which students seemed to enjoy the activity, who got bored, and who wasn't interested or was confused.
# Chapters 1-9 - Cumulative Review Exercises - Page 705: 22 The name of the figure is a rhombus, the perimeter is $36$ ft and the area is $70$ sq ft. #### Work Step by Step 1. Name the figure: Rhombus 2. Find the perimeter by adding all the sides up Let $P =$ the perimeter of the rhombus $P = 8 + 8 + 10 + 10$ $P = 16 + 20$ $P = 36$ ft 3. Find the area of the rhombus Note: It is made of 2 right angled triangles and a rectangle. But first you need to find the height of the triangle using the Pythagorean Formula. $b^{2} = c^{2} - a^{2}$ $b = \sqrt (c^{2} - a^{2})$ $b = \sqrt (8^{2} - 7^{2})$ $b = \sqrt (64 - 49)$ $b = \sqrt 15$ $b = 3.87...$ sq ft Now find the area of the triangle using $base \times height \div 2$ $= (\sqrt 15 \times 7) \div 2$ $= 13.55544...$ sq ft Since there is two right angled triangles, multiply it by 2: $= 13.5544 \times 2$ $= 27.1108...$ sq ft Now, you have to find the area of the rectangle that has a width of $7$ ft and a length of $(10 - \sqrt 15) = 6.127$ ft. $= 7 \times 6.127...$ $= 42.889...$ sq ft Add up the area of the rectangle and the area of the triangles $= 42.889... + 27.110...$ $= 70$ sq ft After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
# COMPLEX ZEROS OF A QUADRATIC FUNCTION ## Presentation on theme: "COMPLEX ZEROS OF A QUADRATIC FUNCTION"— Presentation transcript: COMPLEX ZEROS OF A QUADRATIC FUNCTION SECTION 2.7 COMPLEX ZEROS OF A QUADRATIC FUNCTION SQUARE ROOTS OF NEGATIVE NUMBERS Is a value we have dealt with up to now by simply saying that it is not a real number. And, up to now, we have dealt with the following equation by simply saying it has no solution: x2 + 4 = 0 DEFINITION OF i i 2 = - 1 The number i is called an imaginary number. Imaginary numbers, along with the real numbers, make up a set of numbers known as the complex numbers. COMPLEX NUMBERS Imaginary Real i 2i 5 -1 - 3i 2/3i 1/2 .7 5 -1 1/2 .7 COMPLEX NUMBERS All numbers are complex and should be thought of in the form: a + bi Imaginary Part Real Part COMPLEX NUMBERS a + bi Real Part Imaginary Part When b = 0, the number is a real number. Otherwise, the number is imaginary. OPERATING ON COMPLEX NUMBERS Addition: Example: (3 + 5i) + ( i) Subtraction: (6 + 4i) - ( 3 + 6i) OPERATING ON COMPLEX NUMBERS Multiplication: Example: (5 + 3i) • (2 + 7i) (3 + 4i) • ( 3 - 4i) CONJUGATES 2 + 3i = 2 - 3i Multiplying a complex number by its conjugate always yields a nonnegative real number. THEOREM: If z = a + bi z z = a2 + b2 Writing the reciprocal of a complex number in standard form. Example: Writing the quotient of complex numbers in standard form. Example: Writing the quotient of complex numbers in standard form. Example: POWERS OF i i1 = i i2 = - 1 i3 = - i i4 = 1 i5 = i and so on QUADRATIC EQUATIONS WITH A NEGATIVE DISCRIMINANT Quadratic equations with a negative discriminant have no real solution. But, if we extend our number system to the complex numbers, quadratic equations will always have solutions because we will then be including imaginary numbers. EXAMPLE EXAMPLE Solve the following equations in the complex number system: x2 = 4 x2 = - 9 WARNING! EXAMPLE Solve the following equation in the complex number system: x2 - 4x + 8 = 0 DISCRIMINANT If b2 - 4ac > 0 Two unequal real sol’ns If b2 - 4ac = 0 One double real root If b2 - 4ac < 0 Two imaginary solutions EXAMPLE: Without solving, determine the character of the solution of each equation in the complex number system: 3x2 + 4x + 5 = 0 2x2 + 4x + 1 = 0 9x2 - 6x + 1 = 0 CONCLUSION OF SECTION 2.7
# How do you find the limit of (sqrt(3x - 2) - sqrt(x + 2))/(x-2) as x approaches 2? Mar 11, 2016 ${\lim}_{x \to 2} \frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2} = \frac{1}{2}$ #### Explanation: The first thing we do in limit