Split (1)

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

text stringlengths 22 1.01M
## Section12.4The Two-Body Problem We will consider motion of two bodies interacting with gravitational force only. We have seen before that motion separates into motion of the center of motion and relative position of the two bodies. Let $m_1$ and $m_2$ be the masses and $\vec{r}_1$ and $\vec{r}_2$ their positions. Their separate equations of motions are \begin{align} m_1 \vec{a}_1 \amp = G\dfrac{m_1m_2}{\|\vec{r}_1 - \vec{r}_2\|^2}\, \hat u_{1\rightarrow 2}, \label{eq-vectorized-law-of-grav-m1}\tag{12.4.1}\\ m_2 \vec{a}_2 \amp = -G\dfrac{m_1m_2}{\|\vec{r}_1 - \vec{r}_2\|^2}\, \hat u_{1\rightarrow 2}, \label{eq-vectorized-law-of-grav-m2}\tag{12.4.2} \end{align} where $\hat u_{1\rightarrow 2}$ is a unit vector in the direction of $1\rightarrow 2\text{.}$ The unit vector is needed to express the direction of the vector equation analytically. Here, $\| \vec{r}_1 - \vec{r}_2 \|$ is the distance between the two bodies. Now, from the sum and difference of Eqs. (12.4.1) and (12.4.2) we obtain two useful equations. \begin{align} \amp \vec{A}_\text{cm} = 0, \label{eq-vectorized-law-of-grav-acm}\tag{12.4.3}\\ \amp m_1 \vec{a}_1 - m_2 \vec{a}_2 = 2G\dfrac{m_1m_2}{\|\vec{r}_1 - \vec{r}_2\|^2}\, \hat u_{1\rightarrow 2}, \label{eq-vectorized-law-of-grav-relative}\tag{12.4.4} \end{align} The first of these equations, says that the CM has no acceleration. Therefore, if we use a coordinate system that moves with the CM, i.e., we let the origin to be at the CM. It turns out that this coordinate system also makes the second equation simpler. We say that we are using CM Frame. The coordinates are shown in Figure 12.4.1. Let $\vec r^{\,\prime}_1$ $\vec r^{\,\prime}_2$ be the postions of the two particles with respect to their CM. Then, we have the following relations between the original coordinates and the new ones. \begin{gather} \vec r^{\,\prime}_1 - \vec r^{\,\prime}_2 = \vec{r}_1 - \vec {r}_2, \label{eq-vectorized-law-of-grav-cm-coords-1}\tag{12.4.5}\\ m_1\vec r^{\,\prime}_1 + m_2 \vec r^{\,\prime}_2 = 0. \label{eq-vectorized-law-of-grav-cm-coords-2}\tag{12.4.6} \end{gather} For convenience we give the relative position of particle 1 with respect to particle 2 its own symbol. $$\vec r = \vec r_1 - \vec r_2 = \vec r^{\,\prime}_1 - \vec r^{\,\prime}_2,\tag{12.4.7}$$ and acceleration of this coordinate, i.e., the relative acceleration, $$\vec a = \text{ acceleration of relative position } \vec r.\tag{12.4.8}$$ We also inroduce a combination of masses, called the reduced mass, denoted by $\mu\text{.}$ $$\mu = \dfrac{m_1 m_2 }{m_1 + m_2},\tag{12.4.9}$$ and total mass by $M$ $$M = m_1 + m_2.\tag{12.4.10}$$ After some algebra, which I differ to the end of this section, we can show that Eq. (12.4.4) gives the following for the magnitude of the relative acceleration, $$\mu\,\vec a = -G_N \dfrac{ M\,\mu}{r^2}\,\hat u,\label{eq-vectorized-law-of-grav-relative-cmframe}\tag{12.4.11}$$ where $\hat u$ is a unit vector pointed towards the origin. That is, the original coupled systems of equation of the two-body system in (12.4.1) and (12.4.2), has been decoupled into two separate one-body problems: (1) Eq. (12.4.3) of the CM and (2) Eq. (12.4.11) for the relative coordinate in the center of frame. If you were to look at the original two particles from CM, you would find their relative motion to be described by a single particle of mass $\mu$ moving in the field of a particle of mass $(m_1 + m_2)$ placed at the origin as illustrated in Figure 12.4.2. If $m_1\lt\lt m_2\text{,}$ then, the CM will be very near $m_2$ and $\mu\approx m_1\text{,}$ i.e., in this case, it would appear that $m_1$ is moving in the field of $m_2$ while $m_2$ is fixed, as is the case for Earth-Sun system. Algebra for Equation (12.4.11) \begin{align} \vec a \amp = \vec a_1 - \vec a_2, \\ \amp = \left( \frac{1}{m_1} + \frac{1}{m_2} \right) G_N \frac{ m_1 m_2 }{r^2} \hat u_{1\rightarrow 2}. \end{align} Now, we write this in the CM frame (the right side of Figure 12.4.2), with $\hat u_{1\rightarrow 2}$ is radial unit vector $\hat u_{1\rightarrow 2}= \hat r = (\vec r)/r \text{.}$ \begin{equation} \vec a = G_N \frac{ m_1 + m_2 }{r^2}\, \left( - \frac{\vec r}{r}\right). \end{equation} Now, multiplying by $\mu\text{,}$ we get \begin{equation} \mu\,\vec a = G_N \frac{ \mu M }{r^2}\, \hat u, \end{equation} where $\hat u = \frac{-\vec r}{r}$ is unit vector pointed towards the origin.
### What’s the Next Number? We are all familiar with simple mathematical puzzles that give a short sequence and ask “What is the next number in the sequence”. Simple examples would be $\displaystyle \begin{array}{rcl} && 1, 3, 5, 7, 9, 11, \dots \\ && 1, 4, 9, 16, 25, \dots \\ && 1, 1, 2, 3, 5, 8, \dots \,, \end{array}$ the sequence of odd numbers, the sequence of squares and the Fibonacci sequence. The Pitfalls of Generalising Circle division: ${n}$ points, ${c}$ chords, ${r_G}$ regions. The idea is to spot the rule by which the terms of the sequence are generated. But the rule is never really determined by the first few terms, so the answer is not unique. Indeed, it is simple to construct a rule that matches the given terms and also generates an arbitrary value for the next term. Let us ask “How many regions are formed if each of a set of ${N}$ points on a circle is joined by a chord to all the other ${N-1}$ points?” (see figure). For small ${N}$ it is easy to count the regions. It is tempting to assume that the sequence, which begins ${ 1, 2, 4, 8, 16, \dots }$ reveals the general pattern ${R(n) = n^2}$ for all values of ${n.}$ But this formula breaks down for ${n=6}$, when there are 31 regions. Thus, we may argue that the next number in the sequence ${ 1, 2, 4, 8, 16}$ is 31. Lagrange Polynomials Suppose the given terms are ${\{a_1, a_2, a_3, \dots , a_n\}}$. Linderholm (1971), in his light-hearted book “Mathematics Made Difficult”, argued that we can define a polynomial ${p(x)}$ that passes through all the points ${(k,a_k)}$ and also through an arbitrary value ${(n+1,A)}$. We first define the polynomial $\displaystyle p_{j}(x) := \frac{(x-1)}{(j-1)} \frac{(x-2)}{(j-2)} \dots \frac{(x-(j-1))}{(j-(j-1))} \frac{(x-(j+1))}{(j-(j+1))} \dots \frac{(x-n)}{(j-n)} \,.$ For ${i\ne j}$, we have ${p_j(i) = 0}$, and for ${x = j}$ we have ${p_j(j) = 1}$. Now defining $\displaystyle P(x) := \sum_{j=1}^n a_j p_j(x)$ we see that ${P(k) = a_k}$ for ${k=1, 2, \dots , n}$, so the polynomial ${P(x)}$ interpolates between the given terms ${( a_1, a_2, \dots , a_n )}$. Now we extend the definitions to include an extra term: $\displaystyle q_{j}(x) := \frac{(x-1)}{(j-1)} \frac{(x-2)}{(j-2)} \dots \frac{(x-(j-1))}{(j-(j-1))} \frac{(x-(j+1))}{(j-(j+1))} \dots \frac{(x-(n+1))}{(j-(n+1))}$ and $\displaystyle Q(x) := \sum_{j=1}^n a_j p_j(x) + A\, q_{n+1}(x)$ Then ${Q(k) = P(k) = a_k}$ for ${1\le k\le n}$ and ${Q(n+1) = A}$. Using the polynomial ${Q(x)}$ we can argue that the next term in the sequence ${a_1, a_2, \dots , a_n}$ can take any randomly chosen value ${A}$. In the figure below (left panel) we plot the sequence ${ 1, 4, 9, 16}$ and the interpolating polynomial ${P_4(x)=x^2}$ that fits all four points. Then we extend the sequence in an arbitrary fashion, say ${ 1, 4, 9, 16, 50,30,20,40}$, and plot the interpolating polynomial ${P_8(x)}$ that fits all eight points: $\displaystyle \frac{3427200 - 8614440 x + 8281910 x^2 - 4024839 x^3 + 1083425 x^4 - 163170 x^5 + 12845 x^6 - 411 x^7}{2520}$ The result is shown in the figure (right panel). Sinc Sequences The sinc function is defined thus: $\displaystyle \mathrm{sinc\ } x = \begin{cases} \displaystyle{\frac{\sin x}{x}}, & x \ne 0 \\ \ \ 1, &x = 0 \end{cases}$ We note that ${\mathrm{sinc\ }x}$ oscillates with decaying amplitude as ${\|x\|}$ increases, and has zeros at ${\{\pm \pi, \pm 2\pi, \dots \}}$. Thus, the function ${\mathrm{sinc\ }\pi(x-m)}$ equals ${1}$ for ${x=m}$ and is zero for all other integer values of ${x}$. It is clear that, given a finite sequence ${S = \{a_1, a_2, \dots , a_n \}}$, the function $\displaystyle f_n(x) = \sum_{k=1}^{n} a_k\, \mathrm{sinc\ }\pi(x-k)$ takes the values of the sequence ${S}$ at ${x=1, 2, 3, \dots, n}$. We can extend the sequence ${S}$ by appending an arbitrary value, ${\hat S_k = \{a_1, a_2, \dots , a_k, A \}}$, and we can define an interpolating function that takes the values ${ \{a_1, a_2, \dots , a_n, A \}}$ at ${x=1, 2, 3, \dots, n, n+1}$: $\displaystyle \hat f_n(x) = \sum_{k=1}^{n} [ a_k\,\mathrm{sinc\ }\pi(x-k) ] + A\,\mathrm{sinc\ }\pi(x-(n+1)) \,.$ Using ${\hat f_n(x)}$, we can argue that ${A}$ is the “natural next number” in the sequence ${S = \{a_1, a_2, \dots , a_n \}}$. For example, we plot the sequence ${ 1, 4, 9, 16}$ and the sinc expansion ${s_4(x)}$ that fits all four points (left panel in figure below). Then we extend the sequence arbitrarily to ${ 1, 4, 9, 16, 50,30,20,40}$ and plot the corresponding sinc sum ${s_8(x)}$ (right panel). Conclusion Given the initial terms of a sequence, there are many ways to define a function that fits all the given values and also takes arbitrary subsequent values. This, in purely logical terms, the question “What is the next Number?” does not have a well-defined answer. However, if you apply this logic in your MCQ or your IQ test, you may score badly. Sources ${\bullet}$ Linderholm, Carl E., 1971: Mathematics Made Difficult. Wolfe Publishing, London. 207 pages SBN: 72340415 1.
# Thread: Find sin2x for each given the following condition: 1. ## Find sin2x for each given the following condition: Find sin2x for each given the following condition: a) sin x = 4/5 for 90o < x < 180o b)tan x = -2 for 3pi/2 < x < 2pi 3. ## Re: Find sin2x for each given the following condition: Hello, alejandro! $\displaystyle \text{Find }\sin2x\text{ for each given the following conditions:}$ $\displaystyle a)\;\sin x \,=\,\tfrac{4}{5}\:\text{ for }90^o < x < 180^o$ We are given: .$\displaystyle \sin x \:=\:\tfrac{4}{5} \:=\:\tfrac{opp}{hyp}$ Angle $\displaystyle x$ is in a right triangle with: $\displaystyle opp = 3,\;hyp = 5.$ Pythagorus says: $\displaystyle adj = \pm4 \quad\Rightarrow\quad \cos x \,=\,\pm\tfrac{4}{5}$ Since $\displaystyle x$ in Quadrant 2, $\displaystyle \cos x \,=\,-\tfrac{4}{5}$ Identity: .$\displaystyle \sin2x \,=\,2\sin x\cos x$ Therefore: .$\displaystyle \sin2x \;=\;2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) \;=\; -\frac{24}{25}$ $\displaystyle b)\;\tan x \,=\, -2\:\text{ for }\tfrac{3\pi}{2} < x < 2\pi$ Since $\displaystyle x$ is in Quadrant 4, we have: .$\displaystyle \tan x \,=\,\tfrac{\text{-}2}{1} \,=\,\tfrac{opp}{adj}$ Angle $\displaystyle x$ is in a right triangle with: $\displaystyle opp = \text{-}2,\;adj = 1$ Pythagorus says: .$\displaystyle hyp \,=\,\sqrt{5} \quad\Rightarrow\quad \sin x \,=\,\text{-}\tfrac{2}{\sqrt{5}},\;\cos x \,=\,\tfrac{1}{\sqrt{5}}$ Therefore: .$\displaystyle \sin 2x \:=\:2\left(-\frac{2}{\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}} \right) \;=\; -\frac{4}{5}$
# McGraw Hill My Math Grade 2 Chapter 8 Lesson 5 Answer Key Dollars All the solutions provided inÂ McGraw Hill My Math Grade 2 Answer Key PDF Chapter 8 Lesson 5 Dollars will give you a clear idea of the concepts. ## McGraw-Hill My Math Grade 2 Answer Key Chapter 8 Lesson 5 Dollars Explore and Explain Teacher Directions: Count pennies to 100Â˘. Write the number of quarters it takes to equal 100Â˘. Do the same for the dime and each of the other coins. The number of coins required to make 1 dollar is 100 pennies, 20 nickels, 10 dimes, 4 quarters 1 penny = 1 Â˘, 10 pennies = 100 Â˘ = 1 dollar 1 nickel = 5 Â˘, 20 nickels = 100 Â˘ = 1 dollar 1 dime = 10 Â˘, 10 dimes = 100 Â˘ = 1 dollar 1 quarter = 25 Â˘, 4 quarters = 25 Ă— 4 = 100 Â˘ = 1 dollar. Explanation: See and Show One dollar has a value of 100 cents or 100Â˘. To write one dollar, use a. dollar sign. Use a decimal point to separate the dollars from the cents. Count to find the value of the coins. Circle the combinations that equal $1.00. Question 1. Answer: 3 quarters = 3 Ă— 25 = 75 Â˘. 5 nickels = 5 Ă— 5 = 25 Â˘. Total = 75 + 25 = 100 Â˘. Explanation: Question 2. Answer: 72 Â˘. Explanation: Given, 3 dimes, 3 nickles, 2 pennies, and 1 quarter in the diagram. 1 dime = 10 Â˘, 3 dimes = 3 Ă— 10 = 30 Â˘. 1 nickel = 5 Â˘, 3 nickles = 3 Ă— 5 = 15 Â˘. 1 penny = 1 Â˘, 2 pennies = 2 Ă— 1 = 2 Â˘. 1 quarter = 25 Â˘, Total value of coins = 30 + 15 + 25 + 2 = 72 Â˘. Talk Math How are$ and Â˘ different? How are they alike? The symbol $is called the dollar and the symbol Â˘ is called the cents. Dollars are larger than cents. Explanation: The symbols given are dollars and cents. 1 dollar = 100 cents. Therefore dollars are larger than cents. On My Own Count to find the value of the coins. Circle the combinations that equal$ 1.00. Question 3. Given there are 5 dimes and 5 dimes 1 dime = 10 Â˘, 5 dimes = 50 Â˘ 5 dimes = 50 Â˘ Total value of the coins = 50 Â˘ + 50 Â˘ = 100 Â˘. Explanation: Question 4. The total value of coins = 100 Â˘ = $1.00. Explanation: Given there are 2 quarters, 4 dimes, 2 nickels 50 Â˘Â + 40 Â˘+ 10 Â˘ = 100 Â˘ =$ 1.00. Question 5. The total value of coins given is 50 Â˘. Explanation: Given, 1 quarter, 2 dimes, and 1 nickel. 1 quarter = 25 Â˘, 1 dime = 10 Â˘, 2 dimes = 20 Â˘. 1 nickel = 5 Â˘. 25 + 20 + 5 = 50 Â˘. Question 6. The total value of coins is 100 Â˘ = $1.00. Explanation: Given, There are 1 quarter, 7 dimes, 1 nickel 1 quarter = 25 Â˘ 1 dime = 10 Â˘, 7 dimes = 7 Ă— 10 = 70 Â˘. 1 nickel = 5 Â˘ 25 + 70 + 5 = 100 Â˘. Question 7. Answer: Given there are four quarters 1 quarter = 25 Â˘, 4 quarters = 100 Â˘ =$ 1.00. Explanation: Question 8. Given there are 3 quarters, 2 dimes and1 nickel. The total amount of coins are 100 Â˘ = $1.00. Explanation: 1 quarter = 25 Â˘, 3 quarters = 75 Â˘ 1 dime = 10 Â˘, 2 dimes = 20 Â˘. 1 nickel = 5 Â˘ Total amount of coins are = 75 + 20 + 5 = 100 Â˘. Problem Solving Question 9. Natasha has 1 quarter, 2 dimes, 10 nickels, and 4 pennies. She needs 1 dollar to buy a joke book. How much money does she have? How much more does she need to have one dollar? Answer: She needs 5 Â˘ more to have one dollar. Explanation: Given, Natasha has 1 quarter, 2 dimes, 10 nickels, and 4 pennies 1 quarter = 25 Â˘, 1 dime = 10 Â˘, 1 penny = 1 Â˘ 2 dimes = 20 Â˘, 4 pennies = 4 Â˘, 10 nickels = 50 Â˘ Total value of given coins = 25 + 20 + 4 + 50 = 95 Â˘ But she needs 1 dollar to buy a joke book. 1 dollar = 100 Â˘ 100 Â˘ – 95 Â˘ = 5 Â˘. She needs 5 Â˘ more to have one dollar. Question 10. Chip needs 1 dollar. He has three quarters and one dime. How much does he have? How much more does he need to make 1 dollar? Answer: He needs 15 Â˘ more to make 1 dollar. Explanation: Given, Chip has 1 dollar. He has 3 quarters and 1 dime 1 quarter = 25 Â˘ and 1 dime = 10 Â˘. 3 quarters = 75 Â˘ The total value of coins = 75 + 10 = 85 Â˘. 100 Â˘ – 85 Â˘ = 15 Â˘. Write Math Think of 2 combinations of coins that equal one dollar and write them here. Answer: The 2 combinations of coins are 2 quarters, 4 dimes, 2 nickels and 3 quarters, 2 dimes and1 nickel. Explanation: Given, 2 quarters, 4 dimes, 2 nickels 2 Ă— 25 + 4 Ă— 10 + 2 Ă— 5 50 + 40 + 10 = 100 Â˘. Another combination is 3 quarters, 2 dimes and1 nickel. 3 Ă— 25 + 2 Ă— 10 + 1 Ă— 5 75 + 20 + 5 = 100 Â˘. ### McGraw Hill My Math Grade 2 Chapter 8 Lesson 5 My Homework Answer Key Count to find the value of the coins. Circle combinations that equal$1.00. Question 1. There are 2 quarters, and 5 dimes of coins that make $1.00. Explanation: Given, 2 quarters, 4 dimes 1 quarter = 25 Â˘, 1 dime = 10 Â˘. 2 quarters = 50 Â˘ and 5 dimes = 50 Â˘. Total value of coins = 50 + 50 = 100 Â˘. Question 2. Answer: The total value of the coins = is 68 Â˘. There are 8 pennies and 6 dimes. Explanation: There are 8 pennies and 6 dimes. 1 penny = 1 cent and 1 dime = 10 Â˘. 8 pennies = 8 cents and 6 dimes = 60 cents Total = 68 cents. Question 3. Answer: The total value of the coins is 63 Â˘. Explanation: There are 8 pennies, 1 quarter, 3 dimes. 1 penny = 1 cents, 1 quarter = 25 cents and 1 dime = 10 cents. 8 pennies = 8 cents, 3 dimes = 30 cents Total = 8 + 25 + 30 = 63 Â˘. Question 4. Answer: The total value of coins = 90 Â˘. Explanation: 5 pennies, 3 quarters, 3 nickels. 1 penny = 1 cent, 5 pennies = 5 cents 1 quarter = 25 Â˘, 3 quarters = 75 Â˘. 1 nickel = 5 Â˘, 3 nickels = 15 Â˘. Total value of coins = 75 + 5 + 15 = 90 Â˘. Question 5. Jen has 2 quarters and 4 dimes. She wants to buy 1 bag of pretzels for$ 1. How much more money does she need? 10 Â˘ more to make $1. Explanation: Given, Jen has 2 quarters and 4 dimes 1 quarter = 25 Â˘, 1 dime = 10 Â˘. 2 quarters = 50 Â˘, 4 dimes = 40 Â˘ Total value of coins = 50 + 40 = 90 Â˘. She needs 10 Â˘ more to make$ 1. Question 6. Diego has 3 quarters, 1 dime, and 1 nickel. How many more nickels does he need to have $1 ? _____________ nickels Answer: 1 nickel. Explanation: Given, Diego has 3 quarters, 1 dime, and 1 nickel 1 quarter = 25 Â˘, 3 quarters = 75 Â˘ 1 dime = 10 Â˘ 1 nickel = 5 Â˘ The total value of coins = 75 + 10 + 5 = 95 Â˘. He needs 1 more nickel to have$ 1. Vocabulary Check Circle the correct choices. Question 7. one dollar $1 1$ $1.00 1 Â˘ Answer:$ 1.00 Explanation: Scroll to Top Scroll to Top
Courses Courses for Kids Free study material Offline Centres More Store # The mathematical form for three sinusoidal travelling waves are given by Wave 1 : $y\left( {x,t} \right) = \left( {2cm} \right)\sin \left( {3x - 6t} \right)$Wave 2 : $y\left( {x,t} \right) = \left( {3cm} \right)\sin \left( {4x - 12t} \right)$Wave 3 : $y\left( {x,t} \right) = \left( {4cm} \right)\sin \left( {5x - 11t} \right)$where $x$ is in meters and $t$ is in seconds of these waves(A) Wave 1 has the greatest wave speed and the greatest maximum transverse string speed.(B) Wave 2 has the greatest wave speed and wave 1 has the greatest maximum transverse string speed.(C) Wave 3 has the greatest wave speed and the greatest maximum transverse string speed.(D) Wave 2 has the greatest maximum transverse string speed. Last updated date: 25th Jun 2024 Total views: 395.7k Views today: 11.95k Verified 395.7k+ views Hint: We need to compare each of the wave equations given in the question to the standard wave equation. Then the wave speed will be given by the ratio of $\omega$ and $k$, and the maximum transverse string speed will be the product of the amplitude and $\omega$. Comparing the values for each wave we can find which option is correct. Formula Used: In this solution we will be using the following formula, $\Rightarrow v = \dfrac{\omega }{k}$ where $v$ is the wave speed, $\omega$ is the angular frequency and $k$ is the wave number. $\Rightarrow {u_{\max }} = A\omega$ where ${u_{\max }}$ is the maximum transverse string speed, $A$ is the amplitude. To solve this problem, we need to first compare all the wave equations to the standard wave equation given as $\Rightarrow y = A\sin \left( {kx - \omega t} \right)$ Now for a wave, the wave speed is given as, $\Rightarrow v = \dfrac{\omega }{k}$ and the maximum transverse speed is calculated by taking the derivative of the equation with respect to time, $\Rightarrow {u_{\max }} = {\left. {\dfrac{{dy}}{{dt}}} \right\|_{\max }} = {\left. {\dfrac{d}{{dt}}\left[ {A\sin \left( {kx - \omega t} \right)} \right]} \right\|_{\max }}$ So we get the value as ${u_{\max }} = A\omega$ Now for the first wave we have $y\left( {x,t} \right) = \left( {2cm} \right)\sin \left( {3x - 6t} \right)$ On comparing with $y = A\sin \left( {kx - \omega t} \right)$ we have, $\Rightarrow A = 2cm$, $k = 3$ and $\omega = 6$ So we get for the first wave, $\Rightarrow {v_1} = \dfrac{\omega }{k} = \dfrac{6}{3} = 2cm/s$ and ${u_{\max 1}} = A\omega = 2 \times 6 = 12cm/s$ For the second wave we have $y\left( {x,t} \right) = \left( {3cm} \right)\sin \left( {4x - 12t} \right)$ On comparing with $y = A\sin \left( {kx - \omega t} \right)$ we have, $\Rightarrow A = 3cm$, $k = 4$ and $\omega = 12$ So we get for the second wave, $\Rightarrow {v_2} = \dfrac{\omega }{k} = \dfrac{{12}}{4} = 3cm/s$ and ${u_{\max 2}} = A\omega = 3 \times 12 = 36cm/s$ For the third wave we have $y\left( {x,t} \right) = \left( {4cm} \right)\sin \left( {5x - 11t} \right)$ On comparing with $y = A\sin \left( {kx - \omega t} \right)$ we have, $\Rightarrow A = 4cm$, $k = 5$ and $\omega = 11$ So we get for the third wave, $\Rightarrow {v_3} = \dfrac{\omega }{k} = \dfrac{{11}}{5} = 2.2cm/s$ and ${u_{\max 3}} = A\omega = 4 \times 11 = 44cm/s$ Hence we can say that wave 2 has the greatest wave speed and wave 3 has the greatest maximum transverse string speed. Note The wave speed is the phase velocity of a wave. It is the velocity with which any one frequency component of the wave travels. The phase velocity of the wave can also be written in the terms of the wavelength and time period as $v = \dfrac{\lambda }{T}$.
# Median Of Numbers What is the Median of Numbers? One of the measures of central tendency is median. Median of numbers is the middlemost value of the given set of numbers. It separates the higher half and the lower half of a given data sample. At least half of the observations are smaller than or equal to median and at least half of the observations are greater than or equal to the median. ## How to find the median of numbers? In order to find the median of numbers, we first need to arrange the given numbers either in ascending order or in descending order of their numerical value. It is now easy to find the median after arranging the given numbers. ## Median of Odd Numbers If there are odd number of observations in a data set, then the median can be calculated as below: Step 1: Arrange the data either in ascending or in descending order. Step 2: If the number of observations (say n) is odd, then the middlemost observation is the median of the given data. ## Median of Even Numbers If there are even number of observations in a data set, then the median can be calculated as below: Step 1: Arrange the data either in ascending or in descending order. Step 2: If the number of observations (say n) are even, then identify (n/2)th and [(n/2) + 1]th observations. Step 3: The average of the above two observations (which are identified in step 2) is the median of the given data. Difference Between Mean Median And Mode Median Data Statistics For Class 10 Important Questions Class 9 Maths Chapter 14 Statistics ### Median of Two Numbers If there are only two numbers in a data set, then their median will be the same as their mean (or arithmetic mean or average). For example, the mean and median of 6 and 14 is equal to 10. ### Median of Three Numbers If there are three numbers in a data set, then first arrange them either in ascending order or in descending order. Now the middle-most number is the median of the given data set. For example, 17, 21, and 11 are three numbers of a data set. Now, the descending order of the numbers is 21, 17, 11. Hence,17 is the median. ### Median of a list of Numbers The median is the value which lies at the midpoint of the data set but it is not the midpoint of the values. It doesn’t mean anything for two numbers only. Consider the below illustration to understand in a better way. Let us consider 12 and 6. Mean = Median = (12 + 6)/2 = 9 Adding 2 more values to the above data set. Now, 12, 6, 7, 5 Ascending order: 5, 6, 7, 12 Median = (6 + 7)/2 = 6.5 Mean = (12 + 6 + 7 + 5)/4 = 30/4 = 7.5 Similarly, if we add one more value to the data set, the median will shift. Example 1: The marks scored by 12 students of a class are given by: 40, 50, 60, 55, 70, 45, 60, 80, 65, 50, 82, 58 Find the median marks of the given students. Solution: Given marks are: 40, 50, 60, 55, 70, 45, 60, 80, 65, 50, 82, 58 Ascending order: 40, 45, 50, 50, 55, 58, 60, 60, 65, 70, 80, 82 12 is an even number, i.e. n = 12 and n/2 = 12/2 = 6. 6th observation is 58. 7th observation is 60. Median = (58 + 60)/2 = 118/2 = 59 Example 2: Given below are the ages (in years) of 15 employees in the HR department of a company. 24, 49, 35, 27, 36, 40, 30, 25, 32, 26, 34, 40, 42, 28, 32 Find the median age of the employees in the HR department. Solution: Given data: 24, 49, 35, 27, 36, 40, 30, 25, 32, 26, 34, 40, 42, 28, 32 Ascending order: 24, 25, 26, 27, 28, 30, 32, 32, 34, 35, 36, 40, 40, 42, 49 15 is an odd number, i.e. n = 15 and (n+1)/2 = (15 + 1)/2 = 8 8th observation = 32 Hence, the median age = 32 years
# What is 70 factorial ? Steps to calculate factorial of 70 ## How to calculate the factorial of 70 To find 70 factorial, or 70!, simply use the formula that multiplies the number 70 by all positive whole numbers less than it. Let's look at how to calculate the Factorial of Seventy : 70! is exactly : 11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000000 Factorial of 70 can be calculated as: 70! = 70 x 69 x 68 x 67 x ... x 3 x 2 x 1 The number of trailing zeros in 70! is 16 The number of digits in 70 factorial is 101. ## What Is Factorial? A factorial is displayed by an integer and an exclamation mark. In math concepts, factorial is a multiplication procedure of natural numbers . It multiplies the number by every natural number that is less than it . Symbolically, it is represented as "!". The function is used, among other things, to get the "n" way items can be established . ## Factorial Formula To find the factorial of any given number, substitute the exact value for n in the given solution : n! = n × (n-1) × (n-2) × (n-3) × ….× 3 × 2 × 1 The expansion of the formula provides the numbers to be replicated together to have the factorial of the number. We can also determine a factorial from the prior one. The factorial of any number is that number multiplied the factorial of (that number minus 1). So the rule is : n! = n × (n−1)! Example : 70! Factorial = 70 x 69 x 68 x 67 x ... x 3 x 2 x 1 = 70 × 69! = 11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000000 ## What are Factorials Used For? The best use of factorial is in Combinations and Permutations. Example : Determine how to arrange letters without repeating? There one way for 1 letter "a": 2 ways for two letters "ab": ab, ba. There are 6 ways for 3 letters "abc": abc acb cab bac bca. There are 24 ways for 1234 of the letters "abcd" ## Frequently Asked Questions on Factorial ### Can we have factorials for negative numbers ? Negative integer factorials are undefined ### What Is 0! Zero factorial or Factorial of 0 is simple, and its value is corresponding to 1. So, 0! = 1.
Created By : Abhinandan Kumar Reviewed By : Rajashekhar Valipishetty Last Updated : Apr 21, 2023 Surface Area of Box -The surface area of a box is equal to the sum of the areas of each side of the box. Enter the value of Length(l) Enter the value of Width(w) Enter the value of Height(h) ### What is meant by the Surface Area of a Box? In Mathematics, Rectangular Prism or Box is a three-dimensional figure in which six faces of the solid are rectangles. It is also called cuboid. Opposite faces of the box are equal in measure and have 12 sides and 8 vertices normally. The Surface Area of a Box is equal to the sum of the areas of each side of the box. If "l" is the length, "w" is the width and "h" is the height surface area of the box is given by the formula Surface Area of the box = 2 (lw+lh+hw) square units ### How to Calculate Surface Area of a Box In order to calculate the surface area of a box or rectangular prism all you need to do is find the areas of each side and sum up all those. i.e. A = 2(A1+A2+A3) if "l" is the length "h" is the height and "w" is the width then Areas of all the three sides would be as follows A1 = l w A2 = w * h A3 = l * h Thus, the final formula would be A = 2 (l w+w * h+l * h) A = 2(lw+wh+lh) ∴ Surface area of a Box A = 2(lw+wh+lh) If you aren't Clear about the Surface Area of a Box concept check out the solved examples along with step by step explanation. You will be familiar with how to solve the problems on the Surface Area of a Box by referring to the below examples. Example Calculate the Surface Area of a Box for the length, width, height given as 3Cm, 4 Cm, 5 Cm respectively? Solution: Given Length of a Box = 3 Cm Width of a Box = 4 Cm Height of a Box = 5 Cm Formula to find the Surface Area of a Box is given by A = 2(lw+wh+lh) Substitute the given values in the formula and then we will get the result as under A = 2(34+45+3*5) =2(12+20+15) =2(47) = 94 Cm2 ∴ The surface area of a Box for length 3 cm, width 4 cm, and height 5 cm is 94.0 cm2 ### How to use the Surface Area of a Box Calculator? If you are worried about how to use the Surface Area of a Box Calculator you need not do worry anymore. We will help you out with guidelines and instructions to follow while using the online tool. Following these guidelines, you will arrive at the output easily. They are as under • Firstly, Enter the Length, Width, Height in the input fields provided in the calculator. • Thereafter select the units you want like cm, m, mm, km, in, ft, etc. from the drop-down box. • After that, click on the Area Button. • You will get the output i.e. Surface Area of a Box Calculated showing the detailed explanation in a very short span of time. ### FAQs on Surface Area of a Box 1. How do you find the Surface Area of a Box? In order to calculate the surface area of a box or rectangular prism all you need to do is find the areas of each side and sum up all those. 2. What is the formula for the Surface Area of a Box? The Formula to Calculate the Surface Area of a Box is given by A = 2(lw+wh+lh). 3. How do you Solve Surface Area of a Box Problems Easily? You can Solve the Surface Area of a Box Problems Easily by using our handy tool i.e. Surface Area of a Box Calculator. You just need to provide the input length, width, height, and click on the area button to get the Surface Area of a Box along with detailed explanations. 4. Where can I find the Surface Area of a Box Solved Examples? You can find Various Surface Area of a Box Solved Examples on our page for a better understanding of the concept.
### Theory: There are many techniques to produce more and more irrational numbers between two numbers. But sometimes it is easy to identify the provided number is irrational or not by checking the following properties. Properties of irrational numbers: 1. Addition, subtraction, multiplication and division of two irrational number is may or may not be irrational. 2. Addition of rational and irrational number is always irrational. 3. Subtraction of rational and irrational number is always irrational. 4. Multiplication of rational and irrational is always irrational. 5. Division of rational and irrational is always irrational. 1. The first property says: i) Irrational number $$+$$ Irrational number $$=$$ Irrational$$/$$Rational number. Example: Consider two irrational numbers $$2+\sqrt{3}$$ and $$5+\sqrt{3}$$. $$2+\sqrt{3}+5+\sqrt{3}=7+2\sqrt{3}$$. This is again the irrational number. Now consider another two irrational numbers $$2+\sqrt{3}$$ and $$2-\sqrt{3}$$. $$2+\sqrt{3}+2-\sqrt{3}=4$$. This is a rational number. ii) Irrational number $$-$$ Irrational number $$=$$ Irrational$$/$$Rational number. Example: Consider two irrational numbers $$2+\sqrt{3}$$ and $$5+\sqrt{3}$$. Substract the numbers. $$2+\sqrt{3}+(5+\sqrt{3})=-3$$. It is a rational number. Now consider another two irrational numbers $$2+\sqrt{3}$$ and $$2-\sqrt{3}$$. Substract the numbers. $$2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}$$. It is an irrational number. iii) Irrational number $$×$$ Irrational number $$=$$ Irrational$$/$$Rational number. Example: Consider two irrational numbers $$2\sqrt{3}$$ and $$\sqrt{3}$$. Multiply the numbers. $$2\sqrt{3}\times \sqrt{3}=2\times 3=6$$. It is a rational number. Now consider another two irrational numbers $$2\sqrt{3}$$ and $$\sqrt{2}$$. Multiply the numbers. $$2\sqrt{3}\times \sqrt{2} = 2\sqrt{6}$$. It is an irrational number. iv) Irrational number $÷$ Irrational number $$=$$ Irrational$$/$$Rational number. Example: Consider two irrational numbers $$\sqrt{12}$$ and $$\sqrt{3}$$. Divide the numbers. $$\sqrt{12}\div \sqrt{3} = 2\sqrt{3}\div \sqrt{3} = 2$$. It is a rational number. Now consider another two irrational numbers $$sqrt{15}$$ and $$\sqrt{5}$$. Divide the numbers. $$\sqrt{15}\div \sqrt{5} = \sqrt{3}$$. It is an irrational number. 2. Rational number $$+$$ Irrational number $$=$$ Irrational number. Example: Consider a rational number $$5$$ and an irrational number $$e$$. Adding the rational and irrational numbers becomes $$5+e$$. Adding a rational number with an irrational number will not affect the non-recurring and non-terminating nature of irrational. Thus, the resultant is an irrational number. 3. Rational number $$-$$ Irrational number $$=$$ Irrational number. Example: Consider a rational number $$-8$$ and an irrational number $$\sqrt{19}$$. Substracting the rational and irrational numbers becomes $$-8-\sqrt{19}$$. Subtracting a rational number from the irrational number will not affect the non-recurring and non-terminating nature of irrational. Thus, the resultant is an irrational number. 4. Rational number $$×$$ Irrational number $$=$$ Irrational number. Example: Consider a rational number $$3$$ and an irrational number $$\sqrt{7}$$. Multiplying the rational and irrational numbers becomes, $$3\times\sqrt{7}=3\sqrt{7}$$. Multiplication of a rational number and irrational number will not affect the non-recurring and non-terminating nature of irrational. Thus, the resultant is an irrational number. 5. Rational number $÷$ Irrational number $$=$$ Irrational number. Example: Consider a rational number $$-\frac{3}{2}$$ and an irrational number $$\pi$$. Dividing the rational and irrational numbers becomes, $$-\frac{3}{2}\div \pi = -\frac{3}{2\pi}$$ Division of a rational number and irrational number will not affect the non-recurring and non-terminating nature of irrational. Thus, the resultant is an irrational number.
84 is the amount of pair primes (41 + 43). That is an also composite number that has 2, 3, and also 7 as its prime factors. In this mini lesson, let us learn about the square source of 84, discover out even if it is the square source of 84 is reasonable or irrational, and see how to find the square source of 84 by long division method. Square source of 84√84 = 9.165Square the 84: 842 = 7056 1.You are watching: Square root of 84 in radical form What Is the Square source of 84? 2. Is Square root of 84 reasonable or Irrational? 3. How to discover the Square root of 84? 4. FAQs ~ above Square source of 84 Finding the square root of a number, say n, is finding out what number, speak a, multiplied by itself equals the number n. A × a = n ⇒ a2 = n. Thus a = √n. √84 = √(a × a ) 84 = 9.165 × 9.165 and -9.165 × -9.165√84 = ± 9.165We know that 84 = 2 × 2 × 3 × 7In the most basic radical form √84 = √(2 × 2 × 3 × 7) = 2√21 √84 = 9.1615138991 we cannot compose this as a rational variety of the form p/q. This is a non-terminating decimal. Hence the square root of 84 is irrational. The square root of 84 or any type of number deserve to be calculation in many ways. Two of them are the approximation method and the long department method. ### Square root of 84 by Approximation Method Take 2 perfect square numbers, among which is just smaller sized than 84 and the other is just higher than 84. √81 9 using the mean method, division 84 by 9 or 10.Let us divide through 10⇒ 84 ÷ 10 = 8.4Find the typical of 8.4 and 10.(8.4 + 10) / 2 = 18.4 ÷ 2 = 9.2√84 ≈ 9.2 ### Square source of 84 by the Long division Method The long division method helps us find a much more accurate worth of square root of any type of number. Let"s see just how to discover the square source of by the long department method. Explore square roots utilizing illustrations and interactive examples The square root of any type of number have the right to be assumed come be in between the square source of the two nearest perfect squares of the number. For example, the square source of 108 lies between the square root of 100 and also 121. Therefore, 10 The square root of 84 is evaluated making use of the department method and rounded off to the nearest hundredth. √84 = 9.165. We round it off to the nearest hundredth together 9.17. The square source of 84 is 9.165.The simplified form of radical form is 2√2184 is one irrational number. Example 1: Charlie has actually made 84 cookies. If he needs to arrange castle on the tray as numerous cookies together the variety of rows, how can he arrange them? How numerous cookies will be left out of this arrangement? Solution: Number of cookie per heat × number of rows = total cookies Let cookie per heat = variety of rows = n n × n = 84 n2 = 84 n = 9.1 (approximated to the nearest tenth) He can arrange 81 cookies in 9 rows and also 3 cookies will be left out of the arrangement. Example 2: Sam is playing v his blocks. That has constructed 7 blocks in a row and also extended the form in 12 columns. a) How numerous blocks walk he must remove to make the rectangle come a square? b) exactly how many an ext blocks go he have to make this rectangle come a square? Solution: 7 block in a row × 12 columns = Total number of blocks 7 × 12 = 84 blocks a) He needs to arrange them as a square base. N × n = 84 Since 84 is no a perfect square, let us make it a perfect square. n × n = 81. We subtract 3 native 84 to make it a perfect square. 84 - 3 = 81 Thus he has to remove 3 blocks to build it as a square. b) He needs to arrange them as a square base. N × n = 84 Since 84 is no a perfect square, let us make it a perfect square. n × n = 100. We include 16 to 84 to do it a perfect square. 84 + 16 = 100 Thus he has to add 16 blocks much more to construct it as a square. View much more > go to slidego come slide Breakdown tough principles through basic visuals. See more: Does Mayonnaise Have Dairy In It If You Are Lactose Intolerant? Math will certainly no longer be a hard subject, especially when you know the principles through visualizations.
# Geometric Constructions - Basics ## Objective and Overview In this unit, students make formal geometric constructions with Polypad's geometry tools (utensils) while getting a greater insight into geometrical concepts. In ancient times, geometric constructions of figures and lengths were restricted to the use of only a straightedge and compasses. Today this tradition can be used to create a series of puzzle-like lessons. This unit focuses on the basics of geometric constructions: • Copying a segment • Bisecting a segment • Finding the perpendicular bisector of a line segment • Constructing the center of a circle • Creating perpendicular lines • Constructing a line parallel to a given line through a point • Copying an angle • Bisecting an angle ## Introduction Share this Polypad with students. Ask them to copy the line segment i.e; to draw a line segment congruent to $\overline{AB}$. Students might notice that the line is locked on Polypad so that are not able to select it and just copy it. Invite students to share their methods. Perhaps many used the ruler. Then, ask them to do the same thing without measuring it with the ruler. Invite them to share their thinking. Clarify with the students that it is possible to copy a segment without measuring it. You may use the demo video below or use the construction steps to copy the segment. Copying $\overline{AB}$ means constructing another segment, say $\overline{CD}$, which is congruent to $\overline{AB}$. • Place the compass point on A and the pencil end on point B. • Place the compass point on C and draw an arc without changing the compass width. • Connect any point (D) on the arc with point C. $\overline{CD}$ has the same length as $\overline{AB}$ as long as the compass width is constant. $\overline{AB}≅\overline{CD}$ Discuss with the students the meaning of the words to sketch, draw, and construct using the copying the segment activity. In the first activity, they draw the line using a ruler, whereas in the second one they constructed the segment with the compass. Geometric construction means drawing lines, angles, and shapes accurately without using numbers or equations. To do that, we only need two tools! A straight-edge and a compass. Remind students that a straight-edge is like a ruler but without any markings. A compass allows you to draw a circle of a given size around a point. ## Main Activity The Greeks formulated a great portion of geometry over 2000 years ago. In particular, the mathematician Euclid documented geometric concepts, axioms, logic, and constructions in his book “Elements”. Therefore, these constructions are also known as Euclidean constructions. You may learn more about Euclid’s axioms on Mathigon’s Euclidean Geometry Course. While studying the geometric constructions, students will see the close connection to axiomatic logic used by Euclid to prove his theorems. He proved everything with a minimum of assumptions such as axioms and postulates, and constructed the most complex geometric shapes with a minimum of tools such as a compass and straightedge. From those days to now, using only a straight-edge and compass to construct geometric shapes became almost like a puzzle for mathematicians. Invite students to join you in following in the footsteps of the great mathematicians! Share the following challenges and puzzles students. Options include presenting them one at a time, sharing them all with students at once, or sharing some with groups. ### Bisecting a segment and constructing the perpendicular bisector of a segment First, remind students that there are two rules: • They may only use the compass and the straight-edge tools. • They cannot use the ruler to measure lengths or the protractor to measure the angles. • Place the compass point on point A and open the compass larger than half of the segment's length to draw an arc. • Keep the compass width constant, place the compass point on B to draw the arc. Make sure that the arcs intersect at two points. Label the intersection points as C and D. • Connect C and D with the straight edge. The line segment CD intersects with the segment AB at a point. Label the point as E. • E is the midpoint (median) of $\overlinesegment{AB}$ and $\overlinesegment{CD}$ is the perpendicular bisector of the segment AB. ### Center of a Circle Being able to find the midpoint of a segment without measuring and constructing its perpendicular bisector helps with more complex constructions. One such example is finding the center of a circle. Share this Polypad with students and ask them to construct its center with a compass and straight edge. Let them discuss the possible properties of a circle that can help them to construct (locate) its center. Here is the solution: One of the easiest methods is to find the intersection point of the perpendicular bisectors of any two chords of the circle. This method is based on the fact that the perpendicular bisector of a chord passes through the center of a circle. ### Perpendicular Lines There can be two different options for constructing a perpendicular line to a given line: • Constructing the perpendicular line from a point that is on the line • Constructing the perpendicular line from a point that is not on the line In both cases: • The first step is to place the compass on the given point and draw an arc in a way to intersect with the given segment twice. • Then, place the compass onto the intersection points respectively and open the compass larger than the previous radius to draw the arcs. • Draw the line connecting the intersection point of the arcs and the original point. The newly constructed line is perpendicular to the original line. ### Parallel Lines There are several ways to construct parallel lines. One of the most intuitive ones is using the properties of quadrilaterals with parallel lines. In a rhombus, the congruent sides can be constructed by drawing the radius of a circle. • First place the compass on the given point. • Draw a long arc (almost a full circle) that intersects the original segment twice. • Then, place the compass onto the first intersection point by keeping the compass width constant draw an arc that intersects with the segment. • Now, use this intersection point to draw another arc that intersects with the previous one. That is in fact the fourth vertex of the rhombus. • Finally, connect the original point with the latest constructed point to draw the parallel line. • By this method, one can construct parallel lines as well as a rhombus. ### Copying an angle • Copying $\angle BAC$ means constructing another angle, say $\angle EDF$, congruent to $\angle BAC$. The two angles must have the same measure. • Place the compass point on A and draw an arc. • Keep the compass width constant, place the compass point on D and draw another arc. • Place the compass point on the intersection point of the first arc and ray AB and draw another arc passing through the intersection of the initial arc and the ray AC. • Again, keeping the compass width the same, place the compass point on the intersection of the arc and ray ED, then draw an arc to mark the intersection of both arcs. • Draw a ray from D through the intersection point (F) of the two arcs. The angle EDF is congruent to angle BAC. $\angle EDF\cong \angle BAC$ Ask students to measure the angles with the protractor tool to prove their measures are the same. You can also use the fraction slice of $\frac{1}{5}$ to represent the angle measure. ### Bisecting an angle (Constructing the angle bisector) $\angle ABF\cong \angle FBC$
Courses Courses for Kids Free study material Offline Centres More Store # If a, b, c in the G.P. and ${{a}^{x}}={{b}^{y}}={{c}^{z}}$ then prove that x, y, z are in H.P Last updated date: 15th Jul 2024 Total views: 450.3k Views today: 9.50k Verified 450.3k+ views Hint: use the basic definition of G.P. i.e. if three terms (a, b, c) are in G.P., then, relation between (a, b, c) are ${{b}^{2}}=ac.$ We are given that ${{a}^{x}}={{b}^{y}}={{c}^{z}}...........\left( 1 \right)$ Another information given in the question is that a, b, c are G.P. and hence, we know that, if three terms (a, b, c) in G.P., then can write relation between then as; \begin{align} & {{b}^{2}}=ac \\ & Or \\ & b=\sqrt{ac}........................\left( 2 \right) \\ \end{align} Now we can substitute value of ‘b’ from equation (2) to equation (1), we get; \begin{align} & {{a}^{x}}={{\left( \sqrt{ac} \right)}^{y}}={{c}^{z}} \\ & or \\ & {{a}^{x}}={{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}}...............\left( 3 \right) \\ \end{align} Let us solve the first two terms and second terms individually to get a relation among x, y, z. Now, from first two terms of equation (3), we get; ${{a}^{x}}={{\left( ac \right)}^{\dfrac{y}{2}}}...........\left( 4 \right)$ As we know property of surds that ${{\left( mn \right)}^{r}}={{m}^{r}}{{n}^{r}}$ Hence, we can simplify equation (4), as ${{a}^{x}}={{a}^{\dfrac{y}{2}}}{{c}^{\dfrac{y}{2}}}$ Transferring ${{a}^{\dfrac{y}{2}}}$ to other sides, we get; $\dfrac{{{a}^{x}}}{{{a}^{\dfrac{y}{2}}}}={{c}^{\dfrac{y}{2}}}................\left( 5 \right)$ Now, using property of surds as $\dfrac{{{m}^{r}}}{{{m}^{n}}}={{m}^{r-n}}$ Now, equation (5), becomes ${{a}^{x-\dfrac{y}{2}}}={{c}^{\dfrac{y}{2}}}.................\left( 6 \right)$ Now, taking last two terms of equation (3), we get; ${{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}}$ Now we can simplify the above relation using ${{\left( mn \right)}^{r}}={{m}^{r}}{{n}^{r}}$. Hence, above equation can be written as; \begin{align} & {{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}} \\ & {{a}^{\dfrac{y}{2}}}{{c}^{\dfrac{y}{2}}}={{c}^{z}} \\ \end{align} Transferring ${{c}^{\dfrac{y}{2}}}$ to other sides, we get; ${{a}^{\dfrac{y}{2}}}=\dfrac{{{c}^{z}}}{{{c}^{\dfrac{y}{2}}}}$ Now, using property, $\dfrac{{{m}^{r}}}{{{m}^{n}}}={{m}^{r-n}}$, we can rewrite the given equation as; ${{a}^{\dfrac{y}{2}}}={{c}^{z-\dfrac{y}{2}}}..............\left( 7 \right)$ Now we know the property of surds as, If ${{a}^{m}}={{b}^{n}}$, we can transfer power to other side as $a={{\left( {{b}^{n}} \right)}^{\dfrac{1}{m}}}\text{ or }a={{b}^{\dfrac{n}{m}}}..............\left( 8 \right)$ Using the above property of equation (8), with the equation (7), we get equation (7) as \begin{align} & a={{c}^{\left( z-\dfrac{y}{2} \right)\dfrac{1}{\left( \dfrac{y}{2} \right)}}} \\ & or \\ & a={{c}^{\left( z-\dfrac{y}{2} \right)\dfrac{2}{y}}}..............\left( 9 \right) \\ \end{align} Now putting value of ‘a’ to equation (6) we get; ${{c}^{\left( \left( z-\dfrac{y}{2} \right)\dfrac{2}{y} \right)\left( x-\dfrac{y}{2} \right)}}={{c}^{\dfrac{y}{2}}}..................\left( 10 \right)$ Using the property of surds that if ${{a}^{m}}={{a}^{n}}$ then power should also be equal i.e. m=n. Therefore, we can write from equation (10), $\left( z-\dfrac{y}{2} \right)\dfrac{2}{y}\left( x-\dfrac{y}{2} \right)=\dfrac{y}{2}$ On simplifying the above relation, we get \begin{align} & \left( \dfrac{2z-y}{2} \right)\left( \dfrac{2}{y} \right)\left( \dfrac{2x-y}{2} \right)=\dfrac{y}{2} \\ & \left( 2z-y \right)\left( 2x-y \right)={{y}^{2}} \\ \end{align} Multiplying (2z – y) and (2x – y), we get \begin{align} & 4xz-2yz-2xy+{{y}^{2}}={{y}^{2}} \\ & 4xz-2yz-2xy=0 \\ \end{align} Dividing the whole equation by 2, we get 2xz – yz – xy = 0 Or xy + yz = 2xz Dividing, the whole equation by xyz to both sides, we get, \begin{align} & \dfrac{xy}{xyz}+\dfrac{yz}{xyz}=\dfrac{2xz}{xyz} \\ & \dfrac{1}{z}+\dfrac{1}{x}=\dfrac{2}{y} \\ \end{align} As we know that if three numbers x, y, z are in HP, then $\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{y}$as written in above equation. Hence, it is proved that x, y, z are in H.P. Note: One can go wrong while using the property if ${{a}^{m}}={{c}^{n}}$ then $a={{c}^{\dfrac{n}{m}}}$. One can go wrong while transferring m to other side as If ${{a}^{m}}={{c}^{n}}$then $a={{c}^{\dfrac{n}{m}}}\text{ or }a={{c}^{mn}}$ which are wrong. Hence, be careful while using the above property of surds. Another approach for this question would be that can take log to equation as, ${{a}^{x}}={{b}^{y}}={{c}^{z}}$ Taking log and using property as $\log {{m}^{n}}=n\log m$ \begin{align} & \log {{a}^{x}}=\log {{b}^{y}}=\log {{c}^{z}} \\ & x\log a=y\log b=z\log c \\ \end{align} We know, ${{b}^{2}}=ac$ Taking log to both sides, we get; \begin{align} & \log {{b}^{2}}=\log ac \\ & 2\log b=\log a+\log c \\ \end{align} As, we know log ab = log a + log c Now, using the two equations $x\log a=y\log b=z\log c\text{ and }2\log b=\log a+\log c,$ find relation between x, y and z.
Texas Go Math Grade 7 Lesson 6.2 Answer Key Theoretical Probability of Compound Events Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 6.2 Answer Key Theoretical Probability of Compound Events. Texas Go Math Grade 7 Lesson 6.2 Answer Key Theoretical Probability of Compound Events Finding Probability Using a Table Recall that a compound event consists of two or more simple events. To find the probability of a compound event, you write a ratio of the number of ways the compound event can happen to the total number of equally likely possible outcomes. Jacob rolls two fair number cubes. Find the probability that the numbers he rolls is 8. STEP 1: Use the table to find the sample space for rolling a particular sum on two number cubes. Each cell is the sum of the first number in that row and column. STEP 2: How many possible outcomes are in the sample space? STEP 3: Circle the outcomes that give the sum of 8. ________ STEP 4: How many ways are there to roll a sum of 8? ________ STEP 5: What is the probability of rolling a sum of 8? ________ Reflect Question 1. Give an example of an event that is more likely than rolling a sum of 8. P (rolling a sum of 8) = $$\frac{5}{36}$$ = $$\frac{5}{36}$$ Rolling a sum of 7 is example of an event that is more likely than rolling a sum of 8. There are six ways to roll a sum of 7. P (rolling a sum of 7) = $$\frac{6}{36}$$ = $$\frac{1}{6}$$. Question 2. Give an example of an event that is less likely than rolling a sum of 8. 5 5 P (rolLing a sum of 8) = $$\frac{5}{36}$$ = $$\frac{5}{36}$$ Rolling a sum of 2 is example of an event that is less likely than rolling a sum of 8. Only way to roll a sum of 2 is to roll 1 on each cube and probability for that is: P (rolLing a sum of 2) = $$\frac{1}{36}$$ = $$\frac{1}{36}$$ Example 1 A deli prepares sandwiches with one type of bread (white or wheat), one type of meat (ham, turkey, or chicken), and one type of cheese (cheddar or Swiss). Each combination is equally likely. Find the probability of choosing a sandwich at random and getting turkey and Swiss on wheat bread. STEP 1: Make a tree diagram to find the sample space for the compound event. STEP 2: Find the number of possible outcomes in the sample space: 12 STEP 3: Find the probability of getting turkey and Swiss on wheat bread at random: $$\frac{1}{12}$$ Use the diagram from Example 1 to find the given probabilities. Question 3. P(ham sandwich) _________________________ We have two types of bread for sandwich, and three type of meat, so total number of possible outcomes is 6. Number of favorable outcomes is 2 because we can prepare ham sandwich with one of two types of bread P (ham sandwich) = $$\frac{2}{6}$$ = $$\frac{1}{3}$$ Question 4. P(sandwich containing Swiss cheese) ______________ We have two types of bread for sandwich, three types of meat and two types of cheese, so total number of possible outcomes is 12. Number of favorable outcomes is 6 because we can prepare sandwich with Swiss cheese in 6 ways . P (sandwich containing Swiss cheese) = $$\frac{6}{12}$$ = $$\frac{1}{2}$$ Question 5. Martha types a 4-digit code into a keypad to unlock her car doors. The code uses the numbers 1 and 0. If the digits are selected at random, what is the probability of getting a code with exactly two 0s? Let symbols A. B. C, D represent: A = For first digit in code we have two opportunities: 0 and 1; B = For second digit in code we have two opportunities: 0 and 1; C = For third digit in code we have two opportunities: 0 and 1; D = For forth digit in code we have two opportunities: 0 and 1. Total number of possible outcomes is: A ∙ B ∙ C ∙ D = 2 ∙ 2 ∙ 2 ∙ 2 = 16 For favorable outcomes: Drake rolls two fair number cubes. (Explore Activity) Question 1. Complete the table to find the sample space for rolling a particular product on two number cubes. Each cell is the product of the first number in that row and column. Question 2. What is the probability that the product of the two numbers Drake rolls is a multiple of 4? Count how many numbers in the table are greater than 4. The probability that the product of two numbers Drake rolls is a multiple of 4 is number of the product of two numbers that is greater than 4 divided by total number of possible outcomes: $$\frac{28}{36} \frac{: 4}{: 4} \frac{7}{9}$$ The probability that the product of two numbers Drake rolls is a multiple of 4 is $$\frac{7}{9}$$. Question 3. What is the probability that the product of the two numbers Drake rolls is less than 13? Count how many numbers in the table are less than 13. The probability that the product of two numbers Drake rolls is less than 13 is number of the the product of two numbers that is less than 13 divided by total number of possible outcomes: $$\frac{23}{36}$$ The probability that the product of two numbers Drake rolls is less than 13 is $$\frac{23}{36}$$. You flip three coins and want to explore the probabilities of certain events. (Examples 1 and 2) Question 4. Complete the tree diagram and make a list to find the sample space. List: HHH, HHT, HTH, HTT. List: THH, THT, TTH, TTT. Question 5. How many outcomes are in the sample space? There are 8 outcomes in sample space: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Question 6. List all the ways to get three tails. Only way to flip three tails is that a tail falls on each coin, TTT. Question 7. Complete the expression to find the probability of getting three tails. The probability of getting three tails when three coins are flipped is _________. The probability of getting three tails when three coins are flipped is $$\frac{1}{8}$$. Question 8. What is the probability of getting exactly two heads? There are ________ way(s) to obtain exactly two heads: HHT, __________ There are three ways to obtain exactly two heads. HHT, HTH, THH. P = $$\frac{3}{8}$$ Essential Question Check-In Question 9. There are 6 ways a given compound event can occur. What else do you need to know to find the theoretical probability of the event? Formula for theoretical probability is: P (event) = We know the number of ways the event can occur (6). therefore, to find a theoretical probability we need information for total number of equally likely outcomes. In Exercises 10-12, use the following information. Mattias gets dressed in the dark one morning and chooses his clothes at random. He chooses a shirt (green, red, or yellow), a pair of pants (black or blue), and a pair of shoes (checkered or red). Question 10. Use the space below to make a tree diagram to find the sample space. The sample space has 12 elements. Question 11. What is the probability that Mattias picks an outfit at random that includes red shoes? ___________________________ Total number of possible outcomes is: 3 possibilities for shirt ∙ 2 possibilities for jeans ∙ 2 possibilities for shoes = 3 ∙ 2 ∙ 2 = 12 Number of outfit with red shoes: green-black-red shoes, green-blue-red shoes, red-black-red shoes, red-blue-red shoes, yellow-black-red shoes, yellow-bLue-red shoes = 6 P (outfit with red shoes) = The probability to pick an outfit at random that includes red shoes is $$\frac{1}{2}$$. Question 12. What is the probability that no part of Mattias’s outfit is red? _____________________ Total number of possible outcomes is: 3 possibilities for shirt ∙ 2 possibilities for jeans ∙ 2 possibiLities for shoes = 3 ∙ 2 ∙ 2 = 12 Number of outfit without red clothes or shoes: green-black-checkered, green-blue-checkered, yellow-black checkered, yellow-blue-checkered = 4 P (no part of outfit is red) = The probability that no part of Mattia’s outfit is red is $$\frac{1}{3}$$. Question 13. Rhee and Pamela are two of the five members of a band. Every week, the band picks two members at random to play on their own for five minutes. What is the probability that Rhee and Pamela are chosen this week? Total number of possible outcomes is 5 possibilities to pick one member• 4 possibiLities to pick other member = 5 ∙ 4 = 20 To pick other member we have 4 possibiLities because we already pick one from five for first member so 5 – 1 = 4. Number of possibilities to pick Pamela and Rhee at the same time is 2. P (picking Pamela and Rhee) = The probability that Pamela and Rhee are chosen this week is $$\frac{1}{10}$$ Question 14. Ben rolls two number cubes. What is the probability that the sum of the numbers he rolls is less than 6? Count how many numbers in the table are less than 6. The probability that the sum of two numbers Ben rolls is less than 6 is number of the sum of two numbers that is less than 6 divided by total number of possible outcomes: $$\frac{10}{36}$$ = $$\frac{5}{18}$$ The probability that the sum of two numbers Ben rolls is less than 6 is $$\frac{5}{18}$$. Question 15. Nhan is getting dressed. He considers two different shirts, three pairs of pants, and three pairs of shoes. He chooses one of each of the articles at random. What is the probability that he will wear his jeans but not his sneakers? Number of combination with jeans but without sneakers: collared shirt, jeans and flip-flops; collared shirt, jeans and sandals; t-shirt, jeans and flip-flops; t-shirt, jeans and sandals. P (wearing jeans but not sneakers) = The probability that he will wear his jeans but not his sneakers is $$\frac{2}{9}$$. Question 16. Communicate Mathematical Ideas A ski resort has 3 chair lifts, each with access to 6 ski trails. Explain how you can find the number of possible outcomes when choosing a chair lift and a ski trail without making a list, a tree diagram, or table. Each chair has access to 6 ski trails, therefore, number of possible outcomes is 3 ∙ 6 = 18. Number of possible outcomes is 18. Question 17. Explain the Error For breakfast, Sarah can choose eggs, granola or oatmeal as a main course, and orange juice or milk for a drink. Sarah says that the sample space for choosing one of each contains 32 = 9 outcomes. What is her error? Explain. Sara’s error:She multiply possibilities of her meal which contain milk with eggs, granola and oatmeal and possibilities that contain orange jouce with egg, granola and oatmeal. Right answer 5: With milk Sarah can choose eggs, granola and oatmeal and that is three possibilities. With orange juice Sarah can choose eggs, granola and oatmeal, that also three possibilities. Sample space is: 3 + 3 = 6 The sample space is 6 outcomes. Question 18. Represent Real-World Problems A new shoe comes in two colors, black or red, and in sizes from 5 to 12, including half sizes. If a pair of the shoes is chosen at random for a store display, what is the probability it will be red and size 9 or longer? Total number of possible outcomes: Contain 15 sizes of shoes and two color, so 15 ∙ 2 = 30 P (red and size 9 or larger) = The probability to choose red shoes with size 9 or larger is $$\frac{7}{30}$$. H.O.T. Focus on Higher Order Thinking Question 19. Analyze Relationships At a diner, Sondra tells the server, “Give me one item from each column.” Gretchen says, “Give me one main dish and a vegetable.” Who has a greater probability of getting a meal that includes salmon? Explain. Total number of possible outcomes for Sondra: She order main dish, vegetable and side, so 4 possibilities for mean dish 4 possibilities for vegetable ∙ 2 possibiLities for side = 32. Number of meal with salmon for Sondra: 4 possibilities for vegetable multiply by 2 possibilities for side, so 4 ∙ 2 = 8 The probability of getting a meal that includes salmon for Sondra is: P (getting a salmon) = = $$\frac{8}{32}$$ = $$\frac{1}{4}$$ Total number of possible outcomes for Gretchen: She order main dish and vegetable, so 4 ∙ 4 ∙ 2 = 16 Number of meal with salmon for Gretchen: 2 possibilities for side, so 1 ∙ 2 = 2 The probability of getting a meaL that includes salmon for Gretchen is: P (getting a salmon) = = $$\frac{2}{16}$$ = $$\frac{1}{8}$$ Sondra has greater possibilities than Gretchen. Question 20. The digits 1 through 5 are used for a set of locker codes. a. Look for a Pattern Suppose the digits cannot repeat. Find the number of possible two-digit codes and three-digit codes. Describe any pattern and use it to predict the number of possible five-digit codes. Total number of possible outcomes for two-digits code is: 5 possibilities for first digit ∙ 4 possibilities for second digit = 20 Total number of possible outcomes for three-digits code is: 5 possibilities for first digit ∙ 4 possibilities for second digit ∙ 3 possibilities for third digit = 60 Find the number of possible five-digits codes with even number as first digit. Digits do not repeat, and second, third, forth and fifth digit do not depend on number parity. Total number of possible outcomes for five-digits code S: 2 possibilities for first digit ∙ 4 possibilities for second digit ∙ 3 possibilities for third ∙ 2 possibilities for forth ∙ 1 possibilities for fifth digit = 48 b. Look for a Pattern Repeat part a, but allow digits to repeat. Total number of possible outcomes for two-digits code is: 5 possibilities for first digit ∙ 5 possibilities for second digit = 25 Total number of possible outcomes for three-digits code is: 5 possibilities for first digit ∙ 5 possibilities for second dig it ∙ 5 possibilities for third digit = 125 Find the number of possible five-digits codes with even number as first digit Digits can be repeated, and second, third, forth and fifth digit do not depend on number parity. Total number of possible outcomes for five-digits code is: 2 possibilities for first digit ∙ 5 possibilities for second digit ∙ 5 possibilities for third 5 possibilities for forth ∙ 5 possibilities for fifth digit = 1250 c. Justify Reasoning Suppose that a gym plans to issue numbered locker codes by choosing the digits at random. Should the gym use codes in which the digits can repeat or not? Justify your reasoning.
Question Video: Finding the Result of Adding Three Fractions with Different Denominators and Expressing It as a Mixed Number \| Nagwa Question Video: Finding the Result of Adding Three Fractions with Different Denominators and Expressing It as a Mixed Number \| Nagwa # Question Video: Finding the Result of Adding Three Fractions with Different Denominators and Expressing It as a Mixed Number Mathematics Calculate 1/6 + 5/8 + 1/2. Give your answer as a mixed number. 02:06 ### Video Transcript Calculate one-sixth plus five-eighths plus one-half. Give your answer as a mixed number. To find the sum of these three fractions, we first need to find the least common denominator. In this case, that will be the smallest common multiple of six, eight, and two. We begin by listing a few multiples of six, which includes six, 12, 18, 24, and 30. When we do the same for eight and two, we find the least common multiple is 24. We note that sometimes we need to create longer lists of multiples for smaller denominators before we can find something in common. Now we need to convert the fractions to equivalent fractions with denominator 24. We multiply the numerator and denominator of one-sixth by four. This shows that one-sixth is equivalent to four twenty-fourths. Next, we multiply the numerator and denominator of five-eighths by three. This shows that five-eighths is equivalent to fifteen twenty-fourths. Finally, we multiply the numerator and denominator of one-half by 12. Therefore, we know that one-half is equivalent to twelve twenty-fourths. Now that we have equivalent fractions where the denominators are the same, we can simply add the numerators, four, 15, and 12, to get 31. Altogether, we have thirty-one twenty-fourths, which can be written as a mixed number. The whole number part, one, is equivalent to twenty-four twenty-fourths, so seven out of the 31 parts remain. So we write our mixed number answer as one and seven twenty-fourths. This is the sum of one-sixth plus five-eighths plus one-half. ## 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
## Relations Between Roots Solution4 The page relations between roots solution4 is containing solution of some practice questions from the worksheet relationship between roots and coefficients. (5) If α and β are the roots of 2 x² - 3 x - 5 = 0, form a quadratic equation whose roots are α² and β². Solution: General form of quadratic equation whose roots are α and β x² - (α + β) x + α β = 0 by comparing the given equation with general form of quadratic equation we get a = 2 b = -3 and c = -5 Sum of the roots α + β = -b/a = -(-3)/2 = 3/2 Product of roots α β = c/a = -5/2 here α = α² and β = β² General form of quadratic equation whose roots are α² and β² x² - (α² + β²) x + α² β² = 0 x² - (α² + β²) x + β)² = 0 α² + β² = (α + β)² - 2 α β = (3/2)² - 2 (-5/2) = (9/4) + 5 = (9 + 20)/4 = 29/4 x² - (α² + β²) x + β)² = 0 x² - (29/4) x + (-5/2)² = 0 x² - (29/4) x + (25/4) = 0 (4 x² - 29 x + 25)/4 = 0 4 x² - 29 x + 25 = 0 Therefore the required quadratic equation is 4 x² - 29 x + 25 = 0. (6) If α and β are the roots of x² - 3 x + 2 = 0, form a quadratic equation whose roots are -α and -β. Solution: General form of quadratic equation whose roots are α and β x² - (α + β) x + α β = 0 by comparing the given equation with general form of quadratic equation we get a = 1 b = -3 and c = 2 Sum of the roots α + β = -b/a = -(-3)/1 = 3 Product of roots α β = c/a = 2/1 = 2 here α = - α and β = - β General form of quadratic equation whose roots are α² and β² x² - ( -β) x + (-α) (-β) = 0 x² + (α + β) x + α β = 0 x² - (3) x + 2 = 0 Therefore the required quadratic equations is x² -3 x + 2 = 0. These are the problems solved in the page relations between roots solution4. You can find solution of other problems in the next page. relations between roots solution4
The statement $3 \times 4 = 12$ has two parts: the numbers that are being multiplied, and the answer. The number that are being multiplied are factors, and the answer is the product. We say that $3$ and $4$ are factors of $12$. The factors of a number are those number whose continued product is the number: thus $4$ and $2$ are factors of $8$; $3$ and $6$ or $3$, $3$ and $2$ are the factors of $18$. Factors can also be used to determine how quantities can be divided into smaller groups. In this section we will be learning more about factors of a number. ## Factor of a Number Definition If $a$ and $b$ represent integers, then $a$ is said to be a factor of $b$ if $a$ divides $b$ evenly, that is, if $a$ divides $b$ with no remainder. The prime factors of a number are those prime numbers whose continued product is the number; thus the prime factors of $12$ are $2$, $2$ and $3$. ## Steps to Factoring The factors of a number can be found by two methods: 1. Using division 2. Using multiplication 1. Factorization using division: In this method, start dividing the given number by $1$, $2$, $3$ and so on to determine which number can divide the given number exactly without giving any remainder. When the remainder is zero, both the divisor and the quotient are the factors of the number. When u reach a point where the factors begin to be repeated, there will not be any further factors. For example find the factors of $20$ $20 \div 1 = 20$ … $1$ and $20$ are factors of $20$ $20 \div 2 = 10$ … $2$ and $10$ are factors of $20$ $20 \div 3 =$ Leaves reminder $20 \div 4 = 5$ … $4$ and $5$ are factors of $20$ $20 \div 6 =$ Leaves reminder $20 \div 7 =$ Leaves reminder $20 \div 8 =$ Leaves reminder $20 \div 9 =$ Leaves reminder $20 \div 10 = 2$ … $10$ and $2$ are factors of $20$ Stop here, since $10$ and $2$ are already the factors. $\therefore$ the factors of $20$ are $1$, $2$, $4$, $5$, $10$. 2. Factorization by multiplication: All the numbers used to build a product are the factors of the product. For example find the factors of $18$. $1 \times 18 = 18$ $2 \times 9 = 18$ $3 \times 6 = 18$ $\therefore$, the factors of $18$ are $1, 2, 3, 6, 9$ and $18$. ## Finding Factors of a Number The following are examples of factors of a number. ### Solved Examples Question 1: Is 8 is a factor of 90? Solution: In order to find the whether 8 is a factor of 90 or not, divide 90 by 8. 8) 90 (11 10 8 2 Since there is a reminder, 8 is not a factor of 90. Question 2: What are the factors of 36? Solution: $36 \div 1 = 36$ … $1$ and $36$ are factors of $36$ $36 \div 2 = 18$ … $2$ and $18$ are factors of $36$ $36 \div 3 = 12$ … $3$ and $12$ are factors of $36$ $36 \div 4 = 9$ … $4$ and $9$ are factors of $36$ $36 \div 6 = 6$ … $6$ is a factor of $36$ $36 \div 7 =$ Leaves reminder $36 \div 8 =$ Leaves reminder $36 \div 9 = 4$ … $9$ and $4$ are factors of $36$ Stop here, since 9 and 4 are already the factors. $\therefore$ the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36 Question 3: Find the factors of 16 by multiplication method. Solution: we need to multiply two numbers to get the given number. $1 \times 16 = 16$ $2 \times 8 = 16$ $4 \times 4 = 16$ $\therefore$ the factors of 16 are 1, 2, 4, 8, 16.
