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# How do you solve using the completing the square method x^2 + 10x + 16 = 0? Feb 26, 2016 $x = - 2$ and $x = - 8$ #### Explanation: To solve the equation ${x}^{2} + 10 x + 16 = 0$ using the completing square method, as coefficient of ${x}^{2}$ is $1$ and independent term $16$ positive, we have to identify factors of $16$ whose sum is $10$, coefficient of $x$ term. These are $2$ and $8$ and hence we should split the equation as follows: ${x}^{2} + 2 x + 8 x + 16 = 0$ or $x \left(x + 2\right) + 8 \left(x + 2\right) = 0$ i.e. $\left(x + 2\right) \left(x + 8\right) = 0$. Hence either $x + 2 = 0$ i.e. $x = - 2$ or $x + 8 = 0$ i.e. $x = - 8$ [In general if equation is in form $a {x}^{2} + b x + c = 0$, one should identify two factors whose product is $a \cdot c$ and sum is $b$.]
SITEMAP School-college Physics Notes: Electricity 3.7 Division of p.d. in a circuit UK GCSE level age ~14-16 ~US grades 9-10 Scroll down, take time to study content or follow links Electricity Section 3: 3.7 How the total potential difference is split between resistances in series Doc Brown's Physics exam study revision notes: Explaining and verifying by calculation, how the potential difference (p.d.) is split (divided, shared) between resistances in series i.e. the effect of two (or more) resistors in series 3.7 A little more on potential difference - effect of two resistors in series - explaining and verifying by calculation, how the potential difference is split between resistances in series The circuit 41 shows two resistors wired in series, so how is the total p.d. divided or shared between the two resistance wired in series. This circuit can be used to investigate and confirm the correctness of the following theoretical calculations. On the right is shown what happens to the p.d. going clockwise around the circuit (direction of convention current). The potential store of the battery raises the potential difference of the charge to 12.0 V. The total resistance = 10.0 + 5.0 = 15.0 Ω (you can add resistors up if wired in series) Therefore the current flowing round all of this series circuit = I = V / Rtotal = 12.0 / 15.0 = 0.80 A As the charge passes through the 1st resistor R1, it loses energy and the p.d. falls by 8 V to a p.d. of 4 V. Calculation to confirm this: V1 = I x R1 = 0.80 x 10.0 = 8.0 V As the charge passes through the 2nd resistor R1, it loses energy again and the p.d. falls by 4 V to a p.d. of 0 V. Calculation to confirm this: V2 = I x R2 = 0.80 x 5.0 = 4.0 V As long as there is a complete circuit, the process repeats itself through any number of resistors. What is clear is the total p.d. for the circuit is split into a p.d. ratio identical to the resistor ratio. V1 : V2 is the same ratio as R1 : R2 You can also say that since E = QV, twice as much energy is released by resistor R1 (p.d. 8 V) than R2 (p.d. 4 V) for the same total current, the same total charge transferred, but the lower the p.d. the less potential energy carried by the charge. Keywords, phrases and learning objectives for ? electricity Be able to explain and verifying by calculation (and experiment), how potential difference is split shared divided between resistances in series. Be able to describe the effect of two resistors in series on the p.d. across these resistors wired in series. WHAT NEXT? BIG website and using the [SEARCH BOX] below, maybe quicker than navigating the many sub-indexes for UK KS3 science students aged ~12-14, ~US grades 6-8 ChemistryPhysics for UK GCSE level students aged ~14-16, ~US grades 9-10 for pre-university age ~16-18 ~US grades 11-12, K12 Honors Use your mobile phone in 'landscape' mode? SITEMAP Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries and references to GCSE science course specifications are unofficial. Using SEARCH some initial results may be ad links you can ignore - look for docbrown @import url(https://www.google.co.uk/cse/api/branding.css); ENTER specific physics words or courses e.g. topic, module, exam board, formula, concept, equation, 'phrase', homework question! anything of physics interest! This is a very comprehensive Google generated search of my website TOP OF PAGE
Determining the Shear Force and Bending Moment Equations of Simply Supported Beam ## Example 2. A simply supported beam is loaded as shown in the diagram. Calculate the support reactions and draw the Bending Moment diagram, Shear Force Diagram, Axial Force Diagram. Determine the maximum bending moment. #### Calculate the reactions at the supports of a beam `1. A beam is in equilibrium when it is stationary relative to an inertial reference frame. The following conditions are satisfied when a beam, acted upon by a system of forces and moments, is in equilibrium:ΣFx = 0: HA + P1cos(30) = 0ΣMA = 0: The sum of the moments about a point A is zero: - q14(4/2) - P1sin(30)8 + RB12 + M1 = 0ΣMB = 0: The sum of the moments about a point B is zero: - RA12 + q14(12 - 4/2) + P1sin(30)4 + M1 = 02. Solve this system of equations: HA = - P1cos(30) = - 200.8660 = -17.32 (kN)Calculate reaction of roller support about point B:RB = ( q14(4/2) + P1sin(30)8 - M1) / 12 = ( 64(4/2) + 20sin(30)8 - 42) / 12 = 7.17 (kN)Calculate reaction of pin support about point A:RA = ( q14(12 - 4/2) + P1sin(30)4 + M1) / 12 = ( 64(12 - 4/2) + 20sin(30)4 + 42) / 12 = 26.83 (kN)3. The sum of the forces is zero: ΣFy = 0: RA - q14 - P1sin(30) + RB = 26.83 - 64 - 20sin(30) + 7.17 = 0` #### Draw diagrams for the beam First span of the beam: 0 ≤ x1 < 4 `Determine the equations for the axial force (N):N(x1) = HAN1(0) = 17.32 = 17.32 (kN)N1(4) = 17.32 = 17.32 (kN)Determine the equations for the shear force (Q):Q(x1) = + RA - q1(x1 - 0)Q1(0) = + 26.83 - 6(0 - 0) = 26.83 (kN)Q1(4) = + 26.83 - 6(4 - 0) = 2.83 (kN)Determine the equations for the bending moment (M):M(x1) = + RA(x1) - q1(x1)2/2M1(0) = + 26.83(0) - 6(0 - 0)2/2 = 0 (kNm)M1(4) = + 26.83(4) - 6(4 - 0)2/2 = 59.33 (kNm)` Second span of the beam: 4 ≤ x2 < 8 `Determine the equations for the axial force (N):N(x2) = HAN2(4) = 17.32 = 17.32 (kN)N2(8) = 17.32 = 17.32 (kN)Determine the equations for the shear force (Q):Q(x2) = + RA - q1(4 - 0)Q2(4) = + 26.83 - 6(4 - 0) = 2.83 (kN)Q2(8) = + 26.83 - 6(4 - 0) = 2.83 (kN)Determine the equations for the bending moment (M):M(x2) = + RA(x2) - q1(4 - 0)[(x2 - 4) + (4 - 0)/2]M2(4) = + 26.83(4) - 64(0 + 2) = 59.33 (kNm)M2(8) = + 26.83(8) - 64(4 + 2) = 70.67 (kNm)` Third span of the beam: 8 ≤ x3 < 12 `Determine the equations for the axial force (N):N(x3) = HA - P1cos(30)N3(8) = 17.32 - 200.8660 = 0 (kN)N3(12) = 17.32 - 200.8660 = 0 (kN)Determine the equations for the shear force (Q):Q(x3) = + RA - q1(4 - 0) - P1sin(30)Q3(8) = + 26.83 - 6(4 - 0) - 10.00sin(30) = -7.17 (kN)Q3(12) = + 26.83 - 6(4 - 0) - 10.00sin(30) = -7.17 (kN)Determine the equations for the bending moment (M):M(x3) = + RA(x3) - q1(4 - 0)[(x3 - 4) + (4 - 0)/2] - P1(x3 - 8)sin(30)M3(8) = + 26.83(8) - 64(4 + 2) - 20(8 - 8)sin(30) = 70.67 (kNm)M3(12) = + 26.83(12) - 64(8 + 2) - 20(12 - 8)sin(30) = 42 (kNm)` Solved by BEAMGURU.COM
# Two positive numbers have a sum of 3636. What is the maximum product of one number times the... ## Question: Two positive numbers have a sum of 3636. What is the maximum product of one number times the square of the second number? ## Maximum Value: The maximum and minimum value of a function can be calculated using the first and second derivative of the function. For the critical points, we use the first derivative, and for the maximum and minimum value, we use the second derivative. Let us consider p and q are the two positive numbers. According to the statement in the problem the sum of two positive numbers is equal to: {eq}\begin{align} p + q &= 3636\\ p &= 3636 - q \end{align} {/eq} The maximum product of one number time the square of the second number is: {eq}x = p{q^2}\cdots\cdots\rm{(1)} {/eq} Substitute p into the above equation. {eq}\begin{align} x &= \left( {3636 - q} \right){q^2}\\ x &= 3636{q^2} - {q^3} \end{align} {/eq} We will differentiate the above equation to obtain the critical points. {eq}\begin{align} x' &= 7272q - 3{q^2}\\ &= q\left( {7272 - 3q} \right) \end{align} {/eq} For critical points the first derivative must be equal to zero as: {eq}\begin{align} x' = 0\\ q\left( {7272 - 3q} \right) &= 0\\ q &= 0,\;2424 \end{align} {/eq} Now, we will find the second derivative to obtain the maximum and minimum value. {eq}x'' = 7272 - 6q {/eq} By using critical points in the above equation, we will get maximum and minimum value: {eq}\begin{align} x'' &= 7272 - 6\left( 0 \right)\\ &= 7272 \end{align} {/eq} Also, {eq}\begin{align} x'' &= 7272 - 6\left( {2424} \right)\\ &= - 7272 \end{align} {/eq} According to second derivative, at q=2424 the value of second derivative is negative hence it will be maximum. The value of the number p is calculated as, {eq}\begin{align} p &= 3636 - 2424\\ &= 1212 \end{align} {/eq} Finally, we will substitute the value of the numbers in equation 1. {eq}\begin{align} x &= \left( {1212} \right){\left( {2424} \right)^2}\\ &= 7121440512 \end{align} {/eq} Thus, the maximum product of one number time the square of the second number is 7121440512
## Algebra Showing posts with label two variables. Show all posts Showing posts with label two variables. Show all posts ### Systems of Linear Inequalities (Two Variables) The systems of linear inequalities that we will be solving consist of two linear inequalities and two variables. To solve these we will graph the solution sets of both linear inequalities and then determine where the sets intersect. Any point in the overlap of the graphs will be a solution to the system. Instructional Video: Systems of Linear Inequalities Graph the solution set: The above solution suggests that (−5, 3) is a solution because it is shaded. Check it and others to see if it solves both inequalities. These graphs can sometimes get messy so do your best to think about the solution before actually shading. Use pencil and a good eraser when working on these problems. Given the graph, determine the system. Graph the systems of inequalities. Graph the system: ### Linear Inequalities (Two Variables) When graphing an equation like y = 3x − 6 we know that it will be a line. The graph of a linear inequality such as y >= 3x − 6, on the other hand, gives us a region of ordered pair solutions. Not only do the points on the line satisfy this linear inequality - so does any point in the region that we have shaded. This line is the boundary that separates the plane into two halves - one containing all the solutions and one that does not. Therefore, from the above graph, both (0, 0) and (−2, 4) should solve the inequality. Use a test point not on the boundary to determine which side of the line to shade when graphing solutions to a linear inequality. Usually the origin is the easiest point to test as long as it is not a point on the boundary. Graph the solution set. If the test point yields a true statement shade the region that contains it. If the test point yields a false statement shade the opposite side. When graphing strict inequalities, inequalities without the equal, the points on the line will not satisfy the inequality; hence, we will use a dotted line to indicate this. Otherwise, the steps are the same. Graph the solution set. Given the graph determine the missing inequality.
NSW Syllabuses # Mathematics K–10 - Stage 4 - Number and Algebra Ratios and Rates ## Outcomes #### A student: • MA4-1WM communicates and connects mathematical ideas using appropriate terminology, diagrams and symbols • MA4-2WM applies appropriate mathematical techniques to solve problems • MA4-3WM recognises and explains mathematical relationships using reasoning • MA4-7NA operates with ratios and rates, and explores their graphical representation Related Life Skills outcome: MALS-19NA ## Content • Students: • Recognise and solve problems involving simple ratios (ACMNA173) • use ratios to compare quantities measured in the same units • write ratios using the $$:$$ symbol, eg $$4\!:\!7$$ • express one part of a ratio as a fraction of the whole, eg in the ratio $$4\!:\!7$$, the first part is $$\frac{4}{11}$$ of the whole (Communicating) • simplify ratios, eg $$4\!:\!6 = 2\!:\!3$$, $$\quad \frac{1}{2}\! :\! 2 = 1\!:\!4$$, $$\quad 0.3\!:\!1 = 3\!:\!10$$ • apply the unitary method to ratio problems • divide a quantity in a given ratio • Solve a range of problems involving ratios and rates, with and without the use of digital technologies (ACMNA188) • interpret and calculate ratios that involve more than two numbers • solve a variety of real-life problems involving ratios, eg scales on maps, mixes for fuels or concrete • use rates to compare quantities measured in different units • distinguish between ratios, where the comparison is of quantities measured in the same units, and rates, where the comparison is of quantities measured in different units • convert given information into a simplified rate, eg 150 kilometres travelled in 2 hours = 75 km/h • solve a variety of real-life problems involving rates, including problems involving rate of travel (speed) • Investigate, interpret and analyse graphs from authentic data (ACMNA180) • interpret distance/time graphs (travel graphs) made up of straight-line segments • write or tell a story that matches a given distance/time graph (Communicating) • match a distance/time graph to a description of a particular journey and explain the reasons for the choice (Communicating, Reasoning) • compare distance/time graphs of the same situation, decide which one is the most appropriate, and explain why (Communicating, Reasoning) • recognise concepts such as change of speed and direction in distance/time graphs • describe the meaning of straight-line segments with different gradients in the graph of a particular journey (Communicating) • calculate speeds for straight-line segments of given distance/time graphs (Problem Solving) • recognise the significance of horizontal line segments in distance/time graphs • determine which variable should be placed on the horizontal axis in distance/time graphs • draw distance/time graphs made up of straight-line segments • sketch informal graphs to model familiar events, eg noise level during a lesson • record the distance of a moving object from a fixed point at equal time intervals and draw a graph to represent the situation, eg move along a measuring tape for 30 seconds using a variety of activities that involve a constant rate, such as walking forwards or backwards slowly, and walking or stopping for 10-second increments (Problem Solving) • use the relative positions of two points on a line graph, rather than a detailed scale, to interpret information ### Background Information Work with ratios may be linked to the 'golden rectangle'. Many windows are golden rectangles, as are rectangles used in some of the ancient buildings in Athens, such as the Parthenon. The relationship between the ratios involving the dimensions of the golden rectangle was known to the Greeks in the sixth century BC: $$\frac{\mbox{length}}{\mbox{width}} = \frac{\mbox{length + width}}{\mbox{length}}$$. In Stage 4, the focus is on examining situations where the data yields a constant rate of change. It is possible that some practical situations may yield a variable rate of change. This is the focus in Ratios and Rates in Stage 5.3. It is the usual practice in mathematics to place the independent variable on the horizontal axis and the dependent variable on the vertical axis. This is not always the case in other subjects, such as economics. #### Purpose/Relevance of Substrand As we often need to compare two numbers, amounts or quantities in our daily lives, ratios and rates are important aspects of our study of mathematics. Ratios are used to compare two (or more) numbers, amounts or quantities of the same kind (eg objects, people, weights, heights) and can be expressed as '$$a$$ to $$b$$' or $$a\!:\!b$$. In simple terms, a ratio represents that for a given number or amount of one thing, there is a certain number or amount of another thing (eg 'I have 4 ties for every shirt, so the ratio of ties:shirts is 4:1 and the ratio of shirts:ties is 1:4'). A rate is a particular type of ratio that is used to compare two measurements of different kinds. Speed is a rate in which the distance travelled (by a person, car, etc) is compared to the time taken (to cover the distance travelled). It follows that speed is measured (or expressed) in units such as metres per second and kilometres per hour. Other examples of rates that are important in our everyday lives include interest rates, exchange rates, heart rates, birth rates and death rates. ### Language The use of the word 'per', meaning 'for every', in rates should be made explicit to students. When solving ratio and rate problems, students should be encouraged to write a few key words on the left-hand side of the equals sign to identify what is being found in each step of their working, and to conclude with a statement in words. When describing distance/time graphs (travel graphs), supply a modelled story and graph first, or jointly construct a story with students before independent work is required. When constructing stories and interpreting distance/time graphs, students can use present tense, 'The man travels …', or past tense, 'The man travelled …'. Students should be aware that 'gradient' may be referred to as 'slope' in some contexts. ### National Numeracy Learning Progression links to this Mathematics outcome When working towards the outcome MA4‑7NA the sub-elements (and levels) of Comparing units (CoU2-CoU3) and Positioning and locating (PoL5) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning. The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression.
• Call Now 1800-102-2727 • # Perpendicular axis theorem, definition, mathematical expression, necessary conditions, practice questions, FAQs Sumit is given a planar body (say, a plate). He is told that the moment of inertia of the plate about the axis perpendicular to its plane is 200 kg cm2. He is asked if he can calculate the moment of inertia of the plate about its diametrical axis. Will Sumit be able to calculate this without touching pen-paper-calculator? The perpendicular axis theorem is one of the most important theorems that is used for the calculation of moment of inertias of planar bodies. Since, using the integral method is time consuming and cumbersome, the theorem is developed to reduce the effort in calculation of moment of inertia provided the necessary conditions are met. • Perpendicular axis theorem statement • Mathematical expression • Necessary condition to apply perpendicular axis theorem • Practice questions • FAQs ## Perpendicular axis theorem Statement: For any planar body, the moment of inertia about any of its axes that are perpendicular to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying on the plane of the body, they intersect the first axis in the plane and are perpendicular to the axis as well. ## Mathematical expression Let the x and y axes be chosen in the plane of the body and the z-axis be perpendicular to this plane. The three axes are mutually perpendicular. According to the perpendicular axis theorem : I= I+ Iy This theorem is applicable only for planar objects like circular disc, ring, plane lamina, etc. ${I}_{x}=\sum {x}_{i}^{2}{m}_{i}$ ${I}_{y}=\sum {y}_{i}^{2}{m}_{i}$ ${I}_{z}=\sum {z}_{i}^{2}{m}_{i}$ However ${r}_{i}^{2}={x}_{i}^{2}+{y}_{i}^{2}$ So, ${I}_{z}=\sum \left({x}_{i}^{2}+{y}_{i}^{2}\right){m}_{i}$ $⇒{I}_{z}=\sum {x}_{i}^{2}{m}_{i}+\sum {y}_{i}^{2}{m}_{i}$ $⇒{I}_{z}={I}_{x}+{I}_{y}$ Note: The point of intersection can be any point in the plane of the body, which lies on the body or even outside it. The three axes need not be at the centre of the body. ## Necessary condition to apply perpendicular axis theorem • The body must be planar in nature. This is not applicable for the objects having significant dimensions in all the three dimensions. • The two axes must lie in the major plane of the body and the third axis must be perpendicular to the plane of the body (i.e. perpendicular to those axes as well). • All the axes must be mutually perpendicular. ## Practice problems Q1. For the given uniform square lamina ABCD of side length l, whose centre is O, identify the correct option. A) B) C) D) A. (c) The moment of inertia about an axis perpendicular to the plane of a square lamina of side l and passing through its centre O, ${I}_{O}=\frac{m{l}^{2}}{6}$ The axis GH and EF are perpendicular, hence by perpendicular axis theorem, ${I}_{o}={I}_{EF}+{I}_{GH}$ IEF = IGH due to symmetry of square plate As it is a square shaped planar lamina, the diagonal axes BD and AC are also mutually perpendicular to each other and the axis through O is perpendicular to the plane of the plate. Hence by perpendicular axis theorem, I= IAC + IBD We can say, due to the symmetry of the square plate. From Eq. (i) and (ii), we get Q2. A uniform rectangular lamina is shown in figure. Find the moment of inertia Icom about the z axis which passes through its centre and perpendicular to its plane. Here = 10 m, = 20 m and = 12 kg. A. For a rectangular lamina, the moment of inertia about an axis passing through its COM and perpendicular to the plane of lamina is given by: So, Putting the given values, Q3. Two uniform identical rods each of mass M and length L are arranged as shown in the figure. Find the moment of inertia of the cross about a bisector in the plane of rods as shown by the dotted line in the figure. A. Length of the rod =L Mass of the rod =M As we know that, IZ of a rod is given as $\frac{M{L}^{2}}{12}$ So, moment of inertia of the two rods shown will be 2IZ, which is $\frac{M{L}^{2}}{6}$ Now, using the perpendicular axis theorem, I= I+ Iy and due to symmetry, I= Iy Or Hence, moment of inertia of the cross about the bisector will be Q4. A ring of mass M and radius R has an axis passing through the ring diametrically. If the moment of inertia about the axis passing through its centre of mass and perpendicular to the plane is MR2, then find the moment of inertia about the axis passing diametrically. A. Given, moment of inertia of the ring about the axis passing through the centre of mass and perpendicular to its plane is I= MR2. Now, moment of inertia of ring about the axis passing along its diameter is given by IX or IY Using perpendicular axis theorem we have and due to symmetry, ## FAQs Q1. ABCD is a square plate with centre O. The moment of inertia of the plate about the axes 1,2,3 and 4 are I1, I2, I3, I4 respectively. What will be the relation between these four moments of inertia. A. According to theorem of perpendicular axes, moment of inertia of the plate about an axis passing through the centre O and perpendicular to the plane is, I= I+ I= I+ I4 (Since, the diagonals in a square are mutually perpendicular to each other) I= I= I= I4 This is because the moment of inertia about these axes are the same because of symmetry. Q2. What are the applications of the perpendicular axis theorem? A. The perpendicular axis theorem is very much useful in calculating the moment of inertia for a planar body of complex shapes whose moment of inertia about two mutually perpendicular axes lying on the plane is known. In that case, we don’t have to calculate the moment of inertia by , we can just add the moment of inertia about the axes lying in the plane. Q3. Will the radius of gyration for a planar body be universally fixed? A. The radius of gyration is all about the rotational dynamics of a body. The radius of gyration about the axis of rotation can be defined as the radial distance of a point which would have the same moment of inertia as the body's actual distribution of mass, if the total mass of the body is assumed to be concentrated there. In the case of a body the inertia is dependent on the axis of rotation we chose. Further is the distribution of mass from the axis of rotation, more is the inertia. So a 3D body which is not the same in terms of the radial distribution of mass, the moment of inertia will change according to the axis we choose. A body having the moment of inertia, I , radius of gyration k and mass, m we can say, I = mk2 As I does not remain the same, the radius of gyration, k also changes with the change in axis we chose. Q4. Can we apply the perpendicular axis theorem to a 3D body? A. For a 3D object, we can consider it is made of slices that are all laminar in nature. We can apply the parallel axis theorem independently for each lamina. It does not apply for all of them together i.e., for the whole 3D object. It is because although the laminas have the same z axis, they do not necessarily have the same x and y axes. So we cannot simply add the moments of inertia (MOI) for the x and y axes for every lamina. The x and y axes for each lamina are parallel to each other but they are off-set from each other and from the x, y axes which we are using for the 3D object. Talk to Our Expert Request Call Back Resend OTP Timer = By submitting up, I agree to receive all the Whatsapp communication on my registered number and Aakash terms and conditions and privacy policy
Paul's Online Notes Home / Calculus III / Surface Integrals / Divergence Theorem 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 17.6 : Divergence Theorem 3. Use the Divergence Theorem to evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}$$ where $$\vec F = 2xz\vec i + \left( {1 - 4x{y^2}} \right)\,\vec j + \left( {2z - {z^2}} \right)\,\vec k$$ and $$S$$ is the surface of the solid bounded by $$z = 6 - 2{x^2} - 2{y^2}$$ and the plane $$z = 0$$ . Note that both of the surfaces of this solid included in $$S$$. Show All Steps Hide All Steps Start Solution Let’s start off with a quick sketch of the surface we are working with in this problem. We included a sketch with traditional axes and a sketch with a set of “box” axes to help visualize the surface. The bottom “cap” of the elliptic paraboloid is also included in the surface but isn’t shown. Note as well here that because we are including both of the surfaces shown above that the surface does enclose (or is the boundary curve if you want to use that terminology) the region. Show Step 2 We are going to use Stokes’ Theorem in the following direction. $\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}$ where $$E$$ is just the solid shown in the sketches from Step 1. The region $$D$$ for that we’ll need in converting the triple integral into iterated integrals is just the intersection of the two surfaces from the problem statement. This is, $0 = 6 - 2{x^2} - 2{y^2}\hspace{0.25in} \to \hspace{0.25in} {x^2} + {y^2} = 3$ So, $$D$$ is a disk and so we’ll eventually be doing cylindrical coordinates for this integral. Here are the cylindrical limits for the region $$E$$. $\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le \sqrt 3 \\ 0 \le z \le 6 - 2{x^2} - 2{y^2} = 6 - 2{r^2}\end{array}$ Don’t forget to convert the $$z$$ limits into cylindrical coordinates as well! We’ll also need the divergence of the vector field so here is that. ${\mathop{\rm div}\nolimits} \vec F = \frac{\partial }{{\partial x}}\left( {2xz} \right) + \frac{\partial }{{\partial y}}\left( {1 - 4x{y^2}} \right) + \frac{\partial }{{\partial z}}\left( {2z - {z^2}} \right) = 2 - 8xy$ Show Step 3 Now let’s apply the Divergence Theorem to the integral and get it converted to cylindrical coordinates while we’re at it. \begin{align}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \iiint\limits_{E}{{2 - 8xy\,dV}}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{\left( {2 - 8{r^2}\cos \theta \sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{2r - 8{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align} Don’t forget to pick up the $$r$$ when converting the $$dV$$ to cylindrical coordinates. Show Step 4 All we need to do then in evaluate the integral. \begin{align}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{2r - 8{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\left. {\left( {2r - 8{r^3}\cos \theta \sin \theta } \right)z} \right\|_0^{6 - 2{r^2}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\left( {2r - 8{r^3}\cos \theta \sin \theta } \right)\left( {6 - 2{r^2}} \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{12r - 4{r^3} - \left( {48{r^3} - 16{r^5}} \right)\cos \theta \sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {6{r^2} - {r^4} - \left( {12{r^4} - \frac{8}{3}{r^6}} \right)\cos \theta \sin \theta } \right)} \right\|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{9 - 36\cos \theta \sin \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{9 - 18\sin \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\left( {9\theta - 9\cos \left( {2\theta } \right)} \right)} \right\|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{18\pi }}\end{align}
# How to Determine the Slope of a Line By Maria O'Brien stock.xchng The slope--or slant--of a line can be quantified. You are probably used to referring to inclines as “steep” or “gradual,” but it is possible to give a numerical value to the degree of “steepness” or “gradualness” of a line. Only the basic mathematical operations of subtraction and division are needed to calculate the slope of any line if you have two points on the line. Determining the slope when you are given the formula of a line is simplicity itself, because the slope is part of the formula--or easily derived from the formula. ## Determine the Slope from a Graph or Points on a Line Choose any two points on the line. Designate these points as point 1 and point 2. Name the coordinates of the first point (x1, y1) and designate the ordered pair of the second point (x2, y2). It doesn’t matter which point you call point 1 or point 2. Write the equation for the difference of the y coordinates: y2 - y1. This is the vertical distance between the two points; it is how far the line rises in going from point 1 to point 2. Put down the equation for the difference in the x coordinates: x2 - x1. This represents the horizontal distance the line runs in going from point 1 to point 2. Calculate the slope, the rise per run: y2 - y1 / x2 - x1. This is how far up or down (vertical distance) the line goes per unit of horizontal distance. Label your answer "m." The formula for the slope is thus: m = y2 - y1 / x2 - x1. ## Determine the Slope from the Formula of a Line Pick out the slope immediately if given the equation in the form y = mx + b. Remember, m is the slope. So, if you are asked to find the slope of a line whose equation is y = 3x + 10 , the slope is simply 3, the coefficient of x. Similarly, the slope of y = -½ x -5, is -½. Use ordinary algebraic manipulation to find the slope if given the formula for a line in the form of Ax + By = C. Remember, that in simplifying an equation, you need to respect the equal sign, and whatever you do to one side of the equation, you must also do to the other. Convert the equation above into one in the form of y = mx + b. Here’s how:Ax + By = C[-Ax] + Ax +By = C + [-Ax]By = C -AxBy = -Ax + Cy = -A/B x + C/BThe slope, then is -A/B.For the line, 2x + 3y = 12, solving for y yields y = -2/3 x + 4. The slope, therefore, is -2/3.
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # What Is The Cube Root Of 54? • Simplified Cube Root for ∛54 is 3∛2 • Step by step simplification process to get cube roots radical form and derivative: • First we will find all factors under the cube root: 54 has the cube factor of 9. • Let's check this width ∛92=∛54. As you can see the radicals are not in their simplest form. • Now extract and take out the cube root ∛9 ∛2. Cube of ∛9=3 which results into 3∛2 • All radicals are now simplified. The radicand no longer has any cube factors. ## Determine The Cubed Root Of 54? • The cubed root of fifty-four ∛54 = 3.7797631496846 ## How To Calculate Cube Roots • The process of cubing is similar to squaring, only that the number is multiplied three times instead of two. The exponent used for cubes is 3, which is also denoted by the superscript³. Examples are 4³ = 444 = 64 or 8³ = 888 = 512. • The cubic function is a one-to-one function. Why is this so? This is because cubing a negative number results in an answer different to that of cubing it's positive counterpart. This is because when three negative numbers are multiplied together, two of the negatives are cancelled but one remains, so the result is also negative. 7³ = 777 = 343 and (-7)³ = (-7)(-7)(-7) = -343. In the same way as a perfect square, a perfect cube or cube number is an integer that results from cubing another integer. 343 and -343 are examples of perfect cubes. ## Mathematical Information About Numbers 5 4 • About Number 5. Integers with a last digit as a zero or a five in the decimal system are divisible by five. Five is a prime number. All odd multiples of five border again with the five (all even with zero). The fifth number of the Fibonacci sequence is a five. Five is also the smallest prime number that is the sum of all other primes which are smaller than themselves. The Five is a Fermat prime: 5 = 2 ^ {2 ^ 1} +1 and the smallest Wilson prime. Number five is a bell number (sequence A000110 in OEIS). There are exactly five platonic bodies. There are exactly five tetrominoes. • About Number 4. Four is linear. It is the first composite number and thus the first non-prime number after one. The peculiarity of the four is that both 2 + 2 = 4 and 2 * 2 = 4 and thus 2^2 = 4. Four points make the plane of a square, an area with four sides. It is the simplest figure that can be deformed while keeping it's side lengths, such as the rectangle to parallelogram. Space let's us arrange equidistantly a maximum of four points. These then form a tetrahedron (tetrahedron), a body with four identical triangular faces. Another feature of the four is the impossibility of an algebraic equation of higher degree than four square roots using simple arithmetic and basic operations dissolve. ## What is a cube root? In arithmetic and algebra, the cube of a number n is its third power: the result of the number multiplied by itself twice: n³ = n * n * n. It is also the number multiplied by its square: n³ = n * n². This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also n raised to the one-third power. Both cube and cube root are odd functions: (-n)³ = -(n³). The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 2³ = 8 or (x + 1)³. © Mathspage.com \| Privacy \| Contact \| info [at] Mathspage [dot] com
# Engineering Mathematics: Differential Calculus. Contents Concepts of Limits and Continuity Derivatives of functions Differentiation rules and Higher Derivatives. ## Presentation on theme: "Engineering Mathematics: Differential Calculus. Contents Concepts of Limits and Continuity Derivatives of functions Differentiation rules and Higher Derivatives."— Presentation transcript: Engineering Mathematics: Differential Calculus Contents Concepts of Limits and Continuity Derivatives of functions Differentiation rules and Higher Derivatives Applications Differential Calculus Concepts of Limits and Continuity The idea of limits Consider a function The function is well-defined for all real values of x The following table shows some of the values: x2.92.99 33.0013.013.1 F(x)8.418.948.99499.0069.069.61 The idea of limits Concept of Continuity E.g. is continuous at x=3? The following table shows some of the values: exists as and => f(x) is continuous at x=3! x2.92.99 33.0013.013.1 F(x)8.418.948.99499.0069.069.61 Differential Calculus Derivatives of functions Derivative ( 導數 ) Given y=f(x), if variable x is given an increment  x from x=x 0, then y would change to f(x 0 +  x)  y= f(x 0 +  x) – f(x)  y  x is the slope ( 斜率 ) of triangular ABC x f(x 0 +  x) f(x 0 ) x0+xx0+x x0x0 y xx yy Y=f(x) A B C Derivative What happen with  y  x as  x tends to 0? It seems that  y  x will be close to the slope of the curve y=f(x) at x 0. We defined a new quantity as follows If the limit exists, we called this new quantity as the derivative ( 導數 ) of f(x). The process of finding derivative of f(x) is called differentiation ( 微分法 ). Derivative y f(x 0 +  x) f(x 0 ) x0+xx0+x x0x0 x xx yy Y=f(x) A B C Derivative of f(x) at Xo = slope of f(x) at Xo Differentiation from first principle Find the derivative of with respect to (w.r.t.) x To obtain the derivative of a function by its definition is called differentiation of the function from first principles Differential Calculus Differentiation rules and Higher Derivatives Fundamental formulas for differentiation I Let f(x) and g(x) be differentiable functions and c be a constant. for any real number n Examples Differentiate and w.r.t. x Table of derivative (1) FunctionDerivative Constant k0 x1 kxk Table of Derivatives (2) Function Derivative Angles in radians Differential Calculus Product Rule, Quotient Rule and Chain Rule - The product rule and the quotient rule Form of product function Form of quotient function e.g.1 This is a _________ function with and e.g.2 This is a _________ function with and e.g.3 e.g.4 e.g.5 The product rule Consider the function Using the product rule, e.g.1 Find where Solution: e.g.2 Find where Solution: e.g.3 Find where Solution: The quotient rule Consider the function. Applying the quotient rule, e.g.1 Find where Solution: e.g.2 Find where Solution: e.g.3 Find where Solution: More Example (1) Differentiate w.r.t. x More Example (2) Differentiate w.r.t. x Fundamental formulas for differentiation II Let f(x) and g(x) be differentiable functions Fundamental formulas for differentiation III ln(x) is called natural logarithm ( 自然對數 ) Differentiation of composite functions To differentiate w.r.t. x, we may have problems as we don’t have a formula to do so. The problem can be simplified by considering composite function: Let so and we know derivative of y w.r.t. u (by formula): but still don ’ t know Chain Rule ( 鏈式法則 ) Chain Rule states that : given y=g(u), and u=f(x) So our problem and Example 1 Differentiate w.r.t. x Simplify y by letting so now By chain rule Example 2 Differentiate w.r.t. x Simplify y by letting so now By chain rule Example 3 Differentiate w.r.t. x Simplify y by letting so now By chain rule Higher Derivatives ( 高階導數 ) If the derivatives of y=f(x) is differentiable function of x, its derivative is called the second derivative ( 二階導數 ) of y=f(x) and is denoted by or f ’’(x). That is Similarly, the third derivative = the n-th derivative = Example Find if Differential Calculus Applications Slope of a curve Recall that the derivative of a curve evaluate at a point is the slope of the curve at that point. f(x 0 +  x) f(x 0 ) x0+xx0+x x0x0 x xx yy Y=f(x) A B C Derivative of f(x) at Xo = slope of f(x) at Xo Slope of a curve Find the slope of y=2x+3 at x=0 To find the slope of a curve, we have to compute the derivative of y and then evaluate at a point The slope of y at x=0 equals 2 (y=mx+c now m=2) Slope of a curve Find the slope of at x=0, 2, -2 The slope of y = 2x The slope of y (at x=0) = 2(0) = 0 The slope of y (at x=-2) =2(-2) = -4 The slope of y (at x=2) =2(2) = 4 X=0 X=2X=-2 Local maximum and minimum point For a continuous function, the point at which is called a stationary point. This gives the point local maximum or local minimum of the curve A B C D X1X1 X2X2 First derivative test (Max pt.) Given a continuous function y=f(x) If dy/dx = 0 at x=x o & dy/dx changes from +ve to –ve through x 0, x=x 0 is a local maximum point x=x 0 local maximum point First derivative test (Min pt.) Given a continuous function y=f(x) If dy/dx = 0 at x=x o & dy/dx changes from -ve to +ve through x 0, x=x 0 is a local minimum point x=x 0 local minimum point Example 1 Determine the position of any local maximum and minimum of the function First, find all stationary point (i.e. find x such that dy/dx = 0), so when x=0 By first derivative test x=0 is a local minimum point Example 2 Find the local maximum and minimum of Find all stationary points first: x0x<1x=11<x<3x=3x>3 dy/dx++0-0+ Example 2 (con’d) By first derivative test, x=1 is the local maximum (+ve -> 0 -> -ve) x=3 is the local minimum (-ve -> 0 -> +ve) Second derivative test Second derivative test states: There is a local maximum point in y=f(x) at x=x0, if at x=x 0 and < 0 at x=x 0. There is a local minimum point in y=f(x) at x=x0, if at x=x 0 and > 0 at x=x 0. If dy/dx = 0 and =0 both at x=x 0, the second derivative test fails and we must return to the first derivative test. Example Find the local maximum and minimum point of Solution Find all stationary points first: By second derivative test, x=1 is max and x=3 is min. So, (1,2) is max. point and (3,-2) in min. point. Practical Examples 1 e.g.1 A rectangular block, with square base of side x mm, has a total surface area of 150 mm 2. Show that the volume of the block is given by. Hence find the maximum volume of the block. Solution: Practical Examples 2 A window frame is made in the shape of a rectangle with a semicircle on top. Given that the area is to be 8, show that the perimeter of the frame is. Find the minimum cost of producing the frame if 1 metre costs \$75. Solution: Download ppt "Engineering Mathematics: Differential Calculus. 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GMAT # GMAT math formulas: Essential knowledge for a high quantitative score Updated on: Jun 28, 2023 4.7 rated 85k+ Students counselled 65k+ Courses available The Graduate Management Admission Test (GMAT) is a standardised test that assesses the mathematical and analytical skills of prospective business school candidates. One of the important sections of GMAT is the maths section, also known as the quantitative section, which measures your ability to understand, interpret, and analyse mathematical and data-related problems within a limited time frame. To excel in the GMAT maths section and achieve a high quantitative score, it's essential to have a strong understanding of the fundamental mathematical concepts and formulas that are tested on the exam. In this guide, we will cover the essential GMAT maths formulas that you should know. ## Arithmetic formulas ### 1. Properties of integers: • Even numbers: n is even if n = 2k, where k is an integer. • Odd numbers: n is odd if n = 2k + 1, where k is an integer. • Divisibility rules: • A number is divisible by 2 if its unit digit is even (ends in 0, 2, 4, 6, or 8). • A number is divisible by 3 if the sum of its digits is divisible by 3. • A number is divisible by 4 if its last two digits form a multiple of 4. • A number is divisible by 5 if its unit digit is 0 or 5. • A number is divisible by 6 if it is divisible by both 2 and 3. • A number is divisible by 9 if the sum of its digits is divisible by 9. • Prime numbers: A prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself. ### 2. Percentages • Percentage change: Percentage change = (New Value - Old Value) / Old Value * 100% • Percentage increase: Percentage increase = (Increase/Original Value) * 100% • Percentage decrease: Percentage decrease = (Decrease/Original Value) * 100% ### 3. Ratios and proportions • Ratio: A ratio is a comparison of two quantities. The ratio of a to b is written as a:b or a/b. • Proportion: A proportion is an equation that states two ratios are equal. In a proportion, the product of the means (ad) is equal to the product of the extremes (bc), where a/b = c/d. ### 4. Averages and weighted averages: • Mean (Average): Mean = Sum of all values / Total number of values • Median: The median is the middle value in a set of values when arranged in numerical order. • Mode: The mode is the value that appears most frequently in a set of values. • Weighted Average: Weighted Average = (Sum of (Value * Weight) for all values) / (Sum of Weights) ### 5. Interest • Simple Interest: Simple Interest = (Principal * Rate * Time) / 100 • Compound Interest: Compound Interest = Principal * (1 + Rate/100)^Time - Principal • Effective Interest Rate: Effective Interest Rate = (1 + Nominal Rate/Number of Compounding Periods)^Number of Compounding Periods - 1 ## 1. Linear equations • Formula for solving linear equations with one variable: ax + b = 0 -> x = -b/a • Formula for solving linear equations with two variables: ax + by = c, dx + ey = f -> x = (ce - bf) / (ae - bd), y = (af - cd) / (ae - bd) • Quadratic Formula: For ax^2 + bx + c = 0, x = (-b ± √(b^2 - 4ac)) / (2a) ## 3. Inequalities • Linear Inequalities: ax + b < c, ax + b > c, ax + b ≤ c, ax + b ≥ c -> x > (c - b) / a, x < (c - b) / a, x ≥ (c - b) / a, x ≤ (c - b) / a • Quadratic Inequalities: ax^2 + bx + c < 0 or ax^2 + bx + c > 0, ax^2 + bx + c ≤ 0, ax^2 + bx + c ≥ 0 -> solving quadratic inequalities involves finding the values of x that satisfy the inequality and lie within the specified range. ## 4. Functions • Domain: The domain of a function is the set of all possible input values (x-values) for which the function is defined. • Range: The range of a function is the set of all possible output values (y-values) that the function can produce. • Function notation: If y = f(x), then f(x) is the function notation for the function y. Also Read: GMAT Books for Preparation ## Geometry formulas ### 1. Perimeter and area of basic shapes • Rectangle: Perimeter = 2 * (length + width), Area = length * width • Square: Perimeter = 4 * side length, Area = side length^2 • Circle: Circumference = 2 * π * radius, Area = π * radius^2 • Triangle: Perimeter = sum of lengths of all sides, Area = 0.5 * base * height • Trapezoid: Perimeter = sum of lengths of all sides, Area = 0.5 * (sum of bases) * height ### 2. Properties of triangles • Pythagorean Theorem: In a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. a^2 + b^2 = c^2, where a and b are the lengths of the two shorter sides and c is the length of the hypotenuse. • Triangle Sum Property: The sum of the measures of the angles in a triangle is always 180 degrees. • Area of a Triangle: Area = 0.5 * base * height or Area = √(s * (s-a) * (s-b) * (s-c)), where s is the semi-perimeter of the triangle, and a, b, c are the lengths of the three sides. ## Statistics formulas • Mean: The mean, or average, of a set of numbers is the sum of all the numbers divided by the total number of numbers in the set. Mean = (Sum of all numbers) / (Total number of numbers) • Median: The median of a set of numbers is the middle value when the numbers are arranged in ascending or descending order. If there is an even number of numbers, the median is the average of the two middle values. Median = Middle value(s) • Mode: The mode of a set of numbers is the value(s) that appears most frequently. Mode = Most frequent value(s) ### 1. Exponent rules • Product of Powers: a^m * a^n = a^(m+n) • Quotient of Powers: a^m / a^n = a^(m-n) • Power of a Power: (a^m)^n = a^(mn) • Power of a Product: (ab)^m = a^m b^m • Power of a Quotient: (a/b)^m = a^m / b^m • Negative Exponent: a^(-m) = 1 / a^m • Square Root: √a * √a = a • Cube Root: ∛a * ∛a * ∛a = a • nth Root: √n√a = a^(1/n) ## Probability formulas • Probability: Probability of an event = Number of favourable outcomes / Total number of possible outcomes • Complementary Probability: P(A') = 1 - P(A), where A' represents the complement of event A, and P(A) and P(A') represent the probabilities of event A and its complement, respectively. ## Coordinate geometry formulas • Distance Formula: The distance between two points (x1, y1) and (x2, y2) in a coordinate plane is given by: d = √[(x2 - x1)^2 + (y2 - y1)^2] • Slope of a Line: The slope (m) of a line passing through two points (x1, y1) and (x2, y2) is given by: m = (y2 - y1) / (x2 - x1) • Midpoint Formula: The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is given by: Midpoint = [((x1 + x2)/2), ((y1 + y2)/2)] ## Arithmetic and geometric progression formulas • Arithmetic Progression (AP): An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. The nth term of an arithmetic progression with the first term 'a' and the common difference 'd' is given by: An = a + (n - 1)d where n is the position of the term in the sequence. • Geometric Progression (GP): A geometric progression is a sequence of numbers in which the ratio of any two consecutive terms is constant. The nth term of a geometric progression with the first term 'a' and the common ratio 'r' is given by: Gn = a * r^(n - 1) where n is the position of the term in the sequence. ## Set theory formulas • Union of Sets: The union of two sets A and B, denoted as A ∪ B, is the set of all elements that are in either A or B or in both. • Intersection of Sets: The intersection of two sets A and B, denoted as A ∩ B, is the set of all elements that are in both A and B. • Complement of a Set: The complement of a set A, denoted as A', is the set of all elements that are in the universal set but not in A. ## Profit and loss formulas • Cost Price (CP): The price at which an item is bought is known as the cost price. • Selling Price (SP): The price at which an item is sold is known as the selling price. • Profit (P): The amount earned after selling an item at a price higher than its cost price. P = SP - CP • Loss (L): The amount incurred after selling an item at a price lower than its cost price. L = CP - SP • Profit Percentage (PP): The percentage of profit earned on the cost price. PP = (Profit / CP) * 100 • Loss Percentage (LP): The percentage of loss incurred on the cost price. LP = (Loss / CP) * 100 You might also like ## Conclusion It is important to note that while memorising these formulas is essential, understanding their applications and being able to apply them in various problem-solving scenarios is equally important. Practising GMAT maths problems and using these formulas in real-world scenarios will help you improve your quantitative skills and achieve a high score on the GMAT maths section. How important is it to memorise GMAT maths formulas? While it's important to be familiar with the essential GMAT maths formulas, it's equally important to understand how to apply them in different problem-solving scenarios. The GMAT is not solely about memorization, but also about critical thinking and logical reasoning. So, focus on developing a deep understanding of the concepts and their applications, rather than just memorising formulas. Can I use my own calculator during the GMAT? No, you cannot bring your own calculator to the GMAT test. The GMAT provides an on-screen calculator for certain sections of the test. However, it's important to note that the calculator is basic and only allows for basic arithmetic calculations. How can I effectively prepare for GMAT maths? To effectively prepare for GMAT maths, start by reviewing the fundamental mathematical concepts that are tested on the GMAT. Familiarise yourself with the essential GMAT maths formulas and practice solving a wide range of GMAT maths problems. Additionally, take practice tests and review your mistakes to identify areas where you need to improve. Related tags: 0 Likes
# 3600 in words 3600 in words is written as Three Thousand Six Hundred. In the number 3600, 3 represents the place value of thousand and 6 has the place of hundred. The article on Place Value will get you more details. The number 3600 is used in expressions that relate to count, money, distance, length and others. Let us consider an example for the number 3600. “An advertisement was put up on a hoarding, for a flat that measured 3600 sq feet. The price per foot was Three Thousand Six Hundred rupees” 3600 in words Three Thousand Six Hundred Three Thousand Six Hundred in Numbers 3600 ## How to Write 3600 in Words? We can convert 3600 to words using a place value chart. The number 3600 has 4 digits, so let’s make a chart that shows the place value up to 4 digits. Thousands Hundreds Tens Ones 3 6 0 0 Thus, we can write the expanded form as: 3 × Thousand + 6 × Hundred + 0 × Ten + 0 × One = 3 × 1000 + 6 × 100 + 0 × 10 + 0 × 1 = 3600 = Three Thousand Six Hundred. 3600 is the natural number that is succeeded by 3599 and preceded by 3601. 3600 in words – Three Thousand Six Hundred. Is 3600 an odd number? – No. Is 3600 an even number? – Yes. Is 3600 a perfect square number? – Yes. (60 × 60 ). Is 3600 a perfect cube number? – No. Is 3600 a prime number? – No. Is 3600 a composite number? – Yes. ## Solved Example 1. Write the number 3600 in expanded form Solution: 3 × 1000 + 6 × 100 + 0 × 10 + 0 × 1 We can write 3600 = 3000 + 600 + 0 + 0 = 3 × 1000 + 6 × 100 + 0 × 10 + 0 × 1 ## Frequently Asked Questions on 3600 in words Q1 ### How to write 3600 in words? 3600 in words is written as Three Thousand Six hundred. Q2 ### Is 3600 a perfect square number? Yes. 3600 is a perfect square number. 60 multiplied by itself gives 3600. Q3 ### Is 3600 divisible by 10? Yes. 3600 is divisible by 10.
Active Calculus - Multivariable Section9.8Arc Length and Curvature Given a space curve, there are two natural geometric questions one might ask: how long is the curve and how much does it bend? In this section, we answer both questions by developing techniques for measuring the length of a space curve as well as its curvature. Preview Activity9.8.1. In earlier investigations, we have used integration to calculate quantities such as area, volume, mass, and work. We are now interested in determining the length of a space curve. Consider the smooth curve in 3-space defined by the vector-valued function $$\vr\text{,}$$ where \begin{equation} \vr(t) = \langle x(t), y(t), z(t) \rangle = \langle \cos(t), \sin(t), t \rangle \end{equation} for $$t$$ in the interval $$[0,2\pi]\text{.}$$ Pictures of the graph of $$\vr$$ are shown in Figure 9.8.1. We will use the integration process to calculate the length of this curve. In this situation we partition the interval $$[0,2\pi]$$ into $$n$$ subintervals of equal length and let $$0 = t_0 \lt t_1 \lt t_2 \lt \cdots \lt t_n = b$$ be the endpoints of the subintervals. We then approximate the length of the curve on each subinterval with some related quantity that we can compute. In this case, we approximate the length of the curve on each subinterval with the length of the segment connecting the endpoints. Figure 9.8.1 illustrates the process in three different instances using increasing values of $$n\text{.}$$ 1. Write a formula for the length of the line segment that connects the endpoints of the curve on the $$i$$th subinterval $$[t_{i-1},t_i]\text{.}$$ (This length is our approximation of the length of the curve on this interval.) 2. Use your formula in part (a) to write a sum that adds all of the approximations to the lengths on each subinterval. 3. What do we need to do with the sum in part (b) in order to obtain the exact value of the length of the graph of $$\vr(t)$$ on the interval $$[0,2\pi]\text{?}$$ Subsection9.8.1Arc Length Consider a smooth curve in 3-space that is parametrically described by the vector-valued function $$\vr$$ defined by $$\vr(t) = \langle x(t), y(t), z(t) \rangle.$$ Preview Activity 9.8.1 shows that to approximate the length of the curve defined by $$\vr(t)$$ as the values of $$t$$ run over an interval $$[a,b]\text{,}$$ we partition the interval $$[a,b]$$ into $$n$$ subintervals of equal length $$\Delta t\text{,}$$ with $$a = t_0 \lt t_1 \lt \cdots \lt t_n = b$$ as the endpoints of the subintervals. On each subinterval, we approximate the length of the curve by the length of the line segment connecting the endpoints. The points on the curve corresponding to $$t = t_{i-1}$$ and $$t = t_i$$ are $$(x(t_{i-1}), y(t_{i-1}), z(t_{i-1}))$$ and $$(x(t_i), y(t_i), z(t_i))\text{,}$$ respectively, so the length of the line segment connecting these points is \begin{equation} \sqrt{(x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 + (z(t_i) - z(t_{i-1}))^2}. \end{equation} Now we add all of these approximations together to obtain an approximation to the length $$L$$ of the curve: \begin{equation} L \approx \sum_{i=1}^n \sqrt{(x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 + (z(t_i) - z(t_{i-1}))^2}. \end{equation} We now want to take the limit of this sum as $$n$$ goes to infinity, but in its present form it might be difficult to see how. We first introduce $$\Delta t$$ by multiplying by $$\frac{\Delta t}{\Delta t}\text{,}$$ and see that \begin{align} L \amp \approx \sum_{i=1}^n \sqrt{(x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 + (z(t_i) - z(t_{i-1}))^2}\\ \amp = \sum_{i=1}^n \sqrt{(x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 + (z(t_i) - z(t_{i-1}))^2} \frac{\Delta t}{\Delta t}\\ \amp = \sum_{i=1}^n \sqrt{(x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 + (z(t_i) - z(t_{i-1}))^2} \frac{\Delta t}{\sqrt{(\Delta t)^2}} \end{align} To get the difference quotients under the radical, we use properties of the square root function to see further that \begin{align} L \amp \approx \sum_{i=1}^n \sqrt{\left[(x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 + (z(t_i) - z(t_{i-1})^2\right] \frac{1}{(\Delta t)^2}} \Delta t\\ \amp = \sum_{i=1}^n \sqrt{\left(\frac{x(t_i) - x(t_{i-1})}{\Delta t}\right)^2 + \left(\frac{y(t_i) - y(t_{i-1})}{\Delta t}\right)^2 + \left(\frac{z(t_i) - z(t_{i-1})}{\Delta t}\right)^2} \Delta t. \end{align} Recall that as $$n \to \infty$$ we also have $$\Delta t \to 0\text{.}$$ Since \begin{align} x'(t) \amp = \lim_{\Delta t \to 0} \frac{x(t_i) - x(t_{i-1})}{\Delta t},\\ y'(t) \amp = \lim_{\Delta t \to 0} \frac{y(t_i) - y(t_{i-1})}{\Delta t}, \ \text{ and } \\ z'(t) \amp \lim_{\Delta t \to 0} \frac{z(t_i) - z(t_{i-1})}{\Delta t}, \end{align} we see that \begin{equation} \lim_{n \to \infty} \sum_{i=1}^n \sqrt{\left(\frac{x(t_i) - x(t_{i-1})}{\Delta t}\right)^2 + \left(\frac{y(t_i) - y(t_{i-1})}{\Delta t}\right)^2 + \left(\frac{z(t_i) - z(t_{i-1})}{\Delta t}\right)^2} \Delta t \end{equation} is equal to \begin{equation} \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt. \end{equation} Noting further that \begin{equation} \|\vr'(t)\| = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}, \end{equation} we can rewrite our arclength formula in a more succinct form as follows. The length of a curve. If $$\vr(t)$$ defines a smooth curve $$C$$ on an interval $$[a,b]\text{,}$$ then the length $$L$$ of $$C$$ is given by $$L = \int_a^b \|\vr'(t)\| \, dt.\tag{9.8.1}$$ Note that formula (9.8.1) applies to curves in any dimensional space. Moreover, this formula has a natural interpretation: if $$\vr(t)$$ records the position of a moving object, then $$\vr'(t)$$ is the object’s velocity and $$\|\vr'(t)\|$$ its speed. Formula (9.8.1) says that we simply integrate the speed of an object traveling over the curve to find the distance traveled by the object, which is the same as the length of the curve, just as in one-variable calculus. Activity9.8.2. Here we calculate the arc length of two familiar curves. 1. Use Equation (9.8.1) to calculate the circumference of a circle of radius $$r\text{.}$$ 2. Find the exact length of the spiral defined by $$\vr(t) = \langle \cos(t), \sin(t), t \rangle$$ on the interval $$[0,2\pi]\text{.}$$ We can adapt the arc length formula to curves in 2-space that define $$y$$ as a function of $$x$$ as the following activity shows. Activity9.8.3. Let $$y = f(x)$$ define a smooth curve in 2-space. Parameterize this curve and use Equation (9.8.1) to show that the length of the curve defined by $$f$$ on an interval $$[a,b]$$ is \begin{equation} \int_a^b \sqrt{1+[f'(t)]^2} \, dt. \end{equation} Subsection9.8.2Parameterizing With Respect To Arc Length In addition to helping us to find the length of space curves, the expression for the length of a curve enables us to find a natural parametrization of space curves in terms of arc length, as we now explain. Shown below in Figure 9.8.2 is a portion of the parabola $$y = x^2/2\text{.}$$ Of course, this space curve may be parametrized by the vector-valued function $$\vr$$ defined by $$\vr(t) = \langle t, t^2/2\rangle$$ as shown on the left, where we see the location at a few different times $$t\text{.}$$ Notice that the points are not equally spaced on the curve. A more natural parameter describing the points along the space curve is the distance traveled $$s$$ as we move along the parabola starting at the origin. For instance, the right side of Figure 9.8.2 shows the points corresponding to various values of $$s\text{.}$$ We call this an arc length parametrization. To see that this is a more natural parametrization, consider an interstate highway cutting across a state. One way to parametrize the curve defined by the highway is to drive along the highway and record our position at every time, thus creating a function $$\vr\text{.}$$ If we encounter an accident or road construction, however, this parametrization might not be at all relevant to another person driving the same highway. An arc length parametrization, however, is like using the mile markers on the side of road to specify our position on the highway. If we know how far we’ve traveled along the highway, we know exactly where we are. If we begin with a parametrization of a space curve, we can modify it to find an arc length parametrization, as we now describe. Suppose that the curve is parametrized by the vector-valued function $$\vr = \vr(t)$$ where $$t$$ is in the interval $$[a,b]\text{.}$$ We define the parameter $$s$$ through the function \begin{equation} s=L(t) = \int_a^t \sqrt{(x'(w))^2 + (y'(w))^2 + (z'(w))^2} \,dw, \end{equation} which measures the length along the curve from $$\vr(a)$$ to $$\vr(t)\text{.}$$ The Fundamental Theorem of Calculus shows us that $$\frac{ds}{dt} = L'(t)= \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} = \lvert \vr'(t) \rvert\tag{9.8.2}$$ and so \begin{equation} L(t) = \int_a^t \left\| \frac{d}{dw}\vr(w)\right\| \,dw. \end{equation} If we assume that $$\vr'(t)$$ is never 0, then $$L'(t) > 0$$ for all $$t$$ and $$s=L(t)$$ is always increasing. This should seem reasonable: unless we stop, the distance traveled along the curve increases as we move along the curve. Since $$s=L(t)$$ is an increasing function, it is invertible, which means we may view the time $$t$$ as a function of the distance traveled; that is, we have the relationship $$t=L^{-1}(s)\text{.}$$ We then obtain the arc length parametrization by composing $$\vr(t)$$ with $$t=L^{-1}(s)$$ to obtain $$\vr(s)\text{.}$$ Let’s illustrate this with an example. Example9.8.3. Consider a circle of radius $$5$$ in 2-space centered at the origin. We know that we can parameterize this circle as \begin{equation} \vr(t) = \langle 5\cos(t), 5\sin(t) \rangle, \end{equation} where $$t$$ runs from 0 to $$2\pi\text{.}$$ We see that $$\vr'(t) = \langle -5\sin(t), 5\cos(t) \rangle\text{,}$$ and hence $$\|\vr'(t)\| = 5\text{.}$$ It then follows that \begin{equation} s=L(t) = \int_0^t \|\vr'(w)\|~dw = \int_0^t 5~dw = 5t. \end{equation} Since $$s=L(t) = 5t\text{,}$$ we may solve for $$t$$ in terms of $$s$$ to obtain $$t(s)=L^{-1}(s) = s/5\text{.}$$ We then find the arc length parametrization by composing \begin{equation} \vr(t(s))=\vr(L^{-1}(s)) = \left\langle 5\cos\left(\frac s5\right), 5\sin\left(\frac s5\right)\right\rangle. \end{equation} More generally, for a circle of radius $$a$$ centered at the origin, a similar computation shows that $$\left\langle a\cos\left(\frac sa\right), a\sin\left(\frac sa\right)\right\rangle\tag{9.8.3}$$ is an arc length parametrization. Notice that equation (9.8.2) shows that \begin{equation} \frac{d\vr}{dt} = \frac{d\vr}{ds}\frac{ds}{dt} = \frac{d\vr}{ds}\lvert \vr'(t) \rvert, \end{equation} so \begin{equation} \left\| \frac{d\vr}{ds} \right\| = \left\|\frac{1}{\lvert \vr'(t) \rvert}\frac{d\vr}{dt} \right\| = 1, \end{equation} which means that we move along the curve with unit speed when we parameterize by arc length. This is clearly seen in Example 9.8.3 where $$\|\vr'(s)\| = 1\text{.}$$ It follows that the parameter $$s$$ is the distance traveled along the curve, as shown by: \begin{equation} L(s) = \int_0^s\left\|\frac{d}{ds}\vr(w)\right\|~dw = \int_0^s1~dw = s. \end{equation} Activity9.8.4. In this activity we parameterize a line in 2-space in terms of arc length. Consider the line with parametric equations \begin{equation} x(t) = x_0+at \ \ \ \ \text{ and } \ \ \ \ y(t) = y_0+bt. \end{equation} 1. To write $$t$$ in terms of $$s\text{,}$$ evaluate the integral \begin{equation} s=L(t) = \int_{0}^t \sqrt{(x'(w))^2 + (y'(w))^2} \, dw \end{equation} to determine the length of the line from time 0 to time $$t\text{.}$$ 2. Use the formula from (a) for $$s$$ in terms of $$t$$ to write $$t$$ in terms of $$s\text{.}$$ Then explain why a parameterization of the line in terms of arc length is $$x(s) = x_0+\frac{a}{\sqrt{a^2+b^2}}s \ \ \ \ \text{ and } \ \ \ \ y(s) = y_0+\frac{b}{\sqrt{a^2+b^2}}s.\tag{9.8.4}$$ A little more complicated example is the following. Example9.8.4. Let us parameterize the curve defined by \begin{equation} \vr(t) = \left\langle t^2, \frac{8}{3}t^{3/2}, 4t \right\rangle \end{equation} for $$t \geq 0$$ in terms of arc length. To write $$t$$ in terms of $$s$$ we find $$s$$ in terms of $$t\text{:}$$ \begin{align} s(t) \amp = \int_{0}^t \sqrt{(x'(w))^2 + (y'(w))^2 +(z'(w))^2} \, dw\\ \amp = \int_0^t \sqrt{(2w)^2 + (4w^{1/2})^2 + (4)^2} \, dw\\ \amp = \int_0^t \sqrt{4w^2 + 16w + 16} \, dw\\ \amp = 2\int_0^t \sqrt{(w+2)^2} \, dw\\ \amp = 2\int_0^t w+2 \, dw\\ \amp = \left(w^2+4w\right)\biggm\|_{0}^{t}\\ \amp = t^2+4t. \end{align} Since $$t \geq 0\text{,}$$ we can solve the equation $$s = t^2+4t$$ (or $$t^2+4t-s=0$$) for $$t$$ to obtain $$t = \frac{-4 +\sqrt{16+4s}}{2} = -2 + \sqrt{4+s}\text{.}$$ So we can parameterize our curve in terms of arc length by \begin{equation} \vr(s) = \left\langle \left(-2 + \sqrt{4+s}\right)^2, \frac{8}{3}\left(-2 + \sqrt{4+s}\right)^{3/2}, 4\left(-2 + \sqrt{4+s}\right) \right\rangle. \end{equation} These examples illustrate a general method. Of course, evaluating an arc length integral and finding a formula for the inverse of a function can be difficult, so while this process is theoretically possible, it is not always practical to parameterize a curve in terms of arc length. However, we can guarantee that such a parameterization exists, and this observation plays an important role in the next section. Subsection9.8.3Curvature For a smooth space curve, the curvature measures how fast the curve is bending or changing direction at a given point. For example, we expect that a line should have zero curvature everywhere, while a circle (which is bending the same at every point) should have constant curvature. Circles with larger radii should have smaller curvatures. To measure the curvature, we first need to describe the direction of the curve at a point. We may do this using a continuously varying tangent vector to the curve, as shown at left in Figure 9.8.5. The direction of the curve is then determined by the angle $$\phi$$ each tangent vector makes with a horizontal vector, as shown at right in Figure 9.8.5. Informally speaking, the curvature will be the rate at which the angle $$\phi$$ is changing as we move along the curve. Of course, this rate of change will depend on how we move along the curve; if we move with a greater speed along the curve, then $$\phi$$ will change more rapidly. This is why the speed limit is sometimes lowered when we enter a curve on a highway. In other words, the rate of change of $$\phi$$ will depend on the parametrization we use to describe the space curve. To eliminate this dependence on the parametrization, we choose to work with an arc length parametrization $$\vr(s)\text{,}$$ which means we move along the curve with unit speed. Using an arc length parametrization $$\vr(s)\text{,}$$ we define the tangent vector $$\vT(s) = \vr'(s)\text{,}$$ and note that $$\|\vT(s)\| = 1\text{;}$$ that is, $$\vT(s)$$ is a unit tangent vector. We then have $$\vT(s) = \langle \cos (\phi(s)), \sin(\phi(s)) \rangle\text{,}$$ which means that \begin{equation} \frac{d\vT}{ds} = \left\langle -\sin(\phi(s)) \frac{d\phi}{ds}, ~ \cos(\phi(s)) \frac{d\phi}{ds} \right\rangle = \langle -\sin(\phi(s)),~ \cos(\phi(s)) \rangle \frac{d\phi}{ds}. \end{equation} Therefore \begin{equation} \left\|\frac{d\vT}{ds}\right\| = \left\|\langle -\sin(\phi(s)),~ \cos(\phi(s)) \rangle\right\| ~\left\|\frac{d\phi}{ds}\right\| = \left\|\frac{d\phi}{ds}\right\| \end{equation} Definition9.8.6. If $$C$$ is a smooth space curve and $$s$$ is an arc length parameter for $$C\text{,}$$ then the curvature, $$\kappa\text{,}$$ of $$C$$ is \begin{equation} \kappa = \kappa(s) = \left\lvert \frac{d \vT}{ds} \right\rvert. \end{equation} Note that $$\kappa$$ is the Greek lowercase letter “kappa”. Activity9.8.5. 1. We should expect that the curvature of a line is 0 everywhere. To show that our definition of curvature measures this correctly in 2-space, recall that (9.8.4) gives us the arc length parameterization \begin{equation} x(s) = x_0+\frac{a}{\sqrt{a^2+b^2}}s \ \ \ \ \text{ and } \ \ \ \ y(s) = y_0+\frac{b}{\sqrt{a^2+b^2}}s \end{equation} of a line. Use this information to explain why the curvature of a line is 0 everywhere. 2. Recall that an arc length parameterization of a circle in 2-space of radius $$a$$ centered at the origin is, from (9.8.3), \begin{equation} \vr(s) = \left\langle a \cos\left(\frac{s}{a}\right),~ a \sin\left(\frac{s}{a}\right)\right\rangle. \end{equation} Show that the curvature of this circle is the constant $$\frac{1}{a}\text{.}$$ What can you say about the relationship between the size of the radius of a circle and the value of its curvature? Why does this make sense? The definition of curvature relies on our ability to parameterize curves in terms of arc length. Since we have seen that finding an arc length parametrization can be difficult, we would like to be able to express the curvature in terms of a more general parametrization $$\vr(t)\text{.}$$ To begin, we need to describe the vector $$\vT\text{,}$$ which is a vector tangent to the curve having unit length. Of course, the velocity vector $$\vr'(t)$$ is tangent to the curve; we simply need to normalize its length to be one. This means that we may take $$\vT(t) = \frac{\vr'(t)}{\|\vr'(t)\|}.\tag{9.8.5}$$ Then the curvature of the curve defined by $$\vr$$ is \begin{align} \kappa \amp = \left\lvert \frac{d \vT}{ds} \right\rvert\\ \amp = \left\lvert \frac{d \vT}{dt} \frac{dt}{ds} \right\rvert\\ \amp = \frac{\left\lvert \frac{d \vT}{dt} \right\rvert}{ \left\lvert \frac{ds}{dt} \right\rvert }\\ \amp = \frac{\left\lvert \vT'(t) \right\rvert}{ \left\lvert \vr'(t) \right\rvert}. \end{align} This last formula allows us to use any parameterization of a curve to calculate its curvature. There is another useful formula, given below, whose derivation is left for the exercises. Formulas for curvature. If $$\vr$$ is a vector-valued function defining a smooth space curve $$C\text{,}$$ and if $$\vr'(t)$$ is not zero and if $$\vr''(t)$$ exists, then the curvature $$\kappa$$ of $$C$$ satisfies • $$\displaystyle \kappa = \kappa(t) = \frac{\left\lvert \vT'(t) \right\rvert}{ \left\lvert \vr'(t) \right\rvert}$$ • $$\kappa = \frac{\lvert \vr'(t) \times \vr''(t) \rvert}{\lvert \vr'(t) \rvert^3}\text{.}$$ Activity9.8.6. Use one of the two formulas for $$\kappa$$ in terms of $$t$$ to help you answer the following questions. 1. The ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ has parameterization \begin{equation} \vr(t) = \langle a\cos(t), b\sin(t) \rangle. \end{equation} Find the curvature of the ellipse. Assuming $$0 \lt b \lt a\text{,}$$ at what points is the curvature the greatest and at what points is the curvature the smallest? Does this agree with your intuition? 2. The standard helix has parameterization $$\vr(t) = \cos(t) \vi + \sin(t) \vj + t \vk\text{.}$$ Find the curvature of the helix. Does the result agree with your intuition? The curvature has another interpretation. Recall that the tangent line to a curve at a point is the line that best approximates the curve at that point. The curvature at a point on a curve describes the circle that best approximates the curve at that point. Remembering that a circle of radius $$a$$ has curvature $$1/a\text{,}$$ then the circle that best approximates the curve near a point on a curve whose curvature is $$\kappa$$ has radius $$1/\kappa$$ and will be tangent to the tangent line at that point and has its center on the concave side of the curve. This circle, called the osculating circle of the curve at the point, is shown in Figure 9.8.7 for a portion of a parabola. Subsection9.8.4Summary • The integration process shows that the length $$L$$ of a smooth curve defined by $$\vr(t)$$ on an interval $$[a,b]$$ is \begin{equation} L = \int_a^b \|\vr'(t)\| \, dt. \end{equation} • Arc length is useful as a parameter because when we parameterize with respect to arc length, we eliminate the role of speed in our calculation of curvature and the result is a measure that depends only on the geometry of the curve and not on the parameterization of the curve. • We define the curvature $$\kappa$$ of a curve in 2- or 3-space to be the rate of change of the magnitude of the unit tangent vector with respect to arc length, or \begin{equation} \kappa = \left\lvert \frac{d\vT}{ds} \right\rvert. \end{equation} Exercises9.8.5Exercises The WeBWorK problems are written by many different authors. Some authors use parentheses when writing vectors, e.g., $$(x(t),y(t),z(t))$$ instead of angle brackets $$\langle x(t),y(t),z(t) \rangle\text{.}$$ Please keep this in mind when working WeBWorK exercises. 1. Find the length of the curve \begin{equation} x = -\left(3+3t\right),\quad y = -\left(1+4t\right), z = 4-4t, \end{equation} for $$4 \le t \le 5\text{.}$$ length = (Think of second way that you could calculate this length, too, and see that you get the same result.) 2. Consider the curve $$\displaystyle \mathbf{r} = (e^{-2 t} \cos(6 t), e^{-2 t} \sin(6 t), e^{-2 t})\text{.}$$ Compute the arclength function $$s(t)\text{:}$$ (with initial point $$t=0$$). 3. Find the length of the given curve: \begin{equation} \mathbf{r} \left( t \right) = \left( 3 t, -4 \sin t, -4 \cos t \right) \end{equation} where $$-1 \leq t \leq 1\text{.}$$ 4. Find the curvature of $$y=\sin \left( -3 x \right)$$ at $$x = \frac{\pi}{4}\text{.}$$ 5. Consider the path $$\mathbf{r}(t) = (14 t, 7 t^2, 7\ln t)$$ defined for $$t > 0\text{.}$$ Find the length of the curve between the points $$(14, 7, 0)$$ and $$(70, 175, 7 \ln(5))\text{.}$$ 6. Find the curvature $$\kappa (t)$$ of the curve $$\mathbf{r} (t) = \left( 3 \sin t \right) \mathbf{i} + \left( 3 \sin t \right) \mathbf{j} + \left( -1 \cos t \right) \mathbf{k}$$ 7. A factory has a machine which bends wire at a rate of 2 unit(s) of curvature per second. How long does it take to bend a straight wire into a circle of radius 9? seconds 8. Find the unit tangent vector at the indicated point of the vector function \begin{equation} \mathbf{r}(t) = e^{20 t}\cos t \,\mathbf i + e^{20 t}\sin t \,\mathbf j + e^{20 t} \,\mathbf k \end{equation} $$\mathbf T(\pi/2) = \langle$$ , , $$\rangle$$ 9. Consider the vector function $$\mathbf r(t) = \langle t, t^{2}, t^{5}\rangle$$ Compute $$\mathbf r'(t) = \langle$$ , , $$\rangle$$ $$\mathbf T(1) = \langle$$ , , $$\rangle$$ $$\mathbf r''(t) = \langle$$ , , $$\rangle$$ $$\mathbf r'(t)\times \mathbf r''(t) = \langle$$ , , $$\rangle$$ 10. Starting from the point $$\left( -3, -4, -4 \right)\text{,}$$ reparametrize the curve $$\mathbf{x} (t) = ( -3 + t, -4 + t, -4 + t )$$ in terms of arclength. $$\mathbf{y}(s) = ($$ , , $$)$$ 11. Consider the moving particle whose position at time $$t$$ in seconds is given by the vector-valued function $$\vr$$ defined by $$\vr(t) = 5t \vi + 4\sin(3t) \vj + 4\cos(3t) \vk\text{.}$$ Use this function to answer each of the following questions. 1. Find the unit tangent vector, $$\vT(t)\text{,}$$ to the space curve traced by $$\vr(t)$$ at time $$t\text{.}$$ Write one sentence that explains what $$\vT(t)$$ tells us about the particle’s motion. 2. Determine the speed of the particle moving along the space curve with the given parameterization. 3. Find the exact distance traveled by the particle on the time interval $$[0,\pi/3]\text{.}$$ 4. Find the average velocity of the particle on the time interval $$[0, \pi/3]\text{.}$$ 5. Determine the parameterization of the given curve with respect to arc length. 12. Let $$y = f(x)$$ define a curve in the plane. We can consider this curve as a curve in three-space with $$z$$-coordinate 0. 1. Find a parameterization of the form $$\vr(t) = \langle x(t), y(t), z(t) \rangle$$ of the curve $$y=f(x)$$ in three-space. 2. Use the formula \begin{equation} \kappa = \frac{\lvert \vr'(t) \times \vr''(t) \rvert}{\lvert \vr'(t) \rvert^3} \end{equation} to show that \begin{equation} \kappa = \frac{\lvert f''(x) \rvert}{\left[1+(f'(x))^2\right]^{3/2}}. \end{equation} 13. Consider the single variable function defined by $$y = 4x^2 - x^3.$$ 1. Find a parameterization of the form $$\vr(t) = \langle x(t), y(t) \rangle$$ that traces the curve $$y = 4x^2 - x^3$$ on the interval from $$x = -3$$ to $$x = 3\text{.}$$ 2. Write a definite integral which, if evaluated, gives the exact length of the given curve from $$x = -3$$ to $$x = 3\text{.}$$ Why is the integral difficult to evaluate exactly? 3. Determine the curvature, $$\kappa(t)\text{,}$$ of the parameterized curve. (Exercise 9.8.5.12 might be useful here.) 4. Use appropriate technology to approximate the absolute maximum and minimum of $$\kappa(t)$$ on the parameter interval for your parameterization. Compare your results with the graph of $$y = 4x^2 - x^3\text{.}$$ How do the absolute maximum and absolute minimum of $$\kappa(t)$$ align with the original curve? 14. Consider the standard helix parameterized by $$\vr(t) = \cos(t) \vi + \sin(t) \vj + t \vk\text{.}$$ 1. Recall that the unit tangent vector, $$\vT(t)\text{,}$$ is the vector tangent to the curve at time $$t$$ that points in the direction of motion and has length 1. Find $$\vT(t)\text{.}$$ 2. Explain why the fact that $$\| \vT(t) \| = 1$$ implies that $$\vT$$ and $$\vT'$$ are orthogonal vectors for every value of $$t\text{.}$$ (Hint: note that $$\vT \cdot \vT = \|\vT\|^2 = 1,$$ and compute $$\frac{d}{dt}[\vT \cdot \vT]\text{.}$$) 3. For the given function $$\vr$$ with unit tangent vector $$\vT(t)$$ (from (a)), determine $$\vN(t) = \frac{1}{\|\vT'(t)\|} \vT'(t)\text{.}$$ 4. What geometric properties does $$\vN(t)$$ have? That is, how long is this vector, and how is it situated in comparison to $$\vT(t)\text{?}$$ 5. Let $$\vB(t) = \vT(t) \times \vN(t)\text{,}$$ and compute $$\vB(t)$$ in terms of your results in (a) and (c). 6. What geometric properties does $$\vB(t)$$ have? That is, how long is this vector, and how is it situated in comparison to $$\vT(t)$$ and $$\vN(t)\text{?}$$ 7. Sketch a plot of the given helix, and compute and sketch $$\vT(\pi/2)\text{,}$$ $$\vN(\pi/2)\text{,}$$ and $$\vB(\pi/2)\text{.}$$ 15. In this exercise we verify the curvature formula \begin{equation} \kappa = \frac{\lvert \vr'(t) \times \vr''(t) \rvert}{\lvert \vr'(t) \rvert^3}. \end{equation} 1. Explain why \begin{equation} \lvert \vr'(t) \rvert = \frac{ds}{dt}. \end{equation} 2. Use the fact that $$\vT(t) = \frac{\vr'(t)}{\lvert \vr'(t) \rvert}$$ and $$\lvert \vr'(t) \rvert = \frac{ds}{dt}$$ to explain why \begin{equation} \vr'(t) = \frac{ds}{dt} \vT(t). \end{equation} 3. The Product Rule shows that \begin{equation} \vr''(t) = \frac{d^2s}{dt^2} \vT(t) + \frac{ds}{dt} \vT'(t). \end{equation} Explain why \begin{equation} \vr'(t) \times \vr''(t) = \left(\frac{ds}{dt}\right)^2 (\vT(t) \times \vT'(t)). \end{equation} 4. In Exercise 9.8.5.14 we showed that $$\lvert \vT(t) \rvert = 1$$ implies that $$\vT(t)$$ is orthogonal to $$\vT'(t)$$ for every value of $$t\text{.}$$ Explain what this tells us about $$\lvert \vT(t) \times \vT'(t) \rvert$$ and conclude that \begin{equation} \lvert \vr'(t) \times \vr''(t) \rvert = \left(\frac{ds}{dt}\right)^2 \lvert \vT'(t) \rvert. \end{equation} 5. Finally, use the fact that $$\kappa = \frac{\lvert \vT'(t) \rvert }{\lvert \vr'(t) \rvert}$$ to verify that \begin{equation} \kappa = \frac{\lvert \vr'(t) \times \vr''(t) \rvert}{\lvert \vr'(t) \rvert^3}. \end{equation} 16. In this exercise we explore how to find the osculating circle for a given curve. As an example, we will use the curve defined by $$f(x) = x^2\text{.}$$ Recall that this curve can be parameterized by $$x(t) = t$$ and $$y(t)=t^2\text{.}$$ 1. Use (9.8.5) to find $$\vT(t)$$ for our function $$f\text{.}$$ 2. To find the center of the osculating circle, we will want to find a vector that points from a point on the curve to the center of the circle. Such a vector will be orthogonal to the tangent vector at that point. Recall that $$\vT(s) = \langle \cos(\phi(s)), \sin(\phi(s)) \rangle\text{,}$$ where $$\phi$$ is the angle the tangent vector to the curve makes with a horizontal vector. Use this fact to show that \begin{equation} \vT \cdot \frac{dT}{ds} = 0. \end{equation} Explain why this tells us that $$\frac{dT}{ds}$$ is orthogonal to $$\vT\text{.}$$ Let $$\vN$$ be the unit vector in the direction of $$\frac{dT}{ds}\text{.}$$ The vector $$\vN$$ is called the principal unit normal vector and points in the direction toward which the curve is turning. The vector $$\vN$$ also points toward the center of the osculating circle. 3. Find $$\vT$$ at the point $$(1,1)$$ on the graph of $$f\text{.}$$ Then find $$\vN$$ at this same point. How do you know you have the correct direction for $$\vN\text{?}$$ 4. Let $$P$$ be a point on the curve. Recall that $$\rho = \frac{1}{\kappa}$$ at point $$P$$ is the radius of the osculating circle at point $$P\text{.}$$ We call $$\rho$$ the radius of curvature at point $$P\text{.}$$ Let $$C$$ be the center of the osculating circle to the curve at point $$P\text{,}$$ and let $$O$$ be the origin. Let $$\mathbf{\gamma}$$ be the vector $$\overrightarrow{OC}\text{.}$$ See Figure 9.8.8 for an illustration using an arbitrary function $$f\text{.}$$ Which vector, in terms of $$\rho$$ and $$\vN$$ points from the point $$P$$ to the point $$C\text{?}$$ Use this vector to explain why \begin{equation} \mathbf{\gamma} = \vr + \rho \vN, \end{equation} where $$\vr = \overrightarrow{OP}\text{.}$$ 5. Finally, use the previous work to find the center of the osculating circle for $$f$$ at the point $$(1,1)\text{.}$$ Draw pictures of the curve and the osculating circle to verify your work.
