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# Chapter 13 - Functions of Several Variables - 13.5 Exercises - Page 913: 14 $\partial w/\partial s=-6e^{2s}e^t, \partial w/\partial t=3e^t(e^{2t}-e^{2s})$ For $s=-1$ and $t=2$: $\partial w/\partial s=-6$ and $\partial w/\partial t=3(e^6-1)$ #### Work Step by Step We will find both partial derivatives using the Chain Rule. The partial derivative with respect to $s$ is: $$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial}{\partial x}(y^3-3x^2y)\frac{\partial}{\partial s}(e^s)+\frac{\partial}{\partial y}(y^3-3x^2y)\frac{\partial}{\partial s}(e^t)= -3\cdot2xy\cdot e^s+(3y^2-3x^2)\cdot0=-6xye^s$$ Expressing this in terms of $s$ and $t$ we have: $$\frac{\partial w}{\partial s}=-6xye^s=-6e^{2s}e^t$$ The partial derivative with respect to $t$ is: $$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial}{\partial x}(y^3-3x^2y)\frac{\partial}{\partial t}(e^s)+\frac{\partial}{\partial y}(y^3-3x^2y)\frac{\partial}{\partial t}(e^t)= -3\cdot2xy\cdot0+(3y^2-3x^2)\cdot e^t=3e^t(y^2-x^2)$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial w}{\partial t}=3e^t(y^2-x^2)=3e^t(e^{2t}-e^{2s})$$ Now we will evaluate these partial derivatives at the given values $s=-1$ and $t=2$: $$\frac{\partial w}{\partial s}=-6e^{2s}e^t=-6e^{-1\cdot2}e^2=-6$$ $$\frac{\partial w}{\partial t}=3e^t(e^{2t}-e^{2s})=3e^2(e^{2\cdot2}-e^{2\cdot(-1)})=3e^2(e^4-e^{-2})=3e^6-3\cdot1=3(e^6-1)$$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
# Permutations and Combinations How many ways are there to find the number of permutations of the elements in $\;\{1,2, 3, 4, 5\}\;$ such that it has one cycle of length $2$ and one cycle of length $3$? Please be as descriptive as possible. - There are $\binom{5}{3}$ ways to choose three elements for a cycle of length 3, then $2! = 2$ ways that each of triplets of elements can be permuted, e.g., from the triplet $\{1, 2, 3)\},$ two distinct 3-cycles can be formed $(123)$ and $(132)$, each permutation different from the other. So far, that gives us $\binom 53 \cdot 2$ ways a three cycle can be formed. Then, for the remaining $2$ elements....How many ways are are there to choose $2$ elements from $2$ elements? Only one way: and this is the remaining 2-cycle, whose permutation is itself. That gives us $$\binom{5}{3} \cdot 2 = \binom{5}{2}\cdot 2 = \dfrac{5!}{3!2!}\cdot 2=\frac{5 \cdot 4}{2} \cdot 2 = 20$$ total ways to form a permutation of 5 elements into a disjoint product of a three-cycle and a two-cycle. Another popular way of counting this is as $$\frac{5 \cdot 4}{2} \cdot \frac{3 \cdot 2 \cdot 1}{3}.$$ Here you first choose the two elements of the 2-cycle (in order) in $5 \cdot 4$ ways, then divide by 2 as $(a b) = (b a)$. Then you choose the (remaining) three elements for the 3-cycle (in order) in $3 \cdot 2 \cdot 1$ ways, and divide by 3, as $(abc) = (bca) = (cab)$.
# Frank Solutions for Class 10 Maths Chapter 16 Loci Frank Solutions for Class 10 Maths Chapter 16 Loci are given here. The BYJUâ€™S subject expert team has prepared solutions to help students obtain good marks in Maths. From the exam point of view, the solutions are solved in a simple manner, and students can secure an excellent score by solving Frank Solutions for Class 10 Maths Chapter 16. Solutions that are provided here will help you in getting acquainted with a wide variety of questions and thereby develop problem-solving skills. Chapter 16 – Loci are a set of points with the same property. Loci can be used to construct lines and shapes accurately. These solutions are helpful for the students to cross-check with the exercise answers in the chapter to improve their solving methods. ## Download the PDF of Frank Solutions for Class 10 Maths Chapter 16 Loci ### Access Answers to Frank Solutions for Class 10 Maths Chapter 16 Loci 1. Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement. Solution:- Steps to construction: (1) Draw a line segment AB of 9 cm (2) Then draw perpendicular bisector PQ of AB. PQ is the required locus. Proof: (a) Take any point on PQ i.e. C. (b) Now, join CA and CB. Since, C lies on the perpendicular bisector of line AB. So, C is equidistant from A and B. Therefore, CA = CB Hence, it is proved that perpendicular bisector of AB is the locus of all points which are equidistant from A and B. 2. A and B are fixed points while P is a moving point, moving in a way that it is always equidistant from A and B. What is the locus of the path traced out by the point P? Solution:- The locus of the path traced out by the point P is equidistant from A and B is the perpendicular bisector of the line segment joining the two points. 3. A point moves such that its distance from a fixed line AB is always the same. What is the relation between AB and the path travelled by the point? Solution:- A point moves such that its distance from a fixed straight line AB is the same as a pair of straight lines parallel to the given line, one on each side of it and at the given distance from it. 4. State the locus of a point moving so that its perpendicular distances from two given lines is always equal. Solution:- The locus of a point moving so that its perpendicular distances from two given lines is always equal. A line QS parallel to given lines R and P. 5. AB is fixed line, state the locus of the point P such that âˆ APB = 90o. Solution:- The locus of the point P, such that âˆ APB = 90o, with AB as diameter. 6. Draw and describe the locus in each of the following cases: (a) The locus of points at a distance of 4 cm from a fixed line. Solution:- The locus of points at a distance of 4 cm from a fixed line QS are lines R and P, which are parallel to QS. (b) The locus of points inside a circle and equidistant from two fixed points on the circle. Solution:- The locus of points inside a circle are equidistant from fixed points on the circle will be the diameter which is the perpendicular bisector of the line joining the tow fixed points on the circle. (c) The locus of the mid â€“ points of all parallel chords of a circle. Solution:- The locus of the mid â€“ points of the chords, which are parallel to a given chord, is the diameter perpendicular to the given chords. (d) The locus of a point in rhombus ABCD, which is equidistant from AB and AD. Solution:- 7. Describe completely the locus of points in each of the following cases: (a) Midpoint of radii of a circle. Solution:- (b) Centre of a ball, rolling along a straight line on a level floor. Solution:- (c) Point in a plane equidistant from a given line. Solution:- (d) Centre of a circle of varying radius and touching the two arms of âˆ ABC. Solution:- (e) Centre of a circle of radius 2 cm and touching a fixed circle of radius 3 cm. Solution:- 8. Using only ruler and compasses, construct a triangle ABC, with AB = 5 cm, BC = 3.5 cm and AC = 4 cm. Mark a point P, which is equidistant from AB, BC and also from B and C. Measure the length of PB. Solution:- Steps of construction: 1. Draw a line segment BC = 3.5 cm. 2. With B as a centre and radius 5 cm, draw an arc. 3. With C as a centre and radius 4 cm, draw another arc, cutting the previous arc at A. 4. Join AB and AC. Then, Î”ABC is the required triangle. 5. With B as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 6. With C as centre and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs at P and Q, as shown. Join PQ. Then, PQ is the required perpendicular bisector of BC. Therefore, P is the required point which is equidistant from AB, BC, B and C. Length PB is 2.5 cm 9. Construct a triangle ABC, such that AB = 6 cm, BC = 7.3 cm and CA = 5.2 cm, locate a point which is equidistant from A, B and C. Solution:- Steps of construction: 1. Draw a line segment BC = 7.3 cm. 2. With B as a centre and radius 6 cm, draw an arc. 3. With C as a centre and radius 5.2 cm, draw another arc, cutting the previous arc at A. 4. Join AB and AC. Then, Î”ABC is the required triangle. 5. With B as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 6. With C as centre and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs as shown. Join them. 7. Again with B as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 8. With A as centre and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs as shown. Join them. 9. With C as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 10. With A as centre and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs as shown. Join them So, in triangle ABC, P is the point of intersection of AB, AC and BC. Therefore, PA =PB, PB = PC, PC = PA Hence, P is the point which is equidistant from A, B and C. 10. Two straight roads AB and CD, cross each other at P at an angle of 75o. X is a stone on the road AB, 800m from P towards B, By taking an appropriate scale, draw a figure to locate the position of a pole, which is equidistant from P and X, and is also equidistant from the roads. Solution:- Steps of construction: As per the condition given in the question, 1. Draw two lines AB and CD crossing at an angle of 75o. 2. Then draw an angle bisector of âˆ BPD 3. Now, draw a perpendicular from X on angle bisector meeting at O. 4. From point Y, PX = PY, draw a perpendicular on angle bisector meeting at O. 5. So, O is the point which equidistant from P, X and both the roads. We know that, cos Î¸ = hypotenuse/base cos (75/2) = PO/PX cos (37.5) = PO/800 0.980243 = PO/800 PO = 0.980243 Ã— 800 PO = 784.19 m 11. Draw two intersecting lines to include an angle of 30o. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such points exist? Solution:- As per the condition given in the question, 1. First, draw an angle bisector AB and XY of angles formed by the lines e and f. 2. Now, from centre draw a circle with radius 2 cm, which intersect the angle bisector at p, q, r and s, respectively. Therefore, p, q, r and s are the required four points. 12. AB and CD are two intersecting lines. Find a point equidistant from AB and CD, and also at a distance of 1.8 cm from another given Line EF. Solution:- As per the condition given in the question, 1. First draw an angle bisector AB and CD. 2. Then draw perpendicular from AB and CD on angle bisector i.e. P. Therefore, P is the point equidistant from AB and CD, and also at a distance of 1.8 cm from another given Line EF. 13. Without using set squares or protractor, construct a quadrilateral ABCD in which âˆ BAD = 45o, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm. Locate the point P on BD, which is equidistant from BC and CD. Solution:- Steps of construction: 1. Draw a line segment AB = 6 cm. 2. With B as a centre and radius 6 cm, draw an arc. 3. With C as a centre and radius 5.2 cm, draw another arc, cutting the previous arc at A. 4. Join AB and AC. Then, Î”ABC is the required triangle. 5. With B as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 6. With C as centre and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs as shown. Join them. 7. Again with B as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 8. With A as centre and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs as shown. Join them. 9. With C as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 14. Construct a rhombus ABCD with sides of length 5 cm and diagonal AC of Length 6 cm. Measure âˆ ABC. Find the point R on AD such that RB = RC. Measure the length of AR. Solution:- Steps of construction: 1. Draw a line segment AC = 6 cm. 2. With A as a centre and radius 5 cm, draw two arcs on both sides of line AC. 3. With C as a centre and radius 5 cm, draw another two arcs, cutting the previous arcs at B and D both side. 4. Join AB, AD, BC and CD. Then, ABCD is the required rhombus. 5. With B as centre and radius measuring more than half of BC, draw arcs on both sides of BC. 6. With C as centre and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs as shown. Join them. 7. Then, PQ is the required perpendicular bisector of BC. Therefore, P is the required point which is equidistant from AB, BC, B and C. 15. Construct a rhombus ABCD whose diagonals AC and BD are 8 cm and 6 cm, respectively. Find by construct a point P equidistant from AB and AD and also from C and D. Solution:- Steps of construction: 1. Draw a line segment BC = 6 cm. 2. Then, draw perpendicular AD = 8 cm to BC. 3. With A as a centre and radius 5 cm, draw two arcs on both sides of line AC. 4. With C as a centre and radius 5 cm, draw another two arcs, cutting the previous arcs at B and D both side. 5. Join AB, AD, BC and CD.
# TRANSFORMATIONS AND CONGRUENCE ## About "Transformations and congruence" Transformations and congruence : To translate or reflect or rotate a figure in the coordinate plane, we have to transform each of its vertices. Then, we have to connect the vertices to form the image. When we apply the above mentioned transformations (Translation, Reflection and Rotation), original figure and the image after transformation would have the same size, , just a different orientation. So, the original figure and the image after translation would be congruent. ## Transformations and congruence - Example A triangle has the vertices (3, 4), (5, 4) and (5, 2). Apply the indicated series of transformations to the triangle. Each transformation is applied to the image of the previous transformation, not the original figure. Label each image with the letter of the transformation applied. (i) Reflect across the x-axis. (ii) Translate 3 units to the left. (iii) Reflect across the y-axis. (iv) Translate 4 units up. (v) Rotate of 90 ° clockwise about the origin. Compare the size and shape of the final image to that of the original figure. Solution : Step 1 : (i) Reflect across the x-axis. Since there is a reflection across the x-axis, we have to multiply each y-coordinate by -1. That is, (x, y) -----> (x, -y) So, we have (3, 4) -----> (3, -4) (5, 4) -----> (5, -4) (5, 2) -----> (5, -2) Graph the image. Step 2 : (ii) Translate 3 units to the left. Since there is a translation of 3 units to the left, we have to subtract 3 from each x-coordinate. That is, (x, y) -----> (x-3, y) So, we have (3, -4) -----> (0, -4) (5, -4) -----> (2, -4) (5, -2) -----> (2, -2) Graph the image. Step 3 : (iii) Reflect across the y-axis. Since there is a reflection across the y-axis, we have to multiply each x-coordinate by -1. That is, (x, y) -----> (-x, y) So, we have (0, -4) -----> (0, -4) (2, -4) -----> (-2, -4) (2, -2) -----> (-2, -2) Graph the image. Step 4 : (iv) Translate 4 units up. Since there is a translation of 4 units up, we have to add 4 to each y-coordinate. That is, (x, y) -----> (x, y+4) So, we have (0, -4) -----> (0, 0) (-2, -4) -----> (-2, 0) (-2, -2) -----> (-2, 2) Graph the image. Step 5 : (v) Rotate of 90 ° clockwise about the origin. Since there is a rotation of 90° clockwise about the origin, we have multiply each x-coordinate by -1 and interchange x and y coordinates. That is, (x, y) -----> (y, -x) So, we have (0, 0) -----> (0, 0) (-2, 0) -----> (0, 2) (-2, 2) -----> (2, 2) Graph the image. Compare the size and shape of the final image to that of the original figure. They have the same size and shape, just a different orientation. ## Reflect 1. Which transformations change the orientation of figures ? Reflections and rotations 2. Which transformations do not change the orientation of figures ? Translations 3. Two figures have the same size and shape. What does this indicate about the figures ? One figure is the image of the other, and there is a sequence of transformations that will transform one figure into the other. After having gone through the stuff given above, we hope that the students would have understood "Transformations and congruence" Apart from the stuff given above, if you want to know more about "Transformations and congruence", please click here Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. WORD PROBLEMS HCF and LCM word problems Word problems on simple equations Word problems on linear equations Word problems on quadratic equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6
+0 # area +4 21 1 +180 Two points, A and B, are on the pavement 6 inches apart. Cassidy the caterpillar traces out a path walking along all the points that are twice as far from A as from B. Cassidy has red paint on her feet, so she literally traces out the whole path! What is the area of the region inside Cassidy's path? Nov 18, 2023 #1 +1747 -1 Let C be a point on Cassidy's path. Because C is twice as far from B as it is from A, it must lie on the circle with center B and radius 3 inches. Similarly, it must lie on the circle with center A and radius 1.5 inches. Thus, Cassidy's path traces out the region between two circles. To find the area of this region, we first consider the intersection of the two circles. Let D and E be the points of intersection. From the similarity of triangles ABE and ACB, we have [\frac{BE}{AB} = \frac{BC}{AC}.] From the Pythagorean theorem, we get BC=3−1.5=1.5, so BE=2AB=3. From the Pythagorean theorem again, we get AC=4.52−1.52=3, so [\frac{BE}{AB} = \frac{BC}{AC} = \frac{3}{6} = \frac12.] This shows that △BDE is a 30-60-90 triangle, so BD=21BE=1.5 inches and DE=3⋅BD=1.53 inches. Let the area of the region between the two circles be X. The area of the region inside Cassidy's path is then [X - \frac12 \pi (1.5)^2 - \frac12 \pi (3)^2.] We can divide the region between the two circles into △ABE and four identical regions like △BDE. The area of △ABE is 41⋅6⋅3=4.5 square inches. The area of one of the four regions like △BDE is 21⋅1.5⋅1.53=433 square inches. Therefore, [X = 4.5 + 4 \cdot \frac{3\sqrt{3}}{4} = 6 + 3\sqrt{3}.] Finally, the area of the region inside Cassidy's path is [6 + 3\sqrt{3} - \frac12 \pi (1.5)^2 - \frac12 \pi (3)^2 = 6 + 3\sqrt{3} - \frac{9}{2}\pi = \boxed{\frac{1}{2} \pi + 3 \sqrt{3}}.] The final answer is 1/2pi + 3sqrt(3). Nov 18, 2023
# The Unapologetic Mathematician ## The Exponential Differential Equation So we long ago defined the exponential function $\exp$ to be the inverse of the logarithm, and we showed that it satisfied the exponential property. Now we’ve got another definition, using a power series, which is its Taylor series at ${0}$. And we’ve shown that this definition also satisfies the exponential property. But what really makes the exponential function what it is? It’s the fact that the larger the function’s value gets, the faster it grows. That is, the exponential function satisfies the equation $f(x)=f'(x)$. We already knew this about $\exp$, but there we ultimately had to use the fact that we defined the logarithm $\ln$ to have a specified derivative. Here we use this property itself as a definition. This is our first “differential equation”, which relates a function to its derivative(s). And because differentiation works so nicely for power series, we can use them to solve differential equations. So let’s take our equation as a case in point. First off, any function $f$ that satisfies this equation must by definition be differentiable. And then, since it’s equal to its own derivative, this derivative must itself be differentiable, and so on. So at the very least our function must be infinitely differentiable. Let’s go one step further and just assume that it’s analytic at ${0}$. Since it’s analytic, we can expand it as a power series. So we have some function defined by a power series around ${0}$: $\displaystyle f(x)=\sum\limits_{k=0}^\infty a_kx^k$ We can easily take the derivative $\displaystyle f'(x)=\sum\limits_{k=0}^\infty (k+1)a_{k+1}x^k$ Setting these two power series equal, we find that $a_0=1a_1$, $a_1=2a_2$, $a_2=3a_3$, and so on. In general: $\displaystyle a_k=\frac{1}{k}a_{k-1}=\frac{1}{k}\frac{1}{k-1}a_{k-2}=...=\frac{1}{k}\frac{1}{k-1}...\frac{1}{3}\frac{1}{2}\frac{1}{1}a_0=\frac{a_0}{k!}$ And we have no restriction on $a_0$. Thus we come up with our series solution $\displaystyle f(x)=\sum\limits_{k=0}^\infty\frac{a_0}{k!}x^k=a_0\sum\limits_{k=0}^\infty\frac{x^k}{k!}$ which is just $a_0=f(0)$ times the series definition of our exponential function $\exp$! If we set the initial value $f(0)=a_0=1$, then the unique solution to our equation is the function $\displaystyle\exp(x)=\sum\limits_{k=0}^\infty\frac{x^k}{k!}$ which is our new definition of the exponential function. The differential equation motivates the series, and the series gives us everything else we need. October 10, 2008 - 1. [...] Homogeneous Linear Equations with Constant Coefficients Now that we solved one differential equation, let’s try a wider class: “first-degree homogeneous linear equations with constant [...] Pingback by First-Degree Homogeneous Linear Equations with Constant Coefficients « The Unapologetic Mathematician \| October 10, 2008 \| Reply 2. [...] we can, and we’ll use the same techniques as we did before. We again find that any solution must be infinitely differentiable, and so we will assume that [...] Pingback by Sine and Cosine « The Unapologetic Mathematician \| October 13, 2008 \| Reply 3. Now, I try to verify your statement … “That is, the exponential function satisfies the equation f(x)=f’(x). We already knew this about exp, but there we ultimately had to use the fact that we defined the logarithm ln to have a specified derivative” to prove that: d(e^x)/dx = e^x, (1) Let differentiate the both sides of eq.(1) respect to x, and suppose d(e^x)/dx = u, hence the result can be written as following du/dx = u, (2) Next, separating variables gives du/u = dx, (3) and integrating the both sides of eq.(3) gives ln(u) = x. Here I put the integrating constant c=0, because at x=1, ln(1)=0. From ln(u) = x , I find u = e^x. After u is written into initial form, then I can prove that d(e^x)/dx = e^x.
# Tag Archives: algorithm ## Algorithms, continued Frank Noschese (whose TEDxNYED talk is totally worth your time, by the way) gave me +K for algorithms on Klout the other day. This was presumably in return for the +K he received from me for textbooks a few days prior. To celebrate this achievement (and to demonstrate that it is well deserved), I want to share the common numerator division algorithm. Here’s how it goes. $\frac{a}{b}\div\frac{a}{c}=\frac{c}{b}$ In words, if two fractions have a common numerator, then their quotient is a fraction whose numerator is the denominator of the divisor and whose denominator is the denominator of the dividend. As an example, $\frac{2}{3}\div\frac{2}{6}=\frac{6}{3}=2$ Two thirds compared to two sixths Or, if you start with different numerators, $\frac{2}{3}\div\frac{1}{4}=\frac{2}{3}\div\frac{2}{8}=\frac{8}{3}$ Two thirds compared to one fourth Two thirds compared to two eighths Now, unlike the common numerator algorithm for adding fractions, I can actually describe what’s going on here. Think about a fraction as a part-whole relationship. When we have two fractions with common numerators but different denominators (and assuming each is a fraction of the same-sized whole), then each fraction represents the same number of pieces, but these pieces are different sizes. The quotient, then, is the ratio of the sizes of these pieces. In the first example, thirds are twice as big as sixths. In the second example, thirds are two-and-two-thirds times as big as eighths. The common denominator algorithm for dividing fractions emphasizes a measurement interpretation of division (the quotient answers the question, “How many of this are in that?”) The invert-and-multiply algorithm emphasizes formal numerical/algebraic relationships (the quotient is obtained by using the properties of multiplicative inverses). The common numerator algorithm, though? It emphasizes that fractions and quotients are ratios. The quotient is a multiplicative comparison of the sizes of the pieces. ## This is entirely predictable (Algorithms edition) These errors are usually considered careless errors. They are not careless errors; they come from an inability to think. Oh, and that other one? Algorithms unteach place value Yeah. They’re strong claims. Overly strong. “An inability to think” isn’t really right. “Not thinking in this circumstance” is closer to the truth. Kamii is not a woman of nuance. That’s part of what I enjoy about her work. But it does require interpretation. All of which builds up to an entirely predictable scenario yesterday morning. Griffin (who is seven years old, recall, and whose ability to think has been well documented here) has been well schooled in traditional addition and subtraction algorithms this year, and has learned the lattice algorithm. He loves the lattice, although he struggles to make a lattice neatly enough to do the algorithm (for reasons that will become apparent when you see his handwriting below). All of these are digit-by-digit, ignore-place-value-as-you-work sorts of things. Griffin asked at breakfast, Is 86 divided by 22, 43? and wrote the following: Kamii saw it coming… ## Showing v. Telling: Lattice Continued That last post wasn’t really about the lattice algorithm. This one isn’t either. It’s about supporting claims with evidence. (The last one, in case you’re keeping score, was about crafting tasks that show instead of spiels that tell). Another claim my future elementary teachers like to make is that the lattice is inefficient for problems with large numbers of digits. The idea is that you have to spend a lot of time making that lattice before you can multiply. This stands in contrast to the standard algorithm, which you can just get started with straight away. They make the claim, but they don’t back it up with evidence. (This, after all, is part of what a college education is supposed to teach-how to build arguments-so I’m not complaining that they don’t. I understand that it is my job to teach them to do so.) So I began to wonder whether the lattice-drawing really did set one back. So I put it to the test. Ten digits by ten digits. On your mark, get set, GO! By the way, I was going to do three algorithms head-to-head. I was going to do partial products in the middle of the board. But about a third of the way through, I got fed up and quit. That one really is inefficient for large numbers of digits. ## Place value and the Lattice Algorithm (or A Simple Task, Years in the Making) I’m going to assume you know the lattice algorithm for multidigit multiplication. If you do not, and if you would like a primer, here is one. This post isn’t really about the lattice algorithm, but it’s the context for what I’m really trying to say, which is this: It is worth the time to craft classroom tasks carefully. I have used the lattice algorithm for years with my future elementary teachers. We learn the steps in class, they go off and practice it. And then they write about it, using the ideas of the course to analyze the algorithm. After a number of semesters of this, I became tired of reading in their work some variant of the following claim, The lattice algorithm is very good for teaching place value because you have to pay attention to the places as you work with it. I could not disagree with this claim more strongly. As I work the lattice, I am going digit-by-digit. I am absolutely NOT thinking about the values of those digits. And I suspect most children are not either. This makes it an efficient algorithm. Last semester I decided to put that claim to the test. If these future teachers thought the lattice algorithm exposes important ideas of place value, then what task could I give them to demonstrate that it does not? Well, they have been analyzing the algorithm; they have written papers about it. So if it teaches place value, they should be able to ace any place value task involving the lattice, right? So here’s the task: Perform the lattice algorithm to multiply 7,343 by 1,568. When you are done, use a marker to highlight each and every tens digit in the lattice. No follow-up or clarification questions allowed. If the premise is that the lattice helps us to learn place value, then we should know enough about place value to make a commitment to the meaning of a tens digit. Can you guess which of the answers below is the more popular in my classroom? When both are presented, a really useful discussion of the algorithm and its position with respect to place value ensues. And that discussion helps to explain the really clever “slide trick” for placing the decimal point (as seen about 2:30 into this video). But back to my point. I can tell my students that the lattice doesn’t bring place value understanding along for free. Or I can show them. Showing requires carefully crafted tasks. But I find it’s worth the time. When I have the choice between telling and showing, I nearly always choose to show. Which is why I’m always running behind on content coverage. I made my peace with that years ago.
# Reciprocal of a Rational Number We will learn the reciprocal of a rational number. For every non-zero rational number a/b there exists a rational number b/a such that a/b × b/a = 1 = b/a × a/b The rational number b/a is called the multiplicative inverse or reciprocal of a/b and is denoted by (a/b)-1. The reciprocal of 12 is 1/12 The reciprocal of 5/16 is 16/5. The reciprocal of 3/4 is 4/3 i.e., (3/4)^-1 = 4/3. The reciprocal of -5/12 is 12/-5 i.e., (-5/12)^-1 = 12/-5. The reciprocal of (-14)/17 is 17/-14 i.e., (-17)/14. The reciprocal of -8 is 1/-8 i.e., (-1)/8. The reciprocal of -5 is 1/-5, since -5 × 1/-5 = -5/1 × 1/-5 = -5 × 1/-5 × 1 = 1. Note: The reciprocal of 1 is 1 and the reciprocal of -1 is -1. 1 and -1 are the only rational numbers which are their own reciprocals. No other rational number is its own reciprocal. We know that there is no rational number which when multiplied with 0, gives 1. Therefore, rational number 0 has no reciprocal or multiplicative inverse. Solved example on reciprocal of a rational number: 1. Write the reciprocal of each of the following rational numbers: (i) 5 (ii) -15 (iii) 7/8 (iv) -9/13 (v) 11/-19 Solution: (i) The reciprocal of 5 is 1/5 i.e., (5)^-1 = 1/5. (ii) The reciprocal of -15 is 1/-15 i.e., (-15)^-1 = 1/-15. (iii) The reciprocal of 7/8 is 8/7 i.e., (7/8)^-1 = 8/7. (iv) The reciprocal of -9/13 is 13/-9 i.e., (-9/13)^-1 = 13/-9. (v) The reciprocal of 11/-19 is -19/11 i.e., (11/-19)^-1 = -19/11. 2. Find the reciprocal of 3/7 × 2/11. Solution: 3/7 × 2/11 = (3 × 2)/(7 × 11) = 6/77 Therefore, the reciprocal of 3/7 × 2/11 = Reciprocal of 6/77 = 77/6. 3. Find the reciprocal of -4/5 × 6/-7. Solution: -4/5 × 6/-7 = (-4 × 6)/(5 × -7) = -24/-35 = 24/35 Therefore, the reciprocal of -4/5 × 6/-7 = Reciprocal of 24/35 = 35/24. ` Rational Numbers What is Rational Numbers? Is Every Rational Number a Natural Number? Is Zero a Rational Number? Is Every Rational Number an Integer? Is Every Rational Number a Fraction? Positive Rational Number Negative Rational Number Equivalent Rational Numbers Equivalent form of Rational Numbers Rational Number in Different Forms Properties of Rational Numbers Lowest form of a Rational Number Standard form of a Rational Number Equality of Rational Numbers using Standard Form Equality of Rational Numbers with Common Denominator Equality of Rational Numbers using Cross Multiplication Comparison of Rational Numbers Rational Numbers in Ascending Order Representation of Rational Numbers on the Number Line Addition of Rational Number with Same Denominator Addition of Rational Number with Different Denominator Properties of Addition of Rational Numbers Subtraction of Rational Number with Different Denominator Subtraction of Rational Numbers Properties of Subtraction of Rational Numbers Simplify Rational Expressions Involving the Sum or Difference Multiplication of Rational Numbers Properties of Multiplication of Rational Numbers Rational Expressions Involving Addition, Subtraction and Multiplication Division of Rational Numbers Rational Expressions Involving Division Properties of Division of Rational Numbers To Find Rational Numbers
# Exact Equations Differential equation M(x,y)dx+N(x,y)dy=0 is exact if there exist function f such that df=M(x,y)dx+N(x,y)dy . In this case equation can be rewritten as df=0 which gives solution f=C. Test for exactness: if M(x,y) and N(x,y) are continuous functions and have continuous first partial derivatives on some rectangle of the xy-plane, then differential equation is exact if and only if (partial M(x,y))/(partial y)=(partial N(x,y))/(partial x). To solve this equation we use the facts that (partial f(x,y))/(partial x)=M(x,y) and (partial f(x,y))/(partial y)=N(x,y). We integrate first equation with respect to x to obtain f(x,y) through x and unknown function g(y). We then differentiate result with respect to y and use second equation. After this we find g(y) and thus f(x,y). Solution, as already stated above, is given as f(x,y)=C . Example 1. Solve (x+sin(y))dx+(x cos(y)-2y)dy=0 Here M(x,y)=x+sin(y) , N(x,y)=xcos(y)-2y . Since (partial M)/(partial y)=cos(y) and (partial N)/(partial x)=cos(y) then (partial M)/(partial y)=(partial N)/(partial x) and differential equation is exact. Thus, there exist function f such that (partial f)/(partial x)=M(x,y)=x+sin(y) and (partial f)/(partial y)=N(x,y)=xcos(y)-2y . Integrate first equation with respect to x to obtain that f=int (x+sin(y))dx=1/2 x^2+xsin(y)+g(y) . Note that here constant of integration can depend on y (because we integrate with respect to x). Now diffetiate resulting equation with respect to y: (partial f)/(partial y)=0+xcos(y)+g'(y) . From another side (partial f)/(partial y)=N(x,y)=xcos(y)-2y So, xcos(y)+g'(y)=xcos(y)-2y , or g'(y)=-2y . Integrating it we obtain that g(y)=-y^2+c_1 . Thus, f=1/2 x^2+xsin(y)-y^2+c_1 . Solution is given as 1/2 x^2 +xsin(y)-y^2+c_1=C Or 1/2 x^2 +xsin(y)-y^2=c_2 where (c_2=C-c_1 ). We obtain implicit solution and there is no way to find explicit solution. Example 2. Solve y'=(2+ye^(xy))/(2y-xe^(xy)) . First rewrite it into differential form: (dy)/(dx)=(2+ye^(xy))/(2y-xe^(xy)) or (2+ye^(xy))dx+(xe^(xy)-2y)dy=0 . Here M(x,y)=2+ye^(xy) and N(x,y)=xe^(xy)-2y . Since (partial M)/(partial y)=e^(xy)+xye^(xy) and (partial N)/(partial x)=e^(xy)+xye^(xy) then (partial M)/(partial y)=(partial N)/(partial x) and differential equation is exact. So, exist function f such that (partial f)/(partial x)=M(x,y)=2+ye^(xy) and (partial f)/(partial y)=N(x,y)=xe^(xy)-2y . In previous example we took first equation and integrated it with respect to x. Actuall y we could take second equation and integrate it with respect to y. Final answer is the same. In this example we will take second equation and integrate it with respect to y: f=int (xe^(xy)-2y)dy=e^(xy)-y^2+g(x) . Note that constant of integration depends on x, since we integrate with respect to y. Now, differentiate resulting equation with respect to x: (partial f)/(partial x)=y e^(xy)+g'(x) From another side (partial f)/(partial x)=2+ye^(xy) , so ye^(xy)+g'(x)=2+ye^(xy) or g'(x)=2 . Integrating with respect to x gives g(x)=2x+c_1 . So, f=e^(xy)-y^2+g(x)=e^(xy)-y^2+2x+c_1 . And solution is e^(xy)-y^2+2x+c_1=C or e^(xy)-y^2+2x=c_2 where (c_2=C-c_1) . Sometimes, equation can be not exact, but it can be transformed into exact by multiplying equation by integrating factor. So, I(x,y) is integrating factor for differential equation is exact if I(x,y)(M(x,y)dx+N(x,y)dy)=0 is exact. There are 2 conditions that allow to find integrating factor easily: If 1/N((partial M)/(partial y)-(partial N)/(partial x))=g(x) , a function of x alone, then I(x,y)=e^(int g(x)dx) If 1/M((partial M)/(partial y)-(partial N)/(partial x))=h(y) , a function of x alone, then I(x,y)=e^(-int h(y)dy) Example 3. Solve (y+1)dx-xdy=0 Here M(x,y)=y+1 and N(x,y)=-x . Since (partial M)/(partial y)=1 and (partial N)/(partial x)=-1 then (partial M)/(partial y)!=(partial N)/(partial x) and equation is not exact. However, note that 1/N((partial M)/(partial y)-(partial N)/(partial x))=1/(-x)(1-(-1))=-2/x=g(x) . So, I(x,y)=e^(int -2/xdx)=e^(-2ln(x))=1/x^2 . Multiplying differential equation by integrating factor gives: (y+1)/x^2dx-1/xdy=0. This equation is exact, so exist function f such that df=(y+1)/x^2 dx-1/x dy . Using the fact that (partial f)/(partial x)=(y+1)/x^2, we find that f=int (y+1)/x^2 dx=-(y+1)/x+g(y) . Now, differentiating with respect to y gives: (partial f)/(partial y)=-1/x+g'(y) . From another side (partial f)/(partial y)=-1/x , so -1/x+g'(y)=-1/x or g'(y)=0 . Solving it we obtain that g(y)=C_1 . So, f(x,y)=-(y+1)/x+g(y)=-(y+1)/x+C_1 . Therefore, solution is -(y+1)/x+C_1=C or y=c_2 x-1 where c_2=C_1-C . Example 4. Solve 2xydx+y^2dy=0 Here M(x,y)=2xy and N(x,y)=y^2 . Since (partial M)/(partial y)=2x and (partial N)/(partial x)=0 then (partial M)/(partial y)!=(partial N)/(partial x) and differential equation is not exact. However, note that 1/M((partial M)/(partial y)=(partial N)/(partial x))=1/(2xy)(2x-0)=1/y=h(y) . So, integrating factor is I(x,y)=e^(-int 1/y dy)=e^(-ln(y))=1/y . Multiplying differential equation by integrating factor yields 1/y(2xydx+y^2dy)=0 or 2xdx+ydy=0 This equation is exact. Using the fact that (partial f)/(partial x)=2x we have that f=int (2x)dx=x^2+h(y) . Differentiating last eqaution with respect to y gives: (partial f)/(partial y)=h'(y) . From another side (partial f)/(partial y)=y , so h'(y)=y or h(y)=int ydy=1/2 y^2+c_1 So, f=x^2+h(y)=x^2+1/2 y^2+c_1 and solution is x^2+1/2 y^2+c_1=C or x^2+1/2 y^2=c_2 where (c_2=C-c_1 ) . Another case when there can be easily found integrating factor is case when M=yf(xy) and N=xg(xy) . In this case I(x,y)=1/(xM-yN) . Example 5. Solve y(1-xy)dx+xdy=0 . Equation is not exact, however, note that M(x,y) is in the form y(1-xy) and N(x,y)=x*1 ,so integrating factor is I(x,y)=1/(xy(1-xy)-xy)=-1/(xy)^2 . Multiplying by I(x,y) yields: (xy-1)/(x^2y)dx-1/(xy^2)dy=0 . This equation is exact. Using the fact that (partial f)/(partial x)=(xy-1)/(x^2y) we have that f=int ((xy-1)/(x^2 y))dx=ln(\|x\|)+1/(xy)+g(y) . Differentiating with respect to y gives: (partial f)/(partial y)=-1/(xy^2)+g'(y) . From another side (partial f)/(partial y)=-1/(xy^2) . So, g'(y)=0 or g(y)=c_1 . So, f(x,y)=ln(\|x\|)+1/(xy)+c_1 . Finally, solution is ln(\|x\|)+1/(xy)+c_1=C or ln(\|x\|)+1/(xy)=c_2 where c_2=C-c_1 . Below is the list of common integrating factors: Group of terms I(x,y) Exact differential dg(x,y) ydx-xdy -1/x^2 -(ydx-xdy)/x^2=d(y/x) ydx-xdy 1/y^2 (ydx-xdy)/y^2=d(x/y) ydx-xdy -1/(xy) -(ydx-xdy)/(xy)=d(ln(y/x)) ydx-xdy -1/(x^2+y^2) -(ydx-xdy)/(x^2+y^2)=d(arctan(y/x)) ydy+xdx 1/(xy) (ydy+xdx)/(xy)=d(ln(xy)) ydy+xdx 1/(xy)^n,n>1 (ydy+xdx)/(xy)^n=-d(1/((n-1)(xy)^(n-1))) ydy+xdx 1/(x^2+y^2) (ydy+xdx)/(x^2+y^2)=d(1/2 ln(x^2+y^2)) ydy+xdx 1/(x^2+y^2)^n, n>1 (ydy+xdx)/(x^2+y^2)^n=-d(1/(2(n-1)(x^2+y^2)^(n-1))) aydx+bxdy (a and b constants) x^(a-1)y^(b-1) x^(a-1)y^(b-1)(aydx+bxdy)=d(x^ay^b) In general, integrating factors are difficult to uncover. If a differential equation doesn't have a one of the forms given above, then a search for an integrating factor likely will not be successful, and other methods of solution should be used.
# Complex Numbers – Properties, Graph, and Examples You may have encountered numbers that are said to be imaginary. You may have been asked to disregard the square root of a negative number in the past, but we’ll focus on these types of complex numbers in this article. Along with the rest of the real numbers, these are part of a larger group of complex numbers. Complex numbers are numbers that may contain real and imaginary parts. • The parts that make up complex numbers. • Properties and operation rules for complex numbers. • What the form $a + bi$ represents in a complex plane. Since complex numbers encompass an extensive range of numbers and are crucial elements in higher math, we must familiarize ourselves with all of the concepts mentioned. Why don’t we begin with the most fundamental of them all – the definition of complex numbers? ## What is a complex number? A complex number is a number that belongs to either the imaginary or real number groups. In short, they encompass all numbers belonging to the two mentioned groups. The general form of a complex number is $z = a + bi$, where $i$ represents $\sqrt{-1}$. Let’s inspect the form that we’re given to understand the two crucial parts: real and imaginary. \begin{aligned} \text{Complex Number } &= {\color{red} \boldsymbol{a}} + {\color{blue} \boldsymbol{bi}}\\ {\color{red} \boldsymbol{a}} &: \text{real number part}\\ {\color{blue} \boldsymbol{bi}} &: \text{imaginary number part}\end{aligned} This chart can help you visualize the group of numbers that belong to the entire complex number system. • Real numbers such as $4$, $\sqrt{5}$, $\dfrac{4}{2}$, and $\pi$ all belong to the complex number group as well. • Imaginary numbers such as $\sqrt{-2}$, $-3 + 2i$, and $5 + 2\sqrt{-6}$ also belong under the complex number system group. As long as the number can be expressed in the form $a + bi$, it’s considered part of the complex number group. • For example, $4$ can be expressed as $4 + 0i$, showing that $4$ can be considered a complex number. • Similarly, $\dfrac{2}{3}$ can be rewritten as $\dfrac{2}{3} + 0i$. • Meanwhile, $4i$ can be written as $0 + 4i$, further confirming that imaginary numbers like $4i$ are part of the complex number group. Notice how $i$ is often used to represent complex numbers? Let’s go ahead and review what we know about $i$ – an essential component of imaginary numbers. Understanding the Imaginary Unit, $\boldsymbol{i}$ The number $i$ represents $\sqrt{-1}$ and consequently, $i^2 = -1$. We normally use $i$ to represent the square root of a negative number. This also means that we can rewrite $\sqrt{-36}$ in terms of $i$. • We can factor out $-1$ inside the square root and evaluate $\sqrt{36}$. • Rewrite $\sqrt{-1}$ as $i$. \begin{aligned}\sqrt{-36}&= \sqrt{36 \cdot -1}\\&=\sqrt{36} \cdot \sqrt{-1}\\&=6 \cdot i\\&= 6i\end{aligned} We have another important property to keep in mind when dealing with complex numbers: complex numbers equality. Equality Property The equality property will still apply for complex numbers: $a + bi = m + ni$ only when $a = m$ and $b = n$. What does this mean? Two complex numbers are equal if and only if the real parts are equal and the imaginary part is equal. Now that we have thoroughly discussed the definition of complex numbers let’s go ahead and dive into how the rules of operations may differ or may be alike with the rules we have for real numbers. ## How to solve complex numbers? It’s helpful to think of complex numbers, $z = a + bi$, as binomials when we need to perform arithmetic operations. We consider the real number and imaginary parts as two separate variables. Here are some articles we’ve prepared for a much deeper discussion of the following topics. ### Complex number rules We can start by reviewing how to add and subtract complex numbers – again, and it helps to treat the two parts of the complex numbers as two different variables. When given two complex numbers, we can find their sum or difference by combining the real numbers and the imaginary numbers then simplifying the result. • $(a + bi) + (m + ni) = (a + m) + (b + n)i$ • This means that to add two complex numbers, we add the two real parts and do the same for the imaginary parts. • $(a + bi) – (m + ni) = (a – m) + (b – n)i$ • Similarly, if we want to subtract two complex numbers, we can subtract the real parts and combine them with the difference between the two imaginary parts. Let’s go ahead and try finding the sum and difference of the two complex numbers, $(-3 + 5i)$ and $(4 – 2i)$. • Remove the parenthesis and double-check the signs, especially when finding the difference between the two complex numbers. • Group the real and imaginary number parts and combine each pair of terms. Sum Difference \begin{aligned}(-3 + 5i) + (4 – 2i)&= -3 + 5i+ 4 – 2i\\&= -3 + 4 + 5i -2i\\&= (-3 + 4) + (5 – 2)i\\&= 1 + 3i\end{aligned} \begin{aligned}(-3 + 5i) – (4 – 2i)&= -3 + 5i- 4 + 2i\\&= -3 – 4 + 5i + 2i\\&= (-3 -4) + (5 + 2)i\\&= -7 + 7i\end{aligned} Multiplying Complex Numbers If one of the complex numbers we want to multiply contains a single term, we can distribute this into the remaining complex number. Let’s say we want to multiply $6i$ and $(2 – i)$. We distribute $6i$ into each of the terms inside the parenthesis. Keep in mind that $i^2 = -1$, so make sure to replace $i^2$ with $-1$ whenever possible. \begin{aligned}6i(2 – i)&= (6i)(2) – (6i)(i)\\&=12i – 6i^2\\&= 12i -6(-1)\\&=12i + 6\end{aligned} We use the FOIL method when we want to distribute two binomials, so we also apply the same process if we want to multiply two complex numbers of the form $(a+ bi)$ and $(m + ni)$. Let’s review the FOIL method to multiply the two complex numbers, $(2 – 3i)$ and $(2 + 5i)$. We multiply the first terms, the terms outside and inside the parentheses, and the last term of each binomial. \begin{aligned}(2 – 3i)(2 + 5i)\\\text{First} = 2 \cdot 2 \phantom{xx}\\\text{Outside} = 2 \cdot 5i \phantom{x}\\\text{Inside} = -3i \cdot 2 \\\text{Last} = -3i \cdot 5i \end{aligned} Combine these four terms to find the product of two complex numbers. \begin{aligned}(2 – 3i)(2 + 5i) &= 4 + 10i- 6i – 15i^2\\&= 4 + 10i – 6i – 15(-1)\\&=4 + 10i – 6i + 15\\&= (4 + 15) + (10 – 6)i\\&= 19+ 4i \end{aligned} In general, the product of $(a+ bi)$ and $(m + ni)$ is equal to $(am –bn)(an + bm)i$. Conjugates of Complex Numbers We’ve learned about conjugates in the past, so if you need a quick refresher, make sure to check out this article we wrote about conjugates in math. This means that the conjugate of $a + bi$ will be equal to $a – bi$. Here’s another interesting part – why don’t we multiply the two conjugates? \begin{aligned}(a – bi)(a + bi)&= (a \cdot a – b \cdot b) + (a \cdot b – b \cdot a)i\\&=(a^2 + b^2) + 0i \\&= a^2 + b^2\end{aligned} This means that the product of a complex number and its conjugate is equal to the sum of the squares of their imaginary and real parts. Dividing Complex Numbers When we divide two complex numbers, we might end up with a denominator having an imaginary part. When this happens, we’ll apply a similar process when we want to rationalize a radical expression. Let’s say we want to divide $(a + bi)$ by $(m +ni)$, we’ll end up with a quotient of $\dfrac{a + bi}{m + ni}$. If we want to eliminate the imaginary number found in the denominator, we can: • Multiply both the numerator and denominator by $m – ni$. • Simplify the numerator using the FOIL method. • Simplify the numerator given that the product of a complex number and its conjugate is equal to the sum of the square of their real and imaginary parts. \begin{aligned}\dfrac{a + bi}{m + ni}&= \dfrac{a + bi}{m + ni} \cdot \dfrac{\color{blue} m – ni}{\color{blue} m – ni}\\&= \dfrac{(a + bi){\color{blue}( m – ni)}}{(m + ni){\color{blue}( m – ni)}}\\&= \dfrac{(am -bn)+(mb -an)i}{(m +ni)(m -ni)}, \color{green}\text{FOIL method}\\&= \dfrac{(am -bn)+(mb -an)i}{m^2 + n^2}, \color{green}(a-bi)(a +bi) = a^2 + b^2\\&= \left( \dfrac{am – bn}{m^2 + n^2}\right )+ \left( \dfrac{mb – an}{m^2 + n^2}\right )i \end{aligned} This shows the general form of two complex numbers’ quotient, where $m + ni \neq 0$. We can apply similar steps when finding the quotient of other pairs of complex numbers. We may need to use three or more of these rules for some problems, so make sure to review all these rules and check out the links we’ve provided earlier in this section. ### Solving equations that involve complex numbers Complex numbers can appear in equations. When this happens, it will be helpful to make use of the equality property whenever possible. Keep in mind that we can only equate the real numbers part and the imaginary number parts based on the equality property. Let’s try to solve for $x$ and $y$ in the equation, $5x – (2y + 3)i = 18 – 12i$. Through the equality property, the real and imaginary parts must be equal to each other for the equation to be true. Equation Real Part Imaginary Part $5x – (2y + 3)i$ $5x$ $-(2y + 3)$ $18 – 12i$ $18$ $-2$ Let’s equate the real parts to solve for $x$: \begin{aligned} 5x &= 18\\ x&= \dfrac{18}{5}\end{aligned} We do the same for the imaginary parts so we can solve for $y$: \begin{aligned} -(2y + 3)&= -2\\ 2y + 3 &= 2\\2y &=-1\\y&= -\dfrac{1}{2}\end{aligned} Hence, we have $x = \dfrac{18}{5}$ and $y = -\dfrac{1}{2}$. ## How to graph complex numbers? Now that we’ve learned about complex numbers and how we can manipulate their expressions let’s learn how to represent complex numbers on a graph. When we talk about graphing complex numbers, it’s equally important for us to review what we know about complex planes. Graphing Complex Numbers in Complex Planes The complex plane is a coordinate system where the horizontal coordinate represents the real numbers. The vertical coordinates represent the imaginary parts. This is also why the horizontal and vertical axes are called the real axis and imaginary axis. Comparing this with our regular coordinate system, we can write the coordinate of $a + bi$ as $(x, y)$. This means that the way we plot the complex number will be similar. • We’ll plot the point by moving $a$ units to the right if it’s positive or $a$ units to the left when it’s negative. • Then we also move $b$ upward when it’s positive or downwards when it’s negative. Here are some complex numbers graphed in one complex planes as examples: Let’s check each of the complex numbers graphed: • $4 + 4i$: Graph the point $(4, 4)$ or a point that is $4$ units to the right and upward. • $-2 + i$: Similarly, we can graph $(-2, 1)$ or plotting a point that’s $2$ to the left from the origin and one unit upward. • $3$: Recall that $3 = 3 + 0i$, so that’s $(3, 0)$ or $3$ units along the real axis. • $4i$: We can rewrite $4i$ as $0 + 4i$, so that’s $(0, 4)$. The point is expected to lie $4$ along the positive imaginary axis. Understanding Absolute Values of $\boldsymbol{a + bi}$ The absolute value of a complex number represents the distance of the complex number from $0$. We use Pythagorean theorem ($c^2 = a^2 + b^2$, remember?) to derive the formula for the $a + bi$’s absolute value. The graph above shows the absolute value of the complex number, $\|z\|$, by constructing a right triangle on the complex plane. Using the Pythagorean theorem, we have $\|z\| = \sqrt{a^2 \| b^2}$. This applies to all complex numbers as well. ### Summary of complex numbers definition and properties We’ll learn about complex numbers and more properties involving them in the next articles. For now, we should wrap this up and summarize what we’ve learned so far about complex numbers. • The general form of complex numbers is $a + bi$, where $a$ and $b$ represent the real and imaginary parts, respectively. • The number $i$ represents $\sqrt{-1}$, so $i^2 = -1$. • To solve equations that involve complex numbers, always go back to the equality property. • To graph $a + bi$ on a complex plane, we move $a$ units horizontally and $b$ units vertically. Here’s a summary of the operations that can be done on two complex numbers, $a + bi$ and $m + ni$. Addition $(a + bi) + (m + ni) = (a + m) + (b + n)i$ Subtraction $(a + bi) – (m + ni) = (a – m) + (b – n)i$ Multiplication $(a + bi) \cdot (m + ni) = (am – bn) + (bm + an)i$ Division $\dfrac{a + bi}{m + ni} = \left( \dfrac{am – bn}{m^2 + n^2}\right )+ \left( \dfrac{mb – an}{m^2 + n^2}\right )i$ Product of Two Conjugates $(a + bi)(a – bi) = a^2 – b^2$ Absolute Value $\|a + bi\| = \sqrt{a^2 + b^2}$ ### Example 1 Simplify $(2 – 4i)(3 + i) – (2 – 3i)^2$. Solution Before we can simplify the expression, we’ll have to expand each group of binomials first: • Applying the FOIL method on the first pair of binomials. • Apply the perfect square trinomial property, (a – b)^2 = a^2 -2ab + b^2, to expand the last binomial. \begin{aligned} (2 – 4i)(3 + i) – (2 – 3i)^2 &= [(2\cdot 3 + 4\cdot1) -(-4 \cdot 3 + 2 \cdot 1)i] – (2 – 3i)^2\\&=(10 +12 i) – (2 – 3i)^2\\&= (10 + 12i) -[2^2 – 2 \cdot 3i + (3i)^2]\\&=(10 + 12i)- (4 -6i – 9) \end{aligned} Combine the real and imaginary parts to simplify the expression further. \begin{aligned} (10 + 12i)- (4 -6i – 9) &= 10 + 12i – 4 + 6i + 9\\&= (10 – 4 + 6) + (12 + 6)i\\&=12 + 18i \end{aligned} Hence, we have $(2 – 4i)(3 + i) – (2 – 3ui)^2 = 12 + 18i$. ### Example 2 Simplify $\dfrac{(2 – i)^3}{(1 + i)^2}$. Solution What we can do is expand the numerator and denominator of the rational expression. Let’s begin by expanding $(2 – i)^3$ as shown below. Take note of the blue texts if you need a refresher on the property used in each step. \begin{aligned} (2-i)^3 &= (2- i)(2 -i)^2\\&= (2- i)[2^2 – 2(2)(i) +(i)^2], \phantom{xxx}\color{green} (a -b)^2 = a^2 -2ab + b^2\\&= (2 – i)[4 – 4i – 1], \phantom{xxxxxxxxxxx}\color{green} i^2 = -1\\&= (2 – i)(3 – 4i)\\&=(2\cdot3 – 4\cdot1) + [-1(3)+2(-4)]i, \color{green} \text{FOIL method}\\&= 2 – 11i\end{aligned} We can expand the denominator using the perfect square trinomial property as well. \begin{aligned} (1 + i)^2&= (1)^2 +2(1)(i) + (i)^2, \phantom{xx}\color{green} (a + b)^2 = a^2 + 2ab + b^2 \\&= 1 +- 2i – 1, \phantom{xxxxxxxxxxxxxx}\color{green} i^2 = -1\\&= 2i\end{aligned} Let’s replace the denominator and numerator with these simplified expressions and see what we have. \begin{aligned} \dfrac{(2 – i)^3}{(1 + i)^2} &= \dfrac{2 – 11i}{2i}\end{aligned} To eliminate the imaginary part found in the denominator, we can multiply both denominator and numerator by $i$. \begin{aligned} \dfrac{2 – 11i}{2i} \cdot \dfrac{\color{blue}i}{\color{blue}i}&= \dfrac{2{\color{blue}i} -(11i)({\color{blue}i})}{(2i)({\color{blue}i})}\\&=\dfrac{2i – 11i^2}{2i^2}\\&=\dfrac{2i + 11}{-2}\\&= \dfrac{2i}{-2}- \dfrac{11}{2}\\ &= -i- \dfrac{11}{2}\\&= -\dfrac{11}{2} -i\end{aligned} Hence, $\dfrac{(2 – i)^3}{(1 + i)^2} = -\dfrac{11}{2} -i$. #### Example 3 Let’s observe the power of $i$ by filling in the table shown below. $\boldsymbol{i^1}$ $\boldsymbol{i^5}$ $\boldsymbol{i^2}$ $\boldsymbol{i^6}$ $\boldsymbol{i^3}$ $\boldsymbol{i^7}$ $\boldsymbol{i^4}$ $\boldsymbol{i^8}$ a. What can you observe based on the two columns’ results? b. Using your observation, find $i^k$ if the remainder of $k$ when it’s divided by $4$ is $2$? c. Find the value of $i^{36} – i^{57}$. Solution We can start with {i} and multiply the current value with $i$. Use the fact that $i^2 = -1$ as well. $\boldsymbol{i^1}$ $i^1 = i$ $\boldsymbol{i^5}$ $1 \cdot I = i$ $\boldsymbol{i^2}$ $i \cdot i = -1$ $\boldsymbol{i^6}$ $i \cdot i = -1$ $\boldsymbol{i^3}$ $-1 \cdot i = -i$ $\boldsymbol{i^7}$ $-1 \cdot i = -i$ $\boldsymbol{i^4}$ $-i \cdot i = -i^2 = 1$ $\boldsymbol{i^8}$ $-i \cdot i = -i^2 = 1$ The table above shows that values of $\{i, i^2, i^3, … i^8\}$. The two columns show the same result – meaning, the powers of i will return the same cycle of values shown for $i$, $i^2$, $i^3$, and $i^4$. a. Let’s observe each pair that returns the same values: • $i^5 = i^1 =1$: $5$ and $1$ return a remainder of $1$ when the powers are divided by $4$. • $i^6 = i^2 =-1$: $6$ and $2$ return a remainder of $2$ when the powers are divided by $4$. • $i^7 = i^3 =1$: $7$ and $3$ return a remainder of $3$ when the powers are divided by $4$. • $i^8 = i^4 =1$: $8$ and $4$ are multiples of $4$ and will not return any remainder when the powers arty e divided by $4$. How do we generalize this? When $k$ is divided by $4$ and if the remainder is $1$, $2$ or $3$, $i^k$ will be equal to $i^1 = 1$, $i^2 = -1$, or $i^3 = -i$, respectively. If $k$ is a multiple of $4$, $i^k$ will be equal to $i^4 = 1$. b. Apply what we’ve observed in a, since the remainder of $\dfrac{k}{4}$ is $2$, $i^k = i^2 = -1$. We’ll apply our observations to find the values of $i^{36}$ and $i^{57} •$36$is a multiple of$4$, so$i^{36} = i^4 = 1$. •$57$returns a remainder of$1$when divided by$4$, so$i^{57} = i^1 = i$. c. Hence,$ i^{36} – i^{57} = 1 – i$. #### Example 4 Simplify the expressions and solve for the unknown values. a.$3 – 2i = x – (3 + i)$b.$4 – 6i – x + 3yi = 8 – 3i$c.$\dfrac{1}{y}= \dfrac{1}{2 + 5i}+ \dfrac{1}{5 – 2i}$Solution The first equation requires a straightforward approach of isolating$x$on one side of the equation. We can add$(3 + i)$on both sides of the equation.$ \begin{aligned}3 – 2i&= x -(3 +i)\\3 -2i {\color{blue}+ (3 + i)}&= x- (3 + i){\color{blue}+ (3 + i)}\\6 – i &= x\\x&= 6 – i\end{aligned}$a. Hence, we have$ x = 6 – i$. Let’s isolate$x$and$y$on the left-hand side of the equation.$\begin{aligned}4 – 6i – x + 3yi &= 8 – 3i\\ -x + 3yi &= 4 – 6i + 8 – 3i\\-x + 3yi &= (4 +8) + (-6-3)i\\-x + 3yi &= 12 + (-9)i\end{aligned}$According to the equality property, for two complex numbers to be equal, the real number of parts must be equal to each other. The imaginary parts must be equal as well. Let’s equate$-x$with the real part on the right-hand side and equate$3y$with the imaginary part of the right-hand side’s expression. $\begin{aligned}-x &= 12\\ x&=-12\end{aligned}\begin{aligned}3y &= -9\\y&= \dfrac{-9}{3}\\&=-3\end{aligned}$b. This means that$x$must be$-12$and$y$must be$-3$. Let’s simplify the right-hand expression,$\dfrac{1}{2 + 5i}+ \dfrac{1}{5 – 2i}$, by rewriting each fraction so that they both share a common denominator,$(2 + 5i)(2 – 5i)$. Observe that$ 5- 2i = -(2 + 5i)$, so$\dfrac{1}{2 + 5i}+ \dfrac{1}{5 – 2i} = \dfrac{1}{2 + 5i} – \dfrac{1}{2 – 5i}$.$\begin{aligned}\dfrac{1}{2 + 5i} + \dfrac{1}{2 – 5i}&= \dfrac{1(2 – 5i)}{(2 + 5i)(2 – 5i)} + \dfrac{1(2 + 5i)}{(2 + 5i)(2 – 5i)}\\&= \dfrac{2 – 5i}{(2 + 5i)(2 – 5i)}+ \dfrac{2 + 5i}{(2 + 5i)(2 – 5i)}\\&=\dfrac{2 – 5i + 2 + 5i}{(2 + 5i)(2 – 5i)}\\&= \dfrac{4}{(2 + 5i)(2 – 5i)}\end{aligned}$Simplify the denominator by using the conjugate properties of complex numbers,$(a – bi)(a + bi) = a^2 + b^2$.$\begin{aligned}\dfrac{4}{(2 + 5i)(2 – 5i)} &= \dfrac{4}{(2)^2-(5i)^2}\\&= \dfrac{4}{4 – 25i^2}\\&= \dfrac{4}{4 – 25(-1)}\\&=\dfrac{4}{4 + 25}\\&= \dfrac{4}{29}\end{aligned}$Let’s replace the equation’s right-hand expression to$\dfrac{4}{29}$and solve for$y$.$\begin{aligned}\dfrac{1}{y}&= \dfrac{4}{29}\\y&= \dfrac{29}{4}\end{aligned}$c. This means that with enough manipulation on the complex expression, we could find$ y = \dfrac{24}{9}$. ### Example 5 Graph the following complex numbers on a complex plane and indicate their corresponding absolute values or$\|z\|$. a.$z = -2 + 3i$b.$z = 6 – 12i$c.$z = 6i$d.$z = 12$Solution Let’s begin with$-2 + 3i$. Recall that the absolute value of the complex number,$z = a + bi$can be determined using the property,$\|z\| = \sqrt{a^2 + b^2}$.$ \begin{aligned}\|z\| &= \sqrt{(-2)^2 + (3)^2}\\&= \sqrt{4 + 9}\\&= \sqrt{13}\end{aligned}$To graph$-2 + 3i$, let’s move$2$units to the left and$3$units upward. We’ll apply a similar process when graphing$6 – 12i$and finding its absolute value.$ \begin{aligned}\|z\| &= \sqrt{(6)^2 + (12)^2}\\&= \sqrt{36 + 144}\\&= \sqrt{180}\\&= \sqrt{36 \cdot 5}\\&= 6\sqrt{5}\end{aligned}$The complex number is found$6$units to the imaginary axis’s right and$12$units up the real axis. This graph shows the position of$6 – 12i$on the complex plane and its absolute value,$\|z\|$. The complex number,$6i$, only contains an imaginary part, so its position is expected to lie along the imaginary axis – in fact,$6$units on the positive imaginary axis. From inspection alone, we can see that the distance of$6i$from the origin is$6$. We can also confirm this by using the formula for the absolute value. Since$6i = 0 + 6i$, so we have:$ \begin{aligned}\|z\| &= \sqrt{(6)^2 + (0)^2}\\&= \sqrt{36 + 0}\\&= \sqrt{36}\\&= 6\end{aligned}$Working on the fourth number, we have$12 = 12+ 0i$. This means that we’re expecting the point lying along the real axis –$12$units along the positive real axis. Similarly, we can see that the distance of the point from the origin is$12$and can be confirmed by the traditional method of finding absolute values of complex numbers.$ \begin{aligned}\|z\| &= \sqrt{(0)^2 + (12)^2}\\&= \sqrt{0 + 144}\\&= \sqrt{144}\\&= 12\end{aligned}$### Example 6 Given that$A$,$B$,$C$, and$D$are four complex numbers graphed on one complex plane as shown below. Find the values of the following: a.$A + B$b.$C – D$c.$A \cdot C$d.$\dfrac{B}{C}$Solution We can begin by finding what the four points represent. Take note of the distance of each point from the imaginary and real axes. For example,$A$is$2$units from the right of the imaginary axis and$3$units above the real axis. This means that$A$=$(2, 3)$=$2 + 3i$. We’ll apply the same process for the remaining points and have the following result: •$B = (-4, 5) = -4 + 5i$•$C = (-5, -4) = -5 – 4i$•$D = (6, -6) = 6 – 6i$Let’s go ahead and find the sum of$A$and$B$. Combine the real number parts and imaginary parts to add$(2 + 3i)$and$(-4 + 5i)$.$\begin{aligned}(2 + 3i) + (-4 + 5i)&= (2 + -4) + (3 + 5)i\\&=-2 + 8i\end{aligned}$a. This means that$A + B = -2 + 8i$. We’ll apply a similar process when finding the difference between$C$and$D$.$\begin{aligned}(-5 – 4i) – (6 – 6i)&= (-5 – 6) + (-4 – -6)i\\&=-11 + 2i\end{aligned}$b. Hence,$C – D = -11 + 2i$. To find the product of$A$and$C$, we can apply the FOIL method as shown below.$ \begin{aligned}(2 + 3i) \cdot (-5 – 4i) &= (2\cdot -5 -3 \cdot -4)+(3 \cdot -5 + 2 \cdot -4 )i\\&=(-10+12)+ (-15+-8)i\\&= 2 – 23i\end{aligned}$c. This means that the product of$A$and$C$is$2 – 23i$. The quotient of$B$and$C$is equal to$\dfrac{-4 + 5i}{-5 – 4i}$. Multiply both the numerator and denominator by the conjugate of$C$, which is$-5 + 4i$.$\begin{aligned}\dfrac{-4 + 5i}{-5 – 4i}&= \dfrac{-4 + 5i}{-5 – 4i} \cdot \dfrac{\color{blue}-5 + 4i}{\color{blue}-5+ 4i}\\&= \dfrac{(-4 + 5i)({\color{blue}-5 + 4i})}{(-5 – 4i)({\color{blue}-5 + 4i})}\\&= \dfrac{(-4 + 5i)(-5 + 4i)}{(-5- 4i)(-5 + 4i)}\end{aligned}$Apply the FOIL method on the numerator and the property of the conjugates’ product,$(a -bi)(a + bi) = a^2 + b^2$, to simplify the denominator.$\begin{aligned} \dfrac{(-4 + 5i)(-5 + 4i)}{(-5- 4i)(-5 + 4i)}&= \dfrac{[-4(-5)-5(4)]+ [5(-5)+4(-4)]i}{(-5)^2 + (4)^2}\\&= \dfrac{(20 – 20) + (-25 – 16)i}{25 + 16}\\&= \dfrac{-41i}{41}\\&= -i\end{aligned}$d. This means that$\dfrac{B}{C} = -i$. ### Example 7 Given that$z = a + bi$, show that there is no value for$a$and$b$so that it satisfies$\|z\| – z = i$. Solution Let’s rewrite$\|z\|$and$z$in terms of$a$and$b$. Hence, we have the following: •$\|z\| = \sqrt{a^2 + b^2}$•$z = a + bi \begin{aligned} \|z\| – z &= i \\ \sqrt{a^2 + b^2} – (a + bi)&= i\end{aligned}$Group the real parts and imaginary parts on each side of the equation.$\begin{aligned}(\sqrt{a^2 + b^2} – a) – b i = 0 + i \end{aligned}$For the equation to be true,$\sqrt{a^2 + b^2} – a = 0$and$-b = 1$. This means that$b = -1$, so substitute this value into the first equation and find the value of$a$.$\begin{aligned}\sqrt{a^2 + b^2} – a &=0 \\ \sqrt{a^2 + (-1)^2} – a &=0 \\\sqrt{a^2 + 1} -a&=0\\ \sqrt{a^2 + 1}&=a\\(\sqrt{a^2 + 1})^2 &= a^2\\a^2 + 1 &= a^2 \\1&=0 \Rightarrow \color{red}\text{No solution}\end{aligned}$This shows that when$ b = -1$, the second equation will have no solution. This shows that it will be impossible for the system of equation,$\begin{matrix}\sqrt{a^2 + b^2} – a = 0 \\ b =- 1 \end{matrix}$. Consequently, it’s impossible for a complex number that satisfies the equation$\|z\| – z = i$. ### Practice Questions 1. Simplify$(3 + 4i)(2 + 3i) – (4 – i)^2$. 2. Simplify$\dfrac{(3 + i)^3}{(2 – 5i)^2}$. 3. Determine the values of the following: a.$i^{37}$b.$36i^{23} + 24i^{45}$c.$-5i^{33} + 12i^{45} – 9i^{11}$4. Simplify the expressions and solve for the unknown values. a.$5 – 4i = 2x – (1 + 6i)$b.$12 – 8i – 2x + 4yi = 16 – 3i$c.$\dfrac{1}{y}= \dfrac{1}{12 + 6i}+ \dfrac{1}{12 – 6i}$5. Graph the following complex numbers on a complex plane and indicate their corresponding absolute values or$\|z\|$. a.$z = -6+ 8i$b.$z = 6 – 8i$c.$z = -4i$d.$z = 8$6. Given that$A$,$B$,$C$, and$D$are four complex numbers graphed on one complex plane as shown below. Find the values of the following: a.$A + B$b.$C – D$c.$A \cdot C$d.$\dfrac{B}{C}$### Answer Key 1.$– 21 + 25i$2.$-\dfrac{898}{841} – \dfrac{186}{841}i$3. a.$i$b.$-12i$c.$16i$4. a.$x=3 + i$b.$x=-2$,$y=\dfrac{5}{4}$c.$y =\dfrac{15}{2}$5. 6. a.$-2+5i$b.$11+i$c.$29+3i$d.$-\dfrac{18}{17}-\dfrac{4}{17}i\$ Images/mathematical drawings are created with GeoGebra.