2 Vectors # 2.2 Coordinate Systems and Components of a Vector ### Learning Objectives By the end of this section, you will be able to: • Describe vectors in two and three dimensions in terms of their components, using unit vectors along the axes. • Distinguish between the vector components of a vector and the scalar components of a vector. • Explain how the magnitude of a vector is defined in terms of the components of a vector. • Identify the direction angle of a vector in a plane. • Explain the connection between polar coordinates and Cartesian coordinates in a plane. Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction $37^\circ$ north of east. In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y). In a similar fashion, a vector $\mathbf{\overset{\to }{A}}$ in a plane is described by a pair of its vector coordinates. The x-coordinate of vector $\mathbf{\overset{\to }{A}}$ is called its x-component and the y-coordinate of vector $\mathbf{\overset{\to }{A}}$ is called its y-component. The vector x-component is a vector denoted by ${\mathbf{\overset{\to }{A}}}_{x}$. The vector y-component is a vector denoted by ${\mathbf{\overset{\to }{A}}}_{y}$. In the Cartesian system, the x and y vector components of a vector are the orthogonal projections of this vector onto the x– and y-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components: $\mathbf{\overset{\to }{A}}={\mathbf{\overset{\to }{A}}}_{x}+{\mathbf{\overset{\to }{A}}}_{y}.$ As illustrated in Figure, vector $\mathbf{\overset{\to }{A}}$ is the diagonal of the rectangle where the x-component ${\mathbf{\overset{\to }{A}}}_{x}$ is the side parallel to the x-axis and the y-component ${\mathbf{\overset{\to }{A}}}_{y}$ is the side parallel to the y-axis. Vector component ${\mathbf{\overset{\to }{A}}}_{x}$ is orthogonal to vector component ${\mathbf{\overset{\to }{A}}}_{y}$. It is customary to denote the positive direction on the x-axis by the unit vector $\mathbf{\hat{i}}$ and the positive direction on the y-axis by the unit vector $\mathbf{\hat{j}}$. Unit vectors of the axes, $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$, define two orthogonal directions in the plane. As shown in Figure, the x– and y– components of a vector can now be written in terms of the unit vectors of the axes: $\left\{\begin{array}{c}{\mathbf{\overset{\to }{A}}}_{x}={A}_{x}\mathbf{\hat{i}}\\ {\mathbf{\overset{\to }{A}}}_{y}={A}_{y}\mathbf{\hat{j}}.\end{array}\right.$ The vectors ${\mathbf{\overset{\to }{A}}}_{x}$ and ${\mathbf{\overset{\to }{A}}}_{y}$ defined by Figure are the vector components of vector $\mathbf{\overset{\to }{A}}$. The numbers ${A}_{x}$ and ${A}_{y}$ that define the vector components in Figure are the scalar components of vector $\mathbf{\overset{\to }{A}}$. Combining Figure with Figure, we obtain the component form of a vector: $\mathbf{\overset{\to }{A}}={A}_{x}\mathbf{\hat{i}}+{A}_{y}\mathbf{\hat{j}}.$ If we know the coordinates $b({x}_{b},{y}_{b})$ of the origin point of a vector (where b stands for “beginning”) and the coordinates $e({x}_{e},{y}_{e})$ of the end point of a vector (where e stands for “end”), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates: $\left\{\begin{array}{c}{A}_{x}={x}_{e}-{x}_{b}\\ {A}_{y}={y}_{e}-{y}_{b}.\end{array}\right.$ ### Example #### Displacement of a Mouse Pointer A mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) with respect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what is the displacement vector of the pointer? #### Strategy The origin of the xy-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit vector $\mathbf{\hat{i}}$ on the x-axis points horizontally to the right and the unit vector $\mathbf{\hat{j}}$ on the y-axis points vertically upward. The origin of the displacement vector is located at point b(6.0, 1.6) and the end of the displacement vector is located at point e(2.0, 4.5). Substitute the coordinates of these points into Figure to find the scalar components ${D}_{x}$ and ${D}_{y}$ of the displacement vector $\mathbf{\overset{\to }{D}}$. Finally, substitute the coordinates into Figure to write the displacement vector in the vector component form. #### Solution We identify ${x}_{b}=6.0$, ${x}_{e}=2.0$, ${y}_{b}=1.6$, and ${y}_{e}=4.5$, where the physical unit is 1 cm. The scalar x- and y-components of the displacement vector are $\begin{array}{cc}\hfill {D}_{x}& ={x}_{e}-{x}_{b}=(2.0-6.0)\text{cm}=-4.0\,\text{cm},\hfill \\ \hfill {D}_{y}& ={y}_{e}-{y}_{b}=(4.5-1.6)\text{cm}=+2.9\,\text{cm}.\hfill \end{array}$ The vector component form of the displacement vector is $\mathbf{\overset{\to }{D}}={D}_{x}\mathbf{\hat{i}}+{D}_{y}\mathbf{\hat{j}}=(-4.0\,\text{cm})\mathbf{\hat{i}}+(2.9\,\text{cm})\mathbf{\hat{j}}=(-4.0\mathbf{\hat{i}}+2.9\mathbf{\hat{j}})\text{cm}.$This solution is shown in (Figure 2.17). #### Significance Notice that the physical unit—here, 1 cm—can be placed either with each component immediately before the unit vector or globally for both components, as in Figure. Often, the latter way is more convenient because it is simpler. The vector x-component ${\mathbf{\overset{\to }{D}}}_{x}=-4.0\mathbf{\hat{i}}=4.0(\text{−}\mathbf{\hat{i}})$ of the displacement vector has the magnitude $\|{\mathbf{\overset{\to }{D}}}_{x}\|=\|-4.0\|\|\mathbf{\hat{i}}\|=4.0$ because the magnitude of the unit vector is $\|\mathbf{\hat{i}}\|=1$. Notice, too, that the direction of the x-component is $\text{−}\mathbf{\hat{i}}$, which is antiparallel to the direction of the +x-axis; hence, the x-component vector ${\mathbf{\overset{\to }{D}}}_{x}$ points to the left, as shown in Figure. The scalar x-component of vector $\mathbf{\overset{\to }{D}}$ is ${D}_{x}=-4.0$. Similarly, the vector y-component ${\mathbf{\overset{\to }{D}}}_{y}=+2.9\mathbf{\hat{j}}$ of the displacement vector has magnitude $\|{\mathbf{\overset{\to }{D}}}_{y}\|=\|2.9\|\|\mathbf{\hat{j}}\|=\,2.9$ because the magnitude of the unit vector is $\|\mathbf{\hat{j}}\|=1$. The direction of the y-component is $+\mathbf{\hat{j}}$, which is parallel to the direction of the +y-axis. Therefore, the y-component vector ${\mathbf{\overset{\to }{D}}}_{y}$ points up, as seen in Figure. The scalar y-component of vector $\mathbf{\overset{\to }{D}}$ is ${D}_{y}=+2.9$. The displacement vector $\mathbf{\overset{\to }{D}}$ is the resultant of its two vector components. The vector component form of the displacement vector Figure tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position. A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing. Show Solution $\mathbf{\overset{\to }{D}}=(-5.0\mathbf{\hat{i}}-3.0\mathbf{\hat{j}})\text{cm}$; the fly moved 5.0 cm to the left and 3.0 cm down from its landing site. When we know the scalar components ${A}_{x}$ and ${A}_{y}$ of a vector $\mathbf{\overset{\to }{A}}$, we can find its magnitude A and its direction angle ${\theta }_{A}$. The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis. The angle ${\theta }_{A}$ is measured in the counterclockwise direction from the +x-axis to the vector (Figure). Because the lengths A, ${A}_{x}$, and ${A}_{y}$ form a right triangle, they are related by the Pythagorean theorem: ${A}^{2}={A}_{x}^{2}+{A}_{y}^{2}\enspace\Leftrightarrow\enspace{A}=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}.$ This equation works even if the scalar components of a vector are negative. The direction angle ${\theta }_{A}$ of a vector is defined via the tangent function of angle ${\theta }_{A}$ in the triangle shown in Figure: $\text{tan}\,{\theta }_{A}=\frac{{A}_{y}}{{A}_{x}}\enspace\Rightarrow \enspace{\theta }_{A}={\text{tan}}^{-1}(\frac{{A}_{y}}{{A}_{x}}).$ When the vector lies either in the first quadrant or in the fourth quadrant, where component ${A}_{x}$ is positive (Figure), the angle $\theta$ in Figure is identical to the direction angle ${\theta }_{A}$. For vectors in the fourth quadrant, angle $\theta$ is negative, which means that for these vectors, direction angle ${\theta }_{A}$ is measured clockwise from the positive x-axis. Similarly, for vectors in the second quadrant, angle $\theta$ is negative. When the vector lies in either the second or third quadrant, where component ${A}_{x}$ is negative, the direction angle is ${\theta }_{A}=\theta +180^\circ$ (Figure). ### Example #### Magnitude and Direction of the Displacement Vector You move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What are the magnitude and direction of the displacement vector of the pointer? #### Strategy In Figure, we found the displacement vector $\mathbf{\overset{\to }{D}}$ of the mouse pointer (see Figure). We identify its scalar components ${D}_{x}=-4.0\,\text{cm}$ and ${D}_{y}=+2.9\,\text{cm}$ and substitute into Figure and Figure to find the magnitude D and direction ${\theta }_{D}$, respectively. #### Solution The magnitude of vector $\mathbf{\overset{\to }{D}}$ is $D=\sqrt{{D}_{x}^{2}+{D}_{y}^{2}}=\sqrt{{(-4.0\,\text{cm})}^{2}+{(2.9\,\text{cm})}^{2}}=\sqrt{{(4.0)}^{2}+{(2.9)}^{2}}\,\text{cm}=4.9\,\text{cm}.$ The direction angle is $\text{tan}\,\theta =\frac{{D}_{y}}{{D}_{x}}=\frac{+2.9\,\text{cm}}{-4.0\,\text{cm}}=-0.725\enspace\Rightarrow \enspace\theta ={\text{tan}}^{-1}(-0.725)=-35.9^\circ.$ Vector $\mathbf{\overset{\to }{D}}$ lies in the second quadrant, so its direction angle is ${\theta }_{D}=\theta +180^\circ=-35.9^\circ+180^\circ=144.1^\circ.$ If the displacement vector of a blue fly walking on a sheet of graph paper is $\mathbf{\overset{\to }{D}}=(-5.00\mathbf{\hat{i}}-3.00\mathbf{\hat{j}})\text{cm}$, find its magnitude and direction. Show Solution 5.83 cm, $211^\circ$ In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known. Let us return to the right triangle in Figure. The quotient of the adjacent side ${A}_{x}$ to the hypotenuse A is the cosine function of direction angle ${\theta }_{A}$, ${A}_{x}\text{/}A=\text{cos}\,{\theta }_{A}$, and the quotient of the opposite side ${A}_{y}$ to the hypotenuse A is the sine function of ${\theta }_{A}$, ${A}_{y}\text{/}A=\text{sin}\,{\theta }_{A}$. When magnitude A and direction ${\theta }_{A}$ are known, we can solve these relations for the scalar components: $\left\{\begin{array}{c}{A}_{x}=A\,\text{cos}\,{\theta }_{A}\\ {A}_{y}=A\,\text{sin}\,{\theta }_{A}\end{array}\right. .$ When calculating vector components with Figure, care must be taken with the angle. The direction angle ${\theta }_{A}$ of a vector is the angle measured counterclockwise from the positive direction on the x-axis to the vector. The clockwise measurement gives a negative angle. ### Example #### Components of Displacement Vectors A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction $30^\circ$ west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns $23^\circ$ west of south for 150.0 m. Find the scalar components of Trooper’s displacement vectors and his displacement vectors in vector component form for each leg. #### Strategy Let’s adopt a rectangular coordinate system with the positive x-axis in the direction of geographic east, with the positive y-direction pointed to geographic north. Explicitly, the unit vector $\mathbf{\hat{i}}$ of the x-axis points east and the unit vector $\mathbf{\hat{j}}$ of the y-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use Figure to find the scalar components of the displacements and Figure for the displacement vectors. #### Solution On the first leg, the displacement magnitude is ${L}_{1}=200.0\,\text{m}$ and the direction is southeast. For direction angle ${\theta }_{1}$ we can take either $45^\circ$ measured clockwise from the east direction or $45^\circ+270^\circ$ measured counterclockwise from the east direction. With the first choice, ${\theta }_{1}=-45^\circ$. With the second choice, ${\theta }_{1}=+315^\circ$. We can use either one of these two angles. The components are $\begin{array}{l}{L}_{1x}={L}_{1}\,\text{cos}\,{\theta }_{1}=(200.0\,\text{m})\,\text{cos}\,315^\circ=141.4\,\text{m,}\\ {L}_{1y}={L}_{1}\,\text{sin}\,{\theta }_{1}=(200.0\,\text{m})\,\text{sin}\,315^\circ=-141.4\,\text{m}.\end{array}$ The displacement vector of the first leg is ${\mathbf{\overset{\to }{L}}}_{1}={L}_{1x}\mathbf{\hat{i}}+{L}_{1y}\mathbf{\hat{j}}=(141.4\mathbf{\hat{i}}-141.4\mathbf{\hat{j}})\,\text{m}.$ On the second leg of Trooper’s wanderings, the magnitude of the displacement is ${L}_{2}=300.0\,\text{m}$ and the direction is north. The direction angle is ${\theta }_{2}=+90^\circ$. We obtain the following results: $\begin{array}{ccc}\hfill {L}_{2x}& =\hfill & {L}_{2}\,\text{cos}\,{\theta }_{2}=(300.0\,\text{m})\,\text{cos}\,90^\circ=0.0\,,\hfill \\ \hfill {L}_{2y}& =\hfill & {L}_{2}\,\text{sin}\,{\theta }_{2}=(300.0\,\text{m})\,\text{sin}\,90^\circ=300.0\,\text{m,}\hfill \\ \hfill {\mathbf{\overset{\to }{L}}}_{2}& =\hfill & {L}_{2x}\mathbf{\hat{i}}+{L}_{2y}\mathbf{\hat{j}}=(300.0\,\text{m})\mathbf{\hat{j}}.\hfill \end{array}$ On the third leg, the displacement magnitude is ${L}_{3}=50.0\,\text{m}$ and the direction is $30^\circ$ west of north. The direction angle measured counterclockwise from the eastern direction is ${\theta }_{3}=30^\circ+90^\circ=+120^\circ$. This gives the following answers: $\begin{array}{ccc}\hfill {L}_{3x}& =\hfill & {L}_{3}\,\text{cos}\,{\theta }_{3}=(50.0\,\text{m})\,\text{cos}\,120^\circ=-25.0\,\text{m,}\hfill \\ \hfill {L}_{3y}& =\hfill & {L}_{3}\,\text{sin}\,{\theta }_{3}=(50.0\,\text{m})\,\text{sin}\,120^\circ=+43.3\,\text{m,}\hfill \\ \hfill {\mathbf{\overset{\to }{L}}}_{3}& =\hfill & {L}_{3x}\mathbf{\hat{i}}+{L}_{3y}\mathbf{\hat{j}}=(-25.0\mathbf{\hat{i}}+43.3\mathbf{\hat{j}})\text{m}.\hfill \end{array}$ On the fourth leg of the excursion, the displacement magnitude is ${L}_{4}=80.0\,\text{m}$ and the direction is south. The direction angle can be taken as either ${\theta }_{4}=-90^\circ$ or ${\theta }_{4}=+270^\circ$. We obtain $\begin{array}{ccc}\hfill {L}_{4x}& =\hfill & {L}_{4}\,\text{cos}\,{\theta }_{4}=(80.0\,\text{m})\,\text{cos}\,(-90^\circ)=0\,,\hfill \\ \hfill {L}_{4y}& =\hfill & {L}_{4}\,\text{sin}\,{\theta }_{4}=(80.0\,\text{m})\,\text{sin}\,(-90^\circ)=-80.0\,\text{m,}\hfill \\ \hfill {\mathbf{\overset{\to }{L}}}_{4}& =\hfill & {L}_{4x}\mathbf{\hat{i}}+{L}_{4y}\mathbf{\hat{j}}=(-80.0\,\text{m})\mathbf{\hat{j}}.\hfill \end{array}$ On the last leg, the magnitude is ${L}_{5}=150.0\,\text{m}$ and the angle is ${\theta }_{5}=-23^\circ+270^\circ=+247^\circ$ $(23^\circ$ west of south), which gives $\begin{array}{ccc}\hfill {L}_{5x}& =\hfill & {L}_{5}\,\text{cos}\,{\theta }_{5}=(150.0\,\text{m})\,\text{cos}\,247^\circ=-58.6\,\text{m,}\hfill \\ \hfill {L}_{5y}& =\hfill & {L}_{5}\,\text{sin}\,{\theta }_{5}=(150.0\,\text{m})\,\text{sin}\,247^\circ=-138.1\,\text{m,}\hfill \\ \hfill {\mathbf{\overset{\to }{L}}}_{5}& =\hfill & {L}_{5x}\mathbf{\hat{i}}+{L}_{5y}\mathbf{\hat{j}}=(-58.6\mathbf{\hat{i}}-138.1\mathbf{\hat{j}})\text{m}.\hfill \end{array}$ If Trooper runs 20 m west before taking a rest, what is his displacement vector? Show Solution $\mathbf{\overset{\to }{D}}=(-20\,\text{m})\mathbf{\hat{j}}$ ### Polar Coordinates To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$ along the x-axis and the y-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the polar coordinate system. In the polar coordinate system, the location of point P in a plane is given by two polar coordinates (Figure). The first polar coordinate is the radial coordinate r, which is the distance of point P from the origin. The second polar coordinate is an angle $\phi$ that the radial vector makes with some chosen direction, usually the positive x-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point P. This radial direction is described by a unit radial vector $\mathbf{\hat{r}}$. The second unit vector $\mathbf{\hat{t}}$ is a vector orthogonal to the radial direction $\mathbf{\hat{r}}$. The positive $+\mathbf{\hat{t}}$ direction indicates how the angle $\phi$ changes in the counterclockwise direction. In this way, a point P that has coordinates (x, y) in the rectangular system can be described equivalently in the polar coordinate system by the two polar coordinates $(r,\phi )$. Figure is valid for any vector, so we can use it to express the x– and y-coordinates of vector $\mathbf{\overset{\to }{r}}$. In this way, we obtain the connection between the polar coordinates and rectangular coordinates of point P: $\left\{\begin{array}{c}x=r\,\text{cos}\,\phi \\ y=r\,\text{sin}\,\phi \end{array}\right. .$ ### Example #### Polar Coordinates A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction $20^\circ$ north of east and finds one gold coin at a location 10.0 m away from the well in the direction $20^\circ$ north of west. What are the polar and rectangular coordinates of these findings with respect to the well? #### Strategy The well marks the origin of the coordinate system and east is the +x-direction. We identify radial distances from the locations to the origin, which are ${r}_{S}=20.0\,\text{m}$ (for the silver coin) and ${r}_{G}=10.0\,\text{m}$ (for the gold coin). To find the angular coordinates, we convert $20^\circ$ to radians: $20^\circ=\pi 20\text{/}180=\pi \text{/}9$. We use Figure to find the x– and y-coordinates of the coins. #### Solution The angular coordinate of the silver coin is ${\phi }_{S}=\pi \text{/}9$, whereas the angular coordinate of the gold coin is ${\phi }_{G}=\pi -\pi \text{/}9=8\pi \text{/}9$. Hence, the polar coordinates of the silver coin are $({r}_{S},{\phi }_{S})=(20.0\,\text{m},\pi \text{/}9)$ and those of the gold coin are $({r}_{G},{\phi }_{G})=(10.0\,\text{m},8\pi \text{/}9)$. We substitute these coordinates into (Figure) to obtain rectangular coordinates. For the gold coin, the coordinates are $\left\{\begin{array}{l}{x}_{G}={r}_{G}\,\text{cos}\,{\phi }_{G}=(10.0\,\text{m})\,\text{cos}\,8\pi \text{/}9=-9.4\,\text{m}\\ {y}_{G}={r}_{G}\,\text{sin}\,{\phi }_{G}=(10.0\,\text{m})\,\text{sin}\,8\pi \text{/}9=3.4\,\text{m}\end{array}\right. \enspace\Rightarrow \enspace({x}_{G},{y}_{G})=(-9.4\,\text{m},3.4\,\text{m}).$ For the silver coin, the coordinates are $\left\{\begin{array}{l}{x}_{S}={r}_{S}\,\text{cos}\,{\phi }_{S}=(20.0\,\text{m})\,\text{cos}\,\pi \text{/}9=18.9\,\text{m}\\ {y}_{S}={r}_{S}\,\text{sin}\,{\phi }_{S}=(20.0\,\text{m})\,\text{sin}\,\pi \text{/}9=6.8\,\text{m}\end{array}\right.\enspace\Rightarrow \enspace({x}_{S},{y}_{S})=(18.9\,\text{m},6.8\,\text{m}).$ ### Vectors in Three Dimensions To specify the location of a point in space, we need three coordinates (x, y, z), where coordinates x and y specify locations in a plane, and coordinate z gives a vertical position above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the x-axis $\mathbf{\hat{i}}$ and the unit vector of the y-axis $\mathbf{\hat{j}}$. The third unit vector $\mathbf{\hat{k}}$ is the direction of the z-axis (Figure). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order xyz, which is equivalent to the order $\mathbf{\hat{i}}$ – $\mathbf{\hat{j}}$ – $\mathbf{\hat{k}}$, defines the standard right-handed coordinate system (positive orientation). In three-dimensional space, vector $\mathbf{\overset{\to }{A}}$ has three vector components: the x-component ${\mathbf{\overset{\to }{A}}}_{x}={A}_{x}\mathbf{\hat{i}}$, which is the part of vector $\mathbf{\overset{\to }{A}}$ along the x-axis; the y-component ${\mathbf{\overset{\to }{A}}}_{y}={A}_{y}\mathbf{\hat{j}}$, which is the part of $\mathbf{\overset{\to }{A}}$ along the y-axis; and the z-component ${\mathbf{\overset{\to }{A}}}_{z}={A}_{z}\mathbf{\hat{k}}$, which is the part of the vector along the z-axis. A vector in three-dimensional space is the vector sum of its three vector components (Figure): $\mathbf{\overset{\to }{A}}={A}_{x}\mathbf{\hat{i}}+{A}_{y}\mathbf{\hat{j}}+{A}_{z}\mathbf{\hat{k}}.$ If we know the coordinates of its origin $b({x}_{b},{y}_{b},{z}_{b})$ and of its end $e({x}_{e},{y}_{e},{z}_{e})$, its scalar components are obtained by taking their differences: ${A}_{x}$ and ${A}_{y}$ are given by Figure and the z-component is given by ${A}_{z}={z}_{e}-{z}_{b}.$ Magnitude A is obtained by generalizing Figure to three dimensions: $A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}}.$ This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in Figure, the diagonal in the xy-plane has length $\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}$ and its square adds to the square ${A}_{z}^{2}$ to give ${A}^{2}$. Note that when the z-component is zero, the vector lies entirely in the xy-plane and its description is reduced to two dimensions. ### Example #### Takeoff of a Drone During a takeoff of IAI Heron (Figure), its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower? What is the magnitude of its displacement vector? #### Strategy We take the origin of the Cartesian coordinate system as the control tower. The direction of the +x-axis is given by unit vector $\mathbf{\hat{i}}$ to the east, the direction of the +y-axis is given by unit vector $\mathbf{\hat{j}}$ to the north, and the direction of the +z-axis is given by unit vector $\mathbf{\hat{k}}$, which points up from the ground. The drone’s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector. #### Solution We identify b(300.0 m, 200.0 m, 100.0 m) and e(480.0 m, 370.0 m, 250.0m), and use (Figure) and (Figure) to find the scalar components of the drone’s displacement vector: $\left\{\begin{array}{l}{D}_{x}={x}_{e}-{x}_{b}=1200.0\,\text{m}-300.0\,\text{m}=900.0\,\text{m},\\ {D}_{y}={y}_{e}-{y}_{b}=2100.0\,\text{m}-200.0\,\text{m}=1900.0\,\text{m,}\\ {D}_{z}={z}_{e}-{z}_{b}=250.0\,\text{m}-100.0\,\text{m}=150.0\,\text{m}.\end{array}\right.$ We substitute these components into (Figure) to find the displacement vector: $\mathbf{\overset{\to }{D}}={D}_{x}\mathbf{\hat{i}}+{D}_{y}\mathbf{\hat{j}}+{D}_{z}\mathbf{\hat{k}}=900.0\,\text{m}\mathbf{\hat{i}}+1900.0\,\text{m}\mathbf{\hat{j}}+150.0\,\text{m}\mathbf{\hat{k}}=(0.90\mathbf{\hat{i}}+1.90\mathbf{\hat{j}}+0.15\mathbf{\hat{k}})\,\text{km}.$ We substitute into (Figure) to find the magnitude of the displacement: $D=\sqrt{{D}_{x}^{2}+{D}_{y}^{2}+{D}_{z}^{2}}=\sqrt{{(0.90\,\text{km})}^{2}+{(1.90\,\text{km})}^{2}+{(0.15\,\text{km})}^{2}}=4.44\,\text{km}.$ If the average velocity vector of the drone in the displacement in Figure is $\mathbf{\overset{\to }{u}}=(15.0\mathbf{\hat{i}}+31.7\mathbf{\hat{j}}+2.5\mathbf{\hat{k}})\text{m}\text{/}\text{s}$, what is the magnitude of the drone’s velocity vector? Show Solution 35.1 m/s = 126.4 km/h ### Summary • Vectors are described in terms of their components in a coordinate system. In two dimensions (in a plane), vectors have two components. In three dimensions (in space), vectors have three components. • A vector component of a vector is its part in an axis direction. The vector component is the product of the unit vector of an axis with its scalar component along this axis. A vector is the resultant of its vector components. • Scalar components of a vector are differences of coordinates, where coordinates of the origin are subtracted from end point coordinates of a vector. In a rectangular system, the magnitude of a vector is the square root of the sum of the squares of its components. • In a plane, the direction of a vector is given by an angle the vector has with the positive x-axis. This direction angle is measured counterclockwise. The scalar x-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle, and the scalar y-component can be expressed as the product of its magnitude with the sine of its direction angle. • In a plane, there are two equivalent coordinate systems. The Cartesian coordinate system is defined by unit vectors $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$ along the x-axis and the y-axis, respectively. The polar coordinate system is defined by the radial unit vector $\mathbf{\hat{r}}$, which gives the direction from the origin, and a unit vector $\mathbf{\hat{t}}$, which is perpendicular (orthogonal) to the radial direction. ### Conceptual Questions Give an example of a nonzero vector that has a component of zero. Show Solution a unit vector of the x-axis Explain why a vector cannot have a component greater than its own magnitude. If two vectors are equal, what can you say about their components? Show Solution They are equal. If vectors $\mathbf{\overset{\to }{A}}$ and $\mathbf{\overset{\to }{B}}$ are orthogonal, what is the component of $\mathbf{\overset{\to }{B}}$ along the direction of $\mathbf{\overset{\to }{A}}$? What is the component of $\mathbf{\overset{\to }{A}}$ along the direction of $\mathbf{\overset{\to }{B}}$? If one of the two components of a vector is not zero, can the magnitude of the other vector component of this vector be zero? Show Solution yes If two vectors have the same magnitude, do their components have to be the same? ### Problems Assuming the +x-axis is horizontal and points to the right, resolve the vectors given in the following figure to their scalar components and express them in vector component form. Show Solution a. $\mathbf{\overset{\to }{A}}=+8.66\mathbf{\hat{i}}+5.00\mathbf{\hat{j}}$, b. $\mathbf{\overset{\to }{B}}=+30.09\mathbf{\hat{i}}+39.93\mathbf{\hat{j}}$, c. $\mathbf{\overset{\to }{C}}=+6.00\mathbf{\hat{i}}-10.39\mathbf{\hat{j}}$, d. $\mathbf{\overset{\to }{D}}=-15.97\mathbf{\hat{i}}+12.04\mathbf{\hat{j}}$, f. $\mathbf{\overset{\to }{F}}=-17.32\mathbf{\hat{i}}-10.00\mathbf{\hat{j}}$ Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is horizontal to the right. You drive 7.50 km in a straight line in a direction $15^\circ$ east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (b) Show that you still arrive at the same point if the east and north legs are reversed in order. Assume the +x-axis is to the east. Show Solution a. 1.94 km, 7.24 km; b. proof A sledge is being pulled by two horses on a flat terrain. The net force on the sledge can be expressed in the Cartesian coordinate system as vector $\mathbf{\overset{\to }{F}}=(-2980.0\mathbf{\hat{i}}+8200.0\mathbf{\hat{j}})\text{N}$, where $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$ denote directions to the east and north, respectively. Find the magnitude and direction of the pull. A trapper walks a 5.0-km straight-line distance from her cabin to the lake, as shown in the following figure. Determine the east and north components of her displacement vector. How many more kilometers would she have to walk if she walked along the component displacements? What is her displacement vector? 3.8 km east, 3.2 km north, 2.0 km, $\mathbf{\overset{\to }{D}}=(3.8\mathbf{\hat{i}}+3.2\mathbf{\hat{j}})\text{km}$ The polar coordinates of a point are $4\pi \text{/}3$ and 5.50 m. What are its Cartesian coordinates? Two points in a plane have polar coordinates ${P}_{1}(2.500\,\text{m},\pi \text{/}6)$ and ${P}_{2}(3.800\,\text{m},2\pi \text{/}3)$. Determine their Cartesian coordinates and the distance between them in the Cartesian coordinate system. Round the distance to a nearest centimeter. Show Solution ${P}_{1}(2.165\,\text{m},1.250\,\text{m})$, ${P}_{2}(-1.900\,\text{m},3.290\,\text{m})$, 5.27 m A chameleon is resting quietly on a lanai screen, waiting for an insect to come by. Assume the origin of a Cartesian coordinate system at the lower left-hand corner of the screen and the horizontal direction to the right as the +x-direction. If its coordinates are (2.000 m, 1.000 m), (a) how far is it from the corner of the screen? (b) What is its location in polar coordinates? Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar coordinates. Show Solution 8.60 m, $A(2\sqrt{5}\,\text{m},0.647\pi )$, $B(3\sqrt{2}\,\text{m},0.75\pi )$ A fly enters through an open window and zooms around the room. In a Cartesian coordinate system with three axes along three edges of the room, the fly changes its position from point b(4.0 m, 1.5 m, 2.5 m) to point e(1.0 m, 4.5 m, 0.5 m). Find the scalar components of the fly’s displacement vector and express its displacement vector in vector component form. What is its magnitude? ### Glossary component form of a vector a vector written as the vector sum of its components in terms of unit vectors direction angle in a plane, an angle between the positive direction of the x-axis and the vector, measured counterclockwise from the axis to the vector polar coordinate system an orthogonal coordinate system where location in a plane is given by polar coordinates polar coordinates a radial coordinate and an angle
# A plane goes through $A:(0,1,2), \ B:(3,2,1)$ and $C:(4,-1,0).$ Determine the equation of the plane in affin form. Solution: The vectors $\vec{AB}=(3,2,1)-(0,1,2)=3,1,-1$ and $\vec{AC}=(4,-1,0)-(0,1,2)=(4,-2,-2),$ are two direction vectors of the plane. A normal vector $\vec{n}$ to the plane is then given by $$\vec{n}=\vec{AB}\times\vec{AC}=(-4,2,-10).$$ Since $A$ is a point on the plane, we get $$\vec{n}\cdot (x-0,y-1,z-2)=-4x+2(y-1)-10(z-2)=-4x+2y-10z+18=0.$$ I dont understand the first part of the last equation. 1. What vector is $(x-0,y-1,z-2)$? 2. Why do they take the dot-product of the above with the normal vector? (How does it give the equation of the plane?) The vector $(x-0,y-1,z-2)$ comes directly from the fact that $(0,1,2)$ is a point on the plane. For the second part of your question, if $\vec{v}$ is a vector and $\vec{n}$ is a normal vector to $\vec{v}$, then $\vec{v}\cdot\vec{n}=0$. We use the dot product between the normal vector and the cross product of the displacement vectors because the equation of the plane is $(x-a,y-b,z-c)\cdot \vec{n}=0$, where $(a,b,c)$ is a point on the plane and $\vec{n}$ is normal to the plane. The idea of finding the equation of a plane is to have an equation that can help to locate any point $P(x,y,z)$ on the plane. In order to do that consider the vector $\vec{AP}=(x,y,z)-(0,1,2)$. This vector lies in the given plane. Now that we have a normal vector $\hat{n}$ (which by definition is orthogonal to any vector in the plane) we know that $\hat{n} \cdot \vec{AP}=0$. This gives the equation we are looking for. 1. The vector $\langle x-0,y-1,z-2\rangle$ is a vector connecting a generic point $P(x,y,z)$ lying in the desired plane with the point $A(0,1,2)$, which is also in this plane. Since both of them are in this plane, the vector connecting them $\overrightarrow{AP}=\langle x-0,y-1,z-2\rangle$ is in this plane. 2. The word "normal" means "perpendicular" in this context. So saying "$\overrightarrow{n}$ is normal to the plane" means that $n$ is perpendicular to the plane, and therefore it's perpendicular to any vector that lies in this plane. So we know that $\overrightarrow{n}\perp\overrightarrow{AP}$ must be true. And then there's the property of dot products: two vectors are perpendicular if and only if their dot product is zero. • Thanks. But what I still don't understand is why the equation of the plane is given by THAT dot product? Oct 10, 2017 at 20:07 • Because any point $P(x,y,z)$ lying in the plane has to satisfy this condition. So this equation in terms of $x,y,z$ describes all points $P(x,y,z)$ in the plane, and thus it is an equation of this plane. Oct 10, 2017 at 20:12
# Calculus/Fundamental Theorem of Calculus/Solutions 1. Evaluate ${\displaystyle \int _{0}^{1}x^{6}dx}$. Compare your answer to the answer you got for exercise 1 in section 4.1. ${\displaystyle \int _{0}^{1}x^{6}dx={\frac {x^{7}}{7}}{\biggr \|}_{0}^{1}={\frac {1^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {1}{7}}=0.{\overline {142857}}} }$ This is consistent with the bounds we calculated in exercise 1 in section 4.1. ${\displaystyle \int _{0}^{1}x^{6}dx={\frac {x^{7}}{7}}{\biggr \|}_{0}^{1}={\frac {1^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {1}{7}}=0.{\overline {142857}}} }$ This is consistent with the bounds we calculated in exercise 1 in section 4.1. 2. Evaluate ${\displaystyle \int _{1}^{2}x^{6}dx}$. Compare your answer to the answer you got for exercise 2 in section 4.1. ${\displaystyle \int _{1}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr \|}_{1}^{2}={\frac {2^{7}}{7}}-{\frac {1^{7}}{7}}={\frac {128}{7}}-{\frac {1}{7}}=\mathbf {{\frac {127}{7}}=18.{\overline {142857}}} }$ This is consistent with the bounds we calculated in exercise 2 in section 4.1. ${\displaystyle \int _{1}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr \|}_{1}^{2}={\frac {2^{7}}{7}}-{\frac {1^{7}}{7}}={\frac {128}{7}}-{\frac {1}{7}}=\mathbf {{\frac {127}{7}}=18.{\overline {142857}}} }$ This is consistent with the bounds we calculated in exercise 2 in section 4.1. 3. Evaluate ${\displaystyle \int _{0}^{2}x^{6}dx}$. Compare your answer to the answer you got for exercise 4 in section 4.1. ${\displaystyle \int _{0}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr \|}_{0}^{2}={\frac {2^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {128}{7}}=18.{\overline {285714}}} }$ This is consistent with the bounds we calculated in exercise 4 in section 4.1. ${\displaystyle \int _{0}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr \|}_{0}^{2}={\frac {2^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {128}{7}}=18.{\overline {285714}}} }$ This is consistent with the bounds we calculated in exercise 4 in section 4.1. 4. Compute ${\displaystyle \int \limits _{2}^{3}{\frac {x^{3}+3x^{2}-4}{x^{2}}}dx}$. {\displaystyle {\begin{aligned}\int \limits _{2}^{3}{\frac {x^{3}+3x^{2}-4}{x^{2}}}dx&=\int \limits _{2}^{3}{\frac {x^{3}}{x^{2}}}+{\frac {3x^{2}}{x^{2}}}-{\frac {4}{x^{2}}}dx\\&=\int \limits _{2}^{3}x+3-{\frac {4}{x^{2}}}dx\\&=\left.{\frac {x^{2}}{2}}+3x+{\frac {4}{x}}\right\|_{2}^{3}\\&=\left[{\frac {3^{2}}{2}}+3(3)+{\frac {4}{3}}\right]-\left[{\frac {2^{2}}{2}}+3(2)+{\frac {4}{2}}\right]\\&=\left[{\frac {9}{2}}+9+{\frac {4}{3}}\right]-\left[2+6+2\right]\\&={\frac {89}{6}}-10\\&={\frac {29}{6}}\approx 4.{\overline {833}}\end{aligned}}} {\displaystyle {\begin{aligned}\int \limits _{2}^{3}{\frac {x^{3}+3x^{2}-4}{x^{2}}}dx&=\int \limits _{2}^{3}{\frac {x^{3}}{x^{2}}}+{\frac {3x^{2}}{x^{2}}}-{\frac {4}{x^{2}}}dx\\&=\int \limits _{2}^{3}x+3-{\frac {4}{x^{2}}}dx\\&=\left.{\frac {x^{2}}{2}}+3x+{\frac {4}{x}}\right\|_{2}^{3}\\&=\left[{\frac {3^{2}}{2}}+3(3)+{\frac {4}{3}}\right]-\left[{\frac {2^{2}}{2}}+3(2)+{\frac {4}{2}}\right]\\&=\left[{\frac {9}{2}}+9+{\frac {4}{3}}\right]-\left[2+6+2\right]\\&={\frac {89}{6}}-10\\&={\frac {29}{6}}\approx 4.{\overline {833}}\end{aligned}}} 5. Evaluate ${\displaystyle \int \limits _{1}^{4}2x^{3}+{\frac {1}{3x^{2}}}-4{\sqrt {x}}+5dx}$. {\displaystyle {\begin{aligned}\int \limits _{1}^{4}2x^{3}+{\frac {1}{3x^{2}}}-4{\sqrt {x}}+5dx&=\int \limits _{1}^{4}2x^{3}dx+\int \limits _{1}^{4}{\frac {1}{3x^{2}}}dx-\int \limits _{1}^{4}4{\sqrt {x}}dx+\int \limits _{1}^{4}5dx\\&=2\int \limits _{1}^{4}x^{3}dx+{\frac {1}{3}}\int \limits _{1}^{4}{\frac {1}{x^{2}}}dx-4\int \limits _{1}^{4}{\sqrt {x}}dx+5\int \limits _{1}^{4}dx\\&=\left.{\frac {x^{4}}{2}}-{\frac {1}{3x}}-{\frac {8x^{\frac {3}{2}}}{3}}+5x\right\|_{1}^{4}\\&=\left[{\frac {4^{4}}{2}}-{\frac {1}{3(4)}}-{\frac {8(4)^{\frac {3}{2}}}{3}}+5(4)\right]-\left[{\frac {1}{2}}-{\frac {1}{3}}-{\frac {8}{3}}+5\right]\\&=\left[128-{\frac {1}{12}}-{\frac {64}{3}}+20\right]-\left[{\frac {1}{2}}-{\frac {1}{3}}-{\frac {8}{3}}+5\right]\\&={\frac {1519}{12}}-{\frac {5}{2}}\\&={\frac {1489}{12}}\approx 124.{\overline {083}}\end{aligned}}} 6. Given ${\displaystyle f(x)={\begin{cases}x^{3},&x>1\\x+1,&x\leq 1\end{cases}}}$, then find ${\displaystyle \int \limits _{0}^{6}f(x)dx}$. By using the formula that states the additivity with respect to endpoints as shown in Section 4.1, we can break the integral into two parts based on the conditions provided, for ${\displaystyle f(x)=x^{3},~x>1}$ and ${\displaystyle f(x)=x+1,~x\leq 1}$. Hence, {\displaystyle {\begin{aligned}\int \limits _{0}^{6}f(x)dx&=\int \limits _{0}^{1}x+1dx+\int \limits _{1}^{6}x^{3}dx\\&=\left[{\frac {x^{2}}{2}}+x\right]_{0}^{1}+\left[{\frac {x^{4}}{4}}\right]_{1}^{6}\\&=\left[{\frac {1^{2}}{2}}+1\right]+\left[{\frac {6^{4}}{4}}-{\frac {1^{4}}{4}}\right]\\&={\frac {3}{2}}+{\frac {1295}{4}}\\&={\frac {1301}{4}}=325.25\end{aligned}}} 7. Let ${\displaystyle f(x)=\int \limits _{1}^{x}t^{2}dt}$. Then find ${\displaystyle f'(x)}$. This given function can be found the derivative by using the Fundamental Theorem of Calculus Part I, as stated in Section 4.2. ${\displaystyle f'(x)={\frac {d}{dx}}\left[\int \limits _{1}^{x}t^{2}dt\right]=x^{2}.}$ How is this ${\displaystyle x^{2}}$ coming from? This comes from the definite integral that we find using the second part of the Fundamental Theorem of Calculus. Since we know that the derivative is the opposite of what we do with the antiderivative, {\displaystyle {\begin{aligned}f'(x)={\frac {d}{dx}}\left[\int \limits _{1}^{x}t^{2}dt\right]&={\frac {d}{dx}}\left[\left.{\frac {t^{3}}{3}}\right\|_{1}^{x}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}-{\frac {1^{3}}{3}}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}\right]-{\frac {d}{dx}}\left[{\frac {1}{3}}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}\right]-0\\&={\frac {1}{3}}{\frac {d}{dx}}[x^{3}]\\&={\frac {1}{3}}\cdot 3x^{3-1}\\&={\frac {1}{\cancel {3}}}\cdot {\cancel {3}}x^{2}\\&=x^{2}\end{aligned}}} 8. Given ${\displaystyle A(\theta )=\int \limits _{1}^{\theta ^{2}}2x\cos(4x^{2})dx}$. Then find ${\displaystyle A'(\theta )}$. Here is another form of the function that is defined by the definite integral as a function. To find the derivative, we can apply the Chain Rule of the differentiation as in Section 3.4, which ${\displaystyle u=\theta ^{2}}$, then {\displaystyle {\begin{aligned}A'(\theta )={\frac {d}{d\theta }}\left[\int \limits _{1}^{\theta ^{2}}2x\cos(4x^{2})dx\right]&={\frac {d}{d\theta }}\left[\int \limits _{1}^{u}2x\cos(4x^{2})dx\right]{\frac {du}{d\theta }}\\&=\left(2u\cos(4u^{2})\right){\frac {du}{d\theta }}\\&=\left(2\theta ^{2}\cos \left(4(\theta ^{2})^{2}\right)\right){\frac {d}{d\theta }}\left[\theta ^{2}\right]\\\ &=\left(2\theta ^{2}\cos(4\theta ^{4})\right)(2\theta )\\&=4\theta ^{3}\cos(4\theta )\end{aligned}}} 9. If ${\displaystyle M(x)=\int \limits _{x^{3}}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt}$. Then find ${\displaystyle M'(x)}$. This is quite different from the two exercise problems previously. If previously we looked at the variable as being only at one bound of the integration, then what if the variable was placed on both the upper and lower bounds? By using the additive property of a definite integral, we can break up the integral into two parts and take the derivative to generates ${\displaystyle M'(x)}$ from given function. ${\displaystyle {\frac {d}{dx}}\left[\int \limits _{x^{3}}^{x}\cos ^{4}(t)-\sin ^{2}(\theta )dt\right]={\frac {d}{dx}}\left[\int \limits _{x^{3}}^{0}\cos ^{4}(t)-\sin ^{2}(t)dt+\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]}$ Therefore, {\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[\int \limits _{x^{3}}^{0}\cos ^{4}(t)-\sin ^{2}(t)dt+\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]&={\frac {d}{dx}}\left[\int \limits _{x^{3}}^{0}\cos ^{4}(t)-\sin ^{2}(t)dt\right]+{\frac {d}{dx}}\left[\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]\\&=-{\frac {d}{dx}}\left[\int \limits _{0}^{x^{3}}\cos ^{4}(t)-\sin ^{2}(t)dt\right]+{\frac {d}{dx}}\left[\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]\\&=-\left({\frac {d}{du}}\left[\int \limits _{0}^{u}\cos ^{4}(t)-\sin ^{2}(t)dt\right]{\frac {du}{dx}}\right)+\cos ^{4}(x)-\sin ^{2}(x)\\&=-\left(\left(\cos ^{4}(x^{3})-\sin ^{2}(x^{3})\right){\frac {d}{dx}}\left[x^{3}\right]\right)+\cos ^{4}(x)-\sin ^{2}(x)\\&=-\left(\left[\cos ^{4}(x^{3})-\sin(x^{3})\right](3x^{2})\right)+\cos ^{4}(x)-\sin ^{2}(x)\\&=-3x^{2}\cos ^{4}(x^{3})+3x^{2}\sin ^{2}(x^{3})+\cos ^{4}(x)-\sin ^{2}(x)\end{aligned}}} 10. For the function${\displaystyle {\displaystyle f(x)={\sqrt {x}}}}$ over the given closed interval, ${\displaystyle {\displaystyle [4,9]}}$find the value(s) ${\displaystyle c}$ guaranteed by the mean value theorem for the definite integral. To find the value(s)${\displaystyle c}$ that are guaranteed by the mean value theorem for the definite integral, we can first find the ${\displaystyle f(c)}$ by using the formula shown below. ${\displaystyle f(c)={\frac {1}{b-a}}\int \limits _{a}^{b}f(x)dx}$ Such that, {\displaystyle {\begin{aligned}f(c)&={\frac {1}{9-4}}\int \limits _{4}^{9}{\sqrt {x}}dx\\&={\frac {1}{5}}\left[{\frac {2x^{\frac {3}{2}}}{3}}\right]_{4}^{9}\\&={\frac {1}{5}}\left[{\frac {2(9)^{\frac {3}{2}}}{3}}-{\frac {2(4)^{\frac {3}{2}}}{3}}\right]\\&={\frac {1}{5}}\left[18-{\frac {16}{3}}\right]\\&={\frac {1}{5}}\left[{\frac {38}{3}}\right]\\{\sqrt {c}}&={\frac {38}{15}}\end{aligned}}} Now, in the case of finding the value(s) c that are guaranteed by the mean value theorem for a definite integral, we can give the squared on both sides. {\displaystyle {\begin{aligned}{\sqrt {c}}&={\frac {38}{15}}\\\left({\sqrt {c}}\right)^{2}&=\left({\frac {38}{15}}\right)^{2}\\c&={\frac {1444}{225}}\approx 6.{\overline {417}}\end{aligned}}} For the value of ${\displaystyle c}$ guaranteed by the Mean Value Theorem for Integrals is ${\displaystyle {\frac {1444}{225}}\approx 6.{\overline {417}}}$.
# How do you simplify (3x^2y)(10x^5y^3)? Jan 3, 2017 Multiply each like term in each of the parenthesis. See full explanation below. #### Explanation: We will multiply each common term in parenthesis with the common term in the other parenthesis: $\left(\textcolor{red}{3} \textcolor{b l u e}{{x}^{2}} \textcolor{g r e e n}{y}\right) \left(\textcolor{red}{10} \textcolor{b l u e}{{x}^{5}} \textcolor{g r e e n}{{y}^{3}}\right) \to \left(\textcolor{red}{3} \times \textcolor{red}{10}\right) \left(\textcolor{b l u e}{{x}^{2}} \times \textcolor{b l u e}{{x}^{5}}\right) \left(\textcolor{g r e e n}{y} \times \textcolor{g r e e n}{{y}^{3}}\right)$ We can multiply the constants and use the rules for exponents to multiply the common terms. x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a)+color(blue)(b) and z = z^(color(red)(1) This gives: $\left(\textcolor{red}{3} \times \textcolor{red}{10}\right) \left(\textcolor{b l u e}{{x}^{2}} \times \textcolor{b l u e}{{x}^{5}}\right) \left(\textcolor{g r e e n}{y} \times \textcolor{g r e e n}{{y}^{3}}\right) =$ $\textcolor{red}{30} \textcolor{b l u e}{{x}^{2 + 5}} \textcolor{g r e e n}{{y}^{1 + 3}} =$ $\textcolor{red}{30} \textcolor{b l u e}{{x}^{7}} \textcolor{g r e e n}{{y}^{4}}$
Pre-Calc Exam Notes 98 # Pre-Calc Exam Notes 98 - 0.775 rad ⇒ ∠ CBD = 2(0.775 =... This preview shows page 1. Sign up to view the full content. 98 Chapter 4 Radian Measure §4.3 Example 4.13 The centers of two circles are 7 cm apart, with one circle having a radius of 5 cm and the other a radius of 3 cm. Find the area K of their intersection. Solution: In Figure 4.3.6(a), we see that the intersection of the two circles is the union of the seg- ments formed by the chord CD in each circle. Thus, once we determine the angles CAD and CBD we can calculate the area of each segment and add those areas together to get K . C D 5 4 A B 7 (a) BAC = 1 2 CAD , ABC = 1 2 CBD 5 C 4 7 A B α β (b) Triangle ABC Figure 4.3.6 By symmetry, we see that BAC = 1 2 CAD and ABC = 1 2 CBD . So let α = BAC and β = ABC , as in Figure 4.3.6(b). By the Law of Cosines, we have cos α = 7 2 + 5 2 4 2 2(7)(5) = 0.8286 α = 0.594 rad CAD = 2(0.594) = 1.188 rad cos β = 7 2 + 4 2 5 2 2(7)(4) = 0.7143 β = This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: 0.775 rad ⇒ ∠ CBD = 2(0.775) = 1.550 rad Thus, the area K is K = (Area of segment CD in circle at A ) + (Area of segment CD in circle at B ) = 1 2 (5) 2 (1.188 − sin 1.188) + 1 2 (4) 2 (1.550 − sin 1.550) = 7.656 cm 2 . Exercises For Exercises 1-3, ±nd the area of the sector for the given angle θ and radius r . 1. θ = 2.1 rad, r = 1.2 cm 2. θ = 3 π 7 rad, r = 3.5 ft 3. θ = 78 ◦ , r = 6 m 4. The centers of two belt pulleys, with radii of 3 cm and 6 cm, respectively, are 13 cm apart. Find the total area K enclosed by the belt. 5. In Exercise 4 suppose that both belt pulleys have the same radius of 6 cm. Find the total area K enclosed by the belt.... View Full Document {[ snackBarMessage ]} Ask a homework question - tutors are online
# How do you find critical points of a logarithmic function y = e^x - 2e^(-x) - 3x? Oct 29, 2015 Correction: That is not a logarithmic function, but here is how to find the critical numbers for the function. #### Explanation: $f \left(x\right) = {e}^{x} - 2 {e}^{- x} - 3 x$ Note that "Dom:(f) = (-oo,oo) $f ' \left(x\right) = {e}^{x} - 2 {e}^{- x} \left(- 1\right) - 3$ $= {e}^{x} + 2 {e}^{- x} - 3$ $f ' \left(x\right)$ is never undefined, so we need only find its zeros. We need to solve ${e}^{x} + 2 {e}^{- x} - 3 = 0$ As a first step, try getting rid of negative exponents: ${e}^{x} + \frac{2}{e} ^ x - 3 = 0$ Get a common denomiator, noting that ${e}^{x} \cdot {e}^{x} = {e}^{2 x}$, so we have $\frac{{e}^{2 x} + 2 - 3 {e}^{x}}{e} ^ x = 0$. So now we need to solve: ${e}^{2 x} - 3 {e}^{x} + 2 = 0$ Again note that ${e}^{2 x} = {\left({e}^{x}\right)}^{2}$, so we can factor: $\left({e}^{x} - 1\right) \left({e}^{x} - 2\right) = 0$ ${e}^{x} = 1$ $\text{ }$ OR $\text{ }$ ${e}^{x} = 2$ $x = 0$ $\text{ }$ OR $\text{ }$ $x = \ln 2$ Both are in $\text{Dom} \left(f\right)$, so both are critical numbers.
# How to Find the Square Root of 197? What does square root mean? By definition, the square root of 197 is the number that, multiplied by itself, produces the given number 197. In this case, that number is 14.0357. Square root of 197 = 14.0357 11972 The symbol √ is called radix, or radical sign The number below the radix is the radicand ## Is 197 a Perfect Square Root? No. The square root of 197 is not an integer, hence √197 isn't a perfect square. Previous perfect square root is: 196 Next perfect square root is: 225 ## The Prime Factors of 197 are: 197 is a prime number. It has only two factors: 1 and 197 ## How Do You Simplify the Square Root of 197 in Radical Form? The main point of simplification (to the simplest radical form of 197) is as follows: getting the number 197 inside the radical sign √ as low as possible. 197 is already simplified (prime number). ## Is the Square Root of 197 Rational or Irrational? Since 197 isn't a perfect square (it's square root will have an infinite number of decimals), it is an irrational number. ## The Babylonian (or Heron’s) Method (Step-By-Step) StepSequencing 1 In step 1, we need to make our first guess about the value of the square root of 197. To do this, divide the number 197 by 2. As a result of dividing 197/2, we get the first guess: 98.5 2 Next, we need to divide 197 by the result of the previous step (98.5). 197/98.5 = 2 Calculate the arithmetic mean of this value (2) and the result of step 1 (98.5). (98.5 + 2)/2 = 50.25 (new guess) Calculate the error by subtracting the previous value from the new guess. \|50.25 - 98.5\| = 48.25 48.25 > 0.001 Repeat this step again as the margin of error is greater than than 0.001 3 Next, we need to divide 197 by the result of the previous step (50.25). 197/50.25 = 3.9204 Calculate the arithmetic mean of this value (3.9204) and the result of step 2 (50.25). (50.25 + 3.9204)/2 = 27.0852 (new guess) Calculate the error by subtracting the previous value from the new guess. \|27.0852 - 50.25\| = 23.1648 23.1648 > 0.001 Repeat this step again as the margin of error is greater than than 0.001 4 Next, we need to divide 197 by the result of the previous step (27.0852). 197/27.0852 = 7.2733 Calculate the arithmetic mean of this value (7.2733) and the result of step 3 (27.0852). (27.0852 + 7.2733)/2 = 17.1793 (new guess) Calculate the error by subtracting the previous value from the new guess. \|17.1793 - 27.0852\| = 9.9059 9.9059 > 0.001 Repeat this step again as the margin of error is greater than than 0.001 5 Next, we need to divide 197 by the result of the previous step (17.1793). 197/17.1793 = 11.4673 Calculate the arithmetic mean of this value (11.4673) and the result of step 4 (17.1793). (17.1793 + 11.4673)/2 = 14.3233 (new guess) Calculate the error by subtracting the previous value from the new guess. \|14.3233 - 17.1793\| = 2.856 2.856 > 0.001 Repeat this step again as the margin of error is greater than than 0.001 6 Next, we need to divide 197 by the result of the previous step (14.3233). 197/14.3233 = 13.7538 Calculate the arithmetic mean of this value (13.7538) and the result of step 5 (14.3233). (14.3233 + 13.7538)/2 = 14.0386 (new guess) Calculate the error by subtracting the previous value from the new guess. \|14.0386 - 14.3233\| = 0.2847 0.2847 > 0.001 Repeat this step again as the margin of error is greater than than 0.001 7 Next, we need to divide 197 by the result of the previous step (14.0386). 197/14.0386 = 14.0327 Calculate the arithmetic mean of this value (14.0327) and the result of step 6 (14.0386). (14.0386 + 14.0327)/2 = 14.0357 (new guess) Calculate the error by subtracting the previous value from the new guess. \|14.0357 - 14.0386\| = 0.0029 0.0029 > 0.001 Repeat this step again as the margin of error is greater than than 0.001 8 Next, we need to divide 197 by the result of the previous step (14.0357). 197/14.0357 = 14.0356 Calculate the arithmetic mean of this value (14.0356) and the result of step 7 (14.0357). (14.0357 + 14.0356)/2 = 14.0357 (new guess) Calculate the error by subtracting the previous value from the new guess. \|14.0357 - 14.0357\| = 0 0 < 0.001 Stop the iterations as the margin of error is less than 0.001 Result✅ We found the result: 14.0357 In this case, it took us eight steps to find the result.
# R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 10 Chapter 14 Height and Distance ## Chapter 14 – Height and Distance Exercise Ex. 14 Question 1 A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower. [Take ] Solution 1 Let AB be the tower standing on a level ground and O be the position of the observer. Then OA = 20 m and OAB = 90° and AOB = 60° Let AB = h meters From the right OAB, we have Hence the height of the tower is Question 2 A kite is flying at a height of 75m from the level ground, attached to a string inclined at 60° to the horizontal. Find the length of the string assuming that there is no slack in it. Solution 2 Let OB be the length of the string from the level of ground and O be the point of the observer, then, AB = 75m and OAB = 90° and AOB = 60°, let OB = l meters. From the right OAB, we have Question 3 An observer 1.5 m tall is 30 m away from a chimney. The angle of elevation of the top of the chimney from his eye is 60°. Find the height of the chimney. Solution 3 Question 4 The angles of elevation of the top of a tower from two points at distances of 5 metres and 20 metres from the base of the tower and in the same straight line with it, are complementary. Find the height of the tower. Solution 4 Question 5 The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff. Solution 5 Question 6 From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank (on the top of the tower) is 45 °. Find (i) the height of the tower, (ii) the depth of the tank. Solution 6 Question 7 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30 ° and that of the top of the flagstaff is 60°. Find the height of the tower. Solution 7 Question 8 A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. Solution 8 Let SP be the statue and PB be the pedestal. Angles of elevation of S and P are 60° and 45° respectively. Further suppose AB = x m, PB = h m In right ABS, In right PAB, Thus, height of the pedestal = 2m Question 9 The angle of elevation of the top of an unfinished tower at a distance of 75m from its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°? Solution 9 Let AB be the unfinished tower and let AC be complete tower. Let O be the point of observation. Then, OA = 75 m AOB = 30° and AOC = 60° Let AB = h meters And AC = H meters Hence, the required height is Question 10 On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top and bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it. Solution 10 Let AB be the tower and BC be flagpole, Let O be the point of observation. Then, OA = 9 m, AOB = 30° and AOC = 60° From right angled BOA From right angled OAC Thus Hence, height of the tower= 5.196 m and the height of the flagpole = 10.392 m Question 11 Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of one pole is 60° and the angle of depression from the top of another pole at P is 30°. Find the height of each pole and distances of the point P from the poles. Solution 11 Question 12 Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men. Solution 12 Question 13 From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. Solution 13 Question 14 A straight highway leads to the foot of a tower. A man standing on the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower form this point. Solution 14 Question 15 A TV tower stands vertically on a bank of canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. Solution 15 Question 16 The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building. Solution 16 Question 17 The horizontal distance between two towers is 60metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90metres, find the height of the tower. Solution 17 Let AB and CD be the first and second towers respectively. Then, CD = 90 m and AC = 60 m. Let DE be the horizontal line through D. Draw BF CD, Then, BF = AC = 60 m FBD = EDB = 30° Question 18 The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 metres, find the height of the chimney. According to pollution control norms, the minimum height of a smoke-emitting chimney should be 100 metres. State if the height of the above-mentioned chimney meets the pollution norms. What value is discussed in this question? Solution 18 Question 19 From the top of a 7-metre-high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. Solution 19 Question 20 The angle of depression from the top of a tower of a point A on the ground is 30°. On moving a distance of 20 metres from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60°. Find the height of the tower and its distance from the point A. Solution 20 Question 21 The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower. Solution 21 Question 22 Solution 22 Question 23 A man on the deck of a ship, 16m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60 and 30. Calculate the distance of the cliff from the ship and height of the cliff. Solution 23 Let AB be the height of the deck and let CD be the cliff.. Let the man be at B, then, AB= 16 m Let BE CD and AE CD Then, EBD = 60 and EBC = 30 CE = AB = 16m Let CD = h meters Then, ED = (h 16)m From right BED, we have From right CAB, we have Hence the height of cliff is 64 m and the distance between the cliff and the ship = Question 24 The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of tower PQ. Solution 24 Question 25 The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, find the speed of the aeroplane. Solution 25 Question 26 The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres. Solution 26 Let AB be the tower and let the angle of elevation of its top at C be 30°. Let D be a point at a distance 150 m from C such that the angle of elevation of the top of tower at D is 60°. Let h m be the height of the tower and AD = x m In CAB, we have Hence the height of tower is 129.9 m Question 27 As observed from the top of a lighthouse, 100m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation. Solution 27 Let AB be the light house and let C and D be the positions of the ship. Llet AD =x, CD = y In BDA, The distance travelled by the ship during the period of observation = 115.46 m Question 28 From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find the width of the river. Solution 28 Question 29 The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is 6 metres. Solution 29 Question 30 A ladder of length 6 metres makes an angle of 45° with the floor while leaning against one wall of a room. If the foot of the ladder is kept fixed on the floor and it is made to lean against the opposite wall of the room, it makes an angle of 60° with the floor. Find the distance between two walls of the room. Solution 30 Question 31 From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower. Solution 31 Question 32 An electrician has to repair an electric fault on a pole of height 4 metres. He needs to reach a point 1 metre below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use, which when inclined at an angle of 60° to the horizontal would enable him to reach the required position? Solution 32 Question 33 From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30 ° and 60° respectively. Find (i) the horizontal distance between AB and CD, (ii) the height of the lamp post, (iii) the difference between the heights of the building and the lamp post. Solution 33 ## Chapter 14 – Height and Distance Exercise MCQ Question 1 If the height of a vertical pole is equal to the length of its shadow on the ground, the angle of elevation of the sun is (a) 0 (b) 30 (c) 45 (d) 60 Solution 1 Question 2 (a) 30 (b) 45 (c) 60 (a) 75 Solution 2 Question 3 (a) 45 (b) 30 (c) 60 (d) 90 Solution 3 Question 4 (a) 60 (b) 45 (c) 30 (b) 90 Solution 4 Question 5 The shadow of a 5-m-long stick is 2 m long. At the same time, the length of the shadow of a 12.5-m-high tree is (a) 3 m (b) 3.5 m (c) 4.5 m (d) 5 m Solution 5 Question 6 A ladder makes an angle of 60 with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, the length of the ladder is Solution 6 Question 7 A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60 with the wall then the height of the wall is Solution 7 Question 8 From a point on the ground, 30 m away from the foot of a tower the angle of elevation of the top of the tower is 30°. The height of the tower is Solution 8 Question 9 The angle of depression of a car parked on the road from the top of a 150-m-high tower is 30°. The distance of the car from the tower is Solution 9 Question 10 A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is (a) 45 (b) 30 (c) 60 (a) 90 Solution 10 Question 11 From the top of a cliff 20 m high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is (a) 20 m (b) 40 m (c) 60 m (d) 80 m Solution 11 Question 12 If a 1-5-m-tall girl stands at a distance of 3 m from a lamp post and casts a shadow of length 4.5 m on the ground, then the height of the lamp post is (a) 1.5 m (b) 2 m (c) 2.5 m (d) 2.8 m Solution 12 Question 13 The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun’s elevation is 30 than when it was 45. The height of the tower is Solution 13 Question 14 (a) 30 (b) 45 (c) 60 (d) 90 Solution 14 Question 15 Solution 15 Question 16 Solution 16 Question 17 The tops of two towers of heights x and y, standing on a level ground subtend angles of 30 and 60 respectively at the centre of the line joining their feet. Then, x : y is (a) 1 : 2 (b) 2 : 1 (c) 1 : 3 (d) 3 : 1 Solution 17 Question 18 The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30. The height of the tower is Solution 18 Question 19 The string of a kite is 100 m long and it makes an angle of 60 with the horizontal. If there is no slack in the string, the height of the kite from the ground is Solution 19 Question 20 If the angles of elevation of the top of a tower from two points at distances a and b from the base and in the same straight line with it are complementary then the height of the tower is Solution 20 Question 21 On the level ground, the angle of elevation of a tower is 30. On moving 20 m nearer, the angle of elevation is 60. The height of the tower is Solution 21 Question 22 In a rectangle, the angle between a diagonal and a side is 30 and the length of this diagonal is 8 cm. The area of the rectangle is , Solution 22 Question 23 From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30 and 45. The height of the hill is Solution 23 Question 24 If the elevation of the sun changes from 30 to 60 then the difference between the lengths of shadows of a pole 15 m high, is Solution 24 Question 25 An observer 1.5 m tall is 28.5 m away from a tower and the angle of elevation of the top of the tower from the eye of the observer is 45. The height of the tower is (a) 27 m (b) 30 m (c) 28.5 m (d) none of these Solution 25 error: Content is protected !!
Instasolv IIT-JEE NEET CBSE NCERT Q&A 4.5/5 # NCERT Exemplar Class 6 Maths Chapter 3 Solutions: Integers NCERT Exemplar Solutions Class 6 maths Chapter 3 ‘Integers’ will help you understand what negative numbers and positive numbers are. NCERT Exemplar Class 6 Maths Chapter 3 contains a comprehensive as well as a simple explanation of the concepts of zero, natural numbers, whole numbers, and their relation with integers. This chapter consists of 83 questions which are of diverse variety to help you practice comprehensively. In this chapter, you will also be able to develop a better understanding of the number line by learning to plot the integers on it. The Chapter is designed in such a way that you will not only learn a new set of numbers, that is, Integers but you will also be able to revise the concepts learned in previous classes. The NCERT exemplar solutions will help you develop a strong base of number systems and will enable you to make comparisons between positive or negative integers. You will also learn to perform operations on integers such as addition, and addition with the help of number lines. The subject matter experts at Instasolv have curated NCERT exemplar study material in such a manner that you will be able to quickly grasp the exercises. You will be able to score better marks in all your internal exams and final exams. At Instasolv, the Class 6 Maths Exemplar solutions are prepared to address all your doubts and provide you with a hassle-free, one-stop solution for all your problems. ## Important Topics of NCERT Exemplar Class 6 Maths Solutions Chapter 3 • Introduction: The numbers which lie on the right side of the zero are greater than zero and are positive numbers. The numbers which lie on the left side of zero are smaller than zero and are known as negative numbers. Positive numbers are represented by a + sign before them and negative is represented by a – sign before them. 1. Integers: The collection of numbers that includes the set of positive numbers, zero, and negative numbers is known as integers. The ones which are positive and negative are termed as positive integers and negative integers respectively. Therefore, it may be concluded that all the natural numbers, whole numbers, zero and negative numbers can be termed as an integer whereas all integers may not fall in these categories. 1. Representation of Integer on the number line: Draw a line and mark some equidistant points on it. Mark one point as zero and therefore the points on the right will be marked as +1, +2, +3 etc. or simply 1, 2, 3 and so on. The points on the left of the zero will be marked as -1, -2, -3 and so on. 1. Ordering of Integers: On a number line, the number increases as we move to the right and it decreases as we move to the left. You add when there are two positive integers in the question. You should also add when there are two negative integers in the question, making sure the sum takes a negative sign before the number. When you have one positive and one negative to add, you must always subtract, but the answer will take the sign of the bigger integer (ignoring the signs of the numbers, decide which number is bigger) When a positive integer is added to the given integer, it becomes greater than its original value while when a negative integer is added to the given number, it becomes lesser than its original value. The number which is added to the given number to make its value zero is known as its additive inverse 1. Subtraction of Integers: When a negative integer is subtracted from a given number, we get a greater integer. The subtraction of an integer is the same as the addition of its additive inverse. ## Exercise Discussion of NCERT Exemplar Class 6 Maths Solutions Chapter 3 1. NCERT Class 6 Maths Exemplar Solutions for Chapter 3 are so designed by the experts at Instasolv such that they cover all the segments of the chapter in a simple manner to help you with rigorous practice. 2. The questions in this chapter 3 include exercises which have practical examples to make you understand negative numbers. 3. The questions also make you practice the correct usage of signs for positive and negative integers. There are exercise questions based on all the topics such as representation on the number line, addition and subtraction of integers 4. The syllabus of this chapter is covered in-depth, sticking to the latest trends of CBSE exam pattern with the help of the exercise questions. 5. To score well in the exam, you should only look up to NCERT Solutions and NCERT Exemplar Class 6 Science Solutions as they are self-sufficient and cover the entire syllabus of CBSE. ## Why Use NCERT Exemplar Solutions Class 6 Maths Chapter 3 by Instasolv? 1. NCERT Exemplar Solutions for Class 6 Maths Chapter 3 are very important from the exam perspective as they are designed, keeping in mind the latest CBSE exam patterns. 2. At Instasolv, the exemplar solutions are prepared after comprehensive research on the topic and hence, they are a reliable source for your queries. 3. We have a team designated especially to cater to all your needs related to the subject. 4. The objective of the experts at Instasolv is to make the subject more interactive and interesting for you and therefore prepare a strong foundation for your advanced classes. The NCERT Exemplar for Class 6 maths chapter 3 is designed in a language that is easy to understand and organised systematically so that you approach each topic step by step. More Chapters from Class 6
## How to Multiply Matrices ### Matrices Operations The basic operations on a matrix are addition, subtraction, and multiplication. For matrices to be added or subtracted, they must be in the same order. For matrices to be multiplied, the number of columns in the first Matrix must equal the number of rows in the second Matrix. ### How to Multiply Matrixes If $$A$$ and $$B$$ be any two matrices, then their product $$A \times B$$ will be defined only when the number of columns in $$A$$ is equal to the number of rows in $$B$$. If $$A \ = \ [a_{ij}]_{m \times n}$$ and $$B \ = \ [b_{ij}]_{n \times p}$$ then, their final product will be $$A \times B \ = \ C \ = \ [c_{ij}]_{m \times p}$$: $$C_{ij} \ = \ \sum_{t=1}^n a_{it}b_{tj}$$ Consider $$A \ = \ \begin{bmatrix} a_{11} & a_{12} \\\ a_{21} & a_{22} \end{bmatrix}$$, $$B \ = \ \begin{bmatrix} b_{11} & b_{12} \\\ b_{21} & b_{22} \end{bmatrix}$$ $$⇒ \ A \times B \ = \ C \ = \ \begin{bmatrix} a_{11}b_{11} \ + \ a_{12}b_{21} & a_{11}b_{12} \ + \ a_{12}b_{22} \\\ a_{21}b_{11} \ + \ a_{22}b_{21} & a_{21}b_{12} \ + \ a_{22}b_{22} \end{bmatrix}$$ ### Matrix Multiplication Properties 1- In general, matrix multiplication is not commutative: $$AB \ ≠ \ BA$$ 2- Multiplication of matrices is associative, which means that: $$(AB)C \ = \ A(BC)$$ 3- Multiplication of matrices is distributive over the addition of matrices, which means that $$A(B \ + \ C) \ = \ AB \ + \ AC$$ and $$(A \ + \ B)C \ = \ AC \ + \ BC$$. 4- Is $$A$$ an $$m \times n$$ matrix? If so, then $$I_mA \ = \ A \ = \ AI_n$$. 5- The product of two matrices can be a null matrix even if neither of the two matrices is null. This means that if $$AB \ = \ 0$$, it is not necessary that $$A \ = \ 0$$ or $$B \ = \ 0$$. 6- If $$A$$ is an $$m \times n$$ matrix and $$O$$ is a null matrix, then the product of a matrix and a null matrix is always a null matrix: $$A_{m \times n}O_{n \times p} \ = \ O_{m \times p}$$ 7- If $$AB \ = \ 0$$, it does not necessarily follow that $$A$$ or $$B$$ are zero, and a zero matrix could result from multiplying two non-zero matrices. 8- If $$AB \ = \ AC$$, $$B \ ≠ \ C$$ (Cancellation Law is not applicable). 9- $$tr(AB) \ = \ tr(BA)$$ 10- For every square Matrix, there is a multiplicative identity such that $$AI \ = \ IA \ = A$$ ### Exercises for Matrix Multiplication 1) Find the answer: $$\begin{bmatrix} -6 & 2 \\\ 3 & -5 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\\ 0 & -1 \end{bmatrix}$$ 2) Find the answer: $$\begin{bmatrix} 4 & -8 \\\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\\ 1 & 3 \end{bmatrix}$$ 3) Find the answer: $$\begin{bmatrix} 7 \\\ 3 \\\ 4 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \end{bmatrix}$$ 4) Find the answer: $$\begin{bmatrix} 3 \\\ 9 \end{bmatrix} \times \begin{bmatrix} 7 & 3 & -2 \end{bmatrix}$$ 5) Find the answer: $$\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix}$$ 6) Find the answer: $$\begin{bmatrix} -4 & 1 \\\ -3 & 5 \end{bmatrix} \times \begin{bmatrix} 7 & 9 \end{bmatrix}$$ 7) Find the answer: $$\begin{bmatrix} 4 \\\ -2 \\\ 3 \end{bmatrix} \times \begin{bmatrix} 7 & 5 \end{bmatrix}$$ 8) Find the answer: $$\begin{bmatrix} -9 \\\ 8 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & -4 \end{bmatrix}$$ 9) Find the answer: $$\begin{bmatrix} 5 & -7 \end{bmatrix} \times \begin{bmatrix} 4 & 3 \\\ 7 & -6 \end{bmatrix}$$ 10) Find the answer: $$\begin{bmatrix} 5 & 4 & 9 \end{bmatrix} \times \begin{bmatrix} 7 \\\ -6 \\\ 3 \end{bmatrix}$$ 1) Find the answer: $$\begin{bmatrix} -6 & 2 \\\ 3 & -5 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\\ 0 & -1 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} -6 & 2 \\\ 3 & -5 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\\ 0 & -1 \end{bmatrix} \ =}$$ $$\color{red}{ \begin{bmatrix} (-6 \times 1) \ + \ (2 \times 0) & (-6 \times 3) \ + \ (2 \times -1) \\\ (3 \times 1) \ + \ (-5 \times 0) & (3 \times 3) \ + \ (-5 \times -1) \end{bmatrix} \ =}$$$$\color{red}{ \begin{bmatrix} -6 & -20 \\\ 3 & 14 \end{bmatrix}}$$ 2) Find the answer: $$\begin{bmatrix} 4 & -8 \\\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\\ 1 & 3 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} 4 & -8 \\\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\\ 1 & 3 \end{bmatrix} \ = }$$ $$\color{red}{ \begin{bmatrix} (4 \times 2) \ + \ (-8 \times 1) & (4 \times 0) \ + \ (-8 \times -3) \\\ (-2 \times 2) \ + \ (3 \times 1) & (-2 \times 0) \ + \ (3 \times 3) \end{bmatrix} \ =}$$ $$\color{red}{ \begin{bmatrix} 0 & -24 \\\ -1 & 9 \end{bmatrix}}$$ 3) Find the answer: $$\begin{bmatrix} 7 \\\ 3 \\\ 4 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} 7 \\\ 3 \\\ 4 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \end{bmatrix} \ = \begin{bmatrix} (7 \times (-2)) & (7 \times 1) \\\ (3 \times (-2)) & (3 \times 1) \\\ (4 \times (-2)) & (4 \times 1) \end{bmatrix} \ =}$$ $$\color{red}{ \begin{bmatrix} -14 & 7 \\\ -6 & 3 \\\ -8 & 4 \end{bmatrix}}$$ 4) Find the answer: $$\begin{bmatrix} 3 \\\ 9 \end{bmatrix} \times \begin{bmatrix} 7 & 3 & -2 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} 3 \\\ 9 \end{bmatrix} \times \begin{bmatrix} 7 & 3 & -2 \end{bmatrix} \ = }$$ $$\color{red}{ \begin{bmatrix} (3 \times 7) & (3 \times 3) & (3 \times (-2)) \\\ (9 \times 7) & (9 \times 3) & (9 \times (-2)) \end{bmatrix} \ =}$$ $$\color{red}{\begin{bmatrix} 21 & 9 & -6 \\\ 63 & 27 & -18 \end{bmatrix}}$$ 5) Find the answer: $$\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix} \ =}$$ $$\color{red}{\begin{bmatrix} (7 \times 3) \ + \ (-2 \times 9) \end{bmatrix} \ = \ \begin{bmatrix} 3 \end{bmatrix}}$$ 6) Find the answer: $$\begin{bmatrix} -4 & 1 \\\ -3 & 5 \end{bmatrix} \times \begin{bmatrix} 7 & 9 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} -4 & 1 \\\ -3 & 5 \end{bmatrix} \times \begin{bmatrix} 7 & 9 \end{bmatrix} \ =}$$ $$\color{red}{ \begin{bmatrix} (-4 \times 7) \ + \ (1 \times 9) & (-3 \times 7) \ + \ (5 \times 9) \end{bmatrix} \ =}$$$$\color{red}{ \begin{bmatrix} -19 & 24 \end{bmatrix}}$$ 7) Find the answer: $$\begin{bmatrix} 4 \\\ -2 \\\ 3 \end{bmatrix} \times \begin{bmatrix} 7 & 5 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} 4 \\\ -2 \\\ 3 \end{bmatrix} \times \begin{bmatrix} 7 & 5 \end{bmatrix} \ = \ \begin{bmatrix} (4 \times 7) & (4 \times 5) \\\ ((-2) \times 7) & ((-2) \times 5) \\\ (3 \times 7) & (3 \times 5) \end{bmatrix} \ =}$$ $$\color{red}{ \begin{bmatrix} 28 & 20 \\\ -14 & -10 \\\ 21 & 15 \end{bmatrix}}$$ 8) Find the answer: $$\begin{bmatrix} -9 \\\ 8 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & -4 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} -9 \\\ 8 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & -4 \end{bmatrix} \ =}$$ $$\color{red}{ \begin{bmatrix} (-9 \times 3) & (-9 \times (-2)) & (-9 \times (-4)) \\\ (8 \times 3) & (8 \times (-2)) & (8 \times (-4)) \end{bmatrix} \ = }$$ $$\color{red}{\begin{bmatrix} -27 & 18 & 36 \\\ 24 & -16 & -32 \end{bmatrix}}$$ 9) Find the answer: $$\begin{bmatrix} 5 & -7 \end{bmatrix} \times \begin{bmatrix} 4 & 3 \\\ 7 & -6 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} 5 & -7 \end{bmatrix} \times \begin{bmatrix} 4 & 3 \\\ 7 & -6 \end{bmatrix} \ = }$$ $$\color{red}{ \begin{bmatrix} (5 \times 4) \ + \ (-7 \times 7) & (5 \times 3) \ + \ (-7 \times (-6)) \end{bmatrix} \ = }$$ $$\color{red}{ \begin{bmatrix} -29 & 57 \end{bmatrix}}$$ 10) Find the answer: $$\begin{bmatrix} 5 & 4 & 9 \end{bmatrix} \times \begin{bmatrix} 7 \\\ -6 \\\ 3 \end{bmatrix}$$ $$\color{red}{\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix} \ = \ \begin{bmatrix} (5 \times 7) \ + \ (4 \times (-6)) \ + \ (9 \times 3) \end{bmatrix} \ =}$$ $$\color{red}{ \ \begin{bmatrix} 38 \end{bmatrix}}$$ ## Matrix Multiplication Practice Quiz ### Comprehensive DAT Math Practice BooK $17.99$12.99 ### HiSET Mathematics Formulas $6.99$5.99 ### ParaPro Math Test Prep Bundle $69.99$35.99 ### SSAT Upper Level Mathematics Formulas $6.99$5.99
# Finding the Sum and Difference of Two Matrices To solve a problem like the one described for the soccer teams, we can use a matrix, which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned vertically. Each number is an entry, sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters. For example, three matrices named $A,B,\text{}$ and $C$ are shown below. $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],B=\left[\begin{array}{ccc}1& 2& 7\\ 0& -5& 6\\ 7& 8& 2\end{array}\right],C=\left[\begin{array}{c}-1\\ 0\\ 3\end{array}\begin{array}{c}3\\ 2\\ 1\end{array}\right]$ ## Describing Matrices A matrix is often referred to by its size or dimensions: $\text{ }m\text{ }\times \text{ }n\text{ }$ indicating $m$ rows and $n$ columns. Matrix entries are defined first by row and then by column. For example, to locate the entry in matrix $A$ identified as ${a}_{ij},\text{}$ we look for the entry in row $i,\text{}$ column $j$ . In matrix $A\text{, \hspace{0.17em}}$ shown below, the entry in row 2, column 3 is ${a}_{23}$ . $A=\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right]$ A square matrix is a matrix with dimensions $\text{ }n\text{ }\times \text{ }n,\text{}$ meaning that it has the same number of rows as columns. The $3\times 3$ matrix above is an example of a square matrix. A row matrix is a matrix consisting of one row with dimensions $1\text{ }\times \text{ }n$ . $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]$ A column matrix is a matrix consisting of one column with dimensions $m\text{ }\times \text{ }1$ . $\left[\begin{array}{c}{a}_{11}\\ {a}_{21}\\ {a}_{31}\end{array}\right]$ A matrix may be used to represent a system of equations. In these cases, the numbers represent the coefficients of the variables in the system. Matrices often make solving systems of equations easier because they are not encumbered with variables. We will investigate this idea further in the next section, but first we will look at basic matrix operations. ### A General Note: Matrices A matrix is a rectangular array of numbers that is usually named by a capital letter: $A,B,C,\text{}$ and so on. Each entry in a matrix is referred to as ${a}_{ij}$ , such that $i$ represents the row and $j$ represents the column. Matrices are often referred to by their dimensions: $m\times n$ indicating $m$ rows and $n$ columns. ### Example 1: Finding the Dimensions of the Given Matrix and Locating Entries Given matrix $A:$ 1. What are the dimensions of matrix $A?$ 2. What are the entries at ${a}_{31}$ and ${a}_{22}?$ $A=\left[\begin{array}{rrrr}\qquad 2& \qquad & \qquad 1& \qquad 0\\ \qquad 2& \qquad & \qquad 4& \qquad 7\\ \qquad 3& \qquad & \qquad 1& \qquad -2\end{array}\right]$ ### Solution 1. The dimensions are $\text{ }3\text{ }\times \text{ }3\text{ }$ because there are three rows and three columns. 2. Entry ${a}_{31}$ is the number at row 3, column 1, which is 3. The entry ${a}_{22}$ is the number at row 2, column 2, which is 4. Remember, the row comes first, then the column. We use matrices to list data or to represent systems. Because the entries are numbers, we can perform operations on matrices. We add or subtract matrices by adding or subtracting corresponding entries. In order to do this, the entries must correspond. Therefore, addition and subtraction of matrices is only possible when the matrices have the same dimensions. We can add or subtract a $\text{ }3\text{ }\times \text{ }3\text{ }$ matrix and another $\text{ }3\text{ }\times \text{ }3\text{ }$ matrix, but we cannot add or subtract a $\text{ }2\text{ }\times \text{ }3\text{ }$ matrix and a $\text{ }3\text{ }\times \text{ }3\text{ }$ matrix because some entries in one matrix will not have a corresponding entry in the other matrix. ### A General Note: Adding and Subtracting Matrices Given matrices $A$ and $B$ of like dimensions, addition and subtraction of $A$ and $B$ will produce matrix $C$ or matrix $D$ of the same dimension. $A+B=C\text{ such that }{a}_{ij}+{b}_{ij}={c}_{ij}$ $A-B=D\text{ such that }{a}_{ij}-{b}_{ij}={d}_{ij}$ $A+B=B+A$ It is also associative. $\left(A+B\right)+C=A+\left(B+C\right)$ ### Example 2: Finding the Sum of Matrices Find the sum of $A$ and $B,\text{}$ given $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\text{ and }B=\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]$ ### Solution $\begin{array}{l}A+B=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]\qquad \\ \text{ }=\left[\begin{array}{ccc}a+e& & b+f\\ c+g& & d+h\end{array}\right]\qquad \end{array}$ ### Example 3: Adding Matrix A and Matrix <>B Find the sum of $A$ and $B$ . $A=\left[\begin{array}{cc}4& 1\\ 3& 2\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 9\\ 0& 7\end{array}\right]$ ### Solution ${a}_{11},\text{}$ of matrix $A$ to the entry in row 1, column 1, ${b}_{11}$ , of $B$ . Continue the pattern until all entries have been added. $\begin{array}{l}A+B=\left[\begin{array}{cc}4& 1\\ 3& 2\end{array}\right]+\left[\begin{array}{cc}5& 9\\ 0& 7\end{array}\right]\qquad \\ \text{ }=\left[\begin{array}{ccc}4+5& & 1+9\\ 3+0& & 2+7\end{array}\right]\qquad \\ \text{ }=\left[\begin{array}{cc}9& 10\\ 3& 9\end{array}\right]\qquad \end{array}$ ### Example 4: Finding the Difference of Two Matrices Find the difference of $A$ and $B$ . $A=\left[\begin{array}{cc}-2& 3\\ 0& 1\end{array}\right]\text{ and }B=\left[\begin{array}{cc}8& 1\\ 5& 4\end{array}\right]$ ### Solution We subtract the corresponding entries of each matrix. $\begin{array}{l}A-B=\left[\begin{array}{rr}\qquad -2& \qquad 3\\ \qquad 0& \qquad 1\end{array}\right]-\left[\begin{array}{rr}\qquad 8& \qquad 1\\ \qquad 5& \qquad 4\end{array}\right]\qquad \\ \text{ }=\left[\begin{array}{rrr}\qquad -2 - 8& \qquad & \qquad 3 - 1\\ \qquad 0 - 5& \qquad & \qquad 1 - 4\end{array}\right]\qquad \\ \text{ }=\left[\begin{array}{rrr}\qquad -10& \qquad & \qquad 2\\ \qquad -5& \qquad & \qquad -3\end{array}\right]\qquad \end{array}$ ### Example 5: Finding the Sum and Difference of Two 3 x 3 Matrices Given $A$ and $B:$ 1. Find the sum. 2. Find the difference. $A=\left[\begin{array}{rrr}\qquad 2& \qquad -10& \qquad -2\\ \qquad 14& \qquad 12& \qquad 10\\ \qquad 4& \qquad -2& \qquad 2\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\qquad 6& \qquad 10& \qquad -2\\ \qquad 0& \qquad -12& \qquad -4\\ \qquad -5& \qquad 2& \qquad -2\end{array}\right]$ ### Solution $\begin{array}{l}\qquad \\ A+B=\left[\begin{array}{rrr}\qquad 2& \qquad -10& \qquad -2\\ \qquad 14& \qquad 12& \qquad 10\\ \qquad 4& \qquad -2& \qquad 2\end{array}\right]+\left[\begin{array}{rrr}\qquad 6& \qquad 10& \qquad -2\\ \qquad 0& \qquad -12& \qquad -4\\ \qquad -5& \qquad 2& \qquad -2\end{array}\right]\qquad \\ =\left[\begin{array}{rrr}\qquad 2+6& \qquad -10+10& \qquad -2 - 2\\ \qquad 14+0& \qquad 12 - 12& \qquad 10 - 4\\ \qquad 4 - 5& \qquad -2+2& \qquad 2 - 2\end{array}\right]\qquad \\ =\left[\begin{array}{rrr}\qquad 8& \qquad 0& \qquad -4\\ \qquad 14& \qquad 0& \qquad 6\\ \qquad -1& \qquad 0& \qquad 0\end{array}\right]\qquad \end{array}$ 2. Subtract the corresponding entries. $\begin{array}{l}\qquad \\ A-B=\left[\begin{array}{rrr}\qquad 2& \qquad -10& \qquad -2\\ \qquad 14& \qquad 12& \qquad 10\\ \qquad 4& \qquad -2& \qquad 2\end{array}\right]-\left[\begin{array}{rrr}\qquad 6& \qquad 10& \qquad -2\\ \qquad 0& \qquad -12& \qquad -4\\ \qquad -5& \qquad 2& \qquad -2\end{array}\right]\qquad \\ =\left[\begin{array}{rrr}\qquad 2 - 6& \qquad -10 - 10& \qquad -2+2\\ \qquad 14 - 0& \qquad 12+12& \qquad 10+4\\ \qquad 4+5& \qquad -2 - 2& \qquad 2+2\end{array}\right]\qquad \\ =\left[\begin{array}{rrr}\qquad -4& \qquad -20& \qquad 0\\ \qquad 14& \qquad 24& \qquad 14\\ \qquad 9& \qquad -4& \qquad 4\end{array}\right]\qquad \end{array}$ ### Try It 1 $A$ $B$ $A=\left[\begin{array}{rr}\qquad 2& \qquad 6\\ \qquad 1& \qquad 0\\ \qquad 1& \qquad -3\end{array}\right]\text{ and }B=\left[\begin{array}{rr}\qquad 3& \qquad -2\\ \qquad 1& \qquad 5\\ \qquad -4& \qquad 3\end{array}\right]$
Optimizing Simple Systems An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: Optimizing Complex Systems You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:40 Visualize It • 1:04 Define the Problem • 1:21 Write an Equation • 2:44 Find the Min/Max • 4:54 Lesson Summary Want to watch this again later? Timeline Autoplay Autoplay Recommended Lessons and Courses for You Lesson Transcript Instructor: Sarah Wright Optimization problems may seem overwhelming, but they can actually be quite simple. In this lesson, learn how to use a handy five-step formula to tackle these daunting problems. Five Steps to Solve Optimization Problems Remember that optimization problems are everywhere, but we have a five-step way to solve optimization problems. We must visualize the problem, define the problem, write an equation for the problem, find the minimum or maximum for the problem and then answer the question. With these steps, we can solve most optimization problems. Let's try one. What is the maximum amount of land that you can enclose in a rectangular pen that has a perimeter of 100m? Step 1: Visualize It The first thing we need to do is visualize it. Let's draw this out; I have a rectangular pen. Let's say it's going to hold my turtle (he needs a lot of room). Here's my rectangular pen: it's got a height of h, a width of w and a perimeter of 100m. I don't know what h and w are since they weren't given to me. Instead, I get to pick those to solve this problem. Step 2: Define the Problem Next is the second step, define the problem. I need to maximize the pen area. I'm constrained by my perimeter which has to equal 100m. I don't have any other constraints, so I need to maximize this area which is really the product of my width and height of the pen. Step 3: Write an Equation Let's write this as an equation - step three. Area is height times the width, but I don't know what the height or the width is. I do know that the perimeter of this rectangle is 2(height) + 2(width), and the perimeter has to be 100m. When we plug that in, we get 100 = 2h + 2w. Again, area = hw. I'm almost there, but I really want one equation. I don't know how to optimize two equations. Besides, I have three unknowns here. I'm trying to maximize the area, but I've got h and w that can change. Let's get rid of one of those. Let's solve the perimeter equation for width. So, I've got 100 = 2h + 2w, which I can write as 50 = h + w (I've just divided everything by 2). I can solve that by subtracting h from both sides and I end up with w = 50 - h. Now I can plug my width into my area equation, so that my area just depends on the height. Now, I've got one equation, A = h(50 - h), where h is my height and A is my area. To unlock this lesson you must be a Study.com Member. Register for a free trial Are you a student or a teacher? See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Back What teachers are saying about Study.com Earning College Credit Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
# ANGLE MEASURES IN POLYGONS There are two types of angle measures in polygons. 1. Interior Angles 2. Exterior Angles ## Measures of Interior and Exterior Angles Interior Angle : An interior angle of a polygon is an angle inside the polygon at each of its vertices. Exterior Angle : An exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side. More clearly, ## Polygon Interior Angles Theorem Theorem : The sum of the measures of the interior angles of a convex n-gon is (n - 2) ⋅ 180° Corollary to the above Theorem : The measure of each interior angle of a regular n-gon is 1/n ⋅ (n - 2) ⋅ 180° or [(n - 2) ⋅ 180°]/n ## Polygon Exterior Angles Theorem Theorem : The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is 360° Corollary to the above Theorem : The measure of each exterior angle of a regular n-gon is 1/n ⋅ 360° or 360°/n ## Finding Measures of Interior Angles of Polygons Example 1 : Find the value of x in the diagram shown below. Solution : The polygon shown in the diagram above has 6 sides. So it is hexagon. The sum of the measures of the interior angles of any hexagon is = (6 - 2) ⋅ 180° = 4 ⋅ 180° = 720° We can add the measures of all interior angles of the above hexagon and the sum can be equated to 720°. So, we have 136° + 136° + 88° + 142° + 105° + x° = 720° Simplify. 607 + x = 720 Subtract 607 from each side. x = 113 Hence, the measure of sixth interior angle of the hexagon is 113°. ## Finding the Number of Sides of a Polygon Example 2 : The measure of each interior angle of a regular polygon is 140°. How many sides does the polygon have ? Solution : By Polygon Interior Angles Theorem, we have [(n - 2) ⋅ 180°]/n = 140° Multiply each side by n. (n - 2) ⋅ 180 = 140n Simplify. 180n - 360 = 140n Subtract 140n from each side. 40n - 360 = 0 40n = 360 Divide each side by 40. 40n/40 = 360/40 n = 9 Hence, the polygon has 9 sides and it is a regular nonagon. ## Finding Measures of an Exterior Angle Example 3 : Find the value of x in the regular polygon shown below. Solution : The polygon shown above is regular and it has 7 sides. So, it is a regular heptagon and the measure of each exterior angle is x°. By the Polygon Exterior Angles Theorem, we have x° = 1/7 ⋅ 360° Simplify. x ≈ 51.4 Hence, the measure of each exterior angle of a regular heptagon is about 51.4°. Example 4 : Find the value of x in the diagram shown below. Solution : The polygon shown in the diagram above has 5 sides. So it is pentagon. The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is = 360° We can add the measures of all exterior angles of the above pentagon and the sum can be equated to 360°. So, we have 2x° + x° + 3x° + 4x° + 2x° = 360° Simplify. 12x = 360 Divide each side by 12. 12x/12 = 360/12 x = 30 ## Finding Angle Measures of a Polygon Example 5 : A home plate maker for a soft ball field is a pentagon. Three of the interior angles of the pentagon are right angles. The remaining two interior angles are congruent. What is the measure of each angle ? Solution : Draw a Sketch : Sketch and label a diagram for the above plate maker. It is a non regular pentagon. Let ∠A, ∠B and ∠D be the right angles. Let ∠C and ∠E be the remaining two congruent angles. So, we have ∠C ≅ ∠E The sum of the measures of the interior angles of a pentagon is = (5 - 2) ⋅ 180° = 3 ⋅ 180° = 540° Verbal Model : Labels : Sum of measures of interior angles = 540° Measure of each right angle = 90° Measure of ∠C and ∠E = x° Reasoning : Write the equation. 540° = 3 ⋅ 90° + 2x° Simplify. 540 = 270 + 2x Subtract 270 from each side. 270 = 2x Divide each side by 2. 270/2 = 2x/2 135 = x Hence, the measure of each of the two congruent angles is 135°. ## Using Angle Measures of a Regular Polygon Example 6 : If you were designing the home plate marker for some new type of ball game, would it be possible to make a home plate marker that is a regular polygon with each interior angle having a measure of (a) 135° ? (b) 145° ? Solution : Solution (a) : Let n be the number of sides of the regular polygon. By Polygon Interior Angles Theorem, we have [(n - 2) ⋅ 180°]/n = 135° Multiply each side by n. (n - 2) ⋅ 180 = 135n 180n - 360 = 135n Subtract 135n from each side. 45n - 360 = 0 45n = 360 Divide each side by 45. 45n/45 = 360/45 n = 8 Yes, it would be possible. Because a polygon can have 8 sides. Solution (b) : Let n be the number of sides of the regular polygon. By Polygon Interior Angles Theorem, we have [(n - 2) ⋅ 180°]/n = 145° Multiply each side by n. (n - 2) ⋅ 180 = 145n 180n - 360 = 145n Subtract 145n from each side. 35n - 360 = 0 35n = 360 Divide each side by 35. 35n/35 = 360/35 n ≈ 10.3 No, it would not be possible. Because, a polygon can not have 10.3 sides. Kindly mail your feedback to v4formath@gmail.com ## Recent Articles 1. ### Representation of Set Oct 02, 23 08:32 PM Representation of Set 2. ### Operations on Sets Oct 02, 23 12:38 PM Operations on Sets - Concept - Detailed Explanation with Venn Diagrams 3. ### Introduction to Sets Oct 02, 23 12:00 PM Introduction to Sets
# Vertex of the graph of a quadratic polynomial This is what a website states: Before graphing a quadratic function we rearrange the equation, from this: $f(x) = ax^2 + bx + c$ To this: $f(x) = a(x-h)^2 + k$ Where: $h = -b/2a$ $k = f( h )$ In other words, calculate h$(=-b/2a)$, then find $k$ by calculating the whole equation for $x=h$ The website states the following reason for doing this: Well, the wonderful thing about this new form is that $h$ and $k$ show us the very lowest (or very highest) point, called the vertex: And also the curve is symmetrical (mirror image) about the axis that passes through $x=h$, making it easy to graph I want to know that how is $h$ the x - coordinate and $k$ the y - coordinate of the vertex? The square of any real number is always at least zero, and equals zero only when that number is itself zero. So $(x-h)^2$ is always at least zero, and is equal to zero only when $x-h=0$; that happens only when $x=h$. Since $(x-h)^2$ is always at least zero, we see that any positive multiple of it is also always at least zero. So, if we are told that $a>0$, we can say with certainty that $a(x-h)^2$ is always at least zero and is equal to zero only when $x=h$. Finally, we see that $a(x-h)^2 +k$ is always at least $k$ and is equal to $k$ only when $x=h$. That's because we are adding to $k$ a number that is always at least zero (so, the sum can't be any less than $k$). Clearly $(h,k)$ is a point on the graph of the function $f(x) = a(x-h)^2+k$. Any other point $(x,y)$ on the graph has an $x$-coordinate that is not $h$, so the $y$-coordinate must be greater than $k$. This means that if we move left or right from $(h,k)$, staying on the graph, we can only move up. Done. (The case $a<0$ is handled similarly -- try it yourself! What happens if $a=0$?) Addendum: How to find $h$ and $k$ This is just completing the square to rewrite the original expression. Start with: $$f(x) = ax^2 + bx + c$$ Now suppose there is and $h$ and a $k$ so that this is the same as $$f(x) = a(x-h)^2 + k$$ Let's write this out in full : $$a(x-h)^2 + k=a(x^2-2hx + h^2) + k = ax^2-2ahx + ah^2+k$$ Now if these are to be the same function for all $x$, then $$ax^2 + bx + c = ax^2-2ahx + ah^2+k$$ Move everything to one side (note the $ax^2$ terms add out): $$(\underbrace{b+2ah}_{\textrm{constant}})x + (\underbrace{c-ah^2-k}_{\textrm{constant}})=0$$ This has to be true for all $x$. But this is just a linear function of $x$, and the only linear function which is zero for all $x$ is the linear function whose coefficients are all zero! This means we can write $$\begin{cases}b+2ah = 0\\c-ah^2-k = 0\end{cases}$$ So solve the first equation for $h$ to get $$\boxed{h=-\tfrac b{2a}}$$ Now you could plug this into the second equation and solve for $k$, but it's probably easier to note that, since we assumed $$f(x) = a(x-h)^2+k$$ then just putting $x=h$ into this we get $$\boxed{f(h)} = a(h-h)^2+k=a\cdot 0^2 + k = \boxed{k}$$ In other words, get $k$ by using the fact that $$\boxed{k=f(h)}$$ (and we know $h$ at this point). • I got your point that h and k would be coordinates of vertex. But I still didn't understood how we reached the conclusion that h = -b/2a and k = f(h). – Parth Apr 21 '16 at 17:11 • Okay, I added a section in my answer to show how you know this. – MPW Apr 21 '16 at 20:25 • How is b - 2ah = 0? Shouldn't it be b + 2ah? – Parth Apr 22 '16 at 0:49 • @unknownCoder: Yes, typo. Will fix. Thanks for the catch, +1. – MPW Apr 22 '16 at 12:48 • What is the need of including an a in a(x - h)^2 + k? – Parth Apr 22 '16 at 15:13 The quantity $(x-h)^2$ is always non negative, and the smallest value it can attain is zero when $x=h$. Now we have two cases: 1) $a>0$: in this case the quantity $a(x-h)^2$ is always non negative and has zero as its smallest value when $x=h$. 2) $a<0$: in this case the quantity $a(x-h)^2$ is always non positive and has zero as its largest value when $x=h$. In both cases $x=h$ represents a global minimum or maximum of the function; this property doesn't change if you consider the "shifted" quantity $a(x-h)^2+k$, because adding a constant to a function does not change the position of its maxima or minima. So the x-coordinate of the vertex is at $x=h$. (1) All graphs of quadratic functions are symmetric about some vertical axis. $y = f(x) = ax^2 + bx + c$ (where we assume that $a > 0$ for simplicity) is of no exception. (2) The vertex $V(h, k)$ will lie on that axis. (3) Let us assume that $f(x) = 0$ has roots $\alpha$ and $\beta$. Then, the graph of y = f(x) will cut the x-axis at $A( \alpha, 0)$ and $B( \beta, 0)$. The point $( \dfrac {\alpha + \beta}{2}, 0)$ is midway between A and B. Then, the vertical axis (of symmetry) has the equation $L : x = \dfrac {\alpha + \beta}{2}$. (4) Recall that $\alpha + \beta =$ the sum of roots $= \dfrac {–b}{a}$. Therefore, equation of L is then $L : x = \dfrac {–b}{2a}$. (5) According to (2), $V = (h, k) = (\dfrac {–b}{2a}, k)$; where k is given by $f(\dfrac {–b}{2a})$ or $f(h)$ because $V$ is a point on the curve of $y = f(x)$. In the case that $f(x) = 0$ has no real roots, (ie. the graph of $y = f(x)$ does not cut the x-axis), we can translate the x-axis upward certain suitable units such that the new X-axis will cut $y = f(x)$. The above argument still applies.