# What is row minima method? ## What is row minima method? Row minima method Steps (Rule) Step-1: In this method, we allocate as much as possible in the lowest cost cell of the first row, i.e. allocate min(si,dj). ## What are the steps of Hungarian method? The Hungarian algorithm 1. Step 1: Subtract row minima. For each row, find the lowest element and subtract it from each element in that row. 2. Step 2: Subtract column minima. 3. Step 3: Cover all zeros with a minimum number of lines. 4. Step 4: Create additional zeros. What is Hungarian method in operation research? The Hungarian method is a computational optimization technique that addresses the assignment problem in polynomial time and foreshadows following primal-dual alternatives. In 1955, Harold Kuhn used the term “Hungarian method” to honour two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry. ### What is Russell’s approximation method? The steps of Russell’s Approximation Method are described as follows: 1. Arrange the value matrix of transport costs and value of the capacity of each source into columns and rows 2. Find the highest cost value for each row and column 3. Subtract each cost value on the columns and rows at their highest cost. 4. ### What is stepping stone method? Thus, the stepping stone method is a procedure for finding the potential of any non-basic variables (empty cells) in terms of the objective function. Through Stepping stone method, we determine that what effect on the transportation cost would be in case one unit is assigned to the empty cell. What are the principles of Hungarian method? The Hungarian Method is based on the principle that if a constant is added to every element of a row and/or a column of cost matrix, the optimum solution of the resulting assignment problem is the same as the original problem and vice versa. ## How do you use Hungarian Algorithm? Hungarian Method 1. Subtract the smallest entry in each row from all the entries of its row. 2. Subtract the smallest entry in each column from all the entries of its column. 3. Cover all zero entries by drawing line through appropiate rows and columns, use minimal number of lines. ## What is optimal assignment? The objective function in assigning different jobs to different machines is to find the optimal assignment (allocation) that will minimize the total cost or time taken to finish all the jobs by the machines. What is Modi method in transportation problem? MODI method is an improvement over stepping stone method. This model studies the minimization of the cost of transporting a commodity from a number of sources to several destinations. The supply at each source and the demand at each destination are known. ### What is row and column reduction in Hungarian method? Summary 1. Step 1 – Subtract the row minimum from each row. 2. Step 2 – Subtract the column minimum from each column from the reduced matrix. 3. Step 3 – Assign one “0” to each row & column. 4. Step 4 – Tick all unassigned row. ### What is north west corner method? The North West corner rule is a technique for calculating an initial feasible solution for a transportation problem. In this method, we must select basic variables from the upper left cell, i.e., the North-west corner cell. Begin typing your search term above and press enter to search. Press ESC to cancel.
# The Midpoint Formula05:30 minutes Video Transcript ## TranscriptThe Midpoint Formula Meet Imke, she’s an avid rock climber and a hobby ornithologist. You know - she’s into birds. Imke wants to see a very rare bird that makes its nest in a cave on Mount Massive in Colorado. So she must plan her climb carefully for the arduous journey. The path to the nest is divided into three segments. In order to ensure a safe ascent, Imke will need to use cams to secure her rope to the mountain at different points along the way. ### Midpoint Formula Imke can use the Midpoint Formula to help her. Take a look at this map of the route. Notice the coordinate grid? We're given the ordered pairs for the location of the starting point, and three lookout points. Okay, let’s tackle the first segment. Imke wants to determine three, equidistant locations along the route to place her cams.The starting point is at position (1, 2) and the first lookout point is at (11,8). Or, being more abstract, the coordinates of the two points are (x1, y1) and (x2, y2).,To find the exact midpoint between these two points, you can use the Midpoint Formula, where you add the two x-values and divide by 2 to get the x-coordinate; as well as adding the two y-values and then again dividing by 2 to get the y-coordinate. ### Example 1 Let's plug in our known ordered pairs (1, 2) and (11, 8) for x1, x2, y1 and y2. Now simplify to find the midpoint of the first segment. Imke should place a cam at the location of the ordered pair (6, 5). Now, to determine where Imke should place another cam, we need to find the midpoint between the midpoint we just found and the starting point.We already know the starting point is located at the ordered pair (1,2) so we plug in the x- and y-values of the starting point and the midpoint, respectively: (1, 2) and (6, 5) and we simplify. Imke will need to place a cam at the ordered pair (3.5, 3.5). Now let's figure where Imke should place her last cam on this segment: We already know the first midpoint cam's coordinates and the coordinates of the first lookout point. So we'll need to use the ordered pairs, (6, 5) and (11, 8) to determine the final midpoint. Do the math and the third cam should be placed at the ordered pair (8.5, 6.5). Great. We've calculated the location of three points that are equidistant from each other. We did this by first diving the segment in half and then dividing the resulting segments in half again. Imke knows exactly where to place her cams to ensure a safe climb. ### Example 2 Imke plans to rest for the night at a location smack dab in the middle of the first and second lookout points. The first lookout point is at (11, 8) and the second is located at (-3, 12). Figuring out where Imke needs to set up camp is easy; we just use the midpoint formula with the coordinates of the lookout points to calculate the midpoint. Be careful with the signs! Sweet dreams at (4,10), Imke! After a rejuvenating rest, Imke is ready to tackle the third and last segment. ### Example 3 The bird’s nest is supposed to be right in the middle of the third segment between lookout point (-3,12) and lookout point (-3, 17). Imke can use the midpoint formula again. But this time she sees a similarity in the coordinates. Since the x-values of both the endpoints are the same, she knows the x-value of the midpoint will be the same too, negative 3. Now she just needs to calculate the midpoint of the y-values by finding the average of the sum of 12 and 17, which is 14.5. Finally, Imke is almost there. Just a little bit further. Oh!! Look at the baby birds!! They are just so adorable! Oh no! It's the momma, and lemme tell you, she's one angry bird!! ## Videos in this Topic Radical Expressions / Equations (8 Videos) ## The Midpoint Formula Exercise ### Would you like to practice what you’ve just learned? Practice problems for this video The Midpoint Formula help you practice and recap your knowledge. • #### Determine the midpoint. ##### Hints Just look at the example beside if both points have the same $x$-coordinate. The $y$-coordinate is the arithmetic middle of $-2$ and $4$, namely $1$. For the given points on the left the midpoint is $(-3,14.5)$. You get the $y$-coordinate as follows: $14.5=\frac{12+17}{2}=\frac{29}2$. In general the midpoint formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)$ ##### Solution The first look out point is $(11,8)$ and the second one $(-3,12)$. Use the midpoint formula with the coordinates of the given points: • We sum the $x$-coordinates and divide the sum by $2$. So we get $\frac{11-3}2=\frac82=4$. • We proceed in the same fashion with the $y$-coordinates to get $\frac{8+12}2=\frac{20}2=10$. Thus we get the midpoint $(4,10)$. • #### Calculate the three points. ##### Hints The midpoint formula for two points $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ is given by $\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)$. You have to use this midpoint formula three times. • Once for the midpoint $M$ between $P_1$ and $P_2$. • Next for the midpoint between $P_1$ and $M$. • Lastly for the midpoint between $M$ and $P_2$. The coordinates of $M$ are whole numbers. ##### Solution Imke wants to divide the path starting with $P_1=(1,2)$ and ending with $P_2=(11,8)$ into three equidistant points. So first we have to find the midpoint $M$ of those two points and after two more midpoints: one between the starting point $P_1$ and the midpoint $M$ and the other between the midpoint $M$ and the ending point $P_2$. Let's start with the midpoint between $P_1$ and $P_2$. We use the midpoint formula, $\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)$, and put the coordinates in $M=\left(\frac{1+11}2,\frac{2+8}2\right)=\left(\frac{12}2,\frac{10}2\right)=\left(6,5\right)$. Next we use the midpoint formula again for the starting point and $M=(6,5)$: $\left(\frac{1+6}2,\frac{2+5}2\right)=\left(\frac{7}2,\frac{7}2\right)=\left(3.5,3.5\right)$. And again for $M=(6,5)$ and the ending point: $\left(\frac{6+11}2,\frac{5+8}2\right)=\left(\frac{17}2,\frac{13}2\right)=\left(8.5,6.5\right)$. • #### Find the midpoints of the given points. ##### Hints In general the midpoint of two points $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ is given by the midpoint formula: $\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)$. Pay attention to the sign. If two points have a coordinate in common, then that coordinate stays the same for the midpoint. The order of adding the coordinates doesn't matter at all. ##### Solution The midpoint formula for two points $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ is given by $\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)$. Let's use this formula for a few examples: For $P_1=(-3,2)$ and $P_2=(2,-3)$, we get the midpoint $M=\left(\frac{-3+2}2,\frac{2+(-3)}2\right)=\left(\frac{-1}2,\frac{-1}2\right)=(-0.5,-0.5)$. The points $P_1=(2,2)$ and $P_2=(2,4)$, we get the midpoint $M=\left(\frac{2+2}2,\frac{2+4}2\right)=\left(\frac{4}2,\frac{6}2\right)=(2,3)$. The midpoint of $P_1=(2,-4)$ and $P_2=(6,8)$ is given by $M=\left(\frac{2+6}2,\frac{-4+8}2\right)=\left(\frac{8}2,\frac{4}2\right)=(4,2)$. The midpoint of the last two points $P_1=(3,3)$ and $P_2=(-7,9)$ is given by $M=\left(\frac{3+(-7)}2,\frac{3+9}2\right)=\left(\frac{-4}2,\frac{12}2\right)=(-2,6)$. • #### Calculate the meeting point of Imke and her friend. ##### Hints You have to add the coordinates and divide by $2$. The $x$-coordinate is a rational number while the $y$-coordinate is a natural number. Here you see the calculation of the midpoint of $(3,7)$ and $(7,8)$. ##### Solution Here we have to determine the midpoint of $(2,7)$ and $(9,15)$. To do this, we use the midpoint formula. This means to get a coordinate of the midpoint we add the corresponding coordinates of those points and divide the resulting sum by $2$: $M=\left(\frac{2+9}2,\frac{7+15}2\right)=\left(\frac{11}2,\frac{22}2\right)=(5.5,11)$. • #### Identify the midpoint formula. ##### Hints If both points have the same $x$- or $y$-coordinate, you get the midpoint as follows: • The common coordinate stays the same for the midpoint. • The coordinate not in common is the arithmetic middle of the given (two) coordinates. For example, the midpoint of $(4,4)$ and $(6,10)$ is $(5,7)$. ##### Solution The midpoint of two given points, $(x_1,y_1)$ and $(x_2,y_2)$, is given by the midpoint formula: $\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)$. To get the $x$-coordinate of the midpoint, add the $x$-coordinates of the two points and divide the sum by $2$. To get the $y$-coordinate, proceed in an similar fashion with the $y$-coordinates of the two points. • #### Find the midpoints of the given points. ##### Hints You can calculate the midpoint as follows: • For the $x$-coordinate, add the $x$-coordinates of the points and divide the sum by $2$. • Proceed in a similar fashion for the $y$-coordinate. Keep in mind that only three points are midpoints. The midpoint of $(-2,3)$ and $(4,3)$ is $M=(1,3)$. ##### Solution We use the midpoint formula for each pair of points: we add the coordinates and divide the sum by $2$: Let's start with $A$ and $B$: $M=\left(\frac{3+7}2,\frac{1+3}2\right)=\left(\frac{10}2,\frac{4}2\right)=(5,2)$. For $A$ and $C$ we get the midpoint $M=\left(\frac{3+1}2,\frac{1+5}2\right)=\left(\frac{4}2,\frac{6}2\right)=(2,3)$. Lastly we determine the midpoint of $B$ and $C$: $M=\left(\frac{1+7}2,\frac{5+3}2\right)=\left(\frac{8}2,\frac{8}2\right)=(4,4)$.
# 005 Sample Final A, Question 16 Question Graph the following, ${\displaystyle -x^{2}+4y^{2}-2x-16y+11=0}$ Foundations: 1) What type of function is this? 2) What can you say about the orientation of the graph? 1) Since both x and y are squared it must be a hyperbola or an ellipse. We can conclude that the graph is a hyperbola since ${\displaystyle x^{2}}$ and ${\displaystyle y^{2}}$ have the different signs, one negative and one positive. 2) Since the ${\displaystyle y^{2}}$ is positive, the hyperbola opens up and down. Solution: Step 1: We start by completing the square twice, once for x and once for y. After completing the squares we end up with ${\displaystyle -(x+1)^{2}+4(y-2)^{2}=4}$ Common Mistake: When completing the square we will end up adding numbers inside of parenthesis. So make sure you add the correct value to this other side. In this case we add -1, and 16 for completing the square with respect to x and y, respectively. Step 2: Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1). From the center mark the two points that are 3 units left, and three units right of the center. Then mark the two points that are 2 units up, and two units down from the center. The four vertices are: ${\displaystyle (-3,-1),(3,-1),(0,1){\text{ and }}(0,-3)}$
# Fraction King Shodor > Interactivate > Lessons > Fraction King ### Abstract Through the combination of imagination, block manipulation, and computer applets, students learn about fractions. Using this variety of tools will help grab the interest of all students, while teaching them about fractions. ### Objectives Upon completion of this lesson, students will: • understand naming fractions • work with finding a fraction of a whole number • be able to compare fractions with different denominators using concrete representations such as manipulatives and pictures ### Student Prerequisites • Technological: Students must be able to: • perform basic mouse manipulations such as point, click and drag • use a browser for experimenting with the activities ### Teacher Preparation • a browser • pencil and paper • 50 blocks, 50 chips, or 50 pieces of similarly sized pieces of paper ### Key Terms denominator In a rational number, the number below the fraction bar that indicates how many parts the whole is divided into. fraction A rational number of the form a/b where a is called the numerator and b is called the denominator numerator The number above the fraction bar that indicates the number of parts of the whole there are in a rational number ### Lesson Outline 1. Focus and Review • Review vocabulary • Tell the students that today they will be learning about fractions 2. Objectives Students will be able to demonstrate their knowledge of fractions through the use of manipulatives and computer applets. 3. Teacher Input • Have the students participate in the King Fraction scenario. • Pass 50 blocks or small square similarly sized pieces of paper to each student. • Students should work in pairs. 4. Guided Practice • Instruct partner A to place 30 of his/her blocks in 5 equal piles and to ignore the rest of his/her blocks. Instruct partner B to place 30 of his/her blocks into 3 equal piles and to ignore the rest. • Once all the students have their blocks grouped properly ask them the total number of blocks each person placed in groups. • Begin asking the students questions like: • What number of blocks is equal to 3/5 of 30? • What number of blocks is equal to 1/5 of 30? • What number of blocks is equal to 4/5 of 30? • Once the students no longer have difficulty with this activity begin asking them questions like: • What is 3/4 of 24? • What is 1/6 of 24? • What is 2/8 of 48? • Have the students arrange their blocks to calculate the answer to each of the above questions. • Walk around the class spot checking the students blocks. • Once the students no longer have difficulty with this activity begin asking them questions like: • Which fraction is larger 3/5 or 8/10? Be sure to mention when the students answer these questions they need to be using the same number of blocks to calculate the fractions from each question. You may also want to work through the first question as a class. For example: Have the students arrange 2 sets of 10 blocks. Have them arrange the first set into 10 equal groups and the other set into 5 equal groups. Finally have them compare 3/5 of 10 to 8/10 of 10 and tell you which one is larger. • Which fraction is larger 2/3 or 3/9? • Which fraction is larger 1/2 or 1/5? • Have the students open the Fraction Finder applet. • Walk the students through 1 or 2 of the computer generated problems. 5. Independent Practice • Have the students work in pairs taking turns with the Fraction Finder applet. • You may or may not want to have the students draw and label each of their computer generated problems so that you can have something written to check. 6. Closure • Review pertinent vocabulary such as: fraction, denominator, and numerator • Review what each of the different parts of a fraction represent. • Review that fractions can be part of 1 whole object or part of a number of objects. ### Alternate Outline • For the more advanced students you may want to have the students challenge each other by setting the boundary fractions using the Bounded Fraction Finder applet. • For the students who may not be able to answer the questions provided by the Fraction Finder applet you may want to have them use the Bounded Fraction Finder applet, so that the lower end students can set their own bounding fractions. For example: 1/4 and 3/4. • You may want to extend this lesson over several days in order to slow its pace.
Courses Courses for Kids Free study material Offline Centres More Store # How many litres of water will a hemispherical tank hold whose diameter is $4.2m$ ? Last updated date: 23rd May 2024 Total views: 334.5k Views today: 5.34k Verified 334.5k+ views Hint: First of all find the volume of the hemispherical tank and since measure of diameter is in meter first convert it in radius and resultant volume would be cubic meter and then at last convert the meter cube in liters. Given that: Diameter, $d = 4.2m$ We know that radius is half of the diameter. $\therefore r = \dfrac{d}{2}$ Place value and simplify – $\therefore r = \dfrac{{4.2}}{2} = 2.1m$ Volume of the hemispherical tank is given by $V = \dfrac{2}{3}\pi {r^3}$ Place values in the above equation – $V = \dfrac{2}{3} \times \left( {\dfrac{{22}}{7}} \right) \times {\left( {2.1} \right)^3}$ Simplify the above equation – $V = \dfrac{2}{3} \times \left( {\dfrac{{22}}{7}} \right) \times \left( {2.1} \right) \times \left( {2.1} \right) \times \left( {2.1} \right)$ Common factors from both the denominator and the numerator cancel each other, $V = 2 \times \left( {22} \right) \times \left( {0.1} \right) \times \left( {2.1} \right) \times \left( {2.1} \right)$ Simplify the above expression finding the product of the terms – $V = 19.404{m^3}$ ….. (A) Convert cubic metre in litres. $1{m^3} = 1000{\text{ }}litres$ Use equation (A) and convert it in litres- $\Rightarrow V = 19.404 \times 1000{\text{ litres}} \\ \Rightarrow V = 19404\;litres \;$ Hence, the hemispherical tank can hold $19404$ liters of water. So, the correct answer is “ $19404$ liters of water.”. Note: The most important here is the standard formula and application and also its simplification. Always remember the basic conversion relations and convert as per the need. Meter cube to litres and centimetre cube in liters. Remember the difference among the units such as one meter and one centimetre both gives us the measure of length. Remember all the formulas for the volume and area for the closed and open figures. Know the difference between the volume and the areas of the figures. Volume is measured in cubic units and the area is measured in the square units.
# 2.1 The rectangular coordinate systems and graphs (Page 5/21) Page 5 / 21 ## Using the midpoint formula When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula . Given the endpoints of a line segment, $\text{\hspace{0.17em}}\left({x}_{1},{y}_{1}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left({x}_{2},{y}_{2}\right),$ the midpoint formula states how to find the coordinates of the midpoint $\text{\hspace{0.17em}}M.$ $M=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$ A graphical view of a midpoint is shown in [link] . Notice that the line segments on either side of the midpoint are congruent. ## Finding the midpoint of the line segment Find the midpoint of the line segment with the endpoints $\text{\hspace{0.17em}}\left(7,-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(9,5\right).$ Use the formula to find the midpoint of the line segment. $\begin{array}{l}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)=\left(\frac{7+9}{2},\frac{-2+5}{2}\right)\hfill \\ \phantom{\rule{6.5em}{0ex}}=\left(8,\frac{3}{2}\right)\hfill \end{array}$ Find the midpoint of the line segment with endpoints $\text{\hspace{0.17em}}\left(-2,-1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-8,6\right).$ $\left(-5,\frac{5}{2}\right)$ ## Finding the center of a circle The diameter of a circle has endpoints $\text{\hspace{0.17em}}\left(-1,-4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,-4\right).\text{\hspace{0.17em}}$ Find the center of the circle. The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point. $\begin{array}{c}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\\ \left(\frac{-1+5}{2},\frac{-4-4}{2}\right)=\left(\frac{4}{2},-\frac{8}{2}\right)=\left(2,-4\right)\end{array}$ Access these online resources for additional instruction and practice with the Cartesian coordinate system. ## Key concepts • We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x- axis and displacement from the y- axis. See [link] . • An equation can be graphed in the plane by creating a table of values and plotting points. See [link] . • Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y= _____. See [link] . • Finding the x- and y- intercepts can define the graph of a line. These are the points where the graph crosses the axes. See [link] . • The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment. See [link] and [link] . • The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x -coordinates and the sum of the y -coordinates of the endpoints by 2. See [link] and [link] . ## Verbal Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain. Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants. Describe the process for finding the x- intercept and the y -intercept of a graph algebraically. Describe in your own words what the y -intercept of a graph is. The y -intercept is the point where the graph crosses the y -axis. When using the distance formula $\text{\hspace{0.17em}}d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}},$ explain the correct order of operations that are to be performed to obtain the correct answer. ## Algebraic For each of the following exercises, find the x -intercept and the y -intercept without graphing. Write the coordinates of each intercept. $y=-3x+6$ The x- intercept is $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and the y -intercept is $\text{\hspace{0.17em}}\left(0,6\right).$ $4y=2x-1$ $3x-2y=6$ The x- intercept is $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and the y -intercept is $\text{\hspace{0.17em}}\left(0,-3\right).$ sin^4+sin^2=1, prove that tan^2-tan^4+1=0 what is the formula used for this question? "Jamal wants to save \$54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?" i don't need help solving it I just need a memory jogger please. Kuz A = P(1 + r/n) ^rt Dale how to solve an expression when equal to zero its a very simple Kavita gave your expression then i solve Kavita Hy guys, I have a problem when it comes on solving equations and expressions, can you help me 😭😭 Thuli Tomorrow its an revision on factorising and Simplifying... Thuli ok sent the quiz kurash send Kavita Hi Masum What is the value of log-1 Masum the value of log1=0 Kavita Log(-1) Masum What is the value of i^i Masum log -1 is 1.36 kurash No Masum no I m right Kavita No sister. Masum no I m right Kavita tan20°×tan30°×tan45°×tan50°×tan60°×tan70° jaldi batao Joju Find the value of x between 0degree and 360 degree which satisfy the equation 3sinx =tanx what is sine? what is the standard form of 1 1×10^0 Akugry Evalute exponential functions 30 Shani The sides of a triangle are three consecutive natural number numbers and it's largest angle is twice the smallest one. determine the sides of a triangle Will be with you shortly Inkoom 3, 4, 5 principle from geo? sounds like a 90 and 2 45's to me that my answer Neese Gaurav prove that [a+b, b+c, c+a]= 2[a b c] can't prove Akugry i can prove [a+b+b+c+c+a]=2[a+b+c] this is simple Akugry hi Stormzy x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0 x exposent4+4x exposent3+8x exposent2+4x+1=0 HERVE How can I solve for a domain and a codomains in a given function? ranges EDWIN Thank you I mean range sir. Oliver proof for set theory don't you know? Inkoom find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad factoring polynomial
# Introduction ### Boolean Algebra Share This is an introduction to Boolean algebra tutorial. Mathematician George Boole invented the Boolean Algebra. This algebra deals with the rules by which logical operations are carried out. So, a digital circuit in this system is represented as input and output symbols and the function of the circuit is expressed as a Boolean relationship between the symbols. ## Terms • Binary Number System • Binary Constant • Binary Variable • Basic Logical Operation • Boolean Function ## Binary Number System It is a number system consisting of only two digits 0 and 1. It is also called the base 2 number system. All the numbers in this system is expressed in 0s and 1s. Example: Decimal number (8)10 is written as (1000)2 in binary. ## Binary Constant A binary value that will not change is called a binary constant. ## Binary Variable A binary variable is a symbolic name assigned to a binary value. Example: A = 1010 Here, A is a variable having binary value 1010. ## Basic Logical Operation Following are the basic logical operation. • OR operation (+) • AND operation (.) • NOT operation (') ## The Logical OR operation (+) Logical OR operation of two boolean variables A and B is written as X = A + B It is also called logical addition. The following table contains the input and output of the logical OR operation. So, if any one of the input is 1 then the output is 1, otherwise it is 0. ## The Logical AND operation (.) Logical AND operation of two boolean variables A and B is written as X = A . B It is also called logical multiplication. The following table contains the input and output of the logical AND operation. So, if both the input is 1 then the output is 1, otherwise it is 0. ## The Logical NOT operation (') Logical NOT operation performs inverse operation and converts 1 into 0â€¨and 0 into 1. The following table contains the input and output of the logical NOT operation. So, if the input is 1 then the output is 0 and if the input is 0 then the output is 1. ## Boolean Function A boolean function is an algebraic expression formed using binary constants, binary variables and basic logical operators. Example: Boolean Function F = A + 1 where, A is a binary variable + is a Basic Logical Operator (OR) and, 1 is a binary constant. Share
# Congruent Segments Definition (I) [color=#000000]The following applet illustrates what it means for segments to be classified as [b]congruent segments. [br][br][/b]Interact with the applet below for a few minutes. Be sure to move the locations of the black points around each time [i]before[/i] you drag the slider! [br][br]Then answer the questions that follow. [/color] [color=#000000][b]Questions: [/b][br][br]1) What transformations have we learned about thus far? List them.[br][br]2) What transformations did you observe here while interacting with this applet above? [br][br]3) How would you describe what it means for segments to be classified as [b]congruent [br] segments [/b]with respect to any one (or more) of the transformations you observed here [br] in this applet? Explain. [/color] # Translations and Rotations ##### Direct Isometries Translations and rotations are [i]direct isometries[/i]. If you read the names of the vertices in cyclic order (A-B-C and A'-B'-C'), both would be read in the counterclockwise (or clockwise) direction. # Are the triangles congruent (part 2)? Use the given measurement tools to establish that the corresponding sides of triangle ABS and triangle A'B'C' are parallel and have equal lengths. Use the given transformation tools to establish that triangle ABC is congruent to triangle A'B'C'. # Congruent Figures: Dynamic Illustration [color=#0000ff]Recall an ISOMETRY is a transformation that preserves distance.[/color] So far, we have already explored the following isometries:[br][br][color=#0000ff]Translation by Vector[br]Rotation about a Point[br]Reflection about a Line[br]Reflection about a Point ( same as 180-degree rotation about a point) [/color][br][br]For a quick refresher about [color=#0000ff]isometries[/color], see this [url=https://www.geogebra.org/m/KFtdRvyv]Messing with Mona applet[/url]. ##### CONGRUENT FIGURES [b]Definition: [br][br]Any two figures are said to be CONGRUENT if and only if one can be mapped perfectly onto the other using [color=#0000ff]any 1 or composition of 2 (or more) ISOMETRIES.[/color][/b][br][br]The applet below dynamically illustrates, [b]by DEFINITION[/b], what it means for any 2 figures (in this case, triangles) to be [b]CONGRUENT.[/b] [br][br]Feel free to move the BIG WHITE VERTICES of either triangle anywhere you'd like at any time. # SAS: Dynamic Proof! [color=#000000]The [/color][b][u][color=#0000ff]SAS Triangle Congruence Theorem[/color][/u][/b][color=#000000] states that [/color][b][color=#000000]if 2 sides [/color][color=#000000]and their [/color][color=#ff00ff]included angle [/color][color=#000000]of one triangle are congruent to 2 sides and their [/color][color=#ff00ff]included angle [/color][color=#000000]of another triangle, then those triangles are congruent. [/color][/b][color=#000000]The applet below uses transformational geometry to dynamically prove this very theorem. [br][br][/color][color=#000000]Interact with this applet below for a few minutes, then answer the questions that follow. [br][/color][color=#000000]As you do, feel free to move the [b]BIG WHITE POINTS[/b] anywhere you'd like on the screen! [/color] [color=#000000][b]Questions:[/b][br][br]1) What geometry transformations did you observe in the applet above? List them. [br]2) What common trait do all these transformations (you listed in your response to (1)) have? [br]3) Go to [url=https://www.geogebra.org/m/d9HrmyAp#chapter/74321]this link[/url] and complete the first 5 exercises in this GeoGebra Book chapter. [/color] # CCSS HS GEOMETRY RESOURCES!!! [color=#0000ff]This worksheet contains [/color][color=#980000][b]links[/b][/color][color=#0000ff] to [/color][b]HUNDREDS [/b][b]of dynamic and engaging geometry resources. [/b][b][color=#0000ff]Each worksheet is mapped to 1 (or more) of the standards listed in the [/color][url=http://www.corestandards.org/Math/Content/HSG/introduction/]HIGH SCHOOL: Geometry[/url] [color=#0000ff]section of the [/color][url=http://www.corestandards.org/Math/]Common Core State Standards Initiative for Mathematics[/url][color=#980000]. [/color][color=#274e13] [br][/color][/b][br][color=#980000][b]LINKS: [br][/b][/color][br][url=https://www.geogebra.org/m/z8nvD94T#chapter/0]CCSS High School: Geometry (Congruence) Volume 1[/url][br][br][url=https://www.geogebra.org/m/munhXmzx#chapter/0]CSSS High School: Geometry (Congruence) Volume 2[/url][br][br][url=https://www.geogebra.org/m/dPqv8ACE#chapter/0]CCSS High School: Geometry (Similarity, Right Triangles, & Trigonometry)[/url] [br][br][url=https://www.geogebra.org/m/C7dutQHh#chapter/0]CCSS High School: Geometry (Circles)[/url][br][br][url=https://www.geogebra.org/m/K2YbdFk8#chapter/0]CCSS High School: Geometry (Expressing Geometric Properties with Equations)[/url][br][url=https://www.geogebra.org/m/xDNjSjEK#chapter/0][br]CCSS High School: Geometry (Geometric Measurement & Dimension)[/url][br][br][url=https://www.geogebra.org/m/pptbYhsy#chapter/0]CCSS High School: Geometry (Modeling with Geometry)[/url][br][br][b][color=#0000ff]*NEW![/color][/b] [url=https://www.geogebra.org/m/NjmEPs3t]CCSS High School: Geometry (Higher Level Enrichment Challenges)[/url] ##### SAMPLE 2: What 2 theorems are dynamically being illustrated below? (Feel free to move the white points wherever you'd like.) [color=#000000]Teachers can use these resources as a powerful means to naturally [br][br][/color][b][color=#0000ff]1) Foster Discovery Learning[br][/color][color=#0000ff]2) Provide Meaningful Remediation[br]3) Differentiate Instruction, &[br]4) Assess students' understanding.[/color][color=#000000] [/color][/b][br][br][color=#000000]Since any curriculum is [/color][b][i][color=#980000]always[/color][/i][/b][color=#000000] a fluid document, these books, too, will continue to remain works in progress.[/color][br][br][b][color=#0000ff]Teachers:[/color][/b][color=#000000] [br]It is my hope that these resources help empower your students to actively (and regularly) discover the fascinating world of mathematics around them. [br][/color][br][b][color=#0000ff]Students:[/color][/b][color=#000000] [br]It is my hope that these resources help you discover & help reinforce mathematics concepts in a way that makes sense to you. [/color] [b][color=#980000]These GeoGebra books display the amazing work from several esteemed members of the GeoGebra community. I am truly humbled and amazed by their talents. These comprehensive resources would not have been possible without their contributions. [br][br]I would like to express a [u]HUGE THANK YOU[/u] to[br][br][/color][/b][url=https://www.geogebra.org/orchiming]Anthony C.M. OR[/url][br][url=https://www.geogebra.org/stevephelps]Steve Phelps[/url][br][url=https://www.geogebra.org/jennifer+silverman]Jennifer Silverman[/url][br][url=https://www.geogebra.org/tedcoe]Dr. Ted Coe[/url][br][url=https://www.geogebra.org/scruz10]Samantha Cruz[/url][br][url=https://www.geogebra.org/tlindy]Terry Lee Lindenmuth[br][/url][url=https://www.geogebra.org/ra%C3%BAl+falc%C3%B3n]Raul Manuel Falcon Ganfornina[/url] [br][url=https://www.geogebra.org/walch+education]Walch Education[/url][br][url=https://www.geogebra.org/edc+in+maine#]EDC in Maine[/url][br][br]For questions, suggestions, and/or comments, feel free to e-mail me at any time. [br]I wish you much success in your journey of teaching and/or learning mathematics! [br][br]Best,[br][br][url=https://www.geogebra.org/tbrzezinski]Tim Brzezinski[br][br][/url][color=#1e84cc]Independent Mathematics Education Consultant ([url=http://www.dynamicmathsolutions.com/]Dynamic Math Solutions[/url])[br][/color][color=#1e84cc]Adjunct Mathematics Instructor at Central Connecticut State University[br]Former High School Mathematics Teacher (15 years) at Berlin High School (CT, USA)[/color][br][br]E-Mail: dynamicmathsolutions@gmail.com [br]Twitter: [url=https://twitter.com/dynamic_math]@dynamic_math[/url][br] # CCSS HS FUNCTIONS RESOURCES!!! This worksheet currently contains [color=#0000ff][b]links[/b][/color] to [b]over 150 [/b][b]DYNAMIC and ENGAGING resources pertaining to FUNCTIONS. [/b][b]Each worksheet is mapped to 1 (or more) of the standards listed in the[color=#0000ff] [url=http://www.corestandards.org/Math/Content/HSF/introduction/]CCSS High School: Functions[/url] [/color][/b]section of the[b] [color=#cc0000][url=http://www.corestandards.org/Math/]Common Core State Standards Initiative for Mathematics[/url].[/color][/b][color=#666666][b] [/b][br][br][/color][color=#980000][b]LINKS: [br][/b][/color][br][url=https://www.geogebra.org/m/k6Dvu9f3]CCSS High School: Functions (Interpreting Functions)[/url][br][br][url=https://www.geogebra.org/m/uTddJKRC]CCSS High School: Functions (Building Functions)[/url][br][br][url=https://www.geogebra.org/m/GMvvpwrm]CCSS High School: Functions (Linear, Quadratic, & Exponential Models)[/url][br][br][url=https://www.geogebra.org/m/aWuJMDas]CCSS High School: Functions (Trigonometric Functions)[/url][br][br][br][color=#000000]Teachers can use these resources as a powerful means to naturally [br][br][/color][b][color=#0000ff]1) Foster Discovery Learning[br][/color][color=#0000ff]2) Provide Meaningful Remediation[br]3) Differentiate Instruction, &[br]4) Assess students' understanding.[/color][color=#000000] [br][br][/color][/b][color=#000000]Since any curriculum is [/color][b][i][color=#0000ff]always[/color][/i][/b][color=#000000] a fluid document, these books, too, will continue to remain works in progress. More items will continue to be added to these volumes. [br][/color] ##### Sample 2 - ODD FUNCTION ILLUSTRATOR: Feel free to move any of the BIG points wherever you'd like. [b][color=#0000ff]Teachers:[/color][/b][color=#000000] [br]It is my hope that these resources help empower your students to actively (and regularly) discover the fascinating world of mathematics around them. [br][br][/color][b][color=#0000ff]Students:[/color][/b][color=#000000] [br]It is my hope that these resources help you discover & help reinforce mathematics concepts in a way that makes sense to you.[br][br][/color][b]I would like to express a HUGE THANK YOU to[/b] [url=https://www.geogebra.org/orchiming][color=#0000ff]Anthony C.M.Or[/color][/url] [b]and[/b] [color=#0000ff][url=https://www.geogebra.org/stevephelps]Steve Phelps[/url], [/color][br][b]whose work also appears in this project. [/b][br][br]For questions, suggestions, and/or comments, feel free to e-mail me at any time. [br]I wish you much success in your journey of teaching and/or learning mathematics! [br][br]Best,[br][br][url=https://www.geogebra.org/tbrzezinski]Tim Brzezinski[br][/url][color=#1e84cc][br]Independent Mathematics Education Consultant ([url=http://www.dynamicmathsolutions.com/]Dynamic Math Solutions[/url])[br][/color][color=#1e84cc]Adjunct Mathematics Instructor at Central Connecticut State University[br]Former High School Mathematics Teacher (15 years) at Berlin High School (CT, USA)[/color][br][br]E-Mail: dynamicmathsolutions@gmail.com [br]Twitter: [url=https://twitter.com/dynamic_math]@dynamic_math[/url]
# Second derivative The second derivative of a quadratic function is constant. In calculus, the second derivative, or the second order derivative, of a function f is the derivative of the derivative of f. Roughly speaking, the second derivative measures how the rate of change of a quantity is itself changing; for example, the second derivative of the position of a vehicle with respect to time is the instantaneous acceleration of the vehicle, or the rate at which the velocity of the vehicle is changing with respect to time. In Leibniz notation: $\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d^2\boldsymbol{x}}{dt^2}$ On the graph of a function, the second derivative corresponds to the curvature or concavity of the graph. The graph of a function with positive second derivative curves upwards, while the graph of a function with negative second derivative curves downwards. ## Second derivative power rule The power rule for the first derivative, of solved down a bit, will produce the second derivative power rule. The rule is given below: $\frac{d^2}{dx^2}[x^n]=n(n-1)x^{(n-2)}=(n^2-n)x^{(n-2)}.$ ## Notation For more details on this topic, see Notation for differentiation. The second derivative of a function $f(x)\!$ is usually denoted $f''(x)\!$. That is: $f'' = (f')'\!$ When using Leibniz's notation for derivatives, the second derivative of a dependent variable y with respect to an independent variable x is written $\frac{d^2y}{dx^2}.$ This notation is derived from the following formula: $\frac{d^2y}{dx^2} \,=\, \frac{d}{dx}\left(\frac{dy}{dx}\right).$ ## Example Given the function $f(x) = x^3,\!$ the derivative of f is the function $f'(x) = 3x^2.\!$ The second derivative of f is the derivative of f′, namely $f''(x) = 6x.\!$ ## Relation to the graph A plot of $f(x) = \sin(2x)$ from $-\pi/4$ to $5\pi/4$. The tangent line is blue where the curve is concave up, green where the curve is concave down, and red at the inflection points (0, $\pi$/2, and $\pi$). ### Concavity The second derivative of a function f measures the concavity of the graph of f. A function whose second derivative is positive will be concave up (sometimes referred to as convex), meaning that the tangent line will lie below the graph of the function. Similarly, a function whose second derivative is negative will be concave down (sometimes called simply “concave”), and its tangent lines will lie above the graph of the function. ### Inflection points Main article: Inflection point If the second derivative of a function changes sign, the graph of the function will switch from concave down to concave up, or vice versa. A point where this occurs is called an inflection point. Assuming the second derivative is continuous, it must take a value of zero at any inflection point, although not every point where the second derivative is zero is necessarily a point of inflection. ### Second derivative test The relation between the second derivative and the graph can be used to test whether a stationary point for a function (i.e. a point where $f'(x)=0\!$) is a local maximum or a local minimum. Specifically, • If $\ f^{\prime\prime}(x) < 0$ then $\ f$ has a local maximum at $\ x$. • If $\ f^{\prime\prime}(x) > 0$ then $\ f$ has a local minimum at $\ x$. • If $\ f^{\prime\prime}(x) = 0$, the second derivative test says nothing about the point $\ x$, a possible inflection point. The reason the second derivative produces these results can be seen by way of a real-world analogy. Consider a vehicle that at first is moving forward at a great velocity, but with a negative acceleration. Clearly the position of the vehicle at the point where the velocity reaches zero will be the maximum distance from the starting position – after this time, the velocity will become negative and the vehicle will reverse. The same is true for the minimum, with a vehicle that at first has a very negative velocity but positive acceleration. ## Limit It is possible to write a single limit for the second derivative: $f''(x) = \lim_{h \to 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}.$ The limit is called the second symmetric derivative.[1][2] Note that the second symmetric derivative may exist even when the (usual) second derivative does not. The expression on the right can be written as a difference quotient of difference quotients: $\frac{f(x+h) - 2f(x) + f(x-h)}{h^2} = \frac{\frac{f(x+h) - f(x)}{h} - \frac{f(x) - f(x-h)}{h}}{h}.$ This limit can be viewed as a continuous version of the second difference for sequences. Please note that the existence of the above limit does not mean that the function $f$ has a second derivative. The limit above just gives a possibility for calculating the second derivative but does not provide a definition. As a counterexample look on the sign function $\sgn(x)$ which is defined through $\sgn(x) = \begin{cases} -1 & \text{if } x < 0, \\ 0 & \text{if } x = 0, \\ 1 & \text{if } x > 0. \end{cases}$ The sign function is not continuous at zero and therefore the second derivative for $x=0$ does not exist. But the above limit exists for $x=0$: \begin{align} \lim_{h \to 0} \frac{\sgn(0+h) - 2\sgn(0) + \sgn(0-h)}{h^2} &= \lim_{h \to 0} \frac{1 - 2\cdot 0 + (-1)}{h^2} \\ &= \lim_{h \to 0} \frac{0}{h^2} \\ &= 0 \end{align} Just as the first derivative is related to linear approximations, the second derivative is related to the best quadratic approximation for a function f. This is the quadratic function whose first and second derivatives are the same as those of f at a given point. The formula for the best quadratic approximation to a function f around the point x = a is $f(x) \approx f(a) + f'(a)(x-a) + \tfrac12 f''(a)(x-a)^2.$ This quadratic approximation is the second-order Taylor polynomial for the function centered at x = a. ## Eigenvalues and eigenvectors of the second derivative For many combinations of boundary conditions explicit formulas for eigenvalues and eigenvectors of the second derivative can be obtained. For example, assuming $x \in [0,L]$ and homogeneous Dirichlet boundary conditions, i.e., $v(0)=v(L)=0$, the eigenvalues are $\lambda_j = -\frac{j^2 \pi^2}{L^2}$ and the corresponding eigenvectors (also called eigenfunctions) are $v_j(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{j \pi x}{L}\right)$. Here, $v''_j(x) = \lambda_j v_j(x), \, j=1,\ldots,\infty.$ For other well-known cases, see the main article eigenvalues and eigenvectors of the second derivative. ## Generalization to higher dimensions ### The Hessian Main article: Hessian matrix The second derivative generalizes to higher dimensions through the notion of second partial derivatives. For a function f:R3 → R, these include the three second-order partials $\frac{\part^2 f}{\part x^2}, \; \frac{\part^2 f}{\part y^2}, \text{ and }\frac{\part^2 f}{\part z^2}$ and the mixed partials $\frac{\part^2 f}{\part x \, \part y}, \; \frac{\part^2 f}{\part x \, \part z}, \text{ and }\frac{\part^2 f}{\part y \, \part z}.$ If the function's image and domain both have a potential, then these fit together into a symmetric matrix known as the Hessian. The eigenvalues of this matrix can be used to implement a multivariable analogue of the second derivative test. (See also the second partial derivative test.) ### The Laplacian Main article: Laplace operator Another common generalization of the second derivative is the Laplacian. This is the differential operator $\nabla^2$ defined by $\nabla^2 f = \frac{\part^2 f}{\part x^2}+\frac{\part^2 f}{\part y^2}+\frac{\part^2 f}{\part z^2}.$ The Laplacian of a function is equal to the divergence of the gradient. ## References 1. ^ A. Zygmund (2002). Trigonometric Series. Cambridge University Press. pp. 22–23. ISBN 978-0-521-89053-3. 2. ^ Thomson, Brian S. (1994). Symmetric Properties of Real Functions. Marcel Dekker. p. 1. ISBN 0-8247-9230-0. ### Print • Anton, Howard; Bivens, Irl; Davis, Stephen (February 2, 2005), Calculus: Early Transcendentals Single and Multivariable (8th ed.), New York: Wiley, ISBN 978-0-471-47244-5 • Apostol, Tom M. (June 1967), Calculus, Vol. 1: One-Variable Calculus with an Introduction to Linear Algebra 1 (2nd ed.), Wiley, ISBN 978-0-471-00005-1 • Apostol, Tom M. (June 1969), Calculus, Vol. 2: Multi-Variable Calculus and Linear Algebra with Applications 1 (2nd ed.), Wiley, ISBN 978-0-471-00007-5 • Eves, Howard (January 2, 1990), An Introduction to the History of Mathematics (6th ed.), Brooks Cole, ISBN 978-0-03-029558-4 • Larson, Ron; Hostetler, Robert P.; Edwards, Bruce H. (February 28, 2006), Calculus: Early Transcendental Functions (4th ed.), Houghton Mifflin Company, ISBN 978-0-618-60624-5 • Spivak, Michael (September 1994), Calculus (3rd ed.), Publish or Perish, ISBN 978-0-914098-89-8 • Stewart, James (December 24, 2002), Calculus (5th ed.), Brooks Cole, ISBN 978-0-534-39339-7 • Thompson, Silvanus P. (September 8, 1998), Calculus Made Easy (Revised, Updated, Expanded ed.), New York: St. Martin's Press, ISBN 978-0-312-18548-0
Courses Courses for Kids Free study material Offline Centres More Last updated date: 29th Nov 2023 Total views: 278.1k Views today: 5.78k # Find the cube root of -27000. Verified 278.1k+ views Hint: For solving this question you should know about the cube root of a number. In this problem we will solve the cube root of this number by long division method because it is more accurate and not so much of a lengthy method and it provides us the exact answer whether our number is fully a cube or not. Complete step-by-step solution: According to our question, we have been asked to find the cube root of -27000. As we know for solving using the long division method, we use just simple steps as follows: Step 1: Divide our given term by which it is asked and it will be a simple division. Step 2: Now we will multiply the quotient with the divider and put the value of that at subtraction of the first digit. Step 3: Then we put the remaining single digit with the result of this subtraction and then we get it as a new number. Step 4: Now we get a new number. Apply the same above steps with it and finally get the last remainder which can’t be divisible by anything else or we get a remainder as 0. Step 5: If you want to check this, then you can do so with the formula: Dividend = Divisor $\times$ Quotient + Remainder Now if we look at our question, $\sqrt[3]{-27000}$ by long division method: \begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,30 \\ & \,\,\,\,\,\,\,\,\,9\left\| \!{\overline {\, \begin{align} & 27000 \\ & 27 \\ \end{align} \,}} \right. \\ & 2700\left\| \!{\overline {\, \begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,0 \\ \end{align} \,}} \right. \\ & \,\,\,\,...\left\| \!{\overline {\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \,}} \right. \\ \end{align} $300\times {{3}^{2}}+30\times 3\times 0+{{0}^{2}}=2700$ And as we know if we make a cube of -30, then we get, \begin{align} & =\left( -30 \right)\left( -30 \right)\left( -30 \right)=\left( -30 \right)\left( 900 \right) \\ & =-27000 \\ \end{align} So, the cube root is -30. Note: While solving this question or all other related questions to the long division method, always try to make it fully divided or make the remainder zero. And if the question is asked for the square root, then it is mandatory for the remainder to be zero. If this will be zero, then maybe your calculations are wrong and you will never get the solution right.
Basic Math \| Basic-2 Math \| Prealgebra \| Workbooks \| Glossary \| Standards \| Site Map \| Help # Memory Challenge: Division by Fives This activity was designed to help you practice division with the number five. You're going to see cards in the challenge that ask you to match an equation with a quotient. For example, "15 ÷ 5" would match the "3." If you get stuck, think back to your multiplication tables. The numbers here are very similar. Good luck and have fun. ## Directions Our NumberNut memory challenges are a good break from the usual quizzes. Your goal is to match two (2) cards on the screen. Once you start the game, all of the cards will display. Click on two cards. If they reveal the same value, they will disappear. If they don't match, you will need to click on another two cards. Once you have found all of the matching pairs, the cards will be gone and the activity will be over. If you want to play again, you will match the same values but the cards will be in different places. It's always a random challenge. RELATED LINKS LESSONS: - NumberNut.com: Division ACTIVITIES: - Memory Challenge: Division by One - Memory Challenge: Division by Two - Memory Challenge: Division by Three - Memory Challenge: Division by Four - Memory Challenge: Division by Five - Memory Challenge: Division by Six - Memory Challenge: Division by Seven - Memory Challenge: Division by Eight - Memory Challenge: Division by Nine - Memory Challenge: Division by Ten - QuickQuiz: Single-Digit Division (No Rem.) - QuickQuiz: Single-Digit Division (Rem.) - QuickQuiz: Single-Digit Division (Rem.) - QuickQuiz: One/Two-Digit Division (<10) - QuickQuiz: One/Two-Digit Division (No Rem.) - QuickQuiz: One/Two-Digit Division (No Rem.) - QuickQuiz: One/Two-Digit Division (Rem.) - Pick-a-Card: Division Word Problems - Overview - Shapes-Colors - Numbers - Addition - Subtraction - Multiplication - Division - Operations - Dates & Times > Activities * The custom search only looks at Rader's sites. Go for site help or a list of mathematics topics at the site map!
# What do you learn in elementary statistics? What Do You Learn in Elementary Statistics? # What Do You Learn in Elementary Statistics? Statistics is a branch of mathematics which deals with collecting, analyzing, and interpreting data. It is present in our daily lives, from marking attendance to calculating household expenses to predicting the weather. Statistics is crucial in our society and economy as it allows us to make informed decisions by understanding the data presented to us. In this article, we will discuss what you learn in elementary statistics. ## FAQs ### What are the topics included in elementary statistics? Elementary statistics is a branch of statistics that covers the basics of statistical analysis. Topics covered in this course include: • Descriptive statistics • Probability • Distributions • Hypothesis testing • Inferences • Correlation • Regression ### What is descriptive statistics? Descriptive statistics are used to describe or summarize a set of data. This branch of statistics can be used to calculate measures of central tendency, such as mean, mode, and median. It can also calculate measures of variability, such as range, variance, and standard deviation. Descriptive statistics can help us gain insight into the data and understand its characteristics, such as distribution and outliers. ### What is probability in statistics? Probability is a branch of mathematics that deals with the likelihood of an event occurring. In statistics, probability is used to predict the likelihood of an outcome based on the available data. Probability helps statisticians make better decisions by understanding the chances of certain events happening. For example, in a medical study, data analysts can predict the probability of a certain disease occurring in certain populations based on the prevalence of risk factors. ### What are distributions in statistics? Distributions are used to represent the frequency of data values in a dataset. A distribution gives an idea of how common or uncommon different values are in the dataset. One of the most common distributions is the normal distribution which has a bell-shaped curve. The normal distribution is used in many statistical analyses because it is very common in nature, such as height, weight, and test scores. ### What is hypothesis testing in statistics? Hypothesis testing is a statistical process used to determine whether a hypothesis is true or false based on a sample of data. This process is used in research studies to compare two or more groups and determine if there is a statistically significant difference between them. Statisticians formulate a hypothesis and test it against sample data to determine if it is accurate or not. If the hypothesis is accepted, it means that the data collected supports the hypothesis, and if it is rejected, then it means that there is insufficient data to support the hypothesis. ### What is correlation in statistics? Correlation is used to measure the strength and direction of a relationship between two variables. A correlation coefficient is used to measure the level of correlation, where a value of 0 means no correlation, and a value of 1 means a perfect positive correlation. A value of -1 means a perfect negative correlation. Understanding the correlation between variables helps researchers and analysts to make informed decisions. ### What is regression in statistics? Regression is used to model the relationship between a dependent variable and one or more independent variables. The regression analysis helps to quantify the relationship between the variables and make predictions. There are several types of regression, such as linear regression, which is used to model a linear relationship between the dependent and independent variables.