# 9.8: Area of Circles Difficulty Level: At Grade Created by: CK-12 ## Introduction Do you remember Miguel? He had just finished working on figuring out the circumference of three different on deck pads for the pitchers to use while they warm up. Let’s look at his dilemma again before we look at the area of the pads. Miguel’s latest task is to measure some different “on deck” pads for the pitchers to practice with. An on deck pad is a circular pad that is made up of a sponge and some fake grass. Pitchers practice their warm-ups while standing on them. They work on stretching and get ready to “pitch” the ball prior to their turn on the mound. Miguel has three different on deck pads that he is working with. The coach has asked him to measure each one and find the circumference and the area of each. Miguel knows that the circumference is the distance around the edge of the circle. He decides to start with figuring out the circumference of each circle. He measures the distance across each one. The first one measures 4 ft. across. The second measures 5 ft. across. The third one measures 6 ft across. Miguel has completed the first part of this assignment. He knows the circumference of each pad. Now he has to figure out the area of each. Miguel isn’t sure how to do this. He can’t remember how to find the area of a circle. Miguel needs some help. This lesson will help you learn how to find the area of a circle. When finished, we’ll come back to this problem and you can help Miguel figure out the area of each on deck pad. What You Will Learn By the end of this lesson, you will be able to demonstrate the following skills: • Recognize the formula for the area of a circle. • Find the area of circles given radius or diameter. • Find radius or diameter given area. • Find areas of combined figures involving parts of circles. Teaching Time I. Recognize the Formula for the Area of a Circle In the last lesson you learned how to calculate the circumference of a circle. Let’s take a few minutes to review the terms associated with circles. A circle is a set of connected points equidistant from a center point. The diameter is the distance across the center of the circle and the radius is the distance from the center of the circle to the edge. We also know that the number pi, , is the ratio of the diameter to the circumference. We use 3.14 to represent pi in operations. What does all of this have to do with area? Well, to find the area of a two-dimensional figure, we need to figure out the measurement of the space contained inside. This is the measurement of area. This is also the measurement inside a circle. You learned how to find the radius of a circle, given circumference or diameter. now let’s look at using the radius to find the area of the circle. How do we find the area of a circle? The area of a circle is found by taking the measurement of the radius, squaring it and multiplying it by pi. Here is the formula. Write this formula down in your notebook. II. Find the Area of Circles Given the Radius or Diameter We already know that the symbol represents the number 3.14, so all we need to know to find the area of a circle is its radius. We simply put this number into the formula in place of and solve for the area, . Let’s try it out. Example What is the area of the circle below? We know that the radius of the circle is 12 centimeters. We put this number into the formula and solve for . Remember that squaring a number is the same as multiplying it by itself. The area of a circle with a radius of 12 centimeters is 452.16 square centimeters. Example Some students have formed a circle to play dodge ball. The radius of the circle is 21 feet. What is the area of their dodge ball circle? The dodge ball court forms a circle, so we can use the formula to find its area. We know that the radius of the circle is 21 feet, so let’s put this into the formula and solve for area, . Notice that a circle with a larger radius of 21 feet has a much larger area than the circle with a 12 centimeter radius: 1,384.74 square feet. Sometimes, you will be given a problem with the diameter and not the radius. When this happens, you can divide the measurement of the diameter by two and then use the formula. Example Find the area of a circle with a diameter of 10 in. First, we divide the measurement in half to find the radius. Now we use the formula. 9N. Lesson Exercises Find the area of each circle. 3. Diameter = 8 ft. Take a few minutes to check your work. III. Find the Radius or Diameter Given the Area We have seen that when we are given the radius or the diameter of a circle, we can find its area. We can also use the formula to find the radius or diameter if we know the area. Let’s see how this works. Example The area of a circle is 113.04 square inches. What is its radius? This time we know the area and we need to find the radius. We can put the number for area into the formula and use it to solve for the radius, . Let’s look at what we did to solve this. To solve this problem we needed to isolate the variable . First, we divided both sides by , or 3.14. Then, to remove the exponent, we took the square root of both sides. A square root is a number that, when multiplied by itself, gives the number shown. We know that 6 is the square root of 36 because . The radius of a circle with an area of 113.04 square inches is 6 inches. Example What is the diameter of a circle whose area is ? What is this problem asking us to find? We need to find the diameter (not the radius!). What information is given in the problem? We know the area. Therefore we can use the formula to solve for the radius, . Once we know the radius, we can find the diameter. Let’s give it a try. The radius of a circle with an area of 379.94 square centimeters is 11 centimeters. Remember, this problem asked us to find the diameter, so we’re not done yet. How can we find the diameter? The diameter is always twice the length of the radius, so the diameter of this circle is centimeters. As we have seen, we can use the area formula with lots of kinds of information about a circle. If we know the diameter or radius, we can solve for the area, . If we are given the area, we can solve for the radius, . If we know the radius, we can also find the diameter. 9O. Lesson Exercises Find the radius of each circle. 1. Area = 153.86 sq. in. 2. Area = 354.34 sq. ft. 3. 452.16 sq. m Check your work with a partner and then continue with the next section. IV. Find Areas of Combined Figures Involving Parts of Circles Sometimes we may be asked to find the area of a combined figure. Combined figures often include portions of circles, such as a quarter or semicircle (which is a half circle). We can find the area of combined figures by breaking them down into smaller shapes and finding the area of each piece. We can calculate the area of a portion of a circle. As long as we know the radius of the circle, we can find its area. Then we can divide that area into smaller pieces or subtract a portion to find the area of part of the circle. Let’s try this out. Example What is the area of the figure below? This figure is a semicircle, or half of a circle. Remember that a diameter always divides a circle in half. Therefore the edge measuring 17 inches is the circle’s diameter. Can we use it to find the area of the whole circle? We sure can! The radius of the circle must be inches. Now let’s use the formula to solve for area. We know that a whole circle with a radius of 8.5 inches (and a diameter of 17 inches) is 226.87 square inches. Therefore the semicircle figure has an area of square inches. As long as we can find the area of a whole circle, we can divide or subtract to find the area of a portion of a circle. Now let’s look at a combined figure. Example Find the area of the figure below. First, we have to find the area of the rectangle. We can do this by multiplying the length times the width. Then we can find the area of the circle. If you notice, the width of the rectangle is also the diameter of the circle. This will help us when we want to find the area of the circle. The area is 48 square inches for the rectangle. Now let’s look at the semi-circle. If the diameter is the width which is 6 inches, then the radius is 3 inches. We can find the area of a circle now. Now this is the area of a whole circle. We only need the area of a semi-circle. Let’s divide this value in half. The area of the semi-circle is 14.13 square inches. Now we add the two areas together. 48 +14.13 = 62.13 square inches The area of the entire figure is 62.13 square inches. Now let’s go and use what we have learned to help Miguel figure out the area of each on-deck pad. ## Real–Life Example Completed Here is the original problem once again. Reread it and underline any important information. Do you remember Miguel? He had just finished working on figuring out the circumference of three different on deck pads for the pitchers to use while they warm up. Let’s look at his dilemma again before we look at the area of the pads. Miguel’s latest task is to measure some different “on deck” pads for the pitchers to practice with. An on deck pad is a circular pad that is made up of a sponge and some fake grass. Pitchers practice their warm-ups while standing on them. They work on stretching and get ready to “pitch” the ball prior to their turn on the mound. Miguel has three different on deck pads that he is working with. The coach has asked him to measure each one and find the circumference and the area of each. Miguel knows that the circumference is the distance around the edge of the circle. He decides to start with figuring out the circumference of each circle. He measures the distance across each one. The first one measures 4 ft. across. The second measures 5 ft. across. The third one measures 6 ft across. Miguel has completed the first part of this assignment. He knows the circumference of each pad. Now he has to figure out the area of each. Miguel isn’t sure how to do this. He can’t remember how to find the area of a circle. Miguel needs some help. Now it’s time to help Miguel figure out each area. The first one has a diameter of 4 feet, so it has a radius of 2 ft. Here is the area of the first pad. The second pad has a diameter of 5 feet, so it has a radius of 2.5 feet. The third pad has a diameter of 6 feet, so it has a radius of 3 feet. Miguel is very pleased with his work. He is sure that his coach will be pleased with his efforts as well! ## Vocabulary Here are the vocabulary words that are found in this lesson. Circle a set of connected points that are equidistant from a center point. Diameter the distance across the center of a circle. the distance from the center of the circle to the outer edge. Area the space inside a two-dimensional figure ## Technology Integration Other Videos: 1. http://www.mathplayground.com/mv_area_circles.html – This is a Brightstorm video on finding the area of a circle. ## Time to Practice Directions: Find the area of each circle given the radius or diameter. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Directions: Find each radius given the area of the circle. 13. 12.56 sq. in. 14. 78.5 sq. m 15. 200.96 sq. cm 16. 254.34 sq. in 17. 7.07 sq. ft. 18. 28.26 sq. m Directions: Find each diameter given the area of the circle. 19. 12.56 sq. in. 20. 78.5 sq. m 21. 200.96 sq. cm 22. 254.34 sq. in 23. 7.07 sq. ft. 24. 28.26 sq. m Directions: Solve each problem. 25. Rob is painting large polka dots on a sheet for the backdrop of the school musical. He painted 16 polka dots, each with a radius of 3 feet. What is the total area that the polka dots cover? 26. The librarian is having the library at her school carpeted. The library is a circular room with a diameter of 420 feet. How many square feet of carpet will she need to order? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
# High School Mathematics Extensions/Logic ## Introduction Logic is the study of the way we reason. In this chapter, we focus on the methods of logical reasoning, i.e. digital logic, predicate calculus, application to proofs and the (insanely) fun logical puzzles. ## Boolean algebra In the black and white world of ideals, there is absolute truth. That is to say everything is either true or false. With this philosophical backdrop, we consider the following examples: "One plus one equals two." True or false? That is (without a doubt) true! "1 + 1 = 2 AND 2 + 2 = 4." True or false? That is also true. "1 + 1 = 3 OR Sydney is in Australia" True or false? It is true! Although 1 + 1 = 3 is not true, the OR in the statement made the whole statement to be true if at least one of the statements is true. Now let's consider a more puzzling example "2 + 2 = 4 OR 1 + 1 = 3 AND 1 - 3 = -1" True or false? The truth or falsity of the statements depends on the order in which you evaluate the statement. If you evaluate "2 + 2 = 4 OR 1 + 1 = 3" first, the statement is false, and otherwise true. As in ordinary algebra, it is necessary that we define some rules to govern the order of evaluation, so we don't have to deal with ambiguity. Before we decide which order to evaluate the statements in, we do what most mathematician love to do -- replace sentences with symbols. Let x represent the truth or falsity of the statement 2 + 2 = 4. Let y represent the truth or falsity of the statement 1 + 1 = 3. Let z represent the truth or falsity of the statement 1 - 3 = -1. Then the above example can be rewritten in a more compact way: x OR y AND z To go one step further, mathematicians also replace OR by + and AND by ×, the statement becomes: ${\displaystyle x+y\times z}$ Now that the order of precedence is clear. We evaluate (y AND z) first and then OR it with x. The statement "x + yz" is true, or symbolically ${\displaystyle x+yz=1}$ where the number 1 represents "true". There is a good reason why we choose the multiplicative sign for the AND operation. As we shall see later, we can draw some parallels between the AND operation and multiplication. The Boolean algebra we are about to investigate is named after the British mathematician George Boole. Boolean algebra is about two things -- "true" or "false" which are often represented by the numbers 1 and 0 respectively. Alternative, T and F are also used. Boolean algebra has operations (AND and OR) analogous to the ordinary algebra that we know and love. ### Basic Truth tables We have all had to memorize the 9 by 9 multiplication table and now we know it all by heart. In Boolean algebra, the idea of a truth table is somewhat similar. Let's consider the AND operation which is analogous to the multiplication. We want to consider: x AND y where and x and y each represent a true or false statement (e.g. It is raining today). It is true if and only if both x and y are true, in table form: The AND function x y x AND y F F F F T F T F F T T T We shall use 1 instead of T and 0 instead of F from now on. The AND function x y x AND y 0 0 0 0 1 0 1 0 0 1 1 1 Now you should be able to see why we say AND is analogous to multiplication, we shall replace the AND by ×, so x AND y becomes x×y (or just xy). From the AND truth table, we have: 0 × 0 = 0 0 × 1 = 0 1 × 0 = 0 1 × 1 = 1 To the OR operation. x OR y is FALSE if and only if both x and y are false. In table form: The OR function x y x OR y 0 0 0 0 1 1 1 0 1 1 1 1 We say OR is almost analogous to addition. We shall illustrate this by replacing OR with +: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 1 (like 1 OR 1 is 1) The NOT operation is not a binary operation like AND and OR, but a unary operation, meaning it works with one argument. NOT x is true if x is false and false if x is true. In table form: The NOT function x NOT x 0 1 1 0 In symbolic form, NOT x is denoted x' or ~x (or by a bar over the top of x). Alternative notations: ${\displaystyle x\times y=x\wedge y}$ and ${\displaystyle x+y=x\vee y}$ ### Compound truth tables The three truth tables presented above are the most basic of truth tables and they serve as the building blocks for more complex ones. Suppose we want to construct a truth table for xy + z (i.e. x AND y OR z). Notice this table involves three variables (x, y and z), so we would expect it to be bigger than the previous ones. To construct a truth table, firstly we write down all the possible combinations of the three variables: x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 There is a pattern to the way the combinations are written down. We always start with 000 and end with 111. As to the middle part, it is up to the reader to figure out. We then complete the table by hand computing what value each combination is going to produce using the expression xy + z. For example: 000 x = 0, y = 0 and z = 0 xy + z = 0 001 x = 0, y = 0 and z = 1 xy + z = 1 We continue in this way until we fill up the whole table x y z xyORz 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 The procedure we follow to produce truth tables are now clear. Here are a few more examples of truth tables. Example 1 -- x + y + z x y z x+y+z 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 Example 2 -- (x + yz)' When an expression is hard to compute, we can first compute intermediate results and then the final result. x y z x+yz (x+yz)' 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 0 Example 3 -- (x + yz')w x y z w (x+yz')w 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 #### Exercise Produce the truth tables for the following operations: 1. NAND: x NAND y = NOT (x AND y) 2. NOR: x NOR y = NOT (x OR y) 3. XOR: x XOR y is true if and ONLY if one of x or y is true. Produce truth tables for: 1. xyz 2. x'y'z' 3. xyz + xy'z 4. xz 5. (x + y)' 6. x'y' 7. (xy)' 8. x' + y' ### Laws of Boolean algebra In ordinary algebra, two expressions may be equivalent to each other, e.g. xz + yz = (x + y)z. The same can be said of Boolean algebra. Let's construct truth tables for: xz + yz (x + y)z xz + yz x y z xz+yz 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 (x + y)z x y z (x+y)z 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 By comparing the two tables, you will have noticed that the outputs (i.e. the last column) of the two tables are the same! Definition We say two Boolean expressions are equivalent if the output of their truth tables are the same. We list a few expressions that are equivalent to each other x + 0 = x x × 1 = x xz + yz = (x + y)z x + x' = 1 x × x' = 0 x × x = x x + yz = (x + y)(x + z) Take a few moments to think about why each of those laws might be true. The last law is not obvious but we can prove that it's true using the other laws: ${\displaystyle {\begin{matrix}(x+y)(x+z)&=&x(x+z)+y(x+z)\\&=&xx+xz+xy+yz\\&=&x+xz+xy+yz\\&=&x(1+z+y)+yz\\&=&x+yz\end{matrix}}}$ It has been said: "the only thing to remember in mathematics is that there is nothing to remember. Remember that!". You should not try to commit to memory the laws as they are stated, because some of them are so deadly obvious once you are familiar with the AND, OR and NOT operations. You should only try to remember those things that are most basic, once a high level of familiarity is developed, you will agree there really isn't anything to remember. #### Simplification Once we have those laws, we will want to simplify Boolean expressions just like we do in ordinary algebra. We can all simplify the following example with ease: ${\displaystyle {\begin{matrix}xyzw'+xyzw&=&xyz(w+w')\\&=&xyz\end{matrix}}}$ the same can be said about: ${\displaystyle {\begin{matrix}(x+y)(x'+y')&=&x(x'+y')+y(x'+y')\\&=&xx'+xy'+yx'+yy'\\&=&0+xy'+yx'+0\\&=&xy'+yx'\end{matrix}}}$ From those two examples we can see that complex-looking expressions can be reduced very significantly. Of particular interest are expressions of the form of a sum-of-product, for example: xyz + xyz' + xy'z + x'yz + x'y'z' + x'y'z We can factorise and simplify the expression as follows ${\displaystyle xyz+xyz'+xy'z+x'yz+x'y'z'+x'y'z}$ ${\displaystyle =\ xy(z+z')+xy'z+x'yz+x'y'(z'+z)}$ ${\displaystyle =\ xy+xy'z+x'yz+x'y'}$ ${\displaystyle =\ x(y+y'z)+x'(yz+y')}$ It is only hard to go any further, although we can. We use the identity: x + yz = (x + y)(x + z) If the next step is unclear, try constructing truth tables as an aid to understanding. ${\displaystyle {\begin{matrix}&=&\ x(y+z)+x'(z+y')\\&=&\ xy+xz+x'z+x'y'\\&=&\ xy+(x+x')z+x'y'\\&=&\ xy+z+x'y'\\\end{matrix}}}$ And this is as far as we can go using the algebraic approach (or any other approach). The algebraic approach to simplification relies on the principle of elimination. Consider, in ordinary algebra: x + y - x We simplify by rearranging the expression as follows (x - x) + y = y Although we only go through the process in our head, the idea is clear: we bring together terms that cancel themselves out and so the expression is simplified. #### De Morgan's theorems So far we have only dealt with expressions in the form of a sum of products e.g. xyz + x'z + y'z'. De Morgan's theorems help us to deal with another type of Boolean expressions. We revisit the AND and OR truth tables: { } You would be correct to suspect that the two operations are connected somehow due to the similarities between the two tables. In fact, if you invert the AND operation, i.e. you perform the NOT operations on x AND y. The outputs of the two operations are almost the same: { } The connection between AND, OR and NOT is revealed by reversing the output of x + y by replacing it with x' + y'. { } Now the two outputs match and so we can equate them: (xy)' = x' + y' this is one of de Morgan's laws. The other which can be derived using a similar process is: (x + y)' = x'y' We can apply those two laws to simplify equations: Example 1 Express x in sum of product form ${\displaystyle {\begin{matrix}x&=&(ab'+c)'\\&=&(ab')'c'\\&=&(a'+b)c'\\&=&a'c'+bc'\end{matrix}}}$ Example 2 Express x in sum of product form ${\displaystyle {\begin{matrix}x&=&(a+b+c)'\\&=&(a+b)'c'\\&=&a'b'c'\\\end{matrix}}}$This points to a possible extension of De Morgan's laws to 3 or more variables. Example 3 Express x in sum of product form ${\displaystyle {\begin{matrix}x&=&[(a'+c)\cdot (b+d')]'\\&=&(a'+c)'+(b+d')'\\&=&ac'+b'd\\\end{matrix}}}$ Example 4 Express x in sum of product form ${\displaystyle {\begin{matrix}x&=&[(a+bc)\cdot (d+ef)]'\\&=&(a+bc)'+(d+ef)'\\&=&a'(bc)'+d'(ef)'\\&=&a'(b'+c')+d'(e'+f')\\&=&a'b'+a'c'+d'e'+d'f'\\\end{matrix}}}$ Another thing of interest we learnt is that we can reverse the truth table of any expression by replacing each of its variables by their opposites, i.e. replace x by x' and y' by y etc. This result shouldn't have been a surprise at all, try a few examples yourself. De Morgan's laws ${\displaystyle (x+y)'=x'y'}$ ${\displaystyle (xy)'=x'+y'}$ #### Exercise 1. Express in simplified sum-of-product form: 1. z = ab'c' + ab'c + abc 2. z = ab(c + d) 3. z = (a + b)(c + d + f) 4. z = a'c(a'bd)' + a'bc'd' + ab'c 5. z = (a' + b)(a + b + d)d' 2. Show that x + yz is equivalent to (x + y)(x + z) ## Propositions We have been dealing with propositions since the start of this chapter, although we are not told they are propositions. A proposition is simply a statement (or sentence) that is either TRUE or FALSE. Hence, we can use Boolean algebra to handle propositions. There are two special types of propositions -- tautology and contradiction. A tautology is a proposition that is always TRUE, e.g. "1 + 1 = 2". A contradiction is the opposite of a tautology, it is a proposition that is always FALSE, e.g. 1 + 1 = 3. As usual, we use 1 to represent TRUE and 0 to represent FALSE. Please note that opinions are not propositions, e.g. "42 is an awesome number" is just an opinion, its truth or falsity is not universal, meaning some think it's true, some do not. ### Examples • "It is raining today" is a proposition. • "Sydney is in Australia" is a proposition. • "1 + 2 + 3 + 4 + 5 = 16" is a proposition. • "Earth is a perfect sphere" is a proposition. • "How do you do?" is not a proposition - it's a question. • "Go clean your room!" is not a proposition - it's a command. • "Martians exist" is a proposition. Since each proposition can only take two values (TRUE or FALSE), we can represent each by a variable and decide whether compound propositions are true by using Boolean algebra, just like we have been doing. For example "It is always hot in Antarctica OR 1 + 1 = 2" will be evaluated as true. ### Implications Propositions of the type if something something then something something are called implications. The logic of implications are widely applicable in mathematics, computer science and general everyday common sense reasoning! Let's start with a simple example "If 1 + 1 = 2 then 2 - 1 = 1" is an example of implication, it simply says that 2 - 1 = 1 is a consequence of 1 + 1 = 2. It's like a cause and effect relationship. Consider this example: John says: "If I become a millionaire, then I will donate $500,000 to the Red Cross." There are four situations: 1. John becomes a millionaire and donates$500,000 to the Red Cross 2. John becomes a millionaire and does not donate $500,000 to the Red Cross 3. John does not become a millionaire and donates$500,000 to the Red Cross 4. John does not become a millionaire and does not donate \$500,000 to the Red Cross In which of the four situations did John NOT fulfill his promise? Clearly, if and only if the second situation occurred. So, we say the proposition is FALSE if and only if John becomes a millionaire and does not donate. If John did not become a millionaire then he can't break his promise, because his promise is now claiming nothing, therefore it must be evaluated TRUE. If x and y are two propositions, x implies y (if x then y), or symbolically ${\displaystyle x\Rightarrow y}$ has the following truth table: x y ${\displaystyle x\Rightarrow y}$ 0 0 1 0 1 1 1 0 0 1 1 1 For emphasis, ${\displaystyle x\Rightarrow y}$ is FALSE if and only if x is true and y false. If x is FALSE, it does not matter what value y takes, the proposition is automatically TRUE. On a side note, the two propositions x and y need not have anything to do with each other, e.g. "1 + 1 = 2 implies Australia is in the southern hemisphere" evaluates to TRUE! If ${\displaystyle (x\Rightarrow y)\ {\mbox{AND}}\ (y\Rightarrow x)}$ then we express it symbolically as ${\displaystyle x\Leftrightarrow y}$. It is a two way implication which translates to x is TRUE if and only if y is true. The if and only if operation has the following truth table: x y ${\displaystyle x\Leftrightarrow y}$ 0 0 1 0 1 0 1 0 0 1 1 1 The two new operations we have introduced are not really new, they are just combinations of AND, OR and NOT. For example: ${\displaystyle x\Rightarrow y=x'+y}$ Check it with a truth table. Because we can express the implication operations in terms of AND, OR and NOT, we have open them to manipulation by Boolean algebra and de Morgan's laws. Example 1 Is the following proposition a tautology (a proposition that's always true) ${\displaystyle [(x\Rightarrow y)(y\Rightarrow z)]\Rightarrow (x\Rightarrow z)}$ Solution 1 ${\displaystyle =\ [(x\Rightarrow y)(y\Rightarrow z)]\Rightarrow (x\Rightarrow z)}$ ${\displaystyle =\ [(x'+y)(y'+z)]'+(x'+z)}$ ${\displaystyle =\ (x'+y)'+(y'+z)'+x'+z}$ ${\displaystyle =\ xy'+yz'+x'+z}$ ${\displaystyle =\ y'+y+x'+z}$ ${\displaystyle =\ 1}$ Therefore it's a tautology. Solution 2 A somewhat easier solution is to draw up a truth table of the proposition, and note that the output column are all 1s. Therefore the proposition is a tautology, because the output is 1 regardless of the inputs (i.e. x, y and z). Example 2 Show that the proposition z is a contradiction (a proposition that is always false): z = xy(x + y)' Solution ${\displaystyle {\begin{matrix}z&=&xy(x+y)'\\&=&xy(x'y')\\&=&0\\\end{matrix}}}$ Back to Example 1, :${\displaystyle [(x\Rightarrow y)(y\Rightarrow z)]\Rightarrow (x\Rightarrow z)}$. This isn't just a slab of symbols, you should be able translate it into everyday language and understand intuitively why it's true. #### Exercises 1. Decide whether the following propositions are true or false: 1. If 1 + 2 = 3, then 2 + 2 = 5 2. If 1 + 1 = 3, then boys don't like mud 2. Show that the following pair of propositions are equivalent 1. ${\displaystyle x\Rightarrow y}$ : ${\displaystyle y'\Rightarrow x'}$ ## Logic Puzzles Puzzle is an all-encompassing word, it refers to anything trivial that requires solving. Here is a collection of logic puzzles that we can solve using Boolean algebra. Example 1 We have two type of people -- knights or knaves. A knight always tell the truth but the knaves always lie. Two people, Alex and Barbara, are chatting. Alex says :"We are both knaves" Who is who? We can probably work out that Alex is a knave in our heads, but the algebraic approach to determine Alex 's identity is as follows: Let A be TRUE if Alex is a knight Let B be TRUE if Barbara is a knight There are two situations, either: Alex is a knight and what he says is TRUE, OR he is NOT a knight and what he says is FALSE. There we have it, we only need to translate it into symbols: A(A'B') + A'[(A'B')'] = 1 we simplify: (AA')B' + A'[A + B] = 1 A'A + A'B = 1 A'B = 1 Therefore A is FALSE and B is TRUE. Therefore Alex is a knave and Barbara a knight. Example 2 There are three businessmen, conveniently named Archie, Billy and Charley, who order martinis together every weekend according to the following rules: 1. If A orders a martini, so does B. 2. Either B or C always order a martini, but never at the same lunch. 3. Either A or C always order a martini (or both) 4. If C orders a martini, so does A. 1. ${\displaystyle A\Rightarrow B}$ or ${\displaystyle A'+B=1}$ (simplified from: ${\displaystyle AB+A'B'+A'B=1}$ 2. ${\displaystyle B'C+BC'=1}$ 3. ${\displaystyle A+C=1}$ 4. ${\displaystyle C\Rightarrow A}$ or ${\displaystyle C'+A=1}$ (simplified from: ${\displaystyle CA+C'A'+C'A=1}$ Putting all these into one formula and simplifying: ${\displaystyle {\begin{matrix}1&=&(A'+B)(B'C+BC')(A+C)(C'+A)\\&=&(A'+B)(B'C+BC')(AC'+AA+CC'+AC)\\&=&(A'+B)(B'C+BC')(AC'+A+0+AC)\\&=&(A'+B)(B'C+BC')(AC'+A+AC)\\&=&(A'+B)(B'C+BC')(C'+1+C)A\\&=&(A'+B)(B'C+BC')(1)A\\&=&(A'+B)(B'C+BC')A\\&&{\mbox{Now that we know that }}A=1{\mbox{ we can substitute that in:}}\\&=&(0+B)(B'C+BC')1\\&=&(B)(B'C+BC')\\&&{\mbox{Now that we know that }}B=1{\mbox{ we can substitute that in:}}\\&=&(1)(0C+1C')\\&=&C'\\&&{\mbox{If }}1=C'{\mbox{ then }}C=0\\&&ABC'=1\end{matrix}}}$ ## Problem Set 1. Decide whether the following propositions are equivalent: ${\displaystyle x'\Rightarrow y'}$ ${\displaystyle y\Rightarrow x}$ 2. Express in simplest sum-of-product form the following proposition: ${\displaystyle (x\Leftrightarrow y)\Rightarrow z}$ 3. Translate the following sentences into symbolic form and decide if it's true: a. For all x, if x2 = 9 then x2 - 6x - 3 = 0 b. We can find a x, such that x2 = 9 and x2 - 6x - 3 = 0 are both true. 4. NAND is a binary operation: x NAND y = (xy)' Find a proposition that consists of only NAND operators, equivalent to: (x + y)w + z 5. Do the same with NOR operators. Recall that x NOR y = (x + y)' # Feedback What do you think? Too easy or too hard? Too much information or not enough? How can we improve? Please let us know by leaving a comment in the discussion tab. Better still, edit it yourself and make it better.
Courses Courses for Kids Free study material Offline Centres More Store # In a $\Delta PQR$, right angle at $Q$ , $PQ=4\,cm$ and $RQ=3\,cm$. Find the values of sin P, sec P and sec R. Last updated date: 11th Aug 2024 Total views: 411k Views today: 10.11k Verified 411k+ views Hint: We will first draw the figure from the given details in the question and then we will apply the Pythagoras theorem to find the value of the missing side. And after this we will see the point from which we need to find the values and according to that we will select our perpendicular and hypotenuse. Before proceeding with the question we should know the concept of Pythagoras theorem and right angled triangle. A Right-angled triangle is one of the most important shapes in geometry and is the basics of trigonometry. A right-angled triangle is the one which has 3 sides, “base” “hypotenuse” and “height” with the angle between base and height being 90 degrees. The Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90 degree. So we will first draw the figure from the given details in the question. Now we will use Pythagoras theorem and from the figure we get, $\Rightarrow P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}........(1)$ Now substituting the known values in equation (1) we get, $\Rightarrow P{{R}^{2}}={{4}^{2}}+{{3}^{2}}........(2)$ Now squaring the terms in the right hand side of the equation (2) and then adding we get, $\Rightarrow P{{R}^{2}}=16+9=25........(3)$ Now taking the square root on both sides in equation (3) we get, $\Rightarrow PR=5........(4)$ From the figure we can see that sin C is PQ (perpendicular) divided by PR (hypotenuse). Hence using this information we get, $\Rightarrow \sin P=\dfrac{perpendicular}{hypotenuse}=\dfrac{QR}{PR}=\dfrac{3}{5}$ Now, similarly, we can calculate the value of cos Q as follows $\Rightarrow \cos Q=\dfrac{base}{hypotenuse}=\dfrac{PQ}{PR}=\dfrac{4}{5}$ Now we know that $\text{sec}\,P=\dfrac{1}{\cos P}$. Hence substituting the value of cos P in this we get, $\Rightarrow \text{sec}\,P=\dfrac{5}{4}$ Hence the value of $\text{sec}\,P$ is $\dfrac{5}{4}$. Also, we can calculate the value of cos R as follows $\Rightarrow \cos R=\dfrac{base}{hypotenuse}=\dfrac{QR}{PR}=\dfrac{3}{5}$ Now we know that $\text{sec}\,R=\dfrac{1}{\cos R}$. Hence substituting the value of cos R in this we get, $\Rightarrow \text{sec}\,R=\dfrac{5}{3}$ Hence the value of $\text{sec}\,C$ is $\dfrac{5}{3}$. Note: Always remember that if we have right angle triangle, then using Pythagoras theorem, we can find any of the side of triangle as it states that ${{H}^{2}}={{P}^{2}}+{{B}^{2}}$, where H is hypotenuse, P is perpendicular and B is base of triangle and also $\operatorname{sinA}=\dfrac{Perpendicular}{Hypotenuse}$ , $\operatorname{cosA}=\dfrac{Base}{Hypotenuse}$ and $\tan A=\dfrac{Perpendicular}{Base}$. While simplifying numerical terms do calculation carefully as it will change the final answer and will make the solution more complex.
# Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. Question: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require? Solution: Given that: Mary wants to paste a paper on the outer surface of the wooden block. The quantity of the paper required would be equal to the surface area of the box which is of the shape of a cuboid. The dimensions of the wooden block are: Length (l) = 80 cm Height (h) = 20 cm Surface Area of the wooden box = 2[lb + bh + hl] = 2[(8040) + (4020) + (20*80)] = 2[5600] $=11200 \mathrm{~cm}^{2}$ The Area of each sheet of the paper $=40 \times 40 \mathrm{~cm}^{2}$ $=1600 \mathrm{~cm}^{2}$ Therefore, the number of sheets required $=\frac{\text { Surface area of the box }}{\text { Area of one sheet of paper }}$ =11200/1600 = 7 So, she would require 7 sheets.
Unveiling the Mystery of Interior Angles: How to Easily Calculate the Measure of Each Angle Introduction to Interior Angles: What Are They? Interior angles are those angles found between two sides of a polygon when two adjacent lines or line segments within the polygon form one large angle. The sum of all interior angles within a polygon adds up to 360 degrees for a triangle and 540 for a pentagon. In general, interior angles are used in geometry problems to figure out the number of sides a particular shape may have or its measure in terms of length. It is typical for any closed figure with straight line segments as its sides to contain at least one set of internal angles. When you add the exterior angle (the single angle outside) together with the interior ones, it always adds up to 360° – they are complementary to each other since they “complete” our full rotation around the shape back to the starting point. Using interior angles can help us determine properties of shapes such as size and structure, allowing us to explore geometry in much greater depth and detail than ever before. For example, if we know that an interior angle measures 150° then we can calculate that this shape has three sides since 150 x 3 = 450 which is equal to 360° (we also know it is not a triangle because an internal angle cannot be larger than 180°). Since these principles are essential for math and physics related studies, understanding what an interior angle is and how it should be measured can benefit any student at some point in their study curriculum. How to Calculate the Measure of a Single Angle Calculating the measure of a single angle is an important skill to have when working in geometry and trigonometry. To calculate the measure of a single angle, you will need to first identify the type of triangle it is contained within, then use the appropriate formula to find its measure. To begin, identify if the angle is contained within one of three types of triangles: an acute triangle, an obtuse triangle, or a right triangle. An acute triangle has all three angles that are less than 90 degrees; an obtuse triangle has one angle greater than 90 degrees; and a right triangle has one angle that equals exactly 90 degrees. This step is very important as using the wrong formula could provide inaccurate results. Once you know what type of triangle your angle occupies, use the following formulas to find its measure: Acute Triangle – Add all 3 angles together (A+ B+ C) and subtract this total from 180°. The answer you receive will be your desired angle (X). e.g A +B +C = 180° – X Obtuse Triangle – If two angles are known you can add their totals together (A+B) subtract from 180° and subtract this total from 360° providing your desired angle (X). e.g A +B =180 – 360 = X Right Triangle – When dealing with a right angled-triangle measuring an angle is easy since all other angles must equal 90° so simply take away 450 which will give you your X. e.g A +B =90 – 450 = X Using either method should enable you to easily calculate the measures for any single angle when given other relevant information about its containing triangle! Steps for Finding the Measure of Each Interior Angle Finding the measure of each interior angle of a regular polygon (a 2D shape with sides of equal length and angles of equal measure) is quite simple once you understand the equation involved. The equation to calculate the measure of each interior angle in a regular polygon is: Angle = (n-2)(180/n) Where n represents the number of sides in your shape. Below are a few steps for finding the measure of each interior angle by using this equation: Step 1: Count and record the number of sides your shape has, referring to it as ‘n’. For example, if your shape is a pentagon, ‘n’ would be 5. Step 2: Multiply ‘n’ by 180. For example, if your shape is still a pentagon, you would multiply 5 x 180 = 900. Step 3: Subtract two from ‘n’ and then divide it into the answer from Step 2. You should now have an answer that looks like this: (n-2)(180/n). Using our original example with an n value of 5, this would look like (5-2)(180/5). Solving this gives us an answer of 108 degrees – which is indeed the measure for each interior angle in a pentagon! This same process can be repeated for any regular polygon – follow these steps to find out how many degrees are in each internal angle for equilateral triangles, squares and octagons too! FAQs About Finding the Measure of Each Interior Angle What is the measure of an interior angle? The measure of an interior angle is the sum of the measures of its complementary exterior angles. To find this, you’ll have to know either the number of sides in the polygon or the degree measure of one exterior angle. How do I find the measure of an interior angle if I don’t know any other angle measurements? If you don’t know any other angle measurements, you will need to count each side in order to calculate all angles within that polygon. The formula for calculating an interior angle in a regular polygon is (n − 2)180/n , where n represents the number of sides. For example, in a pentagon there are five sides, so me would use (5-2)180/5 = 108° as our formula for finding each internal angles’ degree measure. How can I find out how many sides a shape has when I’m given just the interior angles? If you are only given the interior angles and not any other information about a shape then it can be difficult to determine exactly how many sides it has – since each individual set will form different kinds of polygons. Generally speaking though, if you have three consecutive angles measuring 60° degrees each then you likely have a triangle; four consecutive angles measuring 90° means you most likely have a square; five consecutive angles measuring 72° likely mean you have a pentagon; six consectutive angles measuring 120° likely mean that your shape is actually hexagonal and so forth. It’s important to note though that not all shapes follow this pattern – some triangles may not add up to 180 degrees while some 3 sided figures may be arranged with unevenly sized intervals between their individual corners. If this happens then it may indicate that each corner forms its own unique shape and isn’t apart from one another – therefore making it impossible to determine directly how many sides are present just by looking at its internal measurements alone. Top 5 Facts About Interior Angles 1. Interior angles are the angles inside a polygon that meet at one vertex. 2. An interior angle is usually equal in measure to its corresponding exterior angle. For example, if an exterior angle measures 30º, then the interior opposite to it will be also 30º, creating a straight line with 60º between them. 3.If all sides of a polygon have equal length, then all of its interior angles are also equal in measure. This makes regular polygons particularly easy to identify due to their even shape and angles. 4.In any triangle (a three-sided polygon), the sum of the three angles must always add up to 180° degrees; therefore, each triangle has three different types of angles: one acute angle (less than 90° degrees), one obtuse angle (more than 90° degrees) and one right angle (exactly 90° degrees). The other two internal angles in a triangle both measure less than 90° as acute angles so their sum equals 180° degrees total for the entire triangle’s interior angles’ measures.. 5. In more complex shapes with more than four sides like pentagons, hexagons and nonagons -all of which have five, six or nine sides respectively- figuring out exactly what type of angles those making up its interior contain requires solving equations based on facts such as number of sides and total sum of measurements.. From here we can then determine which type(s) of shapes make up each individual interior angle like scalene versus isosceles triangles depending on the situation at hand! In conclusion, a blog is a powerful tool that has the potential to reach a large audience. It can be used to communicate ideas, stories and advice or showcase content such as photographs and videos. It offers a platform for discussion, debate and publicity in all areas of life. Blogging is an effective way to stand out from the competition. It allows businesses, individuals and organisations to share their experiences and knowledge with others on the internet and create unique content that resonates with readers across the globe. Building relationships with readers will also help build customer loyalty and attract new customers. 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# Comparing Decimals, Fractions and Percentages II Lesson Now that we know how to convert both fractions and decimals into percentages, it's time to play around with them a bit and compare these different types of numbers. Let's take a look at the following question: #### Examples ##### QUESTION 1 Find the number amongst $4\frac{2}{5}$425%, $1.09$1.09, $\frac{61}{1000}$611000, $43%$43%, $\frac{8}{15}$815 that is closest to $50%$50% Think how can I put them all in the same form so I can compare them easily? Do Let's convert $1.09$1.09$\frac{61}{1000}$611000 and $\frac{8}{15}$815 into percentages. $1.09$1.09 $=$= $1.09\times100%$1.09×100% $=$= $109%$109% $\frac{61}{1000}$611000 $=$= $\frac{61}{1000}\times100%$611000×100% $=$= $\frac{6100%}{1000}$6100%1000 $=$= $6.1%$6.1% $\frac{8}{15}$815 $=$= $\frac{8}{15}\times100%$815×100% $=$= $\frac{800%}{15}$800%15 $=$= $53\frac{1}{3}$5313 $%$% $4\frac{2}{5}$425% < $6.1%$6.1% < $43%$43% < $53\frac{1}{3}$5313% < $109%$109% Therefore we can see that the percentage that is closest to $50%$50% is $53\frac{1}{3}$5313% which comes from $\frac{8}{15}$815. ##### QUESTION 2 Compare: $0.31$0.31 and $45%$45% 1. First convert $0.31$0.31 to a percentage. 2. Which of the two values is greater? $45%$45% A $0.31$0.31 B ##### QUESTION 3 Consider the statement: $\frac{67}{50}$6750 > $154%$154% 1. First convert $\frac{67}{50}$6750 to a percentage 2. Hence, is the statement True or False? True A False B
# Putting the fun into algebra Q I work with a brilliant support teacher who helps with a small group of pupils from my Year 8 set 3 in maths. Soon we will be doing an algebra module from our scheme of work. Have you any suggestions for a fun activity that will help their understanding of substitution into algebraic expressions? A The following activity can be done by individuals or in small groups; I have even done it with whole classes playing in teams. The activity is based on ordering a set of six cards showing different algebraic expressions. The order is determined by substituting a particular value into each expression. The number of copies of the cards for the activity obviously depends on the size of the group. For this article, I have assumed that the support teacher will be working with a group of four pupils. A photocopiable sheet with solutions is available at www.mathagonyaunt.co.ukarticlesindexJan312003.html The pupils play in pairs. Each pair needs four sets of the 6 circle cards (with each set on different colour card) and one set of the 4 rectangular substitution cards. For demonstration purposes you will need another two sets of 6 circle cards and an n = 4 card. Laminate the cards and then cut out. I have found that the organisation of "handing out" and "collecting in" is made easier by placing each set of sets in a re-sealable plastic bag. Pupils also need a calculator and paper for any rough working. First demonstrate the activity using a set of the circle cards and the n = 4 card. Explain that the purpose of this activity is to determine the order of the set of circle cards (from smallest to largest) for substitutions using different values of "n", as indicated on the rectangle cards. Take one demonstration set of six circles and the n = 4 card and ask the pupils, as a group, to decide what order they think the circle cards would be in if the "n" in the expression had a value of 4. In each case ask them why they think the cards should be in that order. Follow this by demonstrating the values that are created by correct substitution for each circle, writing the value in washable pen on the circles. Lay the second set of circles in their correct order beneath the pupils' chosen order, placing the n = 4 card at the end. Discuss the differences, pointing out why these might be different. There is a great deal that pupils can learn from this interactive discussion. Now the pupils work in pairs for each of the values given on the rectangular substitution cards to create the correct order of the expressions on the circle cards. Let your support teacher know that pupils are allowed help as they need it. When they have finished check their solutions with them and discuss those that are incorrect. Your support teacher might find it helpful to have some notes about what pupils might have forgotten or still not understood;I have given some examples of some common problem areas.For example, 8n is 8 x n (that is, 8n = n + n + n + n + n + n + n + n). Pupils sometimes find it useful to see this with numbers, and it helps to solve any misconception. n2 is n x n; pupils often mistake this notation and work out 2 x n instead of multiplying the number by itself. One way to help them is to encourage them to write n x n and then substitute the numbers beneath. For this exercise they can also use a calculator using the x2 button. For example for 22, type then on the calculator. 2n is 2 V n. The mistake pupils tend to make here is to read the calculation bottom to top: "n divided by 2". You should stress that it should be read from the top to the bottom as "two divided by n", and they should write it down in symbols as they say it: 2 V n. n2 is n V 2. The mistake here is similar to that given above. 8 - n. This becomes a problem when the value to be substituted is negative. Sometimes it helps to write the sum out: for example, when n = - 1 we have 8 - (-1) = 8 + 1 = 9. Similarly for - 2n, when n = - 1 this becomes - 2 x - 1 = 2. The support teacher might need to be shown how to use the fraction button on the scientific calculator prior to the session so that they can remind pupils how this is done. Wendy Fortescue-Hubbard is a teacher and game inventor. She has been awarded a three-year fellowship by the National Endowment for Science, Technology and the Arts (NESTA) to spread maths to the masses. Email your questions to Mathagony Aunt at teacher@tes.co.ukOr write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX It only takes a moment and you'll get access to more news, plus courses, jobs and teaching resources tailored to you ## Latest stories Will Hazell 22 June 2018 • ### 'I'm worth it', says £240k academy boss Martin George 22 June 2018 • ### Just 1 in 10 Stem apprentices female, report finds Julia Belgutay 22 June 2018 • ### Exclusive: Money 'biggest barrier' to edtech use Martin George 22 June 2018 Helen Ward 22 June 2018 • ### T levels exam board plans 'fundamentally flawed' Julia Belgutay 22 June 2018 • ### Teachers 'must confront sexism – including theirs' Emma Seith 22 June 2018 Martin George 21 June 2018 • ### First minister grilled over long inspection gaps Henry Hepburn 21 June 2018 John Roberts 21 June 2018
# Tangent Function(KS3, Year 8) ## The Lesson The tangent function relates a given angle to the opposite side and adjacent side of a right triangle. The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side in a right triangle. ## Dictionary Definition The Merriam-Webster dictionary defines the tangent function as "the trigonometric function that for an acute angle is the ratio between the leg opposite to the angle when it is considered part of a right triangle and the leg adjacent." The tangent of an angle is given by the formula below: In this formula, tan denotes the tangent function, θ is an angle of a right triangle, the opposite is the length of the side opposite the angle and the adjacent is the length of the side adjacent the angle. The image below shows what we mean: ## Interactive Widget Use this interactive widget to create a right-angled triangle and then use the tangent function to calculate the hidden element. Start by selecting which element you want to hide (using the green buttons) and then clicking in the shaded area. Angle: 0° Opposite: ? ## A Real Example of the Tangent Function It is easier to understand the tangent function with an example. ## Question Find tan 45° using the right triangle shown below. # 1 tan θ = opposite / adjacent Don't forget: / means ÷ # 2 Substitute the angle θ, the length of the opposite and the length of the adjacent into the formula. In our example, θ = 45°, the opposite is 4 cm and the adjacent is 4 cm. tan (45°) = 4 / 4 tan (45°) = 4 ÷ 4 tan (45°) = 1 tan 45° = 1. ## The Graph of the Tangent Function The tangent function can be plotted on a graph. The tangent function is not defined at 90°, 270° (and any amount of 180° added or subtracted from these angles). This is seen on the graph as vertical red dashed lines, which the tangent function approaches but never touches. These lines are called asymptotes (see Note.) Find the angle along the horizontal axis, then go up until you reach the tangent graph. Go across and read the value of tan θ from the vertical axis. We can see from the graph above that tan 45° = 1. ## The Tangent Function and the Unit Circle The tangent function can be related to a unit circle, which is a circle with a radius of 1 that is centred at the origin in the Cartesian coordinate system. For a point at any angle θ, tan θ is given by the ratio of the y-coordinate to the x-coordinate of the point. ## Lesson Slides The slider below gives more information about the tangent function. Open the slider in a new tab ## Interactive Widget Here is an interactive widget to help you learn about the tangent function on a right triangle. ## Trigonometry and Right Angles The tangent function is a function in trigonometry (called a trigonometric function). The word trigonometry comes from the Greek words 'trigonon' ("triangle") and 'metron' ("measure"). Trigonometry is the branch of mathematics that studies the relationships between the sides and the angles of right triangles. When an angle is defined in a right triangle, the three sides can be defined. • The side next to the angle is called the adjacent. • The side opposite the angle is called the opposite. • The longest side is called the hypotenuse. ## Other Trigonometric Functions The tangent function is only one of the trigonometric functions: ## The Tangent Function and Asymptotes It follows from the definition of the tangent function, and of the sine and cosine functions, that the tangent function can be expressed as: When cos θ = 0 (at θ = 90°, 270°, ...), sin θ is being divided by 0. When any number is divided by 0, the result is undefined. At these values of θ, tan θ is undefined. On the graph, these appear as vertical asymptotes. Either side of these values of θ, cos θ is very close to 0. It is a very small positive or negative number. At these values of θ, sin θ is being divided by a very small positive or negative number. The result is a very large positive or negative number. Either side of the asymptotes, tan θ approaches plus or minus infinity. Help Us To Improve Mathematics Monster • Did you spot a typo? Please tell us using this form
# Step Functions 2 teachers like this lesson Print Lesson ## Objective Students will be able to write and graph step functions #### Big Idea Use real world settings to distinguish between step functions and other general piece wise functions. Can you make the call? ## Warm up and Homework Check 10 minutes I include Warm ups with a Rubric as part of my daily routine. My goal is to allow students to work on Math Practice 3 each day. Grouping students into homogeneous pairs provides an opportunity for appropriately differentiated math conversations. This lesson’s Warm Up- Step Functions asks students to write a piecewise function given a scenario. I also use this time to correct and record the previous day's Homework. ## Introduction to Step Functions 15 minutes This lesson begins with a simple step function scenario: AT&T offers a 10¢ per minute cell phone plan. I ask the students to both graph and write an equation for this situation (Math Practice 4). Most students will want to do f(x)=0.10x along with the corresponding graph. Once they have had some time to work, I put this situation on the board and tell them that there is a problem with this model. I then ask them to do a think-pair-share to figure out the problem. It is important to clarify at this point that talking for 2 minutes and 1 second will be counted as 3 minutes. I have them redraw this graph up through five minutes including the proper open and closed end points. Now that they have a graph of this situation, I show how to write it as a floor function. We also talk about ceiling functions and the difference between the two. I then give them another scenario to model with a function and a graph: The seniors are going on a field trip. Each bus has a capacity of 60 students. This will help solidify this new concept. I also have them find the domain, range and intercepts for this function. It is good to compare the graph for this specific scenario to the general graph of the function. ## Is this a Step Function? 8 minutes The next scenario isn’t really a step function although it does share similarities with one. I took my dog to the groomer to get a bath. Here is the pricing: 0-30 pounds \$15 31-70 pounds \$25 > 70 pounds \$35 I give this to my students without a warning that it is different. Once they have had a chance to work on it a bit, there is a good opportunity for some discussions about the similarities and differences between this and a step function (Math Practice 7). They will then write a piecewise function and graph this scenario. ## Write your own Step Function 15 minutes The last activity in this lesson begins with a brainstorming session on situations that can be modeled as step functions. Students will come up with examples as pairs and then we will make a list as a class. Each student will then create their own step function scenario which they will switch with their partner to graph and write a function (Math Practice 2). This activity will help students deepen their understanding of step functions vs. other piecewise functions. ## Exit Ticket 2 minutes I use an exit ticket each day as a quick formative assessment to judge the success of the lesson. Today’s Exit Ticket asks student to write a step function from a real life scenario. ## Homework This short assignment gives the students 4 scenarios that are either step or piecewise functions. Students need to determine whether they are step or piecewise (Math Practice 7), write a function to match the scenario and graph it on a coordinate plane.
Introductory Statistics 2e Chapter Review 5.1Continuous Probability Functions The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P(a < x < b). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f(x), is used to draw the graph of the probability distribution. The total area under the graph of f(x) is one. The area under the graph of f(x) and between values a and b gives the probability P(a < x < b). Figure 5.35 The cumulative distribution function (cdf) of X is defined by P (Xx). It is a function of x that gives the probability that the random variable is less than or equal to x. 5.2The Uniform Distribution If X has a uniform distribution where a < x < b or axb, then X takes on values between a and b (may include a and b). All values x are equally likely. We write XU(a, b). The mean of X is $μ= a+b 2 μ= a+b 2$. The standard deviation of X is $σ= (b−a) 2 12 σ= (b−a) 2 12$. The probability density function of X is $f(x)= 1 b−a f(x)= 1 b−a$ for axb. The cumulative distribution function of X is P(Xx) = $x−a b−a x−a b−a$. X is continuous. Figure 5.36 The probability P(c < X < d) may be found by computing the area under f(x), between c and d. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. 5.3The Exponential Distribution If X has an exponential distribution with mean μ, then the decay parameter is m = $1 μ 1 μ$, and we write XExp(m) where x ≥ 0 and m > 0 . The probability density function of X is f(x) = me-mx (or equivalently $f(x)= 1 μ e −x/μ f(x)= 1 μ e −x/μ$. The cumulative distribution function of X is P(Xx) = 1 – emx. The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that P(X > x + k\|X > x) = P(X > k). If T represents the waiting time between events, and if TExp(λ), then the number of events X per unit time follows the Poisson distribution with mean λ. The probability density function of X is $P (X=k)= λ k e −k k! P (X=k)= λ k e −k k!$. This may be computed using a TI-83, 83+, 84, 84+ calculator with the command poissonpdf(λ, k). The cumulative distribution function P(Xk) may be computed using the TI-83, 83+,84, 84+ calculator with the command poissoncdf(λ, k).
Suppose we are asked to find the sum of two numbers, say 3 and 4. The sum of 3 and 4 is denoted by 3 + 4. Exactly in the same way, the sum of the literal x and a number 8 is denoted by x + 8 and is read ‘x plus 8’, which can also be read as ‘8 more than x,’ or ‘increase x by 8’. Similarly, y more than a literal x is written as x + y. We can also read x + y as the sum of x and y. (x + y) + z means that the sum of literals x and y is added to the literal z whereas x + ( y + z) means that the literal x is added to the sum of literals y and z. Since literals are used to represent numbers. Therefore, addition of variables obeys all properties of addition of numbers. Here, we list the properties of addition of variables. Commutativity : For any addition of two literals a and b, we have a + b = b + a Associativity : For any three literals a, b and c, we have (a + b) + c = a + (b + c) Identity : For any literal a, we have a + 0 = a = 0 + a, where 0 is known as additive identity Illustration : Write each of the following phrases using numbers, literals and the basic operation of addition: (i) The sum of y and 2. Solution: y + 2 (ii) 4 more than a number y. Solution: y +4 Solution: x + 8 (iv) Increase x by 5 Solution: x + 5 (v) The sum of x and 6 added to z Solution: (x + 6) + z (vi) y added to the sum of z and 4 Solution: y + (z + 4) (vii) The sum of a and b Solution: a + b (viii) The sum of c and 4 Solution: c + 4 Solution: r + 7 Solution: p + 9 Introduction to Algebra
# Solve the following Question: $8 x^{2}-9 x+3=0$ Solution: Given: $8 x^{2}-9 x+3=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=8, b=-9$ and $c=3$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{9+\sqrt{81-4 \times 8 \times 3}}{2 \times 8}$ and $\beta=\frac{9-\sqrt{81-4 \times 8 \times 3}}{2 \times 8}$ $\Rightarrow \alpha=\frac{9+\sqrt{81-96}}{16} \quad$ and $\quad \beta=\frac{9-\sqrt{81-96}}{16}$ $\Rightarrow \alpha=\frac{9+\sqrt{-15}}{16} \quad$ and $\quad \beta=\frac{9-\sqrt{-15}}{16}$ $\Rightarrow \alpha=\frac{9+\sqrt{15 i^{2}}}{16} \quad$ and $\quad \beta=\frac{9-\sqrt{15 i^{2}}}{16}$ $\Rightarrow \alpha=\frac{9+i \sqrt{15}}{16} \quad$ and $\quad \beta=\frac{9-i \sqrt{15}}{16}$ $\Rightarrow \alpha=\frac{9}{16}-\frac{\sqrt{15}}{16} i$ and $\beta=\frac{9}{16}+\frac{\sqrt{15}}{16} i$ Hence, the roots of the equation are $\frac{9}{16} \pm \frac{\sqrt{15}}{16} i$.