Overview The fact that many quadratic equations have two solutions has the potential to be confusing. Bear in mind, however, that we are normally looking for just one solution to a particular problem. Choosing the correct solution from two possible answers is usually a matter of recognising which of the answers is appropriate to the problem. In many cases, a negative answer would make no sense, so we can ignore such an outcome. It is very rare that we are left wondering which of the two answers (should there in fact be two answers) is the correct one. It should be clear from the above that what we are looking for is a value (or values) for x that satisfy the function y = ƒ(x) = ax 2 + bx + c = 0. We can often factorise the left-hand side of the equation to give us two terms for x. Factorising is essentially the reverse process of multiplying out brackets to simplify an expression such as (x + 3)(x - 2), which would give us x 2 + 3x - 2x - 6 x 2 + x - 6. Why do we want to do this? Because it leaves us with an equation like (x + 3)(x - 2) = 0. In order for such an equation to be true, one or both of the expressions inside the brackets must evaluate to zero. Therefore, either x plus three equals zero (x + 3 = 0), or x minus two equals zero (x - 2 = 0). The obvious conclusion is that x is equal to minus three or x is equal to two (x = -3 or x = 2). And that's it. We have solved the quadratic equation. Unfortunately, not all quadratic equations factorise, but many do. The skill involved in using factorisation to solve quadratic equations comes with practice and experience, since you first of all need to be able to determine whether or not the equation will factorise in the first place, and then (assuming it will factorise) you need to find two numbers which would, if the brackets were to be multiplied out again, result in the same quadratic equation we started with. Sometimes the right numbers are easy to find, sometimes they can be somewhat elusive. You can of course check whatever answers you come up with by multiplying the brackets out again to see if you get the original quadratic equation. We can see how this works using an example. Consider the following quadratic equation: x 2 - 4x - 5 = 0 The value of the constant term (the last constant in the equation) is minus five (-5). When the brackets are multiplied out, the constant term will be the product of the integer terms within each bracket. Because five is a prime number, it only has two factors (itself and one). Because its value is minus five, the factors could be minus five and one, or five and minus one (-5 and 1, or 5 and -1). Note that the linear coefficient (that's the number in font of x in the middle of the quadratic equation) is minus four. The integer values we are looking for (both of which will be multiplied by x when the brackets are multiplied out) must therefore sum to minus four, so we have to use minus five and one. Let's write the equation out in its factorised form: (x - 5)(x + 1) = 0 If we now multiply out the brackets again, we get: x 2 + x - 5x - 5 = 0 Now gather like terms together: x 2 - 4x - 5 = 0 It works! So, we have a solution, namely that x must equal five or minus one (x = 5 or x = -1). You can see the graph for this function below (you might even recognise it, since it was used as one of the examples on the page "Quadratic Equations" in this section). If you take a look at this graph, you will see that it does indeed intersect the x-axis at x = 5 and x = -1. The graph of y = ƒ(x) = x 2 - 4x - 5 The method that has emerged here for factorising this type of quadratic equation is relatively straightforward. We are looking for two integer values that (a) multiply together to give the constant term and (b) add together to give the linear coefficient. Let's try another example: x 2 - 7x + 12 = 0 The constant term here is twelve. The factors of twelve are one, two, three, four, six and twelve. We are looking for two factors with a product of twelve, but we can discount one and twelve because they do not sum to minus seven (the linear coefficient) in any combination of positive or negative values. Neither do two and six, which leaves only three and four. In order to sum to minus seven, both numbers must be negative, so we have minus three and minus four. Writing the equation out in its factorised form, we have: (x - 3)(x - 4) = 0 Note that because we are multiplying the brackets together, the order in which they are written does not particularly matter. If we now multiply out the brackets again, we get: x 2 - 4x - 3x + 12 = 0 Now gather like terms together: x 2 - 7x + 12 = 0 Once again we have found two values for x. The graph of the corresponding quadratic function is shown below. The illustration below is a screenshot of the extremely useful "Quadratic Function Explorer" facility at the Math Open Reference website (http://www.mathopenref.com/quadraticexplorer.html). This time, our solution gives us x equals three or x equals four (x = 3 or x = 4). As you get more practice at factoring quadratic equations like the ones above, you will find that you can pick out the numbers you need relatively quickly and painlessly, but (you knew there was a "but" coming, didn't you!) bear in mind that the equations we have looked at so far both had a quadratic coefficient of one. The quadratic equation x 2 - 7x + 12 = 0 has roots of x = 3 and x = 4 When the quadratic coefficient is greater than one When the quadratic coefficient is greater than one, life becomes a tiny bit more complicated. Consider the following quadratic equation: 3x 2 + 5x - 2 = 0 We are still looking for two numbers that add together to give the linear coefficient (which in this case is five). Now, however, instead of looking for factors of minus two (the constant term), we need to find two numbers that, when multiplied together, result in the product of the quadratic coefficient and the constant term. This means we are looking for factors of minus six, i.e. three multiplied by minus two (3 × -2 = -6). The only factors of minus six that actually produce minus six when multiplied together and can be added together to produce five, are six and minus one (6 × -1 = -6, 6 + -1 = 5). This allows us to split the term 5x, and rewrite the equation as follows: 3x 2 + 6x - x - 2 = 0 We can now factorise the first two terms (3x 2 and 6x) by finding the highest common factor, which in this case is 3x: 3x (x + 2) The highest common factor for the last two terms (-x and -2) will be -1, which gives us: -1(x + 2) So we now have the expressions 3x × (x + 2) and -1 × (x + 2) on the left hand side of the equation. Since the term (x + 2) is common to both expressions, we can rewrite the equation as: (3x - 1)(x + 2) = 0 Just to be sure we have factorised correctly, we will multiply out the brackets again: 3x 2 + 6x - x - 2 = 0 Gather like terms together: 3x 2 + 5x - 2 = 0 So, since this is the same as our original equation, our factorisation is correct. This means that we have candidate values for x of one third or minus two (x = 1/3 or x = -2). Let's look at another example: 4x 2 - 6x - 4 = 0 We are looking for two numbers that add together to give the linear coefficient (minus six) and that multiply together to give the product of the quadratic coefficient (four) and the constant term (minus four), which will be minus sixteen (4 × -4 = -16). Looking at factors of minus sixteen, the obvious choices are minus eight and two, since minus eight plus two equals minus six (-8 + 2 = -6). We can now rewrite our quadratic equation to split the term 6x as follows: 4x 2 - 8x + 2x - 4 = 0 Now we factorise the first two terms (4x 2 and -8x) by finding the highest common factor, which in this case is 4x: 4x (x - 2) The highest common factor for the last two terms (2x and -4) will be two (2), which gives us: 2 (x - 2) So we now have the expressions 4x × (x - 2) and 2 × (x - 2) on the left hand side of the equation. Since the term (x - 2) is common to both expressions, we can rewrite the equation as: (4x + 2)(x - 2) = 0 Again, to make sure we have factorised correctly, we will multiply out the brackets again: 4x 2 - 8x + 2x - 4 = 0 Gathering like terms together we get: 4x 2 - 6x - 4 = 0 This is the same as our original equation, so our factorisation is correct. This time we have candidate values for x of one half or two (x = 1/2 or x = 2). The difference of two squares Sometimes you will encounter a quadratic equation in which the linear coefficient is zero (or the term bx is omitted altogether, which amounts to the same thing), and the constant term is itself a perfect square. The following equations fit this scenario, and are equivalent to one another: x 2 - 9 = 0 x 2 + 0x - 9 = 0 Because we are subtracting a number that is a square from another term that is also a square, we call this situation the difference of two squares or sometimes the difference of perfect squares. In the example above, we have nine (which is three squared) being subtracted from x squared (x 2). In order to factorise such an expression, we are looking for two numbers that sum to zero (because the linear coefficient is zero), and which multiply together to give the constant term. In the example above those numbers would be three and minus three. We can use these numbers to split the 0x term into two separate terms as follows: x 2 + 3x - 3x - 9 = 0 If we factorise the first two terms we get: x(x + 3) Factorising the last two terms gives: -3(x + 3) So we now have the expressions x × (x + 3) and -3 × (x + 3) on the left hand side of the equation. Since the term (x + 3) is common to both expressions, we can rewrite the equation as: (x - 3)(x + 3) = 0 To make sure we have factorised correctly, we will multiply out the brackets again: x 2 + 3x - 3x - 9 = 0 Gathering like terms together we get: x 2 - 9 = 0 (or x 2 + 0x - 9 = 0) This is the same as our original equation, so our factorisation is correct. We have candidate values for x of three or minus three (x = 3 or x = -3). In quadratic equations of this type: x 2 - a 2 = 0 The equation will always factorise to (x - a)(x + a). Note that you might also come across an expression such as the following: 2x 2 - 18 = 0 The two terms here have a common factor of two, so we can factorise as follows: 2(x 2 - 9) = 0 The term inside the brackets should look familiar because we have already seen it (see above). The equation therefore factors as: 2(x - 3)(x + 3) = 0 Since two is obviously not equal to zero, we still have a situation where either (x - 3) or (x + 3) must equal zero, so x is either three or minus three (x = 3 or x = -3). An alternative approach of course would be to simply divide both sides of the equation by two in the first place to get x 2 - 9 = 0. Another variation on this theme occurs when the quadratic coefficient is itself a perfect square, as in the following equation: 9x 2 - 16 = 0 Because nine is also a perfect square (9 = 3 2), the term 9x 2 can be written as (3x) 2, so we still have the difference of two squares on the left-hand side of the equation ((3x) 2 - 4 2). We can therefore factorise the equation as follows: 9x 2 - 16 = 0 (3x - 4)(3x + 4) Perfect squares Sometimes the factorisation of a quadratic equation results in a single term multiplied by itself, i.e. squared. Consider the following equation: x 2 - 8x + 16 = 0 We are looking for two numbers that multiply together to give sixteen, and add together to give minus eight. The numbers in this case both turn out to be minus four, because minus four multiplied by itself equals sixteen, and minus four plus minus four equals minus eight (-4 × -4 = 16 and -4 + (-4) = -8). We can therefore rewrite the equation, splitting the -8x term, as follows: x 2 - 4x - 4x + 16 = 0 Factorising the first two terms we get: x (x - 4) And factoring the last two terms we get: -4 (x - 4) We now have the terms x × (x - 4) and -4 × (x - 4) on the left hand side of the equation, and we can rewrite the equation as: (x - 4)(x - 4) = 0 (x - 4) 2 = 0 If we multiply out the brackets again as a double-check we get: (x - 4) 2 = 0 (x - 4)(x - 4) = 0 x 2 - 8x + 16 = 0 When the constant term is missing One more case to consider when factorising quadratic equations is the situation where there is no constant term. In a case like this, you should look for factors common to the quadratic and linear terms in the equation. An example should demonstrate the principle. Consider the following quadratic equation: 3x 2 - 6x = 0 Each of the terms 3x 2 and -6x has a common factor of 3x. The common factor is then written outside the brackets, and the terms that appear inside the brackets are chosen so that multiplying out the brackets again will produce the original equation. in this example, the terms within the bracket will be x and minus two (x and -2): 3x(x - 2) = 0 This means that either 3x is equal to zero, or x - 2 is equal to zero, so x is either zero or two.
Paul's Online Notes Home / Calculus II / Series & Sequences / Convergence/Divergence of Series 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. Assignment Problems Notice Please do not email me to get solutions and/or answers to these problems. I will not give them out under any circumstances nor will I respond to any requests to do so. The intent of these problems is for instructors to use them for assignments and having solutions/answers easily available defeats that purpose. If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above. ### Section 10.4 : Convergence/Divergence of Series For problems 1 – 4 compute the first 3 terms in the sequence of partial sums for the given series. 1. $$\displaystyle \sum\limits_{n = 0}^\infty {\frac{1}{{1 + {3^n}}}}$$ 2. $$\displaystyle \sum\limits_{n = 1}^\infty {\left( {{2^n} - {3^n}} \right)}$$ 3. $$\displaystyle \sum\limits_{n = 1}^\infty {\frac{{1 + n}}{{2n}}}$$ 4. $$\displaystyle \sum\limits_{n = 0}^\infty {10}$$ For problems 5 – 7 assume that the $$n$$th term in the sequence of partial sums for the series $$\displaystyle \sum\limits_{n = 0}^\infty {{a_n}}$$ is given below. Determine if the series $$\displaystyle \sum\limits_{n = 0}^\infty {{a_n}}$$ is convergent or divergent. If the series is convergent determine the value of the series. 1. $${s_n} = \left( {{n^2} + 4n} \right){{\bf{e}}^{ - 2n}}$$ 2. $$\displaystyle {s_n} = \frac{{1 + 2n + 3{n^2}}}{{4{n^2} + 5n + 6}}$$ 3. $$\displaystyle {s_n} = \frac{n}{{\ln \left( {n + 2} \right)}}$$ 4. Let $$\displaystyle {d_n} = \frac{{7 - 8n}}{{4 + 3n}}$$ 1. Does the sequence $$\left\{ {{d_n}} \right\}_{n = 0}^\infty$$ converge or diverge? 2. Does the series $$\displaystyle \sum\limits_{n = 0}^\infty {{d_n}}$$ converge or diverge? 5. Let $${d_n} = 1 + n{{\bf{e}}^{ - n}}$$ 1. Does the sequence $$\left\{ {{d_n}} \right\}_{n = 0}^\infty$$ converge or diverge? 2. Does the series $$\displaystyle \sum\limits_{n = 0}^\infty {{d_n}}$$ converge or diverge? For problems 10 – 12 show that the series is divergent. 1. $$\displaystyle \sum\limits_{n = 1}^\infty {\frac{{9 - 2{n^2}}}{{1 + 4n + {n^2}}}}$$ 2. $$\displaystyle \sum\limits_{n = 0}^\infty {\frac{{{5^n} + 1}}{{{3^n}}}}$$ 3. $$\displaystyle \sum\limits_{n = 1}^\infty {\cos \left( n \right)}$$
How to work out the arc length and perimeter of a sector (part of the circumference of a circle) Updated on March 28, 2013 Arc Length Of A Sector Video If you take a part of the circumference of a circle then the distance along this arc is called the arc length. So since the arc length is part of the circumference of a circle, then the arc length of the sector can be found by using the following formula: l = (x/360) × π × d Where l is the arc length, x is the angle inside the sector and d is the diameter of the sector. So make sure that you use the diameter of the circle if you are calculating the arc length of a sector. Example 1 Work out the arc length of this sector: The angle inside the sector is 41⁰, so x = 41⁰. Now the diameter of the whole circle is 24cm (as the 12cm is the radius), so d = 24cm. All you need to do now is substitute these values into the above formula so you can find the arc length. l = (x/360) × π × d l = (41/360) × π × 24 l = 8.6 cm rounded to 1 decimal place. Example 2 Work out the arc length of this sector: The angle inside the sector is 134⁰, so x = 134⁰. Now the diameter of the whole circle is 18cm (as the 9cm is the radius), so d = 18cm. All you need to do now is substitute these values into the above formula so you can find the arc length. l = (x/360) × π × d l = (134/360) × π × 18 l = 21.0 cm rounded to 1 decimal place. Example 3 A sheet of metal is shaped in the form of a sector with a radius of 6cm and an angle of 308⁰. Work out the perimeter of the sheet of metal. First you need to calculate the arc length of the sector. The diameter of the sector is 12cm (d = 12) and the angle of the sector is 308⁰ (x = 308⁰) l = (x/360) × π × d l = (308/360) × π × 12 l = 32.3 cm rounded to 1 decimal place. Now, the question asks for the perimeter of the sector (the distance around the whole of the shape) So you need to add on two radiuses onto the arc length: P = 32.3 + 6 + 6 = 44.3cm 114 9 4 31 9 8
Prealgebra 2e 4.5Add and Subtract Fractions with Different Denominators Prealgebra 2e4.5 Add and Subtract Fractions with Different Denominators Learning Objectives By the end of this section, you will be able to: • Find the least common denominator (LCD) • Convert fractions to equivalent fractions with the LCD • Add and subtract fractions with different denominators • Identify and use fraction operations • Use the order of operations to simplify complex fractions • Evaluate variable expressions with fractions Be Prepared 4.12 Before you get started, take this readiness quiz. Find two fractions equivalent to $56.56.$ If you missed this problem, review Example 4.14. Be Prepared 4.13 Simplify: $1+5·322+4.1+5·322+4.$ If you missed this problem, review Example 4.48. Find the Least Common Denominator In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators? Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals $2525$ cents and one dime equals $1010$ cents, so the sum is $3535$ cents. See Figure 4.7. Figure 4.7 Together, a quarter and a dime are worth $3535$ cents, or $3510035100$ of a dollar. Similarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is $100.100.$ Since there are $100100$ cents in one dollar, $2525$ cents is $2510025100$ and $1010$ cents is $10100.10100.$ So we add $25100+1010025100+10100$ to get $35100,35100,$ which is $3535$ cents. You have practiced adding and subtracting fractions with common denominators. Now let’s see what you need to do with fractions that have different denominators. First, we will use fraction tiles to model finding the common denominator of $1212$ and $13.13.$ We’ll start with one $1212$ tile and $1313$ tile. We want to find a common fraction tile that we can use to match both $1212$ and $1313$ exactly. If we try the $1414$ pieces, $22$ of them exactly match the $1212$ piece, but they do not exactly match the $1313$ piece. If we try the $1515$ pieces, they do not exactly cover the $1212$ piece or the $1313$ piece. If we try the $1616$ pieces, we see that exactly $33$ of them cover the $1212$ piece, and exactly $22$ of them cover the $1313$ piece. If we were to try the $112112$ pieces, they would also work. Even smaller tiles, such as $124124$ and $148,148,$ would also exactly cover the $1212$ piece and the $1313$ piece. The denominator of the largest piece that covers both fractions is the least common denominator (LCD) of the two fractions. So, the least common denominator of $1212$ and $1313$ is $6.6.$ Notice that all of the tiles that cover $1212$ and $1313$ have something in common: Their denominators are common multiples of $22$ and $3,3,$ the denominators of $1212$ and $13.13.$ The least common multiple (LCM) of the denominators is $6,6,$ and so we say that $66$ is the least common denominator (LCD) of the fractions $1212$ and $13.13.$ Manipulative Mathematics Doing the Manipulative Mathematics activity "Finding the Least Common Denominator" will help you develop a better understanding of the LCD. Least Common Denominator The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators. To find the LCD of two fractions, we will find the LCM of their denominators. We follow the procedure we used earlier to find the LCM of two numbers. We only use the denominators of the fractions, not the numerators, when finding the LCD. Example 4.63 Find the LCD for the fractions $712712$ and $518.518.$ Try It 4.125 Find the least common denominator for the fractions: $712712$ and $1115.1115.$ Try It 4.126 Find the least common denominator for the fractions: $13151315$ and $175.175.$ To find the LCD of two fractions, find the LCM of their denominators. Notice how the steps shown below are similar to the steps we took to find the LCM. How To Find the least common denominator (LCD) of two fractions. 1. Step 1. Factor each denominator into its primes. 2. Step 2. List the primes, matching primes in columns when possible. 3. Step 3. Bring down the columns. 4. Step 4. Multiply the factors. The product is the LCM of the denominators. 5. Step 5. The LCM of the denominators is the LCD of the fractions. Example 4.64 Find the least common denominator for the fractions $815815$ and $1124.1124.$ Try It 4.127 Find the least common denominator for the fractions: $13241324$ and $1732.1732.$ Try It 4.128 Find the least common denominator for the fractions: $928928$ and $2132.2132.$ Convert Fractions to Equivalent Fractions with the LCD Earlier, we used fraction tiles to see that the LCD of $1414$ when $1616$ is $12.12.$ We saw that three $112112$ pieces exactly covered $1414$ and two $112112$ pieces exactly covered $16,16,$ so $14=312 and 16=212.14=312 and 16=212.$ We say that $1414$ and $312312$ are equivalent fractions and also that $1616$ and $212212$ are equivalent fractions. We can use the Equivalent Fractions Property to algebraically change a fraction to an equivalent one. Remember, two fractions are equivalent if they have the same value. The Equivalent Fractions Property is repeated below for reference. Equivalent Fractions Property If $a,b,ca,b,c$ are whole numbers where $b≠0,c≠0,b≠0,c≠0,$ then $ab=a·cb·canda·cb·c=abab=a·cb·canda·cb·c=ab$ To add or subtract fractions with different denominators, we will first have to convert each fraction to an equivalent fraction with the LCD. Let’s see how to change $1414$ and $1616$ to equivalent fractions with denominator $1212$ without using models. Example 4.65 Convert $1414$ and $1616$ to equivalent fractions with denominator $12,12,$ their LCD. Try It 4.129 Change to equivalent fractions with the LCD: $3434$ and $56,56,$ LCD $=12=12$ Try It 4.130 Change to equivalent fractions with the LCD: $−712−712$ and $1115,1115,$ LCD $=60=60$ How To Convert two fractions to equivalent fractions with their LCD as the common denominator. 1. Step 1. Find the LCD. 2. Step 2. For each fraction, determine the number needed to multiply the denominator to get the LCD. 3. Step 3. Use the Equivalent Fractions Property to multiply both the numerator and denominator by the number you found in Step 2. 4. Step 4. Simplify the numerator and denominator. Example 4.66 Convert $815815$ and $11241124$ to equivalent fractions with denominator $120,120,$ their LCD. Try It 4.131 Change to equivalent fractions with the LCD: $13241324$ and $1732,1732,$ LCD $9696$ Try It 4.132 Change to equivalent fractions with the LCD: $928928$ and $2732,2732,$ LCD $224224$ Add and Subtract Fractions with Different Denominators Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators. How To Add or subtract fractions with different denominators. 1. Step 1. Find the LCD. 2. Step 2. Convert each fraction to an equivalent form with the LCD as the denominator. 3. Step 3. Add or subtract the fractions. 4. Step 4. Write the result in simplified form. Example 4.67 Add: $12+13.12+13.$ Try It 4.133 Add: $14+13.14+13.$ Try It 4.134 Add: $12+15.12+15.$ Example 4.68 Subtract: $12−(−14).12−(−14).$ Try It 4.135 Simplify: $12−(−18).12−(−18).$ Try It 4.136 Simplify: $13−(−16).13−(−16).$ Example 4.69 Add: $712+518.712+518.$ Try It 4.137 Add: $712+1115.712+1115.$ Try It 4.138 Add: $1315+1720.1315+1720.$ When we use the Equivalent Fractions Property, there is a quick way to find the number you need to multiply by to get the LCD. Write the factors of the denominators and the LCD just as you did to find the LCD. The “missing” factors of each denominator are the numbers you need. The LCD, $36,36,$ has $22$ factors of $22$ and $22$ factors of $3.3.$ Twelve has two factors of $2,2,$ but only one of $33$—so it is ‘missing‘ one $3.3.$ We multiplied the numerator and denominator of $712712$ by $33$ to get an equivalent fraction with denominator $36.36.$ Eighteen is missing one factor of $22$—so you multiply the numerator and denominator $518518$ by $22$ to get an equivalent fraction with denominator $36.36.$ We will apply this method as we subtract the fractions in the next example. Example 4.70 Subtract: $715−1924.715−1924.$ Try It 4.139 Subtract: $1324−1732.1324−1732.$ Try It 4.140 Subtract: $2132−928.2132−928.$ Example 4.71 Add: $−1130+2342.−1130+2342.$ Try It 4.141 Add: $−1342+1735.−1342+1735.$ Try It 4.142 Add: $−1924+1732.−1924+1732.$ In the next example, one of the fractions has a variable in its numerator. We follow the same steps as when both numerators are numbers. Example 4.72 Add: $35+x8.35+x8.$ Try It 4.143 Add: $y6+79.y6+79.$ Try It 4.144 Add: $x6+715.x6+715.$ Identify and Use Fraction Operations By now in this chapter, you have practiced multiplying, dividing, adding, and subtracting fractions. The following table summarizes these four fraction operations. Remember: You need a common denominator to add or subtract fractions, but not to multiply or divide fractions Summary of Fraction Operations Fraction multiplication: Multiply the numerators and multiply the denominators. $ab·cd=acbdab·cd=acbd$ Fraction division: Multiply the first fraction by the reciprocal of the second. $ab÷cd=ab·dcab÷cd=ab·dc$ Fraction addition: Add the numerators and place the sum over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD. $ac+bc=a+bcac+bc=a+bc$ Fraction subtraction: Subtract the numerators and place the difference over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD. $ac−bc=a−bcac−bc=a−bc$ Example 4.73 Simplify: 1. $−14+16−14+16$ 2. $−14÷16−14÷16$ Try It 4.145 Simplify each expression: 1. $−34−16−34−16$ 2. $−34·16−34·16$ Try It 4.146 Simplify each expression: 1. $56÷(−14)56÷(−14)$ 2. $56−(−14)56−(−14)$ Example 4.74 Simplify: 1. $5x6−3105x6−310$ 2. $5x6·3105x6·310$ Try It 4.147 Simplify: 1. $(27a−32)36(27a−32)36$ 2. $2a32a3$ Try It 4.148 Simplify: 1. $(24k+25)30(24k+25)30$ 2. $24k524k5$ Use the Order of Operations to Simplify Complex Fractions In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. We simplified complex fractions by rewriting them as division problems. For example, $3458=34÷583458=34÷58$ Now we will look at complex fractions in which the numerator or denominator can be simplified. To follow the order of operations, we simplify the numerator and denominator separately first. Then we divide the numerator by the denominator. How To Simplify complex fractions. 1. Step 1. Simplify the numerator. 2. Step 2. Simplify the denominator. 3. Step 3. Divide the numerator by the denominator. 4. Step 4. Simplify if possible. Example 4.75 Simplify: $(12)24+32.(12)24+32.$ Try It 4.149 Simplify: $(13)223+2(13)223+2$. Try It 4.150 Simplify: $1+42(14)21+42(14)2$. Example 4.76 Simplify: $12+2334−16.12+2334−16.$ Try It 4.151 Simplify: $13+1234−1313+1234−13$. Try It 4.152 Simplify: $23−1214+1323−1214+13$. Evaluate Variable Expressions with Fractions We have evaluated expressions before, but now we can also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify. Example 4.77 Evaluate $x+13x+13$ when 1. $x=−13x=−13$ 2. $x=−34.x=−34.$ Try It 4.153 Evaluate: $x+34x+34$ when 1. $x=−74x=−74$ 2. $x=−54x=−54$ Try It 4.154 Evaluate: $y+12y+12$ when 1. $y=23y=23$ 2. $y=−34y=−34$ Example 4.78 Evaluate $y−56y−56$ when $y=−23.y=−23.$ Try It 4.155 Evaluate: $y−12y−12$ when $y=−14.y=−14.$ Try It 4.156 Evaluate: $x−38x−38$ when $x=−52.x=−52.$ Example 4.79 Evaluate $2x2y2x2y$ when $x=14x=14$ and $y=−23.y=−23.$ Try It 4.157 Evaluate. $3ab23ab2$ when $a=−23a=−23$ and $b=−12.b=−12.$ Try It 4.158 Evaluate. $4c3d4c3d$ when $c=−12c=−12$ and $d=−43.d=−43.$ Example 4.80 Evaluate $p+qrp+qr$ when $p=−4,q=−2,p=−4,q=−2,$ and $r=8.r=8.$ Try It 4.159 Evaluate: $a+bca+bc$ when $a=−8,b=−7,a=−8,b=−7,$ and $c=6.c=6.$ Try It 4.160 Evaluate: $x+yzx+yz$ when $x=9,y=−18,x=9,y=−18,$ and $z=−6.z=−6.$ Section 4.5 Exercises Practice Makes Perfect Find the Least Common Denominator (LCD) In the following exercises, find the least common denominator (LCD) for each set of fractions. 316. $2323$ and $3434$ 317. $3434$ and $2525$ 318. $712712$ and $5858$ 319. $916916$ and $712712$ 320. $13301330$ and $25422542$ 321. $23302330$ and $548548$ 322. $21352135$ and $39563956$ 323. $18351835$ and $33493349$ 324. $23,16,23,16,$ and $3434$ 325. $23,14,23,14,$ and $3535$ Convert Fractions to Equivalent Fractions with the LCD In the following exercises, convert to equivalent fractions using the LCD. 326. $1313$ and $14,14,$ LCD $=12=12$ 327. $1414$ and $15,15,$ LCD $=20=20$ 328. $512512$ and $78,78,$ LCD $=24=24$ 329. $712712$ and $58,58,$ LCD $=24=24$ 330. $13161316$ and $-1112,-1112,$ LCD $=48=48$ 331. $11161116$ and $-512,-512,$ LCD $=48=48$ 332. $13,56,13,56,$ and $34,34,$ LCD $=12=12$ 333. $13,34,13,34,$ and $35,35,$ LCD $=60=60$ Add and Subtract Fractions with Different Denominators In the following exercises, add or subtract. Write the result in simplified form. 334. $1 3 + 1 5 1 3 + 1 5$ 335. $1 4 + 1 5 1 4 + 1 5$ 336. $1 2 + 1 7 1 2 + 1 7$ 337. $1 3 + 1 8 1 3 + 1 8$ 338. $1 3 − ( − 1 9 ) 1 3 − ( − 1 9 )$ 339. $1 4 − ( − 1 8 ) 1 4 − ( − 1 8 )$ 340. $1 5 − ( − 1 10 ) 1 5 − ( − 1 10 )$ 341. $1 2 − ( − 1 6 ) 1 2 − ( − 1 6 )$ 342. $2 3 + 3 4 2 3 + 3 4$ 343. $3 4 + 2 5 3 4 + 2 5$ 344. $7 12 + 5 8 7 12 + 5 8$ 345. $5 12 + 3 8 5 12 + 3 8$ 346. $7 12 − 9 16 7 12 − 9 16$ 347. $7 16 − 5 12 7 16 − 5 12$ 348. $11 12 − 3 8 11 12 − 3 8$ 349. $5 8 − 7 12 5 8 − 7 12$ 350. $2 3 − 3 8 2 3 − 3 8$ 351. $5 6 − 3 4 5 6 − 3 4$ 352. $− 11 30 + 27 40 − 11 30 + 27 40$ 353. $− 9 20 + 17 30 − 9 20 + 17 30$ 354. $− 13 30 + 25 42 − 13 30 + 25 42$ 355. $− 23 30 + 5 48 − 23 30 + 5 48$ 356. $− 39 56 − 22 35 − 39 56 − 22 35$ 357. $− 33 49 − 18 35 − 33 49 − 18 35$ 358. $− 2 3 − ( − 3 4 ) − 2 3 − ( − 3 4 )$ 359. $− 3 4 − ( − 4 5 ) − 3 4 − ( − 4 5 )$ 360. $− 9 16 − ( − 4 5 ) − 9 16 − ( − 4 5 )$ 361. $− 7 20 − ( − 5 8 ) − 7 20 − ( − 5 8 )$ 362. $1 + 7 8 1 + 7 8$ 363. $1 + 5 6 1 + 5 6$ 364. $1 − 5 9 1 − 5 9$ 365. $1 − 3 10 1 − 3 10$ 366. $x 3 + 1 4 x 3 + 1 4$ 367. $y 2 + 2 3 y 2 + 2 3$ 368. $y 4 − 3 5 y 4 − 3 5$ 369. $x 5 − 1 4 x 5 − 1 4$ Identify and Use Fraction Operations In the following exercises, perform the indicated operations. Write your answers in simplified form. 370. 1. $34+1634+16$ 2. $34÷1634÷16$ 371. 1. $23+1623+16$ 2. $23÷1623÷16$ 372. 1. $-25−18-25−18$ 2. $-25·18-25·18$ 373. 1. $-45−18-45−18$ 2. $-45·18-45·18$ 374. 1. $5n6÷8155n6÷815$ 2. $5n6−8155n6−815$ 375. 1. $3a8÷7123a8÷712$ 2. $3a8−7123a8−712$ 376. 1. $910·(−11d12)910·(−11d12)$ 2. $910+(−11d12)910+(−11d12)$ 377. 1. $415·(−5q9)415·(−5q9)$ 2. $415+(−5q9)415+(−5q9)$ 378. $− 3 8 ÷ ( − 3 10 ) − 3 8 ÷ ( − 3 10 )$ 379. $− 5 12 ÷ ( − 5 9 ) − 5 12 ÷ ( − 5 9 )$ 380. $− 3 8 + 5 12 − 3 8 + 5 12$ 381. $− 1 8 + 7 12 − 1 8 + 7 12$ 382. $5 6 − 1 9 5 6 − 1 9$ 383. $5 9 − 1 6 5 9 − 1 6$ 384. $3 8 · ( − 10 21 ) 3 8 · ( − 10 21 )$ 385. $7 12 · ( − 8 35 ) 7 12 · ( − 8 35 )$ 386. $− 7 15 − y 4 − 7 15 − y 4$ 387. $− 3 8 − x 11 − 3 8 − x 11$ 388. $11 12 a · 9 a 16 11 12 a · 9 a 16$ 389. $10 y 13 · 8 15 y 10 y 13 · 8 15 y$ Use the Order of Operations to Simplify Complex Fractions In the following exercises, simplify. 390. $( 1 5 ) 2 2 + 3 2 ( 1 5 ) 2 2 + 3 2$ 391. $( 1 3 ) 2 5 + 2 2 ( 1 3 ) 2 5 + 2 2$ 392. $2 3 + 4 2 ( 2 3 ) 2 2 3 + 4 2 ( 2 3 ) 2$ 393. $3 3 − 3 2 ( 3 4 ) 2 3 3 − 3 2 ( 3 4 ) 2$ 394. $( 3 5 ) 2 ( 3 7 ) 2 ( 3 5 ) 2 ( 3 7 ) 2$ 395. $( 3 4 ) 2 ( 5 8 ) 2 ( 3 4 ) 2 ( 5 8 ) 2$ 396. $2 1 3 + 1 5 2 1 3 + 1 5$ 397. $5 1 4 + 1 3 5 1 4 + 1 3$ 398. $2 3 + 1 2 3 4 − 2 3 2 3 + 1 2 3 4 − 2 3$ 399. $3 4 + 1 2 5 6 − 2 3 3 4 + 1 2 5 6 − 2 3$ 400. $7 8 − 2 3 1 2 + 3 8 7 8 − 2 3 1 2 + 3 8$ 401. $3 4 − 3 5 1 4 + 2 5 3 4 − 3 5 1 4 + 2 5$ Mixed Practice In the following exercises, simplify. 402. $1 2 + 2 3 · 5 12 1 2 + 2 3 · 5 12$ 403. $1 3 + 2 5 · 3 4 1 3 + 2 5 · 3 4$ 404. $1 − 3 5 ÷ 1 10 1 − 3 5 ÷ 1 10$ 405. $1 − 5 6 ÷ 1 12 1 − 5 6 ÷ 1 12$ 406. $2 3 + 1 6 + 3 4 2 3 + 1 6 + 3 4$ 407. $2 3 + 1 4 + 3 5 2 3 + 1 4 + 3 5$ 408. $3 8 − 1 6 + 3 4 3 8 − 1 6 + 3 4$ 409. $2 5 + 5 8 − 3 4 2 5 + 5 8 − 3 4$ 410. $12 ( 9 20 − 4 15 ) 12 ( 9 20 − 4 15 )$ 411. $8 ( 15 16 − 5 6 ) 8 ( 15 16 − 5 6 )$ 412. $5 8 + 1 6 19 24 5 8 + 1 6 19 24$ 413. $1 6 + 3 10 14 30 1 6 + 3 10 14 30$ 414. $( 5 9 + 1 6 ) ÷ ( 2 3 − 1 2 ) ( 5 9 + 1 6 ) ÷ ( 2 3 − 1 2 )$ 415. $( 3 4 + 1 6 ) ÷ ( 5 8 − 1 3 ) ( 3 4 + 1 6 ) ÷ ( 5 8 − 1 3 )$ In the following exercises, evaluate the given expression. Express your answers in simplified form, using improper fractions if necessary. 416. $x+12x+12$ when 1. $x=−18x=−18$ 2. $x=−12x=−12$ 417. $x+23x+23$ when 1. $x=−16x=−16$ 2. $x=−53x=−53$ 418. $x+(−56)x+(−56)$ when 1. $x=13x=13$ 2. $x=−16x=−16$ 419. $x+(−1112)x+(−1112)$ when 1. $x=1112x=1112$ 2. $x=34x=34$ 420. $x−25x−25$ when 1. $x=35x=35$ 2. $x=−35x=−35$ 421. $x−13x−13$ when 1. $x=23x=23$ 2. $x=−23x=−23$ 422. $710−w710−w$ when 1. $w=12w=12$ 2. $w=−12w=−12$ 423. $512−w512−w$ when 1. $w=14w=14$ 2. $w=−14w=−14$ 424. $4p2q4p2q$ when $p=−12p=−12$ and $q=59q=59$ 425. $5m2n5m2n$ when $m=−25m=−25$ and $n=13n=13$ 426. $2x2y32x2y3$ when $x=−23 x=−23$ and $y=−12 y=−12$ 427. $8u2v38u2v3$ when $u=−34u=−34$ and $v=−12v=−12$ 428. $u+vwu+vw$ when $u=−4,v=−8,w=2u=−4,v=−8,w=2$ 429. $m+npm+np$ when $m=−6,n=−2,p=4m=−6,n=−2,p=4$ 430. $a+ba−ba+ba−b$ when $a=−3,b=8a=−3,b=8$ 431. $r−sr+sr−sr+s$ when $r=10,s=−5r=10,s=−5$ Everyday Math 432. Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs $316316$ yard of print fabric and $3838$ yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover? 433. Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs $114114$ cups of sugar for the chocolate chip cookies, and $118118$ cups for the oatmeal cookies How much sugar does she need altogether? Writing Exercises 434. Explain why it is necessary to have a common denominator to add or subtract fractions. 435. Explain how to find the LCD of two fractions. Self Check After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. After looking at the checklist, do you think you are well prepared for the next section? Why or why not? Order a print copy As an Amazon Associate we earn from qualifying purchases.
### Is the relation a function? ```Formalizing Relations and Functions Objectives: To determine whether a relation is a function To find domain and range and use function notation Relation: a pairing of numbers in one set with numbers in another set Domain: the set of x-values Range: the set of y-values Problem #1: Identifying Functions Using Mapping Diagram Identify the domain and range of the relation. Represent the relation with a mapping diagram. Is the relation a function? A) {(-2, 0.5), (0, 2.5), (4, 6.5), (5, 2.5)} Problem #1: Identifying Functions Using Mapping Diagram Identify the domain and range of the relation. Represent the relation with a mapping diagram. Is the relation a function? B) {(6,5), (4,3), (6,4), (5,8)} Problem #1 Got It? Identify the domain and range of each relation. Represent the relation with a mapping diagram. Is the relation a function? Vertical Line Test Another method used to determine if a relation is a function. Look at the graph and determine if any vertical line passes through more than one point of the graph, then the relation is not a function. Problem #2: Identifying Functions Using the Vertical Line Test Is the relation a function? Use the Vertical Line Test. A) {(-4, 2), (-3, 1), (0,-2), (-4, -1), (1, 2)} Problem #2: Identifying Functions Using the Vertical Line Test Is the relation a function? Use the Vertical Line Test. B) = − 2 + 3 Problem #2 Got It? Is the relation a function? Use the Vertical Line Test. Function Notation y = -3x + 1 is the same as = −3 + 1 - f(x) replaces y f is the name of the function, not a variable Used to emphasize that the function value f(x) depends on the independent variable x. - Other letters besides f can also be used, such as g and h. Problem #3: Evaluating a Function A) The function w(x) = 250x represents the number of words w(x) you can read in x minutes. How many words can you read in 8 minutes? Problem #3: Evaluating a Function B) The function Y(x) = 1 3 represents the number of yards Y(x) in x feet. How many yards are there in 1 mile? Problem #3 Got It? The function T(x) = 65x represents the number of words T(x) that Rachel can type in x minutes. How many words can she type in 7 minutes? Homework Textbook Page 271; #1 – 3, 5 – 17 All Continued… Objectives: To determine whether a relation is a function To find domain and range and use function notation Problem #4: Finding the Range of a Function A) Multiple Choice The domain of f(x) = 1.5x + 4 is {1, 2, 3, 4}. What is the range? Problem #4: Finding the Range of a Function B) The domain of g(x) = 4x – 12 is {1, 3, 5, 7}. What is the range? Problem #4 Got It? What is the range of f(x) = 3x – 2 with domain {1, 2, 3, 4}? Problem #5: Identifying a Reasonable Domain and Range A) You have 3 qt. of paint to paint the trim in your house. A quart of paint covers 100 2 . The function () = 100 represents the area A(q), in square feet, that q quarts of paint cover. What domain and range are reasonable for the function? What is the graph of the function? Problem #5: Identifying a Reasonable Domain and Range B) Lorena has 4 rolls of ribbon to make party favors. Each roll can be used to make 30 favors. The function F(r) = 30r represents the number of favors F(r) that can be made with r rolls. What are a reasonable domain and range of the function? What is a graph of the function? Problem #5 Got It? A car can travel 32 miles for each gallon of gasoline. The function d(x) = 32x represents the distance d(x), in miles, that the car can travel with x gallons of gasoline. The car’s fuel tank holds 17 gal. Find a reasonable domain and range for each function. Then graph the function. Homework 4 – 6 Worksheet ```
What Do The Stars Say About Lamar Odom? (11/03/2019) How will Lamar Odom fare on 11/03/2019 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is not scientifically verified – take it with a grain of salt. I will first work out the destiny number for Lamar Odom, and then something similar to the life path number, which we will calculate for today (11/03/2019). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology people. PATH NUMBER FOR 11/03/2019: We will consider the month (11), the day (03) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. We’ll show you how it works now. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 03 we do 0 + 3 = 3. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 3 + 12 = 17. This still isn’t a single-digit number, so we will add its digits together again: 1 + 7 = 8. Now we have a single-digit number: 8 is the path number for 11/03/2019. DESTINY NUMBER FOR Lamar Odom: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for Lamar Odom we have the letters L (3), a (1), m (4), a (1), r (9), O (6), d (4), o (6) and m (4). Adding all of that up (yes, this can get tiring) gives 38. This still isn’t a single-digit number, so we will add its digits together again: 3 + 8 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for Lamar Odom. CONCLUSION: The difference between the path number for today (8) and destiny number for Lamar Odom (2) is 6. That is higher than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too worked up about that! As mentioned earlier, this is not at all guaranteed. If you want really means something, 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.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> 11.2: The One-Way ANOVA Test Difficulty Level: At Grade Created by: CK-12 Learning Objectives • Understand the shortcomings of comparing multiple means as pairs of hypotheses. • Understand the steps of the ANOVA method and its advantages. • Compare the means of three or more populations using the ANOVA method. • Calculate the pooled standard deviation and confidence intervals as estimates of standard deviations of the populations. Introduction Previously, we have discussed analysis that allows us to test if the means and variances of two populations are equal. But let’s say that a teacher is testing multiple reading programs to determine the impact on student achievement. There are five different reading programs and her \begin{align}31\end{align} students are randomly assigned to one of the five programs. The mean achievement scores and variances for the groups are recorded along with the means and the variances for all the subjects combined. We could conduct a series of \begin{align}t\end{align}-tests to test that all of the sample means came from the same population. However, this would be tedious and has a major flaw which we will discuss later. Instead, we use something called the Analysis of Variance (ANOVA) that allows us to test the hypothesis that multiple \begin{align}(K)\end{align} population means and variance of scores are equal. Theoretically, we could test hundreds of population means using this procedure. Shortcomings of Comparing Multiple Means Using Previously Explained Methods As mentioned, to test whether pairs of sample means differ by more than we would expect due to chance, we could conduct a series of separate \begin{align}t\end{align}-tests in order to compare all possible pairs of means. This would be tedious, but we could use the computer or TI-83/4 calculator to compute these easily and quickly. However, there is a major flaw with this reasoning. When more than one \begin{align}t\end{align}-test is run, each at its own level of significance ( \begin{align}\alpha =.10, .05, .01\end{align}, etc.) the probability of making one or more Type I errors multiplies exponentially. Recall that a Type I error occurs when we reject the null hypothesis when we should not. The level of significance, \begin{align}\alpha\end{align}, is the probability of a Type I error in a single test. When testing more than one pair of samples, the probability of making at least one Type I error is \begin{align}1-(1-\alpha)^c\end{align} where \begin{align}\alpha\end{align} is the level of significance for each \begin{align}t\end{align}-test and \begin{align}c\end{align} is the number of independent \begin{align}t\end{align}-tests. Using the example from the introduction, if our teacher tested conducted separate \begin{align}t\end{align}-tests to examine the means of the populations, she would have to conduct \begin{align}10\end{align} separate \begin{align}t\end{align}-tests. If she performed these tests with \begin{align}\alpha = .05\end{align}, the probability of committing a Type I error is not \begin{align}.05\end{align} as one would initially expect. Instead, it would be \begin{align}0.40\end{align} – extremely high! The Steps of the ANOVA Method In ANOVA, we are actually analyzing the total variation of the scores including (1) the variation of the scores within the groups and (2) the variation between the group means. Since we are interested in two different types of variation, we first calculate each type of variation independently and then calculate the ratio between the two. We use the \begin{align}F\end{align}-distribution as our sampling distribution and set our critical values and test our hypothesis accordingly. When using the ANOVA method, we are testing the null hypothesis that the means and the variances of our samples are equal. When we conduct a hypothesis test, we are testing the probability of obtaining an extreme \begin{align}F\end{align}-statistic by chance. If we reject the null hypothesis that the means and variances of the samples are equal, then we are saying that there is a small likelihood \begin{align}\alpha\end{align} that we would have obtained such an extreme \begin{align}F\end{align}-statistic by chance. To test a hypothesis using the ANOVA method, there are several steps that we need to take. These include: 1. Calculating the mean squares between groups \begin{align}(MS_B)\end{align}. The \begin{align}MS_B\end{align} is the difference between the means of the various samples. If we hypothesize that the group means are equal \begin{align}(\mu _1 = \mu_2 = \ldots = \mu_k)\end{align}, then they must also equal the population mean. Under our null hypothesis, we state that the means of the different samples are all equal and come from the same population, but we understand that there may be fluctuations due to sampling error. When we calculate the \begin{align}MS_B\end{align} , we must first determine the \begin{align}SS_B\end{align} , which is the sum of the differences between the individual scores and the means in each group. To calculate this difference, we use the formula: \begin{align}{SS_B} = \sum_{k=1}^k n_k(\bar X_k - \bar X)^2\end{align} where: \begin{align}k =\end{align} the group number \begin{align}n_k =\end{align} the sample size in group \begin{align}k\end{align} \begin{align}{\bar X_k} =\end{align} the mean of group \begin{align}k\end{align} \begin{align}{\bar X} =\end{align} mean of all individual observations \begin{align}k =\end{align} the number of groups When simplified, the formula becomes: \begin{align}{SS_B} = \sum_{k=1}^k \frac {T_k^2}{n_k} - \frac {T^2}{N}\end{align} where \begin{align}T_k =\end{align} sum of the observations in group \begin{align}K\end{align} \begin{align}T =\end{align} sum of all observations. Once we calculate this value, we divide by the number of degrees of freedom \begin{align}(K-1)\end{align} to arrive at the \begin{align}MS_B\end{align}. \begin{align}{MS_B} = \frac {SS_B}{K-1}\end{align} 2. Calculating the mean squares within groups \begin{align}(MS_W)\end{align}. The mean squares within groups calculation is also called the pooled estimate of the population variance. Remember that when we square the standard deviation of a sample, we are estimating population variance. Therefore, to calculate this figure, we sum of the squared deviations within each group and then divide by the sum of the degrees of freedom for each group. To calculate the \begin{align}MS_W\end{align} we first find the \begin{align}SS_W\end{align}, which is calculated using the formula: \begin{align}\frac {\sum (X_{i1} - {\bar X_1})^2 + \textstyle \sum (X_{i2} - {\bar X_2})^2 +\ldots + \textstyle \sum (X_{ik} - {\bar X_k})^2} {(n_1-1) + (n_2-1)+\ldots+(n_k-1)}\end{align} Simplified, this formula states: \begin{align}{SS_W} = \sum_{k=1}^k \sum_{i=1}^{n_k} X^2_{ik} - \sum_{k=1}^k \frac {T_k^2}{n_k}\end{align} where \begin{align}T_k =\end{align} sum of the observations in group \begin{align}k\end{align} Essentially, this formula sums the squares of each observation and then subtracts the total of the observations squared divided by the number of observations. Finally, we divide this value by the total number of degrees of freedom in the scenario \begin{align}(N-K)\end{align}. \begin{align}{MS_w} = \frac {SS_w}{N-K}\end{align} 3. Calculate the test statistic. The test statistic is as follows: \begin{align}F = \frac {MS_B}{MS_W}\end{align} 4. Find the critical value on the \begin{align}F\end{align}- distribution. As mentioned above, \begin{align}K-1\;\mathrm{degrees}\end{align} of freedom are associated with \begin{align}MS_B\end{align} and \begin{align}N-K \;\mathrm{degrees}\end{align} of freedom are associated with \begin{align}MS_W\end{align}. The degrees of freedom for \begin{align}MS_B\end{align} are read across the columns and the degrees of freedom for \begin{align}MS_W\end{align} are read across the rows. 5. Interpret the results of the hypothesis test. In ANOVA, the last step is to decide whether to reject the null hypothesis and then provide clarification about what that decision means. The primary advantage to using the ANOVA method is that it takes all types of variation into account so that we have an accurate analysis. In addition, we can use technological tools including computer programs (SAS, SPSS, Microsoft Excel) and the TI-83/4 calculator to easily conduct the calculations and test our hypothesis. We use these technological tools quite often when using the ANOVA method. Let’s take a look at an example to help clarify. Example: Let’s go back to the example in the introduction with the teacher that is testing multiple reading programs to determine the impact on student achievement. There are five different reading programs and her \begin{align}31\end{align} students are randomly assigned to the five programs and she collects the following data: Method \begin{align}& 1 && 2 && 3 && 4 && 5 \\ & 1 && 8 && 7 && 9 && 10 \\ & 4 && 6 && 6 && 10 && 12 \\ & 3 && 7 && 4 && 8 && 9 \\ & 2 && 4 && 9 && 6 && 11 \\ & 5 && 3 && 8 && 5 &&8 \\ & 1 && 5 && 5 &&&&\\ & 6 && && 7 &&&&\\ & &&&& 5 &&&&\end{align} Please (1) compare the means of these different groups by calculating the mean squares between groups and (2) use the standard deviations from our samples to calculate the mean squares within groups and estimate the pooled variance of a population. Solution: To solve for \begin{align}SS_B\end{align} , it is necessary to calculate several summary statistics from the data above. \begin{align}& \text{Number} (n_k) && 7 && 6 && 8 && 5 && 5 && 31\\ & \text{Total} (T_k) && 22 && 33 && 51 && 38 && 50 &&= 194\\ & \text{Mean} (\bar X) && 3.14 && 5.50 && 6.38 && 7.60 && 10.00 && = 6.26\\ & \text{Sum of Squared Obs.} \left (\sum_{i=1}^{n_k} X^2_{ik}\right ) && 92 && 199 && 345 && 306 && 510 && = 1,452\\ & \frac{\text{Sum of Obs. Squared}}{\text{Number of Obs}} \left (\frac {T_k^2}{n_k}\right ) && 69.14 && 181.50 && 325.13 && 288.80 && 500.00 && = 1,364.57\end{align} Using this information, we find that the sum of squares between groups is equal to \begin{align}& {SS_B} = \sum_{k=1}^k \frac {T_k^2}{n_k} - {\frac {T^2}{N}}\\ & \approx 1,364.57 - \frac{(194)^2}{31} \approx {150.5}\end{align} Since there are four Degrees of Freedom for this calculation (the number of groups minus one), the mean squares between groups is \begin{align}MS_B=\frac{SS_B}{K-1}\approx \frac {150.5}{4} \approx 37.6\end{align} Next we calculate the mean squares within groups \begin{align}(MS_W)\end{align} which is also known as the estimation of the pooled variance of a population \begin{align}(\sigma^2)\end{align}. To calculate the mean squares within groups, we use the formula \begin{align}{SS_W} = \sum_{k=1}^k \sum_{i=1}^{n_k}X^2_{ik} - \sum_{k=1}^k \frac {T_k^2}{n_k} \end{align} Using our summary statistics from above, we can calculate that the within groups mean square \begin{align}(MS_W)\end{align} is equal to: \begin{align}{SS_W} & = \sum_{k=1}^k \sum_{i=1}^{n_k}X^2_{ik} - \sum_{k=1}^k \frac {T_k^2}{n_k}\\ & \approx 1,452 - 1,364.57\\ & \approx 87.43\end{align} And so we have \begin{align}{MS_W} = \frac {SS_W}{N-K} \approx \frac {87.43}{26} \approx 3.36\end{align} Therefore, our \begin{align}F\end{align}-Ratio is \begin{align}F = \frac {MS_B}{MS_W}\approx \frac{37.6}{3.36}\approx 11.18\end{align} We would then analyze this test statistic against our critical value (using the \begin{align}F\end{align}-distribution table and a value of \begin{align}(\alpha =.02)\end{align}, we find our critical value equal to \begin{align}4.14\end{align}. Since our test statistic \begin{align}(11.18)\end{align} exceeds our critical value \begin{align}(4.14)\end{align}, we reject the null hypothesis. Therefore, we can conclude that not all of the population means of the five programs are equal and that obtaining an \begin{align}F\end{align}-ratio that extreme by chance is highly improbable. Technology Note - Excel Here is the procedure for performing a One-way ANOVA in Excel using this set of data. 1. Copy and paste the table into an empty Excel worksheet 2. Select Data Analysis from the Tools menu and choose “ANOVA: Single-factor” from the list that appears 3. Place the cursor is in the “Input Range” field and select the entire table. 4. Place the cursor in the “Output Range” and click somewhere in a blank cell below the table. 5. Click “Labels” only if you have also included the labels in the table. This will cause the names of the predictor variables to be displayed in the table 6. Click OK and the results shown below will be displayed. Note: The TI-83/4 also offers a One-way ANOVA test. Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 \begin{align}7\end{align} \begin{align}22\end{align} \begin{align}3.142857\end{align} \begin{align}3.809524\end{align} Column 2 \begin{align}6\end{align} \begin{align}33\end{align} \begin{align}5.5\end{align} \begin{align}3.5\end{align} Column 3 \begin{align}8\end{align} \begin{align}51\end{align} \begin{align}6.375\end{align} \begin{align}2.839286\end{align} Column 4 \begin{align}5\end{align} \begin{align}38\end{align} \begin{align}7.6\end{align} \begin{align}4.3\end{align} Column 5 \begin{align}5\end{align} \begin{align}50\end{align} \begin{align}10\end{align} \begin{align}2.5\end{align} ANOVA Source of Variation \begin{align}SS\end{align} \begin{align}df\end{align} \begin{align}MS\end{align} \begin{align}F\end{align} \begin{align}P-\end{align}value \begin{align}F\end{align} crit Between Groups \begin{align}150.5033\end{align} \begin{align}4\end{align} \begin{align}37.62584\end{align} \begin{align}11.18893\end{align} \begin{align}2.05E-05\end{align} \begin{align}2.742594\end{align} Within Groups \begin{align}87.43214\end{align} \begin{align}26\end{align} \begin{align}3.362775\end{align} Total \begin{align}237.9355\end{align} \begin{align}30\end{align} Lesson Summary 1. When testing multiple independent samples to determine if they come from the same populations, we could conduct a series of separate \begin{align}t\end{align}-tests in order to compare all possible pairs of means. However, a more precise and accurate analysis is the Analysis of Variance (ANOVA). 2. In ANOVA, we analyze the total variation of the scores including (1) the variation of the scores within the groups and (2) the variation between the group means and the total mean of all the groups (also known as the grand mean). 3. In this analysis, we calculate the \begin{align}F\end{align}-ratio, which is the total mean of squares between groups divided by the total mean of squares within groups. 4. The total mean of squares within groups is also known as the estimate of the pooled variance of the population. We find this value by analysis of the standard deviations in each of the samples. Review Questions 1. What does the ANOVA acronym stand for? 2. If we are tested whether pairs of sample means differ by more than we would expect due to chance using multiple \begin{align}t\end{align}-tests, the probability of making a Type I error would ___. 3. In the ANOVA method, we use the ___ distribution. 1. Student’s \begin{align}t\end{align}- 2. normal 3. \begin{align}F\end{align}- 4. In the ANOVA method, we complete a series of steps to evaluate our hypothesis. Put the following steps in chronological order. 1. Calculate the mean squares between groups and the means squares within groups 2. Determine the critical values in the \begin{align}F\end{align}-distribution 3. Evaluate the hypothesis 4. Calculate the test statistic 5. State the null hypothesis A school psychologist is interested whether or not teachers affect the anxiety scores among students taking the AP Statistics exam. The data below are the scores on a standardized anxiety test for students with three different teachers. Teacher's Name Ms. Jones Mr. Smith Mrs. White \begin{align}8\end{align} \begin{align}23\end{align} \begin{align}21\end{align} \begin{align}6\end{align} \begin{align}11\end{align} \begin{align}21\end{align} \begin{align}4\end{align} \begin{align}17\end{align} \begin{align}22\end{align} \begin{align}12\end{align} \begin{align}16\end{align} \begin{align}18\end{align} \begin{align}16\end{align} \begin{align}6\end{align} \begin{align}14\end{align} \begin{align}17\end{align} \begin{align}14\end{align} \begin{align}21\end{align} \begin{align}12\end{align} \begin{align}15\end{align} \begin{align}9\end{align} \begin{align}10\end{align} \begin{align}19\end{align} \begin{align}11\end{align} \begin{align}11\end{align} \begin{align}10\end{align} \begin{align}13\end{align} 1. State the null hypothesis. 2. Using the data above, please fill out the missing values in the table below. Ms. Jones Mr. Smith Mrs. White Totals Number \begin{align}(n_k)\end{align} \begin{align}8\end{align} \begin{align}=\end{align} Total \begin{align}(T_k)\end{align} \begin{align}131\end{align} \begin{align}=\end{align} Mean \begin{align}(\bar X)\end{align} \begin{align}14.6\end{align} \begin{align}=\end{align} Sum of Squared Obs. \begin{align}\textstyle (\sum_{i=1}^{n_k} X^2_{ik})\end{align} \begin{align}=\end{align} Sum of Obs. Squared/Number of Obs. \begin{align}\left (\frac {T_k^2}{n_k}\right )\end{align} \begin{align}= \end{align} 1. What is the mean squares between groups \begin{align}(MS_B)\end{align} value? 2. What is the mean squares within groups \begin{align}(MS_W)\end{align} value? 3. What is the \begin{align}F\end{align}-ratio of these two values? 4. Using a \begin{align}\alpha = .05\end{align}, please use the \begin{align}F\end{align}-distribution to set a critical value 5. What decision would you make regarding the null hypothesis? Why? 1. Analysis of Variance 2. Increase or increase exponentially 3. \begin{align}C\end{align} 4. \begin{align}E, A, D, B, C\end{align} 5. \begin{align}H_0: \mu_1 = \mu_2 = \mu_3\end{align} Ms. Jones Mr. Smith Mrs. White Totals Number \begin{align}(n_k)\end{align} \begin{align}10\end{align} \begin{align}9\end{align} \begin{align}8\end{align} \begin{align}= 27\end{align} Total \begin{align}(T_k)\end{align} \begin{align}109\end{align} \begin{align}131\end{align} \begin{align}137\end{align} \begin{align}= 377\end{align} Mean \begin{align}(\bar X)\end{align} \begin{align}10.9\end{align} \begin{align}14.6\end{align} \begin{align}17.1\end{align} \begin{align}= 5,264\end{align} Sum of Squared Obs. \begin{align}\textstyle (\sum_{i=1}^{n_k} X^2_{ik})\end{align} \begin{align}1,339\end{align} \begin{align}2,113\end{align} \begin{align}2,529\end{align} \begin{align}= 5,981\end{align} Sum of Obs. Squared/Number of Obs. \begin{align}\left (\frac {T_k^2}{n_k}\right )\end{align} \begin{align}1,188\end{align} \begin{align}1,907\end{align} \begin{align}2,346\end{align} \begin{align}= 5,441\end{align} 1. \begin{align}26.35\end{align} 2. \begin{align}4.03\end{align} 3. \begin{align}6.54\end{align} 4. \begin{align}3.40\end{align} 5. The calculated test statistic exceeds the critical value so we would reject the null hypothesis. Therefore, we could conclude that not all the population means are equal. Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
# 2: Vectors in Space Learning Objectives • Describe three-dimensional space mathematically. • Locate points in space using coordinates. • Write the distance formula in three dimensions. • Write the equations for simple planes and spheres. • Perform vector operations in (mathbb{R}^{3}). Vectors are useful tools for solving two-dimensional problems. To expand the use of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space. For example, although a two-dimensional map is a useful tool for navigating from one place to another, in some cases the topography of the land is important. Does your planned route go through the mountains? Do you have to cross a river? To appreciate fully the impact of these geographic features, you must use three dimensions. This section presents a natural extension of the two-dimensional Cartesian coordinate plane into three dimensions. ## Three-Dimensional Coordinate Systems As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontal (x)-axis and the vertical (y)-axis. We can add a third dimension, the (z)-axis, which is perpendicular to both the (x)-axis and the (y)-axis. We call this system the three-dimensional rectangular coordinate system. It represents the three dimensions we encounter in real life. Definition: Three-dimensional Rectangular Coordinate System The three-dimensional rectangular coordinate system consists of three perpendicular axes: the (x)-axis, the (y)-axis, and the (z)-axis. Because each axis is a number line representing all real numbers in (ℝ), the three-dimensional system is often denoted by (ℝ^3). In Figure (PageIndex{1a}), the positive (z)-axis is shown above the plane containing the (x)- and (y)-axes. The positive (x)-axis appears to the left and the positive (y)-axis is to the right. A natural question to ask is: How was this arrangement determined? The system displayed follows the right-hand rule. If we take our right hand and align the fingers with the positive (x)-axis, then curl the fingers so they point in the direction of the positive (y)-axis, our thumb points in the direction of the positive (z)-axis (Figure (PageIndex{1b})). In this text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow a left-hand rule, but the right-hand rule is considered the standard representation. In two dimensions, we describe a point in the plane with the coordinates ((x,y)). Each coordinate describes how the point aligns with the corresponding axis. In three dimensions, a new coordinate, (z), is appended to indicate alignment with the (z)-axis: ((x,y,z)). A point in space is identified by all three coordinates (Figure (PageIndex{2})). To plot the point ((x,y,z)), go (x) units along the (x)-axis, then (y) units in the direction of the (y)-axis, then (z) units in the direction of the (z)-axis. Example (PageIndex{1}): Locating Points in Space Sketch the point ((1,−2,3)) in three-dimensional space. Solution To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in the positive (x) direction, (2) units in the negative (y) direction, and (3) units in the positive (z) direction. Complete the prism to plot the point (Figure). Exercise (PageIndex{1}) Sketch the point ((−2,3,−1)) in three-dimensional space. Hint Start by sketching the coordinate axes. e.g., Figure (PageIndex{3}). Then sketch a rectangular prism to help find the point in space. In two-dimensional space, the coordinate plane is defined by a pair of perpendicular axes. These axes allow us to name any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just as in two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms a coordinate plane: the (xy)-plane, the (xz)-plane, and the (yz)-plane (Figure (PageIndex{3})). We define the (xy)-plane formally as the following set: ({(x,y,0):x,y∈ℝ}.) Similarly, the (xz)-plane and the (yz)-plane are defined as ({(x,0,z):x,z∈ℝ}) and ({(0,y,z):y,z∈ℝ},) respectively. To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished. (Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finished walls meet, facing out into the room, the floor is the (xy)-plane, the wall to your right is the (xz)-plane, and the wall to your left is the (yz)-plane. In two dimensions, the coordinate axes partition the plane into four quadrants. Similarly, the coordinate planes divide space between them into eight regions about the origin, called octants. The octants fill (ℝ^3) in the same way that quadrants fill (ℝ^2), as shown in Figure (PageIndex{4}). Most work in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. In this section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to three dimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space. If two points lie in the same coordinate plane, then it is straightforward to calculate the distance between them. We that the distance (d) between two points ((x_1,y_1)) and ((x_2,y_2)) in the x(y)-coordinate plane is given by the formula [d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2}.] The formula for the distance between two points in space is a natural extension of this formula. The Distance between Two Points in Space The distance (d) between points ((x_1,y_1,z_1)) and ((x_2,y_2,z_2)) is given by the formula [d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}. label{distanceForm}] The proof of this theorem is left as an exercise. (Hint: First find the distance (d_1) between the points ((x_1,y_1,z_1)) and ((x_2,y_2,z_1)) as shown in Figure (PageIndex{5}).) Example (PageIndex{2}): Distance in Space Find the distance between points (P_1=(3,−1,5)) and (P_2=(2,1,−1).) Solution Substitute values directly into the distance formula (Equation ef{distanceForm}): [egin{align} d(P_1,P_2) &=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2} [4pt] &=sqrt{(2−3)^2+(1−(−1))^2+(−1−5)^2} [4pt] &=sqrt{(-1)^2+2^2+(−6)^2} [4pt] &=sqrt{41}. end{align}] Exercise (PageIndex{2}) Find the distance between points (P_1=(1,−5,4)) and (P_2=(4,−1,−1)). Hint (d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}) (5sqrt{2}) Before moving on to the next section, let’s get a feel for how (ℝ^3) differs from (ℝ^2). For example, in (ℝ^2), lines that are not parallel must always intersect. This is not the case in (ℝ^3). For example, consider the line shown in Figure (PageIndex{7}). These two lines are not parallel, nor do they intersect. Figure (PageIndex{7}): These two lines are not parallel, but still do not intersect. You can also have circles that are interconnected but have no points in common, as in Figure (PageIndex{8}). Figure (PageIndex{8}): These circles are interconnected, but have no points in common. We have a lot more flexibility working in three dimensions than we do if we stuck with only two dimensions. ## Writing Equations in (ℝ^3) Now that we can represent points in space and find the distance between them, we can learn how to write equations of geometric objects such as lines, planes, and curved surfaces in (ℝ^3). First, we start with a simple equation. Compare the graphs of the equation (x=0) in (ℝ), (ℝ^2),and (ℝ^3) (Figure (PageIndex{9})). From these graphs, we can see the same equation can describe a point, a line, or a plane. In space, the equation (x=0) describes all points ((0,y,z)). This equation defines the (yz)-plane. Similarly, the (xy)-plane contains all points of the form ((x,y,0)). The equation (z=0) defines the (xy)-plane and the equation (y=0) describes the (xz)-plane (Figure (PageIndex{10})). Understanding the equations of the coordinate planes allows us to write an equation for any plane that is parallel to one of the coordinate planes. When a plane is parallel to the (xy)-plane, for example, the (z)-coordinate of each point in the plane has the same constant value. Only the (x)- and (y)-coordinates of points in that plane vary from point to point. Equations of Planes Parallel to Coordinate Planes 1. The plane in space that is parallel to the (xy)-plane and contains point ((a,b,c)) can be represented by the equation (z=c). 2. The plane in space that is parallel to the (xz)-plane and contains point ((a,b,c)) can be represented by the equation (y=b). 3. The plane in space that is parallel to the (yz)-plane and contains point ((a,b,c)) can be represented by the equation (x=a). Example (PageIndex{3}): Writing Equations of Planes Parallel to Coordinate Planes 1. Write an equation of the plane passing through point ((3,11,7)) that is parallel to the (yz)-plane. 2. Find an equation of the plane passing through points ((6,−2,9), (0,−2,4),) and ((1,−2,−3).) Solution 1. When a plane is parallel to the (yz)-plane, only the (y)- and (z)-coordinates may vary. The (x)-coordinate has the same constant value for all points in this plane, so this plane can be represented by the equation (x=3). 2. Each of the points ((6,−2,9), (0,−2,4),) and ((1,−2,−3)) has the same (y)-coordinate. This plane can be represented by the equation (y=−2). Exercise (PageIndex{3}) Write an equation of the plane passing through point ((1,−6,−4)) that is parallel to the (xy)-plane. Hint If a plane is parallel to the (xy)-plane, the z-coordinates of the points in that plane do not vary. (z=−4) As we have seen, in (ℝ^2) the equation (x=5) describes the vertical line passing through point ((5,0)). This line is parallel to the (y)-axis. In a natural extension, the equation (x=5) in (ℝ^3) describes the plane passing through point ((5,0,0)), which is parallel to the (yz)-plane. Another natural extension of a familiar equation is found in the equation of a sphere. Definition: Sphere A sphere is the set of all points in space equidistant from a fixed point, the center of the sphere (Figure (PageIndex{11})), just as the set of all points in a plane that are equidistant from the center represents a circle. In a sphere, as in a circle, the distance from the center to a point on the sphere is called the radius. The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphere is based on the three-dimensional formula for distance. Standard Equation of a Sphere The sphere with center ((a,b,c)) and radius (r) can be represented by the equation [(x−a)^2+(y−b)^2+(z−c)^2=r^2.] This equation is known as the standard equation of a sphere. Example (PageIndex{4}): Finding an Equation of a Sphere Find the standard equation of the sphere with center ((10,7,4)) and point ((−1,3,−2)), as shown in Figure (PageIndex{12}). Figure (PageIndex{12}): The sphere centered at ((10,7,4)) containing point ((−1,3,−2).) Solution Use the distance formula to find the radius (r) of the sphere: [egin{align} r &=sqrt{(−1−10)^2+(3−7)^2+(−2−4)^2} [4pt] &=sqrt{(−11)^2+(−4)^2+(−6)^2} [4pt] &=sqrt{173} end{align} ] The standard equation of the sphere is [(x−10)^2+(y−7)^2+(z−4)^2=173. onumber] Exercise (PageIndex{4}) Find the standard equation of the sphere with center ((−2,4,−5)) containing point ((4,4,−1).) Hint First use the distance formula to find the radius of the sphere. [(x+2)^2+(y−4)^2+(z+5)^2=52 onumber] Example (PageIndex{5}): Finding the Equation of a Sphere Let (P=(−5,2,3)) and (Q=(3,4,−1)), and suppose line segment (overline{PQ}) forms the diameter of a sphere (Figure (PageIndex{13})). Find the equation of the sphere. Solution: Since (overline{PQ}) is a diameter of the sphere, we know the center of the sphere is the midpoint of (overline{PQ}).Then, [C=left(dfrac{−5+3}{2},dfrac{2+4}{2},dfrac{3+(−1)}{2} ight)=(−1,3,1). onumber] Furthermore, we know the radius of the sphere is half the length of the diameter. This gives [egin{align} r &=dfrac{1}{2}sqrt{(−5−3)^2+(2−4)^2+(3−(−1))^2} [4pt] &=dfrac{1}{2}sqrt{64+4+16} [4pt] &=sqrt{21} end{align}] Then, the equation of the sphere is ((x+1)^2+(y−3)^2+(z−1)^2=21.) Exercise (PageIndex{5}) Find the equation of the sphere with diameter (overline{PQ}), where (P=(2,−1,−3)) and (Q=(−2,5,−1).) Hint Find the midpoint of the diameter first. [x^2+(y−2)^2+(z+2)^2=14 onumber] Example (PageIndex{6}): Graphing Other Equations in Three Dimensions Describe the set of points that satisfies ((x−4)(z−2)=0,) and graph the set. Solution We must have either (x−4=0) or (z−2=0), so the set of points forms the two planes (x=4) and (z=2) (Figure (PageIndex{14})). Exercise (PageIndex{6}) Describe the set of points that satisfies ((y+2)(z−3)=0,) and graph the set. Hint One of the factors must be zero. The set of points forms the two planes (y=−2) and (z=3). Example (PageIndex{7}): Graphing Other Equations in Three Dimensions Describe the set of points in three-dimensional space that satisfies ((x−2)^2+(y−1)^2=4,) and graph the set. Solution The (x)- and (y)-coordinates form a circle in the (xy)-plane of radius (2), centered at ((2,1)). Since there is no restriction on the (z)-coordinate, the three-dimensional result is a circular cylinder of radius (2) centered on the line with (x=2) and (y=1). The cylinder extends indefinitely in the (z)-direction (Figure (PageIndex{15})). Exercise (PageIndex{7}) Describe the set of points in three dimensional space that satisfies (x^2+(z−2)^2=16), and graph the surface. Hint Think about what happens if you plot this equation in two dimensions in the (xz)-plane. A cylinder of radius 4 centered on the line with (x=0) and (z=2). ## Working with Vectors in (ℝ^3) Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow. Three-dimensional vectors can also be represented in component form. The notation (vecs{v}=⟨x,y,z⟩) is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, ((0,0,0)), and terminal point ((x,y,z)). The zero vector is (vecs{0}=⟨0,0,0⟩). So, for example, the three dimensional vector (vecs{v}=⟨2,4,1⟩) is represented by a directed line segment from point ((0,0,0)) to point ((2,4,1)) (Figure (PageIndex{16})). Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If (vecs{v}=⟨x_1,y_1,z_1⟩) and (vecs{w}=⟨x_2,y_2,z_2⟩) are vectors, and (k) is a scalar, then [vecs{v}+vecs{w}=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩] and [kvecs{v}=⟨kx_1,ky_1,kz_1⟩.] If (k=−1,) then (kvecs{v}=(−1)vecs{v}) is written as (−vecs{v}), and vector subtraction is defined by (vecs{v}−vecs{w}=vecs{v}+(−vecs{w})=vecs{v}+(−1)vecs{w}). The standard unit vectors extend easily into three dimensions as well, (hat{mathbf i}=⟨1,0,0⟩), (hat{mathbf j}=⟨0,1,0⟩), and (hat{mathbf k}=⟨0,0,1⟩), and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in (ℝ^3) in the following ways: [vecs{v}=⟨x,y,z⟩=xhat{mathbf i}+yhat{mathbf j}+zhat{mathbf k}]. Example (PageIndex{8}): Vector Representations Let (vecd{PQ}) be the vector with initial point (P=(3,12,6)) and terminal point (Q=(−4,−3,2)) as shown in Figure (PageIndex{17}). Express (vecd{PQ}) in both component form and using standard unit vectors. Solution In component form, [egin{align} vecd{PQ} =⟨x_2−x_1,y_2−y_1,z_2−z_1⟩ [4pt] =⟨−4−3,−3−12,2−6⟩ [4pt] =⟨−7,−15,−4⟩. end{align}] In standard unit form, [vecd{PQ}=−7hat{mathbf i}−15hat{mathbf j}−4hat{mathbf k}. onumber] Exercise (PageIndex{8}) Let (S=(3,8,2)) and (T=(2,−1,3)). Express (vec{ST}) in component form and in standard unit form. Hint Write (vecd{ST}) in component form first. (T) is the terminal point of (vecd{ST}). (vecd{ST}=⟨−1,−9,1⟩=−hat{mathbf i}−9hat{mathbf j}+hat{mathbf k}) As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space (Figure (PageIndex{18})). We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference. Properties of Vectors in Space Let (vecs{v}=⟨x_1,y_1,z_1⟩) and (vecs{w}=⟨x_2,y_2,z_2⟩) be vectors, and let (k) be a scalar. • Scalar multiplication: [kvecs{v}=⟨kx_1,ky_1,kz_1⟩] • Vector subtraction: [vecs{v}−vecs{w}=⟨x_1,y_1,z_1⟩−⟨x_2,y_2,z_2⟩=⟨x_1−x_2,y_1−y_2,z_1−z_2⟩] • Vector magnitude: [\|vecs{v}\|=sqrt{x_1^2+y_1^2+z_1^2}] • Unit vector in the direction of (vecs{v}): [dfrac{1}{\|vecs{v}\|}vecs{v}=dfrac{1}{\|vecs{v}\|}⟨x_1,y_1,z_1⟩=⟨dfrac{x_1}{\|vecs{v}\|},dfrac{y_1}{\|vecs{v}\|},dfrac{z_1}{\|vecs{v}\|}⟩, quad ext{if} , vecs{v}≠vecs{0}] We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions. Example (PageIndex{9}): Vector Operations in Three Dimensions Let (vecs{v}=⟨−2,9,5⟩) and (vecs{w}=⟨1,−1,0⟩) (Figure (PageIndex{19})). Find the following vectors. 1. (3vecs{v}−2vecs{w}) 2. (5\|vecs{w}\|) 3. (\|5 vecs{w}\|) 4. A unit vector in the direction of (vecs{v}) Solution a. First, use scalar multiplication of each vector, then subtract: [egin{align} 3vecs{v}−2vecs{w} =3⟨−2,9,5⟩−2⟨1,−1,0⟩ [4pt] =⟨−6,27,15⟩−⟨2,−2,0⟩ [4pt] =⟨−6−2,27−(−2),15−0⟩ [4pt] =⟨−8,29,15⟩. end{align}] b. Write the equation for the magnitude of the vector, then use scalar multiplication: [5\|vecs{w}\|=5sqrt{1^2+(−1)^2+0^2}=5sqrt{2}. onumber] c. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.: [\|5 vecs{w}\|=∥⟨5,−5,0⟩∥=sqrt{5^2+(−5)^2+0^2}=sqrt{50}=5sqrt{2} onumber] d. Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions: [egin{align} dfrac{vecs{v}}{\|vecs{v}\|} =dfrac{1}{\|vecs{v}\|}⟨−2,9,5⟩ [4pt] =dfrac{1}{sqrt{(−2)^2+9^2+5^2}}⟨−2,9,5⟩ [4pt] =dfrac{1}{sqrt{110}}⟨−2,9,5⟩ [4pt] =⟨dfrac{−2}{sqrt{110}},dfrac{9}{sqrt{110}},dfrac{5}{sqrt{110}}⟩ . end{align}] Exercise (PageIndex{9}): Let (vecs{v}=⟨−1,−1,1⟩) and (vecs{w}=⟨2,0,1⟩). Find a unit vector in the direction of (5vecs{v}+3vecs{w}.) Hint Start by writing (5vecs{v}+3vecs{w}) in component form. (⟨dfrac{1}{3sqrt{10}},−dfrac{5}{3sqrt{10}},dfrac{8}{3sqrt{10}}⟩) Example (PageIndex{10}): Throwing a Forward Pass A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of (30°) (see the following figure). Write the initial velocity vector of the ball, (vecs{v}), in component form. Solution The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector (vecs{w}) extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of (30°) (see the following figure). This vector would have the same direction as (vecs{v}), but it may not have the right magnitude. The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is Dist from QB to receiver(=sqrt{15^2+20^2}=sqrt{225+400}=sqrt{625}=25) yd. We have (dfrac{25}{\|vecs{w}\|}=cos 30°.) Then the magnitude of (vecs{w}) is given by (\|vecs{w}\|=dfrac{25}{cos 30°}=dfrac{25⋅2}{sqrt{3}}=dfrac{50}{sqrt{3}}) yd and the vertical distance from the receiver to the terminal point of (vecs{w}) is Vert dist from receiver to terminal point of (vecs{w}=\|vecs{w}\| sin 30°=dfrac{50}{sqrt{3}}⋅dfrac{1}{2}=dfrac{25}{sqrt{3}}) yd. Then (vecs{w}=⟨20,15,dfrac{25}{sqrt{3}}⟩), and has the same direction as (vecs{v}). Recall, though, that we calculated the magnitude of (vecs{w}) to be (\|vecs{w}\|=dfrac{50}{sqrt{3}}), and (vecs{v}) has magnitude (60) mph. So, we need to multiply vector (vecs{w}) by an appropriate constant, (k). We want to find a value of (k) so that (∥kvecs{w}∥=60) mph. We have (\|k vecs{w}\|=k\|vecs{w}\|=kdfrac{50}{sqrt{3}}) mph, so we want (kdfrac{50}{sqrt{3}}=60) (k=dfrac{60sqrt{3}}{50}) (k=dfrac{6sqrt{3}}{5}). Then (vecs{v}=kvecs{w}=k⟨20,15,dfrac{25}{sqrt{3}}⟩=dfrac{6sqrt{3}}{5}⟨20,15,dfrac{25}{sqrt{3}}⟩=⟨24sqrt{3},18sqrt{3},30⟩). Let’s double-check that (\|vecs{v}\|=60.) We have (\|vecs{v}\|=sqrt{(24sqrt{3})^2+(18sqrt{3})^2+(30)^2}=sqrt{1728+972+900}=sqrt{3600}=60) mph. So, we have found the correct components for (vecs{v}). Exercise (PageIndex{10}) Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of (40) mph and an angle of (45°). Write the initial velocity vector of the ball, (vecs{v}), in component form. Hint Follow the process used in the previous example. (v=⟨16sqrt{2},12sqrt{2},20sqrt{2}⟩) ## Key Concepts • The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples ((x,y,z)) are used to describe the location of a point in space. • The distance (d) between points ((x_1,y_1,z_1)) and ((x_2,y_2,z_2)) is given by the formula [d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}. onumber] • In three dimensions, the equations (x=a,y=b,) and (z=c) describe planes that are parallel to the coordinate planes. • The standard equation of a sphere with center ((a,b,c)) and radius (r) is [(x−a)^2+(y−b)^2+(z−c)^2=r^2. onumber ] • In three dimensions, as in two, vectors are commonly expressed in component form, (v=⟨x,y,z⟩), or in terms of the standard unit vectors, (xi+yj+zk.) • Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let (v=⟨x_1,y_1,z_1⟩) and (w=⟨x_2,y_2,z_2⟩) be vectors, and let (k) be a scalar. Scalar multiplication: [(kvecs{v}=⟨kx_1,ky_1,kz_1⟩ onumber] [vecs{v}+vecs{w}=⟨x_1,y_1,z_1⟩+⟨x_2,y_2,z_2⟩=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩ onumber] Vector subtraction: [vecs{v}−vecs{w}=⟨x_1,y_1,z_1⟩−⟨x_2,y_2,z_2⟩=⟨x_1−x_2,y_1−y_2,z_1−z_2⟩ onumber] Vector magnitude: [‖vecs{v}‖=sqrt{x_1^2+y_1^2+z_1^2} onumber] Unit vector in the direction of (vecs{v}): [dfrac{vecs{v}}{‖vecs{v}‖}=dfrac{1}{‖vecs{v}‖}⟨x_1,y_1,z_1⟩=⟨dfrac{x_1}{‖vecs{v}‖},dfrac{y_1}{‖vecs{v}‖},dfrac{z_1}{‖vecs{v}‖}⟩, vecs{v}≠vecs{0} onumber] ## Key Equations Distance between two points in space: [d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}] Sphere with center ((a,b,c)) and radius (r): [(x−a)^2+(y−b)^2+(z−c)^2=r^2] ## Glossary coordinate plane a plane containing two of the three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the (xy)-plane, (xz)-plane, or the (yz)-plane right-hand rule a common way to define the orientation of the three-dimensional coordinate system; when the right hand is curved around the (z)-axis in such a way that the fingers curl from the positive (x)-axis to the positive (y)-axis, the thumb points in the direction of the positive (z)-axis octants the eight regions of space created by the coordinate planes sphere the set of all points equidistant from a given point known as the center standard equation of a sphere ((x−a)^2+(y−b)^2+(z−c)^2=r^2) describes a sphere with center ((a,b,c)) and radius (r) three-dimensional rectangular coordinate system a coordinate system defined by three lines that intersect at right angles; every point in space is described by an ordered triple ((x,y,z)) that plots its location relative to the defining axes ## Angle Between Two Vectors Calculator With this angle between two vectors calculator, you&aposll quickly learn how to find the angle between two vectors. It doesn&apost matter if your vectors are in 2D or 3D, nor if their representations are coordinates or initial and terminal points - our tool is a safe bet in every case. Play with the calculator and check the definitions and explanations below if you&aposre searching for the angle between two vectors formulas, you&aposll definitely find them there. Since you&aposre here, hunting down solutions to your vector problems, can we assume that you&aposre also interested in vector operations? If you want to start from the basics, have a look at our unit vector calculator. For those who want to dig even more into vector algebra, we recommend the vector projection tool and the cross product calculator. ## Calculate rotation matrix to align two vectors in 3D space? I have two separate vectors of 3D data points that represent curves and I'm plotting these as scatter data in a 3D plot with matplotlib. Both the vectors start at the origin, and both are of unit length. The curves are similar to each other, however, there is typically a rotation between the two curves (for test purposes, I've actually being using one curve and applying a rotation matrix to it to create the second curve). I want to align the two curves so that they line up in 3D e.g. rotate curve b, so that its start and end points line up with curve a. I've been trying to do this by subtracting the final point from the first, to get a direction vector representing the straight line from the start to the end of each curve, converting these to unit vectors and then calculating the cross and dot products and using the methodology outlined in this answer (https://math.stackexchange.com/a/476311/357495) to calculate a rotation matrix. However, when I do this, the calculated rotation matrix is wrong and I'm not sure why? My code is below (I'm using Python 2.7): In my test case, curve 2 is simply curve 1 with the following rotation matrix applied: (just a 60 degree rotation around the x axis). The rotation matrix computed by my code to align the two vectors again is: The plot of the direction vectors for the two original curves (a and b in blue and green respectively) and the result of b transformed with the computed rotation matrix (red) is below. I'm trying to compute the rotation matrix to align the green vector to the blue. ## 2: Vectors in Space For understanding the concept behind Machine Learning, as well as Deep Learning, Linear Algebra principles, are crucial. Linear algebra is a branch of mathematics that allows to define and perform operations on higher-dimensional coordinates and plane interactions in a concise way. Its main focus is on linear equation systems. • Idea behind basis vector? • Defination of basis vector • Properties of basis vector • Basis vectors for a given space • It’s important from a data science viewpoint What’s the idea behind basis vectors? So, the idea here is the following, Let us take an R-squared space which basically means that, we are looking at vectors in 2 dimensions. It means that there are 2 components in each of these vectors as we have taken in the above image. We can take many many vectors. So, there will be an infinite number of vectors, which will be in 2 dimensions. So, the point is can we represent all of these vectors using some basic elements and then some combination of these basic elements. Now, let us consider 2 vectors for example, Now, if you take any vector that given in R squared space, let us say take We can write this vector as some linear combination, of this vector plus this vector as follows. Similarly, if you take We can also write this vector as some linear combination, of this vector plus this vector as follows. Similarly, And that would be true for any vector that you have in this space. So, in some sense what we say is that these 2 vectors(v1 and v2) characterize the space or they form a basis for space and any vector in this space, can simply be written as a linear combination of these 2 vectors. Now you can notice, the linear combinations are actually the numbers themselves. So, for example, if I want vector(2, 1) to be written as a linear combination of the vector(1, 0) and vector(0, 1), the scalar multiples are 2 and 1 which is similarly for vector(4, 4) and so on. So, the key point is while we have an infinite number of vectors here, they can all be generated as a linear combination of just 2 vectors and we have seen here that these 2 vectors are vector(1, 0) and vector(0, 1). Now, these 2 vectors are called the basis for the whole space. Defination of basis vector: If you can write every vector in a given space as a linear combination of some vectors and these vectors are independent of each other then we call them as basis vectors for that given space. 1. Basis vectors must be linearly independent of each other: If I multiply v1 by any scalar, I will never be able to get the vector v2. And that proves that v1 and v2 are linearly independent of each other. We want basis vectors to be linearly independent of each other because we want every vector, that is on the basis to generate unique information. If they become dependent on each other, then this vector is not going to bring in anything unique. 2. Basis vectors must span the whole space: The word span basically means that any vector in that space, I can write as a linear combination of the basis vectors as we see in our previous example. 3. Basis vectors are not unique: One can find many many sets of basis vectors. The only conditions are that they have to be linearly independent and should span the whole space. So let’s understand this property in detail by taking the same example as we have taken before. Let us consider 2 other vectors, which are linearly independent of each other. First we have to check are these 2 vectors obeying the properties of basis vector? You can see that these 2 vectors are linearly independent of each other as multiplying v1 by any scalar never able to get the vector v2. So, for example, if I multiply v1 by -1 I will get vector(-1, -1), but not the vector(1, -1). To verify the second property, let’s take the vector(2, 1). Now, let us see whether we can represent this vector(2, 1) as a linear combination of the vector(1, 1) and vector(1, -1). So, if you take a look at this we have successfully represented this vector(2, 1) as a linear combination of the vector(1, 1) and vector(1, -1). You can notice that in the previous case when we use the vector(1, 0) and vector(0, 1), we said this can be written as 2 times of vector(1, 0) and 1 time of vector(0, 1) however, the numbers have changed now. Nonetheless, I can write this as a linear combination of these 2 basis vectors. Similarly, if you take the vector(1,3) Similarly, if you take the vector(4,4) So, this is another linear combination of the same basis vectors. So, the key point that I want to make here is that the basis vectors are not unique. There are many ways in which you can define the basis vectors however, they all share the same property that, if I have a set of vectors which I call as a basis vector, those vectors have to be independent of each other and they should span the whole space. Point to remember: An interesting thing to note here is that we cannot have 2 basis sets which have a different number of vectors. What I mean here is in the previous example though the basis vectors were v1(1, 0) and v2(0, 1) there were only 2 vectors. Similarly, in this case, the basis vectors are v1(1, 1) and v2(1, -1). However, there are still only 2 vectors. So, while you could have many sets of basis vectors, all of them being equivalent to the number of vectors in each set will be the same, they cannot be different. So something that you should keep in mind that for the same space you can not have 2 basis sets one with n vectors and another one with m vectors that is not possible. So, if it is a basic set for the same space, the number of vectors in each set should be the same. Find basis vectors: • Step 1: To find basis vectors of the given set of vectors, arrange the vectors in matrix form as shown below. • Step 2: Find the rank of this matrix. If you identify the rank of this matrix it will give you the number of linearly independent columns. The rank of the matrix will tell us, how many are fundamental to explaining all of these columns, and how many columns do we need. So, that we can generate the remaining columns as a linear combination of these columns. Explanation: If the rank of the matrix is 1 then we have only 1 basis vector, if the rank is 2 then there are 2 basis vectors if 3 then there are 3 basis vectors and so on. In this case, since the rank of the matrix turns out to be 2, there are only 2 column vectors that I need to represent every column in this matrix. So, the basis set has size 2. So, we can pick any 2 linearly independent columns here and then those could be the basis vectors. So, for example, we could choose v1(6, 5, 8, 11) and v2(1, 2, 3, 4) and say, this is the basis vector for all of these columns or we could choose v1(3, -1, -1, -1) and v2(7, 7, 11, 15) and so on. We can choose any 2 columns as long as they are linearly independent of each other and this is something that we know from above that the basis vectors need not be unique. So, I pick any 2 linearly independent columns that represent this data. Important from a data science viewpoint Now, let me explain to you why this basis vectors concept is very very important from a data science viewpoint. Just take a look at the previous example. We have 10 samples and we want to store these 10 samples since each sample has 4 numbers, we would be storing 4 x 10 = 40 numbers. Now, let us assume we do the same exercise, for these 10 samples and then we find that we have only 2 basis vectors, which are going to be 2 vectors out of this set. What we could do is, we could store these 2 basis vectors that, would be 2 x 4 = 8 numbers and for the remaining 8 samples, instead of storing all the samples and all the numbers in each of these samples, what we could do is for each sample we could just store 2 numbers, which are the linear combinations that we are going to use to construct this. So, instead of storing these 4 numbers, we could simply store those 2 constants and since we already have stored the basis vectors, whenever we want to reconstruct this, we can simply take the first constant and multiply it by v1 plus the second constant multiply it by v2 and we will get this number. We store 2 basis vectors which give me: 4 x 2 = 8 numbers And then for the remaining 8 samples, we simply store 2 constants e.g: 8 x 2 = 16 numbers So, this would give us: 8 + 16 = 24 numbers Hence instead of storing 4 x 10 = 40 numbers, we can store only 24 numbers, which is the approximately half reduction in number. And we will be able to reconstruct the whole data set by storing only 24 numbers. So, for example, if you have a 30-dimensional vector and the basis vectors are just 3, then you can see the kind of reduction, that you will get in terms of data storage. So, this is one viewpoint of data science. • You can identify this basis to identify a model between this data. • You can identify a basis to do noise reduction in the data. Attention reader! Don&rsquot stop learning now. Get hold of all the important Machine Learning Concepts with the Machine Learning Foundation Course at a student-friendly price and become industry ready. In science, mathematics, and engineering, we distinguish two important quantities: scalars and vectors. A scalar is simply a real number or a quantity that has magnitude. For example, length, temperature, and blood pressure are represented by numbers such as 80 m, 20°C, and the systolic/diastolic ratio 120/80, respectively. A vector, on the other hand, is usually described as a quantity that has both magnitude and direction. ### Geometric Vectors Geometrically, a vector can be represented by a directed line segment—that is, by an arrow—and is denoted either by a boldface symbol or by a symbol with an arrow over . Get Advanced Engineering Mathematics, 7th Edition now with O’Reilly online learning. O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers. ## LINEARLY INDEPENDENT SETS OF VECTORS That is, an infinite number of solutions can be constructed in terms of just two vectors, and analysis of the solutions can be performed by considering just these two vectors. To use similar methods of analysis in vector spaces, we will need the concepts of span and linear independence of sets of vectors. Both concepts involve linear combinations of vectors. where the numbers are called the coefficient of the linear combination. Solution Five possibilities are Note that the first and last linear combinations yield the same vector (0,0), even though the coefficients are not the same. The last four linear combinations are called nontrivial because in each at least one coefficient is nonzero. Solution We want to find so that Solution We check to see whether the equation has a solution. This is equivalent to and there is no solution. Hence cannot be written as a linear combination of the given vectors. The span of a set of vectors from is actually a subspace of . which are also in span . Therefore, span is a subspace of . It is called the subspace of spanned by . for all complex and . Another way to say this is that all solutions form the subspace span . In some instances span may be all of . where , and can be any real numbers. Solution We must show that any vector in can be written as a linear combination of the three given vectors. That is, we must show that there are constants so that regardless of what real values , and take. The last equation is equivalent to Therefore , and the span is all of . Solution Let be an arbitrary vector in . We want to know whether it is possible to write The last equation is equivalent to Therefore a solution exists only if but this places a restriction on , and so the very first equation cannot be solved for an arbitrary vector . Therefore the span of the given vectors is not all . The question in Example 10 could have been asked in a slightly different way. Solution Suppose is in span . Then the equation must be solvable. Working as in Example 10, we conclude that . Thus span . That is, the span of is all vectors whose third component is the sum of the first two components. So, for example, and . The vector space is spanned by . It is also spanned by a larger set . As we will see later, and functions on can be analyzed by using spanning sets hence for economy's sake, we want to be able to find the smallest possible spanning sets for vector spaces. To do this, the idea of linear independence is required. is . If the set is not linearly independent, it is called linearly dependent . To determine whether a set is linearly independent or linearly dependent, we need to find out about the solution of If we find (by actually solving the resulting system or by any other technique) that only the trivial solution exists, then is linearly independent. However, if one or more of the 's is nonzero, then the set is linearly dependent. This equation is equivalent to which has only as a solution. Therefore, is linearly independent. This system has solution if , then we have a nontrivial solution, and so is not linearly independent--it is linearly dependent. By collecting terms on the left-hand side, this equation can be rewritten From algebra we know that a polynomial is identically zero only when all the coefficients are zero. So we have which has only the trivial solution. Therefore, is linearly independent. Linear dependence of a set of two or more vectors means that at least one of the vectors in the set can be written as a linear combination of the others. Recall Example 13 and the set . In Fig. 3.4.1 we have shown geometrically the dependence of the vectors in . A general statement of this situation is as follows: Suppose is a nonzero coefficient in the linear combination. Then Therefore, is a linear combination of the other vectors in . Suppose . Then, adding to both sides, we have Because the coefficient of is nonzero, the set is linearly dependent is linearly dependent in . Write one of the vectors as a linear combination of the others. This equation is equivalent to Therefore is a solution, where is arbitrary. Thus the set is linearly dependent. Choosing , we have Of course, we could also write Some Geometry of Spanning Sets in and The span of a single nonzero vector is a line containing the origin. Span is all multiples of v , which is all position vectors in the same direction as v (see Fig. 3.4.2). The terminal points of these vectors form the line with vector equation The span of two independent vectors is a plane containing the origin. To see this in , let v and w be given by and , respectively. The plane containing v and w has normal vector and vector equation If we calculate and write the vector equation, we find where is a vector in the plane. However, if is to be a linear combination of v and w , we must have which has a solution if and only if which holds if and only if the vector equation holds. So the span of two independent vectors is the plane containing the vectors. See Fig. 3.4.3. The span of three nonzero vectors in can be a line, a plane, or all of , depending on the degree of dependence of the three vectors. If all three are multiples of each other, we have only a line. If two of the vectors and are independent but the entire set is linearly dependent, then is a linear combination of and and lies in the plane defined by and . That is, the vectors are coplanar . Lay three pencils on a tabletop with erasers joined for a graphic example of coplanar vectors. If is linearly independent, then the span is all . This can be verified directly in individual cases to show it in general requires methods of the next section. Linear combinations in complex vector spaces have important applications, as the next examples illustrate. Show that is linearly independent in . Discuss the importance of the independence. ## 2 2 -Hilbert spaces 2-vector spaces have to a large extent been motivated by and applied in (2-dimensional) quantum field theory. In that context it is usually not the concept of a plain vector space which needs to be categorified, but that of a Hilbert space. 2-Hilbert spaces as a Hilb Hilb -enriched categories with some extra properties were discribed in In applications one often assumes these 2-Hilbert spaces to be semisimple in which case such a 2-Hilbert space is a Kapranov–Voevodsky 2 2 -vector space equipped with extra structure. A review of these ideas of 2-Hilbert spaces as well as applications of 2-Hilbert spaces to finite group representation theory are in • Bruce Bartlett, On unitary 2-representations of finite groups and topological quantum field theory (arXiv) ## Contents The cross product of two vectors a and b is defined only in three-dimensional space and is denoted by a × b . [1] In physics and applied mathematics, the wedge notation ab is often used (in conjunction with the name vector product), [5] [6] [7] although in pure mathematics such notation is usually reserved for just the exterior product, an abstraction of the vector product to n dimensions. The cross product a × b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a direction given by the right-hand rule [2] and a magnitude equal to the area of the parallelogram that the vectors span. [3] The cross product is defined by the formula [8] [9] • θ is the angle between a and b in the plane containing them (hence, it is between 0° and 180°) • a‖ and ‖b‖ are the magnitudes of vectors a and b • and n is a unit vectorperpendicular to the plane containing a and b, in the direction given by the right-hand rule (illustrated). [3] If the vectors a and b are parallel (i.e., the angle θ between them is either 0° or 180°), by the above formula, the cross product of a and b is the zero vector 0. ### Direction Edit By convention, the direction of the vector n is given by the right-hand rule, where one simply points the forefinger of the right hand in the direction of a and the middle finger in the direction of b. Then, the vector n is coming out of the thumb (see the adjacent picture). Using this rule implies that the cross product is anti-commutative, that is, b × a = −(a × b) . By pointing the forefinger toward b first, and then pointing the middle finger toward a, the thumb will be forced in the opposite direction, reversing the sign of the product vector. As the cross product operator depends on the orientation of the space (as explicit in the definition above), the cross product of two vectors is not a "true" vector, but a pseudovector. See § Handedness for more detail. In 1881, Josiah Willard Gibbs, and independently Oliver Heaviside, introduced both the dot product and the cross product using a period ( a . b ) and an "x" ( a x b ), respectively, to denote them. [10] In 1877, to emphasize the fact that the result of a dot product is a scalar while the result of a cross product is a vector, William Kingdon Clifford coined the alternative names scalar product and vector product for the two operations. [10] These alternative names are still widely used in the literature. Both the cross notation ( a × b ) and the name cross product were possibly inspired by the fact that each scalar component of a × b is computed by multiplying non-corresponding components of a and b. Conversely, a dot product ab involves multiplications between corresponding components of a and b. As explained below, the cross product can be expressed in the form of a determinant of a special 3 × 3 matrix. According to Sarrus's rule, this involves multiplications between matrix elements identified by crossed diagonals. ### Coordinate notation Edit The standard basis vectors i, j, and k satisfy the following equalities in a right hand coordinate system: [2] which imply, by the anticommutativity of the cross product, that The anticommutativity of the cross product (and the obvious lack of linear independence) also implies that These equalities, together with the distributivity and linearity of the cross product (but neither follows easily from the definition given above), are sufficient to determine the cross product of any two vectors a and b. Each vector can be defined as the sum of three orthogonal components parallel to the standard basis vectors: Their cross product a × b can be expanded using distributivity: This can be interpreted as the decomposition of a × b into the sum of nine simpler cross products involving vectors aligned with i, j, or k. Each one of these nine cross products operates on two vectors that are easy to handle as they are either parallel or orthogonal to each other. From this decomposition, by using the above-mentioned equalities and collecting similar terms, we obtain: meaning that the three scalar components of the resulting vector s = s1i + s2j + s3k = a × b are Using column vectors, we can represent the same result as follows: ### Matrix notation Edit The cross product can also be expressed as the formal determinant: [note 1] [2] This determinant can be computed using Sarrus's rule or cofactor expansion. Using Sarrus's rule, it expands to Using cofactor expansion along the first row instead, it expands to [11] which gives the components of the resulting vector directly. ### Using Levi-Civita tensors Edit • In any basis, the cross-product a × b is given by the tensorial formula E i j k a i b j a^b^> where E i j k > is the covariant Levi-Civita tensor (we note the position of the indices). That corresponds to the intrinsic formula given here. • In an orthonormal basis having the same orientation than the space, a × b is given by the pseudo-tensorial formula ε i j k a i b j a^b^> where ε i j k > is the Levi-Civita symbol (which is a pseudo-tensor). That’s the formula used for everyday physics but it’s works only in this special case of basis. • In any orthonormal basis, a × b is given by the pseudo-tensorial formula ( − 1 ) B ε i j k a i b j varepsilon _a^b^> where ( − 1 ) B = ± 1 =pm 1> whether the basis has the same orientation than the space or not. The latest formula avoid to change the orientation of the space when we inverse an orthonormal basis. ### Geometric meaning Edit The magnitude of the cross product can be interpreted as the positive area of the parallelogram having a and b as sides (see Figure 1): [2] Indeed, one can also compute the volume V of a parallelepiped having a, b and c as edges by using a combination of a cross product and a dot product, called scalar triple product (see Figure 2): Since the result of the scalar triple product may be negative, the volume of the parallelepiped is given by its absolute value. For instance, Because the magnitude of the cross product goes by the sine of the angle between its arguments, the cross product can be thought of as a measure of perpendicularity in the same way that the dot product is a measure of parallelism. Given two unit vectors, their cross product has a magnitude of 1 if the two are perpendicular and a magnitude of zero if the two are parallel. The dot product of two unit vectors behaves just oppositely: it is zero when the unit vectors are perpendicular and 1 if the unit vectors are parallel. Unit vectors enable two convenient identities: the dot product of two unit vectors yields the cosine (which may be positive or negative) of the angle between the two unit vectors. The magnitude of the cross product of the two unit vectors yields the sine (which will always be positive). ### Algebraic properties Edit If the cross product of two vectors is the zero vector (i.e. a × b = 0 ), then either one or both of the inputs is the zero vector, ( a = 0 or b = 0 ) or else they are parallel or antiparallel ( ab ) so that the sine of the angle between them is zero ( θ = 0° or θ = 180° and sin θ = 0 ). The self cross product of a vector is the zero vector: and compatible with scalar multiplication so that It is not associative, but satisfies the Jacobi identity: Distributivity, linearity and Jacobi identity show that the R 3 vector space together with vector addition and the cross product forms a Lie algebra, the Lie algebra of the real orthogonal group in 3 dimensions, SO(3). The cross product does not obey the cancellation law: that is, a × b = a × c with a0 does not imply b = c , but only that: This can be the case where b and c cancel, but additionally where a and bc are parallel that is, they are related by a scale factor t, leading to: If, in addition to a × b = a × c and a0 as above, it is the case that ab = ac then As bc cannot be simultaneously parallel (for the cross product to be 0) and perpendicular (for the dot product to be 0) to a, it must be the case that b and c cancel: b = c . From the geometrical definition, the cross product is invariant under proper rotations about the axis defined by a × b . In formulae: More generally, the cross product obeys the following identity under matrix transformations: The cross product of two vectors lies in the null space of the 2 × 3 matrix with the vectors as rows: For the sum of two cross products, the following identity holds: ### Differentiation Edit The product rule of differential calculus applies to any bilinear operation, and therefore also to the cross product: where a and b are vectors that depend on the real variable t. ### Triple product expansion Edit The cross product is used in both forms of the triple product. The scalar triple product of three vectors is defined as It is the signed volume of the parallelepiped with edges a, b and c and as such the vectors can be used in any order that's an even permutation of the above ordering. The following therefore are equal: The vector triple product is the cross product of a vector with the result of another cross product, and is related to the dot product by the following formula The mnemonic "BAC minus CAB" is used to remember the order of the vectors in the right hand member. This formula is used in physics to simplify vector calculations. A special case, regarding gradients and useful in vector calculus, is where ∇ 2 is the vector Laplacian operator. Other identities relate the cross product to the scalar triple product: where I is the identity matrix. ### Alternative formulation Edit The cross product and the dot product are related by: The right-hand side is the Gram determinant of a and b, the square of the area of the parallelogram defined by the vectors. This condition determines the magnitude of the cross product. Namely, since the dot product is defined, in terms of the angle θ between the two vectors, as: the above given relationship can be rewritten as follows: which is the magnitude of the cross product expressed in terms of θ, equal to the area of the parallelogram defined by a and b (see definition above). The combination of this requirement and the property that the cross product be orthogonal to its constituents a and b provides an alternative definition of the cross product. [13] ### Lagrange's identity Edit can be compared with another relation involving the right-hand side, namely Lagrange's identity expressed as: [14] where a and b may be n-dimensional vectors. This also shows that the Riemannian volume form for surfaces is exactly the surface element from vector calculus. In the case where n = 3 , combining these two equations results in the expression for the magnitude of the cross product in terms of its components: [15] The same result is found directly using the components of the cross product found from: In R 3 , Lagrange's equation is a special case of the multiplicativity \| vw \| = \| v \|\| w \| of the norm in the quaternion algebra. It is a special case of another formula, also sometimes called Lagrange's identity, which is the three dimensional case of the Binet–Cauchy identity: [16] [17] If a = c and b = d this simplifies to the formula above. ### Infinitesimal generators of rotations Edit The cross product conveniently describes the infinitesimal generators of rotations in R 3 . Specifically, if n is a unit vector in R 3 and R(φ, n) denotes a rotation about the axis through the origin specified by n, with angle φ (measured in radians, counterclockwise when viewed from the tip of n), then for every vector x in R 3 . The cross product with n therefore describes the infinitesimal generator of the rotations about n. These infinitesimal generators form the Lie algebra so(3) of the rotation group SO(3), and we obtain the result that the Lie algebra R 3 with cross product is isomorphic to the Lie algebra so(3). ### Conversion to matrix multiplication Edit The vector cross product also can be expressed as the product of a skew-symmetric matrix and a vector: [16] where superscript T refers to the transpose operation, and [a]× is defined by: The columns [a]×,i of the skew-symmetric matrix for a vector a can be also obtained by calculating the cross product with unit vectors, i.e.: Also, if a is itself expressed as a cross product: Hence, the left hand side equals Now, for the right hand side, Evaluation of the right hand side gives Comparison shows that the left hand side equals the right hand side. This result can be generalized to higher dimensions using geometric algebra. In particular in any dimension bivectors can be identified with skew-symmetric matrices, so the product between a skew-symmetric matrix and vector is equivalent to the grade-1 part of the product of a bivector and vector. [18] In three dimensions bivectors are dual to vectors so the product is equivalent to the cross product, with the bivector instead of its vector dual. In higher dimensions the product can still be calculated but bivectors have more degrees of freedom and are not equivalent to vectors. [18] This notation is also often much easier to work with, for example, in epipolar geometry. From the general properties of the cross product follows immediately that and from fact that [a]× is skew-symmetric it follows that The above-mentioned triple product expansion (bac–cab rule) can be easily proven using this notation. As mentioned above, the Lie algebra R 3 with cross product is isomorphic to the Lie algebra so(3), whose elements can be identified with the 3×3 skew-symmetric matrices. The map a → [a]× provides an isomorphism between R 3 and so(3). Under this map, the cross product of 3-vectors corresponds to the commutator of 3x3 skew-symmetric matrices. These matrices share the following properties: For other properties of orthogonal projection matrices, see projection (linear algebra). ### Index notation for tensors Edit The cross product can alternatively be defined in terms of the Levi-Civita symbol εijk and a dot product η mi (= δ mi for an orthonormal basis), which are useful in converting vector notation for tensor applications: in which repeated indices are summed over the values 1 to 3. This representation is another form of the skew-symmetric representation of the cross product: In classical mechanics: representing the cross product by using the Levi-Civita symbol can cause mechanical symmetries to be obvious when physical systems are isotropic. (An example: consider a particle in a Hooke's Law potential in three-space, free to oscillate in three dimensions none of these dimensions are "special" in any sense, so symmetries lie in the cross-product-represented angular momentum, which are made clear by the abovementioned Levi-Civita representation). [ citation needed ] ### Mnemonic Edit The word "xyzzy" can be used to remember the definition of the cross product. The second and third equations can be obtained from the first by simply vertically rotating the subscripts, xyzx . The problem, of course, is how to remember the first equation, and two options are available for this purpose: either to remember the relevant two diagonals of Sarrus's scheme (those containing i), or to remember the xyzzy sequence. Since the first diagonal in Sarrus's scheme is just the main diagonal of the above-mentioned 3×3 matrix, the first three letters of the word xyzzy can be very easily remembered. ### Cross visualization Edit Similarly to the mnemonic device above, a "cross" or X can be visualized between the two vectors in the equation. This may be helpful for remembering the correct cross product formula. The cross product has applications in various contexts: e.g. it is used in computational geometry, physics and engineering. A non-exhaustive list of examples follows. ### Computational geometry Edit The cross product appears in the calculation of the distance of two skew lines (lines not in the same plane) from each other in three-dimensional space. The cross product can be used to calculate the normal for a triangle or polygon, an operation frequently performed in computer graphics. For example, the winding of a polygon (clockwise or anticlockwise) about a point within the polygon can be calculated by triangulating the polygon (like spoking a wheel) and summing the angles (between the spokes) using the cross product to keep track of the sign of each angle. which is the signed length of the cross product of the two vectors. The cross product is used in calculating the volume of a polyhedron such as a tetrahedron or parallelepiped. ### Angular momentum and torque Edit The angular momentum L of a particle about a given origin is defined as: where r is the position vector of the particle relative to the origin, p is the linear momentum of the particle. In the same way, the moment M of a force FB applied at point B around point A is given as: In mechanics the moment of a force is also called torque and written as τ > Since position r , linear momentum p and force F are all true vectors, both the angular momentum L and the moment of a force M are pseudovectors or axial vectors. ### Rigid body Edit The cross product frequently appears in the description of rigid motions. Two points P and Q on a rigid body can be related by: ### Lorentz force Edit The cross product is used to describe the Lorentz force experienced by a moving electric charge qe : Since velocity v , force F and electric field E are all true vectors, the magnetic field B is a pseudovector. ### Other Edit In vector calculus, the cross product is used to define the formula for the vector operator curl. The trick of rewriting a cross product in terms of a matrix multiplication appears frequently in epipolar and multi-view geometry, in particular when deriving matching constraints. The cross product can be defined in terms of the exterior product. It can be generalized to an external product in other than three dimensions. [19] This view [ which? ] allows for a natural geometric interpretation of the cross product. In exterior algebra the exterior product of two vectors is a bivector. A bivector is an oriented plane element, in much the same way that a vector is an oriented line element. Given two vectors a and b, one can view the bivector ab as the oriented parallelogram spanned by a and b. The cross product is then obtained by taking the Hodge star of the bivector ab , mapping 2-vectors to vectors: This can be thought of as the oriented multi-dimensional element "perpendicular" to the bivector. Only in three dimensions is the result an oriented one-dimensional element – a vector – whereas, for example, in four dimensions the Hodge dual of a bivector is two-dimensional – a bivector. So, only in three dimensions can a vector cross product of a and b be defined as the vector dual to the bivector ab : it is perpendicular to the bivector, with orientation dependent on the coordinate system's handedness, and has the same magnitude relative to the unit normal vector as ab has relative to the unit bivector precisely the properties described above. ### Consistency Edit When physics laws are written as equations, it is possible to make an arbitrary choice of the coordinate system, including handedness. One should be careful to never write down an equation where the two sides do not behave equally under all transformations that need to be considered. For example, if one side of the equation is a cross product of two polar vectors, one must take into account that the result is an axial vector. Therefore, for consistency, the other side must also be an axial vector. [ citation needed ] More generally, the result of a cross product may be either a polar vector or an axial vector, depending on the type of its operands (polar vectors or axial vectors). Namely, polar vectors and axial vectors are interrelated in the following ways under application of the cross product: • polar vector × polar vector = axial vector • axial vector × axial vector = axial vector • polar vector × axial vector = polar vector • axial vector × polar vector = polar vector • polar × polar = axial • axial × axial = axial • polar × axial = polar • axial × polar = polar Because the cross product may also be a polar vector, it may not change direction with a mirror image transformation. This happens, according to the above relationships, if one of the operands is a polar vector and the other one is an axial vector (e.g., the cross product of two polar vectors). For instance, a vector triple product involving three polar vectors is a polar vector. A handedness-free approach is possible using exterior algebra. ### The paradox of the orthonormal basis Edit Let (i, j,k) be an orthonormal basis. The vectors i, j and k don't depend on the orientation of the space. They can even be defined in the absence of any orientation. They can’t therefore be axial vectors. But if i and j are polar vectors then k is an axial vector for i × j = k or j × i = k. This is a paradox. "Axial" and "polar" are physical qualifiers for physical vectors, that is to say vectors which represent physical quantities such as the velocity or the magnetic field. The vectors i, j and k are mathematical vectors, neither axial nor polar. In mathematics, the cross-product of two vectors is a vector. There is no contradiction. There are several ways to generalize the cross product to the higher dimensions. ### Lie algebra Edit The cross product can be seen as one of the simplest Lie products, and is thus generalized by Lie algebras, which are axiomatized as binary products satisfying the axioms of multilinearity, skew-symmetry, and the Jacobi identity. Many Lie algebras exist, and their study is a major field of mathematics, called Lie theory. ### Quaternions Edit The cross product can also be described in terms of quaternions. In general, if a vector [a1, a2, a3] is represented as the quaternion a1i + a2j + a3k , the cross product of two vectors can be obtained by taking their product as quaternions and deleting the real part of the result. The real part will be the negative of the dot product of the two vectors. ### Octonions Edit A cross product for 7-dimensional vectors can be obtained in the same way by using the octonions instead of the quaternions. The nonexistence of nontrivial vector-valued cross products of two vectors in other dimensions is related to the result from Hurwitz's theorem that the only normed division algebras are the ones with dimension 1, 2, 4, and 8. ### Exterior product Edit In general dimension, there is no direct analogue of the binary cross product that yields specifically a vector. There is however the exterior product, which has similar properties, except that the exterior product of two vectors is now a 2-vector instead of an ordinary vector. As mentioned above, the cross product can be interpreted as the exterior product in three dimensions by using the Hodge star operator to map 2-vectors to vectors. The Hodge dual of the exterior product yields an (n − 2) -vector, which is a natural generalization of the cross product in any number of dimensions. The exterior product and dot product can be combined (through summation) to form the geometric product in geometric algebra. ### External product Edit As mentioned above, the cross product can be interpreted in three dimensions as the Hodge dual of the exterior product. In any finite n dimensions, the Hodge dual of the exterior product of n − 1 vectors is a vector. So, instead of a binary operation, in arbitrary finite dimensions, the cross product is generalized as the Hodge dual of the exterior product of some given n − 1 vectors. This generalization is called external product. [20] ### Commutator product Edit The commutator product could be generalised to arbitrary multivectors in three dimensions, which results in a multivector consisting of only elements of grades 1 (1-vectors/true vectors) and 2 (2-vectors/pseudovectors). While the commutator product of two 1-vectors is indeed the same as the exterior product and yields a 2-vector, the commutator of a 1-vector and a 2-vector yields a true vector, corresponding instead to the left and right contractions in geometric algebra. The commutator product of two 2-vectors has no corresponding equivalent product, which is why the commutator product is defined in the first place for 2-vectors. Furthermore, the commutator triple product of three 2-vectors is the same as the vector triple product of the same three pseudovectors in vector algebra. However, the commutator triple product of three 1-vectors in geometric algebra is instead the negative of the vector triple product of the same three true vectors in vector algebra. Generalizations to higher dimensions is provided by the same commutator product of 2-vectors in higher-dimensional geometric algebras, but the 2-vectors are no longer pseudovectors. Just as the commutator product/cross product of 2-vectors in three dimensions correspond to the simplest Lie algebra, the 2-vector subalgebras of higher dimensional geometric algebra equipped with the commutator product also correspond to the Lie algebras. [22] Also as in three dimensions, the commutator product could be further generalised to arbitrary multivectors. ### Multilinear algebra Edit In the context of multilinear algebra, the cross product can be seen as the (1,2)-tensor (a mixed tensor, specifically a bilinear map) obtained from the 3-dimensional volume form, [note 2] a (0,3)-tensor, by raising an index. In the same way, in higher dimensions one may define generalized cross products by raising indices of the n-dimensional volume form, which is a ( 0 , n ) -tensor. The most direct generalizations of the cross product are to define either: • a ( 1 , n − 1 ) -tensor, which takes as input n − 1 vectors, and gives as output 1 vector – an ( n − 1 ) -ary vector-valued product, or • a ( n − 2 , 2 ) -tensor, which takes as input 2 vectors and gives as output skew-symmetric tensor of rank n − 2 – a binary product with rank n − 2 tensor values. One can also define ( k , n − k ) -tensors for other k. These products are all multilinear and skew-symmetric, and can be defined in terms of the determinant and parity. This formula is identical in structure to the determinant formula for the normal cross product in R 3 except that the row of basis vectors is the last row in the determinant rather than the first. The reason for this is to ensure that the ordered vectors (v1, . vn−1, Λ n–1 i=0 vi) have a positive orientation with respect to (e1, . en). If n is odd, this modification leaves the value unchanged, so this convention agrees with the normal definition of the binary product. In the case that n is even, however, the distinction must be kept. This ( n − 1 ) -ary form enjoys many of the same properties as the vector cross product: it is alternating and linear in its arguments, it is perpendicular to each argument, and its magnitude gives the hypervolume of the region bounded by the arguments. And just like the vector cross product, it can be defined in a coordinate independent way as the Hodge dual of the wedge product of the arguments. In 1773, Joseph-Louis Lagrange introduced the component form of both the dot and cross products in order to study the tetrahedron in three dimensions. [23] In 1843, William Rowan Hamilton introduced the quaternion product, and with it the terms "vector" and "scalar". Given two quaternions [0, u] and [0, v] , where u and v are vectors in R 3 , their quaternion product can be summarized as [−uv, u × v] . James Clerk Maxwell used Hamilton's quaternion tools to develop his famous electromagnetism equations, and for this and other reasons quaternions for a time were an essential part of physics education. In 1878 William Kingdon Clifford published his Elements of Dynamic which was an advanced text for its time. He defined the product of two vectors [24] to have magnitude equal to the area of the parallelogram of which they are two sides, and direction perpendicular to their plane. Oliver Heaviside and Josiah Willard Gibbs also felt that quaternion methods were too cumbersome, often requiring the scalar or vector part of a result to be extracted. Thus, about forty years after the quaternion product, the dot product and cross product were introduced — to heated opposition. Pivotal to (eventual) acceptance was the efficiency of the new approach, allowing Heaviside to reduce the equations of electromagnetism from Maxwell's original 20 to the four commonly seen today. [25] Largely independent of this development, and largely unappreciated at the time, Hermann Grassmann created a geometric algebra not tied to dimension two or three, with the exterior product playing a central role. In 1853 Augustin-Louis Cauchy, a contemporary of Grassmann, published a paper on algebraic keys which were used to solve equations and had the same multiplication properties as the cross product. [26] [27] Clifford combined the algebras of Hamilton and Grassmann to produce Clifford algebra, where in the case of three-dimensional vectors the bivector produced from two vectors dualizes to a vector, thus reproducing the cross product. The cross notation and the name "cross product" began with Gibbs. Originally they appeared in privately published notes for his students in 1881 as Elements of Vector Analysis. The utility for mechanics was noted by Aleksandr Kotelnikov. Gibbs's notation and the name "cross product" later reached a wide audience through Vector Analysis, a textbook by Edwin Bidwell Wilson, a former student. Wilson rearranged material from Gibbs's lectures, together with material from publications by Heaviside, Föpps, and Hamilton. He divided vector analysis into three parts: First, that which concerns addition and the scalar and vector products of vectors. Second, that which concerns the differential and integral calculus in its relations to scalar and vector functions. Third, that which contains the theory of the linear vector function. Two main kinds of vector multiplications were defined, and they were called as follows: • The direct, scalar, or dot product of two vectors • The skew, vector, or cross product of two vectors Several kinds of triple products and products of more than three vectors were also examined. The above-mentioned triple product expansion was also included. The simplest example of a vector space is the trivial one: <0>, which contains only the zero vector (see the third axiom in the Vector space article). Both vector addition and scalar multiplication are trivial. A basis for this vector space is the empty set, so that <0>is the 0-dimensional vector space over F. Every vector space over F contains a subspace isomorphic to this one. The zero vector space is conceptually different from the null space of a linear operator L, which is the kernel of L. (Incidentally, the null space of L is a zero space if and only if L is injective.) The next simplest example is the field F itself. Vector addition is just field addition, and scalar multiplication is just field multiplication. This property can be used to prove that a field is a vector space. Any non-zero element of F serves as a basis so F is a 1-dimensional vector space over itself. The field is a rather special vector space in fact it is the simplest example of a commutative algebra over F. Also, F has just two subspaces: <0>and F itself. The original example of a vector space is the following. For any positive integer n, the set of all n-tuples of elements of F forms an n-dimensional vector space over F sometimes called coordinate space and denoted F n . An element of F n is written where each xi is an element of F. The operations on F n are defined by Commonly, F is the field of real numbers, in which case we obtain real coordinate space R n . The field of complex numbers gives complex coordinate space C n . The a + bi form of a complex number shows that C itself is a two-dimensional real vector space with coordinates (a,b). Similarly, the quaternions and the octonions are respectively four- and eight-dimensional real vector spaces, and C n is a 2n-dimensional real vector space. The vector space F n has a standard basis: where 1 denotes the multiplicative identity in F. Let F ∞ denote the space of infinite sequences of elements from F such that only finitely many elements are nonzero. That is, if we write an element of F ∞ as then only a finite number of the xi are nonzero (i.e., the coordinates become all zero after a certain point). Addition and scalar multiplication are given as in finite coordinate space. The dimensionality of F ∞ is countably infinite. A standard basis consists of the vectors ei which contain a 1 in the i-th slot and zeros elsewhere. This vector space is the coproduct (or direct sum) of countably many copies of the vector space F. Note the role of the finiteness condition here. One could consider arbitrary sequences of elements in F, which also constitute a vector space with the same operations, often denoted by F N - see below. F N is the product of countably many copies of F. By Zorn's lemma, F N has a basis (there is no obvious basis). There are uncountably infinite elements in the basis. Since the dimensions are different, F N is not isomorphic to F ∞ . It is worth noting that F N is (isomorphic to) the dual space of F ∞ , because a linear map T from F ∞ to F is determined uniquely by its values T(ei) on the basis elements of F ∞ , and these values can be arbitrary. Thus one sees that a vector space need not be isomorphic to its double dual if it is infinite dimensional, in contrast to the finite dimensional case. Starting from n vector spaces, or a countably infinite collection of them, each with the same field, we can define the product space like above. Let F m×n denote the set of m×n matrices with entries in F. Then F m×n is a vector space over F. Vector addition is just matrix addition and scalar multiplication is defined in the obvious way (by multiplying each entry by the same scalar). The zero vector is just the zero matrix. The dimension of F m×n is mn. One possible choice of basis is the matrices with a single entry equal to 1 and all other entries 0. When m = n the matrix is square and matrix multiplication of two such matrices produces a third. This vector space of dimension n 2 forms an algebra over a field. ### One variable Edit The set of polynomials with coefficients in F is a vector space over F, denoted F[x]. Vector addition and scalar multiplication are defined in the obvious manner. If the degree of the polynomials is unrestricted then the dimension of F[x] is countably infinite. If instead one restricts to polynomials with degree less than or equal to n, then we have a vector space with dimension n + 1. One possible basis for F[x] is a monomial basis: the coordinates of a polynomial with respect to this basis are its coefficients, and the map sending a polynomial to the sequence of its coefficients is a linear isomorphism from F[x] to the infinite coordinate space F ∞ . The vector space of polynomials with real coefficients and degree less than or equal to n is often denoted by Pn. ### Several variables Edit The set of polynomials in several variables with coefficients in F is vector space over F denoted F[x1, x2, …, xr]. Here r is the number of variables. Let X be a non-empty arbitrary set and V an arbitrary vector space over F. The space of all functions from X to V is a vector space over F under pointwise addition and multiplication. That is, let f : XV and g : XV denote two functions, and let α in F. We define where the operations on the right hand side are those in V. The zero vector is given by the constant function sending everything to the zero vector in V. The space of all functions from X to V is commonly denoted V X . If X is finite and V is finite-dimensional then V X has dimension \|X\|(dim V), otherwise the space is infinite-dimensional (uncountably so if X is infinite). Many of the vector spaces that arise in mathematics are subspaces of some function space. We give some further examples. ### Generalized coordinate space Edit Let X be an arbitrary set. Consider the space of all functions from X to F which vanish on all but a finite number of points in X. This space is a vector subspace of F X , the space of all possible functions from X to F. To see this, note that the union of two finite sets is finite, so that the sum of two functions in this space will still vanish outside a finite set. The space described above is commonly denoted (F X )0 and is called generalized coordinate space for the following reason. If X is the set of numbers between 1 and n then this space is easily seen to be equivalent to the coordinate space F n . Likewise, if X is the set of natural numbers, N, then this space is just F ∞ . A canonical basis for (F X )0 is the set of functions <>x \| xX> defined by The dimension of (F X )0 is therefore equal to the cardinality of X. In this manner we can construct a vector space of any dimension over any field. Furthermore, every vector space is isomorphic to one of this form. Any choice of basis determines an isomorphism by sending the basis onto the canonical one for (F X )0. Generalized coordinate space may also be understood as the direct sum of \|X\| copies of F (i.e. one for each point in X): The finiteness condition is built into the definition of the direct sum. Contrast this with the direct product of \|X\| copies of F which would give the full function space F X . ### Linear maps Edit An important example arising in the context of linear algebra itself is the vector space of linear maps. Let L(V,W) denote the set of all linear maps from V to W (both of which are vector spaces over F). Then L(V,W) is a subspace of W V since it is closed under addition and scalar multiplication. Note that L(F n ,F m ) can be identified with the space of matrices F m×n in a natural way. In fact, by choosing appropriate bases for finite-dimensional spaces V and W, L(V,W) can also be identified with F m×n . This identification normally depends on the choice of basis. ### Continuous functions Edit If X is some topological space, such as the unit interval [0,1], we can consider the space of all continuous functions from X to R. This is a vector subspace of R X since the sum of any two continuous functions is continuous and scalar multiplication is continuous. ### Differential equations Edit The subset of the space of all functions from R to R consisting of (sufficiently differentiable) functions that satisfy a certain differential equation is a subspace of R R if the equation is linear. This is because differentiation is a linear operation, i.e., (a f + b g)′ = a f ′ + b g′, where ′ is the differentiation operator. Suppose K is a subfield of F (cf. field extension). Then F can be regarded as a vector space over K by restricting scalar multiplication to elements in K (vector addition is defined as normal). The dimension of this vector space, if it exists, [a] is called the degree of the extension. For example the complex numbers C form a two-dimensional vector space over the real numbers R. Likewise, the real numbers R form a vector space over the rational numbers Q which has (uncountably) infinite dimension, if a Hamel basis exists. [b] If V is a vector space over F it may also be regarded as vector space over K. The dimensions are related by the formula dimKV = (dimFV)(dimKF) For example C n , regarded as a vector space over the reals, has dimension 2n. Apart from the trivial case of a zero-dimensional space over any field, a vector space over a field F has a finite number of elements if and only if F is a finite field and the vector space has a finite dimension. Thus we have Fq, the unique finite field (up to isomorphism) with q elements. Here q must be a power of a prime (q = p m with p prime). Then any n-dimensional vector space V over Fq will have q n elements. Note that the number of elements in V is also the power of a prime (because a power of a prime power is again a prime power). The primary example of such a space is the coordinate space (Fq) n . These vector spaces are of critical importance in the representation theory of finite groups, number theory, and cryptography. As $Y$ is a $6$ dimensional linear space over $mathbb F$ , it is sufficient to prove that is linearly independent. Suppose that $a_u,a_v,b_u,b_v,c_u,c_v in mathbb F$ are such that $a_u u +a_v v + b_u T(u)+ b_v T(v) +c_u T^2(u) + c_v T^2(v)=0.$ Applying $T^2$ on both sides of this equality, you get $a_u=a_v=0$ according to the hypothesis. Then applying $T$ to the remaining terms, $b_u=b_v=0$ . And applying $T$ again, $c_u=c_v=0$ .
# Question Video: Solving Word Problems by Finding the General Term of a Sequence Mathematics Emma started a workout plan to improve her fitness. She exercised for fourteen minutes on the first day and increased the duration of her exercise plan by six minutes each subsequent day. Find, in terms of π, the πth term of the sequence which represents the number of minutes that Emma spends exercising each day. Assume that π = 1 is the first day of Emmaβs plan. 02:04 ### Video Transcript Emma started a workout plan to improve her fitness. She exercised for 14 minutes on the first day and increased the duration of her exercise plan by six minutes each subsequent day. Find, in terms of π, the πth term of the sequence which represents the number of minutes that Emma spends exercising each day. Assume that π equals one is the first day of Emmaβs plan. Weβre told in this problem that Emma increased her exercise by the same amount every day, which means that the time spent exercising form an arithmetic sequence with a common difference of six. Weβre also told that Emma exercised for 14 minutes on the first day of her plan, which means the first term in the sequence is 14. We are asked to find in terms of π the πth term of this sequence, so we need to recall the general formula for the πth term of an arithmetic sequence. Itβs this: π sub π, the πth term, is equal to π plus π minus one π, where π represents the first term and π represents the common difference. We can therefore substitute the values of π and π, which we were given in the question to find our general term. Itβs π sub π is equal to 14 plus six multiplied by π minus one. Now itβs usual to go on and simplify algebraically. So weβll distribute the parentheses. We have 14 plus six π minus six, which then simplifies to six π plus eight. And it is usual to give the general term of an arithmetic sequence in this form, some multiple of π plus a constant. Notice as well that that common difference of six is the coefficient of π in our general term and that will always be the case for an arithmetic sequence. We found the πth term of this sequence. Itβs six π plus eight. And by substituting any value of π, we can calculate any term in this sequence.
Вы находитесь на странице: 1из 18 # Chapter 8, Part 2 ## Copyright 2010 Pearson Education, Inc. DO NOW PICK 2 1) What is a linear model and what does it help us understand? 2) How do we find the residual value? 3) How do we find the line of best fit? Slide 8 - 3 ## AIM AND HOMEWORK How do we plot and understand linear regressions? Homework 16: Read and Complete Chapter 8 (Page 192-199), #15-27 ODD Check Homework 15 with classmates Absent homework policy Grades Extra Credit Project Slide 8 - 4 Slide 8 - 5 ## Correlation and the Line The figure shows the scatterplot of z-scores for fat and protein. If a burger has average protein content, it should have about average fat content too. Moving one standard deviation away from the mean in x moves us r standard deviations away from the mean in y. Slide 8 - 6 ## Correlation and the Line (cont.) Put generally, moving any number of standard deviations away from the mean in x moves us r times that number of standard deviations away from the mean in y. Slide 8 - 7 ## How Big Can Predicted Values Get? r cannot be bigger than 1 (in absolute value), so each predicted y tends to be closer to its mean (in standard deviations) than its corresponding x was. This property of the linear model is called regression to the mean; the line is called the regression line. Slide 8 - 8 ## The Regression Line in Real Units Remember from Algebra that a straight line can be written as: y = mx + b In Statistics we use a slightly different notation: to emphasize that the points that satisfy this We write y equation are just our predicted values, not the actual data values. This model says that our predictions from our model follow a straight line. If the model is a good one, the data values will scatter closely around it. = b0 + b1 x y Slide 8 - 9 ## The Regression Line in Real Units(cont.) We write b1 and b0 for the slope and intercept of the line. b1 is the slope, which tells us how rapidly y changes with respect to x. b0 is the y-intercept, which tells where the line crosses (intercepts) the y-axis. Slide 8 - 10 ## The Regression Line in Real Units (cont.) In our model, we have a slope (b1): The slope is built from the correlation and the standard deviations: b1 = r sy sx Slide 8 - 11 ## The Regression Line in Real Units (cont.) In our model, we also have an intercept (b0). The intercept is built from the means and the slope: b0 = y - b1 x Slide 8 - 12 ## Fat Versus Protein: An Example The regression line for the Burger King data fits the data well: The equation is The predicted fat content for a BK Broiler chicken sandwich (with 30 g of protein) is 6.8 + 0.97(30) = 35.9 grams of fat. Slide 8 - 13 ## The Regression Line in Real Units (cont.) Since regression and correlation are closely related, we need to check the same conditions for regressions as we did for correlations: Quantitative Variables Condition Straight Enough Condition Outlier Condition ## Copyright 2010 Pearson Education, Inc. Slide 8 - 14 Residuals Revisited The linear model assumes that the relationship between the two variables is a perfect straight line. The residuals are the part of the data that hasnt been modeled. Data = Model + Residual or (equivalently) Residual = Data Model Or, in symbols, e= y- y Slide 8 - 15 ## Residuals Revisited (cont.) Residuals help us to see whether the model makes sense. When a regression model is appropriate, nothing interesting should be left behind. After we fit a regression model, we usually plot the residuals in the hope of findingnothing. Slide 8 - 16 Slide 8 - 17 Conclusion Slide 8 - 18
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Binompdf Function ## Using binompdf(n,p,a) on the TI calculator to evaluate P(X=a) = nCa x p^a x q^(n-a) Estimated11 minsto complete % Progress Practice Binompdf Function Progress Estimated11 minsto complete % Binompdf Function Suppose you were conducting an experiment where you tossed a fair coin 75 times and recorded your results. What is the probability that tails will come up 25 of those times? ### Watch This First watch this video to learn about the binompdf function. Then watch this video to see some examples. ### Guidance The binompdf function on the TI-84 calculator can be used to solve problems involving the probability of a precise number of successes out of a certain number of trials. For example, if you wanted to know the probability of 45 successes out of 100 trials, you could use the binompdf function. Just like when computing the probability manually with paper and pencil, you also have to know the probability of success for any particular trial in order to find the answer. The syntax for the binompdf function is binompdf $(n, p, a)$ , where $n$ is the number of trials, $p$ is the probability of success for any particular trial, and $a$ is the number of successes. #### Example A A local food chain has determined that 40% of the people who shop in the store use an incentive card, such as air miles. If 10 people walk into the store, what is the probability that half of them will be using an incentive card? There are 10 trials, so $n = 10$ . A success is a person using a card. You are interested in 5 successes. Therefore, $a = 5$ . The probability of a success is 40%, or 0.40, and, thus, $p = 0.40$ . Therefore, the probability of a failure is $1 - 0.40$ , or 0.60. From this, you know that $q = 0.60$ . $P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(5 \ \text{people using a card}) & = {_{10}}C_5 \times p^5 \times q^5\\P(5 \ \text{people using a card}) & = {_{10}}C_5 \times (0.40)^5 \times (0.60)^5\\P(5 \ \text{people using a card}) & = 252 \times 0.01024 \times 0.07776\\P(5 \ \text{people using a card}) & = 0.201$ Therefore, the probability of seeing 5 people using a card in a random set of 10 people is 20.1%. You could have also used technology to solve this problem, rather than pencil and paper calculations. However, with technology, it is often very helpful to check our answers using pencil and paper as well. With Example A, you could have used the binompdf function on the TI-84 calculator. Binompdf simply stands for binomial probability. The key sequence for using the binompdf function is as follows: If you used the data from Example A, you would find the following: Notice that you typed in binompdf $(n, p, a)$ to solve the problem. #### Example B Karen and Danny want to have 5 children after they get married. What is the probability that they will have exactly 3 girls? There are 5 trials, so $n = 5$ . A success is when a girl is born, and we are interested in 3 girls. Therefore, $a = 3$ . The probability of a success is 50%, or 0.50, and thus, $p = 0.50$ . Therefore, the probability of a failure is $1 - 0.50$ , or 0.50. From this, you know that $q = 0.50$ . $P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(3 \ \text{girls}) & = {_5}C_3 \times p^3 \times q^2\\P(3 \ \text{girls}) & = {_5}C_3 \times (0.50)^3 \times (0.50)^2\\P(3 \ \text{girls}) & = 10 \times 0.125 \times 0.25\\P(3 \ \text{girls}) & = 0.3125$ Therefore, the probability of having exactly 3 girls from the 5 children is 31.25%. When using technology, you will select the binompdf function, because you are looking for the probability of exactly 3 girls from the 5 children. Using the TI-84 calculator gave us the same result as our calculation (and was a great deal quicker). #### Example C A fair coin is tossed 50 times. What is the probability that you will get heads in 30 of these tosses? There are 50 trials, so $n = 50$ . A success is getting a head, and we are interested in exactly 30 heads. Therefore, $a = 30$ . The probability of a success is 50%, or 0.50, and, thus, $p = 0.50$ . Therefore, the probability of a failure is $1 - 0.50$ , or 0.50. From this, you know that $q = 0.50$ . $P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(30 \ \text{heads}) & = {_{50}}C_{30} \times p^{30} \times q^{20}\\P(30 \ \text{heads}) & = {_{50}}C_{30} \times (0.50)^{30} \times (0.50)^{20}\\P(30 \ \text{heads}) & = (4.713 \times 10^{13}) \times (9.313 \times 10^{-10}) \times (9.537 \times 10^{-7})\\P(30 \ \text{heads}) & = 0.0419$ Therefore, the probability of getting exactly 30 heads from 50 tosses of a fair coin is 4.2%. Using technology to check, you get the following: ### Guided Practice You have a summer job at a jelly bean factory as a quality control clerk. Your job is to ensure that the jelly beans coming through the line are the right size and shape. The probability that any particular shipment of jelly beans passes inspection is 90%. A normal day at the jelly bean factory means 15 shipments are produced. What is the probability that exactly 10 will pass inspection? There are 15 shipments, so $n = 15$ . A success is a shipment passing inspection, and we are interested in exactly 10 passing inspection. Therefore, $a = 10$ . The probability of a success is 90%, or 0.90, and, thus, $p = 0.90$ . Therefore, the probability of a failure is $1 - 0.90$ , or 0.10. From this, you know that $q = 0.10$ . $P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(10 \ \text{shipments passing}) & = {_{15}}C_{10} \times p^{10} \times q^5\\P(10 \ \text{shipments passing}) & = {_{15}}C_{10} \times (0.90)^{10} \times (0.10)^5\\P(10 \ \text{shipments passing}) & = 3003 \times 0.3487 \times (1.00 \times 10^{-5})\\P(10 \ \text{shipments passing}) & = 0.0105$ Therefore, the probability that exactly 10 of the 15 shipments will pass inspection is 1.05%. Using technology to check, you get the following: ### Practice 1. It is determined that because of a particular genetic trend in a family, the probability of having a boy is 60%. Janet and David decide to have 4 children. What is the probability that they will have exactly 2 boys? 2. What is the probability that out of 11 coin flips, 4 will come up tails? 3. The probability of a spinner landing on the color yellow is 65%. What is the probability that when a spinner is spun 75 times, it lands on yellow 45 times? 4. The probability of getting a 4 or greater on one roll of a fair die is 50%. If a fair die is rolled 25 times, what is the probability of getting a 4 or greater 14 times? 5. The probability of Amy's alarm clock going off when it supposed to in the morning is 95%. If Amy sets her alarm all 7 days next week, what is the probability that it goes off when it is supposed to on 6 of those days? 6. The probability of randomly guessing the correct answer to a question on a multiple-choice test is 20%. If the test consists of 20 multiple-choice questions, what is the probability of randomly guessing the correct answer to 8 of the questions? 7. Bob is randomly choosing a card from a deck of cards and then replacing it. If he does this 15 times, what is the probability that he chooses a diamond 5 times? 8. The probability that a game card given out to customers at a fast food restaurant is a winner is 10%. If the restaurant gives out 100 games cards in a day, what is the probability that 10 are winners? 9. 10 people each randomly chose a digit from 0 to 9. What is the probability that none of the people chose a 6? 10. Suppose that each year, the probability of getting your tax return audited is 18%. What is the probability of getting your tax return audited exactly once in the next 5 years? ### Vocabulary Language: English Spanish binompdf function binompdf function The function on the TI-84 calculator that can be used to solve problems involving the probability of a precise number of successes out of a certain number of trials
How to Use the Power Rule for Derivatives. Examples and Interactive Practice Problems. # How to Use the Power Rule for Derivatives ### Quick Overview • Power Rule for Derivatives: $$\displaystyle \frac d {dx}\left( x^n\right) = n\cdot x^{n-1}$$ for any value of $$n$$. • This is often described as "Multiply by the exponent, then subtract one from the exponent." • Works for any function of the form $$x^n$$ regardless of what kind of number $$n$$ is. ### Examples ##### Example 1---A Basic Power Function Suppose $$f(x) = 15x^4$$. Find $$f'(x)$$. Step 1 Use the power rule. \begin{align} f(x) & = 15x^{\blue 4}\\ f'(x) & = 15\left(\blue 4 x^{\blue 4 -1}\right)\\ & = 60x^3 \end{align} $$f'(x) = 60x^3$$ when $$f(x) = 15x^4$$ ##### Example 2---A Polynomial Suppose $$f(x) = 2x^3 + \frac 1 6 x^2 - 5x + 4$$. Find $$f'(x)$$. Step 1 Use the power rule on the first two terms of the function. The last two terms can be differentiated using the basic rules. \begin{align} f(x) & = 2x^{\blue 3} + \frac 1 6 x^{\blue 2} - 5\red{x} + \red 4\\[6pt] f'(x) & = 2(\blue 3 x^{\blue 3 -1}) + \frac 1 6(\blue 2 x^{\blue 2 - 1}) - 5\red{(1)} + \red 0\\[6pt] & = 6x^2 + \frac 1 3 x - 5 \end{align} $$\displaystyle f'(x) = 6x^2 + \frac 1 3 x - 5$$ when $$f(x) = 2x^3 + \frac 1 6 x^2 - 5x + 4$$. ##### Example 3---Fractional and Negative Exponents Suppose $$f(x) = x^{2/3} + 4x^{-6} - 3x^{-1/5}$$. $$f'(x)$$. Step 1 Use the power rule for derivatives on each term of the function. \begin{align} f(x) & = x^{\blue{2/3}} + 4x^{\blue{-6}} - 3x^{\blue{-1/5}}\\[6pt] f'(x) & = \blue{\frac 2 3} x^{\blue{\frac 2 3} -1} + 4\blue{(-6)}x^{\blue{-6}-1} - 3\blue{\left(-\frac 1 5\right)}x^{\blue{-\frac 1 5} - 1}\\[6pt] & = \frac 2 3 x^{\frac 2 3 - \frac 3 3} - 24x^{-7} + \frac 3 5 x^{-\frac 1 5 - \frac 5 5}\\[6pt] & = \frac 2 3 x^{-1/3} - 24x^{-7} + \frac 3 5 x^{-6/5} \end{align} $$\displaystyle f'(x) = \frac 2 3 x^{-1/3} - 24x^{-7} + \frac 3 5 x^{-6/5}$$ when $$f(x) = x^{2/3} + 4x^{-6} - 3x^{-1/5}$$ Suppose $$\displaystyle f(x) = \sqrt[4] x + \frac 6 {\sqrt x}$$. Find $$f'(x)$$. Step 1 Rewrite the function so each term is a power function (i.e., has the form $$ax^n$$). \begin{align} f(x) & = \sqrt[4] x + \frac 6 {\sqrt x}\\[6pt] & = x^{1/4} + \frac 6 {x^{1/2}}\\[6pt] & = x^{1/4} + 6x^{-1/2} \end{align} Step 2 Use the power rule for derivatives to differentiate each term. \begin{align} f(x) & = x^{\blue{1/4}} + 6x^{\red{-1/2}}\\[6pt] & = \blue{\frac 1 4} x^{\blue{\frac 1 4} - 1} + 6\red{\left(-\frac 1 2\right)}x^{\red{-\frac 1 2} -1}\\[6pt] & = \frac 1 4 x^{\frac 1 4 - \frac 4 4} - 3x^{-\frac 1 2 - \frac 2 2}\\[6pt] & = \frac 1 4 x^{-3/4} - 3x^{-3/2} \end{align} Step 3 (Optional) Since the original function was written in terms of radicals, we rewrite the derivative in terms of radicals as well so they match aesthetically. \begin{align} f'(x) & = \frac 1 4 x^{-3/4} - 3x^{-3/2}\\[6pt] & = \frac 1 4\cdot \frac 1 {x^{3/4}} - 3\cdot \frac 1 {x^{3/2}}\\[6pt] & = \frac 1 4 \cdot \frac 1 {\sqrt[4]{x^3}} - \frac 3 {\sqrt{x^3}}\\[6pt] & \frac 1 {4\sqrt[4]{x^3}} - \frac 3 {x\sqrt x} \end{align} If we rationalize the denominators as well we end up with $$f'(x) = \frac{\sqrt[4] x}{4x} - \frac{3\sqrt x}{x^2}$$ $$\displaystyle f'(x) = \frac 1 {4\sqrt[4]{x^3}} - \frac 3 {x\sqrt x} = \frac{\sqrt[4] x}{4x} - \frac{3\sqrt x}{x^2}$$ when $$\displaystyle f(x) = \sqrt[4] x + \frac 6 {\sqrt x}$$. ##### Example 5---Derivatives of $$\frac 1 {x^n}$$ Suppose $$\displaystyle f(x) = \frac 8 {x^{12}} + \frac 2 {x^{1.3}}$$. Find $$f'(x)$$. Notice that $$f$$ is a composition of three functions. This means we will need to use the chain rule twice. Step 1 Rewrite $$f$$ so it is in power function form. $$f(x) = \frac 8 {x^{12}} + \frac 2 {x^{1.3}} = 8x^{-12} + 2 x^{-1.3}$$ Step 2 Use the power rule for derivatives to differentiate each term. \begin{align} f(x) & = 8x^{\blue{-12}} + 2 x^{\red{-1.3}}\\ & = 8(\blue{-12})x^{\blue{-12}-1} + 2(\red{-1.3})x^{\red{-1.3}-1}\\ & = -96x^{-13} - 2.6x^{-2.3} \end{align} Step 3 (Optional) Since the original function was written in fractional form, we write the derivative in the same form. $$f'(x) = -96x^{-13} - 2.6x^{-2.3} = -\frac{96}{x^{13}} - \frac{2.6}{x^{2.3}}$$ $$\displaystyle f'(x) = -\frac{96}{x^{13}} - \frac{2.6}{x^{2.3}}$$ when $$\displaystyle f(x) = \frac 8 {x^{12}} + \frac 2 {x^{1.3}}$$. ### Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!