problems is substitute the $x$ value in question and see what happens. Using $x = 2$, we find: ${\lim}_{x \to 2} \frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2} = \frac{\sqrt{3 \left(2\right) - 2} - \sqrt{\left(2\right) + 2}}{\left(2\right) - 2} = \frac{\sqrt{4} - \sqrt{4}}{0} = \frac{0}{0}$ You may be wondering how this helps us. Well, because we have $\frac{0}{0}$, this problem becomes fair game for an application of L'Hopital's Rule. This rule says that if we evaluate a limit and get $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can find the derivative of the numerator and denominator and try evaluating it then. So, without further ado, let's get to it. Derivative of Numerator We're trying to find $\frac{d}{\mathrm{dx}} \left(\sqrt{3 x - 2} - \sqrt{x + 2}\right)$ here. Using the sum rule, we can simplify this to $\frac{d}{\mathrm{dx}} \sqrt{3 x - 2} - \frac{d}{\mathrm{dx}} \sqrt{x + 2}$. Taking the derivative of the first term, we see: $\frac{d}{\mathrm{dx}} \sqrt{3 x - 2} = {\left(3 x - 2\right)}^{\frac{1}{2}} = \frac{3}{2 \sqrt{3 x - 2}} \to$Using power rule and chain rule For the second term, $\frac{d}{\mathrm{dx}} \sqrt{x + 2} = {\left(x + 2\right)}^{\frac{1}{2}} = \frac{1}{2 \sqrt{x + 2}} \to$Using power rule Thus, our new numerator is: $\frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}$. Derivative of Denominator This one is fairly easy: $\frac{d}{\mathrm{dx}} \left(x - 2\right) = 1$. Yep, that's it. Put it all Together Combining these two results into one, our new fraction is $\frac{\frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}}{1} = \frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}$. We can now evaluate it at $x = 2$ and see if anything changes: ${\lim}_{x \to 2} \left(\frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}\right) = \frac{3}{2 \sqrt{3 \left(2\right) - 2}} - \frac{1}{2 \sqrt{\left(2\right) + 2}} = \frac{3}{2 \sqrt{4}} - \frac{1}{2 \sqrt{4}} = \frac{1}{2}$ And there you have it! We can confirm this result by looking at the graph of $\frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2}$: graph{(sqrt(3x-2)-sqrt(x+2))/(x-2) [-0.034, 3.385, -0.205, 1.504]} Mar 14, 2016 If your calculus course has not yet covered derivatives and l'Hospital's rule, use algebra to rationalize the numerator. #### Explanation: $\frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2} = \frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2} \cdot \frac{\sqrt{3 x - 2} + \sqrt{x + 2}}{\sqrt{3 x - 2} + \sqrt{x + 2}}$ $= \frac{\left(3 x - 2\right) - \left(x + 2\right)}{\left(x - 2\right) \left(\sqrt{3 x - 2} + \sqrt{x + 2}\right)}$ $= \frac{2 x - 4}{\left(x - 2\right) \left(\sqrt{3 x - 2} + \sqrt{x + 2}\right)}$ $= \frac{2 \cancel{\left(x - 2\right)}}{\cancel{\left(x - 2\right)} \left(\sqrt{3 x - 2} + \sqrt{x + 2}\right)}$ So we have ${\lim}_{x \rightarrow 2} \frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2} = {\lim}_{x \rightarrow 2} \frac{2}{\sqrt{3 x - 2} + \sqrt{x + 2}}$ $= \frac{2}{\sqrt{3 \left(2\right) - 2} + \sqrt{\left(2\right) + 2}}$ $= \frac{2}{\sqrt{4} + \sqrt{4}} = \frac{2}{2 + 2} = \frac{1}{2}$
# What Is The Midpoint Theorem? The midpoint theorem states that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side. Whereas, its converse states that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side. Imagine sitting in a baseball stadium and watching your favorite team play!! It was a good game, right? Baseball has one of the most uniquely shaped fields in sports, much different than the most common rectangular or circular playing fields, and it is referred to as a Baseball Diamond. Now, someone with even a bit of mathematical knowledge might wonder how that field was made? Curiosity regarding the construction of the diamond-shaped, field according to the specified dimensions and the theorems involved in its