### Home > CC1 > Chapter 9 > Lesson 9.2.4 > Problem9-78 9-78. Jenna was building a fence for her new sheep’s pen. She needs a total of $40$ linear feet of lumber. Some of her neighbors have agreed to give her lumber, and she wants to know if she has enough. 1. Neighbor Jim will give her a board that he says is $8\frac { 1 } { 2 }$ feet long. Neighbor Malia will give her two boards that she says are each $126$ inches long. What is the total combined length of lumber Jenna has received? The best way to determine the length Jenna has is to put all the board lengths into the same units. Using feet will make it easier to determine if Jenna has enough at the end. Jim has $8.5$ feet to give. Malia has $2\left(126\right)$ inches. How many feet is $2\left(126\right)$inches? Can you do this calculation in your head? There are $12$ inches in $1$ foot, therefore $12$0 inches is $10$ feet. 1. Neighbor Mike called and offered to donate a board that is $400$ centimeters long. Jenna found the conversion information at right. Help her decide if she has enough lumber to make her pen. $100 \text{ cm} = 1 \text{ m}$ Therefore, $400$ centimeters is equal to $4$ meters. How many feet is $4$ meters? $\left. \begin{array} { c } { 1 \; \text{cm} = 0.3937 \; \text{in.} } \\ { 1 \; \text{m} = 3.281 \; \text{ft} } \\ { 100 \; \text{cm} = 1 \; \text{m} } \end{array} \right.$
# Algorithm and Flowchart to check whether a number is twisted prime or not [1905 views] ### Prime Numbers: A number is said to be a prime number when it has only two factors, that is, when the factors of the number are 1 and the number itself. Example: 2, 3, 11, 17, etc. ### What are Twisted Prime Numbers?: A prime number is said to be ‘twisted prime’ when on reversing the number, the new number is also a prime number. Twisted prime numbers are also known as Emirp numbers. For example: 97 and 79. 97 is a prime number. On reversing 97, we get 79, which is also a prime number. Therefore, both 97 and 79 are twisted prime numbers. Other examples include: 2, 3, 5, 7, 13, 17, 37, etc. Now let us have a look at the algorithm and flowchart to check whether a given number is twisted prime or not. To avoid redundancy, we will use the concept of functions. Here, we will use a function to check whether a number is prime or not. While writing the program, we can call this function as many times we want. #### CheckPrime(num): Assumption: For this algorithm, we assume that the function takes a number ‘num’, to be checked and returns true if the number is prime, otherwise, returns false. Step 1: Start Step 2: Initialize number of factors of num, f = 0 Step 3. Initialize i = 1 Step 4. Repeat until i<=num: 4.1: If num % i == 0: 4.2: Increment f by 1 4.3: Increment i by 1 Step 5: If f = 2, then: 5.1: Return True Step 6: Else: 6.1 Return False Step 7. Stop ### Algorithm to check whether a number is twisted prime or not: Step 1: Start Step 2: Read the number to be checked from the user, say n Step 3: original = CheckPrime(n) Step 4: Initialize rev = 0 Step 5. Repeat WHILE n ? 0: 5.1: Extract the last digit by: d = n % 10 5.2: Calculate reverse by: rev = (rev10) + d 5.3: Remove the last digit from the number: n = n / 10 Step 6: reverse = CheckPrime(rev) Step 7: If original = True AND reverse = True, then: 7.1: Display “Given number is twisted prime” Step 8: Else: 8.1: Display “Given number is not twisted prime” Step 9: Stop #### Explanation: We start this algorithm by taking the number to be checked as input from the user. To check whether the number is twisted prime, we first need to check whether the number is prime, and then reverse the number. To check whether the given number is prime, we count the number of factors of that number. We start a loop which runs from 1 to num. If the number is divisible by the loop variable, then the loop variable is a factor of that number. Therefore, we increment the number of factors by one. Once this check is done, we now have to reverse the number. To reverse the number, we initialize a variable, say rev as 0. This variable will store the reverse. We now start a loop which will run until the number is not equal to zero. We extract the last digit of the number, by doing: d = n % 10. This last digit is added to (rev 10); to maintain the place values. Now, the last digit is removed from the number by doing: n = n /10. Once this loop completes its execution, we get the reverse in rev. Now, we will check whether the reversed number is prime or not, using the same method as before. If both the original and reversed numbers are prime, the given number is twisted prime, else, it is not twisted prime. Let us consider an example: n = 71 71 is a prime number. Reverse of the number, rev = 17 17 is also a prime number. Therefore, 71 is twisted prime.
## March 13, 2007 ### Scribe Post Good day ya'll! This is Jann and I'll be your scribe for today. Today, we learned how to find the Average Value for Functions. We started by getting the average value of a set of numbers. 8 +5 + 7 + 7 + 3 = 6 5 Here's the definition of the Average value of a Function: "Let 'f' be a function which is continuous on the closed interval [a,b]. The average value of 'f' from x=a to x=b is the integral... Q: Where did this formula come from? A: Good question! If we examine a graph of any function, we can say that "a" is the lower limit of the graph and "b" is the upper limit of the graph. This is the formula used to find the width of the intervals."x" is the width of the intervals. If we recall, we divide the graph of a curve into rectangular divisions to find the area under the curve. "x" varies because the more the intervals, the width of the intervals will decrease. "n" is the number of intervals used to divide the curve. If we try to solve for "n", we need to rearrange the formula. According to our Average Value rules, we need to add all the outputs and divide it by the number of intervals used. Therefore... We substituted the formula for n in the denominator. The formula boxed in red is the Riemann Sum. The "E" symbolized the summation of all the outputs in the function, "f". "x" is a constant since the width of an interval never changes. The number of intervals will go to infinity to get the area under the curve. The limit and the summation are symbolized by the integral from [a,b]. Example: Find the average value of the given function on the given interval. f(x) = 1 - 2x [0,3] Using the average value formula for functions, we can get the average value... Then, we talked about the Mean Value Theorem for Integrals. If we recall, the Mean Value Theorem for Derivatives state that: Let f be continuous on a closed interval, [a,b]. There lies a value "c" that is equal to the rate of change of the curve. The Mean Value Theorem for Integrals state that: Let f be continuous on a closed interval [a,b]. Then there exists "c" in the closed interval such that... With all that said and done, we practiced 1 problem. Find the i) average value of "f" on the given interval, ii) "c" such that the average of "f" is equal to the f(c), and iii) sketch the graph of f and a rectangle whose area is the same as the area under "f". f(x) = 2x [0,3] I think that's all we did in class. Next scribe is... err.... I'm not sure who to pick. XD I'll just pick during the next class.
mathleaks.com mathleaks.com Start chapters home Start History history History expand_more Community Community expand_more {{ filterOption.label }} {{ item.displayTitle }} {{ item.subject.displayTitle }} arrow_forward {{ searchError }} search {{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} # Describing Domain and Range Sometimes it can be helpful to describe or analyze the set of all inputs and outputs for which a function is defined. These quantities are called domain and range, respectively. ## Domain The domain, D, is the set of all x-values or inputs for which a function is defined. There are two reasons for numbers to be excluded from the domain: • The number gives a forbidden calculation, such as or • The function describes a specific situation. Suppose, for example, f(x) represents the price of x apples. It does not make sense to consider the cost of -5 apples. Thus, -5 would not be in the domain of f(x). ## Range The range, R, is the set of all y-values or outputs a function gives. Since y depends on x, the domain determines the range. Some functions can result in positive and negative y-values, whereas others cannot. For example, consider For any input value, f(x) will show positive and negative outputs. The range of f(x) is all real numbers. Conversely, g(x) will only yield non-negative outputs, since the square of a number is never negative. Then, the range of g(x) is y0. fullscreen Exercise The table describes the function, f(x)=y. x y 2 1 4 5 6 8 8 9 10 9 Determine the domain and the range of the function. Show Solution Solution The domain is the set of all x-values for which the function is defined. We can find them in the left column. The range is the set of all y-values, and we find them in the right column. Thus, the range of the function is fullscreen Exercise Use the graph to determine the domain and the range of the function. Show Solution Solution Since it's impossible to draw an infinitely large coordinate system, we cannot sketch the entire graph. However, it's reasonable to assume it continues in the same manner beyond the drawn region. In this case, the graph will continue infinitely to the right and infinitely upward. Thus, the domain and the range do not have an upper limit. However, we can determine their lower limits. The graph of the function begins at x=-2, so the domain includes all numbers greater than or equal to -2. This is written as x-2. Similarly, we see that the smallest y-value is 0, so the range includes all numbers greater than or equal to 0. This is written y0. Thus, the domain and range of f(x) are as shown. fullscreen Exercise A theater has a square stage, and each side of the stage floor is 5 meters. A circular rug is to be laid out on the stage floor. Create a function that describes the area of the rug, and determine its domain and range. Show Solution Solution Let's start by making a rough sketch of the situation. We're confined to the stage's measurements because the rug cannot be bigger than the stage. We'll name the radius of the rug r. The area of a circle is given by the formula A(r)=πr2, where r is the radius of the circle. Since radius measures the distance from the circle to its center, the radius of the rug must be greater than 0. Additionally, since the length across the entire circle — the diameter — must not be greater than 5 meters, the maximum value of r is 2.5 meters. This gives the domain D:0<r2.5. To find the range, we determine the minimum and the maximum value of the area using the domain above. If the radius is 0 meters the area is also 0. A(0)=π02=0. To find the maximum value of the area, we'll substitute 2.5 for r. A(r)=πr2 A(2.5)=π2.52 The area of the rug can range between 0 and approximately 19.6 square meters. Thus, the range is R:0<A(r)19.6.
This presentation is the property of its rightful owner. 1 / 9 # 9.9: Introduction to Trigonometry PowerPoint PPT Presentation 9.9: Introduction to Trigonometry. Trig ratios. The three basic trigonometric ratios are sine , cosine , and tangent . Each of these is the ratio of two specific sides of a right triangle . SOH – CAH - TOA. sin A = ____________. cos A = ____________. tan A = ____________. Example 1. 9.9: Introduction to Trigonometry Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - ## 9.9: Introduction to Trigonometry ### Trig ratios • The three basic trigonometric ratios are sine, cosine, and tangent. Each of these is the ratio of two specific sides of a right triangle. • SOH – CAH - TOA sin A = ____________ cos A = ____________ tan A = ____________ ### Example 1 • Find each ratio sin A = cos A = tan A = sin B = cos B = tan B = A 25 7 C B 24 ### Example 2 • Find each ratio sin 30 = cos 30 = tan 30 = sin 60 = cos 60 = tan 60 = 30∘ ### Example 3 • Find each ratio sin 45 = cos 45 = tan 45 = 45∘ ### Example 4 • Given: ▭EXPO, EX = 8, EO = 6 • Find: sin ∠XEP E 8 X 6 O P ### Example 5 • If tan ∠W = √3, find m∠W ### Example 6 • Given: sin ∠S= 5/13 • Find: cos ∠S A 10 U S ### Homework • p. 420 1-4, 8-13
AP Statistics Curriculum 2007 Normal Prob (Difference between revisions) Revision as of 21:48, 31 January 2008 (view source)IvoDinov (Talk \| contribs) (→General Normal Distribution)← Older edit Revision as of 21:53, 31 January 2008 (view source)IvoDinov (Talk \| contribs) (→ General Advance-Placement (AP) Statistics Curriculum - Nonstandard Normal Distribution & Experiments: Finding Probabilities)Newer edit → Line 13: Line 13: * See the special case of [[AP_Statistics_Curriculum_2007#The_Standard_Normal_Distribution \| Standard Normal distribution]] where the mean is set to zero and a variance to one. * See the special case of [[AP_Statistics_Curriculum_2007#The_Standard_Normal_Distribution \| Standard Normal distribution]] where the mean is set to zero and a variance to one. + + * The relation between the Standard and the General Normal distribution is provided by these simple linear transformations (Supposed X denotes General and Z denotes Standard Normal random variables): + : $Z = {X-\mu \over \sigma}$ converts general normal scores to standard (Z) values. + : $X = \mu +Z\sigma$ converts standard scores to general normal values. ===Experiments=== ===Experiments=== General Advance-Placement (AP) Statistics Curriculum - Nonstandard Normal Distribution & Experiments: Finding Probabilities Due to the Central Limit Theorem, the normal distribution is perhaps the most important model for studying various quantitative phenomena. Many numerical measurements (e.g., weight, time, etc.) can be well approximated by the normal distribution. While the mechanisms underlying natural processes may often be unknown, the use of the normal model can be theoretically justified by assuming that many small, independent effects are additively contributing to each observation. General Normal Distribution The (general) normal distribution is a continuous distribution that has similar exact areas, bound in terms of its mean, like the Standard Normal distribution and the x-axis on the symmetric intervals around the origin: • The area: μ − σ < x < μ + σ = 0.8413 − 0.1587 = 0.6826 • The area: μ − 2σ < x < μ + 2σ = 0.9772 − 0.0228 = 0.9544 • The area: μ − 3σ < x < μ + 3σ = 0.9987 − 0.0013 = 0.9974 • General Normal density function $f(x)= {e^{{-(x-\mu)^2} \over 2\sigma^2} \over \sqrt{2 \pi\sigma^2}}.$ • The relation between the Standard and the General Normal distribution is provided by these simple linear transformations (Supposed X denotes General and Z denotes Standard Normal random variables): $Z = {X-\mu \over \sigma}$ converts general normal scores to standard (Z) values. X = μ + Zσ converts standard scores to general normal values. Experiments TBD This Distributions help-page may be useful in understanding SOCR Distribution Applet. How many batteries, from the sample of 100, can we expect to have? • Absolute Voltage > 1? P(X>1) = 0.1586, thus we expect 15-16 batteries to have voltage exceeding 1. • \|Absolute Voltage\| > 1? P(\|X\|>1) = 1- 0.682689=0.3173, thus we expect 31-32 batteries to have absolute voltage exceeding 1. • Voltage < -2? P(X<-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than -2. • Voltage <= -2? P(X<=-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than or equal to -2. • -1.7537 < Voltage < 0.8465? P(-1.7537 < X < 0.8465) = 0.761622, thus we expect 76 batteries to have voltage in this range.
# Maximum Volume of Cylinder Inside Cone This question was seen in the GCSE 1985 paper as well as more recently the New South Wales HSC 2015 Mathematics paper. That kinda proves the recycling of questions over time and across different education systems. This is a question I like due to the purely algebraic nature of calculus used. # Question The diagram shows a cylinder of radius $$x$$ and height $$y$$ inscribed in a cone of radius $$R$$ and height $$H$$, where $$R$$ and $$H$$ are constants. 1. Show that the volume, $$V$$ of the cylinder can be written as: $V = \frac{H}{R}\pi x^2(R-x)$ 2. By considering the inscribed cylinder of maximum volume, show that the volume of any inscribed cylinder does not exceed $$\frac{4}{9}$$ of the volume of the cone. # Solution Part 1 Solution 1. By looking at the triangle with sides $$H$$ and $$R$$, and $$y$$ and $$R-x$$, we can easily see that they are similar as they are equiangular. Corresponding sides in similar triangles are in the same ratio, hence: $\frac{H}{R} = \frac{y}{R-x}$ Rearranging this: $y = \frac{H}{R}(R-x)$ Now the volume of a cylinder is $$V = \pi r^2 h$$ and the radius of the cylinder in the question is $$x$$, the height is $$y$$: \begin{align}V &= \pi x^2 y \\ V & = \pi x^2 (\frac{H}{R}(R-x))\\ V & = \frac{H}{R}\pi x^2 (R-x) \end{align} [collapse] Part 2 Solution 2. I don’t want to use product rule as that can potentially make a mess of things, so it’s best to expand the expression for the volume of the cylinder: \begin{align}V &= H\pi x^2 – \frac{H}{R}\pi x^3\\ \frac{dV}{dx} &= 2H\pi x – \frac{3H\pi}{R} x^2\\ \frac{d^2V}{dx^2}& = 2H\pi – \frac{6H\pi }{R}x\end{align} For maximum volume of the cylinder, $$\frac{dV}{dx} = 0$$, which leads to: \begin{align}2H\pi x – \frac{3H\pi}{R} x^2 & =0\\2x – \frac{3x^2}{R} & = 0\\2-\frac{3x}{R} &= 0\\x &= \frac{2R}{3} \end{align} In the second last line above, we can divide by $$x$$ because $$x=0$$ is not a possible radius of the cylinder. Substitute $$x = \frac{2R}{3}$$ into the second derivative to confirm that it is indeed a maximum: \begin{align}\frac{d^2V}{dx^2} & = 2H\pi-\frac{6H\pi}{ R}\cdot\frac{2 R}{3}\\ & = 2H\pi – 4H \pi\\ & = 2H\pi \\ &< 0\end{align} This implies that the graph of $$V$$ at $$x=\frac{2R}{3}$$ is concave down, and is hence a local maximum. Since it is also the only stationary point in the domain, it is also the absolute maximum. Therefore: \begin{align}\mbox{Max }V & = H\pi \left(\frac{2R}{3}\right)^2-\frac{H}{R}\pi\left(\frac{2R}{3}\right)^3\\&=H\pi\frac{4R^2}{9}-\frac{H}{R}\pi \cdot \frac{8R^3}{27}\\& =H\pi\frac{4R^2}{9}-H\pi\frac{8R^2}{27}\end{align} Now we factorise the volume of the cone out of this expression, i.e. $$\frac{1}{3}\pi R^2H$$: \begin{align}&=\frac{1}{3}\pi R^2 H \left( \frac{4}{3} – \frac{8}{9}\right)\\ &=\frac{1}{3}\pi R^2 H \left( \frac{12}{9} – \frac{8}{9}\right)\\ & = \frac{1}{3}\pi R^2 H \left(\frac{4}{9}\right)\end{align} Since the maximum volume of the cylinder is $$\frac{4}{9}$$ of the volume of the cone, then the volume of any inscribed cylinder does not exceed that amount. [collapse]
General Articles # Fractions ### Prime Numbers A whole number greater than one that is divisible by only 1 and itself. The numbers 2, 3, 5, 37, and 101 are some examples of prime numbers. ### Greatest Common Factor The greatest common factor of two or more whole numbers is the largest whole number that divides each of the numbers. There are two methods of finding the greatest common factor of two numbers. Method 1: List all the factors of each number, then list the common factors and choose the largest one. Example: 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 54: 1, 2, 3, 6, 9, 18, 27, 54 The common factors are: 1, 2, 3, 6, 9, and 18. The greatest common factor is: 18. Method 2: List the prime factors, then multiply the common prime factors. Example: 36 = 2 × 2 × 3 × 3 54 = 2 × 3 × 3 × 3 The common prime factors are 2, 3, and 3. The greatest common factor is 2 × 3 × 3 = 18.. ### Least Common Multiple The least common multiple of two or more nonzero whole numbers is the smallest whole number that is divisible by each of the numbers. There are two common methods for finding the least common multiple of 2 numbers. Method 1: List the multiples of each number, and look for the smallest number that appears in each list. Example: Find the least common multiple of 12 and 42. We list the multiples of each number: 12: 12, 24, 36, 48, 60, 72, 84, ... 42: 42, 84, 126, 168, 190, ... We see that the number 84 is the smallest number that appears in each list. Method 2: Factor each of the numbers into primes. For each different prime number in either of the factorizations, follow these steps: 1. Count the number of times it appears in each of the factorizations. 2. Take the largest of these two counts. 3. Write down that prime number as many times as the count in step 2. To find the least common multiple take the product of all of the prime numbers written down in steps 1, 2, and 3. Example: Find the least common multiple of 24 and 90. First, we find the prime factorization of each number. 24 = 2 × 2 × 2 × 3 90 = 2 × 3 × 3 × 5 The prime numbers 2, 3, and 5 appear in the factorizations. We follow steps 1 through 3 for each of these primes. The number 2 occurs 3 times in the first factorization and 1 time in the second, so we will use three 2's. The number 3 occurs 1 time in the first factorization and 2 times in the second, so we will use two 3's. The number 5 occurs 0 times in the first factorization and 1 time in the second factorization, so we will use one 5. The least common multiple is the product of three 2's, two 3's, and one 5. 2 × 2 × 2 × 3 × 3 × 5 = 360 Example: Find the least common multiple of 14 and 49. First, we find the prime factorization of each number. 14 = 2 × 7 49 = 7 × 7 The prime numbers 2 and 7 appear in the factorizations. We follow steps 1 through 3 for each of these primes. The number 2 occurs 1 times in the first factorization and 0 times in the second, so we will use one 2. The number 7 occurs 1 time in the first factorization and 2 times in the second, so we will use two 7's. The least common multiple is the product of one 2 and two 7's. 2 × 7 × 7 = 98 Examples: Some other least common multiples are listed below. The least common multiple of 12 and 9 is 36. The least common multiple of 6 and 18 is 18. The least common multiple of 2, 3, 4, and 5 is 60. ### What is a Fraction? A fraction is a number that expresses part of a group. Fractions are written in the form or a/b, where a and b are whole numbers, and the number b is not 0. For the purposes of these web pages, we will denote fractions using the notation a/b, though the preferred notation is generally . The number a is called the numerator, and the number b is called the denominator. Examples: The following numbers are all fractions 1/2, 3/7, 6/10, 4/99 Example: The fraction 4/6 represents the shaded portion of the circle below. There are 6 pieces in the group, and 4 of them are shaded. Example: The fraction 3/8 represents the shaded portion of the circle below. There are 8 pieces in the group, and 3 of them are shaded. Example: The fraction 2/3 represents the shaded portion of the circle below. There are 3 pieces in the group, and 2 of them are shaded. ### Equivalent Fractions Equivalent fractions are different fractions which name the same amount. Examples: The fractions 1/2, 2/4, 3/6, 100/200, and 521/1042 are all equivalent fractions. The fractions 3/7, 6/14, and 24/56 are all equivalent fractions. We can test if two fractions are equivalent by cross-multiplying their numerators and denominators. This is also called taking the cross-product. Example: Test if 3/7 and 18/42 are equivalent fractions. The first cross-product is the product of the first numerator and the second denominator: 3 × 42 = 126. The second cross-product is the product of the second numerator and the first denominator: 18 × 7 = 126. Since the cross-products are the same, the fractions are equivalent. Example: Test if 2/4 and 13/20 are equivalent fractions. The first cross-product is the product of the first numerator and the second denominator: 2 × 20 = 40. The second cross-product is the product of the second numerator and the first denominator: 4 × 13 = 52. Since the cross-products are different, the fractions are not equivalent. Since the second cross-product is larger than the first, the second fraction is larger than the first. ### Comparing Fractions 1. To compare fractions with the same denominator, look at their numerators. The larger fraction is the one with the larger numerator. 2. To compare fractions with different denominators, take the cross product. The first cross-product is the product of the first numerator and the second denominator. The second cross-product is the product of the second numerator and the first denominator. Compare the cross products using the following rules: a. If the cross-products are equal, the fractions are equivalent. b. If the first cross product is larger, the first fraction is larger. c. If the second cross product is larger, the second fraction is larger. Example: Compare the fractions 3/7 and 1/2. The first cross-product is the product of the first numerator and the second denominator: 3 × 2 = 6. The second cross-product is the product of the second numerator and the first denominator: 7 × 1 = 7. Since the second cross-product is larger, the second fraction is larger. Example: Compare the fractions 13/20 and 3/5. The first cross-product is the product of the first numerator and the second denominator: 5 × 13 = 65. The second cross-product is the product of the second numerator and the first denominator: 20 × 3 = 60. Since the first cross-product is larger, the first fraction is larger. ### Converting and Reducing Fractions For any fraction, multiplying the numerator and denominator by the same nonzero number gives an equivalent fraction. We can convert one fraction to an equivalent fraction by using this method. Examples: 1/2 = (1 × 3)/(2 × 3) = 3/6 2/3 = (2 × 2)/(3 × 2) = 4/6 3/5 = (3 × 4)/(5 × 4) = 12/20 Another method of converting one fraction to an equivalent fraction is by dividing the numerator and denominator by a common factor of the numerator and denominator. Examples: 20/42 = (20 ÷ 2)/(42 ÷ 2) = 10/21 36/72 = (36 ÷ 3)/(72 ÷ 3) = 12/24 9/27 = (9 ÷ 3)/(27 ÷ 3) = 3/9 When we divide the numerator and denominator of a fraction by their greatest common factor, the resulting fraction is an equivalent fraction in lowest terms. ### Lowest Terms A fraction is in lowest terms when the greatest common factor of its numerator and denominator is 1. There are two methods of reducing a fraction to lowest terms. Method 1: Divide the numerator and denominator by their greatest common factor. 12/30 = (12 ÷ 6)/(30 ÷ 6) = 2/5 Method 2: Divide the numerator and denominator by any common factor. Keep dividing until there are no more common factors. 12/30 = (12 ÷ 2)/(30 ÷ 2) = 6/15 = (6 ÷ 3)/(15 ÷ 3) = 2/5 ### Improper Fractions Improper fractions have numerators that are larger than or equal to their denominators. Examples: 11/4, 5/5, and 13/2 are improper fractions. ### Mixed Numbers Mixed numbers have a whole number part and a fraction part. Examples: are mixed numbers also written as 2 3/4 and 6 1/2. In these web pages, we denote mixed numbers in the form a b/c. ### Converting Mixed Numbers to Improper Fractions To change a mixed number into an improper fraction, multiply the whole number by the denominator and add it to the numerator of the fractional part. Examples: 2 3/4 = ((2 × 4) + 3)/4 =11/4 6 1/2 = ((6 × 2) + 1)/2 = 13/2 ### Converting Improper Fractions to Mixed Numbers To change an improper fraction into a mixed number, divide the numerator by the denominator. The remainder is the numerator of the fractional part. Examples: 11/4 = 11 ÷ 4 = 2 r3 = 2 3/4 13/2 = 13 ÷ 2 = 6 r1 = 6 1/2 ### Writing a Fraction as a Decimal Method 1 - Convert to an equivalent fraction whose denominator is a power of 10, such as 10, 100, 1000, 10000, and so on, then write in decimal form. Examples: 1/4 = (1 × 25)/(4 × 25) = 25/100 = 0.25 3/20 = (3 × 5)/(20 × 5) = 15/100 = 0.15 9/8 = (9 × 125)/(8 × 125) = 1125/1000 = 1.125 Method 2 - Divide the numerator by the denominator. Round to the decimal place asked for, if necessary. Example: 13/4 = 13 ÷ 4 = 3.25 Example: Convert 3/7 to a decimal. Round to the nearest thousandth. We divide one decimal place past the place we need to round to, then round the result. 3/7 = 3 ÷ 7 = 0.4285… which equals 0.429 when rounded to the nearest thousandth. Example: Convert 4/9 to a decimal. Round to the nearest hundredth. We divide one decimal place past the place we need to round to, then round the result. 4/9 = 4 ÷ 9 = 0.4444… which equals 0.44 when rounded to the nearest hundredth. ### Rounding a Fraction to the Nearest Hundredth Divide to the thousandths place. If the last digit is less than 5, drop it. This is particularly useful for converting a fraction to a percent, if we want to convert to the nearest percent. 1/3 = 1 ÷ 3 = 0.333… which rounds to 0.33 If the last digit is 5 or greater, drop it and round up. 2/7 = 2 ÷ 7 = 0.285 which rounds to 0.29 If the fractions have the same denominator, their sum is the sum of the numerators over the denominator. If the fractions have the same denominator, their difference is the difference of the numerators over the denominator. We do not add or subtract the denominators! Reduce if necessary. Examples: 3/8 + 2/8 = 5/8 9/2 - 5/2 = 4/2 = 2 If the fractions have different denominators: 1) First, find the least common denominator. 2) Then write equivalent fractions using this denominator. 3) Add or subtract the fractions. Reduce if necessary. Example: 3/4 + 1/6 = ? The least common denominator is 12. 3/4 + 1/6 = 9/12 + 2/12 = 11/12. Example: 9/10 - 1/2 = ? The least common denominator is 10. 9/10 - 1/2 = 9/10 - 5/10 = 4/10 = 2/5. Example: 2/3 + 2/7 = ? The least common denominator is 21 2/3 + 2/7 = 14/21 + 6/21 = 20/21. ### Adding and Subtracting Mixed Numbers To add or subtract mixed numbers, simply convert the mixed numbers into improper fractions, then add or subtract them as fractions. Example: 9 1/2 + 5 3/4 = ? Converting each number to an improper fraction, we have 9 1/2 = 19/2 and 5 3/4 = 23/4. We want to calculate 19/2 + 23/4. The LCM of 2 and 4 is 4, so 19/2 + 23/4 = 38/4 + 23/4 = (38 + 23)/4 = 61/4. Converting back to a mixed number, we have 61/4 = 15 1/4. The strategy of converting numbers into fractions when adding or subtracting is often useful, even in situations where one of the numbers is whole or a fraction. Example: 13 - 1 1/3 = ? In this situation, we may regard 13 as a mixed number without a fractional part. To convert it into a fraction, we look at the denominator of the fraction 4/3, which is 1 1/3 expressed as an improper fraction. The denominator is 3, and 13 = 39/3. So 13 - 1 1/3 = 39/3 - 4/3 = (39-4)/3 = 35/3, and 35/3 = 11 2/3. Example: 5 1/8 - 2/3 = ? This time, we may regard 2/3 as a mixed number with 0 as its whole part. Converting the first mixed number to an improper fraction, we have 5 1/8 = 41/8. The problem becomes 5 1/8 - 2/3 = 41/8 - 2/3 = 123/24 - 16/24 = (123 - 16)/24 = 107/24. Converting back to a mixed number, we have 107/24 = 4 11/24. Example: 92 + 4/5 = ? This is easy. To express this as a mixed number, just put the whole number and the fraction side by side. The answer is 92 4/5. ### Multiplying Fractions and Whole Numbers To multiply a fraction by a whole number, write the whole number as an improper fraction with a denominator of 1, then multiply as fractions. Example: 8 × 5/21 = ? We can write the number 8 as 8/1. Now we multiply the fractions. 8 × 5/21 = 8/1 × 5/21 = (8 × 5)/(1 × 21) = 40/21 Example: 2/15 × 10 = ? We can write the number 10 as 10/1. Now we multiply the fractions. 2/15 × 10 = 2/15 × 10/1 = (2 × 10)/(15 × 1) = 20/15 = 4/3 ### Multiplying Fractions and Fractions When two fractions are multiplied, the result is a fraction with a numerator that is the product of the fractions' numerators and a denominator that is the product of the fractions' denominators. Example: 4/7 × 5/11 = ? The numerator will be the product of the numerators: 4 × 5, and the denominator will be the product of the denominators: 7 × 11. The answer is (4 × 5)/(7 × 11) = 20/77. Remember that like numbers in the numerator and denominator cancel out. Example: 14/15 × 15/17 = ? Since the 15's in the numerator and denominator cancel, the answer is 14/15 × 15/17 = 14/1 × 1/17 = (14 × 1)/(1 × 17) = 14/17 Example: 4/11 × 22/36 = ? In the solution below, first we cancel the common factor of 11 in the top and bottom of the product, then we cancel the common factor of 4 in the top and bottom of the product. 4/11 × 22/36 = 4/1 × 2/36 = 1/1 × 2/9 = 2/9 ### Multiplying Mixed Numbers To multiply mixed numbers, convert them to improper fractions and multiply. Example: 4 1/5 × 2 2/3 = ?. Converting to improper fractions, we get 4 1/5 = 21/5 and 2 2/3 = 8/3. So the answer is 4 1/5 × 2 2/3 = 21/5 × 8/3 = (21 × 8)/(5 × 3) = 168/15 = 11 3/15. Examples: 3/4 × 1 1/8 = 3/4 × 9/8 = 27/32. 3 × 7 3/4 = 3 × 31/4 = (3 × 31)/4 = 93/4 = 23 1/4. ### Reciprocal The reciprocal of a fraction is obtained by switching its numerator and denominator. To find the reciprocal of a mixed number, first convert the mixed number to an improper fraction, then switch the numerator and denominator of the improper fraction. Notice that when you multiply a fraction and its reciprocal, the product is always 1. Example: Find the reciprocal of 31/75. We switch the numerator and denominator to find the reciprocal: 75/31. Example: Find the reciprocal of 12 1/2. First, convert the mixed number to an improper fraction: 12 1/2 = 25/2. Next, we switch the numerator and denominator to find the reciprocal: 2/25. ### Dividing Fractions To divide a number by a fraction, multiply the number by the reciprocal of the fraction. Examples: 7 ÷ 1/5 = 7 × 5/1 = 7 × 5 = 35 1/5 ÷ 16 = 1/5 ÷ 16/1 = 1/5 × 1/16 = (1 × 1)/(5 × 16) = 1/80 3/5 ÷ 7/12 = 3/5 × 12/7 = (3 × 12)/(5 × 7) = 36/35 or 1 1/35 ### Dividing Mixed Numbers To divide mixed numbers, you should always convert to improper fractions, then multiply the first number by the reciprocal of the second. Examples: 1 1/2 ÷ 3 1/8 = 3/2 ÷ 25/8 = 3/2 × 8/25 = (3 × 8)/(2 × 25) = 24/50 1 ÷ 3 3/5 = 1/1 ÷ 18/5 = 1/1 × 5/18 = (1 × 5)/(1 × 18) = 5/18 3 1/8 ÷ 2 = 25/8 ÷ 2/1 = 25/8 × 1/2 = (25 × 1)/(8 × 2) = 25/16 or 1 9/16. ### Simplifying Complex Fractions A complex fraction is a fraction whose numerator or denominator is also a fraction or mixed number. Example of complex fractions: otherwise written as (1/4)/(2/3), (3/7)/100, 11/(2/3), and (23 1/5)/(2/3). To simplify complex fractions, change the complex fraction into a division problem: divide the numerator by the denominator. The first of these examples becomes (1/4)/(2/3) = 1/4 ÷ 2/3 = 1/4 × 3/2 = 3/8. The second of these becomes (3/7)/100 = 3/7 ÷ 100 = 3/7 × 1/100 = 3/700. The third of these becomes 11/(2/3) = 11 ÷ 2/3 = 11 × 3/2 = 33/2 = 16 1/2. The fourth of these becomes (23 1/5)/(2/3) = 23 1/5 ÷ 2/3 = 116/5 ÷ 2/3 = 116/5 × 3/2 = 174/5 = 34 4/5. ## Repeating Decimals Every fraction can be written as a decimal. For example, 1/3 is 1 divided by 3. If you use a calculator to find 1 ÷ 3, the calculator returns 0.333333... This is called a repeating decimal. To represent the idea that the 3's repeat forever, one uses a horizontal bar (overstrike) as shown below: Example: What is the repeating decimal for 1/7 ? Dividing 7 into 1, we get 0.142857142..., and we see the pattern begin to repeat with the second 1, so . # Using Data and Statistics ## Line Graphs A line graph is a way to summarize how two pieces of information are related and how they vary depending on one another. The numbers along a side of the line graph are called the scale. Example 1: The graph above shows how John's weight varied from the beginning of 1991 to the beginning of 1995. The weight scale runs vertically, while the time scale is on the horizontal axis. Following the gridlines up from the beginning of the years, we see that John's weight was 68 kg in 1991, 70 kg in 1992, 74 kg in 1993, 74 kg in 1994, and 73 kg in 1995. Examining the graph also tells us that John's weight increased during 1991 and 1995, stayed the same during 1991, and fell during 1994. Example 2: This line graph shows the average value of a pickup truck versus the mileage on the truck. When the truck is new, it costs \$14000. The more the truck is driven, the more its value falls according to the curve above. Its value falls \$2000 the first 20000 miles it is driven. When the mileage is 80000, the truck's value is about \$4000. ### Pie Charts A pie chart is a circle graph divided into pieces, each displaying the size of some related piece of information. Pie charts are used to display the sizes of parts that make up some whole. Example 1: The pie chart below shows the ingredients used to make a sausage and mushroom pizza. The fraction of each ingredient by weight is shown in the pie chart below. We see that half of the pizza's weight comes from the crust. The mushrooms make up the smallest amount of the pizza by weight, since the slice corresponding to the mushrooms is smallest. Note that the sum of the decimal sizes of each slice is equal to 1 (the "whole" pizza"). Example 2: The pie chart below shows the ingredients used to make a sausage and mushroom pizza weighing 1.6 kg. This is the same chart as above, except that the labels no longer tell the fraction of the pizza made up by that ingredient, but the actual weight in kg of the ingredient used. The sum of the numbers shown now equals 1.6 kg, the weight of the pizza. The size of each slice is still the same, and shows us the fraction of the pizza made up from that ingredient. To get the fraction of the pizza made up by any ingredient, divide the weight of the ingredient by the weight of the pizza. What fraction of the pizza does the sausage make up? We divide 0.12 kg by 1.6 kg, to get 0.075. This is the same value as in the pie chart in the previous example. Example 3: The pie chart below shows the ingredients used to make a sausage and mushroom pizza. The fraction of each ingredient by weight shown in the pie chart below is now given as a percent. Again, we see that half of the pizza's weight, 50%, comes from the crust. Note that the sum of the percent sizes of each slice is equal to 100%. Graphically, the same information is given, but the data labels are different. Always be aware of how any chart or graph is labeled. Example 4: The pie chart below shows the fractions of dogs in a dog competition in seven different groups of dog breeds. We can see from the chart that 4 times as many dogs competed in the sporting group as in the herding group. We can also see that the two most popular groups of dogs accounted for almost half of the dogs in the competition. Suppose 1000 dogs entered the competition in all. We could figure the number of dogs in any group by multiplying the fraction of dogs in any group by 1000. In the toy group, for example, there were 0.12 × 1000 = 120 dogs in the competition. ### Bar Graphs Bar graphs consist of an axis and a series of labeled horizontal or vertical bars that show different values for each bar. The numbers along a side of the bar graph are called the scale. Example 1: The bar chart below shows the weight in kilograms of some fruit sold one day by a local market. We can see that 52 kg of apples were sold, 40 kg of oranges were sold, and 8 kg of star fruit were sold. Example 2: A double bar graph is similar to a regular bar graph, but gives 2 pieces of information for each item on the vertical axis, rather than just 1. The bar chart below shows the weight in kilograms of some fruit sold on two different days by a local market. This lets us compare the sales of each fruit over a 2 day period, not just the sales of one fruit compared to another. We can see that the sales of star fruit and apples stayed most nearly the same. The sales of oranges increased from day 1 to day 2 by 10 kilograms. The same amount of apples and oranges was sold on the second day. ### Mean The mean of a list of numbers is also called the average. It is found by adding all the numbers in the list and dividing by the number of numbers in the list. Example: Find the mean of 3, 6, 11, and 8. We add all the numbers, and divide by the number of numbers in the list, which is 4. (3 + 6 + 11 + 8) ÷ 4 = 7 So the mean of these four numbers is 7. Example: Find the mean of 11, 11, 4, 10, 11, 7, and 8 to the nearest hundredth. (11 + 11 + 4 + 10 + 11 + 7 + 8) ÷ 7 = 8.857… which to the nearest hundredth rounds to 8.86. ### Median The median of a list of numbers is found by ordering them from least to greatest. If the list has an odd number of numbers, the middle number in this ordering is the median. If there is an even number of numbers, the median is the sum of the two middle numbers, divided by 2. Note that there are always as many numbers greater than or equal to the median in the list as there are less than or equal to the median in the list. Example: The students in Bjorn's class have the following ages: 4, 29, 4, 3, 4, 11, 16, 14, 17, 3. Find the median of their ages. Placed in order, the ages are 3, 3, 4, 4, 4, 11, 14, 16, 17, 29. The number of ages is 10, so the middle numbers are 4 and 11, which are the 5th and 6th entries on the ordered list. The median is the average of these two numbers: (4 + 11)/2 = 15/2 = 7.5 Example: The tallest 7 trees in a park have heights in meters of 41, 60, 47, 42, 44, 42, and 47. Find the median of their heights. Placed in order, the heights are 41, 42, 42, 44, 47, 47, 60. The number of heights is 7, so the middle number is the 4th number. We see that the median is 44. ### Mode The mode in a list of numbers is the number that occurs most often, if there is one. Example: The students in Bjorn's class have the following ages: 5, 9, 1, 3, 4, 6, 6, 6, 7, 3. Find the mode of their ages. The most common number to appear on the list is 6, which appears three times. No other number appears that many times. The mode of their ages is 6. # Decimals, Whole Numbers, and Exponents ## Decimal Numbers Decimal numbers such as 3.762 are used in situations which call for more precision than whole numbers provide. As with whole numbers, a digit in a decimal number has a value which depends on the place of the digit. The places to the left of the decimal point are ones, tens, hundreds, and so on, just as with whole numbers. This table shows the decimal place value for various positions: Note that adding extra zeros to the right of the last decimal digit does not change the value of the decimal number. Place (underlined) Name of Position 1.234567 Ones (units) position 1.234567 Tenths 1.234567 Hundredths 1.234567 Thousandths 1.234567 Ten thousandths 1.234567 Hundred Thousandths 1.234567 Millionths Example: In the number 3.762, the 3 is in the ones place, the 7 is in the tenths place, the 6 is in the hundredths place, and the 2 is in the thousandths place. Example: The number 14.504 is equal to 14.50400, since adding extra zeros to the right of a decimal number does not change its value. ## Whole Number Portion The whole number portion of a decimal number are those digits to the left of the decimal place. Example: In the number 23.65, the whole number portion is 23. In the number 0.024, the whole number portion is 0. ## Expanded Form of a Decimal Number The expanded form of a decimal number is the number written as the sum of its whole number and decimal place values. Example: 3 + 0.7 + 0.06 + 0.002 is the expanded form of the number 3.762. 100 + 3 + 0.06 is the expanded form of the number 103.06. To add decimals, line up the decimal points and then follow the rules for adding or subtracting whole numbers, placing the decimal point in the same column as above. When one number has more decimal places than another, use 0's to give them the same number of decimal places. Example: 76.69 + 51.37 1) Line up the decimal points: 76.69 +51.37 76.69 +51.37 128.06 Example: 12.924 + 3.6 1) Line up the decimal points: 12.924 + 3.600 12.924 + 3.600 16.524 ## Subtracting Decimals To subtract decimals, line up the decimal points and then follow the rules for adding or subtracting whole numbers, placing the decimal point in the same column as above. When one number has more decimal places than another, use 0's to give them the same number of decimal places. Example: 18.2 - 6.008 1) Line up the decimal points. 18.2 - 6.008 2) Add extra 0's, using the fact that 18.2 = 18.200 18.200 - 6.008 3) Subtract. 18.200 - 6.008 12.192 ## Comparing Decimal Numbers Symbols are used to show how the size of one number compares to another. These symbols are < (less than), > (greater than), and = (equals). To compare the size of decimal numbers, we compare the whole number portions first. The larger decimal number is the one with the lager whole number portion. If the whole number parts are both equal, we compare the decimal portions of the numbers. The leftmost decimal digit is the most significant digit. Compare the pairs of digits in each decimal place, starting with the most significant digit until you find a pair that is different. The number with the larger digit is the larger number. Note that the number with the most digits is not necessarily the largest. Example: Compare 1 and 0.002. We begin by comparing the whole number parts: in this case 1>0, 0 being the whole number part of 0.002, and so 1>0.002. Example: Compare 0.402 and 0.412. The numbers 0.402 and 0.412 have the same number of digits, and their whole number parts are both 0. We compare the next most significant digit of each number, the digit in the tenths place, 4 in each case. Since they are equal, we go on to the hundredths place, and in this case, 0<1, so 0.402<0.412. Example: Compare 120.65 and 34.999. Comparing the whole number parts, 120>34, so 120.65>34.999. Example: Compare 12.345 and 12.097. Since the whole number parts are both equal, we compare the decimal portions starting with the tenths digit. Since 3>0, we have 12.345>12.097. Note: Remember that adding extra zeros to the right of a decimal does not change its value: 2.4 = 2.40 = 2.400 = 2.4000. ## Rounding Decimal Numbers To round a number to any decimal place value, we want to find the number with zeros in all of the lower places that is closest in value to the original number. As with whole numbers, we look at the digit to the right of the place we wish to round to. Note: When the digit 5, 6, 7, 8, or 9 appears in the ones place, round up; when the digit 0, 1, 2, 3, or 4 appears in the ones place, round down. Examples: Rounding 1.19 to the nearest tenth gives 1.2 (1.20). Rounding 1.545 to the nearest hundredth gives 1.55. Rounding 0.1024 to the nearest thousandth gives 0.102. Rounding 1.80 to the nearest one gives 2. Rounding 150.090 to the nearest hundred gives 200. Rounding 4499 to the nearest thousand gives 4000. ## Estimating Sums and Differences We can use rounding to get quick estimates on sums and differences of decimal numbers. First round each number to the place value you choose, then add or subtract the rounded numbers to estimate the sum or difference. Example: To estimate the sum 119.36 + 0.56 to the nearest whole number, first round each number to the nearest one, giving us 119 + 1, then add to get 120. ## Multiplying Decimal Numbers Multiplying decimals is just like multiplying whole numbers. The only extra step is to decide how many digits to leave to the right of the decimal point. To do that, add the numbers of digits to the right of the decimal point in both factors. Example: 4.032 × 4 We can multiply 4032 by 4 to get 16128. There are three decimal places in 4.032, so place the decimal three digits from the right: 4.032 × 4 = 16.128 Example: 6.74 × 9.063 We can multiply 674 by 9063 to get 6108462. Then there are 5 decimal places: two in the number 6.74 and three in the number 9.063, so place the decimal five digits from the right: 6.74 × 9.063 = 61.08462. ## Dividing Whole Numbers, with Remainders Example: 1400 ÷ 7.. Since 14 ÷ 7 = 2, and 1400 is 100 times greater than 14, the answer is 2 × 100 = 200. Many problems are similar to the above example, where the answer is easily obtained by adding on or taking off an appropriate number of 0's. Others are more complicated. Example: 4934 ÷ 6. Use long division. So the answer is 822 with a remainder of 2, written 822 R2. To double-check that the answer is correct, multiply the quotient by the divisor and add the remainder: (822 × 6) + 2 = 4932 + 2 = 4934. ## Dividing Whole Numbers, with Decimal Portions Example: Find 32 ÷ 6 to the nearest whole number. 32 ÷ 6 = 5 r2. 6 is the divisor; 2 is the remainder. 2 is closer to 0 than 6, so round down. The answer is 5. ## Dividing Decimals by Whole Numbers To divide a decimal by a whole number, use long division, and just remember to line up the decimal points: Example: 13.44 ÷ 12. When rounding an answer, divide one place further than the place you're rounding to, and round the result. Add 0's to the right of the number being divided, if necessary. Example: 1.0 ÷ 6. Round to the nearest thousandth. To round 0.16666 . . . to the nearest thousandth, we take 4 places to the right of the decimal point and round to 3 places. Here, we round 0.1666 to 0.167, the answer. ## Dividing Decimals by Decimals To divide by a decimal, multiply that decimal by a power of 10 great enough to obtain a whole number. Multiply the dividend by that same power of 10. Then the problem becomes one involving division by a whole number instead of division by a decimal. Example: 0.144 ÷ 0.12 Multiplying the divisor (0.12) and the dividend (0.144) by 100, then dividing, gives the same result. Be aware that some problems are less difficult and do not require this procedure. Example: 6 ÷ 2.00 This is the same as 6 ÷ 2! The answer is 3. ## Exponents (Powers of 2, 3, 4, ...) Exponential notation is useful in situations where the same number is multiplied repeatedly. The notation is often shown as "^" The number being multiplied is called the base, and the exponent tells how many times the base is multiplied by itself. Example: 4 ×4 ×4 ×4 ×4 ×4 = 46 The base in this example is 4, the exponent is 6. We refer to this as four to the sixth power, or four to the power of six, written as 4^6. Examples: 2 ×2 ×2 = 2^3 = 8 1.1"2 = 1.1 × 1.1 = 1.21 0.5^3 = 0.5 × 0.5 × 0.5 = 0.125 10^6 = 10 × 10 × 10 × 10 × 10 × 10 = 1000000 Observe that the base may be a decimal. Special Cases: A number with an exponent of two is referred to as the square of a number. The square of a whole number is known as a perfect square. The numbers 1, 4, 9, 16, and 25 are all perfect squares. A number with an exponent of three is referred to as the cube of a number. The cube of a whole number is known as a perfect cube. The numbers 1, 8, 27, 64, and 125 are all perfect cubes. Note: A number written with an exponent of 1 is the same as the given number. 23^1 = 23. ## Factorial Notation n! The product of the first n whole numbers is written as n!, and is the product 1 × 2 × 3 × 4 × … × (n - 1) × n. Examples: 4! = 1 × 2 × 3 × 4 = 24 11! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 = 39916800 Tricks: When dividing factorials, note that many of the numbers cancel out! Note: The number 0! Is defined to be 1. ## Square Roots The square root of a whole number n is the number r with the property that r × r = n. We write this as . We say that the number n is the square of the number r. Examples: The square root of 9 is 3, since 3 × 3 = 9. The square root of 289 is 17, since 17 × 17 = 289. The square root of 2 is close to 1.41421. We say close to because the digits to the right of the decimal point in the square root of 2 continue forever, without any repeating pattern. Such a number is called an irrational number, meaning that it cannot be written as a fraction. Tricks: Since the square root of a whole number n is the number r with the property that r × r = n, we always have That is, the square of the square root of any number is just the original number. We also have, for any number r that the square root of the square of r is the absolute value of r. We say the absolute value, because the notation actually means the positive square root of n. Example: From the example above, we see that each positive number n actually has 2 numbers r that satisfy r × r = n, one is positive, and the other is negative. Whole Numbers and Their Basic Properties Whole Numbers The whole numbers are the counting numbers and 0. The whole numbers are 0, 1, 2, 3, 4, 5, ... Place Value The position, or place, of a digit in a number written in standard form determines the actual value the digit represents. This table shows the place value for various positions: Place (underlined) Name of Position 1 000 Ones (units) position 1 000 Tens 1 000 Hundreds 1 000 Thousands 1 000 000 Ten thousands 1 000 000 Hundred Thousands 1 000 000 Millions 1 000 000 000 Ten Millions 1 000 000 000 Hundred millions 1 000 000 000 Billions Example: The number 721040 has a 7 in the hundred thousands place, a 2 in the ten thousands place, a one in the thousands place, a 4 in the tens place, and a 0 in both the hundreds and ones place. Expanded Form The expanded form of a number is the sum of its various place values. Example: 9836 = 9000 + 800 + 30 + 6. Ordering Symbols are used to show how the size of one number compares to another. These symbols are < (less than), > (greater than), and = (equals.) For example, since 2 is smaller than 4 and 4 is larger than 2, we can write: 2 < 4, which says the same as 4 > 2 and of course, 4 = 4. To compare two whole numbers, first put them in standard form. The one with more digits is greater than the other. If they have the same number of digits, compare the most significant digits (the leftmost digit of each number). The one having the larger significant digit is greater than the other. If the most significant digits are the same, compare the next pair of digits from the left. Repeat this until the pair of digits is different. The number with the larger digit is greater than the other. Example: 402 has more digits than 42, so 402 > 42. Example: 402 and 412 have the same number of digits. We compare the leftmost digit of each number: 4 in each case. Moving to the right, we compare the next two numbers: 0 and 1. Since 0 < 1, 402 < 412. Rounding Whole Numbers To round to the nearest ten means to find the closest number having all zeros to the right of the tens place. Note: when the digit 5, 6, 7, 8, or 9 appears in the ones place, round up; when the digit 0, 1, 2, 3, or 4 appears in the ones place, round down. Examples: Rounding 119 to the nearest ten gives 120. Rounding 155 to the nearest ten gives 160. Rounding 102 to the nearest ten gives 100. Similarly, to round a number to any place value, we find the number with zeros in all of the places to the right of the place value being rounded to that is closest in value to the original number. Examples: Rounding 180 to the nearest hundred gives 200. Rounding 150090 to the nearest hundred thousand gives 200000. Rounding 1234 to the nearest thousand gives 1000. Rounding is useful in making estimates of sums, differences, etc. Example: To estimate the sum 119360 + 500 to the nearest thousand, first round each number in the sum, resulting in a new sum of 119000 + 1000.. Then add to get the estimate of 120000. Divisibility Tests There are many quick ways of telling whether or not a whole number is divisible by certain basic whole numbers. These can be useful tricks, especially for large numbers. Commutative Property of Addition and Multiplication Addition and Multiplication are commutative: switching the order of two numbers being added or multiplied does not change the result. Examples: 100 + 8 = 8 + 100 100 × 8 = 8 × 100 Associative Property Addition and multiplication are associative: the order that numbers are grouped in addition and multiplication does not affect the result. Examples: (2 + 10) + 6 = 2 + (10 + 6) = 18 2 × (10 × 6) = (2 × 10) × 6 =120 Distributive Property The distributive property of multiplication over addition: multiplication may be distributed over addition. Examples: 10 × (50 + 3) = (10 × 50) + (10 × 3) 3 × (12+99) = (3 × 12) + (3 × 99) Adding 0 to a number leaves it unchanged. We call 0 the additive identity. Example: 88 + 0 = 88 The Zero Property of Multiplication Multiplying any number by 0 gives 0. Example: 88 × 0 = 0 0 × 1003 = 0 The Multiplicative Identity We call 1 the multiplicative identity. Multiplying any number by 1 leaves the number unchanged. Example: 88 × 1 = 88 Order of Operations The order of operations for complicated calculations is as follows: 1) Perform operations within parentheses. 2) Multiply and divide, whichever comes first, from left to right. 3) Add and subtract, whichever comes first, from left to right. Example: 1 + 20 × (6 + 2) ÷ 2 = 1 + 20 × 8 ÷ 2 = 1 + 160 ÷ 2 = 1 + 80 = 81. Divisibility by 2 A whole number is divisible by 2 if the digit in its units position is even, (either 0, 2, 4, 6, or 8). Examples: The number 84 is divisible by 2 since the digit in the units position is 4, which is even. The number 333336 is divisible by 2 since the digit in the units position is 6, which is even. The number 1297000 is divisible by 2 since the digit in the units position is 0, which is even. Divisibility by 3 A whole number is divisible by 3 if the sum of all its digits is divisible by 3. Examples: The number 177 is divisible by three, since the sum of its digits is 15, which is divisible by 3. The number 8882151 is divisible by three, since the sum of its digits is 33, which is divisible by 3. The number 162345 is divisible by three, since the sum of its digits is 21, which is divisible by 3. If a number is not divisible by 3, the remainder when it is divided by 3 is the same as the remainder when the sum of its digits is divided by 3. Examples: The number 3248 is not divisible by 3, since the sum of its digits is 17, which is not divisible by 3. When 3248 is divided by 3, the remainder is 2, since when 17, the sum of its digits, is divided by three, the remainder is 2. The number 172345 is not divisible by 3, since the sum of its digits is 22, which is not divisible by 3. When 172345 is divided by 3, the remainder is 1, since when 22, the sum of its digits, is divided by three, the remainder is 1. Divisibility by 4 A whole number is divisible by 4 if the number formed by the last two digits is divisible by 4. Examples: The number 3124 is divisible by 4 since the number formed by its last two digits, 24, is divisible by 4. The number 1333336 is divisible by 4 since the number formed by its last two digits, 36, is divisible by 4. The number 1297000 is divisible by 4 since the number formed by its last two digits, 0, is divisible by 4. If a number is not divisible by 4, the remainder when the number is divided by 4 is the same as the remainder when the last two digits are divided by 4. Example: The number 172345 is not divisible by 4, since the number formed by its last two digits, 45, is not divisible by 4. When 172345 is divided by 4, the remainder is 1, since when 45 is divided by 4, the remainder is 1. Divisibility by 5 A whole number is divisible by 5 if the digit in its units position is 0 or 5. Examples: The number 95 is divisible by 5 since the last digit is 5. The number 343370 is divisible by 5 since the last digit is 0. The number 129700195 is divisible by 5 since the last digit is 5. If a number is not divisible by 5, the remainder when it is divided by 5 is the same as the remainder when the last digit is divided by 5. Example: The number 145632 is not divisible by 5, since the last digit is 2. When 145632 is divided by 5, the remainder is 2, since 2 divided by 5 is 0 with a remainder of 2. The number 7332899 is not divisible by 5, since the last digit is 9. When 7332899 is divided by 5, the remainder is 4, since 9 divided by 5 is 1 with a remainder of 4. Divisibility by 6 A number is divisible by 6 if it is divisible by 2 and divisible by 3. We can use each of the divisibility tests to check if a number is divisible by 6: its units digit is even and the sum of its digits is divisible by 3. Examples: The number 714558 is divisible by 6, since its units digit is even, and the sum of its digits is 30, which is divisible by 3. The number 297663 is not divisible by 6, since its units digit is not even. The number 367942 is not divisible by 6, since it is not divisible by 3. The sum of its digits is 31, which is not divisible by 3, so the number 367942 is not divisible by 3. Divisibility by 8 A whole number is divisible by 8 if the number formed by the last three digits is divisible by 8. Examples: The number 88863024 is divisible by 8 since the number formed by its last three digits, 24, is divisible by 8. The number 17723000 is divisible by 8 since the number formed by its last three digits, 0, is divisible by 8. The number 339122483984 is divisible by 8 since the number formed by its last three digits, 984, is divisible by 8. If a number is not divisible by 8, the remainder when the number is divided by 8 is the same as the remainder when the last three digits are divided by 8. Example: The number 172045 is not divisible by 8, since the number formed by its last three digits, 45, is not divisible by 8. When 172345 is divided by 8, the remainder is 5, since when 45 is divided by 8, the remainder is 5. Divisibility by 9 A whole number is divisible by 9 if the sum of all its digits is divisible by 9. Examples: The number 1737 is divisible by nine, since the sum of its digits is 18, which is divisible by 9. The number 8882451 is divisible by nine, since the sum of its digits is 36, which is divisible by 9. The number 762345 is divisible by nine, since the sum of its digits is 27, which is divisible by 9. If a number is not divisible by 9, the remainder when it is divided by 9 is the same as the remainder when the sum of its digits is divided by 9. Examples: The number 3248 is not divisible by 9, since the sum of its digits is 17, which is not divisible by 9. When 3248 is divided by 9, the remainder is 8, since when 17, the sum of its digits, is divided by 9, the remainder is 8. The number 172345 is not divisible by 9, since the sum of its digits is 22, which is not divisible by 9. When 172345 is divided by 9, the remainder is 4, since when 22, the sum of its digits, is divided by 9, the remainder is 4. Divisibility by 10 A whole number is divisible by 10 if the digit in its units position is 0. Examples: The number 1229570 is divisible by 10 since the last digit is 0. The number 676767000 is divisible by 10 since the last digit is 0. The number 129700190 is divisible by 10 since the last digit is 0. If a number is not divisible by 10, the remainder when it is divided by 10 is the same as the units digit. Examples: The number 145632 is not divisible by 10, since the last digit is 2. When 145632 is divided by 10, the remainder is 2, since the units digit is 2. The number 7332899 is not divisible by 10, since the last digit is 9. When 7332899 is divided by 10, the remainder is 4, since the units digit is 9. Divisibility by 11 Starting with the units digit, add every other digit and remember this number. Form a new number by adding the digits that remain. If the difference between these two numbers is divisible by 11, then the original number is divisible by 11. Examples: Is the number 824472 divisible by 11? Starting with the units digit, add every other number:2 + 4 + 2 = 8. Then add the remaining numbers: 7 + 4 + 8 = 19. Since the difference between these two sums is 11, which is divisible by 11, 824472 is divisible by 11. Is the number 49137 divisible by 11? Starting with the units digit, add every other number:7 + 1 + 4 = 12. Then add the remaining numbers: 3 + 9 = 12. Since the difference between these two sums is 0, which is divisible by 11, 49137 is divisible by 11. Is the number 16370706 divisible by 11? Starting with the units digit, add every other number:6 + 7 + 7 + 6 = 26. Then add the remaining numbers: 0 + 0 + 3 + 1=4. Since the difference between these two sums is 22, which is divisible by 11, 16370706 is divisible by 11. Divisibility by 12 A number is divisible by 12 if it is divisible by 4 and divisible by 3. We can use each of the divisibility tests to check if a number is divisible by 12: its last two digits are divisible by 4 and the sum of its digits is divisible by 3. Examples: The number 724560 is divisible by 12, since the number formed by its last two digits, 60, is divisible by 4, and the sum of its digits is 30, which is divisible by 3. The number 36297414 is not divisible by 12, since the number formed by its last two digits, 14, is not divisible by 4. The number 367744 is not divisible by 12, since it is not divisible by 3. The sum of its digits is 29, which is not divisible by 3, so the number 367942 is not divisible by 3. Divisibility by 15 A number is divisible by 15 if it is divisible by 3 and divisible by 5. We can use each of the divisibility tests to check if a number is divisible by 15: its units digit is 0 or 5, and the sum of its digits is divisible by 3. Example: The number 7145580 is divisible by 15, since its units digit is even, and the sum of its digits is 30, which is divisible by 3. Divisibility by 16 A whole number is divisible by 16 if the number formed by the last four digits is divisible by 16. Examples: The number 898630032 is divisible by 16 since the number formed by its last four digits, 32, is divisible by 16. The number 1772300000 is divisible by 16 since the number formed by its last four digits, 0, is divisible by 16. The number 339122481296 is divisible by 16 since the number formed by its last four digits, 1296, is divisible by 16. If a number is not divisible by 16, the remainder when the number is divided by 16 is the same as the remainder when the last four digits are divided by 16. Example: The number 172411045 is not divisible by 16, since the number formed by its last four digits, 1045, is not divisible by 16. When 172411045 is divided by 16, the remainder is 5, since when 1045 is divided by 16, the remainder is 5. Divisibility by 18 A number is divisible by 18 if it is divisible by 2 and divisible by 9. We can use each of the divisibility tests to check if a number is divisible by 18: its units digit is even and the sum of its digits is divisible by 9. Examples: The number 7145586 is divisible by 18, since its units digit is even, and the sum of its digits is 36, which is divisible by 9. The number 2976633 is not divisible by 18, since its units digit is not even. The number 367942 is not divisible by 18, since it is not divisible by 9. The sum of its digits is 31, which is not divisible by 9, so the number 367942 is not divisible by 9. Divisibility by 20 A number is divisible by 20 if its units digit is 0, and its tens digit is even. In other words, the last two digits form one of the numbers 0, 20, 40, 60, or 80. Examples: The number 3351002760 is divisible by 20, since the number formed by its last two digits is 60. The number 802199730000 is divisible by 20, since the number formed by its last two digits is 0. Divisibility by 22 A number is divisible by 22 if it is divisible by the numbers 2 and 11. We can use each of the divisibility tests to check if a number is divisible by 22: its units digit is even, and the difference between the sums of every other digit is divisible by 11. Example: Is the number 117524 divisible by 22? The units digit is even, so it is divisible by 2. The two sums of every other digit are 4 + 5 + 1 = 10 and 2 + 7 + 1 = 10, which have a difference of 0. Since 0 is divisible by 11, 117524 is divisible by 11. Thus, 117524 is divisible by 22, since it is divisible by both 2 and 11. Divisibility by 25 A number is divisible by 25 if the number formed by the last two digits is any of 0, 25, 50, or 75 (the number formed by its last two digits is divisible by 25). Examples: The number 73224050 is divisible by 25, since its last two digits form the number 50. The number 1008922200 is divisible by 25, since its last two digits form the number 0. This is a help resource for 4th through 8th grades. We have just redesigned and reformatted these pages to enable faster loading and display times, but some pages still load slowly. Different browsers display graphics differently, so some math symbols may not appear to line up perfectly in some sentences. Our educational CD-ROMs contain the complete, searchable, indexed help facility shown below, along with the helpful problem-solving tips section. We hope you find this a helpful resource. Whole numbers and their basic properties Decimals, whole numbers, and exponents Using data and statistics Fractions Geometry Ratio and proportion Percent and probability Integers Metric units and measurement Introduction to algebra Positive and negative numbers # MATHEMATICS LEAGUE 4TH, 5TH, 6TH, 7TH, 8TH GRADE and ALGEBRA COURSE 1 CONTEST ORGANIZATION QUESTIONS (TIME LIMITS AND TOPICS) Each contest is a 30-minute multiple-choice test. Questions may involve any topic appropriate to the grade level of the contest. If, for any reason, a question must be dropped, no replacement will be made. CONTEST COPIES Each school will receive 30 copies of each contest in which they are participating. Schools requiring additional copies of any contest are permitted, on the day of the contest, to make as many additional copies as are required. A separate registration form should be submitted from each participating school. CONTEST MATERIALS PACKAGE You should receive materials for Algebra Course 1 and Grades 4 & 5 by April 1; for Grade 6 by the last Tuesday in February; for Grades 7 & 8 by the next to the last Tuesday in January. If any contest materials have not been received by these dates, the League should be phoned immediately at 1-201-568-6328. Each contest materials package includes 30 copies of the contest and a solution key for the contest. A school needing additional copies of any contest is permitted, on the day of the contest, to make as many additional copies of the contest as are required. Special arrangements for blind or other handicapped students, or for non-English-speaking students, may be made by any school. CONTEST AWARDS GRADES 4 AND 5 and ALGEBRA COURSE 1 In each school, the highest scoring student on each contest receives a book award. Other high scoring students in each school receive certificates of merit. GRADES 6, 7, AND 8 In each school, a certificate of merit is awarded to the highest scoring student on each contest. For each contest, awards are given to the two schools with the highest total scores in the League and also to the two students with the highest total scores in the League. For each contest, additional regional awards are given to the highest scoring school in each region (which may be a county, province, state or other grouping as determined by the League). Counties/Provinces/States with fewer than fifteen participating schools are grouped together into regions for the purposes of awards. No school may win both a regional award and an overall League award on any one grade level in the same school year. The League reserves the right to break ties based upon performance on selected questions, or, at its option, to issue duplicate awards. CONTEST LOCATION Each school may administer the contests on its own premises or any other suitable site. 6TH, 7TH, AND 8TH GRADE CONTEST PROCEDURES (The 4th and 5th Grade and Algebra Course 1 Contests are non-competitive; these procedures do not apply.) CONTEST DATE Except in unusual circumstances, the contests must be held on the scheduled date. In the event of school closings, special testing days, school trips or other administrative functions, severely inclement weather, or similar disruptions of the normal school day, permission is granted to conduct the contest on a proximate school day. STARTING TIME Each contest may be held, on the scheduled date, at any time convenient for the school. All students officially participating in a contest within the same school should take that contest at the same time. Scores of students taking the contest at any later time should not be included on the score report filed with the League. PROCTORING Each contest must be actively proctored at all times by a teacher. Neither the proctor nor anyone else may interpret any question to any student during the contest. ELIGIBILITY Only students officially registered in the same accredited school of record may participate on that school's team. A student may take only a contest designed for a grade the student has not completed or a higher grade (regardless of the math course in which the student is enrolled.) Students taking the 4th or 5th grade contest may also take the 6th, 7th, or 8th grade contest. Students taking the 6th, 7th, or 8th grade contest may take only one such contest (but they may take the Algebra Course 1 Contest). On each contest, all official participants must take the contest in school at the exact same time. A student taking the contest at a later time or period or on a later day must not be included on the score report. Students absent on the contest day may take the contest but must not be listed on the score report. MATERIALS ALLOWED Only plain paper, pencil or pen, and any calculator without a QWERTY keyboard, may be used by the participants. No graph paper, compasses, straight edges, rulers, printed mathematical tables, or other devices shall be allowed, except where special arrangements have been made for handicapped students or when dictionaries are made accessible to non-English-speaking students. START OF THE CONTEST Each contestant should be given a copy of the contest and should complete the information requested on the cover page of the contest. Answers submitted for each question must appear in the appropriate space in the answer column. Answers written elsewhere will receive no credit. After the signal to begin is given by the proctor, the timing of the contest will begin. TIME WARNINGS Warnings that "fifteen minutes remain," that "five minutes remain," and that "one minute remains" should be given at the appropriate times. No other warnings or announcements (relative to the contest) should be made to any contestant during the contest. MARKING THE ANSWERS At the end of the contest, the question papers should be collected by the advisor. The advisor should then open the sealed envelope containing the solution key and should mark each paper, awarding 1 point for each correct answer. All papers should be marked exactly according to the solution key. If you wish to appeal an answer, please follow the appeals procedure, but score your students' papers according to the official answer key. The League has the option to disqualify any school that submits a mismarked paper. SUBMITTING CONTEST RESULTS ONLINE The advisor should score the contests. For each contest, the advisor should submit the scores of the school's participants to the League's Internet Score Report Center. The school score for each contest will be the sum of the scores of the five highest scoring participants from the same school of record. The score report must be submitted to our Internet Score Report Center by Friday of the contest week.Student papers may be returned to the students, except that papers with scores above 30 must be held until June 1. APPEALS PROCEDURE Appeals will be awarded only on the basis of an incorrect official answer or a correct alternative interpretation of a question. Detailed explanations of alternative interpretations should be made in the comments section when filing the score reports. Appeals filed with the League must include the names of all students listed on the score report for whom an appeal is being filed. If you disagree with an official answer, file an appeal. You must use only the official solution key in grading student papers. AUTHENTICATION OF RESULTS League policy is to authenticate scores and eligibility of participants from schools winning major awards. The League reserves the right to authenticate scores and/or to reexamine students or validate student solutions before granting official status to any score. Schools must keep all papers with scores above 30 until June 1. The League has the option to disqualify any school that submits a mismarked paper. A school disqualified for cause on any contest is ineligible for awards in any League contest. Directors: Dan Flegler / Phone: 1-201-568-6328, Fax: 1-201-816-0125 Steve Conrad / Phone: 1-516-365-5656, Fax: 1-516-365-5657 # How to get your school involved in Math League Contests The Math League was formed in 1977 by Steven R. Conrad and Daniel Flegler. In October 1985, Steven R. Conrad and Daniel Flegler were both honored by President Ronald Reagan as recipients of Presidential awards for "Excellence in Mathematics Teaching." Mr. Conrad was the winner from New York, and Mr. Flegler was the winner from New Jersey. Mr. Flegler was the 1977 recipient of Princeton University's award for "Distinguished Secondary School Teaching." Mr. Conrad and Mr. Flegler have been preparing contests for math students across North America since 1977. They have co-authored 18 books. Steven R. Conrad taught at Roslyn High School, Roslyn, New York, from 1980 to 1998. Prior to that, he taught at Benjamin Cardozo High School in Bayside, New York and Francis Lewis High School in Flushing, New York. He began his undergraduate education at Washington University, St. Louis. He received a B.S. from Queens College and an M.S. from Yeshiva University. He has done additional graduate study at the University of San Francisco, Fordham University, and St. John's University, where he earned a certificate in School Administration. More than 60 of Mr. Conrad's students have been named to the honors group of the Intel National Science Talent Search for mathematics papers they have written. Six of them have finished in the top 10 nationwide. His sons are both math professors. He was Problem Editor for The Mathematics Student Journal (an official journal of the National Council of Teachers of Mathematics) from 1972 to 1978 and was an associate editor of The New York State Mathematics Teachers' Journal for 3 years. He has contributed problems to many journals, and has had articles published in Mathematics Magazine, The Mathematics Teacher, , American Mathematical Monthly, and Crux Mathematicorum. He is a reviewer for The Mathematics Teacher. For 4 years, Mr. Conrad was the author for the American Regions Mathematics League. He has also been the contest author for the Association of Mathematics Teachers of New Jersey, and director of contests for New York City. Currently, he is the author for contests sponsored by Bergen County, New Jersey, Suffolk County, New York, and Fairfax County, Virginia. He has also authored contests for Montgomery County, Maryland and Nassau County, New York. County, New York. Mr. Conrad served for six years as a member of the committee which developed the SAT II for the College Board. Daniel Flegler taught at Waldwick High School, Waldwick, New Jersey, from 1965 to 1991, where he served as department chairman for 11 years. Mr. Flegler received his B.A. from Brown University and his M.A. from Columbia University. He has done additional graduate work at Fairleigh Dickinson University, the University of Iowa, and Rutgers University, from which he received his certificate in educational administration. More than 15 of Mr. Flegler's students have been named to the honors group of the Intel National Science Talent Search for mathematics papers they have written. Three of them have finished in the top 40 nationwide. From 1972-78, Mr. Flegler served on the Executive Committee of the Association of Mathematics Teachers of New Jersey, and was assistant editor of the New Jersey Mathematics Teacher. In addition, he has written contest problems for both New York City and Nassau County, New York. He has also directed the contests for Bergen County, New Jersey. ### Books and Copying May I make copies of pages from your books of past contests to use with my students? If you have registered for this year's contest, you may duplicate pages from our books for use with your class. If you are not registered for this year's contest, then you may not duplicate pages from the books. Do the Volume 6 books contain the contests that appeared in previous volumes? Later book volumes do not contain the contests published in previous contest problem books. You can view a complete listing of the contests published in each volume here. ### Math Help I have a math question. Can I e-mail you for help on my math homework from Math League? We are happy to answer letters regarding questions that appear on our contests. You are also welcome to browse our online Help Reference, which contains math reference information, with sample questions and solutions for grades 4-8. We do not respond to math questions that do not appear on our contests. If you need help with homework or another question, visit "Ask Dr. Math" (K-12 grades only). Can you mail or e-mail me the solutions for contest xxx? Sorry, we do not have solutions for our contests in a suitable format for sending by e-mail. Our contest problem books have complete solutions, and can be ordered online. Just use any of the links on the left marked 'Shopping" to browse our list of contest books. ### Fees and Billing I received an invoice for my order but I haven't received the materials yet. Do I have to pay in advance? As indicated on your invoice, payment is due upon receipt of all materials ordered. It is not necessary to pay any part of your order until you have received the complete order, although payment is certainly accepted (and welcomed) at the time you receive the invoice. How much does the High School League cost for the school year? The cost of entering the High School League for a school year is \$90. There are no other fees associated with it. I didn't order the high school contests but I received a set anyway. Must I pay for these? Since the first high school contest occurs early in the school year, the League mails a set of contests to any advisor whose school was enrolled in the League last year. If you wish to participate in the League, please complete the registration form that was included with the contest package. The cost of participation is \$90. If you do not wish to participate, please discard the materials in a secure manner. ### Contests How many contests are in a set? High school contests come with 6 sets of 30 contest copies each: one set for each of the year's six contests. All other contests consist of 30 contest copies per set. You may make additional copies of any contest on the day of the contest. How do I check if my school is registered for contests? A new contest registration index page has been added for teachers, administrators, and parents to check thier school's registration status. The web page is best viewed with a Javascript-enabled browser. The list is updated frequently. School registration listings will appear about 2 weeks after contest registration is mailed to us; about 1 week for schools using our online registration form. I received an invoice for my order but I haven't received the materials yet. Do I have to pay in advance? Payment is due upon receipt of all materials ordered, as indicated on your invoice. It is not necessary to pay any part of your order until you have received the complete order, although payment is certainly accepted (and welcomed) at the time you receive the invoice. How do I obtain additional Certificates of Merit for my top students? If your school needs more Certificates of Merit, send your name, school, and school mailing address to our mailer at: Math League Certificates P.O. Box 17 Tenafly, NJ 07670-0017 Include a self-addressed, stamped envelope (two stamps required) large enough to hold certificates. My school is closed or on a special schedule the day of the contest. May we still participate? You may reschedule the contest for either the alternate testing date listed or, with the permission of the League, for another date close to the official testing date. My package for 4th Grade Contests (or 5th Grade or Algebra Course 1) did not include a score report form. How do I report the scores? Our contests for 4th Grade, 5th Grade, and Algebra Course 1 are intramural contests that do not require the submission of scores to the League. Instead of giving plaques to the top two schools and top two students in the League as we do on all our other contests, we provide each school with a book prize to award to the top student in the school. We also include certificates of merit for other high-scoring students. One of my best students was absent from school on the day of the contest. May this student take the contest later? Although you may administer the contest to this student and award this student a certificate of merit if this student is the highest scoring student in your school, you may not submit the student's score to the League. Only students who take the test during the first testing are eligible for official recognition by the League. I'm scheduled for the 6th, 7th and 8th grade contest. Must all grades take the test at the same time or on the same day? Must all my students take the contest at the same time? For 4th grade, 5th grade, and Algebra Course 1, students may take the contest whenever it is convenient. You may give the 6th, 7th, and 8th grade contests on different dates. For each of these contests, students whose scores are reported to the League must all take the contest at the same time. If you give the 6th (or 7th or 8th) grade contest at several different times and on several different days, only the scores of the students who were present for the first testing session may be reported to the League. I'm a new high school advisor this year, although my school was in the League last year. I have not yet received my high school package. What should I do? Your school package was sent in September to last year's adviser. Please check and see if the package can be located. If not, e-mail us at www.mathleague.com (or call 1-201-568-6328) and we will ship another package immediately. Some of my 8th grade students are studying algebra (or other high school math classes). Are they permitted to participate in the 8th grade contest? Any student who has not yet completed the 8th grade may participate in the 8th grade contest, regardless of the math course in which they may be enrolled. Should my 8th grade students enroll in the Algebra Course 1 contest or the 8th grade contest? Students who take the algebra course 1 contest may also take any other contest the League sponsors. I teach a 6th grade student who is doing 7th grade math. In which contest should this student participate? A student may participate officially in only one of the contests for grades 6, 7, and 8. A 6th grade student taking 7th grade math may participate officially in either the 6th Grade Contest or the 7th Grade Contest, but not both. In this case, the choice of contest is not made by us: it's made by you and/or your student. We only have 1 or 2 students in our school interested in these contests. May we participate? There is no minimum number of students required to participate in any of these contests. Our school wants to participate but we do not want our scores listed. May we participate on an unofficial basis? Any school may choose to be an unofficial participant in the contests. Unofficial schools receive the same materials as official schools, but are not eligible for plaques. For grades 6, 7, and 8, only high scoring schools and students are listed on the score report summary published by the League. For high schools, the scores of all schools are listed in the score report summary unless a school requests not to be listed. My child is being home-schooled. May my child participate in the contest? Your child can participate on either an official or an unofficial basis. To participate unofficially, order our contest subscription package for homeschoolers, which will be sent to you in May for testing at home. You can compare your child's unofficial performance on the contest to scores of official participants by viewing the contest results at our Web site. If you want your child to be an official participant, you need to arrange with an accredited local school to order the official contest package, and proctor the contest for your child on the official contest date. All test materials would be mailed to your local school. Email comments and questions to This email address is being protected from spambots. You need JavaScript enabled to view it.