# Difference between revisions of "2007 AIME I Problems/Problem 6" ## Problem A frog is placed at the origin on the number line, and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A move sequence is a sequence of coordinates which correspond to valid moves, beginning with 0, and ending with 39. For example, $0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39$ is a move sequence. How many move sequences are possible for the frog? ## Solution ### Solution 1 Let us keep a careful tree of the possible number of paths around every multiple of $13$. From $0 \Rightarrow 13$, we can end at either $12$ (mult. of 3) or $13$ (mult. of 13). • Only $1$ path leads to $12$ • Continuing from $12$, there is $1 \cdot 1 = 1$ way to continue to $24$ • There are $1 \cdot \left(\frac{24-15}{3} + 1\right) = 4$ ways to reach $26$. • There are $\frac{12 - 0}{3} + 1 = 5$ ways to reach $13$. • Continuing from $13$, there are $5 \cdot 1 = 5$ ways to get to $24$ • There are $5 \cdot \left(\frac{24-15}{3} + 1 + 1\right) = 25$ ways (the first 1 to make it inclusive, the second to also jump from $13 \Rightarrow 26$) to get to $26$. Regrouping, work from $24 \| 26\Rightarrow 39$ • There are $1 + 5 = 6$ ways to get to $24$ • Continuing from $24$, there are $6 \cdot \left(\frac{39 - 27}{3}\right) = 24$ ways to continue to $39$. • There are $4 + 25 = 29$ ways to reach $26$. • Continuing from $26$, there are $29 \cdot \left(\frac{39-27}{3} + 1\right) = 145$ (note that the 1 is not to inclusive, but to count $26 \Rightarrow 39$). In total, we get $145 + 24 = 169$. In summary, we can draw the following tree, where in $(x,y)$, $x$ represents the current position on the number line, and $y$ represents the number of paths to get there: $(12,1)$ $(24,1)$ $(39,4)$ $(26,4)$ $(39,20)$ $(13,5)$ $(24,5)$ $(39,20)$ $(26,25)$ $(39,125)$ Again, this totals $4 + 20 + 20 + 125 = 169$. ### Solution 2 We divide it into 3 stages. The first occurs before the frog moves past 13. The second occurs before it moves past 26, and the last is everything else. For the first stage the possible paths are $(0,13)$, $(0,3,13)$, $(0,3,6,13)$, $(0,3,6,9,13)$, $(0,3,6,9,12,13)$, and $(0,3,6,9,12)$. That is a total of 6. For the second stage the possible paths are $(26)$, $(15,26)$, $(15,18,26)$, $(15,18,21,26)$, $(15,18,21,24,26)$, and $(15,18,21,24)$. That is a total of 6. For the third stage the possible paths are $(39)$, $(27,39)$, $(27,30,39)$, $(27,30,33,39)$, and $(27,30,33,36,39)$. That is a total of 5. However, we cannot jump from $12 \Rightarrow 26$ (this eliminates 5 paths) or $24 \Rightarrow 39$ (this eliminates 6 paths), so we must subtract $6 + 5 = 11$. The answer is $665 - 11=169$ ### Solution 3 Another way would be to use a table representing the number of ways to reach a certain number $\begin{tabular}{c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\ \hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\ \end{tabular}$ How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$, we can reach it from $13, 15, 18, 21, 24$, so we add all those values to get the value for $26$. For $27$, it is only reachable from $24$ or $26$, so we have $29 + 6 = 35$. The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = \boxed{169}$. ## Solution 4 I believe this is an easier way of organizing the solution to reduce the possibility of mistakes. This is a highly visual solution, so it's much easier to record than a tree or table. Using graph paper, draw a number line from 0-39. On one line, dot every multiple of 3. Then on a line below it, dot every multiple of 13. This way you can clearly see which goes before or after which. To make it easier to understand, I'll compare these jumps to a train system. Imagine that every multiple of 3 is a short transit, while the multiples of 13 are long transits; because of the possibility to skip a large section in one move. As we continue, picture each "transit" of 13 to be an option, like a switch. If you are the frog that are riding these trains, you would probably think like this: "I could use the first long transit, skip the second, and use the third. Or, I could skip both first and second and use only the third. etc.) From now on I'll be calling the multiples of 13: 13, 26, and 39, as stations 1, 2, and 3 for clarity. Thinking like this organizes the problem efficiently into $2^3$ cases, where you could choose which "long transits" to ride. We will start with the harder cases and move downwards. Write out each of the 8 cases to record each one. Once we do the hardest case, which is the first one, every preceding one will be easier with the knowledge we gather. Case #1: 111 You have 5 locations to choose to jump to station 1. From the start (0), 3, 6, 9, or 12. The same goes for station #2, with 13, 15, 18, 21, or 24. However, station #3 is tricky. You can jump from 26, 27, 30, or 33, but if you jumped from 36, then that could qualify as skipping jumping station #3, because station #3 is both a multiple of 3 and a multiple of 13. For example, if you jumped from 36 as your last move, then those moves are essentially the same as case 110. So, $554=100$. Case #2: 110 Start with what you did on case #1, but when you reach station 2 continue only jumping multiples of 3 to 39. $55=25$ Case #3: 101 We have the 5 original choices, and when you reach station #1, move to 27 on the number line, slightly ahead of station #2. Here is another tricky part. Since we did not start on station 2 like we did in case #1, there are only 3 cases instead of 4. $53=15$ Case #4: 100 This one is simple. $5$ Case #5: 011 Jump multiples of 3 to get to 15, where you will have 4 locations to jump to station 2, and another 4 choices to reach station 3. $4*4=16$ Case #6: 010 This one is also simple. Once you reach 15 there are only 4 choices. $4$ Case #7: 001 It only gets simpler. You can only jump at 27, 30, or 33. $3$ Case #8: 000 Simple jumps. $1$. You have reached the end! $100+25+15+5+16+4+3+1=\boxed{169}$. -jackshi2006
Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Sonia.trakroo S Sonia.trakroo Community Contributor Quizzes Created: 1 \| Total Attempts: 251 Questions: 10 \| Attempts: 251 Settings . • 1. 5674 more than in 1287. • A. 6961 • B. 6691 • C. 8732 • D. 6960 A. 6961 Explanation The correct answer is 6961 because it is 5674 more than 1287. By adding 5674 to 1287, we get 6961. Rate this question: • 2. Multiply 264 by 987 using a suitable strategy. • A. 260,000 • B. 602,865 • C. 260,568 • D. 266,588 C. 260,568 Explanation To multiply 264 by 987, we can use the standard long multiplication method. We multiply 264 by each digit of 987, starting from the rightmost digit. We get 4 multiplied by 264 equals 1056, 8 multiplied by 264 equals 2112, and 9 multiplied by 264 equals 2376. Then, we add these three products together: 1056 + 2112 + 2376 = 5544. Therefore, the correct answer is 260,568. Rate this question: • 3. Estimate the sum of 9708 and 4302. • A. 14023 • B. 14000 • C. 14010 • D. 15000 B. 14000 Explanation The correct answer is 14000 because when estimating the sum of 9708 and 4302, we round each number to the nearest thousand. 9708 rounds down to 9000 and 4302 rounds up to 5000. Adding these rounded numbers together gives us 9000 + 5000 = 14000. Rate this question: • 4. Estimate the product of 54 and 56. • A. 3024 • B. 3042 • C. 2500 • D. 3000 D. 3000 Explanation To estimate the product of 54 and 56, we can round both numbers to the nearest tens place. 54 rounds to 50 and 56 rounds to 60. Multiplying 50 and 60 gives us 3000, which is the closest estimate to the actual product. Therefore, the answer is 3000. Rate this question: • 5. Reggie bought some toys for his new puppy to play with. The puppy played with the rope toy for 5 minutes. The puppy played with the frisbee for 5 times as long as he played with the rope toy. For how many minutes did he play with the frisbee? • A. 30 • B. 10 • C. 25 • D. 15 C. 25 Explanation The puppy played with the rope toy for 5 minutes. According to the information given, the puppy played with the frisbee for 5 times as long as he played with the rope toy. Therefore, if the puppy played with the rope toy for 5 minutes, he played with the frisbee for 5 * 5 = 25 minutes. Rate this question: • 6. Over the weekend, Tess sold boxes of cookies in her neighborhood. She sold 9 boxes of cookies on Saturday. She sold 3 times as many boxes of cookies on Sunday as she sold on Saturday. How many boxes of cookies did Tess sell on Sunday? • A. 27 • B. 10 • C. 30 • D. 20 A. 27 Explanation Tess sold 9 boxes of cookies on Saturday. The question states that she sold 3 times as many boxes of cookies on Sunday as she sold on Saturday. Therefore, to find out how many boxes she sold on Sunday, we need to multiply the number of boxes sold on Saturday (9) by 3. 9 multiplied by 3 equals 27, so Tess sold 27 boxes of cookies on Sunday. Rate this question: • 7. In the winter, Mr. Murphy's bee hives had 759 worker bees. Now that spring has come, the number of worker bees has grown by 388. About how many worker bees does he have now? Choose a better estimate. • A. 1200 • B. 200 • C. 1500 • D. 1147 A. 1200 Explanation Mr. Murphy's bee hives had 759 worker bees in the winter. Now that spring has come, the number of worker bees has grown by 388. To estimate the current number of worker bees, we can round 388 to the nearest hundred, which is 400. Adding 400 to the initial count of 759 gives us a total of 1159 worker bees. The closest option to this estimate is 1200, which is the correct answer. Rate this question: • 8. To water her crops, a farmer used 746 liters of water from a river. In addition, she used 593 liters from a well. In all, how many liters of water did she use? • A. 1340 • B. 1339 • C. 1430 • D. 1437 B. 1339 Explanation The farmer used 746 liters of water from the river and 593 liters from the well. To find the total amount of water used, we add these two quantities together: 746 + 593 = 1339. Therefore, the correct answer is 1339 liters. Rate this question: • 9. Martin makes a fruit smoothie for breakfast every day. This morning, he put 5 strawberries in the smoothie. He also put twice as many mango chunks in the smoothie as strawberries. How many mango chunks did Martin put in the smoothie? • A. 10 • B. 5 • C. 15 • D. 20 A. 10 Explanation Martin put 10 mango chunks in the smoothie. Since he put twice as many mango chunks as strawberries, and he put 5 strawberries, he put 2 * 5 = 10 mango chunks in the smoothie. Rate this question: • 10. A shipping company delivered 4,580 letters and 5,856 packages. How many items did the company deliver in all? • A. 10,000 • B. 10,643 • C. 10,436 • D. 10,017 C. 10,436 Explanation The company delivered a total of 4,580 letters and 5,856 packages. To find the total number of items, we add these two numbers together. Therefore, the correct answer is 10,436. Rate this question: Related Topics
Newton-Raphson Method I was recently asked by a class to go over the Newton-Raphson method for solving non-linear equations. Specifically in this case it was to solve 1D gas dynamics equations. Here I will just do a brief overview of the method, and how its used. I will solve two cases, one where the derivative of the function is known, and one where the derivative of the function should be approximated. Contents Here are the contents you’ll find in this post: 1. Introduction to the Newton-Raphson Method 2. Using the Method 1. Example 1 (derivative of the function is known) • Calculating Mach number from critical area ratio (M from A/A) 2. Example 2 (derivative of the function is unknown… or to annoying to derive) • Calculating incident shock pressure ratio from diaphragm pressure ratio 3. Conclusions and Useful References Introduction to the Newton-Raphson Method What is the Newton-Raphson method? Basically it is an iterative approach for solving the roots of functions. There are tons of these. You probably don’t need to know all of them (just pick a few that work for you!) Typically I stick to the Newton-Raphson method and the bisection method and I rarely have problems. The Newton-Raphson method is used when you have some function f(x) and you want to find the value of the dependent variable (x) when the function equals zero. AKA you want to find the roots of the equation. If you have an initial guess at some point, $x_i$, the tangent can be extended to some point that crosses 0 at an easily calculable point $x_{i+1}$. This point gives an improved estimation of the root. The basics of the method come from the fact that the first derivative is equivalent to the slope and therefore if you know it, you can calculate the tangent intersection: $f'(x_i)=\frac{f(x_i)-0}{x_i-x_{i+1}}$ (1) Which we can rearrange to get the “Newton-Raphson Formula”: $x_{i+1}=x_{i}-\frac{f(x_{i})}{\frac{df}{dx}(x_{i})}$ (2) The method is depicted in the figure below: Equation 2 pretty much sums up the method. If you start with a known derivative and function value $x_i$ you can calculate a new prediction $x_{i+1}$. By repeating this process (perform iterations) better and better approximations of the value for x are obtained. Using the Method Example 1 (derivative of the function is known): Lets say for this example that we know a critical area ratio (let’s say A/A=3.3, with $\gamma=1.4$) at a point in the flow field, and we want to use it calculate Mach number. For this we will solve the following equation: $\frac{A}{A^}=\frac{1}{M}\left[\left( \frac{2}{\gamma+1}\right) \left(1 +\frac{\gamma -1}{2} M^2\right)\right]^{\frac{\gamma+1}{2\left(\gamma-1\right)}}$ (3) This equation is equation for the critical area ratio for a given Mach number. Sometimes however the A/A is known and the Mach number is desired. Let’s say A/A* =3.3 what is the Mach number? To do this, this equation can be easily solved using the Newton-Raphson approach. Essentially the Newton-Raphson method is a root finding method. This means we are finding where an equation crosses zero. So how do we apply this to the problem above? We aren’t interested in where the equation above crosses zero, we are interested in where it crosses 3.3! Well it’s easy to change the above equation to give: $f(M)= \left[\left( \frac{2}{\gamma+1}\right) \left(1 +\frac{\gamma -1}{2} M^2\right)\right]^{\frac{\gamma+1}{2\left(\gamma-1\right)}} - M\frac{A}{A^} = 0$ (4) So now the problem is in a form that we can use the Newton-Raphson approach. In general, the approach is written as: $x_{new}=x_{old}-\frac{f(x_{old})}{\frac{df}{dx}(x_{old})}$ (5) Here, f(x) is the equation that is to be solved (in our case Equation 4), x is the variable to be solved for (in our case Mach number) and df/dx is the derivative of the equation with respect to the unknown variable (in our case the derivative of Equation 4 with respect to Mach number). So at this point we know f(M), but we must calculate the derivative of f(M). The result is the following: $\frac{df}{dM}(M)= \frac{A}{A^}-(\frac{2}{\gamma+1})^{\frac{\gamma+1}{2\left(\gamma-1\right)}-1}M\left(1+\frac{(\gamma-1)M^2}{2}\right)^{\frac{\gamma+1}{2\left(\gamma-1\right)}-1}$ (6) The next step is to apply the iterative algorithm. Essentially, you have to make an intelligent initial guess for the unknown variable (Mach number). In our case let’s make our first guess M=3. In our first step this is $M_{old}$. Using this value we can get a guess for $M_{new}$. Then at our next iteration $M_{new}$ becomes $M_{old}$ and this process is repeated until convergence ($M_{old} \approx M_{new}$). The following is a table showing the convergence of this problem: So the Mach number for an A/A* of 3.3 is 2.738! But what if we had chosen a different starting point? Say… 0.5? Well then the answer looks like… Wait why is the answer different?? Dont forget! There are two solutions to this problem, a subsonic and a supersonic solution. When we started at M=3 we converged on the supersonic solution (M=2.73805). When we start at 0.5 we converged on the subsonic solution (M=0.17875). Example 2 (derivative df/dx is unknown… or too annoying to derive): For example 2 we are again going to solve a simple 1D gas dynamics problem using the Newton-Raphson method. For this problem, let’s say that we are given a diaphragm pressure ratio for a shock tube ($P_4/P_1=40$). We want to solve for the pressure ratio that will exist across the normal shock that is created when the diaphragm ruptures. For this problem see John and Keith (2006) for an excellent chapter on the shock-tube problem. Basically this problem boils down to solving the following equation $\frac{P_4}{P_1}=\frac{P_2}{P_1}\left[1-\frac{(\gamma_4-1)(\frac{a_1}{a_4})(\frac{P_2}{P_1}-1)}{\sqrt{2\gamma_1}\sqrt{2\gamma_1+(\gamma_1+1)(\frac{P_2}{P_1}-1)}}\right]^{-\frac{2\gamma_4}{\gamma_4-1}}$ (7) for a $\frac{P_2}{P_1}$ with a known shock tube pressure ratio $\frac{P_4}{P_1}$. Equation 5 is not the type of equation that is very attractive to differentiate, so in this example we will use a simple central different to approximate the derivative. Essentially what this means is that instead of directly calculating the derivative at each of our iterations, we will use a point slightly to the right and slightly to the left in order to approximate the slope. First of all we rearrange (5) to give us an equation that equals zero (so that we can find its roots). Demonstrating this is easier if we set our unknown $p=P_2/P_1$. $f(p)=\frac{P_4}{P_1}\left[1-\frac{(\gamma_4-1)(\frac{a_1}{a_4})(p-1)}{\sqrt{2\gamma_1}\sqrt{2\gamma_1+(\gamma_1+1)(p-1)}}\right]^{\frac{2\gamma_4}{\gamma_4-1}}-p=0$ (8) For the Newton-Raphson the above equation would be solved using: $p_{new}=p_{old}-\frac{f(p_{old})}{\frac{df}{dp}(p_{old})}$ (9) Since we do not know the derivative of f approximate it with a central difference (centered approximation of the local slope at that point): $\frac{df}{dp}(p_{old})=\frac{f(p_{old}+\Delta p) -f(p_{old}-\Delta p)} {2 \Delta p}$ (10) so: $p_{new}=p_{old}-2\Delta p \frac{f(p_{old})}{f(p_{old}+\Delta p) -f(p_{old}-\Delta p)}$ (11) With the above iterative equation, the problem is solved in the same fashion as Example 1. See the following table: Using Newton-Raphson we have now calculated that for a diaphragm pressure ratio of 40, the pressure ratio across the shock ($P_2/P_1$) should be 4.7726. Conclusion and Useful References Hopefully this was helpful. The Newton-Raphson method is extremely useful. However be careful! There are some cases where this method doesn’t work very well. For example if an equation has multiple roots, your initial guess must be fairly close to the answer you are looking for or you could get the completely wrong root. It often helps to plot your function beforehand or after so that you know what it looks like, and it also helps to never blindly trust the answer. If it isn’t physical… it probably isn’t the one you were looking for. Here are some useful references [1] https://en.wikipedia.org/wiki/Newton%27s_method [2] John, J. E. A., & Keith, T, G. (2006) Gas Dynamics, 3rd Edition, Pearson Prentice Hall [3] Chapra, S. C., Canale, R. P. (2010) Numerical Methods for Engineers (6th Ed.), McGraw-Hill Categories Uncategorized
# Teaching Properties of Numbers to 7th Graders Would You Rather Listen to the Lesson? Properties of real numbers is one of the earliest lessons students learn in pre-algebra in 7th grade. By using the properties of numbers, 7th graders are prepared to easily simplify expressions and solve linear equations. To help students achieve mastery of properties of real numbers, math teachers can benefit from a number of tips. Read on to learn more about such tips that are guaranteed to keep your students engaged throughout the lesson! ## What Are Properties of Numbers (7th Grade)? For starters, you can explain to children how we define properties of real numbers. Properties of real numbers are simply “laws” of numbers that apply to certain math operations. Point out that there are four basic properties: • Commutative • Associative • Identity • Distributive Make sure to highlight that these properties don’t apply to all operations, but only when we add or multiply numbers. In other words, the properties of real numbers don’t work for division or subtraction. You can illustrate this later with examples as you go through each property. ## How to Teach Properties of Numbers (7th Grade) After providing a brief introduction, you can provide an overview of each property separately, alongside examples of how it’s used. You can also rely on using videos to teach this lesson in your class, such as this fun video that features a ‘Mathemagician’ showing the properties of real numbers by using magical math tricks. ### Commutative Property The commutative property states that we can change the order of the numbers that we’re adding, but it won’t affect the answer. That is, we can swap out numbers but the sum remains the same. We can present this in the following way: a + b = b + a Provide a few examples, such as: 3 + 2 = 2 + 3 = 5 4 + 5 = 5 + 4 = 9 A useful way of demonstrating this is with the help of marbles. Bring a few marbles to class and share them between two student groups – group one should have four marbles and group two should have five marbles. Let children visualize the commutative property by swapping the groups and counting the sum of marbles, which stays the same even when the groups are swapped. #### Multiplication The commutative property also holds true for multiplications, stating that we can swap over the numbers that are being multiplied without making any difference to the product. That is, regardless of the order of numbers being multiplied, the product remains the same. We can present this in the following way: a x b = b x a Provide a few examples, such as: 4 x 2 = 2 x 4 = 8 3 x 5 = 5 x 3 = 15 #### Subtraction and Division Finally, remind children that the commutative property doesn’t work when we’re dividing or subtracting numbers. Demonstrate this with a few examples, such as: 20 ÷ 4 = 5 But: 4 ÷ 20 = 0.2 5 – 1 = 4 But: 1 – 5 = – 4 ### Associative Property Explain that the associative property refers to number grouping, that is, which one we calculate first. It states that when adding three or more numbers, we can change the way in which we group these numbers, but this won’t affect the sum. We can present it like this: (a + b) + c = a + (b + c) Provide an example, such as: (2 + 3) + 5 = 2 + (3 + 5) = 10 #### Multiplication Add that the associative property also applies to multiplication. It states that when we’re multiplying three or more numbers, we can change the grouping of the numbers being multiplied, but this won’t affect the product. We can present it like this: (a × b) × c = a × (b × c) Provide an example, such as: (2 x 3) x 5 = 2 x (3 x 5) = 30 #### Subtraction and Division Same as with the commutative property, the associative property does not work for subtraction or division. Demonstrate this with an example, such as: (8 ÷ 4) ÷ 2 = 1 But: 8 ÷ (4 ÷ 2) = 4 ### Identity Property Explain that the identity property states that for any number a, the sum of 0 and the number a is the number itself. That is, by adding any real number to zero, the sum is equal to the number itself. We can present this property for addition in the following way: a + 0 = a 0 + a = a Provide a few examples to demonstrate how this works, for instance: – 4 + 0 = – 4 And also: 0 + (-4) = – 4 3 + 0 = 3 And also: 0 + 3 = 3 #### Multiplication Add that the identity property also works for multiplication. That is, the identity property states that for any number a, the product of 1 and the number a is the number itself. We can present this in the following way: a x 1 = a Provide a few examples, such as: 7 x 1 = 7 – 23 x 1 = – 23 ### Distributive Property As the name suggests, the distributive property helps us distribute a given expression. This property states that if we add two or more numbers and multiply this sum to an outside number, – the result will be the same if we multiply every number in the parenthesis by the outside number and then add the products. We can present this in the following way: a x (b + c) = a x b + a x c Provide an example of this, such as: 3 x (4 + 5) = 3 x 4 + 3 x 5 = 27 ## Activities to Practice Properties of Numbers (7th Grade) ### Group Work Use this simple activity at the end of the lesson to reinforce students’ knowledge of the properties of real numbers. To prepare for this activity, simply print out this Assignment Worksheet. Make sure you make enough copies for each student. The worksheet contains exercises with equations where children are asked to evaluate an expression and then identify the type of property that is used in each equation. Writing and evaluating expressions is something that children should already master, but you can refresh their skills by referring to this article. Divide children into groups of 3, 4. Hand out the worksheets and provide instructions. All members of a given group work together to solve the exercises. In the end, each group presents their answers in front of the class. If there’s time left, you can also use this Assignment Worksheet that focuses on solving math problems by applying the distributive property of numbers. As these exercises are more advanced, you can leave them for the end to provide a challenge to students. Parents who are homeschooling their children can easily transform this activity into an individual activity. Or if you have an older child, you can simply involve them in reviewing the properties of numbers! ## Before You Leave… If you liked these tips on teaching properties of numbers to 7th graders, we have a whole free lesson that’s dedicated to this topic! So make sure you check out our worksheets and resources below. These are all PDFs, so they’ll print out easily: To get the Editable versions of these files, join us in the Math Teacher Coach community!
# How do you integrate int sqrt(x^2-25) by trigonometric substitution? Sep 5, 2016 $\frac{x \sqrt{{x}^{2} - 25} - 25 \ln \left(\left\mid \sqrt{{x}^{2} - 25} + x \right\mid\right)}{2} + C$ #### Explanation: We have: $I = \int \sqrt{{x}^{2} - 25} \mathrm{dx}$ Let $x = 5 \sec \left(u\right)$. Note that this implies $\mathrm{dx} = 5 \sec \left(u\right) \tan \left(u\right) \mathrm{du}$. Thus: $I = 5 \int \sec \left(u\right) \tan \left(u\right) \sqrt{25 {\sec}^{2} \left(u\right) - 25} \mathrm{du}$ Factoring $\sqrt{25} = 5$: $I = 25 \int \sec \left(u\right) \tan \left(u\right) \sqrt{{\sec}^{2} \left(u\right) - 1} \mathrm{du}$ Note that since ${\tan}^{2} \left(u\right) + 1 = {\sec}^{2} \left(u\right)$, we can say that $\sqrt{{\sec}^{2} \left(u\right) - 1} = \tan \left(u\right)$, which is why we chose the original subsitution of $x = 5 \sec \left(u\right)$. This yields: $I = 25 \int \sec \left(u\right) {\tan}^{2} \left(u\right) \mathrm{du}$ Now, rewrite this using ${\tan}^{2} \left(u\right) = {\sec}^{2} \left(u\right) - 1$: $I = 25 \int \sec \left(u\right) \left({\sec}^{2} \left(u\right) - 1\right) \mathrm{du}$ $I = 25 \int {\sec}^{3} \left(u\right) \mathrm{du} - 25 \int \sec \left(u\right) \mathrm{du}$ The second integral is a common integral: $I = 25 \int {\sec}^{3} \left(u\right) \mathrm{du} - 25 \ln \left(\left\mid \tan \left(u\right) + \sec \left(u\right) \right\mid\right) \text{ } \textcolor{red}{\overline{\underline{\left\mid \star \right\mid}}}$ Letting ${I}_{1} = \int {\sec}^{3} \left(u\right) \mathrm{du}$, we will solve for it using integration by parts, in the form: $\int s \mathrm{dt} = s t - \int t \mathrm{ds}$ So, let: $\left\{\begin{matrix}s = \sec \left(u\right) \text{ "=>" "ds=sec(u)tan(u)du \\ dt=sec^2(u)du" "=>" } t = \tan \left(u\right)\end{matrix}\right.$ Thus: ${I}_{1} = \sec \left(u\right) \tan \left(u\right) - \int \sec \left(u\right) {\tan}^{2} \left(u\right)$ Letting $\tan \left(u\right) = {\sec}^{2} \left(u\right) - 1$: ${I}_{1} = \sec \left(u\right) \tan \left(u\right) - \int \sec \left(u\right) \left({\sec}^{2} \left(u\right) - 1\right) \mathrm{du}$ ${I}_{1} = \sec \left(u\right) \tan \left(u\right) - \int {\sec}^{3} \left(u\right) + \int \sec \left(u\right)$ Here, we see that $\int {\sec}^{3} \left(u\right) = {I}_{1}$, the integral we're currently solving for. Additionally, we've already integrated $\sec \left(u\right)$ once: ${I}_{1} = \sec \left(u\right) \tan \left(u\right) - {I}_{1} + \ln \left(\left\mid \tan \left(u\right) + \sec \left(u\right) \right\mid\right)$ Adding ${I}_{1}$ to both sides: $2 {I}_{1} = \sec \left(u\right) \tan \left(u\right) + \ln \left(\left\mid \tan \left(u\right) + \sec \left(u\right) \right\mid\right)$ ${I}_{1} = \frac{1}{2} \sec \left(u\right) \tan \left(u\right) + \frac{1}{2} \left(\left\mid \tan \left(u\right) + \sec \left(u\right) \right\mid\right)$ Returning to $\textcolor{red}{\overline{\underline{\left\mid \star \right\mid}}}$, we see that: $I = 25 \left(\frac{1}{2} \sec \left(u\right) \tan \left(u\right) + \frac{1}{2} \left(\left\mid \tan \left(u\right) + \sec \left(u\right) \right\mid\right)\right) - 25 \ln \left(\left\mid \tan \left(u\right) + \sec \left(u\right) \right\mid\right)$ $I = \frac{25}{2} \sec \left(u\right) \tan \left(u\right) - \frac{25}{2} \ln \left(\left\mid \tan \left(u\right) + \sec \left(u\right) \right\mid\right)$ Since we have $x = 5 \sec \left(u\right)$, write this all in terms of $\sec \left(u\right)$: $I = \frac{25}{2} \sec \left(u\right) \sqrt{{\sec}^{2} \left(u\right) - 1} - \frac{25}{2} \ln \left(\left\mid \sqrt{{\sec}^{2} \left(u\right) - 1} + \sec \left(u\right) \right\mid\right)$ Note that $\frac{x}{5} = \sec \left(u\right)$ and ${\sec}^{2} \left(u\right) = {x}^{2} / 25$: $I = \frac{25}{2} \left(\frac{x}{5}\right) \sqrt{{x}^{2} / 25 - 1} - \frac{25}{2} \ln \left(\left\mid \sqrt{{x}^{2} / 25 - 1} + \frac{x}{5} \right\mid\right)$ $I = \frac{5}{2} x \sqrt{\frac{{x}^{2} - 25}{25}} - \frac{25}{2} \ln \left(\left\mid \sqrt{\frac{{x}^{2} - 25}{25}} + \frac{x}{5} \right\mid\right)$ Factoring out $\sqrt{\frac{1}{25}} = \frac{1}{5}$: $I = \frac{1}{2} x \sqrt{{x}^{2} - 25} - \frac{25}{2} \ln \left(\left\mid \frac{1}{5} \sqrt{{x}^{2} - 25} + \frac{x}{5} \right\mid\right)$ $I = \frac{1}{2} x \sqrt{{x}^{2} - 25} - \frac{25}{2} \ln \left(\left\mid \frac{1}{5} \left(\sqrt{{x}^{2} - 25} + x\right) \right\mid\right)$ $I = \frac{1}{2} x \sqrt{{x}^{2} - 25} - \frac{25}{2} \ln \left(\left\mid \sqrt{{x}^{2} - 25} + x \right\mid\right) - \frac{25}{2} \ln \left(\frac{1}{5}\right) + C$ Note that $- \frac{25}{2} \ln \left(\frac{1}{5}\right)$ is absorbed into $C$: $I = \frac{1}{2} x \sqrt{{x}^{2} - 25} - \frac{25}{2} \ln \left(\left\mid \sqrt{{x}^{2} - 25} + x \right\mid\right) + C$ And, finally: $I = \frac{x \sqrt{{x}^{2} - 25} - 25 \ln \left(\left\mid \sqrt{{x}^{2} - 25} + x \right\mid\right)}{2} + C$
Get here NCERT Solutions for Class 6 Maths Chapter 2. These NCERT Solutions for Class 6 of Maths subject includes detailed answers of all the questions in Chapter 2 – Whole Numbers provided in NCERT Book which is prescribed for class 6 in schools. Resource: National Council of Educational Research and Training (NCERT) Solutions Class: 6th Class Subject: Maths Chapter: Chapter 2 – Whole Numbers ## NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers Class 6th Maths Chapter 2 Whole Numbers NCERT Solution is given below. Exercise – 2.1 Question 1: Write the next three natural numbers after 10999. Next three natural numbers after 10999 are 11000, 11001, 11002 Question 2: Write the three whole numbers occurring just before 10001. 3 whole numbers just before 10001 are 10000, 9999, 9998 Question 3: Which is the smallest whole number? The smallest whole number is 0. Question 4: How many whole numbers are there between 32 and 53? Whole numbers between 32 and 53 = 20 (53 − 32 − 1 = 20) (33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52) Question 5: Write the successor of: (a) 2440701 (b) 100199 (c) 1099999 (d) 2345670 (a) 2440701 + 1 = 2440702 (b) 100199 + 1 = 100200 (c) 1099999 + 1 = 1100000 (d) 2345670 + 1 = 2345671 Question 6: Write the predecessor of: (a) 94 (b) 10000 (c) 208090 (d) 7654321 (a) 94 − 1 = 93 (b) 10000 − 1 = 9999 (c) 208090 − 1 =208089 (d) 7654321 − 1 = 7654320 Question 7: In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them. (a) 530, 503 (b) 370, 307 (c) 98765, 56789 (d) 9830415, 10023001 (a) 530, 503 As 530 > 503, 503 is on the left side of 530 on the number line. (b) 370, 307 As 370 > 307, 307 is on the left side of 370 on the number line. (c) 98765, 56789 As 98765 > 56789, 56789 is on the left side of 98765 on the number line. (d) 9830415, 10023001 Since 98, 30, 415 < 1, 00, 23, 001, 98,30,415 is on the left side of 1,00,23,001 on the number line. Question 8: Which of the following statements are true (T) and which are false (F)? (a) Zero is the smallest natural number. (b) 400 is the predecessor of 399. (c) Zero is the smallest whole number. (d) 600 is the successor of 599. (e) All natural numbers are whole numbers. (f) All whole numbers are natural numbers. (g) The predecessor of a two digit number is never a single digit number. (h) 1 is the smallest whole number. (i) The natural number 1 has no predecessor. (j) The whole number 1 has no predecessor. (k) The whole number 13 lies between 11 and 12. (l) The whole number 0 has no predecessor. (m) The successor of a two digit number is always a two digit number. (a) False, 0 is not a natural number. (b) False, as predecessor of 399 is 398 (399 − 1 = 398). (c) True (d) True, as 599 + 1 = 600 (e) True (f) False, as 0 is a whole number but it is not a natural number. (g) False, as predecessor of 10 is 9. (h) False, 0 is the smallest whole number. (i) True, as 0 is the predecessor of 1 but it is not a natural number. (j) False, as 0 is the predecessor of 1 and it is a whole number. (k) False, 13 does not lie in between 11 and 12. (l) True, predecessor of 0 is −1, which is not a whole number. (m) False, as successor of 99 is 100. Exercise 2.2 Question 1: Find the sum by suitable rearrangement: (a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647 (a) 837 + 208 + 363 = (837 + 363) + 208 = 1200 + 208 = 1408 (b) 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600 Question 2: Find the product by suitable rearrangement: (a) 2 × 1768 × 50 (b) 4 × 166 × 25 (c) 8 × 291 × 125 (d) 625 × 279 × 16 (e) 285 × 5 × 60 (f) 125 × 40 × 8 × 25 (a) 2 × 1768 × 50 = 2 × 50 × 1768 = 100 × 1768 = 176800 (b) 4 × 166 × 25 = 4 × 25 × 166 = 100 × 166 = 16600 (c) 8 × 291 × 125 = 8 × 125 × 291 = 1000 × 291 = 291000 (d) 625 × 279 × 16 = 625 × 16 × 279 = 10000 × 279 = 2790000 (e) 285 × 5 × 60 = 285 × 300 = 85500 (f) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25 = 1000 × 1000 = 1000000 Question 3: Find the value of the following: (a) 297 × 17 + 297 × 3 (b) 54279 × 92 + 8 × 54279 (c) 81265 × 169 − 81265 × 69 (d) 3845 × 5 × 782 + 769 × 25 × 218 (a) 297 × 17 + 297 × 3 = 297 × (17 + 3) = 297 × 20 = 5940 (b) 54279 × 92 + 8 × 54279 = 54279 × 92 + 54279 × 8 = 54279 × (92 + 8) = 54279 × 100 = 5427900 (c) 81265 × 169 − 81265 × 69 = 81265 × (169 − 69) = 81265 × 100 = 8126500 (d) 3845 × 5 × 782 + 769 × 25 × 218 = 3845 × 5 × 782 + 769 × 5 × 5 × 218 = 3845 × 5 × 782 + 3845 × 5 × 218 = 3845 × 5 × (782 + 218) = 19225 × 1000 = 19225000 Question 4: Find the product using suitable properties. (a) 738 × 103 (b) 854 × 102 (c) 258 × 1008 (d) 1005 × 168 (a) 738 × 103 = 738 × (100 + 3) = 738 × 100 + 738 × 3 (Distributive property) = 73800 + 2214 = 76014 (b) 854 × 102 = 854 × (100 + 2) = 854 × 100 + 854 × 2 (Distributive property) = 85400 + 1708 = 87108 (c) 258 × 1008 = 258 × (1000 + 8) = 258 × 1000 + 258 × 8 (Distributive property) = 258000 + 2064 = 260064 (d) 1005 × 168 = (1000 + 5) × 168 = 1000 × 168 + 5 × 168 (Distributive property) = 168000 + 840 = 168840 Question 5: A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs Rs 44 per litre, how much did he spend in all on petrol? Quantity of petrol filled on Monday = 40 l Quantity of petrol filled on Tuesday = 50 l Total quantity filled = (40 + 50) l Cost of petrol (per l) = Rs 44 Total money spent = 44 × (40 + 50) = 44 × 90 = Rs 3960 Question 6: A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs Rs 15 per litre, how much money is due to the vendor per day? Quantity of milk supplied in the morning = 32 l Quantity of milk supplied in the evening = 68 l Total of milk per litre = (32 + 68) l Cost of milk per litre = Rs 15 Total cost per day = 15 × (32 + 68) = 15 × 100 = Rs 1500 Question 7: Match the following: (i) 425 × 136 = 425 × (6 + 30 + 100) (a) Commutativity under multiplication (ii) 2 × 49 × 50 = 2 × 50 × 49 (iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition (i) 425 × 136 = 425 × (6 + 30 + 100) [Distributivity of multiplication over addition] Hence, (c) (ii) 2 × 49 × 50 = 2 × 50 × 49 [Commutativity under multiplication] Hence, (a) (iii) 80 + 2005 + 20 = 80 + 20 + 2005 [Commutativity under addition] Hence, (b) Exercise 2.3 Question 1: Which of the following will not represent zero? (a) 1 + 0 (b) 0 × 0 Question 2: If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples. If the product of 2 whole numbers is zero, then one of them is definitely zero. For example, 0 × 2 = 0 and 17 × 0 = 0 If the product of 2 whole numbers is zero, then both of them may be zero. 0 × 0 = 0 However, 2 × 3 = 6 (Since numbers to be multiplied are not equal to zero, the result of the product will also be non-zero.) Question 3: If the product of two whole numbers is 1, can we say that one of both of them will be 1? Justify through examples. If the product of 2 numbers is 1, then both the numbers have to be equal to 1. For example, 1 × 1 = 1 However, 1 × 6 = 6 Clearly, the product of two whole numbers will be 1 in the situation when both numbers to be multiplied are 1. Question 4: Find using distributive property: (a) 728 × 101 (b) 5437 × 1001 (c) 824 × 25 (d) 4275 × 125 (e) 504 × 35 (a) 728 × 101= 728 × (100 + 1) = 728 × 100 + 728 × 1 = 72800 + 728 = 73528 (b) 5437 × 1001 = 5437 × (1000 + 1) = 5437 × 1000 + 5437 × 1 = 5437000 + 5437 = 5442437 (c) 824 × 25 = (800 + 24) × 25 = (800 + 25 − 1) × 25 = 800 × 25 + 25 × 25 − 1 × 25 = 20000 + 625 − 25 = 20000 + 600 = 20600 (d) 4275 × 125 = (4000 + 200 + 100 − 25) × 125 = 4000 × 125 + 200 × 125 + 100 × 125 − 25 × 125 = 500000 + 25000 + 12500 − 3125 = 534375 (e) 504 × 35 = (500 + 4) × 35 = 500 × 35 + 4 × 35 = 17500 + 140 = 17640 Question 5: Study the pattern: 1 × 8 + 1 = 9 1234 × 8 + 4 = 9876 12 × 8 + 2 = 98 12345 × 8 + 5 = 98765 123 × 8 + 3 = 987 Write the next two steps. Can you say how the pattern works? (Hint: 12345 = 11111 + 1111 + 111 + 11 + 1). 123456 × 8 + 6 = 987648 + 6 = 987654 1234567 × 8 + 7 = 9876536 + 7 = 9876543 Yes, the pattern works. As 123456 = 111111 + 11111 + 1111 + 111 + 11 + 1, 123456 × 8 = (111111 + 11111 + 1111 + 111 + 11 + 1) × 8 = 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8 = 888888 + 88888 + 8888 + 888 + 88 + 8 = 987648 123456 × 8 + 6 = 987648 + 6 = 987654 « Previous Next » ### NCERT Solutions for Class 6 Maths All Chapters We hope that our NCERT Solutions for class 6 Maths helped with your studies! If you liked our NCERT Solutions for Class 6 Maths, please share this post. ### NCERT Solutions for Class 6 All Subjects Maths English Science Hindi Social Science Sanskrit
Geometric Distribution: Definition & Example What is a Geometric Distribution? The geometric distribution represents the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function: f(x) = (1 − p)x − 1p For example, you ask people outside a polling station who they voted for until you find someone that voted for the independent candidate in a local election. The geometric distribution would represent the number of people who you had to poll before you found someone who voted independent. You would need to get a certain number of failures before you got your first success. If you had to ask 3 people, then X=3; if you had to ask 4 people, then X=4 and so on. In other words, there would be X-1 failures before you get your success. If X=n, it means you succeeded on the nth try and failed for n-1 tries. The probability of failing on your first try is 1-p. For example, if p = 0.2 then your probability of success is .2 and your probability of failure is 1 – 0.2 = 0.8. Independence (i.e. that the outcome of one trial does not affect the next) means that you can multiply the probabilities together. So the probability of failing on your second try is (1-p)(1-p) and your probability of failing on the nth-1 tries is (1-p)n-1. If you succeeded on your 4th try, n = 4, n – 1 = 3, so the probability of failing up to that point is (1-p)(1-p)(1-p) = (1-p)3. Example Sample question: If your probability of success is 0.2, what is the probability you meet an independent voter on your third try? Inserting 0.2 as p and with X = 3, the probability density function becomes: f(x) = (1 − p)x − 1p P(X=3) = (1 − 0.2)3 − 1(0.2) P(X=3) = (0.8)20.2 = 0.128. Theoretically, there are an infinite number of geometric distributions. The value of any specific distribution depends on the value of the probability p. Assumptions for the Geometric Distribution The three assumptions are: • There are two possible outcomes for each trial (success or failure). • The trials are independent. • The probability of success is the same for each trial. ------------------------------------------------------------------------------ Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! Statistical concepts explained visually - Includes many concepts such as sample size, hypothesis tests, or logistic regression, explained by Stephanie Glen, founder of StatisticsHowTo.
# Show that $(x-5)^2+\frac{9(4-x)}{4x}>1$ on the interval $(0,4)$ Consider the function $$f$$ given by $$f(x)=(x-5)^2+\frac{9(4-x)}{4x},$$ for $$x\in (0,4)$$. I'm asked to show that $$f(x)>1$$ on the interval $$(0,4)$$. I've started by recognising that $$(x-5)^2>0$$ on this interval, in which case $$f(x)>\frac{9(4-x)}{4x}.$$ How can I now show that this is greater than $$1$$? In the interval $$(0,4)$$, $$4x>0$$ and $$9(4-x)>0$$. So $$\frac{9(4-x)}{4x}>0$$ on $$(0,4)$$. And $$(x-5)^2>1$$ for $$\forall x\in (0,4)$$ so $$f(x)>1$$ $$\forall x\in (0,4).$$ Note that $$(x-5)^2+\frac{9(4-x)}{4x}-1=\frac{(x-4)(4x^2-24x-9)}{4x}$$ We need to prove that $$(x-6)(x-4)+\frac{9(4-x)}{4x}>0$$ or $$6-x+\frac{9}{4x}>0,$$ which is obvious. To prove the stated condition, it is enough to prove its equivalent i.e. $$$$4x^3-40x^2+87x+36>0$$$$ To find if the above equation is an increasing function i.e. $$>0$$, we find its critical points. Its critical points are: $$$$\frac{20\pm\sqrt{139}}{6}$$$$ Thus, this leads to two intervals viz. $$I=\overbrace{\Big(0, \frac{20-\sqrt{139}}{6}\Big]}^{I_1} \cup \overbrace{\Big[\frac{20-\sqrt{139}}{6},4\Big)}^{I_2}$$ since $$4<\frac{20+\sqrt{139}}{6}$$ (enough for the proof). Now, substituting the extremities of $$I_1$$, we get $$90.39947522$$ for $$x=\frac{20-\sqrt{139}}{6}$$ and $$36$$ for $$x=0$$. Hence, the above reduced equation is strictly increasing in the interval $$I_1$$. In the interval $$I_2$$, we get, $$0$$ for $$x=4$$. Hence the function is a decreasing function in the interval $$I_2$$. But since $$4$$ is not included in the interval, it is a guarantee that the function is always strictly greater than zero. Hence proved.
Probability 18.01.2019 Episode #9 of the course Everyday math by Jenn Schilling Welcome to a quick dive into probability! Probability is a complex field of mathematics, but an introduction to it will help you better understand chance events such as weather, games of chance, sports, and more! The Math Probability tells us the likelihood of an event occurring. For example, a 60% chance of rain means there is a 60% likelihood that it will rain. There are a few important terms to understand around probability. The first is outcome, which is the result of an experiment or event. For example, in a coin flip, an outcome might be heads. The second is favorable outcome, which is the result that we are looking for. So, in a coin flip, perhaps we want to get tails; tails is the favorable outcome. The third is sample space, which encompasses the total possible outcomes of an experiment. For example, when flipping a coin, you can either get heads or tails, so the sample space consists of two outcomes: heads and tails. To determine the probability of an event, we divide the number of favorable outcomes by the number of total outcomes in the sample space. For example, the probability of rolling a 3 on a six-sided die is 1/6 because there are 6 possible outcomes, but only 1 outcome in which we get a 3. In more complicated experiments, there are multiple ways of obtaining the favorable outcome. For instance, suppose we are going to roll two dice, and we want to know the probability of getting a sum of 5. There are multiple ways of getting a sum of 5 from two dice: 1 + 4 4 + 1 2 + 3 3 + 2 We have a total of 4 favorable outcomes. There are 36 total possible outcomes for this experiment because each die has six sides that are all equally likely to occur (assuming we are being fair and not using weighted dice). So, the probability of rolling a sum of 5 is 4 (the number of favorable outcomes) divided by 36 (the number of total possible outcomes), which is 4/36, or 1/9 if we simplify. So, the probability of rolling a sum of 5 is 1 in 9. Probabilities can also be turned into percents by dividing out the fraction. In our example above, the probability (or chance) of rolling a sum of 5 is 11%. The higher the percent, the greater the probability, and the more likely it is that the desired event will occur. Everyday Applications Probability appears frequently in our everyday lives. Whether it’s the weather forecast for the day, the lottery, or sports, probability is everywhere! In weather forecasting, probability is used to indicate the chance of precipitation. This probability is calculated by using a sample space of all days with similar weather characteristics and evaluating the number of those days on which there was precipitation. For example, if the chance of rain is 60%, that means that out of all the days with similar weather conditions, it rained on 60% of them. Games of chance and the lottery are another area in which probability is prevalent. In any game that involves dice, coin flips, cards, or spinners, the probability of a favorable outcome can be calculated as above in the dice sum example. Understanding the probabilities of certain cards or rolls can help anyone make better choices in games of chance. In sports, probabilities appear in the likelihood of certain outcomes (who will win the game?), as well as in individual player statistics (how likely is it a certain player will hit the ball or make a field goal?). Knowing that the probability represents the likelihood of an outcome happening, based on all past events and whether the desired outcome occurred, helps us interpret and evaluate probabilities in various fields. I hope you enjoyed this overview of probability. In tomorrow’s lesson, we will wrap up this course with an explanation of statistics! We will cover important statistical terms, and you will learn how to interpret statistics in sports, popular science, and the news. Share with friends
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Beat Frequencies ## When waves of two frequencies interfere a third wave is perceived by the observer. Estimated8 minsto complete % Progress Practice Beat Frequencies Progress Estimated8 minsto complete % Beat Frequencies Students will learn the concept of destructive and constructive interference in the context of beat frequencies. ### Key Equations ; two interfering waves create a beat wave with frequency equal to the difference in their frequencies Guidance • Constructive interference occurs when two waves combine to create a larger wave. This occurs when the peaks of two waves line up. • Destructive interference occurs when two waves combine and cancel each other out. This occurs when a peak in one wave lines up with a trough in the other wave. • When waves of two different frequencies interfere, a phenomenon known as beating occurs. The frequency of a beat is the difference of the two frequencies. #### Example 1 You want to find out the frequency of a tuning fork. When you strike the unknown fork and a fork which is known to create a sound at 100 Hz at the same time, you hear a beat frequency of 3 Hz. You strike the unknown fork a second time, but this time with a tuning fork rated for 105 Hz and you hear a beat frequency of 2 Hz. What is the frequency of the unknown fork? ##### Solution Each test gives us two possibilities for a frequency. Based on the first test, the unknown fork could either be rated for 97 Hz or 103 Hz. Based on the second test, the frequency of the unknown fork could either be 103 Hz or 107 Hz. The two tests agree on 103 Hz, so that must be the frequency of the unknown tuning fork. ### Explore More 1. At the Sunday drum circle in Golden Gate Park, an Indian princess is striking her drum at a frequency of 2 Hz. You would like to hit your drum at another frequency, so that the sound of your drum and the sound of her drum “beat” together at a frequency of 0.1 Hz. What frequencies could you choose? 2. You and your friend play the same note. Your friend's instrument is in tune and is playing a note at 256 Hz, but you are slightly out of tune and are playing a note at 258 Hz. What is the beat frequency? 1. 1.9 Hz or 2.1 Hz 2. 2 Hz
Question laura has saved her babysitting money for the past two weeks. currently has 5 time her weekly pay. this week she spent a less then 2 times her weekly pay and now has 156. how much does laura earn each week To find out how much Laura earns each week, you can set up an equation using the information given. Let’s call Laura’s weekly pay “W”. You know that Laura has saved 5 times her weekly pay, so you can write this as 5W. You also know that she spent less than 2 times her weekly pay this week, so you can write this as 2W – X, where X is the amount that she spent less than 2 times her weekly pay. Finally, you know that she has 156 left after spending some money, so you can write this as 156. You can now set up the equation: 5W + (2W – X) = 156 Combining like terms, you get: 7W – X = 156 Adding X to both sides, you get: 7W = 156 + X Dividing both sides by 7, you get: W = (156 + X) / 7 Thus, Laura’s weekly pay is equal to (156 + X) / 7. You know that she spent less than 2 times her weekly pay this week, so X must be less than 2W. However, you don’t know the exact value of X, so you can’t find the exact value of Laura’s weekly pay. :
# Problems on Principle of Mathematical Induction Solved Problems on Principle of Mathematical Induction are shown here to prove Mathematical Induction. ### Problems on Principle of Mathematical Induction 1. Using the principle of mathematical induction, prove that 1² + 2² + 3² + ..... + n² = (1/6){n(n + 1)(2n + 1} for all n ∈ N. Solution: Let the given statement be P(n). Then, P(n): 1² + 2² + 3² + ..... +n² = (1/6){n(n + 1)(2n + 1)}. Putting n =1 in the given statement, we get LHS = 1² = 1 and RHS = (1/6) × 1 × 2 × (2 × 1 + 1) = 1. Therefore LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, P(k): 1² + 2² + 3² + ..... + k² = (1/6){k(k + 1)(2k + 1)}. Now, 1² + 2² + 3² + ......... + k² + (k + 1)² = (1/6) {k(k + 1)(2k + 1) + (k + 1)² = (1/6){(k + 1).(k(2k + 1)+6(k + 1))} = (1/6){(k + 1)(2k² + 7k + 6}) = (1/6){(k + 1)(k + 2)(2k + 3)} = 1/6{(k + 1)(k + 1 + 1)[2(k + 1) + 1]} ⇒ P(k + 1): 1² + 2² + 3² + ….. + k² + (k+1)² = (1/6){(k + 1)(k + 1 + 1)[2(k + 1) + 1]} ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N. Problems on Principle of Mathematical Induction 2. Using the principle of mathematical induction, prove that 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}. Solution: Let the given statement be P(n). Then, P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}. Thus, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k + 1) = (1/3){k(k + 1)(k + 2)}. Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +...+ k(k + 1) + (k + 1)(k + 2) = (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ....... + k(k + 1)) + (k + 1)(k + 2) = (1/3) k(k + 1)(k + 2) + (k + 1)(k + 2) [using (i)] = (1/3) [k(k + 1)(k + 2) + 3(k + 1)(k + 2) = (1/3){(k + 1)(k + 2)(k + 3)} ⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +......+ (k + 1)(k + 2) = (1/3){k + 1 )(k + 2)(k +3)} ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all values of ∈ N. Problems on Principle of Mathematical Induction 3. Using the principle of mathematical induction, prove that 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 +.....+ (2n - 1)(2n + 1) = (1/3){n(4n² + 6n - 1). Solution: Let the given statement be P(n). Then, P(n): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 +...... + (2n - 1)(2n + 1)= (1/3)n(4n² + 6n - 1). When n = 1, LHS = 1 ∙ 3 = 3 and RHS = (1/3) × 1 × (4 × 1² + 6 × 1 - 1) = {(1/3) × 1 × 9} = 3. LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, P(k): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ….. + (2k - 1)(2k + 1) = (1/3){k(4k² + 6k - 1) ......(i) Now, 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + …….. + (2k - 1)(2k + 1) + {2k(k + 1) - 1}{2(k + 1) + 1} = {1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ………… + (2k - 1)(2k + 1)} + (2k + 1)(2k + 3) = (1/3) k(4k² + 6k - 1) + (2k + 1)(2k + 3) [using (i)] = (1/3) [(4k³ + 6k² - k) + 3(4k² + 8k + 3)] = (1/3)(4k³ + 18k² + 23k + 9) = (1/3){(k + 1)(4k² + 14k + 9)} = (1/3)[k + 1){4k(k + 1) ² + 6(k + 1) - 1}] ⇒ P(k + 1): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ..... + (2k + 1)(2k + 3) = (1/3)[(k + 1){4(k + 1)² + 6(k + 1) - 1)}] ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N. More Problems on Principle of Mathematical Induction 4. Using the principle of mathematical induction, prove that 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{n(n + 1)} = n/(n + 1) Solution: Let the given statement be P(n). Then, P(n): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{n(n + 1)} = n/(n + 1). Putting n = 1 in the given statement, we get LHS= 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2. LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, P(k): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{k(k + 1)} = k/(k + 1) ..…(i) Now 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)} [1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)} = k/(k + 1)+1/{ (k + 1)(k + 2)}. {k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii) = {k(k + 2) + 1}/{(k + 1)(k + 2} = {(k + 1)² }/{(k + 1)(k + 2)} = (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1) ⇒ P(k + 1): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)} = (k + 1)/(k + 1 + 1) ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N. Problems on Principle of Mathematical Induction 5. Using the principle of mathematical induction, prove that {1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ….... + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3)}. Solution: Let the given statement be P(n). Then, P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3). Putting n = 1 in the given statement, we get and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15. LHS = RHS Thus , P(1) is true. Let P(k) be true. Then, P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i) Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3 = {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)} = k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)] = {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)} = (2k² + 5k + 3)/[3(2k + 3)(2k + 5)] = {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)} = (k + 1)/{3(2k + 5)} = (k + 1)/[3{2(k + 1) + 3}] = P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}] = (k + 1)/{3{2(k + 1) + 3}] ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for n ∈ N. Problems on Principle of Mathematical Induction 6. Using the principle of mathematical induction, prove that 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} for all n ∈ N. Solution: Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} . Putting n = 1 in the given statement, we get LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6. Therefore LHS = RHS. Thus, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……... + 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}. …….(i) Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)} = [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1)(k + 2}] + 1/{(k + 1)(k + 2)(k + 3)} = [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)] = {k(k + 3)² + 4}/{4(k + 1)(k + 2)(k + 3)} = (k³ + 6k² + 9k + 4)/{4(k + 1)(k + 2)(k + 3)} = {(k + 1)(k + 1)(k + 4)}/{4 (k + 1)(k + 2)(k + 3)} = {(k + 1)(k + 4)}/{4(k + 2)(k + 3) ⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1)(k + 2)(k + 3)} = {(k + 1)(k + 2)}/{4(k + 2)(k + 3)} ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N. Problems on Principle of Mathematical Induction 7. Using the Principle of mathematical induction, prove that {1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... {1 - 1/(n + 1)} = 1/(n + 1) for all n ∈ N. Solution: Let the given statement be P(n). Then, P(n): {1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... {1 - 1/(n + 1)} = 1/(n + 1). When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½. Therefore LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, P(k): {1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... [1 - {1/(k + 1)}] = 1/(k + 1) Now, [{1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... [1 - {1/(k + 1)}] ∙ [1 – {1/(k + 2)}] = [1/(k + 1)] ∙ [{(k + 2 ) - 1}/(k + 2)}] = [1/(k + 1)] ∙ [(k + 1)/(k + 2)] = 1/(k + 2) Therefore p(k + 1): [{1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... [1 - {1/(k + 1)}] = 1/(k + 2) ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N. Problems on Principle of Mathematical Induction Mathematical Induction
# What must be added to each of the following expressions to make it a whole square? Question: What must be added to each of the following expressions to make it a whole square? (i) 4x2 − 12x + 7 (ii) 4x2 − 20x + 20 Solution: (i) Let us consider the following expression: $4 x^{2}-12 x+7$ The above expression can be written as: $4 x^{2}-12 x+7=(2 x)^{2}-2 \times 2 x \times 3+7$ It is evident that if 2x is considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square. Therefore, 7 must become 9. Therefore, adding and subtracting 2 in the above expression, we get: $\left(4 x^{2}-12 x+7\right)+2-2=\left\{(2 x)^{2}-2 \times 2 x \times 3+7\right\}+2-2=\left\{(2 x)^{2}-2 \times 2 x \times 3+9\right\}-2=(2 x+3)^{2}-2$ (ii) Let's consider the following expression: $4 x^{2}-20 x+20$ The above expression can be written as: $4 x^{2}-20 x+20=(2 x)^{2}-2 \times 2 x \times 5+20$ It is evident that if 2x is considered as the first term and 5 is considered as the second term, 5 is required to be added to the above expression to make it a perfect square. Therefore, number 20 must become 25. Therefore, adding and subtracting 5 in the above expression, we get: $\left(4 x^{2}-20 x+20+5\right)-5=\left\{(2 x)^{2}-2 \times 2 x \times 5+20\right\}+5-5=\left\{(2 x)^{2}-2 \times 2 x \times 5+25\right\}-5=(2 x+5)^{2}-5$
Dividing a Quantity in Three Given Ratios – Definition, Rules, Examples \| How do you Divide a Quantity in Three Given Ratios? In this platform, you will learn about the rules of dividing a quantity into three given ratios. As we know, a ratio indicates their relative sizes and always expresses the ratios in simplest form. It is a comparison of two or more different quantities having the same units of measure and it can even be denoted as a fraction. In this dividing a quantity in three given Ratios word problems, students can get the Dividing a Quantity into Three Given Ratios Questions with Answers. In this article, you will even find the definition of dividing a quantity into three given ratios, rules, and solved example problems on dividing a quantity into three given ratios. Dividing a Quantity in Three Given Ratios – Definition Dividing a quantity into three given ratios is stated as taking a common multiple that is k and then finding each part in terms of k. We have to divide a number into three parts having a ratio of 1:3:4 then, let the common multiple be k. Therefore, the three parts are 1k, 3k, and 4k. You can also download the worksheet on dividing a quantity in a given ratios pdf from this page. Rules for Dividing a Quantity in a Three Given Ratios The rules of dividing a quantity into three given ratios are explained below. If a quantity K is divided into three parts in the ratio of X:Y: Z, then (i) The first part is X/(X +Y+Z) Ã— K. (ii) The second part is Y/(X +Y+Z) Ã— K, and (iii) The third part is Z/(X+Y+Z) x K. Problems on Dividing a Quantity into Three Given Ratios Problem 1:Â If $162 is divided among three people in the ratio of 2 : 3: 4. Find the share of each person. Solution: As given in the question, the value is$162 and the ratio is 2:3:4. Now, we will find the share of each person. First, find the sum of the terms of the ratio. The sum is 2 + 3 + 4 = 9. The share of the first boy isÂ 2/9 Ã— 162 = $36. Next, the share of the second boy is 3/9 Ã— 162 =$54. The share of the third boy is 4/9 Ã— 162= $72. Therefore, the required shares are$ 36, $54, and$72 respectively. Problem 2: How to divide 88 into three parts in the ratio of 2:4:5. Write the divided values. Solution:Â As given in the question, the value is 88 and the ratio is 2:4:5. Now, we need to find the sum of the given ratios. The sum is 2+4+5 = 11. So, the first part value is 2/11 x 88 = 16. The second part is 4/11×88 = 32. The third part is 5/11×88 = 40. Hence, after dividing the values are 16, 32, and 40. Problem 3: If the angles of a triangle are in the ratio of 2:3:4. What is the value of each angle? Solution:Â Given in the question, Let the common ratio be 2x,3x, and 4x. So, the common ratio is x and the angles are 2x, 3x, and 4x. 2x+3x + 4x is180 degrees. So, 9x = 180 degrees. X = 180/9 = 20 degrees. The 2x angle is, 2(20) = 40 degrees. The angle at 3xÂ is, 3(20) = 60 degees. The angle at 4x is , 4(20) = 80 degrees. Therefore, in the triangle, each angle value is 40 degrees, 60 degrees, and 80 degrees. Problem 4: The sides of a triangle are in the ratio of 1:2:3 and the perimeter of the triangle is 36cm, find its sides? Solution: As given in the question, the sides are in the ratio of 1:2:3 and the perimeter is 36cm. First, we can assume that the sides of the triangle are 1 x t, 2 x t, 3 x t = t, 2t, 3t. The perimeter is 36cm. So, the value is t+2t+3t = 36cm. i . e., 6t = 36 cm t = 36 cm/6 = 6cm. The side of the triangle t is 6 cm. The side of the triangle 2t is 2×6 = 12 cm. The side of the triangle 3t is 3×6 = 18 cm. Therefore, the sides of the triangle are 6 cm, 12 cm, and 18 cm respectively. Problem 5: Divide Rs.1500 among A, B, and C in the ratio of 3:5:2. Solution: Given in the question, the value is Rs.1500 and in the ratio of 3:5:2. Now, find the sum of the given ratios. The sum is 3+5+2 = 10. So, the A value is 3/10 x 1500 = Rs.450. The B value is 5/10 x 1500 = Rs.750. The C value is 2/10 x 1500 = Rs.300. The total amount is First amount + Second amount + Third amount = Rs.450 + Rs.750 + Rs.300 = Rs.1500 Thus, the dividing values areÂ Rs.450, Rs.750, and Rs.300. Problem 6: A bag contains 3 dollars, 50 cents, and 4 dollars in the ratio of 5: 4: 2. The total amount is \$ 2450. Find the number of each denomination? Solution: In the given question, the ratios are 5:4:2. So the number for each denomination is 5x, 4x, and 2x respectively. The amount of 3 dollars is 5x Ã— 300 cents that is 1500x cents The amount of 50 cents is 4x Ã— 50 cents = 200x cents The amount of 4 dollars is 2x Ã— 400 cents = 800x cents The total amount given is 2450 Ã— 100 cents =245000 cents. Now, 1500x + 200x + 800x = 245000 â‡’ 2500x = 245000 â‡’ x = 2450002500 â‡’ x = 98 Now, we have to substitute the x value in each denomination. We get, The number of 3 dollars i.e., 5x = 5Ã—98 = 490 The number for 50cents is 4x = 4Ã—98 = 392 The number of 4 dollars i.e., 2x = 2Ã—98 = 196 Hence, each number denominations are 490, 392, and 196. Problem 7: A certain sum of money is divided into three parts in the ratio of 4: 3: 2.Â If the first part is â‚¹448, then find the total amount, the second part, and the third part respectively? Solution: Given that, the three parts in the ratio are 4: 3: 2. Assume that, the amount of money is 4x, 3x, and 2x. Using the given first part value. We will find the X value. The first part value is â‚¹448. i.e., 4x = â‚¹448 â‡’ x = 4484 = 112 Thus, x = 112 Next, 3x = 3Ã—112 = 336 and 2x = 2Ã—112 = 224 Therefore, the second amount = â‚¹336 the third amount = â‚¹224 So, the total amount of money is, First amount + Second amount + Third amount i.e., â‚¹448 + â‚¹336 + â‚¹224 = â‚¹1008. Therefore, the total amount of money is â‚¹1008, â‚¹336, and â‚¹224 respectively. FAQs on Dividing a Quantity into Three given Ratios 1. What is meant by dividing a quantity into three given ratios? Dividing a quantity into three given ratios is stated as taking a common multiple that is x and then finding each part in terms of x. We have to divide a number into three parts having a ratio of 1:2:4 then, let the common multiple be x. Therefore, the three parts are 1x, 2x, and 4k. 2. How many rules are there for finding the three given ratios quantity? There are 3 rules for dividing a quantity into three given ratios. 3. How do you find the ratio of three quantities? The following are the steps, for calculating a ratio of 3 numbers. The steps are: • Step 1: First, find the total number of parts in the ratio by adding the numbers in the ratio together. • Step 2: Next, find the value of each part in the ratio by dividing the given amount by the total number of parts. • Step 3: Then, multiply the original ratio by the value of each part. 4. What happens when you divide a ratio? When dividing ratios, we are essentially dividing a whole number into a number of smaller numbers and assigning those in proportion to the specified ratio by multiplying. Scroll to Top Scroll to Top
# Problem of the Week Problem A and Solution Enough Money ## Problem Rosa has six dimes, seven quarters, three $$\5$$ bills, and one toonie. She would like to purchase some LegoTM for $$\20.00$$. Does Rosa have enough money? Justify your answer. In Canada, a dime is worth $$10$$¢, a quarter is worth $$25$$¢, a toonie is worth $$\2.00$$, and $$100$$¢ is equal to $$\1.00$$. ## Solution We will determine how much money Rosa has, in dollars. \begin{aligned} \text{six dimes} & = 6 \times 10¢ = 60¢ = \0.60 \\ \text{seven quarters} & = 7 \times 25¢ = 175¢ = \1.75 \\ \text{three} ~\5~ \text{bills} & = 3 \times \5.00 = \15.00 \\ \text{one toonie} & = 1 \times \2.00 = \2.00 \end{aligned} Thus, in total, Rosa has $$\0.60+\1.75+\15.00+\2.00=\19.35$$. This is less than $$\20.00$$, which is the cost of the LegoTM. Therefore, Rosa does not have enough money to buy the LegoTM. ### Teacher’s Notes The value of modern coins is not normally connected to the value of the material with which they are made. In fact, if the metal used to mint a coin is actually worth more than the coin’s face value, usually there is an effort to remove the coin from circulation. For example, the penny in Canada (1 cent coin) was removed in 2012. At that time, it cost approximately 1.6 cents to mint each penny. In 1965, the United States stopped circulating the silver dollar since the material was worth more than a dollar. Most of the dollar coins that had been minted were then melted to recover the base metal. In history, coins did not always have a value imprinted on them. The purity of the metal that composed the coin could be tested on a touchstone, and the monetary value was calculated based on its weight and purity. Eventually gold and silver coins were stamped with their weights instead of a fixed monetary value. Governments would mint the coins as a way of guaranteeing the weight and purity of the material was correct.