construction, must make one feel inquisitive. If you take a closer look towards the field or see a birds’ eye view of the field, you will notice that the diamond looks like the sector of a circle with a triangle in it (joining the endpoints of the grass line with the help of a straight line). With that image in mind, let’s explore the mathematics behind the construction of the legendary baseball diamond. Recommended Video for you: ## Midpoint Theorem The midpoint theorem states that “For a given triangle ∆ABC, let D and E be the midpoints of AC and AB, respectively. Then the segment DE is parallel to BC and its length is one half the length of segment BC.” Or, in simple words, it can be stated as The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side. ### Midpoint Theorem Proof Any theorem must have a mathematical proof for it to be valid and the midpoint theorem also has one. To Prove- DE = (1/2) BC and DE\|\|BC In the above figure, extend the line segment DE to a point F in such a way that DE = EF and also joins F to point C. In triangle ADE and ECF, we have – DE = EF (by construction), ∠ AED = ∠ CEF (since they are vertically opposite angles) and EC = AE (since E is the midpoint of AC). According to the above results, we can say that the triangles AED and CEF are congruent. Therefore, we can say that ∠ ADE = ∠CFE (alternate interior angles), and similarly, ∠DAE = ∠FCE (alternate interior angles) and AD=CF. Therefore, we can say that CF\|\|AB, so CF\|\|BD. Since opposite sides of the quadrilateral BDFC are parallel and equal, BDFC is a parallelogram, hence BC\|\|DF, i.e., BC\|\|DE and DE = (1/2) BC. Hence, the midpoint theorem is proved. This is the general textbook explanation that students tend to understand, but never question in terms of its application to real-world problems. Now, before this gets boring, we’ll shift back into baseball to make this concept more interesting and easy to understand. Also Read: What Is Ptolemy’s Theorem? ## Practical Understanding To understand any theorem, it’s essential to understand its practical importance and application. So, we’re coming back to the baseball field for a practical understanding of the theorem. Below are the dimensions of a baseball field (listing only the important/relevant dimensions to prove the practical application of the midpoint theorem). • Home to first base – 27.43 m • Third base to home – 27.43 m • Home to left-field foul pole – 99.06 m • Home to right field foul pole – 99.06 m The known distance between the two foul poles is 140.09 meters; we can now use the midpoint theorem to calculate this and find out whether the theorem is practically valid or not. Considering the triangle formed by the two foul poles and home plate, we have two sides of the triangle, both having their length equal to 99.06 m, and the third side, i.e., the distance between the foul poles, which is 140.09 m. The midpoints of the equal sides (from home plate to the left and right foul poles) are at a distance of 49.53 m from home and poles. Now, if you join the two midpoints with the help of a line segment, the length of the line segment is unknown, but can be easily determined using basics or trigonometry and triangle congruency. Here we have A (home plate), B (right foul pole) and C (left foul pole). O is the perpendicular dropped from A to line segment DE. We will consider AB = AC since in a baseball field, the distance of the two foul poles from home plate is the same. Now, we know that in the triangle AOD, we can calculate DO by – Cosine = Base/Hypotenuse, ∠ ADO = 45 (since a baseball field is symmetrical) Therefore, cos 45= Base/49.53, which gives us the length of the base DO, i.e., 35.02 m. Also, since the triangles ADO and AEO are congruent by RHS congruency, line segment MN = 70.04 m, which is one half of LR, hence proving the midpoint theorem!