# Exponential Decay – Formulas and Examples We can model real-life situations with exponential functions. One of these situations is population decline. There are “standard” formulas that can be used to easily calculate the population or any other quantity by knowing some facts about the situation. Here, we will look at what the population decrease means and we will learn about its formulas. In addition, we will see several examples with answers of exponential decay that apply these formulas. ##### ALGEBRA Relevant for Learning to solve exponential decay problems with examples. See examples ##### ALGEBRA Relevant for Learning to solve exponential decay problems with examples. See examples ## Exponential decay summary and formulas Exponential decay describes the process of reducing an amount by a consistent percentage over a period of time. Exponential decay is very useful for modeling a large number of real-life situations. Most notably, we can use exponential decay to monitor inventory that is used regularly in the same amount, such as food for schools or cafeterias. The following is the formula used to model exponential decay. It is important to recognize this formula and each of its elements: Recall that the exponential function has the basic form . Therefore, in the exponential decay formula, we have replaced with . Then, we have: • initial amount. Initial amount before decrement. • decay factor. Represented as a decimal. • time interval. Many natural events can be modeled using the exponential number e. We can think of as a universal constant that can be used to represent the growth or decrease that occurs with continuous processes. Furthermore, using we can also represent exponential growth or decay measured periodically over time. Therefore, if we have quantities that continuously increase or decrease with a fixed percentage, we can model these scenarios with the following formula: In this formula, we have: • final amount. Amount after decrement. • initial amount. Amount before decrement. • exponential. e is approximately equal to 2.718… • rate of continuous growth or decline. It is also called the constant of proportionality. • time interval. ## Exponential decay – Examples with answers The following examples are solved using what we have learned about exponential decay. Each example has its respective solution, but try to solve the problems yourself before looking at the answer. ### EXAMPLE 1 A restaurant served 5,000 customers on Monday. There was a health inspection and the restaurant scored low, so the restaurant served 2,500 customers on Tuesday. On Wednesday, the restaurant served 1,250 customers, and on Thursday only 625 customers. How many clients will there be after five days starting on Monday? We can see that the number of clients decreases by 50% every day. This type of decrement is different from the linear decrease. In linear decrease, the amount that decreases each day would be the same each day. In this case, the starting quantity is 5000 and the decay factor (r) would be 0.5. x would represent the amount of time in days. Therefore, we can plug these values into the exponential decay formula: The result is 312.5, but since we cannot have a half client, we round this up to have 313 clients. ### EXAMPLE 2 A forest has a population of 1000 birds. Due to deforestation, the bird population is decreasing at a rate of 5% per year. Calculate the size of the bird population after 10 years. We have to start by finding the exponential decay function that models the bird population. Then, we use the general form with the following data: Therefore, we have: The bird population after 10 years is 599. ### EXAMPLE 3 The cost of a car model decreases in value at a continuous rate of 8% per year. If a car costs 20,000 USD when new, what will it be worth after 5 years? This is a case of continuous decay, so we have to use the second formula given above. We use the formula with the following data: Therefore, after 5 years the car will cost 13,406.4 USD. ### EXAMPLE 4 Rewrite the function from the previous problem in the form . To write the function in the form , we can use the substitution . Therefore, we have: ⇒ We can also find the value of r using . Therefore, we have: This means that the annual decay rate of the car price is 7.69%. ### EXAMPLE 5 Carbon-14 is a radioactive isotope of carbon. Its presence in organic materials is the basis of the radiocarbon dating method. The amounts of carbon-14 present in materials allow for the age of the specimens to be determined. One artifact had 12 grams of carbon-14. How many grams of carbon-14 will be present in the artifact after 10,000 years if the model describes the amount of carbon-14 present after t years? In this case, we need to find the amount of carbon-14 present after 10,000 years. Therefore, we have to solve for variable A by substituting the value : Therefore, after 10,000 years, the artifact will have roughly 3.58 grams of carbon-14. ### EXAMPLE 6 An artifact found at an archaeological site contained 20% of the original carbon-14. Determine the age of the artifact using the exponential decay model for carbon-14 . In this case, we want to find the age of the artifact, so we have to solve for t. We do not have specific values of and . However, we know that the carbon-14 found is 20% of the original, so we can use . Therefore, we have: If we divide both sides by , we have: We take the natural logarithm of both sides to eliminate the exponential: Therefore, the age of the artifact is approximately 13301 years. ## Exponential decay – Practice problems Put into practice what you have learned about exponential decay with the following problems. Solve the problems and select an answer. Check your answer to verify that you selected the correct one.
# Average Visibility of Moviegoers ### Solution 1 Observe that the $i^{th}$ tallest person is viewable if and only if he is in front of all the people $1,2,\ldots,i-1.$ This happens with probability $\displaystyle \frac{1}{i}.$ Therefore, the answer is $\displaystyle \sum_{i=1}^{n}\frac{1}{i}\approx \ln n+\gamma,$ where $\gamma=0.5772157\ldots,$ Euler's constant. ### Solution 2 For $n$ patrons, let the number of viewable patrons summed over all permutations be $g(n)$. Thus, the average number of patrons over the permutations is $g(n)/n!$ (denote this by $f(n))$. There are $(n-1)!$ permutations where the tallest person is last in line. For this case, the sum of viewable patrons over all permutations is $g(n-1)+(n-1)!$ ($g(n-1)$ for the first $n-1$ in line and $(n-1)!$ because the tallest patron is a viewable patron for all the permutations). If a patron other than the tallest patron is last in line, the sum of viewable patrons over all permutations where the last patron is held fixed is $g(n-1)$ (the last patron is not a viewable patron in this case). However, the last patron can be chosen in $(n-1)$ ways for this case. Thus, \displaystyle \begin{align} g(n)&=g(n-1)+(n-1)!+(n-1)g(n-1) \\ &=(n-1)!+ng(n-1) \\ \Rightarrow\; f(n)&=\frac{1}{n}+f(n-1)~\text{(Dividing both sides byn!)}. \end{align} Noting that $f(1)=1$, the solution of this recurrence relation is $\displaystyle f(n)=\sum_{k=1}^{n} \frac{1}{k}=H_n~\text{($n^{th}$harmonic number)}.$ ### Acknowledgment This is problem 2 from the 1992 Indiana College Mathematics Competition, see A Friendly Mathematics Competition by Rick Gillman (ed.) Solution 2 is by Amit Itagi.
Courses Courses for Kids Free study material Offline Centres More Store # Show that the middle term in the expansion of ${\left( {1 + x} \right)^n}$is $6{x^2}$ if n = 4. Last updated date: 13th Jul 2024 Total views: 450.6k Views today: 7.50k Verified 450.6k+ views Hint - There can be two methods to solve this problem, one is based upon the general formula of expansion of ${\left( {1 + x} \right)^n}$ using the binomial expansion and the other one focuses on direct formula to find the middle term in expansion of ${\left( {1 + x} \right)^n}$ depending upon whether n is even or odd. Now let’s use the direct formula for finding the middle term in the expansion of ${\left( {1 + x} \right)^n}$. Here n =4 (given in question) Clearly n is even thus the middle term in expansion of ${\left( {1 + x} \right)^n}$is $^n{C_{\dfrac{n}{2}}}{x^{\dfrac{n}{2}}}$……………… (1) So let’s directly put the value of n in equation (1) we get Middle term will be $^4{C_{\dfrac{4}{2}}}{x^{\dfrac{4}{2}}} = {{\text{ }}^4}{C_2}{x^2}$………………….. (2) Now The formula for $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$…………………. (3) Using equation three we get $^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \\ \Rightarrow \dfrac{{4!}}{{2!\left( 2 \right)!}} \\$ Now $n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right).........\left( {n - r} \right)$ such that $r < n$ Using this concept we get $^4{C_2} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \\ \Rightarrow 6 \\$ Hence putting it in equation (2) we get Middle term of ${\left( {1 + x} \right)^n}$is $6{x^2}$ Note – Now let’s talk about a method in which we can use the entire expansion of ${\left( {1 + x} \right)^n}$using binomial theorem. The expansion of ${\left( {1 + x} \right)^n}$is$1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ............\infty$. Hence after completely expanding till the given value of n, find the middle term in that series, this will also give the right answer.
<meta http-equiv="refresh" content="1; url=/nojavascript/"> # 8.7: Cone: Base Area, Lateral Area, Surface Area and Volume Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Find the surface area of a cone using a net or a formula. • Find the volume of a cone. ## Cones A cone is a three-dimensional figure with a single curved base that tapers to a single point called an apex. The base of a cone can be a circle or an oval of some type. In this chapter, we will only use circular cones. • The ______________________ is the point on top of a cone. • The base of the cones we will study is in the shape of a ____________________. You can remember the name “cone” of this shape because it looks like an upside-down ice cream cone. The apex of a right cone lies above the center of the cone’s circle. In an oblique cone, the apex is not in the center: • The apex of a right cone is the point directly above the _____________________ of the cone’s circular base. The height of a cone h\begin{align}h\end{align} is the perpendicular distance from the center of the cone’s base to its apex. • The height of a cone is just like an altitude: a _______________________ line from the center of the circular base to the apex. ## Surface Area of a Cone Using Nets Most three-dimensional figures are easy to deconstruct into a net. The cone is different in this regard. Can you predict what the net for a cone looks like? In fact, the net for a cone looks like a small circle and a sector, or part of a larger circle. The diagram below shows how the half-circle sector folds to become a cone: Note that the circle that the sector is cut from is much larger than the base of the cone. • The net for a cone is a circular base plus a _____________________. Example 1 Which sector will give you a taller cone—a half circle or a sector that covers three-quarters of a circle? Assume that both sectors are cut from congruent circles. Make a model of each sector: 1. The half circle makes a cone that has a height that is about equal to the radius of the semi-circle. 2. The three-quarters sector gives a cone that has a wider base (greater diameter) but its height as not as great as the half-circle cone. Example 2 Predict which will be greater in height—a cone made from a half-circle sector or a cone made from a one-third-circle sector. Again, assume that both sectors are cut from congruent circles. The relationship in Example #1 on the previous page holds true—the greater (in degrees) the sector, the smaller the height of the cone. In other words, the fraction 13\begin{align}\frac{1}{3}\end{align} is less than 12\begin{align}\frac{1}{2}\end{align}, so a one-third sector will create a cone with greater height than a one-half sector. • The larger the sector, the __________________________ the height of its cone. Example 3 Predict which will be greater in diameter—a cone made from a half-circle sector or a cone made from a one-third-circle sector. Assume that the sectors are cut from congruent circles. Here you have the opposite relationship—the larger (in degrees) the sector, the greater the diameter of the cone. In other words, 12\begin{align}\frac{1}{2}\end{align} is greater than 13\begin{align}\frac{1}{3}\end{align}, so a one-half sector will create a cone with greater diameter than a one-third sector. • The larger the sector, the ________________________ the diameter of its cone. ## Surface Area of a Regular Cone The surface area of a regular pyramid is given by: A=(12lP)+B where l\begin{align}l\end{align} is the slant height of the figure, P\begin{align}P\end{align} is the perimeter of the base, and B\begin{align}B\end{align} is the area of the base. Imagine a series of pyramids in which n\begin{align}n\end{align}, the number of sides of each figure’s base, increases. As you can see, as n\begin{align}n\end{align} increases, the figure more and more resembles a circle. You can also think of this as: a circle is like a polygon with an infinite number of sides that are infinitely small. Similarly, a cone is like a pyramid that has an infinite number of sides that are infinitely small in length. As a result, the formula for finding the total surface area of a cone is similar to the pyramid formula. The only difference between the two is that the pyramid uses P\begin{align}P\end{align}, the perimeter of the base, while a cone uses C\begin{align}C\end{align}, the circumference of the base. A(pyramid)=12lP+BandA(cone)=12lC+B Since the circumference of a circle is 2πr\begin{align}2 \pi r\end{align}: A(cone)=12lC+B=12l(2πr)+B=πrl+B You can also express B\begin{align}B\end{align} as πr2\begin{align}\pi r^2\end{align} to get: ## Surface Area of a Right Cone A(cone)=πrl+B=πrl+πr2=πr(l+r) Any of these forms of the equation can be used to find the surface area of a right cone. There are a few different formulas to find the surface area of a cone. Pick one formula and describe what every variable represents. \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} Example 4 Find the total surface area of a right cone with a radius of 8 cm and a slant height of 10 cm. Use the formula: A(cone)=πr(l+r)=π(8)(10+8)=8π18=144π cm2or(144)(3.14)=452.16 cm2 The exact area of the cone is 144π cm2\begin{align}144 \pi \ cm^2\end{align} and the approximate area is 452.16 cm2\begin{align}452.16 \ cm^2\end{align}. Example 5 Find the total surface area of a right cone with a radius of 3 feet and an altitude (not slant height) of 6 feet. Use the Pythagorean Theorem to find the slant height l\begin{align}l\end{align}: r2+h232+629+36454535=l2=l2=l2=l2=l=l Now use the area formula: A(cone)=πr(l+r)=π(3)(35+3)=3π(35+3) If we leave this as an exact answer, we cannot simplify anymore. This would be an ideal time to use a decimal approximation with a calculator: 3π(35+3)3(3.14)(35+3)=91.45 cm2 The surface area of the cone is approximately 91.45 cm2\begin{align}91.45 \ cm^2\end{align}. ## Volume of a Cone Which has a greater volume, a pyramid, cone, or cylinder if the figures have bases with the same "diameter" (i.e., distance across the base) and the same altitude? To find out, compare pyramids, cylinders, and cones that have bases with equal diameters and the same altitude. Here are three figures that have the same dimensions—cylinder, a right regular hexagonal pyramid, and a right circular cone. Which figure appears to have a greater volume? It seems obvious that the volume of the cylinder is greater than the other two figures. This is because the pyramid and cone taper off to a single point, while the cylinder’s sides stay the same width. Determining whether the pyramid or the cone has a greater volume is not so obvious. If you look at the bases of each figure you see that the apothem of the hexagon is congruent to the radius of the circle. You can see the relative size of the two bases by superimposing one onto the other: From the diagram you can see that the hexagon is slightly larger in area than the circle. Therefore, the volume of the right hexagonal regular pyramid would be greater than the volume of a right circular cone. It is, but only because the area of the base of the hexagon is slightly greater than the area of the base of the circular cone. • When comparing the volumes of a cylinder, a pyramid, and a cone, the __________________________ has the largest volume and the __________________________ has the smallest volume. The __________________________ has a volume in between the other two shapes. The formula for finding the volume of each figure is virtually identical. Both formulas follow the same basic form: V=13Bh Since the base of a circular cone is, by definition, a circle, you can substitute the area of a circle, πr2\begin{align}\pi r^2\end{align} for the base of the figure. This is expressed as a volume postulate for cones. • Instead of using B\begin{align}B\end{align} for base area, we use the area of a _____________________, πr2\begin{align}\pi r^2\end{align}, in the formula for volume of a cone. Volume of a Right Circular Cone Given a right circular cone with height h\begin{align}h\end{align} and a base that has radius r\begin{align}r\end{align}: VV=13Bh=13πr2h Example 6 Find the volume of a right cone with a radius of 9 cm and a height of 16 cm. Use the formula: V=13πr2h\begin{align}V = \frac{1}{3} \pi r^2 h\end{align} Substitute the values for r=\begin{align}r =\end{align} ___________ and h=\begin{align}h =\end{align} _____________ : VV=13π(92)(16)=1296π3=432π cm3or(432)(3.14)=1356.48 cm3 The cone has an exact volume of \begin{align}432 \pi\end{align} cubic centimeters and an approximate volume of 1356.48 cubic centimeters. By now, you have seen the units \begin{align}cm^2\end{align} or \begin{align}in^2\end{align} and \begin{align}cm^3\end{align} or \begin{align}in^3\end{align} in the examples. When we calculate area, we use a “square” unit, such as \begin{align}cm^2\end{align} (square centimeters) or \begin{align}in^2\end{align} (square inches) When we calculate volume, we use a “cubic” unit, such as \begin{align}cm^3\end{align} (cubic centimeters) or \begin{align}in^3\end{align} (cubic inches) Example 7 Find the volume of a right cone with a radius of 10 feet and a slant height of 13 feet. Use the Pythagorean theorem to find the height \begin{align}h\end{align}: Now use the volume formula: \begin{align}V = \frac{1}{3} \pi r^2 h\end{align} Substitute the values for \begin{align}r =\end{align} ___________ and \begin{align}h =\end{align} _____________ : The cone’s volume can be written as \begin{align}277 \pi \ ft^3\end{align} or \begin{align}869.78 \ ft^3\end{align}. 1. What type of units are used to express volume? What type for area? \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} 2. What shape is the base of a right circular cone? \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} 3. When calculating the volume of a cone, what information do you need? \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} \begin{align}{\;}\end{align} 8 , 9 , 10 ## Date Created: Feb 23, 2012 May 12, 2014 You can only attach files to None which belong to you If you would like to associate files with this None, please make a copy first.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use. # Distributive Property for Multi-Step Equations ## a(x + b) = c Estimated22 minsto complete % Progress Practice Distributive Property for Multi-Step Equations MEMORY METER This indicates how strong in your memory this concept is Progress Estimated22 minsto complete % Candy Teacher Contributed ## Real World Applications – Algebra I Selling Candy ### Student Exploration One of the most common ways that teenagers start to make money is by selling candy. If you buy a case of candy, it can cost you about $26 for a box of 30 pieces of big candy bars, which means that if you sell each candy bar for$1, you can get a 4 profit. If you and your friend Jay are both selling candy, we can represent this relationship by using and proving the distributive property. You and Jay sold a lot of candy bars. Let’s say you sold 15 boxes in one week and she sold 8 boxes. We can represent the total amount that the two of you earned in two different ways. The first way to represent how much money you both made total is by adding the total number of boxes sold and then multiplying this by 4. So, we would get the sum, which is 15+8\begin{align}15 + 8\end{align} and then multiplying that by 4, which is 4(15+8)\begin{align}4(15 + 8)\end{align}. Another way to represent this relationship is if you figured out how much you made first, before Jay. Since you knew that you sold 15 bars, you multiplied your earnings by 4, so 4×15\begin{align}4 \times 15\end{align} or60, and then added Jay’s earnings. Then, your total earnings would be (4×15)+(4×8)\begin{align}(4 \times 15) + (4 \times 8)\end{align}. These two methods would yield the same answer, right? Do the calculations to check. Since these are two different ways of representing the total profit, what does this mean? These two ways actually help prove the distributive property, and why the distributive property holds true. If we were to distribute the 4 to 15 and 8, we would get the second expression above. But, let’s say that you and Jay both decided that 4 is not a very good profit, and wanted to see what kinds of profit you can make by changing how much you want your profit to be. We can also represent this using the distributive property! Let’s say that “x\begin{align}x\end{align}” will represent the number of dollars you want to add to your4 current profit. We can use the expression (x+4)(15+8)\begin{align}(x + 4)(15 + 8)\end{align} to represent the total amount of profit between you and your friend. We can also represent this as the sum of both of your profits individually. Your profit would be (x+4)(15)\begin{align}(x + 4)(15)\end{align}, which represents the amount added to your current profit and then the sum multiplied by 15, or the number of boxes of candy you sold. Jay’s profit would be (x+4)(8)\begin{align}(x + 4)(8)\end{align}, which represents the amount added to her current profit and then the sum multiplied by 8, or the number of boxes of candy she sold. To find the total profit between the two of you, you’d take the sum of the two quantities, or adding 15(x+4)+8(x+4)\begin{align}15(x + 4) + 8(x + 4)\end{align}. Since a common factor in this expression is (x+4)\begin{align}(x + 4)\end{align}, we can also rewrite this as (x+4)(15+8)\begin{align}(x + 4)(15 + 8)\end{align}. We just factored this expression by finding the common factor, or by grouping! ### Extension Investigation Check out other ways that you can represent the distributive property and factoring in your world. When would this be useful to you? If we went back to the example above about selling candy, could you figure out a way to sell something and find a way to make the maximum profit? If so, how? What type of relationship would this represent? (Linear? Quadratic?) If it’s quadratic, what would the point of that would yield the maximum profit be called? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes
Lesson Objectives • Demonstrate an understanding of exponents and logarithms • Learn how to solve exponential equations with different bases ## How to Solve Exponential Equations with Logarithms Before we jump in and start solving exponential equations, let's look at some of the properties we will be using in this lesson. ### Properties for Solving Exponential and Logarithmic Equations • b, x, and y are real numbers, b > 0, b ≠ 1 • If x = y, then bx = by • If bx = by, then x = y • If x = y and x > 0, y > 0, then logb(x) = logb(y) • If x > 0, y > 0 and logb(x) = logb(y), then x = y ### Solving Exponential Equations with Different Bases Previously, we learned how to solve exponential equations with the same base. For that scenario, we can set the exponents equal to each other and solve for the unknown. We can't do this with every scenario since it is not always possible to have the same base on each side of an equation. When we try to solve an exponential equation and different bases are involved, we use logarithms to obtain our solution. We will first isolate the exponential expression and then we can take logarithms to the same base on both sides. This will allow us to use our product rule for logarithms to isolate the variable. Let's look at a few examples. Example 1: Solve each equation $$6^x=15$$ To solve this equation, we can take the log of each side. $$log(6^x)=log(15)$$ Using our power rule for logarithms, we can bring the exponent (x) down in front of the logarithm on the left side of the equation. $$xlog(6)=log(15)$$ To get x by itself, we divide each side of the equation by log(6). $$x=\frac{log(15)}{log(6)}$$ This is our answer. If the teacher or assignment calls for a decimal approximation, we can divide using a calculator and then round to the appropriate number of decimal places. One thing to note, the textbook may give you an alternative method for solving this type of equation. Notice that our base of the exponential expression is a 6. Recall the property of logarithms that tells us: logb(bx) = x Therefore, we can also solve the equation in the following way: $$6^x=15$$ $$log_{6}(6^x)=log_{6}(15)$$ $$x=log_{6}(15)$$ Since using the change of base rule gives us our answer above, we can see that both answers are the same. It is just a matter of personal preference. Example 2: Solve each equation $$10 \cdot 19^{4 - x}- 5=77$$ To solve this equation, we want to isolate the exponential expression. This will allow us to use logarithms and bring the variable down out of the exponent. We will begin by adding 5 to each side: $$10 \cdot 19^{4 - x}=82$$ Now we will divide each side by 10: $$19^{4 - x}=\frac{82}{10}$$ We can reduce our fraction on the right: $$19^{4 - x}=\frac{41}{5}$$ Now, we can take the log of each side: $$log(19^{4 - x})=log\left(\frac{41}{5}\right)$$ Bring the exponent (4 - x) out in front: $$(4 - x)log(19)=log\left(\frac{41}{5}\right)$$ Divide each side by log(19): $$4 - x=\frac{log\left(\frac{41}{5}\right)}{log(19)}$$ Subtract 4 away from each side: $$-x=\frac{log\left(\frac{41}{5}\right)}{log(19)}- 4$$ Multiply each side by -1: $$x=-\frac{log\left(\frac{41}{5}\right)}{log(19)}+ 4$$ Example 3: Solve each equation $$e^{x - 3}- 4=10$$ Let's begin by isolating our exponential expression. We will add 4 to each side of the equation: $$e^{x - 3}=14$$ When e is involved, we want to use ln. Remember ln is loge $$ln(e^{x - 3})=ln(14)$$ $$x - 3=ln(14)$$ We will add 3 to each side of the equation: $$x=ln(14) + 3$$ #### Skills Check: Example #1 Solve each equation. $$4 \cdot 12^{5x + 2}=23$$ A No Solution B $$x=\frac{log_{2}\left(\frac{11}{4}\right)}{5}$$ C $$x=\frac{log_{12}\left(\frac{23}{4}\right) - 2}{5}$$ D $$x=\frac{log_{23}(21) - 4}{5}$$ E $$x=\frac{log_{14}(61) + 9}{9}$$ Example #2 Solve each equation. $$5 \cdot 11^{3x + 10}=11$$ A $$x=\frac{log_{5}(20)}{3}$$ B $$x=\frac{log_{11}\left(\frac{11}{5}\right) - 10}{3}$$ C $$x=\frac{log_{5}\left(\frac{10}{11}\right) - 11}{3}$$ D $$x=\frac{log_{11}(5) - 3}{11}$$ E No Solution Example #3 Solve each equation. $$14^{9x - 9}+ 3=64$$ A No Solution B $$x=\frac{15}{7}$$ C $$x=\frac{log_{14}\left(61\right) + 9}{9}$$ D $$x=\frac{log_{14}\left(\frac{13}{64}\right) - 3}{9}$$ E $$x=\frac{-log_{14}\left(\frac{55}{14}\right) + 3}{9}$$
# Applications of Methods of Proof Save this PDF as: Size: px Start display at page: ## Transcription 1 CHAPTER 4 Applications of Methods of Proof 1. Set Operations 1.1. Set Operations. The set-theoretic operations, intersection, union, and complementation, defined in Chapter 1.1 Introduction to Sets are analogous to the operations,, and, respectively, that were defined for propositions. Indeed, each set operation was defined in terms of the corresponding operator from logic. We will discuss these operations in some detail in this section and learn methods to prove some of their basic properties. Recall that in any discussion about sets and set operations there must be a set, called a universal set, that contains all others sets to be considered. This term is a bit of a misnomer: logic prohibits the existence of a set of all sets, so that there is no one set that is universal in this sense. Thus the choice of a universal set will depend on the problem at hand, but even then it will in no way be unique. As a rule we usually choose one that is minimal to suit our needs. For example, if a discussion involves the sets {1, 2, 3, 4} and {2, 4, 6, 8, 10}, we could consider our universe to be the set of natural numbers or the set of integers. On the other hand, we might be able to restrict it to the set of numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. We now restate the operations of set theory using the formal language of logic Equality and Containment. Definition Sets A and B are equal, denoted A = B, if x[x A x B] Note: This is equivalent to x[(x A x B) (x B x A)]. Definition Set A is contained in set B (or A is a subset of B), denoted A B,if x[x A x B]. 96 2 1. SET OPERATIONS 97 The above note shows that A = B iff A B and B A Union and Intersection. Definitions The union of A and B, The intersection of A and B, A B = {x (x A) (x B)}. A B = {x (x A) (x B)}. If A B =, then A and B are said to be disjoint Complement. Definition The complement of A A = {x U (x A)} = {x U x A}. There are several common notations used for the complement of a set. For example, A c is often used to denote the complement of A. You may find it easier to type A c than A, and you may use this notation in your homework Difference. Definition The difference of A and B, or the complement of B relative to A, A B = A B. Definition The symmetric difference of A and B, A B = (A B) (B A) = (A B) (A B). The difference and symmetric difference of two sets are new operations, which were not defined in Module 1.1. Notice that B does not have to be a subset of A for 3 1. SET OPERATIONS 98 the difference to be defined. This gives us another way to represent the complement of a set A; namely, A = U A, where U is the universal set. The definition of the difference of two sets A and B in some universal set, U, is equivalent to A B = {x U (x A) (x B)}. Many authors use the notation A \ B for the difference A B. The symmetric difference of two sets corresponds to the logical operation, the exclusive or. The definition of the symmetric difference of two sets A and B in some universal set, U, is equivalent to A B = {x U [(x A) (x B)] [ (x A) (x B)]} Product. Definition The (Cartesian) Product of two sets, A and B, is denoted A B and is defined by 1.7. Power Set. A B = {(a, b) a A b B} Definition Let S be a set. The power set of S, denoted P(S) is defined to be the set of all subsets of S. Keep in mind the power set is a set where all the elements are actually sets and the power set should include the empty set and itself as one of its elements Examples. Example Assume: U = {a, b, c, d, e, f, g, h}, A = {a, b, c, d, e}, B = {c, d, e, f}, and C = {a, b, c, g, h}. Then (a) A B = {a, b, c, d, e, f} (b) A B = {c, d, e} (c) A = {f, g, h} (d) B = {a, b, g, h} (e) A B = {a, b} (f) B A = {f} (g) A B = {a, b, f} 4 1. SET OPERATIONS 99 (h) (A B) C = {a, b, c} (i) A B = {(a, c), (a, d), (a, e), (a, f), (b, c), (b, d), (b, e), (b, f), (c, c), (c, d), (c, e), (c, f), (d, c), (d, d), (d, e), (d, f), (e, c), (e, d), (e, e), (e, f)} (j) P(A) = {, {a}, {b}, {c}, {d}, {e}, {a, b}, {a, c}, {a, d}, {a, e}, {b, c}, {b, d}, {b, e}, {c, d}, {c, e}, {d, e}, {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e}, {b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}, {a, b, c, d}, {a, b, c, e}, {a, b, d, e}, {a, c, d, e}, {b, c, d, e}, {a, b, c, d, e}} (k) P(A) = 32 Exercise Use the sets given in Example to find (1) B A (2) P(B) (3) P(U) Example Let the universal set be U = Z + the set of all positive integers, let P be the set of all prime (positive) integers, and let E be the set of all positive even integers. Then (a) P E = {n Z + n is prime or even}, (b) P E = {2}, (c) P is the set of all positive composite integers, (d) E is the set of all positive odd integers, {2n + 1 n N}, (e) P E is the set of all positive odd prime numbers (all prime numbers except 2), (f) E P = {4, 6, 8, 10,... } = {2n n Z + n 2}, (g) E P = {n Z + (n is prime or even) n 2} Exercise (1) If A = n and B = m, how many elements are in A B? (2) If S is a set with S = n, what is P(S)? Exercise Does A B = B A? Prove your answer Venn Diagrams. A Venn Diagram is a useful geometric visualization tool when dealing with three or fewer sets. The Venn Diagram is generally set up as follows: The Universe U is the rectangular box. A set are represented by a circle and its interior. In the absence of specific knowledge about the relationships among the sets being represented, the most generic relationships should be depicted. 5 1. SET OPERATIONS 100 Venn Diagrams can be very helpful in visualizing set operations when you are dealing with three or fewer sets (not including the universal set). They tend not to be as useful, however, when considering more than three sets. Although Venn diagrams may be helpful in visualizing sets and set operations, they will not be used for proving set theoretic identities Examples. Example The following Venn Diagrams illustrate generic relationships between two and three sets, respectively. U U A B A B C Example This Venn Diagram represents the difference A B (the shaded region). U A A - B B The figures in the examples above show the way you might draw the Venn diagram if you aren t given any particular relations among the sets. On the other hand, if you knew, for example, that A B, then you would draw the set A inside of B. 6 1. SET OPERATIONS Set Identities. Example Prove that the complement of the union is the intersection of the complements: A B = A B. Proof 1. One way to show the two sets are equal is to use the fact that A B = A B iff A B A B and A B A B. Step 1. Show A B A B. Assume x is an arbitrary element of A B (and show x A B). Since x A B, x A B. This means x A and x B (De Morgan s Law). Hence x A B. Thus, by Universal Generalization, so that, by definition, x[x (A B) x (A B)] A B A B. Step 2. Show A B A B. Suppose x is an arbitrary element of A B. Then x A and x B. Therefore, x A B (De Morgan s Law). This shows x A B. Thus, by Universal Generalization, so that, by definition, x[x (A B) x (A B)] A B A B. Proof 2. The following is a second proof of the same result, which emphasizes more clearly the role of the definitions and laws of logic. We will show x[x A B x A B]. 7 1. SET OPERATIONS 102 Assertion x: x A B x [A B] [x A B] [(x A) (x B)] (x A) (x B) (x A) (x B) x A B Reason Definition of complement Definition of Definition of union De Morgan s Law Definition of complement Definition of intersection Hence x[x A B x A B] is a tautology. (In practice we usually omit the formality of writing x in the initial line of the proof and assume that x is an arbitrary element of the universe of discourse.) Proof 3. A third way to prove this identity is to build a membership table for the sets A B and A B, and show the membership relations for the two sets are the same. The 1 s represent membership in a set and the 0 s represent nonmembership. A B A B A B A B A B Compare this table to the truth table for the proof of De Morgan s Law: (p q) ( p q) A set identity is an equation involving sets and set operations that is true for all possible choices of sets represented by symbols in the identity. These are analgous to identities such as (a + b)(a b) = a 2 b 2 8 that you encounter in an elementary algebra course. 1. SET OPERATIONS 103 There are various ways to prove an identity, and three methods are covered here. This is a good place to be reminded that when you are proving an identity, you must show that it holds in all possible cases. Remember, giving an example does not prove an identity. On the other hand, if you are trying to show that an expression is not an identity, then you need only provide one counterexample. (Recall the negation of xp (x) is x P (x)). Proof 1 establishes equality by showing each set is a subset of the other. This method can be used in just about any situation. Notice that in Proof 1 we start with the assumption, x is in A B, where x is otherwise an arbitrary element in some universal set. If we can show that x must then be in A B, then we will have established x, [x A B] [x A B]. That is, the modus operandi is to prove the implications hold for an arbitrary element x of the universe, concluding, by Universal Generalization, that the implications hold for all such x. Notice the way De Morgan s Laws are used here. For example, in the first part of Proof 1, we are given that x (A B). This means [x (A B)] [(x A) (x B)] [(x A) (x B)]. Proof 2 more clearly exposes the role of De Morgan s Laws. Here we prove the identity by using propositional equivalences in conjunction with Universal Generalization. When using this method, as well as any other, you must be careful to provide reasons. Proof 3 provides a nice alternative when the identity only involves a small number of sets. Here we show two sets are equal by building a member table for the sets. The member table has a 1 to represent the case in which an element is a member of the set and a 0 to represent the case when it is not. The set operations correspond to a logical connective and one can build up to the column for the set desired. You will have proved equality if you demonstrate that the two columns for the sets in question have the exact same entries. Notice that all possible membership relations of an element in the universal set for the sets A and B are listed in the first two columns of the membership table. For example, if an element is in both A and B in our example, then it satisfies the conditions in the first row of the table. Such an element ends up in neither of the two sets A B nor A B. This is very straight forward method to use for proving a set identity. It may also be used to prove containment. If you are only trying to show the containment M N, you would build the membership table for M and N as above. Then you would look in every row where M has a 1 to see that 9 1. SET OPERATIONS 104 N also has a 1. However, you will see examples in later modules where a membership table cannot be created. It is not always possible to represent all the different possibilities with a membership table. Example Prove the identity (A B) C = (A C) (B C) Proof 1. Suppose x is an arbitrary element of the universe. Assertion x (A B) C [x (A B)] [x C] [(x A) (x B)] [x C] [(x A) (x C)] [(x B) (x C)] [x (A C)] [(x (B C)] x [(A C) (B C)] Reason definition of intersection definition of union distributive law of and over or definition of intersection definition of union Proof 2. Build a membership table: A B C A B (A C) (B C) (A B) C (A C) (B C) Since the columns corresponding to (A B) C and (A C) (B C) are identical, the two sets are equal. 10 Example Prove (A B) C A (B C). 1. SET OPERATIONS 105 Proof. Consider the membership table: A B C A B B C (A B) C A (B C) Notice the only 1 in the column for (A B) C is the fourth row. The entry in the same row in the column for A (B C) is also a 1, so (A B) C A (B C). Exercise Prove the identity A B = A B using the method of Proof 2 in Example Exercise Prove the identity A B = A B using the method of Proof 3 in Example Exercise Prove the identity (A B) C = (A C) (B C) using the method of Proof 1 in Example Exercise Prove the identity (A B) C = (A C) (B C) using the method of Proof 2 in Example Exercise Prove the identity (A B) C = (A C) (B C) using the method of Proof 3 in Example Union and Intersection of Indexed Collections. Definition The the union and intersection of an indexed collection of sets {A 1, A 2, A 3,..., A n } can be written as n A i = A 1 A 2 A 3 A n 11 1. SET OPERATIONS 106 and respectively. n A i = A 1 A 2 A 3 A n, Infinite Unions and Intersections. Definition The union and intersection of an indexed collection of infinitely many sets {A 1, A 2, A 3,..., } can be written as and A i = {a a A i for some i in Z + } A i = {a a A i for all i in Z + } If you have a collection of more than two sets, you can define the intersection and the union of the sets as above. (Since the operations are associative, it isn t necessary to clutter the picture with parentheses.) The notation is similar to the Σ notation used for summations. The subscript is called an index and the collection of sets is said to be indexed by the set of indices. In the example, the collection of sets is {A 1, A 2,..., A n }, and the set of indices is the set {1, 2,..., n}. There is no requirement that sets with different indices be different. In fact, they could all be the same set. This convention is very useful when each of the sets in the collection is naturally described in terms of the index (usually a number) it has been assigned. An equivalent definition of the union and intersection of an indexed collection of sets is as follows: and n A i = {x i {1, 2,..., n} such that x A i } n A i = {x i {1, 2,..., n}, x A i }. 12 1. SET OPERATIONS 107 Another standard notation for unions over collections of indices is A i = A i. i Z + More generally, if I is any set of indices, we can define A i = {x i I such that x A i }. i I Example Example Let A i = [i, i + 1), where i is a positive integer. Then n A i = [1, n + 1), and n A i =, if n > 1. A i = [1, ) A i = This is an example of a collection of subsets of the real numbers that is naturally indexed. If A i = [i, i + 1), then A 1 = [1, 2), A 2 = [2, 3), A 3 = [3, 4), etc. It may help when dealing with an indexed collection of sets to explicitly write out a few of the sets as we have done here. Example Suppose C i = {i 2, i 1, i, i + 1, i + 2}, where i denotes an arbitrary natural number. Then C 0 = { 2, 1, 0, 1, 2}, C 1 = { 1, 0, 1, 2, 3}, C 2 = {0, 1, 2, 3, 4}, n C i = { 2, 1, 0, 1,..., n, n + 1, n + 2} i=0 4 C i = {2} i=0 13 n C i = if n > 4. i=0 C i = { 2, 1, 0, 1, 2, 3,... } i=0 C i = i=0 1. SET OPERATIONS 108 Exercise For each positive integer k, let A k = {kn n Z}. For example, Find A 1 = {n n Z} = Z A 2 = {2n n Z} = {..., 2, 0, 2, 4, 6,...} A 3 = {3n n Z} = {..., 3, 0, 3, 6, 9,...} A k k=1 m A k, where m is an arbitrary positive integer. k=1 Exercise Use the definition for A k in exercise to answer the following questions. (1) (2) A i A i Computer Representation of a Set. Here is a method for storing subsets of a given, finite universal set: Order the elements of the universal set and then assign a bit number to each subset A as follows. A bit is 1 if the element corresponding to the position of the bit in the universal set is in A, and 0 otherwise. Example Suppose U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, with the obvious ordering. Then The bit string corresponding to A = {2, 4, 6, 8, 10} is The bit string corresponding to B = {1, 2, 3, 4} is 14 1. SET OPERATIONS 109 There are many ways sets may be represented and stored in a computer. One such method is presented here. Notice that this method depends not only on the universal set, but on the order of the universal set as well. If we rearrange the order of the universal set given in Example to U = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1}, then the bit string corresponding to the subset {1, 2, 3, 4} would be The beauty of this representation is that set operations on subsets of U can be carried out formally using the corresponding bitstring operations on the bit strings representing the individual sets. Example For the sets A, B U in Example : A = ( ) = = {1, 3, 5, 7, 9} A B = = = {1, 2, 3, 4, 6, 8, 10} A B = = = {2, 4} A B = = = {1, 3, 6, 8, 10} Exercise Let U = {a, b, c, d, e, f, g, h, i, j} with the given alphabetical order. Let A = {a, e, i}, B = {a, b, d, e, g, h, j}, and C = {a, c, e, g, i}. (1) Write out the bit string representations for A, B, and C. (2) Use these representations to find (a) C 15 (b) A B (c) A B C (d) B C 1. SET OPERATIONS 110 ### A set is a Many that allows itself to be thought of as a One. (Georg Cantor) Chapter 4 Set Theory A set is a Many that allows itself to be thought of as a One. (Georg Cantor) In the previous chapters, we have often encountered sets, for example, prime numbers form a set, domains ### Introduction Russell s Paradox Basic Set Theory Operations on Sets. 6. Sets. Terence Sim 6. Sets Terence Sim 6.1. Introduction A set is a Many that allows itself to be thought of as a One. Georg Cantor Reading Section 6.1 6.3 of Epp. Section 3.1 3.4 of Campbell. Familiar concepts Sets can ### Sets. A set is a collection of (mathematical) objects, with the collection treated as a single mathematical object. 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Keep in mind the following facts about regular languages: CPS 2: Discrete Mathematics Instructor: Bruce Maggs Assignment Due: Wednesday September 2, 27 A Tool for Proving Irregularity (25 points) When proving that a language isn t regular, a tool that is often ### We give a basic overview of the mathematical background required for this course. 1 Background We give a basic overview of the mathematical background required for this course. 1.1 Set Theory We introduce some concepts from naive set theory (as opposed to axiomatic set theory). The ### Congruences. Robert Friedman Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition ### Lecture 12. More algebra: Groups, semigroups, monoids, strings, concatenation. V. Borschev and B. Partee, November 1, 2001 p. 1 Lecture 12. More algebra: Groups, semigroups, monoids, strings, concatenation. 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Solving and Graphing Inequalities We know 4 is greater than 3, so is 5, so is 6, so is 7 and 3.1 works also. So does 3.01 ### The Mathematics Driving License for Computer Science- CS10410 The Mathematics Driving License for Computer Science- CS10410 Venn Diagram, Union, Intersection, Difference, Complement, Disjoint, Subset and Power Set Nitin Naik Department of Computer Science Venn-Euler ### Matrix Algebra. Some Basic Matrix Laws. Before reading the text or the following notes glance at the following list of basic matrix algebra laws. Matrix Algebra A. Doerr Before reading the text or the following notes glance at the following list of basic matrix algebra laws. Some Basic Matrix Laws Assume the orders of the matrices are such that ### We now explore a third method of proof: proof by contradiction. CHAPTER 6 Proof by Contradiction We now explore a third method of proof: proof by contradiction. This method is not limited to proving just conditional statements it can be used to prove any kind of statement ### 2. Propositional Equivalences 2. PROPOSITIONAL EQUIVALENCES 33 2. Propositional Equivalences 2.1. Tautology/Contradiction/Contingency. Definition 2.1.1. A tautology is a proposition that is always true. Example 2.1.1. p p Definition ### Sample Induction Proofs Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Sample Induction Proofs Below are model solutions to some of the practice problems on the induction worksheets. The solutions given ### Mathematical Induction. Mary Barnes Sue Gordon Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by ### 9.2 Summation Notation 9. Summation Notation 66 9. Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a ### Chapter I Logic and Proofs MATH 1130 1 Discrete Structures Chapter I Logic and Proofs Propositions A proposition is a statement that is either true (T) or false (F), but or both. s Propositions: 1. I am a man.. I am taller than ### Mathematical Induction Chapter 2 Mathematical Induction 2.1 First Examples Suppose we want to find a simple formula for the sum of the first n odd numbers: 1 + 3 + 5 +... + (2n 1) = n (2k 1). How might we proceed? The most natural ### WUCT121. Discrete Mathematics. Logic WUCT121 Discrete Mathematics Logic 1. Logic 2. Predicate Logic 3. Proofs 4. Set Theory 5. Relations and Functions WUCT121 Logic 1 Section 1. Logic 1.1. Introduction. In developing a mathematical theory, ### Fundamentele Informatica II Fundamentele Informatica II Answer to selected exercises 1 John C Martin: Introduction to Languages and the Theory of Computation M.M. Bonsangue (and J. Kleijn) Fall 2011 Let L be a language. It is clear ### Problem Set. Problem Set #2. Math 5322, Fall December 3, 2001 ANSWERS Problem Set Problem Set #2 Math 5322, Fall 2001 December 3, 2001 ANSWERS i Problem 1. [Problem 18, page 32] Let A P(X) be an algebra, A σ the collection of countable unions of sets in A, and A σδ the collection ### 1 Introduction to Counting 1 Introduction to Counting 1.1 Introduction In this chapter you will learn the fundamentals of enumerative combinatorics, the branch of mathematics concerned with counting. While enumeration problems can ### Finite Sets. Theorem 5.1. Two non-empty finite sets have the same cardinality if and only if they are equivalent. MATH 337 Cardinality Dr. Neal, WKU We now shall prove that the rational numbers are a countable set while R is uncountable. This result shows that there are two different magnitudes of infinity. But we ### CHAPTER 7 GENERAL PROOF SYSTEMS CHAPTER 7 GENERAL PROOF SYSTEMS 1 Introduction Proof systems are built to prove statements. They can be thought as an inference machine with special statements, called provable statements, or sometimes ### Classical Sets and Fuzzy Sets Classical Sets Operation on Classical Sets Properties of Classical (Crisp) Sets Mapping of Classical Sets to Functions Classical Sets and Fuzzy Sets Classical Sets Operation on Classical Sets Properties of Classical (Crisp) Sets Mapping of Classical Sets to Functions Fuzzy Sets Notation Convention for Fuzzy Sets Fuzzy ### Matthias Beck Ross Geoghegan. The Art of Proof. Basic Training for Deeper Mathematics Matthias Beck Ross Geoghegan The Art of Proof Basic Training for Deeper Mathematics ! "#\$\$%&#'!()+!!,-''!.)-/%)/#0! 1)2#3\$4)0\$!-5!"#\$%)4#\$&'!! 1)2#3\$4)0\$!-5!"#\$%)4#\$&#6!7&)0)'! 7#0!83#0&'-!7\$#\$)!90&:)3'&\$;! ### The Foundations: Logic and Proofs. Chapter 1, Part III: Proofs The Foundations: Logic and Proofs Chapter 1, Part III: Proofs Rules of Inference Section 1.6 Section Summary Valid Arguments Inference Rules for Propositional Logic Using Rules of Inference to Build Arguments ### Mathematics Review for MS Finance Students Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions,
Search # What is the formula for calculating the area of the square? What is the formula for calculating the area of the square? The area of the square is the measure of the surface between the sides of the square, and is calculated by raising the measure of the side to the second, that is, by multiplying the measure of the side by itself. ## How do you calculate the area of a formula square? A = lxl = l2. In other words, the AREA of the SQUARE is obtained BY MULTIPLYING the measurement of the SIDE for ITSELF. ## How do you find the area of the square inverse formula? Now we know that the AREA of the SQUARE is obtained by MULTIPLICATING the measurement of the SIDE by ITSELF, that is, by squaring the measurement of the side. We also know that the inverse operation to squaring is the extraction of the square root. ## How do you discover the side of the square from the area? If the area of the square is known, to obtain the measurement of the side it is sufficient to extract the square root of the area. ## How do you find the height of a square? We only know the base, but by applying the Pythagorean theorem we can calculate the height. In fact: h = lx 0,866 = 18 x 0,866 = 15,59 cm. ## Find 28 related questions ### How is the area of the elementary school square calculated? In practice, to calculate the area of the square we only need the measurement of one side and it will be enough to multiply this value by itself. The area of the square is equal to the product of the side by the side. ### How do you calculate the perimeter and area of a square? therefore to find the perimeter knowing the side of the square you have to multiply the measurement of the side by 4. A square has a side of 4,5 centimeters. What is its perimeter? To calculate the perimeter of the square from the area, the square root of the area must be extracted and the result multiplied by 4. ### How is the area and perimeter of the square calculated? Conceptually it is: for the area, to take the units of one side as many times as there are units on the other; for the perimeter, to add the lengths of all sides, which in the square are 4 and all have the same size. ### How do you calculate the area of a figure whose perimeter is known? The perimeter, in this case, will be found by calculating the sum of the length of all sides, while the area will be calculated through the product between the longest side, called base, and the shortest side, called height. ### How do you calculate the perimeter of a rectangle having only the area? The formula used to find the area of the rectangle is "A = bxh". The formula for the perimeter of the rectangle is "P = 2 x (b + h)". In the previous formulas "A" is the area, "P" is the perimeter, "b" is the base of the rectangle and "h" its height. ### How is the area and perimeter of the triangle calculated? The formula for the perimeter is the general one: P = a + b + c. As well as that for the area: A = (base x height) / 2 = (axh) / 2. ### How do you calculate the dimensions of a rectangle knowing the area? A = bxh = 6 cm x 4 cm = 24 cm2. Now let's imagine we only know the area and the base of the rectangle. It is evident that we can calculate the height as follows: h = A / b = 24/6 = cm 4. ### How is the perimeter and area of the isosceles trapezium calculated? ISOSCELE KEYSTONE: PERIMETER FORMULA The perimeter of the isosceles trapezium is obtained by adding the two bases (major and minor) and the two sides of the trapezium. ### What is the perimeter of a square that has a side of 4 cm? To calculate the perimeter of the square you can: add the measurements of the sides 3 + 3 + 3 + 3 = 12 cm. multiply the measure of the side by four 3 × 4 = 12 cm in fact AB = BC = CD = DA so the perimeter of the square is four times AB. ### How is the area of the rectangle and square calculated? So let's calculate the surface: 5 cm2 repeated 5 times = 5 x 5 = 25 cm2. But since 5 cm is the measure of the base and the height, we can find out the formula to calculate the area of each square: bxh = lxl = l2. ### How do you calculate the area of a square starting from the diagonal? Example 1: calculate the area of a square whose diagonal measures 7 cm. We apply the formula: A = d2 / 2 = 72/2 = 2 cm24,5. ### How to explain the area in primary school? The area indicates the extent of the surface enclosed by the polygon. The area is a quantity of two dimensions: to calculate the area, in fact, we take a “unit” figure as a reference and see how many times it is contained in the figure we are analyzing. ### How to calculate the area of a trapezoid with the perimeter? Formulas of the area and perimeter of the trapezoid and rhombus. (Major base + minor base) by height divided by two. ### How do you find the area of the trapezius? The area of the trapezoid is calculated by multiplying the sum of the bases by the height and dividing the result by 2. ### How are the sides of an isosceles trapezoid located? oblique side (knowing only bases and height) = √ height2 + [(major base-minor base): 2] height (knowing the area) = (2xarea): (major base + minor base) height (knowing only bases and side oblique) = √ oblique side2 - [(major base-minor base): 2] ### How do you find the catheti having the area? The area of the right triangle can be calculated by multiplying the measurements of the two legs and dividing the result by 2, or by dividing the product between the measurements of the hypotenuse and the height of the right triangle by 2. ### How do you find the width of the rectangle? The length of the two longer opposite sides is called the length or base of the rectangle, while the length of the two shorter sides is called the width or height. ### What is the formula for calculating the area of the triangle? TRIANGLE AREA ANY The formula for calculating the area of the triangle in general is: Base for Height / 2. You can also use another formula, that of Heron, which foresees the knowledge, however, of the measure of the three sides of the triangle. ##### add a comment of What is the formula for calculating the area of the square? Comment sent successfully! We will review it in the next few hours.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 8 Go to the latest version. # 1.5: Connect Variable Expressions and the Order of Operations with Real-World Problems Difficulty Level: At Grade Created by: CK-12 0% Progress Practice Expressions for Real-Life Situations Progress 0% Have you ever been to a ballet? Sara is going to see the "Nutcracker", but she needs to figure out the cost for tickets. She isn't sure if three or four of her friends are going to join her at the performance. The cost for one ticket is $35.00, and there is a$2.00 one time fee for the ticket purchase. Can you write a variable expression to explain this situation? Can you figure out the total cost if three people attend the performance? Can you figure out the total cost if four people attend the performance? Pay attention to this Concept, and you will be able to help Sara at the end of the Concept. ### Guidance When we have an unknown quantity in a problem, we can use a variable to help us to find the value of our problem. You can find many real-world scenarios where variable expressions would be helpful in solving problems. Many different places sell tickets. There are also prices for adults and prices for children. The number of tickets can vary, or they can be the same. If we know the number of tickets, then we can figure out the total amount of revenue earned based on the cost of the ticket for an adult and for a child. Take a look at this situation. An amusement park charges eight dollars admission and one dollar and fifty-cents per ride. Write an expression to find the cost of admission and five ride tickets. First, notice that we have eight dollars admission. That is the first part of the expression. 8 Next, they charge 1.50 per ride. 8+1.50x\begin{align}8 + 1.50x\end{align} We used the variable for the number of rides since this is the variable or changeable facet. The number of rides can change. In this example, we were asked to figure out the cost for 5 rides. That is the value that we can substitute into our expression for x\begin{align}x\end{align}. 8+1.50(5)\begin{align}8 + 1.50(5)\end{align} Now we can use the order of operations to solve this problem. 8+7.5015.50 The cost of admission plus five ride tickets is15.50. Using variable expressions can help you solve for unknown quantities. Just remember to use the order of operations so that your work is accurate!! #### Example C How much would it cost for five cones with sprinkles? Solution: 19.25\begin{align}19.25\end{align} Now let's go back to the dilemma at the beginning of the Concept. Let's think about what we know. Sara knows the cost for one ticket.35.00\begin{align}35.00\end{align} She isn't sure how many people are going. That is our variable. Let's use x\begin{align}x\end{align}. 35x\begin{align}35x\end{align} But there is also a 2.00\begin{align}2.00\end{align} one time charge. Let's add that to our expression. 35x+2\begin{align}35x + 2\end{align} Now we can figure out the cost for three people by substituting three in for x\begin{align}x\end{align}. 35(3)+2=107\begin{align}35(3) + 2 = 107\end{align} We can also figure out the cost for four people. 35(4)+2=142\begin{align}35(4) + 2 = 142\end{align} Our work is complete. ### Vocabulary Evaluate to figure out the value of a numerical or variable expression. Equation a mathematical statement with an equals sign where one side of the equation has the same value as the other side. Expression a group of numbers, symbols and variables that represents a quantity. Numerical Expression a group of numbers and operations. Variable Expression a group of numbers, operations and at least one variable. Variable a letter used to represent an unknown quantity. Grouping Symbols parentheses or brackets used to group numbers and operations. ### Guided Practice Here is one for you to try on your own. Harriet is making a cake. She needs to go to the store and purchase some almonds and a can of evaporated milk. When she arrives at the store, she picks up the can of evaporated milk first. It costs1.99. Then she goes to get the almonds. Almonds are $3.99 per pound. Can you write a variable expression so that Harriet can calculate her total cost based on how many pounds she purchases? Solution To accomplish this task, first let's figure out the variable. The variable is going to be the number of pounds purchased since that is the changeable amount. The price per pound is$3.99. Now let's write the first part of the variable expression. 3.99x\begin{align}3.99x\end{align} Harriet isn't only purchasing almonds, she is also buying a can of evaporated milk. We can add that cost to our variable expression. 3.99x+1.99\begin{align}3.99x + 1.99\end{align} ### Practice Directions:Write a variable expression for each situation described below. 1. A pound of apples costs $4.50. Kelly isn't sure how many pounds she is going to purchase. 2. Each touchdown at a football game is worth 7 points. Write a variable expression where the number of touchdowns scored can change. 3. Alex bought several bunches of bananas. Each bunch costs$.89. Write a variable expression to describe this situation. 4. If an ice cream cone costs $3.20 and hot fudge is an additional$.45, write a variable expression where the number of ice cream cones can change. 5. There are six children in the Smith family. If each child gets a haircut, write a variable expression to show the cost of the haircut so that the total cost can be calculated. 6. A turkey costs $6.75 per pound. Write a variable expression where the weight of the turkey is the changeable amount. 7. The car wash charges$15.00 per car. Write a variable expression that can be used to calculate the number of cars washed in one hour. 8. Kelly bought a pair of sneakers for $35.00. She also bought a pile of different laces. Each set of laces costs$3.00. Write a variable expression to show how Kelly could calculate her total cost. Directions:Now go back to each variable expression that you wrote in numbers 1 - 8 and evaluate each expression as if the given value for the variable was 4. These are your answers for numbers 11 - 16. ### Vocabulary Language: English Algebraic Expression Algebraic Expression An expression that has numbers, operations and variables, but no equals sign. Equation Equation An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs. Evaluate Evaluate To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value. Expression Expression An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols. Grouping Symbols Grouping Symbols Grouping symbols are parentheses or brackets used to group numbers and operations. Numerical expression Numerical expression A numerical expression is a group of numbers and operations used to represent a quantity. Dec 19, 2012 Aug 10, 2015
# 3.6 Function operations Page 1 / 2 The form of “function as a rule” suggests that we may think of carrying out arithmetic operations like addition, multiplication etc with two functions. If we limit ourselves to real function, then we can attach meaning to equivalent of arithmetic operations with predictable domain intervals. We should, however, clearly understand that function operations with real functions involve new domain for the resulting function. In general, function operation results in contraction of intervals in which new rule formed from algebraic operation is valid. As pointed out, function operations are defined for a new domain, depending on the type of operations - we carry out. In the nutshell, we may keep following aspects in mind, while describing function operations : • Function operations are defined for real functions. • The result of function operation is itself a real function. • New function resulting from function operation is defined in new interval(s) of real numbers as determined by the nature of operation involved. ## Domain of resulting function The function operations, like addition, involve more than one function. Each function has its domain in which it yields real values. The resulting domain will depend on the way the domain intervals of two or more functions interact. In order to understand the process, let us consider two functions “ ${y}_{1}$ ” and “ ${y}_{2}$ ” as given below. ${y}_{1}=\sqrt{\left({x}^{2}-3x+2\right)}$ ${y}_{2}=\frac{1}{\sqrt{\left({x}^{2}-3x-4\right)}}$ Let " ${D}_{1}$ " and" ${D}_{2}$ " be their respective domains. Now, the expressions in the square roots need to be non-negative. For the first function : ${x}^{2}-3x+2\ge 0$ $⇒\left(x-1\right)\left(x-2\right)\ge 0$ The sign diagram is shown here. The domain for the function is the intervals in which function value is non-negative. $⇒{D}_{1}=-\infty Note that domain, here, includes end points as equality is permitted by the inequality "greater than or equal to" inequality. In the case of second function, square root expression is in the denominator. Thus, we exclude end points corresponding to roots of the equation. ${x}^{2}-3x-4>0$ $⇒\left(x+1\right)\left(x-4\right)>0$ $⇒{D}_{2}=-\infty Now, let us define addition operation for the two functions as, $y={y}_{1}+{y}_{2}$ The domain, in which this new function is defined, is given by the common interval between two domains obtained for the individual functions. Here, domain for each function is shown together one over other for easy comparison. For new function defined by addition operation, values of x should be such that they simultaneously be in the domains of two functions. Consider for example, x = 0.75. This falls in the domain of first function but not in the domain of second function. It is, therefore, clear that domain of new function is intersection of the domains of individual functions. The resulting domain of the function resulting from addition is shown in the figure. $⇒D={D}_{1}\cap {D}_{2}$ This illustration shows how domains interact to form a new domain for the new function when two functions are added together. where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good What is power set Period of sin^6 3x+ cos^6 3x Period of sin^6 3x+ cos^6 3x
Enable contrast version # Tutor profile: Christine A. Christine A. Tutor for over 9 years ## Questions ### Subject:Pre-Algebra TutorMe Question: Solve $$4(x+6)+2(3x-5)$$. Christine A. Answer: To solve this, we work from left to right in the equation. The first step is to distribute the $$4$$ to what is inside of the parenthesis. To do this, we first multiply $$4$$ by $$x$$, and then we multiply $$4$$ by $$6$$. So the first part of the equation becomes $$4(x)+4(6)$$. Next we simplify the equation. We cannot simplify $$4x$$ beyond what we have already done, because we do not know what the variable $$x$$ means here, so it stays as $$4x$$. However, we can simplify $$4(6)$$: $$4$$ times $$6$$ is $$24$$. So the first part of the equation is $$4x+24$$. The same thing we did to the first part of this equation, we do to the second part of the equation as well. Remember that the equation, although it looks like $$3x-5$$, it is actually $$3x+(-5)$$. Now we multiply $$2$$ by $$3x$$, and we multiply $$2$$ by $$-5$$. $$2$$ times $$3x$$ becomes $$6x$$. $$2$$ times $$-5$$ is $$-10$$. So the second part of the equation becomes $$6x+(-10)$$. To simplify that even more, it becomes $$6x-10$$. Now we combine what we did to find the final answer. $$4x+24$$ plus $$6x-10$$. This equation becomes $$4x+24+6x-10$$. Now we combine like terms - any terms that have $$x$$ in it can be combined, and any terms that that are normal numbers can also be combined. Here, we have two terms that have an $$x$$: $$4xx$$ and $$6x$$. So these terms can be added together: $$4x+6x$$, which becomes $$10x$$. We also have two numbers that can be added together: $$24$$ and $$-10$$. This equation becomes $$24+(-10)$$, which is $$24-10=14$$. Now we bring it all together to find the final answer: $$10x+14$$. So the answer to the question asking us to solve $$4(x+6)+2(3x-5)$$ is $$10x+14$$. ### Subject:English TutorMe Question: Please provide an example of a metaphor and an example of a simile. Christine A. Answer: A metaphor is a figure of speech that compares one thing to another, but the sentence does not use the words "like" or "as". An example is: "She is a couch potato." A simile is similar to a metaphor, but the main difference is that it uses the words "like" or "as". An example is "She ran like the wind." ### Subject:Algebra TutorMe Question: What is the inverse of the function $$f(x)=x^2+4$$? Christine A. Answer: When you see a question that asks you to find the inverse to a function, the easiest thing to do is to switch the variables around. It sounds more complicated than it is. Remember that $$f(x)$$ is just another way to say "y" in an equation. So putting that understanding in, the question is asking what the inverse of $$y=x^2+4$$ is. Now that we know what is "x" and "y" here, remember that an inverse means to switch the variables. So to find the inverse of the equation that they gave us, switch around the "x" and "y" variables. When you switch those variables around, the equation becomes $$x=y^2+4$$. Now, like most other algebra equations, we solve for y. To solve for "y", you have to get "y" by itself. So the first thing we can do is subtract 4 from each side - we want to do that so that we can start to get the "y" variable by itself, and whatever we do to one side of an equation we have to do to the other side as well. So when we subtract 4 from both sides, we get $$x-4=y^2$$. In the equation, we see that "y" is squared. To finish the problem and get "y" by itself, we have to find the square root of "y". And remember, what we do to one side of an equation, we have to do to the other as well. So we have to find the square root of both sides of the equation. That leaves us with the answer: $$\sqrt{x-4}=y$$. Remember, "y" is another way to say $$f(x)$$. To signify a function is the inverse of another, we use the term $$f^{-1}(x)$$ [and the original function remains $$f(x)$$]. So the answer is $$\sqrt{x-4}=f^{-1}(x)$$. To put it in an order that is more familiar, the answer is $$f^{-1}(x)=\sqrt{x-4}$$. ## FAQs What is a lesson? A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard. How do I begin a lesson? If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson. Who are TutorMe tutors? Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you. BEST IN CLASS SINCE 2015 TutorMe homepage Made in California by GoGuardian Copyright © 2022. Zorro Holdco, LLC doing business as TutorMe. All Rights Reserved. High Contrast Mode On Off
MEASURING MAGNITUDES - Numbers and Counting - Numbers: Their Tales, Types, and Treasures ## Chapter 1: Numbers and Counting ### 1.14.MEASURING MAGNITUDES Natural numbers are “counting numbers.” They can be used to count every finite collection of arbitrary things; they measure the size of a set. But there is an even more important aspect of numbers that we have not covered yet, and that goes beyond just natural numbers. In everyday life, we use numbers not so much to measure sets but to measure and describe quantities and magnitudes. These entities cannot be counted in a strict sense. For example, the length of a line segment is a magnitude—in itself it is something quite different from a natural number, which, as we have just seen, describes the cardinality of a set. How is it that we can express lengths through numbers? The key idea here is to measure a length in multiples of unit lengths. The choice of units is a mere matter of convention. In practice, the unit is provided by the measuring device that we use to measure the given quantity. For measuring lengths we have, for example, a measuring tape or a ruler with the units printed on it. By comparing a given length with the ruler and its units, we determine a measuring number that describes the length. For example, in figure 1.10, we find that seven unit lengths in a row add up to give us the length of a stick. We see that measuring lengths is just another form of counting. Figure 1.10: Measuring lengths is done by counting units. What do we do when the length of the object does not fit exactly to a whole number of units, as in figure 1.11? In this case, we first count the number of whole units that measure the length of the stick. This gives two inches. Now there is a remainder that is shorter than one inch. The given length is not a multiple of the unit. In that case, we divide the unit into suitable subunits and count the remainder in these subunits. A subunit is always a fraction of the given unit, which means the subunit is given as one nth of the unit. For example, one could take inch as a subunit for one inch, as in figure 1.11, where we count seven of these subunits. This means that the length of the stick is two inches plus seven times of an inch, which is usually written as 2 ". Figure 1.11: Measuring lengths with units and subunits. Metric units are usually divided into ten parts. One tenth of a centimeter is a millimeter. The same stick would then have a length of 7 cm and 3 mm. For the division of units into ten parts, we have a convenient notation that is an extension of the place-value system described in theprevious section: We write 7.3 (seven point three) when the magnitude is given by seven units and three tenths of that unit. And if we need a higher accuracy, we could divide smaller subunits even further. With more precise methods, we would find that a stick of 2 and inches would measure 7.3025 cm, which means 7 units + 3 tenths + 2 thousandths + 5 ten-thousandths. Notice that we have the role of the zero symbol as a place holder, denoting the absence of hundredths in this case. We call a number that measures a quantity by comparison with a given unit a real number. An important special case is a rational number. A rational number describes a quantity as a multiple of a unit plus a multiple of a subunit, as, for example, in 2 plus inches or 7 plus centimeters or 5 plus gallons. In all these cases, it is possible to express the quantity as an integer multiple of a suitably chosen subunit: It has come as a surprise that this cannot be done in all cases. Not every length can be described as an integer multiple of a unit plus an integer multiple of a subunit. Such a number would be called irrational—as distinguished from the rational numbers shown in the examples above. In other words, an irrational number cannot be expressed as a fraction. An example of an irrational number is the length of the diagonal of a unit square—one whose side length is one. The length d of the diagonal of the unit square is the square root of two or, written symbolically, d = = 1.41421356…. This means that the length of the diagonal equals the length of the side plus four tenths of the side, plus one hundredth of the side, plus four thousandths, plus two ten-thousandths, and so on. The chain of digits behind the dot will never end. This alone would not make d irrational, as the example with 5.33333…= 5 shows. This number has also infinitely many digits in the base-10 representation, but these digits are all the same. In contrast, there is no repetitive pattern in the digits of d = . And one can show that it is not possible to write d as a multiple of a certain fraction of the unit, which makes it irrational. Another famous example of an irrational number is π = 3.14159… The measure of the circumference of a circle, using the diameter of that circle as the unit of length equals three times the diameter of the circle, plus one tenth of the diameter, plus four hundredths of the diameter, plus…. Again, the chain of digits would never end and follows no regular pattern (no periodic repetitions). Thus the number π is irrational: it cannot be written as a multiple of a fraction of the unit. More about this amazing number π will be discussed in chapter 10. Provided that a unit (a “gauge”) is fixed, magnitudes of all kinds can be measured by the same type of real number, and in quite a similar way as we measure lengths. We measure areas by square meters and volumes by cubic meters or similar units. We measure time by counting hours, minutes, and seconds. It is quite interesting that hours and minutes have sixty subunits, which reminds us of the ancient Babylonian numeral system (see chapter 3). Thus, we now have a better understanding of the nature and development of how numbers were represented over time to where we now have a rather sophisticated system for representing quantities.