## MP Board Class 6th Maths Solutions Chapter 4 Basic Geometrical Ideas Ex 4.1 Question 1. Use the figure to name : (a) Five points (b) A line (c) Four rays (d) Five line segments Solution: (a) Five points : O, B, C, D, E (b) A line $$\overleftrightarrow { DB }$$ (c) Four rays $$\overrightarrow{O D}$$, $$\overrightarrow{O E}$$, $$\overrightarrow{O C}$$, $$\overrightarrow{O B}$$ (d) Five line segments : $$\overline{D E}$$, $$\overline{O E}$$, $$\overline{O C }$$, $$\overline{O B}$$, $$\overline{O D}$$. Question 2. Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given. Solution: Possible lines are $$\overleftrightarrow { AB }$$, $$\overleftrightarrow { AC }$$, $$\overleftrightarrow { AD }$$, $$\overleftrightarrow { BC }$$, $$\overleftrightarrow { BD }$$, $$\overleftrightarrow { CD }$$, $$\overleftrightarrow { BA }$$, $$\overleftrightarrow { CA }$$, $$\overleftrightarrow { DA }$$, $$\overleftrightarrow { CB }$$, $$\overleftrightarrow { DB }$$, $$\overleftrightarrow { DC }$$. Question 3. Use the figure to name : (a) Line containing point E. (b) Line passing through A. (c) Line on which O lies (d) Two pairs of intersecting lines. Solution: (a) A line containing point E is $$\overleftrightarrow { AE }$$. (b) A line passing through A is $$\overleftrightarrow { AE }$$. (c) A line on which O lies is $$\overleftrightarrow { CO }$$ or $$\overleftrightarrow { OC }$$. (d) Two pairs of intersecting lines are $$\overleftrightarrow { AD }$$, $$\overleftrightarrow { CO }$$ and $$\overleftrightarrow { AE }$$, $$\overleftrightarrow { FE }$$. Question 4. How many lines can pass through (a) one given point? (b) two given points? Solution: (a) Infinite number of lines can pass through one given point. (b) Only one line can pass through two given points. Question 5. Draw a rough figure and label suitably in each of the following cases: (a) Point Plies on $$\overline{A B}$$. (b) $$\overleftrightarrow { XY }$$ and $$\overleftrightarrow { PQ }$$ intersect at M. (c) Line l contains E and F but not D. (d) $$\overleftrightarrow { OP }$$ and $$\overleftrightarrow { OQ }$$ meet at O. Solution: Question 6. Consider the following figure of line $$\overleftrightarrow { MN }$$. Say whether following statements are true or false in context of the given figure. (a) Q, M, O, N, P are points on the line $$\overleftrightarrow { MN }$$. (b) M, O, N are points on a line segment $$\overline{M N}$$. (c) M and N are end points of line segment $$\overline{M N}$$. (d) O and N are end points of line segment $$\overline{O P}$$. (e) M is one of the end points of line segment $$\overline{Q O}$$. (f) M is point on ray $$\overrightarrow { OP }$$ (g) Ray $$\overrightarrow { OP }$$ is different from ray $$\overrightarrow { OP }$$. (h) Ray OP is same as ray $$\overrightarrow { OM }$$. (i) Ray $$\overrightarrow { OM }$$ is not opposite to ray $$\overrightarrow { OP }$$. (j) O is not an initial point of $$\overrightarrow { OP }$$. (k) N is the initial point of $$\overrightarrow { NP }$$ and $$\overrightarrow { NM }$$. Solution: (a) True (b) True (c) True (d) False (e) False (f) False (g) True (h) False (i) False (j) False (k) True
# Rounding Numbers to the 100 Thousands Place ## Presentation on theme: "Rounding Numbers to the 100 Thousands Place"— Presentation transcript: Rounding Numbers to the 100 Thousands Place FOCUS QUESTION: What does it mean to round a number? When you round a number, you are replacing it with another number that tells about how much or how many. You’re saying what the number is closest to. Copyright © 2013 Kelly Mott Round to the Nearest Ten What is 13 rounded to the nearest ten? Find 13 on the number line. Put a dot. Is 13 closer to 10 or 20? __________________ Answer: 13 rounded to the nearest ten is ____ Copyright © 2013 Kelly Mott Round to the Nearest Hundred What is 180 rounded to the nearest hundred? Find 180 on the number line. Put a dot. Is 180 closer to 100 or 200? _______________ Answer: 180 rounded to the nearest hundred is _____________________ Copyright © 2013Kelly Mott Round to the Nearest Thousand What is 1,300 rounded to the nearest thousand? Find 1,300 on the number line. Put a dot. Is 1,300 closer to 1,000 or 2,000? ___________ Answer: 1,300 rounded to the nearest thousand is _____________________ 1,000 1,100 1,200 1,300 1,400 1,500 1,600 1,700 1,800 1,900 2,000 Copyright © 2013 Kelly Mott Rounding to the Nearest 10 15 Rounding Practice 1: Round to the nearest ten: RULES FOR ROUNDING Round to the nearest ten: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 15 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 33 Rounding Practice 2: Round to the nearest ten: RULES FOR ROUNDING Round to the nearest ten: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 33 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 228 Rounding Practice 3: Round to the nearest ten: RULES FOR ROUNDING Round to the nearest ten: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 228 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 3,614 Rounding Practice 4: Round to the nearest ten: RULES FOR ROUNDING Round to the nearest ten: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 3,614 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 32,576 Rounding Practice 5: Round to the nearest ten: RULES FOR ROUNDING Round to the nearest ten: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 32,576 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott Does this make sense so far? Rounding to the Nearest 100 215 Rounding Practice 1: Round to the nearest hundred: RULES FOR ROUNDING Round to the nearest hundred: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 215 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 563 Rounding Practice 2: Round to the nearest hundred: RULES FOR ROUNDING Round to the nearest hundred: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 563 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 3,481 Rounding Practice 3: Round to the nearest hundred: RULES FOR ROUNDING Round to the nearest hundred: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 3,481 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 13,854 Rounding Practice 4: Round to the nearest hundred: RULES FOR ROUNDING Round to the nearest hundred: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 13,854 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 51,318 Rounding Practice 5: Round to the nearest hundred: RULES FOR ROUNDING Round to the nearest hundred: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 51,318 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott Does this make sense so far? Rounding to the Nearest 1,000 2,415 Rounding Practice 1: Round to the nearest thousand: RULES FOR ROUNDING Round to the nearest thousand: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 2,415 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 1,233 Rounding Practice 2: Round to the nearest thousand: RULES FOR ROUNDING Round to the nearest thousand: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 1,233 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 13,581 Rounding Practice 3: Round to the nearest thousand: RULES FOR ROUNDING Round to the nearest thousand: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 13,581 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 39,150 Rounding Practice 4: Round to the nearest thousand: RULES FOR ROUNDING Round to the nearest thousand: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 39,150 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott 103,718 Rounding Practice 5: Round to the nearest thousand: RULES FOR ROUNDING Round to the nearest thousand: Underline the number in the place value you will round. 2. Circle the number to the right of the underlined number. 103,718 Use the rounding pyramid: Underlined number rounds up one Add the zeros Underlined number stays the same Add the zeros 9 4 8 3 2 7 1 6 5 4. Any digit to the left of the underlined digit stays the same. Copyright © 2013 Kelly Mott Does this make sense so far? Rounding to the Nearest 10,000 Round to the Nearest Ten Thousand Round to the Nearest Ten Thousand Does this make sense so far? Rounding to the Nearest 100,000 Round to the Nearest 100 Thousand Round to the Nearest 100 Thousand
Courses Courses for Kids Free study material Offline Centres More Store # How do you graph the line $y = - 4x + 4$ ? Last updated date: 22nd Jun 2024 Total views: 374.4k Views today: 3.74k Verified 374.4k+ views Hint: In the above question, we need to plot the above line on the graph. Since, the given equation is a line where the equation of the straight line is $y = mx + c$so we can compare with this equation and find out the coordinates of the line on the cartesian plane. Complete step by step solution: In the above equation, we have to plot the line on the graph. To find the coordinates of the line what we can do is to select some values of x coordinate and then evaluate those values in the given equation. So, now we will have to select a value for x i.e., zero and we substitute the value in $y = mx + c$ where m is the slope and c is the constant term y-intercept. So, substitute the value in equation we get, The equation is $y = - 4x + 4$ .So when $x = 0$ we get So when $x = 0$we get, $\Rightarrow y = 0 + 4 \\ \Rightarrow y = 4 \\$ Therefore, we get the y-intercept as 4. The line will cut the Y-axis at point 4. Now we will select the point y=0. $\Rightarrow 0 = - 4x + 4 \\ \Rightarrow 4x = 4 \\ \Rightarrow x = 1 \\$ Therefore, we get x-intercept as 1. The line will cut x-axis at 1. Hence, we plot the points (0,4) and (1,0) on the graph and the line is plotted by joining these points. Note: An important thing to note is that slope contains the direction in which how you go from one point to another where $m = \dfrac{y}{x} = \dfrac{{rise}}{{run}}$. The numerator tells you many steps to go up and down and the denominator tell us how much steps to move left or right.
# 7.4: Percent - Mathematics Learning Objectives • understand the relationship between ratios and percents • be able to make conversions between fractions, decimals, and percents ## Ratios and Percents Ratio, Percent We defined a ratio as a comparison, by division, of two pure numbers or two like denominate numbers. A most convenient number to compare numbers to is 100. Ratios in which one number is compared to 100 are called percents. The word percent comes from the Latin word "per centum." The word "per" means "for each" or "for every," and the word "centum" means "hundred." Thus, we have the following definition. Percent means “for each hundred," or "for every hundred." The symbol % is used to represent the word percent. Sample Set A The ratio 26 to 100 can be written as 26%. We read 26% as "twenty-six percent." Sample Set A The ratio (dfrac{165}{100}) can be written as 165%. We read 165% as "one hundred sixty-five percent." Sample Set A The percent 38% can be written as the fraction (dfrac{38}{100}). Sample Set A The percent 210% can be written as the fraction (dfrac{210}{100}) or the mixed number (2dfrac{1)}{100}) or 2.1. Sample Set A Since one dollar is 100 cents, 25 cents is (dfrac{25}{100}) of a dollar. This implies that 25 cents is 25% of one dollar. Practice Set A Write the ratio 16 to 100 as a percent. Answer 16% Practice Set A Write the ratio 195 to 100 as a percent. Answer 195% Practice Set A Write the percent 83% as a ratio in fractional form. Answer (dfrac{83}{100}) Practice Set A Write the percent 362% as a ratio in fractional form. Answer (dfrac{362}{100}) or (dfrac{181}{50}) ## The Relationship Between Fractions, Decimals, and Percents – Making Conversions Since a percent is a ratio, and a ratio can be written as a fraction, and a fraction can be written as a decimal, any of these forms can be converted to any other. Before we proceed to the problems in Sample Set B and Practice Set B, let's summarize the conversion techniques. To Convert a Fraction To Convert a Decimal To Convert a Percent To a decimal: Divide the numerator by the denominator To a fraction: Read the decimal and reduce the resulting fraction To a decimal: Move the decimal point 2 places to the left and drop the % symbol To a percent: Convert the fraction first to a decimal, then move the decimal point 2 places to the right and affix the % symbol. To a percent: Move the decimal point 2 places to the right and affix the % symbol To a fraction: Drop the % sign and write the number “over” 100. Reduce, if possible. Sample Set B Convert 12% to a decimal. Solution (12\% = dfrac{12}{100} = 0.12) Note that The % symbol is dropped, and the decimal point moves 2 places to the left. Sample Set B Convert 0.75 to a percent. Solution (0.75 = dfrac{75}{100} = 75\%) Note that The % symbol is affixed, and the decimal point moves 2 units to the right. Sample Set B Convert (dfrac{3}{5}) to a percent. Solution We see in Example above that we can convert a decimal to a percent. We also know that we can convert a fraction to a decimal. Thus, we can see that if we first convert the fraction to a decimal, we can then convert the decimal to a percent. (dfrac{3}{5} o egin{array} {r} {.6} {5overline{)3.0}} {underline{3 0}} {0} end{array} ext{ or } dfrac{3}{5} = 0.6 = dfrac{6}{10} = dfrac{60}{100} = 60\%) Sample Set B Convert 42% to a fraction. Solution (42\% = dfrac{42}{100} = dfrac{21}{50}) or (42\% = 0.42 = dfrac{42}{100} = dfrac{21}{50}) Practice Set B Convert 21% to a decimal. Answer 0.21 Practice Set B Convert 461% to a decimal. Answer 4.61 Practice Set B Convert 0.55 to a percent. Answer 55% Practice Set B Convert 5.64 to a percent. Answer 564% Practice Set B Convert (dfrac{3}{20}) to a percent. Answer 15% Practice Set B Convert (dfrac{11}{8}) to a percent. Answer 137.5% Practice Set B Convert (dfrac{3}{11}) to a percent. Answer (27.overline{27})% ## Exercises For the following 12 problems, convert each decimal to a percent. Exercise (PageIndex{1}) 0.25 Answer 25% Exercise (PageIndex{2}) 0.36 Exercise (PageIndex{3}) 0.48 Answer 48% Exercise (PageIndex{4}) 0.343 Exercise (PageIndex{5}) 0.771 Answer 77.1% Exercise (PageIndex{6}) 1.42 Exercise (PageIndex{7}) 2.58 Answer 258% Exercise (PageIndex{8}) 4.976 Exercise (PageIndex{9}) 16.1814 Answer 1,618.14% Exercise (PageIndex{10}) 533.01 Exercise (PageIndex{11}) 2 Answer 200% Exercise (PageIndex{12}) 14 For the following 10 problems, convert each percent to a decimal. Exercise (PageIndex{13}) 15% Answer 0.15 Exercise (PageIndex{14}) 43% Exercise (PageIndex{15}) 16.2% Answer 0.162 Exercise (PageIndex{16}) 53.8% Exercise (PageIndex{17}) 5.05% Answer 0.0505 Exercise (PageIndex{18}) 6.11% Exercise (PageIndex{19}) 0.78% Answer 0.0078 Exercise (PageIndex{20}) 0.88% Exercise (PageIndex{21}) 0.09% Answer 0.0009 Exercise (PageIndex{22}) 0.001% For the following 14 problems, convert each fraction to a percent. Exercise (PageIndex{23}) (dfrac{1}{5}) Answer 20% Exercise (PageIndex{24}) (dfrac{3}{5}) Exercise (PageIndex{25}) (dfrac{5}{8}) Answer 62.5% Exercise (PageIndex{26}) (dfrac{1}{16}) Exercise (PageIndex{27}) (dfrac{7}{25}) Answer 28% Exercise (PageIndex{28}) (dfrac{16}{45}) Exercise (PageIndex{29}) (dfrac{27}{55}) Answer (49.overline{09})% Exercise (PageIndex{30}) (dfrac{15}{8}) Exercise (PageIndex{31}) (dfrac{41}{25}) Answer 164% Exercise (PageIndex{32}) (6 dfrac{4}{5}) Exercise (PageIndex{33}) (9 dfrac{9}{20}) Answer 945% Exercise (PageIndex{34}) (dfrac{1}{200}) Exercise (PageIndex{35}) (dfrac{6}{11}) Answer (54.overline{54})% Exercise (PageIndex{36}) (dfrac{35}{27}) For the following 14 problems, convert each percent to a fraction. Exercise (PageIndex{37}) 80% Answer (dfrac{4}{5}) Exercise (PageIndex{38}) 60% Exercise (PageIndex{37}) 25% Answer (dfrac{1}{4}) Exercise (PageIndex{38}) 75% Exercise (PageIndex{37}) 65% Answer (dfrac{13}{20}) Exercise (PageIndex{38}) 18% Exercise (PageIndex{37}) 12.5% Answer (dfrac{1}{8}) Exercise (PageIndex{38}) 37.5% Exercise (PageIndex{37}) 512.5% Answer (dfrac{41}{8}) or (5 dfrac{1}{8}) Exercise (PageIndex{38}) 937.5% Exercise (PageIndex{37}) (9.overline{9})% Answer (dfrac{1}{10}) Exercise (PageIndex{38}) (55.overline{5})% Exercise (PageIndex{37}) (22.overline{2})% Answer (dfrac{2}{9}) Exercise (PageIndex{38}) (63.overline{6})% #### Exercises for Review Exercise (PageIndex{39}) Find the quotient. (dfrac{40}{54} div 8 dfrac{7}{21}). Answer (dfrac{2}{9}) Exercise (PageIndex{40}) (dfrac{3}{8}) of what number is (2dfrac{2}{3})? Exercise (PageIndex{41}) Find the value of (dfrac{28}{15} + dfrac{7}{10} - dfrac{5}{12}). Answer (dfrac{129}{60}) or (2 dfrac{9}{60} = 2 dfrac{3}{20}) Exercise (PageIndex{42}) Round 6.99997 to the nearest ten thousandths. Exercise (PageIndex{43}) On a map, 3 inches represent 40 miles. How many inches represent 480 miles? Answer 36 inches ## How to Change Any Number to a Percent, With Examples Understanding and calculating percentages can help you in many ways: working out the correct tip at a restaurant, knowing how much you are saving on that mega sale or allowing you to interpret data from mathematical and scientific research. In short, learning more about percentages is important for chemistry and all other areas. A percentage is a way of expressing one number as a portion or share of a whole number, and percentages are always based on their relation to 100, which represents the whole number or object. For example, 75% is the same as 75 out of 100. Any percentage lower than 100 is just part of the whole or total. Percentages are ratios, they can, therefore, be written as fractions and then decimals. Converting answers from percentages to fractions to decimals can be a good exercise to check the accuracy of your work.You can convert any number to a percent. ## 9.3. Mathematical Functions and Operators Mathematical operators are provided for many PostgreSQL types. For types without common mathematical conventions for all possible permutations (e.g., date/time types) we describe the actual behavior in subsequent sections. Table 9-2 shows the available mathematical operators. Table 9-2. Mathematical Operators The bitwise operators are also available for the bit string types bit and bit varying, as shown in Table 9-3. Bit string operands of &, \|, and # must be of equal length. When bit shifting, the original length of the string is preserved, as shown in the table. Table 9-3. Bit String Bitwise Operators Table 9-4 shows the available mathematical functions. In the table, dp indicates double precision. Many of these functions are provided in multiple forms with different argument types. Except where noted, any given form of a function returns the same data type as its argument. The functions working with double precision data are mostly implemented on top of the host system's C library accuracy and behavior in boundary cases may therefore vary depending on the host system. Table 9-4. Mathematical Functions Function Return Type Description Example Result abs (x) (same as x) absolute value abs(-17.4) 17.4 cbrt (dp) dp cube root cbrt(27.0) 3 ceil (dp or numeric) (same as input) smallest integer not less than argument ceil(-42.8) -42 degrees (dp) dp radians to degrees degrees(0.5) 28.6478897565412 exp (dp or numeric) (same as input) exponential exp(1.0) 2.71828182845905 floor (dp or numeric) (same as input) largest integer not greater than argument floor(-42.8) -43 ln (dp or numeric) (same as input) natural logarithm ln(2.0) 0.693147180559945 log (dp or numeric) (same as input) base 10 logarithm log(100.0) 2 log (b numeric, x numeric) numeric logarithm to base b log(2.0, 64.0) 6.0000000000 mod (y, x) (same as argument types) remainder of y/x mod(9,4) 1 pi () dp "π" constant pi() 3.14159265358979 pow (a dp, b dp) dp a raised to the power of b pow(9.0, 3.0) 729 pow (a numeric, b numeric) numeric a raised to the power of b pow(9.0, 3.0) 729 radians (dp) dp degrees to radians radians(45.0) 0.785398163397448 random () dp random value between 0.0 and 1.0 random() round (dp or numeric) (same as input) round to nearest integer round(42.4) 42 round (v numeric, s integer) numeric round to s decimal places round(42.4382, 2) 42.44 setseed (dp) int32 set seed for subsequent random() calls setseed(0.54823) 1177314959 sign (dp or numeric) (same as input) sign of the argument (-1, 0, +1) sign(-8.4) -1 sqrt (dp or numeric) (same as input) square root sqrt(2.0) 1.4142135623731 trunc (dp or numeric) (same as input) truncate toward zero trunc(42.8) 42 trunc (v numeric, s integer) numeric truncate to s decimal places trunc(42.4382, 2) 42.43 Finally, Table 9-5 shows the available trigonometric functions. All trigonometric functions take arguments and return values of type double precision. ## 7.4: Percent - Mathematics Introducing Proportional Relationships • Representing Proportional Relationships with Tables • Representing Proportional Relationships with Equations • Comparing Proportional and Nonproportional Relationships • Representing Proportional Relationships with Graphs • Let's Put it to Work ### Unit 4 Proportional Relationships and Percentages • Proportional Relationships with Fractions • Percent Increase and Decrease • Applying Percentages • Let's Put it to Work ### Unit 5 Rational Number Arithmetic • Interpreting Negative Numbers • Adding and Subtracting Rational Numbers • Multiplying and Dividing Rational Numbers • Four Operations with Rational Numbers • Solving Equations When There Are Negative Numbers • Let's Put It to Work ### Unit 6 Expressions, Equations, and Inequalities • Representing Situations of the Form $px+q=r$ and $p(x+q)=r$ • Solving Equations of the Form $px+q=r$ and $p(x+q)=r$ and Problems That Lead to Those Equations • Inequalities • Writing Equivalent Expressions • Let's Put it to Work ### Unit 7 Angles, Triangles, and Prisms • Angle Relationships • Drawing Polygons with Given Conditions • Solid Geometry • Let's Put It to Work ### Unit 8 • Probabilities of Single Step Events • Probabilities of Multi-step Events • Sampling • Using Samples • Let's Put it to Work ### Unit 9 IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at https://openupresources.org/math-curriculum/. Adaptations and updates to IM 6–8 Math are copyright 2019 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Spanish translation of the "B" assessments are copyright 2020 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. This site includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information. ## Percentage Calculator Use Alcula's percentage calculator to compute percentages and answer questions such as: • How much is 7% of 25000? • What percentage of 10000 is 120? • 250 is 8 percent of what amount? • How much is 12000+8% In the calculator window, choose the question you need answered and enter the 2 quantities that you already know. The calculated result will automatically display on the right of the question you chose, along with the answers to all the other questions. To calculate percentage change, use one of the three calculators at the bottom. The percentage increase calculator calculates the chosen percentage from the initial quantity and adds it to the initial quantity to calculate the quantity after the increase. Similarly, the percentage decrease calculator subtracts the chosen percentage of the initial quantity from the initial quantity. Finally, the percent change calculator takes as input an initial quantity and a final quantity and calculates the difference as a percentage. ## Quartiles Another related idea is Quartiles, which splits the data into quarters: ### Example: 1, 3, 3, 4, 5, 6, 6, 7, 8, 8 The numbers are in order. Cut the list into quarters: In this case Quartile 2 is half way between 5 and 6: The Quartiles also divide the data into divisions of 25%, so: • Quartile 1 (Q1) can be called the 25th percentile • Quartile 2 (Q2) can be called the 50th percentile • Quartile 3 (Q3) can be called the 75th percentile ### Example: (continued) For 1, 3, 3, 4, 5, 6, 6, 7, 8, 8: • The 25th percentile = 3 • The 50th percentile = 5.5 • The 75th percentile = 7 ## Practice questions 1. A survey of American car buyers indicates that if a person buys a Ford, there is a 60% chance that their next purchase will be a Ford, while owners of a GM will buy a GM again with a probability of 0.80. Express the buying habits of these consumers in a transition matrix. 2. A hockey player decides to either shoot the puck (S) or pass it to a teammate (P) according to the following transition matrix. a. If the player shot on the first play, what is the probability that he will pass on the third play? b. What is the long-term shoot vs. pass distribution of this player? 3. The local police department conducts a campaign to reduce the rates of texting and driving in the community. The effects of the campaign are summarized in the transition matrix below: If 35% of people in the community reported texting and driving before the campaign: a. What is the percentage of people in the community that reported texting and driving after the campaign? b. If the campaign were to be repeated multiple times, what is the long-range trend in terms of the lowest rate that texting and driving can be reduced to in this community? 4. A large company conducted a training program with their employees to reduce the incidence of slips, trips and falls in the workplace. About 15% of workers reported a slip, trip or fall accident the previous year (year 1). After the training program (year 2), 75% of those who previously reported an accident reported no further accidents, while 5% of those who didn’t report a previous accident reported one this year. a. Create a transition matrix for this scenario. b. If the company employs 8500 workers, how many slip, trip and fall accidents were reported in year 2? c. If the program continued for another year, how many accidents would be reported in year 3? d. If the training program were to be repeated for many years, what is the lowest prevalence of slip, trip or fall accidents that could be achieved? ## 3 Simple Ways to Calculate Percentages (Math) How to calculate percentages is easier than you think. Learning this can help you to easily calculate tips at restaurants and how to use percentages to easily calculate sales prices when shopping. If you’re not sure how to perform any of those handy calculations, or if you’re just in need of a general percentage refresher, check out our guide on how to calculate percentages below. ### 1. Calculating the Percentage of a Whole To calculate a percentage, the whole amount must be known. This is in addition to the percentage or portion amount. You may be asked “what percentage of W is P,” where W is the whole amount and P is the portion amount. Or the question may be “how much is X percent of W,” where X represents a percentage figure. ##### 1. What is a percentage? A percentage is a way to express a number as a part of a whole. To calculate a percentage, we look at the whole as equal to 100%. For example, say you have 10 apples (=100%). If you eat 2 apples, then you have eaten 2/10 × 100% = 20% of your apples and you are left with 80% of your original apples. The term “percent” in English comes from the Italian per cento or the French pour cent, which literally mean per hundred. ##### 2. What is the value of the whole? For instance, let’s say we have a jar containing 1199 red marbles and 485 blue marbles, making it 1684 marbles in total. In this case, 1684 makes up a whole jar of marbles and will be set equal to 100%. ##### 3. Turn the value into percentage Let’s say we want to find out the percentage of the jar that is taken up by the 485 blue marbles. ##### 4. Put the two values into a fraction In our example, we need to find out what percent 485 (number of blue marbles) is of 1684 (total number of marbles). Therefore the fraction, in this case, is 485/1684. ##### 5. Convert the fraction to a decimal. To turn 485/1684 to a decimal, divide 485 by 1684. This comes to 0.288. ##### 6. Convert the decimal into a percent Multiply the result obtained in the step above by 100. For this example, 0.288 multiplied by 100 equals 28.8 or 28.8%. Formula: 0.288 x 100 = 28.8 or 28.8% A simple way to multiply a decimal by 100 is to move the decimal to the right two places. ### 2. Reverse Percentage You may come across a question that will ask you to work backward and find the original price of something after the price has increased. If you are given a quantity after a percentage increase or decrease, you may need to find the original amount. ##### 1. When to do reverse percentage? Sometimes you’re given the percentage of an amount and need to know the numerical value of the percent. Examples include calculating taxes, tips, and loan interest. Say you borrowed money from a friend who is going to charge you interest. The amount borrowed was initially $15 and the interest rate is 3% per day. These are the only two numbers you need for the calculation. ##### 3. Convert the percent into a decimal Multiply the percent by .01 or simply move the decimal to the left two places. This turns 3% into .03. ##### 4. Multiply your initial total by the new decimal In this case, multiply 15 by .03. This comes to 0.45. In this example, .45 is the amount of interest accrued each day that you do not pay your friend back. Formula: 0.3 x 15 = .45 (amount of interest accrued) ### 3. Calculating Discounts Doing your shopping but want to save money by picking discounted items? Learning how to solve discount percentages will help you learn if you are actually saving money or wasting money. ##### 1. Price and the discount amount This is a very simple way to calculate a discounted price, but you must begin with an accurate percent off. Double check what your item is on sale for. ##### 2. Opposite of the discount percent. The opposite of a percent is 100% minus the percent you are working with. If you want to buy a shirt that is 30% off, the opposite of this is 70%. Formula: 100% – 30% (discount) = 70 % ##### 3. Convert the opposite percent to a decimal. To convert a percent to a decimal, multiply it by .01 or move the decimal two places to the left. In this example, 70% becomes .7. ##### 4. Multiply the price by the new decimal If the shirt you want is$20, multiply 20 by .7. This comes to 14. This means the shirt is on sale for $14. Formula:$20 (item price) x .7 = 14 (discounted item price) Did you find this article helpful? Share it with your friends! ## Lesson 7 Summary We can use a double number line diagram to show information about percent increase and percent decrease: The initial amount of cereal is 500 grams, which is lined up with 100% in the diagram. We can find a 20% increase to 500 by adding 20% of 500: egin500+(0.2)oldcdot 500 &= (1.20)oldcdot 500&=600end In the diagram, we can see that 600 corresponds to 120%. If the initial amount of 500 grams is decreased by 40%, we can find how much cereal there is by subtracting 40% of the 500 grams: egin500−(0.4)oldcdot 500 &= (0.6)oldcdot 500&=300end So a 40% decrease is the same as 60% of the initial amount. In the diagram, we can see that 300 is lined up with 60%. To solve percentage problems, we need to be clear about what corresponds to 100%. For example, suppose there are 20 students in a class, and we know this is an increase of 25% from last year. In this case, the number of students in the class last year corresponds to 100%. So the initial amount (100%) is unknown and the final amount (125%) is 20 students. Looking at the double number line, if 20 students is a 25% increase from the previous year, then there were 16 students in the class last year. ## How to calculate percentages Learn how to calculate percentages in this easy lesson! When you're asked to calculate an (unknown) percentage ("What percentage. "), you need to first write the fraction PART/TOTAL, and then simply write that fraction as a decimal and as a percentage. See many examples below. The concepts and ideas of this lesson are also explained in this video: What percentage of the height of a 15-ft tree is a 3-ft sapling? A choir has 22 women and 18 men. Find what percentage of the choir&rsquos members are men. One pair of jeans costs $25 and another costs$28. How many percent is the price of cheaper jeans of the price of the more expensive jeans? Look carefully at the questions above. Notice that the problems don&rsquot tell you the percentage in other words, there is no number in the problem written as x%. Instead, they ask you to find it! Asking &ldquoWhat percentage?&rdquo or "How many percent?" is the same as asking &ldquoHow many hundredth parts?&rdquo We can solve these questions in a two-part process: 1. First find out the part that is being asked for as a fraction. The denominator probably won&rsquot be 100. 2. Convert that fraction to a decimal. Then you can easily convert the decimal to a percentage! Example 1. A choir has 22 women and 18 men. Find what percentage of the choir&rsquos members are men. 1. Find out what part (fraction) of the choir&rsquos members are men. That is 18/40, or 9/20. 2. Write 9/20 as a percent. Use equivalent fractions: 9/20 = 45/100 = 45%. Example 2. One pair of jeans costs $25 and the other costs$28. How many percent is the price of cheaper jeans of the price of the more expensive jeans? 1. Write what part the cheaper price is of the more expensive price. The answer is 25/28. 2. Write 25/28 as a percentage. A calculator gives 25/28 = 0.8928. Rounded to the nearest whole percent, that&rsquos 89%. 1. a. What percentage of a 15-ft tree is a little 3-ft sapling? b. How many percent is $12 of$16? 2. Find how many percent the smaller object&rsquos height is of the taller object&rsquos height. 3. A 2-year old child measures 32 inches tall and weighs 24 pounds. A 10-year old child measures 52 inches tall and weighs 96 pounds. a. How many percent is the smaller child&rsquos age of the older child&rsquos age? b. How many percent is the smaller child&rsquos height of the older child&rsquos height? 4. Write the percentages into the sectors in the circle graphs Think of fractions! 5. The circle graph at the right gives the angle measure of each sector of the circle. Find what percentage each sector is of the whole circle, and write the percentage in the sector. Remember, the whole circle is 360°.
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # What Are The Prime Factors Of 92 What are the prime factors of 92? Answer: 2, 23 The number 92 has 2 prime factors. Primes can only have two factors(1 and itself) and only be divisible by those two factors. Any figure where this rule applies can be called a prime factor. The biggest prime factor of 92 is 23. The smallest prime factor of 92 is 2. ## What Is The Factor Tree Of 92 How to use a factor tree to find the prime factors of 92? A factor tree is a diagram that organizes the factoring process. • First step is to find two numbers that when multiplied together equal the number we start with. 92 ↙ ↘ 2 × 46 • Second step is to check the multiplication(of the first step) for numbers that are not primes. 46 ↙ ↘ 23 × 2 We found 2 prime factors(2, 23) using the factor tree of 92. Now let us explain the process to solving factor trees in more detail. Our goal is to find all prime factors of a given whole number. In each step of our factor tree diagram for 92 we always checked both multiplication numbers if they were primes or not. If one or both of the integers are not prime numbers then this means that we will have to make diagrams for them too. This process continues until only prime numbers are left. Remember that often a factor tree for the same integer can be solved in more than one correct way! An example of this is the figure 12 where 26=12 and 43=12. The primes of a factor tree for 12 are the same regardles if we start the factor tree with 26 or 43. ## How To Verify If Prime Factors Of 92 Are Correct Answers To know if we got the correct prime factors of 92 we have to get the prime factorization of 92 which is 2 * 2 * 23. Because when you multiply the primes of the prime factorization the answer has to be equal with 92. After having checked the prime factorization we can now safely say that we got all prime factors. ## General Mathematical Properties Of Number 92 92 is a composite number. 92 is a composite number, because it has more divisors than 1 and itself. This is an even number. 92 is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 92 is not an odd number. When we simplify Sin 92 degrees we get the value of sin(92)=-0.7794660696158. Simplify Cos 92 degrees. The value of cos(92)=-0.62644444791034. Simplify Tan 92 degrees. Value of tan(92)=1.2442700581287. When converting 92 in binary you get 1011100. Converting decimal 92 in hexadecimal is 5c. The square root of 92=9.5916630466254. The cube root of 92=4.514357435474. Square root of √92 simplified is 2√23. All radicals are now simplified and in their simplest form. Cube root of ∛92 simplified is 92. The simplified radicand no longer has any more cubed factors. ## Determine Prime Factors Of Numbers Smaller Than 92 Learn how to calculate primes of smaller numbers like: ## Determine Prime Factors Of Numbers Bigger Than 92 Learn how to calculate primes of bigger numbers such as: ## Single Digit Properties For Number 92 Explained • Integer 9 properties: 9 is odd and the square of 3. It is a composite, with the following divisors:1, 3, 9. Since the quantity of the divisors(excluding itself) is 4<9, it is a defective number. In mathematics nine is a perfect total, suitable and a Kaprekar figure. Any amount is divisible by 9 if and only if the quantity of its digits is. Being divisible by the count of its divisors, it is refactorizable. Each natural is the sum of at most 9 cubes. If any sum of the digits that compose it is subtracted from any natural, a multiple of 9 is obtained. The first odd square and the last single-digit quantity. In the binary system it is a palindrome. Part of the Pythagorean triples (9, 12, 15), (9, 40, 41). A repeated number in the positional numbering system based on 8. Nine is a Colombian digit in the numerical decimal system. If multiplied 9 always leads back to itself: 2×9=18 → 1+8=9, 3×9=27 → 2+7=9 in the same way if you add a number to 9, the result then refers to the initial digit: 7+9=16 → 1+6=7, 7+9+9=25 → 2+5=7, 7+9+9+9=34 → 3+4=7. If you put 111111111 in the square (ie 1 repeated 9 times) you get the palindrome 12345678987654321, also if you add all the numbers obtained: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 you get 81, and in turn 8 + 1 = 9. • Integer 2 properties: 2 is the first of the primes and the only one to be even(the others are all odd). The first issue of Smarandache-Wellin in any base. Goldbach's conjecture states that all even numbers greater than 2 are the quantity of 2 primes. It is a complete Harshad, which is a integer of Harshad in any expressed base. The third of the Fibonacci sequence, after 1 and before 3. Part of the Tetranacci Succession. Two is an oblong figure of the form n(n+1). 2 is the basis of the binary numbering system, used internally by almost all computers. Two is a number of: Perrin, Ulam, Catalan and Wedderburn-Etherington. Refactorizable, which means that it is divisible by the count of its divisors. Not being the total of the divisors proper to any other arithmetical value, 2 is an untouchable quantity. The first number of highly cototent and scarcely totiente (the only one to be both) and it is also a very large decimal. Second term of the succession of Mian-Chowla. A strictly non-palindrome. With one exception, all known solutions to the Znam problem begin with 2. Numbers are divisible by two (ie equal) if and only if its last digit is even. The first even numeral after zero and the first issue of the succession of Lucas. The aggregate of any natural value and its reciprocal is always greater than or equal to 2. ## Finding All Prime Factors Of A Number We found that 92 has 2 primes. The prime factors of 92 are 2, 23. We arrived to this answer by using the factor tree. However we could have also used upside down division to get the factorization primes. There are more that one method to factorize a integer. ## List of divisibility rules for finding prime factors faster Rule 1: If the last digit of a figure is 0, 2, 4, 6 or 8 then it is an even number. All even numbers are divisible by 2. Rule 2: If the sum of digits of a integer is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 12 are 1 and 2 so 1+2=3 and 3 is divisible by 3, meaning that 12 is divisible by 3). The same logic also works for 9. Rule 3: If the last two digits of a number are 00 then this integer is divisible by 4(example: we know that 124=100+24 and 100 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 24). In order to use this rule to it's fullest it is best to know multiples of 4. Rule 4: If the last digit of a number is 0 or 5 then 5 it is divisible by 5. Rule 5: All integers that are divisible by both 2 and 3 are also divisible by 6. This is logical because 2*3=6. ## What Are Prime Factors Of A Number? All numbers that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself).
Math1011_Week07_3_678 # Math1011_Week07_3_678 - Math1011 Learning Strategies Center... This preview shows pages 1–5. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Math1011 Learning Strategies Center Section 3.6 10/24/2009 Find the slope of circle x 2 + y 2 = 25 at point (3,-4). Use implicit differentiation to show that the circle x 2 + y 2 =25 is concave down in 2nd quadrant. Note: to show f(x) is concave down, show f’’(x) < 0. Math1011 Learning Strategies Center Section 3.6 10/24/2009 Find the slope of circle x 2 + y 2 = 25 at point (-3,4). 2 2 d d d dx dx dx d d dx dx 2 2 3 3 4 4 x 25 Implicit Differentiation and Chain Rule 2 ( x ) 2 ( ) 2 2 2 2 (-3, 4), The slope on the circle on (-3, 4) is 3/4. dy dx dy dx dy x x y y dx dy x y dx y x y y x y y x At − − − − − + = + = + = = − = = = = = Use implicit differentiation to show that the circle x 2 + y 2 =25 is concave down in 2nd quadrant. Note: to show f(x) is concave down, show f’’(x) < 0. 2 2 2-x y ( 1) ' '' ( ) Take 2nd derivative ( ) ( ) '' Quotient Rule ( 1) ' '' 2nd quadrant: y > 0 and x < 0 so y'= '' As y''< 0, y is concave down in 2nd qua d x y dx d d dx dx y x y y y y x x y y y y x y y y y − − − + = − − = − + = > = < drant (Doc #011w.36.01t) Math1011 Section 3.6 2 2 2 Consider the curve x y - 6y + 2 = 0 Find the equation of the tangent line to the curve at point (2,1) Math1011 Section 3.6 2 2 2 Consider the curve x y - 6y + 2 = 0 Find the equation of the tangent line to the curve at point (2,1) 2 2 2 2 2 2 d d d d dx dx dx dx 2 2 2 2 2 2 2 2 2 2 2 2... View Full Document ## This note was uploaded on 11/22/2009 for the course CHEM 2070 at Cornell. ### Page1 / 16 Math1011_Week07_3_678 - Math1011 Learning Strategies Center... This preview shows document pages 1 - 5. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# Statistics Solution of TS & AP Board Class 9 Mathematics ###### Exercise 9.1 Question 1. Write the mark wise frequencies in the following frequency distribution table. The mark wise frequency representation indicates the upper range of the marks and their corresponding frequencies. Let’s form a frequency table showing mark wise frequencies. We can also represent marks and their corresponding frequencies on a graph, which will show relationship between them. Question 2. Write the mark wise frequencies in the following frequency distribution table. The mark wise frequency representation indicates the upper range of the marks and their corresponding frequencies. Let’s form a frequency table showing mark wise frequencies. We can also represent marks and their corresponding frequencies on a graph, which will show relationship between them. Question 3. The blood groups of 36 students of IX class are recorded as follows. Represent the data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students? Given the blood groups of 36 students can be represented in the form of frequency distribution table. Just count the occurrence of blood groups in the data and formulate a table. Count of blood group: A = 10 B = 9 O = 15 AB = 2 Representing the same on a graph to show the relationship between blood group and number of students (frequency), From the frequency table as well as from the graph, notice that: Most common blood group = O [As O has the greatest no. of occurrence, i.e., frequency = 15] Rarest blood group = AB [As AB has the least no. of occurrence, i.e., frequency = 2] Question 4. The blood groups of 36 students of IX class are recorded as follows. A O A O A B O A B A B O B O B O O A B O B AB O A O O O A AB O A B O A O B Represent the data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students? Given the blood groups of 36 students can be represented in the form of frequency distribution table. Just count the occurrence of blood groups in the data and formulate a table. Count of blood group: A = 10 B = 9 O = 15 AB = 2 Representing the same on a graph to show the relationship between blood group and number of students (frequency), From the frequency table as well as from the graph, notice that: Most common blood group = O [As O has the greatest no. of occurrence, i.e., frequency = 15] Rarest blood group = AB [As AB has the least no. of occurrence, i.e., frequency = 2] Question 5. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows; Prepare a frequency distribution table for the data given above. Given the number of heads on tossing three coins can be represented in the form of frequency distribution table. Just count the occurrence of number of heads in the data and formulate a table. Count of occurrence of head by 3 coins: 0 (head occurred 0 times) = 3 1 (head occurred 1 times) = 10 2 (head occurred 2 times) = 10 3 (head occurred 3 times) = 7 It can also be analyzed and represented graphically: Question 6. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows; 1 2 3 2 3 1 1 1 0 3 2 1 2 2 1 1 2 3 2 0 3 0 1 2 3 2 2 3 1 1 Prepare a frequency distribution table for the data given above. Given the number of heads on tossing three coins can be represented in the form of frequency distribution table. Just count the occurrence of number of heads in the data and formulate a table. Count of occurrence of head by 3 coins: 0 (head occurred 0 times) = 3 1 (head occurred 1 times) = 10 2 (head occurred 2 times) = 10 3 (head occurred 3 times) = 7 It can also be analyzed and represented graphically: Question 7. A TV channel organized a SMS(Short Message Service) poll on prohibition on smoking, giving options like A – complete prohibition, B – prohibition in public places only, C – not necessary. SMS results in one hour were Represent the above data as grouped frequency distribution table. How many appropriate answers were received? What was the majority of peoples’ opinion? Given the poll options (A, B & C) can be represented in the form of frequency distribution table. Just count the occurrence of each option in the data and formulate a table. Count of occurrence of options are: A (complete prohibition) = 19 B (prohibition in public places only) = 36 C (not necessary) = 10 It can also be analyzed and represented graphically: That is, 19 + 36 + 10 = 65 To analyze majority votes on poll, mark the highest frequency. Note that, highest frequency is 36, that is of option B. And hence, majority of people’s opinion is B, i.e., prohibition in public places only. Question 8. A TV channel organized a SMS(Short Message Service) poll on prohibition on smoking, giving options like A – complete prohibition, B – prohibition in public places only, C – not necessary. SMS results in one hour were A B A B C B A B B A C C B B A B B A B C B A B C B A B B A B B C B A B A B C B B A B C B B A B B A B B A B C B A B B A B C A B B A Represent the above data as grouped frequency distribution table. How many appropriate answers were received? What was the majority of peoples’ opinion? Given the poll options (A, B & C) can be represented in the form of frequency distribution table. Just count the occurrence of each option in the data and formulate a table. Count of occurrence of options are: A (complete prohibition) = 19 B (prohibition in public places only) = 36 C (not necessary) = 10 It can also be analyzed and represented graphically: That is, 19 + 36 + 10 = 65 To analyze majority votes on poll, mark the highest frequency. Note that, highest frequency is 36, that is of option B. And hence, majority of people’s opinion is B, i.e., prohibition in public places only. Question 9. Represent the data in the adjacent bar graph as frequency distribution table. Analyzing a bar graph is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph. There are 4 parameters for vehicles namely, cycles, autos, bikes and cars. So now, create a table consisting of these vehicles and their corresponding frequencies. Thus, this is the required frequency distribution table. Question 10. Represent the data in the adjacent bar graph as frequency distribution table. Analyzing a bar graph is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph. There are 4 parameters for vehicles namely, cycles, autos, bikes and cars. So now, create a table consisting of these vehicles and their corresponding frequencies. Thus, this is the required frequency distribution table. Question 11. Identify the scale used on the axes of the adjacent graph. Write the frequency distribution from it. Observe that, on x-axis we have classes from I Class to VI Class. The scale used in x-axis is class on a unit interval. And on y-axis, we have number of students. The scale used on y-axis is from 0 to 90 with an interval of 10 units. Analyzing a histogram is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph. There are 6 parameters for classes namely, I Class, II Class, III Class, IV Class, V Class and VI Class. So now, create a table consisting of these classes and their corresponding frequencies (number of students). Thus, this is the required frequency distribution table. Question 12. Identify the scale used on the axes of the adjacent graph. Write the frequency distribution from it. Observe that, on x-axis we have classes from I Class to VI Class. The scale used in x-axis is class on a unit interval. And on y-axis, we have number of students. The scale used on y-axis is from 0 to 90 with an interval of 10 units. Analyzing a histogram is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph. There are 6 parameters for classes namely, I Class, II Class, III Class, IV Class, V Class and VI Class. So now, create a table consisting of these classes and their corresponding frequencies (number of students). Thus, this is the required frequency distribution table. Question 13. The marks of 30 students of a class, obtained in a test (out of 75), are given below: 42, 21, 50, 37, 42, 37, 38, 42, 49, 52, 38, 53, 57, 47, 29 59, 61, 33, 17, 17, 39, 44, 42, 39, 14, 7, 27, 19, 54, 51. Form a frequency table with equal class intervals. (Hint : one of them being 0-10)
# Iowa - Grade 1 - Math - Numbers and Operations in Base Ten - Comparing Numbers - 1.NBT.3 ### Description Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <. • State - Iowa • Standard ID - 1.NBT.3 • Subjects - Math Common Core ### Keywords • Math • Numbers and Operations in Base Ten ## More Iowa Topics Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.) Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
Question 16 # In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through Band C. Then the area, in sq. cm, of the overlapping region between the two circles is Solution BC is the diameter of circle C2 so we can say that $$\angle BAC=90^{\circ\ }$$ as angle in the semi circle is $$90^{\circ\ }$$ Therefore overlapping area = $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1 AB= AC = 8 cm and as $$\angle BAC=90^{\circ\ }$$, so we can conclude that BC= $$8\sqrt{2}$$ cm Radius of C2 = Half of length of BC = $$4\sqrt{2}$$ cm Area of C2 = $$\pi\left(4\sqrt{2}\right)^2=32\pi$$ $$cm^2$$ A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm Area of the minor sector made be BC in C1 = $$\frac{1}{4}$$(Area of circle C1) - Area of triangle ABC = $$\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$$ $$cm^2$$ Therefore, Overlapping area between the two circles= $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1 = $$\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$$ $$cm^2$$
# How To Solve a Magic Square The Magic Square has been around for over 4000 years - possibly over 5000! We'll see How to Solve a Magic Square! Let's look at a 4 x 4 magic square. Basic Rules to solve the magic Square: Fill in the small squares inside the large square with the consecutive integers from 1 to 16 so that the sum of all rows, columns, and diagonal are the same number - which is the "Magic Number". This is where we use of Algebra in real life • Determine the Magic Sum of your 4 x 4 magic square. Below is the an "Algebra" formula that we can use to help determine what our "Magic Sum" can be. Since we're working with a 4 x4 magic square, the Order of our magic square is 4. See formula below: • Place the number 1 in the center box on the top row. This same thing is done for any magic square with an odd number of columns or rows. • Follow an up-one, right-one pattern to fill in the remaining numbers. Start at the middle box in the top row where you placed the 1. Move up one row and to the right one column and write in the next number sequentially. Whenever a move takes you above the top row (like the first move will), go to the bottom row instead. If you need to move a square to the right but are already in the rightmost column, move to the leftmost column instead. Continue filling in the square following the pattern until it’s complete. 5. Numerical Patterns: Magic squares contain intriguing numerical patterns that captivate mathematicians and enthusiasts alike. 6. Cultural Diversity: Different cultures have developed their own variations of magic squares, each with its unique properties and rules. 7. Mystery and Challenge: The challenge of constructing and solving magic squares adds an element of mystery, making them a lifelong pursuit for math aficionados. 8. Educational Value: Working with magic squares can enhance mathematical skills, including addition, problem-solving, and pattern recognition. 9. Mind-Bending: The concept of magic squares is mind-bending, as the mathematical rules governing them are not immediately obvious. 10. Versatile Application: Magic squares find applications in various fields, from recreational math to art, cryptography, and even some areas of science. 11. Cognitive Benefits: Solving magic squares can improve cognitive functions such as memory, attention, and logical reasoning. In summary, magic squares are a captivating fusion of mathematics, history, art, and culture, making them an endlessly fascinating topic for exploration and enjoyment. 1. Let's say we are trying to solve a 4 x 4 magic square... The Magic Sum for the magic Square is 34. This means all the rows, columns, and diagonals of the magic square have to add up to 34 using the consecutive integers: 1 through 16.
# Highest Common Factor (HCF) Tnpsc ## Highest Common Factor The highest common factor, also known as the greatest common divisor, is the largest positive integer that divides two or more numbers without leaving a remainder. It is commonly denoted as HCF or GCD. To find the highest common factor of two numbers, you can use various methods such as prime factorization, division method, or the Euclidean algorithm. For example, let’s find the highest common factor of 24 and 36 using the prime factorization method: ### Prime factorization 1. Find the prime factors of each number: • 24 = 2 * 2 * 2 * 3 = 2^3 * 3 • 36 = 2 * 2 * 3 * 3 = 2^2 * 3^2 2. Identify the common prime factors and their lowest powers: • Common factors: 2^2 * 3 3. Multiply the common factors to find the highest common factor: • HCF(24, 36) = 2^2 * 3 = 12 Therefore, the highest common factor of 24 and 36 is 12. Finding the highest common factor is important in various mathematical operations, simplifying fractions, and solving mathematical problems. ### Division Method To find the highest common factor (HCF) of 24 and 36 using the division method, follow these steps: 1. Start by dividing the larger number (36) by the smaller number (24): 36 ÷ 24 = 1, with a remainder of 12 2. Now divide the divisor (24) by the remainder (12): 24 ÷ 12 = 2 3. Divide the remainder from the previous step (12) by the new remainder (0): 12 ÷ 0 = undefined Since the remainder has become zero, the HCF is the divisor from the last non-zero remainder. In this case, the HCF of 24 and 36 using the division method is 12. Therefore, the highest common factor of 24 and 36 is 12 when using the division method. * * All the Notes in this blog, are referred from Tamil Nadu State Board Books and Samacheer Kalvi Books. Kindly check with the original Tamil Nadu state board books and Ncert Books.
Book Allocation Problem Introduction In this article, we will solve the problem: Book Allocation. We will start with a quite naive and intuitive approach and then proceed towards a more optimal binary search solution. Problem Statement Given an array ‘pages’ of integer numbers, where ‘pages[i]’ represents the number of pages in the ‘i-th’ book. There are ‘m’ number of students, and the task is to allocate all the books to their students. Allocate books in a way such that: 1. Each student gets at least one book. 2. Each book should be allocated to a student. 3. Book allocation should be in a contiguous manner. You have to allocate the books to ‘m’ students such that the maximum number of pages assigned to a student is minimum. Let’s understand the problem statement through an example. Example: Number of books = 4 and Number of students = 2 pages[] = { 10,20,30,40} Important points to consider • Assign at least one book to every student, so there can’t be any allocation such that a student gets no books assigned. • While allocating the books, no book should be left out. In other words, you have to allocate each and every book given. • Allocate in a contiguous manner. Let's say you have to allocate 3 books to a student from pages[] = { 10,20,30,40} Then, the possible allocations can be - {10,20,30} and {20,30,40}. You can’t allocate {10,30,40} as it is not contiguous. All possible ways of book allocation are shown in the below figure- The minimum of the maximum number of pages assigned = min{90,70,60} = 60. Hence, the required answer is 60. Brute Force Approach Let be the number of books and the number of students. In the case of M > N: Simply return -1 as the number of students is greater than the number of books available, and according to the given constraints, it is not possible to allocate at least one book to each student. Let’s see the algorithm for all other cases where M<=N: • We have to minimize the value of the maximum number of pages assigned to a student in an allocation. • If the maximum number of pages assigned to a student in a book allocation is max_pages, then this implies that the number of pages assigned to every student is less than or equal to max_pages. • Let’s focus on the possible values of the number of pages that can be assigned to a student? • The minimum value can be equal to 0 (Though not in this problem because you have to assign each student at least one book, so the number of pages can't be zero) • The maximum number of pages will be the sum of the number of pages in all books. (This will happen when you assign all the books to one student). • The range of the maximum number of pages we obtained is - (0, sum of all the values of pages array]. Open interval on 0 because at least one book needs to be assigned to every student. So, the interval we get is [1, sum of pages array]. Steps of algorithm 1. Iterate over all values ranging from numPages=1 to numPages=sum of pages array. 2. Check if it is possible to allocate the books such that the value of the number of pages assigned to any student is less than or equal to numPages in each iteration. 3. If the maximum number of pages can be equal to numPages, then return numPages as the answer. We don’t need to iterate further because after this the value of numPages will increase, but we are interested in finding the minimum value of the maximum number of pages that can be allocated. How to check if it is possible to allocate the books such that the maximum number of pages assigned to any student is numPages? 1. Initialize count of students with 1. 2. Keep allocating the books to a student until the sum of the pages assigned is less than numPages. 3. If at any point the number of pages assigned to a student exceeds numPages, then allocate the current book to the next student and increment the count of students. 4. If the count of students becomes greater than M, then return false. 5. In the end, if the count of students is equal to M, return true. Implementation in C++ ``````/* C++ code for Book Allocation problem to find the minimum value of the maximum number of pages/ #include <bits/stdc++.h> using namespace std; /function to check if it is possible to allocate the books such that the maximum number of pages assigned to any student is numPages/ bool isPossible(int pages[], int n, int m, int numPages) { int cntStudents = 1; int curSum = 0; for (int i = 0; i < n; i++) { if (pages[i] > numPages) { return false; } if (curSum + pages[i] > numPages) { / Increment student count by '1'/ cntStudents += 1; / assign current book to next student and update curSum / curSum = pages[i]; / If count of students becomes greater than given no. of students, return False/ if (cntStudents > m) { return false; } } else { / Else assign this book to current student and update curSum / curSum += pages[i]; } } return true; } int allocateBooks(int pages[], int n, int m) { / If number student is more than number of books / if (n < m) { return -1; } / Count total number of pages / int sum = 0; for (int i = 0; i < n; i++) { sum += pages[i]; } / Check for every possible value / for (int numPages = 1; numPages <= sum; numPages++) { if (isPossible(pages, n, m, numPages)) { return numPages; } } return -1; } int main() { int n = 4; int m = 2; int pages[] = {10, 20, 30, 40}; cout << "The minimum value of the maximum number of pages in book allocation is: " << allocateBooks(pages, n, m) << endl; }`````` Output: ``The minimum value of the maximum number of pages in book allocation is: 60`` Implementation in Python ``````def isPossible(arr, n, m, curr_min): cntStudents = 1 curSum = 0 # iterate over all the books for i in range(n): if (arr[i] > curr_min): return False # Increment student count by '1' if (curSum + arr[i] > curr_min): cntStudents += 1 # assign current book to next student and update curSum curSum = arr[i] # If count of students becomes greater than # given no. of students, return False # update curSum if (cntStudents > m): return False else: curSum += arr[i] return True # function to find minimum pages def allocateBooks(arr, n, m): sum = 0 # If number student is more than number of books if (n < m): return -1 # Count total number of pages for i in range(n): sum += arr[i] for numPages in range(1, sum+1): ans = isPossible(arr, n, m, numPages) if (ans): return numPages return -1 # Number of pages in books arr = [10, 20, 30, 40] n = len(arr) m = 2 # No. of students print("The minimum value of the maximum number of pages in book allocation is", allocateBooks(arr, n, m))`````` Output: ``The minimum value of the maximum number of pages in book allocation is 60`` Implementation in Java ``````public class Main { //function to check if it is possible to allocate the books such that the //maximum number of pages assigned to any student is numPages static boolean isPossible(int arr[], int n, int m, int curr_min) { int cntStudents = 1; int curSum = 0; // iterate over all the books for (int i = 0; i < n; i++) { if (arr[i] > curr_min) return false; if (curSum + arr[i] > curr_min) { //Increment student count by '1' cntStudents++; / assign current book to next student and update curSum / curSum = arr[i]; / If count of students becomes greater than given no. of students, return False/ if (cntStudents > m) return false; } / Else assign this book to current student and update curSum / else curSum += arr[i]; } return true; } // method to find minimum pages static int findPages(int arr[], int n, int m) { long sum = 0; / If number student is more than number of books / if (n < m) return -1; / Count total number of pages / for (int i = 0; i < n; i++) sum += arr[i]; for (int numPages = 1; numPages <= sum; numPages++) { if (isPossible(arr, n, m, numPages)) { return numPages; } } return -1; } public static void main(String[] args) { int arr[] = {10, 20, 30, 40}; int m = 2; //No. of students int n = 4; System.out.println("The minimum value of the maximum number of pages in book allocation is " + findPages(arr, n, m)); } }`````` Output: ``The minimum value of the maximum number of pages in book allocation is 60`` Complexity Analysis Time Complexity - O(nSum) O(nSum), where ‘n’ is the number of integers in the array ‘pages’ and ‘Sum’ is the sum of all the elements of ‘pages’. We are using two nested loops of size ‘Sum’ and ‘n’. So, the time complexity is O(nSum). Space Complexity - O(1) O(1) as we are using constant space. Binary Search Approach The idea is to use binary search over the search space [1, Sum of pages array] to improve the time complexity. Steps of algorithm • Initially ‘start = 0’ and ‘end = sum of all pages’ • Find mid = (start+end)/2 • Check if it is possible to allocate books such that the number of pages allocated to each student is less than or equal to mid. • If possible then: • Update the minimum answer found so far and put end mid-1. • Since we need the minimum value of the maximum number of pages, the end is set as mid-1 so that our search space now becomes [start,mid-1] to get a value less than mid. • Else • Update start mid+1. Because if it is not possible to allocate the books with maximum pages equal to mid then it won’t be possible for any value less than mid. So, the search space becomes [mid+1,end]. • Repeat the above steps until start <= end. Implementation in C++ ``````/* C++ code for Book Allocation problem to find the minimum value of the maximum number of pages/ #include <bits/stdc++.h> using namespace std; /function to check if it is possible to allocate the books such that the maximum number of pages assigned to any student is numPages/ bool isPossible(int pages[], int n, int m, int numPages) { int cntStudents = 1; int curSum = 0; for (int i = 0; i < n; i++) { if (pages[i] > numPages) { return false; } if (curSum + pages[i] > numPages) { / Increment student count by '1'/ cntStudents += 1; / assign current book to next student and update curSum / curSum = pages[i]; / If count of students becomes greater than given no. of students, return False/ if (cntStudents > m) { return false; } } else { / Else assign this book to current student and update curSum / curSum += pages[i]; } } return true; } int allocateBooks(int pages[], int n, int m) { / If number student is more than number of books / if (n < m) { return -1; } / Count total number of pages / int sum = 0; for (int i = 0; i < n; i++) { sum += pages[i]; } int start = 0, end = sum; int result = INT_MAX; / Traverse until start <= end , binary search / while (start <= end) { / Check if it is possible to distribute books by using mid as current maximum / int mid = start + (end - start) / 2; if (isPossible(pages, n, m, mid)) { result = min(result, mid); /update the result/ end = mid - 1; } else { start = mid + 1; } } return result; } int main() { int n = 4; int m = 2; int pages[] = {10, 20, 30, 40}; cout << "The minimum value of the maximum number of pages in book allocation is: " << allocateBooks(pages, n, m) << endl; }`````` This video gives a detailed tutorial of the “Book Allocation Problem”, so you can watch this video to understand it better. Output: ``The minimum value of the maximum number of pages in book allocation is: 60`` Implementation in Python ``````# function to check if it is possible to allocate the books such that the # maximum number of pages assigned to any student is numPages def isPossible(arr, n, m, curr_min): cntStudents = 1 curSum = 0 # iterate over all the books for i in range(n): if (arr[i] > curr_min): return False if (curSum + arr[i] > curr_min): # Increment student count by '1' cntStudents += 1 # assign current book to next student and update curSum # If count of students becomes greater than # given no. of students, return False curSum = arr[i] if (cntStudents > m): # update curSum return False else: curSum += arr[i] return True # function to find minimum pages def allocateBooks(arr, n, m): sum = 0 # If number student is more than number of books if (n < m): return -1 # Count total number of pages for i in range(n): sum += arr[i] start, end = 0, sum result = 109 # Traverse until start <= end , binary search while (start <= end): mid = (start + end) // 2 if (isPossible(arr, n, m, mid)): result = mid end = mid - 1 else: start = mid + 1 return result # Number of pages in books arr = [10, 20, 30, 40] n = len(arr) m = 2 # No. of students print("The minimum value of the maximum number of pages in book allocation is", allocateBooks(arr, n, m))`````` Output: ``The minimum value of the maximum number of pages in book allocation is 60`` Implementation in Java ``````public class Main { //function to check if it is possible to allocate the books such that the //maximum number of pages assigned to any student is numPages static boolean isPossible(int arr[], int n, int m, int curr_min) { int cntStudents = 1; int curSum = 0; // iterate over all the books for (int i = 0; i < n; i++) { if (arr[i] > curr_min) return false; if (curSum + arr[i] > curr_min) { //Increment student count by '1' cntStudents++; / assign current book to next student and update curSum / curSum = arr[i]; / If count of students becomes greater than given no. of students, return False/ if (cntStudents > m) return false; } / Else assign this book to current student and update curSum / else curSum += arr[i]; } return true; } // method to find minimum pages static int findPages(int arr[], int n, int m) { long sum = 0; / If number student is more than number of books / if (n < m) return -1; / Count total number of pages / for (int i = 0; i < n; i++) sum += arr[i]; int start = 0, end = (int) sum; int result = Integer.MAX_VALUE; / Traverse until start <= end , binary search / while (start <= end) { / Check if it is possible to distribute books by using mid as current maximum / int mid = (start + end) / 2; if (isPossible(arr, n, m, mid)) { result = mid; end = mid - 1; } else start = mid + 1; } return result; } public static void main(String[] args) { int arr[] = {10,20,30,40}; int m = 2; //No. of students int n=4; System.out.println("The minimum value of the maximum number of pages in book allocation is " +findPages(arr,n, m)); } }`````` Output: ``The minimum value of the maximum number of pages in book allocation is 60`` Complexity Analysis Time Complexity - O(Nlog(Sum)) O(N*log(Sum)), where ‘N’ is the number of integers in the array ‘pages’ and ‘Sum’ is the sum of all the elements of ‘pages’ as for every number ‘mid’ we have an iteration loop of size ‘N’ and binary search takes ‘log(Sum)’ time. Space Complexity - O(1) O(1) as we are using constant space. Does a greedy algorithm always work? No, a greedy algorithm does not always work. To solve a problem via a greedy algorithm, you need to have intuitive proof in your mind, at least to lead to the correct solution. To show when it doesn’t work, you can think of a counter-example that will help rule out the greedy algorithm. What is a comparator? Comparator is an important concept. It is a function object that helps to achieve custom sorting depending on the problem requirements. What are the benefits of the greedy approach? The Benefits of a Greedy Approach 1. The algorithm is simpler to explain. 2. This algorithm has the potential to outperform other algorithms (but, not in all cases). Conclusion In this article, we discussed the Book Allocation problem to find the minimum value of the maximum number of pages assigned to a student. This question is one of the excellent applications of Binary Search. We discussed the naive approach first and then optimized the solution by binary search as the problem had monotonic property. With this problem, you must have got an idea of when and how to use binary search. Some of the problems which you can practice based on a similar concept are- Are you planning to ace the interviews of reputed product-based companies like Amazon, Google, Microsoft, and more? Attempt our Online Mock Test Series on CodeStudio now!