Share # Find the Term of the Arithmetic Progression 9, 12, 15, 18, ... Which is 39 More than Its 36th Term. - CBSE Class 10 - Mathematics #### Question Find the term of the arithmetic progression 9, 12, 15, 18, ... which is 39 more than its 36th term. #### Solution In the given problem, let us first find the 36st term of the given A.P. A.P. is 9, 12, 15, 18 … Here, First term (a) = 9 Common difference of the A.P. (d) = 12 - 9 = 3 a_n = a + (n - 1)d So for 36th term (n = 36) a_36 = 9 + (36 - 1)(3) = 9 + 35(3) = 9 + 105 = 114 Let us take the term which is 39 more than the 36th term as an. So, a_n = 39 + a_36 = 39 + 114 = 153 Also a_n = a + (n -1)d 153 = 9 + (n -1)3 153 = 9 + 3n - 3 153 = 6 + 3n 153 - 6 = 3n Further simplifying, we get, 147 = 3n n = 147/3 n = 49 Therefore, the 49 th term if the given A.P. is 39 more than the 36 th term Is there an error in this question or solution? #### Video TutorialsVIEW ALL [6] Solution Find the Term of the Arithmetic Progression 9, 12, 15, 18, ... Which is 39 More than Its 36th Term. Concept: Arithmetic Progression. S
# 7.4: Dividing Fractions Difficulty Level: At Grade Created by: CK-12 ## Introduction The Rainforest Game As part of her project, Julie has decided to create a game about the rainforest. She will play the game with her classmates as part of her presentation, so the students can learn some information in a new way. Besides, Julie loves games! To create the game, Julie is going to create question cards on strips of paper. She wants each strip of paper to be . That way she will have enough room to write the questions but the strips won’t be too wide either. Julie takes the big chart paper and cuts off a piece 20” wide. She is sure that she will have enough paper to cut her question strips. There are 25 students in Julie’s class. She wants each student to have one question to answer. Given the size of the chart paper and the size Julie wants each strip to be, does she have enough paper? Julie isn’t sure. She needs your help. To figure out the problem, Julie will need to divide fractions. You can help her. Pay attention in this lesson and you will learn all that you need to know about dividing fractions. What You Will Learn By the end of this lesson you will be able to complete the following: • Identify and write reciprocal fractions. • Divide a fraction by a whole number. • Divide a whole number by a fraction. • Divide a fraction by a fraction. • Solve real-world problems involving quotients of fractions. Teaching Time I. Identify and Write Reciprocal Fractions This lesson focuses on dividing fractions. But before we dive into the mechanics of dividing fractions, let’s think about some division facts. We know that division is the opposite of multiplication, in fact we could say that multiplication is the inverse operation of division. What is an inverse operation? An inverse operation is the opposite operation. The word “inverse” is a fancy way of saying opposite. If the opposite of addition is subtraction, then subtraction is the inverse operation of addition. We can also say that division is the inverse of multiplication. What do inverse operations have to do with dividing fractions? Well, when we divide fractions, we need to perform the inverse operation. To divide a fraction, we have to multiply by the reciprocal of the second fraction. What is a reciprocal? A reciprocal is the inverse or opposite form of a fraction. When we change the division to its inverse, multiplication, we also change the second fraction to its reciprocal. We can make any fraction a reciprocal by simply flipping the numerator and the denominator. Example The reciprocal of four-fifths is five-fourths. We simply flipped the numerator and the denominator of the fraction to form its reciprocal. Example Notice that if we multiply a fraction and it’s reciprocal that the product is 1. Example We will begin dividing fractions in the next section, but for right now it is important that you understand that a reciprocal is the inverse of a fraction and know how to write a reciprocal of a fraction. Try a few of these on your own. Write a reciprocal for each fraction. Take a few minutes to check your work with a peer. II. Divide a Fraction by a Whole Number You have learned a couple of things about dividing fractions. The first is that to divide fractions we are actually use the inverse operation, multiplication. The second is that the second fraction is going to become its reciprocal or opposite. These are a few basic notes, but we haven’t applied them to actually dividing yet. Let’s begin. How do we divide a fraction by