# How do you factor 16+ 22x - 3x ^ { 2}? Jul 25, 2017 $- \left(3 x + 2\right) \left(x - 8\right)$ #### Explanation: $f \left(x\right) = - y = - \left(3 {x}^{2} - 22 x - 16\right)$ Factor y by the new AC Method (Socratic Search) $y = 3 {x}^{2} - 22 x - 16 =$ 3( x + p)(x + q) Converted trinomial: $y ' = {x}^{2} - 22 x - 48 =$ (x + p')(x + q') Proceeding: Find p' and q', then, divide them by a = 3. Find p' and q', knowing the sum (b = -22) and the product (ac = -48). They are: p' = 2 and q' = - 24. Back to y --> $p = \frac{p '}{a} = \frac{2}{3}$, and $q = - \frac{24}{3} = - 8$ Factored form: $y = 3 \left(x + \frac{2}{3}\right) \left(x - 8\right) = \left(3 x + 2\right) \left(x - 8\right)$ Finally, $f \left(x\right) = - \left(3 x + 2\right) \left(x - 8\right)$
# What's a trillion times a trillion ### Just count it one two three four,... how far can you count? Can you count in other languages too? Can you write the numbers too? As a word and / or as a number? ### Numbers, numbers and more numbers All around you you will find numbers: your age, your phone number, the number of your siblings, the time, the odometer in the car, prices, population figures, etc. Numbers bring order to the world. Mathematicians call these numbers natural numbers . The set of natural numbers is unlimited: \$\$ NN = {0, 1, 2, 3, 4,…} \$\$ Some books differentiate between \$\$ NN = {1, 2, 3, 4,…} \$\$ and \$\$ NN_0 = {0, 1, 2, 3, 4,…} \$\$. ### Predecessor and successor The predecessor is the number that came directly before the given number when counting, i.e. the number one smaller. Example: 3 is the predecessor of 4, 56 is the predecessor of 57, 1 345 678 is the predecessor of 1 345 679 The successor is the number that comes next after the given number when counting, i.e. the number that is one larger. Example: 5 is the successor to 4, 58 is the successor to 57, 1 345 680 is the successor to 1 345 679 Predecessor - number - successor All natural numbers have exactly one successor. All natural numbers except the smallest natural number have exactly one predecessor. ### The system of ten For writing numbers let's use the Digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. With this you can write any number. Why? We write our numbers in Decimal system. There are 10 One to a Tens combined, 10 tens to one Hundreds, 10 hundreds to one Thousands etc. The number So 3 333 consists of 3 thousands, 3 hundreds, 3 tens and 3 ones, but is only used with the Digit 3 written. Image: Druwe & Polastri kapiert.decan do more: • interactive exercises and tests • individual classwork trainer • Learning manager ### Names for large numbers 1 million (million) = 1,000 thousand = 1,000,000 1 billion (billion) = 1,000 million = 1,000,000,000 1 trillion (trillion) = 1,000 billion = 1,000,000,000,000 1 trillion = 1,000 trillion = 1,000,000,000,000,000 1 trillion = 1 000 trillion = 1 000 000 000 000 000 000 It's best to remember the number of zeros in large numbers: Thousand: 3 zeros Million: 6 zeros Billion: 9 zeros Billion: 12 zeros Be careful when translating English numerals: German English million million billion billion (also: thousand million) trilliontrillion (also: million million) ### The place value table You can write large numbers on a place value board for a better overview: Example: The number word fifteen million four hundred and sixty thousand as a number: 15,460,000 In order to be able to read large numbers better without a place value table, you write them in packs of three and leave a small space in between. 1345789 looks like this in a pack of three: 1 345 789 Sometimes you also see points: 1,345,789 You write numbers up to a million in one word. Example: three hundred and sixty thousand Numbers over a million are written separately. Example: two million sixty thousand The place value table is also called the place value table.
# 4.5: Intercepts by Substitution Difficulty Level: Basic Created by: CK-12 Estimated9 minsto complete % Progress Practice Intercepts by Substitution MEMORY METER This indicates how strong in your memory this concept is Progress Estimated9 minsto complete % Estimated9 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Suppose the number of gallons of gas left in the tank of a car depends on the number of miles driven and can be represented by a linear equation. You know how much gas is in the tank when 0 miles have been driven, and you also know after how many miles you will have 0 gallons left in the tank. Could you graph the linear equation? In this Concept, you'll learn how to graph equations in situations like this one, where you know the intercepts. ### Guidance As you may have seen in the previous Concept, graphing solutions to an equation of two variables can be time-consuming. Fortunately, there are several ways to make graphing solutions easier. This Concept will focus on graphing a line by finding its intercepts. A future Concept will show you how to graph a line using its slope and \begin{align}y-\end{align}intercept. In geometry, there is a postulate that states, “Two points determine a line.” Therefore, to draw any line, all you need is two points. One way is to find its intercepts. An intercept is the point at which a graphed equation crosses an axis. The \begin{align}x-\end{align}intercept is an ordered pair at which the line crosses the \begin{align}x-\end{align}axis (the horizontal axis). Its ordered pair has the form \begin{align}(x,0)\end{align}. The \begin{align}y-\end{align}intercept is an ordered pair at which the line crosses the \begin{align}y-\end{align}axis (the vertical axis). Its ordered pair has the form \begin{align}(0,y)\end{align}. By finding the intercepts of an equation, you can quickly graph all the possible solutions to the equation. Finding Intercepts Using Substitution Remember that the Substitution Property allows the replacement of a variable with a numerical value or another expression. You can use this property to help find the intercepts of an equation. #### Example A Graph \begin{align}2x+3y=-6\end{align} using its intercepts. Solution: The \begin{align}x-\end{align}intercept has the ordered pair \begin{align}(x,0)\end{align}. Therefore, the \begin{align}y-\end{align}coordinate has a value of zero. By substituting zero for the variable of \begin{align}y\end{align}, the equation becomes: \begin{align}2x+3(0)=-6\end{align} Continue solving for \begin{align}x\end{align}: \begin{align}2x+0& =-6\\ 2x& =-6\\ x& =-3\end{align} The \begin{align}x-\end{align}intercept has an ordered pair of (–3, 0). Repeat the process to find the \begin{align}y-\end{align}intercept. The ordered pair of the \begin{align}y-\end{align}intercept is \begin{align}(0,y)\end{align}. Using substitution: \begin{align}2(0)+3y& =-6\\ 3y& =-6\\ y& =-2\end{align} The \begin{align}y-\end{align}intercept has the ordered pair (0, –2). To graph the line formed by the solutions of the equation \begin{align}2x+3y=-6\end{align}, graph the two intercepts and connect them with a straight line. #### Example B Graph \begin{align}4x-2y=8\end{align} using its intercepts. Solution: Determine the \begin{align}x-\end{align}intercept by substituting zero for the variable \begin{align}y\end{align}. \begin{align}4x-2(0)& =8\\ 4x& =8\\ x& =2\end{align} The ordered pair of the \begin{align}x-\end{align}intercept is (2, 0). By repeating this process, you'll find the \begin{align}y-\end{align}intercept has the ordered pair (0, –4). Graph these two ordered pairs and connect them with a line. #### Example C Graph \begin{align}x-2y=10\end{align} using intercepts. Solution: First, find the intercepts by substituting in zero for \begin{align}x\end{align} and \begin{align}y\end{align}: \begin{align} \text{Substitute in zero for x.} && (0)-2y &=10 \\ \text{Simplify.} && -2y&=10 \\ \text{Solve for y.}&& y&=-5 \end{align} This means that the \begin{align}y\end{align}-intercept is at the point (0,-5). \begin{align} \text{Substitute in zero for y.} && x-6(0) &=10 \\ \text{Solve for x.}&& x&=10 \end{align} This means that the \begin{align}x\end{align}-intercept is at the point (10,0). ### Guided Practice Find the intercepts and use them to graph \begin{align}y=-2x+8\end{align}. Solution: Substitute in zero for \begin{align}x\end{align} and \begin{align}y\end{align}, in order to find the intercepts. \begin{align} \text{Substitute in zero for x.} && y&=-2(0)+8 \\ \text{Solve for y.}&& y&=8 \end{align} This means that the \begin{align}y\end{align}-intercept is at the point (0,8). \begin{align} \text{Substitute in zero for y.} && 0&=-2x+8 \\ \text{Solve for x.}&& x&=4 \end{align} This means that the \begin{align}x\end{align}-intercept is at the point (4,0). ### Practice Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Graphing Using Intercepts (12:18) 1. Define intercept. 2. What is the ordered pair for an \begin{align}x-\end{align}intercept? Find the intercepts for the following equations using substitution. 1. \begin{align}y=3x-6\end{align} 2. \begin{align}y=-2x+4\end{align} 3. \begin{align}y=14x-21\end{align} 4. \begin{align}y=7-3x\end{align} 1. Which intercept does a vertical line have? 2. Does the equation \begin{align}y=5\end{align} have both an \begin{align}x-\end{align}intercept and a \begin{align}y-\end{align}intercept? Explain your answer. 3. Write an equation having only an \begin{align}x-\end{align}intercept at (–4, 0). 4. How many equations can be made with only one intercept at (0, 0)? Hint: Draw a picture to help you. Mixed Review For 11-12, Determine whether each ordered pair is a solution to the equation. 1. \begin{align}5x+2y=23;(7,-6)\end{align} and (3, 4) 2. \begin{align}3a-2b=6;(0,3)\end{align} and \begin{align}\left (\frac{5}{3},\frac{-1}{2}\right )\end{align}. 3. Graph the solutions to the equation \begin{align}x=-5\end{align}. 4. Solve: \begin{align}\frac{4}{5} k-16=-\frac{1}{4}\end{align}. 5. Is the following relation a function? \begin{align}\left \{(-1,1),(0,0),(1,1),(2,3),(0,6)\right \}\end{align} 6. Using either a finite set of counting numbers, whole numbers, or integers, how would you describe the domain of the following situation: The number of donuts purchased at a coffee shop on a particular day? 7. Find the percent of change: Old price = $1,299; new price =$1,145. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish TermDefinition $X$-intercept An ordered pair at which the line crosses the $x-$axis (the horizontal axis). Its ordered pair has the form $(x,0)$. $Y$-intercept The $y-$intercept is an ordered pair at which the line crosses the $y-$axis (the vertical axis). Its ordered pair has the form $(0,y)$. Intercept The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis. Intercept Method The intercept method is a way of graphing a linear function by using the coordinates of the $x$ and $y$-intercepts. The graph is drawn by plotting these coordinates on the Cartesian plane and then joining them with a straight line. Show Hide Details Description Difficulty Level: Basic Tags: Subjects:
Courses Courses for Kids Free study material Offline Centres More Store # A certain amount of money is distributed among A, B and C. A gets $\dfrac{3}{16}$ and B gets $\dfrac{1}{4}$ of the whole amount. If C gets Rs.81; then what does B get?(a) Rs.36(b) Rs.56(c) Rs.46(d) Rs.63 Last updated date: 13th Jun 2024 Total views: 392.4k Views today: 4.92k Verified 392.4k+ views Hint: Assume the total money that is distributed as ‘x’. Find the amount A and B get by finding ${{\left( \dfrac{3}{16} \right)}^{th}}$ part of x and ${{\left( \dfrac{1}{4} \right)}^{th}}$ of part of x respectively. Now, calculate the amount of money C gets by subtracting the sum of money of A and B from total money. Equate the amount of money C gets with 81 and find the value of x. Finally, substitute this value of x in the expression of money obtained by B to get the answer. Here, we have been given that a certain amount of money is distributed among A, B and C. Let us assume the total amount of money distributed is ‘x’. Now, it is given that A gets $\dfrac{3}{16}$ and B gets $\dfrac{1}{4}$ of the whole amount. Therefore, we have, Money obtained by A = $\dfrac{3}{16}\times x=\dfrac{3x}{16}$ Money obtained by B = $\dfrac{1}{4}\times x=\dfrac{x}{4}$ Therefore, the amount of money obtained by C will be the difference of total money and sum of money obtained by A and B. So, we have, $\Rightarrow$ Money obtained by C = $x-\left( \dfrac{3x}{16}+\dfrac{x}{4} \right)$ $\Rightarrow$ Money obtained by C = $x-\left( \dfrac{3x+4x}{16} \right)$ $\Rightarrow$ Money obtained by C = $x-\dfrac{7x}{16}$ $\Rightarrow$ Money obtained by C = $\dfrac{16x-7x}{16}$ $\Rightarrow$ Money obtained by C = $\dfrac{9x}{16}$ It is given that money obtained by C is Rs.81. Therefore, equating it with the expression of money obtained by C, we get, \begin{align} & \Rightarrow \dfrac{9x}{16}=81 \\ & \Rightarrow x=\dfrac{16\times 81}{9} \\ \end{align} $\Rightarrow x=$ Rs.144 Therefore, the total amount of money that was distributed is Rs.144. Now, we have to find the amount of money obtained by B. So, substituting the obtained value of x in the expression of money obtained by B, we get, Money obtained by B = $\dfrac{144}{4}$ = Rs.36 So, the correct answer is “Option a”. Note: One may note that we do not have to assume the amount of money obtained by A and B as different variables. It may confuse us. We just have to assume one variable and carry out our calculation using that assumption. Note that the amount of money obtained by C was given to us. That is why we equated it with the obtained expression for money obtained by C. Remember that the value of x is not our solution, we have to substitute it in the expression $\dfrac{x}{4}$. Sometimes in a hurry students just write the value of ‘x’ as the answer. So, the question must be read carefully.
Brickwork Calculation Formula Contents How to calculate Number of bricks, cement and Sand for brickwork? Any construction works requires a pre estimation of the quantities of the materials to be used, which is necessary to predict the cost and availability of materials. So, With brickwork calculation formula we will estimate how much quantity of the Bricks, Cement and Sand are Required. Lets Consider, Standard Size of a Brick = 190 x 90 x 90 mm Mortar Thickness = 10 mm Brickwork Volume = 1 x 1 x 1 = 1 Cu.m Number of Bricks Step 1: Lets find volume of a single Brick without mortar Volume of Brick without mortar = 0.19 x .09 x .09 = 0.001539 Cu.m Step 2: Now find the volume of single brick with mortar Volume of Brick without mortar = 0.2 x 0.1 x 0.1 = 0.002 Cu.m Step 3: Now lets find the number of bricks required for the required brickwork volume. Wall size = 1 cu.m Number of bricks with mortar =1/0.002 = 500 No’s. Related : Quantity of Mortar Total Quantity of Mortar = Volume of wall -(Volume of brick without mortar x Number of bricks required) = 1 – ( 0.001539 x 500 ) = 1 – 0.7695 = 0.2305 Cu.m. Now lets find the quantity of mortar i.e Cement and Sand with ratio of 1: 6 Mortar Quantity in wet Condition = 0.2305 Cu.m Mortar Quantity in dry Condition = 0.2305 x 1.33 = 0.306565 Cu.m. Cement Calculation Cement Quantity = Volume of Mortar x (Ratio of cement / Sum of the ratio) = 0.306565 x (1/7) = 0.0438 cu.m Density of Cement = 1440 kg/m3 No of Bags = 0.0438 x 1440 = 63.06 kg Sand Calculation Volume = Volume of Mortar x (Ratio of cement / Sum of the ratio) = 0.306565 x (6/7) = 0.2627 Cu.m. Cubic Feet = 0.2627 x 35.31 = 9.27841 Total Material quantity required for 1 cu.m wall brick work No’s Of Brick = 500 No’s Cement Quantity = 63 kg Sand Quantity = 0.2627 Cu.m Found Useful, Share with others
# How do you solve 2x² - 7x + 15 = 0? Jun 26, 2015 Note that this equation has no Real roots. Complex roots can be determined using the quadratic formula. $\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{7 + \sqrt{71} i}{4}$ or $x = \frac{7 - \sqrt{71} i}{4}$ #### Explanation: The roots of a quadratic of the form $\textcolor{w h i t e}{\text{XXXX}}$$a {x}^{2} + b x + c = 0$ can be determined using the quadratic formula: $\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ In this case: $\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{7 \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(2\right) \left(15\right)}}{2 \left(2\right)}$ $\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{7 \pm \sqrt{- 71}}{4}$ $\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{7 + \sqrt{71} i}{4}$ or $x = \frac{7 - \sqrt{71} i}{4}$
top Chapter 7—Percents 7-2 Expanding the Scope of Percent Notation The previous lesson introduced you you percent notation and their equivalent fraction and decimal forms. There is more to the percent number line that integers between 0 and 100. Here is what you will be studying in this section: Percents that have fraction/decimal parts, such as: 56½%, ¼%, 37.8%, 3.625% Percents that are greater than 100, such as: 110%, 200%, 102.4% Percents with Fraction or Decimal Parts Sometimes you will find percents that include fractions and decimal values. Financial reports and mortgage rates, for instance, sometimes include percentages written as 6½ % or 12.3 %. There is nothing tricky about these fractional parts of a percent. They are simply parts of one-hundred percent. You will also find percentages that are less than 1. Examples are 0.1%, 0.125%, and so on. These are nothing more than expressions of small percentages—amounts that are less than 1 out of 100. Recall To rewrite a percent to decimal: Drop the % sign and multiply by 0.01 Examples Rewrite 12½ % as a decimal value: 12½ x 0.01 = 12.5 x 0.01 = 0.125 Rewrite ½ % as a decimal value: 12½ x 0.01 = 12.5 x 0.01 = 0.125 Rewrite 44.25% as a decimal value 44.25 x 0.01 = 0.4425 Examples & Exercises Percent to Decimal Use these Examples & Exercises to check your understanding of how to rewrite percents with fraction parts to decimal. Of course, you should also be able to rewrite any fraction or decimal value to percent. Recall To rewrite a decimal or fraction to percent: Multiply by 100 and attach the % sign. Examples Rewrite 0.0625 as a percent: 0.0625 x 100 = 6.25% or 6¼% Rewrite 0.005 as a percent: 0.005 x 100 = 0.5% or ½% Rewrite 0.00021 as a percent: 0.0002 x 100 = 0.021% Rewrite 1/8 as a percent: 1/8 x 100 = 100/8 = 12.5% Examples & Exercises Decimal to Percent Use these Examples & Exercises to check your understanding of how to rewrite any decimal/fraction to percent. Percents Greater than 100% Examples Rewrite 1.0 as a percent 1.0 x 100 = 100% Rewrite 1.125 as a percent 1.125 x 100 = 112.5% Rewrite 220% as a decimal 220 x 0.01 = 2.2 Rewrite 505% as a decimal 505 x 0.01 = 5.05 Examples & Exercises Percents Larger than 100% Use these Examples & Exercises to check your understanding of percents larger than 100.
# Find the equation of a circle whose centre is (3, –1) Question: Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0. Solution: Given equation of the chord is, $2 x-5 y+18=0$ $5 y=2 x+18$ $y=\frac{2}{5} x+\frac{18}{5}$ As we have, $y=m x+C$ Where, $\mathrm{m}$ is the slope of the line, $m=\frac{2}{5}$ Slope of the line perpendicular to the chord, $m^{\prime}=-\frac{5}{2}$ As the product of slope of perpendicular lines $=-1$, $y-y_{1}=m^{\prime}\left(x-x_{1}\right)$ $y-(-1)=-\frac{5}{2}(x-3)$ $2 y+2=-5 x+15$ $5 x+2 y=13$ $.2$ [Equation of line passing from centre and cutting the chord] Solving both the equations, $2 x-5 y=-18 \& 5 x+2 y=13$ Multiplying the equation $1 \&$ equation 2 by $2 \& 5$ respectively, we get $4 x-10 y=-36$ $25 x+10 y=65$ $29 x=29$ $x=1$ $2(1)-5 y=-18$ $2-5 y+18=0$ $5 y=20$ $y=4$ Point of intersection at chord and radius $=(1,4)$ Distance between the point of intersection \& centre Distance formula $=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ $=\sqrt{(1-3)^{2}+(4-(-1))^{2}}$ $=\sqrt{(-2)^{2}+(4+1)^{2}}$ $=\sqrt{4+(5)^{2}}$ $=\sqrt{4+25}$ $=\sqrt{2} 9$ units Using Pythagoras Theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 = (3)2 + (√29)2 = 29 + 9 = √38 Since, the radius bisects the chord into two equal halves, Since, the equation of a circle having centre (h, k), having radius as r units, is (x – h)2 + (y – k)2 = r2 (x – 3)2 + (y – (-1))2 = (√38)2 x2 – 6x + 9 + (y + 1)2 = 38 x2 – 6x + y2 + 2y + 1 + 9- 38 = 0 x2 – 6x + y2 + 2y – 28 = 0 Hence, the required equation of the circle is x2 – 6x + y2 + 2y – 28 = 0.
Question Video: Finding a Root of a Quadratic Equation given the Other Root \| Nagwa Question Video: Finding a Root of a Quadratic Equation given the Other Root \| Nagwa # Question Video: Finding a Root of a Quadratic Equation given the Other Root Mathematics • First Year of Secondary School ## Join Nagwa Classes Given that β10 is a root of the equation 2π₯Β² + 13π₯ β 70 = 0, what is the other root? 02:57 ### Video Transcript Given that negative 10 is a root of the equation two π₯ squared plus 13π₯ minus 70 equals zero, what is the other root? Weβre told that negative 10 is a root of our equation, which means that our quadratic must be equal to zero when π₯ is equal to negative 10. Essentially, itβs a solution to the equation two π₯ squared plus 13π₯ minus 70 equals zero. Now, what this actually means is that π₯ plus 10 must be a factor of two π₯ squared plus 13π₯ minus 70. Two π₯ squared plus 13π₯ minus 70 can therefore be written as π₯ plus 10 times some other binomial. Now, letβs give that binomial a form. Letβs say itβs in the form ππ₯ plus π, where π and π are real constants. What weβre going to do is distribute the parentheses on the right-hand side of this equation and see what we get. We begin by multiplying π₯ by ππ₯. Thatβs ππ₯ squared. We then multiply the outer terms. Thatβs π₯ times π, which is ππ₯. Next, we multiply the inner terms. Thatβs 10 times ππ₯, which is 10ππ₯. And finally, we multiply 10 by π, to give us 10π. So we find that this is equal to two π₯ squared plus 13π₯ minus 70. And we now use a process called comparing coefficients. We look at the coefficients of our various terms. Letβs begin by comparing our coefficients of π₯ squared. On the left-hand side, we have a two. And on the right-hand side, the coefficient of π₯ squared is π. So when we compare coefficients of π₯ squared, we find π is equal to two. We could next compare coefficients of π₯. In fact though, weβre going to compare constants. We might say these are the coefficients of the π₯ to the power of zero terms. On the left-hand side, our constant is negative 70. And on the right-hand side, we have 10π. So negative 70 equals 10π. And so weβre going to divide by 10 to solve for π. π is therefore equal to negative seven. This means that our quadratic expression can be written as π₯ plus 10 times two π₯ minus seven. Weβve replaced π and π with their solutions. But we know that weβre using this to solve the equation two π₯ squared plus 13π₯ minus 70 equals zero. We already know that we have one root of negative 10. Thatβs found by setting π₯ plus 10 equal to zero. Weβre now going to set two π₯ minus seven equal to zero and solve for π₯. Weβll add seven to both sides of this equation so that two π₯ is equal to seven. And then weβll divide through by two. So π₯ is equal to seven over two or 3.5. And we could check this solution by substituting π₯ equals seven over two into our original equation, making sure that it is indeed equal to zero. ## 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
Courses Chapter Notes - Perimeter and Area Class 7 Notes \| EduRev Class 7 : Chapter Notes - Perimeter and Area Class 7 Notes \| EduRev The document Chapter Notes - Perimeter and Area Class 7 Notes \| EduRev is a part of the Class 7 Course Mathematics (Maths) Class 7. All you need of Class 7 at this link: Class 7 11. Perimeter and Area Plane Figures The area of the parallelogram is given by base x height. Triangle: A triangle is a polygon with three vertices, and three sides or edges that are line segments. A triangle with vertices A, B, and C is denoted as ABC The perimeter of a triangle is the sum of the lengths of its sides. If the three sides are a, b, and c, then perimeter Page 43 The area of a triangle is the space enclosed by its three sides. It is given by the formula, where b is the base and h is the altitude. A simple closed figure bounded by four line segments is called a quadrilateral. â€¢ Rectangle â€¢ Square â€¢ Parallelogram â€¢ Rhombus A rectangle is a quadrilateral with opposite sides equal, and each angle of measure 90o. Page 44 The perimeter of a rectangle is twice the sum of the lengths of its adjacent sides. In the figure, the perimeter of rectangle ABCD = 2(AB + BC). The area of a rectangle is the product of its length and breadth. In the figure, the area of rectangle ABCD = AB x BC. A square is a quadrilateral with four equal sides, and each angle of measure 90o. The perimeter of a square with side s units is 4s. In the figure, the perimeter of square ABCD = 4AB or 4BC or 4CD or 4DA. The area of a square with side s is s2 In the figure, the perimeter of square ABCD = AB2 or BC2 or CD2 or DA2. Page 45 A quadrilateral in which both the pairs of opposite sides are parallel is called a parallelogram. The perimeter of a parallelogram is twice the sum of the lengths of the adjacent sides. In the figure, the perimeter of parallelogram ABCD = 2(AB + BC) The area of a parallelogram is the product of its base and perpendicular height or altitude. Any side of a parallelogram can be taken as the base. The perpendicular dropped on that side from the opposite vertex is known as the height (altitude). In the figure, the area of parallelogram ABCD = AB x DE or AD x BF. A parallelogram in which the adjacent sides are equal is called a rhombus. The perimeter and area of a rhombus can be calculated using the same formulae as that for a parallelogram. Page 46 Circles A circle is defined as a collection of points on a plane that are at an equal distance... Circle: A circle is defined as a collection of points on a plane that are at an equal distance from a fixed point on the plane. The fixed point is called the centre of the circle. Circumference: The distance around a circular region is known as its circumference. Diameter: Any straight line segment that passes through the centre of a circle and whose end points are on the circle is called its diameter. Any line segment from the centre of the circle to its circumference. Page 47 Circumference of a circle = 2Ï€r, where r is the radius of the circle or , where Ï€d is the diameter of the circle. Î is an irrational number, whose value is approximately equal to . Circumference = Diameter x 3.14 Diameter(d) is equal to twice radius(r). d=2r Circles with the same centre but different radii are called concentric circles. Circle: The area of a circle is the region enclosed in the circle. The area of a circle can be calculated by using the formula: â€¢ if radius r is given â€¢ if diameter D is given â€¢ if circumference C is given Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity! , , , , , , , , , , , , , , , , , , , , , ;
## Year 4 Homelearning (Roman Numerals pre-learning) 20/4/18 Year 4 Home Learning – Friday 20th April 2018 Spellings Due to being away on residential for much of next week, there will be no spelling homework. Please use the time to aid your child in practising to put a duvet cover on their duvet! Maths (Pre-learning) Please have a go at the Roman Numerals task provided. As this is the only piece of homework this week, please hand it to your teacher on Tuesday. Roman Numerals pre-learning – Friday 20th April 2018 These are the building blocks of the Roman Numerals. I= 1 V= 5 X= 10 L= 50 C= 100 D= 500 M= 1000 In the our system (the Arabic numeral system), there are ten different digits, (0,1,2,3,4,5,6,7,8,9) and the place of these digits in the number determines its value. For example, 2 on its own means “two”, but in 3240, the “2” now means “two hundred”. In this way, any number can be written down, using only ten digits. The Roman numerals have a similar set of “rules”, which allow you to work out the number that is displayed: 1. If a smaller numeral follows a larger numeral, add the smaller number to the larger number 2. If a smaller numeral precedes a larger numeral, subtract the smaller number from the larger number 3. Do not use the same symbol more than three times in a row Following these rules, you should be able to construct and decipher Roman Numerals. It should help if you break the numbers down into thousands, hundreds, tens and units. Let us now look at some examples. What number is “XII”? Using rule number one above, and breaking the number down into tens and units: X= ten II= two the two comes after the ten, and so we add two to ten, and get the number 12. This time, we use rule number two: X=ten and I= one. The one comes before the ten, and so we subtract one from ten, and get the number 9. The same principle works for all numbers. Can you work out how to write “1984” in Roman Numerals? Have a go at this now, and then show … This number written in Roman numerals has more digits than when it is written in Arabic Numerals. However, this will not always be the case. Can you think of some examples when the number of Roman Numerals is less than the number of Arabic numerals for the same number? Once you feel confident, have a go at the following questions. These involve converting Roman numerals to Arabic numerals and vice versa. Try converting the following Roman numerals into Arabic numerals: 1. III 2. I 3. IV 4. VI 5. X 6. XIII 7. XX 8. XVIII 9. XIX 10. MCMLXXVI 11. MMVII 12. MCMLXXXI 13. MMXXIII 14. MCMLXII Now try converting the following Arabic numerals into Roman numerals: 15. 55 16. 86 17. 31 18. 40 19. ninety-six 20. two thousand five hundred and ninety-two 21. 1372 22. 913 23. 2107
# Lesson 2 Describing Patterns These materials, when encountered before Algebra 1, Unit 6, Lesson 2 support success in that lesson. ## 2.1: Continue the Pattern (5 minutes) ### Warm-up In earlier units, students worked with linear and exponential functions. In this unit, they encounter quadratic functions, which are neither linear nor exponential but change in a predictable way. This warm-up helps students recall what makes a pattern linear or exponential. This will be useful later in this lesson when students need to identify what type of pattern they see, and explain why. This activity is a good opportunity to practice some mental math involving fractions. The number in the list is given in fraction form so that it’s easier to compute the next values in the list mentally. If desired, this could be changed to 2.5 and students could use a calculator to find answers in the form of a decimal. ### Student Facing Consider a list that starts $$1, \frac52, \dots$$ What would be the next three numbers in the list, if it followed a pattern that grew: 1. exponentially? 2. linearly? ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis Reiterate that in a pattern that grows exponentially, successive terms change by a common factor. In a pattern that grows linearly, successive terms change by a common difference. Some questions for discussion: • “What strategies did you use to find the next three numbers when the list grew exponentially?” • “What strategies did you use to find the next three numbers when the list grew linearly?” ## 2.2: Patterns of Sticks (20 minutes) ### Activity In this activity, students have an opportunity to encounter patterns as they do in the associated Algebra 1 lesson. There are three linear patterns, and one that is neither linear nor exponential. This work is more scaffolded than the work in the Algebra 1 lesson, giving students an opportunity to step through the same type of work. When students express the $$n$$th term of a pattern using a variable after figuring out several numerical values in the pattern, they are expressing regularity in repeated reasoning (MP8). ### Launch Consider arranging students into groups of 2. Present the image for all to see, and give students quiet think time to describe how they see the pattern changing and generate the next two steps. Have students discuss their observations with a partner before working on the rest of this task. Before moving on to the second pattern, you may need to clarify for the whole class how each table represents a different aspect of how the figures are growing. ### Student Facing 1. Here’s a pattern. 1. How do you see the pattern changing? 2. Extend the pattern to show your prediction of the next two steps. 2. Here are tables that represent the pattern. step 0 1 2 3 6 11 $$n$$ 3 5 7 step 0 1 2 3 6 11 $$n$$ 3 4 5 9 1. In each pattern, what quantity is represented in the second row? 2. Complete each table. 3. Describe each pattern as linear, exponential, or neither. Be prepared to explain how you know. 3. Here is another pattern. 1. Lin says that step 3 will have 8 segments. Andre says that step 3 will have 7 segments. How does each student see the pattern growing? 2. Complete the tables to show the relationship between step number and number of segments, as Lin and Andre would see it. 3. Describe each pattern as linear, exponential, or neither. Be prepared to explain how you know. Lin step number of segments 0 1 2 3 6 9 $$n$$ 1 2 4 Andre step number of segments 0 1 2 3 6 9 1 2 4 ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis Ensure students understand why Andre’s pattern is neither growing linearly nor exponentially. (There is no need to talk about second differences or introduce the word quadratic at this time.) • “How did you complete the table for Andre’s pattern?” (The number of sticks increased by 1 and then by 2. To continue the pattern, I added 3 sticks for step 4 and 4 sticks for step 5.) • “How did you recognize linear patterns?” (Linear patterns have a constant rate of change so you add (or subtract) the same amount each time the step increases by 1 to find the next number.) • “How do you recognize exponential patterns?” (Exponential patterns have a constant growth factor so you multiply by the same number each time to find the next number in the pattern, or, when you divide sequential numbers in the pattern, you always get the same quotient.) ## 2.3: Patterns of Dots (20 minutes) ### Activity This activity provides a scaffolded opportunity to describe the way two different patterns change. ### Student Facing 1. Here is a pattern of dots. 1. Describe how you see the pattern growing. 2. Draw the next step. 3. Complete the table to continue the pattern. step number of dots 0 1 2 3 4 6 $$n$$ 3 6 4. Is the relationship between step number and number of dots linear, exponential, or neither? Explain how you know. 2. Here is another pattern of dots. 1. Describe how you see the pattern growing. 2. Draw the next step. 3. Complete the table to continue the pattern. step number of dots 0 1 2 3 4 6 $$n$$ 5 7 4. Is the relationship between step number and number of dots linear, exponential, or neither? Explain how you know. ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis The purpose of this discussion is for students to share strategies they used to complete the last few rows of each table and to clarify how they knew the relationship was linear, exponential, or neither. If time permits, highlight connections between the visual patterns and how the total number of dots is increasing in the tables. Display a completed table for both patterns and refer to it when discussing the following questions: • “How did you extend the pattern to figure out how many dots were in step _____?” • “How could you predict the number of dots in step 10? Or step 20?” • “What advice would you give a classmate to help them recognize a linear relationship? An exponential relationship?” • “Did anyone see the pattern in question 1 growing in a different way?”
### Increasing and decreasing functions This is my third post in the series of “Applications of derivatives”. The previous two were based on “Tangent and Normal” and “Maxima and Minima”.In this post, IB Maths Tutors will teach about increasing and decreasing functions. That is one more application of derivatives. Increasing and Decreasing Functions- We shall first learn about increasing functions Increasing Function- (a) Strictly increasing function- A function f (x) is said to be a strictly increasing function on (a, b) if x1< x2 f(x1) < f (x2) for all xl, x2(a, b).Thus, f(x) is strictly increasing on (a, b) if the values of f(x) increase with the increase in the values of x.Refer to the graph in below-given figure Graphically, f (x) is increasing on (a, b) if the graph y = f (x) moves up as x moves to the right. The graph in Fig.1 is the graph of a strictly increasing function on (a, b). Classification of Strictly Increasing Functions on the basis of Shape- (i) Concave up– When f'(x)>0 and f”(x)>0 for all x domain Refer to the graph in below-given figure (ii) When f'(x)>0 and f”(x)=0 for all x domain Refer to the graph in below-given figure (iii) Concave down or convex up – When f'(x)>0 and f”(x)<0 for all x domain Refer to the graph in below-given figure (b) Only Increasing or non-decreasing functions- A function is said to be non-decreasing if for as shown in the graph, for AB and CD; and for the portion BC ; thus overall we can say that so it is obvious for increasing or non-decreasing functions, f'(x) 0 with equality holding in interval like BC. Refer to the graph in below-given figure Decreasing Functions- (a) Strictly Decreasing Function-A function f (x) is said to be a strictly increasing function on (a, b) if x1< x2 f (x1) > f (x2) for all xl, x2(a, b) Thus, f (x) is strictly decreasing on (a, b) if the values of f (x) decrease with the increase in the values of x. Graphically, f (x) is a decreasing function on (a, b) if the graph y = f (x) moves down as x moves to the right. The graph in Fig. is the graph of a strictly decreasing function on (a, b). Refer to the graph in below-given figure Classification of Strictly Decreasing Functions on the basis of Shape- (i) Concave up or Convex down-When f'(x)<0 and f”(x)>0 for all x domain Refer to the graph in below-given figure (ii) When f'(x)<0 and f”(x)=0 for all x domain Refer to the graph in below-given figure (iii) Concave down or convex up – When f'(x)<0 and f”(x)<0 for all x domain Refer to the graph in below-given figure (b) Only decreasing or non-increasing functions- A function is said to be non-increasing if for as shown in the graph, for AB and CD ; and for the portion BC ; thus overall we can say that Obviously for this where equality holds for horizontal part of the graph i.e, in the interval of BC. Refer to the graph in below-given figure Monotonic Function: A function is said to be monotonic in the given interval (a,b) If it is either increasing or decreasing in the given interval. Definition: A function is said to be increasing (decreasing) at a point x0 if there is an interval (x0-h,x0+h) containing x0 such that f(x) is increasing(decreasing) on (x0-h,x0+h) Definition: A function is said to be increasing or decreasing on [a,b] if it is increasing(decreasing) at (a,b) and it is also increasing (decreasing) at x=a and x=b Conditions for Increasing and Decreasing Functions- We can easily identify increasing and decreasing functions with the help of differentiation. (i) If f'(x)>0 for all x (a,b), then f(x) is increasing on (a,b) (ii) If f'(x)<0 for all x (a,b), then f(x) is decreasing on (a,b) Properties of Monotonic Functions- (a) If f(x) is a function that is strictly increasing in the interval [a,b] then inverse of given function (f-1) exists and f-1 is also strictly increasing function (b) If f(x) and g(x) are functions that are strictly increasing or decreasing in the interval [a,b] then composite of given functions gof(x) is also strictly increasing or decreasing (c) If f(x) increasing function then its reciprocal 1/f(x) is decreasing function (d) If f(x) and g(x) are both increasing functions then (f+g)(x) is also increasing (e) If f(x) is increasing and g(x) is decreasing or vice-versa then gof(x) is decreasing
ma006 ยป ## Expressions Map In the previous lessons we worked with polynomial expressions. The variables had non negative integer exponents. And our expressions could contain any number of terms. We learned how to add, subtract, multiply and even divide these polynomials. In the upcoming lessons, we'll make use of our new factoring and look at performing these same operations for rational expressions. A rational expression is a ratio of two polynomials. It is a fraction in which the numerator and the denominator are both polynomials, 1 over x is an example of a rational expression, since 1 is a mynomial and x is also a mynomial. The second example contains the polynomial x plus 2 in the numerator, and 3x minus 1 in the denominator. So this is also a rational expression. And this course will prepare you to handle any sort of polynomial or rational expression. You'll be able to add, subtract, multiply and even divide these rational expressions by the end of this unit. ## Simplify Fractions Before we go into more depth about rational expressions, let's recall our knowledge about fractions. A fraction is a rational number or a ratio of two integers. In this case, we have the ratio of five to eight or five eighths. A student showed the following work when trying to simplify five eighths. Do you think this work is correct, and why? Choose yes or no and then check off either reasons that support your answer. ## Simplify Fractions This work would not be correct, since we know that four-eighths would be equal to one-half. More importantly, we know that this step cannot be correct. The student tried to simplify terms in the numerator and in the denominator. Remember terms are separated by addition or subtraction. We cannot cancel terms, but we can cancel factors. Factors are numbers or variables that are multiplied in the numerator or denominator. For example, if we started with six-12ths. We can rewrite the 6 as 2 times 3, and 12 as 2 times 6. We can simplify the factors of 2, since these cancel out to equal 1. This would leave us with 3 6. We repeat this process one more time by rewriting 3 and 6 with factors. We can cancel the whenever we're working throughout this unit keep in mind that we can't cancel terms, we can only cancel factors, things that are multiplied in the numerator and in the denominator. And finally, we know 5/8 cannot be simplified since 5 and 8 do not share a common factor. So we also need to include this reason. ## Simplify Rationals Expressions 1 We can use our knowledge of canceling factors to help us simplify rational expressions. We simplify a rational expression the same way that we simply fractions. We just need to cancel the factors that appear in the numerator and the denominator since the factors reduced to 1. We can use these steps for any rational expression we encounter. So let's start off with this one. I want you to factor the numerator and the denominator in the set here. ## Simplify Rational Expressions 1 Well we know we can factor a 3 from both the 1st term and the 2nd term leaving us with x minus 4 for our 2nd factor. For the denominator we know that they share a common factor of 5 so we put that out in front which leaves us with x minus 4 for our 2nd factor. So, this would be our factor rational expression great job if you found it. ## Simplified Form Now that we have factored each part of our fraction, we can remove the common factors or simplify them since they equal one. For example, we know that this quantity that X minus four divided by the quantity X minus four would equal one. This would leave us with three times one which is three and five times one which is five, or three 5th. We know we can do this because X really represents a number, so if X were five we would be left with three times one and five times one, which is 3 5th. But x doesn't have to be 5, it could be any number. If we let x equal 6 then we plug in 6 for x and we get 3 time 6 minus 4 divided by 5 times 6 minus 4. This is really the same as 3 times 2 divided by 5 times 2. The twos are factors that reduce to 1, so we really have 3 fifths. It wouldn't matter what the value of x was. The quantity x minus 4 divided by the quantity x minus 4 would always equal 1. ## Excluded Values of x But this simplification to 3 5th wont be true for every value of X. There is actually one value of X that makes it not true. What value do you think it is? You'll want to think back to what you know about fractions and what you know about their denominators. Take some time to think, and when you're ready, put your answer here. ## Excluded Values of x We know that the denominator of a fraction cannot equal 0. So, if we solve this for x, we know x cannot take on the value of 4. Now, I didn't show the steps of adding 20 to both sides and then dividing by 5, but I think that you can easily see that if x were 4, we would get 0, which can't happen for the denominator. This idea of excluded values of x is really important. Later, when we solve equations with rational expressions, we want to make sure we don't allow certain values for the variable or for x. We don't want the denominators to be 0. ## Simplify Rational Expressions 2 Let's try simplifying this rational expression. First, I just want you to factor the numerator and the denominator. Write your factors here. ## Simplify Rational Expressions 2 To factor the first trinomial, we want to find factors of 30 that sum to negative 11. These factors are negative 5 and negative 6, and since our quadratic term has a 1 in front of it, we can just use these factors in our factored form here. X minus 5 times x minus 6. For the denominator we want to find factors of 1 times 20, or 20. That sum to negative 9. Those would be negative 4 and negative 5. And again since our quadratic term has a 1 as the coefficient, we'll use these factors in our factored form. Great job factoring if you got one or both of those correct. ## Simplified Form Now, that our fraction has been factored, we can cancel the factors that appear in the numerator and in the denominator. You can write your answers here. ## Simplified Form We can see that there's a common factor of x minus 5 in the numerator and in the denominator. So we know those simplify to 1. This would leave us with x minus 6 in the numerator and x minus 4 in the denominator. Great job on that simplification. ## Are We Done Now, that we have this fraction, can we make this simplification to get 3 halves, yes or no? ## Are We Done Well that answer would be no. We can't do this simplification. We cancelled terms here. Remember we can only cancel factors, or things that are multiplied in the numerator and in the denominator. X minus 6 divided by the quantity x minus 4, would be our final answer. If you're ever trying to simplify a fraction further, ask yourself if you can factor the numerator and the denominator, or if you can cancel factors that appear in both. In this case, we can't factor anything besides 1 out of the numerator and the denominator, and these 2 factors aren't the same. So, this is our final result. ## Simplify Rational Expressions 3 Try simplifying this rational expression. Remember to only cancel factors that appear in the numerator and in the denominator ## Simplify Rational Expressions 3 The first thing we want to do is factor our numerator and denominator. To factor the numerator, we find factors of negative 8 that sum to positive 2. And since the coefficient in front of m squared is 1, we can use these two factors in our binomials. For the denominator, we want to find factors of negative 12 that sum to negative 1. These factors are negative 4 and positive 3. Now, we can't just use these factors in our binomials since the m squared has a coefficient of 2. Instead, we'll need to use factoring by grouping. Using that method, we get a denominator of 2 m plus 3 times m minus 2. Now that we have our factored form, we can cancel factors that appear in the numerator and denominator. Since they equal 1. And this leaves us with our solution. We know we can't simplify the m's anymore since we have a sum and the numerator and the denominator. We don't have anymore factors to cancel. ## Opposite Integers in Fractions In the last few examples, we saw when rational expressions had a factor in the numerator and the denominator that equaled 1. This is true for the quantity m minus 2 divided by m minus 2. We know this fraction really equals 1. But what do you think about for these expressions? What do you think these would equal? Write your answer in the boxes. ## Opposite Integers in Fractions Well it turns out all of these are negative 1. For the first fraction we can think about simplifying the common factor of 6. Leaving us with negative 1 divided by 1 or just negative 1. We can use the same reasoning for the rest of the examples. They will all equal negative 1 since we'll have one factor that's positive in the numerator. And one factor that's negative in the denominator. It doesn't matter where the negative sign falls, we just need one of them to be negative in order for our answer to be negative one. ## Opposite Polynomials in Fractions Instead of just looking at numbers, now, let's look at polynomials. What do you think these expressions would equal? Write your answers here. ## Opposite Polynomials in Fractions The second one might have been tricky but they both equal negative 1. We want to remember that the negative sign is really a negative 1 being multiplied by this expression. We know x plus 3 divided by x plus 3 equals 1. So now we can simplify our fraction. This just leaves us with negative 1. For our second rational expression, we can think of a 1 being outside of the expression a plus negative 1 a plus 4. If we were to distribute this negative 1 back inside the parenthesis we'd get our original denominator. We know this step is correct. We can simplify the a plus 4s since they equal 1leaving us with 1 times 1 in the numerator and negative 1 times 1 in the denominator. This just simplifies to negative 1. You might have spotted this first one easily. Great work if you. If you did. It's okay if this one gave you trouble. ## Opposite Polynomials We're going to focus in on this idea of opposite polynomials. For two polynomial expressions to be opposite, every corresponding pair of terms must opposite in sign. We know addition is commutative so we can rewrite these terms by adding the opposite. I'll move the negative x squared to the front, the 3x to the middle, and the negative 8 to the end. Notice that when we move these terms around, we just keep the sign with it. The three x's are positive, the eight is still negative and the x squared is also still negative. Now we can definitely see that these two polynomials are opposites. The x squared terms are opposites since one is positive and the other is negative. The three x's are opposites since one is negative and the other is positive And the eights are opposites, since the first one's positive and the second one's negative. And the neat part about opposite polynomials and rational expressions is that they simplify to negative1 . We can think about factoring a negative 1 from each of these terms. So we'll have negative 1. Times the opposite sign of all those terms in the denominator. The numerator will just stay the same and we can think about multiplying it by a positive 1. Here we have a common factor in the numerator and a denominator, so this all equals 1. And finally, since we have this negative sign down in the denominator, our answer simplifies to negative 1. ## Which One Part 1 For two polynomial expressions to be opposites every corresponding pair of terms must be opposite in sign. Knowing this, I want you to determine whether each of these rational expressions is equal to positive one, negative one or something else. Choose one of these for each expression. Good luck. ## Which One Part 1 For the first expression, we can switch the terms in the denominator, since addition is commutative. So we'll have x plus 7 divided by the quantity x plus the second rational expression, we know it equals negative 1. This b squared is positive and this b squared is negative, whereas this 5 is negative and this 5 is positive. So, we definitely have opposite polynomials. We see that here, if we just reverse the terms of the denominator and keep the signs in front of them. We factor a negative 1 from the denominator and then the b squared minus have thought it was negative 1. You might have cancelled the terms which are a and then simplified negative 5 over positive 5. But, remember, we can't cancel terms. We can only cancel factors. A is positive in the numerator and positive in the denominator, so, these aren't opposite polynomials. The 5s have different signs which mean this polynomial can't equal positive 1. It must be something else. And for the last one, we could switch the terms in the denominator, and then, factor a negative 1 from the numerator. The common factors of 2x minus 9 simplify to 1, leaving us with negative 1 as our result. Great work if you even got two of those correct. These were pretty tough. ## Which One Part 2 What about these rational expressions? Are they equal to positive 1, negative 1, or something else? ## Which One Part 2 The first two are equal to negative 1, the 3rd one is equal to positive 1, and the last one is something else. We know these first two are equal to negative 1 since each rational expression contains opposite polynomials. The x terms are opposite in sign here, and the y terms are also opposite in sign. So we know that this fraction can simplify to negative 1. The same is true for this fraction. The b squared terms are opposite in sign, the 5b terms are opposite in sign and the constant of 1 is also opposite in sign. I didn't show the factoring here, but we know that it will equal negative 1 if we remove a factor of negative 1 from the denominator and cancel our trinomial. In our third example, the y terms have the same sign, and the x terms have the same sign. So really, this fraction is the same expression. In this third rational expression, the y terms have the same sign, as do the x terms. If we switch the order of addition in the numerator, we'll have x minus y divided by the quantity x minus y. We know these expressions are exactly the same, so we really have 1 divided by 1, or just positive 1. For this last expression, we know it needs to equal something else. We can't just square each term and call it a day. We know that this is really x plus 2 times x plus 2. When we square a binomial, we get a perfect square trinomial. Looking at this numerator and denominator, we know that our fraction doesn't equal positive 1 or negative 1. ## The Different Expression 1 Let's continue our exploration of rational expressions. This time I'm going to show you 4 rational expressions and I want you to find the one expression that is not equal to the other 3. In other words you want to try and either to simplify these to be positive 1 or negative 1 or you want to rearrange them to make them look like each other. Be sure to chose the odd one out. The one that looks different. ## The Different Expression 1 This expression is the odd one out. The other ones equal positive 1. This expression equals 1 since we have the same expression in the numerator and the denominator. We've seen something similar like this before if we just switch the terms in the denominator we can see that numerator and the denominator have the same factors. Notice that I'm not cancelling an individual term. I'm cancelling an entire numerator with an entire denominator. So I have one divided by one or just one. This one might have been a little bit tricky. Here we have a negative one times our rational expression. Well we can just distribute this negative sign or this negative one to each term and the numerator. This would give us x plus 3 Divided by 3 plus x. This fraction is identical to this rational expression, so we know it equals 1. For this rational expression, we know that these factors can reduce to 1, but we're left with a negative sign out in front, or a negative 1 times 1, which is just negative 1. ## The Different Expression 2 What about for these four expressions? Which one of these is different? ## The Different Expression 2 We know this is the answer. This rational expression is the odd one out, since it equals positive 1. This expression over here is really equal to negative 1, since x minus 7 divided by the quantity x minus 7 simplifies to 1. We get a negative 1 since we have a negative sign sitting out here. This is really negative 1 times 1. For this first rational expression, we can switch the order of the denominator. I can have the negative x come first, and then followed by the positive 7. We can see that this expression is the same as this last expression down here. But this doesn't quite look like negative 1. But these two rational expressions don't look like negative 1, but in fact, they are. We know these expressions in the numerator and denominator are opposite polynomials. The x terms are opposite in sign, as are the constant terms. We can factor a negative 1 from our denominator to get negative 1 times x minus 7. Then we can simplify our common factors in the numerator and denominator, leaving us with negative 1. Nice thinking if you found that one. ## The Different Expression 3 What about for these four expressions? Which one is different? And as a hint, these won't simplify to positive 1 or negative 1. You'll need to manipulate each rational expression to see if they're the same. ## The Different Expression 3 We can see that these two rational expressions are equal. If we switch the order of the terms in the numerator of this expression, we'll get x plus 9 divided by the quantity x minus 9. Now we need to decide between these two expressions. We can switch the order of the terms in the numerator and in the denominator. The x stays negative here and the 9 stays positive. For this expression, we don't want to switch the order of the numerator, since it's already x plus 9. It matches the other ones. And here's where this negative sign comes in handy. We can distribute this negative sign to each term in the numerator, or to each term in the denominator. I'm going to multiply our denominator by negative 1. Negative 1 times 9 is negative 9. And negative 1 times negative x is positive x. Finally, we commute this edition to get x plus 9, divided by the quantity x minus 9. So we know this rational expression is equal to these two. This one is definitely different. ## Simplify Rational Expressions Practice 1 Now that we've seen different types of rational expressions that equal 1 and negative 1, let's try and define some others. What do you think this would equal? Write your numerator here and your denominator here. Good luck. ## Simplify Rational Expressions Practice 1 This first trinomial can be factored as x plus 4 times x plus 6. We want factors of positive 24 that sum to positive 10. The second trinomial can be factored into x plus 5 times x plus 4. We want factors of positive 20 that sum to positive 9. Once we have our factored form, we cancel the factors that appear in the numerator and the denominator so we're left with x plus 6. Divided by the quantity, x plus 5. Great work if you've got this as your final rational expression. ## Simplify Rational Expressions Practice 2 Try factoring this expression. Remember to use all your factoring skills and start by looking for a GCF. Good luck. ## Simplify Rational Expressions Practice 2 This is our simplified expression, 8a divided by the quantity a plus 5. We can start factoring the numerator by removing a common factor of 8a from each term. This leaves us with a second factor of a plus 4. This denominator is actually identical to the last problem we looked at. The x has been replaced with a. So we have the same factors, a plus 5 and a plus 4. We just have a instead of x. We cancel the common factor of a plus four in the numerator and denominator, leaving us with eight a divided by a plus five. Keep in mind these are both positive ones, so we wouldn't need to change anything in terms of signs. Sometimes our two factors might reduce to negative one. In that case we would need to multiply our expression by a negative one. We would have a negative sign in front. ## Simplify Rational Expressions Practice 3 Here's our answer. Great work if you got that one correct. We can factor the numerator to be x plus 3 times x minus 3, since this is the difference of two squares. Our denominator factors into x plus 3 squared. This is a perfect square trinomial. The first term is a square. The second term is a square, and the middle terms is twice the square root of the first term and the last term. We know for the square exponent, we really write this factor twice. From here, it's easy to see we have a common factor of x plus 3, in the numerator and the denominator. So those simplify to equal 1. Another way to think about simplifying this is to do it over here. We have an exponent of 1 for this x plus with the 1 and the numerator. So our power of 2 drops down to a power of 1 and we lose this factor altogether. Keep in mind that things aren't equal in 0 here, it's really that we have an equivalence of 1, x plus 3 divided by x plus 3. This leaves us with our final answer. ## Simplify Rational Expressions Practice 4 Try simplifying this expression. Write your answer here and be sure to factor out a greatest common factor first. ## Simplify Rational Expressions Practice 4 We can remove a 4 from the first 3 terms of out numerator, leaving us with 4 times x squared plus 4x plus 4. We can also remove a 4 as a greatest common factor from 4x and 8. This leaves us with 4 times x plus 2 in our denominator. We have a quadratic trinomial here and not only that it's a perfect square trinomial, so we can factor that as x plus 2 times x plus 2. Finally, we cancel the factors in the numerator and the denominator. So x plus two divided by x plus two equals one, and four divided by four equals one. Remember, we're only simplifying factors, or things that are being multiplied in our numerator and our denominator. This leaves us with x plus two divided by one, or just x plus two. Notice that not every rational expression simplifies to another fraction. Sometimes we'll just get another polynomial. ## Simplify Rational Expressions Practice 5 We can factor this numerator as the difference of two squares. We can write it as 6 minus x times 6 plus x. For this denominator we want to find the factors of negative 48 that sum to positive 2. These factors are negative 6 and positive 8. When I look at my factors it doesn't seem as if anything can simplify out to equal 1 but if i look at these two factors i can see that these are opposite polynomials. The constant term 6 is positive while this one's negative and the x term is negative here while this x term is positive. This means I'll have a factor of negative 1. We can see this if we were to just switch the order of the terms for these two, we'll have negative 6 plus x, then I can factor a negative signs of both of these terms since i factored a negative 1 out. These factors of I thought here. Whenever you see opposite polynomial factors you can simplify them to be negative 1. We don't need to show all this work. So we're left with this expression. Now, I want to acknowledge there are a lot of different answers here. This negative sign could have been distributed to either the numerator or the denominator. If we distribute the negative 1 to the numerator, we could have gotten this expression. Or switch the order of these terms and you got this expression. These are all equal. If we distribute this negative one instead of the denominator, we can wind up with any of these expressions. And in fact, all of these are correct. In general, it's easiest to list the negative sign out in front of a fraction and then write your rational expressions. ## Simplify Rational Expressions Practice 6 What about this rational expression? What would you get? Notice we've changed one thing here, the sign. This is a negative 2x instead of a positive 2x. ## Simplify Rational Expressions Practice 6 Our solution is 6 minus x, divided by the quantity x minus 8. Great job if you got this rational expression. We factor the numerator the same as before, but this time in the denominator we want factors of -8 and positive 6. They multiply to give us -48, and a sum to get us -2. I noticed that x plus 6 and 6 plus x are really the same factors.So I can cancel those since they equal 1. This leaves me with the remaining factors, so I have 6 minus x divided by x minus 8. You might have switched the order of these two terms, giving you negative x plus 6 for the numerator. This would be okay. You might even have gone a step further, and factored negative 1 from the numerator, and that would have been okay too. All of these answers are correct. And if you even followed my advice from the first time, a negative sign out in front would look like this. ## Simplify Rational Expressions Practice 7 For our last problem, try simplifying this expression. ## Simplify Rational Expressions Practice 7 Our numerator is a difference of two perfect squares. So we can factor it as 2x plus 5y times 2x minus 5y. This denominator isn't identical to one of our factors, but it is an opposite polynomial for this factor. We can see that the factors simplify to be negative 1 and were left with negative 1. Times 2x plus Fantastic work if you got that expression.