# lecture 3 Slides.pptx 28 de May de 2023 1 de 13 ### lecture 3 Slides.pptx • 1. Lecture 3 Topics: Outlier in Box Plot Mean, Median and Spread data SADAF SALEEM Department of Computer Sciences GIFT University, Pakistan Course Title: Probability Theory Course Code: MATH-313 Program: BS MATH Semester: Spring 2023 • 2. Box-and-Whisker Plot Outlier: • Sometime in box and whisker plot we mark another point called outlier. • In a box plot, an asterisk (*) identifies an outlier. • An outlier is a value that is inconsistent with the rest of the data. It is defined as a value that is more than 1.5 times the interquartile range smaller than Q1 or larger than Q3. • 3. Box-and-Whisker Plot Example 1: 10.2, 14.1, 14.4. 14.4, 14.4, 14.5, 14.5, 14.6, 14.7, 14.7, 14.7, 14.9, 15.1, 15.9, 16.4 Mark any outlier if exist. Solution: Step 1: arrange data Step 2: Find Q2, Q1, Q3 • Q2 = 14.6 • Q1 = 14.4 • Q3 = 14.9 Step 3: Find IQR (inter quartile range) IQR = 14.9 – 14.4 = 0.5 Step 4: Calculated Lower limit and Upper limit • Q1 – 1.5 ×IQR = 14.4 – 0.75 = 13.65 • Q3 + 1.5×IQR = 14.9 + 0.75 = 15.65. Step 5: Mark outliers Then the outliers are at: 10.2, 15.9, and 16.4 • 4. Box-and-Whisker Plot Example 1: 21, 23, 24, 25, 29, 33, 49 Mark any outlier if exist. Solution: Step 1: arrange data Step 2: Find Q2, Q1, Q3 • Q2 = 25 • Q1 = 23; • Q3 = 33 Step 3: Find IQR (inter quartile range) IQR = 33 – 23 = 10 Step 4: Calculated Lower limit and Upper limit • Q1 – 1.5 ×IQR = 23 – 1.5×10 = 23 – 15 = 8 • Q3 + 1.5×IQR = 33 + 1.5×10 = 33 + 15 = 48 Step 5: Mark outliers • 5. Box-and-Whisker Plot Example 3: Let the data range be 199, 201, 236, 269,271,278,283,291, 301, 303, and 341. Mark any outlier if exist. Solution: • Q2 = 278 • Q1 = 236 • Q3 = 301 Step 3: Find IQR (inter quartile range) IQR = 301-236=65 Step 4: Calculated Lower limit and Upper limit • Q1 – 1.5 ×IQR = 138.5 • Q3 + 1.5×IQR =398.5 No outlier exist • 6. Box-and-Whisker Plot Example 3: The box plot below shows the amount spent for books and supplies per year by students at four-year public colleges. a. Estimate the median amount spent. b. Estimate the first and third quartiles for the amount spent. c. Estimate the interquartile range for the amount spent. d. Beyond what point is a value considered an outlier? e. Identify any outliers and estimate their value. • 9. Sample Mean • Average value of the sample • Influenced by extreme values • Represented by 𝑥 • Formula: • Example 1: • 1.7, 2.2, 3.9, 3.11, 14.7 • 𝒙 = 1.7+2.2+3.9+3.11+14.7 5 • 𝒙 = 5.12 • 10. Sample Mean: Class Practice • 11. Sample Median • Central tendency of the Sample • Don't influenced by extreme values • Represented by 𝑥 • Formula: • Arrange observation in increasing order • Example 1: 1.7, 2.2, 3.9, 3.11, 14.7 • 1.7, 2.2, 3.11, 3.90, 14.7 • 𝒙 = 𝟑. 𝟏𝟏
## Monday, October 25, 2021 ### Projectile Motion When Vertical Displacement Is Not Zero Where are we going with this? The information on this page relates to the skills needed to investigate and evaluate the graphical and mathematical relationship (using either manual graphing or computers) of one-dimensional kinematic parameters (distance, displacement, speed, velocity, acceleration) with respect to an object's position, direction of motion, and time. Projectile Motion When Vertical Displacement Is Not Zero Why is this even a thing! Source 2021-10-25 What happens if a projectile is set into motion, but from a position above the horizontal? Like in shot putting. The shot is launched from a distance of around 2.25 meters above the ground. How does this change the calculations? The problem starts out just as if the projectile began on the ground. The difference occurs in step 4 (below). Because the vertical displacement is NOT ZERO, the projectile has further to go before hitting the ground. tdown is now NOT EQUAL to tup. To find tdown, you will need to use the distance equation: df = di + vit + 1/2at2 Since we are calculating time from the point that the object stops moving up (and therefore is not moving at in the vertical direction) vi is zero and the middle term goes away. Hence, we can simplify the equation to: df = di + 1/2at2 Obviously, a is acceleration do to gravity, but what are di and df? The most accurate way to think about this is to say df is the ground and di is the total vertical displacement (dy) plus the initial vertical displacement (dyi). Since these would be measured from the ground to the top of the trajectory, they would be positive meaning a is in the opposite direction and is negative. Further df would be when it hits the ground and would be zero. Therefore: df = di + 1/2at2 where di = dyi + dy a = -9.81 Solving for t we find that tdown = √ (dyi + dy) / 1/2(9.81) So, if you are given theta and the velocity, here's a checklist sort of process… • #1 Draw the diagram and label everything. • #2 Find the component velocities in the x and y directions: v • cosθ = vx v • sinθ = vy • #3 Find the time up using vy and acceleration due to gravity (probably 9.81 m/s/s) tup = v / 9.81 • #4 Find the total time where… t = tup + tdown and (different for dy ≠ zero as follows) Find the maximum vertical displacement: dy = vytup + 1/2atup2 where a = -9.81 m/s/s, vy was found in step 2 and tup was found in step 3. df = di + 1/2at2 0 = (dyi + dy) + 1/2 (-9.81)t2 tdown = √ (dyi + dy) / 1/2(9.81) • #5 Use the distance equation to find the horizontal displacement: d= vx•t where vx was found in step 2 and was found in step 4. Something like this…
GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 26 May 2019, 19:35 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # What is the area of the shaded region? Author Message TAGS: ### Hide Tags Intern Joined: 01 Mar 2014 Posts: 1 ### Show Tags 01 Mar 2014, 20:44 1 00:00 Difficulty: (N/A) Question Stats: 100% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics Hello everyone, First time poster here just wanted to ask you how to solve this question quicker than the way I just did. Provided info: Given the small circle has radius of 3, what is the area of the shaded region? The triangle inscribed in smaller circle is split into 3 isosceles triangles with 2 sides equaling 3. Using the formula x : x : xsqrt2 You get the side of the inscribed triangle is 3sqrt2 Once you have that you can find the area of the equilateral triangle by splitting it into 2 right triangles and using the pythagorean theorem. Where x = height ((3sqrt2)/2)^2 + x^2 = (3sqrt2)2^2 18/4 + x^2 = 18 9/2 + x^2 = 18 subtract the 9/2 x^2 = 18 - 4.5 = 13.5 x=sqrt13.5 Then you find the area of the triangle by formula (bh)/2 3sqrt2sqrt13.5 = 3sqrt27 = 9sqrt3 lastly divide by 2 = 4.5sqrt3 A=4.5sqrt3 Now you know the area of the original smaller triangle. The area of the smaller circle using A=pi r^2 = 9pi Now given the ratio of the larger triangle to the smaller triangle 4x You can find the area of the shaded region of the top part by using the two answers above and divide by 3. 4(4.5sqrt3) - 9pi (18sqrt3 - 9pi)/3 6sqrt3 -3pi Since the smaller triangle is 1/4 to the larger triangle, we know the larger circle will be 36pi and we already know the area of the larger triangle = 18sqrt3 So in knowing that we can determine the larger shaded region (at the bottom) by subtracting the area of the larger circle by the larger triangle and then divide it all by 3 (36pi - 18srqt3) / 3 which equals 12pi - 6sqrt3 6sqrt3 -3pi + 12pi - 6sqrt3 9pi Can someone tell me if my math is correct and how can you solve this question in under ~2mins? Attachments gmat.jpg [ 32.56 KiB \| Viewed 6709 times ] Intern Status: FTW Joined: 01 Apr 2012 Posts: 9 Location: India GMAT Date: 09-27-2014 GPA: 3.5 WE: Consulting (Venture Capital) ### Show Tags 02 Mar 2014, 01:46 Let r= Radius of smaller circle a= side of smaller triangle A= side of bigger triangle Since, r=3, using the formula Radius of inscribed circle= √3a/6 (where a= side of the triangle), we get a= 6√3 Now, using the formula Radius of circumscribed circle= √3A/3 (where A= side of the triangle), we get R= 6 Now, we can see that the Area of bigger circle = Area of bigger triangle + Area of 3 big sections => ⊼ (6)^2 = √3 (6√3)^2 / 4 + Area of 3 Big sections => Area of 3 big sections = 36⊼ - 27√3 => Area of 1 big section = 12 ⊼ - 9√3.............(i) Now, similarly, Area of bigger triangle = Area of smaller circle + Area of 3 small sections =>√3(6√3)^2 / 4 = ⊼ (3)^2 + Area of 3 small sections => Area of 3 small sections = 27√3 - 9⊼ => Area of 1 small section = 9√3 - 3⊼...........(ii) Now, as required by the question, the area of shaded portion = area of 1 small section + area of 1 big section Adding (i) and (ii), we get the required area = 9⊼ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9246 Location: Pune, India ### Show Tags 02 Mar 2014, 23:42 joshoowa wrote: Hello everyone, First time poster here just wanted to ask you how to solve this question quicker than the way I just did. Provided info: Given the small circle has radius of 3, what is the area of the shaded region? The triangle inscribed in smaller circle is split into 3 isosceles triangles with 2 sides equaling 3. Using the formula x : x : xsqrt2 You get the side of the inscribed triangle is 3sqrt2 Once you have that you can find the area of the equilateral triangle by splitting it into 2 right triangles and using the pythagorean theorem. Where x = height ((3sqrt2)/2)^2 + x^2 = (3sqrt2)2^2 18/4 + x^2 = 18 9/2 + x^2 = 18 subtract the 9/2 x^2 = 18 - 4.5 = 13.5 x=sqrt13.5 Then you find the area of the triangle by formula (bh)/2 3sqrt2sqrt13.5 = 3sqrt27 = 9sqrt3 lastly divide by 2 = 4.5sqrt3 A=4.5sqrt3 Now you know the area of the original smaller triangle. The area of the smaller circle using A=pi r^2 = 9pi Now given the ratio of the larger triangle to the smaller triangle 4x You can find the area of the shaded region of the top part by using the two answers above and divide by 3. 4(4.5sqrt3) - 9pi (18sqrt3 - 9pi)/3 6sqrt3 -3pi Since the smaller triangle is 1/4 to the larger triangle, we know the larger circle will be 36pi and we already know the area of the larger triangle = 18sqrt3 So in knowing that we can determine the larger shaded region (at the bottom) by subtracting the area of the larger circle by the larger triangle and then divide it all by 3 (36pi - 18srqt3) / 3 which equals 12pi - 6sqrt3 6sqrt3 -3pi + 12pi - 6sqrt3 9pi Can someone tell me if my math is correct and how can you solve this question in under ~2mins? It's good to understand the relations between circles and inscribed polygons. Check these two posts: http://www.veritasprep.com/blog/2013/07 ... relations/ http://www.veritasprep.com/blog/2013/07 ... other-way/ When a circle is inscribed in an equilateral triangle, Side of the triangle = $$2\sqrt{3} $$ Radius of the circle Side of the larger triangle = $$2\sqrt{3} $$ Radius of the smaller circle = $$2\sqrt{3} * 3$$ = $$6\sqrt{3}$$ When an equilateral triangle is inscribed in a circle, Side of the triangle = $$\sqrt{3} $$ Radius of the circle Side of the larger triangle = $$\sqrt{3} $$Radius of the larger circle $$6\sqrt{3} = \sqrt{3}$$ * Radius of larger circle Radius of larger circle = 6 Area of the top shaded region = (Area of larger triangle - Area of smaller circle)/3 Area of the lower shaded region = (Area of larger circle - Area of larger triangle)/3 Total shaded area = (Area of larger circle - Area of smaller circle)/3 =$$(\pi6^2 - \pi 3^2)/3 = 9\pi$$ _________________ Karishma Veritas Prep GMAT Instructor Non-Human User Joined: 09 Sep 2013 Posts: 11029 ### Show Tags 07 Sep 2017, 00:11 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: What is the area of the shaded region? [#permalink] 07 Sep 2017, 00:11 Display posts from previous: Sort by
Paul's Online Notes Home / Algebra / Common Graphs / Parabolas 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 4-2 : Parabolas 7. Sketch the graph of the following parabola. The graph should contain the vertex, the $$y$$‑intercept, $$x$$-intercepts (if any) and at least one point on either side of the vertex. $f\left( x \right) = - 3{x^2} + 6x + 3$ Show All Steps Hide All Steps Start Solution Let’s find the vertex first. In this case the equation is in the form $$f\left( x \right) = a{x^2} + bx + c$$. And so we know the vertex is the point $$\left( { - \frac{b}{{2a}},f\left( { - \frac{b}{{2a}}} \right)} \right)$$. The vertex is then, $\left( { - \frac{6}{{2\left( { - 3} \right)}},f\left( { - \frac{6}{{2\left( { - 3} \right)}}} \right)} \right) = \left( {1,f\left( 1 \right)} \right) = \left( {1,6} \right)$ Also note that $$a = - 3 < 0$$ for this parabola and so the parabola will open downwards. Show Step 2 The $$y$$-intercept is just the point $$\left( {0,f\left( 0 \right)} \right)$$. A quick function evaluation gives us that $$f\left( 0 \right) = 3$$ and so for our equation the $$y$$-intercept is $$\left( {0,3} \right)$$. Show Step 3 For the $$x$$-intercepts we just need to solve the equation $$f\left( x \right) = 0$$. So, let’s solve that for our equation. $- 3{x^2} + 6x + 3 = 0\hspace{0.25in} \to \hspace{0.25in} x = \frac{{ - 6 \pm \sqrt {{6^2} - 4\left( { - 3} \right)\left( 3 \right)} }}{{2\left( { - 3} \right)}} = \frac{{ - 6 \pm \sqrt {72} }}{{ - 6}} = \frac{{ - 6 \pm 6\sqrt 2 }}{{ - 6}} = 1 \pm \sqrt 2$ The two $$x$$-intercepts for this parabola are then : $$\left( {1 - \sqrt 2 ,0} \right) = \left( { - 0.4142,0} \right)$$ and $$\left( {1 + \sqrt 2 ,0} \right) = \left( {2.4142,0} \right)$$. Show Step 4 Because we had two $$x$$-intercepts for this parabola we already have at least one point on either side of the vertex and so we don’t really need to find any more points for our graph. However, just for the practice let’s find the point corresponding to the $$y$$-intercept on the other side the vertex. The $$y$$-intercept is a distance of 1 to the left of the vertex and so there will be a corresponding point at the same $$y$$ value to the right and it will be a distance of 1 to the right of the vertex. Therefore, the point to the right of the vertex corresponding to the y‑intercept is $$\left( {2,3} \right)$$. Show Step 5 Here is a sketch of the parabola including all the points we found above.
## Mathematics that I learned – By Angela NATURAL NUMBERS Natural numbers – We know , we use 1, 2, 3, 4,……when we begin to count. They come naturally when we start counting . Hence, mathematicians call the counting numbers as natural numbers. To add a natural number of any digit write the number below one another. We can write the number in any order. Place the numbers , in such a way that the ones place is matching.Then add till the number of places. Look at the picture below to add a natural number. How to subtract natural numbers To subtract a natural number of 1 digit , 2 digit , 3 digit or any digit write the biggest number first then the smallest number. Write the number below one another .Place the numbers , in such a way that the ones place is matching. Now subtract till the number of places. FACTORS WHAT IS FACTOR If we divide a number with any number and if we get the reminder 0 that number is the factor of the first number. HOW TO FIND A FACTOR OF A NUMBER To find a factor of a number we should divide the number with 1 . Then we should divide that number with 2 and if the remainder is 0 , that number is the factor of 2. Continue this process till the half of the number. WHAT IS PRIME NUMBER Prime number are numbers that have 2 factors. Those factors are 1 and the number itself. WHAT IS PRIME FACTOR The factors of a number with the prime number is called the prime factors of the number. WHAT IS PRIME FACTORIZATION ? HOW TO FIND OUT ? The method using to find the prime factors of a number is called prime factorization of the prime factors of a number . There are two methods to do prime factorization . They are 1 . Factor tree 2. Prime factorization FACTOR TREE PRIME FACTORIZATION HOW TO FIND FACTORS OF A NUMBER USING PRIME FACTORS To find all factors of a number multiply all possible combination. Now look at the examples shown below. LCM – LOWEST COMMON MULTIPLE Method 1 – In this method, first we have to write the numbers as shown in the fig. . Now we can select any smallest number which should be at least divisible with any one of the number. Now we can divide each number with the divisor. If the number is divisible with divisor, then we can write the quotient bellow that number. If the number is not divisible then we can directly bring down the number itself. Now repeat the process as shown in the figure until all the numbers reach 1. Then by multiply all the divisors we will get the LCM. HCF – HIGHEST COMMON FACTOR Method 1 – In this method first write the numbers as shown in the given fig.Now we should write the least possible number which should be divisible with all the numbers. Repeat this process till there is no more possible number that should be divisible with all the number.Then multiply all the divisor to get the HCF. If there is only one divisor , that divisor is the HCF of all the number. CO – PRIME NUMBER Two numbers that have 1 as the only common factor is known as co – prime numbers. TWIN PRIME NUMBERS Two consecutive prime numbers with only 1 composite number is called twin prime numbers Example : 11 , 13 PRIME TRIPLE Three consecutive prime numbers with only 1 composite prime number is called prime triple. Example : 3, 5, 7 # Divisibility rule of numbers 1 to 11 ###### Divisibility rule of numbers 1 Number 1 is divisible with all numbers. ###### Divisibility rule of numbers 2 If a number is an even number then it is divisible by 2. Or if the last digit (ones place) of the number is 0,2,4,6,8, then it is divisible by 2. Example : 24 , 32 , 50 . ###### Divisibility rule of numbers 3 The given number is divisible by 3, then the sum of the all digits if the number will be the multiple of 3. Example : 54 = 5 + 4 =9 ###### Divisibility rule of numbers 4 A number with 3 or more digits is divisible with 4, then the number formed by it last 2 digits (once and tens place) will be divisible with 4. Example : 524 ###### Divisibility rule of numbers 5 The number which has either 0 or 5 in its once place is divisible with 5. Example : 235 , 350 ###### Divisibility rule of numbers 6 If a number is divisible with 2 and 3 both then the number is divisible with 6 also. Example : 24 Last digit is 4 . It is divisible by 2. 2+4 = 6 . 6 is divisible with 3. therefore , 24 is divisible by 6 ###### Divisibility rule of numbers 8 A number with 4 or more digit is divisible by 8, if the number formed by its last 3 digit is divisible by 8. Example : 3248 ###### Divisibility rule of numbers 9 If the sum of the digits of a number is divisible by 9 then the number itself is divisible by 9. Example : 99 = 9+9=18 ###### Some more divisibility rules • If a number is divisible by another number, then it is divisible by each of the factors of that number. Ex : 24 is divisible by 8 . It is divisible by each of the factors of 8. Factors of 8 are 1 , 2 , 4 and 8. So 1,2,4 and 8 are also the factors of 24. • If a number is divisible by two co-prime numbers, then it is divisible by it product also. Ex :The number 80 is divisible by 4 and 5 . It is also divisible by 4 multiplied by 5 = 20 . 4 and 5 are co – prime numbers. • If two given number are divisible by a number, then their sum is also divisible by that number. Ex : The numbers 16 and 20 are both divisible by 4 . The number 16 +20 = 36 is also divisible by 4. • If two given number are divisible by a number, then their difference is also divisible by that number. Ex : The numbers 35 and 20 are both divisible by 5 . There difference is 35 – 20 = 15 . It is divisible by 5. BASIC GEOMETRICAL IDEAS Point Point is the basic unit of geometry. it have no length breadth and no thickness. That means point have no dimensions. It can be represented by dot and can be called as dotA. LINE a line is a collection of points going endlessly in both directions along a straight path. Plane A plane is a smooth flat surface which extends endlessly in all directions. It does not have boundaries. A plane have only length and breadth but no thickness. Line segment A line segment is the part of the line with 2 endpoints. It can also define as the shortest distance between 2 points. (line segment PQ). Ray Ray is the part of the line with one start point and extended endlessly in one direction. Angle Interior of an angle The space with in the arms of an angle is called interior of the angle. Exterior of an angle The space outside the arms of an angle is called exterior of an angle. Two angles which have a common arm,a common vertex, and the non common arm lie on either side of common arm are called adjacent angles. curves a curve is a collection of points going endlessly in both directions along a straight or curved path. Simple curve Any curve which does not cross itself and can be draw without taking the pencil is called simple curve. Open curve Any curve which have a start point and an end point then it is called open curve. Closed curve Any curve which have no start point and no end point then it is called closed curve. Different spaces in a closed curve 1. Interior of the curve (inside ) 2. Boundary (on the curve) 3. Exterior of the curve (outside) The interior and boundary together in a closed curve is called the region Polygons The simple closed curve made up with line segment is called polygon Side, Vertex, Diagonals of a polygon The line segments used to form a polygon is called side. The common points of two sides (line segments) in a polygon is called vertex. The line segment between two opposite vertex of a polygon is called diagonal of the polygon. If two vertex of a polygon formed by at least one common side is called adjacent vertex. Opposite vertex If two vertex of a polygon formed by no common side is called opposite vertex. ## Addition, multiplication, division and subtraction of rational numbers / fractional numbers Division If we have decimal places in the dividend, we can divide the dividend with divisor with out decimal places and move the decimal point from right to left as per the number of places in the dividend. If we have the decimal places in the divisor, we can divide the dividend with divisor without decimal places and move the decimal point in the quentiont from left to right as per the number of decimal places in the divisor. example 652/3.2 ## Different types of angles An Angle can formed when two rays originate from the same endpoint. The rays making angle are called the arm of the angle. angle can also form by intersecting two lines or line segments. We can see the various type of angle in the fig-1. Angle can be represent by ∠ABC. Where AB and BC are the arms of the angles and B is the vertex of the angle. Complementary angle : If the sum of 2 angles are equal to 90°, then that angles are called complementary angles. Supplementary angles : If the sum of 2 angles are equal to 180°, then that angles are called Supplementary angles. Adjacent angles: If two angles are adjacent angles then they have a common vertex .a common arm and there non -common arm are the different side of the common arm. In the below figure-2 we can see a adjacent angles. Here the angle ∠ABD and ∠DBC are adjacent angle. But also note that ∠ ABC is not adjacent angle. More over when there is two adjacent angle, then the sum is always equal to the angle formed by two non-common arms that is ∠ABD+∠CBD=∠ABC ###### Linear pair of Angle If the non-common arm of an adjacent angle are opposite ray the angles are called linear pair of angle. As shown in the fig3, AB and BC are non common arm and the angle ABD and angle CBD are linear pair of angle. From the fig3 we can see that ∠ABD+∠CBD=∠ABC (adjacent angle). And ∠ABC =180° as it is straight angle. So from the above statement we can make the axiom 6.1 as follows If a ray stand on a line, then the sum of two adjacent angle so formed is 180° The opposite (converse of each other) of the axiom 6.1 can be state like bellow. If sum of two adjacent angle is 180° then the non common arm of the angle form a line. This two axioms together called linear pair axiom. ## Facts about lines / definitions of lines ###### Lines/(straight line) Eventually distributed points in any direction can called line. Eventually distributed points in particular direction can called straight line. In this section we are discussing about the straight line only. As shown in the fig1-a A line has no end. It can be extended in either side endlessly. It can denoted by as shone in the fig. As shown in the figure 1-b a line segment have a start and end point. It can be denoted as shown in the figure. A ray have a start point but don’t have an end point. it can also be denoted by as shown in the figure 1-c. But in normal case the line, line segment and the ray are denoted by AB only. And the length (measure) of a line segment is also denoted by AB. Parallel lines are two lines whose every point have equal perpendicular distance each other. it can be see in the figure 1-d. Intersetting ling are two lines which a common point. An intersecting line will also create angles on intersecting point. We can see an example of intersecting line as shown in the fig-e. Line can also denoted by a single letter like m,n or l. If three or more points lie on a same line, then this points are called collinear points. Otherwise they are called non collinear points. ## Algebraic identities or equations 1. (x+y)(x+y) = x²+2xy+y² 2. (x-y)(x-y) = x²-2xy+y² 3. (x+y)(x-y) = x²-y² 4. (x+a)(x+b) = x²+(a+b)x+ab 5. (x+y+z)² = x²+y²+z²+2xy+2xz+2yz 6. (x+y)³ = x³+y³+3x²y+3y²x or x³+y³+3xy(x+y) 7. (x-y)³ = x³-y³-3x²y+3y²x or x³-y³-3xy(x-y) 8. x³+y³+z³-3xyz = (x+y+Z)(x²+y²+z²-xy-xz-yz) 9. x³+y³ = (x+y)(x²-xy+y²) 10. x³-y³ = (x-y)(x²+xy+y²) 11. if x+y+z = 0 then x³+y³+z³ = 3xyz ## Equations for surface area and volume of geometrical shapes ###### surface area Area of rectangle = l x b -length x birth Area of the circle = πr² Surface area of cuboid = 2( lb+bh+hl) where l= length, b= breth h = height Surface area of cube = 6a² where a= edge of the cube Lateral Surface area of cube = 4a² where a= edge of the cube Curved surface area of a cylinder = 2πrh where r= radius of the cynder, h= height of the cylinder. Total surface area of the cylinder = 2πr(r+h) where r= radius of the cynder, h= height of the cylinder. Curved surface area of cone = πrl, where r= radius of the cone, l is slant height of cone. Total surface area of a cone = πr(l+r), where r radius, l= slant height if we know height and radius of the cone we can find the slant height (l) of the cone by l= √(r²+h²) Surface area of a sphere = 4πr², where r is the radius of the sphere. Curved surface area of a hemisphere (half sphere) = 2πr² Total surface area of a hemisphere (half sphere) = 3πr² ##### Volume volume of cuboid = l x b x h where l= length, b= breth, h= height Volume of the cube = a³ where a is the edge of the cube Volume of the cylinder = πr²h where r= radius, h = height Volume of the cone = 1/3(πr²h) where r= radius of the cone , h= height of the cone Volume of the sphere = 4/3(πr³) where r = radius of the sphere
### Math Notes Subjects #### Algebra Solutions ##### Topics \|\| Problems If an object is shot vertically from the surface of the earth with an initial velocity of $$v$$ feet per second, and if air resistance and other disturbing factors are neglected, it is found that $$s = vt - 0.5gt^2$$, where $$s$$ feet is the height of the object above the surface at the end of $$t$$ seconds and $$g = 32$$, approximately. (a) Solve for $$t$$ in terms of $$s$$ (b) If $$v=300 \text{fps}$$, use (a) to find where $$s=450$$ and $$s=0$$. Solution: (a) $$s = vt -0.5gt^2$$ $$2s =2vt -gt^2$$ $$gt^2 -2vt = -2s$$ $$t^2 - \frac{2v}{g}t = \frac{-2s}{g}$$ $$t^2 -\frac{2v}{g}t + (\frac{v}{g})^2 = \frac{-2s}{g} + (\frac{v}{g})^2$$ $$(t-\frac{v}{g})^2 = \frac{v^2 - 2sg}{g^2}$$ $$t - \frac{v}{g} = \frac{\sqrt{v^2-2sg}}{g}$$ $$t = \frac{v\pm\sqrt{v^2-2sg}}{g}$$ (b) When $$s = 450$$ $$t = \frac{300 + \sqrt{300^2-2(450)(32)}}{32}$$ $$t = 17.11 \text{sec}$$ $$t = \frac{300 - \sqrt{300^2-2(450)(32)}}{32}$$ $$t = 1.64 \text{sec}$$ When $$s = 0$$ $$t = \frac{300 + \sqrt{300^2-2(0)(32)}}{32}$$ $$t = 18.75 \text{sec}$$ $$t = \frac{300 - \sqrt{300^2-2(0)(32)}}{32}$$ $$t = 0 \text{sec}$$
# Lesson Video: Equation of a Straight Line: Vector Form Mathematics In this video, we will learn how to find the equation of a straight line in vector form. 18:28 ### Video Transcript In this video, we will learn how to find the equation of a straight line in vector form. There are many different ways that we can write the equation of a straight line in the π₯π¦-plane. Letβs recall a few of these. Firstly, in slope intercept form, we have π¦ equals ππ₯ plus π, where π is the slope or gradient and π is the π¦-intercept. In point slope form, we have π¦ minus π¦ sub zero is equal to π multiplied by π₯ minus π₯ sub zero, where once again π is the slope and the line passes through the point π₯ sub zero, π¦ sub zero. Thirdly, we have the standard form, which is π΄π₯ plus π΅π¦ equals πΆ, where π΄, π΅, and πΆ are integers and π΄ is nonnegative. It is also important to note that only one of π΄ and π΅ can be zero. All three of these methods have advantages and disadvantages. As both slope intercept and point slope form do not allow for vertical lines, weβll use the vector notation for a straight line to overcome this. We can find the position vector of any point on a straight line by using a known point π with coordinates π₯ sub zero, π¦ sub zero on the line that has a position vector π« sub zero together with any nonzero vector π that is parallel to the line. Vector π is called the direction vector of the line. This leads us to the following definition. The position vector π« of any point on a line containing the point π with position vector π« sub zero is given by π« is equal to π« sub zero plus π‘ multiplied by π, where π is the direction vector of the line and π‘ is any scalar value. This is because if the point π lies on the line, which is parallel to the nonzero vector π, then we can find the position vector of any point on the line by adding a scalar multiple of π to the position vector of π. Letβs consider the line with direction vector one, two that passes through the point with coordinates one, one such that π« sub zero is equal to one, one. The vector equation of this line will therefore be equal to one, one plus π‘ multiplied by one, two. We can represent this on the coordinate plane. We know that the line passes through the point with coordinates one, one. The direction vector is one, two. And we recall that the first component tells us the horizontal displacement and the second component, the vertical displacement. This means that the vector one, two represents movement one unit to the right and two units upwards. We move along the line from the point one, one in scalar multiples of the direction vector one, two. If the scalar multiple was two, then our direction vector becomes two, four. In fact, any nonzero vector parallel to the line is an equivalent direction vector of the line since weβre allowed to take any scalar multiples of the direction vector. We will now consider how we can find the π₯- and π¦-intercepts using the vector form of the equation of a line. To find the π₯- and π¦-intercepts, we set each component of the vector equation equal to zero and then solve for π‘. Recalling the vector equation of a line that we saw earlier, we know that this vector will have π₯- and π¦-components. When the line intersects the π¦-axis, we know that π₯ is equal to zero. Simplifying the right-hand side of our equation, we have the components one plus π‘ and one plus two π‘. We can then equate the π₯-components. And solving this gives us π‘ is equal to negative one. As π¦ is equal to one plus two π‘, π¦ is also equal to negative one. We can therefore conclude that the π¦-intercept occurs at π¦ equals negative one. Using the same method and setting the π¦-component equal to zero, we see that the line intersects the π₯-axis at π₯ equals one-half. We noted earlier that there is one big advantage for using the vector equation of a line; that is, we can find the vector equation of any line, including vertical ones. Before moving on to some specific examples, letβs consider this. The line π₯ equals negative one passes through the point negative one, zero and its direction is vertical. This means that it has no horizontal direction at all. This means that the first component of the direction vector will be equal to zero. Using the vector form of the equation of a line, we see that π« is equal to negative one, zero plus π‘ multiplied by zero, one. This can be seen on the π₯π¦-plane as shown. We have a position vector negative one, zero and a direction vector zero, one. Letβs now look at some specific questions involving the vector equation of a straight line. Write the vector equation of the straight line that passes through the point six, negative nine with direction vector nine, negative two. Is it (A) π« is equal to nine, negative two plus π multiplied by six, negative nine? (B) π« is equal to six, negative nine plus π multiplied by nine, negative two. Is it option (C) π is equal to six, negative nine plus vector π« multiplied by nine, negative two? Or option (D) π is equal to nine, negative two plus vector π« multiplied by six, negative nine. We begin by recalling that the vector equation of a straight line is written in the form π« is equal to π« sub zero plus π‘ multiplied by π, where π« sub zero is a position vector of a point on the line and π is the direction vector of the line. In this question, we are given both of these. As the line passes through the point six, negative nine, π« sub zero is equal to six, negative nine. We are also told that the direction vector is nine, negative two. Therefore, π« is equal to six, negative nine plus π‘ multiplied by nine, negative two. Noting that the value of π‘ can be any scalar quantity, to match the options, we will let this equal π. The correct answer is therefore option (B). Before moving on to our next example, we will consider how we can write the equation of a line given its slope. We know that the slope or gradient of any line is equal to the change in π¦ over the change in π₯. Letβs begin by considering the line drawn with slope negative eight over three. This means that for every three units we move horizontally to the right, we need to move eight units vertically down. This can also be written as the vector three, negative eight. It is important to note that this is equivalent to the direction vector negative three, eight. This demonstrates a useful property for finding the direction vector when given the slope of a line. If a line has slope π, then the line will have direction vector one, π. When the slope is equal to a fraction π over π as in this case, the line will have direction vector π, π. We will now look at an example where we will use this information. Find the vector equation of the straight line passing through the points six, negative seven and negative four, six. Is it (A) π« is equal to six, negative seven plus π multiplied by 10, negative 13? (B) π« is equal to negative four, six plus π multiplied by negative 13, 10. (C) π« is equal to six, negative four plus π multiplied by negative seven, six. Or (D) π« is equal to negative four, six plus π multiplied by 10, 13. We begin by recalling that the vector equation of a straight line is written in the form π« is equal to π« sub zero plus π multiplied by π, where π« sub zero is the position vector of any point that lies on the line, π is the direction vector of the line, and π is any scalar. We are given the coordinates of two points, six, negative seven and negative four, six, that both lie on the line. And we can use either of these as the position vector to help find the vector equation of the line. In option (A), the position vector is six, negative seven. And in options (B) and (D), the position vector is negative four, six. Letβs begin by letting the position vector π« sub zero be six, negative seven. We now need to find the direction vector given the two points that lie on the line. We recall that the slope of any line is equal to the change in π¦ over the change in π₯. Substituting in the points given, we see that the slope is equal to negative seven minus six over six minus negative four. This is equal to negative 13 over 10. Recalling that any line with slope π equal to π over π has direction vector π, π, we see that the direction vector here is equal to 10, negative 13. As this is one possible direction vector of our straight line, we have π« is equal to six, negative seven plus π multiplied by 10 negative 13. We noticed that this corresponds to option (A), proving that this is the vector equation of the straight line passing through the points six, negative seven and negative four, six. Looking at the other options, we recall that options (B) and (D) do indeed pass through the point negative four, six. However, they do not have a direction vector which is equal to or parallel to 10, negative 13. We can therefore rule out options (B) and (D). The direction vector of option (C) is negative seven, six. And this too is not parallel to the direction vector 10, negative 13. At this point, it is worth recalling that the vector equation of a straight line is not unique. Other possible vector equations of the straight line from the information given are π« is equal to negative four, six plus π multiplied by 10, negative 13; π« is equal to six, negative seven plus π multiplied by negative 10, 13; and π« is equal to negative four, six plus π multiplied by negative 10, 13. The direction vectors in the last two options move in the opposite direction. Any of these four solutions are valid. However, the only one that was given as one of the options is π« is equal to six, negative seven plus π multiplied by 10, negative 13. This leads us to a useful result about finding the direction vector of a line when we are given two distinct points on the line. Given two distinct points π΄ and π΅ with coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two, the vector equation of the line that passes through them is equal to π₯ sub one, π¦ sub one plus π multiplied by π₯ sub two minus π₯ sub one, π¦ sub two minus π¦ sub one. We can also use this information to determine whether three points in the π₯π¦-plane are collinear. We will see an example of this in our final question. Using the vector form of the equation of a straight line, identify whether the points negative seven, five; negative one, two; and five, negative one are collinear. We recall that a set of points are said to be collinear if all of the points lie on the same straight line. There are several ways of checking this. One way is to begin by finding the equation between a pair of points and then checking whether the third point satisfies this. We will begin by finding the vector equation of the line which passes through negative seven, five and negative one, two. We recall that this line has equation π« is equal to π₯ sub one, π¦ sub one plus π multiplied by π₯ sub two minus π₯ sub one, π¦ sub two minus π¦ sub one. Substituting in the given values, the right-hand side becomes negative seven, five plus π multiplied by negative one minus negative seven, two minus five. This, in turn, simplifies to negative seven, five plus π multiplied by six, negative three. Grouping the corresponding components on the right-hand side, we have π« is equal to negative seven plus six π, five minus three π. If the three points are collinear, our third point five, negative one will lie on this line. We can check this by substituting its position vector five, negative one into the vector equation of the line. Equating the components, we have five is equal to negative seven plus six π and negative one is equal to five minus three π. We can solve the first equation by adding seven to both sides and then dividing through by six. This gives us π is equal to two. The second equation also gives us a solution of π equals two. As this value of π is the same, we can therefore conclude that the point five, negative one lies on the line. The correct answer is therefore yes, the three points are collinear. This can also be shown graphically on the π₯π¦-plane. We will now finish this video by summarizing the key points. The vector equation of a line can be written in the form π« is equal to π« sub zero plus π‘ multiplied by π, where π« sub zero is the position vector of any point that lies on the line, π is the direction vector of the line, and π‘ is any scalar. If we have two distinct points π΄ and π΅ that lie on the line, we can find its direction vector as shown, where the coordinates of the two points are π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two. We also saw in this video that a line of slope π has direction vector one, π. And if this value of π is equal to the fraction π over π, then the direction vector can be written π, π. We also saw that the vector equation of a line is not unique as we can choose any point that lies on the line as the position vector and any nonzero vector parallel to the line as the direction vector. Finally, we saw that any three or more points are collinear if the direction vectors between each pair of points are equivalent.
# How many times is 3 75? The number that multiplies by 3 to give 75 is 25. To answer the question of what times 3 equals 75, we need to find a number x, such that the See full answer below. ## What multiplication answer is 75? 75 = 1 x 75, 3 x 25, or 5 x 15. ## What times 15 gives you 75? Thus, the answer to “15 times what equals 75?” is 5. To double-check our work, multiply 5 by 15 to see that it equals 75. (Note that our answer on this page is rounded to 4 digits if necessary.) ## How do you work out 15 x 15? How many times is 3 75? – Related Questions ## How do you work out 12 x 18? Here is step-by-step how to calculate 18 times 12 using long multiplication: 1. Step 1: Setup. x. 2. Step 2: Multiply. 8 x 2 = 16. 3. Step 3: Multiply. 2 x 1 = 2. 4. Step 4: Add 0. x. 5. Step 5: Multiply. 1 x 8 = 8. 6. Step 6: Multiply. 1 x 1 = 1. ## How do you calculate 15 tax from total? How to Calculate Sales Tax 1. Find list price and tax percentage. 2. Divide tax percentage by 100 to get tax rate as a decimal. 3. Multiply list price by decimal tax rate to get tax amount. 4. Add tax amount to list price to get total price. ## How do you work out 15 x 14? Here is step-by-step how to calculate 15 times 14 using long multiplication: 1. Step 1: Setup. x. 2. Step 2: Multiply. 5 x 4 = 20. 3. Step 3: Multiply. 4 x 1 = 4. 4. Step 4: Add 0. x. 5. Step 5: Multiply. 1 x 5 = 5. 6. Step 6: Multiply. 1 x 1 = 1. ## How do you get 30% of 100? Answer: 30% of 100 is 30. ## What score is 8 out of 15? First, you need to calculate your grade in percentages. The total answers count 15 – it’s 100%, so we to get a 1% value, divide 15 by 100 to get 0.15. Next, calculate the percentage of 8: divide 8 by 1% value (0.15), and you get 53.33% – it’s your percentage grade. ## How do you calculate 20 percent off 100? How do I take 20 % off a price? 1. Take the original price. 2. Divide the original price by 5. 3. Alternatively, divide the original price by 100 and multiply it by 20. 4. Subtract this new number from the original one. 5. The number you calculated is the discounted value. ## What is 20% of a \$1000? Solution. After multiplying, you see that 20% of 1,000 is 200. Thus, 20% of a \$1,000 bill is \$200. ## What number is 20% out of 200? Answer: 20% of 200 is 40. Let’s find 20% of 200. ## How can I find 20% of 120? Answer: 20% of 120 is 24. ## What is 18 out of 20 on a quiz? First, you need to calculate your grade in percentages. The total answers count 20 – it’s 100%, so we to get a 1% value, divide 20 by 100 to get 0.20. Next, calculate the percentage of 18: divide 18 by 1% value (0.20), and you get 90.00% – it’s your percentage grade. ## What score is 120 out of 200? The number 120 is 60% of the number 200. To find our percentage, we first divide 120 by 200 as a fraction. ## How do you get 30% of 90? Answer: 30% of 90 is 27. ## What grade is 20 right out of 30? For the USA grading system, a value of 66.67% corresponds to the letter mark D.
Education.com Try Brainzy Try Plus # Derivatives Study Guide (not rated) By Updated on Oct 1, 2011 ## Derivatives Straight lines may be ideal to human beings, but most functions have curved graphs. This does not stop us from projecting straight lines on them! For example, at the point marked x on the graph in Figure 6.1, the function is clearly increasing. However, exactly how fast is the function increasing at that point? Since "how fast" refers to a slope, we draw in the tangent line, the line straight through the point that heads in the same direction as the curve (see Figure 6.2). The slope of the tangent line tells us how fast the function is increasing at the given point. We can figure out the y-value of this point by plugging x into f and getting (x, f(x)). However, we can't get the slope of the tangent line when we have just one point. To get a second point, we go ahead a little further along the graph (see Figure 6.3). If we go ahead by distance a, the second point will have an x-value of x + a and a y-value of f(x + a). Because this second point is on the curve and not on the tangent line, we get a line that is not quite the tangent line. Still, its slope will be close to the one we want, so we calculate as follows: To make things more accurate, we pick a second point that is closer to the first one by using a smaller a. This is depicted in Figure 6.4. In fact, if we take the limit as a goes to zero, we will get the slope of the tangent line exactly. This is called the derivative of f(x) and is written f'(x). #### Example 1 What is the derivative of f(x) = x2? #### Solution 1 Use f(x) = x2. Multiply out and simplify. Factor and simplify. Plug in for the limit. The derivative is f'(x) = 2x. This means that the slope at any point on the curve y = x2 is exactly twice the x-coordinate. The situation at x = –2, x = 0, and x = 1 is shown in Figure 6.5. #### Example 2 What is the slope of the line tangent to g(x) =√x at x = 9? #### Solution 2 Use g(x) =√x. Rationalize the numerator. Multiply and simplify. Simplify. Plug in to evaluate the limit. The derivative of g(x) =√x is thus .This means that at x = 9, the slope of the tangent line is . This is illustrated in Figure 6.6. #### Example 3 Find the equation of the tangent line to h(x) = 2x2 – 5x + 1 at x = 3. #### Solution 3 To find the equation of the tangent line, we need a point and a slope. The y-value at x = 3 is h(3) = 2(3)2 – 5(3) + 1 = 4, so the point is (3,4). And to get the slope, we need the derivative. Start with the definition of the derivative. Use h(x) = 2x2 – 5x + 1. Multiply out and simplify. Factor out and simplify. Evaluate the limit. Thus, the derivative of h(x) = 2x2 – 5x + 1 is h'(x) = 4x – 5. The slope at x = 3 is h'(3) = 4(3) – 5 = 7. The equation of the tangent line is therefore y = 7(x – 3) + 4 = 7x – 17. This is shown in Figure 6.7. Find practice problems and solutions for these concepts at Derivatives Practice Questions 150 Characters allowed ### Related Questions #### Q: See More Questions ### Today on Education.com Top Worksheet Slideshows
Courses Courses for Kids Free study material Offline Centres More Store # The V - I graph for a conductor at temperature ${T_1}$ and ${T_2}$ are as shown in the figure. The term $({T_2} - {T_1})$ is proportional to:A) $\dfrac{{\sin 2\theta }}{{{{\sin }^2}\theta }}$B) $\dfrac{{\cot 2\theta }}{{{{\sin }^2}\theta }}$C) $\dfrac{{\cos 2\theta }}{{\sin 2\theta }}$D) $\dfrac{{\tan 2\theta }}{{{{\sin }^2}\theta }}$ Last updated date: 13th Sep 2024 Total views: 79.5k Views today: 0.79k Verified 79.5k+ views Hint: Recall that the slope of $V - I$ graph gives $\operatorname{R}$ . Revise the trigonometric formulae for the ease of calculations in the question. Also, we must know how the resistance varies with respect to temperature. Complete step by step solution: Here we are given a $V - I$ graph. By Ohm’s law, we know that: $V \propto I$ $\Rightarrow V = IR$ $\Rightarrow \dfrac{V}{I} = R$ $\Rightarrow$ The slope of $V - I$ graph gives $R$ $\therefore$ Consider $\dfrac{{{V_1}}}{{{I_1}}} = {R_1}$ resistance at temperature ${T_1}$ and $\dfrac{{{V_2}}}{{{I_2}}} = {R_2}$ resistance at temperature ${T_2}$ . self-made diagram But, we also know that $\tan \theta = \dfrac{{opposite}}{{adjacent}}$ $\Rightarrow \tan \theta = \dfrac{{{V_1}}}{{{I_1}}}$ $\Rightarrow {R_1} = \tan \theta$ And ${R_2} = \tan ({90^ \circ } - \theta )$ $\Rightarrow {R_2} = \cot \theta$ The resistance of a conductor always depends on the temperature. As the temperature increases the resistance of the conductor also increases. For small temperatures, the resistance of the conductor increases linearly with temperature, which is given by the equation: $R = {R_o}(1 + \alpha T)$ Where $R$ is resistance at temperature $T$ in $Ohms(\Omega )$ ${R_o}$ is resistance at absolute temperature in $\Omega$ $T$ is temperature in $Kelvin(K)$ $\alpha$ is temperature coefficient of resistance $\therefore {R_1} = {R_o}(1 + \alpha {T_1})$ and ${R_2} = {R_o}(1 + \alpha {T_2})$ Now, ${R_2} - {R_1} = {R_o}[1 + \alpha ({T_2} - {T_1})]$ But ${R_2} = \cot \theta$ and ${R_1} = \tan \theta$ Substituting these values in the above equation, we get $\cot \theta - \tan \theta = {R_o}[1 + \alpha ({T_2} - {T_1})]$ $\Rightarrow {T_2} - {T_1} \propto \cot \theta - \tan \theta$ $equation(1)$ Now, we need to simply the equation $\cot \theta - \tan \theta$ $\cot \theta - \tan \theta = \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}$ $\Rightarrow \cot \theta - \tan \theta = \dfrac{{\cos \theta \times \cos \theta - \sin \theta \times \sin \theta }}{{\sin \theta \times \cos \theta }}$ $\Rightarrow \cot \theta - \tan \theta = \dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{\sin \theta \cos \theta }}$ Now, we can substitute ${\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta$ $\Rightarrow \cot \theta - \tan \theta = \dfrac{{\cos 2\theta }}{{\sin \theta \cos \theta }}$ Now multiplying the numerator and denominator by $2$ , we get: $\Rightarrow \cot \theta - \tan \theta = \dfrac{{2\cos 2\theta }}{{2\sin \theta \cos \theta }}$ We know that $2\sin \theta \cos \theta = \sin 2\theta$ $\therefore \cot \theta - \tan \theta = \dfrac{{2\cos 2\theta }}{{\sin 2\theta }}$ Substituting this value in $equation(1)$ $\Rightarrow {T_2} - {T_1} \propto \dfrac{{2\cos 2\theta }}{{\sin 2\theta }}$ $\Rightarrow {T_2} - {T_1} \propto \dfrac{{\cos 2\theta }}{{\sin 2\theta }}$ ($\because 2$ is an integer we can ignore it) $\therefore$ Option $(C), \dfrac{{\cos 2\theta }}{{\sin 2\theta }}$ is the correct option. Note: One must know that the resistance of the conductor for small temperatures increases with increase in temperature. One is very likely to forget the trigonometric formulas. Do not confuse or make mistakes in the trigonometric formulas of $\cos 2\theta,\sin 2\theta$.
Categories ## Largest Area of Triangle \| AIME I, 1992 \| Question 13 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Largest Area of Triangle. ## Area of Triangle – AIME I, 1992 Triangle ABC has AB=9 and BC:AC=40:41, find the largest area that this triangle can have. • is 107 • is 820 • is 840 • cannot be determined from the given information ### Key Concepts Ratio Area Triangle AIME I, 1992, Question 13 Coordinate Geometry by Loney ## Try with Hints First hint Let the three sides be 9, 40x, 41x area = $\frac{1}{4}\sqrt{(81^2-81x^2)(81x^2-1)} \leq \frac{1}{4}\frac{81^2-1}{2}$ Second Hint or, $\frac{1}{4}\frac{81^2-1}{2}=\frac{1}{8}(81-1)(81+1)$ Final Step =(10)(82) =820. Categories ## Problem on Ratio \| PRMO 2017 \| Question 12 Try this beautiful problem from the Pre-RMO, 2017 based on ratio. ## Problem on Ratio – PRMO 2017 In a sports team, the total number of boys and girls are in the ratio 4:3. On one day it was found that 8 boys and 14 girls were absent from the team and that the number of boys was the square of the number of girls, find the total number of members in the sports team. • is 107 • is 42 • is 840 • cannot be determined from the given information ### Key Concepts Ratio Integers Algebra PRMO, 2017, Question 12 Elementary Algebra by Hall and Knight ## Try with Hints First hint here ratio = 4:3 $\Rightarrow boys =4x, girls=3x$ given $(4x-8)=(3x-14)^{2}$ Second Hint $\Rightarrow 4x-8=9x^{2}-84x+196$ $\Rightarrow 9x^{2}+196-84x=4x-8$ $\Rightarrow 9x^{2}-88x+204=0$ Final Step $x=\frac{88\pm\sqrt{88^{2}-4(9)(204)}}{18}$ $\Rightarrow x=6 or \frac{34}{9}$ $\Rightarrow 7x=42$ or non-integer $\Rightarrow 42 members.$ Categories ## What is Ratio ? In mathematics, a ratio indicates how many times one number contains another. For example, if there are eight oranges and six lemons in a bowl of fruit, then the ratio of oranges to lemons is eight to six which is equivalent to 4:3. ## Try this Problem from AMC 10B – 2020 – Problem No.-3 The ratio of w to x, y to z and z to x are 4:3, 3:2 and 1:6 respectively . What is the ratio of w to y ? A) 4:3 B) 3:2 C) 8:3 D) 4:1 E) 16 :3 American Mathematics Competition 10 (AMC 10B), 2020, Problem Number 3 Ratio 3 out of 10 Mathematics can be fun ## Use some hints This one is an easy sum to solve but those who are confused right now can use the first hint : The ratio of w : x = 4 : 3 ; so we can write it $\frac {w}{x} = \frac {4}{3}$. Similarly we can do it for the other given ratios. Try to do it…… I think you have already got the answer . If not then try this out ….. z : x = 1 : 6 i.e $\frac {z}{x} = \frac {1}{6}$ y : z = 3 : 2 i.e $\frac {y}{z} = \frac {3}{2}$ This hint is the final hint as already mentioned in header : Lets multiply each ratios : $\frac {w}{x} \times \frac {x}{z} \times \frac {z}{y} = \frac {4}{3} \times 6 \times \frac {2}{3}$ After canceling out all the similar terms $\frac {w}{y} = \frac {16}{3}$ Categories ## Ratio and Proportion The given problem is based upon calculating the number of marbles in jars of specific color, to do so we have to use ratios of the marbles of different colors and use the ratio to calculate the actual number of marbles of required color. ## Try the problem Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$, and the ratio of blue to green marbles in Jar $2$ is $8:1$. There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$? $\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$ 2019 AMC 10B Problem 11 Ratio and proportion 6 out of 10 Secrets in Inequalities. ## Use some hints Let $2x$ is the total no of marbles in both the jars. so each of the jar have $x$ marbles. Thus, $\frac{x}{10}$ is the number of green marbles in Jar $1$, and $\frac{x}{9}$ is the number of green marbles in Jar $2$. Since $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$, we have $\frac{19x}{90}=95$, so there are $x=450$ marbles in each jar. Since $\frac{9}{10}th$ of the jar 1 marbles are of blue color and $\frac{8}{9}th$ of the jar 2 marbles are of blue color. Now we can easily find the no of blue marbles in both the jars and then we can subtract them to get the amount by which one exceed the other.
# How do you graph f(x)=3^(x - 2)? May 4, 2018 $\text{ }$ #### Explanation: $\text{ }$ Given the exponential function: color(red)(f(x)=3^(x-2) Before graphing this function, create a data table: The table should contain values for color(red)(x, corresponding values for color(red)(y=3^x) and color(red)(y=3^(x-2) We include the base function: color(red)(y=3^x, since it provides an opportunity to examine the behavior of both the graphs by comparing them. The table shows $\textcolor{red}{x}$ and the corresponding color(red)(y values: Construct both the graphs: $\textcolor{g r e e n}{\text{Graph 1:}}$ color(blue)("Graph of "y = f(x) = 3^x Domain : $\left(- \infty , \infty\right)$ Range : $\left(0 , \infty\right)$ y-intercept : $\left(0 , 1\right)$ Horizontal Asymptote :$y = 0$ $\textcolor{g r e e n}{\text{Graph 2:}}$ color(blue)("Graph of "y = f(x) = 3^(x-2 Domain : $\left(- \infty , \infty\right)$ Range : $\left(0 , \infty\right)$ y-intercept : $\left(0 , \frac{1}{0}\right)$ Horizontal Asymptote :$y = 0$ $\textcolor{g r e e n}{\text{Graph 3:}}$ color(blue)("Graph of "y = f(x) = 3^(x) and y = f(x) = 3^(x-2) Compare the behavior of both graphs: Translation is horizontal for $y = f \left(x\right) = {3}^{x - 2}$ by color(red)(2 units. Hope it helps.
Last time I focused on some basics about learning the number bonds (combinations) of 10 as well as adding 10 to any number. Today I want to show the benefits of making a 10 when adding numbers with sums greater than 10 (such as 8 + 5). Then I’ll show how to help students add up to apply that to addition and subtraction of larger numbers. I’ll model this using concrete and pictorial representations (which are both important before starting abstract forms). Using a 10 Frame: A ten frame is an excellent manipulative for students to experience ways to “Make a 10.” I am attaching a couple of videos I like to illustrate the point. • Model this process with your students using 2 ten frames. • Put 8 counters on one ten frame. (I love using 2-color counters.) • Put 5 counters (in another color) on the second ten frame. • Determine how many counters to move from one ten frame to the other to “make a 10.” In this example, I moved 2 to join the 8 to make a 10. That left 3 on the second ten frame. 10 + 3 = 13 (and 8 + 5 = 13). The example below shows the same problem, but this time move 5 from the first ten frame to the second ten frame to “make a 10.” That left 3 on the first ten frame. 3 + 10 = 13 (and 8 + 5 = 13). Continue reading Math Problem Solving Part 1: Join (aka Some and Some More) I have 1 penny in one pocket. I have 6 more pennies in another pocket. This is a Join or “some and some more” story structure. Many teachers I work with have asked for advice on problem solving in math. Students often don’t know how to approach them or know what operation to use. Should teachers help students focus on key words or not? What about the strategies such as CUBES, draw pictures, make a list, guess and check, work backwards, find a pattern? While all of those strategies definitely have their purpose, I find we often give kids so many steps to follow (underline this, circle this, highlight that, etc.) that they lose sight of what the problem is basically about. In this post, I will focus on two basic questions (who and what) and a simple graphic organizer that will help students think about (and visualize) the actions in a one-step Join story problem. KG and first grade students can act out these actions using story mats or ten frames. Late first through 5th grade can use a part-part-whole box. There are two FREE items offered. These are the types of problems I will focus on in the next few posts. 1. Join (also referred to as SSM – Some and Some More) 2. Separate (also referred to as SSWA – Some, Some Went Away) 3. Part-Part-Whole 4. Comparing 5. Equal groups JOIN problems have 3 versions: • a + b = ___ (The result is unknown.) • a + ____ = c (How the story changed is unknown / missing addend.) • ____ + b = c (The start is unknown / missing addend.) They can also be referred to as “Some and Some More” stories (SSM). This means, you have some and you get some more for a total amount. These present themselves as additive stories, but it doesn’t necessarily mean you have to add to solve them. The second and third version above are often referred to as missing addend problems. Continue reading Addition and Subtraction Part 5: Separate and Comparison Problem Solving Structures KG-4th This week I will focus on subtraction problem structures. There are two types: separate and compare. I suggest teaching these models separately. Also, some part-part-whole problems can be solved using subtraction. I will refer to the same terms as in addition: start, change, result. You can also use the same materials used with addition problems: part-part-whole templates, bar models, ten frames, two-color counters, number lines, and connecting cubes. The goal is for students to see that subtraction has different models (separate vs. comparison) and an inverse relationship with addition — we can compose as well as decompose those numbers. Knowledge of number bonds will support the addition / subtraction relationship. Here is the same freebie I offered last week you can download for your math files: Addition and Subtraction Story Structure Information The six color anchor charts shown below are also attached here free for your use: Subtraction structure anchor charts Separate: Result Unknown • Example: 10 – 4 = ____; There were 10 cookies on the plate. Dad ate 4 of them. How many are left on the plate? • Explanation: The problem starts with 10. It changes when 4 of the cookies are eaten. The result in this problem is the answer to the question (how many are left on the plate). • Teaching and practice suggestions: • Ask questions such as: Do we know the start? (Yes, it is 10.) Do we know what changed? (Yes, 4 cookies were eaten so we take those away.) How many cookies are left on the plate now? (Result is 6.) • Reinforce the number bonds of 10: What goes with 4 to make 10? (6) • Draw a picture to show the starting amount. Cross out the items to symbolize removal. • Show the problem in this order also: ____ = 10 – 4. Remember the equal sign means the same as — what is on the left matches the amount on the right of the equal sign. Addition and Subtraction Part 4: The Equal Sign and Join Problem Solving Structures KG-4th Ask your students this question: “What does the equal sign mean?” I venture many or most of them will say, “It’s where the answer goes,” or something to that effect. You will see them apply this understanding with the type of problem in which the answer blank comes after the equal sign such as in 5 + 6 = _____ or 14 – 9 = _____. This is the most common type of problem structure called Result Unknown. I was guilty of providing this type of equation structure a majority of the time because I didn’t know there actually were other structures, and I didn’t realize the importance of teaching students the other structures. So . . . here we go. (Be sure to look for freebies ahead.) Teach your students the equal sign means “the same as.” Think of addition and subtraction problems as a balance scale: what’s on one side must be the same as the other side — to balance it evenly. Let’s look at the equal sign’s role in solving the different addition / subtraction structures. I will be using the words start to represent how the problem starts, the word change to represent what happens to the starting amount (either added to or subtracted), and the result. In this post I will be addressing the join / addition models of Result Unknown, Change Unknown, Start Unknown and Part-Part-Whole. I will highlight the subtraction models (Separate and Comparison) next time. Knowing these types of structures strengthens the relationship between addition and subtraction. Three of these addition models use the term join because that is the action taken. There are some, then some more are joined together in an additive model. One addition model, however, is the part-part-whole model. In this type there is no joining involved – only showing that part of our objects have this attribute and the other part have a different attribute (such as color, type, size, opposites). Helpful materials to teach and practice these strategies are bar models, part whole templates, a balance scale, and ten frames using cubes and/or two color counters. Here are 2 free PDF attachments. First one: Addition and Subtraction Story Structure Information. Second one is a copy of each of the 4 anchor charts shown below: Join and part-part-whole story structure anchor charts Join: Result Unknown • Example: 5 + 4 = ____; A boy had 5 marbles and his friend gave him 4 more marbles. How many marbles does the boy have now? • Explanation: The boy started with 5 marbles. There was a change in the story because he got 4 more. The result in this problem is the action of adding the two together. • Teaching and practice suggestions: • Ask questions such as: Do we know the start? (Yes, it is 5.) Do we know what changed? (Yes, his friend gave him more, so we add 4.) What happens when we put these together?(Result is 9.) • Show the problem in this order also (with result blank first instead of last) : ____ = 5 + 4 • Common questions: How many now? How many in all? How many all together? What is the sum? Join: Change Unknown • Example: 5 + ____ = 9; A boy had 5 marbles. His friend gave him some more. Now he has 9 marbles. How many marbles did his friend give him? You could also call this a missing addend structure. • Explanation: This problem starts with 5. There is a change of getting some more marbles, but we don’t know how many yet. The result after the change is (9). It is very likely students will solve like this: 5 + 14 = 9. Why? Because they see the equal sign and a plus sign and perform that operation with the 2 numbers showing. Their answer of 14 is put in the blank because of their misunderstanding of the meaning of the equal sign. • Teaching and practice suggestions: • Ask questions such as: Do we know the start? (Yes, he started with 5 marbles.) Do we know what changed in the story? (Yes, his friend gave him some marbles, but we don’t know how many.) Do we know the result? (Yes, he ended up with 9 marbles.) So the mystery is “How many marbles did his friend give him?” • Count up from the start amount to the total amount. This will give you the change involved in the story. • Reinforce number bonds by asking, “What goes with 5 to make 9?” or “What is the difference between 5 and 9?” Join: Start Unknown • Example: ____ + 4 = 9; A boy had some marbles. His friend gave him 4 more. Now he has 9. How many did he have to start with? • Explanation: The boy had some marbles to start with, but the story does not tell us. Then there is a change in the story when his friend gives him some more (4). The result is that he has 9. • Teaching and practice suggestions: • Ask questions such as: Do we know how many marbles the boy started with? (No, the story didn’t tell us.) Do we know what changed in the story? (Yes, his friend have him 4 marbles.) Do we know the result? (Yes, he had 9 marbles at the end.) • Count up from 4 to 9. • Reinforce knowledge of number bonds by asking, “What goes with 4 to make 9?” Addition and Subtraction Part 3: Facts Strategies KG-3rd This is part three in a series of strategies regarding addition and subtraction strategies. This part will focus on a variety of strategies to help toward memorization of facts, meaning automatic computation. While children are learning their number bonds (building up to 5 in KG, to 10 in first grade, and to 20 in second grade), there are other facts which cross several number bonds that students can work towards. These strategies to build mental math automaticity are highlighted below. Get some freebies in the section on doubles / near doubles. Identity (or Zero) Property: • The value of the number does not change when zero is added or subtracted. • 3 + 0 = 3 • 9 – 0 = 9 Subtracting All: • The answer is always zero when you take away / subtract all. • 9 – 9 = 0 • 50 – 50 = 0 • Adding 1 results in the next number in the counting sequence. • Subtracting 1 means naming the number that comes right before it in the counting sequence. • With manipulatives, lay out an amount for student to count. Slide one more and see if he/she can name the amount without recounting. • Do the same as above, but take one away from the group to see if he/she can name the amount without recounting. • Show this concept using a number line. • 6 + 1 = 7; 26 + 1 = 27 • 7 – 1 = 6; 37 – 1 = 36 • After +1 or -1 strategies are in place, then go for +2 or -2 for automatic processing. Next-Door Neighbor Numbers: • If subtracting two sequential numbers (ie 7 subtract 6), the answer is always one because you are taking away almost all of the original amount. • Help students identify these types of problems: 8-7; 10-9; 98-97; 158-157 • Guide students to writing these types of problems. • Relate these to subtracting 1 problems. If 10-1 = 9; then 10 – 9 = 1. • Show on a number line. Addition and Subtraction Part 2: Part-Part-Whole Models KG-2nd In Part 1 I focused on a numerical fluency continuum, which defines the stages a child goes through to achieve number sense. After a child has a firm grasp of one-to-one correspondence, can count on, and understands concepts of more and less, he/she is ready to explore part-part-whole relationships which lead to the operations of addition and subtraction. That will be the focus of this post. Read on for free number bond activities and a free number bond assessment! One way to explore part-part-whole relationships is through various number bonds experiences. Number Bonds are pairs of numbers that combine to total the target or focus number. When students learn number bonds they are applying the commutative, identity, and zero properties. Do you notice from the chart below that there are 4 number bonds for the number 3; 5 number bonds for the number 4; 6 number bonds for the number 5, etc? And . . . half of the number bonds are actually just the commutative property in action, so there really aren’t as many combinations for each number to learn after all. • KG students should master number bonds to 5. • First graders should master number bonds to 10. • Second graders should master number bonds to 20.Teaching Methods for Number Bonds • Ideally, students should focus on the bonds for one number at a time, until mastery is achieved. In other words, if working on the number bond of 3, they would learn 0 and 3, 3 and 0, 1 and 2, 2 and 1 before trying to learn number bonds of 4. See the end of this post for assessment ideas. • Ten Frame cards: Use counters to show different ways to make the focus number. (See above example of 2 ways to show 6.) Shake and Spill games are also great for this: Using 2-color counters, shake and spill the number of counters matching your focus number. See how many spilled out red and how many spilled out yellow. Record results on a blank ten-frame template. Repeat 10 times. • Number Bond Bracelets: Use beads and chenille stems to form bracelets for each number 2-10. Slide beads apart to see different ways to make the focus number. • Reckenreck: Slide beads on the frame to show different combinations. • Part-Part-Whole Graphic Organizers: Here are two templates I like. Start with objects matching the focus number in the “whole” section. Then move “part” of them to one section and the rest to the other section. Rearrange to find different bonds for the same focus number. Start students with manipulalatives before moving to numbers. Or use numbers as a way for students to record their findings. Once students have a good concept of number bonds, these part-part-whole organizers are very helpful when doing addition and subtraction problems (including story problems) using these structures: Result Unknown, Change Unknown, and Start Unknown. Children should use manipulatives at first to “figure out” the story. • Here is an example of a change unknown story: “I have 5 pennies in one pocket and some more in my other pocket. I have 7 pennies all together. How many pennies in my other pocket?” To do this, put 5 counters in one “part” section. Count on from 5 to 7 by placing more counters in the second “part” section (2). Then move them all to the whole section to check that there are 7 all together. Students are determining “What goes with 5 to make 7?” 5 + ___ = 7 • Here is an example of a result unknown subtraction story: “Mom put 7 cookies on a plate. I ate 2 of them. How many cookies are still on the plate?” To do this start with the whole amount (7) in the large section. Then move the 2 that were eaten to a “part” section. Count how many are remaining in the “whole section” to find out how many are still on the plate? 7 – 2 = ____. • How are number bonds related to fact families? A fact family is one number bond shown with 2 addition and 2 subtraction statements. Ex: With number bonds 3 and 4 for the number 7, you can make 4 problems: 3 + 4 = 7; 4 + 3 = 7; 7-3 = 4; and 7-4=3. Addition and Subtraction Part 1: Numerical Fluency To be able to add and subtract, students normally pass through several phases as they build readiness for these operations with numbers. As teachers, we know oral counting does not necessarily indicate an understanding of numbers and sets, just like reciting the alphabet doesn’t necessarily mean a child can recognize letters and sounds. Read ahead for freebies in the Part-Part-Whole section. Numerical Fluency Continuum: There are 7 steps to numerical fluency. If a child gets stuck on any of these steps, it may very likely halt their progress. Hopefully children move through these by the end of 2nd grade, but many students beyond that level have a breakdown which is likely because they missed one of these stages. Can you determine which of these stages your students are in? 1. One-to-one correspondence: The ability to count objects so each object counted is matched with one number word. 2. Inclusion of set: Does a child realize that the last number counted names the number of objects in the set? A child counts 5 objects. When you ask how many, can they state “5.” If you mix them up after they just counted them, do they realize there are still 5? 3. Counting on: If a child counts 5 objects and the teacher then puts 2 more objects for the child to count, do they start all over or continue counting from 5? 5 . . . 6, 7. 4. Subitizing:Recognize an amount without physically counting (ie on dice, dot cards, fingers). 5. More Than / Less Than / Equal To: Can a child look at two sets of objects and tell whether the second set is more, less, or equal to the first set. Can a child build a second set with one more, one less, or equal to the first set? 6. Part / Part / Whole: Compose and decompose sets by looking at the whole and the parts that make up the whole. 7. Unitizing: The child is able to move from counting by ones to count by sets / groups: fives, tens, etc. Number Talks Part 3: Computational Strategies 3rd-5th grades This is the Part 3 of Number Talks. If you are just tuning in, please refer to NT Parts 1 and 2. As I mentioned before, conducting a Number Talk session with your students is a chance for them to explain different ways to solve the same problem. This is meant to highlight strategies which have already been taught. Click below to watch 2 videos of how to conduct a Number Talk session with intermediate students. You will see many strategies being used. Number Talk 3rd grade 90-59 = ____ Number Talk 5th grade 12 x 15 = ___ Addition and Subtraction Strategies: I like using the methods listed below before teaching the standard algorithm. This is because they build on a solid knowledge of place value (and number bonds 1-10). If your students are adding and subtracting using the standard algorithm and can’t adequately explain the meaning of the regrouping process in terms of place value, then try one of the following methods. In many cases, I will ask a student the meaning of the “1” that has been “carried” over in double-digit addition. About 85% of the time, the student cannot explain that the “1” represents a group of 10. When adding the tens’ column, they often forget they are adding groups of 10 and not single digits. So they get caught up in the steps and don’t always think about the magnitude of the number (which is part of number sense). You will notice teachers write the problems horizontally in order to elicit the most strategies possible. • Partial Sums • Place Value Decomposition • Expanded Notation • Compensation • Open Number Line (to add or subtract) Here are some possible Number Talk problems and solutions: Multiplication and Division Strategies: I like using these methods before teaching the standard algorithms. Again, they build a solid understanding of place value, the use of the distributive property, and how knowledge of doubling and halving increases the ability to compute problems mentally. Once these methods have been learned, then it is easy to explain the steps in the standard algorithm. • Area Model • Partial Products • Distributive Property • Doubling and Halving • Partial Quotients Here are some possible Number Talk problems and solutions: Number Talks Part 2: Strategies and decomposing with 1st-3rd grade For 1st -3rd grade students: Refer to “Number Talks Part I” (posted Nov. 12, 2016) for ways to conduct a Number Talk with KG and early 1st grade students (focusing on subitizing and number bonds). For students in 1st – 3rd grade, place extra emphasis on number bonds of 10. Write a problem on the board, easel, or chart tablet with students sitting nearby to allow for focused discussion. Have the following available for reference and support: ten frame, part-part-whole template, base ten manipulatives, and a 0-100 chart. Present addition and subtraction problems to assist with recall of the following strategies. If time allows, post another similar problem so students can relate previous strategy to new problem. Students show thumbs up when they have an answer in mind. The teacher checks with a few on their answer. Then he/she asks, “How did you solve this problem?” The teacher writes how each student solved the problem. Number Bonds are pairs of numbers that combine to total the target or focus number. When students learn number bonds they are applying the commutative, identity, and zero properties. PLUS, the information can be applied to both addition and subtraction problems. Number bonds of 10 are very critical to our place value system, and will enhance a student’s success with future addition and subtraction strategies such as use of an open number line. • KG students should master number bonds to 5. • First graders should master number bonds to 10. • Second graders should master number bonds to 20. Daily Math Meeting Part 1: Ways to Build Number Sense K-5 To build number sense, students need frequent exposure or review of concepts you have previously introduced. There are many ways to build number sense on an on-going, informal basis – especially when you can squeeze in 10-15 minutes daily: • During morning meeting time • During a Number Talks session • At the beginning of your math lesson • At the end of your math lesson • End of day closure time I have included several of my power point slides on this topic as a PDF file (daily-practice-to-build-number-sense-pdf). Continue reading
#### need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (v) Answer: $\frac{\pi }{5}$ Hint: The range of principal value of $\sec ^{-1}$ is $\left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]$ Given: $\sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)$ Explanation: First we solve $\sec \left(\frac{9 \pi}{5}\right)$ \begin{aligned} &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(2 \pi-\frac{\pi}{5}\right) \\ &\therefore[\sec (2 \pi-\theta)]=\sec \theta \end{aligned} \begin{aligned} &\sec \left(2 \pi-\frac{\pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \\ &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \end{aligned} By substituting these value in $\sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)$ we get, \begin{aligned} &\sec ^{-1}\left(\sec \frac{\pi}{5}\right) \\ &\therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \\ &\sec ^{-1}\left(\sec \frac{\pi}{5}\right)=\frac{\pi}{5} \end{aligned} Hence, $\sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)=\frac{\pi}{5}$
### Theory: In this section, let us discuss different types of matrices. 1. Row matrix A matrix is a row matrix when it is made up of just one row and '$$n$$' number of columns. Row matrices are also called row vectors. For a matrix $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & … & a_{1n} \end{bmatrix}$$, the order of the matrix is $$1$$ $$\times$$ $$n$$. Example: Let us look at a few row matrices. 1. $$A = \begin{bmatrix} \sqrt{2} & \frac{\sqrt 7}{2} & 12 \end{bmatrix}$$, where the order of the matrix is $$1$$ $$\times$$ $$3$$. 2. $$A = \begin{bmatrix} 7 & 14 & 21 & 28 & 35 & 42 & 49 \end{bmatrix}$$, where the order of the matrix is $$1$$ $$\times$$ $$7$$. 2. Column matrix A matrix is a column matrix when it is made up of 'm' number of rows and just one column. Column matrices are also called column vectors. For a matrix $$A = \begin{bmatrix} a_{11}\\ a_{21}\\ a_{31}\\ \vdots\\ a_{m1} \end{bmatrix}$$, the order of the matrix is $$m$$ $$\times$$ $$1$$. Example: Let us look at a few column matrices. 1. $$A = \begin{bmatrix} 2x\\ 3x\\ x \end{bmatrix}$$, where the order of the matrix is $$3$$ $$\times$$ $$1$$. 2. $$A = \begin{bmatrix} 4\\ 8\\ 12\\ 16\\ 20 \end{bmatrix}$$, where the order of the matrix is $$5$$ $$\times$$ $$1$$. 3. Square matrix A matrix is a square matrix when the number of rows equals the number of columns. In other words, $$m$$ $$=$$ $$n$$. The order of a square matrix is $$m$$. Example: Let us look at a few square matrices. 1. $$A = \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$$, where the order of the matrix is $$2$$ $$\times$$ $$2$$. 2. $$A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}$$, where the order of the matrix is $$3$$ $$\times$$ $$3$$. Leading diagonal of a square matrix: In a square matrix, the entries $$a_{ij}$$ where $$i$$ $$=$$ $$j$$ form the leading diagonal of a matrix. Example: Let us consider the matrix given below. $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ Here, the elements $$a_{11}$$, $$a_{22}$$ and $$a_{33}$$ form the elements of the leading diagonal. 4. Diagonal matrix In a square matrix, when all the entries except the leading diagonal is zero, then it is called a diagonal matrix. In other words, $$a_{ij}$$ $$=$$ $$0$$ for $$i$$ $$\neq$$ $$j$$. Example: Let us look at a few diagonal matrices. 1. $$A = \begin{bmatrix} 1 & 0\\ 0 & 4 \end{bmatrix}$$ 2. $$A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 9 \end{bmatrix}$$ 5. Scalar matrix In a square matrix, when all the elements of the leading diagonal is the same, then it is a scalar matrix The general representation of a scalar matrix $$A$$ $$=$$ $$(a_{ij})_{m \times n}$$ is $$a_{ij} = \begin{cases} 0 & \text{ when } i \neq j \\ k & \text{ when } i = j \end{cases}$$, where $$k$$ is a constant. Example: Let us look at a few scalar matrices. 1. $$A = \begin{bmatrix} 7 & 0 & 0\\ 0 & 7 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ 2. $$A = \begin{bmatrix} \sqrt{3} & 0 & 0\\ 0 & \sqrt{3} & 0\\ 0 & 0 & \sqrt{3} \end{bmatrix}$$ 6. Identity or unit matrix In a square matrix, when all the leading diagonal elements are $$1$$, then it is an identity matrix or a unit matrix. The general representation of a unit matrix $$A$$ $$=$$ $$(a_{ij})_{m \times n}$$ is $$a_{ij} = \begin{cases} 0 & \text{ when } i \neq j \\ 1 & \text{ when } i = j \end{cases}$$. Example: Let us look at a few identity matrices. 1. $$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ 2. $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ 7. Zero matrix or null matrix A matrix is a zero matrix or a null matrix when all the matrix elements are zero. Example: Let us look at a few null matrices. 1. $$\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$$ 2. $$\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$$ 8. Transpose of a matrix The transpose of a matrix is obtained by interchanging the elements in the rows and columns. The transpose of a matrix is denoted by $$A^T$$, and $$A^T$$ is read as '$$A$$ $$\text{transpose}$$'. If the order of matrix $$A$$ is $$m \times n$$, then the order of $$A^T$$ is $$n \times m$$. Example: Let us now look at a few examples. 1. If $$A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}$$, then $$A^T = \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix}$$. 2. If $$B = \begin{bmatrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12 \end{bmatrix}$$, then $$B^T = \begin{bmatrix} 1 & 5 & 9\\ 2 & 6 & 10\\ 3 & 7 & 11\\ 4 & 8 & 12 \end{bmatrix}$$. 9. Triangular matrix If all the entries above the leading diagonal are zero in a square matrix, then it is a lower triangular matrix. If all the entries below the leading diagonal are zero, then it is an upper triangular matrix. A matrix $$A$$ is a upper triangular matrix if $$a_{ij}$$ $$=$$ $$0$$ for $$i$$ $$>$$ $$j$$. Similarly, matrix $$A$$ is a lower triangular matrix if $$a_{ij}$$ $$=$$ $$0$$ for $$i$$ $$<$$ $$j$$. Example: Let us look at a few examples. Upper triangular matrices: $$A = \begin{bmatrix} 1 & 0 & 0\\ 2 & 3 & 0\\ 4 & 5 & 6 \end{bmatrix}$$, $$B = \begin{bmatrix} 7 & 0 & 0\\ 8 & 9 & 0\\ 10 & 11 & 12 \end{bmatrix}$$ Lower triangular matrices: $$C = \begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \end{bmatrix}$$, $$D = \begin{bmatrix} 7 & 8 & 9\\ 0 & 10 & 11\\ 0 & 0 & 12 \end{bmatrix}$$
# Algebra and General Mathematics: Set Theory Review notes in set theory topic in Algebra and General Mathematics. This will help you prepare in taking the Engineering Board Exam or simply reviewing for your Algebra subject. Set theory is the branch of mathematical logic that studies sets, which informally are collections of objects. The modern study of set theory was initiated by Georg Cantor and Richard Dedekind in the 1870s. A set is any well defined collection of symbols or objects. The objects comprising the set are called elements or members. ### REPRESENTATION OF A SET Tabular Form – all the elements of the set are written in braces ( ), each one of them are separated by a comma. Example: If set A is a set of counting numbers, then: A = { 1, 2, 3, 4, 5, 6, …} Set Builder Form – the elements are not written explicitly but their characteristic properties are stated and written within braces. Example: If set A is a set of natural numbers less than 12, then: A = { x \| x is a natural number < 12 } where: “\|” is read as “such that” ### KINDS OF SETS Finite Set – a set containing a finite number of elements. Example: A = { x \| x is a natural number ≤ 6 } or A = { 1, 2, 3, 4, 5, 6 } Infinite Set – a set containing an infinite number of elements. Example: A = { x \| x is a natural number } or A = { 1, 2, 3, 4, … } Singleton – a set having only one element (unit set). Example: A = { x \| x is a natural number > 5 but < 7 } or A = { 6 } Null Set – a set having no element. Also called empty set or vid. Example: A = { x \| x is a natural number between 2 and 3 } or A = Ï† Subset – a set “A” is a subset of another set B, if every element of the set “A” belongs to the set B, denoted by A ⊆ B which is read as “ a is a subset of B” or “A is contained in B”. Properties of Subset: • 1. Every set is a subset of itself : A ⊆ A • 2. Null set is a subset of every set : Ï† ⊆ A Proper Subset – a set “A” is said to be a proper subset of a set B if A is a subset of B but is not a subset of A, denoted by A ⊂ B. Example: A = { x \| x is an odd integer }, B = { x \| x is an integer } Then, A ⊂ B Universal Set – a set U is called a universal set, if all the sets under discussion are subsets of the set U. Example: A = { a, c, e, g, i, k } B = {b, d, f, h, j, l, } C = { a, b, c, d, e, f, g, h, I, j, k, l, … } Since, set A and B are subset of these set C, the set C is the Universal Set. Equal Sets – two sets A and B are said to be equal if: A ⊆ B and B ⊆ A Example: A = { 1, 3, 5, 7, 9 } and B = { 1, 3, 5, 7, 9 } Then, A = B Equivalent Sets – two sets A and B having equal number of distinct elements are said to be equivalent sets. Example: A = { 2, 3, 4, 5, 6, 7 } B = {a, b, c, d, e, f } Since, both sets have the same number of distinct elements, the two sets are said to be equivalent sets. Notes to Remember: • Equal sets are always equivalent sets but not all equivalent sets are equals • Infinite sets are always equivalent • Two sets are equivalent if their Cardinal Number or Order is the same. ### CARDINAL NUMBER OF A SET The number of distinct elements in a set A is called Cardinal Number or its Order and is denoted by n(A). Example: A = { 1, 3, 5, 7, 9, 11, 13 } n(A) = 7 ; since there are 7 distinct elements ### SET OPERATIONS Union – the union of sets A and B, denoted by A ⋃ B, is the set of elements which belong to A or B or both A and B. It is one of the fundamental operations through which sets can be combined and related to each other. Example: A = { x is an even integer larger than 1 } B = { x is an odd integer larger than 1 } A ⋃ B = { 2, 3, 4, 5, 6, … } Sets cannot have duplicate elements, so the union of the sets { 1, 2, 3 } and { 2, 3, 4 } is { 1, 2, 3, 4 }. Multiple occurrences of identical elements have no effect on the cardinality of a set or its contents. The number 9 is not contained in the union of the set of prime numbers {2 , 3, 5, 7, 11, …} and the set of even numbers { 2, 4, 6, 8, 10, …}, because 9 is neither prime nor even. Intersection – the intersection of sets A and B, denoted by A ⋂ B, is the set of elements which belong to both A and B. Example: A = { e, f, r, e, n } B = { k, a, r, e, n } A ⋂ B = { r, e, n } If a and B do not have any element in common, that is A ⋂ B = ⌀, then A and B are said to disjoint. Difference – the difference of sets A and B, is the set of elements which belong to A but not B. Example: A = { e, f, r, e, n } B = { k, a, r, e, n } A – B = { e, f } and also, B – A = { k, a } Complement – the complement of set A, denoted by Ac, is the set of elements which belong to the universal set but not to the set A. Example: A = { 1, 3, 5, 7, 9, 11, … } and U = { 1, 2, 3, 4, … } Ac = { 2, 4, 6, 8, 10, 12, … } ### VENN DIAGRAM – John Venn, (1834 – 1883) a graphical representation defining set of operations. Sample Problem: ECE Board November 1998 A club of 40 executives, 33 likes to smoke Marlboro and 20 like to smoke Philip Morris. How many like both? Solution: Let: x = number of executives who likes to smoke both Marlboro and Philip Morris. Using Venn – diagram (33 – x ) + (20 – x) + x = 40 x = 53 – 40 x = 13, number of executives who likes to smoke both Marlboro and Philip Morris. ### LAWS OF THE ALGEBRA OF SETS Idempotent Laws • A ⋃ A = A • A ⋂ A = A Associative Laws • (A ⋃ B) ⋃ C = A ⋃ (B ⋃ C) • (A ⋂ B) ⋂ C = (A ⋂ B) ⋂ C Commutative Laws • A ⋃ B = B ⋃ A • A ⋂ B = B ⋂ A Distributive Laws • A ⋃ (B ⋂ C) = (A ⋃ B) ⋂ (A ⋃ C) • A ⋂ (B ⋃ C) = (A ⋂ B) ⋃ (A ⋂ C) Identity Laws • A ⋃ ⌀ = A • A ⋂ ⌀ = A Complement Laws • A ⋃ Ac = U • (Ac)c = A • A ⋂ Ac = ⌀ • Uc = ⌀, ⌀c = U De Morgan’s Law • (A ⋃ B)c = Ac ⋂ Bc • (A ⋂ B)c = Ac ⋃ Bc Labels:
Set Theory: Venn Diagrams And Subsets Free Roman Numeral Games Set Operations & Venn Diagrams Venn Diagram Examples Set Worksheets What Is A Venn Diagram? A Venn Diagram is a pictorial representation of the relationships between sets. We can represent sets using Venn diagrams. In a Venn diagram, the sets are represented by shapes; usually circles or ovals. The elements of a set are labeled within the circle. The following diagrams show the set operations and Venn Diagrams for Complement of a Set, Disjoint Sets, Subsets, Intersection and Union of Sets. Scroll down the page for more examples and solutions. The set of all elements being considered is called the Universal Set (U) and is represented by a rectangle. • The complement of A, A’, is the set of elements in U but not in A. A’ ={x \| x ∈ U and x ∉ A} • Sets A and B are disjoint sets if they do not share any common elements. • B is a proper subset of A. This means B is a subset of A, but B ≠ A. • The intersection of A and B is the set of elements in both set A and set B. A ∩ B = {x \| x ∈ A and x ∈ B} • The union of A and B is the set of elements in set A or set B. A ∪ B = {x \| x ∈ A or x ∈ B} • A ∩ ∅ = ∅ • A ∪ ∅ = A Set Operations And Venn Diagrams Example: 1. Create a Venn Diagram to show the relationship among the sets. U is the set of whole numbers from 1 to 15. A is the set of multiples of 3. B is the set of primes. C is the set of odd numbers. 2. Given the following Venn Diagram determine each of the following set. a) A ∩ B b) A ∪ B c) (A ∪ B)’ d) A’ ∩ B e) A ∪ B' Venn Diagram Examples Example: Given the set P is the set of even numbers between 15 and 25. Draw and label a Venn diagram to represent the set P and indicate all the elements of set P in the Venn diagram. Solution: List out the elements of P. P = {16, 18, 20, 22, 24} ← ‘between’ does not include 15 and 25 Draw a circle or oval. Label it P. Put the elements in P. Example: Draw and label a Venn diagram to represent the set R = {Monday, Tuesday, Wednesday}. Solution: Draw a circle or oval. Label it R . Put the elements in R. Example: Given the set Q = { x : 2x – 3 < 11, x is a positive integer }. Draw and label a Venn diagram to represent the set Q. Solution: Since an equation is given, we need to first solve for x. 2x – 3 < 11 ⇒ 2x < 14 ⇒ x < 7 So, Q = {1, 2, 3, 4, 5, 6} Draw a circle or oval. Label it Q. Put the elements in Q. Venn Diagram Videos What’s a Venn Diagram, and What Does Intersection and Union Mean? Venn Diagram and Subsets Learn about Venn diagrams and subsets. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