<meta http-equiv="refresh" content="1; url=/nojavascript/"> You are viewing an older version of this Concept. Go to the latest version. # Sums of Fractions with Like Denominators ## Result of adding numerators over denominator 0% Progress Practice Sums of Fractions with Like Denominators Progress 0% Sums of Fractions with Like Denominators Have you ever had to add small measurements or fractions to put something together? Having successfully completed the estimation project, Travis is off to do some more measuring for his uncle. Uncle Larry has told Travis that he needs to make some measurements on a wall in what will be the kitchen. Uncle Larry shows Travis which wall to mark on and hands him a ruler and a pencil. “I need you to make a small mark at $\frac{1}{8}''$ , another small mark at $\frac{2}{8}''$ past the first, and a large mark at $\frac{3}{8}''$ past the second mark,” says Uncle Larry. “Then continue that pattern across the wall. The most important marks are the large ones, please be sure that those marks are in the correct place. The large marks will indicate where I need to put brackets later.” “Okay,” says Travis, smiling. He is confident that he knows what he is doing. Uncle Larry goes off to work on another project and leaves Travis to his work. “Hmmm,” thinks Travis to himself. “If I write in all of the large marks first, I will be done a lot quicker. Then I can go back and do the small ones. I can add these fractions to figure out at what measurement I need to draw in the large marks.” Travis has a plan, but will his plan work? If Travis adds up the fractions, at what measurement will the large marks be drawn? This Concept will teach you all that you need to know to answer each of these questions. ### Guidance You have already learned how to add whole numbers and how to add decimals, now you are going to learn how to add fractions. In this Concept, you will learn all about adding fractions with like or common denominators . What is a like denominator? A like denominator is a denominator that is the same. This means that the whole has been divided up into the same number of parts. If the denominator of two fractions is a five, then both of those fractions have been divided into five parts. The numerators may be different, but the denominators are the same. This picture shows two different fractions with like denominators. Now let’s say that we want to add these two fractions. Because the denominators are common, we are adding like parts. We can simply add the numerators and we will have our new fraction. $\frac{2}{6} + \frac{4}{6} = \frac{6}{6}$ Here it is as a picture. We combined both of these fractions together to have a fraction we can call six-sixths. We must simplify or reduce all of our answers. In this example, when we have six out of six parts, we have one whole. You can see that one whole figure is shaded in. We simplify our answer and then our work is complete. Our final answer is $\frac{6}{6} = 1$ . Let’s look at another one. We can work on this one without looking at a picture. $\frac{2}{8} + \frac{4}{8} = \underline{\;\;\;\;\;\;\;\;\;}$ The first step is to make sure that you have like denominators. In this example, both denominators are 8, so we can add the numerators because the denominators are alike. Our next step is to add the numerators. $2 + 4 = 6$ We put that number over the common denominator. $\frac{6}{8}$ Our last step is to see if we can simplify our answer. In this example, 6 and 8 have the greatest common factor of 2. We divide both the numerator and the denominator by 2 to simplify the fraction. $\frac{6 \div 2}{8 \div 2} = \frac{3}{4}$ Our final answer is $\frac{3}{4}$ . Now it is time for you to try a few of these on your own. Be sure that your answer is in simplest form. #### Example A $\frac{1}{7} + \frac{2}{7} = \underline{\;\;\;\;\;\;\;\;\;}$ Solution: $\frac{3}{7}$ #### Example B $\frac{3}{9} + \frac{1}{9} = \underline{\;\;\;\;\;\;\;\;\;}$ Solution: $\frac{4}{9}$ #### Example C $\frac{2}{10} + \frac{3}{10} = \underline{\;\;\;\;\;\;\;\;\;}$ Solution: $\frac{5}{10} = \frac{1}{2}$ Now let's go back and help Travis with his dilemma. For Travis to follow his plan, he needs to add up the fractions to figure out what fraction of an inch should be between the large marks for the brackets. $\frac{1}{8} + \frac{2}{8} + \frac{3}{8} = \underline{\;\;\;\;\;\;\;\;\;}$ These fractions all have common denominators, so Travis can simply add the numerators. $1 + 2 + 3 = 6$ Next, we can put this answer over the common denominator. $\frac{6}{8}''$ Travis needs to make a large mark every six-eighths of an inch. It will be a lot simpler to measure the marks if Travis simplifies this fraction. $\frac{6}{8} = \frac{3}{4}$ Travis needs to make a large mark every $\frac{3}{4}''$ of an inch. Confident in his calculations, he gets right to work. ### Vocabulary Like Denominators when the denominators of fractions being added or subtracted are the same. Simplifying dividing the numerator and the denominator of a fraction by its greatest common factor. The result is a fraction is simplest form. ### Guided Practice Here is one for you to try on your own. $\frac{3}{10} + \frac{2}{10} + \frac{2}{10} = \underline{\;\;\;\;\;\;\;\;\;}$ Our answer is $\frac{7}{10}$ . ### Practice 1. $\frac{1}{3} + \frac{1}{3} = \underline{\;\;\;\;\;\;\;\;\;}$ 2. $\frac{2}{5} + \frac{2}{5} = \underline{\;\;\;\;\;\;\;\;\;}$ 3. $\frac{4}{7} + \frac{2}{7} = \underline{\;\;\;\;\;\;\;\;\;}$ 4. $\frac{5}{11} + \frac{4}{11} = \underline{\;\;\;\;\;\;\;\;\;}$ 5. $\frac{6}{10} + \frac{1}{10} = \underline{\;\;\;\;\;\;\;\;\;}$ 6. $\frac{4}{10} + \frac{1}{10} = \underline{\;\;\;\;\;\;\;\;\;}$ 7. $\frac{3}{4} + \frac{1}{4} = \underline{\;\;\;\;\;\;\;\;\;}$ 8. $\frac{5}{6} + \frac{3}{6} = \underline{\;\;\;\;\;\;\;\;\;}$ 9. $\frac{4}{9} + \frac{2}{9} = \underline{\;\;\;\;\;\;\;\;\;}$ 10. $\frac{5}{10} + \frac{1}{10} = \underline{\;\;\;\;\;\;\;\;\;}$ 11. $\frac{6}{13} + \frac{4}{13} = \underline{\;\;\;\;\;\;\;\;\;}$ 12. $\frac{9}{10} + \frac{1}{10} = \underline{\;\;\;\;\;\;\;\;\;}$ 13. $\frac{6}{9} + \frac{1}{9} = \underline{\;\;\;\;\;\;\;\;\;}$ 14. $\frac{8}{12} + \frac{1}{12} = \underline{\;\;\;\;\;\;\;\;\;}$ 15. $\frac{10}{20} + \frac{4}{20} = \underline{\;\;\;\;\;\;\;\;\;}$ 16. $\frac{11}{18} + \frac{5}{18} = \underline{\;\;\;\;\;\;\;\;\;}$ ### Vocabulary Language: English Like Denominators Like Denominators Two or more fractions have like denominators when their denominators are the same. "Common denominators" is a synonym for "like denominators". Simplify Simplify To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions.
# ESTIMATING MULTIPLICATION WITH DECIMALS When we want to do the multiplications of "Three digit number x One digit number" "Three digit number" x Two digit number" "Decimal number x Decimal number" we have to calculate step by step. Sometimes we may have to estimate the above products without step by step calculation. Even though estimation is just approximation of the exact answer, we can get quickly. ## Rule to be Followed in Estimating the Products When we estimate multiplication with decimal, we have to follow the rule given below as much as possible. If one number is increased to round off, then decrease the other number (if it is needed to round off). Sometimes we may have to either increase both the numbers or decrease both the numbers to round off. Examples : In 48 x 29, we have 48 ----> 50 and 29 ---->30 In 52 x 31, we have 52 ----> 50 and 31 ---->30 ## Practice Problems Problem 1 : Estimate the product of : 28.3 x 3.2 Solution : In the above product, the numbers can be rounded as given below. 28.3 -----> 30 3.2 -----> 3 Now, the above product will become 30 x 3 Then, the estimated product is = 3 x 3 = 9 -----> 90 Problem 2 : Estimate the product of : 34.3 x 3.5 Solution : In the above product, the numbers can be rounded as given below. 34.3 -----> 30 3.5 -----> 4 Now, the above product will become 30 x 4 Then, the estimated product is = 3 x 4 = 12 -----> 120 Problem 3 : Estimate the product of : 74.8 x 5.7 Solution : In the above product, the numbers can be rounded as given below. 74.8 -----> 70 5.7 -----> 6 Now, the above product will become 70 x 6 Then, the estimated product is = 7 x 6 = 42 -----> 420 Problem 4 : Estimate the product of : 328 x 42 Solution : This is bit different case. In the above product, the numbers can be rounded as given below. 328 -----> 300 42 -----> 40 But both the numbers are decreased. (Because 328 is close to 300 and 42 is close to 40) Now, the estimated product is = 3 x 4 = 12 -----> 12000 But the exact product is 13776. Here, there is bit higher difference between the exact product and estimation. So, we can round the given numbers as given below. 328 -----> 350 42 -----> 40 Now, the product will become 350 x 40 Then, the estimate of the product is = 35 x 4 = 140 -----> 14000 Problem 5 : Estimate the product of : 428.5 x 57.3 Solution : In the above product, the numbers can be rounded as given below. 428.5 -----> 400 57.3 -----> 60 Now, the above product will become 400 x 60 Then, the estimated product = 4 x 6 = 24 -----> 24000 Problem 6 : Estimate the product of : 5283 x 78 Solution : In the above product, the numbers can be rounded as given below. 5283 -----> 5000 78 -----> 80 Now, the above product will become 5000 x 80 Then the estimated product is = 5 x 8 = 40 -----> 400000 Problem 7 : Estimate the product of : 6523.56 x 73.569 Solution : In the above product, the numbers can be rounded as given below. 6523.56 -----> 7000 73.569 -----> 70 Now, the above product will become 7000 x 70 Then, the estimated product is = 7 x 7 = 49 -----> 490000 Apart from the stuff given above, if you need any other stuff, please use our google custom search here. You can also visit the following web pages on different stuff in math. WORD PROBLEMS Word problems on simple equations Word problems on linear equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6
##### Objectives 1. Learn to view a matrix geometrically as a function. 2. Learn examples of matrix transformations: reflection, dilation, rotation, shear, projection. 3. Understand the vocabulary surrounding transformations: domain, codomain, range. 4. Understand the domain, codomain, and range of a matrix transformation. 5. Pictures: common matrix transformations. 6. Vocabulary words: transformation / function, domain, codomain, range, identity transformation, matrix transformation. In this section we learn to understand matrices geometrically as functions, or transformations. We briefly discuss transformations in general, then specialize to matrix transformations, which are transformations that come from matrices. Informally, a function is a rule that accepts inputs and produces outputs. For instance, is a function that accepts one number as its input, and outputs the square of that number: In this subsection, we interpret matrices as functions. Let be a matrix with rows and columns. Consider the matrix equation (we write it this way instead of to remind the reader of the notation ). If we vary then will also vary; in this way, we think of as a function with independent variable and dependent variable • The independent variable (the input) is which is a vector in • The dependent variable (the output) is which is a vector in The set of all possible output vectors are the vectors such that has some solution; this is the same as the column space of by this note in Section 2.3. At this point it is convenient to fix our ideas and terminology regarding functions, which we will call transformations in this book. This allows us to systematize our discussion of matrices as functions. ##### Definition A transformation from to is a rule that assigns to each vector in a vector in • is called the domain of • is called the codomain of • For in the vector in is the image of under • The set of all images is the range of The notation means is a transformation from to It may help to think of as a “machine” that takes as an input, and gives you as the output. The points of the domain are the inputs of this simply means that it makes sense to evaluate on vectors with entries, i.e., lists of numbers. Likewise, the points of the codomain are the outputs of this means that the result of evaluating is always a vector with entries. The range of is the set of all vectors in the codomain that actually arise as outputs of the function for some input. In other words, the range is all vectors in the codomain such that has a solution in the domain. ##### Definition The identity transformation is the transformation defined by the rule In other words, the identity transformation does not move its input vector: the output is the same as the input. Its domain and codomain are both and its range is as well, since every vector in is the output of itself. # Subsection3.1.3Matrix Transformations Now we specialize the general notions and vocabulary from the previous subsection to the functions defined by matrices that we considered in the first subsection. ##### Definition Let be an matrix. The matrix transformation associated to is the transformation This is the transformation that takes a vector in to the vector in If has columns, then it only makes sense to multiply by vectors with entries. This is why the domain of is If has rows, then has entries for any vector in this is why the codomain of is The definition of a matrix transformation tells us how to evaluate on any given vector: we multiply the input vector by a matrix. For instance, let and let be the associated matrix transformation. Then Suppose that has columns If we multiply by a general vector we get This is just a general linear combination of Therefore, the outputs of are exactly the linear combinations of the columns of the range of is the column space of See this note in Section 2.3. Let be an matrix, and let be the associated matrix transformation. • The domain of is where is the number of columns of • The codomain of is where is the number of rows of • The range of is the column space of In the case of an square matrix, the domain and codomain of are both In this situation, one can regard as operating on it moves the vectors around in the same space.
# Question Video: Differentiating Functions Involving Reciprocal Trigonometric Ratios and Roots Using the Chain Rule Mathematics • Higher Education If π¦ = β(19 csc π₯ + 18), find dπ¦/dπ₯. 03:22 ### Video Transcript If π¦ is equal to the square root of 19 csc of π₯ plus 18, find dπ¦ by dπ₯. Weβre given π¦ as a function in π₯, and weβre asked to find dπ¦ by dπ₯. Thatβs the first derivative of π¦ with respect to π₯. So we need to differentiate our expression for π¦ with respect to π₯. Looking at this, we can see this is the square root of a trigonometric function. In other words, π¦ is the composition of functions. And to differentiate the composition of functions, we can do this by using the chain rule, and this would work. However, because our outer function is a power function, we can actually do this by using the general power rule. It doesnβt matter which method we would use; both would give us the same answer; itβs personal preference. In this case, weβre going to use the general power rule. To use the general power rule, weβre first going to need to use our laws of exponents to rewrite π¦ as 19 csc of π₯ plus 18 all raised to the power of one-half. Letβs now recall the general power rule. This tells us for any real constant π and differentiable function π of π₯, if π¦ is equal to π of π₯ all raised to the πth power, then dπ¦ by dπ₯ would be equal to π times π prime of π₯ multiplied by π of π₯ all raised to the power of π minus one. And we can see that π¦ is exactly written in this form, with our exponent π equal to one-half and π of π₯, our inner trigonometric function, 19 csc of π₯ plus 18. However, to use the general power rule, we need to first find an expression for π prime of π₯. And π prime of π₯ will be the derivative of 19 csc of π₯ plus 18 with respect to π₯. And thereβs a few different ways we could evaluate this derivative. For example, we could rewrite the csc of π₯ by using our Pythagorean identity to be one over the sin of π₯. We could then differentiate this by using the general power rule or the chain rule, or we could do this by using the quotient rule. And any of those methods would work. However, we can also recall the following result for differentiating reciprocal trigonometric functions. The derivative of the csc of π₯ with respect to π₯ is equal to negative the cot of π₯ multiplied by the csc of π₯. We can then use this to differentiate π of π₯ term by term. First, the derivative of 19 csc of π₯ with respect to π₯ will be negative 19 cot of π₯ times the csc of π₯. Next, we know the derivative of the constant 18 with respect to π₯ will just be equal to zero. Now that weβve found an expression for π prime and we know the value of π and an expression for π of π₯, we can use the general power rule to find dπ¦ by dπ₯. Substituting in our expressions for π of π₯, π prime of π₯, and π is equal to one-half into our general power rule, we get dπ¦ by dπ₯ is equal to one-half times negative 19 cot of π₯ times the csc of π₯ multiplied by 19 csc of π₯ plus 18 all raised to the power of one-half minus one. And we can simplify this. First, in our exponent, one-half minus one is equal to negative one-half. We could then simplify our coefficient. However, recall from our laws of exponents, π to the power of negative one-half is equal to one divided by the square root of π. So instead of multiplying by 19 csc of π₯ plus 18 all raised to the power of negative one-half, we can instead divide by the square root of 19 csc of π₯ plus 18. So by doing this and rearranging slightly, we get our final answer. Therefore, by using the general power rule, we were able to show if π¦ is equal to the square root of 19 csc of π₯ plus 18, then dπ¦ by dπ₯ will be equal to negative 19 cot of π₯ times the csc of π₯ all divided by two times the square root of 19 csc of π₯ plus 18.
# Using Phasors in AC Circuits concept Dealing with sines and cosines is mathematically a pain in the ass. Which is unfortunate because in AC it's mostly what we do. Luckily it's possible to use a new mathematical tool called "phasors" to remove the trigonometry from our problems and replace it with normal multiplication, division, addition and subtraction. Phasors are a way of representing sine waves as well as complex numbers and operating on them without needing to do any complex algebra or trigonometry. Before we go into how to use phasors we'll need to do a quick refresher on complex numbers. fact A complex number is a point on a 2D graph written $$x + iy$$ where $$x$$ is the horizontal coordinate and $$y$$ is the vertical coordinate. $$x + iy$$ is the same as point (x, y) in geometry. fact The $$i$$ in complex numbers happens to be: $$i = \sqrt{-1}$$ fact In a complex number the $$x$$ is called the "Real" part and the $$y$$ is called the "Imaginary" part (hence the use of the letter $$i$$ to denote the imaginary part). fact Two complex numbers can be added together by adding their real and imaginary parts separately. example If $$Z_1 = 1 + 2i$$ and $$Z_2 = -3 + i$$ find $$Z_1 + Z_2$$ The real part of $$Z_1 + Z_2$$ is going to be the real part of $$Z_1$$ added to the real part of $$Z_2$$. So the real part of $$Z_1 + Z_2$$ is $$-2$$. Likewise the imaginary part of $$Z_1 + Z_2$$ is going to be $$2 + 1 = 3$$ So $$Z_1 + Z_2 = -2 + 3i$$ It's useful when doing multiplication or division to write complex numbers in their polar form instead, this uses polar coordinates to mark the point rather than rectangular coordinates like we used above. fact Polar form is given by $$Z = re^{i\theta}$$ To convert $$Z = x + iy$$ into polar form we use the formulas: $$r = \sqrt{x^2 + y^2}$$ $$\theta = \tan^{-1}(\frac{y}{x})$$ example Convert $$1 + i$$ into polar form We just use the above formulas to get: $$r = \sqrt{2}$$ $$\theta = \tan^{-1}(1) = \frac{\pi}{4}$$ So $$i + i = \sqrt{2}e^{i\frac{\pi}{4}}$$ fact To convert a polar form to a rectangular form use the following formulas: For $$Z = Ae^{i\theta}$$: $$Z = A\cos(\theta) + iA\sin(\theta)$$ fact To multiply two complex numbers use their polar forms and the following formulas: $$Z_3 = Z_1Z_2$$ $$r_{Z3} = r_{Z1}\cdot r_{Z2}$$ $$\theta_{Z3} = \theta_{Z1}+\theta_{Z2}$$ fact To divide two complex numbers use their polar forms and the following formulas: $$Z_3 = \frac{Z_1}{Z_2}$$ $$r_{Z3} = \frac{r_{Z1}}{r_{Z2}}$$ $$\theta_{Z3} = \theta_{Z1}-\theta_{Z2}$$ Okay, that's enough of the complex numbers for us to get into phasors. A phasor is really just a complex number in polar form. We can use it to represent a thing called "Impedance" which replaces resistance in AC circuits. We can also use phasors to represent sine waves so we can convert trigonometry problems into simpler algebra problems fact To write the polar complex number $$re^{i\theta}$$ as a phasor we use the slightly different form: $$r \angle \theta$$ fact If all of the AC sources in a circuit have the same frequency then the voltage and current everywhere in the circuit has that one frequency. fact If all of the AC frequencies in a circuit are the same we can write all the voltages and currents as phasors. fact To write a cosine wave $$A\cos(\omega t + \theta)$$ as a phasor it's simply: $$A \angle \theta$$ fact To write a sine wave $$A\sin(\omega t + \theta)$$ as a phasor we write: $$A \angle (\theta - 90^\circ)$$ example Find $$12\angle 30^\circ \cdot 2\angle 11^\circ$$ $$12 \cdot 2 = 24$$ $$30 + 11 = 41$$ So $$12\angle 30^\circ \cdot 2\angle 11^\circ = 24\angle 41^\circ$$ example Find $$\frac{12\angle 30^\circ}{2\angle 11^\circ}$$ $$\frac{12}{2} = 6$$ $$30 - 11 = 19$$ so $$\frac{12\angle 30^\circ}{2\angle 11^\circ} = 6\angle 19^\circ$$ fact To add two phasors we must first change them into rectangular form, add them like we normally add two complex numbers, then turn the result back into phasor form. example Find $$5\angle 30^\circ + 2\angle 45^\circ$$ $$5\angle 30^\circ = 5\cos(30^\circ) + 5i\sin(30^\circ) \approx 4.33 + 2.5i$$ $$2\angle 45^\circ = 2\cos(45^\circ) + 2i\sin(45^\circ) \approx 1.41 + 1.41i$$ So $$5\angle 30^\circ + 2\angle 45^\circ \approx 5.74 + 3.91i$$ Now we return it to phasor form: $$5.74 + 3.91i = 6.95\angle 34.26^\circ$$ practice problems
# 4.6: Data Transformations Skills to Develop • To learn how to use data transformation if a measurement variable does not fit a normal distribution or has greatly different standard deviations in different groups. ## Introduction Many biological variables do not meet the assumptions of parametric statistical tests: they are not normally distributed, the standard deviations are not homogeneous, or both. Using a parametric statistical test (such as an anova or linear regression) on such data may give a misleading result. In some cases, transforming the data will make it fit the assumptions better. To transform data, you perform a mathematical operation on each observation, then use these transformed numbers in your statistical test. For example, as shown in the first graph above, the abundance of the fish species Umbra pygmaea (Eastern mudminnow) in Maryland streams is non-normally distributed; there are a lot of streams with a small density of mudminnows, and a few streams with lots of them. Applying the log transformation makes the data more normal, as shown in the second graph. Here are $$12$$ numbers from the mudminnow data set; the first column is the untransformed data, the second column is the square root of the number in the first column, and the third column is the base-$$10$$ logarithm of the number in the first column. Untransformed Square-root transformed Log transformed 38 6.164 1.580 1 1.000 0.000 13 3.606 1.114 2 1.414 0.301 13 3.606 1.114 20 4.472 1.301 50 7.071 1.699 9 3.000 0.954 28 5.292 1.447 6 2.449 0.778 4 2.000 0.602 43 6.557 1.633 You do the statistics on the transformed numbers. For example, the mean of the untransformed data is $$18.9$$; the mean of the square-root transformed data is $$3.89$$; the mean of the log transformed data is $$1.044$$. If you were comparing the fish abundance in different watersheds, and you decided that log transformation was the best, you would do a one-way anova on the logs of fish abundance, and you would test the null hypothesis that the means of the log-transformed abundances were equal. ## Back transformation Even though you've done a statistical test on a transformed variable, such as the log of fish abundance, it is not a good idea to report your means, standard errors, etc. in transformed units. A graph that showed that the mean of the log of fish per $$75m$$ of stream was $$1.044$$ would not be very informative for someone who can't do fractional exponents in their head. Instead, you should back-transform your results. This involves doing the opposite of the mathematical function you used in the data transformation. For the log transformation, you would back-transform by raising 10 to the power of your number. For example, the log transformed data above has a mean of $$1.044$$ and a $$95\%$$ confidence interval of $$\pm 0.344$$ log-transformed fish. The back-transformed mean would be $$10^{1.044}=11.1$$ fish. The upper confidence limit would be $$10^{(1.044+0.344)}=24.4$$ fish, and the lower confidence limit would be $$10^{(1.044-0.344)}=5.0$$ fish. Note that the confidence interval is not symmetrical; the upper limit is $$13.3$$ fish above the mean, while the lower limit is $$6.1$$ fish below the mean. Also note that you can't just back-transform the confidence interval and add or subtract that from the back-transformed mean; you can't take $$10^{0.344}$$ and add or subtract that. ## Choosing the right transformation Data transformations are an important tool for the proper statistical analysis of biological data. To those with a limited knowledge of statistics, however, they may seem a bit fishy, a form of playing around with your data in order to get the answer you want. It is therefore essential that you be able to defend your use of data transformations. There are an infinite number of transformations you could use, but it is better to use a transformation that other researchers commonly use in your field, such as the square-root transformation for count data or the log transformation for size data. Even if an obscure transformation that not many people have heard of gives you slightly more normal or more homoscedastic data, it will probably be better to use a more common transformation so people don't get suspicious. Remember that your data don't have to be perfectly normal and homoscedastic; parametric tests aren't extremely sensitive to deviations from their assumptions. It is also important that you decide which transformation to use before you do the statistical test. Trying different transformations until you find one that gives you a significant result is cheating. If you have a large number of observations, compare the effects of different transformations on the normality and the homoscedasticity of the variable. If you have a small number of observations, you may not be able to see much effect of the transformations on the normality and homoscedasticity; in that case, you should use whatever transformation people in your field routinely use for your variable. For example, if you're studying pollen dispersal distance and other people routinely log-transform it, you should log-transform pollen distance too, even if you only have $$10$$ observations and therefore can't really look at normality with a histogram. ## Common transformations There are many transformations that are used occasionally in biology; here are three of the most common: ### Log transformation This consists of taking the log of each observation. You can use either base-$$10$$ logs (LOG in a spreadsheet, LOG10 in SAS) or base-$$e$$ logs, also known as natural logs (LN in a spreadsheet, LOG in SAS). It makes no difference for a statistical test whether you use base-$$10$$ logs or natural logs, because they differ by a constant factor; the base-$$10$$ log of a number is just $$2.303…\times \text{the\; natural\; log\; of\; the\; number}$$. You should specify which log you're using when you write up the results, as it will affect things like the slope and intercept in a regression. I prefer base-$$10$$ logs, because it's possible to look at them and see the magnitude of the original number: $$log(1)=0,\; log(10)=1,\; log(100)=2$$, etc. The back transformation is to raise $$10$$ or $$e$$ to the power of the number; if the mean of your base-$$10$$ log-transformed data is $$1.43$$, the back transformed mean is $$10^{1.43}=26.9$$ (in a spreadsheet, "=10^1.43"). If the mean of your base-e log-transformed data is $$3.65$$, the back transformed mean is $$e^{3.65}=38.5$$ (in a spreadsheet, "=EXP(3.65)". If you have zeros or negative numbers, you can't take the log; you should add a constant to each number to make them positive and non-zero. If you have count data, and some of the counts are zero, the convention is to add $$0.5$$ to each number. Many variables in biology have log-normal distributions, meaning that after log-transformation, the values are normally distributed. This is because if you take a bunch of independent factors and multiply them together, the resulting product is log-normal. For example, let's say you've planted a bunch of maple seeds, then $$10$$ years later you see how tall the trees are. The height of an individual tree would be affected by the nitrogen in the soil, the amount of water, amount of sunlight, amount of insect damage, etc. Having more nitrogen might make a tree $$10\%$$ larger than one with less nitrogen; the right amount of water might make it $$30\%$$ larger than one with too much or too little water; more sunlight might make it $$20\%$$ larger; less insect damage might make it $$15\%$$ larger, etc. Thus the final size of a tree would be a function of $$\text{nitrogen}\times \text{water}\times \text{sunlight}\times \text{insects}$$, and mathematically, this kind of function turns out to be log-normal. ### Square-root transformation This consists of taking the square root of each observation. The back transformation is to square the number. If you have negative numbers, you can't take the square root; you should add a constant to each number to make them all positive. People often use the square-root transformation when the variable is a count of something, such as bacterial colonies per petri dish, blood cells going through a capillary per minute, mutations per generation, etc. ### Arcsine transformation This consists of taking the arcsine of the square root of a number. (The result is given in radians, not degrees, and can range from $$-\pi /2\; to\; \pi /2$$.) The numbers to be arcsine transformed must be in the range $$0$$ to $$1$$. This is commonly used for proportions, which range from $$0$$ to $$1$$, such as the proportion of female Eastern mudminnows that are infested by a parasite. Note that this kind of proportion is really a nominal variable, so it is incorrect to treat it as a measurement variable, whether or not you arcsine transform it. For example, it would be incorrect to count the number of mudminnows that are or are not parasitized each of several streams in Maryland, treat the arcsine-transformed proportion of parasitized females in each stream as a measurement variable, then perform a linear regression on these data vs. stream depth. This is because the proportions from streams with a smaller sample size of fish will have a higher standard deviation than proportions from streams with larger samples of fish, information that is disregarded when treating the arcsine-transformed proportions as measurement variables. Instead, you should use a test designed for nominal variables; in this example, you should do logistic regression instead of linear regression. If you insist on using the arcsine transformation, despite what I've just told you, the back-transformation is to square the sine of the number. ## How to transform data In a blank column, enter the appropriate function for the transformation you've chosen. For example, if you want to transform numbers that start in cell $$A2$$, you'd go to cell $$B2$$ and enter =LOG(A2) or =LN(A2) to log transform, =SQRT(A2) to square-root transform, or =ASIN(SQRT(A2)) to arcsine transform. Then copy cell $$B2$$ and paste into all the cells in column $$B$$ that are next to cells in column $$A$$ that contain data. To copy and paste the transformed values into another spreadsheet, remember to use the "Paste Special..." command, then choose to paste "Values." Using the "Paste Special...Values" command makes Excel copy the numerical result of an equation, rather than the equation itself. (If your spreadsheet is Calc, choose "Paste Special" from the Edit menu, uncheck the boxes labeled "Paste All" and "Formulas," and check the box labeled "Numbers.") To back-transform data, just enter the inverse of the function you used to transform the data. To back-transform log transformed data in cell $$B2$$, enter =10^B2 for base-$$10$$ logs or =EXP(B2) for natural logs; for square-root transformed data, enter =B2^2; for arcsine transformed data, enter =(SIN(B2))^2 ### Web pages I'm not aware of any web pages that will do data transformations. ### SAS To transform data in SAS, read in the original data, then create a new variable with the appropriate function. This example shows how to create two new variables, square-root transformed and log transformed, of the mudminnow data. DATA mudminnow; INPUT location $banktype$ count; countlog=log10(count); countsqrt=sqrt(count); DATALINES; Gwynn_1 forest 38 Gwynn_2 urban 1 Gwynn_3 urban 13 Jones_1 urban 2 Jones_2 forest 13 LGunpowder_1 forest 20 LGunpowder_2 field 50 LGunpowder_3 forest 9 BGunpowder_1 forest 28 BGunpowder_2 forest 6 BGunpowder_3 forest 4 BGunpowder_4 field 43 ; The dataset "mudminnow" contains all the original variables ("location", "banktype" and "count") plus the new variables ("countlog" and "countsqrt"). You then run whatever PROC you want and analyze these variables just like you would any others. Of course, this example does two different transformations only as an illustration; in reality, you should decide on one transformation before you analyze your data. The SAS function for arcsine-transforming X is ARSIN(SQRT(X)). You'll probably find it easiest to backtransform using a spreadsheet or calculator, but if you really want to do everything in SAS, the function for taking $$10$$ to the $$X$$ power is 10X; the function for taking $$e$$ to a power is EXP(X); the function for squaring $$X$$ is X2; and the function for backtransforming an arcsine transformed number is SIN(X)**2. ## Reference Picture of a mudminnow from The Virtual Aquarium of Virginia. ## Contributor • John H. McDonald (University of Delaware)
5 Q: # Two boys and a girl can do a work in 5 days, while a boy and 2 girls can do it in 6 days. If the boy is paid at the rate of 28$a week, what should be the wages of the girl a week ? A) 24$ B) 22 $C) 16$ D) 14 $Answer: C) 16$ Explanation: Let the 1 day work of a boy=b and a girl=g, then 2b + g = 1/5 ---(i) and b + 2g = 1/6 ---(ii) On solving (i) & (ii), b=7/90, g=2/45 As payment of work will be in proportion to capacity of work and a boy is paid $28/week, so a girl will be paid $28×245790$ = 16$. Q: 5 boys and 3 girls can together cultivate a 23 acre field in 4 days and 3 boys and 2 girls together can cultivate a 7 acre field in 2 days. How many girls will be needed together with 7 boys, if they cultivate 45 acres of field in 6 days? A) 4 B) 3 C) 2 D) 1 Explanation: Let workdone 1 boy in 1 day be b and that of 1 girl be g From the given data, 4(5b + 3g) = 23 20b + 12g = 23 .......(a) 2(3b + 2g) = 7 6b + 4g = 7 ........(b) Solving (a) & (b), we get b = 1, g = 1/4 Let number og girls required be 'p' 6(7 x 1 + p x 1/4) = 45 => p = 2. Hence, number of girls required = 2 1 59 Q: 70000 a year is how much an hour? A) 80 B) 8 C) 0.8 D) 0.08 Explanation: Given for year = 70000 => 365 days = 70000 => 365 x 24 hours = 70000 => 1 hour = ? 70000/365x24 = 7.990 = 8 0 160 Q: A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained? A) 1/5 B) 1/6 C) 1/7 D) 1/8 Explanation: Total work is given by L.C.M of 72, 48, 36 Total work = 144 units Efficieny of A = 144/72 = 2 units/day Efficieny of B = 144/48 = 3 units/day Efficieny of C = 144/36 = 4 units/day According to the given data, 2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100 3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54 p = 198/16.5 p = 12 days. Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day (C+D) in p days = (4 + 6) x 12 = 120 unit Remained part of work = (144-120)/144 = 1/6. 5 605 Q: 10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work? A) 215 days B) 225 days C) 235 days D) 240 days Explanation: Given that (10M + 15W) x 6 days = 1M x 100 days => 60M + 90W = 100M => 40M = 90W => 4M = 9W. From the given data, 1M can do the work in 100 days => 4M can do the same work in 100/4= 25 days. => 9W can do the same work in 25 days. => 1W can do the same work in 25 x 9 = 225 days. Hence, 1 woman can do the same work in 225 days. 8 1036 Q: A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work? Given A,B,C can complete a work in 15,20 and 30 respectively. The total work is given by the LCM of 15, 20, 30 i.e, 60. A's 1 day work = 60/15 = 4 units B's 1 day work = 60/20 = 3 units C's 1 day work = 60/30 = 2 units (A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units Let B + C worked for x days = (3 + 2) x = 5x units C worked for 2 days = 2 x 2 = 4 units Then, 18 + 5x + 4 = 60 22 + 5x = 60 5x = 38 x = 7.6 Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days. 553 Q: M, N and O can complete the work in 18, 36 and 54 days respectively. M started the work and worked for 8 days, then N and O joined him and they all worked together for some days. M left the job one day before completion of work. For how many days they all worked together? A) 4 B) 5 C) 3 D) 6 Explanation: Let M, N and O worked together for x days. From the given data, M alone worked for 8 days M,N,O worked for x days N, O worked for 1 day But given that M alone can complete the work in 18 days N alone can complete the work in 36 days O alone can complete the work in 54 days The total work can be the LCM of 18, 6, 54 = 108 units M's 1 day work = 108/18 = 6 units N's 1 day work = 108/36 = 3 units O's 1 day work = 108/54 = 2 units Now, the equation is 8 x 6 + 11x + 5 x 1 = 108 48 + 11x + 5 = 108 11x = 103 - 48 11x = 55 x = 5 days. Hence, all M,N and O together worked for 5 days. 2 544 Q: P, Q, and R can do a job in 12 days together. If their efficiency of working be in the ratio 3 : 8 : 5, Find in what time Q can complete the same work alone? A) 36 days B) 30 days C) 24 days D) 22 days Explanation: Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5 Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively They can do work for 12 days. => Total work = 12 x 16x = 192x Now, the required time taken by Q to complete the job alone = days. 5 746 Q: 5 men and 3 boys can together cultivate a 23 acre field in 4 days and 3 men and 2 boys together can cultivate a 7 acre field in 2 days. How many boys will be needed together with 7 men, if they cultivate 45 acre of field in 6 days. A) 6 B) 4 C) 2 D) 3 Explanation: Let work done by 1 man in i day be m and Let work done by 1 boy in 1 day be b From the given data, 4(5m + 3b) = 23 20m + 12b = 23....(1) 2(3m + 2b) = 7 6m + 4b = 7 ....(2) By solving (1) & (2), we get m = 1, b = 1/4 Let the number of required boys = n 6(7 1 + n x 1/4) = 45 => n = 2.
Parametric Equations This section covers: Note that there is information on the parametric form of the equation of a line in space here in the Vectors section. Introduction to Parametric Equations So far, we’ve dealt with Rectangular Equations, which are equations that can be graphed on a regular coordinate system, or Cartesian Plane. Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. But sometimes we need to know what both $$x$$ and $$y$$ are, for example, at a certain time, so we need to introduce another variable, say $$\boldsymbol{t}$$ (the parameter). Parametric equations are also referred to as plane curves. Note that $$t$$ can be negative; this means before motion has started (I won’t get into the Physics part of parameters here!). Here is a simple set of parametric equations that represent a cubic $$y={{x}^{3}}$$ for $$t$$ in $$[0, 3]$$: $$x\left( t \right)=t,\,\,y\left( t \right)={{t}^{3}},\,\,0\le t\le 3$$. Many write this simply as $$x=t,\,\,y={{t}^{3}},\,\,0\le t\le 3$$. Do you see how when we introduce the parametric variable $$t$$, we can see how the curve is being drawn for certain values of $$t$$? For example, for $$t=0$$, we are at the point $$(0,0)$$, for $$t=1$$, we are at the point $$(1,1)$$, for $$t=2$$, we are at the point $$(2,8)$$, and so on. We can even put arrows on a graph to show the direction, or orientation of the set of parametric equations. Here is a t-chart and graph for this parametric equation, as well as some others. Note that the domain is the lowest $$x$$ value to the highest $$x$$ value, regardless what the value for $$t$$ is. The range is the lowest $$y$$ value to the highest $$y$$ value, again regardless what the value for $$t$$ is. The end points are the points with the lowest $$t$$ value and the highest $$t$$ value. Note how the domains and ranges aren’t necessarily the same as the order of the points in the parametric t-charts, but they are always from low to high: Parametric T-Chart Parametric Graph $$\begin{array}{l}x=t\\y={{t}^{3}}\end{array}$$ for $$t$$ in $$\left[ {0,3} \right]$$ t x y 0 0 0 1 1 1 2 2 8 3 3 27 Domain: $$\left[ {0,3} \right]$$ Range: $$\left[ {0,27} \right]$$ End Points: $$\left( {0,0} \right)$$ and $$\left( {3,27} \right)$$ $$\begin{array}{l}x=-t+2\\y={{t}^{2}}-4\end{array}$$ for $$t$$ in $$\left[ {-1,4} \right]$$ t x y –1 3 –3 0 2 –4 1 1 –3 2 0 0 3 –1 5 4 –2 12 Domain: $$\left[ {-2,3} \right]$$ Range: $$\left[ {-4,12} \right]$$ End Points: $$\left( {3,-3} \right)$$ and $$\left( {-2,12} \right)$$ $$\begin{array}{l}x=4-2t\\y=\sqrt{{{{t}^{2}}+1}}\end{array}$$ for $$t$$ in $$\left[ {-1,2} \right]$$ t x y –1 6 $$\sqrt{2}$$ 0 4 1 1 2 $$\sqrt{2}$$ 2 0 $$\sqrt{5}$$ Domain: $$\left[ {0,6} \right]$$ Range: $$\left[ {1,\sqrt{5}} \right]$$ End Points: $$\left( {6,\sqrt{2}} \right)$$ and $$\left( {0,\sqrt{5}} \right)$$ Parametric Equations in the Graphing Calculator We can graph the set of parametric equations above by using a graphing calculator: < First change the MODE from FUNCTION to PARAMETRIC, and enter the equations for X and Y in “Y =”. For the WINDOW, you can put in the min and max values for $$t$$, and also the min and max values for $$x$$ and $$y$$ if you want to. Tstep will determine how many points are graphed; the smaller the Tstep, the more points will be graphed (smoother curve); you can play around with this. Then hit GRAPH to see the graph: < Let’s put Trigonometry parametric equations in the calculator. Make sure the calculator is in radians. I like to use a Tstep of $$\displaystyle \frac{\pi }{{12}}$$, with $$t$$ from 0 to 2π, and you might want to use ZOOM Zsquare to make the screen square. Note that when the coefficients of $$cos(t)$$and $$sin(t)$$ are the same, we get a circle; we will show this below algebraically. Note: If you are putting two sets of parametrics equations in the calculator and want to see if the paths collide at the same time, put the calculator in MODE SIMUL for simultaneous. You can put the Tstep in Window low, like 1, and you can the parametrics in “real time”. Converting Parametric Equations to Rectangular: Eliminating the Parameter Sometimes we want to get a set of parametric equations back to its simplest form – without the parameter (usually if we don’t care about extra variable, which in many cases is time). There’s a trick to do this; you have to solve for $$t$$ one of the equations (typically the simplest one), and then plug what you get into the other equation, so you only are left with $$x$$’s and $$y$$’s. This new equation is called a rectangular equation. When dealing with parametric equations with trig functions (when trig functions are in both equations), you typically don’t want to solve for $$t$$, but solve for the trig functions with argument $$t$$. Then you can plug this expression in the other parametric equation and many times a Trigonometric Identity can be used to simplify. In these cases, we sometimes get equations for a circle, ellipse, or hyperbola (found in the Conics section). But if we don’t have the trig functions in both parametric equations, we’ll want to get the $$t$$ by itself by taking the inverse of the trig function. Here are some examples; let’s do problems without trig first. Do you see how our goal is to not have $$t$$ in our equation at all? Parametric Equations Rectangular Equations Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=4t-2\\y=2+4t\end{array} \right.$$ Solve for $$t$$ in one of the equations and then substitute this in for the $$t$$ in the other equation: \displaystyle \begin{align}x&=4t-2\\x+2&=4t\\t&=\frac{{x+2}}{4}\end{align} Plug this into the second equation: \displaystyle \begin{align}y&=2+4t\\y&=2+4\left( {\frac{{x+2}}{4}} \right)\,\\y&=x+4\,\,\,\text{(line)}\end{align} Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=t-3\\y={{t}^{2}}-6t+9\end{array} \right.$$ Solve for $$t$$ in the simplest equation and then substitute this in for the $$t$$ in the other equation: $$\begin{array}{l}x=t-3\\t=x+3\end{array}$$ Plug this into the equation: $$\begin{array}{l}y={{t}^{2}}-6t+9\\y={{\left( {x+3} \right)}^{2}}-6\left( {x+3} \right)+9\\y={{x}^{2}}+6x+9-6x-18+9\\y={{x}^{2}}\,\,\,\text{(parabola)}\end{array}$$ Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=5t\\y=3{{e}^{t}}\end{array} \right.$$ Solve for $$t$$ in the simplest equation and then substitute this in for the $$t$$ in the other equation: \displaystyle \begin{align}x&=5t\\t&=\frac{x}{5}\end{align} Plug this into the second equation: $$\begin{array}{l}y=3{{e}^{t}}\\y=3{{e}^{{\frac{x}{5}}}}\,\,\,\text{(exponential)}\end{array}$$ Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=\sqrt{{t-3}}\\y=2+t\end{array} \right.$$ Solve for $$t$$ in the simplest equation and then substitute this in for the $$t$$ in the other equation. Note that the second equation is simpler, since it doesn’t have the square root. $$\begin{array}{l}y=2+t\\t=y-2\end{array}$$ Plug this into the second equation: $$\begin{array}{l}x=\sqrt{{y-3}}\\x=\sqrt{{\left( {y-2} \right)-3}}\\x=\sqrt{{y-5}}\\{{x}^{2}}=y-5\\y={{x}^{2}}+5\,\,\,\text{(parabola)}\end{array}$$ Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=3t\\y=-2{{t}^{2}}-1\end{array} \right.$$ $$t\,\,\text{on}\,\,\left[ {-1,3} \right]$$ Solve for $$t$$ in the simplest equation and then substitute this in for the $$t$$ in the other equation. $$\displaystyle t=\frac{x}{3},\,\,\,\,y=-2{{\left( {\frac{x}{3}} \right)}^{2}}-1$$ $$\displaystyle y=-\frac{{2{{x}^{2}}}}{9}-1\,\,\,\,\,\,\text{(parabola)}$$ Note that we are given an interval for $$t$$, so we are expected to find the domain and range for the rectangular equations. To find the domain and range, make a t-chart: t x y –1 –3 –3 0 0 –1 1 3 –3 2 6 –9 3 9 –19 Domain is$$\left[ {-3,9} \right]$$ and Range is $$\left[ {-19,-1} \right]$$. Here are more problems where you have to eliminate the parameter with trig. Notice that when we have trig arguments in both equations, we can sometimes use a Pythagorean Trig Identity to eliminate the parameter (and we end up with a Conic): Parametric Equations Rectangular Equation Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=3\sin t+2\\y=3\cos t-1\end{array} \right.$$ $$\displaystyle t\,\,\text{on}\,\,\left[ {-\frac{\pi }{2},\frac{\pi }{2}\,} \right]$$ Solve for $$\sin t$$ in the first equation and $$\cos t$$ in the second (since it’s too complicated to solve for $$t$$). Then use the Pythagorean identity $${{\sin }^{2}}t+{{\cos }^{2}}t=1$$, and substitute: \begin{align}x&=3\sin t+2\\3\sin t&=x-2\\\sin t&=\frac{{x-2}}{3}\end{align} \begin{align}y&=3\cos t-1\\3\cos t&=y+1\\\cos t&=\frac{{y+1}}{3}\end{align} \displaystyle \begin{align}{{\sin }^{2}}t+{{\cos }^{2}}t&=1\\{{\left( {\frac{{x-2}}{3}} \right)}^{2}}+{{\left( {\frac{{y+1}}{3}} \right)}^{2}}&=1\\{{\left( {x-2} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}&=9\,\text{ }\left( {\text{circle}} \right)\end{align} This looks like a circle, but since the interval for $$t$$ is $$\displaystyle \left[ {-\frac{\pi }{2},\frac{\pi }{2}\,} \right]$$, we have a semi-circle. Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=4\sec t+1\\y=3\tan t\end{array} \right.$$ Solve for $$\sec t$$ in the first equation $$\tan t$$ and in the second. Then use the Pythagorean identity $${{\sec }^{2}}t-{{\tan }^{2}}t=1$$, and substitute: \begin{align}x&=4\sec t+1\\4\sec t&=x-1\\\sec t&=\frac{{x-1}}{4}\end{align} \begin{align}y&=3\tan t\\\tan t&=\frac{y}{3}\end{align} \begin{align}{{\sec }^{2}}t-{{\tan }^{2}}t&=1\\{{\left( {\frac{{x-1}}{4}} \right)}^{2}}-{{\left( {\frac{y}{3}} \right)}^{2}}&=1\\\frac{{{{{\left( {x-1} \right)}}^{2}}}}{{16}}-\frac{{{{y}^{2}}}}{9}&=1\,\,\,\,\,\text{(hyperbola)}\end{align} Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=5\cot t-4\\y=4\csc t\end{array} \right.$$ Solve for $$\cot t$$ in the first equation and $$\csc t$$ in the second. Then use the Pythagorean identity $${{\csc }^{2}}t-{{\cot }^{2}}t=1$$, and substitute: \begin{align}x&=5\cot t-4\\5\cot t&=x+4\\\cot t&=\frac{{x+4}}{5}\end{align} \begin{align}y&=4\csc t\\\csc t&=\frac{y}{4}\end{align} \begin{align}{{\csc }^{2}}t-{{\cot }^{2}}t&=1\\{{\left( {\frac{y}{4}} \right)}^{2}}-{{\left( {\frac{{x+4}}{5}} \right)}^{2}}&=1\\\frac{{{{y}^{2}}}}{{16}}-\frac{{{{{\left( {x+4} \right)}}^{2}}}}{{25}}&=1\,\,\,\,\text{(hyperbola)}\end{align} Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=\sin t-4\\y=2\cos t+2\end{array} \right.$$ Solve for $$\sin t$$ in the first equation and $$\cos t$$ in the second (since it’s too complicated to solve for $$t$$). Then use the Pythagorean identity $${{\sin }^{2}}t+{{\cos }^{2}}t=1$$, and substitute: \begin{align}x&=\sin t-4\\\sin t&=x+4\end{align} \begin{align}y&=2\cos t+2\\\cos t&=\frac{{y-2}}{2}\end{align} \displaystyle \begin{align}{{\sin }^{2}}t+{{\cos }^{2}}t&=1\\{{\left( {x+4} \right)}^{2}}+{{\left( {\frac{{y-2}}{2}} \right)}^{2}}&=1\\\frac{{{{{\left( {x+4} \right)}}^{2}}}}{1}+\frac{{{{{\left( {y-2} \right)}}^{2}}}}{4}&=1\,\text{ }\,\,\left( {\text{ellipse}} \right)\end{align} Eliminate the parameter and describe the resulting equation: $$\left\{ \begin{array}{l}x=3t\\y=\sin t\end{array} \right.$$ Solve for $$t$$ in the simplest equation and then substitute in the $$t$$ in the other equation. Since we don’t have a trig function in both equations, we can’t use the Pythagorean identity like we did above. \begin{align}x&=3t\\t&=\frac{x}{3}\end{align} Plug this into the second equation: \begin{align}y&=\sin t\\y&=\sin \left( {\frac{x}{3}} \right)\,\,\,\,\left( {\text{sin}\,\text{graph}} \right)\end{align} Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. This seems to be a bit tricky, since technically there are an infinite number of these parametric equations for a single rectangular equation. And remember, you can convert what you get back to rectangular to make sure you did it right! If no other requirements are given (such as what $$t$$ value ranges should be), the easiest way is to always let $$x\left( t \right)=t$$ and then $$y\left( t \right)$$ is just the function with a $$t$$ instead of an $$x$$. For example, one way to write equation $$y=4x-5$$ into a set of parametric equations is $$\left\{ \begin{array}{l}x\left( t \right)=t\\y\left( t \right)=4t-5\end{array} \right.$$. As another example, to convert $$f\left( x \right)=8{{x}^{2}}+4x-2$$ into a set of parametric equations, we have $$\left\{ \begin{array}{l}x=t\\y=8{{t}^{2}}+4t-2\end{array} \right.$$. (I purposely left out the $$t$$’s on the left side of the equations; you see parametrics written both way). Work these the other way (from parametric to rectangular) to see how they work! And remember that this is just one way to write the set of parametric equations; there are many! You may also be asked come up with parametric equations from a rectangular equation, given an interval or range for $$t$$, or write one from a set of points or a point and a slope (these last three would be linear). < Here are some examples: Problem Solution Write a set of parametric equations for the line segment starting at $$\left( {-1,2} \right)$$ and ending at $$\left( {3,8} \right)$$. Graph: (The “regular” rectangular equation for this line is $$y=\frac{3}{2}x+\frac{7}{2}$$ .) Since we are looking for a line segment, let’s find the set of parametric equations $$\left\{ \begin{array}{l}x\left( t \right)=at+b\\y\left( t \right)=ct+d\end{array} \right.$$, where we need to find $$a,\,b,\,c,$$ and $$d$$. Since we are starting at $$\left( {-1,2} \right)$$ and ending at $$\left( {3,8} \right)$$, one way to create parametric equations is with $$t=0$$ at the first point, and $$t=1$$ at the second. Then we have: $$\left\{ \begin{array}{l}-1=a\left( 0 \right)+b\\\,\,\,\,2=c\left( 0 \right)+d\end{array} \right.$$, or $$b=-1$$ and $$d=2$$. Now, the second point: $$\left\{ \begin{array}{l}3=a\left( 1 \right)-1\\8=c\left( 1 \right)+2\end{array} \right.$$, or $$a=4$$ and $$c=6$$. The parametric equations are $$\displaystyle \left\{ \begin{array}{l}x\left( t \right)=4t-1\\y\left( t \right)=6t+2\end{array} \right.\,\,\,\,\,\,0\le t\le 1$$. Easier way using vectors: Start with the point $$\left( {-1,2} \right)$$, and “add” the vector $$\left\langle {3–1,8-2} \right\rangle$$ to get the next point, so we have $$\left\langle {-1,2} \right\rangle +\left\langle {4,6} \right\rangle t=\left\langle {-1+4t,\,2+6t} \right\rangle =\left\langle {4t-1,\,6t+2} \right\rangle$$. Write a set of parametric equations for the line segment starting at $$\left( {2,-4} \right)$$ with slope $$-3$$. (The “regular” rectangular equation for this line is $$y=-3x+2$$ .) Since we are looking for a line segment, let’s find the set of parametric equations $$\left\{ \begin{array}{l}x\left( t \right)=at+b\\y\left( t \right)=ct+d\end{array} \right.$$, where we need to find $$a,\,b,\,c,$$ and $$d$$. Since we are starting at $$\left( {2,-4} \right)$$ and with slope $$-3$$, one line could have parametric equations with $$t=0$$ at the first point, and $$t=1$$ at the point where we go down $$3$$ and over $$1$$ (thus a slope of $$-3$$; this point is $$\left( {3,\,-7} \right)$$). Then we have: $$\left\{ \begin{array}{l}\,\,\,\,2=a\left( 0 \right)+b\\-4=c\left( 0 \right)+d\end{array} \right.$$, or $$b=2$$ and $$d=-4$$. Now, the second point: $$\left\{ \begin{array}{l}\,\,\,\,3=a\left( 1 \right)+2\\-7=c\left( 1 \right)-4\end{array} \right.$$, or $$a=1$$ and $$c=-3$$. The parametric equations are .$$\displaystyle \left\{ \begin{array}{l}x\left( t \right)=t+2\\y\left( t \right)=-3t-4\end{array} \right.\,\,\,\,\,\,0\le t\le 1$$. Easier way using vectors: Start with the point $$\left( {2,\,-4} \right)$$, and “add” the vector $$\left\langle {1,\,-3} \right\rangle$$ (slope –3) to get to the next point, so we have $$\left\langle {2,\,-4} \right\rangle +\left\langle {1,\,-3} \right\rangle t=\left\langle {2+t,\,\,-4-3t} \right\rangle =\left\langle {t+2,\,\,-3t-4} \right\rangle$$. VECTOR METHOD EXAMPLE: Write a set of parametric equations for the line segment between points $$\left( {2,6} \right)$$ and $$\left( {4,–6} \right)$$, so that when $$t=0$$, we are at $$\left( {2,6} \right)$$, and at $$t=2$$, we are at $$\left( {4,–6} \right)$$. Let’s do one more, using the vector method. Start with the point $$\left( {2,\,6} \right)$$, and “add” the vector $$\left\langle {4-2,-6-6\,} \right\rangle$$ to get the next point. Since that point is at $$t=2$$, we’ll multiply the vector by $$\displaystyle \frac{t}{2}$$ (not sure exactly how this works!), so we have $$\displaystyle \left\langle {2,\,6} \right\rangle +\left\langle {2,\,-12} \right\rangle \frac{t}{2}=\left\langle {2+t,\,\,6-6t} \right\rangle =\left\langle {t+2,-6t+6\,} \right\rangle$$. We have $$\displaystyle \left\{ \begin{array}{l}x\left( t \right)=t+2\\y\left( t \right)=-6t+6\end{array} \right.\,\,\,\,\,\,0\le t\le 2$$. Try it; it works! Simultaneous Solutions Sometimes we need to find the $$x$$ and $$y$$ coordinates of any intersections of parametric sets of equations. To do this, we want to set the $$x$$’s together and solve for $$t$$, and then set the $$y$$’s together and also solve for $$t$$. Where we have the same $$t$$ when setting both the $$x$$ and $$y$$ equations together, we have an intersection. Then we have to put the $$\boldsymbol{t}$$ back in either $$x$$ equation and either $$y$$ equation to get the intersection. Here are some examples; find the $$x$$ and $$y$$ coordinates of any intersections: Simultaneous Solution Problem Solution Find the $$x$$ and $$y$$ coordinates of any intersections: $$\left\{ \begin{array}{l}x={{t}^{2}}+1\\y=-5t+6\end{array} \right.$$ $$\left\{ \begin{array}{l}x=2t\\y={{t}^{2}}\end{array} \right.$$ Solve for $$t$$ by setting the “$$x$$” equations together, and do the same for the “$$y$$” equations. If any $$t$$ values are the same in both, we have a solution; we then solve for $$x$$ and $$y$$ in either equation for that $$t$$. $$\begin{array}{c}{{t}^{2}}+1=2t\\{{t}^{2}}-2t+1=0\\\left( {t-1} \right)\left( {t-1} \right)=0\\t=1\end{array}$$ $$\begin{array}{c}-5t+6={{t}^{2}}\\{{t}^{2}}+5t-6=0\\\left( {t-1} \right)\left( {t+6} \right)=0\\t=1,\,-6\end{array}$$ Since $$t=1$$ works for both sets of equations, we have a solution! Plug in $$t$$ for $$x$$ and $$y$$ in either set of equations to get $$(2,1)$$. Find the $$x$$ and $$y$$ coordinates of any intersections: $$\displaystyle \left\{ \begin{array}{l}x=\sin \left( t \right)\\y=\frac{{\sin \left( {2t} \right)}}{2}\end{array} \right.$$ $$\displaystyle \left\{ \begin{array}{l}x=\frac{{\sqrt{3}}}{2}\\y=-\sqrt{3}{{\cos }^{2}}\left( t \right)\end{array} \right.$$ Solve for $$t$$ by setting the “$$x$$” equations together, and do the same for the “$$y$$” equations. If any $$t$$ values are the same in both, we have a solution; we then solve for $$x$$ and $$y$$ in either equation for that $$t$$. Note that we used the identity $$\sin \left( {2t} \right)=2\sin \left( t \right)\cos \left( t \right)$$ in the second set: $$\require {cancel} \displaystyle \begin{array}{c}\sin \left( t \right)=\frac{{\sqrt{3}}}{2}\\t=\frac{\pi }{3},\,\,\frac{{2\pi }}{3}\end{array}$$ $$\begin{array}{c}\frac{{\sin \left( {2t} \right)}}{2}=-\sqrt{3}{{\cos }^{2}}\left( t \right)\\\frac{{\cancel{2}\sin \left( t \right)\cos \left( t \right)}}{{\cancel{2}}}=-\sqrt{3}{{\cos }^{2}}\left( t \right)\\\sqrt{3}{{\cos }^{2}}\left( t \right)+\sin \left( t \right)\cos \left( t \right)=0\\\color{#2E8B57}{{\cos \left( t \right)}}\left( {\color{blue}{{\sqrt{3}\cos \left( t \right)+\sin \left( t \right)}}} \right)=0\end{array}$$ $$\begin{array}{c}\color{#2E8B57}{{\cos \left( t \right)}}=0\\\\\\t=\frac{\pi }{2},\,\,\frac{{3\pi }}{2}\,\end{array}$$ $$\begin{array}{c}\color{blue}{{\sqrt{3}\cos \left( t \right)=-\sin \left( t \right)}}\\\,\frac{{\sin }}{{\cos }}\left( t \right)=-\sqrt{3}\\\,\tan \left( t \right)=-\sqrt{3}\\t=\,\frac{{2\pi }}{3},\,\,\frac{{5\pi }}{3}\end{array}$$ Since $$\displaystyle t=\frac{{2\pi }}{3}$$ works for both sets of equations, we have a solution! Plug in $$t$$ for $$x$$ and $$y$$ in either set of equations to get $$\displaystyle \left( {\frac{{\sqrt{3}}}{2},\,\,\frac{{-\sqrt{3}}}{4}} \right)$$. Find the $$x$$ and $$y$$ coordinates of any intersections: $$\left\{ \begin{array}{l}x={{e}^{{2t}}}\\y=\ln t\end{array} \right.$$ $$\left\{ \begin{array}{l}x={{e}^{{t-4}}}\\y=2\ln \left( {t-4} \right)\end{array} \right.$$ Solve for $$t$$ by setting the “$$x$$” equations together, and do the same for the “$$y$$” equations. If any $$t$$ values are the same in both, we have a solution; we then solve for $$x$$ and $$y$$ in either equation for that $$t$$. Note that we used the quadratic formula in the second set: $$\begin{array}{c}{{e}^{{2t}}}={{e}^{{t-4}}}\\2t=t-4\\t=-4\end{array}$$ $$\displaystyle \begin{array}{c}\ln t=2\ln \left( {t-4} \right)\\\ln t=\ln {{\left( {t-4} \right)}^{2}}\\t={{\left( {t-4} \right)}^{2}}\\t={{t}^{2}}-8t+16\\{{t}^{2}}-9t+16=0\\t=\frac{{9\pm \sqrt{{81-4\left( 1 \right)\left( {16} \right)}}}}{2}=\frac{{9\pm \sqrt{{17}}}}{2}\end{array}$$ Since we don’t have the same $$t$$ value when we put the $$x$$ equations together and the $$y$$ equations together, we have no solution. Applications of Parametric Equations Parametric Equations are very useful applications, including Projectile Motion, where objects are traveling on a certain path at a certain time. Let’s first talk about Simultaneous Solution examples, where we might find out whether or not certain objects collide (are at the same place at the same time). Simultaneous Solution Examples Here is an example of type of Parametric Simultaneous Solution problem you might see: Problem: A hiker in the woods travels along the path described by the parametric equations $$\left\{ \begin{array}{l}x=80-.7t\\y=.3t\end{array} \right.$$. A bear leaves another area of the woods to the west and travels along the path described by the parametric equations $$\left\{ \begin{array}{l}x=.2t\\y=20+.1t\end{array} \right.