a whole number? To divide a fraction by a whole number we have to think about what we are actually being asked to do. We are being asked to take a part of something and split it up into more parts. Let’s look at an example so that we can make sense of this. Example This problem is asking us to take one-half and divide into three parts. Here is a picture of what this would look like. This is one half. If we were going to divide one-half into three parts, how much would be in each part? Here we divided the one-half into three sections. But we couldn’t just do that with one part of the whole so we divided the other half into three sections too. Each part is of the whole. How can we do this without drawing a lot of pictures? That is where multiplying by the reciprocal comes in handy. Example First, change the division to multiplication. Next, invert the second fraction, which is a whole number 3, make it to the fraction then make that into its reciprocal . Now, we can find the product. Notice that the answer is the same as when we divided using the pictures!! Practice solving these on your own. Remember to simplify the quotient (the answer) if you can. Take a few minutes to check your work with a peer. III. Divide a Whole Number by a Fraction We can also divide a whole number by a fraction. When we divide a whole number by a fraction we are taking a whole and dividing it into new wholes. Let’s look at an example to understand this. Example Now at first glance, you would think that this answer would be one-half, but it isn’t. We aren’t asking for of one we are asking for 1 divided by one-half. Let’s look at a picture. Now we are going to divide one whole by one-half. Now we have two one-half sections. We can test this out by using the rule that we learned in the last section. Example Our answer is the same as when we used the pictures. It’s time for you to try a few of these on your own. Take a few minutes to check your work with a friend. Are your answers all whole numbers? IV. Divide a Fraction by a Fraction We have been dividing whole numbers by fractions and fractions by whole numbers. We can also use what we have learned when dividing a fraction by another fraction. Here is the rule. Let’s apply these rules to dividing a fraction by another fraction. Example Start by applying the first rule and change the sign to multiplication. Then apply the second rule, the reciprocal of one-third is three over one. Next, we multiply across and simplify. As long as we apply the rules, the problem is very straightforward and simple to figure out. Let’s try another one. Example We started with a fraction divided by a fraction, so we multiplied by the reciprocal. Our product was an improper fraction which we converted to a mixed number. Our quotient is . It’s time for you to practice a few of these on your own. Be sure that your answer is in simplest form. ## Real Life Example Completed The Rainforest Game Now that you have learned about how to divide fractions and whole numbers, let’s help Julie figure out her problem. Here it is once again. As part of her project, Julie has decided to create a game about the rainforest. She will play the game with her classmates as part of her presentation, so the students can learn some information in a new way. Besides, Julie loves games! To create the game, Julie is going to create question cards on strips of paper. She wants each strip of paper to be . That way she will have enough room to write the questions but the strips won’t be too wide either. Julie takes the big chart paper and cuts off a piece 20” wide. She is sure that she will have enough paper to cut her question strips. There are 25 students in Julie’s class. She wants each student to have one question to answer. Given the size of the chart paper and the size Julie wants each strip to be, does she have enough paper? First, let’s go back and underline any important information or questions. Next, let’s look at what we are trying to figure out. Julie needs to figure out if she can cut at least 25 strips of paper that are wide from the large sheet of paper that is 20” wide. To figure this out, we can set up a division problem. We are dividing the 20” into as many strips as possible. Our first step is to change the operation to multiplication and to multiply 20 by the reciprocal of three-fourths. Notice that we also made 20 into a fraction over one. Now we are ready to multiply and simplify. Julie can cut 26 strips of paper from her large sheet. She will have enough strips for each student to have a question. There is also of another strip left over. ## Vocabulary Here are the vocabulary words that are found in this lesson. Inverse Operation opposite operation. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction. Reciprocal the inverse of a fraction. We flip a fraction’s numerator and denominator to write a reciprocal. The product of a fraction and its reciprocal is one. ## Technology Integration Other Videos: http://www.mathplayground.com/howto_divide_fractions.html – This is a great basic video on dividing fractions. ## Time to Practice Directions: Divide each fraction and whole number. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. Directions: Divide each pair of fractions. 21. 22. 23. 24. 25. 26. 27. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
You are on page 1of 10 # Karin Dickensheets Module 2 Written Assignment Linear Algebra ## Lesson 2.1 Problem 11 The problem asks that I use the arrow method to find the determinant of the above matrix. I will begin by rewriting the matrix with the first two colums repeated to the right (making the arrow method possible). Now I will multiply each diagonal descending from left to right and find the sum of those products: This equals 45. Next I will multiply each diagonal descending from right to left and find the sum of those products: This is eual to 110. Finally, I will subtract my second sum from my first sum to find my determinant. 45-110=-65 My determinant is -65. Check work (1.1.1.1) (1.1.1.2) Problem 25 This problem wants me to evaluate det(A) by a cofactor expansion along a row or column of my choice. Because the third column has the most zeroes, I believe that will be the easiest column to work with. After making this choice, I will rewrite the matrix as a difference between the two matrices I will create using my two non-zero elements: Since I have no more zeroes, I will go ahead and use the first row for my next expansion in the same manner as before (now having 3 non-zero elements). I have used red to help demonstrate where these numbers are coming from: I want to show this a piece at a time, but then I will put it all together: ## I will put these pieces together with my original -3 as so: I will then do the same process to the second matrix listed above in my determinant equation and subtract that sum from my first sum: ## From here, I will use ad-cb to further simplify this problem: Check Work (1.2.1.1) (1.2.1.2) Lesson 2.2 Problem 13 The problem asks me to evaluate the determinant of the matrix by using reduced row echelon form. Here is the matrix: Reduced row echelon form creates an upper right triangle of numbers, while the lower left triangle is zeroes with a diagonal of ones down from left to right (but not necessarily in the far right bottom). I will begin by first dividing the first row by 3. (2.1.1) ## Because, I divided by three, I would rewrite my matrix as: Next I want to add two times the first row to the second row: (2.1.2) Written as: Now I want to swap the second and third row (so that I can have that 1 where I want it). (2.1.3) ## Because I swapped rows, the sign on my 3 will change as so: In my final reduction step, I want to multiply row two by negative three and add it to my third row. (2.1.4) Written as: -3 ## Now I just need to multiply -3 by -11 to find my determinant. =33 Check Work 33 (2.1.1.1) Lesson 2.3 Problem 6 The problem asks that I verify that det(AB) = det(BA) and to determine whether the equality det (A+B) = det(A)+det(B) holds. ## (3.1.1) Then I will find (BA). (3.1.2) Then I will find the determinant of each. I prefer using the arrow method as I demonstrated in Lesson 2.1. I will do that now for (AB). = 322 - 388 det(AB)= -66 ## [(5)(14)(-5)+(8)(7)(5)+(-3)(-6)(-2)] - [(-3)(14)(5)+(5)(7)(-2)+(8)(-6)(-5)] = (-350 + 280 - 36) - (-210 - 70 + 240) = -106 - (-40) det(BA)= -66 Next I want to find det(A+B). I first need to find A+B by adding like elements of both matrices and putting their sums into one: ## det(A) = 2 det(A) + det(B) = -31 det(B) = -33 Therefore, while det(AB) and det(BA) are the same, det(A+B) does not equal det(A) + det(B) in this case. Check Work ## (3.1.1.1) (3.1.1.2) (3.1.1.3) (3.1.1.4) Problem 10 For this problem, I am to use determinants to decide wehther the given matrix is invertible. The given matrix is: ## First I will use the arrow method to find the determinant: [(-3)(0)(3)+(0)(6)(8)+(1)(5)(0)]-[(1)(0)(8)+(-3)(6)(0)+(0)(5)(3)] =(0+0+0) - (0+0+0) =0 According to theorem 2.3.3 in our text on page 108, a square matrix is invertible if and only if its determinite does not equal zero. (3.2.1.1)

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