Open in App Not now # Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.6 • Last Updated : 07 Nov, 2021 ### (i) Solution: Lets, take 1/x = a and 1/y = b Here, the two given equation will be as follows: + = 2 Multiply it by 6, we get 3a + 2b = 12 -(1) and, + = Multiply it by 6, we get 2a + 3b = 13 -(2) Now, by using Elimination method, Multiply eq(1) by 2 and multiply eq(2) by 3, and then subtract them 5b = 15 b = 3 Now putting b = 3 in eq(1), we get 3a + 2(3) = 12 a = 6/3 a = 2 So, Now As a = 1/x = 2 x = 1/2 b = 1/y = 3 y = 1/3 ### (ii) Solution: Lets, take 2/âˆšx = a and 3/âˆšy = b Here, the two given equation will be as follows: a + b = 2 -(1) and, 2a – 3b =-1 -(2) Now, by using Elimination method, Multiply eq(1) by 3, and then add them 5a = 5 a = 1 Now putting a = 1 in eq(1), we get 1 + b = 2 b = 1 So, Now As a = 2/âˆšx = 1 âˆšx = 2 x = 4 b = 3/âˆšy = 1 âˆšx = 3 y = 9 ### – 4y = 23 Solution: Lets, take 1/x = a Here, the two given equation will be as follows: 4a + 3y = 14 -(1) and, 3a – 4y = 23 -(2) Now, by using Elimination method, Multiply eq(1) by 3 and multiply eq(2) by 4, and then subtract them -25y = 50 y = -2 Now putting y = -2 in eq(1), we get 4a + 3(-2) = 14 4a = 20 a = 5 So, Now As a = 1/x = 5 x = 1/5 y = -2 ### (iv) Solution: Lets, take = a and, = b Here, the two given equation will be as follows: 5a + b = 2 -(1) and, 6a – 3b = 1 -(2) Now, by using Elimination method, Multiply eq(1) by 3, and then add them 21a = 7 a = 1/3 Now putting a = 1/3 in eq(1), we get 5(1/3) + b = 2 b = 2 – 5/3 b = 1/3 So, Now As a = x – 1 = 3 x = 4 b = y – 2 = 3 y = 5 ### (v) Solution: = 5 = 15 Lets, take 1/x = a and 1/y = b Here, the two given equation will be as follows: 7b – 2a = 5 -(1) and, 8b + 7a = 15 -(2) Now, by using Elimination method, Multiply eq(1) by 7, multiply eq(2) by 2 and then add them 65b = 65 b = 1 Now putting b = 1 in eq(1), we get 7(1) – 2a = 5 2a = 7 – 5 a = 1 So, Now As a = 1/x = 1 x = 1 b = 1/y = 1 y = 1 ### 2x + 4y = 5xy Solution: Divide both the equations by xy, we get = 6 = 5 Lets, take 1/x = a and, 1/y = b Here, the two given equation will be as follows: 6b + 3a = 6 Divide the above equation by 2, 2b + a = 2 -(1) and, 2b + 4a = 5 -(2) Now, by using Elimination method, Subtract eq(1) from eq(2), we get 3a = 3 a = 1 Now putting a = 1 in eq(1), we get 2b + 1 = 2 b = 1/2 So, Now As a = 1/x = 1 x = 1 b = 1/y = 1/2 y = 2 ### (vii) Solution: Lets, take = a and = b Here, the two given equation will be as follows: 10a + 2b = 4 Divide the above equation by 2, 5a + b = 2 -(1) and, 15a – 5b = -2 -(2) Now, by using Elimination method, Multiply eq(1) by 3 and subtract them, 8b = 8 b = 1 Now putting b = 1 in eq(1), we get 5a + 1 = 2 a = 1/5 So, Now As a = = x + y = 5 -(3) b = = 1 x – y = 1 -(4) By adding eq(3) and (4), we get 2x = 6 x = 3 and y = 2 ### (viii) Solution: Lets, take = a and, = b Here, the two given equation will be as follows: a + b = 3/4 -(1) and, Multiply it by 2, we get a – b = -1/4 -(2) Now, by using Elimination method, Add eq(1) and eq(1), we get 2a = 1/2 a = 1/4 Now putting a = 1/4 in eq(1), we get + b = b = 1/2 So, Now As a = 3x + y = 4 -(3) b = 3x – y = 2 -(4) By adding eq(3) and eq(4), we get 6x = 6 x = 1 and y = 1 ### (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. Solution: Let us consider, Speed of Ritu in still water = x km/hr Speed of Stream = y km/hr Now, speed of Ritu during, Downstream = (x + y) km/h Upstream = (x â€“ y) km/h As Speed = According to the given question, x + y = 20/2 x + y = 10 -(1) and, x – y = 4/2 x – y = 2 -(2) Add eq(1) and eq(2), we get 2x = 12 x = 6 and y = 4 Hence, speed of Ritu rowing in still water = 6 km/hr Speed of Stream = 4 km/hr ### (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. Solution: Let’s take, The total number of days taken by women to finish the work = x The total number of days taken by men to finish the work = y Work done by women in one day will be = 1/x Work done by women in one day will be = 1/y So, according to the question 4() = 1 And, 3() = 1 Lets, take 1/x = a and, 1/y = b Here, the two given equation will be as follows: 4(2a + 5b) = 1 8a + 20b = 1 -(1) and, 3(3a + 6b) = 1 9a + 18b = 1 -(2) Now, by using Cross multiplication method, a = b = 1/36 So, Now As a = x = 18 b = y = 36 Hence, number of days taken by women to finish the work = 18 days Number of days taken by men to finish the work = 36 days. ### (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. Solution: Lets, take Speed of the train = x km/h Speed of the bus = y km/h According to the given question, = 4 Lets, take 1/x = a and 1/y = b Here, the two given equation will be as follows: 60a + 240b = 4 Divide it by 4, we get 15a + 60b = 1 -(1) and, 100a + 200b = 25/6 Divide it by 25/6, we get 24a + 48b = 1 -(2) Now, by using Cross multiplication method, a = = b = = So, Now As a = = x = 60 b = = y = 80 Hence, speed of the train = 60 km/h Speed of the bus = 80 km/h My Personal Notes arrow_drop_up Related Articles
Associated Topics \|\| Dr. Math Home \|\| Search Dr. Math ### Volume of Soil ``` Date: 02/16/99 at 09:05:17 From: Amy Atteberry Subject: Volume of soil needed for a trapezoidal bed I am building some planting beds in a trapezoidal shape. The beds will be 8" deep. The dimensions of the trapezoidal bed are 2 x 6 x 4 x 4' by 8". Therefore, I need to be able to calculate how many cubic feet of soil will be needed to fill these beds. ``` ``` Date: 02/16/99 at 12:41:14 From: Doctor Rick Subject: Re: Volume of soil needed for a trapezoidal bed Okay, your 3-dimensional shape might be called a "right prism with a trapezoidal base." Your question might still need more clarification. I am guessing that the dimensions of the trapezoid you give are the lengths of the sides, and in this order: 2 +-----+ 4 / \ 4 / \ +-----------+ 6 and that the 2' and 6' sides are parallel. The formula for the area of a trapezoid is A = (L1 + L2) * H / 2 where L1 and L2 are the lengths of the two parallel sides, and H is the height, that is, the perpendicular distance between the parallel sides. We need to find the height. If you draw a line from the left corner of the 2' side, perpendicular to the 6' side, you get a right triangle with hypotenuse 4' and bottom leg 2'. Using the Pythagorean Theorem, we find that the height (the other side of the triangle) is H = sqrt(4^2 - 2^2) = sqrt(12) = 3.464' Now the area of the trapezoid is A = (2' + 6') * 3.464' / 2 = 13.856 ft^2 To get the volume of the prism, we multiply the area of the base by the height; this time the height is the truly vertical dimension, 8" = 2/3 foot. So the volume is V = A * h = 13.856 ft^2 * 2/3 ft = 9.237 ft^3 That's how many cubic feet of soil you need, if I interpreted your description correctly. If not, perhaps I have given you enough information to figure out the volume yourself. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ ``` Associated Topics: High School Geometry High School Higher-Dimensional Geometry High School Practical Geometry Search the Dr. Math Library: Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words Submit your own question to Dr. Math Math Forum Home \|\| Math Library \|\| Quick Reference \|\| Math Forum Search
# What is the LCM of 10 and 14? 70 #### Explanation: LCM, or Lowest Common Multiple, is the number which is both a multiple of 10 and 14 and is the first number that is. Many students will look at the two numbers, in this case 10 and 14 and say: "If I multiply 10 and 14, I'll get 140 and that is for sure a multiple of 10 and 14 and so I'll just use that." And it is true that 140 is a multiple of 10 and 14. But it isn't the lowest. So let's find the lowest. When I do these, I like to break the numbers in question down to their prime factors: For 10, the prime factors are 5 and 2 - our LCM will need a 5 and a 2 For 14, the prime factors are 7 and 2 - our LCM will need a 7 and a 2 So we'll need a 2, a 5, and a 7 to make our LCM: $2 \times 5 \times 7 = 70$ And we can check this by saying: $10 \times 7 = 70$ $14 \times 5 = 70$ The first attempt we made, with 140, wasn't the lowest because it had an extra 2 that wasn't needed: $2 \times 5 \times 2 \times 7 = 140$ Sep 29, 2016 70 #### Explanation: Consider the prime factors of 10 and 14. $10 = 2 \times 5$ and $14 = 2 \times 7$ Hence the factors of the lowest common multiple must be $2 \times 5 \times 7$ - Since 2 appears in both it must occur only once in the LCM. $\therefore L C M \left(10 , 14\right) = 2 \times 5 \times 7 = 70$ Sep 29, 2016 $70$ #### Explanation: multiples of $10 : 10 , 20 , 30 , 40 , 50 , 60 , \textcolor{red}{70} , 80 \ldots$ multiples of $14 : 14 , 28 , 42 , 56 , \textcolor{red}{70} , 84 \ldots \ldots$ Hence, LCM of $10 \mathmr{and} 14 = 70$
# Properties of a quadratic equation In the quadratic equation form ax2+bx+c=0, x stands for the unknown value and a, b and c the numbers that are known such that a will never be equal to zero. If in case a is zero, then that equation is not a quadratic equation but a linear equation. a, b and c are the coefficients of the equation which can be distinguished by referring to them as the constant, the linear or the quadratic coefficient or the free term. A quadratic equation can be solved using methods like factoring by the quadratic formula, by graphing or by completing the square. To solve a quadratic equation in one variable can be done through the factoring technique together with the zero factor property as below: • Begin by writing the quadratic equation in a standard form. • Factor the quadratic polynomial to a product of linear factors. • Use Zero Factor Property to make each factor equal to zero. • Find the solution to each of the resulting linear equations. The resulting solution is the solution of the original quadratic equation. The quadratic equation is also known as a univariate because it involves only one unknown. The quadratic equation only has powers of x which are non-negative integers and this makes it a polynomial equation and in particular, it is a second-degree polynomial equation because its greatest power is two. A quadratic equation which has real or complex coefficients has two solutions known as roots. These solutions can or cannot be distinct and can also be or not be real. The first method of solving a quadratic equation is through factoring by inspection. It is possible to write the quadratic equation ax2 + bx + c = 0 as a product (px + q)(rx + s) = 0. And with simple inspection, it is also possible to know the values of p, q, r and s which make the two forms equivalent to each other. If you write the quadratic equation in a second form, the zero factor property states that a quadratic equation is satisfied if px + q = 0 or rx + s = 0. The solution of these two linear equations gives you the roots of the quadratic. Factoring by inspection method is the first method of solving a quadratic equation that students learn. If you are given a quadratic equation in the form x2 + bx + c = 0, the factorization will have the form(x + q)(x + s), and you will have to find q and s that add up to b and whose product is c. For example, x2 + 5x + 6 factors as (x + 3)(x + 2). In the case where a is not equal to 1, you will need more guesses in order to find the solution if at all it can be factored by inspection. Except for special cases like where b = 0 or c = 0, factoring by inspection works for a quadratic equation that has rational roots. This means that most quadratic equations in practical cases cannot be computed through factoring by inspection. The second method of solving a quadratic equation is by completing the square. The completing by square method is used to come up with a new formula for solving a quadratic equation. The formula is known as the quadratic formula. The quadratic formula’s mathematical proof is. One property of this equation form is that it gives one valid root when a = 0, while the other root has division by zero, since when a = 0, the quadratic equation changes to a linear equation which has one root. By contrast, the formula contains division by zero in both cases. With the polynomial expansion, you can easily tell that this equation is equivalent to the quadratic equation. The third method of solving a quadratic equation is by graphing. The graph of a quadratic equation is a parabola that opens up when the leading coefficient is positive and it opens down when the leading is a negative. The x intercepts are crucial and they should be gotten, plotted and labeled. In any function, the x intercepts are gotten by getting the real zeros of that function and the zeros of any given function can be gotten by solving that equation that results from f(x)=0. In a quadratic function, the function f of the equation that results from f(x) =0 can be solved by factoring along with the zero factor property or with the quadratic formula. The vertex of a parabola is also crucial and can be gotten, plotted and labeled. In the case of a quadratic function f, the equation resulting from f(x) = 0 is always solvable with the quadratic formula or by factoring in along with the Zero Factor Property. The first step in graphing a quadratic equation is determining the form of the equation given. A quadratic equation can be expressed in three forms. These forms are the quadratic form, vertex form and standard form. Any of the forms can be used in graphing but the process of graphing each of them is a bit different. The standard form is the form where the equation is written as f(x) = ax2+bx+c whereby a, b and c are real numbers and a is not a zero. Examples of standard form equations are f(x) = x2 +2x+1 and f(x) =9 x2+10x-8. A vertex form is the form whereby the equation is expressed as f(x) =a(x-h) 2 +k whereby a, h and k are real numbers and a is not zero. This equation form is known as a vertex form because h and k will directly give you a central point of the parabola at point (h, k). Examples of vertex form equations are -3(x-5) 2+1 and 9(x-4) 2+18. To graph any of this quadratic equation forms you will first have to get the vertex of the parabola which is the middle point (h, k) at the tip of the curve. The vertex coordinates in the standard form are given by k=f(h) and h=-b/2a. In the vertex equation form, h and k are gotten directly from the equation. The second step in graphing a quadratic equation is to define the variables. To solve a quadratic equation, the variables a, b and c or a, h and k must be defined. Common algebra problems will give you a quadratic equation with variables already filled in a vertex or standard form. An example of a standard equation form with variables is f(x)= 2×2+16x+39, so a is 2, b is 16 and c is 39. An example of a vertex equation form with variables is f(x)=4(x-5) 2+12, a is 4, b is 16 and k is 12. The third step is to find h. In the vertex form, h is already provided. In standard form h is calculated by h =-b/2a and so in f(x)= 2×2+16x+39 h will be -16/2(2). After solving you will get h=-4. In the vertex form f(x)=4(x-5) 2+12, h is equal to 5. The third step is to find k. In the vertex form, k is already known just like h but for standard form k is equal to f(h). This means that you can get k in the standard form equation by replacing every x with the value of h. So k=2(-4) 2 +16(-4)+39, k=2(16)-64+39, k=32-64+39 =7. The value of k in the vertex quadratic equation form is 12. The fifth step is to plot the vertex. The parabola vertex is at point (h, k). The h represents the x coordinate and the k represents the y coordinate. The vertex is the middle point of the parabola. In the standard equation form, the vertex will be at point (-4, 7). This point will be plotted on the graph and labeled. In the vertex equation form, the vertex is at (5, 12). The sixth step is to draw the parabola axis. The axis of symmetry in a parabola is a line that runs through the middle and divides the parabola into half. Across the axis, the right side of a parabola will mirror the left side of the parabola. For the quadratic equation of the form ax2+bx+c or a(x-h) 2+k, the axis is the line which is vertical and passes through the vertex. In the standard form equation, the axis is the line that is parallel to the y axis and passes through the point (-4, 7). This line is not part of the parabola but it shows you how a parabola curves symmetrically. The seventh step is to get the direction of opening. After knowing the axis and vertex of the parabola determine if the parabola opens downwards or upwards. If a is positive, the parabola opens upwards but if a is negative the parabola will open downwards. For the standard form equation f(x)= 2×2+16x+39, the parabola will open upwards since a=2(positive), and in the vertex equation form f(x)=4(x-5) 2+12, the parabola will also open upwards since a=4(positive). The eight step if necessary, you can find and plot the x intercepts. The x intercepts are the two points where the parabola meets the x axis. Not all parabolas have x intercepts. If the parabola is a vertex that opens upwards and has the vertex above the x axis or if it opens downwards and has a vertex below the x axis, it will not have x intercepts. You can also find and plot the y intercept too. To find the y intercept, you will set x to zero and then solve the equation for y or f(x). This will give you the value of y when the parabola passes through the y axis. And unlike the x intercept, standard parabolas can have one y intercept and for the standard quadratic equation forms, the y intercept is at y=c. # Our Service Charter 1. ### Excellent Quality / 100% Plagiarism-Free We employ a number of measures to ensure top quality essays. The papers go through a system of quality control prior to delivery. We run plagiarism checks on each paper to ensure that they will be 100% plagiarism-free. So, only clean copies hit customers’ emails. We also never resell the papers completed by our writers. So, once it is checked using a plagiarism checker, the paper will be unique. Speaking of the academic writing standards, we will stick to the assignment brief given by the customer and assign the perfect writer. By saying “the perfect writer” we mean the one having an academic degree in the customer’s study field and positive feedback from other customers. 2. ### Free Revisions We keep the quality bar of all papers high. But in case you need some extra brilliance to the paper, here’s what to do. First of all, you can choose a top writer. It means that we will assign an expert with a degree in your subject. And secondly, you can rely on our editing services. Our editors will revise your papers, checking whether or not they comply with high standards of academic writing. In addition, editing entails adjusting content if it’s off the topic, adding more sources, refining the language style, and making sure the referencing style is followed. 3. ### Confidentiality / 100% No Disclosure We make sure that clients’ personal data remains confidential and is not exploited for any purposes beyond those related to our services. We only ask you to provide us with the information that is required to produce the paper according to your writing needs. Please note that the payment info is protected as well. Feel free to refer to the support team for more information about our payment methods. The fact that you used our service is kept secret due to the advanced security standards. So, you can be sure that no one will find out that you got a paper from our writing service. 4. ### Money Back Guarantee If the writer doesn’t address all the questions on your assignment brief or the delivered paper appears to be off the topic, you can ask for a refund. Or, if it is applicable, you can opt in for free revision within 14-30 days, depending on your paper’s length. The revision or refund request should be sent within 14 days after delivery. The customer gets 100% money-back in case they haven't downloaded the paper. All approved refunds will be returned to the customer’s credit card or Bonus Balance in a form of store credit. Take a note that we will send an extra compensation if the customers goes with a store credit. We have a support team working 24/7 ready to give your issue concerning the order their immediate attention. If you have any questions about the ordering process, communication with the writer, payment options, feel free to join live chat. Be sure to get a fast response. They can also give you the exact price quote, taking into account the timing, desired academic level of the paper, and the number of pages. Excellent Quality Zero Plagiarism Expert Writers or ## Custom Writing Service Instant Quote Subject: Please Select... Art Architecture Dance Design Analysis Drama Movies Music Paintings Theatre Biology Business Chemistry Communications and Media Advertising Communication Strategies Journalism Public Relations Creative writing Economics Accounting Case Study Company Analysis E-Commerce Finance Investment Logistics Trade Education Application Essay Education Theories Pedagogy Teacher's Career Engineering English Ethics History African-American Studies American History Asian Studies Canadian Studies East European Studies Holocaust Latin-American Studies Native-American Studies West European Studies Law Criminology Legal Issues Linguistics Literature American Literature Antique Literature Asian Literature English Literature Shakespeare Studies Management Marketing Mathematics Medicine and Health Alternative Medicine Healthcare Nursing Nutrition Pharmacology Sport Nature Agricultural Studies Anthropology Astronomy Environmental Issues Geography Geology Philosophy Physics Political Science Psychology Religion and Theology Sociology Technology Aeronautics Aviation Computer Science Internet IT Management Web Design Tourism Type: Please Select... Essay Term Paper Research Paper Coursework Book Report Book Review Movie Review Dissertation Thesis Thesis Proposal Research Proposal Dissertation Chapter - Abstract Dissertation Chapter - Introduction Chapter Dissertation Chapter - Literature Review Dissertation Chapter - Methodology Dissertation Chapter - Results Dissertation Chapter - Discussion Dissertation Services - Editing Dissertation Services - Proofreading Formatting Admission Services - Admission Essay Admission Services - Scholarship Essay Admission Services - Personal Statement Admission Services - Editing Editing Proofreading Case Study Lab Report Speech Presentation Math Problem Article Article Critique Annotated Bibliography Reaction Paper PowerPoint Presentation Statistics Project Multiple Choice Questions (None-Time-Framed) Other (Not listed) Pages/Words: 1 page/approx 275 words 2 pages/approx 550 words 3 pages/approx 825 words 4 pages/approx 1100 words 5 pages/approx 1375 words 6 pages/approx 1650 words 7 pages/approx 1925 words 8 pages/approx 2200 words 9 pages/approx 2475 words 10 pages/approx 2750 words 11 pages/approx 3025 words 12 pages/approx 3300 words 13 pages/approx 3575 words 14 pages/approx 3850 words 15 pages/approx 4125 words 16 pages/approx 4400 words 17 pages/approx 4675 words 18 pages/approx 4950 words 19 pages/approx 5225 words 20 pages/approx 5500 words 21 pages/approx 5775 words 22 pages/approx 6050 words 23 pages/approx 6325 words 24 pages/approx 6600 words 25 pages/approx 6875 words 26 pages/approx 7150 words 27 pages/approx 7425 words 28 pages/approx 7700 words 29 pages/approx 7975 words 30 pages/approx 8250 words 31 pages/approx 8525 words 32 pages/approx 8800 words 33 pages/approx 9075 words 34 pages/approx 9350 words 35 pages/approx 9625 words 36 pages/approx 9900 words 37 pages/approx 10175 words 38 pages/approx 10450 words 39 pages/approx 10725 words 40 pages/approx 11000 words 41 pages/approx 11275 words 42 pages/approx 11550 words 43 pages/approx 11825 words 44 pages/approx 12100 words 45 pages/approx 12375 words 46 pages/approx 12650 words 47 pages/approx 12925 words 48 pages/approx 13200 words 49 pages/approx 13475 words 50 pages/approx 13750 words 51 pages/approx 14025 words 52 pages/approx 14300 words 53 pages/approx 14575 words 54 pages/approx 14850 words 55 pages/approx 15125 words 56 pages/approx 15400 words 57 pages/approx 15675 words 58 pages/approx 15950 words 59 pages/approx 16225 words 60 pages/approx 16500 words 61 pages/approx 16775 words 62 pages/approx 17050 words 63 pages/approx 17325 words 64 pages/approx 17600 words 65 pages/approx 17875 words 66 pages/approx 18150 words 67 pages/approx 18425 words 68 pages/approx 18700 words 69 pages/approx 18975 words 70 pages/approx 19250 words 71 pages/approx 19525 words 72 pages/approx 19800 words 73 pages/approx 20075 words 74 pages/approx 20350 words 75 pages/approx 20625 words 76 pages/approx 20900 words 77 pages/approx 21175 words 78 pages/approx 21450 words 79 pages/approx 21725 words 80 pages/approx 22000 words 81 pages/approx 22275 words 82 pages/approx 22550 words 83 pages/approx 22825 words 84 pages/approx 23100 words 85 pages/approx 23375 words 86 pages/approx 23650 words 87 pages/approx 23925 words 88 pages/approx 24200 words 89 pages/approx 24475 words 90 pages/approx 24750 words 91 pages/approx 25025 words 92 pages/approx 25300 words 93 pages/approx 25575 words 94 pages/approx 25850 words 95 pages/approx 26125 words 96 pages/approx 26400 words 97 pages/approx 26675 words 98 pages/approx 26950 words 99 pages/approx 27225 words 100 pages/approx 27500 words 101 pages/approx 27775 words 102 pages/approx 28050 words 103 pages/approx 28325 words 104 pages/approx 28600 words 105 pages/approx 28875 words 106 pages/approx 29150 words 107 pages/approx 29425 words 108 pages/approx 29700 words 109 pages/approx 29975 words 110 pages/approx 30250 words 111 pages/approx 30525 words 112 pages/approx 30800 words 113 pages/approx 31075 words 114 pages/approx 31350 words 115 pages/approx 31625 words 116 pages/approx 31900 words 117 pages/approx 32175 words 118 pages/approx 32450 words 119 pages/approx 32725 words 120 pages/approx 33000 words 121 pages/approx 33275 words 122 pages/approx 33550 words 123 pages/approx 33825 words 124 pages/approx 34100 words 125 pages/approx 34375 words 126 pages/approx 34650 words 127 pages/approx 34925 words 128 pages/approx 35200 words 129 pages/approx 35475 words 130 pages/approx 35750 words 131 pages/approx 36025 words 132 pages/approx 36300 words 133 pages/approx 36575 words 134 pages/approx 36850 words 135 pages/approx 37125 words 136 pages/approx 37400 words 137 pages/approx 37675 words 138 pages/approx 37950 words 139 pages/approx 38225 words 140 pages/approx 38500 words 141 pages/approx 38775 words 142 pages/approx 39050 words 143 pages/approx 39325 words 144 pages/approx 39600 words 145 pages/approx 39875 words 146 pages/approx 40150 words 147 pages/approx 40425 words 148 pages/approx 40700 words 149 pages/approx 40975 words 150 pages/approx 41250 words 151 pages/approx 41525 words 152 pages/approx 41800 words 153 pages/approx 42075 words 154 pages/approx 42350 words 155 pages/approx 42625 words 156 pages/approx 42900 words 157 pages/approx 43175 words 158 pages/approx 43450 words 159 pages/approx 43725 words 160 pages/approx 44000 words 161 pages/approx 44275 words 162 pages/approx 44550 words 163 pages/approx 44825 words 164 pages/approx 45100 words 165 pages/approx 45375 words 166 pages/approx 45650 words 167 pages/approx 45925 words 168 pages/approx 46200 words 169 pages/approx 46475 words 170 pages/approx 46750 words 171 pages/approx 47025 words 172 pages/approx 47300 words 173 pages/approx 47575 words 174 pages/approx 47850 words 175 pages/approx 48125 words 176 pages/approx 48400 words 177 pages/approx 48675 words 178 pages/approx 48950 words 179 pages/approx 49225 words 180 pages/approx 49500 words 181 pages/approx 49775 words 182 pages/approx 50050 words 183 pages/approx 50325 words 184 pages/approx 50600 words 185 pages/approx 50875 words 186 pages/approx 51150 words 187 pages/approx 51425 words 188 pages/approx 51700 words 189 pages/approx 51975 words 190 pages/approx 52250 words 191 pages/approx 52525 words 192 pages/approx 52800 words 193 pages/approx 53075 words 194 pages/approx 53350 words 195 pages/approx 53625 words 196 pages/approx 53900 words 197 pages/approx 54175 words 198 pages/approx 54450 words 199 pages/approx 54725 words 200 pages/approx 55000 words Single spaced approx 275 words per page Urgency (Less urgent, less costly): Please Select... 6 hours 12 hours 24 hours 48 hours 3 days 4 days 5 days 7 days 10 days 20 days 30 days Level: Choose High School Undergraduate Master Ph. D. Currency: USD GBP CAD AUD EUR Total Cost: NaN Get 10% Off on your 1st order!
# How to use the Z Table (With Examples) A z-table is a table that tells you what percentage of values fall below a certain z-score in a standard normal distribution. A z-score simply tells you how many standard deviations away an individual data value falls from the mean. It is calculated as: z-score = (x – μ) / σ where: • x: individual data value • μ: population mean • σ: population standard deviation This tutorial shows several examples of how to use the z table. ### Example 1 The scores on a certain college entrance exam are normally distributed with mean μ = 82 and standard deviation σ = 8. Approximately what percentage of students score less than 84 on the exam? Step 1: Find the z-score. First, we will find the z-score associated with an exam score of 84: z-score = (x – μ) / σ = (84 – 82) / 8 = 2 / 8 = 0.25 Step 2: Use the z-table to find the percentage that corresponds to the z-score. Next, we will look up the value 0.25 in the z-table: Approximately 59.87% of students score less than 84 on this exam. ### Example 2 The height of plants in a certain garden are normally distributed with a mean of μ = 26.5 inches and a standard deviation of σ = 2.5 inches. Approximately what percentage of plants are greater than 26 inches tall? Step 1: Find the z-score. First, we will find the z-score associated with a height of 26 inches. z-score = (x – μ) / σ = (26 – 26.5) / 2.5 = -0.5 / 2.5 = -0.2 Step 2: Use the z-table to find the percentage that corresponds to the z-score. Next, we will look up the value -0.2 in the z-table: We see that 42.07% of values fall below a z-score of -0.2. However, in this example we want to know what percentage of values are greater than -0.2, which we can find by using the formula 100% – 42.07% = 57.93%. Thus, aproximately 59.87% of the plants in this garden are greater than 26 inches tall. ### Example 3 The weight of a certain species of dolphin is normally distributed with a mean of μ = 400 pounds and a standard deviation of σ = 25 pounds. Approximately what percentage of dolphins weigh between 410 and 425 pounds? Step 1: Find the z-scores. First, we will find the z-scores associated with 410 pounds and 425 pounds z-score of 410 = (x – μ) / σ = (410 – 400) / 25 = 10 / 25 = 0.4 z-score of 425 = (x – μ) / σ = (425 – 400) / 25 = 25 / 25 = 1 Step 2: Use the z-table to find the percentages that corresponds to each z-score. First, we will look up the value 0.4 in the z-table: Then, we will look up the value 1 in the z-table: Lastly, we will subtract the smaller value from the larger value: 0.8413 – 0.6554 = 0.1859. Thus, approximately 18.59% of dolphins weigh between 410 and 425 pounds. ## 9 Replies to “How to use the Z Table (With Examples)” 1. Loralee says: Thank you for all of the information on this site! You are a lifesaver! 2. hana says: Hello perfect 3. Coster Itayi Mukusha says: Your calculation of 1.645 was help. But I need to know how to calculate z alpha 98% cofidence interval of 2.326 using tables manually 4. Shireen says: What will be the alpha value or tabulated value of z=-17.5 tabulated value or alpha value,pls send the answer to shrnfatima@gmail.com 5. Gene says: Well understood thanks 6. Mohammed Abdul Luqmaan says: Super explanation with out ads 😍😍 7. Hyacinth Seymour says: thank you your example help me to understand this topics 8. Basil Okorie says: Thanks for the examples, they really helped me in understanding the use of Z tables. 1. James Carmichael says: Hi Basil…You are very welcome!
# denominator 74 results denominator - The bottom portion of a fraction. For a/b, b is the denominator 1, 1/2, 1/4, 1/8, 1/16 The next number in the sequence is 1/32. What is the function machine you wou 1, 1/2, 1/4, 1/8, 1/16 The next number in the sequence is 1/32. What is the function machine you would use to find the nth term of this sequence? Hint: look at the denominators We notice that 1/2^0 = 1/1 = 1 1/2^1 = 1/2 1/2^2 = 1/4 1/2^3 = 1/8 1/2^4 = 1/32 So we write our explicit formula for term n: f(n) = [B]1/2^(n - 1)[/B] 1, 1/2, 1/4, 1/8, 1/16, ... At first glance, we see powers of 2 in the denominator of every term except the first one. But if we remember 2^0 = 1, we get our breakthrough. 1/2^0 = 1/1 = 1 Therefore, we stagger the powers of 2 by 1 less than the term we are on: a(n) = [B]1/2^(n - 1) [MEDIA=youtube]Ua-arUukOew[/MEDIA][/B] 1/a + 1/b = 1/2 for a 1/a + 1/b = 1/2 for a Subtract 1/b from each side to solve this literal equation: 1/a + 1/b - 1/b = 1/2 - 1/b Cancel the 1/b on the left side, we get: 1/a = 1/2 - 1/b Rewrite the right side, using 2b as a common denominator: 1/a = (b - 2)/2b Cross multiply: a(b - 2) = 2b Divide each side by (b - 2) a = [B]2b/(b - 2)[/B] 14 is the what of 1/14? 14 is the what of 1/14? [B]Denominator[/B] 18 divided by the difference of t and 9 The difference of t and 9 is: t - 9 Now, we set up a quotient with 18 as the numerator and t - 9 as the denominator 18 ------ t - 9 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9 x 9/10 x 10/11 x 11/12 x 12/13 x 13/14 Do this without a calculator. Notice, all numerators have a matching denominator except 2 in the first term and 14 in the last term. So we have 2/14 Using 2 as a common factor, we have: 2/2 / 14/2 [B]1/7[/B] 3abc^4/12a^3(b^3c^2)^2 * 8ab^-4c/4a^2b 3abc^4/12a^3(b^3c^2)^2 * 8ab^-4c/4a^2b Expand term 1: 3abc^4/12a^3(b^3c^2)^2 3abc^4/12a^3b^6c^4 Now simplify term 1: 3/12 = 1/4 c^4 terms cancel Subtract powers from variables since the denominator powers are higher: b^(6 - 1) = b^5 a^(3 - 1) = a^2 1/4a^2b^5 Now simplify term 2: 8ab^-4c/4a^2b 8/4 = 2 2c/a^(2 - 1)b^(1 - -4) 2c/ab^5 Now multiply simplified term 1 times simplified term 2: 1/4a^2b^5 * 2c/ab^5 (1 * 2c)/(4a^2b^5 * ab^5) 2c/4a^(2 + 1)b^(5 + 5) 2c/4a^3b^10 2/4 = 1/2, so we have: [B]c/2a^3b^10[/B] 5 times a number increased by 4 is divided by 6 times the same number 5 times a number increased by 4 is divided by 6 times the same number Take this algebraic expression in parts. Part 1: 5 times a number increased by 4 [LIST] []The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x []5 times the number means multiply x by 5: 5x [][I]Increased by 4[/I] means we add 4 to 5x: 5x + 4 [/LIST] Part 2: 6 times the same number [LIST] []From above, [I]a number[/I] is x: x []6 times the number means we multiply x by 6: 6x [/LIST] The phrase [I]is divided by[/I] means we have a quotient, where 5x + 4 is the numerator, and 6x is the denominator. [B](5x + 4)/6x[/B] A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an ho A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat's speed and how long was the upstream journey? [U]Set up the relationship of still water speed and downstream speed[/U] Speed down stream = Speed in still water + speed of the current Speed down stream = x+2 Therefore: Speed upstream =x - 2 Since distance = rate time, we rearrange to get time = Distance/rate: 15/(x+ 2) + 15 /(x- 2) = 3 Multiply each side by 1/3 and we get: 5/(x + 2) + 5/(x - 2) = 1 Using a common denominator of (x + 2)(x - 2), we get: 5(x - 2)/(x + 2)(x - 2) + 5(x + 2)/(x + 2)(x - 2) (5x - 10 + 5x + 10)/5(x - 2)/(x + 2)(x - 2) 10x = (x+2)(x-2) We multiply through on the right side to get: 10x = x^2 - 4 Subtract 10x from each side: x^2 - 10x - 4 = 0 This is a quadratic equation. To solve it, [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2-10x-4%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type it in our search engine[/URL] and we get: Speed of the boat in still water =X=5 +- sq. Root of 29 kmph We only want the positive solution: x = 5 + sqrt(29) x = 10.38 [U]Calculate time for upstream journey:[/U] Time for upstream journey = 15/(10.38 - 2) Time for upstream journey = 15/(8.38) Time for upstream journey = [B]1.79[/B] [U]Calculate time for downstream journey:[/U] Time for downstream journey = 15/(10.38 + 2) Time for downstream journey = 15/(12.38) Time for downstream journey = [B]1.21[/B] a = v^2/r for r a = v^2/r for r Start by cross multiplying to get r out of the denominator: ar = v^2 Divide each side of the equation by a to isolate r: ar/a = v^2/a Cancel the a's on the left side, and we get: r = [B]v^2/a[/B] A boy spent one-half of his money for a book and one-third of his money for a pen. The remaining \$2. A boy spent one-half of his money for a book and one-third of his money for a pen. The remaining \$2.25 he saved. How much money did he originally have? Find out what percent of money was spent Using a common denominator of 6, we have 1/2 + 1/3 = 3/6 + 2/6 = 5/6. Therefore, 1/6 of his money is left to save. Let the boy's original money be x. We have: x/6 = 2.25 Cross multiply, we get x = [B]13.50[/B] a class has 24 people and 1/6 of them have blue eues. what fraction of the class has blue eyes, usin a class has 24 people and 1/6 of them have blue eues. what fraction of the class has blue eyes, using 24 as the denominator Blue eyes = 1/6 * 24 Blue Eyes = 24/6 Blue Eyes = 4 Since 64 = 24, we have: 24/6 4/4 = [B]96/24[/B] A company has 12,600 employees. Of these, 1/4 drive alone to work, 1/6 car pool, 1/8 use public tran A company has 12,600 employees. Of these, 1/4 drive alone to work, 1/6 car pool, 1/8 use public transportation, 1/9 cycle, and the remainder use other methods of transportation. How many employees use each method of transportation? Find the remainder fraction: Remainder = 1 - (1/4 + 1/6 + 1/8 + 1/9) The least common multiple of 4, 6, 8, 9 is 72. So we divide 72 by each fraction denominator to get our multiplier: 1/4 = 18/72 1/6 = 12/72 1/8 = 9/72 1/9 = 8/72 Add those all up: (18 + 12 + 9 + 8)/72 47/72 Now subtract the other methods out from 1 to get the remainder of who use other methods: Remainder = 1 - 47/72 Since 1 = 72/72, we have: (72 - 47)/72 [B]25/72[/B] A fraction has a value of 3/4. If 7 is added to the numerator, the resulting fraction is equal to th A fraction has a value of 3/4. If 7 is added to the numerator, the resulting fraction is equal to the reciprocal of the original fraction. Find the original fraction. Let the fraction be x/y. We're given two equations: [LIST=1] []x/y = 3/4 [](x + 7)/y = 4/3. [I](The reciprocal of 3/4 is found by 1/(3/4)[/I] [/LIST] Cross multiply equation 1 and equation 2: [LIST=1] []4x = 3y []3(x + 7) = 4y [/LIST] Simplifying, we get: [LIST=1] []4x = 3y []3x + 21 = 4y [/LIST] If we divide equation 1 by 4, we get: [LIST=1] []x = 3y/4 []3x + 21 = 4y [/LIST] Substitute equation (1) into equation (2) for x: 3(3y/4) + 21 = 4y 9y/4 + 21 = 4y Multiply the equation by 4 on both sides to eliminate the denominator: 9y + 84 = 16y To solve this equation for y, we type it in our math engine and we get: y = [B]12 [/B] We then substitute y = 12 into equation 1 above: x = 3 * 12/4 x = 36/4 x = [B]9 [/B] So our original fraction x/y = [B]9/12[/B] A lottery offers 1 \$1000 prize and 5 \$100 prizes. 1000 tickets are sold. Find the expectation if a p A lottery offers 1 \$1000 prize and 5 \$100 prizes. 1000 tickets are sold. Find the expectation if a person buys 1 ticket for \$5. Set up the expected values E(x): for the 1,000 price: E(x) = (1000 - 5) * 1/1000 = 995/1000 For the 5 \$100 prizes: E(x) = (100 - 5) * 5/1000 = 475/1000 For the losing ticket. With 6 winning tickets, we have 1000 - 6 = 994 losing tickets: E(x) = -3 * 994/1000 = -2982/1000 We get our total expected value by adding all of these expected values up. Since they all have the same denominator, we add numerators: E(x) = (995 + 475 - 2982)/1000 E(x) = -1512/1000 E(x) = [B]-1.51[/B] A quarter of the learners in a class have blond hair and two thirds have brown hair. The rest of the A quarter of the learners in a class have blond hair and two thirds have brown hair. The rest of the learners in the class have black hair. How many learners in the class if 9 of them have blonde hair? Total learners = Blond + Brown + Black Total Learners = 1/4 + 2/3 + Black Total Learners will be 1, the sum of all fractions 1/4 + 2/3 + Black = 1 Using common denominators of 12, we have: 3/12 + 8/12 + Black = 12/12 11/12 + Black = 12/12 Subtract 11/12 from each side: Black = 1/12 Let t be the total number of people in class. We are given for blondes: 1/4t = 9 Multiply each side by 4 [B]t = 36[/B] Brown Hair 2/3(36) = 24 Black Hair 1/12(36) = 3 A rational expression is undefined when what is 0? A rational expression is undefined when what is 0? The [B]denominator[/B]. Because division by zero is undefined. A rational number is such that when you multiply it by 7/3 and subtract 3/2 from the product, you ge A rational number is such that when you multiply it by 7/3 and subtract 3/2 from the product, you get 92. What is the number? Let the rational number be x. We're given: 7x/3 - 3/2 = 92 Using a common denominator of 32 = 6, we rewrite this as: 14x/6 - 9/6 = 92 (14x - 9)/6 = 92 Cross multiply: 14x - 9 = 92 6 14x - 9 = 552 To solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=14x-9%3D552&pl=Solve']type this equation into our search engine [/URL]and we get: x = [B]40.07[/B] A recipe calls for 2 ½ cup of flour, but Paul only has 2 ⅙. How much more flour does he need? A recipe calls for 2 ½ cup of flour, but Paul only has 2 ⅙. How much more flour does he need? Convert to improper fractions [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%261%2F2&frac2=3%2F8&pl=Simplify']2 & 1/2 [/URL]= 5/2 [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%261%2F6&frac2=3%2F8&pl=Simplify']2 & 1/6[/URL] = 13/6 [URL='https://www.mathcelebrity.com/equivalent-fractions.php?num=5%2F2&pl=Equivalent+Fractions']Convert 5/2 to an equivalent fraction with a denominator[/URL] of 6: 15/6 [URL='https://www.mathcelebrity.com/fraction.php?frac1=15%2F6&frac2=13%2F6&pl=Subtract']We subtract 13/6 from 15/6[/URL] 15/6 - 13/6 = 2/6 2/6 = [B]1/3[/B] a recipe of 20 bread rolls requires 5 tablespoons of butter. How many tablespoons of butter are need a recipe of 20 bread rolls requires 5 tablespoons of butter. How many tablespoons of butter are needed for 30 bread rolls? Set up a proportion of bread rolls per tablespoons of butter where t is the amount of tablespoons of butter needed for 30 bread rolls: 20/5 = 30/t Cross multiply our proportion: Numerator 1 * Denominator 2 = Denominator 1 * Numerator 2 20t = 30 * 5 20t = 150 Divide each side of the equation by 20: 20t/20 = 150/20 Cancel the 20's on the left side and we get: t = [B]7.5[/B] a son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age. a son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age. Declare variables: [LIST] []Let f be the father's age []Let s be the son's age [/LIST] We're given two equations: [LIST=1] []s = f/4 []f - s = 30. [I]The reason why we subtract s from f is the father is older[/I] [/LIST] Using substitution, we substitute equaiton (1) into equation (2) for s: f - f/4 = 30 To remove the denominator/fraction, we multiply both sides of the equation by 4: 4f - 4f/4 = 30 4 4f - f = 120 3f = 120 To solve for f, we divide each side of the equation by 3: 3f/3 = 120/3 Cancel the 3's on the left side and we get: f = [B]40[/B] a stone mason builds 7 houses in 3 days. How many days does it take to build 11 houses? a stone mason builds 7 houses in 3 days. How many days does it take to build 11 houses? The build rate of houses per days is proportional. Set up a proportion of [I]houses to days[/I] where d is the number of days it takes to build 11 houses: 7/3 = 11/d Cross multiply: Numerator 1 Denominator 2 = Denominator 1 * Numerator 2 7d = 11 * 3 7d = 33 Divide each side of the equation by 7: 7d/7 = 33/7 d = [B]4.7142857142857[/B] Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 45 minutes in a week, how much time did she devote to homework in the same week If Amanda spent 2/5 of her time after school doing homework, she has 1 - 2/5 time left over. We convert 1 to a fraction using a denominator of 5, we get: 5/5 - 2/5 = 3/5 And Amanda spent 1/4 of 3/5 of her time bike riding, which means she spent: 1(3)/4(5) = 3/20 of her time. If the total time after school is t, we have: 3t/20 = 45 [URL='https://www.mathcelebrity.com/prop.php?num1=3t&num2=45&den1=20&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Typing in 3t/20 = 45 to our search engine[/URL], we get t = 300. So Amanda has 300 total minutes after school, which means she spent 2/5(300) = [B]120 minutes (2 hours)[/B] doing homework. An ancient Greek was said to have lived 1/4 of his live as a boy, 1/5 as a youth, 1/3 as a man, and An ancient Greek was said to have lived 1/4 of his live as a boy, 1/5 as a youth, 1/3 as a man, and spent the last 13 years as an old man. How old was he when he died? Set up his life equation per time lived as a boy, youth, man, and old man 1/4 + 1/5 + 1/3 + x = 1. Using our [URL='http://www.mathcelebrity.com/gcflcm.php?num1=4&num2=3&num3=5&pl=LCM']LCM Calculator[/URL], we see the LCM of 3,4,5 is 60. This is our common denominator. So we have 15/60 + 12/60 + 20/60 + x/60 = 60/60 [U]Combine like terms[/U] x + 47/60 = 60/60 [U]Subtract 47/60 from each side:[/U] x/60 = 13/60 x = 13 out of the 60 possible years, so he was [B]60 when he died[/B]. An experienced accountant can balance the books twice as fast as a new accountant. Working together An experienced accountant can balance the books twice as fast as a new accountant. Working together it takes the accountants 10 hours. How long would it take the experienced accountant working alone? Person A: x/2 job per hour Person B: 1/x job per hour Set up our equation: 1/x + 1/(2x) = 1/10 Multiply the first fraction by 2/2 to get common denominators; 2/(2x) + 1/(2x) = 1/10 Combine like terms 3/2x = 1/10 Cross multiply: 30 = 2x Divide each side by 2: [B]x = 15[/B] Anna painted 1/6 of a wall, Eric painted 1/5 of the wall, and Meadow painted 1/4 of the wall. There Anna painted 1/6 of a wall, Eric painted 1/5 of the wall, and Meadow painted 1/4 of the wall. There are now 3910 square feet left to paint. How many square feet did Anna paint? [URL='https://www.mathcelebrity.com/gcflcm.php?num1=4&num2=5&num3=6&pl=LCM']Using 60 as a common denominator through least common multiple[/URL], we get: 1/6 = 10/60 1/5 = 12/60 1/4 = 15/60 10/60 + 12/60 + 15/60 = 37/60 Remaining part of the wall is 60/60 - 37[B]/[/B]60 = 23/60 3910/23 = 170 for each 1/60 of a wall Anna painted 1/6 or 10/60 of the wall. So we multiply 170 * 10 = [B]1,700 square feet[/B] at a party, there are 72 people. The ratio of men to ladies to kids is 4 to 3 to 2. at a party, there are 72 people. The ratio of men to ladies to kids is 4 to 3 to 2. [LIST] []How many men at the party? []How many ladies at the party? []How many kids at the party? [/LIST] Our total ratio denominator is 4 + 3 + 2 = 9. To find the number of each type of person, we take their ratio divided by their ratio numerator times 72 people at the party [U]Calculate ratios:[/U] [LIST] []Men: [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F9&frac2=72&pl=Multiply']4/9 * 72[/URL] = [B]32[/B] []Ladies: [URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F9&frac2=72&pl=Multiply']3/9 72[/URL] = [B]24[/B] []Kids: [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F9&frac2=72&pl=Multiply']2/9 72[/URL] = [B]16[/B] [/LIST] [U]Check our work:[/U] Men + Ladies + Kids = 32 + 24 + 16 Men + Ladies + Kids = 72 <-- This checks out! Cards in a pack are either orange or purple. 80% of the cards are orange. Write the ratio of orange Cards in a pack are either orange or purple. 80% of the cards are orange. Write the ratio of orange cards to purple cards. [URL='https://www.mathcelebrity.com/perc.php?num=+5&den=+8&num1=+16&pct1=+80&pct2=+90&den1=+80&pct=80&pcheck=4&decimal=+65.236&astart=+12&aend=+20&wp1=20&wp2=30&pl=Calculate']80% as a fraction [/URL]is 4/5. Fractions to ratios can be written as numerator : denominator, so we have: [B]4:5[/B] Conjugates Free Conjugates Calculator - Simplifies a fraction with conjugates in the denominator difference between 2 positive numbers is 3 and the sum of their squares is 117 difference between 2 positive numbers is 3 and the sum of their squares is 117 Declare variables for each of the two numbers: [LIST] []Let the first variable be x []Let the second variable be y [/LIST] We're given 2 equations: [LIST=1] []x - y = 3 []x^2 + y^2 = 117 [/LIST] Rewrite equation (1) in terms of x by adding y to each side: [LIST=1] []x = y + 3 []x^2 + y^2 = 117 [/LIST] Substitute equation (1) into equation (2) for x: (y + 3)^2 + y^2 = 117 Evaluate and simplify: y^2 + 3y + 3y + 9 + y^2 = 117 Combine like terms: 2y^2 + 6y + 9 = 117 Subtract 117 from each side: 2y^2 + 6y + 9 - 117 = 117 - 117 2y^2 + 6y - 108 = 0 This is a quadratic equation: Solve the quadratic equation 2y2+6y-108 = 0 With the standard form of ax2 + bx + c, we have our a, b, and c values: a = 2, b = 6, c = -108 Solve the quadratic equation 2y^2 + 6y - 108 = 0 The quadratic formula is denoted below: y = -b ± sqrt(b^2 - 4ac)/2a [U]Step 1 - calculate negative b:[/U] -b = -(6) -b = -6 [U]Step 2 - calculate the discriminant Δ:[/U] Δ = b2 - 4ac: Δ = 62 - 4 x 2 x -108 Δ = 36 - -864 Δ = 900 <--- Discriminant Since Δ is greater than zero, we can expect two real and unequal roots. [U]Step 3 - take the square root of the discriminant Δ:[/U] √Δ = √(900) √Δ = 30 [U]Step 4 - find numerator 1 which is -b + the square root of the Discriminant:[/U] Numerator 1 = -b + √Δ Numerator 1 = -6 + 30 Numerator 1 = 24 [U]Step 5 - find numerator 2 which is -b - the square root of the Discriminant:[/U] Numerator 2 = -b - √Δ Numerator 2 = -6 - 30 Numerator 2 = -36 [U]Step 6 - calculate your denominator which is 2a:[/U] Denominator = 2 * a Denominator = 2 * 2 Denominator = 4 [U]Step 7 - you have everything you need to solve. Find solutions:[/U] Solution 1 = Numerator 1/Denominator Solution 1 = 24/4 Solution 1 = 6 Solution 2 = Numerator 2/Denominator Solution 2 = -36/4 Solution 2 = -9 [U]As a solution set, our answers would be:[/U] (Solution 1, Solution 2) = (6, -9) Since one of the solutions is not positive and the problem asks for 2 positive number, this problem has no solution Divide 10 by the difference of z and y [U]The difference of z and y means we subtract y from z[/U] z - y [U]Now, we form a fraction, where 10 is the numerator and z - y is the denominator[/U] 10/(z - y) double v, add u, then divide t by what you have double v, add u, then divide t by what you have Double v means we multiply the variable v by 2: 2v Add u: 2v + u We build a fraction, with t as the numerator, and 2v + u as the denominator [B]t/(2v + u)[/B] Explain the steps you would take to find an equation for the line perpendicular to 4x - 5y = 20 and Explain the steps you would take to find an equation for the line perpendicular to 4x - 5y = 20 and sharing the same y-intercept Get this in slope-intercept form by adding 5y to each side: 4x - 5y + 5y = 5y + 20 Cancel the 5y's on the left side and we get: 5y + 20 = 4x Subtract 20 from each side 5y + 20 - 20 = 4x - 20 Cancel the 20's on the left side and we get: 5y = 4x - 20 Divide each side by 5: 5y/5 = 4x/5 - 4 y = 4x/5 - 4 So we have a slope of 4/5 to find our y-intercept, we set x = 0: y = 4(0)/5 - 4 y = 0 - 4 y = -4 If we want a line perpendicular to the line above, our slope will be the negative reciprocal: The reciprocal of 4/5 is found by flipping the fraction making the numerator the denominator and the denominator the numerator: m = 5/4 Next, we multiply this by -1: -5/4 So our slope-intercept of the perpendicular line with the same y-intercept is: [B]y = -5x/4 - 4[/B] Factorials Free Factorials Calculator - Calculates the following factorial items: * A factorial of one number such as n! * A factorial of a numerator divided by a factorial of a denominator such as n!m!/a!b! * Double Factorials such as n!! * Stirlings Approximation for n! Fraction with variable x in numerator and 6 in the denominator. Fraction with variable x in numerator and 6 in the denominator. The numerator is the top of the fraction. The denominator is the bottom of the fraction. [B]x/6[/B] Greatest Common Factor and Least Common Multiple Free Greatest Common Factor and Least Common Multiple Calculator - Given 2 or 3 numbers, the calculator determines the following: * Greatest Common Factor (GCF) using Factor Pairs * Rewrite Sum using the Distributive Property and factoring out the GCF * Least Common Multiple (LCM) / Least Common Denominator (LCD) using Factor Pairs * GCF using the method of Successive Division * GCF using the Prime Factorization method * Determine if the numbers are coprime and twin prime Harry got 42 out of 49 correct in his test. What fraction of the marks did he get correct? Harry got 42 out of 49 correct in his test. What fraction of the marks did he get correct? The fraction correct is: 42/49 Both the numerator and denominator [URL='http://www.mathcelebrity.com/gcflcm.php?num1=42&num2=49&num3=&pl=GCF']have a common factor[/URL] of 7 Reducing top and bottom by 7, we get: [B]6/7[/B] Help Plz Nick's age: x John's age: x/2 Pip's age = 2/3 * x/2 = x/3 The sum is 26, so we have: x + x/2 + x/3 = 26 Common denominator is (1 * 2 * 3) = 6 6x/6 + 3x/6 + 2x/6 = 26 Combine like terms: 11x/6 = 26 Cross multiply: 11x = 156 x = 14.1818 This doesn't make sense for age. Are you sure you wrote out the problem right? The formula is: Numerator = One color Denominator - Numerator = Another Color [MEDIA=youtube]mJD2acCpol4[/MEDIA] I only own blue blankets and red blankets. 8 out of every 15 blankets I have are red. I only own blue blankets and red blankets. 8 out of every 15 blankets I have are red. If have i 45 blankets, how many are blue? If 8 out of 15 blankets are red, then 15 - 8 = 7 are blue So 7 out of every 15 blankets are blue. Set up a proportion of blue blankets to total blankets where b is the number of blue blankets in 45 blankets 7/15 = b/45 Cross multiply: If 2 proportions are equal, then we can do the following: Numerator 1 * Denominator 2 = Denominator 1 * Numerator 2 15b = 45 * 7 15b = 315 To solve for b, divide each side of the equation by 15: 15b/15 = 315/15 Cancel the 15's on the left side and we get: b = [B]21[/B] If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numer If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions. Convert 2 to a fraction with a denominator of 10: 20/2 = 10, so we multiply 2 by 10/10: 210/10 = 20/10 Add 2 to the numerator and denominator: (n + 2)/(d + 2) = 9/10 Cross multiply and simplify: 10(n + 2) = 9(d + 2) 10n + 20 = 9d + 18 Move constants to right side by subtracting 20 from each side and subtracting 9d: 10n - 9d = -2 Subtract 3 from the numerator and denominator: (n - 3)/(d - 3) = 4/5 Cross multiply and simplify: 5(n - 3) = 4(d - 3) 5n - 15 = 4d - 12 Move constants to right side by adding 15 to each side and subtracting 4d: 5n - 4d = 3 Build our system of equations: [LIST=1] []10n - 9d = -2 []5n - 4d = 3 [/LIST] Multiply equation (2) by -2: [LIST=1] []10n - 9d = -2 []-10n + 8d = -6 [/LIST] Now add equation (1) to equation (2) (10 -10)n (-9 + 8)d = -2 - 6 The n's cancel, so we have: -d = -8 Multiply through by -1: d = 8 Now bring back our first equation from before, and plug in d = 8 into it to solve for n: 10n - 9d = -2 10n - 9(8) = -2 10n - 72 = -2 To solve for n, we [URL='https://www.mathcelebrity.com/1unk.php?num=10n-72%3D-2&pl=Solve']plug this equation into our search engine[/URL] and we get: n = 7 So our fraction, n/d = [B]7/8[/B] if 200 is divided in the ratio of 1:3:4 , what is the greatest number if 200 is divided in the ratio of 1:3:4 , what is the greatest number Determine the ratio denominator by adding up the ratio amounts: 1 + 3 + 4 = 8 So we have the following ratios and ratio amounts with our greatest number in bold: [LIST] []1/8 * 200 = 25 []3/8 200 = 75 []4/8 200 = [B]100[/B] [/LIST] If 9/20 of a salad is eaten, how much is leftover? If 9/20 of a salad is eaten, how much is leftover? The full salad is 1. Using a common denominator, we have 1 = 20/20 [URL='https://www.mathcelebrity.com/fraction.php?frac1=1&frac2=9%2F20&pl=Subtract']So the leftover is 20/20 - 9/20[/URL] = [B]11/20[/B] In Trina's desk drawer, there are 15 paper clips and 18 rubber bands. In Kirk's office supply tray, In Trina's desk drawer, there are 15 paper clips and 18 rubber bands. In Kirk's office supply tray, there are 13 paper clips and 16 rubber bands. Who has a higher ratio of paper clips to rubber bands? Trina: 15/18 Kirk: 13/16 We want common denominators to compare, so we get a greatest common factor (GCF) for 16 and 18. [URL='https://www.mathcelebrity.com/gcflcm.php?num1=16&num2=18&num3=&pl=GCF+and+LCM']Running this through our search engine[/URL], we get GCF(16, 18) = 144 For Trina, 144/18 = 8 For Kirk, 144/16 = 9 We multiply Trina's fraction, top and bottom by 8: 15 * 8 / 18 * 8 120/144 We multiply Trina's fraction, top and bottom by 8: 13 * 8 / 16 * 8 104/144 [B]Trina[/B] has more in her numerator, so her ratio of paper clips to rubber bands is greater. John mows 3 lawns in 4 hours, Paul mows 5 lawns in 6 hours. Who mows faster? John mows 3 lawns in 4 hours, Paul mows 5 lawns in 6 hours. Who mows faster? To see who mows faster, we set up fractions with a common denominator. You can see this by running this statement in the calculator: [URL='https://www.mathcelebrity.com/fraction.php?frac1=3/4&frac2=5/6&pl=Compare']3/4 or 5/6[/URL] You'll see that 5/6 is larger, so Paul mores more lawns per hour. Kyle can walk ½ mile in ¼ of an hour. What is Kyle’s speed in miles per hour? Kyle can walk ½ mile in ¼ of an hour. What is Kyle’s speed in miles per hour? We write this in terms of miles per hour as: 1/2 / 1/4 We want 1 for the denominator to represent an hour, so we multiply top and bottom of the fraction by 4: 4/2 / 4/4 2 / 1 [B]2 miles per hour[/B] Laura found a roll of fencing in her garage. She couldn't decide whether to fence in a square garden Laura found a roll of fencing in her garage. She couldn't decide whether to fence in a square garden or a round garden with the fencing. Laura did some calculations and found that a circular garden would give her 1380 more square feet than a square garden. How many feet of fencing were in the roll that Laura found? (Round to the nearest foot.) Feet of fencing = n Perimeter of square garden = n Each side of square = n/4 Square garden's area = (n/4)^2 = n^2/16 Area of circle garden with circumference = n is: Circumference = pi * d n = pi * d Divide body tissues by pi: d = n/pi Radius = n/2pi Area = pi * n/2pi * n/2pi Area = pin^2/4pi^2 Reduce by canceling pi: n^2/4pi n^2/4 * 3.14 n^2/12.56 The problem says that the difference between the square's area and the circle's area is equal to 1380 square feet. Area of Circle - Area of Square = 1380 n^2/12.56 - n^2/16 = 1380 Common denominator = 200.96 (16n^2 - 12.56n^2)/200.96 = 1380 3.44n^2/200.96 = 1380 Cross multiply: 3.44n^2 = 277,324.8 n^2 = 80,617.7 n = 283.9 Nearest foot = [B]284[/B] Leilani can read 20 pages in 2 minutes. if she can maintain this page, how many pages can she read in an hour? We know that 1 hour is 60 minutes. Let p be the number of pages Leilani can read in 1 hour (60 minutes) The read rate is constant, so we can build a proportion. 20 pages /2 minutes = p/60 We can cross multiply: Numerator 1 * Denominator 2 = Denominator 1 * Numerator 2 [SIZE=5][B]Solving for Numerator 2 we get:[/B][/SIZE] Numerator 2 = Numerator 1 * Denominator 2/Denominator 1 [SIZE=5][B]Evaluating and simplifying using your input values we get:[/B][/SIZE] p = 20 * 60/ 2 p = 1200/2 p = [B]600[/B] m=u/k-r/k for k m=u/k-r/k for k Multiply both sides by k to eliminate the k denominator: km = uk/k - rk/k Cancel the k's on the right side and we get km = u - r Divide each side by m: km/m = (u - r)/m Cancel the m on the left side: [B]k = (u - r)/m[/B] n + n/2 + n/4 + n/8 + n/16 = 19,375 n + n/2 + n/4 + n/8 + n/16 = 19,375 Convert to like fractions with a denominator of 16: 16n/16 + 8n/16 + 4n/16 + +2n/16 + n/16 = 19,375 31n/16 = 19,375 Cross multiply: 31n = 19,375 * 16 31n = 310000 Divide each side by 1: 31n/31 = 310000/31 n = [B]10,000[/B] numerator of a fraction is 5 less than its denominator. if 1 is added to the numerator and to the de numerator of a fraction is 5 less than its denominator. if 1 is added to the numerator and to the denominator the new fraction is 2/3. find the fraction. Let n be the numerator. Let d be the denominator. We're given 2 equations: [LIST=1] []n = d - 5 [](n + 1)/(d + 1) = 2/3 [/LIST] Substitute equation (1) into equation (2) for n: (d - 5 + 1) / (d + 1) = 2/3 (d - 4) / (d + 1) = 2/3 Cross multiply: 3(d - 4) = 2(d + 1) To solve this equation for d, we type it in our search engine and we get: d = 14 Substitute d = 14 into equation (1) to solve for n: n = 14 - 5 n = 9 Therefore, our fraction n/d is: [B]9/14[/B] One positive number is one-fifth of another number. The difference between the two numbers is 192, f One positive number is one-fifth of another number. The difference between the two numbers is 192, find the numbers. Let the first number be x and the second number be y. We're given two equations: [LIST=1] []x = y/5 []x + y = 192 [/LIST] Substitute equation 1 into equation 2: y/5 + y = 192 Since 1 equals 5/5, we rewrite our equation like this: y/5 = 5y/5 = 192 We have fractions with like denominators, so we add the numerators: (1 + 5)y/5 = 192 6y/5 = 192 [URL='https://www.mathcelebrity.com/prop.php?num1=6y&num2=192&den1=5&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Type this equation into our search engine[/URL], and we get: [B]y = 160[/B] Substitute this value into equation 1: x = 160/5 x = [B]32[/B] Percentage-Decimal-Fraction Relations Free Percentage-Decimal-Fraction Relations Calculator - Calculates the relational items between a fraction, a decimal (including repeating decimal and terminating decimal), a percentage, and the numerator and denominator piece of that fraction. Also calculates the percentage change going from one number to another or the amount increase or decrease of a percentage above/below a number. Round decimals. decimals into fractions Pete ate 1/2 a pizza, Carol ate 1/3, and Joe ate 1/4. How much pizza was eaten? Pete ate 1/2 a pizza, Carol ate 1/3, and Joe ate 1/4. How much pizza was eaten? Total pizza eaten is: 1/2 + 1/3 + 1/4 We need a common denominator. List out factors: 2: 1, 2, 4, 6, 8, 10, 12 3: 1, 3, 6, 9, 12 4: 1, 4, 8, 12 12 is our least common multiple: 1/2 = 6/12 1/3 = 4/12 1/4 = 3/12 So our addition becomes: (6 + 4 + 3)/12 = 13/12 of a pizza Prove the sum of any two rational numbers is rational Take two integers, r and s. We can write r as a/b for integers a and b since a rational number can be written as a quotient of integers We can write s as c/d for integers c and d since a rational number can be written as a quotient of integers Add r and s: r + s = a/b + c/d With a common denominator bd, we have: r + s = (ad + bc)/bd Because a, b, c, and d are integers, ad + bc is an integer since rational numbers are closed under addition and multiplication. Since b and d are non-zero integers, bd is a non-zero integer. Since we have the quotient of 2 integers, r + s is a rational number. [MEDIA=youtube]0ugZSICt_bQ[/MEDIA] raise t to the 10th power, then find the quotient of the result and s raise t to the 10th power, then find the quotient of the result and s Raise t to the 10th power means we use t as our variable and 10 as our exponent: t^10 The quotient means a fraction, where the numerator is t^10 and the denominator is s: [B]t^10/s[/B] Sergeant U has 360 rounds of ammunition to distribute to Lance Corporal (LCpl) F, Lance Corporal (LC Sergeant U has 360 rounds of ammunition to distribute to Lance Corporal (LCpl) F, Lance Corporal (LCpl) M and Lance Corporal (LCpl) Z in the ratio 3:5:7. How many rounds did Lance Corporal (LCpl) M receive? Our ratio denominator is: 3 + 5 + 7 = 15 Lance Corporal (LCpl) M gets 5:15 of the ammunition. [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F15&frac2=3%2F8&pl=Simplify']Using our fraction simplifier[/URL], we see that 5/15 = 1/3 So we take 360 rounds of ammunition times 1/3: 360/3 = [B]120[/B] Simplify 2/sqrt(5) To do this, we need rationalize the denominator. This means get rid of the radical: Multiply top and bottom by sqrt(5) 2sqrt(5)/sqrt(5) * sqrt(5) sqrt(5) * sqrt(5) = sqrt(25) so we have: 2sqrt(5)/sqrt(25) sqrt(25) = 5, so we have: [B]2sqrt(5)/5[/B] [MEDIA=youtube]jearVN9LhBE[/MEDIA] Solve for x Expand the right side: 1/3x + 1/2 = 6/4x - 10 Simplify as 6/4 is 3/2 x/3 + 1/2 = 3x/2 - 10 Common denominator of 2 and 3 is 6. So we have: 2x/6 + 1/2 = 9x/6 -10 Subtract 2x/6 from each side 7x/6 - 10 = 1/2 Add 10 to each side. 10 is 20/2 7x/6 = 21/2 Using our [URL='http://www.mathcelebrity.com/prop.php?num1=7x&num2=21&den1=6&den2=2&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator[/URL], we get: [B]x = 9[/B] The denominator of a fraction is 4 more than the numerator. If 4 is added to the numerator and 7 is The denominator of a fraction is 4 more than the numerator. If 4 is added to the numerator and 7 is added to the denominator, the value of the fraction is 1/2. Find the original fraction. Let the original fraction be n/d. We're given: [LIST=1] []d = n + 4 [](n + 4) / (d + 7) = 1/2 [/LIST] Cross multiply Equation 2: 2(n + 4) = d + 7 2n + 8 = d + 7 Now substitute equation (1) into tihs: 2n + 8 = (n + 4) + 7 2n + 8 = n + 11 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B8%3Dn%2B11&pl=Solve']Type this equation into our search engine[/URL], and we get: n = 3 This means from equation (1), that: d = 3 + 4 d = 7 So our original fraction n/d = [B]3/7[/B] The fraction has a value of 3/5. The sum of the numerator and the denominator was 40. What was the f The fraction has a value of 3/5. The sum of the numerator and the denominator was 40. What was the fraction? We're given two equations with a fraction with numerator (n) and denominator (d): [LIST=1] []n + d = 40 []n/d = 3/5 [/LIST] Cross multiply equation 2, we get: 5n = 3d Divide each side by 5: 5n/5 = 3d/5 n = 3d/5 Substitute this into equation 1: 3d/5 + d = 40 Multiply through both sides of the equation by 5: 5(3d/5) = 5d = 40 * 5 3d + 5d =200 To solve this equation for d, we [URL='https://www.mathcelebrity.com/1unk.php?num=3d%2B5d%3D200&pl=Solve']type it in our search engine and we get[/URL]: d = [B]25 [/B] Now substitute that back into equation 1: n + 25 = 40 Using [URL='https://www.mathcelebrity.com/1unk.php?num=n%2B25%3D40&pl=Solve']our equation solver again[/URL], we get: n = [B]15[/B] The quotient of 49 and n squared n squared is written as n to the power of 2, n^2 We have a fraction, where 49 is the numerator, and n^2 is the denominator 49 ----- n^2 the quotient of 77 and x the quotient of 77 and x Quotient means we have a fraction where 77 is the numerator and x is the denominator: [B]77/x[/B] the quotient of 8 and the difference of x and m The difference of x and m means we subtract: x - m Quotient means a fraction. 8 is the numerator, and x - m is the denominator: [B] 8 ------ x - m[/B] the quotient of a variable and 7 the quotient of a variable and 7. A variable means an arbitrary number, let's call it x. A quotient means a fraction, where x is the numerator and 7 is the denominator: [B] x --- 7[/B] the quotient of the cube of a number x and 5 the quotient of the cube of a number x and 5 [LIST] []A number means an arbitrary variable, let's call it x []The cube of a number means raise it to the 3rd power, so we have x^3 []Quotient means we have a fraction, so our numerator is x^3, and our denominator is 5 [/LIST] [B]x^3 ---- 5[/B] The sum of three numbers is 171. The second number is 1/2 of the first and the third is 3/4 of the f The sum of three numbers is 171. The second number is 1/2 of the first and the third is 3/4 of the first. Find the numbers. We have three numbers, x, y, and z. [LIST=1] []x + y + z = 171 []y = 1/2x []z = 3/4x [/LIST] Substitute (2) and (3) into (1) x + 1/2x + 3/4x = 171 Use a common denominator of 4 for each x term 4x/4 + 2x/4 + 3x/4 = 171 (4 + 2 + 3)x/4 = 171 9x/4 = 171 [URL='https://www.mathcelebrity.com/prop.php?num1=9x&num2=171&den1=4&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Plug this equation into our search engine[/URL], and we get [B]x = 76[/B] So y = 1/2(76) --> [B]y = 38[/B] Then z = 3/4(76) --> [B]z = 57[/B] The sum of x and one half of x The sum of x and one half of x To write this algebraic expression correctly, we have (1 + 1/2)x To get common denominators, we write 1 as 2/2. So we have: (2/2 + 1/2)x [B]3/2x[/B] What fraction lies exactly halfway between 2/3 and 3/4? What fraction lies exactly halfway between 2/3 and 3/4? A) 3/5 B) 5/6 C) 7/12 D) 9/16 E) 17/24 Halfway means taking the average, which is dividing the sum of the fractions by 2 for 2 fractions: 1/2(2/3 + 3/4) 1/2(2/3) + 1/2(3/4) 1/3 + 3/8 We need common denominators, so [URL='https://www.mathcelebrity.com/fraction.php?frac1=1%2F3&frac2=3%2F8&pl=Add']we type this fraction sum into our search engine[/URL] and get: [B]17/24 - Answer E[/B] Which of the followings can increase the value of t? (select all the apply) a. Increase the standar Which of the followings can increase the value of t? (select all the apply) a. Increase the standard deviation of difference scores b. Decrease the standard deviation of difference scores c. Increase the difference between means d. Decrease the difference between means [B]b. Decrease the standard deviation of difference scores c. Increase the difference between means[/B] [I]Increase numerator or decrease denominator of the t-value formula[/I] Write a fraction with a denominator of 9. The fraction should be less than 1/2 Write a fraction with a denominator of 9. The fraction should be less than 1/2 Let n be the numerator. We have: n/9 < 1/2 multiply each side by 9: 9n/9 < 9/2 n < 9/2 Examples are 8/2, 7/2, 6/2, 5/2, 4/2, 3/2, x/y + 9 = n for x x/y + 9 = n for x Subtract 9 from each side to isolate the x term: x/y + 9 - 9 = n - 9 Cancel the 9's on the left side and we get: x/y = n - 9 Because we have a fraction on the left side, we can cross multiply the denominator y by n - 9 [B]x =[/B] [B]y(n - 9)[/B] You have 30 DONUTS. 1/6 of them are Boston Cream. 2/5 of them are Maple. 3/10 of them are Chocolate You have 30 DONUTS. 1/6 of them are Boston Cream. 2/5 of them are Maple. 3/10 of them are Chocolate Dip and 1/3 are Sprinkled. IS THIS EVEN POSSIBLE?? How many donuts OVER or UNDER am I? (Show your work and use EQUIVALENTS.) We use 30 as our common denominator. Let's get [I]equivalent fraction[/I]s for each donut type with a denominator of 30: [LIST] [][URL='https://www.mathcelebrity.com/equivalent-fractions.php?num=1%2F6&pl=Equivalent+Fractions']1/6[/URL] = 5/30 [][URL='https://www.mathcelebrity.com/equivalent-fractions.php?num=2%2F5&pl=Equivalent+Fractions']2/5 [/URL]= 12/30 [][URL='https://www.mathcelebrity.com/equivalent-fractions.php?num=3%2F10&pl=Equivalent+Fractions']3/10[/URL] = 9/30 [][URL='https://www.mathcelebrity.com/equivalent-fractions.php?num=1%2F3&pl=Equivalent+Fractions']1/3[/URL] = 10/30 [/LIST] Add up our numerators of the common denominator of 30: 5 + 12 + 9 + 10 = 36 So our fraction is 36/30. This makes our scenario [B]impossible[/B]. Fractions of the donut should add up to 1. Which would mean our numerators need to sum to 1 or less. Since 36 > 30, this scenario is [B]impossible.[/B] z/w=x+z/x+y for z z/w=x+z/x+y for z This is a literal equation. Let's isolate z on one side. Subtract z/x from each side. z/w - z/x = x + y Factor our z on the left side: z(1/w - 1/x) = x + y Divide each side by (1/w - 1/x) z = x + y/(1/w - 1/x) To remove reciprocals in the denominator, we rewrite 1/w - 1/x with a common denominator xw (x - w)/xw Then multiply x + y by the reciprocal z = [B](x + y)xw/(x - w)[/B]
# Linear Algebra/Introduction to Matrices and Determinants (Redirected from Linear Algebra/Introduction to Determinants) The determinant is a function which associates to a square matrix an element of the field on which it is defined (commonly the real or complex numbers). ## Matrices Organization of a matrix Informally an m×n matrix (plural matrices) is a rectangular table of entries from a field (that is to say that each entry is an element of a field). Here m is the number of rows and n the number of the columns in the table. Those unfamiliar with the concept of a field, can for now assume that by a field of characteristic 0 (which we will denote by F) we are referring to a particular subset of the set of complex numbers. An m×n matrix (read as m by n matrix), is usually written as: $A=\left(\begin{matrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \end{matrix}\right)$ The $i^{th}$ row is an element of $F^n$, showing the n components $\begin{pmatrix}a_{i1} & a_{i2} &\cdots a_{in} \end{pmatrix}$. Similary the $j^{th}$ column is an element of $F^m$ showing the m components $\begin{pmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{mj}\end{pmatrix}$. Here m and n are called the dimensions of the matrix. The dimensions of a matrix are always given with the number of rows first, then the number of columns. It is also said that an m by n matrix has an order of m×n. Formally, an m×n matrix M is a function $M:A\rightarrow F$ where A = {1,2...m} × {1,2...n} and F is the field under consideration. It is almost always better to visualize a matrix as a rectangular table (or array) then as a function. A matrix having only one row is called a row matrix (or a row vector) and a matrix having only one column is called a column matrix (or a column vector). Two matrices of the same order whose corresponding entries are equal are considered equal. The (i,j)-entry of the matrix (often written as $A_{ij}$ or $A_{i,j}$) is the element at the intersection of the $i^{th}$ row (from the top) and the $j^{th}$ column (from the left). For example, $\begin{pmatrix} 3 & 4 & 8 \\ 2 & 7 & 11 \\ 1 & 1 & 1 \end{pmatrix}$ is a 3×3 matrix (said 3 by 3). The 2nd row is $\begin{pmatrix}2 & 7 & 11 \end{pmatrix}$ and the 3rd column is $\begin{pmatrix} 8\\ 11 \\ 1\end{pmatrix}$. The (2,3) entry is the entry at intersection of the 2nd row and the 3rd column, that is 11. Some special kinds of matrices are: • A square matrix is a matrix which has the same number of rows and columns. A diagonal matrix is a matrix with non zero entries only on the main diagonal (ie at $A_{i,i}$ positions). • The unit matrix or identity matrix In, is the matrix with elements on the diagonal set to 1 and all other elements set to 0. Mathematically, we may say that for the identity matrix $I_{i,j}$ (which is usually written as $\delta_{i,j}$ and called Kronecker's delta) is given by: $\delta_{i,j} = \begin{cases} 1, & i=j \\ 0, & i \neq j \end{cases}$ For example, if n = 3: $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} .$ • The transpose of an m-by-n matrix A is the n-by-m matrix AT formed by turning rows into columns and columns into rows, i.e. $A_{i,j} = A^T_{j,i} \forall i,j$. An example is $\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}^{\mathrm{T}} \!\! \;\! = \, \begin{bmatrix} 1 & 3 & 5\\ 2 & 4 & 6 \end{bmatrix} \;$ • A square matrix whose transpose is equal to itself is called a symmetric matrix; that is, A is symmetric if $A^{\mathrm{T}} = A.\,$. An example is $\begin{bmatrix} 1 & 2 & 3\\ 2 & 4 & -5\\ 3 & -5 & 6\end{bmatrix}$ • A square matrix whose transpose is equal to its negative is called skew-symmetric matrix; that is, A is skew-symmetric if $A^{\mathrm{T}} = -A.\,$. An example is $\begin{bmatrix} 0 & -3 & 4\\ 3 & 0 & -5\\ -4 & 5 & 0\end{bmatrix}$ Properties of these matrices are developed in the exercises. ## Determinants To define a determinant of order n, suppose there are n2 elements of a field sij where i and j are less than or equal to n. Define the following function (this function is important in the definition): S(a1,a2,a3,...,an)=# of reversals, meaning the number of times an1<an2 when n1>n2, for each possible combination. Suppose you have a permutation of numbers from 1 to n {a1,a2,a3,...,an). Then define a term of the determinant to be equal to (-1)S(a1,a2,a3,...,an)s1a1,s2a2,s3a3,...,snan. The sum of all possible terms (i. e. through all possible permutations) is called the determinant. ## Theorem Definition: The transpose of a matrix A, AT is the matrix resulting when the columns and rows are interchanged i. e. the matrix sji when A is the matrix sij A matrix and its transpose have the same determinant: $\det(A^\top) = \det(A). \,$ ### Proof All terms are the same, and the signs of the terms are also unchanged since all reversals remain reversals. Thus, the sum is the same. ## Theorem Interchanging two rows (or columns) changes the sign of the determinant: $\det \begin{bmatrix} \cdots \\ \mbox{row A} \\ \cdots \\ \mbox{row B} \\ \cdots \end{bmatrix} = - \det \begin{bmatrix}\cdots \\ \mbox{row B} \\ \cdots \\ \mbox{row A} \\ \cdots \end{bmatrix}$. ### Proof To show this, suppose two adjacent rows (or columns) are interchanged. Then any reversals in a term would not be affected except for the reversal of the elements of that term within that row (or column), in which case adds or subtracts a reversal, thus changing the signs of all terms, and thus the sign of the matrix. Now, if two rows, the ath row and the (a+n)th are interchanged, then interchange successively the ath row and (a+1)th row, and then the (a+1)th row and (a+2)th row, and continue in this fashion until one reaches the (a+n-1)th row. Then go backwards until one goes back to the ath row. This has the same effect as switching the ath row and the (a+n)th rows, and takes n-1 switches for going forwards, and n-2 switches for going backwards, and their sum must then be an odd number, so it multiplies by -1 an odd number of times, so that its total effect is to multiply by -1. ### Corollary A determinant with two rows (or columns) that are the same has the value 0. Proof: This determinant would be the additive inverse of itself since interchanging the rows (or columns) does not change the determinant, but still changes the sign of the determinant. The only number for which it is possible is when it is equal to 0. ## Theorem It is linear on the rows and columns of the matrix. $\det \begin{bmatrix} \ddots & \vdots & \ldots \\ \lambda a_1 + \mu b_1 & \cdots & \lambda a_n + \mu b_n \\ \cdots & \vdots & \ddots \end{bmatrix} = \lambda \det \begin{bmatrix} \ddots & \vdots & \cdots \\ a_1 & \cdots & a_n \\ \cdots & \vdots & \ddots \end{bmatrix} + \mu \det \begin{bmatrix} \ddots & \vdots & \cdots \\ b_1 & \cdots & b_n \\ \cdots & \vdots & \ddots \end{bmatrix}$ ### Proof The terms are of the form a1...$\lambda a + \mu b$...an. Using the distributive law of fields, this comes out to be a1...$\lambda a$...an + a1...$\mu b$...an, an thus its sum of such terms is the sum of the two determinants: $\lambda \det \begin{bmatrix} \ddots & \vdots & \cdots \\ a_1 & \cdots & a_n \\ \cdots & \vdots & \ddots \end{bmatrix} + \mu \det \begin{bmatrix} \ddots & \vdots & \cdots \\ b_1 & \cdots & b_n \\ \cdots & \vdots & \ddots \end{bmatrix}$ ### Corollary Adding a row (or column) times a number to another row (or column) does not affect the value of a determinant. #### Proof Suppose you have a determinant A with the kth column added by another column times a number: $\begin{bmatrix} a_{11} & a_{12} & a_{13} & \ldots & a_{1k}+\mu a_{1b} & \ldots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \ldots & a_{2k}+\mu a_{2b} & \ldots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \ldots & a_{3k}+\mu a_{3b} & \ldots & a_{3n} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \ldots & a_{nk}+\mu a_{nb} & \ldots & a_{nn} \end{bmatrix}$ where akb are elements of another column. By the linear property, this is equal to $\begin{bmatrix} a_{11} & a_{12} & a_{13} & \ldots & a_{1k} & \ldots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \ldots & a_{2k} & \ldots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \ldots & a_{3k} & \ldots & a_{3n} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \ldots & a_{nk} & \ldots & a_{nn} \end{bmatrix} + \begin{bmatrix} a_{11} & a_{12} & a_{13} & \ldots & \mu a_{1b} & \ldots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \ldots & \mu a_{2b} & \ldots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \ldots & \mu a_{3b} & \ldots & a_{3n} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \ldots & \mu a_{nb} & \ldots & a_{nn} \end{bmatrix}$ The second number is equal to 0 because it has two columns that are the same. Thus, it is equal to $\begin{bmatrix} a_{11} & a_{12} & a_{13} & \ldots & a_{1k} & \ldots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \ldots & a_{2k} & \ldots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \ldots & a_{3k} & \ldots & a_{3n} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \ldots & a_{nk} & \ldots & a_{nn}\end{bmatrix}$ which is the same as the matrix A. • It is easy to see that $\det(rI_n) = r^n \,$ and thus $\det(rA) = \det(rI_n \cdot A) = r^n \det(A) \,$ for all $n$-by-$n$ matrices $A$ and all scalars $r$. • A matrix over a commutative ring R is invertible if and only if its determinant is a unit in R. In particular, if A is a matrix over a field such as the real or complex numbers, then A is invertible if and only if det(A) is not zero. In this case we have $\det(A^{-1}) = \det(A)^{-1}. \,$ Expressed differently: the vectors v1,...,vn in Rn form a basis if and only if det(v1,...,vn) is non-zero. The determinants of a complex matrix and of its conjugate transpose are conjugate: $\det(A^) = \det(A)^. \,$
Question # If ${{\text{S}}_1}$ and ${{\text{S}}_2}$ are respectively the sets of local minimum and local maximum points of the functions $f\left( x \right) = 9{x^4} + 12{x^3} - 36{x^2} + 25,x \in R$, then write the elements contained by the two sets ${{\text{S}}_1}$ and ${{\text{S}}_2}$.${\text{A}}{\text{. }}{{\text{S}}_1} = \left\{ { - 2,1} \right\};{{\text{S}}_2} = \left\{ 0 \right\} \\ {\text{B}}{\text{. }}{{\text{S}}_1} = \left\{ { - 2,0} \right\};{{\text{S}}_2} = \left\{ 1 \right\} \\ {\text{C}}{\text{. }}{{\text{S}}_1} = \left\{ { - 2} \right\};{{\text{S}}_2} = \left\{ {0,1} \right\} \\ {\text{D}}{\text{. }}{{\text{S}}_1} = \left\{ { - 1} \right\};{{\text{S}}_2} = \left\{ {0,2} \right\} \\$ Hint- Here, we will proceed by differentiating the given function once and then putting $f'\left( x \right) = 0$ in order to obtain the values of x where local maxima and local minima can occur according to the sign of $f''\left( x \right)$. The given function in x is $f\left( x \right) = 9{x^4} + 12{x^3} - 36{x^2} + 25{\text{ }} \to {\text{(1)}},x \in R$ It is also given that ${{\text{S}}_1}$ corresponds to the set of values of x where local minima occurs and ${{\text{S}}_2}$ corresponds to the set of values of x where local maxima occurs. Let us differentiate the given function with respect to x, we get $\Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = \dfrac{d}{{dx}}\left[ {9{x^4} + 12{x^3} - 36{x^2} + 25} \right] \\ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {9{x^4}} \right] + \dfrac{d}{{dx}}\left[ {12{x^3}} \right] - \dfrac{d}{{dx}}\left[ {36{x^2}} \right] + \dfrac{d}{{dx}}\left[ {25} \right] \\ \Rightarrow f'\left( x \right) = 9\dfrac{d}{{dx}}\left[ {{x^4}} \right] + 12\dfrac{d}{{dx}}\left[ {{x^3}} \right] - 36\dfrac{d}{{dx}}\left[ {{x^2}} \right] + 0 \\ \Rightarrow f'\left( x \right) = 9\left( {4{x^3}} \right) + 12\left( {3{x^2}} \right) - 36\left( {2x} \right) \\ \Rightarrow f'\left( x \right) = 36{x^3} + 36{x^2} - 72x \\ \Rightarrow f'\left( x \right) = 36x\left( {{x^2} + x - 2} \right){\text{ }} \to {\text{(2)}} \\$ Since, we know that local maxima or local minima are the points where local maximum and local minimum values will be occurring. At local maxima and local minima, for any function f(x) the necessary condition is $f'\left( x \right) = 0$. By putting $f'\left( x \right) = 0$ in equation (2), we get $\Rightarrow 0 = 36x\left( {{x^2} + x - 2} \right) \\ \Rightarrow x\left( {{x^2} + x - 2} \right) = 0 \\ \Rightarrow x\left( {{x^2} - x + 2x - 2} \right) = 0 \\ \Rightarrow x\left[ {x\left( {x - 1} \right) + 2\left( {x - 1} \right)} \right] = 0 \\ \Rightarrow x\left( {x - 1} \right)\left( {x + 2} \right) = 0 \\$ From the above equation, we have x=0 or $x - 1 = 0 \\ \Rightarrow x = 1 \\$ or $x + 2 = 0 \\ \Rightarrow x = - 2 \\$ So, the points where maxima or minima can occur are x=0,1,-2 Also we know that for any function f(x) to attain local maxima at a point x=a, $f''\left( a \right) < 0$ and for this function f(x) to attain local minima at a point x=b, $f''\left( b \right) > 0$. By differentiating the equation (2) both sides with respect to x, we get $\Rightarrow \dfrac{d}{{dx}}\left[ {f'\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {36x\left( {{x^2} + x - 2} \right)} \right] \\ \Rightarrow f''\left( x \right) = 36\dfrac{d}{{dx}}\left[ {{x^3} + {x^2} - 2x} \right] \\ \Rightarrow f''\left( x \right) = 36\left[ {\dfrac{d}{{dx}}\left[ {{x^3}} \right] + \dfrac{d}{{dx}}\left[ {{x^2}} \right] - 2\dfrac{{dx}}{{dx}}} \right] \\ \Rightarrow f''\left( x \right) = 36\left[ {3{x^2} + 2x - 2} \right]{\text{ }} \to {\text{(3)}} \\$ Put x=0 in equation (3), we get $\Rightarrow f''\left( 0 \right) = 36\left[ {3{{\left( 0 \right)}^2} + 2\left( 0 \right) - 2} \right] = - 72$ Clearly, $f''\left( 0 \right) < 0$ so x=0 is a point of local maxima i.e., corresponding to point x=0, the given function f(x) has local maximum value. So, x=0 is a value in the set ${{\text{S}}_2}$. Put x=1 in equation (3), we get $\Rightarrow f''\left( 1 \right) = 36\left[ {3{{\left( 1 \right)}^2} + 2\left( 1 \right) - 2} \right] = 36\left[ 3 \right] = 108$ Clearly, $f''\left( 1 \right) > 0$ so x=1 is a point of local minima i.e., corresponding to point x=1, the given function f(x) has local minimum value. So, x=1 is a value in the set ${{\text{S}}_1}$. Put x=-2 in equation (3), we get $\Rightarrow f''\left( -2 \right) = 36\left[ {3{{\left( { - 2} \right)}^2} + 2\left( { - 2} \right) - 2} \right] = 36\left[ 6 \right] = 216$ Clearly, $f''\left( -2 \right) > 0$ so x=-2 is a point of local minima i.e., corresponding to point x=-2, the given function f(x) has local minimum value. So, x=-2 is a value in the set ${{\text{S}}_1}$. So, set ${{\text{S}}_1} = \left\{ { - 2,1} \right\};{{\text{S}}_2} = \left\{ 0 \right\}$ Hence, option A is correct. Note- In this particular problem, we have used second derivative test i.e., if $f''\left( a \right) < 0$, then x=a is a point of local maxima and if $f''\left( a \right) > 0$, then x=a is a point of local minima. Also, if $f''\left( a \right) = 0$occurs then x=a is a point of inflection. Here, the local maximum value of function f(x) is obtained by substituting x=0 in the function and the local minimum values of f(x) is obtained by substituting x=-2 and x=1 in the function.