$$. (a) Do the pathways of the hiker and the bear intersect? (b) Do the hiker and bear collide? Solution: (a) The reason we might want to have the paths of the hiker and the bear represented by parametric equations is because we are interested in where they are at a certain time. It appears that each of the set of parametric equations form a line, but we need to make sure the two lines cross, or have an intersection, to see if the paths of the hiker and the bear intersect. To do this, we’ll want to eliminate the parameter in both cases, by solving for $$t$$ in one of the equations and then substituting this in for the $$t$$ in the other equation. We can eliminate the parameter in this case, since we don’t care about the time. < Hiker: \displaystyle \begin{align}x&=80-.7t\\-.7t&=x-80\\t&=\frac{{x-80}}{{-.7}}=\frac{{80-x}}{{.7}}\end{align} Plug this into the second equation: \displaystyle \begin{align}y&=.3t\\y&=.3\left( {\frac{{80-x}}{{.7}}} \right)\,\\y&=-\frac{3}{7}x+\frac{{240}}{7}\,\,\,\,\,\text{(a line)}\end{align} Bear: \displaystyle \begin{align}x&=.2t\\t&=\frac{x}{{.2}}=5x\end{align} Plug this into the second equation: $$\displaystyle \begin{array}{l}y=20+.1t\\y=20+.1\left( {5x} \right)\\y=.5x+20\,\,\,\,\,\,\text{(a line)}\end{array}$$ We see that that the two lines are not parallel, so they must intersect! Thus, the answer to (a) above is yes, the pathways of the hiker and bear intersect. We can see where the two lines intersect by solving the system of equations: $$\displaystyle \left\{ \begin{array}{l}y=-\frac{3}{7}x+\frac{{240}}{7}\\y=.5x+20\end{array} \right.;\,\,\,\,-\frac{3}{7}x+\frac{{240}}{7}=.5x+20;\,\,\,\,x=\frac{{200}}{{13}};\,\,y=\frac{{360}}{{13}}\,\,\,\,\,\,\,\,\text{Intersection:}\,\,\left( {\frac{{200}}{{13}},\,\,\frac{{360}}{{13}}} \right)$$ (We could also find these intersections by putting the two equations in the graphing calculator without using parametrics). < Here’s what it looks like in a graphing calculator using parametric equations (I had to play around with the WINDOW to get the graph to display properly): (b) We can’t really tell from the graph whether or not the hiker and and bear collide, although we might be able to by looking at the TABLE in the graphing calculator. But the easiest way would be to set the two $$x$$ equations together, and then set the two $$y$$ equations together, and see if we have the same $$t$$ (like we did above in the Simultaneous Solutions section): Solve for $$t$$ by setting the “$$x$$” equations together, and do the same for the “$$y$$” equations. If any $$t$$ values are the same in both, we have a solution; we then solve for $$x$$ and $$y$$ in either equation for that $$t$$. $$\begin{array}{c}80-.7t=.2t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.3t=20+.1t\\\,\,\,\,\,80=.9t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.2t=20\,\,\,\,\,\,\,\,\\t=88.89\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=100\end{array}$$ Since the $$t$$ values aren’t the same for the $$x$$ and $$y$$, the hiker and bear won’t be at the same place at the same time. Whew! Problem: At noon, Julia starts out from Austin and starts driving towards Dallas; she drives at a rate of 50 mph. Marie starts out in Dallas and starts driving towards Austin; she leaves two hours later Julia (at 2pm), and drives at a rate of 60 mph. The cities are roughly 200 miles apart. When will Julia and Marie pass each other? How far will they be from Dallas when they pass each other? Solution: Let’s first draw this situation and then try to come up with a pair of parametric equations. Remember that $$\text{distance}=\text{rate}\times \text{time}$$. Let $$t=$$ the time they travel (starting at noon), so Julia’s driving time is $$t$$, and Marie’s driving time is $$t-2$$, since she leaves two hours later. Julia’s distance from Austin is $$50t$$, and Marie’s distance from Dallas is $$60\left( {t-2} \right)$$. We can make the $$x$$ equations the distance from Austin for each of the girls, and the $$y$$ equations the paths of the two girls, so we can randomly assign $$y=0$$ to Julia, and $$y=1$$ to Marie. Note that since Marie’s distance from Dallas is $$60\left( {t-2} \right)$$, her distance from Austin is $$200-60\left( {t-2} \right)$$, since both the distances add up to 200 miles. We’ll set the two distances from Austin together (the $$x$$ part of the equations) and solve for $$t$$ to get the time that they meet, measured in the time from Julia leaving Austin (noon). We really don’t need to use the $$y$$ equations, but it’s important to see how we can model a situation with them. In almost 3 hours from noon (about 3pm), they will pass each other. They will be $$50t$$ miles from Austin, or about 145 miles from Austin. This will make them about 200 – 145 = 54 miles from Dallas. Projectile Motion Applications Again, parametric equations are very useful for projectile motion applications. With parametric equations and projectile motion, think of $$x$$ as the distance along the ground from the starting point, $$y$$ as the distance from the ground up to the sky, and $$t$$ as the time for a certain $$x$$ value and $$y$$ value. This is called the trajectory, or path of the object. If we remember from the Quadratic Applications section here (Quadratic Projectile Problem), we can define a parabolic curve of an object going up into the sky and back down as $$h\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{h}_{0}}$$, where, in simplistic terms, the $$-16$$ is the gravity (in feet per seconds per seconds), the $${{v}_{0}}$$ is the initial velocity (in feet per seconds) and the $${{h}_{0}}$$ is the initial height (in feet). (With a quadratic equation, we could also model the height of an object, given a certain distance from where it started; don’t confuse these two types of models.) Now we can model both distance and time of this object using parametric equations to get the trajectory of an object. Note that we’re using trigonometry again: (Note that the $$y$$ equation includes an initial height $${{h}_{0}}$$; also note that we assume the object starts at $$x=0$$; if not, we have to add an initial value $${{x}_{0}}$$ to the $$x$$ equation). These equations make sense since the horizontal and vertical distances use the “famous” equation $$\text{distance}=\text{rate}\times \text{time}$$, where rate is the initial velocity at a certain angle, and time is $$x$$. To solve these problems, we’ll typically want to use one equation first to get the time $$t$$, depending on what we know about either the distance from the starting point ($$x$$) or how high up the object is ($$y$$). We then want to see either how far away the object is from the starting point ($$x$$), or how high up it is ($$y$$). When the problems ask how long the object is in the air, we typically want to set the $$y$$ equation to 0, since this is when the ball is on the ground. Then we can solve for $$t$$, or time. When the problem asks how far the object travels, we typically want to find when the ball hits the ground (from the $$y$$ equation) and then plug that $$t$$ into the $$x$$ equation to see how far it traveled. The $$x$$ part of the equation is typically linear. When the problem asks the maximum height of the object, and when it hits that height, we typically want to find the vertex of the $$y$$ equation, since this is the height curve (parabola) for the object. Remember that we can use $$\displaystyle \left( {-\frac{b}{{2a}},\,\,f\left( {-\frac{b}{{2a}}} \right)} \right)$$ to find the vertex of the quadratic $$a{{x}^{2}}+bx+c$$. Here are some examples: Projectile Motion Problems Solutions The following parametric equations model the path of a soccer ball, where $$t$$ is in seconds, and distances are in feet: $$\left\{ \begin{array}{l}x\left( t \right)=120t\\y\left( t \right)=64t-16{{t}^{2}}\end{array} \right.$$ (Notice that $${{h}_{0}}=0$$, since the ball starts from the ground). (a) When will the ball hit the ground? How far does the ball travel? (b) Find the maximum height of the ball. When does this occur? (a) The ball hits the ground when the height of the ball is 0; this is when the $$y$$ equation equals 0. Notice that it is also at the ground at 0 seconds (before it leaves the ground). $$\displaystyle \begin{array}{c}64t-16{{t}^{2}}=0\\16t\left( {4-t} \right)=0\end{array}$$ $$\displaystyle \begin{array}{c}16t=0\,\,\,\,\,\,\,\,\,\,4-t=0\\t=0, 4\end{array}$$ The ball hits the ground after 4 seconds. To get how far the ball travels, we plug this value into the $$x$$ equation, which is distance: $$x\left( 4 \right)=120\left( 4 \right)=480$$. Therefore, the ball travels 480 feet before it hits the ground again. (b) The maximum height of the ball occurs at the vertex of the height curve of the ball ($$y$$). Using the standard equation $$y=a{{x}^{2}}+bx+c$$, we can use $$\displaystyle -\frac{b}{{2a}}$$ to find the $$t$$ part of the equation (when the maximum height occurs), and then use this value to get the $$y$$ (the actual maximum height): $$\displaystyle -\frac{b}{{2a}}=-\frac{{64}}{{2\left( {-16} \right)}}=2\,\,\,\,\,\,\,\,y\left( 2 \right)=64\left( 2 \right)-16{{\left( 2 \right)}^{2}}=\,\,64$$ The maximum height of the ball is 64 feet, and this happens 2 seconds after it leaves the ground the first time. This makes sense since the ball starts from the ground, and this is half the time for the ball to hit the ground again. Jillian, a pro golfer, hits the golf ball with an initial velocity of 50 ft./sec and at an angle of 35° from the horizon. (a) Find when and where the ball will hit the ground. (b) Find the maximum height of the ball. When does this occur? First, let’s set up the parametric equations that models the distance ($$x$$) and height ($$y$$) at a time $$t$$: $$\left\{ \begin{array}{l}x\left( t \right)=\left( {{{v}_{0}}\cos \alpha } \right)t\\y\left( t \right)={{h}_{0}}+\left( {{{v}_{0}}\sin \alpha } \right)t-16{{t}^{2}}\end{array} \right.$$ or $$\left\{ \begin{array}{l}x\left( t \right)=\left( {50\cos 35} \right)t\\y\left( t \right)=0+\left( {50\sin 35} \right)t-16{{t}^{2}}\end{array} \right.$$ (a) The ball hits the ground when the height of the ball is 0; this is when the $$y$$ equation equals 0. Notice that it is also at the ground at 0 seconds (this makes sense). $$\begin{array}{c}\left( {50\sin 35} \right)t-16{{t}^{2}}=0\\t\left( {50\sin \left( {35} \right)-16t} \right)=0\end{array}$$ $$\begin{array}{c}t=0\,\,\,\,\,\,\,\,50\sin \left( {35} \right)-16t=0\\t=0, \approx 1.792\end{array}$$ The ball hits the ground in 1.792 seconds. To get how far the ball travels, we plug this value into the $$x$$ equation, which is distance: $$x\left( {1.792} \right)=\left( {50\cos 35} \right)\left( {1.792} \right)\approx 73.396$$. The ball travels about 73.4 feet. (b) The maximum height of the ball occurs at the vertex of the height curve of the ball ($$y$$). We can use $$\displaystyle -\frac{b}{{2a}}$$ to find the $$t$$ part of the equation (when the maximum height occurs), and then use this value to get the $$y$$ (the actual maximum height): $$\displaystyle -\frac{b}{{2a}}=-\frac{{50\sin 35}}{{2\left( {-16} \right)}}\approx .896$$ $$\displaystyle y\left( {.896} \right)=50\sin 35\left( {.896} \right)-16{{\left( {.896} \right)}^{2}}\approx 12.851$$ The maximum height of the ball is 12.851 feet, and this happens .896 seconds after it leaves the ground the first time. This makes sense since the ball starts from the ground, and this is half the time for the ball to hit the ground again. Here are some projectile motion problems with wind; we’ll have to add or subtract vectors. Remember that the wind against the object will have to subtracted from the $$x$$ equation, and the wind in the same direction of the object will have to be added. For any straight-line wind (or if the wind is in a horizontal direction), we can use (or 180°, depending on the direction) for the trig arguments, since it comes straight across. When the wind is straight line, it turns out that we’re not adding or subtracting anything from the $$y$$ equation, only the $$x$$. Problems Solutions Lisa hits a golf ball off the ground with a velocity of 60 ft./sec at an angle of 45°. The wind is blowing against the path of the ball at 10 ft./sec with an angle of depression of 15°. (a) Write a set of parametric equations to model this situation. (b) How long does the ball stay in the air (hang time)? (c) How far does Lisa hit the ball (in the horizontal direction)? (a) Note that wind at an angle of depression affects both the $$x$$ and $$y$$ equations, and since it is blowing against the path of the ball, we need to subtract the wind from both equations: $$\displaystyle \left\{ \begin{array}{l}x\left( t \right)=\left( {60\cos 45} \right)t-\left( {10\cos 15} \right)t\\y\left( t \right)=\left( {60\sin 45} \right)t-16{{t}^{2}}-\left( {10\sin 15} \right)t\end{array} \right.$$ (If the wind were blowing in the same direction as the ball, we would add to both). (b) To find out how long the ball stays in the air, we have to set the $$y$$ equation to 0 and solve for $$t$$. (This is because when the $$y$$ equation hits 0, the ball hits the ground again.) To solve, either use quadratic formula, or put in graphing calculator (degree mode): $$\begin{array}{c}\left( {60\sin 45} \right)t-16{{t}^{2}}-\left( {10\sin 15} \right)t=0\\42.4264t-16{{t}^{2}}-2.5882t=0\\-16{{t}^{2}}+39.8382t=0\\t=0,\,\approx 2.49\,\,\sec \end{array}$$ (c) To find how long the ball stays in the air, we’ll use the positive $$t$$ that we just got in the $$x$$ equation: $$\displaystyle \left( {60\cos 45} \right)\left( {2.49} \right)-\left( {10\cos 15} \right)\left( {2.49} \right)\approx 81.59$$. Lisa hits the golf ball 81.59 feet in the horizontal direction. Pretty good! Jade, a pro softball player, hits the ball when it is 3 feet off the ground with an initial velocity of 100 ft./sec and at an angle of 35° from the horizon. A straight-line wind is blowing at 14 ft./sec in the direction the ball is traveling. She hits the ball towards a 40-foot fence that is 120 feet from the plate; if it clears this fence, the ball is a home run. (a) Will the hit be a home run? (b) If so, how much does it clear the fence; if not, how much does it miss the fence? First, let’s set up the parametric equations that models the distance ($$x$$) and height ($$y$$) at a time $$t$$: $$\displaystyle \left\{ \begin{array}{l}x\left( t \right)=\left( {100\cos 35} \right)t+\left( {14\cos 0} \right)t\\y\left( t \right)=3+\left( {100\sin 35} \right)t-16{{t}^{2}}+\left( {14\sin 0} \right)t\end{array} \right.\,$$ or $$\displaystyle \left\{ \begin{array}{l}x\left( t \right)=\left( {100\cos 35} \right)t+14t\\y\left( t \right)=3+\left( {100\sin 35} \right)t-16{{t}^{2}}\end{array} \right.$$ Note we had to add the wind expressions, and use $$\cos \left( 0 \right)$$ and $$\sin \left( 0 \right)$$, since the wind is blowing in the direction of the ball, and it’s a straight-line wind. Also note that $$\cos \left( 0 \right)=1$$ and $$\sin \left( 0 \right)=0$$ (so we’re not adding any wind to the vertical equation, which makes sense). Note also that we had to add the initial height of 3. (If the wind were blowing in against the ball, we would subtract it). (a) We need to find the height of the ball when it is 120 feet from the plate; if this height is over 40 feet, then the ball will be home run. Let’s plug in 120 for the $$x$$ and solve back for $$t$$: . \displaystyle \begin{align}120&=\left( {100\cos 35} \right)t+14t\\120&\approx 95.92t\\t&\approx 1.25\end{align} At $$t=1.25$$ seconds, the ball reaches the fence horizontally. Now let’s see how tall the ball is at this time (height); we’ll use the $$y\left( {1.25} \right)=3+\left( {100\sin 35} \right)\left( {1.25} \right)-16{{\left( {1.25} \right)}^{2}}\approx 49.7$$. At 120 feet from the home plate, the ball has a height of about 49.7 feet. (b) Since the ball has a height of 49.7 feet when it is 120 feet from the home plate, it will clear the 40-foot fence by 9.7 feet; Jade will make a home run! Learn these rules, and practice, practice, practice! For Practice: Use the Mathway widget below to try an Eliminate the Parameter problem. Click on Submit (the blue arrow to the right of the problem), and then click on Eliminate the Parameter to see the answer. You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic. If you click on “Tap to view steps”, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!). On to Sequences and Series – you are ready! <
# The correct way you should calculate probability Probability has been used in many different fields over the years and it has been something that helped people in many situations. As the new coronavirus spread across the world, many have started talking about the probability of how many people it will infect, kill, etc. The simplest way to define probability is to say how likely something is to happen. Whenever you are unsure of the outcome, you can use probability to understand the possible outcomes. The easiest example that comes to mind is flipping the coin. There are two possible outcomes when it comes to flipping the coin: heads and tails. What is the probability of the coin landing on one of the sides? Since we are looking for one outcome, and there are a total of two, all we gonna do is to divide the number of ways it can happen to the total number of possible outcomes. In this situation, simply divide 1 to 2. So, the chance for the coin to land on heads is 50%. Now, let’s talk about something that has more than 2 possible outcomes, For example, dice. In this case, there are 6 possible outcomes, now, let’s say that you want to know what is the probability of rolling 4 and 6. In this situation, we have two different outcomes that we want to know the probability of, and in total, there are 6 possible outcomes. This means that the number of ways it can happen is two, and the total possible outcomes, 6. In this situation, we will have to divide 2 to 6. There are many fields actively using probability and without it, it would be impossible for many of them to exist at all. ## How is probability used in sports In sports, it is very important to have a good gaming strategy in order to be successful. For different types of games and competitions, best athletes and coaches frequently use probability to come up with different types of strategies that would work the best. For example, in baseball, the coach has the ability to create a lineup according to the player’s batting average. If a player has a 200 batting average, for example, it means that he has gotten a base hit two out of every 10. Someone with a 400 batting average will be higher in the lineup since he has gotten 4 base hits out of every 10. ## The world of gambling and probability Gambling is one of those fields where people are using probability very actively. There are many games in the world of gambling where probability can be used, for example, one could use it in every single one of the card games. Also, it could be very actively used in Roulette. Let’s say that you are playing Roulette with 37 digits, you choose a number and wait for an outcome. The probability of you winning would be 1 divided by 37. But there are many other options of betting in roulette, for example, the probability of the ball landing on a red number is 18 divided by 38. However, it seems to be very hard for many to trust probabilities in the world of gambling. In 2019, there was a study about roulette probability. During the study, some people were asked if they would use probability to play casino games such as roulette and many of them said that they would not. The study focused mostly on gamblers of online casinos, who said that strategies that they had never worked before. Still, probability can be used in many different games, what types of cards you will receive while playing poker, where the roulette stops spinning, and many other things heavily depend on probability in the world of gambling. ## Probability is actively used in weather forecasting People use probability every day of their lives without even realizing that they are using it. We enjoy some of the greatest examples of it while checking the weather every day. Meteorologists can not predict the exact weather, so they are using probability to say what is going to happen. For example, if your phone says that there is a 60 percent chance that it will rain, it means that in the case of 60 out of 100 days with the same conditions, it has rained. Professionals also take a look at historical data in order to forecast for the day of the week and talk about possible weather trends. The greatest thing about probability is that people are using it in their everyday lives without even realizing it. It helps us to make many different decisions by trusting the numbers, every time you decide to use an umbrella because of the weather forecast or are working on a new strategy for sports, you are using probability. There are many fields where probability is being used on an everyday basis and it makes our lives easier. This mathematical chance of something happening has influenced many of our decisions, and it continues to help us make some of the hardest decisions every day. ## GOING BACK TO UNIVERSITY? HERE’S A CHECKLIST FOR YOU! #### Get more stuff Subscribe to our mailing list and get interesting stuff and updates to your email inbox. Thank you for subscribing. Something went wrong.
# K12MATH007: Math Grade 7 Unit 5: Probability and Statistics Imagine that you would like to open a new business, but you need data on what the local people would think and whether the business would be successful. Your work on probability and statistics will help you build an appropriate sample for surveying and show you how to use your data to analyze the results and make predictions. In this unit, you will begin by finding the probability of single events. This will lead to the probability of compound events. You will draw inferences based on random sampling, compare multiple samples, and build representations of events using tables and tree diagrams. From the models, you will make predictions and calculate the measure of center and variability. Unit 5 Time Advisory Completing this unit should take approximately 11 hours and 30 minutes. ☐ Subunit 5.1: 5 hours and 30 minutes ☐ Subunit 5.2: 2 hours and 15 minutes ☐ Subunit 5.3: 3 hours and 45 minutes Unit5 Learning Outcomes Upon successful completion of this course, you will be able to: - Solve problems involving basic and compound probabilities. - Make predictions about an event based on theoretical and experimental probabilities. - Use data from random samples to make inferences about a population. - Calculate the measures of center and variability. - Create tree diagrams and organized lists to represent sample spaces. - Develop models to represent the probability of different events occurring. 5.1 Probability “There is a 30% chance of rain for Friday.” The chance of something happening is the probability of an event occurring. Probabilities are used to help make decisions. A 30% chance of rain is a fairly low probability of occurring. A 90% chance of rain tells us that it is most likely going to rain. We may need to cancel plans or at least remember to bring an umbrella. In this subunit, you will take a look at how to compute probabilities and how some events can affect the probability of other events. 5.1.1 Basic Probability - Explanation: Khan Academy’s “Basic Probability” Link: Khan Academy’s “Basic Probability” (YouTube) Instructions: Watch the video to see the notation that is used when writing probabilities and the three different forms that can be used to write the probabilities. Viewing this video should take approximately 15 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.C.5](http://www.corestandards.org/Math/Content/7/SP/C/5) attributed to Khan Academy. `````` • Activity: XP Math: “Plinko Probability - The Probability is Right” Link: XP Math: “Plinko Probability - The Probability is Right” (Flash) Instructions: In this game, you will answer probability questions about the amounts you are able to win. You will place your answer in the maroon box at the bottom right. The game will tell you whether your answer is correct. If the sound is on when you play the game, you will hear a noise for a wrong answer and you will not be given a chip. If you are correct, you will be given a chip to drop down the plinko board. The hint button places a triangle under each of the given values you need to answer the question. This way, you can count up the total to get the probability. Playing this game should take approximately 15 minutes. Standards Addressed (Common Core): • Did I Get This? Activity: orgLib: “Preview Questions for Test: 7.SP.5 - Worksheet” Link: orgLib: “Preview Questions for Test: 7.SP.5 - Worksheet” (HTML) Instructions: Scroll down to the five questions and click on “Consume” beneath them. You will be brought to a new screen where you can type in your answers. After each question you will need to click on “Submit Answer.” When you have all five questions done, you will click on “Submit Test.” At this point, you will be brought to a screen that will show whether your answers are correct. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): 5.1.2 Compound Events - Explanation: CK-12: “Probability of Compound Events” Link: CK-12: “Probability of Compound Events” (PDF) Instructions: Read the section entitled “Guidance” to find the difference between independent and dependent events. Be sure to take note of how calculating the probability for a dependent event is different from calculating the probability of an independent event. Then solve the 17 practice questions at the bottom of the page. Completing this activity should take approximately 45 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.C.8](http://www.corestandards.org/Math/Content/7/SP/C/8) attributed to CK-12 Foundation. The original version can be found [here](http://www.ck12.org/probability/Probability-of-Compound-Events/lesson/Probability-of-Compound-Events/). `````` 5.1.2.1 Tree Diagrams, Organized Lists and Tables - Explanation: Dr. Carol Fisher Burns’s “Probability Tree Diagrams” Link: Dr. Carol Fisher Burns’s “Probability Tree Diagrams” (HTML) Explanation: Read through the examples of probability questions and take a look at the tree diagrams formed to answer the questions. Then go back up to the top of the page and click on “Jump right to the exercises!” This will bring you to a new screen where you can practice making your own tree diagrams. Making tree diagrams should take approximately 15 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.C.8](http://www.corestandards.org/Math/Content/7/SP/C/8) `````` • Did I Get This? Activity: orgLib: “Preview Questions for Test: 7.SP.8 - Worksheet” Link: orgLib: “Preview Questions for Test: 7.SP.8 - Worksheet” (HTML) Instructions: Scroll down to the four questions and click on “Consume” beneath them. You will be brought to a new screen where you can type in your answers. After each question you will need to click on “Submit Answer.” When you have all four questions done, you will click on “Submit Test.” At this point, you will be brought to a screen that will show whether your answers are correct. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): 5.1.3 Experimental Probability - Explanation: CK-12: “Experimental Probability” Link: CK-12: “Experimental Probability” (HTML) Instructions: Read through the entire lesson on experimental probability. Focus on the differences and similarities between your hypothesis (theoretical probability) and the actual results (experimental probability). You do not need to watch the video clip after the reading. Instead, work through the questions in the “Time to Practice” section. Completing this activity should take approximately 1 hour and 15 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.C.6](http://www.corestandards.org/Math/Content/7/SP/C/6) `````` • Activity: Shodor: “Experimental Probability” Link: Shodor: “Experimental Probability” (HTML) Instructions: Now is your chance to try out your own experiments. On this site there are several different interactive spinners. For three of the spinners, do the following: • 1. Calculate the theoretical probability of the spinner landing on red. • 2. Complete 20 spins and keep track of the results. • 3. Use the results to calculate the experimental probability of landing on red. • 4. Compare the theoretical to the experimental probability. • 5. Make a prediction: If you were to continue spinning the spinner 1,000 times, how many times would you expect the spinner to land on red? Completing this activity should take approximately 45 minutes. Standards Addressed (Common Core): - CCSS.Math.Content.7.SP.C.6 • Activity: Dan Meyer’s Three-Act Math Tasks: “Yellow Starbursts” Link: Dan Meyer’s Three-Act Math Tasks: “Yellow Starbursts” (HTML) Instructions: Follow the directions for the activity. There are videos and images to help you answer the questions. In Act 3, the answer will be provided and you will be given a chance to reflect on your work. Completing this activity should take approximately 30 minutes. Standards Addressed (Common Core): • Did I Get This? Activity: orgLib: “Preview Questions for Test: 7.SP.6 - Worksheet” Link: orgLib: “Preview Questions for Test: 7.SP.6 - Worksheet” (HTML) Instructions: Scroll down to the five questions and click on “Consume” beneath them. You will be brought to a new screen where you can type in your answers. After each question you will need to click on “Submit Answer.” When you have all five questions done, you will click on “Submit Test.” At this point, you will be brought to a screen that will show whether your answers are correct. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): 5.1.4 Geometric Probability - Activity: GeoGebra: “A Look at Geometric Probability” Link: GeoGebra: “A Look at Geometric Probability” (HTML) Instructions: The probability of landing in the inside shape can be calculated by finding the area of the inside shape and the area of the outside shape and writing the results as a fraction. This activity allows you to manipulate the lengths of the sides and therefore change the areas. Follow the directions given in the activity. Use the dots below to change the values of the lengths. Then, on separate paper, calculate the probability to see if you get the same answer as the program. Practice with five different lengths. Completing this activity should take approximately 30 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.C.7](http://www.corestandards.org/Math/Content/7/SP/C/7) - [CCSS.ELA-Literacy.RST.6-8.3](http://www.corestandards.org/ELA-Literacy/RST/6-8/3) - [CCSS.ELA-Literacy.RST.6-8.7](http://www.corestandards.org/ELA-Literacy/RST/6-8/7) - [CCSS.ELA-Literacy.WHST.6-8.1b](http://www.corestandards.org/ELA-Literacy/WHST/6-8/1/b) - [CCSS.ELA-Literacy.WHST.6-8.4](http://www.corestandards.org/ELA-Literacy/WHST/6-8/4) `````` • Did I Get This? Activity: orgLib: “Preview Questions for Test: 7.SP.7 - Worksheet” Link: orgLib: “Preview Questions for Test: 7.SP.7 - Worksheet” (HTML) Instructions: Scroll down to the four questions and click on “Consume” beneath them. You will be brought to a new screen where you can type in your answers. After each question you will need to click on “Submit Answer.” When you have all four questions done, you will click on “Submit Test.” At this point, you will be brought to a screen that will show whether your answers are correct. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): 5.2 Samples Have you ever heard a commercial claim “9 out of 10 dentists recommend this toothpaste”? The claim that 9 out of 10 dentists like the toothpaste isn’t very meaningful if 9 of the 10 dentists also work for that toothpaste company. In that case, the sample that was questioned does not truly represent the entire population of dentists. In this subunit, we will take a look at samples and see how a well-chosen sample can represent the population and help make informed decisions. 5.3 Compare Data Sets Oftentimes we look at one set of data in a graph or table. However, there are times when we might want to compare two different sets of data. Maybe Mr. Smith’s math class is in a competition with Ms. Sanchez’s class. Both classes have a set of data that will need to be compared to see who wins the competition. This subunit looks at different ways of comparing multiple sets of data. It starts by looking at different types of graphs and measures of center and variability. Then it continues on to discuss multiple sets of data. 5.3.1 Mean Deviation - Interactive Lab: GeoGebra: “Calculate Mean Absolute Deviation (M.A.D.)” Link: GeoGebra: “Calculate Mean Absolute Deviation (M.A.D.)” (HTML) Instructions: There is a detailed list of the steps that need to be taken to get the MAD at the bottom of the worksheet. Follow them to calculate the MAD. Then use the option to show the steps in #1 to check your work. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.B.3](http://www.corestandards.org/Math/Content/7/SP/B/3) - [CCSS.Math.Content.7.SP.B.4](http://www.corestandards.org/Math/Content/7/SP/B/4) `````` 5.3.2 Box and Whisker Plots - Explanation: CK-12: “Box-and-Whisker Plots” Link: CK-12: “Box-and-Whisker Plots” (PDF) Instructions: Read the lesson on creating a box-and-whisker plot until you reach the videos from Khan Academy. While reading, take note of what the following terms mean and where they appear in a box-and-whisker plot: median, upper quartile, and lower quartile. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.B.3](http://www.corestandards.org/Math/Content/7/SP/B/3) - [CCSS.Math.Content.7.SP.B.4](http://www.corestandards.org/Math/Content/7/SP/B/4) - [CCSS.ELA-Literacy.RST.6-8.4](http://www.corestandards.org/ELA-Literacy/RST/6-8/4) text is attributed to CK-12 and the video is attributed to Khan Academy. The original version can be found [here](http://www.ck12.org/user%3AbXN0b2RlcmxAcGVyaGFtLmsxMi5tbi51cw../book/Mrs.-Stoderl%2527s-Pre-Algebra-book/r20/section/13.4/). `````` 5.3.3 Stem and Leaf Plots - Explanation: CK-12: “Stem-and-Leaf Plots” Link: CK-12: “Stem-and-Leaf Plots” (PDF) Instructions: Start by reading the example of using data from the Iditarod in Alaska. Continue reading until you reach the videos. Take note of the following while reading. ``````- 1. How is the stem different than a leaf? - 2. How can I write a three-digit number as a stem and leaf? - 3. What does the key for a stem and leaf plot look like? - 4. How can I get the mean, median, and mode from the stem and leaf plot? Next, watch the video from Khan Academy in order to see a stem and leaf plot made and why it is important. Finally, using the information in the reading and the video, answer the questions in the “Time to Practice” section. Completing these activities should take approximately 1 hour. Standards Addressed (Common Core): - [CCSS.Math.Content.7.SP.B.3](http://www.corestandards.org/Math/Content/7/SP/B/3) - [CCSS.Math.Content.7.SP.B.4](http://www.corestandards.org/Math/Content/7/SP/B/4) - [CCSS.ELA-Literacy.RST.6-8.2](http://www.corestandards.org/ELA-Literacy/RST/6-8/2) text is attributed to CK-12 and the video is attributed to Khan Academy. The original version can be found [here](http://www.ck12.org/user%3AbXN0b2RlcmxAcGVyaGFtLmsxMi5tbi51cw../book/Mrs.-Stoderl%2527s-Pre-Algebra-book/r20/section/13.3/). `````` 5.3.4 Comparing Multiple Sets of Data - Explanation: CK-12: “Data Analysis Using Dot Plots, Measures of Central Tendency, and Interquartile Range” Link: CK-12: “Data Analysis Using Dot Plots, Measures of Central Tendency, and Interquartile Range” (PDF) Instructions: Read through the lesson, taking note of how measures of center (mean, median, mode) and measures of variability (MAD, interquartile range, range) can be used to compare two sets of data. Reading this lesson should take approximately 30 minutes. Standards Addressed (Common Core): ``````- [CCSS.Math.Content.7.SP.B.3](http://www.corestandards.org/Math/Content/7/SP/B/3) - [CCSS.Math.Content.7.SP.B.4](http://www.corestandards.org/Math/Content/7/SP/B/4) attributed to CK-12 Foundation. The original version can be found [here](http://www.ck12.org/na/Data-Analysis-Using-Dot-Plots,-Measures-of-Central-Tendency,-and-Interquartile-Range---7.SP.3,4-1/lesson/user:c2ZveDJAb3N3ZWdvLm9yZw../Data-Analysis-Using-Dot-Plots%252C-Measures-of-Central-Tendency%252C-and-Interquartile-Range---7.SP.3%252C4/). `````` • Did I Get This? Activity: orgLib: “Preview Questions for Test: 7.SP.3 - Worksheet” Link: orgLib: “Preview Questions for Test: 7.SP.3 - Worksheet” (HTML) Instructions: Scroll down to the four questions and click on “Consume” beneath them. You will be brought to a new screen where you can type in your answers. After each question you will need to click on “Submit Answer.” When you have all four questions done, you will click on “Submit Test.” At this point, you will be brought to a screen that will show whether your answers are correct. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): • Did I Get This? Activity: orgLib: “Preview Questions for Test: 7.SP.4 - Worksheet” Link: orgLib: “Preview Questions for Test: 7.SP.4 - Worksheet” (HTML) Instructions: Scroll down to the four questions and click on “Consume” beneath them. You will be brought to a new screen where you can type in your answers. After each question you will need to click on “Submit Answer.” When you have all four questions done, you will click on “Submit Test.” At this point, you will be brought to a screen that will show whether your answers are correct. Completing this activity should take approximately 15 minutes. Standards Addressed (Common Core): • Checkpoint: Illustrative Mathematics: “College Athletes” Link: Illustrative Mathematics: “College Athletes” (PDF) Instructions: Complete parts a though e. You may use a calculator when completing this checkpoint. Then check your work on page 3. Completing this checkpoint and checking your work should take approximately 1 hour. Standards Addressed (Common Core): • Checkpoint: Illustrative Mathematics: “Waiting Times” Link: Illustrative Mathematics: “Waiting Times” (PDF) Instructions: Determine the number of boxes of cereal that would need to be purchased in parts A and B. Then scroll to page 2 to check your work. Completing this checkpoint and checking your work should take approximately 15 minutes. Standards Addressed (Common Core): `````` Instructions: Complete this exam.
# The Twelve Facts of Christmas: Pascal’s Triangle At this time of year, you’ll see a lot of decorations around and a lot of shapes – stars, trees, snowmen, and so on. One shape you might not see many of though, is the triangle, and I think that’s a shame. So I’m going to share with you my 12 facts of Christmas, about one of the greatest triangles in mathematics: Pascal’s triangle. 1. ## It can be created simply by adding numbers together To generate Pascal’s triangle, start with a one (the number of partridges in a pear tree) and imagine it’s sitting in an infinite row of zeroes going off to either side. Then you can generate the next row by writing the sum of each pair of numbers in the row in the gap underneath them. This will give you an infinite row of zeroes with two ones in the middle, then repeating again will give you that with 1, 2, 1 in the middle, and so on. The rest of the triangle is obtained by simply repeating this process for the rest of time. 1. ## It’s named after Blaise Pascal, but someone else discovered it first Although Pascal’s triangle is named after mathematician Blaise Pascal, many other mathematicians knew about the triangle hundreds of years earlier. Chinese mathematician Jia Xian (c. 1050) supposedly “[used] the triangle to extract square and cube roots of numbers,” and Persian mathematician Omar Khayyam (c. 1048–1113) seemed to also have knowledge of the structure. 1. ## The numbers in the triangle are the coefficients of binomial expansions If you have a bracket containing a sum, (a+b), and you want to raise that bracket to a power n, you will find that the different powers of a and b in the resulting expansions have different coefficients in front of them. (a + b)2 = (1)a2 + 2ab + (1)b2 (a + b)3 = (1)a3 + 3a2b + 3ab2 + (1)b3 (a + b)4 = (1)a4 + 4a3b + 6a2b2 + 4ab3 + (1)b4 These coefficients are exactly the numbers in the nth row of Pascal’s triangle! (The first row of the triangle is called row 0, so when I say the nth row, I mean the one that starts ‘1, n’.) This is connected to the fact that the triangle also encodes the ‘choose’ function in mathematics: If you have a pile of n things, and you want to choose k of them, the number of possible ways to do this will be the kth entry in the nth row of Pascal’s triangle (again starting from the 0th in rows and columns). For example, if I have 5 gold rings, and I want to give one of them to each of my two true loves, the number of different ways to choose two rings from five is 10 – which can be found in the fifth row, second column. 1. ## The diagonals in the triangle are interesting number sequences Apart from the outside edge, which is all made up of 1s, the other diagonals all have nice properties – the first stripe is just the whole numbers counting upwards; the next is the triangular numbers (sums like 1, 1+2, 1+2+3, 1+2+3+4 etc) – if you had a triangular number of Christmas puddings, you could arrange them into a triangle on the table. The next stripe is the 3D triangle numbers, or tetrahedral numbers: the number of spherical Christmas puddings you can stack in a triangular-based pyramid, and each such number is a sum of the first triangular numbers. In fact, this pattern continues – the next set is 4D tetrahedral numbers, called pentatope numbers, and so on. 1. ## It’s also got Fibonacci Numbers in there Taking a less steep diagonal slice through the triangle, and adding the resulting numbers, you’ll find they each sum to a Fibonacci number – each one is the sum of the previous two. These numbers have many mysterious and deep connections in mathematics, and also occur in hypothetical rabbit population dynamics. 1. ## There’s a pattern in the sums of each row If you add the numbers in each horizontal row together, you’ll find there’s an interesting pattern in the numbers you get. Bonus points if you can work out why! 1. ## The products of the rows also do something interesting If you multiply together all the numbers in each row, that’ll give you a sequence of numbers. If you write Sn for the nth number in that sequence, then the following is true:This says that if you take the product of a row, multiply it by the product of the row two below that, and then divide that by the square of the product of the row in between, this value gets closer and closer to the exponential constant e (as your value for n gets bigger and bigger). It gets everywhere, the number e… 1. ## It also gives you the number of paths in a grid If you have a square grid, and are allowed to move from one box to another vertically or horizontally adjacent box, you can challenge yourself to find a route from anywhere in the grid to the top left square. If you write the numbers of Pascal’s triangle diagonally across a square grid, you’ll find that the number in each square gives the number of different routes you can take from that square to the top left square. 1. ## If you colour it in, you get a fractal Colouring in the even numbers of Pascal’s triangle one colour, and the odd numbers another, results in an exact copy of Sierpinski’s triangle, the triangular fractal. The more rows you colour down the triangle, the more detailed a fractal you get (not much can be seen in this example, but this video shows the process nicely). 1. ## There are still open mathematical questions about it Even though we know many things about the numbers in Pascal’s triangle, there are still mysteries to be solved. An open conjecture, called Singmaster’s Conjecture, muses on the question of how many times, at most, any given number N > 1 can occur in the triangle. The most frequently occurring number known is 3003, which is in there 8 times, but there’s no exact value known for a maximum, if one even exists. I wrote about it for The Aperiodical a few years ago, if you’d like to read more. 1. ## There’s a higher dimensional version called Pascal’s pyramid The 3D triangle is a tetrahedron of numbers, and encodes trinomial coefficients – found when you raise a bracket of the form (a+b+c) to the nth power. The numbers on each layer of the tetrahedron are the sum of the three adjacent numbers in the layer above it. Makes you wonder what a 4D tetrahedron looks like, if you weren’t already from Fact 4. 1. ## There’s such a thing as the Christmas stocking identity, and it’s about Pascal’s triangle In case you were worried this post was veering away from being mildly Christmas-themed, don’t worry – if you draw the shape of a Christmas stocking and overlay it on Pascal’s triangle, there’s a nice pattern to be found there as well. Also called the hockey-stick identity, it states that the a sequence of numbers moving diagonally in towards the middle of the triangle will add up to the final number just off the diagonal at the bottom. Festive! ### Posted by Katie Steckles is a mathematician based in Manchester, who gives talks and workshops on different areas of maths. She finished her PhD in 2011, and since then has talked about maths in schools, at science festivals, on BBC radio, at music festivals, as part of theatre shows and on the internet. Katie writes blog posts and editorials for The Aperiodical, a semi-regular maths news site.
# Understanding Limits and Continuity Limits and continuity of function is where calculus topics starts, however, to know calculus you must complete all lessons of pre- calculus with examples. Limit of a function is a certain value that a function approaches, whereas a function is said to be continuous when the function has no breaks or drawn without taking the pen away from the paper. ### How to understand a limit of a function? Limit of a function is a certain value on x- axis that a function of x approaches from both the sides. When a function of x approaches to n (where n is an integer), then all the values of x are very close to n from both the side of the number line. For example, when function of x approaches to 2, then the the values near to the limit of a function from the left if 1.9, 1.99, 1.999 etc. and from the right is 2.1, 2.01, 2.001 etc. ### Limit with an example Let us say $f(x) = x^{2} + 2x - 6$ is a function whose limit is 2, then, the limit of a function can be written as $\displaystyle \lim_{x \to 2}$. This can be showed in the form of a graph. While finding the limit of a function, what should we observe? Finding the limit of a function is all about observing the particular value a function approaches from left and right. From figure 1, it can be observed that both the right hand limit and left hand limit of a function approaches to the same number. In the above graph, the limit approaches from both right and left to 2. $\displaystyle \lim_{x \to c^-}$ means that x approaches c from left and reach all numbers smaller than c. Similarly, $\displaystyle \lim_{x \to c^-}$ means that x approaches c from right and reach all numbers greater than c. Note: the value of x is never equal to c. ### Continuity of a function The limits applies to a continuous function. If the function is continuous, then a limit exists, otherwise not. What is continuity of a given function? A function is said to be continuous, when graph for the given function shows no breaks or discontinuity at a point or at least in a given interval. How to find whether a function is continuous? Continuity of a function can be examined in two ways. • Continuity of a function at a given point • Continuity over a given interval For finding the continuity of is given function is, the given function should have no breaks at the neighbourhood to a given point. How to determine the continuity of a function at a given point? For checking the continuity of a function at a given point, then it must satisfy the following conditions viz., • The point at which continuity of a function is examined should be within the domain of the function. • Both the left handed and right handed limit of the function approaching to the point should be equal. ### Summary • Limit of a function is approached from both the sides. • If the difference between the nearest value and the limit n is $h, h_{1}, h_{2}$ etc., then the values approaching from the left is $n - h, n - h_{1}, n - h_{2}$. • In the same case, the values approaching from the right is $n + h, n + h_{1}, n + h_{2}$ . • Continuity is defined as a function without any breaks or jumps.
# 12.02 Time intervals Lesson We work with time in almost every area of our life, so knowing how to add and subtract time, calculate the time between events, or even use a $24$24 hour clock are important. Let's look at some common ways we may need to work with time. ### Adding and subtracting time When we want to find the difference between times we can construct a subtraction expression in a similar way to how we usually subtract two numbers - but it's important to remember that there are $60$60 minutes in an hour, so we have to adjust our counting accordingly. People often forgot this in the calculations which can lead to errors. For example, the difference between $6$6 pm and $5:30$5:30 pm is $30$30 minutes. This is not the same as $6-5.3$65.3 which people commonly (and mistakenly) write, as this results in $0.7$0.7 (which they then interpret as $70$70 minutes - again, mistakenly). Similarly, when adding time, once the total number of minutes reaches $60$60 minutes we add $1$1 to the hours instead. #### Worked example ##### Example 1 Add $2$2 hours and $45$45 minutes to $5:25$5:25 pm. Do: We could start by first adding the minutes $25+45$25+45 which is equal to $70$70 minutes. We can think of this as $60$60 minutes $+10$+10 minutes, which is $1$1 hour and $10$10 minutes. Now add the hours $5+2+1$5+2+1 and we get $8$8. So adding $2$2 hours and $45$45 minutes onto $5:25$5:25 pm takes us to $8:10$8:10 pm. Reflect: Alternatively, we could view the problem more visually. Starting at $5:25$5:25 there are $35$35 minutes until $6:00$6:00. This leaves us with $2$2 hours and $10$10 minutes to add. First add the $10$10 minutes to reach $6:10$6:10 and then $2$2 hours to reach our final answer of $8:10$8:10 pm. Remember! $60$60 minutes makes $1$1 hour #### Practice questions ##### Question 1 Find the value of $3$3 hours $3$3 minutes $+$+ $3$3 hours $29$29 minutes 1. $\editable{}$ hours $\editable{}$ minutes ##### Question 2 James went to a movie at $11:50$11:50 am. The movie went for $1$1 hour and $10$10 minutes. 1. Would the movie finish in the am or pm? am A pm B am A pm B 2. Complete the statement: The movie finishes at $\editable{}$$::\editable{}$$\editable{}$ pm. ##### Question 3 Evaluate $6$6 hours $10$10 minutes $-$ $2$2 hours $30$30 minutes. 1. $\editable{}$ hour(s) $\editable{}$ minute(s) ### Time between events #### Worked example ##### Example 2 If a bus departs at $9:20$9:20 am and arrives at its final destination at $12:55$12:55 pm calculate the the length of the journey. Think: We can do this by breaking the time down into smaller parts. This can be done visually as follows: Do: Add the minutes together, remembering that once the minutes reach $60$60, we can add another hour. From $9:20$9:20 am until $10:00$10:00 am $=$= $40$40 min From $10:00$10:00 am to $12:00$12:00 pm $=$= $2$2 h From $12:00$12:00 pm to $12:55$12:55 pm $=$= $55$55 min Total: $=$= $2$2 h $95$95 min $95$95 minutes is $60$60 minutes + $35$35 minutes. ($1$1 hour and $35$35 minutes) Thus, the journey was $3$3 hours and $35$35 minutes in duration. #### Practice questions ##### question 4 A song clip starts playing at $7$7:$55$55 pm and finished at $8$8:$11$11 pm. How long is the song clip in minutes? ##### question 5 A taxi departs at 13:32 from Sydney and arrives at 20:14 at Melbourne. How long did the taxi take for the trip? 1. The trip took $\editable{}$ hour(s) and $\editable{}$ minute(s) ### Using the calculator for time calculations We can also use our calculators to help with some of these calculations. #### Worked example ##### Example 3 Calculate $3$3 hours $40$40 minutes - $1$1 hour $55$55 minutes. Think: We can put it in the calculator using the DMS button like we would if we were to calculate angles. Do: Hence, the answer is $1$1 hour $45$45 minutes. Reflect: Try the practice questions above again, this time using the calculator in this way. Do you get the same answers as before? ### Time zones in Australia Australia is divided into $3$3 time zones (shown below on the left-hand map). They are: Eastern Standard Time (AEST): covering Queensland, New South Wales, Australian Capital Territory, Victoria and Tasmania. Western Standard Time (AWST): covering Western Australia. ($2$2 hours behind AEST) Central Standard Time (ACST): covering South Australia, Northern Territory. ($\frac{1}{2}$12 an hour behind AEST) The time zones become a bit more tricky during Daylight Savings because not all the states use it. Queensland, the Northern Territory and Western Australia stay on Standard Time. The map on the right shows that during Daylight Savings, Australia actually has $5$5 time zones! During daylight savings, clocks are moved forward $1$1 hour from the standard time, so $8$8 am standard time would be $9$9 am in daylight savings time, $6$6 pm AEST would become $7$7 pm AEDT and so on. Similarly, when we go from daylight savings time to standard time, clocks are moved back $1$1 hour, so $12$12 pm AEDT would be $11$11 am AEST and $14:22$14:22 AEDT would be $13:22$13:22 AEST. #### Practice questions ##### Question 6 If it is $16$16:$14$14 in Sydney in Daylight Savings Time, what time would it be in Eastern Standard Time? 1. $16$16:$14$14 AEDT would be $\editable{}$:$\editable{}$ AEST. ##### Question 7 If it is $14$14:$35$35 in Sydney in Eastern Standard Time, what time would it be in Daylight Savings Time? 1. $14$14:$35$35 AEST would be $\editable{}$:$\editable{}$ AEDT. ##### Question 8 Complete the table to show the time in each time zone. 1. WST CST EST $\editable{}$$::\editable{} 55::2525 \editable{}$$:$:$\editable{}$ $5$5$:$:$11$11 $\editable{}$$::\editable{} \editable{}$$:$:$\editable{}$ $\editable{}$$::\editable{} \editable{}$$:$:$\editable{}$ $12$12$:$:$25$25 $\editable{}$$::\editable{} 77::1717 \editable{}$$:$:$\editable{}$ ##### Question 9 Luke and Maria board a train at $7$7:$58$58 PM on Saturday from Sydney, New South Wales towards Adelaide, South Australia. 1. If the trip to Adelaide took $20$20 hours and $22$22 mins then they arrived on $\editable{}$:$\editable{}$ PM Sunday in Adelaide local time. 2. Luke boards a train from Adelaide, South Australia at $5$5:$45$45 PM on Sunday to Darwin, Northern Territory. If his trip took $28$28 hours and $36$36 mins, then he arrived at $\editable{}$:$\editable{}$ PM Monday Darwin local time. 3. Maria boards a train from Adelaide, South Australia at $4$4:$55$55 PM Sunday towards Perth, Western Australia. If the trip took her $19$19 hours and $20$20 mins, then she arrived at $\editable{}$:$\editable{}$ AM in Perth local time. ### Outcomes #### ACMEM078 calculate time intervals, such as time between, time ahead, time behind
# In the same ratio In this lesson, we will learn to scale up from one group to 'many' groups and use multiplicative relationships to calculate unknown values in the 'many' groups. #### Unit quizzes are being retired in August 2023 Why we're removing unit quizzes from the website > Quiz: # Intro quiz - Recap from previous lesson Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz! Q1.Fill in the gap: We can describe the ____________ between two or more quantities by comparing them with each other. 1/5 Q2.Fill in the gap: We can describe the relationship between two values using __________ notation. 2/5 Q3.I make groups of grey and white counters in the ratio 3:4. How many grey counters are there in two groups? 3/5 Q4.I make groups of grey and white counters in the ratio 3:4. How many white counters are there in five groups? 4/5 Q5.I make groups of grey and white counters in the ratio 3:4. How many grey counters and white counters are there in 100 groups? 5/5 #### Unit quizzes are being retired in August 2023 Why we're removing unit quizzes from the website > Quiz: # Intro quiz - Recap from previous lesson Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz! Q1.Fill in the gap: We can describe the ____________ between two or more quantities by comparing them with each other. 1/5 Q2.Fill in the gap: We can describe the relationship between two values using __________ notation. 2/5 Q3.I make groups of grey and white counters in the ratio 3:4. How many grey counters are there in two groups? 3/5 Q4.I make groups of grey and white counters in the ratio 3:4. How many white counters are there in five groups? 4/5 Q5.I make groups of grey and white counters in the ratio 3:4. How many grey counters and white counters are there in 100 groups? 5/5 # Video Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should: • Click "Close Video" • Click "Next" to view the activity Your video will re-appear on the next page, and will stay paused in the right place. # Worksheet These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below. #### Unit quizzes are being retired in August 2023 Why we're removing unit quizzes from the website > Quiz: # In the same ratio quiz Apply your understanding to this quiz. Q1.Fill in the gap: Two quantities are said to be in the same ratio, if there is an equivalent _____________ relationship between them. 1/5 Q2.Fill in the gap: 2 litres of red paint and 5 litres of white paint are in the same_______ as 5 litres of red paint and 12.5 litres of white paint. 2/5 Q3.I mix 2 litres of squash with 3 litres of water. I need more that tastes the same. How much squash should I mix with 9 litres of water? 3/5 Q4.Fill in the missing ratio: I mix 2 litres of squash with 7 litres of water. The ratio of squash to water is …. : .... 4/5 Q5.Fill in the missing number: I mix 4 litres of yellow paint with 10 litres of red paint. The ratio of yellow to red is …. : 5. 5/5 #### Unit quizzes are being retired in August 2023 Why we're removing unit quizzes from the website > Quiz: # In the same ratio quiz Apply your understanding to this quiz. Q1.Fill in the gap: Two quantities are said to be in the same ratio, if there is an equivalent _____________ relationship between them. 1/5 Q2.Fill in the gap: 2 litres of red paint and 5 litres of white paint are in the same_______ as 5 litres of red paint and 12.5 litres of white paint. 2/5 Q3.I mix 2 litres of squash with 3 litres of water. I need more that tastes the same. How much squash should I mix with 9 litres of water? 3/5 Q4.Fill in the missing ratio: I mix 2 litres of squash with 7 litres of water. The ratio of squash to water is …. : .... 4/5 Q5.Fill in the missing number: I mix 4 litres of yellow paint with 10 litres of red paint. The ratio of yellow to red is …. : 5. 5/5 # Lesson summary: In the same ratio ## Time to move! Did you know that exercise helps your concentration and ability to learn? For 5 mins... Move around: Climb stairs On the spot: Chair yoga
# QTS Numeracy Model Solutions – Test 2, Q23 In this series of blog posts, we look at some of the detailed methods that you could use to tackle the questions on the QTS Numeracy practice papers from the Department for Education. In this post, we look at question 23 from practice test 2. The full practice paper can be downloaded here. ## The Question A teacher presented the following box-and-whisker diagram as part of a staff discussion on pupils’ performance. The diagram shows the percentage test marks in mathematics for a revision test for two class groups. Tick all the true statements: • The range of percentage marks was greatest in class A. • The median percentage mark in Class A was 15 percentage points less than the median percentage in Class B. • The interquartile range was the same in both classes. ## Worked Solution This is a question on box plots, which is a type of graph that you might come across that conveys some specific pieces of information to you. In particular, the upper and lower ends of the ‘whiskers’ show the highest and lowest percentage scores achieved. The line in the middle of the box shows the median (the score that half of students achieved above and half below), and the two ends of the box show the upper and lower quartiles (the scores the one-quarter and three-quarters of students respectively scored above). To answer this question, consider each of the statements in turn. The range of percentage marks was greatest in class A. The range is the difference between the highest score and the lowest score, so you will need to read these values off the box plot for each graph and subtract them. For class A, the highest score was 60% and the lowest score was 30%, so the range is 60% – 30% = 30%. For class A, the highest score was 60% and the lowest score was 35%, so the range is 60% – 35% = 25%. The range for class A is greater than the range for class B, so the statement is TRUE. The median percentage mark in class A was 15 percentage points less than the median percentage in class B. To find the median, read off the value of the line in the middle of the box. For class A, this is at 45% and for class B, it is at 50%. The difference between these values is 50% – 45% = 5%, so the statement is FALSE. The interquartile range is the same in both classes. The interquartile range means the difference between the value of the upper quartile and the lower quartile. Read these values off the box plot by looking at the top and bottom of the box, and then subtract the values. For Class A, the upper quartile is 50% and the lower quartile is 40%, so the interquartile range is 50% – 40% = 10%. For Class B, the upper quartile is 55% and the lower quartile is 45%, so the interquartile range is 55% – 45% = 10%. These values are the same, so the statement is true. ## Final Answer: True, False, True Note – I have made a point of being thorough in this method to ensure you understand each point as we go. When you sit your QTS Skills Tests, your process will probably be much quicker as these techniques start to become second nature to you. ## Further Help If you require any further help with questions like this, we have created a selection of resources to provide all the help you need. Revision Book – The Guide to the QTS Skills Tests book devotes a whole chapter to each QTS numeracy topic, with detailed methods, worked examples and plenty of practice questions for you to have a go at. The book also includes three practice tests and fully worked solutions to every question. Practice Tests – Want more practice tests to see how ready you are? Take a look at our selection of practice papers for the QTS numeracy test (including some totally free papers as our gift to you). Revision Cheat Sheets – Our cheat sheets boil down everything you need to know to just the key points on the topic. They are a perfect resource for your last minute revision!