What Astrology Experts Say About Alicia Silverstone (11/10/2019) How will Alicia Silverstone fare on 11/10/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is just for fun – do not take this too seriously. I will first work out the destiny number for Alicia Silverstone, and then something similar to the life path number, which we will calculate for today (11/10/2019). 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 experts. PATH NUMBER FOR 11/10/2019: We will analyze the month (11), the day (10) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 10 we do 1 + 0 = 1. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 1 + 12 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 11/10/2019. DESTINY NUMBER FOR Alicia Silverstone: 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 Alicia Silverstone we have the letters A (1), l (3), i (9), c (3), i (9), a (1), S (1), i (9), l (3), v (4), e (5), r (9), s (1), t (2), o (6), n (5) and e (5). Adding all of that up (yes, this can get tedious) gives 76. This still isn’t a single-digit number, so we will add its digits together again: 7 + 6 = 13. This still isn’t a single-digit number, so we will add its digits together again: 1 + 3 = 4. Now we have a single-digit number: 4 is the destiny number for Alicia Silverstone. CONCLUSION: The difference between the path number for today (6) and destiny number for Alicia Silverstone (4) is 2. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too happy yet! As mentioned earlier, this is not scientifically verified. 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.
Contact The Learning Centre # Simultaneous equations ### Substitution method In this method, we substitute one variable from one equation into the other. The steps involved are: 1. Using either of the equations, express one variable in terms of the other. 2. This expression is then substituted into the other equation to form an equation in one variable only. 3. Solve this equation to find the value of one of the variables. 4. Substitute the value of this variable into the equation formed in the first step to find the value of the other variable. For example, solve for $$x$$ and $$y$$:$\begin{eqnarray} 2x+y &=& 21 \label{eqn:sub1} \\ 3x+4y &=& 44\ \label{eqn:sub2} \end{eqnarray}$ Rearranging $$(\ref{eqn:sub1})$$ to give \begin{eqnarray} y&=& 21-2x \label{eqn:sub3} \end{eqnarray} Substituting $$(\ref{eqn:sub3})$$ into $$(\ref{eqn:sub2})$$ and solving for $$x$$ gives: \begin{eqnarray} 3x+4(21-2x) &=& 44 \nonumber \\ 3x+84-8x &=& 44 \nonumber \\ -5x + 84 &=& 44 \nonumber \\ -5x &=& -40 \nonumber \\ x&=& 8 \label{eqn:sub4} \end{eqnarray} Finally, substituting $$(\ref{eqn:sub4})$$ into $$(\ref{eqn:sub3})$$ to solve for $$y$$ gives $$y = 21-2\times 8 = 5$$. Therefore the solution is $$(8,5)$$.
# Pre-Algebra : Commutative Property of Multiplication ## Example Questions ### Example Question #1 : Commutative Property Of Multiplication Which property is illustrated by the example: Multiplicative identity property Distributive property Associative property of multiplication Commutative property of multiplication Commutative property of multiplication Explanation: The commutative property of mulitplication states that the order in which 2 numbers are multiplied does not affect the result: Or in this case, ### Example Question #2 : Identities And Properties Use the Commutative Property of Multiplication to write the below expression in a different way. Explanation: The rule for Commutative Property of Multiplication is Using this rule, the expression can be written as ### Example Question #2 : Commutative Property Of Multiplication Which of the following equations correctly shows the commutative property of multiplication? Explanation: The commutative property states that factors can be multiplied in any order and the product is always the same. While each of our answer choices are real equations, only one correctly displays this property: ### Example Question #1 : Commutative Property Of Multiplication Evaluate the following expression without using a calculator Explanation: This problem could easily turn into some super heavy calculations, but nobody has time for that. Instead, we can make this problem very easy by utilizing the commutative property of multiplication, which in essence says that numbers can be multiplied in any order. In other words, and That means that rather multiplying our numbers in their orginal order, which can actually change the order of the numbers without affecting the final product. At this point we should notice that one of our numbers is while another is . Similarly the fractions and are each accompanied by their recripocals and respecitively. At this point we must also remember the multiplicative inverse property, which tells us that the product of a number and its reciprocal will always be 1. With that in mind, the strategic way to order the multiplication is as follows: Doing this pairs each number up with its reciprocal. Finally, we can utlize the associative properpty of multiplication, which tells us that we can multiply any two numbers at a time. In other words, . This allows us to pair off the multiplication as follows. Within each parentheses we now have a number multiplied by its reciprocal. Remembering that this always equals 1, we can simplify, giving Our answer is 1. ### Example Question #3 : Commutative Property Of Multiplication Which answer shows the communitive property of multiplication to the equation below? Explanation: The communitive property of multiplication is a property that states that the numbers can be multiplied in any combination. To show which answer is correct simply regroup the numbers in a different or and you will have your answer. ### Example Question #4 : Commutative Property Of Multiplication Which of the following displays the commutative property of multiplication?
• https://me.yahoo.com # Variable Cross Section Time of emptying a tank of variable cross-section through an orifice ## Overview Previously the time of emptying of geometrical tanks (i.e., rectangular, hemispherical and circular) was discussed. But, sometimes, we come across tanks, which have variable cross-section. In such cases, there are two variables instead of one, as in the case of tanks of uniform cross-section. Since a single relation cannot be derived for different cross-sections, it is therefore essential that such problems should be solved from the first principles i.e., from the equation. $dt&space;=&space;\frac{-A.dh}{C_{d}.a\sqrt&space;{2gh}}$ This can be best understood from the following examples. Example: [metric] ##### Example - Time of emptying a tank of variable cross-section Problem A rectangular tank of 20m$\inline&space;\times$12m at the top and 10m$\inline&space;\times$6m at the bottom is 3m deep as shown in figure. There is an orifice of 450mm diameter at the bottom of the tank. Determine the time taken to empty the tank completely, if coefficient of discharge is 0.64. Workings Given, • Top length = 20m • Top width = 12m • Bottom length = 10m • Bottom width = 6m • Depth of water = 3m • Diameter of orifice, $\inline&space;d$ = 450mm = 0.45m • $\inline&space;C_{d}$ = 0.64 We know that the area of the orifice, $a&space;=&space;\frac{\pi}{4}\times&space;d^2&space;=&space;\frac{\pi}{4}\times&space;(0.45)^2&space;=&space;0.159m^2$ First of all, let us consider a small strip of water of thickness (dh) at a height (h) from the bottom of the tank. From the geometry of the figure, we find that the length of the strip of water, $l&space;=&space;10&space;+&space;\frac{10h}{3}$ and the breadth of the strip of water, $b&space;=&space;6&space;+&space;\frac{6h}{3}&space;=&space;6&space;+&space;2h$ So, the area of the strip, $A&space;=&space;l\times&space;b&space;=(10&space;+&space;\frac{10h}{3})(6&space;+&space;2h)&space;=&space;6.67&space;h^2&space;+&space;40h&space;+&space;60$ Now let us use the general equation for the time to empty a tank, $\therefore&space;dt&space;=&space;\frac{-A.dh}{C_{d}.a.\sqrt&space;{2gh}}$ The total time required to empty the tank may be found by integrating the above equation between the limits 3 and 0 (because initial head of water is 3m and final head of water is 0m). $T&space;=&space;\int_{3}^{0}&space;\frac{-A.dh}{C_{d}.a.\sqrt&space;{2gh}}$ $\;\;\;\;\;=&space;\frac{-1}{C_{d}.a.\sqrt&space;{2g}}\int_{3}^{0}A.h^{-\frac{1}{2}}.dh$ $\;\;\;\;\;=&space;\frac{1}{C_{d}.a.\sqrt&space;{2g}}\int_{0}^{3}(6.67&space;h^2&space;+&space;40h&space;+&space;60).h^{-\frac{1}{2}}.dh$ $\;\;\;\;\;=&space;\frac{1}{C_{d}.a.\sqrt&space;{2g}}\int_{0}^{3}(6.67&space;h^{\frac{3}{2}}&space;+&space;40h^{\frac{1}{2}}&space;+&space;60h^{-\frac{1}{2}}).dh$ $\;\;\;\;\;=&space;\frac{1}{0.64\times&space;0.159\times&space;\sqrt&space;{2\times&space;9.81}}\times&space;[2.668\times&space;(3)^{\frac{5}{2}}+26.67\times&space;(3)^{\frac{3}{2}}+120\times&space;(3)^{\frac{1}{2}}]$ $\;\;\;\;\;=&space;860\;s&space;=&space;14\;min\;20\;s$ Solution Time taken to empty the tank = 14 min 20 s
# How to know if the angle is positive or negative in inverse trigonometry I have been doing this problem this problem: $$Cos[Tan^{-1}(-\frac{2}{3})]$$ So I was instructed to draw a triangle to guide me so I did Now once I drew my triangle I found the hypotenuse, which is $$\sqrt{13}$$ And then I was able to obtain the answer to this expression which I got: $$\frac{2\sqrt{13}}{13}$$ However, I am told I drew the triangle wrong, it is actually -2 (negative) and (3) positive. Why is it that the triangle is wrong? I was told the actual answer is $$\frac{3\sqrt{13}}{13}$$ Arctan has a range of $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$. Now let $arctan\frac{-2}{3}=y$. This implies $tany=\frac{-2}{3}$. Because tan is negative, we know y must lie in quadrant IV. It cannot lie in quadrant I because tan is positive in quadrant I. Therefore we draw our angle as you have above. Now, note that $tan\theta=\frac{opposite}{adjacent}$. Therefore, you should have $-2$ where $3$ is in your picture, and you should have $3$ where $-2$ is. Your line should be drawn to the coordinate $(3,-2)$. Now, we find the hypotenuse as you have already done using the Pythagorean Theorem. We find that it is $\sqrt{13}$ as you've noted. Now, because we are finding $cos(arctan\frac{-2}{3})$, we will use the fact that $cos\theta=\frac{adjacent}{hypotenuse}$. So, this gives $\frac{3}{\sqrt{13}}$. Rationalizing we have $\frac{3\sqrt{13}}{13}$ • Yes, tangent is negative in quadrant IV and II, but the range for arctan is limited to $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$ so our angle cannot lie in quadrant II. It must lie in either quadrant I or IV. And since it is negative, it must lie in quadrant IV. – MathGuy Jan 27 '17 at 3:04 • Yes, the ranges are different though depending on the inverse trig function. For arcsin the range is $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$, for arccos the range is $0\le{y}\le\pi$ and for arctan the range is $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$ – MathGuy Jan 27 '17 at 3:12
Course Content Class 7th Science 0/36 Class 7th Math 0/71 Class 7 Social Science Geography 0/20 Class 7 Social Science History 0/20 Class 7 Social Science Civics 0/19 Class 7 English Honeycomb 0/20 Class 7 English Honeycomb Poem 0/18 Class 7 English An Alien Hand Supplementary 0/20 Online Class For 7th Standard Students (CBSE) About Lesson ## Introduction ### Triangle • A triangle is a closed curve made of three line segments. • It has three:Sides: (i) Sides: ¯¯¯¯¯¯¯¯AB, ¯¯¯¯¯¯¯¯BC and ¯¯¯¯¯¯¯¯CA (ii) Angles: ∠BAC, ∠ACB and ∠CBA (iii) Vertices: A, B and C ## Important Lines in a Triangle ### Median • Median is the line that connects a vertex of a triangle to the mid-point of the opposite side. • In the given figure, ¯¯¯¯¯¯¯¯¯AD is the median, joining the vertex A to the midpoint of ¯¯¯¯¯¯¯¯BC. ### Altitude • An altitude is a line segment through a vertex of the triangle and perpendicular to a line containing the opposite side. ## Sides Also Have Constraints ### Sum of the lengths of two sides of a triangle • The sum of the lengths of any two sides of a triangle is greater than the third side. • In the above triangle, 9+11=20 >14 11+14=25 >9 9+14=23 >11 ### Difference between lengths of two sides of a triangle • The difference between lengths of any two sides is smaller than the length of the third side. • In the above triangle, 11 – 9 = 2 < 14 14 – 11 = 3 < 9 14 – 9 = 5 < 11 • Using the concept of sum of two sides and difference of two sides, it is possible to determine the range of lengths that the third side can take. ## Triangle Properties ### Angle sum property of a triangle • The total measure of the three angles of a triangle is 1800. • In △PQR, ∠RPQ+∠PQR+∠QRP =70+ 60+ 50= 1800 ### Exterior angle of a triangle and its property • An exterior angle of a triangle is equal to the sum of its interior opposite angles. In the given figure, ∠1+∠2=∠3. ### Pythagoras Theorem • The side opposite to the right angle in a right-angled triangle is called the hypotenuse. • The other two sides are known as legs of the right-angled triangle. • In a right-angled triangle, square of hypotenuse is equal to the sum of squares of legs. AC2=AB2+BC2 ⇒52=42+32 ### Properties of isosceles and equilateral triangles Properties of Isosceles Triangle • Two sides are equal in length. • Base angles opposite to the equal sides are equal. Properties of Equilateral Triangle • All three sides are equal in length. • Each angle equals to 600. ## Classification of Triangles ### Classification of triangles based on sides • Equilateral triangle: A triangle in which all the three sides are of equal lengths. • Isosceles triangle: A triangle in which two sides are of equal lengths. • Scalene Triangle: A triangle in which all three sides are of different length. ### Classification of triangles based on angles • Acute-angled: A triangle with three acute angles. • Right-angled: A triangle with one right angle. • Obtuse-angled: A triangle with one obtuse angle. Join the conversation
Going Around in Circles If we pick any two distinct points on a circle, and connect them with a chord, the chord will divide the interior of the circle into two distinct, nonintersecting regions. If we pick three distinct points on a circle and connect each pair of them with a chord, we can form four distinct, nonintersecting regions. What is the maximum number of nonintersecting regions that we can form by selecting four distinct points on a circle and connecting each pair of them with a chord? By selecting five distinct points? Six distinct points? And of course, then, $n$ distinct points? Source: NCTM Problem to Ponder 11/2011 SOLUTION One point: $0$ chord, $0$ intersection, $1$ region Two points: $1$ chord, $0$ intersection, $2$ regions Three points: $3$ chords, $0$ intersection, $4$ regions Four points: $6$ chords, $1$ intersection, $8$ regions Five points: $10$ chords, $5$ intersections, $16$ regions Six points: $15$ chords, $15$ intersections, $31$ regions Seven points: $21$ chords, $35$ intersections, $57$ regions We tabulate the previous data in the following table Looking at the table, we easily notice a pattern: $Regions=Chords+Intersections+1$ But, is it true for all values of $n$? When we add a new point, the number of chords, intersections, and regions increase. But by how much? We will now examine in depth the transition from a 5-point circle to a 6-point circle in order to gain insights into the increase of these numbers. Let $R$ denote the number of regions; $C$ the number of chords; $I$ the number of intersections. We start from a 5-point circle as shown in the figure below $C=10,\; I=5,\; R=16$ Now, let’s add a sixth point and draw chord $A$. Chord $A$ does not intersect any old chord and splits an old region into two. The number of regions $R$ increases by $1$ from $16$ to $17$. Chord $B$ intersects $3$ old chords and splits $4$ old regions into $8$. $R$ increases by $4$ from $17$ to $21$. Chord $C$ intersects $4$ old chords and splits $5$ old regions into $10$. $R$ increases by $5$ from $21$ to $26$. Chord $D$ intersects $3$ old chords and splits $4$ old regions into $8$. $R$ increases by $4$ from $26$ to $30$. Chord $E$ does not intersect any old chord and splits an old region into two. $R$ increases by $1$ from $30$ to $31$. Summary When we add a point, we add several new chords. A new chord may intersect no old chord or may intersect several old chords at several intersection points. When a new chord intersects an old chord it leaves an old region and enters another old region splitting each old region it passes through into two. Hence, it follows that the increase in $R$ will be one more than the increase in $I$ including the case where the new chord does not intersect any chord. In other words, $\Delta R=\Delta I+1$ Proof by induction Show that $R=C+I+1$ for all positive integers $n=1,2,3,\cdots$ $n=1$ $1=0+0+1$ The formula is true for $n=1$ Suppose it is true for $n$. We write $R_n=C_n+I_n+1$ We want to show that it is true for $n+1$, that is $R_{n+1}=C_{n+1}+I_{n+1}+1$ When we add a point, we add a number of new chords. If the new chord does not intersect any old chord, $\Delta I=0$ $\Delta C=1$ $\Delta R=\Delta I+1$ $=0+1$ $=1$ $R_{n+1}=R_n+\Delta R$ $=R_n+1$ $C_{n+1}+I_{n+1}+1=\left (C_n+\Delta C\right )+\left (I_n+\Delta I\right )+1$ $=\left (C_n+1\right )+\left (I_n+0\right )+1$ $=\left (C_n+I_n+1\right )+1$ $=R_n+1$ Comparing the equations we conclude that $R_{n+1}=C_{n+1}+I_{n+1}+1$ If the new chord intersects $k$ old chords at $k$ intersection points $\Delta I=k$ $\Delta C=1$ $\Delta R=\Delta I+1$ $=k+1$ $R_{n+1}=R_n+\Delta R$ $=R_n+k+1$ $C_{n+1}+I_{n+1}+1=\left (C_n+\Delta C\right )+\left (I_n+\Delta I\right )+1$ $=\left (C_n+1\right )+\left (I_n+k\right )+1$ $=\left (C_n+I_n+1\right )+k+1$ $=R_n+k+1$ Comparing the equations we conclude that $R_{n+1}=C_{n+1}+I_{n+1}+1$ DONE Now that we have proved the formula $R=C+I+1$, it is time to express it in terms of $n$. A chord connects two points. How many ways can we select two points from a collection of $n$ points? $C=\binom {n}{2}$ Two chords (connecting four points) intersect at one point. How many ways can we select four points from a collection of $n$ points? $I=\binom{n}{4}$ The formula expressed in terms of the number of points $n$ is $R=\binom{n}{2}+\binom{n}{4}+1$ Verification $R_1=\binom{1}{2}+\binom{1}{4}+1$ $=0+0+1$ $=1$ $R_2=\binom{2}{2}+\binom{2}{4}+1$ $=1+0+1$ $=2$ $R_3=\binom{3}{2}+\binom{3}{4}+1$ $=3+0+1$ $=4$ $R_4=\binom{4}{2}+\binom{4}{4}+1$ $=6+1+1$ $=8$ $R_5=\binom{5}{2}+\binom{5}{4}+1$ $=10+5+1$ $=16$ $R_6=\binom{6}{2}+\binom{6}{4}+1$ $=15+15+1$ $=31$ $R_7=\binom{7}{2}+\binom{7}{4}+1$ $=21+35+1$ $=57$ $R_8=\binom{8}{2}+\binom{8}{4}+1$ $=28+70+1$ $=99$ Answer: $R=\binom{n}{2}+\binom{n}{4}+1$.
• # General Course Information We will continue following the Florida Standards. ## Duration: Aug. 24th – Sept. 10th (14 Days) MAFS.5.NBT.1.1- Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left. (DOK 1) MAFS.5.NBT.1.2 - Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10. (DOK 2) MAFS.5.NBT.1.3 - Read, write, and compare decimals to thousandths. a. Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). b. Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. (DOK 2) MAFS.5.NBT.1.4 - Use place value understanding to round decimals to any place. (DOK 1) Standards for Mathematical Practice Academic Vocabulary: Unit 3: Focus of the Unit/Lesson: Multiply and Divide Whole Numbers (Until Oct 20th) Standards: MAFS.5.NBT.2.5 Fluently multiply multi-digit whole numbers using the standard algorithm. MAFS.5.NBT.2.6 Find whole number quotients of whole numbers with up to four digit dividends and two digit divisors using strategies based on place value, the properties of operations and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. ## Inquiry Mondays Focus Standards (Every Monday we will work on Nature of Science concepts.) SC.5.N.1.1 (DOK 3) - Define a problem, use appropriate reference materials to support scientific understanding, plan and carry out scientific investigations of various types such as: systematic observations, experiments requiring the identification of variables, collecting and organizing data, interpreting data in charts, tables, and graphics, analyze information, make predictions, and defend conclusions. SC.5.N.1.2 (DOK 2) - Explain the difference between an experiment and other types of scientific investigation. SC.5.N.1.3 (DOK 2) - Recognize and explain the need for repeated experimental trials SC.5.N.1.4 (DOK 2) - Identify a control group and explain its importance in an experiment. SC.5.N.1.5 (DOK 2) - Recognize and explain that authentic scientific investigation frequently does not parallel the steps of "the scientific method." SC.5.N.1.6 (DOK 2) - Recognize and explain the difference between personal opinion/interpretation and verified observation. SC.5.N.2.1 (DOK 2) - Recognize and explain that science is grounded in empirical observations that are testable; explanation must always be linked with evidence. SC.5.N.2.2 (DOK 2) - Recognize and explain that when scientific investigations are carried out, the evidence produced by those investigations should be replicable by others. ## Unit 1: Nature of Science Focus Standards (First Week) SC.5.N.1.1 (DOK 3) - Define a problem, use appropriate reference materials to support scientific understanding, plan and carry out scientific investigations of various types such as: systematic observations, experiments requiring the identification of variables, collecting and organizing data, interpreting data in charts, tables, and graphics, analyze information, make predictions, and defend conclusions. SC.5.N.1.2 (DOK 2) - Explain the difference between an experiment and other types of scientific investigation. SC.5.N.1.3 (DOK 2) - Recognize and explain the need for repeated experimental trials SC.5.N.1.4 (DOK 2) - Identify a control group and explain its importance in an experiment. SC.5.N.1.5 (DOK 2) - Recognize and explain that authentic scientific investigation frequently does not parallel the steps of "the scientific method." SC.5.N.1.6 (DOK 2) - Recognize and explain the difference between personal opinion/interpretation and verified observation. SC.5.N.2.1 (DOK 2) - Recognize and explain that science is grounded in empirical observations that are testable; explanation must always be linked with evidence. SC.5.N.2.2 (DOK 2) - Recognize and explain that when scientific investigations are carried out, the evidence produced by those investigations should be replicable by others. # Unit 2: Matter (Tested Mid Sept) Focus of the Unit/Lesson: Matter can be described (classified) by physical and chemical properties and is susceptible to change. Unit Essential Question: How can matter and changes to matter be described? Standards: Essential SC 5.P.8.1 – Compare and contrast the basic properties of solids, liquids, and gases, such as mass, volume, color, texture and Temperature. SC 5.P.8.3 – Demonstrate and explain that mixtures of solids can be separated based on observable properties of their parts Such as particle size, shape, color, and magnetic attraction. Important: SC 5.P.9.1 – Investigate and describe that many physical and chemical changes are affected by temperature. # Unit 3: Force and Motion (tested Mid Oct) Focus of the Unit/Lesson: Force directly affects the motion of an object. Unit Essential Question: How do forces directly affect the motion of an object? Standards: SC 5.P.13.1 – Identify familiar forces that cause objects to move, such as pushes or pulls, including gravity acting on falling objects. SC 5.P.13.2 – Investigate and describe that the greater the force applied to it, the greater the change in motion of a given Object. SC 5.P.13.3 – Investigate and describe that the more the mass an object has, the less effect a force will have on the objects’ Motion. SC 5.P.13.4 – Investigate and explain that when a force is applied to an object, but it does not move, it is because another Opposing force is being applied by something in the environment so that the forces are balanced. Focus of the Unit/Lesson: Energy exists in many forms and has the ability to cause motion or create change. Unit Essential Question: How do the different forms of energy cause motion and create change? # Unit 3: Energy (tested Nov) Standards: SC 5.P.10.1 – Investigate and describe some basic forms of energy, including light, heat, sound, electrical, chemical and Mechanical. SC 5.P.10.2 – Investigate and explain that energy has the ability to cause motion or create change. SC 5.P.10.4 – Investigate and explain that electrical energy can be transformed into heat, light, and sound energy as well as The energy of motion. ## Language Arts ### Focus Standard for Reading Comprehension LAFS.5.RI.1.1- Quote accurately from a text when explaining what the text says explicitly and when drawing inferences from the text. (DOK 2) LAFS.5.RI.1.2- Determine two or more main ideas of a text and explain how they are supported by key details; summarize the text. (DOK 2) LAFS.5.RI.1.3- Explain the relationships or interactions between two or more individuals, events, ideas, or concepts in a historical, scientific, or technical text based on specific information in the text. ### Focus Standard for Language Arts LAFS.5.L.1.1e - Recognize and correct inappropriate shifts in verb tense. (DOK 2) LAFS.5.L.1.2a - Use punctuation to separate items in a series. (DOK 1) LAFS.5.L.2.3a - Expand, combine, and reduce sentences for meaning, reader/listener interest, and style. (DOK 3) LAFS.5.L.2.3b - Compare and contrast the varieties of English (e.g., dialects, registers) used in stories, dramas, or poems. (DOK 3) LAFS.5.L.3.4b - Use common, grade-appropriate Greek and Latin affixes and roots as clues to the meaning of a word (e.g., photograph, photosynthesis). (DOK 2) LAFS.5.L.3.4c - Consult reference materials (e.g., dictionaries, glossaries, thesauruses), both print and digital, to find the pronunciation and determine or clarify the precise meaning of key words and phrases. (DOK 2) ## Social Studies Unit 1: The World in Spatial Terms Duration: 9/7-9/18 (9 days) Standards: SS.5.G.1.1 – Interpret current and historical information using a variety of geographic tools. SS.5.G.1.2 – Use latitude and longitude to locate places. SS.5.G.1.3 – Identify major United States physical features on a map of North America. SS.5.G.1.4 – Construct maps, charts, and graphs to display geographic information. SS.5.G.1.6 – Locate and identify states, capitals, and United States Territories on a map. Unit 2: Focus of the Unit/Lesson: Geographic features influenced cultural aspects of Native American life. Essential Question: How did geographic features influence cultural aspects of Native American life? Standards: SS 5.A.2.1 – Compare cultural aspects of ancient American civilizations. (Aztecs/Mayas; Mound Builders/Anasazi/Inuit) SS 5.A.2.2 – Identify Native American tribes from different geographical regions of North America (cliff dwellers and Pueblo People of the desert Southwest, coastal tribes of the Pacific Northwest, nomadic nations of the Great Plains, woodland Tribes east of the Mississippi River. major SS 5A.2.3 – Compare cultural aspects of Native American tribes from different geographic regions of North America including But not limited to clothing, shelter, food, major beliefs and practices, music, art, and interactions with the environment. Unit 3: Focus of the Unit/Lesson: European explorers have contributed to the development of Florida Essential Question: How did European exploration impact the development of Florida? Standards: SS 4.A.3.1 – Identify explorers who came to Florida and the motivations for their expeditions SS 4.A.3.2 – Describe causes and effects of European colonization on the Native American tribes of Florida SS 4.A.3.3 – Identify the significance of St. Augustine as the oldest permanent European settlement in the United States. SS 4.A.3.4 – Explain the purpose of and daily life on missions (San Luis de Tailmail in present day Tallahassee) SS 4.A.3.5 – Identify the significance of Fort Mose as the first free African community in the United States. SS 4.A.3.6 – Identify the effects of Spanish rule in Florida. SS 4.A.3.7 – Identify nations (Spain, France, England), that controlled Florida before it became a United States territory. SS 4.A.6.3 – Describe the contributions of significant individuals to Florida
# Common Core: 3rd Grade Math : Dividing Within 100 to Solve Word Problems ## Example Questions ### Example Question #1 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #2 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #3 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #1 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #5 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #6 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #7 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #8 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #9 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is ### Example Question #1 : Dividing Within 100 To Solve Word Problems I have oranges and boxes. I am going to put the same number of oranges in each box. How many oranges will be in each box? Possible Answers: Correct answer: Explanation: Looking at the picture below, we have a total of oranges and boxes. Each box has oranges in it. Because we are splitting the oranges up into equal groups, this is a division problem. Let's let equal the total number of oranges per box. Our equation is
## Using the Simple Interest Formula to Calculate Interest Earned ### Learning Outcomes • Use the simple interest formula Do you know that banks pay you to let them keep your money? The money you put in the bank is called the principal, $P$, and the bank pays you interest, $I$. The interest is computed as a certain percent of the principal; called the rate of interest, $r$. The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable for time, $t$, represents the number of years the money is left in the account. ### Simple Interest If an amount of money, $P$, the principal, is invested for a period of $t$ years at an annual interest rate $r$, the amount of interest, $I$, earned is $I=Prt$ where $\begin{array}{ccc}\hfill I& =& \text{interest}\hfill \\ \hfill P& =& \text{principal}\hfill \\ \hfill r& =& \text{rate}\hfill \\ \hfill t& =& \text{time}\hfill \end{array}$ Interest earned according to this formula is called simple interest. The formula we use to calculate simple interest is $I=Prt$. To use the simple interest formula we substitute in the values for variables that are given, and then solve for the unknown variable. It may be helpful to organize the information by listing all four variables and filling in the given information. ### example Find the simple interest earned after $3$ years on $\text{\500}$ at an interest rate of $\text{6%.}$ Solution Organize the given information in a list. $I = ?$ $P = 500$ $r = 6%$ $t= 3 \text{ years}$ We will use the simple interest formula to find the interest. Write the formula. $I=Prt$ Substitute the given information. Remember to write the percent in decimal form. $I=\left(500\right)\left(0.06\right)\left(3\right)$ Simplify. $I=90$ Check your answer. Is $\text{\90}$ a reasonable interest earned on $\text{\500}$ in $3$ years? In $3$ years the money earned $18\text{%}$. If we rounded to $20\text{%}$, the interest would have been $500(0.20)$ or $\text{\100}$. Yes, $\text{\90}$ is reasonable. Write a complete sentence that answers the question. The simple interest is $\text{\90}$. ### try it In the next example, we will use the simple interest formula to find the principal. ### example Find the principal invested if $\text{\178}$ interest was earned in $2$ years at an interest rate of $\text{4%.}$ ### try it Now we will solve for the rate of interest. ### example Find the rate if a principal of $\text{\8,200}$ earned $\text{\3,772}$ interest in $4$ years.
# Section 6.2 Calculus 2 Taylor and Maclaurin Series ## 6.2 Taylor and Maclaurin Series ### 6.2.1 Power Series from Functions • Let $$f(x)$$ have derivatives of all orders nearby $$a$$. Then the Taylor series generated by $$f$$ at $$a$$ is given by $$\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k$$, where $$f^{(k)}(a)$$ is the $$k^{th}$$ derivative of $$f$$ at $$a$$. • A Maclaurin series is a Taylor series where $$a=0$$. • A Taylor/Maclaurin series is said to converge to its generating function if it is equal to it for all members of its domain. • Example Let $$f(x)=\frac{1}{1+x}$$ with the domain $$-1<x<1$$, and guess a formula for $$f^{(k)}(0)$$ by computing its first few terms. Then show that the Maclaurin series generated by $$f$$ converges to $$f$$. • Example Let $$g(x)=\frac{2}{x}$$ with the domain $$0<x<4$$, and guess a formula for $$g^{(k)}(2)$$ by computing its first few terms. Then show that the Taylor series generated by $$g$$ at $$2$$ converges to $$g$$. • It can be shown that $$f$$ defined by $$f(0)=0$$ and $$f(x)=e^{-1/x^2}$$ otherwise satisfies $$f^{(k)}(0)=0$$, giving an example of a function which doesn’t converge to its Taylor series. • If a power series of the form $$\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k$$ converges to $$f(x)$$, then that power series is the Taylor series generated by $$f(x)$$ at $$a$$. ### 6.2.2 Maclaurin Series for $$e^x$$, $$\sin x$$, $$\cos x$$ • The following Maclaurin Series can be shown to converge to their generating functions: • $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$$ • $$\cos x = \sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k)!} = 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\dots$$ • $$\sin x = \sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!} = x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\dots$$ • Example Show how to generate the Maclaurin series for $$e^x$$. • Example Show how to generate the Maclaurin series for $$\sin x$$. ### Review Exercises 1. Let $$f(x)=\frac{1}{1-x}$$ with the domain $$-1<x<1$$, and guess a formula for $$f^{(k)}(0)$$ by computing its first few terms. Then show that the Maclaurin series generated by $$f$$ converges to $$f$$. 2. Let $$g(x)=\frac{3}{x}$$ with the domain $$0<x<6$$, and guess a formula for $$g^{(k)}(3)$$ by computing its first few terms. Then show that the Taylor series generated by $$g$$ at $$3$$ converges to $$g$$. 3. Let $$h(x)=\frac{1}{x^2+1}$$ with the domain $$-1<x<1$$. It may be shown that the first few terms of $$\<h^{(k)}(0)\>_{k=0}^\infty$$ are $$\<1,0,-2,0,24,0,-720,\dots\>$$. Show that the Maclaurin series generated by $$h$$ converges to $$h$$. 4. Show how to generate the Maclaurin series $$\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k)!}$$ for $$\cos x$$. 5. Find the Maclaurin series for $$\sinh x$$. 6. Find the Maclaurin series for $$\cosh x$$. 7. Find the Maclaurin series for $$e^{-x}$$. 8. Find the Maclaurin series for $$x^3+3x-7$$. Solutions ### Textbook References • University Calculus: Early Transcendentals (3rd Ed) • 9.8
# What is a discontinuity in math? ## What is a discontinuity in math? Discontinuity in Maths Definition The function of the graph which is not connected with each other is known as a discontinuous function. A function f(x) is said to have a discontinuity of the first kind at x = a, if the left-hand limit of f(x) and right-hand limit of f(x) both exist but are not equal. ## What is an example of a discontinuity? Discontinuous Function Examples Example 1: Identify if the function f(x) = (x – 2)/(x – 4) is a discontinuous function. Solution: As we can see, the function f(x) = (x – 2)/(x – 4) is not defined at x = 4. Hence it is discontinuous at x = 4. Answer: f(x) = (x – 2)/(x – 4) is a discontinuous function. What are the 4 types of discontinuity? There are four types of discontinuities you have to know: jump, point, essential, and removable. What is simple discontinuity? 1: 1.4 Calculus of One Variable … ►A simple discontinuity of ⁡ at occurs when ⁡ and ⁡ exist, but ⁡ ( c + ) ≠ f ⁡ . If ⁡ is continuous on an interval save for a finite number of simple discontinuities, then ⁡ is piecewise (or sectionally) continuous on . For an example, see Figure 1.4. ### How do you find a discontinuity? Explanation: Start by factoring the numerator and denominator of the function. A point of discontinuity occurs when a number is both a zero of the numerator and denominator. Since is a zero for both the numerator and denominator, there is a point of discontinuity there. ### How do you write a discontinuity? Start by factoring the numerator and denominator of the function. A point of discontinuity occurs when a number is both a zero of the numerator and denominator. Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation. What is another term for discontinuity? In this page you can discover 20 synonyms, antonyms, idiomatic expressions, and related words for discontinuity, like: , divergence, perturbation, space/time, circularity, polarisation, asymmetry, break, discontinuance, discontinuation and disruption. What is the difference between continuity and discontinuity? Discontinuity in human development usually signifies some form of change, whereas continuity implies maintaining the status quo (Lerner, 2002). Continuity and discontinuity include descriptions of and explanations for behavior, which are not necessarily undivided. #### What is a point discontinuity? A point of discontinuity occurs when a number is both a zero of the numerator and denominator. Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation. #### How to find discontinuity calculator? Start by factoring the numerator and denominator of the function. A point of discontinuity occurs when a number is both a zero of the numerator and denominator. Since is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation. What are the different types of discontinuities in calculus? Quick Overview. Discontinuities can be classified as jump,infinite,removable,endpoint,or mixed. • Jump Discontinuities. The graph of f ( x) below shows a function that is discontinuous at x = a. • Infinite Discontinuities. • Removable Discontinuities. • Removable Discontinuities can be Fixed. • Endpoint Discontinuities. • Mixed Discontinuities. • How to classify discontinuities? On graphs,the open and closed circles,or vertical asymptotes drawn as dashed lines help us identify discontinuities. • As before,graphs and tables allow us to estimate at best. • When working with formulas,getting zero in the denominator indicates a point of discontinuity. • ## What are the three types of discontinuity? Discontinuities can be classified as jump, infinite, removable, endpoint, or mixed. Removable discontinuities are characterized by the fact that the limit exists. Removable discontinuities can be “fixed” by re-defining the function. The other types of discontinuities are characterized by the fact that the limit does not exist.
Courses Courses for Kids Free study material Offline Centres More Store # Find the factors of 330.A. $2 \times 4 \times 5 \times 11$ B. $2 \times 3 \times 7 \times 13$ C. $2 \times 3 \times 5 \times 13$ D. $2 \times 3 \times 5 \times 11$ Last updated date: 20th Jun 2024 Total views: 386.1k Views today: 7.86k Verified 386.1k+ views Hint: Here, we will find the factors of the given number using the concept of prime factorization. Prime factorization is a method in which a number is factorized to get factors as prime numbers. Factors are defined as the whole number multiplied by a number to get another number. We are given a number 330. We will find the factors of 330 by using the method of prime factorization. We will follow the following steps has to perform doing prime factorization: First, we will divide the number by the smallest prime number which divides the number exactly. We will divide the quotient again by the smallest or the next smallest prime number if it is not exactly divisible by the smallest prime number. We divide the process again and again till the quotient becomes 1. We multiply the factors. So, using the steps, we get Thus, the factors of 330 are 2, 3, 5, and 11. As the factors are prime numbers, therefore we cannot factorize it further. Therefore, the factors of 330 are $2 \times 3 \times 5 \times 11$. Note: We should remember that we should use only the prime factors. the product should be the number itself. We can also use the factor tree method. In factor tree method, the given number has to be multiplied by a prime factor and a composite factor. The composite factor has to be factorized by a prime factor and a composite factor. This has to be continued till all the factors become the prime factors. Each number is a factor of itself and The number 1 is a factor of every number. Thus the number 1 can be neglected. The product should be the number itself. Thus the numbers become the factors.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 7.4: Solving Multiplication and Division Equations Difficulty Level: At Grade Created by: CK-12 ## Introduction Figuring Fares While visiting their grandparents, Kara has been handling the money for their rides on the subway. For the past two days, Marc has been watching her calculating and handing out money. Then when they were coming back from the mall, Marc noticed something on a sign about a day pass and a month long pass. “How much money do we have set aside for the subway?” Marc asked Kara the day after he saw the sign. “We have $45.00 each allotted for the subway,” she said. “Well, a day pass is$9.00, but a month long pass is $20.00. I think that is a better deal,” he told her. “Let’s figure out how many days we can ride for$9.00 per day.” Kara wrote the following equation on a piece of paper. \begin{align}9x=45\end{align} Solving equations is essential for Kara and Marc at this point in time. Then they will know whether day passes or a month long pass is the way to go. You can figure this out too. Solving equations is what you will learn in this lesson. Take what you learn and then apply it back to this problem. At the end of the lesson, you will know which train option makes the most sense. What You Will Learn In this lesson you will learn the following skills: • Solve single-variable multiplication equations. • Solve single-variable division equations. • Recognize and apply the multiplication and division properties of equality and the inverse relationship of multiplication and division. • Model and solve real-world problems using multiplication and division equations. Teaching Time I. Solve Single-Variable Multiplication Equations As you saw in the last lesson, we can have equations with addition and subtraction in them. We can also have equations with multiplication and division in them. These single-variable equations can be solved for the value of the variable. We want to solve the equation and figure out what the variable is equal to. How does this work? In the algebraic equation below, the variable \begin{align}z\end{align} represents only one possible number. \begin{align}z \times 2=8\end{align} What number does \begin{align}z\end{align} represent? We can find out by asking ourselves: what number, when multiplied by 2, equals 8? Since \begin{align}4 \times 2=8, z\end{align} must be equal to 4. We solved this equation using mental math. When we determine the value for a variable in an equation, we are solving that equation. If the equation involves a simple number fact, such as \begin{align}4 \times 2=8\end{align}, then we can solve the problem as we did above using mental math. What happens when we have a more challenging equation to solve? Some equations are more complex and mental math might not get the job done. To solve a more complex equation, such as \begin{align}z \times 7=105\end{align}, we might need to use a different strategy. Let's take a look at one strategy for solving an equation now. One strategy for solving an equation is to use inverse operations to isolate the variable. Remember, isolating the variable means getting the variable by itself on one side of the equal (=) sign. To solve an equation in which a variable is multiplied by a number, we can use the inverse operation of multiplication: division. We can divide both sides of the equation by that number to find the value of the variable. We must divide both sides of the equation by that number because of the Division Property of Equality, which states: if \begin{align}a=b\end{align} and \begin{align}c \neq 0\end{align}, then \begin{align}\frac{a}{c}=\frac{b}{c}\end{align}. This is much simpler than it looks. If you divide one side of an equation by a nonzero number, \begin{align}c\end{align}, you must divide the other side of the equation by that same number, \begin{align}c\end{align}, to keep the values on both sides equal. Example \begin{align}z \times 7=105\end{align}. In the equation, \begin{align}z\end{align} is multiplied by 7. So, we can divide both sides of the equation by 7 to solve for \begin{align}z\end{align}. \begin{align}z \times 7 &= 105\\ \frac{z \times 7}{7} &= \frac{105}{7}\\ z \times \frac{7}{7} &= 15\\ z \times 1 &= 15\\ z &= 15\end{align} It may help to separate out the factors like we did above. Having divided both sides by 7, we can see that the value of \begin{align}z\end{align} is 15. Example \begin{align}-8r=128\end{align} In the equation, -8 is multiplied by \begin{align}r\end{align}. So, we can divide both sides of the equation by -8 to solve for \begin{align}r\end{align}. \begin{align}-8r &= 128\\ \frac{-8r}{-8} &= \frac{128}{-8}\end{align} We need to use what we know about dividing integers to help us solve this problem. \begin{align}\frac{-8r}{-8} &= \frac{128}{-8}\\ 1r &= -16\\ r &= -16\end{align} The value of \begin{align}r\end{align} is -16 . Remember the integer rules for dividing positive and negative numbers? Let’s review. Now you can practice a few of these on your own. 7I. Lesson Exercises Solve each problem for the missing variable. 1. \begin{align}-4x=12\end{align} 2. \begin{align}8a=64\end{align} 3. \begin{align}9b=81\end{align} II. Solve Single-Variable Division Equations Sometimes, you will see equations that have division in them. Remember, that we can use a fraction bar to show division. To solve an equation in which a variable is divided by a number, we can use the inverse of division––multiplication. We can multiply both sides of the equation by that number to solve it. We must multiply both sides of the equation by that number because of the Multiplication Property of Equality, which states: if \begin{align}a=b\end{align}, then \begin{align}a \times c=b \times c\end{align}. So, if you multiply one side of an equation by a number, \begin{align}c\end{align}, you must multiply the other side of the equation by that same number, \begin{align}c\end{align}, to keep the values on both sides equal. Example \begin{align}k \div (-4)=12\end{align}. In the equation, \begin{align}k\end{align} is divided by -4. So, we can multiply both sides of the equation by -4 to solve for \begin{align}k\end{align}. You will need to use what you know about multiplying integers to help you solve this problem. It may help to rewrite \begin{align}k \div (-4)\end{align} as \begin{align}\frac{k}{-4}\end{align}. \begin{align}k \div (-4) &= 12\\ \frac{k}{-4} &= 12\\ \frac{k}{-4} \times (-4) &= 12 \times (-4)\\ \frac{k}{-4} \times \frac{-4}{1} &= -48\\ \frac{k}{\cancel{-4}} \times \frac{\cancel{-4}}{1} &= -48\\ \frac{k}{1} &= -48\\ k &= -48\end{align} The –4's will cancel each other out when they are divided. Then we multiply. The value of \begin{align}k\end{align} is –48. Remember the rules for multiplying integers will apply when working with these equations!! Think back and use them as you work. Example \begin{align}\frac{n}{1.5}=10\end{align} In the equation, \begin{align}n\end{align} is divided by 1.5. So, we can multiply both sides of the equation by 1.5 to solve for \begin{align}n\end{align}. \begin{align}\frac{n}{1.5} &= 10\\ \frac{n}{1.5} \times 1.5 &= 10 \times 1.5\\ \frac{n}{1.5} \times \frac{1.5}{1} &= 15\\ \frac{n}{\cancel{1.5}} \times \frac{\cancel{1.5}}{1} &= 15\\ \frac{n}{1} &= 15\\ n &= 15\end{align} The value of \begin{align}n\end{align} is 15. Now it is time for you to practice solving a few of these equations. 7J. Lesson Exercises Solve each equation for the missing variable. 1. \begin{align}\frac{x}{-2}=5\end{align} 2. \begin{align}\frac{y}{5}=6\end{align} 3. \begin{align}\frac{b}{-4}=-3\end{align} Take a few minutes to check your work with a peer. III. Using and Understanding Properties In the last two sections, we have been using properties to help us in solving equations. Remember that a property is a rule for performing operations in mathematics that makes our lives simpler. These properties are no exception. If you can look at an equation, see that it is a multiplication equation and know that division is the way to solve it, then the property of the Division Property of Equality is being utilized. In the same way if you use the Multiplication Property of Equality to solve a division problem, then the property has been useful. Always remember to think about the inverse operations and associate them with the different properties. This will help you to keep it all straight and not get mixed up. The Multiplication and Division Properties of Equality in combination with inverse operations are the best ways to solve equations correctly. IV. Model and Solve Real-World Problems Using Multiplication and Division Equations Equations are very useful when working with real-world problems. In fact, an equation is often the best way to sift through the information in a problem, select the correct information and work to solve an equation to find an answer. Let’s apply what we have learned to some real-life dilemmas. Example Three friends evenly split the total cost of the bill for their lunch. The amount each friend paid for his share was 4.25. a. Write an equation to represent \begin{align}c\end{align}, the total cost, in dollars, of the bill for lunch. b. Determine the total cost of the bill. Consider part \begin{align}a\end{align} first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. Because the total cost was evenly split by 3 friends, write a division equation to represent the problem. \begin{align}& \underline{Three \ friends} \ \underline{evenly \ split} \ the \ \underline{total \ cost} \ldots amount \ each \ friend \ paid \ \underline{was} \ \underline{\4.25} \ldots\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ c \qquad \qquad \qquad \ \div \qquad \qquad \quad 3 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ = \quad 4.25\end{align} This equation, \begin{align}c \div 3=4.25\end{align}, represents \begin{align}c\end{align}, the total cost of the lunch bill. Next, consider part \begin{align}b\end{align}. Solve the equation to find the total cost, in dollars, of the lunch bill. \begin{align}c \div 3 &= 4.25\\ \frac{c}{3} &= 4.25\\ \frac{c}{3} \times 3 &= 4.25 \times 3\\ \frac{c}{\cancel{3}} \times \frac{\cancel{3}}{1} &= 12.75\\ c &= 12.75\end{align} The total cost of the lunch bill was12.75. Example Sarvenaz earns $8 for each hour she works. She earned a total of$168 last week. a. Write an equation to represent \begin{align}h\end{align}, the number of hours she worked last week. b. Determine how many hours Sarvenaz worked last week. Consider part \begin{align}a\end{align} first. Use a number, an operation sign, a variable, or an equal sign to represent each part of that problem. She earns $8 for each hour she works, so you could multiply the number of hours she worked by$8 to find the total amount she earned. Write a multiplication equation. \begin{align}& \underline{\8 \ for \ each \ hour}\ldots She \ \underline{earned} \ a \ \underline{total \ of \ \168} \ last \ week.\\ & \qquad \downarrow \qquad \qquad \qquad \qquad \qquad \downarrow \qquad \qquad \ \downarrow\\ & \qquad 8h \qquad \qquad \qquad \qquad \quad \ = \qquad \quad \ 168\end{align} So, this equation, \begin{align}8h=168\end{align}, represents \begin{align}h\end{align}, the number of hours she worked last week. Next, consider part \begin{align}b\end{align}. Solve the equation to find \begin{align}h\end{align}, the number of hours she worked last week. \begin{align}8h &= 168\\ \frac{8h}{8} &= \frac{168}{8}\\ 1h &= 21\\ h &= 21\end{align} Sarvenaz worked 21 hours last week. Now let’s go back and apply what we have learned to the introduction problem. ## Real Life Example Completed Figuring Fares Here is the original problem once again. Reread it and underline any important information. While visiting their grandparents, Kara has been handling the money for their rides on the subway. For the past two days, Marc has been watching her calculating and handing out money. Then when they were coming back from the mall, Marc noticed something on a sign about a day pass and a month long pass. “How much money do we have set aside for the subway?” Marc asked Kara the day after he saw the sign. We have $45.00 each allotted for the subway,” she said. “Well, a day pass is$9.00, but a month long pass is $20.00. I think that is a better deal,” he told her. “Let’s figure out how many days we can ride for$9.00 per day.” Kara wrote the following equation on a piece of paper. \begin{align}\underline{9x=45}\end{align} Solving equations is essential for Kara and Marc at this point in time. Then they will know whether day passes or a month long pass is the way to go. First, let’s solve the equation and figure out how many day passes Kara and Marc can purchase day passes given their budget. \begin{align}9x &= 45\\ \frac{9x}{9} &= \frac{45}{9}\\ x &= 5\end{align} According to this, they can purchase day passes for only five days. A month long pass is 20.00 each. This is a much better deal for them. \begin{align}20(2) = \40.00\end{align} The total subway fares will cost them40.00 together instead of \$45.00 each. Kara and Marc purchase month long T passes the very next morning. ## Vocabulary Here are the vocabulary words that are found in this lesson. Isolate the Variable get the variable alone on one side of the equals sign. Inverse Operation Opposite operation Division Property of Equality states that if both sides of an equation are divided by the same number, they will stay equal. Multiplication Property of Equality states that if both sides of an equation are multiplied by the same number, they will stay equal. ## Technology Integration Other Videos: 1. http://www.mathplayground.com/mv_solving_single_step_equations.html – This is a Brightstorm video about solving single step equations. ## Time to Practice Directions: Solve each single-variable multiplication equation for the missing value. 1. \begin{align}4x=16\end{align} 2. \begin{align}6x=72\end{align} 3. \begin{align}-6x=72\end{align} 4. \begin{align}-3y=24\end{align} 5. \begin{align}-3y=-24\end{align} 6. \begin{align}-5x=-45\end{align} 7. \begin{align}-1.4x=2.8\end{align} 8. \begin{align}3.5a=7\end{align} 9. \begin{align}7a=-49\end{align} 10. \begin{align}14b=-42\end{align} Directions: Solve each single-variable division equation for the missing value. 11. \begin{align}\frac{x}{5}=2\end{align} 12. \begin{align}\frac{y}{7}=3\end{align} 13. \begin{align}\frac{b}{9}=-4\end{align} 14. \begin{align}\frac{b}{8}=-10\end{align} 15. \begin{align}\frac{x}{3}=-3\end{align} 16. \begin{align}\frac{x}{5}=-8\end{align} 17. \begin{align}\frac{x}{1.3}=3\end{align} 18. \begin{align}\frac{x}{2.4}=4\end{align} 19. \begin{align}\frac{x}{6}=1.2\end{align} 20. \begin{align}\frac{y}{1.5}=3\end{align} Directions: Solve each problem. 21. Mrs. Kowalski divided the students in the seventh grade into 7 equal-sized groups. There were 25 students in each group. a. Write an equation to represent \begin{align}s\end{align}, the total number of students in the seventh grade. b. Determine the total number of students in the seventh grade. 22. Evan needs to buy a total of 180 hot dogs for a school event. Hot dogs are sold in packages of 12 each. a. Write an equation to represent \begin{align}p\end{align}, the number of packages Evan must buy. b. Determine the number of hot dog packages Evan must buy. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
# What is the sum of 7+77+777+7777+... to n terms ? Nov 7, 2016 ${\sum}_{k = 1}^{n} {a}_{k} = \frac{70}{81} \left({10}^{n} - 1\right) - \frac{7}{9} n$ #### Explanation: Note that $\frac{7}{9} = 0. \overline{7}$ Hence we can write a formula for the $k$th term: ${a}_{k} = \frac{7}{9} \left({10}^{k} - 1\right) \text{ }$ for $k = 1 , 2 , 3. . .$ Let: ${b}_{k} = \frac{7}{9} \left({10}^{k}\right) = \frac{70}{9} \cdot {10}^{k - 1}$ This is in the form ${b}_{k} = b \cdot {r}^{k - 1}$, the general term of a geometric series with initial term $b = \frac{70}{9}$ and common ratio $r = 10$ The sum to $n$ terms is given by the formula: ${S}_{n} = \frac{b \left({r}^{n} - 1\right)}{r - 1} = \frac{70}{9} \cdot \frac{{10}^{n} - 1}{10 - 1} = \frac{70}{81} \cdot \left({10}^{n} - 1\right)$ Hence: ${\sum}_{k = 1}^{n} {a}_{k} = \frac{70}{81} \left({10}^{n} - 1\right) - \frac{7}{9} n$
Courses Courses for Kids Free study material Offline Centres More Store # The figure shows an irregular quadrilateral and the lengths of its individual sides. Which of the following equations best represents the perimeter of the quadrilateral.A) $4{m^2} + 5$ B) 5m+5C) $2{m^4} + 5$ D) ${m^4} + 5$ Last updated date: 12th Sep 2024 Total views: 429.9k Views today: 10.29k Verified 429.9k+ views Hint: Let us first try to understand the meaning of the perimeter of any polygon. It stands for the length of the sum of all the sides of the polygon. Thus, in order to find the perimeter of any given polygon, all you need to do is add all the lengths of the sides. In the question given to us, we have a quadrilateral. All you need to do is add the given lengths to find its perimeter. Complete step by step solution: Given to us is a quadrilateral having the length of the sides as m, 2m, m+2 and m+3 and we need to find the perimeter of the quadrilateral. For that all we need to do is add the lengths of all the sides. Let: ${\text{Length of the 1st side}} = m + 3 \\ {\text{Length of the 2nd side}} = m \\ {\text{Length of the 3rd side}} = 2m \\ {\text{Length of the 4th side}} = m + 2 \\$ Add all of them to find the perimeter of the quadrilateral given in the question. ${\text{Perimeter}} = m + 2m + m + 2 + m + 3 \\ = 5m + 5 \\$ Thus, we get the perimeter of the given figure as 5 m + 5. Thus, the correct option is option (b). Note: If in any given question, we have the opposite sides of the 4-sided polygon are equal, then the perimeter of that polygon can be calculated using the formula, ${\text{Perimeter}} = 2 \times \left( {{\text{sid}}{{\text{e}}_1} + {\text{sid}}{{\text{e}}_2}} \right)$ And if in any polygon all the 4 sides are equal then, ${\text{Perimeter}} = 4 \times \left( {\text{s}} \right)$, where ‘s’ stands for the side of the polygon.
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 16.8: Stokes's Theorem $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ Recall that one version of Green's Theorem (see equation 16.5.1) is $$\int_{\partial D} {\bf F}\cdot d{\bf r} =\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf k}\,dA.\] Here $$D$$ is a region in the $$x$$-$$y$$ plane and $$\bf k$$ is a unit normal to $$D$$ at every point. If $$D$$ is instead an orientable surface in space, there is an obvious way to alter this equation, and it turns out still to be true: Stoke's Theorem Provided that the quantities involved are sufficiently nice, and in particular if $$D$$ is orientable,$$\int_{\partial D} {\bf F}\cdot d{\bf r}=\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS,\] if $$\partial D$$ is oriented counter-clockwise relative to $$\bf N$$. Note how little has changed: $$\bf k$$ becomes $$\bf N$$, a unit normal to the surface, and $$dA$$ becomes $$dS$$, since this is now a general surface integral. The phrase "counter-clockwise relative to $$\bf N$$" means roughly that if we take the direction of $$\bf N$$ to be "up", then we go around the boundary counter-clockwise when viewed from "above''. In many cases, this description is inadequate. A slightly more complicated but general description is this: imagine standing on the side of the surface considered positive; walk to the boundary and turn left. You are now following the boundary in the correct direction. Example $$\PageIndex{2}$$: Let $${\bf F}=\langle e^{xy}\cos z,x^2z,xy\rangle$$ and the surface $$D$$ be $$x=\sqrt{1-y^2-z^2}$$, oriented in the positive $$x$$ direction. It quickly becomes apparent that the surface integral in Stokes's Theorem is intractable, so we try the line integral. The boundary of $$D$$ is the unit circle in the $$y$$-$$z$$ plane, $${\bf r}=\langle 0,\cos u,\sin u\rangle$$, $$0\le u\le 2\pi$$. The integral is $$\int_0^{2\pi} \langle e^{xy}\cos z,x^2z,xy\rangle\cdot\langle 0,-\sin u,\cos u\rangle\,du= \int_0^{2\pi} 0\,du = 0,\] because $$x=0$$. Example $$\PageIndex{3}$$: Consider the cylinder $${\bf r}=\langle \cos u,\sin u, v\rangle$$, $$0\le u\le 2\pi$$, $$0\le v\le 2$$, oriented outward, and $${\bf F}=\langle y,zx,xy\rangle$$. We compute$$\iint_\limits{D} \nabla\times{\bf F}\cdot {\bf N}\,dS= \int_{\partial D}{\bf F}\cdot d{\bf r}\] in two ways. First, the double integral is $$\int_0^{2\pi}\int_0^2 \langle 0,-\sin u,v-1\rangle\cdot \langle \cos u, \sin u, 0\rangle\,dv\,du= \int_0^{2\pi}\int_0^2 -\sin^2 u\,dv\,du = -2\pi. \] The boundary consists of two parts, the bottom circle $$\langle \cos t,\sin t, 0\rangle$$, with $$t$$ ranging from $$0$$ to $$2\pi$$, and $$\langle \cos t,\sin t, 2\rangle$$, with $$t$$ ranging from $$2\pi$$ to $$0$$. We compute the corresponding integrals and add the results:$$ \int_0^{2\pi} -\sin^2 t\,dt+\int_{2\pi}^0 -\sin^2t +2\cos^2t =-\pi-\pi=-2\pi, \] as before. An interesting consequence of Stokes's Theorem is that if $$D$$ and $$E$$ are two orientable surfaces with the same boundary, then $$\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS =\int_{\partial D} {\bf F}\cdot d{\bf r} =\int_{\partial E} {\bf F}\cdot d{\bf r} =\iint_\limits{E}(\nabla\times {\bf F})\cdot{\bf N}\,dS. \] Sometimes both of the integrals$$\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS\qquad\hbox{and}\qquad\int_{\partial D} {\bf F}\cdot d{\bf r}\] are difficult, but you may be able to find a second surface $$E$$ so that $$\iint_\limits{E}(\nabla\times {\bf F})\cdot{\bf N}\,dS\] has the same value but is easier to compute. Example $$\PageIndex{4}$$: In example 16.8.2 the line integral was easy to compute. But we might also notice that another surface $$E$$ with the same boundary is the flat disk $$y^2+z^2\le 1$$. The unit normal $$\bf N$$ for this surface is simply $${\bf i}=\langle 1,0,0\rangle$$. We compute the curl:$$\nabla\times{\bf F}=\langle x-x^2,-e^{xy}\sin z-y,2xz-xe^{xy}\cos z\rangle.\] Since $$x=0$$ everywhere on the surface, $$(\nabla\times{\bf F})\cdot {\bf N}=\langle 0,-e^{xy}\sin z-y,2xz-xe^{xy}\cos z\rangle\cdot\langle 1,0,0\rangle=0,$$ so the surface integral is $$\iint_\limits{E}0\,dS=0,\] as before. In this case, of course, it is still somewhat easier to compute the line integral, avoiding $$nabla\times{\bf F}$$ entirely. Example $$\PageIndex{5}$$: Let $${\bf F}=\langle -y^2,x,z^2\rangle$$, and let the curve $$C$$ be the intersection of the cylinder $$x^2+y^2=1$$ with the plane $$y+z=2$$, oriented counter-clockwise when viewed from above. We compute $$\int_C {\bf F}\cdot d{\bf r}$$ in two ways. First we do it directly: a vector function for $$C$$ is({\bf r}=\langle \cos u,\sin u, 2-\sin u\rangle\), so $${\bf r}'=\langle -\sin u,\cos u,-\cos u\rangle$$, and the integral is then$$\int_0^{2\pi} y^2\sin u+x\cos u-z^2\cos u\,du =\int_0^{2\pi} \sin^3 u+\cos^2 u-(2-\sin u)^2\cos u\,du =\pi.\] To use Stokes's Theorem, we pick a surface with $$C$$ as the boundary; the simplest such surface is that portion of the plane $$y+z=2$$ inside the cylinder. This has vector equation $${\bf r}=\langle v\cos u,v\sin u,2-v\sin u\rangle$$. We compute $${\bf r}_u= \langle -v\sin u,v\cos u,-v\cos u\rangle$$, $${\bf r}_v= \langle \cos u,\sin u, -\sin u\rangle$$, and $${\bf r}_u\times{\bf r}_v=\langle 0,-v,-v\rangle$$. To match the orientation of $$C$$ we need to use the normal $$\langle 0,v,v\rangle$$. The curl of $$\bf F$$ is $$\langle 0,0,1+2y\rangle= \langle 0,0,1+2v\sin u\rangle$$, and the surface integral from Stokes's Theorem is $$\int_0^{2\pi}\int_0^1 (1+2v\sin u)v\,dv\,du=\pi.\] In this case the surface integral was more work to set up, but the resulting integral is somewhat easier. Proof of Stokes's Theorem We can prove here a special case of Stokes's Theorem, which perhaps not too surprisingly uses Green's Theorem. Suppose the surface $$D$$ of interest can be expressed in the form $$z=g(x,y)$$, and let $${\bf F}=\langle P,Q,R\rangle$$. Using the vector function $${\bf r}=\langle x,y,g(x,y)\rangle$$ for the surface we get the surface integral$$\eqalign{\iint_\limits{D} \nabla\times{\bf F}\cdot d{\bf S}&= \iint_\limits{E} \langle R_y-Q_z,P_z-R_x,Q_x-P_y\rangle\cdot \langle -g_x,-g_y,1\rangle\,dA\cr &=\iint_\limits{E}-R_yg_x+Q_zg_x-P_zg_y+R_xg_y+Q_x-P_y\,dA.\cr}\] Here $$E$$ is the region in the $$x$$-$$y$$ plane directly below the surface$$D$$. For the line integral, we need a vector function for $$\partial D$$. If $$\langle x(t),y(t)\rangle$$ is a vector function for $$\partial E$$ then we may use $${\bf r}(t)=\langle x(t),y(t),g(x(t),y(t))\rangle$$ to represent $$\partial D$$. Then $$\int_{\partial D}{\bf F}\cdot d{\bf r}=\int_a^b P{dx\over dt}+Q{dy\over dt}+R{dz\over dt}\,dt=\int_a^b P{dx\over dt}+Q{dy\over dt}+R\left({\partial z\over\partial x}{dx\over dt}+{\partial z\over\partial y}{dy\over dt}\right)\,dt.\] using the chain rule for $$dz/dt$$. Now we continue to manipulate this:$$\eqalign{\int_a^b P{dx\over dt}+Q{dy\over dt}+&R\left({\partial z\over\partial x}{dx\over dt}+{\partial z\over\partial y}{dy\over dt}\right)\,dt\cr &=\int_a^b \left[\left(P+R{\partial z\over\partial x}\right){dx\over dt}+ \left(Q+R{\partial z\over\partial y}\right){dy\over dt}\right]\,dt\cr &=\int_{\partial E} \left(P+R{\partial z\over\partial x}\right)\,dx+\left(Q+R{\partial z\over\partial y}\right)\,dy,\cr}\] which now looks just like the line integral of Green's Theorem, except that the functions $$P$$ and $$Q$$ of Green's Theorem have been replaced by the more complicated $$P+R(\partial z/\partial x)$$ and $$Q+R(\partial z/\partial y)$$. We can apply Green's Theorem to get $$\int_{\partial E} \left(P+R{\partial z\over\partial x}\right)\,dx+\left(Q+R{\partial z\over\partial y}\right)\,dy= \iint_\limits{E} {\partial\over \partial x}\left(Q+R{\partial z\over\partial y}\right)-{\partial\over \partial y}\left(P+R{\partial z\over\partial x}\right)\,dA.\] Now we can use the chain rule again to evaluate the derivatives inside this integral, and it becomes$$\eqalign{\iint_\limits{E} &Q_x+Q_zg_x+R_xg_y+R_zg_xg_y+Rg_{yx}-\left(P_y+P_zg_y+R_yg_x+R_zg_yg_x+Rg_{xy}\right)\,dA\cr&=\iint_\limits{E} Q_x+Q_zg_x+R_xg_y-P_y-P_zg_y-R_yg_x\,dA,\cr}\] which is the same as the expression we obtained for the surface integral. $$\square$$
# Rules for Deductive Reasoning – Syllogism in Mathematics Share Cognitive Science Implications for Teaching Mathematics • • • • • • A less precise, but correct, statement of The Transitive Property of Equality which is quite useful for constructing mathematical models is: It two expressions represent the same quantity, the two expressions must be equal. The key to every “word” problem, or “application” problem, or “modeling” problem is this interpretation of the The Transitive Property of Equality. However, whenever the model is a well established formula it is wise to use the formula (what was a key step in deriving that formula – make a guess). I will address this topic in detail in future posts. Example The number of man days to complete the job is (6)(14) The number of man days to complete the job is 21x Therefore (by the Transitive Property) these two expressions must be equal. This observation yields the model 21x = (6)(14). Example Some real estate will increase in value. Anything that will increase in value is a good investment. Therefore, some real estate is a good investment. Example Every odd natural number can be written in the form 2k + 1 for some integer k. 23875 is an odd natural number. Therefore 23875 can be written in the form 2k + 1 for some integer k. Every beginning algebra textbook presents some form or another the following First Two Properties of Equations. (1) : If any expression is added to both sides of an equation the resulting equation is equivalent to the original equation. (2) : If both sides of an equation are multiplied by the same non-zero real number, the resulting equation is equivalent to the original equation. The solution of every linear equation is now solved with the same simple application of syllogism. For example Every linear equation in one variable can be solved using the first two properties of equations. 3x + 14 = 11x – 9 is a linear equation in one variable. Therefore 3x + 14 = 11x – 9 can be solved using the first two properties of equations. Another example Every linear equation in one variable can be solved using the first two properties of equations. is a linear equation in one variable. Therefore can be solved using the first two properties of equations. If you like the posts of this blog, share them with your friends and associates. Encourage them to read the post every Tuesday and Thursday. Use the Social Media buttons to share this essay with friends and associates. Use the like button to indicate your approval. ## 2 thoughts on “Rules for Deductive Reasoning – Syllogism in Mathematics” 1. M P HODGES says: I love rules. I am most acceptable to all rules without cost! At what age in general do our young agree that rules are good or beneficial?