### 1 Changing variables in multiple integrals When the method of substitution is used to solve an integral of the form ${\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx$ three parts of the integral are changed, the limits, the function and the infinitesimal $dx$ . So if the substitution is of the form $x=x\left(u\right)$ the $u$ limits, $c$ and $d$ , are found by solving $a=x\left(c\right)$ and $b=x\left(d\right)$ and the function is expressed in terms of $u$ as $f\left(x\left(u\right)\right)$ . Figure 28 Figure 28 shows why the $dx$ needs to be changed. While the $\delta u$ is the same length for all $u$ , the $\delta x$ change as $u$ changes. The rate at which they change is precisely $\frac{d}{\phantom{\rule{0.3em}{0ex}}du}x\left(u\right)$ . This gives the relation $\phantom{\rule{2em}{0ex}}\delta x=\frac{dx}{\phantom{\rule{0.3em}{0ex}}du}\delta u$ Hence the transformed integral can be written as $\phantom{\rule{2em}{0ex}}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.3em}{0ex}}dx={\int }_{c}^{d}f\left(x\left(u\right)\right)\frac{dx}{du}du$ Here the $\frac{dx}{du}$ is playing the part of the Jacobian that we will define. Another change of coordinates that you have seen is the transformations from cartesian coordinates $\left(x,y\right)$ to polar coordinates $\left(r,\theta \right)$ . Recall that a double integral in polar coordinates is expressed as $\phantom{\rule{2em}{0ex}}\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int f\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dxdy=\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int g\left(r,\theta \right)\phantom{\rule{0.3em}{0ex}}rdrd\theta$ Figure 29 We can see from Figure 29 that the area elements change in size as $r$ increases. The circumference of a circle of radius $r$ is $2\pi r$ , so the length of an arc spanned by an angle $\theta$ is $2\pi r\frac{\theta }{2\pi }=r\theta$ . Hence the area elements in polar coordinates are approximated by rectangles of width $\delta r$ and length $r\delta \theta$ . Thus under the transformation from cartesian to polar coordinates we have the relation $\phantom{\rule{2em}{0ex}}\delta x\delta y\to r\delta r\delta \theta$ that is, $r\delta r\delta \theta$ plays the same role as $\delta x\delta y$ . This is why the $r$ term appears in the integrand. Here $r$ is playing the part of the Jacobian.
## Asymptotic Expansion of an integral Problem: find the asymptotic expansion of the following integral: $$I_n=\int_0^1\exp(x^n)dx$$ as $n \to \infty$. This expansion, while at first pedestrian-seeming, turns out to have a very interesting set of coefficients. You can expand the exponential in a Taylor series quite accurately: $$\exp{\left ( x^n \right )} = 1 + x^n + \frac12 x^{2 n} + \ldots$$ Because $x \in [0,1]$, this series converges rapidly as $n \to \infty$. Then the integral is $$1 + \frac{1}{n+1} + \frac12 \frac{1}{2 n+1} + \ldots = \sum_{k=0}^{\infty}\frac{1}{k!} \frac{1}{k n+1}$$ We can rewrite this as \begin{align}I_n&=1+\frac{1}{n} \sum_{k=1}^{\infty} \frac{1}{k \cdot k!} \left ( 1+\frac{1}{k n} \right )^{-1}\\ &= 1+\frac{1}{n} \sum_{m=0}^{\infty} \frac{(-1)^m}{n^m} \: \sum_{k=1}^{\infty} \frac{1}{k^{m+1} k!}\\ &= 1+\sum_{m=1}^{\infty} (-1)^{m+1}\frac{K_m}{n^m} \end{align} where $$K_m = \sum_{k=1}^{\infty} \frac{1}{k^{m+1} k!}$$ To first order in $n$: $$I_n \sim 1+\frac{K_1}{n} \quad (n \to \infty)$$ where $$K_1 = \sum_{k=1}^{\infty} \frac{1}{k\, k!} = \text{Ei}(1) – \gamma \approx 1.3179$$ This checks out numerically in Mathematica. BONUS As a further check, I computed the following asymptotic approximation: $$g(n) = 1+\frac{K_1}{n} -\frac{K_2}{n^2}$$ where $$K_2 = \sum_{k=1}^{\infty} \frac{1}{k^2 k!} \approx 1.1465$$ I computed $$\log_2{\left[\frac{\left\|g\left(2^m\right)-I_{2^m}\right\|}{I_{2^m}}\right]}$$ for $m \in \{1,2,\ldots,9\}$ The results are as follows $$\left( \begin{array}{cc} 1 & -4.01731 \\ 2 & -6.56064 \\ 3 & -9.26741 \\ 4 & -12.0963 \\ 5 & -15.0028 \\ 6 & -17.9538 \\ 7 & -20.9287 \\ 8 & -23.916 \\ 9 & -26.9096 \\ \end{array} \right)$$ Note that the difference between successive elements is about $-3$; because this is a log-log table, that means that this error is $O(1/n^3)$ and that the approximation is correct.
# 2.21: Conversions of Length, Mass, Capacity in Metric Units Difficulty Level: At Grade Created by: CK-12 Estimated16 minsto complete % Progress Practice Conversions of Length, Mass, Capacity in Metric Units MEMORY METER This indicates how strong in your memory this concept is Progress Estimated16 minsto complete % Estimated16 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Charlotte is going on vacation with her parents and her 3 month old baby brother for a week. Her baby brother drinks 750 milliliters of formula each day. Charlotte’s mom has 6 liters of formula that she is bringing on the trip. How can Charlotte help her mom to figure out if she has enough formula for the trip? In this concept, you will learn how to convert among metric units of measure using powers of 10. ### Converting Metric Units of Measure Recall that the metric system is the system of measurement primarily used in science and in countries outside of the United States. In the metric system, the base unit of length is the meter (m). The base unit of mass is the gram (g). The base unit of capacity is the liter (L). The same prefixes are used throughout the metric system. The prefix of each unit of measurement indicates how that unit relates to the base unit. • “milli” means one thousandth. For example, there are 1000 millimeters in 1 meter. • “centi” means one hundredth. For example, there are 100 centimeters in 1 meter. • “kilo” means one thousand. For example, there are 1000 meters in 1 kilometer. The metric system is based on powers of 10 just like our number system. To move between different units of length, mass, and capacity all you need to do is move the decimal point. • Any time you are going from a smaller unit of measure to a larger unit of measure you will need to divide or move the decimal point to the left. • Any time you are going from a larger unit of measure to a smaller unit of measure you will need to multiply or move the decimal point to the right. The table below shows the most common metric conversions. Kilometers (km) to Meters (m) Move decimal point 3 to the right Meters (m) to Centimeters (cm) Move decimal point 2 to the right Meters (m) to Millimeters (mm) Move decimal point 3 to the right Centimeters (cm) to Millimeters (mm) Move decimal point 1 to the right Millimeters (mm) to Centimeters (cm) Move decimal point 1 to the left Millimeters (mm) to Meters (m) Move decimal point 3 to the left Centimeters (cm) to Meters (m) Move decimal point 2 to the left Meters (m) to Kilometers (km) Move decimal point 3 to the left The same conversions will work for grams and liters. Just use the prefixes to find the correct conversion. Here is an example. Convert 525 meters to centimeters. First, notice that you are going from a larger unit to a smaller unit. This means you will need to multiply and move the decimal point to the right. Next, remember that there are 100 centimeters in a meter. This means you will need to multiply by 100 or move the decimal point 2 to the right. Insert zeros into the blank spaces. Here is another example. Convert 95,231 milligrams to kilograms. First, notice that you are going from a smaller unit to a larger unit. This means you will need to divide and move the decimal point to the left. Next, remember that there are 1000 milligrams in a gram and 1000 grams in a kilogram. This means there are 1,000,000 milligrams in a kilogram. You will need to divide by 1,000,000 or move the decimal point 6 to the left. ### Examples #### Example 1 Earlier, you were given a problem about Charlotte and her family trip with her baby brother. Her brother drinks 750 milliliters of formula a day and her mom has 6 liters of formula packed for the 7 day trip. Charlotte wants to make sure they have enough formula. First, Charlotte should figure out how many milliliters of formula her brother will need for the 7 days. He drinks 750 milliliters a day, so multiply 750 times 7. Her brother will need 5250 milliliters of formula for the trip. Next, convert 5250 milliliters to liters. Charlotte should notice that she is going from a smaller unit to a larger unit. This means she will need to divide and move the decimal point to the left. Now, Charlotte needs to remember that there are 1000 milliliters in a liter. This means she will need to divide by 1000 or move the decimal point 3 to the left. The answer is that because Charlotte’s mom has 6 liters of formula and Charlotte’s brother will need 5.25 liters of formula, Charlotte’s mom has enough formula for the trip. #### Example 2 Convert 150 grams to centigrams. First, notice that you are going from a larger unit to a smaller unit. This means you will need to multiply and move the decimal point to the right. Next, remember that there are 100 centigrams in a gram. This means you will need to multiply by 100 or move the decimal point 2 to the right. Insert zeros into the blank spaces. #### Example 3 Convert 500 meters to centimeters. First, notice that you are going from a larger unit to a smaller unit. This means you will need to multiply and move the decimal point to the right. Next, remember that there are 100 centimeters in a meter. This means you will need to multiply by 100 or move the decimal point 2 to the right. Insert zeros into the blank spaces. #### Example 4 Convert 120 meters to kilometers. First, notice that you are going from a smaller unit to a larger unit. This means you will need to divide and move the decimal point to the left. Next, remember that there are 1000 meters in a kilometer. This means you will need to divide by 1000 or move the decimal point 3 to the left. #### Example 5 Convert 50 centiliters to liters. First, notice that you are going from a smaller unit to a larger unit. This means you will need to divide and move the decimal point to the left. Next, remember that there are 100 centiliters in a liter. This means you will need to divide by 100 or move the decimal point 2 to the left. ### Review Convert the following metric units of length. 1. 10 cm to millimeters 2. 100 kilometers to meters 3. 453 meters to kilometers 4. 1,567 kilometers to meters 5. 6,700 centimeters to meters 6. 7.8 meters to centimeters Convert the following measurements into milliliters. 1. 65.57 liters 2. 28.203 centiliters 3. 0.009761 kiloliters Convert the following measurements into centigrams. 1. 29.467 grams 2. 0.0562 milligrams 3. 0.0450584 kilograms Convert the following measurements into kiloliters. 1. 89.96 liters 2. 45,217 milliliters 3. 3,120,700 centiliters ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Customary System The customary system is the measurement system commonly used in the United States, including: feet, inches, pounds, cups, gallons, etc. Equivalence Equivalence is the condition of being equal in value or meaning. Estimate To estimate is to find an approximate answer that is reasonable or makes sense given the problem. Measurement A measurement is the weight, height, length or size of something. Proportion A proportion is an equation that shows two equivalent ratios. Ratio A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:
# Quick Answer: What Is Greatest To Least? ## Are fractions with different denominators? If the denominators are not the same, then you have to use equivalent fractions which do have a common denominator . To do this, you need to find the least common multiple (LCM) of the two denominators. To add fractions with unlike denominators, rename the fractions with a common denominator. Then add and simplify. ## Which list of numbers is in order from the smallest to the largest? Answer. Answer: List of numbers which is in order from the smallest to the largest is Ascending order whereas List of numbers which is in order from the largest to the smallest is Descending order. ## Which fraction is least to greatest? Use the top number to order the fractions. Now that they all have the same denominator, the fractions are easy to compare. Use their top number, or numerator, to rank them from least to greatest. Ranking the fractions we found above, we get: 6/18, 12/18, 15/18. ## Is ascending greatest to least? In general terms, Ascending means smallest to largest, 0 to 9, and/or A to Z and Descending means largest to smallest, 9 to 0, and/or Z to A. Ascending order means the smallest or first or earliest in the order will appear at the top of the list: For numbers or amounts, the sort is smallest to largest. ## What is the greatest fraction? The larger fraction is the one with the greater numerator. Compare two fractions at a time. Look at the denominators. The fraction with the smaller denominator is the larger fraction. ## How do you determine which fraction is greater? Step 1: Compare denominators. If they are different, rewrite one or both fractions with a common denominator. Step 2: Check the numerators. If the denominators are the same, then the fraction with the greater numerator is the greater fraction. ## Is 0.1 or 0.01 greater? 2 Answers. 0.1 is one tenth and 0.01 is one hundredth, which is a tenth of a tenth, so 0.1 is bigger. ## What fraction is bigger 7/8 or 910? The numerator of the first fraction 35 is less than the numerator of the second fraction 36 , which means that the first fraction 3540 is less than the second fraction 3640 and that 78 is less than 910 . ## How do you know which fraction is greater cross multiplication? You can use cross-multiplication to compare fractions and find out which is greater. When you do so, make sure that you start with the numerator of the first fraction. To find out which of two fractions is larger, cross-multiply and place the two numbers you get, in order, under the two fractions. ## How do you know which decimal is bigger? We can use this method to see which decimals are bigger:Set up a table with the decimal point in the same place for each number.Put in each number.Fill in the empty squares with zeros.Compare using the first column on the left.If the digits are equal move to the next column to the right until one number wins. ## Is zero is a positive integer? Zero is defined as neither negative nor positive. The ordering of integers is compatible with the algebraic operations in the following way: if a < b and c < d, then a + c < b + d. ## What is the greatest integer? The Greatest Integer Function is denoted by y = [x]. For all real numbers, x, the greatest integer function returns the largest integer. less than or equal to x. In essence, it rounds down a real number to the nearest integer. For example: [1] = 1 [1.5] = 1 [3.7] = 3 [4.3] = 4. ## Which is the smallest integer? zeroThe smallest integer is zero. ## Is 0.5 or 0.25 bigger? 0.5 because 0.5 is the same as 0.50, which is larger than 0.25. ## Is ascending order smallest to largest? Ascending Order. Arranging numbers (or other items) in ascending order means to arrange them from smallest to largest.
# How do you solve abs(-2r-1)=11? Oct 30, 2017 $r = - 6 , r = 5$ #### Explanation: This is an absolute value equation meaning that we need to split this into two equations; the original equation and the opposite of it: $- 2 r - 1 = 11$ and $- 2 r - 1 = - 11$ Notice that the $11$ became $- 11$. Now we can solve this by simplifying. Let's simplify the first one: $- 2 r - 1 = 11$ $- 2 r = 12$ $r = - 6$ Second one: $- 2 r - 1 = - 11$ $- 2 r = - 10$ $r = 5$ Now let's check our work by plugging what we got for r back into the original equation $\| - 2 r - 1 \| = 11$. $\| - 2 \left(- 6\right) - 1 \| = 11$ $\| 12 - 1 \| = 11$ $11 = 11$ :) $\| - 2 \left(5\right) - 1 \| = 11$ $\| - 10 - 1 \| = 11$ $\| - 11 \| = 11$ $11 = 11$ :) So our two solutions are $r = - 6$ and $r = 5$.
# Divide by 10, 100 and 1000 Divisors We will learn how to divide by 10, 100 and 1000 divisors to find the quotient and remainder. Let us consider some examples: So, 3458 ÷ 10 = 345 with remainder 8 So, 9396 ÷ 10 = 939 with remainder 6 When we divided a number by 10 then the very first digit from the extreme right of the dividend forms the remainder and the remaining digits from the quotient. For example: (i) Divide 2542 ÷ 100 So, 2542 ÷ 100 = 25 with remainder 42 (ii) Divide 4978 ÷ 100 So, 4978 ÷ 100 = 49 with remainder 78 When we divide a number by 100 then the first two digits from the extreme right of the dividend from the remainder and remaining digits from the quotient. For example: (i) Divide 7943 ÷ 1000 So, 7943 ÷ 1000 = 7 with reminder 94 (ii) Divide 9869 ÷ 1000 So, 9869 ÷ 1000 = 9 with remainder 869 When we divide a number by 1000 then the first three digits from the extreme right of the dividend from the remainder and the remaining digits form the quotient. Observe the following table carefully to divide by 10, 100 and 1000 divisors. Dividend Divisor Quotient Remainder 3536 ÷ 10 = 353 6 3536 ÷ 100 = 35 36 3536 ÷ 1000 = 3 536 13536 ÷ 1000 = 13 536 213536 ÷ 1000 = 213 536 Word Problems: 1. 42 boys collected $4662 to donate in a fund. If each boy contributed equally, how much did each boy contribute? Solution: 42 boys contributed together =$4662 Each boy contributed = $4662 ÷ 42. So, each boy contributed$111. 2. A dealer bought 25 chairs for $12350. Find the cost of one chair. Solution: 25 chairs cost =$12350 1 chair costs = $12350 ÷ 25. So, the price of one chair =$494. `
# Sides of a Right-angled Triangle $$(2n + 1)^2 + (2n^2 + 2n)^2 = (2n^2 +2n +1)^2$$ It can be used to generate infinitely many sides of right-angled triangles with integer lengths by putting values of $$n = 1, 2, 3, ...$$ I wanted to know that how we came to this equation. How do we know that putting n = 1, 2, 3, … into this we'll get all the sides of a right-angled triangle. I'm trying to find more about this on the internet, if you can help me what should I find about. If you start with the sequence of square numbers, and take all the differences between consecutive terms, you get all the odd numbers in sequence. For instance, $$2^2 - 1^2 = 3\\ 3^2 - 2^2 = 5\\ 4^2-3^3 = 7$$ Some times, that odd number happens to be a square itself. For instance, we have $$5^2 - 4^2 = 9 = 3^2\\ 13^2 - 12^2 = 25 = 5^2$$ Rearranging these, we get Pythagorean triples: $$3^2 + 4^2 = 5^2\\ 5^2 + 12^2 = 13^2$$ If we want to describe all the different Pythagorean triples that appear this way, we end up with exactly your expression. Your formula notes that often Pythagorean Triples are of the form $$(t_1, t_2, t_2+1)$$. It attempts to see when this is the case and this formula generates that, we have: $$(2n^2+2n+1)^2-(2n^2+2n)^2=4n^2+4n+1=(2n+1)^2$$ Since the difference here is just a square, it's a valid triple. However, this misses many triples, particularly multiples of existing triples, but an example is $$(8,15,17)$$. I prefer the usage of: $$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2$$ This is most notable for its link to complex numbers, it is $$\Re(z^2)+\Im(z^2)=\|z^2\|$$ and this can be used to generate them. Set your calculator to Complex mode and use $$(1000\text{Ran#}+\text{1000Ran#}i)^2=$$ and every result will be the base and height of a Pythagoras triangle. (Ran# is my calculators random number generator, it might be different to yours. Also it generates from $$0.001$$ to $$1$$, max $$3$$dp, hence my multiplying by $$1000$$) Consider the Primitive Pythagorean triple $$(a, b, c)$$. Consider $$b=4T_n$$ where $$T_n$$ is the $$n^{\text{th}}$$ triangular number. Notice that if this is so, then the value of $$c$$ is $$4T_n+1$$. Note that the formula for the $$n^{\text{th}}$$ triangular number is given by $$\frac{n(n+1)}{2}$$. All that we are left to do is solve for $$a$$ from the following equation: $$a^2+[4T_n]^2=[4T_n+1]^2$$. Solving this gives the value of $$a$$ as $$2n+1$$. Hence, $$(a, b, c)=(2n+1, 4T_n, 4T_n+1)$$ is a Pythagorean triple.
# Integral of the absolute Value of $$x$$: $$\int \|x\| dx$$ Evaluate the integral $\int \|x\| \; dx$ Rewrite as $\int \|x\| \; dx = \int 1 \cdot \|x\| \; dx \quad (I)$ Note that $$\|x\| = \sqrt {x^2}$$ and hence $$\dfrac{d(\|x\|)}{dx} = \dfrac{d( \sqrt {x^2})}{dx} = \dfrac{x}{\sqrt {x^2}} = \dfrac{x}{\|x\|} \quad (II)$$ Apply the integration by parts: $$\displaystyle \int u' v \; dx = u v - \int u v' \; dx$$ to the integral on the right side of (I) above. Let $$u' = 1$$ , $$v = \|x\|$$ which gives $$u = x$$ and $$v' = \dfrac{x}{\|x\|}$$ , see (II) above. We substitute all the above in the formula of the integration by parts given above. $= x \|x\| - \int x \dfrac{x}{\|x\|} dx \quad (III)$ Multiply the numerator and denominator of the integrand $$x \dfrac{x}{\|x\|}$$ in the above integral $x \dfrac{x}{\|x\|} = x \dfrac{x}{\|x\|} \dfrac{\|x\|}{\|x\|}$ Simplify noting that $$\|x\| \|x\| = x^2$$ $= \|x\|$ We now substitute the integrand $$x \dfrac{x}{\|x\|}$$ in (III) by $$\|x\|$$ and write $\int \|x\| dx = x \|x\| - \int \|x\| dx$ Add $$\displaystyle \int \|x\| dx$$ to both sides $\int \|x\| dx + \int \|x\| dx = x \|x\| - \int \|x\| dx + \int \|x\| dx$ Simplify by grouping $2 \int \|x\| \; dx = x \|x\|$ The final answer is given by $\boxed {\int \|x\| \; dx = \dfrac{1}{2} x \|x\| }$ ## More References and Links 1. Table of Integral Formulas 2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540 3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824 4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8
# Identify Parallel Lines Related Topics: Lesson Plans and Worksheets for Grade 4 Lesson Plans and Worksheets for all Grades Examples, solutions, and videos to help Grade 4 students learn how to identify, define, and draw parallel lines. Common Core Standards: 4.G.1 ### New York State Common Core Math Grade 4, Module 4, Lesson 4 The following diagram shows a worksheet to identify parallel lines in given figures. Scroll down the page for more examples of parallel lines. NYS Math Module 4 Grade 4 Lesson 4 Problem Set 4. Determine which of the following figures have lines that are parallel by using a straightedge and the right angle template that you created. Circle the letter of the shapes that have at least one pair of parallel lines. Mark each pair of parallel lines with arrows and then identify the parallel lines with a statement modeled after the one in 4(a). NYS Math Module 4 Grade 4 Lesson 4 Concept Development Problem 1: Define and identify parallel lines. Two lines that never touch no matter how far you extend them are parallel. Problem 3: Represent parallel lines with symbols. NYS Math Module 4 Grade 4 Lesson 4 Homework 5. True or false? All shapes with a right angle have sides that are parallel. Explain your thinking. 7. Draw a line using your straightedge. Now use your right angle template and straightedge to construct a line parallel to the first line you drew. NYS Math Module 4 Grade 4 Lesson 4 Homework 1. On each object, trace at least one pair of lines that appear to be parallel. 2. How do you know if two lines are parallel? 3. In the square and triangular grids below, use the given segments in each grid to draw a line that is parallel using a straightedge. 4. Determine which of the following figures have lines that are parallel by using a straightedge and the right angle template that you created. Circle the letter of the shapes that have at least one pair of parallel lines. Mark each pair of parallel lines with arrows and then identify the parallel lines with a statement modeled after the one in 4(a). 5. True or false? All shapes with a right angle have sides that are parallel. Explain your thinking. 6. Explain why AB and CD are parallel but EF and GH are not. 7. Draw a line using your straightedge. Now use your right angle template and straightedge to construct a line parallel to the first line you drew. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
Ex 11.1 Chapter 11 Class 10 Constructions [Deleted] Serial order wise Β This video is only available for Teachoo black users Get live Maths 1-on-1 Classs - Class 6 to 12 ### Transcript Ex 11.1, 3 Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle. Letβs first construct Ξ ABC with sides 5 cm, 6 cm, 7 cm Steps to draw Ξ ABC Draw base AB of side 5 cm With A as center, and 6 cm as radius, draw an arc With B as center, and 7 cm as radius, draw an arc 3. Let C be the point where the two arcs intersect. Join AC & BC Thus, Ξ ABC is the required triangle Now, letβs make a similar triangle with Scale factor = 7/5 Steps of construction Draw any ray AX making an acute angle with AB on the side opposite to the vertex C. Mark 7 (the greater of 7 and 5 in 7/5 ) points π΄_1, π΄_2, π΄_3, π΄_4, π΄_5, π΄_6, π΄_7 on AX so that γπ΄π΄γ_1=π΄_1 π΄_2=π΄_2 π΄_3 β¦ and so on Join π΄_5 π΅ (5th point as 5 is smaller in 7/5) and draw a line through π΄_7 parallel to π΄_5 π΅, to intersect AB extended at Bβ². Draw a line through Bβ² parallel to the line BC to intersect AC extended at Cβ². Thus, Ξ ABβCβ² is the required triangle Justification Since scale factor is 7/5, we need to prove (π¨π©^β²)/π¨π©=(π¨πͺ^β²)/π¨πͺ=(π©^β² πͺ^β²)/π©πͺ = π/π By construction, (π΄B^β²)/π΄π΅=(π΄π΄_7)/(π΄π΄_5 )= 7/5 Also, BβCβ is parallel to BC So, the will make the same angle with line AB β΄ β ABβCβ = β ABC Now, In Ξ ABβCβ and Ξ ABC β A = β A β ABβCβ = β ABC Ξ ABβCβ βΌ Ξ ABC Since corresponding sides of similar triangles are in the same ratio (π΄π΅^β²)/π΄π΅=(π΄πΆ^β²)/π΄πΆ=(π΅^β² πΆ^β²)/π΅πΆ So,(π¨π©^β²)/π¨π©=(π¨πͺ^β²)/π¨πͺ=(π©^β² πͺ^β²)/π©πͺ =π/π. Thus, our construction is justified
Free Algebra Tutorials! Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: # Solving Systems of Equations by Substitution Solving a system by graphing is certainly limited by the accuracy of the graph. If the lines intersect at a point whose coordinates are not integers, then it is difficult to determine those coordinates from the graph. The method of solving a system by substitution does not depend on a graph and is totally accurate. For substitution we replace a variable in one equation with an equivalent expression obtained from the other equation. Our intention in this substitution step is to eliminate a variable and to give us an equation involving only one variable. Example 1 An independent system solved by substitution Solve the system by substitution: 2x + 3y = 8 y + 2x = 6 Solution We can easily solve y + 2x = 6 for y to get y = -2x + 6. Now replace y in the first equation by -2x + 6: 2x + 3y = 8 2x + 3(-2x + 6) = 8 Substitute -2x + 6 for y. 2x - 6x + 18 = 8 -4x = -10 x To find y, we let in the equation y = -2x + 6: The next step is to check and y = 1 in each equation. If and y = 1 in 2x + 3y = 8, we get If and y = 1 in y + 2x = 6, we get Because both of these equations are true, the solution set to the system is . The equations of this system are independent. Example 2 An inconsistent system solved by substitution Solve by substitution: x - 2y = 3 2x - 4y = 7 Solution Solve the first equation for x to get x - 2y = 3 . Substitute 2y + 3 for x in the second equation: 2x - 4y = 7 2(2y + 3) = 7 4y + 6 - 4y = 7 6 = 7 Because 6 = 7 is incorrect no matter what values are chosen for x and y, there is no solution to this system of equations. The equations are inconsistent. To check, we write each equation in slope-intercept form: x - 2y = 3 2x - 4y = 7 -2y = -x + 3 -4y = -2x + 7 y y The graphs of these equations are parallel lines with different y-intercepts. The solution set to the system is the empty set, Ã˜.
# Lagrange multipliers ## Lagrange multipliers : Lagrange multipliers are a mathematical tool used to find the maximum or minimum value of a function subject to constraints. Essentially, they allow us to find the optimal solution to a problem by combining multiple equations and constraints. For example, let’s say we have a rectangular field that we want to fence off. We have a limited amount of fencing material and want to find the dimensions of the field that will give us the maximum area. We can use Lagrange multipliers to solve this problem. First, we set up our objective function, which is the area of the rectangular field. This is given by the formula A = lw, where l is the length and w is the width of the field. Next, we set up our constraint equation, which is the amount of fencing material we have available. Let’s say we have 100 feet of fencing material. This means that the perimeter of the field must be equal to 100 feet, or 2l + 2w = 100. Now, we can use Lagrange multipliers to find the optimal dimensions of the field that will give us the maximum area. We do this by combining our objective function and constraint equation into a single equation: A – λ(2l + 2w – 100) = 0. Solving this equation, we find that the optimal dimensions of the field are l = 25 and w = 20, which gives us an area of 500 square feet. Another example of using Lagrange multipliers is in finding the shortest distance between two points. Let’s say we want to find the shortest distance between the points (2,3) and (5,7). We can use Lagrange multipliers to solve this problem. First, we set up our objective function, which is the distance between the two points. This is given by the formula d = √((x1-x2)^2+(y1-y2)^2), where x1 and y1 are the coordinates of the first point, and x2 and y2 are the coordinates of the second point. Next, we set up our constraint equation, which is the fact that the two points are fixed. This means that x1 and y1 are fixed at the coordinates (2,3), and x2 and y2 are fixed at the coordinates (5,7). Now, we can use Lagrange multipliers to find the shortest distance between the two points. We do this by combining our objective function and constraint equation into a single equation: d – λ(x1 – 2) – λ(y1 – 3) – λ(x2 – 5) – λ(y2 – 7) = 0. Solving this equation, we find that the shortest distance between the two points is d = √(3^2 + 4^2) = 5. In both of these examples, Lagrange multipliers allowed us to find the optimal solution to the problem by combining multiple equations and constraints. In the first example, we found the dimensions of a rectangular field that would give us the maximum area, and in the second example, we found the shortest distance between two points. Lagrange multipliers are a powerful tool that can help us find the optimal solutions to a wide range of problems.
What Is 11 Out Of 15? Are you curious to know what is 11 out of 15? You have come to the right place as I am going to tell you everything about 11 out of 15 in a very simple explanation. Without further discussion let’s begin to know what is 11 out of 15? In various contexts, you might encounter expressions like “11 out of 15,” leaving you wondering about its significance. In this detailed exploration, we will unravel the meaning behind this numerical ratio, addressing common questions about what it represents and how it is interpreted. What Is 11 Out Of 15? When you see the expression “11 out of 15,” it denotes a ratio or fraction representing the relationship between the two numbers. In this case, it signifies that you have 11 items or parts out of a total of 15. Understanding Ratios • Numerical Representation: The ratio “11 out of 15” is written mathematically as 11/15, indicating the proportion or fraction of the whole. • Visualizing the Ratio: Imagine having 15 items, and out of those, 11 are of a specific category or meet a certain condition. What Is A 11 Out Of 15? The expression “a 11 out of 15” is essentially another way of saying the same thing as “11 out of 15.” The term “a” is used to represent any variable or unknown quantity. Variable Representation • Mathematical Flexibility: Using “a” allows for mathematical flexibility, enabling you to solve for the unknown quantity if other parts of the ratio are provided. • Example Scenario: If you know that “a 11 out of 15” equals 7, you can set up the equation (a/15) * 11 = 7 and solve for “a.” What Grade Is 11 Out Of 15? In an academic context, you might encounter the question, “What grade is 11 out of 15?” This scenario often relates to assessments or evaluations where a numerical score is converted into a percentage or letter grade. If you love learning new things then you then you can read interesting topics here at squareroott. • Percentage Calculation: To determine the percentage, divide 11 by 15 and multiply by 100. In this case, (11/15) * 100 = 73.33%. • Letter Grade Interpretation: The corresponding letter grade might depend on the grading scale used by the educational institution. For example, a score of 73.33% could translate to a letter grade of “C” on a standard grading scale. What Is An 11 Out Of 15? When you encounter the phrase “an 11 out of 15,” it generally refers to a specific instance or occurrence where 11 out of 15 elements meet a certain criteria or condition. Contextual Interpretation • Example Scenario: For instance, in a survey context, “an 11 out of 15” positive responses might indicate a favorable sentiment. • Performance Evaluation: In a performance review, “an 11 out of 15” could represent the number of successfully completed tasks or goals. Conclusion In summary, “11 out of 15” is a numerical ratio or fraction that represents the relationship between two quantities. Whether you encounter this expression in a mathematical, academic, or contextual setting, understanding its interpretation allows for informed decision-making and analysis. From grade calculations to performance evaluations, the expression “11 out of 15” serves as a versatile tool for conveying proportions and assessing outcomes. FAQ What Is The Grade 11 Out Of 15? First, you need to calculate your grade in percentages. The total answers count 15 – it’s 100%, so we to get a 1% value, divide 15 by 100 to get 0.15. Next, calculate the percentage of 11: divide 11 by 1% value (0.15), and you get 73.33% – it’s your percentage grade. What Is A 10 Out Of 15 Grade? 10 out of 15 is equivalent 2/3 or 66.7 %. 10/15 = 2/3 = 0.6667 = 66.67 % What Is A 12 Out Of 15 Grade? 12/15 as a percent is 80% Is 95 An A Or A+? Common examples of grade conversion are: A+ (97–100), A (93–96), A- (90–92), B+ (87–89), B (83–86), B- (80–82), C+ (77–79), C (73–76), C- (70–72), D+ (67–69), D (65–66), D- (below 65). I Have Covered All The Following Queries And Topics In The Above Article What Is 11 Out Of 15 What Is A 11 Out Of 15 What Grade Is 11 Out Of 15 What Is An 11 Out Of 15 What Is 11 Out Of 15
Lesson: Input/Output Tables: 2 Step Rules 10 Views 0 Favorites Lesson Objective SWBAT use a rule to complete an input/output table Lesson Plan Materials Needed: white board, dry erase markers, DN worksheet, Example Problem, scrap paper, pencils, GP worksheet, IND practice worksheet Vocabulary: input, output, numerical relationship, solve, equations ………. Do Now (2 - 3 min): Teacher gives each student a copy of the DN worksheet. Students should complete the worksheet quickly and quietly. Opening (2 -3 min): Teacher reviews the answers to the DN worksheet. Teacher says, “Yesterday, we learned how to find the rule when looking at an input/output table. Today, we are going to continue talking about input/output tables. By the end of the lesson, you will be able to use a rule to complete an input/output table on your own.” Direct Instruction (10 - 12 min): Teacher posts the Example Problem on the board and says, “When we look at this problem, we should be reminded about what we learned yesterday when we were determining the rule. I want you all to quickly pull out a piece of scratch paper and see if you can figure out what the rule is with this number machine. Ok, does anyone have a rule? [Answers will vary, teacher is looking for multiply by 2, then add 4 to the answer] Very good, some of you figured out that we need to multiply by 2 and then add 4 to the answer. That is the rule for this input/output table. Now that we know the rule, we need to understand how to use the rule. We need to be able to complete the input/output table using the rule.” Teacher points to the first row of the table and says, “If I put 8 into the number machine, I would multiply 8 by 2 to get 16 and then add 4 to get 20.” Teacher writes 8x2=16+4=20 on the board and says, “Really, you just have to follow the rule. If you use the rule, then you can solve the problem in order to get the output of any number. Lets practice with the next number in our example. I have the number 5, what is the first thing I will do to this 5 based on the rule? [Multiply it by 2] Right, and 5x2 is 10, and then what do I do? [Add 4] Right, and 10 + 4 is 14. So if I put 5 into the number machine and follow the rule, I will get an output of 14. Lets look at the last row in our example; can anyone use their scratch paper or their mental math skills and tell me what the output will be if the input is 12? [26] Awesome, the answer would be 26. You all are ready to practice following the rules and complete an input/output table in small groups!” Guided Practice (10 -12 min): Teacher gives each math group a GP worksheet and circulates the room to answer any questions. When most students are finished, the teacher reviews the answers and clarifies any points of confusion. Independent Practice (10 min): The teacher hands out the IND Practice worksheet. Students are asked to complete the worksheet independently and turn it in. Closing (2-3 min): Teacher calls the attention of the students back toward the front of the class to quickly review the answers to the Independent Practice worksheet/ ask what we learned about. Lesson Resources IND Lesson 14 Classwork 11 DN Lesson 14 Starter / Do Now Chart with Strategies to find Rule Exemplar 2
# Factors Of 40 \| With Easy Division and Prime Factorization Contents In maths, factors of 40 are the numbers that divide the original number 40 and produce the whole number in quotient form. You can find the original number by multiplying any two factors in pairs. Likewise, multiples of 40 are the lengthened versions, such as 40, 80, 120, 160, 200, 240, and so on. Factors and multiples have different properties. A composite number is a number that has more than two factors, such as 36, 24, 18, 60, 45, etc. Here I will discuss factors in pairs and prime factors of number 40. I will analyze 40, its pair factors, and its prime factors by the prime factorization method. ## What are the Factors of 40? A factor of 40 is a number that divides 40 without leaving a remainder. To explain this further, the factors of 40 are the numbers multiplied in pairs, resulting in a new number. It has many factors other than 1 and 40 because it is an even composite number. Accordingly, the factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. Factors of 40: 1, 2, 4, 5, 8, 10, 20 and 40. Prime Factorization of 40: 2×2×2×5 or 23 × 5 ## What are the Factors of -40? Factors of -40: -1, -2, -4, -5, -8, -10, -20 and -40. We can get -40 by having (-1, 40), (-2, 20), (-4, 10), and (-5, 8)as a pair of factors. Similarly, we can also get -40 by having (1, -40), (2, -20), (4, -10), and (5, -8) as a pair of factors. ## Pair Factors of 40 A pair factor of 40 is the result of multiplying two numbers together that result in an original number of 40. They can either be positive or negative pairs. This will give us the original number 40 when we multiply the two negative numbers. Here are the positive and negative factors of 40: Positive Pair Factors of 40: Negative Pair Factors of 40: ## Factors of 40 by Division Method You can also find the factors of 40 by using the division method. To divide 40 using the division method, we must divide it by successive integers. Factors of 40 are the integers that divide 40 exactly without leaving any remainder. Let’s now divide 40 by 1. • 40/1 = 40 • 40/2 = 20 • 40/4 = 10 • 40/5 = 8 • 40/8 = 5 • 40/10 = 4 • 40/20 = 2 • 40/40 = 1 As a result, the factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. Note: When we divide 40 by any number other than 1, 2, 4, 5, 8, 10, 20, and 40, it leaves the remainder, and therefore, it is not a factor of 40. ## Prime Factorization of 40 Now that we know 40 is a composite number let’s find its prime factors. First, we need to divide 40 by the smallest prime factor, i.e., 2. 40 ÷ 2 = 20 Divide 20 by 2, once more. 20 ÷ 2 = 10 Divide until you get an odd number because you get a fraction if you divide by 2. Fractions cannot be factors. Therefore, 10 ÷ 2 = 5 Dividing 5 by 2 gives us a fraction. Next, consider prime numbers 3, 5, 7, and so on. 5 ÷ 3 = 1.67; not a factor Next, we will move to the number 5. If we divide 5 by 5, we get: 5 ÷ 5 = 1 Moreover, we received one at the end, so we cannot use the division method. As a result, the prime factorization of 40 is 2 x 2 x 2 x 5 or 23 x 5, where 2 and 5 are prime numbers. ## Examples ### Example 1: Find the factors that are common to 35 and 41. Solution: The factors of 35 are 1, 5, 7, and 35. 1 and 41 are the factors of 41. Since 41 is a prime number, the common factor between 35 and 41 is 1. ### Example 2: Find the common factor between 40 and 24. Solution: The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, and 24. Therefore, 40 and 39 have a common factor of 1,2,4 and 8. ### Example 3: Calculate the common factors of 40 and 64. Solution: 40 has the following factors: 1, 2, 4, 5, 8, 10, 20, and 40. Factors of 64 = 1, 2, 4, 8, 16, 32, and 64. As a result, 40 and 64 share these common factors: 1, 2, 4, 8, and 10. ## FAQs ### How many factors are there in 40? 1, 2, 4, 5, 8, 10, 20, and 40 are the factors of 40. ### The prime factorization of 40 is? The factorization of 40 is 2 × 2 × 2 × 5 or 23 × 5. ### How many positive pair factors does 40 have? Among the positive pair factors of 40, there are (1, 40), (2, 20), (4, 10), and (5, 8). ### How many negative pair factors does 40 have? (-1, -40), (-2, -20), (-4, -10), and (-5, -8) are the negative pair factors of 40. ### 16 is a factor of 40; is it true? 16 cannot be a factor of 40. As 16 leaves a remainder when divided by 40, it is not a factor of 40.
Polynomials (Chapter 2, Exercise 2.2) – Class 10 Maths Solutions NCERT This page consists of the solutions of the Exercise 2.2, Chapter 2 of Class 10 Mathematics under NCERT Syllabus. The Chapter 2 of Class 10 Mathematics is about the Polynomials. The Exercise 2.2 deals with the questions based on the Relationship between Zeros and Coefficients of a Polynomial. The Exercise 2.2 consists of 2 questions only. The solution of all the questions of Exercise 2.2 of Class 10 Mathematics is given below. Question 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. i.$x^2-2x-8$ Solution: Given, the quadratic polynomial $x^2-2x-8$ Factorising, $x^2-2x-8$ $=x^2-(4-2)x-8$ $=x^2-4x+2x-8$ $=x(x-4)+2(x-4)$ $=(x-4)(x+2)$ The zeros of the quadratic polynomial are, $x-4=0$ $\Rightarrow x=4$ And $x+2=0$ $\Rightarrow x=-2$ Sum of the zeros$=4+(-2)=4-2=2$ Product of the zeros$=4\times (-2)=-8$ Verification: Here, $a=1,\, b=-2$ and $c=-8$ Sum of zeros$=\frac{-b}{a}=\frac{-(-2)}{1}=2$ Product of zeros$=\frac{c}{a}=\frac{-8}{1}=-8$ Hence verified. ii. $4s^2-4s+1$ Solution: Given, the quadratic polynomial $4s^2-4s+1$ Factorising, $4s^2-4s+1$ $=4s^2-(2+2)s+1$ $=4s^2-2s-2s+1$ $=2s(2s-1)-1(2s-1)$ $=(2s-1)(2s-1)$ $=(2s-1)^2$ The zeros of the quadratic polynomial are, $(2s-1)^2=0$ $2s-1=0$ and $2s-1=0$ $s=\frac{1}{2},\,\frac{1}{2}$ Sum of the zeros$=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1$ Product of the zeros$=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$ Verification: Here, $a=4,\, b=-4$ and $c=1$ Sum of zeros$=\frac{-b}{a}=\frac{-(-4)}{4}=1$ Product of zeros$=\frac{c}{a}=\frac{1}{4}=\frac{1}{4}$ Hence verified. iii. $6x^2-3-7x$ Solution: Given, the quadratic polynomial $6x^2-3-7x=6x^2-7x-3$ Factorising, $6x^2-7x-3$ $=6x^2-(9-2)x-3$ $=6x^2-9x+2x-3$ $=3x(2x-3)+1(2x-3)$ $=(2x-3)(3x+1)$ The zeros of the quadratic polynomial are, $2x-3=0$ and $3x+1=0$ $2x=3$ and $3x=-1$ $x=\frac{3}{2}$ and $x=\frac{-1}{3}$ Sum of the zeros$=\frac{3}{2}+\frac{-1}{3}=\frac{7}{6}$ Product of the zeros$=\frac{3}{2}\times \frac{-1}{3}=-\frac{1}{2}$ Verification: Here, $a=6,\, b=-7$ and $c=-3$ Sum of zeros$=\frac{-b}{a}=\frac{-(-7)}{6}=\frac{7}{6}$ Product of zeros$=\frac{c}{a}=\frac{-3}{6}=-\frac{1}{2}$ Hence verified. iv. $4u^2+8u$ Solution: Given, the quadratic polynomial $4u^2+8u$ Equating the polynomial to zero, $4u^2+8u=0$ $\Rightarrow 4u(u+2)=0$ The zeros of the quadratic polynomial are, $4u=0$ and $u+2=0$ $u=0$ and $u=-2$ Sum of the zeros$=0+(-2)=-2$ Product of the zeros$=0\times (-2)=0$ Verification: Here, $a=4,\, b=8$ and $c=0$ Sum of zeros$=\frac{-b}{a}=\frac{-8}{4}=-2$ Product of zeros$=\frac{c}{a}=\frac{0}{4}=0$ Hence verified. v. $t^2-15$ Solution: Given, the quadratic polynomial $t^2-15$ Equating the polynomial to zero, $t^2-15=0$ $\Rightarrow t^2=15$ $\Rightarrow t=\pm \sqrt{15}$ $\Rightarrow t=+\sqrt{15},\,-\sqrt{15}$ The zeros of the quadratic polynomial are $+\sqrt{15}$ and $-\sqrt{15}$ Sum of the zeros$=+\sqrt{15}-\sqrt{15}=0$ Product of the zeros$=+\sqrt{15}\times -\sqrt{15}=-15$ Verification: Here, $a=1,\, b=0$ and $c=-15$ Sum of zeros$=\frac{-b}{a}=\frac{-0}{1}=0$ Product of zeros$=\frac{c}{a}=\frac{-15}{1}=-15$ Hence verified. vi. $3x^2-x-4$ Solution: Given, the quadratic polynomial $3x^2-x-4$ Factorising, $3x^2-x-4$ $=3x^2-(4-3)x-4$ $=3x^2-4x+3x-4$ $=x(3x-4)+1(3x-4)$ $=(3x-4)(x+1)$ The zeros of the quadratic polynomial are, $3x-4=0$ and $x+1=0$ $3x=4$ and $x=-1$ $x=\frac{4}{3}$ and $x=-1$ Sum of the zeros$=\frac{4}{3}+(-1)=\frac{1}{3}$ Product of the zeros$=\frac{4}{3}\times (-1)=-\frac{4}{3}$ Verification: Here, $a=3,\, b=-1$ and $c=-4$ Sum of zeros$=\frac{-b}{a}=\frac{-(-1)}{3}=\frac{1}{3}$ Product of zeros$=\frac{c}{a}=\frac{-4}{3}=-\frac{4}{3}$ Hence verified. Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. i.$\frac{1}{4},\, -1$ Solution: Given, $α+β=\frac{1}{4}\,\,\,\,\,……(i)$ and $α\times β=-1\,\,\,\,\,……(ii)$ We know that, $α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and $α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$ Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$ $a=4,\, b=-1$ and $c=-4$ Therefore, the required quadratic polynomial will be, $ax^2+bx+c$ $4x^2+(-1)x+(-4)$ $4x^2-x-4$ ii. $\sqrt{2},\,\frac{1}{3}$ Solution: Given, $α+β=\sqrt{2}\,\,\,\,\,……(i)$ and $α\times β=\frac{1}{3}\,\,\,\,\,……(ii)$ We know that, $α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and $α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$ Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$ $a=3,\, b=-3\sqrt{2}$ and $c=1$ Therefore, the required quadratic polynomial will be, $ax^2+bx+c$ $3x^2+(-3\sqrt{2})x+(1)$ $3x^2-2\sqrt{2}x+1$ iii. $0,\,\sqrt{5}$ Solution: Given, $α+β=0\,\,\,\,\,……(i)$ and $α\times β=\sqrt{5}\,\,\,\,\,……(ii)$ We know that, $α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and $α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$ Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$ $a=1,\, b=0$ and $c=\sqrt{5}$ Therefore, the required quadratic polynomial will be, $ax^2+bx+c$ $1x^2+(0)x+\sqrt{5}$ $x^2+\sqrt{5}$ iv. $1,\, 1$ Solution: Given, $α+β=1\,\,\,\,\,……(i)$ and $α\times β=1\,\,\,\,\,……(ii)$ We know that, $α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and $α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$ Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$ $a=1,\, b=-1$ and $c=1$ Therefore, the required quadratic polynomial will be, $ax^2+bx+c$ $1x^2+(-1)x+1$ $x^2-x+1$ v. $-\frac{1}{4},\,\frac{1}{4}$ Solution: Given, $α+β=-\frac{1}{4}\,\,\,\,\,……(i)$ and $α\times β=\frac{1}{4}\,\,\,\,\,……(ii)$ We know that, $α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and $α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$ Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$ $a=4,\, b=1$ and $c=1$ Therefore, the required quadratic polynomial will be, $ax^2+bx+c$ $4x^2+1x+1$ $4x^2-x+1$ vi. $4,\,1$ Solution: Given, $α+β=4\,\,\,\,\,……(i)$ and $α\times β=1\,\,\,\,\,……(ii)$ We know that, $α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and $α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$ Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$ $a=1,\, b=-4$ and $c=1$ Therefore, the required quadratic polynomial will be, $ax^2+bx+c$ $1x^2+(-4)x+1$ $x^2-4x+1$ I hope the solutions of Class 10, Exercise 2.2, Chapter 2 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 2.2, feel free to contact me at [email protected] or [email protected] or fill the form here.
Tips & Tricks: Probability # Probability Tips and Tricks for Government Exams ## Introduction • Probability is the section of mathematics, that deals with the results of random events. The word probability means chance or possibility of an outcome. It explains the possibility of a particular event occurring. We often use sentences like - ‘It will probably rain today, ‘he will probably pass the test’, ‘there is very less probability of getting a storm tonight’, and ‘most probably the price of onion will go high again. In all these sentences, we replace words like chance, doubt, maybe, likely, etc., with the word probability. ### Theory • Probability defines the likelihood of occurrence of an event. There are many real-life situations in which we may have to predict the outcome of an event. We may be sure or not sure of the results of an event. In such cases, we say that there is a probability of this event to occur or not occur. Probability generally has great applications in games, in business to make probability-based predictions, and also probability has extensive applications in this new area of artificial intelligence. • The probability of an event can be calculated by probability formula by simply dividing the favorable number of outcomes by the total number of possible outcomes. The value of the probability of an event to happen can lie between 0 and 1 because the favorable number of outcomes can never cross the total number of outcomes. Also, the favorable number of outcomes cannot be negative. ### Formula's • The probability formula defines the likelihood of the happening of an event. It is the ratio of favorable outcomes to the total favorable outcomes. The probability formula can be expressed as, P(E) = Number of favorable outcomes/Number of total outcomes Or, P(E) = n(A)/n(S) where, • P(B) is the probability of an event 'B'. • n(B) is the number of favorable outcomes of an event 'B'. • n(S) is the total number of events occurring in a sample space. ### Tricks • When we throw a coin on the air, the coin appears either a Head (H) or a Tail (T). • A dice is a solid cube having 6 faces, marked as 1, 2, 3, 4, 5, 6 gradually. When we throw a dice the outcome is the number that appears on it’s upper face of the dice. ### Solved Examples Question for Tips & Tricks: Probability Try yourself: The probability of two persons of passing the interview are 1/3 and 3/5. Then calculate the probability that only one of them pass the interview ? Question for Tips & Tricks: Probability Try yourself: Among a group of 5 boys and 6 girls, 4 members is to be selected for an event. Find the probability that at least one boy is selected ? Question for Tips & Tricks: Probability Try yourself: The probability that the problem will be solved by three persons are 1/2, 1/3 and 1/6. Find the probability that the problem is solved ? Question for Tips & Tricks: Probability Try yourself: A urn contains 4 red balls, 5 green balls and 6 white balls, If one ball is drawn at random, find the probability that it is neither red nor white ? The document Probability Tips and Tricks for Government Exams is a part of the Bank Exams Course Tips & Tricks for Government Exams. All you need of Bank Exams at this link: Bank Exams ## Tips & Tricks for Government Exams 66 videos\|66 docs ## FAQs on Probability Tips and Tricks for Government Exams 1. What is probability and how is it calculated? Ans. Probability is a measure of the likelihood of an event occurring. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. 2. What is the difference between theoretical probability and experimental probability? Ans. Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability, on the other hand, is based on actual observations or experiments and may vary from the theoretical probability. 3. How do you find the probability of two independent events occurring? Ans. To find the probability of two independent events occurring, you multiply the probabilities of each event. This can be calculated using the formula: P(A and B) = P(A) * P(B). 4. What is the concept of complementary events in probability? Ans. Complementary events in probability are two events that together encompass all possible outcomes. The probability of either event occurring is equal to 1. For example, if the probability of event A is 0.4, then the probability of event A not occurring (complementary event) is 0.6. 5. How do you calculate the probability of dependent events? Ans. To calculate the probability of dependent events, you multiply the probability of the first event by the conditional probability of the second event given that the first event has already occurred. This can be calculated using the formula: P(A and B) = P(A) * P(B\|A), where P(B\|A) denotes the probability of event B given that event A has occurred. ## Tips & Tricks for Government Exams 66 videos\|66 docs ### Up next Explore Courses for Bank Exams exam Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;
# 2012 AMC 12B Problems/Problem 17 ## Problem Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$? $\textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8$ ## Solutions $[asy] size(7cm); pair A=(0,0),B=(1,1.5),D=Bdir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);[/asy]$ (diagram by [i] MSTang [/i]) ### Solution 1 $[asy] size(14cm); pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8); dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H); draw(A--SS--D--cycle); draw(P--Q--R^^B--Q--C); draw(EE--M--F^^G--B^^C--H,dotted); label("A",A,SW); label("B",B,S); label("C",C,S); label("D",D,SE); label("E",EE,S); label("F",F,S); label("P",P,W); label("Q",Q,NW); label("R",R,NE); label("S",SS,N); label("M",M,S); label("G",G,W); label("H",H,NE);[/asy]$ Construct the midpoints $E=(4,0)$ and $F=(10,0)$ and triangle $\triangle EMF$ as in the diagram, where $M$ is the center of square $PQRS$. Also construct points $G$ and $H$ as in the diagram so that $BG\parallel PQ$ and $CH\parallel QR$. Observe that $\triangle AGB\sim\triangle CHD$ while $PQRS$ being a square implies that $GB=CH$. Furthermore, $CD=6=3\cdot AB$, so $\triangle CHD$ is 3 times bigger than $\triangle AGB$. Therefore, $HD=3\cdot GB=3HC$. In other words, the longer leg is 3 times the shorter leg in any triangle similar to $\triangle AGB$. Let $K$ be the foot of the perpendicular from $M$ to $EF$, and let $x=EK$. Triangles $\triangle EKM$ and $\triangle MKF$ also have legs in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$, so $10x=EF=6$. It follows that $EK=0.6$ and $MK=1.8$, so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $\boxed{\mathbf{(C)}\ 6.4}$. ### Solution 2 $[asy] size(7cm); pair A=(0,0),B=(1,1.5),D=Bdir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);[/asy]$ Let the four points be labeled $P_1$, $P_2$, $P_3$, and $P_4$, respectively. Let the lines that go through each point be labeled $L_1$, $L_2$, $L_3$, and $L_4$, respectively. Since $L_1$ and $L_2$ go through $SP$ and $RQ$, respectively, and $SP$ and $RQ$ are opposite sides of the square, we can say that $L_1$ and $L_2$ are parallel with slope $m$. Similarly, $L_3$ and $L_4$ have slope $-\frac{1}{m}$. Also, note that since square $PQRS$ lies in the first quadrant, $L_1$ and $L_2$ must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: $L_1: y = m(x-3)$, $L_2: y = m(x-5)$, $L_3: y = -\frac{1}{m}(x-7)$, $L_4: y = -\frac{1}{m}(x-13)$. Since $PQRS$ is a square, it follows that $\Delta x$ between points $P$ and $Q$ is equal to $\Delta y$ between points $Q$ and $R$. Our approach will be to find $\Delta x$ and $\Delta y$ in terms of $m$ and equate the two to solve for $m$. $L_1$ and $L_3$ intersect at point $P$. Setting the equations for $L_1$ and $L_3$ equal to each other and solving for $x$, we find that they intersect at $x = \frac{3m^2 + 7}{m^2 + 1}$. $L_2$ and $L_3$ intersect at point $Q$. Intersecting the two equations, the $x$-coordinate of point $Q$ is found to be $x = \frac{5m^2 + 7}{m^2 + 1}$. Subtracting the two, we get $\Delta x = \frac{2m^2}{m^2 + 1}$. Substituting the $x$-coordinate for point $Q$ found above into the equation for $L_2$, we find that the $y$-coordinate of point $Q$ is $y = \frac{2m}{m^2+1}$. $L_2$ and $L_4$ intersect at point $R$. Intersecting the two equations, the $y$-coordinate of point $R$ is found to be $y = \frac{8m}{m^2 + 1}$. Subtracting the two, we get $\Delta y = \frac{6m}{m^2 + 1}$. Equating $\Delta x$ and $\Delta y$, we get $2m^2 = 6m$ which gives us $m = 3$. Finally, note that the line which goes though the midpoint of $P_1$ and $P_2$ with slope $3$ and the line which goes through the midpoint of $P_3$ and $P_4$ with slope $-\frac{1}{3}$ must intersect at at the center of the square. The equation of the line going through $(4,0)$ is given by $y = 3(x-4)$ and the equation of the line going through $(10,0)$ is $y = -\frac{1}{3}(x-10)$. Equating the two, we find that they intersect at $(4.6, 1.8)$. Adding the $x$ and $y$-coordinates, we get $6.4$. Thus, answer choice $\boxed{\textbf{(C)}}$ is correct. ### Solution 3 Note that the center of the square lies along a line that has an $x-$intercept of $\frac{3+5}{2}=4$, and also along another line with $x-$intercept $\frac{7+13}{2}=10$. Since these 2 lines are parallel to the sides of the square, they are perpendicular (since the sides of a square are). Let $m$ be the slope of the first line. Then $-\frac{1}{m}$ is the slope of the second line. We may use the point-slope form for the equation of a line to write $l_1:y=m(x-4)$ and $l_2:y=-\frac{1}{m}(x-10)$. We easily calculate the intersection of these lines using substitution or elimination to obtain $\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)$ as the center or the square. Let $\theta$ denote the (acute) angle formed by $l_1$ and the $x-$axis. Note that $\tan\theta=m$. Let $s$ denote the side length of the square. Then $\sin\theta=s/2$. On the other hand the acute angle formed by $l_2$ and the $x-$axis is $90-\theta$ so that $\cos\theta=\sin(90-\theta)=s/6$. Then $m=\tan\theta=3$. Substituting into $\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)$ we obtain $\left(\frac{23}{5},\frac{9}{5}\right)$ so that the sum of the coordinates is $\frac{32}{5}=6.4$. Hence the answer is $\framebox{C}$. ### Solution 4 (Fast) Suppose $$SP: y=m(x-3)$$ $$RQ: y=m(x-5)$$ $$PQ: -my=x-7$$ $$SR: -my=x-13$$ where $m >0$. Recall that the distance between two parallel lines $Ax+By+C=0$ and $Ax+By+C_1=0$ is $\|C-C_1\|/\sqrt{A^2+B^2}$, we have distance between $SP$ and $RQ$ equals to $2m/\sqrt{1+m^2}$, and the distance between $PQ$ and $SR$ equals to $6/\sqrt{1+m^2}$. Equating them, we get $m=3$. Then, the center of the square is just the intersection between the following two "mid" lines: $$L_1: y=3(x-4)$$ $$L_2: -3y = x-10$$ The solution is $(4.6,1.8)$, so we get the answer $4.6+1.8=6.4$. $\framebox{C}$.
# Variance of a Data Set ## The mean of the squares of the deviation of data values Estimated20 minsto complete % Progress Practice Variance of a Data Set MEMORY METER This indicates how strong in your memory this concept is Progress Estimated20 minsto complete % Calculating Variance If you were told that the mean income at a certain company was 35,000, you wouldn’t really know much about the actual income of the majority of the employees, since there could be a few upper-level managers or owners whose income might skew the mean badly. However, if you were also given the variance of the incomes, how would that help? ### Calculating Variance Variance (commonly denoted \begin{align}\sigma ^2\end{align}) is a very useful measure of the relative amount of ‘scattering’ of a given set. In other words, knowing the variance can give you an idea of how closely the values in a set cluster around the mean. The greater the variance, the more the data values in the set are spread out away from the mean. Variance is an important calculation to become familiar with because, like the arithmetic mean, variance is used in many other more complex statistical evaluations. The calculation of variance is slightly different depending on whether you are working with a population (you do not intend to generalize the results back to a larger group) or a sample (you do intend to use the sample results to predict the results of a larger population). The difference is really only at the end of the process, so let’s start with the calculation of the population. To calculate the variance of a population: 1. First, identify the arithmetic mean of your data by finding the sum of the values and dividing it by the number of values. 2. Next, subtract each value from the mean and record the result. This value is called the deviation of each score from the mean. 3. For each value, square the deviation. 4. Finally, divide the sum of the squared deviations by the number of values in the set. The resulting quotient is the variance \begin{align}(\sigma^2)\end{align} of the set. To calculate the variance of a sample, the only difference is that in step 4, you divide the sum of squared deviations by the number of values in the sample minus 1. By dividing the sum of squared deviations by one less than the number of values, you help reduce the effect of outliers in the sample and increase the calculated variance of the sample by a small amount to allow more ‘room’ for the unknown values in the population. #### Calculating the Variance 1. Calculate the variance of set \begin{align}x\end{align}: \begin{align}x=\left \{12, 7, 6, 3, 10, 5, 18, 15\right \}\end{align} Follow the steps from above to calculate the variance: • First, calculate the arithmetic mean: \begin{align}\mu =\frac{12+7+6+3+10+5+18+15}{8}=9.5\end{align} • Subtract each value from the mean to get the deviation of each value, square the deviation of each value: \begin{align}\text{Value} - \text{Mean} = \text{Deviation}\end{align} \begin{align}\text{Deviation}^2\end{align} \begin{align}12-9.5=2.5\end{align} 6.25 \begin{align}7-9.5=-2.5\end{align} 6.25 \begin{align}6-9.5=-3.5\end{align} 12.25 \begin{align}3-9.5=-6.5\end{align} 42.25 \begin{align}10-9.5=.5\end{align} .25 \begin{align}5-9.5=-4.5\end{align} 20.25 \begin{align}18-9.5=8.5\end{align} 72.25 \begin{align}15-9.5=5.5\end{align} 30.25 TOTAL (sum of deviation2): 190.00 • Finally, divide the sum of the squared deviations by the count of values in the data set: \begin{align}\frac{190}{8} & =23.75\\ \therefore \ The \ variance \ & of \ set \ x \ is \ 23.75\end{align} 2. Find the variance of set \begin{align}z\end{align}: \begin{align}z=\left \{1, 2, 3, 4, 5, 6, 7, 9\right \}\end{align} Divide the squared deviation of each value from the mean by the total number of values in the set: \begin{align}& \qquad \qquad \mu =\frac{1+2+3+4+5+6+7+9}{8}=4.625 \\ &(1-4.625)^2+(2-4.625)^2+(3-4.625)^2+(4-4.625)^2\\ &\qquad +(5-4.625)^2+(6-4.625)^2+(7-4.625)^2+(9-4.625)^2 =49.875 \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \frac{49.875}{8} =6.234\\ & \qquad \qquad \qquad \ \therefore \ Variance \ (\sigma^2) \ of \ set \ z = 6.234\end{align} 3. Find \begin{align}\sigma^2 \ of \ y\end{align}: \begin{align}y=\left \{13, 14, 15, 16, 17, 18, 19, 20, 21\right \}\end{align} Let’s do this one differently, using a nifty trick known as the “mean of the squares minus the square of the mean.” Start, as before, by finding the arithmetic mean: \begin{align}\mu =\frac{13+14+15+16+17+18+19+20+21}{9}=17\end{align} Then, to find the variation, divide the sum of the squares of each value by the number of values (this is the “mean of the squares”), then square the mean we calculated above, 17 (the “square of the mean”), and subtract it from the mean of the squares: \begin{align}&\sigma ^2 = \frac{13^2+14^2+15^2+16^2+17^2+18^2+19^2+20^2+21^2}{9}-17^2=6.6\overline{6} \\ &\qquad \qquad \qquad \qquad \qquad \quad \therefore \ \sigma ^2 \ of \ y=6.6\overline{6}\end{align} #### Earlier Problem Revisited If you were told that the mean income at a certain company was35,000, you wouldn’t really know much about the actual income of the majority of the employees, since there could be a few upper-level managers or owners whose income might skew the mean badly. However, if you were also given the variance of the incomes, how would that help? By learning the variance of the set of incomes, you could get a feel for how representative the \$35,000 figure was of the likely salary of a common employee. ### Examples #### Example 1 Find \begin{align}\mu\end{align} and \begin{align}\sigma ^2\end{align} of set \begin{align}z\end{align} Let’s use the “mean of the squares minus the square of the mean” method: First find the mean of the set: \begin{align}\frac{3.25+3.5+2.85+3.4+2.95+3.02+3.17}{7}=3.16286\end{align} Now divide the sum of each of the values squared by the number of values: \begin{align}\frac{3.25^2+3.5^2+2.85^2+3.4^2+2.95^2+3.02^2+3.17^2}{7}-10.0036=10.0524-10.0036=0.049\end{align} is the variance. #### Example 2 If all values of set \begin{align}z\end{align}, above, were increased by 5, what would the new mean and variance be? Find the mean of the new set: \begin{align}\frac{8.25+8.5+7.85+8.4+7.95+8.02+8.17}{7}=8.16286\end{align} Divide the sum of the values squared by the number of values: \begin{align}\frac{466.7668}{7}=66.681\end{align} Subtract the squared mean from the mean of the squares: \begin{align}66.681-66.632=0.049\end{align} is the variance. The variance is the same as before! Does that surprise you? It should, because they actually aren’t the same, it just appears that way due to rounding. The new set actually has a variance closer to 0.048688, and the original is more accurately 0.04873469. Obviously they are very close, but not exactly the same. #### Example 3 If all values of set \begin{align}z\end{align} from question #1 were doubled, how would that affect \begin{align}\mu\end{align} and \begin{align}\sigma ^2\end{align}? The question is what would happen if all of the values were doubled. Do the mean and variance also double? Let’s see: The mean of the new set is \begin{align}\frac{6.5+7+5.7+6.8+5.9+6.04+6.34}{7}=\frac{44.28}{7}=6.326\end{align}, which is twice the mean of the original set. So far so good. The “mean of the squares” is \begin{align}\frac{6.5^2+7^2+5.7^2+6.8^2+5.9^2+6.04^2+6.34^2}{7}=\frac{281.47}{7}=40.21\end{align}, which is four times the original mean of the squares, not double after all (which makes sense, given that each doubled value was squared). Finally, subtract the two values: \begin{align}40.21-6.326^2 = .192\end{align} is the variance. If we compare this to the original: \begin{align}\frac{.192}{.049}\approx 4\end{align}, we can see that doubling the original values quadruples the variance. ### Review Questions 1-12: find \begin{align}\sigma ^2\end{align} 1. \begin{align}y=\left \{4, 50, 63, 2, 82, 99\right \}\end{align} 2. Set \begin{align}x\end{align} is a random sample from a population with 38 members: \begin{align}x=\left \{8, 13, 5, 10\right \}\end{align} 3. Set \begin{align}z\end{align} is a random sample from a larger population: \begin{align}z=\left \{4,3,5,15,5\right \}\end{align} 4. \begin{align}y=\left \{3,26,5,1,1\right \}\end{align} 5. 22, 21, 13, 19, 16, 18 6. Sample: 1, 2, 5, 1 7. Sample: 10, 6, 3, 4 8. 8, 11, 17, 7, 19 9. 15, 17, 19, 21, 23, 25, 27, 29 10. Sample: 15, 17, 19, 21, 23, 25, 27, 29 11. .25, .35, .45, .55, .26, .75 12. Find the variance of the data in the table: HEIGHTS (rounded to the nearest inch) FREQUENCY OF STUDENTS 60 35 61 33 62 45 63 4 64 3 65 4 66 7 67 4 ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition absolute deviation The absolute deviation is the sum total of how different each number is from the mean. deviation Deviation is a measure of the difference between a given value and the mean. Mean The mean of a data set is the average of the data set. The mean is found by calculating the sum of the values in the data set and then dividing by the number of values in the data set. mean absolute deviation The mean absolute deviation is an alternate measure of how spread out the data is. It involves finding the mean of the distance between each data value and the mean. While this method might seem more intuitive, in statistics it has been found to be too limited and is not commonly used. Population In statistics, the population is the entire group of interest from which the sample is drawn. Sample A sample is a specified part of a population, intended to represent the population as a whole. Skew To skew a given set means to cause the trend of data to favor one end or the other standard deviation The square root of the variance is the standard deviation. Standard deviation is one way to measure the spread of a set of data. variance A measure of the spread of the data set equal to the mean of the squared variations of each data value from the mean of the data set.