+0 # Help with graphing 0 7 1 +806 help with this graph Find the area of the region enclosed by the graph of x^2 + y^2 = 2x - 6y + 6 + 12x - 8y + 22. Dec 17, 2023 #1 +222 0 Step 1: Rewrite the equation First, let's rewrite the equation to a more recognizable form: x^2 + y^2 + 14x - 14y = 28 Step 2: Complete the square for both x and y terms To find the area enclosed by the graph, we can try to rewrite the equation as the standard equation of a circle: (x - h)^2 + (y - k)^2 = r^2 We can achieve this by completing the square for both the x and y terms: (x^2 + 14x) + (y^2 - 14y) = 28 (x^2 + 14x + 49) + (y^2 - 14y + 49) = 28 + 49 + 49 (x + 7)^2 + (y - 7)^2 = 126 Now, we see that the equation represents a circle centered at (-7, 7) with a radius of √126. Step 3: Calculate the area The area enclosed by the circle is simply the area of a circle minus the areas of the four sectors cut off by the coordinate axes: Area = πr^2 - 4 * (1/4) * πr^2 Area = π * √126^2 - 4 * (1/4) * π * √126^2 Area = 126π - 31.5π Area = 94.5π square units Dec 17, 2023
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 7.2: Solving Linear Systems by Substitution Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Solve systems of equations with two variables by substituting for either variable. • Manipulate standard form equations to isolate a single variable. • Solve real-world problems using systems of equations. • Solve mixture problems using systems of equations. ## Introduction In this lesson, we will learn to solve a system of two equations using the method of substitution. ## Solving Linear Systems Using Substitution of Variable Expressions Let’s look again at the problem involving Peter and Nadia racing. Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter? In that example, we came up with two equations. \begin{align}& \text{Nadia’s equation} && d = 6t \\ & \text{Peter’s equation} && d = 5t + 20\end{align} We have seen that each relationship produces its own line on a graph, but that to solve the system we find the point at which the lines intersect (Lesson 1). At that point the values for \begin{align}d\end{align} and \begin{align}t\end{align} satisfy both relationships. In this simple example, this means that the \begin{align}d\end{align} in Nadia’s equation is the same as the \begin{align}d\end{align} in Peter’s. We can set the two equations equal to each other to solve for \begin{align}t\end{align}. \begin{align}6t & = 5t + 20 && \text{Subtract}\ 5t\ \text{from both sides.}\\ t & = 20 && \text{Substitute this value for}\ t\ \text{into Nadia’s equation.}\\ d & = 6\cdot 20 = 120\end{align} Even if the equations are not so obvious, we can use simple algebraic manipulation to find an expression for one variable in terms of the other. We can rearrange Peter’s equation to isolate \begin{align}t\end{align}. \begin{align} d & = 5t + 20 && \text{Subtract}\ 20\ \text{from both sides}.\\ d - 20 & = 5t && \text{Divide by}\ 5.\\ \frac{d - 20}{5} & = t \end{align} We can now substitute this expression for \begin{align}t\end{align} into Nadia’s equation \begin{align}(d = 6t)\end{align} to solve it. \begin{align} d & = 6 \left (\frac {d-20}{5} \right ) && \text{Multiply both sides by}\ 5.\\ 5d & = 6(d - 20) && \text{Distribute the}\ 6.\\ 5d & = 6d - 120 && \text{Subtract}\ 6d\ \text{from both sides.}\\ -d & = -120 && \text{Divide by}\ -1.\\ d & = 120 && \text{Substitute value for}\ d\ \text{into our expression for}\ t.\\ t & = \frac {120 - 20}{5}=\frac {100}{5}=20\end{align} We find that Nadia and Peter meet 20 seconds after they start racing, at a distance of 120 yards away. The method we just used is called the Substitution Method. In this lesson, you will learn several techniques for isolating variables in a system of equations, and for using the expression you get for solving systems of equations that describe situations like this one. Example 1 Let us look at an example where the equations are written in standard form. Solve the system \begin{align}2x+3y & = 6 \\ -4x + y & = 2\end{align} Again, we start by looking to isolate one variable in either equation. If you look at the second equation, you should see that the coefficient of \begin{align}y\end{align} is 1. It makes sense to use this equation to solve for \begin{align}y\end{align}. Solve the second equation for the \begin{align}y\end{align} variable: \begin{align}-4x + y & = 2 && \text{Add}\ 4x\ \text{to both sides}.\\ y & = 2 + 4x \end{align} Substitute this expression into the second equation. \begin{align}2x + 3(2 + 4x) & = 6 && \text{Distribute the}\ 3.\\ 2x+6+12x & = 6 && \text{Collect like terms.}\\ 14x + 6 & = 6 && \text{Subtract}\ 6\ \text{from both sides.}\\ 14x & = 0 \\ x & = 0\end{align} Substitute back into our expression for \begin{align}y\end{align}. \begin{align}y= 2 + 4\cdot 0 = 2 \end{align} As you can see, we end up with the same solution \begin{align}(x = 0, y = 2)\end{align} that we found when we graphed these functions (Lesson 7.1). As long as you are careful with the algebra, the substitution method can be a very efficient way to solve systems. Next, consider a more complicated example. In the following example the solution gives fractional answers for both \begin{align}x\end{align} and \begin{align}y\end{align}, and so would be very difficult to solve by graphing alone! Example 2 Solve the system \begin{align}2x + 3y & = 3 \\ 2x - 3y & = -1\end{align} Again, we start by looking to isolate one variable in either equation. Right now it doesn’t matter which equation we use or which variable we solve for. Solve the first equation for \begin{align}x\end{align} \begin{align}2x+ 3y & = 3 && \text{Subtract}\ 3y\ \text{from both sides.}\\ 2x & = 3 - 3y && \text{Divide both sides by} \ 2.\\ x & = \frac{3-3y}{2}\end{align} Substitute this expression into the second equation. \begin{align}\cancel{2}. \frac{1} {\cancel{2}} (3 - 3y) - 3y & = -1 && \text{Cancel the fraction and rewrite terms.}\\ 3 - 3y - 3y & = -1 && \text{Collect like terms.}\\ 3 - 6y & = -1 && \text{Subtract}\ 3\ \text{from both sides.}\\ -6y & = -4 && \text{Divide by}\ -6.\\ y & = \frac{2}{3}\end{align} Substitute into the expression and solve for \begin{align}x\end{align}. \begin{align}x & = \frac{1} {2} \left (3 - \cancel{3} \frac{2} {\cancel{3}}\right )\\ x & = \frac {1}{2}\end{align} So our solution is, \begin{align} x = \frac {1}{2}, y = \frac {2}{3}\end{align}. You can see why the graphical solution \begin{align} \left(\frac {1}{2}, \frac {2}{3} \right )\end{align} might be difficult to read accurately. ## Solving Real-World Problems Using Linear Systems There are many situations where we can use simultaneous equations to help solve real-world problems. We may be considering a purchase. For example, trying to decide whether it is cheaper to buy an item online where you pay shipping or at the store where you do not. Or you may wish to join a CD music club, but do not know if you would really save any money by buying a new CD every month in that way. One example with which we are all familiar is considering phone contracts. Let’s look at an example of that now. Example 3 Anne is trying to choose between two phone plans. The first plan, with Vendafone costs $20 per month, with calls costing an additional 25 cents per minute. The second company, Sellnet, charges$40 per month, but calls cost only 8 cents per minute. Which should she choose? Anne’s choice will depend upon how many minutes of calls she expects to use each month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the number of minutes is the independent variable, it will be our \begin{align}x\end{align}. Cost is dependent on minutes. The cost per month is the dependent variable and will be assigned \begin{align}y\end{align}. \begin{align}& \text{For Vendafone} && y = 0.25x + 20\\ & \text{For Sellnet} && y = 0.08x + 40\end{align} By writing the equations in slope-intercept form \begin{align}(y = mx + b)\end{align} you can visualize the situation in a simple sketched graph, shown right. The line for Vendafone has an intercept of 20 and a slope of 0.25. The Sellnet line has an intercept of 40 and a slope of 0.08 (which is roughly a third of the Vendafone line). In order to help Anne decide which to choose, we will determine where the two lines cross, by solving the two equations as a system. Since equation one gives us an expression for \begin{align}y\end{align} \begin{align}(0.25x + 20)\end{align}, we can substitute this expression directly into equation two. \begin{align}0.25x + 20 & = 0.08x + 40 && \text{Subtract}\ 20\ \text{from both sides.}\\ 0.25x & = 0.08x + 20 && \text{Subtract}\ 0.08x\ \text{from both sides.}\\ 0.17x & = 20 && \text{Divide both sides by}\ 0.17.\\ x & = 117.65\ minutes && \text{Rounded to two decimal places.}\end{align} We can now use our sketch, plus this information to provide an answer: If Anne will use 117 minutes or less every month, she should choose Vendafone. If she plans on using 118 or more minutes, she should choose Sellnet. ## Mixture Problems Systems of equations crop up frequently when considering chemicals in solutions, and can even be seen in things like mixing nuts and raisins or examining the change in your pocket! Let’s look at some examples of these. Example 4 Nadia empties her purse and finds that it contains only nickels (worth 5 cents each) and dimes (worth 10 cents each). If she has a total of 7 coins and they have a combined value of 55 cents, how many of each coin does she have? Since we have two types of coins, let’s call the number of nickels \begin{align}x\end{align} and the number of dimes will be our \begin{align}y\end{align}. We are given two key pieces of information to make our equations, the number of coins and their value. \begin{align}& \text{Number of coins equation} && x + y = 7 && \text{(number of nickels)} + \text{(number of dimes)}\\ & \text{The value equation} && 5x + 10y = 55 && \text{Since nickels are worth five cents and dimes ten cents}\end{align} We can quickly rearrange the first equation to isolate \begin{align}x\end{align}. \begin{align}x & = 7 -y && \text{Now substitute into equation two}.\\ 5(7 -y) + 10y & = 55 && \text{Distribute the}\ 5.\\ 35 - 5y + 10y & = 55 && \text{Collect like terms.}\\ 35 + 5y & = 55 && \text{Subtract}\ 35\ \text{from both sides.}\\ 5y & = 20 && \text{Divide by}\ 5.\\ y & = 4 && \text{Substitute back into equation one.}\\ + 4 & = 7 && \text{Subtract}\ 4\ \text{from both sides}.\\ x & =3\end{align} Solution Nadia has 3 nickels and 4 dimes. Sometimes the question asks you to determine (from concentrations) how much of a particular substance to use. The substance in question could be something like coins as above, or it could be a chemical in solution, or even heat. In such a case, you need to know the amount of whatever substance is in each part. There are several common situations where to get one equation you simply add two given quantities, but to get the second equation you need to use a product. Three examples are below. Type of Mixture First Equation Second Equation Coins (items with value) Total number of items \begin{align}(n_1 \ n_2)\end{align} Total value (item value \begin{align}\times\end{align} no. of items) Chemical solutions Total solution volume \begin{align}(V_1 + V_2)\end{align} Amount of solute (vol \begin{align}\times\end{align} concentration) Density of two substances Total amount or volume of mix Total mass (volume \begin{align}\times\end{align} density) For example, when considering mixing chemical solutions, we will most likely need to consider the total amount of solute in the individual parts and in the final mixture. A solute is simply the chemical that is dissolved in a solution. An example of a solute is salt when added to water to make a brine. Even if the chemical is more exotic, we are still interested in the total amount of that chemical in each part. To find this, simply multiply the amount of the mixture by the fractional concentration. To illustrate, let’s look at an example where you are given amounts relative to the whole. Example 5 A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. In order to do this, he wishes to use a high concentration solution (60%) and dilute it with a low concentration solution (5%). How much of each solution should he use? To set this problem up, we first need to define our variables. Our unknowns are the amount of concentrated solution \begin{align}(x)\end{align} and the amount of dilute solution \begin{align}(y)\end{align}. We will also convert the percentages (60%, 15% and 5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information we need is the final volume (500 ml) and the final amount of solute (15% of \begin{align}500 \ ml = 75 \ ml\end{align}). Our equations will look like this. \begin{align}\text{Volume equation} && x + y &= 500 \\ \text{Solute equation} && 0.6x + 0.05y &= 75\end{align} You should see that to isolate a variable for substitution it would be easier to start with equation one. \begin{align}x + y & = 500 && \text{Subtract}\ y\ \text{from both sides.}\\ x & = 500 -y && \text{Now substitute into equation two.}\\ 0.6(500 - y) + 0.05y & = 75 && \text{Distribute the}\ 6.\\ 300-0.6y + 0.05y & = 75 && \text{Collect like terms.}\\ 300 - 0.55y & = 75 && \text{Subtract}\ 300\ \text{from both sides.}\\ -0.55y & = -225 && \text{Divide both sides by}\ -0.55.\\ y & = 409\ ml && \text{Substitute back into equation for}\ x.\\ x & = 500 - 409 = 91\ ml\end{align} Solution The chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution. ## Review Questions 1. Solve the system: \begin{align}x + 2y = 9\!\\ 3x + 5y= 20\end{align} 2. solve the system. \begin{align}x -3y = 10\!\\ 2x + y = 13\end{align} 3. Of the two non-right angles in a right angled triangle, one measures twice that of the other. What are the angles? 4. The sum of two numbers is 70. They differ by 11. What are the numbers? 5. A rectangular field is enclosed by a fence on three sides and a wall on the fourth side. The total length of the fence is 320 yards. If the field has a total perimeter of 400 yards, what are the dimensions of the field? 6. A ray cuts a line forming two angles. The difference between the two angles is \begin{align}18^{\circ}\end{align}. What does each angle measure? 7. I have15 and wish to buy five pounds of mixed nuts for a party. Peanuts cost $2.20 per pound. Cashews cost$4.70 per pound. How many pounds of each should I buy? 8. A chemistry experiment calls for one liter of sulfuric acid at a 15% concentration, but the supply room only stocks sulfuric acid in concentrations of 10% and in 35%. How many liters of each should be mixed to give the acid needed for the experiment? 9. Bachelle wants to know the density of her bracelet, which is a mix of gold and silver. Density is total mass divided by total volume. The density of gold is 19.3 g/cc and the density of silver is 10.5 g/cc. The jeweler told her that the volume of silver used was 10 cc and the volume of gold used was 20 cc. Find the combined density of her bracelet. 1. \begin{align}x=-5, y=7\end{align} 2. \begin{align}x= 7, y = -1\end{align} 3. \begin{align}x = 30^{\circ}, y = 60^{\circ}\end{align} 4. 29.5 and 40.5 5. \begin{align}x = 120 \ yards, y = 80 \ yards\end{align} 6. \begin{align}x = 81^{\circ}, y = 99^{\circ}\end{align} 7. 3.4 pounds of peanuts, 1.6 pounds of cashews 8. 0.8 liters of 10%, 0.2 liters of 35% 9. 16.4 g/cc ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
Survey Thank you for your participation! * Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project Document related concepts no text concepts found Transcript ```MCT4C1 Special Angles and the CAST RULE Many realworld applications of trigonometry deal with angle measurements of 30, 45, and 60 degrees in a right triangle. ie) Construction. Because we see these angles so much it is worth memorizing their EXACT trigonometric values! Special Triangles: To help learn these values, try memorizing the following special triangles! 60 45 2 1 1 45 30 1 Example: Using the special triangles provide EXACT answers for the following. θ 0 30 45 60 90 sinθ cosθ tanθ Related Angles: y Terminal Arm Rotational Angle θ Related/Reference Angle x Rotational Angle θ The angle formed between the positive xaxis, counterclockwise to the terminal arm. Related/Reference Angle the acute angle between the teminal arm and the xaxis. Example: Sketch the following angles in standard position, and state the corresponding RELATED angle. o a) 240 c) 330 o b) 135 o o d) 210 CAST rule: The CAST rule is used to determine the "sign" of a trigonometric ratio, depending on the location of the terminal arm. Each letter determines the POSITIVE ratio. y S (sine) A (all) x T (tan) C (cos) Evaluating Trig Ratios using RELATED angles and the CAST rule: Since Trigonometric Functions are PERIODIC, the same trig ratios, repeat at regular intervals!!! Because of this we can use RELATED angles to evaluate trigonometric ratios larger than 90 degrees. Steps: Example: Find the exact value of sin210o 1.) Rewrite the trig ratio using the RELATED angle. 1.) sin210o = sin30o 2.) sin 30o = 2.) Using the special triangles, state the EXACT value. 1 2 3.) Since 210o is in Quadrant III, 3.) Use the CAST rule to determine the sign. (sine is Negative) o ∴ sin210 = 1 2 Example: Find the EXACT Value of each trigonometric function. a) cos300o b) tan120o c) sin330o d) tan225o Homework: Worksheet and p. 2 3 # 4 6, 8 12 ``` Related documents
# Question Video: Selecting a Data Set with the Given Mode and Median Mathematics • 6th Grade Which of the following sets of data has a mode of 48 and a median of 20. [A] 48, 21, 11, 48, 20, 17 [B] 21, 48, 19, 48, 17, 11 [C] 47, 47, 11, 48, 20, 17 [D] 10, 16, 19, 21, 47, 47 [E] 20, 48, 48, 11, 11, 19 03:16 ### Video Transcript Which of the following sets of data has a mode of 48 and a median of 20. Is it A) 48, 21, 11, 48, 20, 17? B) 21, 48, 19, 48, 17, 11? C) 47, 47, 11, 48, 20, 17? D) 10, 16, 19, 21, 47, 47? Or E) 20, 48, 48, 11, 11, 19? We recall that the mode is the most frequently occurring value. Therefore, we need to find a set of data where 48 is the most common or most frequently occurring number. 48 does not occur in set D. Therefore, this cannot be the correct answer. Whilst there is a 48 in set C, there are two 47s. Therefore, the mode of set C is 47. We can, therefore, rule this out as the correct answer. Set E has two 48s, but it also has two 11s. This means that it has two modes, or it’s bimodal. It has a mode of 11 and 48. This means that option E is also incorrect. Both option A and option B have two 48s. This is the most frequently occurring value in both data sets. This means that the mode of both of these is 48. Let’s now consider our second piece of information. The median of the data set has to be 20. We know that the median is the middle value once the numbers are in ascending or descending order. Once we have put both of these data sets in order, we notice that they have an even number of values, in this case, six. This means that there are two middle members, in set A, 20 and 21 and in set B, 19 and 21. The median can be calculated by finding the mean of these two values. This is the same as finding the midpoint of the two values. The mean of 20 and 21 is 20.5. And the mean of 19 and 21 is 20. This means that set A has a median of 20.5 and set B has a median of 20. We can, therefore, rule out set A. The set of data that has a mode of 48 and a median of 20 is set B, 21, 48, 19, 48, 17, and 11.
# 1.15 Rectilinear motion (Page 2/6) Page 2 / 6 This attribute of rectilinear motion allows us to do away with the need to use vector notation and vector algebra for quantities with directional attributes like position vector, displacement and velocity. Instead, the vectors are treated simply as scalars with one qualification that vectors in the direction of chosen reference is considered positive and vectors in the opposite direction to the chosen reference is considered negative. The most important aspect of the sign convention is that a vector like velocity can be expressed by a scalar value say, 5 m/s. Though stated without any aid for specifying direction like using unit vector, the direction of the velocity is indicated, which is in the positive x-direction. If the velocity of motion is -5 m/s, then the velocity is in the direction opposite to the direction of reference. To illustrate the construct, let us consider a motion of a ball which transverses from O to A to B to C to O along x-axis as shown in the figure. The velocities at various points of motion in m/s (vector form) are : $\begin{array}{l}{\mathbf{v}}_{\mathbf{O}}=2\mathbf{i};\phantom{\rule{4pt}{0ex}}{\mathbf{v}}_{\mathbf{A}}=3\mathbf{i};\phantom{\rule{4pt}{0ex}}{\mathbf{v}}_{\mathbf{B}}=-4\mathbf{i};\phantom{\rule{4pt}{0ex}}{\mathbf{v}}_{\mathbf{C}}=3\mathbf{i}\end{array}$ Going by the scalar construct, we can altogether drop use of unit vector like " i " in describing all vector quantities used to describe motion in one dimension. The velocities at various points of motion in m/s (in equivalent scalar form) can be simply stated in scalar values for rectilinear motion as : $\begin{array}{l}{v}_{O}=2;\phantom{\rule{4pt}{0ex}}{v}_{A}=3;\phantom{\rule{4pt}{0ex}}{v}_{B}=-4;\phantom{\rule{4pt}{0ex}}{v}_{C}=3\end{array}$ Similarly, we can represent position vector simply by the component in one direction, say x, in meters, as : $\begin{array}{l}{x}_{O}=0;\phantom{\rule{4pt}{0ex}}{x}_{A}=5;\phantom{\rule{4pt}{0ex}}{x}_{B}=10;\phantom{\rule{4pt}{0ex}}{x}_{C}=-5\end{array}$ Also, the displacement vector (in meters) is represented with scalar symbol and value as : $\begin{array}{l}\mathrm{OA}=5;\phantom{\rule{4pt}{0ex}}\mathrm{OB}=10;\phantom{\rule{4pt}{0ex}}\mathrm{OC}=5\end{array}$ Following the same convention, we can proceed to write defining equations of speed and velocity in rectilinear motion as : $\begin{array}{l}\|v\|=\|\frac{đx}{đt}\|\end{array}$ and $\begin{array}{l}v=\frac{đx}{đt}\end{array}$ ## Rectilinear motion Problem : If the position of a particle along x – axis varies in time as : $\begin{array}{l}x=2{t}^{2}-3t+1\end{array}$ Then : • What is the velocity at t = 0 ? • When does velocity become zero? • What is the velocity at the origin ? • Plot position – time plot. Discuss the plot to support the results obtained for the questions above. Solution : We first need to find out an expression for velocity by differentiating the given function of position with respect to time as : $\begin{array}{l}v=\frac{đ}{đt}\left(2{t}^{2}-3t+1\right)=4t-3\end{array}$ (i) The velocity at t = 0, $\begin{array}{l}v=4x0-3=-3\mathrm{m/s}\end{array}$ (ii) When velocity becomes zero : For v = 0, $\begin{array}{l}4t-3=0\\ ⇒t=\frac{3}{4}=\mathrm{0.75 s.}\end{array}$ (iii) The velocity at the origin : At origin, x = 0, $\begin{array}{l}x=2{\mathbf{t}}^{2}-3t+1=0\\ ⇒2{\mathbf{t}}^{2}-2t-t+1=0\\ ⇒2t\left(t-1\right)-\left(t-1\right)=0\\ ⇒t=\mathrm{0.5 s, 1 s.}\end{array}$ This means that particle is twice at the origin at t = 0.5 s and t = 1 s. Now, $\begin{array}{l}v\left(t=\mathrm{0.5 s}\right)=4t-3=4x0.5-3=\mathrm{-1 m/s.}\end{array}$ Negative sign indicates that velocity is directed in the negative x – direction. $\begin{array}{l}v\left(t=\mathrm{1 s}\right)=4t-3=4x1-3=\mathrm{1 m/s.}\end{array}$ We observe that slope of the curve from t = 0 s to t<0.75 s is negative, zero for t = 0.75 and positive for t>0.75 s. The velocity at t = 0, thus, is negative. We can realize here that the slope of the tangent to the curve at t = 0.75 is zero. Hence, velocity is zero at t = 0.75 s. The particle arrives at x = 0 for t = 0.5 s and t = 1 s. The velocity at first arrival is negative as the position falls on the part of the curve having negative slope, whereas the velocity at second arrival is positive as the position falls on the part of the curve having positive slope. pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them what are scalars show that 1w= 10^7ergs^-1 what's lamin's theorems and it's mathematics representative if the wavelength is double,what is the frequency of the wave What are the system of units A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained 58asagravitasnal firce Amar water boil at 100 and why what is upper limit of speed what temperature is 0 k Riya 0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale Mustapha How MKS system is the subset of SI system? which colour has the shortest wavelength in the white light spectrum if x=a-b, a=5.8cm b=3.22 cm find percentage error in x x=5.8-3.22 x=2.58 what is the definition of resolution of forces
# Place Value Pursuit: The Quest for Decimal Dominance Greetings, math educators and enthusiasts! In the pursuit of mathematical mastery, an understanding of decimal numbers and their place values is essential. Today, we bring to you an innovative, engaging activity that turns learning about decimals into a thrilling quest: “Place Value Pursuit: The Quest for Decimal Dominance”. ## Aims and Objectives This dynamic card-based game encourages students to delve into: • Enhancing logical reasoning and decision-making skills. • Acquiring proficiency in comparing, contrasting, and ordering decimal numbers. • Developing a deep understanding of the place value system, particularly its application to decimal numbers. ## Gathering the Tools for Our Quest To facilitate a learning environment that is both inclusive and stimulating, this game utilizes readily available classroom resources: • A standard deck of playing cards (retain only the face cards – Jack, Queen, King, Ace). • A whiteboard or a piece of paper for each participant. • A coin for flipping. • An optional scorekeeping sheet which can be drawn on the whiteboard or paper. ## The Journey Begins: Gameplay Instructions Step 1: Map the Territory First, assign numerical values to the face cards as follows: Jack (11), Queen (12), King (13), Ace (14). This elevates the gameplay with higher numbers, requiring more strategic decision-making. Step 2: Divide the Deck Shuffle the deck thoroughly and distribute the cards equally among the players. Step 3: Set the Challenge Flip a coin. If it shows heads, the players strive to create the largest possible decimal number. If it lands on tails, they aim for the smallest decimal number. Step 4: Deal the Cards Players, in turns, deal out one card to each participant. Step 5: Construct the Decimal Each player records the value of their card in one of the five positions meant for creating a decimal number (__ __ . __ __). Importantly, once a number is written down, it cannot be moved or replaced, encouraging strategic thinking. Step 6: The Great Reveal After all five positions are filled, each player reads their decimal number aloud. The player who fulfills the round’s objective (creating the largest or smallest decimal number) wins the round and collects the played cards. Step 7: Crowning the Decimal Champion Continue in this manner until all cards have been played. The player who has amassed the most cards by the end is hailed as the Decimal Champion! ## Gameplay Scenarios To provide a clearer understanding, let’s visualize a couple of scenarios: Scenario 1: The coin lands on heads (make the largest number). Player A draws a King (13) and decides to place it in the tens position. Player B receives a Queen (12) and follows the same strategy. As the round unfolds, players continue to make calculated choices to build the largest possible decimal number. Scenario 2: The coin shows tails (make the smallest number). Both players receive a Jack (11). However, Player A chooses to place it in the ones position, while Player B selects the tenths position. This initial choice could drastically change the final outcome, showcasing the importance of strategic thinking in this game. ## Accommodations and Modifications “Place Value Pursuit” can easily be adapted to accommodate diverse learning needs and abilities: • For students who find decimals challenging, start with a simpler version of the game focusing only on the ones and tens place (i.e., __ . __). Gradually introduce tenths and hundredths as they become more comfortable. • Visual aids such as base-ten blocks or place value charts can be provided to support visual learners. • To increase the difficulty for advanced students, consider adding a thousandths place or introducing negative numbers. • For promoting cooperative learning, allow students to form pairs or small groups, facilitating discussions and collaborative decision-making. ## Aligning with Common Core State Standards (CCSS) This game is in sync with key math standards defined under CCSS: • 4.NF.C.7 – Students compare two decimals to hundredths by reasoning about their size. • 5.NBT.A.3 – Students read, write, and compare decimals to thousandths. • 5.NBT.A.4 – Students use place value understanding to round decimals to any place. To wrap up, “Place Value Pursuit: The Quest for Decimal Dominance” is a unique, engaging way to make learning decimals fun. It promotes active learning, encourages strategic thinking, and increases mathematical confidence. So let’s gather our deck of cards and embark on this exciting decimal adventure. Happy Teaching! ## Transform Your Child into a Math Genius Overnight with the Ultimate Place Value Learning Bundle – Guaranteed Fun & Success! Master the world of place value with the Ultimate Focus on Place Value Bundle! Perfect for kids in the 2nd to 6th grades, this bundle includes 28 stimulating games, assorted place value cards, and 132 decimal place value mazes/worksheets. Transform math struggles into victories by building a solid foundation of place value understanding. Enjoy long-term use with activities designed to grow with your child’s learning level over five years. Games like “Speed” Place Value Yahtzee and Place Value Battleship turn learning into exciting challenges. Don’t wait! Enhance your child’s math journey today. Click ‘Buy Now’ and make math fun with the Ultimate Focus on Place Value Bundle!
# Operations With Polynomials Worksheet Operations With Polynomials Worksheet. 2.1 Operations on polynomials key.pdf – Polynomial… You will see that on this case, the middle time period will disappear. Hopefully this video will assist you to determine what step to do next when you’re using polynomial lengthy division. In this part we will present examples of the means to use two completely different strategies to multiply to binomials. Also notice that you have introduced down all of the terms in the dividend, and that the quotient extends to the right edge of the dividend. These are other ways to verify whether or not you might have accomplished the problem. Bring down the relaxation of the expression within the dividend. It’s useful to bring down the entire remaining terms. Divide every term in the polynomial by the monomial. Combine like terms, paying close consideration to the signs. Now the diploma of the polynomial is given by the utmost of the above levels. The fixed values present in a polynomial are knows as its coefficients / coefficient values. The constants used in the above polynomial are 1, 5 and -3. There are frequent trends that apply to polynomials on a coordinate graph. An example is that the each of the ends of the graph will level up if the polynomial is even and the lead coefficient is constructive. [toc] ## Ultimate Math Solver Free Scholars should graph, clear up, and justify quadratic issues. Now let’s look at an instance of dividing a trinomial with more than one variable by a monomial with a couple of variable. This follows the identical procedure as when you have one variable, but you should take observe of distinguishing between the variables. Pay attention to the signs when subtracting. When there is not an identical like term for each time period in each polynomial, there will be empty places within the vertical association of the polynomials. This structure makes it straightforward to check that you are combining like terms only. In this Algebra I tutorial activity, ninth graders add, subtract, and multiply polynomials. Students factor polynomials and solve quadratic equations. The six page educational exercise contains thirty problems. So far, we now have shown two methods for multiplying two binomials collectively. They are one of the most nicely studied and extensively used polynomials, so there might be plenty of data on the market about them. Perform the division of each term by dividing the coefficients and dividing the variables by subtracting the exponents of variables with like bases. [newline]To multiply a monomial by a binomial, you use the distributive property in the same way as multiplying polynomials with one variable. In the following instance, you will see that the identical ideas apply when you are dividing a trinomial by a monomial. You can distribute the divisor to each term in the trinomial and simplify using the rules for exponents. ### Math Worksheets By Subject Regroup to match like phrases, remember to examine the signal of every term. Distribute −1[/latex] to each time period in the polynomial. Write one polynomial beneath the other, lining up like phrases vertically. The subtraction of polynomials is as simple as the addition of polynomials. Using columns would help us to match the right terms together in a complicated subtraction. While subtracting polynomials, separate the like terms and simply subtract them. Keep two rules in mind whereas performing the subtraction of polynomials. A polynomial is made out of a quantity of terms. Term is a smaller expression consisting of variables and coefficients sure with multiplication. ## Math Worksheets Click any of the pictures or words beneath to see all the worksheets within that subtopic. Here we take away parenthesis by altering the sign of every time period within the second bracket. Input polynomials $p$ and $q$ and choose an operation to be performed. Since 6 has a variable with exponent zero, and our divisor has exponent 1, we can’t divide anymore. The most necessary factor is to be careful for exponents, all the time put them in order from best exponent to lowest. This will stop you from making errors, and you don’t have to think much about what is dividing what. • That means you can solely combine squared variable phrases with squared variable phrases, cubed variable terms with cubed variable phrases, and so forth. • The other methodology was to put the polynomial being subtracted beneath the other after altering the signs of every time period. • Understanding polynomial merchandise is a vital step in studying to resolve algebraic equations involving polynomials. • For those of you that use footage to study, you can draw an space model to help make sense of the method. In the previous example, we noticed the outcomes of squaring a binomial that was a sum of two phrases. In the subsequent example we will find the product of squaring a binomial that is the distinction of two phrases. Students will apply including and subtracting, multiplying, and dividing polynomials. By following these steps we are ready to clear up adding and subtracting polynomials. Complete interactive PDF worksheets online, with no printing! ### Follow Questions On Including Polynomials Always start by discovering that GCF to get this snowball a rolling. If it’s a trinomial, you will want to use the FOIL methodology. Each degree gets progressively harder. Level 3 problems are catered in the course of larger college students. The polynomial coefficients may be any actual numbers. These formulas can’t be simplified and also you just should learn them by heart. As we’ve all through the course, simplifying with exponents includes rewriting unfavorable exponents as constructive. Pay consideration to the indicators of the terms in the next example, we are going to divide by a negative monomial. So far we now have seen examples of binomials with variable terms on the left and fixed phrases on the best, such as this binomial \left(2r-3\right)[/latex]. Variables may be on the right of the fixed term, as on this binomial \left(5+r\right)[/latex]. In the following example, we’ll present that multiplying binomials on this type requires one extra step on the end. In the final part we completed with an instance of multiplying two binomials,\left(x+4\right)\left(x+2\right)[/latex]. When adding combine the likes, when subtracting just flip the indicators of the phrases and that pretty much sums it up. Worksheets that work with polynomial operations, polynomial word problems, and even factoring polynomials. The algebraic expressions having adverse or irrational power of the variable are not polynomials. Factor in numerous strategies in a lesson plan for factoring quadratics. [ssba-buttons] ## Related posts of "Operations With Polynomials Worksheet" Parent Function Worksheet Answers. Many mother and father select to allow Internet filtering or supervise their youngsters's on-line activities in an try to guard their children from inappropriate material on the Internet. If a person adjustments the name of a sheet and their macro stops working, that is on them. However, the World Wide Web... #### Solving Absolute Value Inequalities Worksheet Solving Absolute Value Inequalities Worksheet. Hard math divisions on-line, free trigonometry worksheets for dummies, integer guidelines practice sheets, how do you remedy (x+y)10th energy, Unknown Variables and Exponents. A quadratic with a term lacking is called an incomplete quadratic (as lengthy because the ax 2 time period isn't missing). To remedy an absolute value inequality,... #### Soh Cah Toa Worksheet Soh Cah Toa Worksheet. 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# How do you figure out 70 percent off? ## How do you figure out 70 percent off? How to calculate percent off? 1. Divide the number by 100 (move the decimal place two places to the left). 2. Multiply this new number by the percentage you want to take off. 3. Subtract the number from step 2 from the original number. This is your percent off number. ### How do you work out 70% of 200? Working out 70% of 200 If you are using a calculator, simply enter 70÷100×200 which will give you 140 as the answer. #### What is the best way to calculate percentage? Generally, the way to figure out any percentage is to multiply the number of items in question, or X, by the decimal form of the percent. To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1. How do you work out 70 of 250? Percentage Calculator: What is 70 percent of 250.? = 175. What is the meaning of up to 70% off? A sale up to a 70% discount means that there may be multiple sales, some being 70% and lower. So there could be 10%, 20%, 30%, 40%, 50%, 60%, and 70% discount, but no higher. So they could have any sort of sale that is 70% and under. See a translation. 1 like. ## What number is 70 percent of 500? Percentage Calculator: What is 70 percent of 500? = 350. ### How do you work out 70 of 300? Working out 70% of 300 1. Write 70% as 70100. 2. 70100 of 300 = 70100 × 300. 3. Therefore, the answer is 210. If you are using a calculator, simply enter 70÷100×300 which will give you 210 as the answer. #### How do you calculate percentages online on a calculator? To calculate a percentage of a percentage, convert both percentages to fractions of 100, or to decimals, and multiply them. For example, 50% of 40% is calculated; (50/100) x (40/100) = 0.50 x 0.40 = 0.20 = 20/100 = 20%. How do I calculate what percentage one number is of another? Answer: To find the percentage of a number between two numbers, divide one number with the other and then multiply the result by 100. Let us see an example of finding the percentage of a number between two numbers. How do you find 60 percent of a number? You have learned that to find 1% of a number means finding 1/100 of it. Similarly, finding 60% of a number means finding 60/100 (or 6/10) of it. 60% of \$700 → 60% × \$700. 60% of \$700 → 0.6 × \$700.
# What are the points of inflection, if any, of f(x)=e^(2x) - e^x ? Dec 24, 2015 Crap. #### Explanation: Was utter crap so forget I said anything. Dec 24, 2015 There is an inflection point at $x = - 2 \ln \left(2\right)$ #### Explanation: To find inflection points, we apply the second derivative test. $f \left(x\right) = {e}^{2 x} - {e}^{x}$ $f ' \left(x\right) = 2 {e}^{2 x} - {e}^{x}$ $f ' ' \left(x\right) = 4 {e}^{2 x} - {e}^{x}$ We apply the second derivative test by setting $f ' ' \left(x\right)$ equal to $0$. $4 {e}^{2 x} - {e}^{x} = 0$ $4 {e}^{2 x} = {e}^{x}$ $\ln \left(4 {e}^{2 x}\right) = \ln \left({e}^{x}\right)$ One property of logarithms is that terms being multiplied in a single logarithm can be turned into a sum of logarithms for each term: $\ln \left(4 {e}^{2 x}\right) = \ln \left({e}^{x}\right)$ $\ln \left(4\right) + \ln \left({e}^{2 x}\right) = \ln \left({e}^{x}\right)$ $\ln \left(4\right) + 2 x = x$ $x = - \ln \left(4\right)$ $x = - \ln \left({2}^{2}\right)$ $x = - 2 \ln \left(2\right) \approx - 1.3863 \ldots$ Although you usually don't see inflection points with exponentials, the fact that one is being subtracted from the other means that there is the possibility of them "affecting" the graph in ways that offers the possibility for an inflection point. graph{e^(2x) - e^(x) [-4.278, 1.88, -1.63, 1.447]} graph: $f \left(x\right) = {e}^{2 x} - {e}^{x}$ You can see that the portion of the line left of the point appears to be concave down, whereas the portion to the right changes and becomes concave up.
# Solving Quadratic Equations by Completing the Square Created by Anna Szczepanek, PhD Reviewed by Rijk de Wet Last updated: Nov 15, 2023 Here, we discuss how you can quickly and easily solve a quadratic equation with the method of completing the square. Examples of solutions are included! ## What is completing the square? We've already introduced this method in our completing the square calculator, but it's always good to have a quick refresher, isn't it? OK. Completing the square is a method in mathematics (in algebra, to be precise) that we use to solve quadratic equations (or equivalently, to factor quadratic trinomials). It's an alternative method to using the quadratic formula calculator. The goal is to obtain a perfect square (this notion is explained in depth in our perfect square trinomial calculator) on the left side of the equation (where we have the unknown $x$) and a number on the right side. That is, we transform the equation $ax^2 +bx + c = 0$ to $a(x+d)^2 = e$ and further to $(x+d)^2 = e/a$ where $a$, $b$, $c$, $d$, and $e$ are real coefficients. Next, we take a look at the right-hand side: • If $e/a \ge 0$ (i.e., $e/a$ is non-negative), we simplify and solve the equation by taking the square root of both sides. • If $e/a < 0$ (i.e., $e/a$ is negative), then our equation has no solutions. ## What is a quadratic equation? In math, a quadratic equation is an equation where a value of $x$ is desired so that a quadratic polynomial (a polynomial of degree $2$) is equal to zero: $ax^2+bx+c=0$ where $a$, $b$, and $c$ are real coefficients. If the right side is not zero (i.e. $ax^2+bx+c = w$ and $w \ne 0$), you can always transfer to the left side to get the form given above: $ax^2 + bx + (c-w) = 0$ ## Why 'complete the square'? This method is called "complete the square" because we are hunting for perfect square trinomials. Formally, we want to transform the expression $x^2+bx+c$ so as to obtain $(x+d)^2$, which is a trinomial that arises from squaring a linear binomial $x+d$. We can then apply the square root to both sides of the equation to solve the initial equation. ## Completing the square formula In order to solve a quadratic equation by completing the square, follow these steps: 1. If the leading coefficient of your quadratic equation is not $1$ (i.e., if the polynomial is not monic), then divide both sides by $a$. 2. Assume we have the expression $x^2 + bx + c = 0$. Observe that $(x+b/2)^2 = x^2 + bx +b^2/4$. The first two terms are the same, but the last terms differ. 3. To get from $c$ to $b^2/4$, we have to subtract $c$ and add $b^2/4$. Observe that $(x^2 + bx + c) - c + b^2/4 = (x+b/2)^2$. 4. We have to perform the same operation on the right side! Finally, our equation is equivalent to $(x+b/2)^2 = -c + b^2/4$. 5. The next step depends on the sign of the right-hand side: • If $b^2/4 > c$, then there's no solution. • If $b^2/4 = c$, then we have one solution, and it is equal to $-b/2$. • If $b^2/4 < c$, then there are two solutions and you can find them by taking the square root of both sides. ## How do you know when to apply complete the square formula? Completing the square is a method of solving quadratic equations that always works β even if the coefficients are irrational or if the equation does not have real roots! It's up to you to decide whether you want to deal with a given quadratic expression by using the quadratic formula, or by the method of completing the square. There are many quadratic equations for which the latter is much faster and more elegant β you just need to gain a bit of experience to be able to quickly choose the best method. ## Examples of completing the square Let's discuss a few examples of solving quadratic equations by completing the square. Example 1. Solve by completing the square: $x^2 + 4x + 4 = 0$. We immediately recognize the short multiplication formula working in reverse: $(x+2)^2 =x^2 + 4x + 4$. Thus, our problem can be rewritten as $(x+2)^2 = 0$. Since $\sqrt{0} = 0$, we get $x+2=0$ and therefore $x = -2$. In fact, in this example we didn't have to complete the square, because the perfect square trinomial was already there, staring at us defiantly! Example 2. Solve using the completing the square method: $x^2 + 6x + 5 = 0$. Let's take a look at the part containing the unknown $x$ we have $x^2 + 6x$. To produce these terms by squaring a linear binomial, we can use: $(x + 3)^2 = x^2 + 6x + 9$. As you can see, the third term doesn't agree with what we have in our equations, so we need to complete the square. We have $5$ in the original equation and $9$ in the perfect square. So let's add $4$ to both sides of the initial equation: $\begin{split} x^2 + 6x + 5 + 4 &= 0 + 4 \\ x^2 + 6x + 9 &= 4 \\ (x + 3)^2 &= 4 \\ \|x + 3\| &= 2 \\ x + 3 &= \pm 2 \\ \therefore\qquad x &= -1 \\ \text{ or } x &= -5 \\ \end{split}$ Example 3. Solve by completing the square: $x^2 - 2x + 4 = 0$. At the left side, we easily recognize $x^2 - 2x$ as part of the perfect square trinomial $x^2 - 2x + 1 = (x-1)^2$. However, we have $4$ in our equation while we need $1$. So let's subtract $3$ from both sides: $\begin{split} x^2 - 2x + 4 - 3 &= -3 \\ x^2 - 2x + 1 &= -3 \\ (x - 1)^2 &= -3 \\ \end{split}$ Ouch. The equation says that some number squared (represented by the left-hand side of the equation) should be equal to $-3$. The problem is, squaring always leads to non-negative numbers! Therefore, we can deduce that our equation has no solutions (in real numbers). Anna Szczepanek, PhD Related calculators We will solve the quadratic equation ax2 + bx + c = 0 a b c People also viewedβ¦ ### BMR - Harris-Benedict equation Harris-Benedict calculator uses one of the three most popular BMR formulas. Knowing your BMR (basal metabolic weight) may help you make important decisions about your diet and lifestyle. ### Ellipse perimeter Use the ellipse perimeter calculator to determine the circumference and area of an ellipse. ### Titration Use our titration calculator to determine the molarity of your solution. ### Triangle side angle Determine the sides of a triangle or the angles between them with our triangle side angle calculator.
Scale Invariance By exploring the concept of scale invariance, find the probability that a random piece of real data begins with a 1. Into the Exponential Distribution Get into the exponential distribution through an exploration of its pdf. PCDF When can a pdf and a cdf coincide? PDF Stage: 5 Challenge Level: Here is another excellent solution from Andrei at Tudor Vianu National College, Romania: We have the probability distribution: $$\rho(x) = {2\over 27}\big(6+x-x^2\big)$$ for $\{0 \leq x \leq 3\}.$ The mean of this distribution is $m$, where $$m=\int_0^3x\,\rho(x)\,dx$$ Thus $$m = {2\over 27}\int_0^3x\big(6+x-x^2\big)\, dx ={2\over 27}\int_0^3(6x+x^2-x^3)\,dx$$ hence $$m={2\over 27}\Big[3x^2 + x^3/3 - x^4/4\Big]_0^3 ={2\over 27}[27 +9 -81/4] = {7\over 6}$$ To calculate the median $t$ of the distribution I must have: $$\text{Prob}\{0\leq X\leq t\} = \text{Prob} \{t \leq X \leq 3\}$$ But the total probability is 1, so each of the two probabilities is 0.5. Then I have to solve the equation $$\int_0^t\rho(x)\,dx = {1\over 2}$$ which gives: $${2\over 27}\int_0^t\left(6+x-x^2\right)dx = {1\over 2}.$$ I obtain the following equation $$4t^3-6t^2-72t+81=0$$ which has the solutions: $$t_1 = {9\over 2},\ t_2 = {3\over 2}\left(-1-\sqrt 3\right),\ {3\over 2}\left(-1 + \sqrt 3\right).$$ The only solution in the interval $[0,3]$ is $t_3$ which is approximately $1.098$. In a discrete distribution the mode is the value that occurs most frequently. In a continuous distribution it is the value where the probability density function takes its maximum value. To find this maximum I shall calculate the derivative $\rho\prime(x)$ and then $\rho \prime\prime(x)$: $$\rho^{\prime}(x)={2\over 27}(1 - 2x)$$ $$\rho^{\prime\prime}(x) = {-4\over 27}.$$ As the second derivative is negative the function has indeed a maximum. The maximum value of $\rho$ occurs at $x = 1/2$ so the mode is $1/2$.
# Quick Answer: What Happens When You Take The Square Root Of A Square Root? ## Can you subtract a square root from a square root? We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. However, it is often possible to simplify radical expressions, and that may change the radicand. ## Can you square a square root? Note: Remember that addition and subtraction are opposite operations and multiplication and division are opposite operations? Turns out, squaring and taking the square root are opposite operations too! See why in this tutorial! ## What happens when you take the square root of a perfect square? Perfect Squares are the resultant of whole numbers being multiplied together. So, if you take the square root of a perfect square, you get a nice whole number as your answer, instead of a long scary decimal. Otherwise, you ‘ll get a decimal, and one that goes on forever at that. ## How do I multiply square roots? Correct answer: To multiply square roots, we multiply the numbers inside the radical. Any numbers outside the radical are also multiplied. ## What is the root square of 169? The square root of 169 is 13. You might be interested: FAQ: How To Put Square Root In Word 2016? ## Is 3 a square root? The process of multiplying a number times itself is called squaring. Numbers whose square roots are whole numbers, (or more accurately positive integers) are called perfect square numbers. List of Perfect Squares. NUMBER SQUARE SQUARE ROOT 3 9 1.732 4 16 2.000 5 25 2.236 6 36 2.449 ## IS 400 a perfect square? 400 is a perfect square. Because 20 * 20 = 400. ## How do you know if a square root is a perfect square? You can also tell if a number is a perfect square by finding its square roots. Finding the square root is the inverse (opposite) of squaring a number. If you find the square root of a number and it’s a whole integer, that tells you that the number is a perfect square. ## What is the square root of a square? (see ± shorthand). Although the principal square root of a positive number is only one of its two square roots, the designation “the square root ” is often used to refer to the principal square root. As decimal expansions. n truncated to 50 decimal places 10 3.16227766016837933199889354443271853371955513932521 ## What are the positive and negative square roots of 4? But the roots could be positive or negative or we can say there are always two roots for any given number. Hence, root 4 is equal to ±2 or +2 and -2 ( positive 2 and negative 2). You can also find square root on a calculator. Square Root From 1 to 50. Number Square Root Value 4 2 5 2.236 6 2.449 7 2.646
Dear Uncle Colin, I recently came across a problem in which I had to integrate $\cos^3(x)$. Somewhere in my mind, I recall that the thing to do is to make it into something involving $\cos(3x)$, but I couldn’t put the details together. Could you help? -- Not A Very Inspired Expression Rearranger Dear NAVIER, But of course! In fact, I can show you two ways to accomplish it: the cloggerish way I learnt, and a much nicer way. The cloggerish way The standard way is to say $\cos(3x) = \cos(2x+x) = \cos(2x)\cos(x) - \sin(2x)\sin(x)$. From there, you replace $\cos(2x)$ with $2\cos^2(x) -1$ and $\sin(2x)$ with $2\sin(x)\cos(x)$, leaving you with: $\cos(3x) = \left(2 \cos^2(x) -1\right)\cos(x) - 2\cos(x)\sin^2(x)$ We can replace $\sin^2(x)$ with $1 - \cos^2(x)$: $\cos(3x) = \left(2 \cos^2(x) -1\right)\cos(x) - 2\cos(x)\left(1-\cos^2(x)\right)$ Then expand everything: $\cos(3x) = 2\cos^3(x) - \cos(x) - 2\cos(x) + 2 \cos^3(x) = 4\cos^3(x) - 3\cos(x)$ With that identity in place, you can rearrange to get $\cos^3(x)$ on its own: $\cos^3(x) = \frac{1}{4}\left( \cos(3x) + 3\cos(x)\right)$. The nicer way In your formula book, assuming you’re doing a qualification that has such a thing, you’ll find a group of trig identities that are rarely, if ever, examined. These include $2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) = \cos(A) + \cos(B)$, and this is the one we’re interested in: it turns a product of cosines into a sum. For example, if you consider $2\cos(x)\cos(x)$, you find that if $A=2x$ and $B=0$, you get $\cos(2x) + 1$ – which rearranges, if you want it to, as $\cos(2x) = 2\cos^2(x) - 1$. In general, you can turn $\cos(P)\cos(Q)$ into $\frac 12 \left( \cos(P+Q) + \cos(P-Q) \right)$, which is extremely handy here. $\cos^3(x) = \left[\cos(x)\cos(x)\right] \cos(x)$, obviously, so it’s $\frac 12 \left( \cos(2x) + 1 \right) \cos(x)$, or $\frac{1}{2} \cos(2x)\cos(x) + \frac{1}{2} \cos(x)$. Applying the same rule to the first term gives $\cos^3(x) = \frac{1}{4} \left(\cos(3x) + \cos(x)\right) + \frac{1}{2}\cos(x) = \frac{1}{4} \left( \cos(3x) + 3 \cos(x) \right)$. The Ninja Way “Who let you in?” “Oh, of course. You don’t need letting in.” ”$\cos^3(x) = \left(1 - \sin^2(x)\right)\cos(x)$”, he said. “Or $\cos(x) - \sin^2(x) \cos(x)$.” “First term’s trivial. Second one’s function-derivative.” And he was gone.
Smartick is an online platform for children to master math in only 15 minutes a day Dec23 # Geometric Plane Shapes: Circles, Triangles, Rectangles, Squares, and Trapezoids What are geometric plane shapes? What characteristics do they have? These are the questions that we will answer in this post. The principal geometric plane shapes are: ## The Circle The circle is a shape that can be made by tracing a curve that is always the same distance from a point that we call the center. The distance around a circle is called the circumference of the circle. ## The Triangle The triangle is a shape that is formed by 3 straight lines that are called sides. There are different ways of classifying triangles, according to their sides or angles. 1. According to their angles: • Right triangle: the largest of the 3 angles is a right angle. • Acute Triangle: the largest of the 3 angles is an acute angle (less than 90 degrees). • Obtuse Triangle: the largest of the 3 angles is an obtuse angle (more than 90 degrees). 2. According to their sides: • Equilateral Triangle: all 3 sides are the same length. • Isosceles Triangle: it has 2 (or more) sides that are of equal length. (An equilateral triangle is also isosceles.) • Scalene Triangle: no 2 sides are of equal measure. ## The Rectangle The rectangle is a shape that has 4 sides. The distinguishing characteristic of a rectangle is that all 4 angles measure 90 degrees. ## The Rhombus The rhombus is a shape formed by 4 straight lines. Its 4 sides measure the same length but, unlike the rectangle, any of all 4 angles measure 90 degrees. ## The Square The square is a type of rectangle, but also a type of rhombus. It has characteristics of both of these. That is to say, all 4 angles are right angles, and all 4 sides are equal in length. ## The Trapezoid The trapezoid also has 4 sides. It has two sides that are parallel but the other 2 are not. You can practice with the geometric plane shapes by registering in Smartick. Latest posts by Smartick (see all) • Kwabena Nyamekye Adjepong.Jun 02 2020, 10:42 AM Like the site. • GloryJun 02 2020, 7:19 AM Nice and easy to understand,my kids enjoyed it!!! • Ugwu DestinyMay 25 2020, 5:23 AM Thank you Mrs Mercy • Ugwu DestinyMay 25 2020, 5:17 AM Nice lesson,Thank you • Kelly SimpsonApr 24 2020, 9:01 AM Please can you put more shapes???!!!!! • Abeer MizanApr 15 2020, 3:50 AM thank you for the lesson! • AlinaSep 18 2019, 3:38 PM Thank the lord. I couldn’t tell the difference between square and trapezoid • HerobrineMay 13 2019, 1:04 AM Thank you for my lesson • Yola BingaliMar 28 2019, 11:16 AM Nice! • Sai KumarDec 22 2018, 11:24 AM 2D square problem • NewcomerJan 17 2019, 1:54 PM There are other shapes worth considering. • AmnaFeb 17 2018, 3:32 AM I enjoyed learning different geometrical shapes • SanikaOct 28 2017, 9:58 AM You should tell more about each • HemanthJul 17 2019, 9:34 AM Nice
The Purplemath ForumsHelping students gain understanding and self-confidence in algebra powered by FreeFind Conics: Parabolas: Introduction (page 1 of 4) Sections: Introduction, Finding information from the equation, Finding the equation from information, Word problems & Calculators In algebra, dealing with parabolas usually means graphing quadratics or finding the max/min points (that is, the vertices) of parabolas for quadratic word problems. In the context of conics, however, there are some additional considerations. To form a parabola according to ancient Greek definitions, you would start with a line and a point off to one side. The line is called the "directrix"; the point is called the "focus". The parabola is the curve formed from all the points (x, y) that are equidistant from the directrix and the focus. The line perpendicular to the directrix and passing through the focus (that is, the line that splits the parabola up the middle) is called the "axis of symmetry". The point on this axis which is exactly midway between the focus and the directrix is the "vertex"; the vertex is the point where the parabola changes direction. "regular", or vertical, parabola (in blue), with the focus (in green) "inside" the parabola, the directrix (in purple) below the graph, the axis of symmetry (in red) passing through the focus and perpendicular to the directrix, and the vertex (in orange) on the graph "sideways", or horizontal, parabola (in blue), with the focus (in green) "inside" the parabola, the directrix (in purple) to the left of the graph, the axis of symmetry (in red) passing through the focus and perpendicular to the directrix, and the vertex (in orange) on the graph The name "parabola" is derived from a New Latin term that means something similar to "compare" or "balance", and refers to the fact that the distance from the parabola to the focus is always equal to (that is, is always in balance with) the distance from the parabola to the directrix. In practical terms, you'll probably only need to know that the vertex is exactly midway between the directrix and the focus. In previous contexts, your parabolas have either been "right side up" or "upside down" graph, depending on whether the leading coefficient was positive or negative, respectively. In the context of conics, however, you will also be working with "sideways" parabolas, parabolas whose axes of symmetry parallel the x-axis and which open to the right or to the left. A basic property of parabolas "in real life" is that any light or sound ray entering the parabola parallel to the axis of symmetry and hitting the inner surface of the parabolic "bowl" will be reflected back to the focus. "Parabolic dishes", such as "bionic ears" and radio telescopes, take advantage of this property to concentrate a signal onto a receiver. The focus of a parabola is always inside the parabola; the vertex is always on the parabola; the directrix is always outside the parabola. The "general" form of a parabola's equation is the one you're used to, y = ax2 + bx + c — unless the quadratic is "sideways", in which case the equation will look something like x = ay2 + by + c. The important difference in the two equations is in which variable is squared: for regular (vertical) parabolas, the x part is squared; for sideways (horizontal) parabolas, the y part is squared. The "vertex" form of a parabola with its vertex at (h, k) is: regular: y = a(xh)2 + k sideways: The conics form of the parabola equation (the one you'll find in advanced or older texts) is: regular: 4p(yk) = (xh)2 sideways: 4p(xh) = (yk)2 Why "(h, k)" for the vertex? Why "p" instead of "a" in the old-time conics formula? Dunno. The important thing to notice, though, is that the h always stays with the x, that the k always stays with the y, and that the p is always on the unsquared variable part. The relationship between the "vertex" form of the equation and the "conics" form of the equation is nothing more than a rearrangement: y = a(xh)2 + k y k = a(xh)2 (1/a)(yk) = (xh)2 4p(yk) = (xh)2 In other words, the value of 4p is actually the same as the value of 1/a; they're just two ways of saying the exact same thing. But this new variable p is one you'll need to be able to work with when you're doing parabolas in the context of conics: it represents the distance between the vertex and the focus, and also the same (that is, equal) distance between the vertex and the directrix. And 2p is then clearly the distance between the focus and the directrix. • State the vertex and focus of the parabola having the equation (y – 3)2 = 8(x – 5). • Comparing this equation with the conics form, and remembering that the h always goes with the x and the k always goes with the y, I can see that the center is at (h, k) = (5, 3). The coefficient of the unsquared part is 4p; in this case, that gives me 4p = 8, so p = 2. Since the y part is squared and p is positive, then this is a sideways parabola that opens to the right. The focus is inside the parabola, so it has to be two units to the right of the vertex: vertex: (5, 3); focus: (7, 3) • State the vertex and directrix of the parabola having the equation (x + 3)2 = –20(y – 1). • The temptation is to say that the vertex is at (3, 1), but that would be wrong. The conics form of the equation has subtraction inside the parentheses, so the (x + 3)2 is really (x – (–3))2, and the vertex is at (–3, 1). The coefficient of the unsquared part is –20, and this is also the value of 4p, so p = –5. Since the x part is squared and p is negative, then this is a regular parabola that opens downward. This means that the directrix, being on the outside of the parabola, is five units above the vertex. vertex: (–3, 1); directrix: y = 6 Top \| 1 \| 2 \| 3 \| 4 \| Return to Index Next >> Cite this article as: Stapel, Elizabeth. "Conics: Parabolas: Introduction." Purplemath. Available from http://www.purplemath.com/modules/parabola.htm. Accessed [Date] [Month] 2016 Purplemath: Linking to this site Printing pages School licensing Reviews of Internet Sites: Free Help Practice Et Cetera The "Homework Guidelines" Study Skills Survey Tutoring from Purplemath Find a local math tutor This lesson may be printed out for your personal use.
# How to Divide Exponents Dividing expressions that have exponents is easier than it looks. As long as you're working with the same base, all you have to do is subtract the values of the exponents from each other and keep the base the same. If you're struggling with this, jump down to Step 1 for easy instructions on getting through the process. Part 1 Part 1 of 2: ### Understanding the Basics 1. 1 Write down the problem. The most simple version of this problem will be in the form of ma ÷ mb. In this case, you're working with the problem m8 ÷ m2. Write it down. 2. 2 Subtract the second exponent from the first. The second exponent is 2 and the first is 8. So, you can rewrite the problem as m8-2.[1] 3. 3 State your final answer. Since 8 - 2 = 6, your final answer is m6. It's that simple. If you're not working with a variable and have an actual number in the base, such as 2, then you would have to do the math (26 = 64) to finish solving the problem.[2] Part 2 Part 2 of 2: ### Going the Extra Mile 1. 1 Make sure that each expression has the same base. If you're working with different bases, then you cannot divide the exponents. Here's what you need to know: • If you're working with a problem with variables, such as m6 ÷ x4, then there's nothing more you can do to simplify it. • However, if the bases are numbers and not variables, you may be able to manipulate them so you end up with the same base. For example, in the problem, 23 ÷ 41, you first have to make both bases be "2." All you do is rewrite 4 as 22 and do the math: 23 ÷ 22 = 21, or 2. • You can only do this, however, if you can turn the larger base into an expression of squared numbers to make it have the same base as the first. 2. 2 Divide expressions with multiple variables. If you have an expression with multiple variables, then you just have to divide the exponents from each identical base to get your final answer. Here's how you do it:[3] • x6y3z2 ÷ x4y3z = • x6-4y3-3z2-1 = • x2z 3. 3 Divide expressions with coefficients. As long as you're working with the same base, it's no problem if each expression has a different coefficient. Just divide the exponents as you normally would and divide the first coefficient by the second. Here's how:[4] • 6x4 ÷ 3x2 = • 6/3x4-2 = • 2x2 4. 4 Divide expressions with negative exponents. To divide expressions with negative exponents, all you have to do is move the base to the other side of the fraction line. So, if you have 3-4 in the numerator of a fraction, you'll have to move it to the denominator. Here are two examples:[5] • Example 1: • x-3/x-7 = • x7/x3 = • x7-3 = • x4 • Example 2: • 3x-2y/xy = • 3y/(x2 * xy) = • 3y/x3y = • 3/x3 ## Community Q&A Search • Question How can I divide a by b, if both have exponent c? Donagan It would be the fraction (a/b) raised to the power of c. • Question How do I divide a positive number with a positive exponent by a positive number with a negative exponent? First of all, the two positive numbers (the bases) have to be the same. If they are, you subtract the exponent in the denominator from the exponent in the numerator. If the denominator's exponent is negative, you treat it as if it were positive and add it to the numerator's exponent. Thus, x^3 ÷ x^(-1) = x^4. • Question What is 400 divided by 10^3? 400 / 10³ = 400 / 1000 = 0.4. 200 characters left ## Tips • Don't worry if you get it wrong! Keep trying! ⧼thumbs_response⧽ • If you have a calculator it is usually a good idea to check your answer. Calculate both the sum and your answer to it to check that they are the same. ⧼thumbs_response⧽ • If you are confused, you can always ask your parents or someone that knows the subject well enough to assist you. ⧼thumbs_response⧽ Co-authored by: Co-authors: 19 Updated: March 9, 2023 Views: 160,561 Article SummaryX To divide exponents with the same base, start by subtracting the second exponent from the first. For example, if your problem is m to the 4th power divided by m to the 2nd power, then you would subtract 2 from 4 in order to get 2. That means your final answer is m to the 2nd power. Alternatively, if your problem is 2 to the 5th power divided by 2 to the 2nd power, subtract 2 from 5 to get an answer of 2 to the 3rd power. To learn how to divide expressions with coefficients, keep reading! Thanks to all authors for creating a page that has been read 160,561 times.