Enable contrast version # TutorMe Blog ## How To Use the Leading Coefficient Test To Graph End Behavior Charlotte Taylor January 14, 2021 You can use the leading coefficient test to figure out end behavior of the graph of a polynomial function. This isn’t some complicated theorem. There’s no factoring or x-intercepts. Here are two steps you need to know when graphing polynomials for their left and right end behavior. ## Defining the Terms Let’s review some common precalculus terms you’ll need for the leading coefficient test: A polynomial is a fancy way of saying "many terms." A polynomial function is a function (a statement that describes an output for any given input) that is composed of many terms. 2x3+8-4 is a polynomial. y=2x3+8-4 is a polynomial function. A leading term in a polynomial function f is the term that contains the biggest exponent. A coefficient is the number in front of the variable. End behavior is another way of saying whether the graph ascends or descends in either direction. The leading coefficient test is a quick and easy way to discover the end behavior of the graph of a polynomial function by looking at the term with the biggest exponent. ## Step 1: The Coefficient of the Leading Term Determines Behavior to the Right The behavior of the graph is highly dependent on the leading term because the term with the highest exponent will be the most influential term. Let’s step back and explain these terms. 2x3 is the leading term of the function y=2x3+8-4. 2 is the coefficient of the leading term. When graphing a function, the leading coefficient test is a quick way to see whether the graph rises or descends for either really large positive numbers (end behavior of the graph to the right) or really large negative numbers (end behavior of the graph to the left). Let's start with the right side of the graph, where only positive numbers are in the place of x. If the leading coefficient is positive, bigger inputs only make the leading term more and more positive. The graph will rise to the right. If the leading coefficient is negative, bigger inputs only make the leading term more and more negative. The graph will descend to the right. ## Step 2: The Degree of the Exponent Determines Behavior to the Left The variable with the exponent is x3. When you replace x with positive numbers, the variable with the exponent will always be positive. So you only need to look at the coefficient to determine right-hand behavior. When you replace x with negative numbers, the variable with the exponent can be either positive or negative depending on the degree of the exponent. A negative number multiplied by itself an odd number of times will remain negative. A negative number multiplied by itself an even number of times will become positive. The leading coefficient test tells us that the graph rises or falls depending on whether the leading terms are positive or negative, so for left-hand behavior (negative numbers), you will need to look at both the coefficient and the degree of the component together. Let’s look at the following examples of when x is negative: A trick to determine end graphing behavior to the left is to remember that "Odd" = "Opposite." If the degree is odd, the end behavior of the graph for the left will be the opposite of the right-hand behavior. If the degree is even, the variable with the exponent will be positive and, thus, the left-hand behavior will be the same as the right. ## Using the Leading Coefficient Test When graphing a polynomial function, look at the coefficient of the leading term to tell you whether the graph rises or falls to the right. Look at the exponent of the leading term to compare whether the left side of the graph is the opposite (odd) or the same (even) as the right side. When in doubt, split the leading term into the coefficient and the variable with the exponent and see what happens when you substitute either a negative number (left-hand behavior) or a positive number (right-hand behavior) for x. Finally, here are some complete examples illustrating the leading coefficient test: BEST IN CLASS SINCE 2015 TutorMe homepage
# 0.4 8.5 - rotational kinetic energy: work and energy revisited Page 1 / 8 • Derive the equation for rotational work. • Calculate rotational kinetic energy. • Demonstrate the Law of Conservation of Energy. In this module, we will learn about work and energy associated with rotational motion. [link] shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy . Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in [link] ) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled: $\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(\text{net}\phantom{\rule{0.25em}{0ex}}F\right)\text{Δ}s.$ To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by $r$ , and gather terms: $\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(r\phantom{\rule{0.25em}{0ex}}\text{net}\phantom{\rule{0.25em}{0ex}}F\right)\frac{\text{Δ}s}{r}.$ We recognize that $r\phantom{\rule{0.25em}{0ex}}\text{net}\phantom{\rule{0.25em}{0ex}}F=\text{net τ}$ and $\Delta s/r=\theta$ , so that $\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(\text{net τ}\right)\theta .$ This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation $\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(\text{net τ}\right)\theta$ is valid in general, even though it was derived for a special case. To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that $\text{net τ}=\mathrm{I\alpha }$ , so that $\text{net}\phantom{\rule{0.25em}{0ex}}W=I\text{αθ}.$ ## Making connections Work and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation . Now, we solve one of the rotational kinematics equations for $\text{αθ}$ . We start with the equation ${{\omega }_{}}^{2}={{\omega }_{\text{0}}}^{2}+2\text{αθ}.$ Next, we solve for $\text{αθ}$ : $\text{αθ}=\frac{{\omega }^{2}-{{\omega }_{\text{0}}}^{2}}{2}.$ Substituting this into the equation for net $W$ and gathering terms yields $\text{net}\phantom{\rule{0.25em}{0ex}}W=\frac{1}{2}{\mathrm{I\omega }}^{2}-\frac{1}{2}I{{\omega }_{\text{0}}}^{2}.$ This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term $\left(\frac{1}{2}\right){\mathrm{I\omega }}^{2}$ to be rotational kinetic energy ${\text{KE}}_{\text{rot}}$ for an object with a moment of inertia $I$ and an angular velocity $\omega$ : are nano particles real yeah Joseph Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master? no can't Lohitha where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine has a lot of application modern world Kamaluddeen yes narayan what is variations in raman spectra for nanomaterials ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? 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# Illustrative Mathematics Unit 6.1, Lesson 7: From Parallelograms to Triangles Related Topics: Math Worksheets Learn about comparing the area of parallelograms and the area of triangles. After trying the questions, click on the buttons to view answers and explanations in text or video. Same Parallelograms, Different Bases Let’s compare parallelograms and triangles. Illustrative Math Unit 6.1, Lesson 7 (printable worksheets) 7.1 - Same Parallelograms, Different Bases Here are two copies of a parallelogram. Each copy has one side labeled as the base b and a segment drawn for its corresponding height and labeled h. 1. The base of the parallelogram on the left is 2.4 centimeters; its corresponding height is 1 centimeter. Find its area in square centimeters. 2. The height of the parallelogram on the right is 2 centimeters. How long is the base of that parallelogram? Explain your reasoning. 7.2 - A Tale of Two Triangles (Part 1) Two polygons are identical if they match up exactly when placed one on top of the other. 1. Open the applet. Draw one line segment to decompose each of the following polygons into two identical triangles, if possible. If you choose to, you can also draw the triangles. 2. Which quadrilaterals can be decomposed into two identical triangles? 3. Study the quadrilaterals that were, in fact, decomposable into two identical triangles. What do you notice about them? Write a couple of observations about what these quadrilaterals have in common. Open the next applet. Draw some other types of quadrilaterals that are not already shown. Try to decompose them into two identical triangles. Can you do it? Come up with a general rule about what must be true if a quadrilateral can be decomposed into two identical triangles. 1. These are examples of how the quadrilaterals can be decomposed into triangles by connecting opposite vertices. However, triangles from the same quadrilateral are not always identical. See the answers to the following questions for more detail. 2. A, B, D, F, and G can be decomposed into two identical triangles. C and E cannot. 3. A, B, D, F, and G have two pairs of parallel sides, equal opposite sides, and equal opposite angles, while C and E do not. These are examples of additional quadrilaterals that can be decomposed into two identical triangles. Based on this and the previous answers, we can conclude that for a quadrilateral to be decomposable into two identical triangles it must be a parallelogram. 7.3 - A Tale of Two Triangles (Part 2) 3. This applet has eight pairs of triangles. Choose 1–2 pairs of triangles. Use them to help you answer the following questions. 1. Which pair(s) of triangles do you have? _________________ Can each pair of triangles be composed into: a. a rectangle? b. a parallelogram? 2. Check the other pairs. Complete each of the following statements with the words "all", "some", or "none". Sketch 1–2 examples to illustrate each completed statement. a. ________________ of these pairs of identical triangles can be composed into a rectangle. b. ________________ of these pairs of identical triangles can be composed into a parallelogram. Pair P can be composed into a square, which has all the properties of a rectangle or a parallelogram. (See this link for more information about how squares, rectangles, and parallelograms are related.) Pair Q can be composed into a parallelogram. Some of these pairs of identical triangles can be composed into a rectangle. All of these pairs of identical triangles can be composed into a parallelogram. Lesson 7 Summary A parallelogram can always be decomposed into two identical triangles by a segment that connects opposite vertices. Going the other way around, two identical copies of a triangle can always be arranged to form a parallelogram, regardless of the type of triangle being used. To produce a parallelogram, we can join a triangle and its copy along any of the three sides, so the same pair of triangles can make different parallelograms. Here are examples of how two copies of both Triangle A and Triangle F can be composed into three different parallelograms. This special relationship between triangles and parallelograms can help us reason about the area of any triangle. Practice Problems 1. To decompose a quadrilateral into two identical shapes, Clare drew a dashed line as shown in the diagram. A: Clare said the that two resulting shapes have the same area. Do you agree? Explain your reasoning. B: Did Clare partition the figure into two identical shapes? Explain your reasoning. A: The two shapes do have the same area. The area of the rectangle is 4 × 2 = 8 square units, while the area of the triangle is half the area of a square that is 4 by 4 units, as shown below, so its area is ½ × (4 × 4) = 8 square units. B: These are not two identical shapes. One is a triangle and the other is a rectangle. The original quadrilateral is not a parallelogram either, so it may or may not be possible to divide the original quadrilateral into identical halves. 2. Triangle R is a right triangle. Can we use two copies of Triangle R to compose a parallelogram that is not a square? If so, explain how or sketch a solution. If not, explain why not. It is possible to use two copies of Triangle R to compose a parallelogram that is not a square. 3. Two copies of this triangle are used to compose a parallelogram. Which parallelogram cannot be a result of the composition? If you get stuck, consider using tracing paper. A, B, and D can all be composed out of copies of this triangle, as seen by the triangle covering exactly half of each of these parallelograms. C cannot be composed out of copies of this triangle, as the remaining unshaded area is not a triangle. 4. A: On the grid, draw at least three different quadrilaterals that can each be decomposed into two identical triangles with a single cut (show the cut line). One or more of the quadrilaterals should have non-right angles. B: Identify the type of each quadrilateral. All parallelograms are quadrilaterals that can be decomposed into two identical triangles with a single cut. Squares and rectangles have all the properties of parallelograms. A is a rectangle. B is a parallelogram with non-right angles. C is a square. 5. A: A parallelogram has a base of 9 units and a corresponding height of ⅔ units. What is its area? B: A parallelogram has a base of 9 units and an area of 12 square units. What is the corresponding height for that base? C: A parallelogram has an area of 7 square units. If the height that corresponds to a base is ¼ unit, what is the base? A: A = b · h A = 9 units · ⅔ units A = 6 square units B: A = b · h 12 square units = 9 units · h h = 12 square units ÷ 9 units h = 1⅓ units C: A = b · h 7 square units = b · ¼ unit b = 7 square units ÷ ¼ unit b = 28 units 6. List all segments that could represent a corresponding height if the side n is the base. g and h are perpendicular to the base n and could represent its corresponding height. The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics. 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.
# Trigonometry solver with steps Trigonometry solver with steps can be a helpful tool for these students. So let's get started! ## The Best Trigonometry solver with steps One instrument that can be used is Trigonometry solver with steps. There are many ways to solve simultaneous equations, and the best method depends on the specific equations involved. However, there are some general tips that can be useful when solving simultaneous equations. First, it can be helpful to rearrange the equations so that one variable is isolated on one side. Next, one of the equations can be divided by a constant to simplify it. Finally, the two equations can be solved by substitution or elimination. A direct variation is a proportionality between two variables, in which one variable is a multiple of the other. For example, if y varies directly as x, and we know that y = 10 when x = 5, then we can find the value of y when x = 2 by multiplying 5 by 2, which equals 10. In general, if y varies directly as x, and y = k when x = a, then we can find the value of y when x = b by Log equations can be difficult to solve, but there are a few basic steps that can be followed to make solving them easier. First, identify the base of the logs and rewrite the equation so that all logs are of the same base. Next, use the properties of logs to simplify the equation as much as possible. Finally, solve the equation using traditional algebraic methods. Maths learning apps are a great way to improve your child's maths skills. By using an app, your child can learn at their own pace and work on specific areas that they need to improve. There are many different maths learning apps available, so it is important to choose one that is suitable for your child's age and ability. You can also find apps that focus on specific areas of maths, such as addition, subtraction, multiplication, or division. By using a maths learning app This process involves using the square root of a number to isolate a variable that is being solving for. While this process can be complicated, it is often a useful tool for solving equations that would otherwise be difficult or impossible to solve. ## We cover all types of math problems I- I have no words. I simply love it. It works amazingly. It helps with all my math hawks and I understand better. Thanks to this app. I recommend to all person out there having troubles with math cause this app is just wonderful. You just need a good camera for this app and it may take some time to scan a work but it's worth it. I simply love this app. Great job the developers. ### Layla Alexander Leaps and bounds better than Microsoft Math Solver. Didn't seem to ask for unnecessary permissions either. The free version is very helpful, but the paid version will really help you understand math. ### Raquel Jones How to solve fractions over fractions How to solve vertical asymptote Help with math How to cheat in math Live math help Math answers to all problems for free
# Differences Between d^2y/dx^2 and (dy/dx)^2 in Calculus Derivatives have many uses outside of mathematics and everyday life, including in subjects like science, engineering, physics, and others. You must have mastered the ability to calculate the derivative of various functions in earlier courses, including trigonometric, implicit, logarithm, etc. d2y/dx2 and (dydx)^2 are two derivative equations. But to understand them, first, you need to understand what exactly is the second derivative. The derivative of a function in calculus is known as the second derivative, sometimes known as the second-order derivative. The second derivative, roughly speaking, measures how a quantity’s rate of change is itself changing. For instance, the second derivative of an object’s position with respect to time is the object’s instantaneous acceleration or the rate at which the object’s velocity is changing with respect to time. In this article, I’ll tell you what the difference is between d2y/dx2=(dy/dx)^2 and what exactly the derivative means. Page Contents ## D2y/dx2 Vs (dy/dx)^2 Derivative ofÂ dy/dxÂ (These 2s may look like index notation, but they arenâ€™t). (dy/dx)2, on the other hand, is the square of the first derivative. Example: Take Y=3????3+6????2y=3×3+6×2 The first derivative: dy/dx=9????2+12????dydx=9×2+12x The second derivative: d2yd????2=18????+12d2ydx2=18x+12 The square of the first derivative:Â (dy/dx)2=(9????2+12????)2=(81????4+216????3+144 ## What Is the Second Derivative? When you differentiate the derivative, you get the second derivative. Remember that dy/dx is the derivative of y with respect to x. The second derivative, pronounced “dee two y by d x squared,” is represented as d2y/dx2. The nature of stationary points can be more easily ascertained using the second derivative (whether they’re maximum points, minimum points, or points of inflection). When dy/dx = 0, a curve reaches a stationary point. The type of stationary point (maximum, minimum, or point of inflection) can be determined using the second derivative once the location of the stationary point has been established. ## What Is Derivative? The derivative of a function of a real variable in mathematics quantifies the sensitivity of the function’s value (output value) to changes in its argument (input value). Calculus’s core tool is the derivative. The velocity of an item, for instance, is the derivative of its position with respect to time. It quantifies how quickly the object’s position varies as time passes. When it occurs, the slope of the tangent line to the function’s graph at a given input value is the derivative of a function of a single variable. The function closest to that input value is best approximated linearly by the tangent line. Because of this, the derivative is frequently referred to as the “instantaneous rate of change,” which is the ratio of the instantaneous change in the dependent variable to that in the independent variable. To include functions of several real variables, derivatives can be generalized. This generalization reinterprets the derivative as a linear transformation whose graph, after a suitable translation, is the best linear approximation to the graph of the original function. Regarding the foundation provided by the selection of independent and dependent variables, the Jacobian matrix is the matrix that represents this linear transformation. ### Understanding Differentiation, Antidifferentiation, and Integration in Calculus It can be computed using the partial derivatives of the independent variables. The gradient vector replaces the Jacobian matrix for a real-valued function with several variables. Differentiation is the action of locating a derivative. Antidifferentiation is the term for the opposite process. Antidifferentiation and integration are related in the calculus fundamental theorem. The two fundamental operations of single-variable calculus are differentiation and integration. ## Different Notations ### Leibniz’s notation In 1675, Gottfried Wilhelm Leibniz introduced the letters dx, dy, and dy/dx. Even today, it is frequently employed when the relationship between the dependent and independent variables in the equation y = f(x) is considered to be functional. The variable for differentiation (in the denominator) can be specified using Leibniz’s notation, which is important for partial differentiation. ### Lagrange’s notation One of the most popular modern differentiation notations, sometimes known as prime notation, uses the prime mark and is credited to Joseph-Louis Lagrange. It denotes the derivative of a function f as f1. The latter notation generalizes to provide the notation f(n) for the nth derivative of f, which is more convenient when discussing the derivative as a function rather than a function of itself because the Leibniz notation can be complicated in this situation. ### Newton’s notation A dot is placed over the function name in Newton’s differentiation notation, often known as the “dot notation,” to signify a time derivative. Only derivatives with regard to time or arc length are represented using this notation. Usually, it’s applied to differential equations in differential geometry and physics. However, the dot notation is inapplicable to several independent variables and high-order derivatives (order 4 or more). ### Euler’s notation The first derivative, Df, is obtained using the differential operator D in Euler’s notation by applying it to a function f. Dnd stands for the nth derivative.Â If y = f(x) is a dependent variable, the independent variable x is frequently clarified by adding the subscript x to the D. Although when the variable x is understood, such as when this is the sole independent variable contained in the equation, this subscript is frequently left off. For expressing and resolving linear differential equations, Euler’s notation is helpful. ## Application of Derivatives in Maths Derivatives are frequently used in mathematics. They can be used to determine a function’s maximum or minimum, the slope of a curve, or even the inflection point. Below are a few instances where we will use the derivative. And the following sections go into great detail about each of them. The application of derivatives is most frequently found in: • Calculating a quantity’s Rate of change • Getting a good estimate of the value • Finding the equation for a curve’s tangent and normal • Identifying the point of inflection, maxima, and minima • Assessing the increasing and decreasing functions ## Application of Derivatives in Real Life Derivatives can be used in many situations in real life. Here is a list of a few situations in which you can use derivation: • To calculate profit and loss in the business. • In order to measure temperature variation. • To calculate the rate of travel, such as miles per hour, kilometers per hour, etc. • Numerous physics equations are derived using derivatives. • Finding the earthquake magnitude range is a favorite task in seismology research. ## Conclusion • d^2y/dx^2 is the second derivative. (dy/dx)^2 is the square of the first derivative. • Derivatives matter in business, physics, and temperature measurement. • Finding highs and lows in math relies on derivatives. • The second derivative spots still points and turning points. • Notations like Leibniz’s, Lagrange’s, Newton’s, and Euler’s help show derivatives. • Derivatives gauge change, locate slopes, and analyze functions. • They’re used for profits and temperature readings. Also, for physics formulas. • Knowing derivatives matters for accurate real-world math. Scroll to Top
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis : An additive inverse of a number is defined as the value, which on adding with the original number results in zero value. It is the value we add to a number to yield zero. Suppose, a is the original number, then its additive inverse will be minus of a i.e.,-a, such that; a+(-a) = a – a = 0 Example: • Additive inverse of 10 is -10, as 10 + (-10) = 0 • Additive inverse of -9 is 9, as (-9) + 9 = 0 Check: Multiplicative Inverse Additive inverse is also called the opposite of the number, negation of number or changed sign of original number. Fact: Additive inverse of zero is zero only. How to Find the Additive inverse? The additive inverse of any given number can be found by changing the sign of it. The additive inverse of a positive number will be a negative, whereas the additive inverse of a negative number will be positive. However, there will be no change in the numerical value except the sign. For example, the additive inverse of 8 is -8, whereas the additive inverse of -6 is 6. Properties Additive inverse simply means changing the sign of the number and adding it to the original number to get an answer equal to 0. The properties of additive inverse are given below, based on negation of the original number. For example, x is the original number, then its additive inverse is -x. So, here we will see the properties of -x. • −(−x) = x • (-x)2 = x2 • −(x + y) = (−x) + (−y) • −(x – y) = y − x • x − (−y) = x + y • (−x) × y = x × (−y) = −(x × y) • (−x) × (−y) = x × y We have understood that an additive inverse is added to a value to make it zero. Now this value can be a natural number, integer, rational number, irrational number, complex number, etc. Let us find the additive inverse of different types of numbers. Additive inverse of Natural or Whole Numbers As we know, natural numbers are the positive integers. Therefore, the additive inverse of positive integers will be negative. Whole numbers/Natural numbers Additive Inverse Result 0 0 0+0 = 0 1 -1 1+(-1) = 0 2 -2 2+(-2) = 0 3 -3 3+(-3) = 0 4 -4 4+(-4) = 0 5 -5 5+(-5) = 0 10 -10 10+(-10) = 0 20 -20 20+(-20) = 0 50 -50 50+(-50) = 0 100 -100 100+(-100) =0 Complex numbers are the combination of real numbers and imaginary numbers. A + iB is a complex number, where A is the real number and B is the imaginary number. Now the additive inverse of A + iB should be a value, that on adding it with a given complex number, we get a result as zero. Therefore, it will be -(A + iB) Example: Additive inverse of 2 + 3i is -(2+3i) 2+3i + [-(2+3i)] = 2+3i -2-3i = 0 Suppose a/b is a rational number such that the additive inverse of a/b is -a/b and vice versa. Fraction Additive Inverse Result 1/2 -1/2 (½) + (-½) = 0 1/4 -1/4 (¼) + (-¼) = 0 3/4 -3/4 (¾) + (-¾) = 0 2/5 -2/5 ⅖ + (-⅖) = 0 10/3 -10/3 10/3 + (-10/3) = 0 Difference Between Additive Inverse and Multiplicative Inverse Additive inverse and multiplicative inverse, both have different properties. See the below table to know the differences. Additive Inverse Multiplicative Inverse It is added to the original number to get 0 It is multiplied to the original number to get 1 Results in 0 Results in 1 Sign of the original number is changed and added Reciprocal of the original number is multiplied Example: 66 + (-66) = 0 Example: 66 × (1/66) = 1 Solved Examples Example 1: What is the additive inverse of 2/3? Solution: Given number is 2/3. We know that the additive inverse of a/b is -a/b. Hence, the additive inverse of 2/3 is -2/3. Example 2: What is the additive inverse of -5/9? Solution: Given number is -5/9. We know that the additive inverse of -a/b is a/b. Hence, the additive inverse of -5/9 is 5/9. Practice Problems 1. Find the additive inverse of 4/5. 2. What is the additive inverse of -7/9? 3. Write the additive inverse of 5 and -7. An additive inverse of a number is the value, which on adding with the original number results in zero value. No, both are different. Additive inverse is the added to get the result as zero.Additive identity is the value that is added to get the original number, which is zero. Examples: 3 + (-3) = 0 3 + 0 = 3 -17 17+(-17) = 0 What is the additive inverse of 23? -23 23 + (-23) = 23 -23 = 0 What is the additive inverse of ab? (-ab) ab + (-ab) = ab – ab = 0
How to solve Irrational Numbers How to solve Irrational Numbers The best way of understanding how can we solve irrational number is given below. Friends First we discuss about irrational number:- An irrational number is any number that is real but not rational and cannot be expressed as a simple fraction or non repeating decimal. Most of irrational number the set of all rational number and take more space in column or decimal. Some Example of irrational numbers are:- 2, 5,6,7, ?3.14). The square root of any prime number is irrational. Irrational number cannot be obtained by dividing one integer by another. So -1/3=0.333 is not a irrational because it is obtained by the ratio of two integer. 1 and 3. Irrational number cant have a finite decimal expression. Now, we discuss on some equation to solve the irrational number. We assuming that 3 is an rational number i.e 3=a/b equation Know More About Worksheet on Scientific Notation and Rational Number Tutorcircle.com Page No. : 1/4 (1) Where a and b are integers having no common factor (b`0) on squaring both side (3)2= (a/b)2 3= a2/b2 equation (2) 3b2=a2 equation (3) Where a and b are both odd number and a/b reduce to smallest possible terms. It is not possible that a and b are even because if a and b are even one can always be divided by 2 as we assume a/b is an rational numbers a=2m+1 Assuming a and b are odd b=2n+1 by putting the value of a and b in eq 3 : 3(2n+1)2= (2m+1)2 3(4n2+1+4n)= (2m2+1+4m) 12n2+3+12n=4m2+1+4m 12n2+12n+2=4m2+4m 6n2+6n+1=2m2+2m 6n2+6n+1=2(m2+n) In this equation all the value of m is always odd and the value of n is always n for all values s so this equation has no solution. Our assumption a and b are odd in invalid so we can say that root of 3 is an irrational number. In above articles we discuss about how to solve irrational numbers. What are irrational Numbers Irrational numbers are the numbers which are not rational numbers. In other words we can say that any number that cannot be expressed in the form of p/q are termed as irrational numbers. If any floating point number (that is a number that has an integer part as well as an decimal part is termed as floating point number.) cannot expressed as the ratio of two integers that floating point number is termed as irrational numbers. Let us take some of the examples of Irrational numbers Now if we take the value of “ pi (π ) “ that is π = 3.1415926535897932384626433832795 This value of π is Read More About Rational Numbers Properties Worksheets Tutorcircle.com Page No. : 2/4 impossible to express as the simple ratio of two numbers or two integers instead. Thus the value of π is an irrational number. Let us take some more examples to clearly get an image about the irrational numbers Let us take a value 3.2. Now 3.2 is not an irrational number, it is a rational number as 3.2 can be expressed as a ratio of two integers that is A square root of every non perfect square is an irrational number and similarly, a cube root of non-perfect cube is also an example of the irrational number. When we multiply any two irrational numbers and the result is rational number, then each of these irrational numbers is called rationalizing factor of the other one. Here is a general irrational number which is frequently used in mathematics. “Pi” is a best example of an irrational number. The value of 'pi' is solved to over one million decimal places and still there is no pattern found. Tutorcircle.com Page No. : 2/3 Page No. : 3/4 ThankÂ You TutorCircle.com How to solve Irrational Numbers Know More About Worksheet on Scientific Notation and Rational Number Now, we discuss on some equation to solve the irrational number. We ass... How to solve Irrational Numbers Know More About Worksheet on Scientific Notation and Rational Number Now, we discuss on some equation to solve the irrational number. We ass...
# E3 Questions on Gradually varied flow - including integration of the backwater curve. 1. E3.1 A Rectangular channel is 3.0m wide, has a 0.01 slope, discharge of $5.3m^{3}/s$, and n=0.011. Find $y_{n}$ and $y_{c}$. If the actual depth of flow is 1.7m, what type of profile exists? (Answer: $y_{n}$ = 0.4m, $y_{c}$ = 0.683m, S1 Curve) We need to solve Manning’s equation for $y_{n}$ for a rectangular channel. For a rectangular channel, we have: $A=by$ and $P=b+2y$ so $R=\frac{A}{P}=\frac{by}{b+2y}$ The Manning’s equation Eq-2 is thus: $Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{% 2/3}}So^{1/2}$ This equation must be solved for $y_{n}$ by a suitable numerical solution (e.g. trial and error or secant method). This gives for $Q=5.3m^{3}/s$, $b=3.0m$, $So=0.01$ and $n=0.011$: $\text{Normal depth}\quad y_{n}=0.412m$ At critical depth $y_{c}$, $Fr=1$ and also $Fr^{2}=1$, so we can use equation Eq-20 $Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}$ For a rectangular channel, this simplifies (using $A=By$) to $1=\frac{Q^{2}}{gB^{2}y_{c}^{3}}$ This can be formulated to be solved explicitly for $y_{c}$, eqn Eq-22: $y_{c}=\sqrt[3]{\frac{Q^{2}}{gB^{2}}}=\left(\frac{Q^{2}}{gB^{2}}\right)^{1/3}$ To give $\text{Critical depth}\quad y_{c}=0.683m$ As $y_{n} then the flow is super-critical. We can calculate the Froude number of normal depth (eqn Eq-16) to confirm this: $Fr=\frac{V}{\sqrt{gy_{n}}}=2.13$ As $Fr>1$ this confirms that uniform flow is super-critical. So we have a steep channel, and all surface profiles will be "S" curves. Figure 24 shows the relative positions of the critical and normal depth lines for a steep slope. If at a position in the channel, the depth is $1.7m$, as this is above the critical depth then the profile is in region 1. The surface profile will be an $S1$ profile. This takes the form as shown in figure 25 2. E3.2 A rectangular channel with a bottom width of $4.0m$ and a bottom slope of $0.0008$ has a discharge of $1.50m^{3}/s$. In a gradually varied flow in this channel, the depth at a certain location is found to be $0.30m$ assuming $n=0.016$, determining the type of GVF profile, the critical depth, and the normal depth. (Answer: M2, $y_{c}=0.24m$, $y_{n}=0.43m$) Following example E3.1, the Manning’s equation Eq-2 is: $Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{% 2/3}}So^{1/2}$ This equation must be solved for $y_{n}$ by a suitable numerical solution (e.g. trial and error or secant method). This gives for $Q=1.5m^{3}/s$, $b=4.0m$, $So=0.0008$ and $n=0.016$: $\text{Normal depth}\quad y_{n}=0.426m$ At critical depth $y_{c}$, $Fr=1$ and also $Fr^{2}=1$, so we can use equation Eq-14 equated to 1 $Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}$ For a rectangular channel, this simplifies (using $A=By$) to $1=\frac{Q^{2}}{gB^{2}y_{c}^{3}}$ This can be formulated to be solved explicitly for $y_{c}$ (eqn Eq-22): $y_{c}=\sqrt[3]{\frac{Q^{2}}{gB^{2}}}$ To give $\text{Critical depth}\quad y_{c}=0.243m$ As $y_{n}>y_{c}$ then the flow is sub-critical. We can calculate the Froude number of normal depth to confirm this: $Fr=\frac{V}{\sqrt{gy_{n}}}=0.43$ As $Fr<1$ this confirms that uniform flow is sub-critical. So we have a mild channel, and all surface profiles will be "M" curves. Figure 26 shows the relative positions of the critical and normal depth lines for a mild slope. If at a position in the channel, the depth is $0.3m$, as this is above the critical depth and below the normal depth the profile is in region 2. The surface profile will be an $M2$ profile. This takes the form as shown in figure 27 3. E3.3 The figure below shows a backwater curve in a long rectangular channel. Determine using a numerical integration method, the profile for the following high flow conditions: $Q=10m^{3}/s$, $b=3m$, $n=0.022$, and a bed slope of $0.001$. Take the depth just upstream of the dam as the control point equal to $5m$. At what distance is the water level not affected by the dam? perform your integration using a) 2-steps and b) 10-steps (Answer: $y_{n}=2.437m$. Using $y_{0}$ (Euler) 2-step $x=-3490m$, 10-step $x=-4724m$. Using $y_{1/2}$ 2-step $x=-4480m$, 10-step $x=-5906m$. Using a forth-order Runge Kutta method gave 2-step $x=-3893m$, 10-step $x=-5461m$) We will integrate using a numerical method that solves to obtain a distance from depth change. There are various methods. We will demonstrate 3 here. We must solve (integrate) this backwater, or gradually varied flow, equation Eq-42: $\frac{dy}{dx}=\frac{S_{o}-S_{f}}{\-Fr^{2}}$ Solving for distance, $x$, given a change in depth $y$. So in total differential form, we will solve: $\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)$ There are options now as to the way the right-hand side is approximated (or discretised). Some options are 1. i) As in equation i), the first-order Euler method (where the subscript ${}_{o}$ indicates at the known, or initial, point), i.e. $y_{0}$: $\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{o}$ 2. ii) Or equation ii), at the averaged depth $y_{1/2}=(y_{0}+y_{1})/2$ $\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{mean}=\Delta y% \left(\frac{1-Fr^{2}_{y{{}_{1/2}}}}{S_{o}-S_{f_{y_{1/2}}}}\right)=\Delta y% \left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{{1/2}}$ 3. iii) or, as in equation iii), the whole function averaged between the initial and subsequent point: $\displaystyle\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{mean}$ $\displaystyle=\Delta y\left.\left[\left(\frac{1-Fr_{0}^{2}}{S_{o}-S_{f_{y_{0}}% }}\right)+\left(\frac{1-Fr_{1}^{2}}{S_{o}-S_{f_{y_{1}}}}\right)\right]\middle/% 2\right.$ $\displaystyle=\Delta y\left.\left[\left(\frac{1-Fr}{S_{o}-S_{f}}\right)_{0}+% \left(\frac{1-Fr}{S_{o}-S_{f}}\right)_{1}\right]\middle/2\right.$ For all methods, we calculate $Fr^{2}$ and $S_{f}$ in the same way using the appropriate depth $y$ and other channel properties. The equations to use are, for the Froude number, in a rectangular channel (equation Eq-14): $Fr^{2}=\frac{Q^{2}b}{A^{3}g}=\frac{Q^{2}b}{(by)^{3}\;g}=\frac{Q^{2}}{b^{2}y^{3% }\;g}$ And remembering that $S_{f}$ is determined from the Manning’s equation (Eq-2), assuming that over the short distance $S_{f}=S_{o}$ then: $Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}AR^{2/3}S_{f}^{1/2}$ rearranging for $S_{f}$ and substituting appropriately for a rectangular channel we get from equation Eq-45: $S_{f}=\frac{Q^{2}n^{2}}{A^{2}R^{4/3}}=\frac{Q^{2}n^{2}}{(by)^{2}\left(\frac{by% }{b+2y}\right)^{4/3}}$ The question asks to find the distance to where the depth is not affected by the dam - this is asking, when does the depth reach normal depth?. So we must first calculate $y_{n}$. This is done via a straightforward iteration of the Manning’s equation for a rectangular channel (see previous examples). In this case using $Q=10m^{3}/s$, $b=3m$, $n=0.022$ and $S_{o}=0.001$ then the solution is, normal depth $y_{n}=2.437m$. The level just upstream of the dam is $5m$, so we need to integrate to find $x$ when $y$ has dropped to $y_{n}=2.437$. That is a drop, (so negative), of $dy=2.437-5.0=-2.563$ We are asked to calculate the distance using a 2-step integration which would give $\Delta y=-2.563/2=-1.2815m$ and also for a 10-step integration which would give $\Delta y=-2.563/10=-0.2563m$ We can now take each of the three discretisation methods and construct an iteration table (e.g. in Excel) for each to calculate the two solutions using the two $\Delta y$ values. Integration method E3.3i) - First-order Euler Figure 29 shows the integration table for the 2-step integration and a distance of $3490m$ upstream of the dam to normal depth. Figure 30 shows the integration table for the 10-step integration and a distance of $4724m$ upstream of the dam to normal depth. Integration method E3.3ii) - Average $y$ Figure 31 shows the integration table for the 2-step integration and a distance of $4480m$ upstream of the dam to normal depth. Figure 32 shows the integration table for the 10-step integration and a distance of $5906m$ upstream of the dam to normal depth. Integration method E3.3iii) - Average $(1-Fr^{2})/(S_{o}-S_{f})$ Figure 33 shows the integration table for the 2-step integration and a distance of $1712420m$ upstream of the dam to normal depth. This seems incorrect (unlikely), however, let’s have a look a the 10-step solution. Figure 34 shows the integration table for the 10-step integration and a distance of $346510m$ upstream of the dam to normal depth. This again seems incorrect. The reason why this method does not give expected answers is that it fails when depth is equal to $y_{n}$. The reason is that $(S_{o}-S_{f})$ would is zero at $y_{n}$. In the numerical calculation, it is close to zero, $\approx 10^{-7}$, and gives a very long last $\Delta x$. We can say that this method should not be used when the depth of $y_{n}$ is to be encountered. A hack that could allow us to use this method would be to perform an Euler step for the last integration step (i.e. avoid the use of $y_{n}$). [We might call this a hybrid method.]. This would give, for the 2-step $x=-2718m$ and for the 10-step $x=-4570m$. A fourth-order Runge-Kutta method is usually considered as accurate as practically possible for a numerical method. Using this on the above problem gave for the 2-step $x=-3893m$, and for the 10-step $x=-5461m$. In summary, the solutions are shown in the table 1 below 4. E3.4 A trapezoidal, concrete-lined, channel has a constant bed slope of $0.0015$, a bed width of $3m$, and side slopes of $1:1$. A control gate increased the depth immediately upstream to $4m$. When the discharge is $19m^{3}/s$ compute the water surface profile upstream and identify the distance when the water depth is $1.8m$. ($n=0.017$) (Answer: Using $y_{0}$ (Euler) 2-step $x=-1540m$, 10-step $x=-1695m$. Using $y_{1/2}$ 2-step $x=-1670m$, 10-step $x=-1809m$. Using mean GVF function: 2-step $x=-2729m$, 10-step $x=-1933m$. Using a forth-order Runge Kutta method gave 2-step $x=-2023m$, 10-step $x=-1850m$) This solution follows similarly to that of example question E3.3. We have been given an initial depth of $y_{0}=4m$ and asked to find the distance upstream to where the depth is $y=1.8$. We should first determine the normal and critical depths. Although these are not directly used in the question, they do inform what solutions is expected and whether the depths rise or fall. The calculations must be done iteratively, as seen in previous examples. Undertaking the iterative calculation we get $y_{n}=1.725m$ and $y_{c}=1.364m$. So flow is sub-critical, and the slope is mild. The depths given for the calculation of the surface are above the normal depth so the surface profile will be a $M1$ profile. It will slope downwards from 4m toward the target of 1.8m. The depth fall is a drop, (so negative), of $dy=4.0-1.8=-2.2m$ We are asked to calculate the distance using a 2-step integration which would give $\Delta y=-2.2/2=-1.1m$ And also for a 10-step integration which would give $\Delta y=-2.1/10=-0.22m$ Here we demonstrate results from the three integration methods identified in example E3.3 above. Integration method E3.3i) - First-order Euler Figure 35 shows the integration table for the 2-step integration and a distance of $1540m$ upstream of the dam to the target depth. Figure 36 shows the integration table for the 10-step integration and a distance of $1696m$ upstream of the dam to the target depth. Integration method E3.3ii) - Average $y$ Figure 37 shows the integration table for the 2-step integration and a distance of $1670m$ upstream of the dam to the target depth. Figure 38 shows the integration table for the 10-step integration and a distance of $1809m$ upstream of the dam to the target depth. Integration method E3.3iii) - Average $(1-Fr^{2})/(S_{o}-S_{f})$ Figure 39 shows the integration table for the 2-step integration and a distance of $2279m$ upstream of the dam to the target depth. This seems incorrect (unlikely), however, let’s have a look a the 10-step solution. Figure 40 shows the integration table for the 10-step integration and a distance of $1934m$ upstream of the dam to the target depth. This again seems incorrect. A fourth-order Runge-Kutta method is usually considered as accurate as practically possible for a numerical method. Using this on the above problem gave for the 2-step $x=-2020m$, and for the 10-step $x=-1850m$. A 100-step Runge Kutta give $x=-1844m$, so the 10-step can probably be considered sufficiently accurate. In summary, the solutions are shown in the table 1 below 5. E3.5 Using the figure below, determine the profile for the channel conditions using a step length of $\Delta x=100m$. $Q=600m^{3}/s$, $n=0.04$, the bed slope of the rectangular channel is $S_{o}=0.002$ and has a width of $B=50m$. The sill height of the weir is $2.5m$ and the water depth over the weir is $4m$. Compare results from each method. The question is asking to integrate upstream from a depth of $6.5m(=4.0+2.5)$ at the weir enough distance to understand and compare the flow profile for each method. It suggests an integration step of $\Delta x=100m$ which should be used for the depth from distance integration. It does not specify that this should be the method so the method integration distance from depth will also be used and the results compared. We should first understand the flow regime by calculation normal and critical depths and by comparison examine the expected flow profile. We have a rectangular channel and Manning’s n, so will use the following uniform flow equation: $Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{% 2/3}}So^{1/2}$ This equation must be solved for $y_{n}$ by a suitable numerical solution (e.g. trial and error or secant method). This gives for $Q=600.0m^{3}/s$, $b=50.0m$, $So=0.002$ and $n=0.04$: $\text{Normal depth}\quad y_{n}=4.434m$ At critical depth $y_{c}$, $Fr=1$ and also $Fr^{2}=1$, so we can use the equation Eq-19, written for a rectangular channel $Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}=\frac{Q^{2}}{gB^{2}y_{c}^{3}}$ And formulated to solve for critical depth $y_{c}$, giving from equation Eq-22 $y_{c}=\sqrt[3]{\frac{Q^{2}}{gB^{2}}}=2.448m$ As $y_{n}>y_{c}$ then uniform flow is sub-critical. The slope of the channel is mild. The depth at the weir is higher that normal depth and moving upstream must fall gradually toward normal depth. This is the region 1 (above normal fro a mild slope), so the profile must be an M1 profile. The first integration will be the following depth from distance, using $\Delta x=100m$ and we will use am Euler method i.e. use the conditions at the known point to calculate the derivative. Equation Eq-53 $\Delta y=\Delta x\left(\frac{S_{o}-S_{f}}{1-Fr^{2}}\right)_{i}$ Where the friction slope, $S_{f}$, term is, as previously, equation Eq-45: $S_{f}=\frac{n^{2}Q^{2}}{R^{4/3}A^{2}}$ And the Froude number squared, equation Eq-14 $Fr^{2}=\frac{Q^{2}B}{A^{3}g}$ See previous questions for a greater description of the Euler integration methodology. Tabulating the calculation and produces the integration table in figure 42. The integration has been performed for 20 steps and depth has fallen to $4.64m$. Which indicates that normal depth has still not been achieved $2km$ upstream of the weir. As we wish to compare results from a distance from depth integration we should select appropriate parameters. The integration above used 20 steps and reached as depth of $y=4.636m$. Using these as parameters would give a $\Delta y=(6.5-4.636)/20=0.0932m$ and 20 steps of integration. Results in an integration table using a derivative approximated at $y_{1/2}$ is shown in figure 43. We see from the table that there is a slight difference in the distance to that same depth of $75m$. To investigate any further differences by plotting a graph of the two profiles. Such a graph is shown in figure 44. It is clear that the two integrations are very similar - differing by only millimeters along the whole profile length. The shape of the profile is as expected that of an M1 profile.
# Lesson Planning of Multiplication of Fraction Students` Learning Outcomes • Verify the Commutative property of multiplication of fractions. • Verify the Associative property of multiplication of fractions. ### Information for Teachers • There are properties involving multiplication that will help make problems easier to solve. • Commutative Property: Change the place in multiplication of two fraction does n`t change the product. • Associative Property: When three or more fractions are multiplied, the product is the same regardless of the grouping of the factors. • While teaching the lesson, the teacher should consult with textbook where and when it is required. #### Material / Resources Writing board, chalk, marker, duster, chart paper, textbook ##### Introduction • Write following statements on the board and recall the properties of commutative and associative for addition, as; 4 + 2 = 2 + 4, (5 + 3) + 2 = 5 + (2 + 3) • Which properties are being used? (Expected answer would be as; Commutative and Associative Properties) • B + a = a + b, (a + b) + c = a + (b + c) • Now tell that these properties are also verified for multiplication. • 4 x 2 = 2 x 4 (Expected answer would be as; Commutative Properties) • (5 x 3) x 2 = 5 x (2 x 3) (Expected answer would be as; Associative Properties) • Announce that in this lesson we will verify these properties for the multiplication of fractions. ##### Development Activity 1 Individual & Pair Group Task: Ask students to note down the following on their notebook and find the answer Ask the students to pair up now. Rearrange the fractions of each question, as; 3/5 x 7/7 will turn as 4/7 x 3/5 Do the same for all the five Apply multiplication What do you find about them? They will conclude that :it makes no difference on the answer if we change the position or the order of the fractions” Hence like addition, multiplication is also commutative. Activity 2 Group Task: Divide the students in 3 groups. Call one student from each group. Ask each of them to write one simple fraction on the board and they decide who will write first, then second and then third, and so on. Ask them to multiply these fractions as they are written. Now call other three students and ask them to swap the places of the fractions and then multiply. Is there any change in the answer (Expected answer would be as; no) Repeat this activity 3 to 4 times and students will deduce that changing the order of multiplication does n`t change the answer, hence multiplication of fraction is Associative. Refer students to do textbook exercise questions for further practice. Keep checking students work and help them if they make any mistake. ##### Sum up / Conclusion • Recap the main points of the lesson and conclude by giving one example of each that multiplication of fractions is commutative as well as associative. ##### Assessment • Write following questions of fractions on the board and ask students to work in pairs, and groups. • Select any 2 questions and verify both properties. • Each student in pair or group must select different questions. • Then swap their copies for peer checking.
# 6.2 Graphs of exponential functions (Page 6/6) Page 6 / 6 ## Verbal What role does the horizontal asymptote of an exponential function play in telling us about the end behavior of the graph? An asymptote is a line that the graph of a function approaches, as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small. What is the advantage of knowing how to recognize transformations of the graph of a parent function algebraically? ## Algebraic The graph of $\text{\hspace{0.17em}}f\left(x\right)={3}^{x}\text{\hspace{0.17em}}$ is reflected about the y -axis and stretched vertically by a factor of $\text{\hspace{0.17em}}4.\text{\hspace{0.17em}}$ What is the equation of the new function, $\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$ State its y -intercept, domain, and range. $g\left(x\right)=4{\left(3\right)}^{-x};\text{\hspace{0.17em}}$ y -intercept: $\text{\hspace{0.17em}}\left(0,4\right);\text{\hspace{0.17em}}$ Domain: all real numbers; Range: all real numbers greater than $\text{\hspace{0.17em}}0.$ The graph of $\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{2}\right)}^{-x}\text{\hspace{0.17em}}$ is reflected about the y -axis and compressed vertically by a factor of $\text{\hspace{0.17em}}\frac{1}{5}.\text{\hspace{0.17em}}$ What is the equation of the new function, $\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$ State its y -intercept, domain, and range. The graph of $\text{\hspace{0.17em}}f\left(x\right)={10}^{x}\text{\hspace{0.17em}}$ is reflected about the x -axis and shifted upward $\text{\hspace{0.17em}}7\text{\hspace{0.17em}}$ units. What is the equation of the new function, $\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$ State its y -intercept, domain, and range. $g\left(x\right)=-{10}^{x}+7;\text{\hspace{0.17em}}$ y -intercept: $\text{\hspace{0.17em}}\left(0,6\right);\text{\hspace{0.17em}}$ Domain: all real numbers; Range: all real numbers less than $\text{\hspace{0.17em}}7.$ The graph of $\text{\hspace{0.17em}}f\left(x\right)={\left(1.68\right)}^{x}\text{\hspace{0.17em}}$ is shifted right $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units, stretched vertically by a factor of $\text{\hspace{0.17em}}2,$ reflected about the x -axis, and then shifted downward $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units. What is the equation of the new function, $\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$ State its y -intercept (to the nearest thousandth), domain, and range. The graph of $\text{\hspace{0.17em}}f\left(x\right)=2{\left(\frac{1}{4}\right)}^{x-20}$ is shifted left $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ units, stretched vertically by a factor of $\text{\hspace{0.17em}}4,$ reflected about the x -axis, and then shifted downward $\text{\hspace{0.17em}}4\text{\hspace{0.17em}}$ units. What is the equation of the new function, $\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$ State its y -intercept, domain, and range. $g\left(x\right)=2{\left(\frac{1}{4}\right)}^{x};\text{\hspace{0.17em}}$ y -intercept: Domain: all real numbers; Range: all real numbers greater than $\text{\hspace{0.17em}}0.$ ## Graphical For the following exercises, graph the function and its reflection about the y -axis on the same axes, and give the y -intercept. $f\left(x\right)=3{\left(\frac{1}{2}\right)}^{x}$ $g\left(x\right)=-2{\left(0.25\right)}^{x}$ y -intercept: $\text{\hspace{0.17em}}\left(0,-2\right)$ $h\left(x\right)=6{\left(1.75\right)}^{-x}$ For the following exercises, graph each set of functions on the same axes. $f\left(x\right)=3{\left(\frac{1}{4}\right)}^{x},$ $g\left(x\right)=3{\left(2\right)}^{x},$ and $\text{\hspace{0.17em}}h\left(x\right)=3{\left(4\right)}^{x}$ $f\left(x\right)=\frac{1}{4}{\left(3\right)}^{x},$ $g\left(x\right)=2{\left(3\right)}^{x},$ and $\text{\hspace{0.17em}}h\left(x\right)=4{\left(3\right)}^{x}$ For the following exercises, match each function with one of the graphs in [link] . $f\left(x\right)=2{\left(0.69\right)}^{x}$ B $f\left(x\right)=2{\left(1.28\right)}^{x}$ $f\left(x\right)=2{\left(0.81\right)}^{x}$ A $f\left(x\right)=4{\left(1.28\right)}^{x}$ $f\left(x\right)=2{\left(1.59\right)}^{x}$ E $f\left(x\right)=4{\left(0.69\right)}^{x}$ For the following exercises, use the graphs shown in [link] . All have the form $\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x}.$ Which graph has the largest value for $\text{\hspace{0.17em}}b?$ D Which graph has the smallest value for $\text{\hspace{0.17em}}b?$ Which graph has the largest value for $\text{\hspace{0.17em}}a?$ C Which graph has the smallest value for $\text{\hspace{0.17em}}a?$ For the following exercises, graph the function and its reflection about the x -axis on the same axes. $f\left(x\right)=\frac{1}{2}{\left(4\right)}^{x}$ $f\left(x\right)=3{\left(0.75\right)}^{x}-1$ $f\left(x\right)=-4{\left(2\right)}^{x}+2$ For the following exercises, graph the transformation of $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.\text{\hspace{0.17em}}$ Give the horizontal asymptote, the domain, and the range. $f\left(x\right)={2}^{-x}$ $h\left(x\right)={2}^{x}+3$ Horizontal asymptote: $\text{\hspace{0.17em}}h\left(x\right)=3;$ Domain: all real numbers; Range: all real numbers strictly greater than $\text{\hspace{0.17em}}3.$ $f\left(x\right)={2}^{x-2}$ For the following exercises, describe the end behavior of the graphs of the functions. $f\left(x\right)=-5{\left(4\right)}^{x}-1$ As $x\to \infty$ , $f\left(x\right)\to -\infty$ ; As $x\to -\infty$ , $f\left(x\right)\to -1$ $f\left(x\right)=3{\left(\frac{1}{2}\right)}^{x}-2$ $f\left(x\right)=3{\left(4\right)}^{-x}+2$ As $x\to \infty$ , $f\left(x\right)\to 2$ ; As $x\to -\infty$ , $f\left(x\right)\to \infty$ For the following exercises, start with the graph of $\text{\hspace{0.17em}}f\left(x\right)={4}^{x}.\text{\hspace{0.17em}}$ Then write a function that results from the given transformation. Shift $f\left(x\right)$ 4 units upward Shift $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ 3 units downward $f\left(x\right)={4}^{x}-3$ Shift $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ 2 units left Shift $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ 5 units right $f\left(x\right)={4}^{x-5}$ Reflect $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ about the x -axis Reflect $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ about the y -axis $f\left(x\right)={4}^{-x}$ For the following exercises, each graph is a transformation of $\text{\hspace{0.17em}}y={2}^{x}.\text{\hspace{0.17em}}$ Write an equation describing the transformation. $y=-{2}^{x}+3$ For the following exercises, find an exponential equation for the graph. $y=-2{\left(3\right)}^{x}+7$ ## Numeric For the following exercises, evaluate the exponential functions for the indicated value of $\text{\hspace{0.17em}}x.$ $g\left(x\right)=\frac{1}{3}{\left(7\right)}^{x-2}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}g\left(6\right).$ $g\left(6\right)=800+\frac{1}{3}\approx 800.3333$ $f\left(x\right)=4{\left(2\right)}^{x-1}-2\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}f\left(5\right).$ $h\left(x\right)=-\frac{1}{2}{\left(\frac{1}{2}\right)}^{x}+6\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}h\left(-7\right).$ $h\left(-7\right)=-58$ ## Technology For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth. $\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x}+d.$ $-50=-{\left(\frac{1}{2}\right)}^{-x}$ $116=\frac{1}{4}{\left(\frac{1}{8}\right)}^{x}$ $x\approx -2.953$ $12=2{\left(3\right)}^{x}+1$ $5=3{\left(\frac{1}{2}\right)}^{x-1}-2$ $x\approx -0.222$ $-30=-4{\left(2\right)}^{x+2}+2$ ## Extensions Explore and discuss the graphs of $\text{\hspace{0.17em}}F\left(x\right)={\left(b\right)}^{x}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}G\left(x\right)={\left(\frac{1}{b}\right)}^{x}.\text{\hspace{0.17em}}$ Then make a conjecture about the relationship between the graphs of the functions $\text{\hspace{0.17em}}{b}^{x}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\left(\frac{1}{b}\right)}^{x}\text{\hspace{0.17em}}$ for any real number $\text{\hspace{0.17em}}b>0.$ The graph of $\text{\hspace{0.17em}}G\left(x\right)={\left(\frac{1}{b}\right)}^{x}\text{\hspace{0.17em}}$ is the refelction about the y -axis of the graph of $\text{\hspace{0.17em}}F\left(x\right)={b}^{x};\text{\hspace{0.17em}}$ For any real number $\text{\hspace{0.17em}}b>0\text{\hspace{0.17em}}$ and function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ the graph of $\text{\hspace{0.17em}}{\left(\frac{1}{b}\right)}^{x}\text{\hspace{0.17em}}$ is the the reflection about the y -axis, $\text{\hspace{0.17em}}F\left(-x\right).$ Prove the conjecture made in the previous exercise. Explore and discuss the graphs of $\text{\hspace{0.17em}}f\left(x\right)={4}^{x},$ $\text{\hspace{0.17em}}g\left(x\right)={4}^{x-2},$ and $\text{\hspace{0.17em}}h\left(x\right)=\left(\frac{1}{16}\right){4}^{x}.\text{\hspace{0.17em}}$ Then make a conjecture about the relationship between the graphs of the functions $\text{\hspace{0.17em}}{b}^{x}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(\frac{1}{{b}^{n}}\right){b}^{x}\text{\hspace{0.17em}}$ for any real number n and real number $\text{\hspace{0.17em}}b>0.$ The graphs of $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ are the same and are a horizontal shift to the right of the graph of $\text{\hspace{0.17em}}f\left(x\right);\text{\hspace{0.17em}}$ For any real number n , real number $\text{\hspace{0.17em}}b>0,$ and function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ the graph of $\text{\hspace{0.17em}}\left(\frac{1}{{b}^{n}}\right){b}^{x}\text{\hspace{0.17em}}$ is the horizontal shift $\text{\hspace{0.17em}}f\left(x-n\right).$ Prove the conjecture made in the previous exercise. factoring polynomial find general solution of the Tanx=-1/root3,secx=2/root3 find general solution of the following equation Nani the value of 2 sin square 60 Cos 60 0.75 Lynne 0.75 Inkoom when can I use sin, cos tan in a giving question depending on the question Nicholas I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks. John I want to learn the calculations where can I get indices I need matrices Nasasira hi Raihany Hi Solomon need help Raihany maybe provide us videos Nasasira Raihany Hello Cromwell a Amie What do you mean by a Cromwell nothing. I accidentally press it Amie you guys know any app with matrices? Khay Ok Cromwell Solve the x? x=18+(24-3)=72 x-39=72 x=111 Suraj Solve the formula for the indicated variable P=b+4a+2c, for b Need help with this question please b=-4ac-2c+P Denisse b=p-4a-2c Suddhen b= p - 4a - 2c Snr p=2(2a+C)+b Suraj b=p-2(2a+c) Tapiwa P=4a+b+2C COLEMAN b=P-4a-2c COLEMAN like Deadra, show me the step by step order of operation to alive for b John A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes. The sequence is {1,-1,1-1.....} has how can we solve this problem Sin(A+B) = sinBcosA+cosBsinA Prove it Eseka Eseka hi Joel yah immy June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler? 7.5 and 37.5 Nando how would this look as an equation? Hayden 5x+x=45 Khay
# Texas Go Math Grade 3 Lesson 14.2 Answer Key Subtraction Strategies Refer to ourÂ Texas Go Math Grade 3 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 3 Lesson 14.2 Answer Key Subtraction Strategies. ## Texas Go Math Grade 3 Lesson 14.2 Answer Key Subtraction Strategies Unlock the Problem A sunflower can grow to be very tall. Dylan is 39 inches tall. She watered a sunflower that grew to be 62 inches tall. How many inches shorter was Dylan than the sunflower? One Way Use a number line to find 62 – 39. A. Count up by tens and then ones. Add the lengths of the jumps to find the difference. 10 + 10 + 3 = ____________ Explanation: 10 + 10 + 3 = 23 B. Take away tens and ones. Take away lengths of jumps to end on the difference. Complete the equation. 62 – 39 = ___________ So, Dylan was _________ inches shorter than the sunflower. Complete the equation. 62 – 39 = 23 So, Dylan was 23 inches shorter than the sunflower. Try This! Find 74 – 38. Draw jumps and label the number line to show your thinking. Then complete the equation. So, 74 – 38 = _________. Explanation: The jumps and label are described on number line and completed the equation. 74 – 38 = 36 Another Way Use strip diagrams to solve a subtraction problem. At Mr. Cruzâ€™s vegetable farm, there are 348 tomato plants. There are 136 fewer pepper plants than tomato plants. There are 92 fewer carrot plants than pepper plants. How many carrot plants are at the farm? STEP 1: Complete the strip diagram to find the number of pepper plants. Complete the equations. STEP 2: Complete the strip diagram to find the number of carrot plants. Complete the equations. So, there are _________ carrot plants at the farm. STEP 1: Complete the strip diagram to find the number of pepper plants. Complete the equations. STEP 2: Complete the strip diagram to find the number of carrot plants. Complete the equations. So, there are 120 carrot plants at the farm. Share and Show Yumi has 61 plant stickers. She has 24 fewer flower stickers than plant stickers. How many stickers does Yumi have in all? Use the number lines to represent and solve. Then complete the equations. So, Yumi has _________ stickers in all. So, Yumi has 98 stickers in all. Complete the strip diagram to represent and solve. Then complete the equation for the strip diagram. Question 2. On Monday, a large flower shop sold 425 roses. The flower shop sold Monday 123 fewer roses on Friday than on Monday. How many roses did the Friday shop sell on Friday? Explanation: Problem Solving Use the table for 3-6. You may use models, number lines, and equations to represent and solve. Question 3. Analyze How many more people attended the museum on Saturday than on Thursday? Explanation: 130 more people attended the museum on Saturday than on Thursday, Since 585 – 450 =130 Question 4. Apply How many fewer people attended the museum on Monday than on Friday? Explanation: 220 fewer people attended the museum on Monday than on Friday, Since 526 – 306 =220 Question 5. H.O.T. Multi-Step How many fewer people attended the museum on Tuesday and Wednesday combined than on Thursday and Friday combined? Explanation: 216 fewer people attended the museum on Tuesday and Wednesday combined than on Thursday and Friday combined, Since 450 + 526 = 976 415 +345 = 760 976 – 760 = 216 H.O.T. Pose a Problem Write a subtraction word problem about the data in the table. Then solve your problem. Answer: How many more people attended the museum on Thursday than on Wednesday Explanation: 105 more people attended the museum on Thursday than on Wednesday, Since 450-345 =105 Fill in the bubble for the correct answer choice. You can use models, number lines, and equations to represent and solve. Question 7. Multi-Step Jan and Dave are having a penny race. Dave will win the race if he finds at least 40 more pennies than Jan. Dave finds 213 pennies. Jan finds 172 pennies. Which is a true statement about Jan and Daveâ€™s penny race? (A) Jan wins because she finds 41 more pennies than Dave. (B) Jan wins because Dave finds 31 more pennies than she does. (C) Dave wins because he finds more pennies than Jan. (D) Dave wins because he finds 41 more pennies than Jan. Answer: (D) Dave wins because he finds 41 more pennies than Jan. Explanation: Dave finds 213 pennies. Jan finds 172 pennies. So Dave finds 41 more pennies than Jan. So Dave won the race since he finds at least 40 more pennies than Jan. 213-172 = 41. So Dave wins because he finds 41 more pennies than Jan. Question 8. Kyle’s basketball team scores 62 points. Kyle’s team members score 51 of the points. Kyle scores the rest of the points. How many points does Kyle score? (A) 21 (B) 13 (C) 11 (D) 23 Explanation: Kyle’s basketball team scores 62 points. Kyle’s team members score 51 of the points. Kyle scores the rest of the points. Then 11 points Kyle scored Since 62 -51 = 11. Question 9. Multi-Step Suzi has 86 songs on her digital music player. Xavier has 112 songs on his digital music player. Then he deletes 18 songs. How many more songs does Xavier have now than Suzi? (A) 8 (B) 94 (C) 26 (D) 18 Explanation: Suzi has 86 songs on her digital music player. Xavier has 112 songs on his digital music player. Then he deletes 18 songs. Then 112 – 18 = 94 Songs in Xavier’s music player. Then 8 more songs does Xavier have now than Suzi. Since 94 -86= 8. Texas Test Prep 3rd Grade Go Math Subtraction Strip DiagramÂ Question 10. There were 87 sunflowers at the flower shop in the morning. There were 56 sunflowers left at the end of the day. How many sunflowers were sold? (A) 143 (B) 31 (C) 13 (D) 41 Explanation: There were 87 sunflowers at the flower shop in the morning. There were 56 sunflowers left at the end of the day. 31Â sunflowers were sold. Since 87 – 56 =31 ### Texas Go Math Grade 3 Lesson 14.2 Homework and Practice Answer Key Complete the strip diagram to represent and solve. Then complete the equation for the strip diagram. Question 1. There were 285 visitors at the art show on Sunday. There were 173 visitors at the art show on Friday. How many more visitors were at the art show on Sunday than on Friday? Explanation: There were 285 visitors at the art show on Sunday. There were 173 visitors at the art show on Friday. 