# Degree (graph theory) explained In graph theory, the degree (or valency) of a vertex of a graph is the number of edges that are incident to the vertex; in a multigraph, a loop contributes 2 to a vertex's degree, for the two ends of the edge.[1] The degree of a vertex v is denoted \deg(v) or \degv . The maximum degree of a graph G is denoted by \Delta(G) , and is the maximum of G 's vertices' degrees. The minimum degree of a graph is denoted by \delta(G) , and is the minimum of G 's vertices' degrees. In the multigraph shown on the right, the maximum degree is 5 and the minimum degree is 0. In a regular graph, every vertex has the same degree, and so we can speak of the degree of the graph. A complete graph (denoted Kn , where n is the number of vertices in the graph) is a special kind of regular graph where all vertices have the maximum possible degree, n-1 . In a signed graph, the number of positive edges connected to the vertex v is called positive deg (v) and the number of connected negative edges is entitled negative deg (v) .[2] [3] ## Handshaking lemma See main article: Handshaking lemma. The degree sum formula states that, given a graph G=(V,E) , \sumv\deg(v)=2\|E\| . The formula implies that in any undirected graph, the number of vertices with odd degree is even. This statement (as well as the degree sum formula) is known as the handshaking lemma. The latter name comes from a popular mathematical problem, which is to prove that in any group of people, the number of people who have shaken hands with an odd number of other people from the group is even.[4] ## Degree sequence The degree sequence of an undirected graph is the non-increasing sequence of its vertex degrees; for the above graph it is (5, 3, 3, 2, 2, 1, 0). The degree sequence is a graph invariant, so isomorphic graphs have the same degree sequence. However, the degree sequence does not, in general, uniquely identify a graph; in some cases, non-isomorphic graphs have the same degree sequence. The degree sequence problem is the problem of finding some or all graphs with the degree sequence being a given non-increasing sequence of positive integers. (Trailing zeroes may be ignored since they are trivially realized by adding an appropriate number of isolated vertices to the graph.) A sequence which is the degree sequence of some graph, i.e. for which the degree sequence problem has a solution, is called a graphic or graphical sequence. As a consequence of the degree sum formula, any sequence with an odd sum, such as (3, 3, 1), cannot be realized as the degree sequence of a graph. The inverse is also true: if a sequence has an even sum, it is the degree sequence of a multigraph. The construction of such a graph is straightforward: connect vertices with odd degrees in pairs (forming a matching), and fill out the remaining even degree counts by self-loops.The question of whether a given degree sequence can be realized by a simple graph is more challenging. This problem is also called graph realization problem and can be solved by either the Erdős–Gallai theorem or the Havel–Hakimi algorithm. The problem of finding or estimating the number of graphs with a given degree sequence is a problem from the field of graph enumeration. More generally, the degree sequence of a hypergraph is the non-increasing sequence of its vertex degrees. A sequence is k -graphic if it is the degree sequence of some k -uniform hypergraph. In particular, a 2 -graphic sequence is graphic. Deciding if a given sequence is k -graphic is doable in polynomial time for k=2 via the Erdős–Gallai theorem but is NP-complete for all k\ge3 .[5] ## Special values • A vertex with degree 0 is called an isolated vertex. • A vertex with degree 1 is called a leaf vertex or end vertex or a pendant vertex, and the edge incident with that vertex is called a pendant edge. In the graph on the right, is a pendant edge. This terminology is common in the study of trees in graph theory and especially trees as data structures. • A vertex with degree n - 1 in a graph on n vertices is called a dominating vertex. ## Global properties • If each vertex of the graph has the same degree k, the graph is called a k-regular graph and the graph itself is said to have degree k. Similarly, a bipartite graph in which every two vertices on the same side of the bipartition as each other have the same degree is called a biregular graph. • An undirected, connected graph has an Eulerian path if and only if it has either 0 or 2 vertices of odd degree. If it has 0 vertices of odd degree, the Eulerian path is an Eulerian circuit. • A directed graph is a directed pseudoforest if and only if every vertex has outdegree at most 1. A functional graph is a special case of a pseudoforest in which every vertex has outdegree exactly 1. • By Brooks' theorem, any graph G other than a clique or an odd cycle has chromatic number at most Δ(G), and by Vizing's theorem any graph has chromatic index at most Δ(G) + 1. • A k-degenerate graph is a graph in which each subgraph has a vertex of degree at most k. ## References • P.. Erdős. Paul Erdős. T.. Gallai. Tibor Gallai. Gráfok előírt fokszámú pontokkal. hu. Matematikai Lapok. 11. 1960. 264 - 274. . • Havel. Václav. V. J. Havel. 1955. A remark on the existence of finite graphs. cs. Časopis Pro Pěstování Matematiky. 80. 4. 477–480. 10.21136/CPM.1955.108220. free. • Hakimi . S. L. . S. L. Hakimi . Journal of the Society for Industrial and Applied Mathematics . 0148049 . 496–506 . On realizability of a set of integers as degrees of the vertices of a linear graph. I . 10 . 1962. 3 . 10.1137/0110037 . . • Sierksma . Gerard . Hoogeveen . Han . 10.1002/jgt.3190150209 . 2 . . 1106533 . 223–231 . Seven criteria for integer sequences being graphic . 15 . 1991. . ## Notes and References 1. Book: Diestel . Reinhard . Graph Theory . Springer-Verlag . Berlin, New York . 3rd . 978-3-540-26183-4 . 2005 . 5, 28. 2. Ciotti . Valerio . Bianconi . Giestra . Capocci . Andrea . Colaiori . Francesca . Panzarasa . Pietro . Degree correlations in signed social networks . Physica A: Statistical Mechanics and Its Applications . 2015 . 422 . 25–39 . 10.1016/j.physa.2014.11.062 . 1412.1024 . 2015PhyA..422...25C . 4995458 . 2021-02-10 . 2021-10-02 . https://web.archive.org/web/20211002175332/https://www.sciencedirect.com/science/article/abs/pii/S0378437114010334 . live . 3. Saberi . Majerid . Khosrowabadi . Reza . Khatibi . Ali . Misic . Bratislav . Jafari . Gholamreza . Topological impact of negative links on the stability of resting-state brain network . Scientific Reports . January 2021 . 11 . 1 . 2176 . 33500525 . 7838299 . 10.1038/s41598-021-81767-7 . 2021NatSR..11.2176S . 4. Book: Grossman, Peter . Discrete Mathematics for Computing. 2009. 185. Bloomsbury. 978-0-230-21611-2. 5. Deza . Antoine . Levin . Asaf . Meesum . Syed M. . Onn . Shmuel . January 2018 . Optimization over Degree Sequences . SIAM Journal on Discrete Mathematics . en . 32 . 3 . 2067–2079 . 10.1137/17M1134482 . 0895-4801 . 1706.03951 . 52039639.
# How do you find an ordered pair for 3x - 4y = 18 and 5y = x - 28? #### Answer: Solve for x in one equation in order to substitute it into the other equation and you'll work through to get (-2,-6) #### Explanation: Let's start with the original equations: $3 x - 4 y = 18$ $5 y = x - 28$ We can solve the 2nd equation for x, then substitute that into the first equation. So let's first take the 2nd equation: $5 y = x - 28$ and add 28 to both sides to get: $x = 5 y + 28$ We can now substitute that into the first equation: $3 x - 4 y = 18$ $3 \left(5 y + 28\right) - 4 y = 18$ now let's solve for y: $15 y + 84 - 4 y = 18$ $11 y = - 66$ $y = - 6$ We can now substitute in the y value into one of our equations to find the x value. I'm going to use $x = 5 y + 28$ $x = 5 \left(- 6\right) + 28$ $x = - 2$ Now let's check the answer: $3 x - 4 y = 18$ $3 \left(- 2\right) - 4 \left(- 6\right) = 18$ $- 6 + 24 = 18$ check! $5 y = x - 28$ $5 \left(- 6\right) = \left(- 2\right) - 28$ $- 30 = - 2 - 28$ check! So the final answer is (-2,-6)
# The IntMath Newsletter - 21 Sep 2008 By Murray Bourne, 21 Sep 2008 In this Newsletter: 1. Math tip - Hexadecimal Numbers 2. Encryption - and a challenge 3. Current IntMath Poll - Math Anxiety 4. From the math blog 5. Final thought ## 1. Math tip - Hexadecimal Numbers First, so we can better understand hexadecimal numbers, let's talk about our normal number system: 0, 1, 2, 3, 4, ... This system is called the Hindu-Arabic number system because it was developed in India (by the Hindus) and was spread throughout the Middle East and Europe by Muslim scholars around 1000 years ago. Hindu-Arabic numerals finally replaced the Roman Numeral system (I, II, III, IV, V,...) in Europe around 1500. The biggest contribution of the new number system was that it was positional. That is, if I have the number 439, the "4" really means "4x100 = 400", the "3" means "3x10 = 30" and the "9" means "9x1 = 9 units". The position of the digit in the number determines its value. Our number system uses base 10. This means that after every 10 digits, we need to add "1" to the column on the left. Consider 36, 37, 38, 39. The next number is 40. We have added "1" to the column on the left, and the 9 "ticks over" to 0 once again. Using exponents, base 10 numbers are based on the following values in each column (right to left): 100 = 1 101 = 10 102 = 100 103 = 1,000 104 = 10,000 Now let's look at an alternative number system that has important applications in computers. Here we learn that 100 is not always "one hundred"! Hexadecimal Numbers Hexadecimal numbers use base 16 (not base 10 like our normal number system). "Hexa-" means "6" and "deci-" is "ten". In hexadecimal numbers, the digits 0 to 9 are used and the next number is written using the letter "A" (to mean "ten"). Continuing, "B" means "eleven", "C" means "twelve", "D" means "thirteen", "E" means "fourteen" and "F" means "fifteen". The next number is written "10" (which actually means "sixteen" in our normal system), and the following number is "11" (which is really "seventeen"). Next is "12" which means "16+2=18", and so on. Like the Hindu-Arabic decimal system, the hexadecimal system is positional. For example, in the hexadecimal number "34A", • The "3" means "3x162 = 768", • The "4" means "4x16 = 64", and • The "A" means "ten". Altogether, the number "34A" has decimal value 768+64+10 = 842. The hexadecimal system is based on powers of 16, as follows: 160 = 1 161 = 16 162 = 256 163 = 4,096 164 = 65,536 So in the hexadecimal system, the number "100" is actually equal to 256 in the ordinary decimal system. Similar to the decimal system, we add "1" to the column on the left as we count up. In the hexadecimal system, "E + 1 = 10" (which means "fifteen plus one equals sixteen"). Or we could have the counting sequence 39, 3A, 3B, 3C, 3D, 3E, 3F, 40. That last number is obtained by adding one to the left column (giving "4") and the "F" ticks over back to "0". For an example of how hexadecimal numbers are used in computing, the colors of the Web page that you are looking at right now are coded using hexadecimal values. A white background will be written #FFFFFF, whereas the light blue used as the background for Interactive Mathematics is #E5EDEF. ## 2. Encryption [This topic was requested by a reader. It involves ETAOIN, computers and Facebook.] One of the hot areas in current mathematical research is data encryption. With so much sensitive data floating around the Internet, encryption has become vital to protect online banking, e-commerce and even our FaceBook profiles. 🙂 I guess you’re all familiar with simple encryption, like the following: K N O W L E D G E I S P O W E R 11 14 15 23 12 5 4 7 5 9 19 16 15 23 5 18 I have simply used A=1, B=2, C=3,..., Z=26 to turn my text message "knowledge is power" into a set of numbers. Such encoding of a message is called a "cipher" (or "cypher"). However, this cipher is too simple and very easy to crack. One way to make it more secure is to use what is called a "shift cipher", where we add the same number to each of the digits in our code. Let's add 9. So the first number (representing "K") will become 11+9=20, and N will become 14+9=23. When we get to "W", we have a little problem, since 23+9=32. There is no 32nd letter of the alphabet, so we just use modulo arithmetic. (What is the remainder when we divide 32 by 26? Answer: 6. So we say "32 is equivalent to 6, modulo 26"). By applying our shift cipher and modulo arithmetic, our coded message is now: K N O W L E D G E I S P O W E R 20 23 24 6 21 11 13 16 14 13 2 25 24 6 14 1 We could write our encrypted message using letters by translating the original letter-number equivalents: K N O W L E D G E I S P O W E R T W X F U N M P N R B Y X F N A This is better, but it is still easy to decode. You could study the frequency of each letter and figure out what each represents. (The letters E, T, A, O, I, N are the most common English letters, while J, X, Q and Z are the least common.) Using Hexadecimal Numbers I could have made our message harder to crack by using the hexadecimal number system that we learned about above. If I use hexadecimal numbers with A=1, B=2,..., 8=H, 9=J, A=K, B=L, etc (and without the shift cipher), then my encrypted message is: K N O W L E D G E I S P O W E R B E F 17 C 5 4 7 5 9 13 10 F 17 5 12 We could make it harder to crack by applying the shift cipher as before. Ready for a Challenge? I have used hexadecimal numbers and a shift cipher to encode a message. Can you decrypt it? 18 C 9 16 9 17 5 B 9 12 D 19 17 D 12 5 10 10 13 A 19 17 The answer will appear in the next Newsletter. ## 3. Current IntMath Poll - Math Anxiety There has been a big response to the current IntMath Poll where I ask users "How is your math anxiety?" How about you? Are you math anxious right now? You can respond to the poll on any page in Interactive Mathematics (you'll see it in the right column). ## 4. From the Math Blog 1) Friday Math Movie - The Math in MP3 Audio Files Remember LP records? Some purists still love them. This movie takes a look at the differences between digital and analog music recording. 2) Another misleading credit card advertisement Here's another case where a bank offers what seems to be a very good deal - 0% interest on a loan. But let's have a closer look. 3) Review: How Computer Games Help Children Learn "How Computer Games Help Children Learn" is an interesting look at the world of learning games. 4) Tools for making a math website A reader asked what tools I used to produce the Interactive Mathematics site. ## 5. Final Thought Here's something to think about from A.P. Gouthey: To get profit without risk, experience without danger, and reward without work is as impossible as it is to live without being born. Have a good week. See the 7 Comments below. ### 7 Comments on “The IntMath Newsletter - 21 Sep 2008” 1. Li-sa says: "In hexadecimal numbers, the digits 0 to 9 are used and the next number is written using the letter “A” (to mean “ten”). Continuing, “B” means “eleven”, “C” means “twelve”, “D” means “thirteen”, “E” means “fourteen” and “E” means “fifteen”." A minor typing mistake: “F” means fifteen, not “E”. 2. Murray says: Doh! Thanks, Li-sa. I have corrected it in the post above. 3. Felix Muvangua says: Mr. Murray Thanks for your maths newsletter, it is very informative and educative. Today I learn something new about "hexadecimal numbers. Keep up the good work, and may the Lord Almighty give you more wisedom. 4. lyn says: Thanks for te INTMATH Newsletter. It's educational and I learned new things espcially the ENCPRYPTION. I will tell this to my friends. 5. Myke says: What you are doing is incredibly fantastic. keep up the good work. 6. olanipekun christianah says: I really apprieciate your effort on this chapter,you have made me known what i did not know before.It was very understandable. Thank you. 7. will says: No comment on this my first newsletter. I enjoy learning more about peripheral skills in electronics. I'm grateful for simplification as you present math. ### Comment Preview HTML: You can use simple tags like <b>, <a href="...">, etc. To enter math, you can can either: 1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone): a^2 = sqrt(b^2 + c^2) (See more on ASCIIMath syntax); or 2. Use simple LaTeX in the following format. Surround your math with $$ and $$. $$\int g dx = \sqrt{\frac{a}{b}}$$ (This is standard simple LaTeX.) NOTE: You can't mix both types of math entry in your comment.
# square root Graph of the square root function ${\ displaystyle y = {\ sqrt {x}}}$ In the logarithmic representation , the graph of the square root function becomes a straight line with a slope of 12 . The square root (colloquially root; English square root , just sqrt ) a non-negative number is that (uniquely determined) non-negative number whose square is equal to the given number is. The symbol for the square root is the square root sign , so the square root of the number is represented by. The number or the term under the root is called a radicand . The more detailed notation is less common. In addition, the square root can be expressed as a power : is equivalent to For example, because and the square root of is equal . ${\ displaystyle y}$${\ displaystyle y}$ ${\ displaystyle {\ sqrt {}}}$${\ displaystyle y}$${\ displaystyle {\ sqrt {y}}}$ ${\ displaystyle y}$${\ displaystyle {\ sqrt {y}}}$${\ displaystyle {\ sqrt [{2}] {y}}.}$${\ displaystyle y ^ {\ frac {1} {2}}}$${\ displaystyle {\ sqrt {y}}.}$${\ displaystyle 3 ^ {2} = 3 \ cdot 3 = 9}$${\ displaystyle 3 \ geq 0}$${\ displaystyle 9}$${\ displaystyle 3}$ Since the equation has for two solutions , the square root is usually defined as the nonnegative of the two solutions; That is , it always applies. This means that the concept of the square root is unique. The two solutions to the equation are thus and${\ displaystyle x ^ {2} = y}$${\ displaystyle y> 0}$${\ displaystyle {\ sqrt {y}} \ geq 0.}$${\ displaystyle x_ {1} = {\ sqrt {y}}}$${\ displaystyle x_ {2} = - {\ sqrt {y}}.}$ ## Preliminary note on the definitions There are two problems to consider when formally defining the square root: • If you limit yourself to nonnegative rational numbers , then in many cases the square root is undefined. Even in antiquity it was found out that the number, for example, can not be a rational number (see Euclid's proof of the irrationality of the root of 2 ).${\ displaystyle {\ sqrt {2}}}$ • In general, there are two different numbers whose squares match a given number. For example, the number would also be a possible candidate for the square root .${\ displaystyle (-3) ^ {2} = (- 3) \ cdot (-3) = 9}$${\ displaystyle -3}$${\ displaystyle 9}$ The symbol for the square root was first used during the 16th century . It is believed that the character is a modified form of the lowercase r , which is an abbreviation for the Latin word "radix" (root). Originally the symbol was placed in front of the radicand; the horizontal extension was missing. Even Carl Friedrich Gauss therefore used brackets for more complicated root expressions and wrote, for example, instead of${\ displaystyle {\ sqrt {}} (b ^ {2} -4ac)}$${\ displaystyle {\ sqrt {b ^ {2} -4ac}}.}$ In English, the square root is called "square root", which is why the term "sqrt" is used for the square root function in many programming languages . ## Square roots of real numbers Graph of the square function (red and blue). By reflecting only the blue half at the bisector of the 1st quadrant, the diagram of the square root function (green) is created. Definition: The square root of a nonnegative real number is the nonnegative real number whose square is equal . ${\ displaystyle {\ sqrt {y}}}$ ${\ displaystyle y}$${\ displaystyle x}$${\ displaystyle x ^ {2} = x \ cdot x}$${\ displaystyle y}$ Equivalent to this, the real square root can be defined as a function: Let {\ displaystyle {\ begin {aligned} q \ colon [0; \ infty {[} & \ rightarrow [0; \ infty {[} \\ x & \ mapsto y = x ^ {2} \ end {aligned}}} the (bijective) restriction of the square function to the set of nonnegative real numbers. The inverse of this function is called the square root function${\ displaystyle q}$${\ displaystyle y \ mapsto x = {\ sqrt {y}}.}$ ### Remarks • It should be noted that the square function explained by is defined for all real numbers, but is not reversible. It is neither injective nor surjective .${\ displaystyle \ mathbb {R} \ rightarrow \ mathbb {R}; x \ mapsto x ^ {2}}$ • The constraint of the square function is reversible and is reversed by the real root function. Since only nonnegative real numbers appear as images of , the real root function is only defined for these numbers.${\ displaystyle q}$${\ displaystyle q}$ • Because of the restriction made before the inversion from to nonnegative real numbers, the values of the square root function are nonnegative numbers. Restricting the square function to other subsets of , in which different real numbers always have different squares, would lead to other inverse functions, but these are not called real square root functions.${\ displaystyle q}$${\ displaystyle \ mathbb {R}}$ ### Examples Square numbers and their square roots 1 1 121 11 4th 2 144 12 9 3 169 13 16 4th 196 14th 25th 5 225 15th 36 6th 256 16 49 7th 289 17th 64 8th 324 18th 81 9 361 19th 100 10 400 20th ### Properties and rules of calculation The properties of the square root function result from the properties of the square function restricted to the set of non-negative real numbers: • ${\ displaystyle {\ sqrt {a \ cdot b}} = {\ sqrt {a}} \ cdot {\ sqrt {b}} \;}$for .${\ displaystyle \; 0 \ leq a, \, 0 \ leq b}$ • ${\ displaystyle {\ sqrt {a \ cdot b}} = {\ sqrt {-a}} \ cdot {\ sqrt {-b}} \;}$for .${\ displaystyle \; a \ leq 0, \, b \ leq 0}$ • ${\ displaystyle 0 \ leq a , d. i.e., the square root function is strictly increasing. • ${\ displaystyle {\ sqrt {a ^ {2}}} = \| a \|}$holds with the real amount for any real numbers .${\ displaystyle a}$ • In contrast, only applies to non-negative .${\ displaystyle ({\ sqrt {a}}) ^ {2} = a}$${\ displaystyle a}$ • The square root function is differentiable , there applies .${\ displaystyle \ mathbb {R} _ {+}}$ ${\ displaystyle {\ frac {\ mathrm {d} {\ sqrt {x}}} {\ mathrm {d} x}} = {\ frac {1} {2 {\ sqrt {x}}}}}$ • At the point 0 it is not differentiable, its diagram has a perpendicular tangent with the equation .${\ displaystyle x = 0}$ • It can be Riemann-integrable on every closed subinterval of its domain , one of its antiderivatives is .${\ displaystyle [a, b]}$${\ displaystyle F (x) = {\ tfrac {2} {3}} \ cdot {\ sqrt {x ^ {3}}}}$ ## Calculating square roots from real numbers Rational (approximate) values of some square roots ${\ displaystyle {\ begin {array} {ccccr} {\ sqrt {2}} & \ approx & {\ sqrt {\ frac {49} {25}}} & = & {\ frac {7} {5}} \\ {\ sqrt {2}} & \ approx & {\ sqrt {\ frac {289} {144}}} & = & {\ frac {17} {12}} \\ {\ sqrt {2}} & \ approx & {\ sqrt {\ frac {100} {49}}} & = & {\ frac {10} {7}} \\ {\ sqrt {3}} & \ approx & {\ sqrt {\ frac { 49} {16}}} & = & {\ frac {7} {4}} \\ {\ sqrt {4}} &&& = & 2 \\ {\ sqrt {5}} & \ approx & {\ sqrt {\ frac {81} {16}}} & = & {\ frac {9} {4}} \\ {\ sqrt {6}} & \ approx & {\ sqrt {\ frac {2401} {400}}} & = & {\ frac {49} {20}} \\ {\ sqrt {7}} & \ approx & {\ sqrt {\ frac {64} {9}}} & = & {\ frac {8} {3 }} \\ {\ sqrt {8}} & \ approx & {\ sqrt {\ frac {289} {36}}} & = & {\ frac {17} {6}} \\ {\ sqrt {9} } &&& = & 3 \\ {\ sqrt {10}} & \ approx & {\ sqrt {\ frac {361} {36}}} & = & {\ frac {19} {6}} \\ {\ sqrt { 11}} & \ approx & {\ sqrt {\ frac {100} {9}}} & = & {\ frac {10} {3}} \ end {array}}}$ Even if the square root of a natural number is to be taken, the result is often an irrational number , the decimal fraction expansion of which is therefore a non-periodic, non-terminating decimal fraction (namely precisely when the result is not natural). The calculation of a square root, which is not a rational number, therefore, is to determine an approximate value sufficient accuracy. There are a number of ways to do this: Written rooting This is an algorithm similar to the common method of written division. Interval nesting This procedure is quite easy to understand, albeit very cumbersome in practice. Example (approximate value for ):${\ displaystyle {\ sqrt {2}}}$ From and it follows that lies between 1 and 2. Therefore you try , etc. From and you can see that it must be between 1.4 and 1.5. Continuing this procedure with more and more decimal places finally delivers an approximate value with the desired accuracy: ${\ displaystyle 1 ^ {2} = 1 <2}$${\ displaystyle 2 ^ {2} = 4> 2}$${\ displaystyle {\ sqrt {2}}}$${\ displaystyle 1 {,} 1 ^ {2}}$${\ displaystyle 1 {,} 2 ^ {2}}$${\ displaystyle 1 {,} 4 ^ {2} = 1 {,} 96 <2}$${\ displaystyle 1 {,} 5 ^ {2} = 2 {,} 25> 2}$${\ displaystyle {\ sqrt {2}}}$ ${\ displaystyle 1 {,} 41421356 ^ {2} <2 <1 {,} 41421357 ^ {2} \; \ Rightarrow \; {\ sqrt {2}} \ approx 1 {,} 41421356}$ Babylonian root pulling or Heron method This iteration method is often used when programming the calculation of the roots for calculators because it converges quickly . It is Newton's method of finding zeros applied to the function .${\ displaystyle x \ mapsto x ^ {2} -a}$ Taylor series development The Taylor series expansion of the root function with an expansion point can be used as a Taylor expansion of around the point as a binomial series${\ displaystyle t \ mapsto {\ sqrt {t}}}$${\ displaystyle t = 1}$${\ displaystyle x \ mapsto (1 + x) ^ {1/2}}$${\ displaystyle x = 0}$ ${\ displaystyle \ sum _ {n = 0} ^ {\ infty} {\ binom {1/2} {n}} \, x ^ {n} = \ sum _ {n = 0} ^ {\ infty} { \ binom {2n} {n}} \, {\ frac {(-1) ^ {n + 1}} {(2n-1) \, 4 ^ {n}}} \, x ^ {n} = 1 + {\ frac {1} {2}} x - {\ frac {1} {8}} x ^ {2} + {\ frac {1} {16}} x ^ {3} - {\ frac {5 } {128}} x ^ {4} \ pm \ dotsb}$ be found because this series for pointwise to converge . With that results ${\ displaystyle \| x \| \ leq 1}$${\ displaystyle {\ sqrt {1 + x}}}$ ${\ displaystyle x: = t-1}$ ${\ displaystyle {\ sqrt {t}} = 1 + {\ frac {1} {2}} (t-1) - {\ frac {1} {8}} (t-1) ^ {2} + { \ frac {1} {16}} (t-1) ^ {3} - {\ frac {5} {128}} (t-1) ^ {4} \ pm \ dotsb}$ For ${\ displaystyle 0 \ leq t \ leq 2.}$ Calculation using the CORDIC algorithm This process is mainly used in calculators, FPUs and microcontrollers . ### Determination of the square root graphically One possibility is the Kathetensatz : The number whose square root is sought, is on a number line of coated out. A semicircle with a radius is drawn over the distance between and ( Thales circle ). At , a perpendicular to the baseline is established that intersects the semicircle (height of a right triangle). The distance from this point of intersection to the zero point is the square root of (cathetus). ${\ displaystyle n}$${\ displaystyle 0}$${\ displaystyle 0}$${\ displaystyle n}$${\ displaystyle r = {\ tfrac {n} {2}}}$${\ displaystyle 1}$${\ displaystyle n}$ ## Square roots of complex numbers The extraction of the roots corresponds to bisecting the angle in the complex plane. Example:${\ displaystyle {\ sqrt {\ mathrm {i}}}}$ If a complex number other than zero, then the equation has ${\ displaystyle z}$ ${\ displaystyle w ^ {2} = z}$ exactly two solutions for , also known as the roots or square roots of . These lie in the Gaussian plane of numbers on the two points of intersection of the circle around 0 with the radius and the bisector of the angle between the rays emanating from or through . The one of the two roots that lies in the right half-plane is called the principal value of the root. For negative (real) values , the root with the positive imaginary part is the main value. ${\ displaystyle w}$${\ displaystyle z}$${\ displaystyle {\ sqrt {\| z \|}}}$${\ displaystyle 0}$${\ displaystyle 1}$${\ displaystyle z}$${\ displaystyle z}$ If you write the complex number in the form ${\ displaystyle z}$ ${\ displaystyle z = r \ cdot {\ rm {e ^ {\ mathrm {i} \ varphi},}}}$ where and are real with and , then the following applies to the principal value of the root: ${\ displaystyle \ varphi}$${\ displaystyle r}$${\ displaystyle r> 0}$${\ displaystyle - \ pi <\ varphi \ leq \ pi}$ ${\ displaystyle w_ {1} = {\ sqrt {r}} \ cdot {\ rm {e ^ {\ mathrm {i} \ varphi / 2}}}}$ The second square root value (the secondary value) results from point reflection (180 ° rotation) at the zero point: ${\ displaystyle w_ {2} = {\ sqrt {r}} \ cdot {\ rm {e ^ {\ mathrm {i} (\ varphi / 2 + \ pi)}}}}$ ### definition The complex function “ square z ”, like the real square function, does not have an inverse function because it is not injective , but unlike real numbers it is surjective, that is, every complex number is the square of a complex number. Analogous to the real (non-negative) square roots, one can therefore define complex square root functions by restricting the domain of definition to a subset of the complex numbers on which is injective and remains surjective. Depending on which subset you choose, you get different branches of the square root function as the inverse. ${\ displaystyle q \ colon \ mathbb {C} \ rightarrow \ mathbb {C}; z \ mapsto z ^ {2}}$${\ displaystyle q}$${\ displaystyle D}$${\ displaystyle q}$ The main branch of the complex square root function results if one considers the domain of${\ displaystyle q}$ ${\ displaystyle D_ {H}: = \ {x + \ mathrm {i} \, y \ in \ mathbb {C} \ mid x> 0 {\ text {or}} (x = 0 {\ text {and}} y \ geq 0) \}}$ is based on, this is the right half plane of the complex number plane , whereby only the numbers with nonnegative imaginary part belong from the edge . The restriction of to is a bijective mapping of to to the complex numbers, so its inverse function, the main branch of the square root of to, is entirely defined. The value of this inverse function is called the principal value of the square root of . If by a specific complex number is meant, then it is that main value. ${\ displaystyle D_ {H}}$${\ displaystyle q}$${\ displaystyle D_ {H}}$${\ displaystyle D_ {H}}$${\ displaystyle \ mathbb {C}}$${\ displaystyle {\ sqrt {z}}}$${\ displaystyle z}$${\ displaystyle {\ sqrt {z}}}$ Is given in Cartesian coordinates, i.e. with real numbers and , then results ${\ displaystyle z}$${\ displaystyle z = x + {\ rm {iy}}}$${\ displaystyle x}$${\ displaystyle y}$ ${\ displaystyle {\ sqrt {z}} = {\ sqrt {x + {\ rm {iy}}}} = {\ sqrt {\ tfrac {\| z \| + x} {2}}} + \ mathrm {i} \ cdot \ operatorname {sgn ^ {+}} (y) \ cdot {\ sqrt {\ tfrac {\| z \| -x} {2}}}}$ for the main value of the square root, where the function has the value −1 for negative and the value 1 otherwise (also for and thus different from the sign function ): ${\ displaystyle \ operatorname {sgn ^ {+}}}$${\ displaystyle y}$${\ displaystyle y = 0}$ ${\ displaystyle \ operatorname {sgn}}$ ${\ displaystyle {\ text {sgn}} ^ {+} (y) = {\ begin {cases} +1 & {\ text {for}} y \ geq 0 \\ - 1 & {\ text {for}} y < 0 \ end {cases}}}$ The only side branch of is${\ displaystyle q}$${\ displaystyle - {\ sqrt {z}}.}$ Is in polar coordinates given with , the main value of the square root of ${\ displaystyle z}$${\ displaystyle z = \| z \| \ cdot \ mathrm {e} ^ {\ mathrm {i} \ cdot \ arg (z)}}$${\ displaystyle \ arg (z) \ in (- \ pi, \ pi]}$ ${\ displaystyle {\ sqrt {z}} = {\ sqrt {\| z \|}} \ mathrm {e} ^ {\ mathrm {i} \ cdot \ arg (z) / 2}}$ where is the real (nonnegative) square root of . The secondary value results again as . ${\ displaystyle {\ sqrt {\| z \|}}}$${\ displaystyle \| z \|}$${\ displaystyle - {\ sqrt {z}} = {\ sqrt {\| z \|}} \ mathrm {e} ^ {\ mathrm {i} \ cdot (\ arg (z) / 2 + \ pi)}}$ The magnitude of the two roots is therefore the root of the magnitude of the complex number. For the main value, the argument ("the angle of z ", see below) is halved. The other solution results geometrically by point reflection of this main value at the origin . ${\ displaystyle \ arg (z)}$ The argument of a complex number is the oriented angle in the complex plane of numbers , which are points and in real coordinates. In the example below, the argument from and the argument from are highlighted in color. ${\ displaystyle z = x + \ mathrm {i} \, y}$${\ displaystyle \ angle (EOZ)}$${\ displaystyle E (1 \| 0),}$ ${\ displaystyle O (0 \| 0)}$${\ displaystyle Z (x \| y)}$${\ displaystyle z}$${\ displaystyle w_ {1}}$ ### Example: Computing a complex square root Find the square roots of First, the amount of the radicand is determined: ${\ displaystyle z = -1 + \ mathrm {i} \, {\ sqrt {3}}.}$ ${\ displaystyle \| z \| = \ left \| -1+ \ mathrm {i} {\ sqrt {3}} \ right \| = {\ sqrt {(-1) ^ {2} + ({\ sqrt {3}} ) ^ {2}}} = {\ sqrt {1 + 3}} = {\ sqrt {4}} = 2}$ This gives the main value of the square root to {\ displaystyle {\ begin {aligned} w_ {1} & = {\ sqrt {\ tfrac {2 + (- 1)} {2}}} + \ mathrm {i} \ cdot \ operatorname {sgn ^ {+} } ({\ sqrt {3}}) \ cdot {\ sqrt {\ tfrac {2 - (- 1)} {2}}} \\ [0.3em] & = {\ sqrt {\ tfrac {1} {2 }}} + \ mathrm {i} \ cdot (+1) \ cdot {\ sqrt {\ tfrac {3} {2}}} = {\ sqrt {2}} \ cdot \ left ({\ tfrac {1} {2}} + \ mathrm {i} \ cdot {\ tfrac {1} {2}} {\ sqrt {3}} \ right) \ end {aligned}}} The other root is obtained by reversing the sign: ${\ displaystyle w_ {2} = - w_ {1} = {\ sqrt {2}} \ cdot \ left (- {\ tfrac {1} {2}} - \ mathrm {i} \ cdot {\ tfrac {1 } {2}} {\ sqrt {3}} \ right)}$ ### Power law The power law ${\ displaystyle (a \ cdot b) ^ {r} = a ^ {r} \ cdot b ^ {r} \ qquad \ qquad \ qquad \ qquad \ qquad \ qquad \ quad {\ text {(P)}}}$ does not apply to everyone , not even to the main values of the roots. This can already be seen in the special case resulting from the further specification${\ displaystyle r = 1/2}$ ${\ displaystyle a, b \ in \ mathbb {C}}$ ${\ displaystyle a = b =: z}$ ${\ displaystyle {\ sqrt {z ^ {2}}} = \ left ({\ sqrt {z}} \ right) ^ {2},}$ who because of identity too ${\ displaystyle \ left ({\ sqrt {z}} \ right) ^ {2} = z}$ ${\ displaystyle {\ sqrt {z ^ {2}}} = z}$ can be simplified, according to which apparently every negative number provides a counterexample, for example : ${\ displaystyle z = -1}$ Because of and the main value of has the argument , while the main value of has the argument .${\ displaystyle (-1) ^ {2} = 1}$${\ displaystyle \ arg (1) = 0}$${\ displaystyle {\ sqrt {(-1) ^ {2}}}}$${\ displaystyle \ arg ({\ sqrt {1}}) = 0/2 = 0}$${\ displaystyle -1}$${\ displaystyle \ arg (-1) = \ pi}$ Remarks 1. Since principal values of roots from positive radicands must be positive, the counterexample shows that there cannot be a square root function for which the power law applies to all .${\ displaystyle {\ text {(P)}}}$${\ displaystyle a, b \ in \ mathbb {C}}$ 2. For and any one can freely choose the “sign” of two of the three roots, after which exactly one possibility remains for the “sign” of the last third.${\ displaystyle r = 1/2}$${\ displaystyle a, b \ in \ mathbb {C}}$${\ displaystyle {\ text {(P)}}}$ ## Square roots modulo n Square roots can also be defined in the remainder class ring . Analogously to the real and complex numbers, a square root of is called if: ${\ displaystyle \ mathbb {Z} / n \ mathbb {Z}}$${\ displaystyle q}$${\ displaystyle x}$ ${\ displaystyle q ^ {2} \ equiv x {\ bmod {n}}}$ However, one has to use different methods to calculate square roots modulo than to calculate real or complex square roots. To find the square roots of modulo , one can proceed as follows: ${\ displaystyle n}$${\ displaystyle x}$${\ displaystyle n}$ First you determine the prime factorization ${\ displaystyle n = p_ {1} ^ {m_ {1}} \ cdot p_ {2} ^ {m_ {2}} \ dotsm p_ {k} ^ {m_ {k}}}$ of the module and then the solutions modulo of the individual prime powers . These solutions are finally put together to form the desired solution using the Chinese remainder theorem . ${\ displaystyle n}$${\ displaystyle p ^ {m}}$ ### Calculation of square roots modulo a prime number p The case is simple: Because of and modulo 2, every number has a uniquely determined square root, namely itself. For prime numbers not equal to 2, the square roots of : ${\ displaystyle p = 2}$${\ displaystyle 0 ^ {2} = 0, \, 1 ^ {2} = 1}$${\ displaystyle 1 \ not \ equiv 0 {\ bmod {2}}}$ ${\ displaystyle p}$${\ displaystyle x}$ To test whether it even has a square root in , one computes the value of the Legendre symbol${\ displaystyle x}$${\ displaystyle \ mathbb {Z} / p \ mathbb {Z}}$ ${\ displaystyle \ left ({\ frac {x} {p}} \ right) \ equiv x ^ {\ frac {p-1} {2}} {\ bmod {p}}}$, because it applies: ${\ displaystyle \ left ({\ frac {x} {p}} \ right) = {\ begin {cases} -1, & {\ text {if}} x {\ text {quadratic non-remainder modulo}} p {\ text {is}} \\ 0, & {\ text {if}} x {\ text {and}} p {\ text {are not prime}} \\ 1, & {\ text {if}} x {\ text {a quadratic remainder modulo}} p {\ text {is}} \ end {cases}}}$ In the first case there is no square root in and in the second case it only has the square root 0. The interesting case is the third case, and therefore we assume in the following that it holds. ${\ displaystyle x}$${\ displaystyle \ mathbb {Z} / p \ mathbb {Z}}$${\ displaystyle {\ bigl (} {\ tfrac {x} {p}} {\ bigr)} = 1}$ #### Calculation for the case p mod 4 = 3 If the Legendre symbol is 1, then are ${\ displaystyle {\ bigl (} {\ tfrac {x} {p}} {\ bigr)}}$ ${\ displaystyle q \ equiv \ pm x ^ {\ frac {p + 1} {4}} {\ bmod {p}}}$ the two square roots of modulo . ${\ displaystyle x}$${\ displaystyle p}$ #### Calculation for the case p mod 4 = 1 If the Legendre symbol is 1, then are ${\ displaystyle {\ bigl (} {\ tfrac {x} {p}} {\ bigr)}}$ ${\ displaystyle q \ equiv \ pm {\ frac {x} {2r}} \ left (W _ {\ frac {p-1} {4}} + W _ {\ frac {p + 3} {4}} \ right ) {\ bmod {p}}}$ the two square roots of modulo . Here one chooses so that ${\ displaystyle x}$${\ displaystyle p}$${\ displaystyle r}$ ${\ displaystyle \ left ({\ frac {r ^ {2} -4x} {p}} \ right) = - 1}$ applies. You can simply test different values of . The consequence is recursively through ${\ displaystyle r}$${\ displaystyle W_ {n}}$ ${\ displaystyle W_ {n} = {\ begin {cases} r ^ {2} / x-2, & {\ text {if}} n = 1 \\ W_ {n / 2} ^ {2} -2, & {\ text {if}} n {\ text {even}} \\ W _ {(n + 1) / 2} W _ {(n-1) / 2} -W_ {1}, & {\ text {if }} n> 1 {\ text {odd}} \ end {cases}}}$ Are defined. Sample calculation for and : ${\ displaystyle x = 3}$${\ displaystyle p = 37}$ According to the above formula, the square roots of through are ${\ displaystyle x}$ ${\ displaystyle q \ equiv \ pm {\ frac {x} {2r}} \ left (W_ {9} + W_ {10} \ right) {\ bmod {3}} 7}$ given. For one finds the value by trying , because the following applies: ${\ displaystyle r}$${\ displaystyle r = 2}$ ${\ displaystyle \ left ({\ frac {r ^ {2} -4x} {p}} \ right) \ equiv (r ^ {2} -4x) ^ {\ frac {p-1} {2}} \ equiv (-8) ^ {18} \ equiv 36 \ equiv -1 {\ bmod {3}} 7}$ The values for and result as follows: ${\ displaystyle W_ {9}}$${\ displaystyle W_ {10}}$ ${\ displaystyle {\ begin {matrix} W_ {1} & \ equiv & r ^ {2} / x-2 & \ equiv & 4 / 3-2 & \ equiv & 24 & {\ bmod {3}} 7 \\ W_ {2} & \ equiv & W_ {1} ^ {2} -2 & \ equiv & 24 ^ {2} -2 & \ equiv & 19 & {\ bmod {3}} 7 \\ W_ {3} & \ equiv & W_ {1} W_ {2} - W_ {1} & \ equiv & 24 \ cdot 19-24 & \ equiv & 25 & {\ bmod {3}} 7 \\ W_ {4} & \ equiv & W_ {2} ^ {2} -2 & \ equiv & 19 ^ {2} -2 & \ equiv & 26 & {\ bmod {3}} 7 \\ W_ {5} & \ equiv & W_ {2} W_ {3} -W_ {1} & \ equiv & 19 \ cdot 25-24 & \ equiv & 7 & {\ bmod {3}} 7 \\ W_ {9} & \ equiv & W_ {4} W_ {5} -W_ {1} & \ equiv & 26 \ cdot 7-24 & \ equiv & 10 & {\ bmod {3}} 7 \\ W_ {10} & \ equiv & W_ {5} ^ {2} -2 & \ equiv & 7 ^ {2} -2 & \ equiv & 10 & {\ bmod {3}} 7 \\\ end {matrix}}}$ Substituting these values gives ${\ displaystyle q \ equiv \ pm {\ frac {x} {2r}} \ left (W_ {9} + W_ {10} \ right) \ equiv \ pm {\ frac {3} {4}} (10+ 10) \ equiv \ pm 15 {\ bmod {3}} 7.}$ That means: 15 and 22 are the two square roots of 3 modulo 37. ## Square roots from matrices As the root of a square matrix is any matrix , which multiplied by itself result: ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle A}$ ${\ displaystyle A = B \ cdot B \ Leftrightarrow B {\ text {is the root of}} A}$ As with the square root of real or complex numbers, the square root of matrices is not necessarily unique. However, if one only looks at positively definite symmetric matrices, then the formation of the roots is unambiguous: Every positively definite symmetric matrix has a positively definite symmetric root which is clearly defined.It is obtained by diagonalizing with the help of an orthogonal matrix (this is always possible according to the spectral theorem ) and then replaced the diagonal elements with their roots; however, the positive root must always be chosen. See also Cholesky decomposition . The uniqueness follows from the fact that the exponential mapping is a diffeomorphism from the vector space of the symmetric matrices to the subset of the positively definite symmetric matrices. ${\ displaystyle A}$${\ displaystyle A ^ {\ frac {1} {2}}.}$${\ displaystyle A}$ ## Square root of an approximate integral operator You can numerically approximate the specific integral function from 0 to with and a given function , which assumes the values at the equidistant interpolation points , as a matrix multiplication as follows (for ): ${\ displaystyle G, \, g_ {i}: = g (x_ {i})}$${\ displaystyle x_ {i}}$${\ displaystyle x_ {i} = i \ Delta x}$${\ displaystyle i = 0.1, \ dotsc, n-1}$${\ displaystyle F, \, f_ {i}: = f (x_ {i})}$${\ displaystyle x_ {i}}$${\ displaystyle f_ {i}}$ ${\ displaystyle G = FI}$${\ displaystyle n = 4}$ ${\ displaystyle G = FI = {\ begin {pmatrix} g_ {0} & g_ {1} & g_ {2} & g_ {3} \\ 0 & g_ {0} & g_ {1} & g_ {2} \\ 0 & 0 & g_ {0} & g_ {1} \\ 0 & 0 & 0 & g_ {0} \ end {pmatrix}} = {\ begin {pmatrix} f_ {0} & f_ {1} & f_ {2} & f_ {3} \\ 0 & f_ {0} & f_ {1} & f_ { 2} \\ 0 & 0 & f_ {0} & f_ {1} \\ 0 & 0 & 0 & f_ {0} \ end {pmatrix}} {\ begin {pmatrix} \ Delta x & \ Delta x & \ Delta x & \ Delta x \\ 0 & \ Delta x & \ Delta x & \ Delta x \\ 0 & 0 & \ Delta x & \ Delta x \\ 0 & 0 & 0 & \ Delta x \ end {pmatrix}}}$ It is vividly clear that one can repeat this operation and thus obtain the double integral : ${\ displaystyle H, \, h_ {i}: = h (x_ {i})}$ ${\ displaystyle H = GI = FII = FI ^ {2}}$ So one can understand the matrix as a numerically approximated integral operator. ${\ displaystyle I}$ The matrix cannot be diagonalized and its Jordanian normal form is: ${\ displaystyle I}$ ${\ displaystyle {\ begin {pmatrix} \ Delta x & 1 & 0 & 0 \\ 0 & \ Delta x & 1 & 0 \\ 0 & 0 & \ Delta x & 1 \\ 0 & 0 & 0 & \ Delta x \ end {pmatrix}}}$ To get a square root from it, one could proceed as described for the non-diagonalizable matrices. However, there is a more direct formal solution in this case as follows: ${\ displaystyle I ^ {\ beta} = {\ begin {pmatrix} \ alpha _ {0} & \ alpha _ {1} & \ alpha _ {2} & \ alpha _ {3} \\ 0 & \ alpha _ { 0} & \ alpha _ {1} & \ alpha _ {2} \\ 0 & 0 & \ alpha _ {0} & \ alpha _ {1} \\ 0 & 0 & 0 & \ alpha _ {0} \ end {pmatrix}}}$ with , and . ${\ displaystyle \ alpha _ {0} = (\ Delta x) ^ {\ beta}}$${\ displaystyle \ alpha _ {k} = \ sum _ {j = 1} ^ {k} {\ frac {\ Gamma (\ beta +1) (- 1) ^ {j + 1} \ alpha _ {kj} } {\ Gamma (j + 1) \ Gamma (\ beta -j + 1)}}}$${\ displaystyle k = 1,2, \ dotsc, n-1}$ The indices denote the subdiagonals (0 is the diagonal) and the exponent is equal . If one assumes real and positive, then real and by definition is positive. ${\ displaystyle \ alpha}$${\ displaystyle \ beta}$${\ displaystyle {\ tfrac {1} {2}}}$${\ displaystyle \ Delta x}$${\ displaystyle (\ Delta x) ^ {\ frac {1} {2}}}$ With this one can numerically approximate a "half" certain integral from 0 to the function as follows: ${\ displaystyle L, \, l_ {i}: = l (x_ {i})}$${\ displaystyle x_ {i}}$${\ displaystyle f (x)}$ ${\ displaystyle L = FI ^ {\ beta} = {\ begin {pmatrix} l_ {0} & l_ {1} & l_ {2} & l_ {3} \\ 0 & l_ {0} & l_ {1} & l_ {2} \\ 0 & 0 & l_ {0} & l_ {1} \\ 0 & 0 & 0 & l_ {0} \ end {pmatrix}} = {\ begin {pmatrix} f_ {0} & f_ {1} & f_ {2} & f_ {3} \\ 0 & f_ {0} & f_ {1} & f_ {2} \\ 0 & 0 & f_ {0} & f_ {1} \\ 0 & 0 & 0 & f_ {0} \ end {pmatrix}} {\ begin {pmatrix} \ alpha _ {0} & \ alpha _ {1} & \ alpha _ {2} & \ alpha _ {3} \\ 0 & \ alpha _ {0} & \ alpha _ {1} & \ alpha _ {2} \\ 0 & 0 & \ alpha _ {0} & \ alpha _ {1 } \\ 0 & 0 & 0 & \ alpha _ {0} \ end {pmatrix}}}$ If you are looking for all operators which, when multiplied by themselves , result in the approximate integral operator , you also have to insert the negative sign, that is, there are two solutions . ${\ displaystyle I}$${\ displaystyle \ pm I ^ {\ frac {1} {2}}}$ To derive the formula, you can first invert, raise the result to the power and then invert it again. ${\ displaystyle I}$${\ displaystyle \ beta}$ 1. The validity of the power law for square roots is not assumed in the quoted passage, but is sometimes assumed in the literature (for negative real radicands): Klaus Fritzsche: Tutorium Mathematik für Einsteiger . Springer-Verlag, 2016, ISBN 978-3-662-48910-9 ( limited preview in Google Book Search [accessed July 4, 2017]).${\ displaystyle {\ text {(P)}}}$
# Ascending Order Ascending order is a method of arranging numbers from smallest value to largest value. The order goes from left to right. Ascending order is also sometimes named as increasing order. For example, a set of natural numbers are in ascending order, such as 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8… and so on. The less than symbol (<), is used to denote the increasing order. The inverse method of increasing order is descending order, where the numbers are arranged in decreasing order of values. Learn the ascending order definition, symbol/sign, examples, representation on a number line, ascending order of fractions, solved problems, etc., in this article. ## What is Ascending Order? In Mathematics, the process of arranging the numbers from smallest value to largest value is called ascending order. The numbers are arranged from left to right in increasing order. A real-life example of ascending order is an arrangement of English alphabets from A to Z. The word ‘ascending’ means going up. Hence, in the case of Mathematics, if the numbers are going up, then they are arranged in ascending order. The other terms used for ascending order are: • Lowest value to highest value • Bottom value to Top value Note: It is not necessary that the numbers are available in a pattern, to be arranged in ascending order. ## Ascending order symbol Ascending order is denoted by the less than symbol ‘<‘. For example, 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 The above numbers are arranged in ascending order from 1 to 9. The symbol represents that the succeeding number is greater than the preceding number in the arrangement. The sign of ascending order represents numbers from lowest to highest, which is opposite to the concept of descending Order, where numbers are arranged from largest to smallest. ## Examples of Ascending Order Some of the examples of numbers arranged in ascending order are given below: • 1 < 2 < 3 • 10 < 11 < 12 < 13 • 43 < 55 < 78 < 98 < 101 • 100 < 1000 < 10000 • -10 < -9 < -8 < -7 ## Ascending Order on Number Line The concept of ascending order is been introduced in primary classes for students of Class 1. With the help of the number line, students can easily understand the arrangement of numbers in increasing order. In the above number lines numbers from -6 to 6 are arranged in ascending order. It states that numbers on the left side of 0 are smaller than the numbers on the right side of 0. As we go from left to right on the number line the value of numbers increases. ## How to Arrange Numbers in Ascending Order? To arrange the numbers in ascending order, first, we need to compare the values and then order them in ascending order. We can arrange here different numbers such as: • Integers • Negative numbers • Fractions • Decimals ### Arranging Integers in Ascending Order As we know integers are numbers that can be negative, positive or zero. But they are not fractions. Let us learn how to arrange positive integers in ascending order. • Firstly compare the number of digits in each number • The number with lesser digits is the smallest number • The number with the largest digits is the largest number • If the number of digits is the same, then compare the left-most digits of the numbers • Compare the numbers in the same manner and arrange them all from smallest to largest. Example: 2 < 4 < 12 < 23 < 451 < 541 ### Negative Numbers in Ascending Order Arranging negative numbers can be a little challenging for the students at the beginning. But once they have understood the logic, it will be very easy for them to arrange the numbers in ascending order. If a bigger number is having a negative sign, then it becomes the smallest value. For example, 3 is greater than 2, but -3 is smaller than 2. Similarly, a two-digit negative number is smaller than a single-digit negative number. -43 < -8 -50 < -25 < -10 < -1 ### Fractions in Ascending order There are two methods involved in finding the ascending order of the fractional numbers. Both the method will give the same solution Method 1: • Step 1: For a given series of fractions, first convert it into decimal numbers. • Step 2: Find the increasing order of the decimal numbers. • Step 3: Finally, replace the decimal values with the respective fractional numbers. Method 2 : • Step 1: Find the L.C.M of the denominators. • Step 2: Divide the L.C.M value by the denominator of the fraction. • Step 3: Multiply both the numerator and denominator of the fraction with the resultant value of step 2. • Step 4: As a result of step 2 and step 3, compare the like fractions. • Step 5: Since the denominators are the same, compare the numerator values of like fractions. • Step 6: Finally, arrange the fractions in increasing order with their respective fractions given in the problem. ### Decimals in Ascending Order To arrange the list of decimal numbers, first, check which decimal has the least whole number part. Now arrange the decimals with increasing order of whole number part. If the whole number part is the same for two or more decimals, then arrange the decimals according to the decimal part after th decimal point, in ascending order. Example: 1.2 < 2.3 < 3.5 < 3.6 < 3.8 < 4.3 ### Ascending order of Alphabets Same as numbers you can also arrange alphabets in ascending order and descending orders. For example : a < b < c < d < e < f < g < h < i < j < k < l < m < n < o < p < q < r < s < t < u < v < w < x < y < z (For small alphabets). You can reverse the order of the alphabets in the case of descending order. ## Ascending Order and Descending Order Descending order is the contradiction of ascending order. That means it is the opposite process of writing the numbers in increasing order. Therefore, it can be mentioned as a decreasing order. In the case of descending order, for a given set of numbers, the highest valued number is written first, and the lowest valued number is written at last. It is denoted by the symbol ‘>’. Ascending Order Descending Order Numbers are arranged in increasing order Numbers are arranged in decreasing order Smallest to largest Largest to smallest It is represented by less than symbol: ‘<’ It is represented by greater than symbol: ‘>’ Example: 3<6<7<9<10 Example: 10>9>7>6>3 ## Solved Problems on Ascending Order Question 1: Arrange the numbers 34, 10, 6, 4, 45, 25 in ascending order. Solution: The increasing order of the numbers are 4, 6, 10, 25, 34 and 45 Question 2: Write these numbers in ascending order: 8, 10, 92, 1, 27. Solution: The following is the increasing order of the given numbers: 1 < 8 < 10 < 27 < 92 Question 2: Arrange the following numbers in increasing order. 63 , 72 , 42 , 26 ,34 and 102 Solution: Given , 63 , 72 , 42 , 26 ,34 and 102 63 = 6 x 6 x 6 = 216 72= 7 x 7 =49 42 = 4 x 4 = 16 26 =2 x 2 x 2 x 2 x 2 x 2 = 64 34 = 3 x 3 x 3 x 3 = 81 102 = 10 x 10 = 100 Therefore, the increasing order of the numbers are 16, 49, 64, 81, 100 and 216 and will be represented as : 42 < 72 < 26 < 34 < 102 < 63 Question 4: Find the increasing order of the following fractions: 8/6, 2/3, 10/12 and 9/6. Solution: Given series: 8/6, 2/3, 10/12 and 9/6. Step 1: Converting fractions into decimals 8/6 = 1.33 2/3 = 0.67 10/12 = 0.83 9/6 = 1.5 Step 2: Arranging the increasing order of the decimal values 0.67 < 0.83 < 1.33 < 1.5 Step 3 : Replacing the decimal values with fraction values 2/3 < 10/12 < 8/6 < 9/6 Therefore, the increasing order of the given fractions is 2/3, 10/12, 8/6 and 9/6. Question 5: Write 3, 7, 8, 2, 10, 28, 15 in descending order. Solution: 28 > 15 > 10 > 8 > 7 > 3 > 2 is the descending order of the given set of numbers. ## Practice Problems on Ascending Order These problems are the worksheet for students to practice more in ascending order. 1. Arrange in ascending order: 2, 1, 7, 3, 4. 2. Write the numbers in increasing order: 90, 34, 92, 1, 35. 3. Write the numbers in ascending order using the symbol: 80, 1, 12, 10, 72. 4. Rearrange the numbers in increasing order: 18, 11, 67, 19, 07. 5. Write the numbers in increasing order: 7, 15, 90, 81, 56. ## Frequently Asked Questions on Ascending Order ### What is the ascending order and descending order? When the numbers are written in increasing order, smallest to largest value, then it is said to be ascending, for example, 3<5<7<9<11<13 When the numbers are written in decreasing order, i.e. largest to smallest value, then it is said to be descending order. For example, 13>11>9>7>5>3 ### How can we arrange the numbers in increasing order? Suppose a set of values are given: 23, 11, 15, 9, 55, 43. By arranging them in ascending order, we have to figure the smallest value first, which is 9. 9 Now the number greater than 9 here is 11. Therefore, 9<11 It indicates that 11 is greater than 9. In the same way, we can arrange all the numbers as; 9<11<15<23<43<55 ### What is ascending order in A to Z? Ascending order means going up from small value to high value and text from A to Z. Descending order means arranging the numbers from largest to smallest and text from Z to A. When the names are arranged for a list, then it is usually arranged in A to Z order, alphabetically. ### How to arrange negative numbers in ascending order? To arrange negative numbers, we should always remember that the highest number with a minus sign (-) or negative symbol, is the smallest value.For example, -3, -8, -10, -4, -9 are the integers. Then on arranging them in increasing order, we get; -10, -9, -8, -4, -3 Hence, -10 is the smallest value here, and -3 is the largest one. ### What is the sign of ascending and descending orders? Ascending order is represented by < (less than) symbol, whereas descending order is represented by > (greater than) symbol.
# Difference between revisions of "1997 AHSME Problems/Problem 26" ## Problem Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$, angle $APB$ is twice angle $ACB$, and $\overline{AC}$ intersects $\overline{BP}$ at point $D$. If $PB = 3$ and $PD= 2$, then $AD\cdot CD =$ $[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("A",A,SW);label("B",B,SE);label("C",C,N);label("D",D,NE + N);label("P",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]$ $\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ ## Solution The product of two lengths with a common point brings to mind the Power of a Point Theorem. Since $PA = PB$, we can make a circle with radius $PA$ that is centered on $P$, and both $A$ and $B$ will be on that circle. Since $\angle APB = \widehat {AB} = 2 \angle ACB$, we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the masure of inscribed angle $\angle ACB$, which is true for all inscribed angles. $[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3.06,0.9); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); pair E = (0,4.5); dot(A);dot(B);dot(C);dot(P);dot(D);dot(E); label("A",A,SW);label("B",B,SE);label("C",C,NW);label("D",D,NE + N);label("P",P,N);label("E", E, NW); draw(A--B--P--cycle); draw(A--C--B--cycle); draw(circle(P, 2.46)); draw(P--E);[/asy]$ Since $PDB$ is a line, we have $PD + DB = PB$, which gives $3 = DB + 2$, or $DB = 1$. We now extend radius $PB = 3$ to diameter $EB = 6$. Since $EDB$ is a line, we have $ED + DB = EB$, which gives $ED + 1 = 6$, or $ED = 5$. Finally, we apply the power of a point theorem to point $D$. This states that $AD \cdot DC = DB \cdot BE$. Since $DB = 1$ and $BE = 5$, the desired product is $5$, which is $\boxed{A}$. ## See also 1997 AHSME (Problems • Answer Key • Resources) Preceded byProblem 25 Followed byProblem 27 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions Invalid username Login to AoPS
# Area Of Square Using Diagonals The area of a square can be calculated using its diagonals. Apart from the traditional method of finding the area of a square by squaring the sides, this is one of the most useful methods to compute the area of a square if the diagonal length is given. ## Formula to Calculate Area of a Square Using Diagonal It is not always mandatory to have the measure of the side for computing a square’s area. If we have the length of the diagonal, then the area can be calculated as: Area of Square = ½ × d2 Square units • Here, “d” is the length of any of the diagonal (in a square, diagonals are equal) ### Derivation for Area of Square using Diagonal Formula The formula to find the area of any square if its diagonals are given can be derived using Pythagoras theorem as explained below: Consider a square of sides “a” units and diagonal as “d” units. In the above figure, the square of the side “a” unit, has been divided into two right triangles with the help of diagonal of length “d” units. Thus, the diagonal of the square divides it into two right triangles. Consider any right triangle and apply Pythagoras theorem. According to Pythagoras theorem, for a right-angled triangle, Hypotenuse2 = Perpendicular2 + Base2 In the above diagram, • Perpendicular = a • Base = a • Hypotenuse = d So, a2 + a2 = d2 ⇒ 2a2 = d2 Or, a2 = d2/2 Now, area of a square = a2 = d2/2 So, area of a square using diagonals = ½ × dSquare units. ### Example Question to Solve Area of Square When Diagonal is Given: Question: Find the area of a square having a diagonal of length 10 cm. Solution: Given, d = 10 cm Now, the area of square = ½ × (diagonal)2 So, A = ½ × 102 Or, area = 50 cm2 Note: Using the diagonal, the perimeter of the square can also be found as explained below. ### Finding Perimeter of Square Using Diagonal It is known that the area of a square in terms of diagonal = ½ × d2 Now, a × a = ½ × d2 Or, a = √(½ × d2) As the perimeter of square = 4a, P = 4 × √(d2/2) units.
# Vertex Form Pronunciation: /ˈvɜr.tɛks fɔɹm/ Explain Click on the blue points in the figure and drag them to change the figure. How can you change the figure so that the parabola opens downwards? Manipulative 1 - Vertex Form of a Quadratic Equation Created with GeoGebra. The vertex form of a quadratic equation is f(x) = a(x - h)2 + k. This form is called the vertex form because the point (h, k) is the vertex of the parabola described by the equation. ### Graphing a Quadratic Equation in Vertex Form An advantage of transforming a quadratic equation into vertex form is that it is easier to graph. Since the vertex can be read from the equation, it can quickly be identified and plotted. 1. Step 1 In the equation y = -1 ( x - 2 ) ^ 2 + 1, the vertex is (2, 1). 2. Step 2 Plot ( h + 1, k + a ). General CaseExampleDescription y = a((h + 1) - h)2 + k y = -1((2 + 1) - 2)2 + 1 Substitute h + 1 in for x. y = a(h - h + 1)2 + k y = -1(3 - 2)2 + 1 Simplify the innermost parenthesis. y = a · 12 + k y = -1 · 12 + 1 Simplify the remaining parenthesis. y = a · 1 + k y = -1 · 1 + 1 Simplify the exponent. y = a + k y = -1 + 1 Simplify the multiplication. y = k + a y = 0 Simplify the addition and subtraction. So, (h + 1, k + a) is a point on the graph. So, (3, 0) is a point on the graph. Plot the point. Table 1: Plot (h + 1, k + a). 3. Step 3 Plot (h - 1, k + a). General CaseExampleDescription y = a((h - 1) - h)2 + k y = -1((2 - 1) - 2)2 + 1 Substitute h + 1 in for x. y = a(h - h - 1)2 + k y = -1(1 - 2)2 + 1 Simplify the innermost parenthesis. y = a(-1)2 + k y = -1(-1)2 + 1 Simplify the remaining parenthesis. y = a · 1 + k y = -1 · 1 + 1 Simplify the exponent. y = a + k y = -1 + 1 Simplify the multiplication. y = k + a y = 0 Simplify the addition and subtraction. So, (h - 1, k + a) is a point on the graph. So, (1, 0) is a point on the graph. Plot the point. Table 1: Plot (h - 1, k + a). 4. Step 4 Sketch the parabola. 5. Click on the points on the sliders and drag them to change the figure. Click on the check boxes to see each step. Manipulative 2 - How to Graph a Quadratic Equation in Vertex Form Created with GeoGebra. ### Converting a Quadratic Equation from Standard Form to Vertex Form Often it is easier to sketch a quadratic equation by converting it to vertex form. This can be done using complete the square. ### References 1. McAdams, David E.. All Math Words Dictionary, vertex form. 2nd Classroom edition 20150108-4799968. pg 189. Life is a Story Problem LLC. January 8, 2015. Buy the book • McAdams, David E.. Complete the Square. allmathwords.org. All Math Words Encyclopedia. Life is a Story Problem LLC. 12/13/2009. http://www.allmathwords.org/en/c/completethesquare.html. McAdams, David E. Vertex Form. 5/13/2019. All Math Words Encyclopedia. Life is a Story Problem LLC. http://www.allmathwords.org/en/v/vertexform.html. ### Revision History 5/13/2019: Changed equations and expressions to new format. (McAdams, David E.) 12/21/2018: Reviewed and corrected IPA pronunication. (McAdams, David E.) 12/17/2018: Removed broken links, updated license, implemented new markup, update to new GeoGebra app. (McAdams, David E.)
# If $A (-3, 5), B (-2, -7), C (1, -8)$ and $D (6, 3)$ are the vertices of a quadrilateral $ABCD$, find its area. Given: $A (-3, 5), B (-2, -7), C (1, -8)$ and $D (6, 3)$ are the vertices of a quadrilateral $ABCD$. To do: We have to find the area of the quadrilateral. Solution: Join $A$ and $C$ to get two triangles $ABC$ and $ADC$. This implies, Area of quadrilateral $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$. We know that, Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$ Therefore, Area of triangle $ABC=\frac{1}{2}[-3(-7+8)+(-2)(-8-5)+1(5+7)]$ $=\frac{1}{2}[-3(1)+(-2)(-13)+1(12)]$ $=\frac{1}{2}[-3+26+12]$ $=\frac{1}{2} \times (35)$ $=\frac{35}{2}$ sq. units. Area of triangle $ADC=\frac{1}{2}[-3(3+8)+6(-8-5)+1(5-3)]$ $=\frac{1}{2}[-3(11)+6(-13)+1(2)]$ $=\frac{1}{2}[-33-78+2]$ $=\frac{1}{2} \times (-109)$ $=\frac{109}{2}$ sq. units. Therefore, The area of the quadrilateral $ABCD=\frac{35}{2}+\frac{109}{2}=\frac{35+109}{2}=72$ sq. units. The area of the quadrilateral $ABCD$ is $72$ sq. units. Tutorialspoint Simply Easy Learning Updated on: 10-Oct-2022 32 Views Get certified by completing the course
Graphing 3.1 Understanding Proportional Relationships (Extra Lesson)UNIT 3 • LESSON 1 UNDERSTANDING PROPORTIONAL RELATIONSHIPS adybug and ant move at constant speeds. The diagrams with tick marks show their positions at different times. Each tick mark represents 1 centimeter. 1. Lines uu and v also show the positions of the two bugs. Which line shows the ladybug’s movement? Which line shows the ant’s movement? Explain your reasoning. 2. How long does it take the ladybug to travel 12 cm? The ant? 3. Scale the vertical and horizontal axes by labeling each grid line with a number. You will need to use the time and distance information shown in the tick-mark diagrams. 4. Mark and label the point on line u and the point on line v that represent the time and position of each bug after travelling 1 cm. Refer to the tick-mark diagrams and graph in the earlier activity when needed. 1. Imagine a bug that is moving twice as fast as the ladybug. On each tick-mark diagram, mark the position of this bug. 2. Plot this bug’s positions on the coordinate axes with lines uu and vv, and connect them with a line. 3. Write an equation for each of the three lines. Here is a graph that could represent a variety of different situations. 1. Write an equation for the graph. 2. Sketch a new graph of this relationship. You teacher will give you 12 graphs of proportional relationships. 1. Sort the graphs into groups based on what proportional relationship they represent. 2. Write an equation for each different proportional relationship you find. Two large water tanks are filling with water. Tank A is not filled at a constant rate, and the relationship between its volume of water and time is graphed on each set of axes. Tank B is filled at a constant rate of 12 liters per minute. The relationship between its volume of water and time can be described by the equation v=12t, where t is the time in minutes and v is the total volume in liters of water in the tank. The graph of a very slight curve representing the volume of the tank in liters versus time in minutes The graph of a very slight curve representing the volume of the tank in liters versus time in minutes Sketch and label a graph of the relationship between the volume of water v and time t for Tank B on each of the axes. Answer the following questions and say which graph you used to find your answer. After 30 seconds, which tank has the most water? At approximately what times do both tanks have the same amount of water? At approximately what times do both tanks contain 1 liter of water? 20 liters?
# Statistics Outlier Function – lesscss By \| April 24, 2019 # Statistics – Outlier Function An outlier in a probability distribution function is a number that is more than 1.5 times the length of the data set away from either the lower or upper quartiles. Specifically, if a number is less than \${Q_1 – 1.5 times IQR}\$ or greater than \${Q_3 + 1.5 times IQR}\$, then it is an outlier. Outlier is defined and given by the following probability function: ## Formula \${Outlier datas are, lt Q_1 – 1.5 times IQR (or) gt Q_3 + 1.5 times IQR }\$ Where − • \${Q_1}\$ = First Quartile • \${Q_2}\$ = Third Quartile • \${IQR}\$ = Inter Quartile Range ### Example Problem Statement: Consider a data set that represents the 8 different students 15, 3, 16, 25, 12 and 14. Discover the outlier data from the Solution: Given data set is: 11 13 15 3 16 25 12 14 Arrange it in ascending order: 3 11 12 13 14 15 16 25 First Quartile Value() \${Q_1}\$ \${ Q_1 = frac{(11 + 12)}{2} \[7pt] = 11.5 }\$ Third Quartile Value() \${Q_3}\$ \${ Q_3 = frac{(15 + 16)}{2} \[7pt] = 15.5 }\$ Lower Outlier Range (L) \${ Q_1 – 1.5 times IQR \[7pt] = 11.5 – (1.5 times 4) \[7pt] = 11.5 – 6 \[7pt] = 5.5 }\$ Upper Outlier Range (L) \${ Q_3 + 1.5 times IQR \[7pt] = 15.5 + (1.5 times 4) \[7pt] = 15.5 + 6 \[7pt] = 21.5 }\$ In the given information, 5.5 and 21.5 is more greater than the other values in the given data set i.e. except from 3 and 25 since 3 is greater than 5.5 and 25 is lesser than 21.5. In this way, we utilize 3 and 25 as the outlier values. ‘; (vitag.displayInit = window.vitag.displayInit \|\| []).push(function () { viAPItag.display(ad_id); }); }())
Game Development Reference In-Depth Information equation using what are known as the inverse trig functions. The names of these functions match their counterparts, but prefixed with the word arc . In this case, we need to use arctangent to find the value of each of these angles. B A = arctan 15/20 ð Þ 90 ° A B = arctan 20/15 ð Þ Based on these equations, angle A would be ~37 ° and B would Figure 11.6 A triangle where we know just two of the sides, but no angles and no hypotenuse. be ~53 ° . If you add these together with the right angle of 90 ° ,you can see that we indeed have a proper triangle of 180 . For our final theoretical example, look back again to Fig. 11.6 . Suppose all you needed was the hypotenuse and you weren ° t inter- ested in the angles at all. You could do what we did previously, using arctangent to get the values of the angles and then use those angles with either sine or cosine to determine the hypotenuse. However, as this is a multiple-step process, it is inefficient when we have a much quicker way. In addition to the standard trig functions, there is another equation to determine the third side of a triangle when you know the other two, which is known as the Pythagorean theorem. The theorem states that the hypotenuse of a triangle, squared, is equal to the sum of the squares of the other two sides. Let ' s look at this as an equation, calling the two shorter sides a and b and the hypotenuse c . ' a 2 b 2 c 2 + = Finding any one side when you know the other two is just a simple permutation of this equation as follows: q ð q ð q ð c = a 2 + b 2 , b = c 2 a 2 , a = c 2 b 2 Þ Þ Þ For our purposes, we know sides a and b to be 15 and 20 (or 20 and 15; it doesn t really matter). From these values, the hypotenuse would therefore be equal to ' (15 2 +20 2 ), or 25. Now that we have defined these functions and have seen how to use them, let s look at a couple of practical examples in Flash and how to apply the functions there. A fairly common use of the trig functions is finding the angle of the mouse cursor relative to another point. This angle can then be applied to the rotation of a DisplayObject to make the object ' look at the mouse. If you open the MousePointer.fla file, you ll find just such an example setup. It consists of a triangle MovieClip called ' on the Stage. One of the corners of the triangle is colored differently to differentiate the direction it is pointing. For simplicity, the ActionScript to perform this math is on the timeline; if you were using this code as part of something larger, it would make sense to put it in a class. Let pointer ' s look at this code now.
# How do you solve for x: 6(x+1)=12(x-3)? Feb 4, 2017 See the entire solution process below: #### Explanation: First, expand all the terms in exponents: $\left(6 \times x\right) + \left(6 \times 1\right) = \left(12 \times x\right) - \left(12 \times 3\right)$ $6 x + 6 = 12 x - 36$ Next, subtract $\textcolor{red}{6 x}$ and add $\textcolor{b l u e}{36}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced: $6 x + 6 - \textcolor{red}{6 x} + \textcolor{b l u e}{36} = 12 x - 36 - \textcolor{red}{6 x} + \textcolor{b l u e}{36}$ $6 x - \textcolor{red}{6 x} + 6 + \textcolor{b l u e}{36} = 12 x - \textcolor{red}{6 x} - 36 + \textcolor{b l u e}{36}$ $0 + 42 = 6 x - 0$ $42 = 6 x$ Now, divide each side of the equation by $\textcolor{red}{6}$ to solve for $x$ while keeping the equation balanced: $\frac{42}{\textcolor{red}{6}} = \frac{6 x}{\textcolor{red}{6}}$ $7 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} x}{\cancel{\textcolor{red}{6}}}$ $7 = x$ $x = 7$ Feb 4, 2017 $x = 7$ #### Explanation: $6 \left(x + 1\right) = 12 \left(x - 3\right)$ $\therefore 6 x + 6 = 12 x - 36$ $\therefore 6 x - 12 x = - 36 - 6$ $\therefore - 6 x = - 42$ multiply both sides by$- 1$ $\therefore 6 x = 42$ $\therefore x = {\cancel{42}}^{7} / {\cancel{6}}^{1}$ $\therefore x = 7$ substitute $x = 7$ $\therefore 6 \left(7 + 1\right) = 12 \left(7 - 3\right)$ $\therefore 6 \times 8 = 12 \times 4$ $\therefore 48 = 48$
Courses Courses for Kids Free study material Offline Centres More Store # The vertex of a variable $\Delta ABC$ are $a(3,4)$, $b(5\cos t,5\sin t)$, and $c(5\sin t, - 5\cos t)$. Find the locus of orthocentres of $\Delta ABC$. Last updated date: 22nd Jun 2024 Total views: 404.7k Views today: 6.04k Verified 404.7k+ views Hint: The given problem has to be solved with distance formula and orthocentre formula which is also known as centroid. Centroid means the midpoint of a triangle. By also circumcenter and basic calculation solved by complete step-by-step explanation. Formulas used: Distance between two points ${\text{ = }}\sqrt {{{({x_2} - {x_1})}^2} - {{({y_2} - {y_1})}^2}}$ Centroid${\text{ = }}\left( {\dfrac{{{x_1} + {x_2} + x{}_3}}{3},\dfrac{{{y_1} + {y_2} + y{}_3}}{3}} \right)$ ${\sin ^2}\theta + {\cos ^2}\theta = 1$ ${(a + b)^2} = {a^2} + {b^2} + 2ab$ It is the given vertex of a variable $a(3,4)$, $b(5\cos t,5\sin t)$, and $c(5\sin t, - 5\cos t)$. We have to find the locus of the orthocentre of $\Delta ABC$. First, we need to find the distance of the vertex $ABC$, and also check the circumcenter of the circle. Let the origin be $\left( {0,0} \right)$ Then finding distance formula for $OA$, $OB$, $OC$ $\Rightarrow {\text{OA = }}\sqrt {{{(3 - 0)}^2} + {{(4 - 0)}^2}}$ Simplifying we get, $\Rightarrow {\text{OA = }}\sqrt {{{(3)}^2} + {{(4)}^2}}$ Now, squaring numbers and taking square root, we get $\Rightarrow {\text{OA = }}\sqrt {9 + 16}$ $\Rightarrow OA = \sqrt {25}$ Since, $25$ is a multiple of $5$, $\Rightarrow OA = \sqrt {{5^2}}$ Taking square root we get, $\Rightarrow OA = 5$ Likewise, we are going to find the other distance $OB\& OC$, we get, $\Rightarrow {\text{OB = }}\sqrt {{{(5\cos t - 0)}^2} + {{(5\sin t - 0)}^2}}$ Simplifying we get, $\Rightarrow {\text{OB = }}\sqrt {{{(5\cos t)}^2} + {{(5\sin t)}^2}}$ Squaring the terms, $\Rightarrow {\text{OB = }}\sqrt {25{{\cos }^2}t + 25{{\sin }^2}t}$ Taking Commonly $25$, we get, $\Rightarrow {\text{OB = }}\sqrt {25({{\cos }^2}t + {{\sin }^2}t)}$ By applying formula mentioned in formula used that is ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we get $\Rightarrow {\text{OB = }}\sqrt {25}$ Since, $25$ is a multiple of $5$, $\Rightarrow OB = \sqrt {{5^2}}$ Taking square root we get, $\Rightarrow OB = 5$ Likewise we find ${\text{OC}}$, applying distance formula $\Rightarrow {\text{OC = }}\sqrt {{{(5\sin t - 0)}^2} + {{( - 5\cos t - 0)}^2}}$ Simplifying we get, $\Rightarrow {\text{OC = }}\sqrt {{{(5\sin t)}^2} + {{( - 5\cos t)}^2}}$ Squaring the terms, $\Rightarrow {\text{OC = }}\sqrt {25{{\cos }^2}t + 25{{\sin }^2}\theta }$ Taking Commonly $25$, we get $\Rightarrow {\text{OC = }}\sqrt {25({{\cos }^2}t + {{\sin }^2}t)}$ By applying formula mentioned in formula used that is ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we get $\Rightarrow {\text{OC = }}\sqrt {25}$ Since, $25$ is a multiple of $5$, $\Rightarrow OC = \sqrt {{5^2}}$ Taking square root we get, $\Rightarrow OC = 5$ Thus the distance of $OA$, $OB$, $OC$ is the same so $\left( {0,0} \right)$ is the circumcenter of $\Delta ABC$. Let $H$ be the orthocentre of $\Delta ABC$ Now, Solve using centroid $G$ divides using section formula in the ratio $2:1$ Then, Centroid $G$ is using vertex, $\Rightarrow G = \left( {\dfrac{{3 + 5\cos t + 5\sin t}}{3},\dfrac{{4 + 5\sin t - 5\cos t}}{3}} \right) - - - - - (1)$ Now, Centroid $G$ using Section formula, We know that the centroid, $H\left( {h,k} \right)$ and circumcenter $\left( {0,0} \right)$ in the ratio $2:3$ $\Rightarrow G = \left( {\dfrac{h}{3},\dfrac{k}{3}} \right) - - - - (2)$ Now, equating equation (1) & (2), we get $\Rightarrow \dfrac{h}{3} = \dfrac{{3 + 5\cos t + 5\sin t}}{3}$ $\Rightarrow \dfrac{k}{3} = \dfrac{{4 + 5\sin t - 5\cos t}}{3}$ Now, canceling denominator value because same on both sides, $\Rightarrow h = 3 + 5\cos t + 5\sin t$ $\Rightarrow k = 4 + 5\sin t - 5\cos t$ Now, changing the constant term towards the left-hand side, $\Rightarrow k - 4 = 5\sin t - 5\cos t - - - - (3)$ $\Rightarrow h - 3 = 5\cos t + 5\sin t - - - - - - (4)$ Now, squaring on both sides and adding equation (3) & (4), we get $\Rightarrow {(h - 3)^2} + {(k - 4)^2} = {(5\cos t + 5\sin t)^2} + {(5\sin t - 5\cos t)^2}$ Now, after squaring the RHS side only, and applying ${(a+b)}^2$ formula mention in the formula used, we get $\Rightarrow {(h - 3)^2} + {(k - 4)^2} = 25{\cos ^2}t + 25{\sin ^2}t + 50\cos t\sin t + 25{\cos ^2}t + 25{\sin ^2}t - 50\cos t\sin t$ Canceling the same positive and negative values, we get $\Rightarrow {(h - 3)^2} + {(k - 4)^2} = 25{\cos ^2}t + 25{\sin ^2}t + 25{\cos ^2}t + 25{\sin ^2}t$ By adding the right-hand side of the equation, we get $\Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50{\cos ^2}t + 50{\sin ^2}t$ By taking 50 commonly, we get $\Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50({\cos ^2}t + {\sin ^2}t)$ Once again we applying the formula mentioned in the formula used, we get $\Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50(1)$ $\Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50$ $\therefore$ The locus of orthocentre is ${(h - 3)^2} + {(k - 4)^2} = 50$ Note: This kind of problem has special attention to points for the values we substitute to that we get things right in the manner. By knowing formula and calculation more than we want to know about the content of the question. This may be given in the form of orthocentre but no clear explanation regarding centroid. So, we want to create an idea to make a centroid value and section form and compare those we solved. Like this, we must try to find out in many ways.

Subsets and Splits

No community queries yet

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