173 more visitors were at the art show on Sunday than on Friday. Since 285 -175 = 112 Problem Solving The table shows how many of each kind of fruit was sold at the Farmers’ Market in one week. Use the table for 2-4. You may use models, number lines, and equations to represent and solve. Question 2. How many more oranges than melons were sold? Explanation: 223 more oranges than melons were sold, Since 435 – 212 = 223 How many fewer peaches than grapefruits and melons combined were sold at the market? Explanation: 341 fewer peaches than grapefruits and melons combined were sold at the market. Since 347+212 = 559 559-218= 341 Question 4. Write a subtraction word problem about the data in the table. Then solve your problem. Answer: How many more peaches than melons were sold? Explanation: 6 more peaches than melons were sold Since 218-212 =6 Lesson Check Fill in the bubble completely to show your answer. You may use models, number lines, and equations to represent and solve. Question 5. Jenna scores 87 points bowling. Karl scores 35 fewer points than Jenna. How many points does Karl score? (A) 122 (B) 112 (C) 52 (D) 42 Explanation Jenna scores 87 points bowling. Karl scores 35 fewer points than Jenna. then, 87-35=52 points Karl scored Question 6. A toy store has 78 action figures. At the end of the day, 52 action figures are left. How many action figures were sold? (A) 130 (B) 25 (C) 26 (D) 120 Explanation: A toy store has 78 action figures. At the end of the day, 52 action figures are left. Then, 78 – 52 = 26 action figures were sold. Question 7. There are 278 people in a movie theater. Of those people, 145 are adults, and the rest are children. How many children are in the movie theater? (A) 133 (B) 122 (C) 100 (D) 103 Explanation: There are 278 people in a movie theater. Of those people, 145 are adults, and the rest are children. Then 278 – 145 = 133children are in the movie theater There are 295 people at a softball game. Of those people, 153 are adults, and the rest are children. How many children are at the softball game? (A) 300 (B) 124 (C) 142 (D) 150 Explanation: There are 295 people at a softball game. Of those people, 153 are adults, and the rest are children. Then 295- 153 = 142 children are at the softball game Question 9. (A) 58 (B) 39 (C) 27 (D) 29 Explanation: If the market sells a total of 29 baskets, then 58 – 29 = 29 baskets are left. Question 10. Multi-Step There are 27 girls and 19 boys at the library. Fifteen students leave to go to class. How many students are still at the library? (A) 46 (B) 61 (C) 34 (D) 31
We think you are located in United States. Is this correct? 3.2 Describing sequences 3.2 Describing sequences (EMAY) A sequence is an ordered list of items, usually numbers. Each item which makes up a sequence is called a “term”. Sequences can have interesting patterns. Here we examine some types of patterns and how they are formed. Examples: 1. $$1; 4; 7; 10; 13; 16; 19; 22; 25; \ldots$$ There is difference of $$\text{3}$$ between successive terms. The pattern is continued by adding $$\text{3}$$ to the previous term. 2. $$13; 8; 3; -2; -7; -12; -17; -22; \ldots$$ There is a difference of $$-\text{5}$$ between successive terms. The pattern is continued by adding $$-\text{5}$$ to (i.e. subtracting $$\text{5}$$ from) the previous term. 3. $$2; 4; 8; 16; 32; 64; 128; 256; \ldots$$ This sequence has a factor of $$\text{2}$$ between successive terms. The pattern is continued by multiplying the previous term by 2. 4. $$3; -9; 27; -81; 243; -729; 2187; \ldots$$ This sequence has a factor of $$-\text{3}$$ between successive terms. The pattern is continued by multiplying the previous term by $$-\text{3}$$. 5. $$9; 3; 1; \frac{1}{3}; \frac{1}{9}; \frac{1}{27}; \ldots$$ This sequence has a factor of $$\frac{1}{3}$$ between successive terms. The pattern is continued by multiplying the previous term by $$\frac{1}{3}$$ which is equivalent to dividing the previous term by 3. Some learners may see example 3 as $$2^{1}; 2^{2}; 2^{3}; \ldots$$ and see a pattern with the powers. You may choose to discuss this in class as a precursor to geometric series which will be introduced in Grade 12. temp text Worked example 1: Study table You and $$\text{3}$$ friends decide to study for Maths and are sitting together at a square table. A few minutes later, $$\text{2}$$ other friends arrive and would like to sit at your table. You move another table next to yours so that $$\text{6}$$ people can sit at the table. Another $$\text{2}$$ friends also want to join your group, so you take a third table and add it to the existing tables. Now $$\text{8}$$ people can sit together. Examine how the number of people sitting is related to the number of tables. Is there a pattern? Make a table to see if a pattern forms Number of tables, n Number of people seated $$\text{1}$$ $$4=4$$ $$\text{2}$$ $$4+2=6$$ $$\text{3}$$ $$4+2+2=8$$ $$\text{4}$$ $$4+2+2+2=10$$ $$\vdots$$ $$\vdots$$ n $$4+2+2+2+\cdots +2$$ Describe the pattern We can see that for $$\text{3}$$ tables we can seat $$\text{8}$$ people, for $$\text{4}$$ tables we can seat $$\text{10}$$ people and so on. We started out with $$\text{4}$$ people and added two each time. So for each table added, the number of people increased by $$\text{2}$$. So the pattern formed is $$4; 6; 8; 10; \ldots$$. temp text To describe terms in a number pattern we use the following notation: The first term of a sequence is $${T}_{1}$$. The fourth term of a sequence is $${T}_{4}$$. The tenth term of a sequence is $${T}_{10}$$. The general term is often expressed as the $$n^{\text{th}}$$ term and is written as $${T}_{n}$$. A sequence does not have to follow a pattern but, when it does, we can write down the general formula to calculate any term. For example, consider the following linear sequence: $$1; 3; 5; 7; 9; \ldots$$ The $$n^{\text{th}}$$ term is given by the general formula: $${T}_{n}=2n-1$$ You can check this by substituting values into the formula: \begin{align} {T}_{1}& = 2\left(1\right)-1=1 \\ {T}_{2}& = 2\left(2\right)-1=3 \\ {T}_{3}& = 2\left(3\right)-1=5 \\ {T}_{4}& = 2\left(4\right)-1=7 \\ {T}_{5}& = 2\left(5\right)-1=9 \end{align} If we find the relationship between the position of a term and its value, we can find a general formula which matches the pattern and find any term in the sequence. Common difference (EMAZ) Consider the following sequence: $6; 1; -4; -9; ...$ We can see that each term is decreasing by 5 but how would we determine the general formula for the $$n^{\text{th}}$$ term? Let us try to do this with a table. Term number $$T_{1}$$ $$T_{2}$$ $$T_{3}$$ $$T_{4}$$ $${T}_{n}$$ Term $$\text{6}$$ $$\text{1}$$ $$-\text{4}$$ $$-\text{9}$$ $${T}_{n}$$ Formula $$6 - 0 \times 5$$ $$6 - 1 \times 5$$ $$6-2 \times 5$$ $$6-3 \times 5$$ $$6 - (n-1) \times 5$$ You can see that the difference between the successive terms is always the coefficient of $$n$$ in the formula. This is called a common difference. Therefore, for sequences with a common difference, the general formula will always be of the form: $$T_{n}=dn+c$$ where $$d$$ is the difference between each term and $$c$$ is some constant. Sequences with a common difference are called linear sequences. Common difference The common difference is the difference between any term and the term before it. The common difference is denoted by $$d$$. For example, consider the sequence $$10; 7; 4; 1; \ldots$$ To calculate the common difference, we find the difference between any term and the previous term. Let us find the common difference between the first two terms. \begin{align} d& = {T}_{2}-{T}_{1} \\ & = 7-10 \\ & = -3 \end{align} Let us check another two terms: \begin{align} d& = {T}_{4}-{T}_{3} \\ & = 1-4 \\ & = -3 \end{align} We see that $$d$$ is constant. In general, $$d=T_{n}-T_{n-1}$$ $$d \ne T_{n-1}-T_{n}$$ for example, $$d={T}_{2}-{T}_{1}$$, not $${T}_{1}-{T}_{2}$$. Worked example 2: Study table, continued As before, you and $$\text{3}$$ friends are studying for Maths and are sitting together at a square table. A few minutes later $$\text{2}$$ other friends arrive so you move another table next to yours. Now $$\text{6}$$ people can sit at the table. Another $$\text{2}$$ friends also join your group, so you take a third table and add it to the existing tables. Now $$\text{8}$$ people can sit together as shown below. 1. Find an expression for the number of people seated at $$n$$ tables. 2. Use the general formula to determine how many people can sit around $$\text{12}$$ tables. 3. How many tables are needed to seat $$\text{20}$$ people? Make a table to see the pattern Number of Tables, $$n$$ Number of people seated Pattern $$\text{1}$$ $$4=4$$ $$=4+2\left(0\right)$$ $$\text{2}$$ $$4+2=6$$ $$=4+2\left(1\right)$$ $$\text{3}$$ $$4+2+2=8$$ $$=4+2\left(2\right)$$ $$\text{4}$$ $$4+2+2+2=10$$ $$=4+2\left(3\right)$$ $$\vdots$$ $$\vdots$$ $$\vdots$$ n $$4+2+2+2+\cdots +2$$ $$=4+2\left(n-1\right)$$ Note: There may be variations in how you think of the pattern in this problem. For example, you may view this problem as the person on one end fixed, two people seated opposite each other per table and one person at the other end fixed. This results in $$1 + 2n + 1 = 2n + 2$$. Your formula for $$T_n$$ will still be correct. Describe the pattern The number of people seated at $$n$$ tables is $${T}_{n}=4+2\left(n-1\right)$$ Calculate the $$12^{\text{th}}$$ term, in other words, find $${T}_{n}$$ if $$n=12$$ \begin{align} {T}_{12}& = 4+2\left(12-1\right) \\ & = 4+2\left(11\right) \\ & = 4 + 22 \\ & = 26 \end{align} Therefore $$\text{26}$$ people can be seated at $$\text{12}$$ tables. Calculate the number of tables needed to seat $$\text{20}$$ people, in other words, find $$n$$ if $${T}_{n}=20$$ \begin{align} {T}_{n}& = 4+2\left(n-1\right) \\ 20& = 4+2\left(n-1\right) \\ 20& = 4+2n-2 \\ 20-4+2& = 2n \\ 18 & = 2n \\ \frac{18}{2} & = n \\ n& = 9 \end{align} Therefore $$\text{9}$$ tables are needed to seat $$\text{20}$$ people. temp text It is important to note the difference between $$n$$ and $${T}_{n}$$. $$n$$ can be compared to a place holder indicating the position of the term in the sequence, while $${T}_{n}$$ is the value of the place held by $$n$$. From our example above, the first table holds $$\text{4}$$ people. So for $$n=1$$, the value of $${T}_{1}=4$$ and so on: $$n$$ $$\text{1}$$ $$\text{2}$$ $$\text{3}$$ $$\text{4}$$ $$\ldots$$ $${T}_{n}$$ $$\text{4}$$ $$\text{6}$$ $$\text{8}$$ $$\text{10}$$ $$\ldots$$ Worked example 3: Data plans Raymond subscribes to a limited data plan from Vodacell. The limited data plans cost $$\text{R}\,\text{120}$$ for $$\text{1}$$ gigabyte (GB) per month, $$\text{R}\,\text{135}$$ for $$\text{2}$$ $$\text{GB}$$ per month and $$\text{R}\,\text{150}$$ for $$\text{3}$$ $$\text{GB}$$ per month. Assume this pattern continues indefinitely. 1. Use a table to set up the pattern of the cost of the data plans. 2. Find the general formula for the sequence. 3. Use the general formula to determine the cost for a $$\text{30}$$ $$\text{GB}$$ data plan. 4. The cost of an unlimited data plan is $$\text{R}\,\text{510}$$ per month. Determine the amount of data Raymond would have to use for it to be cheaper for him to sign up for the unlimited plan. Make a table to see the pattern Number of GB $$(n)$$ $$\text{1}$$ $$\text{2}$$ $$\text{3}$$ $$\text{4}$$ Cost (in Rands) $$\text{120}$$ $$\text{135}$$ $$\text{150}$$ $$\text{165}$$ Pattern $$\text{120}$$ $$120+(1)(15)=135$$ $$120+(2)(15)=150$$ $$120+(3)(15)=165$$ Use the observed pattern to determine the general formula. The price of $$n$$ GB of data is $${T}_{n}=120+15\left(n-1\right)$$ Determine the cost of $$\text{30}$$ $$\text{GB}$$ of data. This question requires us to determine the value of the $$30^{\text{th}}$$ term, in other words, find $${T}_{n}$$ if $$n=30$$. Using the general formula, we get: \begin{align} {T}_{n}& = 120+15\left(n-1\right) \\ \therefore {T}_{30}& = 120+15\left(30-1\right) \\ & = 120+15\left(29\right) \\ & = 120 + 435 \\ & = 555 \end{align} Therefore the cost of a $$\text{30}$$ $$\text{GB}$$ data package is $$\text{R}\,\text{555}$$. Determine when it is cheaper to purchase the unlimited data plan The final question of this worked example requires us to determine when it would be cheaper for Raymond to purchase an unlimited data plan instead of a limited plan. In other words, we need to find $$n$$ where $$T_n$$ is less than $$\text{R}\,\text{510}$$. We know that: ${T}_{n} = 120+15\left(n-1\right)$ Therefore, if $$T_n=510= 120+15\left(n-1\right)$$ Solving for $$n$$, we get: \begin{align} 510& = 120+15\left(n-1\right) \\ 510& = 120+15n-15 \\ 510& = 105+15n \\ 405& = 15n \\ \frac{405}{15} & = n \\ n& = 27 \end{align} Therefore it is cheaper for Raymond to purchase the unlimited data plan if he uses more than $$\text{27}$$ $$\text{GB}$$ per month. temp text Video: 2F72 Textbook Exercise 3.1 Use the given pattern to complete the table below. Figure number $$\text{1}$$ $$\text{2}$$ $$\text{3}$$ $$\text{4}$$ $$n$$ Number of dots Number of lines Total Figure number $$\text{1}$$ $$\text{2}$$ $$\text{3}$$ $$\text{4}$$ $$n$$ Number of dots $$\text{3}$$ $$\text{4}$$ $$\text{5}$$ $$\text{6}$$ $$n + 2$$ Number of lines $$\text{3}$$ $$\text{5}$$ $$\text{7}$$ $$\text{9}$$ $$2n + 1$$ Total $$\text{6}$$ $$\text{9}$$ $$\text{12}$$ $$\text{15}$$ $$3(n + 1)$$ Consider the sequence shown here: $$-4 \; ; \; -1 \; ; \; 2 \; ; \; 5 \; ; \; 8 \; ; \; 11 \; ; \; 14 \; ; \;17 \; ; \; \ldots$$ If $$T_{n} = \text{2}$$ what is the value of $$T_{n-1}$$? \begin{align} T_{3} & = \text{2} \\ \therefore T_{n-1} & = -1 \end{align} Consider the sequence shown here: $$C \; ; \; D \; ; \; E \; ; \; F \; ; \; G \; ; \; H \; ; \; I \; ; \; J \; ; \; \ldots$$ If $$T_{n} = G$$ what is the value of $$T_{n-4}$$? \begin{align} T_{5} & = G \\ \therefore T_{n-4} & = C \end{align} For each of the following sequences determine the common difference. If the sequence is not linear, write “no common difference”. $$9 \; ; \; -7 \; ; \; -8 \; ; \; -25 \; ; \; -34 \; ; \; \ldots$$ \begin{align} d & = T_{2} - T_{1} = (-7) - (9) = -16 \\ d & = T_{3} - T_{2} = (-8) - (-7) = -1 \\ d & = T_{4} - T_{3} = (-25) - (-8) = -17 \end{align} You can see that the results are not the same - the difference is not 'common.' That means that this sequence of numbers in not linear, and it has no common difference. $$5 \; ; \; 12 \; ; \; 19 \; ; \; 26 \; ; \; 33 \; ; \; \ldots$$ \begin{align} d & = T_{2} - T_{1} = (12) - (5) = 7 \\ d & = T_{3} - T_{2} = (19) - (12) = 7 \\ d & = T_{4} - T_{3} = (26) - (19) = 7 \end{align} All of the results are the same, which means we have found the common difference for these numbers: $$d = 7$$. $$\text{2,93} \; ; \; \text{1,99} \; ; \; \text{1,14} \; ; \; \text{0,35} \; ; \; \ldots$$ \begin{align} d & = T_{2} - T_{1} = \left(\text{1,99}\right) - \left(\text{2,93}\right) = -\text{0,94} \\ d & = T_{3} - T_{2} = \left(\text{1,14}\right) - \left(\text{1,99}\right) = -\text{0,85} \end{align} In this case the sequence is not linear. Therefore the final answer is that there is no common difference. $$\text{2,53} \; ; \; \text{1,88} \; ; \; \text{1,23} \; ; \; \text{0,58} \; ; \; \ldots$$ \begin{align} d & = T_{2} - T_{1} = \left(\text{1,88}\right) - \left(\text{2,53}\right) = -\text{0,65} \\ d & = T_{3} - T_{2} = \left(\text{1,23}\right) - \left(\text{1,88}\right) = -\text{0,65} \end{align} The common difference is $$d = -\text{0,65}$$. Write down the next three terms in each of the following sequences: $$5 \; ; \; 15 \; ; \; 25 \; ; \; \ldots$$ The common difference is: \begin{align} d & = T_{2} - T_{1} \\ & = 15 - 5 \\ & = 10 \end{align} Therefore we add $$\text{10}$$ each time to get the next term in the sequence. The next three numbers are: $$\text{35}$$, $$\text{45}$$ and $$\text{55}$$ and the sequence becomes: $$5 \; ; \; 15 \; ; \; 25 \; ; \; 35 \; ; \; 45 \; ; \; 55 \; ; \; \ldots$$ $$-8 \; ; \; -3 \; ; \; 2 \; ; \; \ldots$$ The common difference is: \begin{align} d & = T_{2} - T_{1} \\ & = -3 - (-8) \\ & = 5 \end{align} Therefore we add $$\text{5}$$ each time to get the next term in the sequence. The next three numbers are: $$\text{7}$$, $$\text{12}$$ and $$\text{17}$$ and the sequence becomes: $$-8 \; ; \; -3 \; ; \; 2 \; ; \; 7 \; ; \; 12 \; ; \; 17 \; ; \; \ldots$$ $$30 \; ; \; 27 \; ; \; 24 \; ; \; \ldots$$ The common difference is: \begin{align} d & = T_{2} - T_{1} \\ & = 27 - 30 \\ & = -3 \end{align} Therefore we subtract $$\text{3}$$ each time to get the next term in the sequence. The next three numbers are: $$\text{21}$$, $$\text{18}$$ and $$\text{15}$$ and the sequence becomes: $$30 \; ; \; 27 \; ; \; 24 \; ; \; 21 \; ; \; 18 \; ; \; 15 \; ; \; \ldots$$ $$-\text{13,1}\; ; \;-\text{18,1}\; ; \;-\text{23,1}\; ;\ldots$$ \begin{align} d &= T_{2}-T_{1} \text{ or } T_{3}-T_{2}\\ &= (-\text{18,1}) - (-\text{13,1}) \text{ or } (-\text{23,1}) - (-\text{18,1}) \\ &= -\text{5} \\ \text{Therefore } T_4 &= -\text{28,1} \\ T_5 &= -\text{33,1} \\ T_6 &= -\text{38,1} \end{align} $$- 9 x\; ;- 19 x\; ;- 29 x\; ;\ldots$$ \begin{align} d &= T_{2}-T_{1} \text{ or } T_{3}-T_{2}\\ &= (- 19 x) - (- 9 x) \text{ or } (- 29 x) - (- 19 x) \\ &= -10 x \\ \text{Therefore } T_4 &= - 39 x \\ T_5 &= - 49 x \\ T_6 &= - 59 x \end{align} $$-\text{15,8}\; ; \;\text{4,2}\; ; \;\text{24,2}\; ;\ldots$$ \begin{align} d &= T_{2}-T_{1} \text{ or } T_{3}-T_{2}\\ &= (\text{4,2}) - (-\text{15,8}) \text{ or } (\text{24,2}) - (\text{4,2}) \\ &= \text{20} \\ \text{Therefore } T_4 &= \text{44,2} \\ T_5 &= \text{64,2} \\ T_6 &= \text{84,2} \end{align} $$30 b\; ;34 b\; ;38 b\; ;\ldots$$ \begin{align} d &= T_{2}-T_{1} \text{ or } T_{3}-T_{2}\\ &= (34 b) - (30 b) \text{ or } (38 b) - (34 b) \\ &= 4 b \\ \text{Therefore } T_4 &= 42 b \\ T_5 &= 46 b \\ T_6 &= 50 b \end{align} Given a pattern which starts with the numbers: $$3 \; ; \; 8 \; ; \; 13 \; ; \; 18 \; ; \; \ldots$$ determine the values of $$T_{6}$$ and $$T_{9}$$. $3 \; ; \; 8 \; ; \; 13 \; ; \; 18 \; ; \; 23 \; ; \; \underline{28} \; ; \;33 \; ; \; 38 \; ; \; \underline{43} \; ; \; \ldots$ $$T_{6} = 28 \text{ and } T_{9} = 43$$ Given a sequence which starts with the letters: $$C \; ; \; D \; ; \; E \; ; \; F \; ; \; \ldots$$ determine the values of $$T_{5}$$ and $$T_{8}$$. $C \; ; \; D \; ; \; E \; ; \; F \; ; \; \underline{G} \; ; \; H \; ; \; I \; ; \; \underline{J} \; ; \; \ldots$ $$T_{5} = G \text{ and } T_{8} = J$$ Given a pattern which starts with the numbers: $$7 \; ; \; 11 \; ; \; 15 \; ; \; 19 \; ; \; \ldots$$ determine the values of $$T_{5}$$ and $$T_{8}$$. $7 \; ; \; 11 \; ; \; 15 \; ; \; 19 \; ; \; \underline{23} \; ; \;27 \; ; \; 31 \; ; \; \underline{35} \; ; \;\ldots$ $$T_{5} = 23 \text{ and } T_{8} = 35$$ The general term is given for each sequence below. Calculate the missing terms (each missing term is represented by $$\ldots$$). $$0 \; ; \; 3 \; ; \; \ldots \; ; \; 15 \; ; \; 24 \; \qquad T_n = n^2 - 1$$ The third term is: \begin{align} T_n & = n^{2} - 1 \\ T_{3} & = (3)^{2} - 1 \\ & = 9 - 1 \\ & = 8 \end{align} The fourth term is: \begin{align} T_n & = n^{2} - 1 \\ T_{4} & = (4)^{2} - 1 \\ & = 16 - 1 \\ & = 15 \end{align} Therefore the only missing term is the third one, which is $$\text{8}$$. The full sequence is: $$0 \; ; \; 8 \; ; \; 15 \; ; \; 24$$ $$3 \; ; \; 2 \; ; \; 1 \; ; \; 0 \; ; \; \ldots; -2 \qquad T_{n} = -n + 4$$ The fifth term is: \begin{align} T_n & = -n + 4 \\ T_{5} & = -(5) + 4 \\ & = -1 \end{align} The sixth term is: \begin{align} T_n & = -n + 4 \\ T_{6} & = -(6) + 4 \\ & = -2 \end{align} Therefore the only missing term is the fifth one, which is $$-\text{1}$$. The full sequence is: $$3 \; ; \; 2 \; ; \; 1 \; ; \; 0 \; ; \; -1 \; ; \; -2$$ $$-11 \; ; \; \ldots \; ; \; -7 \; ; \; \ldots \; ; \; -3 \qquad T_{n} = -13 + 2n$$ The second term is: \begin{align} T_n & = -13 + 2n \\ T_{2} & = -13 + 2(2) \\ & = -13 + 4 \\ & = -9 \end{align} The third term is: \begin{align} T_n & = -13 + 2n \\ T_{3} & = -13 + 2(3) \\ & = -13 + 6 \\ & = -7 \end{align} The fourth term is: \begin{align} T_n & = -13 + 2n \\ T_{4} & = -13 + 2(4) \\ & = -13 + 8 \\ & = -5 \end{align} The fifth term is: \begin{align} T_n & = -13 + 2n \\ T_{5} & = -13 + 2(5) \\ & = -13 + 10 \\ & = -3 \end{align} Therefore the two missing terms are the second and fourth ones, which are $$-\text{9}$$ and $$-\text{5}$$. The full sequence is: $$-11 \; ; \; -9 \; ; \; -7 \; ; \; -5 \; ; \; -3$$ $$\text{1} \; ; \; \text{10} \; ; \; \text{19} \; ; \; \ldots \; ; \; \text{37} \qquad T_n = 9 n - 8$$ \begin{align} T_n & = 9 n - 8\\ T_4 & = \text{9} (\text{4}) -\text{8}\\ &= \text{28} \end{align} $$\text{9} \; ; \; \ldots \; ; \; \text{21} \; ; \; \ldots \; ; \; \text{33} \qquad T_n = 6 n + 3$$ To find the two missing terms, we use the equation for the general term: \begin{align} T_n & = \text{6}n\text{+3} \\ T_2 & = \text{6}(\text{2})\text{+3} \\ &= \text{15} \\ T_4 & = \text{6}(\text{4})\text{+3} \\ &= \text{27} \end{align} Find the general formula for the following sequences and then find $$T_{10}$$, $$T_{50}$$ and $$T_{100}$$ $$2; 5; 8; 11; 14; \ldots$$ We first need to find $$d$$: \begin{align} d & = T_{2} - T_{1} \\ & = 5 - 2 \\ & = 3 \end{align} Next we note that for each successive term we add $$d$$ to the last term. We can express this as: \begin{align} T_{1} & = a = 2 \\ T_{2} & = a + d = 2 + 3 \\ & = 2 + 1(3)\\ T_{3} & = T_{2} + d = 2 + 3 + 3 \\ & = 2 + 2(3) \\ T_{4} & = T_{3} + d = 2 + 3 + 3 + 3 \\ & = 2 + 3(3)\\ T_{n} & = T_{n-1} + d = 2 + 3(n-1) \\ & = 3n - 1 \end{align} The general formula is $$T_n = 3n - 1$$. $$T_{10}$$, $$T_{50}$$ and $$T_{100}$$ are: \begin{align} T_{10} & = 3(10) - 1 \\ & = 29 \\ T_{50} & = 3(50) - 1 \\ & = 149 \\ T_{100} & = 3(100) - 1 \\ & = 299 \end{align} $$0; 4; 8; 12; 16; \ldots$$ We first need to find $$d$$: \begin{align} d & = T_{2} - T_{1} \\ & = 4 - 0 \\ & = 4 \end{align} Next we note that for each successive term we add $$d$$ to the last term. We can express this as: \begin{align} T_{1} & = a = 0 \\ T_{2} & = a + d = 0 + 4 \\ & = 4(1)\\ T_{3} & = T_{2} + d = 0 + 4 + 4 \\ & = 4(2)\\ T_{4} & = T_{3} + d = 0 + 4 + 4 + 4 \\ & = 4(3)\\ T_{n} & = T_{n-1} + d = 0 + 4(n-1) \\ & = 4n - 4 \end{align} The general formula is $$T_n = 4n - 4$$. $$T_{10}$$, $$T_{50}$$ and $$T_{100}$$ are: \begin{align} T_{10} & = 4(10) - 4 \\ & = 36 \\ T_{50} & = 4(50) - 4 \\ & = 196 \\ T_{100} & = 4(100) - 4 \\ & = 396 \end{align} $$2; -1; -4; -7; -10; \ldots$$ We first need to find $$d$$: \begin{align} d & = T_{2} - T_{1} \\ & = -1 - 2 \\ & = -3 \end{align} Next we note that for each successive term we add $$d$$ to the last term. We can express this as: \begin{align} T_{1} & = a = 2 \\ T_{2} & = a + d = 2 + (-3) \\ & = 2 + (-3)(1)\\ T_{3} & = T_{2} + d = 2 + (-3) + (-3) \\ & = 2 + (-3)(2)\\ T_{4} & = T_{3} + d = 2 + (-3) + (-3) + (-3) \\ & = 2 + (-3)(3)\\ T_{n} & = T_{n-1} + d = 2 + (-3)(n-1) \\ & = 5 - 3n \end{align} The general formula is $$T_n = 5 - 3n$$. $$T_{10}$$, $$T_{50}$$ and $$T_{100}$$ are: \begin{align} T_{10} & = 5 - 3(10) \\ & = -25 \\ T_{50} & = 5 - 3(50) \\ & = -145 \\ T_{100} & = 5 - 3(100) \\ & = -295 \end{align} The diagram below shows pictures which follow a pattern. How many triangles will there be in the $$5^{\text{th}}$$ picture? $$5 \; ; \; 7 \; ; \; 9 \; ; \; 11 \; ; \; \ldots$$ Therefore two triangles are added each time and the fifth picture will have $$\text{13}$$ triangles. Determine the formula for the $$n^{\text{th}}$$ term. The general term of the pattern is: \begin{align} T_{n} & = T_{1} + d = 5 + (2)(n-1) \\ & = 2n + 3 \end{align} Use the formula to find how many triangles are in the $$25^{\text{th}}$$ picture of the diagram. \begin{align} T_{n} & = 2n + 3 \\ T_{25} & = 2(25) + 3 \longleftarrow \text{substitute n } = 25 \\ & = 53 \end{align} Study the following sequence: $$15 \; ; \; 23 \; ; \; 31 \; ; \; 39 \; ; \; \ldots$$ Write down the next $$\text{3}$$ terms. We note that we add 8 to each term to get the next term. Therefore the next three terms are $$47 \; ; \; 55 \; ; \; 63$$. Find the general formula for the sequence \begin{align} T_n & = T_1 + d(n - 1) \\ & = 15 + 8(n-1) \\ & = 8n + 7 \end{align} Find the value of $$n$$ if $$T_n$$ is $$\text{191}$$. \begin{align} 191 &= 8n + 7 \\ 184 &= 8n \\ n &= 23 \end{align} Study the following sequence: $$-44 \; ; \; -14 \; ; \; 16 \; ; \; 46 \; ; \; \ldots$$ Write down the next $$\text{3}$$ terms. We note that we add 30 to each term to get the next term. Therefore the next three terms are $$76 \; ; \; 106 \; ; \; 136$$. Find the general formula for the sequence \begin{align} T_n & = T_1 + d(n - 1) \\ & = -44 + 30(n - 1) \\ & = 30n -74 \end{align} Find the value of $$n$$ if $$T_n$$ is $$\text{406}$$. \begin{align} 406 &= 30n - 74 \\ 480 &= 30n \\ n &= 16 \end{align} Consider the following list: $- z - 5 \; ; \; - 4 z - 5 \; ; \; - 6 z - 2 \; ; \; - 8 z - 5 \; ; \; - 10 z - 5 \; ; \; \ldots$ Find the common difference for the terms of the list. If the sequence is not linear (if it does not have a common difference), write “no common difference”. \begin{align} d & = T_{2} - T_{1} = (- 4 z - 5) - (- z - 5) = - 3 z \\ & = T_{3} - T_{2} = (- 6 z - 2) - (- 4 z - 5) = - 2 z + 3 \\ & = T_{4} - T_{3} = (- 8 z - 5) - (- 6 z - 2) = - 2 z - 3 \end{align} No common difference. If you are now told that $$z = -2$$, determine the values of $$T_{1}$$ and $$T_{2}$$. \begin{align} T_{1} & = - z - 5 \\ & = - (-2)-5 \\ & = -3 \\ T_{2} & = - 4 z - 5 \\ & = - 4 (-2)-5 \\ & = 3 \end{align} Consider the following pattern: $2n + 4 \; ; \; 1 \; ; \; - 2n - 2 \; ; \; - 4n - 5 \; ; \; - 6n - 8 \; ; \;\ldots$ Find the common difference for the terms of the pattern. If the sequence is not linear (if it does not have a common difference), write “no common difference”. \begin{align} d & = T_{2} - T_{1} = (1) - (2n + 4) = - 2n - 3 \\ & = T_{3} - T_{2} = (- 2n - 2) - (1) = - 2n - 3 \\ & = T_{4} - T_{3} = (- 4n - 5) - (- 2n - 2) = - 2n - 3 \end{align} The common difference for these numbers: $$d = - 2n - 3$$. If you are now told that $$n = -1$$, determine the values of $$T_{1}$$ and $$T_{3}$$. \begin{align} T_{1} & = 2 n + 4 \\ & = 2 (-1)+4 \\ & = 2 \\ T_{3} & = - 2 n - 2 \\ & = - 2 (-1)-2 \\ & = 0 \end{align} If the following terms make a linear sequence: $\frac{k}{3} - 1 \; ; \;- \frac{5k}{3} + 2 \; ; \;- \frac{2k}{3} + 10 \; ; \;\ldots$ Determine the value of $$k$$. If the answer is a non-integer, write the answer as a simplified fraction. \begin{align} T_{2} - T_{1} & = T_{3} - T_{2} \\ \left(- \frac{5k}{3} + 2\right) - \left(\frac{k}{3} - 1\right) & = \left(- \frac{2k}{3} + 10\right) - \left(- \frac{5k}{3} + 2\right) \\ \\ 3\left(- \frac{5k}{3} + 2\right) - 3\left(\frac{k}{3} - 1\right) & = 3\left(- \frac{2k}{3} + 10\right) - 3\left(- \frac{5k}{3} + 2\right) \\ - 5k + 6 - \left(k - 3\right) & = - 2k + 30 - \left(- 5k + 6\right) \\ - 6k + 9 & = 3k + 24 \\ -15 & = 9k \\ k & = - \frac{5}{3} \end{align} Now determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions. \begin{align} \text{First term: } T_{1} & = \frac{k}{3} - 1 \\ & = \frac{\left(- \frac{5}{3}\right)}{3}-1 \\ & = - \frac{14}{9} \\ \text{Second term: }T_{2} & = - \frac{5 k}{3} + 2 \\ & = - \frac{5 \left(- \frac{5}{3}\right)}{3}+2 \\ & = \frac{43}{9} \\ \text{Third term: }T_{3} & = - \frac{2 k}{3} + 10 \\ & = - \frac{2 \left(- \frac{5}{3}\right)}{3}+10 \\ & = \frac{100}{9} \end{align} The first three terms of this sequence are: $$- \frac{14}{9} , \frac{43}{9}$$ and $$\frac{100}{9}$$. If the following terms make a linear sequence: $y - \frac{3}{2} \; ; \; - y - \frac{7}{2} \; ; \; - 7 y - \frac{15}{2} \; ; \; \ldots$ find $$y$$. If the answer is a non-integer, write the answer as a simplified fraction. \begin{align} T_{2} - T_{1} & = T_{3} - T_{2} \\ \left(- y - \frac{7}{2}\right) - \left(y - \frac{3}{2}\right) & = \left(- 7 y - \frac{15}{2}\right) - \left(- y - \frac{7}{2}\right) \\ \\ 2\left(- y - \frac{7}{2}\right) - 2\left(y - \frac{3}{2}\right) & = 2\left(- 7y - \frac{15}{2}\right) - 2\left(- y - \frac{7}{2}\right) \\ - 2y - 7 - \left(2y - 3\right) & = - 14y - 15 - \left(- 2y - 7\right) \\ - 4y - 4 & = - 12y - 8 \\ 8y & = -4 \\ y & = - \frac{1}{2} \end{align} Now determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions. \begin{align} \text{First term: } T_{1} & = y - \frac{3}{2} \\ & = \left(- \frac{1}{2}\right)- \frac{3}{2} \\ & = -2 \\ \text{Second term: }T_{2} & = - y - \frac{7}{2} \\ & = - \left(- \frac{1}{2}\right)- \frac{7}{2} \\ & = -3 \\ \text{Third term: }T_{3} & = - 7 y - \frac{15}{2} \\ & = - 7 \left(- \frac{1}{2}\right)- \frac{15}{2} \\ & = -4 \end{align} The first three terms of this sequence are: $$-2 , -3 \text{ and } -4$$. What is the $$649^{\text{th}}$$ letter of the sequence: PATTERNPATTERNPATTERNPATTERNPATTERNPATTERNPATTE.............? The word “PATTERN” is 7 letters long, so: $\frac{649}{7} = 92\text{ r } 5$ The remainder of 5 shows us that the $$649^{\text{th}}$$ letter is the $$5^{\text{th}}$$ letter in the word, which is E
Minima/ maxima (statement questions) - Number questions Chapter 6 Class 12 Application of Derivatives Concept wise ### Transcript Ex 6.3, 13 Find two numbers whose sum is 24 and whose product is as large as possible. Let first number be š„ Now, given that First number + Second number = 24 š„ + second number = 24 Second number = 24 ā š„ Product = (šššš š” šš¢šššš ) Ć (š ššššš šš¢šššš) = š„ (24āš„) Let P(š„) = š„ (24āš„) We need product as large as possible Hence we need to find maximum value of P(š„) Finding Pā(x) P(š„)=š„(24āš„) P(š„)=24š„āš„^2 Pā(š„)=24ā2š„ Pā(š„)=2(12āš„) Putting Pā(š„)=0 2(12āš„)=0 12 ā š„ = 0 š„ = 12 Finding Pāā(š„) Pā(š„)=24ā2š„ Pāā(š„) = 0 ā 2 = ā 2 Thus, pāā(š„) < 0 for š„ = 12 š„ = 12 is point of maxima & P(š„) is maximum at š„ = 12 Finding maximum P(x) P(š„)=š„(24āš„) Putting š„ = 12 p(12)= 12(24ā12) = 12(12) = 144 ā“ First number = x = 12 & Second number = 24 ā x = (24 ā 12)= 12
## How To Find The Sum Of Consecutive Integers #### How to Add Consecutive Integers from 1 to 100 9 Steps find the two integers The sum of two numbers is 56 . The smaller number is 18 less than the larger number. #### the sum of the reciprocals of two consecutive integers is If four consecutive odd integers greater than 9 are added together, what is the smallest possible sum of those four integers? The 4 consecutive of integers greater than 9 (but not including 9) are 11, 13, 15, 17. Added together, we get 56. to find the integers you can guess and check (you know #### Find the smallest possible sum of six consecutive integers Sum of the last three integers = 21 The sum of the first three is 9 less than the last three. Since the sum of the last 3 integers is 624, the sum of the first 3 integers = 624 - 9 = 615 #### What three consecutive integers have a sum of 225? Valeur Sum of integers squared from 1 to N is also called "Square Pyramid Numbers" because each layer of the balls makes a square pattern. It is expressed as Pyr n = 1 2 + 2 2 + 3 2 + + N 2 Instead ,if the cross section pattern is a triangle, then it makes the following number sequence. How to find the sum of consecutive integers #### In how many ways can a number be expressed as a sum of 30/04/2018 · To find the sum of consecutive numbers 1 to 100, you multiply the number of sets (50) by the sum of each set (101): () = So, the sum of consecutive number 1 through 100 is … #### Sums of Consecutive Integers maa.org 7/09/2016 · Consider the two integers as n and n+1 the sum of two integers is 111 that is n+n+1=111 2n+1=111 take 1 to the RHS 2n=111-1 2n=110 n=110/2 n=55 #### Find 4 consecutive integers with a sum of 106Please show A sum of consecutive numbers is a difference of triangular numbers. The paper below gives a solution for the case of nonconsecutive triangular numbers. 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Wearing a light jacket and not just a top is a big improvement but not usually do-able in my hot climate. 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# Sum of k How to use the sum of k formula to solve summation problems: formula, proof, example, and its solution. ## Formula The sum of k as k goes from 1 to n is n(n + 1)/2. ## Proof Previouly, you've proved this formula. Mathematical induction So let's see the proof not only to learn about the formula, but also to remember how to write mathematical induction. Show that 'if n = 1, the given statement is true'. Assume that 'if n = k, the given statement is true'. So assume that 1 + 2 + 3 + ... + k = k(k + 1)/2 is true. Show that 'if n = k + 1, the given statement is true'. For n = k + 1, (left side) = 1 + 2 + 3 + ... + k + (k + 1). So add (k + 1) on both sides: 1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1). Then change the right side to (k + 1)[(k + 1) + 1]/2. By showing this, you can prove that 'if n = k + 1, the given statement is true'. This part is the most tricky part when solving mathematical induction problems. 'If n = 1, the given statement is true' makes 'if n = 2, the given statement is true'. 'If n = 2, the given statement is true' makes 'if n = 3, the given statement is true'. 'If n = 3, the given statement is true' makes 'if n = 4, the given statement is true'. ... By this way, for all n, the given statement is true. So the given statement is true. ## Example Rewrite [sum of (4k - 1)] into 4⋅[sum of k] - [sum of 1]. Basic properties of summation As as k goes from 1 to n, [sum of k] = n(n + 1)/2, [sum of 1] = n (= 1⋅n). So (given) = 4⋅[n(n + 1)/2] - n.
# Sample 1 has 17 "yes" responses out of 97 in the sample, sample 2 has "yes" 46 responses out of 131 responses in the sample. Calculate the point estimate for the difference between the population proportions of "yes" responses. Question Probability Sample 1 has 17 "yes" responses out of 97 in the sample, sample 2 has "yes" 46 responses out of 131 responses in the sample. Calculate the point estimate for the difference between the population proportions of "yes" responses. 2021-02-01 Given that Sample 1 has 17 "yes" responses out of 97 responses in the sample, and Sample 2 has 46 "yes" responses out of 131 responses in the sample. We need to calculate the point estimane for the difference between the population proportions of "yes" responses. Point estimate: We define p is the point estimate for population proportions. p=x/need where x is the number of successes and n is the population size. Now, for the Sample 1 the point estimate for the population proportion is PSKp1=x/n =17/97 =0.17525ZSK and for the Sample 2 the point estimate for the population proportion is PSKp2=x/n =46/131 =0.35114ZSK Therefore, the point estimate for the difference between the population proportion of "yes" responses is PSKp=p1-p2 =0.17525-0.35114 =-0.17589ZSK ### Relevant Questions A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age. $$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25 $$\begin{array}{\|c\|c\|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$ A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old. $$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50 $$\begin{array}{\|c\|c\|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$ (i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.) (ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$ (a) What is the level of significance? State the null and alternate hypotheses. $$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$ $$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$ $$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$ $$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$ (b) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations, The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations, The Student's t. We assume that both population distributions are approximately normal with known standard deviations, What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.) What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.) (c) Find (or estimate) the P-value. P-value $$\displaystyle>{0.250}$$ $$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$ $$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$ $$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$ $$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$ P-value $$\displaystyle<{0.005}$$ Sketch the sampling distribution and show the area corresponding to the P-value. P.vaiue Pevgiue P-value f P-value Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$ $$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$ $$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$ $$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Money reports that the average annual cost of the first year of owning and caring for a large dog in 2017 is $1,448. The Irish Red and White Setter Association of America has requested a study to estimate the annual first-year cost for owners of this breed. A sample of 50 will be used. Based on past studies, the population standard deviation is assumed known with $$\displaystyle\sigma=\{230}.$$ $$\begin{matrix} 1,902 & 2,042 & 1,936 & 1,817 & 1,504 & 1,572 & 1,532 & 1,907 & 1,882 & 2,153 \\ 1,945 & 1,335 & 2,006 & 1,516 & 1,839 & 1,739 & 1,456 & 1,958 & 1,934 & 2,094 \\ 1,739 & 1,434 & 1,667 & 1,679 & 1,736 & 1,670 & 1,770 & 2,052 & 1,379 & 1,939\\ 1,854 & 1,913 & 2,163 & 1,737 & 1,888 & 1,737 & 2,230 & 2,131 & 1,813 & 2,118\\ 1,978 & 2,166 & 1,482 & 1,700 & 1,679 & 2,060 & 1,683 & 1,850 & 2,232 & 2,294 \end{matrix}$$ (a) What is the margin of error for a $$95\%$$ confidence interval of the mean cost in dollars of the first year of owning and caring for this breed? (Round your answer to nearest cent.) (b) The DATAfile Setters contains data collected from fifty owners of Irish Setters on the cost of the first year of owning and caring for their dogs. Use this data set to compute the sample mean. Using this sample, what is the $$95\%$$ confidence interval for the mean cost in dollars of the first year of owning and caring for an Irish Red and White Setter? (Round your answers to nearest cent.)$_______ to $________ asked 2020-11-22 List the assumptions necessary for each of the following inferential techniques: a. Large-sample inferences about the difference $$\displaystyle{\left(\mu_{{1}}-\mu_{{2}}\right)}$$ between population means using a two-sample z-statistic b. Small-sample inferences about $$\displaystyle{\left(\mu_{{1}}-\mu_{{2}}\right)}$$ using an independent samples design and a two-sample t-statistic c. Small-sample inferences about $$\displaystyle{\left(\mu_{{1}}-\mu_{{2}}\right)}$$ using a paired difference design and a single-sample t-statistic to analyze the differences d. Large-sample inferences about the differences $$\displaystyle{\left(\mu_{{1}}-\mu_{{2}}\right)}$$ between binomial proportions using a two sample z-statistic e. Inferences about the ratio $$\displaystyle{\frac{{{\sigma_{{{1}}}^{{{2}}}}}}{{{\sigma_{{{2}}}^{{{2}}}}}}}$$ of two population variances using an F-test. asked 2021-01-28 Indicate true or false for the following statements. If false, specify what change will make the statement true. a) In the two-sample t test, the number of degrees of freedom for the test statistic increases as sample sizes increase. b) When the means of two independent samples are used to to compare two population means, we are dealing with dependent (paired) samples. c) The $$\displaystyle{x}^{{{2}}}$$ distribution is used for making inferences about two population variances. d) The standard normal (z) score may be used for inferences concerning population proportions. e) The F distribution is symmetric and has a mean of 0. f) The pooled variance estimate is used when comparing means of two populations using independent samples. g) It is not necessary to have equal sample sizes for the paired t test. asked 2021-02-09 A two-sample inference deals with dependent and independent inferences. In a two-sample hypothesis testing problem, underlying parameters of two different populations are compared. In a longitudinal (or follow-up) study, the same group of people is followed over time. Two samples are said to be paired when each data point in the first sample is matched and related to a unique data point in the second sample. This problem demonstrates inference from two dependent (follow-up) samples using the data from the hypothetical study of new cases of tuberculosis (TB) before and after the vaccination was done in several geographical areas in a country in sub-Saharan Africa. Conclusion about the null hypothesis is to note the difference between samples. The problem that demonstrates inference from two dependent samples uses hypothetical data from the TB vaccinations and the number of new cases before and after vaccination. PSK\begin{array}{\|c\|c\|} \hline Geographical\ regions & Before\ vaccination & After\ vaccination\\ \hline 1 & 85 & 11\\ \hline 2 & 77 & 5\\ \hline 3 & 110 & 14\\ \hline 4 & 65 & 12\\ \hline 5 & 81 & 10\\\hline 6 & 70 & 7\\ \hline 7 & 74 & 8\\ \hline 8 & 84 & 11\\ \hline 9 & 90 & 9\\ \hline 10 & 95 & 8\\ \hline \end{array}ZSK Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided $$\displaystyle{95}\%$$ confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance. asked 2021-01-02 Geographical Analysis (Jan, 2010) presented a study of Emergency Medical Services (EMS) ability to meet the demand for an ambulance. In one example, the researchers presented the following scenario. An ambulance station has one vehicle and two demand locations, A and B. The probability that the ambulance can travel to a location in under eight minutes is .58 for location A and .42 for location B. The probability that the ambulance is busy at any point in time is .3. a. Find the probability that EMS can meet demand for an ambulance at location A. b. Find the probability that EMS can meet demand for an ambulance at location B. asked 2020-10-23 1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are$42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is \$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance. The owner of a large equipment rental company wants to make a rather quick estimate of the average number of days a piece of ditch-digging equipment is rented out per person per time. The company has records of all rentals, but the amount of time required to conduct an audit of all accounts would be prohibitive. The owner decides to take a random sample of rental invoices. Fourteen different rentals of ditch-diggers are selected randomly from the files, yielding the following data. She wants to use these data to construct a $$99\%$$ confidence interval to estimate the average number of days that a ditch-digger is rented and assumes that the number of days per rental is normally distributed in the population. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 2 decimal places, using conventional rounding rules.
Question Video: Finding the Measure of the Angle between a Straight Line and a Coordinate Axis \| Nagwa Question Video: Finding the Measure of the Angle between a Straight Line and a Coordinate Axis \| Nagwa # Question Video: Finding the Measure of the Angle between a Straight Line and a Coordinate Axis Mathematics • Third Year of Secondary School ## Join Nagwa Classes Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher! Find, to the nearest second, the measure of the angle between the straight line (π₯ + 1)/2 = (π¦ β 2)/β4 = (π§ + 2)/5 and the positive direction of the π₯-axis. 04:12 ### Video Transcript Find, to the nearest second, the measure of the angle between the straight line π₯ plus one over two is equal to π¦ minus two over negative four is equal to π§ plus two over five and the positive direction of the π₯-axis. In this question, weβre asked to determine the measure of the angle between two straight lines. And we need to give this angle to the nearest second. To do this, letβs start by recalling how we find the measure of the angle between two straight lines. We know if we have two straight lines with direction vectors π sub one and π sub two, then the angle π between the two lines will satisfy the equation cos of π is equal to the dot product of π sub one and π sub two divided by the magnitude of π sub one times the magnitude of π sub two. Therefore, to find the angle between the two lines, we need to find the direction vector of each line. Letβs start by finding the direction vector of the first line. We can do this by recalling if a straight line is given in Cartesian form, then the denominators of these fractions represents the components of its direction vector. And when using this method, it is important to check the signs of the variables. They all need to be positive. So, in this case, the direction vector π sub one is the vector two, negative four, five. Our second straight line is just the positive direction of the π₯-axis. So the π¦- and π§-coordinates of every point on this line remain constant at zero. Only the π₯-coordinate changes. So weβll choose the unit directional vector π’, which is the vector one, zero, zero, to be the direction vector of this line. We could now substitute these vectors into our equation involving the angle between the two lines. However, itβs easier to evaluate both the numerator and denominator of the right-hand side of this equation first. Letβs start by finding the dot product of the two vectors. We can do this by recalling to find the dot product of two vectors of the same dimension, we just need to find the sum of the products of the corresponding components. However, in this case, our second vector has two components equal to zero. So, in this case, itβs just equal to the product of the first components of the two vectors. Two times one is two. Next, we need to determine the magnitude of the two vectors. Letβs start with the magnitude of vector π sub one. Thatβs the square root of the sum of the squares of its components. Thatβs the square root of two squared plus negative four squared plus five squared, which we can evaluate is root 45, which we could simplify to give three root five. And we could apply the same process to determine the magnitude of vector π sub two. However, we chose this to be the unit vector π’, so we already know its magnitude is one. We can now substitute these values into our equation. We get that cos of π is equal to two divided by root 45 times one, which simplifies to root 45. We can now solve for π by taking the inverse cosine of both sides of the equation. This then gives us that π is the inverse cos of two over root 45. And making sure our calculator is in degrees mode, we get 72.65 and this expansion continues degrees. But the question wants us to give our answer to the nearest second. So weβre going to need to convert this angle into degrees, minutes, and seconds. To do this, we start by noting there are 72 degrees in this angle. This leaves us with a remaining angle of 0.65 and this expansion continues degrees. Since there are 60 minutes in a degree, we can multiply this angle by 60 to convert it into minutes. Evaluating this and making sure we use the exact values, we get 39.23 and this expansion continues minutes. So we can see that this angle has 39 minutes and then a remaining angle of 0.23 and this expansion continues minutes. And once again, weβre going to want to know what this remaining angle is in seconds. And since there are 60 seconds in a minute, we can do this by multiplying our angle by 60. And by using the exact angle, we get 14.16 and this expansion continues seconds. We want to give our answer to the nearest second. So we need to look at the first decimal digit to determine if we need to round up or round down. This is one, so we need to round down. This then gives us our final answer. The measure of the angle between the straight line π₯ plus one over two is equal to π¦ minus two over negative four is equal to π§ plus two over five and the positive direction of the π₯-axis to the nearest second is 72 degrees, 39 minutes, and 14 seconds. ## 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 Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
## Engage NY Eureka Math Grade 6 Module 5 Lesson 2 Answer Key ### Eureka Math Grade 6 Module 5 Lesson 2 Exploratory Challenge Answer Key Exploratory Challenge: Question a. Use the shapes labeled with an X to predict the formula needed to calculate the area of a right triangle. Explain your prediction. Formula for the area of right triangles: Area of the given triangle: ________ Formula for the area of right triangles: A = $$\frac{1}{2}$$ Ã— base Ã— height or A = $$\frac{\text { base } \times \text { height }}{2}$$ Area of the given triangle: A = $$\frac{1}{2}$$ Ã— 3 in. Ã— 2 in. = 3 in2 Question b. Use the shapes labeled with a Y to determine if the formula you discovered in part (a) is correct. Does your area formula for triangle Y match the formula you got for triangle X? Answers will vary; however, the area formulas should be the same if students discovered the correct area formula. If so, do you believe you have the correct formula needed to calculate the area of a right triangle? Why or why not? If not, which formula do you think is correct? Why? Area of the given triangle: A = $$\frac{1}{2}$$ Ã— 3 in. Ã— 3 in. = 4.5 im <sup>2</sup> ### Eureka Math Grade 6 Module 5 Lesson 2 Exercise Answer Key Exercises: Calculate the area of each right triangle below. Each figure is not drawn to scale. Question 1. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (8 ft.) (15 ft.) = 60 ft2 Question 2. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (11.4 cm) (17.7 cm) = 100.89 cm2 Question 3. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (6 in.) (8 in.) = 24 in2 Question 4. Question 5. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (32.7 km) (21.4 km) = 349.89 km2 Question 6. Mr. Jones told his students they each need half of a piece of paper. Calvin cut his piece of paper horizontally, and Matthew cut his piece of paper diagonally. Which student has the larger area on his half piece of paper? Explain. After cutting the paper, both Calvin and Matthew have the same area. Calvin cut his into two rectangles that are each half the area of the original piece of paper. Matthew cut his paper into two equivalent right triangles that are also half the area of the original piece of paper. Question 7. Ben requested that the rectangular stage be split into two equal sections for the upcoming school play. The only instruction he gave was that he needed the area of each section to be half of the original size. If Ben wants the stage to be split into two right triangles, did he provide enough information? Why or why not? Ben did not provide enough information because the stage may be split horizontally or vertically through the middle of the rectangle. This would result in two equal pieces, but they would not be right triangles. Question 8. If the area of a right triangle is 6. 22 sq. in. and its base is 3. 11 in., write an equation that relates the area to the height, h, and the base. Solve the equation to determine the height. 6.22 = $$\frac{1}{2}$$ (3. 11)h 6.22 = (1.555)h 6.22 Ã· 1.555 = (1. 555)h Ã· 1.555 4 = h Therefore, the height of the right triangle is 4 in. ### Eureka Math Grade 6 Module 5 Lesson 2 Problem Set Answer Key Calculate the area of each right triangle below. Note that the figures are not drawn to scale. Question 1. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (31.2 cm) (9.1 cm) = 141.96 cm2 Question 2. Question 3. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (2.4 in.) (3.2 in.) = 3.84 in2 Question 4. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (11 mm) (60 mm) = 330 mm2 Question 5. Question 6. Elania has two congruent rugs at her house. She cut one vertically down the middle, and she cut diagonally through the other one. After making the cuts, which rug (labeled A, B, C, or D) has the larger area? Explain. All of the rugs are the same size after making the cuts. The vertical line goes down the center of the rectangle, making two congruent parts. The diagonal line also splits the rectangle into two congruent parts because the area of a right triangle is exactly half the area of the rectangle. Question 7. Give the dimensions of a right triangle and a parallelogram with the same area. Explain how you know. Question 8. If the area of a right triangle is $$\frac{9}{16}$$ sq. ft. and the height is $$\frac{3}{4}$$ ft., write an equation that relates the area to the base, b, and the height. Solve the equation to determine the base. Therefore, the base of the right triangle is 1$$\frac{1}{2}$$ ft. ### Eureka Math Grade 6 Module 5 Lesson 2 Exit Ticket Answer Key Question 1. Calculate the area of the right triangle. Each figure is not drawn to scale. A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (8 in.)(6 in.) = 24 in2 A = $$\frac{1}{2}$$ Ã— base Ã— height.
How to Master the Average Rate of Change Understanding the average rate of change is crucial across various fields of study, including calculus, where it represents the slope of the secant line between two points on a function. Here’s a step-by-step guide to grasp this concept fully and apply it in different contexts: Step-by-step Guide to Master the Average Rate of Change Step 1: Understanding the Concept Definition • The average rate of change is a measure of how much a quantity changes, on average, between two points. • In mathematical terms, for a function $$f(x)$$, the average rate of change from $$x=a$$ to $$x=b$$ is $$\frac{f(b)−f(a)}{b−a}$$. Graphical Representation • It’s the slope of the straight line (secant line) connecting two points on a curve. Step 2: Calculating the Average Rate of Change Identify the Points • Choose two points on the graph of the function or in your data set, labeled as $$(a,f(a))$$ and $$(b,f(b))$$. Apply the Formula • Subtract the $$y$$-values: $$f(b)−f(a)$$. • Subtract the $$x$$-values: $$b−a$$. • Divide the difference in $$y$$-values by the difference in $$x$$-values to find the average rate of change. Step 3: Interpreting the Average Rate of Change Positive or Negative • A positive average rate of change indicates an increasing function in the interval, while a negative one indicates a decreasing function. Magnitude • The greater the magnitude of the average rate of change, the steeper the line and the more significant the change over the interval. Step 4: Applying the Average Rate of Change in Different Fields Calculus • Motion: It can represent the average velocity of an object over a time interval. • Functions: Helps in understanding the behavior of functions over an interval before delving into instantaneous rates of change (derivatives). Economics • Market Analysis: Calculate the average rate of change of stock prices to gauge overall market trends. • Growth Rates: Determine the average growth rate of a company’s revenue or profit over time. Biology • Population Dynamics: Measure the average growth rate of a population over a given time period. • Biochemical Processes: Calculate the rate of change of reactant or product concentration in a reaction. Physics • Thermodynamics: Analyze the average rate of temperature change in a substance. • Kinematics: Use it to find the average acceleration when velocity changes over time. Step 5: Advanced Considerations in Calculus • Secant Line to Tangent Line: As the interval between $$a$$ and $$b$$ gets smaller, the average rate of change approaches the instantaneous rate of change (the derivative). • Curve Analysis: Use the average rate of change to approximate the behavior of curves before using more advanced calculus techniques. Step 6: Communicating Results • When presenting your findings, contextualize the average rate of change within the problem’s framework, explaining what the change represents in real-world terms. • Use graphs to illustrate the average rate of change visually for a more impactful presentation. Final Word The average rate of change is a fundamental concept that serves as a stepping stone to more advanced calculus ideas like derivatives. Its utility spans across various disciplines, making it a versatile tool for analyzing changes and trends in a myriad of contexts. By following this guide, you can harness this concept to extract meaningful insights from a range of data sets and functions. Examples: Example 1: Determine the average rate of change of the function $$g(x)=3x^2−4x+1$$ from $$x=2$$ to $$x=5$$. Solution: • Calculate $$g(2)=3(2)^2−4(2)+1=5$$. • Calculate $$g(5)=3(5)^2−4(5)+1=56$$. • Apply the average rate of change formula: $$\frac{g(5)−g(2)}{5−2}=\frac{56−5}{3}=\frac{51}{3}=17$$. The average rate of change from $$x=2$$ to $$x=5$$ is $$17$$. Example 2: Determine the average rate of change of the function $$h(x)=2x^3−3x^2+x−5$$ from $$x=3$$ to $$x=6$$. Solution: • Calculate $$h(3)=2(3)^3−3(3)^2+3−5=25$$. • Calculate $$h(6)=2(6)^3−3(6)^2+6−5=325$$. • Apply the average rate of change formula: $$\frac{h(6)−h(3)}{6-3}=\frac{325−25}{3}=\frac{300}{3}=100$$. The average rate of change from $$x=3$$ to $$x=6$$ is $$100$$. What people say about "How to Master the Average Rate of Change - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE $5 It was$16.99 now it is \$11.99
Select Page As I like to do before getting down to some hardcore maths, I’m going to take an artistic break and present to you a Haiku that I wrote a couple of days ago: The flower withers Petal falls on wilted earth New shoot takes its place There, that’s enough of the artsy fartsy stuff for now. I will firstly explain briefly what cost curves are, and why they look like they do. I will then go on to prove the fact that the Marginal Cost curve intersects the Average Cost curve at its lowest point. Using some simple but fancy maths, of course. ##### Cost Curves I’m sure you’ve noticed the 3 curves in the graph above. These generic graphs show us the various costs incurred by a typical firm when they produce some good (e.g a box of chocolates). The horizontal axis represents the quantity (q) of the item produced, and the vertical axis represents the cost (C) of producing q amounts of the good. Let’s roll with the boxes of chocolate example. The first curve we’re interested in is the total cost (TC) curve. It has 2 key features: 1. The curve is always increasing because it obviously costs you more money in total to produce more boxes of chocolate (and also the cost is never negative!) 2. The more boxes of chocolate you produce, the more expensive each additional item becomes to produce. In other words, we can say the marginal cost (MC) is increasing. In the graph, this means that it gets steeper as q increases. (A reason for this could be because the machines used to produce the chocolate need additional maintenance the more chocolate is produced.) The second point does not always apply, but it makes our example neat, because we can represent the TC curve with a quadratic equation: TC = aq2 + bq + c In this general function a, b and c are just constants – any 3 fixed numbers. The average cost (AC) curve represents the average cost of producing q boxes of chocolate (Easy isn’t it?). How do you work out the average? By taking the total and dividing it by how many you have: AC = (TC/q) = aq + b + c/q You can see that the AC curve is sort of ‘U’ shaped – it starts off high, then goes down until it hits a minimum point, and finally it goes up and keeps going up. The marginal cost (MC) curve represents the marginal cost of q – which is the cost of producing each extra box of chocolates. Remember when I said the marginal cost is increasing because the slope of the TC curve is increasing? Well this is because the MC curve is just the slope of the TC curve. We work this out by differentiating the TC function: MC = (dTC/dq) = 2aq + b This is the equation of a straight line, which explains the graph above. This tells us, for example, that the 3rd box of chocolates is more expensive for you to produce than the 2nd, which is more expensive than the 1st, and so on. ##### The thing we want to prove Just from looking at the curves, you might notice that the MC curve crosses the AC curve at exactly its minimum point. That is no coincidence my friend, and we are going to prove that it is always true. I’ll explain why this fact should be true intuitively. When the average cost is falling, it means that the cost of producing each extra good must be less than the average (otherwise it wouldn’t go down). Conversely, when the average is rising, the cost of each extra good must be greater than the average. Only when the cost of the next good is exactly equal to the average, will the average remain constant. And when the average remains constant, it is at the stationary point of the graph (its lowest point in the ‘U’). To simplify this, take a list of numbers 1,2,3. The average is 2 because (1+2+3)/3 = 6/3 = 2. If the next number we added to the list was 6 (higher than the average) then the average would rise to 12/4 = 3. If instead we added a 1 to the list, the average would fall to 7/4 = 1.75. But if we added a 2 to the list (which is exactly the average), then the new average would be the same as the number we added – 8/4 = 2. ##### The Proof That explanation was nice, but we need to do some maths to show that the MC always crosses that AC at its lowest point. First, we find the lowest point of the AC curve, by differentiating AC (denoted by AC’) and setting it equal to 0: AC’ = a – c/q2 = 0 so a = c/q2 which means that q = √(c/a) at the lowest point of AC. Now we need to show that the point where AC = MC has the same value of q as we have found above. Let AC = MC. Then: aq + b + c/q = 2aq + b aq = c/q q2 = c/a q = √(c/a) And as if by magic, the point where AC and MC cross is the same as the lowest point of AC. Hence this proves our statement. (As a disclaimer, this only strictly proves that the statement is true for quadratic cost functions. But the statement is true for all functions, and this can be proved pretty easily. However, the notation becomes a bit advanced and abstract, which is why I’ve only shown the proof for the neat quadratic case.)
#### Ads Blocker Detected!!! We have detected that you are using extensions to block ads. Please support us by disabling these ads blocker. # Mathematics Class 7 Chapter 3 – Data Handling – Exercise 3.4 – NCERT Exercise Solution 1. Tell whether the following is certain to happen, impossible, can happen but not certain. (i) You are older today than yesterday. Solution: Chance: It is certain to happen. (ii) A tossed coin will land heads up. Solution: Chance: It can happen but not certain. (iii) A die when tossed shall land up with 8 on top. Solution: Chance: It is impossible because, there are only six faces on a die that is marked as 1, 2, 3, 4, 5, and 6 on it. (iv) The next traffic light seen will be green. Solution: Chance: It can happen but not certain. (v) Tomorrow will be a cloudy day. Solution: Chance: It can happen but not certain. 1. There are 6 marbles in a box with numbers from 1 to 6 marked on each of them. (i) What is the probability of drawing a marble with number 2? Solution: Given, Total number of marbles marked with the number from 1 to 6 = 6 Now, Probability of drawing a marble with number 2 = Number of favourable outcomes/ Number of possible outcomes = (1/6) Hence, the required probability is (1/6) (ii) What is the probability of drawing a marble with number 5? Solution: Given, There are 6 marbles in the box with numbers from 1 to 6 are marked. Now, number or marble marked with 5 = 1 Now, Probability of drawing a marble with number 5 = Number of favourable outcomes/ Number of possible outcomes = (1/6) Hence, the required probability is (1/6) 1. A coin is flipped to decide which team starts the game. What is the probability that your team will start? Solution: We know that a coin has two faces one is Head (H) and another one is Tail (T). Now, one team can choose either Head or Tail. So, Sample space (Sn) = 2 The probability of our team starts first= Number of favourable outcomes/ Number of possible outcomes = ½ Hence, the required probability is (1/2) 👍👍👍 error:
# 3.4 \| Exponential and Logarithmic Equations To solve an exponential equation of the form $$a b^{k x -c} + d = 0$$ • Isolate the exponential. • Take logarithm base $$b$$ of both sides. If you obtain an expression involving taking the logarithm of a negative number, then no solution exists. Otherwise continue to the next step. • Use the inverse property to obtain a linear equation. • Now solve for $$x$$. Strategy for Solving Exponential Equations Given an exponential equation, rewrite the equation to obtain equations of the type $$a b^{kx-c} +d = 0.$$ This may require factoring, combining fractions, etc… Find the exact solution to $$68e^{4x+1}+14 = 16.$$ • Isolate the exponential. $$e^{4x+1} = \frac{1}{34}$$ • Take natural logarithm of both sides and apply the inverse property for logarithms. \begin{aligned} \ln e^{4x+1} &= \ln(1/34) \\ 4x+1 & = \ln(1/34) \end{aligned} • Solve for $$x$$. $$x = \frac{\ln(1/34) -1}{4}$$ Solve $$\left( 4 – \frac{2.471}{40} \right)^{9t} = 21.$$ \begin{aligned} \log \left( 4 – \frac{2.471}{40} \right)^{9t} & = \log 21 \\ 9t \log \left( 4 – \frac{2.471}{40} \right) & = \log 21 \\ t & = \frac{\log 21}{9 \log \left( 4 – \frac{2.471}{40} \right)} \\ t & \approx 0.246788 \end{aligned} Solve $$e^{2x} -2 e^{x} – 3 = 0$$ • Factor. $$(e^x +1) (e^x-3) = 0$$ • Set each factor equal to zero. $$e^x +1 = 0 \Rightarrow x = \ln(-1)$$ or $$e^x – 3 = 0 \Rightarrow x = \ln 3$$ Thus, the only solution to the given equation is $$x = \ln 3$$ since $$\ln(-1)$$ is undefined. To solve an logarithmic equation of the form $$a \log_b (k x -c) + d = 0$$ • Consolidate and Isolate the logarithms. • Exponentiate both sides base $$b$$. • Use the inverse property to obtain a linear equation. • Now solve for $$x$$. Strategy for Solving Logarithmic Equations Given a logarithmic equation, rewrite the equation to obtain equations of the type $$a \log_b (k x -c) + d = 0.$$ This may require factoring, combining fractions, etc… Find the exact solution to $$-2 +2 \ln(3x) = 17.$$ • Isolate the logarithm. $$\ln(3x) = \frac{19}{2}$$ • Exponentiate both sides base $$e$$ and apply the inverse property. \begin{aligned} e^{\ln(3x)} &= e^{19/2} \\ 3x &= e^{19/2} \end{aligned} • Solve for $$x$$. $$x = \frac{1}{3} e^{19/2}$$ Solve $$\log_3(x+5) = \log_3(x-1) – \log_3(x+1)$$ • Consolidate and Isolate the logarithm. $$\log_3\left( \frac{x+5}{x-1} \cdot (x+1) \right) = 0$$ • Exponentiate both sides base $$3$$ and apply the inverse property. \begin{aligned} 3^{\log_3\left( \frac{x+5}{x-1} \cdot (x+1) \right)} &= 3^0 \\ \frac{x+5}{x-1} \cdot (x+1) &= 1 \end{aligned} • Solve for $$x$$. \begin{aligned} (x+5)(x+1) & = x-1 \\ x^2 +6x+5 & = x-1 \\ x^2 +5x +6 & = 0 \\ (x +2)(x+3) & = 0 \end{aligned} So, $$x = -2 \text{ or } x =-3$$ • But $$x = -2,$$ and $$x = -3$$ both make the original equation undefined. Therefore, the given equation has no solutions.

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