# Introduction To Differentiation In this article, we will introduce the concepts of differentiation as per the Secondary 4 A Math coursework. We will learn about: • The gradient of the curve at a point. • Differentiation of power functions • Differentiation of composite functions using Chain Rule. ## Gradient Of A Curve At A Point The gradient of a curve at a point on the curve can be estimated by finding the gradient of the tangent at that point. Differentiation is a more accurate method of finding the gradient of a curve at any point. There are three different ways to represent as follows, The first row represents the equation form before differentiation. The second row represents the equation form after differentiation. ## Differentiation Of Power Functions Find the derivative of the following with respect to $\displaystyle{x}$ 1. $\displaystyle{y=a^n}$ • In this type, the power $\mathrm{n}$ needs to be multiplied by the coefficient. • Then subtract 1 from the power. Therefore after differentiation the equation becomes, $\displaystyle{\frac { dy } { dx } =anx^{n-1}}$ 1. $\displaystyle{f(x) = ax^n}$ • In this type, add a prime (i.e $f'$) in between $f(x)$. • The power $n$ needs to be multiplied with the coefficient. • Then subtract 1 from the power. Therefore after differentiation, the equation becomes as below. $f'(x)=anx^{n-1}$ 1. $\displaystyle{ax^n}$ • In this type, write $\displaystyle{\frac {d}{dx} }$ on the left hand side of the expression followed by the value in the question in brackets. • The power $\displaystyle{n}$ needs to be multiplied with the coefficient. • Then subtract 1 from the power. Therefore after differentiation the equation becomes, $\displaystyle{\frac { d } { dx } (a^n) = anx^{n-1}}$ The above method works if the power $\displaystyle{n}$ is a real number and $\displaystyle{a}$ is not $0$. Example 1: Differentiate the following with respect to $x$. 1. $\displaystyle{y=3x^2}$ 2. $\displaystyle{f(x)={{1}\over 2}{x^3}}$ 3. $\displaystyle{-\frac25x^{-2}}$ 4. $\displaystyle{y=-2x^{\frac{1}{3}}}$ Solution: • Following Type 1, mention the LHS as $\displaystyle{\frac { dy } { dx }}$. • Then, multiply the power $2$ with the coefficient and subtract 1 from the power. $\displaystyle{\frac { dy } { dx } =(2\times3)x^{2-1}}$ $\displaystyle{\frac { dy } { dx } =6x}$ 1. $\displaystyle{f(x)={{1}\over 2}{x^3}}$ Solution: • For the LHS, add a prime (i.e $f'$) in between $f$ and $(x)$. • For the RHS, multiply the power $\displaystyle{\frac {1}{2}}$ with the co-efficient and subtract 1 from the power. $\displaystyle{f'(x) = (3 \times \frac12)x^{3-1}}$ $\displaystyle{f'(x) = \frac32x^{2}}$ 1. $\displaystyle{-\frac25x^{-2}}$ Solution: Add $\displaystyle{\frac {d}{dx}}$ and in the brackets, multiply the power $-2$ with the co-efficient and subtract 1 from the power. $\displaystyle{\frac { d } { dx } (-\frac {2}{5}x^{-2}) =\frac45x^{-3}}$ D. $\displaystyle{y=-2x^{\frac{1}{3}}}$ Solution: • Following Type 1, mention the LHS as $\displaystyle{\frac { dy } { dx } }$. • Then, multiply the power $\displaystyle{\frac{1}{3}}$ with the co-efficient and subtract 1 from the power. \begin{align} \displaystyle{\frac { dy } { dx }} &= \displaystyle{\bigg(-2 \times \frac13\bigg)x^{\big(-\frac {1}{3}-1\big)}}\\[2ex] &=\displaystyle{-\frac23x^{-\frac {2}{3}}} \end{align} ## Laws Of Indices These laws have a lot of usage in differentiation. ### Positive indices 1. $\displaystyle{a^m \times a^n =a^{m+n}}$ 2. $\displaystyle{a^m \div a^n =a^{m - n}}$ 3. $\displaystyle{(a^m)^n = a^{mn}}$ 4. $\displaystyle{(ab)^n = a^{n} b^{n}}$ 5. $\displaystyle{\bigg(\frac {a}{b} \bigg)^n = \bigg(\frac {a^n} {b^n} \bigg)}$ ### Negative and Zero indices 1. $\displaystyle{a^0 = 1}$ 2. $\displaystyle{\bigg(\frac {1}{a^n} \bigg) = a^{-n}}$ ### Fractional indices 1. $\displaystyle{\sqrt [n]a = a^{\frac1n}}$ 2. $\displaystyle{\sqrt [n]{a^m} = a^{\frac mn}}$ ### Example 2: Differentiate of $ax$ with respect to $x$ 1. Differentiate $\displaystyle{y=5x^1}$ with respect to $x$ Solution: • Following the first type of differentiation, the LHS becomes $\displaystyle{\frac { dy } { dx }}$. • In the RHS, the power 1 is multiplied with the coefficient 5 and subtracting the power 1 by 1. $\displaystyle{\frac { dy } { dx } = 5x^0}$ The value of $x^0$ is 1. $\displaystyle{\frac { dy } { dx } =5}$ In summary, $\displaystyle{\frac { d } { dx } (ax) = a}$ , where $a$ is a constant. 1. Differentiate $\displaystyle{y=5}$ with respect to $x$. Solution: Since there is no $x$ in the question, it is difficult to differentiate. You need to add $x^0$ to it. Therefore, $\displaystyle{y=5x^0}$ Since, the value of $x^0$ is 1, it does not change the question. Hence, when you differentiate, \begin{align} \frac { dy } { dx } &= (5 \times 0)x^{-1}\\[2ex] &= 0 \end{align} In summary, $\displaystyle{\frac{d}{dx} (a)=0}$, where $a$ is a constant. ### Example 3: Differentiate $\displaystyle{y = \frac {2} {x^2} + 5x + 3}$ with respect to $x$. Solution: According to the laws of indices (negative and zero indices law 2), we are going to change the denominator to numerator $\displaystyle{\frac {2} {x^2}}$ , Hence, you get $\displaystyle{y = {2} {x^{-2}} + 5x + 3}$ Now you can differentiate it like you do normally, • Multiply the power $-2$ with the coefficient 2 and subtract the power $-2$ with 1. • As we saw in the last two examples, when you differentiate $5x$ you get 5 and when you differentiate 3 you get 0. $\displaystyle{\frac { dy } { dx } =-4x^{-3} +5 +0}$ Since, the final answer cannot be written in negative power, so we change the $\displaystyle{x^{-3}}$ to the fractional form. Hence, the answer is $\displaystyle{\frac{dy}{dx} =-\frac {4}{x^{3}} +5}$ ### Example 4: Differentiate $\displaystyle{\frac {4}{\sqrt {x}}}$ with respect to $x$. Solution: Firstly the question needs to be converted to the $\displaystyle{\frac { d } { dx } (ax^n)}$ form. Square root can also be written as power $\displaystyle{\frac{1}{2}}$, which gives us $\displaystyle{\frac{d}{dx} \bigg[\frac {4}{x^{\frac12}}\bigg]}$ The denominator $\displaystyle{x^{\frac12}}$ is converted to numerator, thus becoming $\displaystyle{\frac { d } { dx } (4x^{-\frac12})}$. Then, differentiating this equation via multiplying the power $\displaystyle{-\frac12}$ with the coefficient 4, and subtracting by 1, the equation becomes $\displaystyle{-2x^{-\frac 32}}$. Since the answer cannot be written in negative powers, $\displaystyle{x^{^{-{3/_2}}}}$ is converted to the fractional form , i.e. $\displaystyle{- \frac {2}{x^{^3/_2}}}$ $\displaystyle{x^{^{3/_2}}}$ can also be written as $\displaystyle{\sqrt {x^3}}$. Therefore, the answer is $\displaystyle{- \frac {2} {\sqrt {x^3} }}$. ### Example 5: Differentiate $\displaystyle{y= 2x^3 - {\sqrt {3} } - { \frac {5}{\sqrt {x}} }}$ , with respect to $x$. Solution: Rewrite the question by converting the denominator at $\sqrt{x}$ to $x^{-1/2}$ , $\displaystyle{y= 2x^3 - {\sqrt {3} } - { 5x^{-\frac12} }}$ Now differentiating, The coefficient is multiplied with the powers and then the powers are subtracted by 1, $\displaystyle{\frac { dy } { dx } =6x^2 + \frac {5}{2}x^{-\frac 32}}$ The $x^{-\frac32}$ is converted into denominator. $\displaystyle{\frac { dy } { dx } = 6x^2 + {\frac {5}{2x^{\frac32}}}}$ The $\displaystyle{x^{\frac32}}$ in the above equation is written as $\sqrt{x^{3}}$. $\displaystyle{\frac { dy } { dx } = 6x^2 + {\frac {5}{2\sqrt {x^{3}}}}}$ ## Differentiation Of Composite Functions Now that we have learnt how to differentiate power function, have you wondered how can we find the derivative of composite functions like $\displaystyle{y=(3x^2+5)^{\frac12} }$? For this type of differentiation we need to use the chain rule. Let us take the equation, $\displaystyle{\frac { d } { dx } \big(a[f(x)]^n\big)}$ where $a$ is a constant and $n$ is a real number. The power $n$ outside the brackets should be multiplied with the coefficient outside the brackets and subtract 1 from the power, let's assume this as Equation 1. $an[f(x)]^{n-1}$ ---------- Equation (1) Then, differentiate the equation in the bracket and multiply it with the equation 1. Therefore, it becomes $\displaystyle{an[f(x)]^{n-1} \times f'(x)}$ ### Example 6: Differentiate the following with respect to $x$. 1. $\displaystyle{y=(3x^2+5)^{12}}$ Solution: Multiply the power 12 with the coefficient 1. Then, differentiate the equation in brackets and multiply it. $\displaystyle{\frac { dy } { dx } =12(3x^2+5)^{11} \times6x}$ Simplifying the above equation, $\displaystyle{\frac { dy } { dx } =72x(3x^2+5)}$ Therefore, the answer to the question (a) is, $\displaystyle{\frac { dy } { dx } =72x(3x^2+5)}$ Differentiate the following with respect to $x$. 1. $\displaystyle{f(x)=\sqrt {2x-3}}$ Solution: Let us rewrite the question by rewriting the $\displaystyle{\sqrt {2x-3}}$ as $\displaystyle{(2x-3)^{\frac12}}$, Therefore, the equation becomes $\displaystyle{f(x)=(2x-3)^{\frac12}}$ Multiply the power $\displaystyle{\frac12}$ with the coefficient 1 and subtract the power with 1. Then, differentiate the equation in the bracket and multiply it. Therefore, the equation becomes, $\displaystyle{f'(x)=\frac {1}{2}(2x-3)^{-\frac12} \times 2}$ The 2 in numerator and 2 in denominator in the above equation cancels each other, $\displaystyle{f'(x)=(2x-3)^{-\frac12}}$ Since the answers cannot have negative powers, $\displaystyle{f'(x)= \frac{1}{\sqrt {(2x-3)} }}$ Therefore the answer for the question (b) is $\displaystyle{f'(x)= \frac{1}{\sqrt {(2x-3)} }}$ Differentiate the following with respect to $x$. 1. $\displaystyle{\frac {1}{(2-3x)^2}}$ Solution Rewriting the question by converting the denominator $\displaystyle{(2-3x)^2}$ into the negative power form, we get $\displaystyle{(2-3x)^{-2}}$ . Differentiating the above equation by multiplying the power with the coefficient and subtracting the power with 1, and following by differentiating the function in the bracket. we get, $$$-2(2-3x)^{-3} \times ({-3}) = 6(2-3x)^{-3}$$$ Since the answers cannot be written in negative powers, converting $\displaystyle{(2-3x)^{-3}}$ into denominator. $\displaystyle{f'(x) = \frac {6}{(2-3x)^3}}$ Therefore the answer for the question (c) is. $\displaystyle{f'(x) = \frac {6}{(2-3x)^3}}$ ## Conclusion In the article, we learned about differentiation as per the Secondary 4 A Maths coursework. We understood the basics of differentiation, how to differentiate using the law of indices and chain rule. 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# 9.8 Rational exponents (Page 4/7) Page 4 / 7 Simplify: ${\left(16{m}^{\frac{1}{3}}\right)}^{\frac{3}{2}}$ ${\left(81{n}^{\frac{2}{5}}\right)}^{\frac{3}{2}}$ . $64{m}^{\frac{1}{2}}$ $729{n}^{\frac{3}{5}}$ Simplify: ${\left({m}^{3}{n}^{9}\right)}^{\frac{1}{3}}$ ${\left({p}^{4}{q}^{8}\right)}^{\frac{1}{4}}$ . ## Solution 1. $\begin{array}{cccc}& & & {\left({m}^{3}{n}^{9}\right)}^{\frac{1}{3}}\hfill \\ \\ \\ \begin{array}{c}\text{First we use the Product to a Power}\hfill \\ \text{Property.}\hfill \end{array}\hfill & & & {\left({m}^{3}\right)}^{\frac{1}{3}}{\left({n}^{9}\right)}^{\frac{1}{3}}\hfill \\ \\ \\ \begin{array}{c}\text{To raise a power to a power, we multiply}\hfill \\ \text{the exponents.}\hfill \end{array}\hfill & & & m{n}^{3}\hfill \end{array}$ 2. $\begin{array}{cccc}& & & {\left({p}^{4}{q}^{8}\right)}^{\frac{1}{4}}\hfill \\ \\ \\ \begin{array}{c}\text{First we use the Product to a Power}\hfill \\ \text{Property.}\hfill \end{array}\hfill & & & {\left({p}^{4}\right)}^{\frac{1}{4}}{\left({q}^{8}\right)}^{\frac{1}{4}}\hfill \\ \\ \\ \begin{array}{c}\text{To raise a power to a power, we multiply}\hfill \\ \text{the exponents.}\hfill \end{array}\hfill & & & p{q}^{2}\hfill \end{array}$ We will use both the Product and Quotient Properties in the next example. Simplify: $\frac{{x}^{\frac{3}{4}}·{x}^{-\frac{1}{4}}}{{x}^{-\frac{6}{4}}}$ $\frac{{y}^{\frac{4}{3}}·y}{{y}^{-\frac{2}{3}}}$ . ## Solution 1. $\begin{array}{cccc}& & & \frac{{x}^{\frac{3}{4}}·{x}^{-\frac{1}{4}}}{{x}^{-\frac{6}{4}}}\hfill \\ \\ \\ \begin{array}{c}\text{Use the Product Property in the numerator,}\hfill \\ \text{add the exponents.}\hfill \end{array}\hfill & & & \frac{{x}^{\frac{2}{4}}}{{x}^{-\frac{6}{4}}}\hfill \\ \\ \\ \begin{array}{c}\text{Use the Quotient Property, subtract the}\hfill \\ \text{exponents.}\hfill \end{array}\hfill & & & {x}^{\frac{8}{4}}\hfill \\ \text{Simplify.}\hfill & & & {x}^{2}\hfill \end{array}$ 2. $\begin{array}{cccc}& & & \frac{{y}^{\frac{4}{3}}·y}{{y}^{-\frac{2}{3}}}\hfill \\ \\ \\ \begin{array}{c}\text{Use the Product Property in the numerator,}\hfill \\ \text{add the exponents.}\hfill \end{array}\hfill & & & \frac{{y}^{\frac{7}{3}}}{{y}^{-\frac{2}{3}}}\hfill \\ \\ \\ \begin{array}{c}\text{Use the Quotient Property, subtract the}\hfill \\ \text{exponents.}\hfill \end{array}\hfill & & & {y}^{\frac{9}{3}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & {y}^{3}\hfill \end{array}$ Simplify: $\frac{{m}^{\frac{2}{3}}·{m}^{-\frac{1}{3}}}{{m}^{-\frac{5}{3}}}$ $\frac{{n}^{\frac{1}{6}}·n}{{n}^{-\frac{11}{6}}}$ . ${m}^{2}$ ${n}^{3}$ Simplify: $\frac{{u}^{\frac{4}{5}}·{u}^{-\frac{2}{5}}}{{u}^{-\frac{13}{5}}}$ $\frac{{v}^{\frac{1}{2}}·v}{{v}^{-\frac{7}{2}}}$ . ${u}^{3}$ ${v}^{5}$ ## Key concepts • Summary of Exponent Properties • If $a,b$ are real numbers and $m,n$ are rational numbers, then • Product Property ${a}^{m}·{a}^{n}={a}^{m+n}$ • Power Property ${\left({a}^{m}\right)}^{n}={a}^{m·n}$ • Product to a Power ${\left(ab\right)}^{m}={a}^{m}{b}^{m}$ • Quotient Property : $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n},\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}a\ne 0,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}m>n$ $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}},\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}a\ne 0,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}n>m$ • Zero Exponent Definition ${a}^{0}=1$ , $a\ne 0$ • Quotient to a Power Property ${\left(\frac{a}{b}\right)}^{m}=\frac{{a}^{m}}{{b}^{m}},\phantom{\rule{0.5em}{0ex}}b\ne 0$ ## Practice makes perfect Simplify Expressions with ${a}^{\frac{1}{n}}$ In the following exercises, write as a radical expression. ${x}^{\frac{1}{2}}$ ${y}^{\frac{1}{3}}$ ${z}^{\frac{1}{4}}$ ${r}^{\frac{1}{2}}$ ${s}^{\frac{1}{3}}$ ${t}^{\frac{1}{4}}$ $\sqrt{r}$ $\sqrt[3]{s}$ $\sqrt[4]{t}$ ${u}^{\frac{1}{5}}$ ${v}^{\frac{1}{9}}$ ${w}^{\frac{1}{20}}$ ${g}^{\frac{1}{7}}$ ${h}^{\frac{1}{5}}$ ${j}^{\frac{1}{25}}$ $\sqrt[7]{g}$ $\sqrt[5]{h}$ $\sqrt[25]{j}$ In the following exercises, write with a rational exponent. $-\sqrt[7]{x}$ $\sqrt[9]{y}$ $\sqrt[5]{f}$ $\sqrt[8]{r}$ $\sqrt[4]{t}$ ${r}^{\frac{1}{8}}$ ${s}^{\frac{1}{10}}$ ${t}^{\frac{1}{4}}$ $\sqrt[3]{a}$ $\sqrt{c}$ $\sqrt[5]{u}$ $\sqrt{v}$ ${u}^{\frac{1}{5}}$ ${v}^{\frac{1}{2}}$ ${w}^{\frac{1}{16}}$ $\sqrt[3]{7c}$ $\sqrt[7]{12d}$ $3\sqrt[4]{5f}$ $\sqrt[4]{5x}$ $\sqrt[8]{9y}$ $7\sqrt[5]{3z}$ ${\left(5x\right)}^{\frac{1}{4}}$ ${\left(9y\right)}^{\frac{1}{8}}$ $7{\left(3z\right)}^{\frac{1}{5}}$ $\sqrt{21p}$ $\sqrt[4]{8q}$ $4\sqrt[6]{36r}$ $\sqrt[3]{25a}$ $\sqrt{3b}$ ${\left(25a\right)}^{\frac{1}{3}}$ ${\left(3b\right)}^{\frac{1}{2}}$ ${\left(40c\right)}^{\frac{1}{10}}$ In the following exercises, simplify. ${81}^{\frac{1}{2}}$ ${125}^{\frac{1}{3}}$ ${64}^{\frac{1}{2}}$ ${625}^{\frac{1}{4}}$ ${243}^{\frac{1}{5}}$ ${32}^{\frac{1}{5}}$ 5 3 2 ${16}^{\frac{1}{4}}$ ${16}^{\frac{1}{2}}$ ${3125}^{\frac{1}{5}}$ ${216}^{\frac{1}{3}}$ ${32}^{\frac{1}{5}}$ ${81}^{\frac{1}{4}}$ 6 2 3 ${\left(-216\right)}^{\frac{1}{3}}$ $\text{−}{216}^{\frac{1}{3}}$ ${\left(216\right)}^{-\frac{1}{3}}$ ${\left(-243\right)}^{\frac{1}{5}}$ $\text{−}{243}^{\frac{1}{5}}$ ${\left(243\right)}^{-\frac{1}{5}}$ $-3$ $-3$ $\frac{1}{3}$ ${\left(-1\right)}^{\frac{1}{3}}$ ${-1}^{\frac{1}{3}}$ ${\left(1\right)}^{-\frac{1}{3}}$ ${\left(-1000\right)}^{\frac{1}{3}}$ $\text{−}{1000}^{\frac{1}{3}}$ ${\left(1000\right)}^{-\frac{1}{3}}$ $-10$ $-10$ $\frac{1}{10}$ ${\left(-81\right)}^{\frac{1}{4}}$ $\text{−}{81}^{\frac{1}{4}}$ ${\left(81\right)}^{-\frac{1}{4}}$ ${\left(-49\right)}^{\frac{1}{2}}$ $\text{−}{49}^{\frac{1}{2}}$ ${\left(49\right)}^{-\frac{1}{2}}$ not a real number $-7$ $\frac{1}{7}$ ${\left(-36\right)}^{\frac{1}{2}}$ $-{36}^{\frac{1}{2}}$ ${\left(36\right)}^{-\frac{1}{2}}$ ${\left(-1\right)}^{\frac{1}{4}}$ ${\left(1\right)}^{-\frac{1}{4}}$ $\text{−}{1}^{\frac{1}{4}}$ not a real number $1$ $-1$ ${\left(-100\right)}^{\frac{1}{2}}$ $\text{−}{100}^{\frac{1}{2}}$ ${\left(100\right)}^{-\frac{1}{2}}$ ${\left(-32\right)}^{\frac{1}{5}}$ ${\left(243\right)}^{-\frac{1}{5}}$ $\text{−}{125}^{\frac{1}{3}}$ $-2$ $\frac{1}{3}$ $-5$ Simplify Expressions with ${a}^{\frac{m}{n}}$ In the following exercises, write with a rational exponent. $\sqrt{{m}^{5}}$ $\sqrt[3]{{n}^{2}}$ $\sqrt[4]{{p}^{3}}$ $\sqrt[4]{{r}^{7}}$ $\sqrt[5]{{s}^{3}}$ $\sqrt[3]{{t}^{7}}$ ${r}^{\frac{7}{4}}$ ${s}^{\frac{3}{5}}$ ${t}^{\frac{7}{3}}$ $\sqrt[5]{{u}^{2}}$ $\sqrt[5]{{v}^{8}}$ $\sqrt[9]{{w}^{4}}$ $\sqrt[3]{a}$ $\sqrt{{b}^{5}}$ $\sqrt[3]{{c}^{5}}$ ${a}^{\frac{1}{3}}$ ${b}^{\frac{1}{5}}$ ${c}^{\frac{5}{3}}$ In the following exercises, simplify. ${16}^{\frac{3}{2}}$ ${8}^{\frac{2}{3}}$ ${10,000}^{\frac{3}{4}}$ ${1000}^{\frac{2}{3}}$ ${25}^{\frac{3}{2}}$ ${32}^{\frac{3}{5}}$ 100 125 8 ${27}^{\frac{5}{3}}$ ${16}^{\frac{5}{4}}$ ${32}^{\frac{2}{5}}$ ${16}^{\frac{3}{2}}$ ${125}^{\frac{5}{3}}$ ${64}^{\frac{4}{3}}$ 64 3125 256 ${32}^{\frac{2}{5}}$ ${27}^{-\frac{2}{3}}$ ${25}^{-\frac{3}{2}}$ ${64}^{\frac{5}{2}}$ ${81}^{-\frac{3}{2}}$ ${27}^{-\frac{4}{3}}$ 32,768 $\frac{1}{729}$ $\frac{1}{81}$ ${25}^{\frac{3}{2}}$ ${9}^{-\frac{3}{2}}$ ${\left(-64\right)}^{\frac{2}{3}}$ ${100}^{\frac{3}{2}}$ ${49}^{-\frac{5}{2}}$ ${\left(-100\right)}^{\frac{3}{2}}$ 1000 $\frac{1}{16,807}$ not a real number $\text{−}{9}^{\frac{3}{2}}$ $\text{−}{9}^{-\frac{3}{2}}$ ${\left(-9\right)}^{\frac{3}{2}}$ $\text{−}{64}^{\frac{3}{2}}$ $\text{−}{64}^{-\frac{3}{2}}$ ${\left(-64\right)}^{\frac{3}{2}}$ $-512$ $-\frac{1}{512}$ not a real number $\text{−}{100}^{\frac{3}{2}}$ $\text{−}{100}^{-\frac{3}{2}}$ ${\left(-100\right)}^{\frac{3}{2}}$ $\text{−}{49}^{\frac{3}{2}}$ $\text{−}{49}^{-\frac{3}{2}}$ ${\left(-49\right)}^{\frac{3}{2}}$ $-343$ $-\frac{1}{343}$ not a real number Use the Laws of Exponents to Simplify Expressions with Rational Exponents In the following exercises, simplify. ${4}^{\frac{5}{8}}·{4}^{\frac{11}{8}}$ ${m}^{\frac{7}{12}}·{m}^{\frac{17}{12}}$ ${p}^{\frac{3}{7}}·{p}^{\frac{18}{7}}$ ${6}^{\frac{5}{2}}·{6}^{\frac{1}{2}}$ ${n}^{\frac{2}{10}}·{n}^{\frac{8}{10}}$ ${q}^{\frac{2}{5}}·{q}^{\frac{13}{5}}$ 216 $n$ ${q}^{3}$ ${5}^{\frac{1}{2}}·{5}^{\frac{7}{2}}$ ${c}^{\frac{3}{4}}·{c}^{\frac{9}{4}}$ ${d}^{\frac{3}{5}}·{d}^{\frac{2}{5}}$ ${10}^{\frac{1}{3}}·{10}^{\frac{5}{3}}$ ${x}^{\frac{5}{6}}·{x}^{\frac{7}{6}}$ ${y}^{\frac{11}{8}}·{y}^{\frac{21}{8}}$ 100 ${x}^{2}$ ${y}^{4}$ ${\left({m}^{6}\right)}^{\frac{5}{2}}$ ${\left({n}^{9}\right)}^{\frac{4}{3}}$ ${\left({p}^{12}\right)}^{\frac{3}{4}}$ ${\left({a}^{12}\right)}^{\frac{1}{6}}$ ${\left({b}^{15}\right)}^{\frac{3}{5}}$ ${\left({c}^{11}\right)}^{\frac{1}{11}}$ ${a}^{2}$ ${b}^{9}$ $c$ ${\left({x}^{12}\right)}^{\frac{2}{3}}$ ${\left({y}^{20}\right)}^{\frac{2}{5}}$ ${\left({z}^{16}\right)}^{\frac{1}{16}}$ ${\left({h}^{6}\right)}^{\frac{4}{3}}$ ${\left({k}^{12}\right)}^{\frac{3}{4}}$ ${\left({j}^{10}\right)}^{\frac{7}{5}}$ ${h}^{8}$ ${k}^{9}$ ${j}^{14}$ $\frac{{x}^{\frac{7}{2}}}{{x}^{\frac{5}{2}}}$ $\frac{{y}^{\frac{5}{2}}}{{y}^{\frac{1}{2}}}$ $\frac{{r}^{\frac{4}{5}}}{{r}^{\frac{9}{5}}}$ $\frac{{s}^{\frac{11}{5}}}{{s}^{\frac{6}{5}}}$ $\frac{{z}^{\frac{7}{3}}}{{z}^{\frac{1}{3}}}$ $\frac{{w}^{\frac{2}{7}}}{{w}^{\frac{9}{7}}}$ $s$ ${z}^{2}$ $\frac{1}{w}$ $\frac{{t}^{\frac{12}{5}}}{{t}^{\frac{7}{5}}}$ $\frac{{x}^{\frac{3}{2}}}{{x}^{\frac{1}{2}}}$ $\frac{{m}^{\frac{13}{8}}}{{m}^{\frac{5}{8}}}$ $\frac{{u}^{\frac{13}{9}}}{{u}^{\frac{4}{9}}}$ $\frac{{r}^{\frac{15}{7}}}{{r}^{\frac{8}{7}}}$ $\frac{{n}^{\frac{3}{5}}}{{n}^{\frac{8}{5}}}$ $u$ $r$ $\frac{1}{n}$ ${\left(9{p}^{\frac{2}{3}}\right)}^{\frac{5}{2}}$ ${\left(27{q}^{\frac{3}{2}}\right)}^{\frac{4}{3}}$ ${\left(81{r}^{\frac{4}{5}}\right)}^{\frac{1}{4}}$ ${\left(64{s}^{\frac{3}{7}}\right)}^{\frac{1}{6}}$ $3{r}^{\frac{1}{5}}$ $2{s}^{\frac{1}{14}}$ ${\left(16{u}^{\frac{1}{3}}\right)}^{\frac{3}{4}}$ ${\left(100{v}^{\frac{2}{5}}\right)}^{\frac{3}{2}}$ ${\left(27{m}^{\frac{3}{4}}\right)}^{\frac{2}{3}}$ ${\left(625{n}^{\frac{8}{3}}\right)}^{\frac{3}{4}}$ $9{m}^{\frac{1}{2}}$ $125{n}^{2}$ Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles. LeBron needs 150 milliliters of a 30% solution of sulfuric acid for a lab experiment but only has access to a 25% and a 50% solution. How much of the 25% and how much of the 50% solution should he mix to make the 30% solution? 5% Michael hey everyone how to do algebra Felecia answer 1.5 hours before he reaches her I would like to solve the problem -6/2x 12x Andrew how Christian Does the x represent a number or does it need to be graphed ? latonya -3/x Venugopal -3x is correct Atul Arnold invested $64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received$4,500 in interest in one year? Tickets for the community fair cost $12 for adults and$5 for children. On the first day of the fair, 312 tickets were sold for a total of $2204. How many adult tickets and how many child tickets were sold? Alpha Reply 220 gayla Three-fourths of the people at a concert are children. If there are 87 children, what is the total number of people at the concert? Tsimmuaj Reply Erica earned a total of$50,450 last year from her two jobs. The amount she earned from her job at the store was $1,250 more than four times the amount she earned from her job at the college. How much did she earn from her job at the college? Tsimmuaj Erica earned a total of$50,450 last year from her two jobs. The amount she earned from her job at the store was $1,250 more than four times the amount she earned from her job at the college. How much did she earn from her job at the college? Tsimmuaj ? Is there anything wrong with this passage I found the total sum for 2 jobs, but found why elaborate on extra If I total one week from the store 4 would = the month than the total is = x than x can't calculate 10 month of a year candido what would be wong candido 87 divided by 3 then multiply that by 4. 116 people total. Melissa the actual number that has 3 out of 4 of a whole pie candido was having a hard time finding Teddy use Matrices for the 2nd question Daniel One number is 11 less than the other number. If their sum is increased by 8, the result is 71. Find the numbers. Tsimmuaj Reply 26 + 37 = 63 + 8 = 71 gayla 26+37=63+8=71 ziad 11+52=63+8=71 Thisha how do we know the answer is correct? Thisha 23 is 11 less than 37. 23+37=63. 63+8=71. that is what the question asked for. gayla 23 +11 = 37. 23+37=63 63+8=71 Gayla by following the question. one number is 11 less than the other number 26+11=37 so 26+37=63+8=71 Gayla your answer did not fit the guidelines of the question 11 is 41 less than 52. gayla 71-8-11 =52 is this correct? Ruel let the number is 'x' and the other number is "x-11". if their sum is increased means: x+(x-11)+8 result will be 71. so x+(x-11)+8=71 2x-11+8=71 2x-3=71 2x=71+3 2x=74 1/2(2x=74)1/2 x=37 final answer tesfu just new Muwanga Amara currently sells televisions for company A at a salary of$17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of$29,000 plus a $20 commission for each television she sells. How televisions would Amara need to sell for the options to be equal? Tsimmuaj Reply yes math Kenneth company A 13 company b 5. A 17,000+13×100=29,100 B 29,000+5×20=29,100 gayla need help with math to do tsi test Toocute me too Christian have you tried the TSI practice test **tsipracticetest.com gayla DaMarcus and Fabian live 23 miles apart and play soccer at a park between their homes. DaMarcus rode his bike for 34 of an hour and Fabian rode his bike for 12 of an hour to get to the park. Fabian’s speed was 6 miles per hour faster than DaMarcus’s speed. Find the speed of both soccer players. gustavo Reply ? Ann DaMarcus: 16 mi/hr Fabian: 22 mi/hr Sherman Joy is preparing 20 liters of a 25% saline solution. She has only a 40% solution and a 10% solution in her lab. How many liters of the 40% solution and how many liters of the 10% solution should she mix to make the 25% solution? Wenda Reply 15 and 5 32 is 40% , & 8 is 10 % , & any 4 letters is 5%. Karen It felt that something is missing on the question like: 40% of what solution? 10% of what solution? Jhea its confusing Sparcast 3% & 2% to complete the 25% Sparcast because she already has 20 liters. Sparcast ok I was a little confused I agree 15% & 5% Sparcast 8,2 Karen Jim and Debbie earned$7200. Debbie earned \$1600 more than Jim earned. How much did they earned 5600 Gloria 1600 Gloria Bebbie: 4,400 Jim: 2,800 Jhea A river cruise boat sailed 80 miles down the Mississippi River for 4 hours. It took 5 hours to return. Find the rate of the cruise boat in still water and the rate of the current. A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic equation A=x(100−2x) gives the area, A , of the dog run for the length, x , of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run. ggfcc Mike
# Video: Evaluating Numerical Expressions by Multiplying and Dividing Rational Numbers Evaluate ((1 1/6) ÷ (3/4))((−4/3) ÷ 3) giving the answer in its simplest form. 03:23 ### Video Transcript Evaluate one and one-sixth divided by three-quarters multiplied by negative four-thirds divided by three, giving the answer in its simplest form. Let’s begin by looking at the first parentheses or bracket, one and one-sixth divided by three-quarters. Our first step here is to convert one and one-sixth into an improper or top-heavy fraction. We do this by multiplying the whole number by the denominator and then adding the numerator. This gives us seven. Therefore, one and one-sixth is the same as seven-sixths. We need to divide this by three-quarters. When dividing two fractions, we can use the acronym KCF. We keep the first fraction the same, in this case, seven over six. We change the division sign to a multiplication sign. We flip or find the reciprocal of the second fraction, in this case, four over three. Before multiplying these two fractions, we notice that we can cross simplify or cross cancel by dividing the four and six by two. Multiplying the numerators at this stage gives us 14. And multiplying the denominators gives us nine. One and one-sixth divided by three-quarters is 14 over nine or fourteen-ninths. Let’s now consider our second parentheses or bracket. We need to divide negative four-thirds by three. As any integer can be written as this number over one, we can rewrite this as negative four-thirds divided by three over one. We now need to repeat the process from the first parentheses. This gives us negative four over three multiplied by one over three or one-third. Multiplying the numerators gives us negative four, and multiplying the denominators gives us nine. Recalling that multiplying a negative number by a positive number gives a negative answer and also that we can let the numerator or denominator of a negative fraction be negative. Alternatively, we could consider the whole fraction as negative. Negative four-thirds divided by three is negative four-ninths. The two parentheses have therefore simplified to 14 over nine and negative four over nine. Once again, we need to multiply the numerators and then multiply the denominators. 14 multiplied by negative four is negative 56. Nine multiplied by nine is 81. The answer written as a fraction in its simplest form is negative 56 over 81.
# 8-1 The Pythagorean Theorem and Its Converse. Parts of a Right Triangle In a right triangle, the side opposite the right angle is called the hypotenuse. ## Presentation on theme: "8-1 The Pythagorean Theorem and Its Converse. Parts of a Right Triangle In a right triangle, the side opposite the right angle is called the hypotenuse."— Presentation transcript: 8-1 The Pythagorean Theorem and Its Converse Parts of a Right Triangle In a right triangle, the side opposite the right angle is called the hypotenuse. – It is the longest side. The other two sides are called the legs. The Pythagorean Theorem Pythagorean Theorem: If a triangle is a right triangle, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. a 2 + b 2 = c 2 Pythagorean Triples A Pythagorean triple is a set of nonzero whole numbers that satisfy the Pythagorean Theorem. Some common Pythagorean triples include: 3, 4, 55, 12, 138, 15, 177, 24, 25 – If you multiply each number in the triple by the same whole number, the result is another Pythagorean triple! Finding the Length of the Hypotenuse What is the length of the hypotenuse of  ABC? Do the sides form a Pythagorean triple?  The legs of a right triangle have lengths 10 and 24. What is the length of the hypotenuse? Do the sides form a Pythagorean triple? Finding the Length of a Leg What is the value of x? Express your answer in simplest radical form.  The hypotenuse of a right triangle has length 12. One leg has length 6. What is the length of the other leg? Express your answer in simplest radical form. Triangle Classifications Converse of the Pythagorean Theorem: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. – If c 2 = a 2 + b 2, than  ABC is a right triangle. Theorem 8-3: If the square of the length of the longest side of a triangle is great than the sum of the squares of the lengths of the other two sides, then the triangle is obtuse. – If c 2 > a 2 + b 2, than  ABC is obtuse. Theorem 8-4: If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then the triangle is acute. – If c 2 < a 2 + b 2, than  ABC is acute. Classifying a Triangle  Classify the following triangles as acute, obtuse, or right. – 85, 84, 13 – 6, 11, 14 – 7, 8, 9 Download ppt "8-1 The Pythagorean Theorem and Its Converse. Parts of a Right Triangle In a right triangle, the side opposite the right angle is called the hypotenuse." Similar presentations
Introduction Measurement is the basis of science. It gives the idea of comparison between the sizes of different objects. We use scientific notation to express any measurement in a simple form and order of magnitude to find how magnified an object is. Both are excellent concepts for approximation of measurement. In this article, we will discuss more about them along with their differences. What is Scientific Notation? It is a way of writing very small or large numbers in the simplest form. The main purpose of writing the numbers in the simplest form is to compare them to find nearest approximations. Scientific notation has made it easier to write very large or small numbers. You can also use the scientific notation calculator. Therefore, it is used by scientists to write experimental measurements. For example, Earth has a surface area of 5,101,000,000 miles2, which is a very large number. In scientific notation we can write it as: 5.101 × 109 miles 2 In scientific notation, every non-zero number can be written in the form of $$m\;×\;10^n$$ The non-zero number is also called a significant figure. What is the Order of Magnitude? The order of magnitude gives the idea about a measurement that tells how big or small it is from another measurement. This concept is based on the scientific notation and it is defined as the power of 10 which is closest to the magnitude of that measurement. To express a number N in the form of order of magnitude then there will be number x such that 0.5 < n ≤ 5 so, $$N\;=\;n\;×\;10^x$$ For example, the surface area of the Earth is written as $$5,101,000,000\;=\;5.101\;×\;10^9\;miles^2$$ It is the scientific notation of surface area and we say that it has 9 orders of magnitude. The power of 10 refers to the order of magnitude. What is the difference between scientific notation and order of magnitude? Scientific notation and the order of magnitude are used to approximate and compare two or more measurements. Here we will discuss the difference between scientific notation and order of magnitude in the following difference table. Difference Table Scientific Notation Order of Magnitude It is a way of writing very large and small numbers in a simple organised form. It uses the scientific notation to give the idea how much an object is magnified. The mathematical form of scientific notation $$N\;=\;m\;×\;10^n$$ The mathematical form of order of magnitude is $$N\;=\;n\;×\;10^x$$ The range of m in scientific notation is $$1\;<\;m\;≤\;10$$ The range of order of magnitude is $$0.5\;<\;n\;≤\;5$$ Scientific Notation Rules There are some rules of writing a number by using scientific notation. • The base should be always 10. • The exponent must be a non-zero integer, that means it can be either positive or negative. • The absolute value of the coefficient is greater than or equal to 1 but it should be less than 10. • Coefficients can be positive or negative numbers including whole and decimal numbers. • The mantissa carries the rest of the significant digits of the number. Let’s discuss an example of scientific notation by following the above rules. For example, The population of the Earth is 7,403,000,000. We will write it in scientific notation. $$7,403,000,000\;=\;7.403\;×\;10^9$$ Order of Magnitude Rules There are some rules of finding the order of magnitude for any number. The online order of magnitude calculator also follow the same rules. • The base should be always 10. • The exponent must be a non-zero integer, that means it can be either positive or negative. • The absolute value of the coefficient is greater than or equal to 0/5 but it should be less than 5. • Coefficients can be positive or negative numbers including whole and decimal numbers. • The mantissa carries the rest of the significant digits of the number. Consider the following example to understand the order of magnitude. For example, The population of the Earth is 7,403,000,000. We will find its order of magnitude by using scientific notation. So, $$7,403,000,000\;=\;7.403\;×\;10^9$$ The order of magnitude of Earth’s surface area is 9. FAQ’s What do you mean by Order of Magnitude? The order of magnitude of a measurement is an estimation about it that tells how much it is bigger or smaller from the other measurement. It uses scientific notation to find the order of magnitude. How do you convert from standard form to scientific notation? To convert from standard form to scientific notation, you need to place the decimal point after first non-zero number. Count the number of digits that you moved. Write the number of digits as the power of 10 such as, $$m\;×\;10^n$$. Why do we use Scientific Notation? Scientific notation allows us to express very small and large numbers in a simple and understandable way. It helps to compare two or more experimental measurements in order to find an approximate measurement. What is meant by scientific notation in physics? In physics, it is a scientific way of writing very large and small numbers using significant figures. It converts a number as the power of 10 to express it in a manageable way. What are the 3 Parts in Scientific Notation? There are three main parts of scientific notation. These are 1. Coefficient 2. Base 3. Exponent In the formula of scientific notation, N = m × 10n , m is coefficient, 10 is base value and its power is the exponent value.
# How do I find the base angles without a vertex angle in a isosceles triangle? How can I find the base angles in a isosceles triangle if the vertex angle is missing? Normally, I would go: 2x + vertex angle = 180, but now even the vertex is missing, the only thing I have is a line in the middle, with 90 degree angle. Maybe the 90 degree should be used to find one of the base angles and one of the vertex angles, any help is appreciated, thanks a lot, I am still learning this, any tips are very useful. All you need to do is find one base angle, since in an isosceles triangle, both those angles are opposite equal sides, and are therefore equal. Edit: Now that I see your image, You have enough information to find everything. • You can use the Pythagorean Theorem to obtain the altitude (the unknown side in the middle.) • The vertex, B, is split into two equal angles, but you can easily find the vertex angle by first finding one of the base angles. • You can find the base angle B by using the fact that $$\cos B = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}\implies B = \cos^{-1}\left(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\right) = \cos^{-1}\left(\frac{10.9}{25.6}\right) = 64.8$$ Each base angle, let's call them each of measure $x$, plus the vertex angle add up to $180$. So if you need to solve for the vertex angle $\theta$, we have that $\theta = 180 - 2x$. Also note, to find an angle, given none of the angles are known, you can to use the lengths of the sides of the triangle to solve for the base angles and/or vertex, and can use the Law of Cosines to do that. • Yes, I know that, but the Q is how, I do not have the vertex angle either, I only have a 90 degree angle, I can subtract 90 degree from 180, but that would still not allow me to find the one of the base angles or the vertex angle. Oct 27 '13 at 18:52 • Yes, John, you can use the Pythagorean Theorem to obtain the altitude (the unknown side in the middle.) The vertex, B, is split into two equal angles. But you can find the base angle B by using the fact that $$\cos B = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}\implies B = \cos^{-1}\left(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\right)$$ Oct 27 '13 at 19:53 • Exactly! That's correct! Oct 28 '13 at 19:05 • No, the base angles are both $64.8$, the vertex of the largest triangle is $50.4$, and half of that at the top is the angle $25.2.$ The line in the middle of the largest triangle divided the big triangle exactly in half. Oct 28 '13 at 19:09 • Not currently, but I have taught both at a community college and at a university. Oct 28 '13 at 19:13
# Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2. Text Solution Verified by Experts ## (i) As the function f(x)=[x] is neither continuous nor derivable , conditions of mean value theorem is not satisfied hence not verified .(ii) Given function is, f(x)=[x],x∈[−2,2]From the above equation (1), it is clear that the function f(x) is not continuous at x=−2andx=2.The value of function at point −2 is,f(−2)=[−2]=−2The value of function at point 2 is,f(2)=[2]=2It can be observed that f(−2)inf(2).The differentiability of the function can be check as follows. Let, b is an integer.The left hand limit of fat x=bis,limh→0f(b+h)−f(b)h=limh→0[b+h]−[b]h=limh→0b−1−bh=limh→0−−1h=∞The right hand limit of f at x=b is,limh→0f(b+h)−f(b)h=limh→0[b+h]−[b]h=limh→0b−bh=0Since, right hand limit is not equal to the left hand limit therefore function is not differentiable in the given interval. Therefore, Mean value theorem is not satisfied for the given function f(x)=[x] for x∈[−2,2].(iii)Given function is,f(x)=x2−1,x∈[1,2]The first derivative of the function f'(c)=0 for some value, where c∈(a,b).From the above equation (1), it is clear that the function f(x) is continuous at x=1andx=2.The value of function at point 1 is,f(1)=(1)2−1=0The value of function at point 2 is,f(2)=(2)2−1=3It can be observed that f(1)∈f(2).The differentiability of the function can be check as follows. Let, b is an integer.The left hand limit of the function at x=b is,limh→0(f(b+h)−fb))h=limh→0{(b+h)2−1}−{b2−1}h=limh→0{b2+2bh+h2−1}−(b2+1)h=limh→02bhh=2bThe right hand limit of fat x=b is,limh→0f(b+h)−f(b)h=limh→0{(b+h)2−1}−{b2−1}h=limh→02bhh=2bSince, right hand limit is equal to the left hand limit therefore function is differentiable in the given interval.Therefore, Mean value theorem is satisfied for the given function f(x)=x2−1 for x∈[1,2]. \| Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation
# Approximate number An approximate number is a number that is close but not exactly equal to another number. It is the counterpart to exact numbers. There is no uncertainty in an exact number, while the definition of an approximate number is one in which uncertainty exists. One way to think of approximate numbers is as numbers that arise through measurement or calculation. An exact number on the other hand, can be thought of as one that arises from counting. For example, there are 4 clovers on a four-leaf clover. 4 is an exact number. If the leaf of a four-leaf clover measured 7 cm, the 7 cm is an approximate number. This is because the measured length, while closer to 7 cm than it is to 7.1 cm or 7.2 cm, is still only an approximate measure. There is no way to exactly measure the length of the leaf. With highly precise tools, the measured value can be very close to the exact value, but we can never be certain that the measured value is exact. Example Chad saw a sign at the grocery store that stated: 3 candy bars for \$1.00. At this rate, what was the cost of each candy bar? \$1.00 ÷ 3 = 0.3, where the line over the 3 indicates that it repeats indefinitely. This means that the real value is somewhere between \$0.33 and \$0.34, since 0.3 cannot be expressed exactly in decimals. Both \$0.33 and \$0.34 are approximate numbers for the exact cost of each bar. It is not possible to charge Chad \$0.3, so instead we could say that he was charged \$0.33 for two candy bars, and \$0.34 for the third candy bar. ## Rounding There are many different ways to round numbers. Rounding to an integer, as we did above, is the most common. ### Rounding down to an integer Rounding down to an integer is referred to as taking the floor, which means rounding towards negative infinity. 25.9 → 25 -25.1 → -26 Half values are rounded towards negative infinity. 25.5 → 25 -25.5 → -26 ### Rounding up to an integer Rounding up to an integer is referred to as taking the ceiling which means rounding towards positive infinity. 25.1 → 26 -25.9 → -25 Half values are rounded towards infinity. 25.5 → 26 -25.5 → -25 ### Rounding towards zero Rounding towards zero is referred to as truncating, which means rounding away from infinity. 25.9 → 25 -25.1 → -25 Half values are rounded away from infinity. 25.5 → 25 -25.5 → -25 ### Rounding away from zero Rounding away from zero means rounding towards infinity. 25.1 → 26 -25.1 → -26 Half values are rounded towards infinity. 25.5 → 26 -25.5 → -26 There are also other rounding methods, but these are some of the most common.
NCERT Solutions: Basic Geometrical Ideas (Exercise 4.4, 4.5 & 4.6) # NCERT Solutions for Class 6 Maths Chapter 4 - Basic Geometrical Ideas (Exercise 4.4, 4.5 and 4.6) ### Exercise 4.4 (Old NCERT) Ques 1: Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its interior? Ans: A is neither interior of the figure nor exterior of triangle. It is a vertex Ques 2: (a) Identify three triangles in the figure. (b) Write the names of seven angles. (c) Write the names of six line segments. (d) Which two triangles have ∠B as common? Ans: (a) The three triangles are: ΔABC, ΔABD, ΔADC (c) Line segments are: (d) Triangles having common ∠B: ΔABC, ΔABD, ### Exercise 4.5 (Old NCERT) Ques 1: Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonals in the interior or exterior of the quadrilateral? Ans: Diagonal PR and diagonal SQ meet at O, which is inside the quadrilateral. Ques 2: Draw a rough sketch of a quadrilateral KLMN. State, (a) Two pairs of opposite sides. (b) Two pairs of opposite angles. (c) Two pairs of adjacent sides. (d) Two pairs of adjacent angles. Ans: (a) Pair of opposite sides: KL and MN, KN and LM (b) Pair of opposite angles: ∠K and ∠M, ∠L and ∠N (c) Pair of adjacent sides: KN and NM, KL and LM (d) Pair of adjacent angles: ∠K and ∠N, ∠L and ∠M Ques 3: Investigate: Use strip and fasteners to make a triangle and a quadrilateral. Try to push inward at any one vertex of the triangle. Do the same to the quadrilateral. Is the triangle distorted? Is the quadrilateral distorted? Is the triangle rigid? Why is it that structures like electric towers make use of triangular shapes and not quadrilateral? Ans: O is common to both the angles ∠AOC and ∠BOC. No, the triangle is not distorted but the quadrilateral is distorted and also the triangle is rigid. Structures like electric towers make use of triangular shape so that they could not be distorted and they could be rigid. ### Exercise 4.6 (Old NCERT) Ques 1: From the figure, identify: (a) The centre of circle (c) a diameter (d) a chord (e) Two points in the interior (f) a point in the exterior (g) a sector (h) a segment Ans: (a) O is the centre. (b) Three radii: OA, OB and OC (c) A diameter: AC (d) A chord: ED (e) Interior points: O, P (f) Exterior point: Q (g) A sector: OAB (h) A segment: ED Ques 2: (a) Is every diameter of a circle also a chord? (b) Is every chord of circle also a diameter? Ans: (a) Yes, every diameter of a circle is also a chord. (b) No, every chord of a circle is not a diameter. Ques 3: Draw any circle and mark: (a) Its centre. (c) A diameter. (d) A sector. Ans: (a) Its centre is O. (c) A diameter is AC. (d) A sector OAB. Ques 4: Say true or false: (a) Two diameters of a circle will necessarily intersect. (b) The centre of a circle is always in its interior. Ans: (a) True (b) True The document NCERT Solutions for Class 6 Maths Chapter 4 - Basic Geometrical Ideas (Exercise 4.4, 4.5 and 4.6) is a part of the Class 6 Course NCERT Textbooks & Solutions for Class 6. All you need of Class 6 at this link: Class 6 494 docs ## FAQs on NCERT Solutions for Class 6 Maths Chapter 4 - Basic Geometrical Ideas (Exercise 4.4, 4.5 and 4.6) 1. What are some basic geometric shapes? Ans. Some basic geometric shapes are point, line, line segment, ray, angle, triangle, quadrilateral, circle, square, rectangle, and polygons. 2. What is the difference between a line segment and a ray? Ans. A line segment is a part of a line that has two endpoints, whereas a ray is a part of a line that has only one endpoint and extends infinitely in one direction. 3. What is an acute angle? Ans. An acute angle is an angle that measures less than 90 degrees. 4. What are the properties of a rectangle? Ans. A rectangle is a quadrilateral with four right angles, opposite sides that are parallel and congruent, and diagonals that are congruent and bisect each other. 5. What is a polygon? Ans. A polygon is a closed figure made up of line segments that are joined together. It has three or more straight sides and angles. ## NCERT Textbooks & Solutions for Class 6 494 docs ### Up next Explore Courses for Class 6 exam ### Top Courses for Class 6 Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , , , , ;
Credit:The Washington Post/Getty ImagesThe Washington PostGetty Images Q: # How do you find the sum or difference of fractions? A: To find the sum or difference of fractions, first find the lowest common denominator (LCD) of each fractions. Once you find the LCD, add or subtract the numerators to discover your answer. Know More 1. ### Find the LCD When the fractions have the same denominator, or bottom number, continue on to solve the problem. When the fractions have different denominators, convert them to equivalent fractions using the LCD rule. To find the LCD, calculate the least common multiple of all of the denominators. For example, you are given two-thirds and four-sixths. The least common multiple is six. Convert two-thirds to an equivalent fraction by multiplying the numerator and the denominator by two. The equivalent fraction is four-sixths. 2. ### Find the sum Perform addition to find the sum of two or more fractions. When the denominators are the same, add the numerators, or top numbers, to calculate the answer. The denominator doesn't change. For example, two-thirds plus four-thirds equals six-thirds. 3. ### Find the difference Subtract the smaller fraction from the larger fraction to calculate the difference. When the denominators are the same, subtract the smallest numerator from the largest. Write the answer over the original denominator for the correct answer. For example, five-eighths minus three-eighths equals two-eighths. Sources: ## Related Questions • A: Fractions were created by the ancient Egyptians in 1800 B.C. Egyptians created a number system on a base 10 concept, which is similar to the numerical systems used in the United States and around the world for creating fractions. Originally, Egyptians used pictures as well as numerals to display and differentiate numbers. • A: To add fractions and whole numbers, convert a whole number into a fraction with the same denominator as the fraction that needs to be added. Add up the numerators and keep the denominator the same.
# Solve my inequalities These sites allow users to input a Math problem and receive step-by-step instructions on how to Solve my inequalities. We can solving math problem. ## Solving my inequalities When you try to Solve my inequalities, there are often multiple ways to approach it. Math can be a tough subject for a lot of students. Word problems in particular can be tricky, since they often require students to use a variety of Math concepts in order to solve them. Luckily, there are a number of online Math word problem solvers that can help. These websites allow students to enter a word problem and receive step-by-step instructions on how to solve it. In addition, many of these websites also provide helpful Math tools, such as calculators and conversion charts. As a result, Math word problem solver websites can be a valuable resource for students who are struggling with Math word problems. To solve a factorial, simply multiply the given number by every number below it until you reach one. So, to solve 5!, you would multiply 5 by 4, then 3, then 2, and then 1. The answer would be 120. It is important to start with the given number and work your way down, rather than starting with one and working your way up. This is because the factorial operation is not commutative - that is, 5! is not the same as 1 x 2 x 3 x 4 x 5. When solving factorials, always start with the given number and work your way down to one. This method is based on the Taylor expansion of a function, which states that a function can be approximated by a polynomial if it is differentiable. The Taylor series method involve taking the derivative of the function at each point and then adding up all of the terms to get the sum. This can be a very tedious process, but it is often the only way to find the sum of an infinite series. There are some software programs that can help to automate this process, but they can be expensive. In other words, x would be equal to two (2). However, if x represented one third of a cup of coffee, then solving for x would mean finding the value of the whole cup. In this case, x would be equal to three (3). The key is to remember that, no matter what the size of the fraction, solving for x always means finding the value of the whole. With a little practice, solving for x with fractions can become second nature. ## Math solver you can trust Everyone is saying bad things about the app. But honestly, it's such a life saver. I haven't had that ad. And they have that feature "Plus" which I understand. They have to earn money. But some people think that everything should be free. They're asking too much. If they want that feature, they should buy it. But. As I said the app has saved me and thought me math. And now that they gave everyone the "plus" feature for free until 20th April. That's amazing. 6/5 Esmeralda Garcia This is wonderful, it gives solutions and step by step solving. It has helped me to understand math problems. But make all the steps to be explained, "why". Very helpful, and reliable app! It helps me solve most of my math problems and shows solving step by step. Isabella Gonzales How to solve elimination method Pre algebra help calculator System of two equations solver System solver calculator Solving complex numbers
# Lines and Angles: A Comprehensive Overview 1. A Level Maths Tutorials 2. Geometry Tutorials 3. Lines and Angles Lines and angles are two of the most fundamental concepts in geometry, and understanding them is essential for success in mathematics. With this comprehensive overview, you will gain an in-depth look at what lines and angles are, how they are related, and how to use them in problem solving. We'll cover the basics of lines and angles, as well as their more advanced applications, so you can apply your knowledge to geometry and beyond. From the basic definitions of lines and angles to more challenging applications, this overview will provide the essential background knowledge to help you understand the fundamentals of mathematics. We'll discuss properties of lines and angles, how to calculate measurements, and how to apply your knowledge to more complex problems. By the end of this overview, you'll be better equipped to tackle any geometry problem that comes your way. Lines and angles are two of the fundamental building blocks of geometry. Lines are defined as straight paths connecting two points, while angles are formed when two lines meet at a common point. Understanding the properties of lines and angles is essential for problem-solving in many different scenarios. #### Basic Properties of Lines and Angles Lines can be categorized as vertical, horizontal, or diagonal. They can also be classified as parallel or intersecting. When two lines intersect, they form an angle. Angles are measured in degrees, with a full circle having 360 degrees. A right angle has 90 degrees, while a straight angle has 180 degrees. There are also other types of angles, including complementary angles (two angles that add up to 90 degrees), supplementary angles (two angles that add up to 180 degrees), vertical angles (two angles that are opposite each other and have the same measure), adjacent angles (two angles that share a side and a vertex), and linear pairs (two angles that add up to 180 degrees). #### Calculating Lines and Angles The most basic way to calculate lines and angles is to use a ruler or protractor. With these tools, you can measure the length of a line or the degree of an angle. There are also more complex methods for calculating lines and angles, such as using trigonometry or the Pythagorean Theorem. Knowing how to use these methods can be useful in more advanced mathematics problems. #### Uses of Lines and Angles in Everyday Life Lines and angles are used in many everyday scenarios. Architects use them to construct buildings, engineers use them to design bridges, and navigators use them to determine direction. In mathematics, they are used to solve equations and calculate distances. Even in sports, athletes use lines and angles to perfect their technique. #### Parallel Lines Parallel lines are lines that never intersect, no matter how far they extend. They have the same slope and the same distance between them at all points. Parallel lines are used in architecture to create strong foundations for buildings, as well as in navigation to mark distances on a map. Angles of Elevation and DepressionAngles of elevation and depression are used in navigation and surveying to measure the height of an object in relation to the observer’s position. These angles can be calculated using trigonometry or other more complex mathematical methods. #### Angle Bisectors An angle bisector is a line or ray that divides an angle into two equal parts. This concept is used in construction to ensure that walls and other structures are built at the correct angle. It is also used in geometry problems to calculate the area of shapes or triangles. ## Advanced Topics Related to Lines and Angles Beyond the basics, there are a variety of advanced topics related to lines and angles that are important to know. These include the angle sum theorem, the parallel line theorem, the alternate interior angle theorem, and the properties of perpendicular lines. Each of these topics is essential for understanding more complex principles of geometry. #### Angle Sum Theorem The angle sum theorem states that the sum of the angles in a triangle is equal to 180 degrees. This theorem can be applied to other shapes as well, such as quadrilaterals, where the sum of the angles will be 360 degrees. Knowing this theorem is important for understanding how to calculate interior and exterior angles in polygons. #### Parallel Line Theorem The parallel line theorem states that if two parallel lines are cut by a transversal, then the corresponding angles are equal. This can be used to find missing angles when working with parallel lines and transversals. Additionally, it can be used to prove that two lines are parallel if certain angles are known. #### Alternate Interior Angle Theorem The alternate interior angle theorem states that if two parallel lines are cut by a transversal, then the alternate interior angles are equal. This theorem is similar to the parallel line theorem but applies to the interior angles rather than the corresponding angles. #### Properties of Perpendicular Lines The last advanced topic related to lines and angles is the properties of perpendicular lines. Perpendicular lines are two lines that intersect at a 90-degree angle. There are many properties associated with perpendicular lines including the fact that opposite angles formed by intersecting perpendicular lines are equal. ## How Lines and Angles Are Used in Everyday Life Lines and angles are essential for understanding the world around us. From the layout of a room to the construction of a bridge, lines and angles are used in almost every aspect of everyday life. We can see lines and angles in the simple task of mowing the lawn. The mower follows a straight line with the blade at a fixed angle. This allows for an even cut across the grass. In construction, lines and angles are especially important. Many structures require precise measurements and calculations to ensure stability and safety. Architects use lines and angles to measure the size, shape, and orientation of their buildings. The same principles apply to roads and bridges. Engineers must consider the correct angles and lengths when designing roads to make sure they can handle the traffic load. Similarly, bridges must be able to support a certain amount of weight while staying structurally sound. Surveying is another area where lines and angles play an important role. Surveyors use precise instruments to measure and calculate distances, angles, and elevations. This information is then used to create maps and plans for construction projects. Finally, lines and angles are used in navigation. By understanding how the Earth is oriented, navigators can determine their position on a map or chart. They must also be able to accurately measure angles to ensure they are traveling in the right direction. ## The Basics of Lines and Angles Lines and angles are essential components of geometry that are used to describe the shape and size of objects. A line is a straight path that extends in two directions without ending, while an angle is the difference between two lines that meet at a common point. Lines and angles can be measured using various units, such as degrees or radians. Lines can be categorized into different types, such as horizontal lines, vertical lines, oblique lines, parallel lines, and perpendicular lines. Horizontal lines extend from left to right without any change in elevation, while vertical lines extend from top to bottom without any change in elevation. Oblique lines have a slant and may be angled in any direction. Parallel lines are two lines that never intersect, while perpendicular lines are two lines that intersect at a right angle. Angles are typically measured in degrees, with the most common angles being right angles (90°), acute angles (less than 90°), and obtuse angles (more than 90°). Angles can also be classified as complementary angles (two angles that add up to 90°) or supplementary angles (two angles that add up to 180°). Lines and angles are used in a variety of everyday situations, such as constructing buildings and measuring distances. They are also integral components of more complex mathematical operations, such as trigonometry and calculus. By understanding the basics of lines and angles, students can better comprehend more advanced math concepts. In conclusion, lines and angles are two of the most fundamental building blocks of geometry. They have many uses in everyday life, from forming the corners of buildings to the angles of a triangle. Knowing how to calculate them and understanding their properties is essential for anyone studying geometry or mathematics. By understanding the basics of lines and angles, as well as more advanced topics, you can gain an appreciation for the importance of these shapes in our world.
# 7th Grade Math Circles A circle is a simple closed shape where all points have the same distance from the centre. The centre of a circle is the centre point in a circle, from which all the distances to the points on the circle are equal. The radius is the distance from the centre to any point on the circle. The diameter of the circle is defined as the distance across the circle. The length of any chord passing through the centre is called the diameter of the circle. It is twice the radius. ## Important points of the circle • The distance around the outside of the circle is called the circumference of the circle. • Circumference of the circle = 2πr. • Area of circle = π * (radius)2 = π * r= πr2 • The circles are said to be congruent if they have equal radii. • Equal chords of a circle subtend equal angles at the centre. • The radius drawn perpendicular to the chord bisects the chord. • Circles having different radii are similar. • The chords that are equidistant from the centre are equal in length. • The distance from the centre of the circle to the longest chord (diameter) is zero. ## Circle Example Here are a few examples of grade 7 circle problems. 1. Find the circumference of the circles with a radius of 14 cm (Take π = 22/7) Ø Circumference of the circle = 2πr Ø = 2 * 22/7 * 14 Ø = 2 * 22 * 2 Ø = 88 cm 2. Find the area of the following circles with a radius of 5 cm (Take π = 22/7) Ø Area of circle = πr2 Ø = π * r * r = 22/7 * 5 * 5 Ø = (22 * 25)/7 Ø = 550/7mm2 Teachers are looking for lessons, activities, worksheets, quiz, mock test papers that are suitable for 7th grade math. Here's your compilation of lessons suitable for 7th grade to share with your students. ## Sum Up Studying grade 7 circles help students to develop a robust understanding of basic concepts related to circles and their components. ### What teachers are saying about BytelearnWhat teachers are saying Stephen Abate 19-year math teacher Carmel, CA Any math teacher that I know would love to have access to ByteLearn. Jennifer Maschino 4-year math teacher Summerville, SC “I love that ByteLearn helps reduce a teacher’s workload and engages students through an interactive digital interface.” Rodolpho Loureiro Dean, math program manager, principal Miami, FL “ByteLearn provides instant, customized feedback for students—a game-changer to the educational landscape.”
# Understanding 1/8th As A Decimal: Conversion Methods, Examples, And Importance // Thomas Discover the steps to convert 1/8th to a decimal using division methods, explore real-world examples, and understand the importance of mastering this concept for everyday life, academics, and problem-solving skills. Plus, get valuable tips for working with decimals to avoid common mistakes and improve your accuracy. ## Understanding 1/8th as a Decimal ### Explanation of Fractions and Decimals Fractions and decimals are two ways to represent numbers that are not whole numbers. Fractions are a way of expressing a part of a whole or a ratio between two numbers, while decimals are a way of expressing numbers in a base-10 system. Both fractions and decimals can be used to represent the same value, but they are written and used in slightly different ways. ### Converting Fractions to Decimals Converting fractions to decimals is a useful skill that allows us to work with numbers in different formats. To convert a fraction to a decimal, we divide the numerator (the top number) by the denominator (the bottom number). In the case of 1/8th, we divide 1 by 8. The result is 0.125, which is the of 1/8th. ### Simplifying Fractions Simplifying fractions involves reducing them to their simplest form. In the case of 1/8th, it is already in its simplest form because 1 and 8 do not have any common factors other than 1. However, it is important to note that simplifying fractions is necessary when working with larger and more complex fractions. ### Representing 1/8th as a Decimal Representing 1/8th as a decimal is straightforward. As mentioned earlier, the decimal equivalent of 1/8th is 0.125. This means that if we have one whole divided into eight equal parts, each part represents 0.125 of the whole. Understanding 1/8th as a decimal allows us to work with numbers in decimal form and perform calculations more easily. It is important to grasp the concept of fractions and decimals, as they are used in various real-life situations and academic fields. In the following sections, we will explore different methods for calculating and representing 1/8th as a decimal, as well as its importance in different contexts. ## How to Calculate 1/8th as a Decimal Calculating fractions as decimals is an essential skill in math, and understanding how to convert 1/8th to a decimal is no exception. There are several methods you can use to perform this calculation, including the division method, long division method, using a decimal equivalent chart, or even utilizing a calculator. ### Division Method The division method is a straightforward way to calculate 1/8th as a decimal. Simply divide the numerator (1) by the denominator (8). In this case, 1 divided by 8 equals 0.125. Therefore, 1/8th as a decimal is 0.125. ### Long Division Method Another method you can use to calculate 1/8th as a decimal is the long division method. Begin by writing 1 as the dividend and 8 as the divisor. Then, perform the long division process, dividing 1 by 8. The quotient you will obtain is 0.125, which represents 1/8th as a decimal. ### Decimal Equivalent Chart If you prefer a visual reference, you can consult a decimal equivalent chart to find the decimal representation of 1/8th. These charts list common fractions and their corresponding decimal values. Locate 1/8th on the chart, and you’ll find that it is represented by the decimal 0.125. ### Calculator Use Using a calculator is another option for calculating 1/8th as a decimal. Most calculators have a division function that allows you to input the fraction and obtain the decimal result. Enter 1 divided by 8, and the calculator will display the decimal equivalent of 0.125. By utilizing any of these methods – division, long division, decimal equivalent chart, or calculator – you can confidently calculate 1/8th as a decimal. Each method offers its advantages, so choose the one that suits your preference and helps you understand the concept better. ## Examples of 1/8th as a Decimal ### 1/8th as a Decimal in Numerical Form When we want to represent 1/8th as a decimal, we can do so by performing a simple division. By dividing 1 by 8, we find that the of 1/8th is 0.125. This means that if we have one whole item divided into 8 equal parts, each part would be 0.125 of the whole. ### 1/8th as a Decimal in Written Form In written form, 1/8th as a decimal is expressed as 0.125. This can be read as “zero point one two five”. It is important to note that the zero before the decimal point is typically omitted when reading the decimal aloud. ### Real-world Examples of 1/8th as a Decimal Understanding how to represent 1/8th as a decimal can be useful in various real-world scenarios. Here are a few examples: 1. Cooking and Baking: Recipes often call for measurements in fractions, and being able to convert them to decimals can make it easier to follow the instructions. For instance, if a recipe requires 1/8th of a cup of an ingredient, you would know that it is equal to 0.125 cups. 2. Construction and Carpentry: Measurements in construction and carpentry often involve fractions. Being able to convert fractions to decimals helps in accurately measuring and cutting materials. For example, if you need to cut a piece of wood that is 1/8th of an inch long, converting it to a decimal would give you a length of 0.125 inches. 3. Financial Calculations: Understanding decimals is crucial in financial calculations, such as calculating interest rates or determining percentages. For instance, if you need to calculate 1/8th of a dollar, you would know that it is equal to 0.125 dollars. By being able to convert 1/8th to a decimal, you can confidently apply this knowledge in various everyday situations. ## Importance of Understanding 1/8th as a Decimal ### Everyday Applications Understanding the concept of 1/8th as a decimal is crucial in various everyday situations. Whether you are cooking, measuring ingredients, or calculating proportions, knowing how to work with fractions and decimals is essential. For example, if a recipe calls for 1/8th of a cup of an ingredient, you need to convert that fraction into a decimal to accurately measure it. By understanding how to represent 1/8th as a decimal, you can confidently tackle daily tasks that involve measurements and calculations. Proficiency in working with fractions and decimals, including 1/8th, is not only applicable in everyday life but also in academic and professional settings. In mathematics, fractions and decimals are fundamental concepts that form the basis for more complex mathematical operations. Understanding how to convert fractions to decimals and vice versa, including 1/8th, is a crucial skill for students studying mathematics, science, engineering, and various other disciplines. Additionally, many professions require the ability to work with decimals, such as finance, accounting, and statistics. ### Financial Implications The ability to understand and work with 1/8th as a decimal has significant financial implications. In personal finance, knowing how to calculate percentages and decimals accurately is vital for budgeting, saving, and making informed financial decisions. For example, if you need to calculate 1/8th of a dollar amount, understanding the decimal representation of 1/8th enables you to calculate discounts, interest rates, and determine appropriate allocations of funds. Moreover, in business and economics, decimals are used extensively in financial statements, investment analysis, and profit calculations. Proficiency in working with 1/8th as a decimal is essential for individuals and professionals to navigate the financial landscape effectively. ### Problem-Solving Skills Understanding 1/8th as a decimal develops problem-solving skills that are applicable in various contexts. By learning how to convert fractions to decimals, including 1/8th, individuals enhance their analytical thinking and reasoning abilities. Problem-solving often requires breaking down complex situations into more manageable components, and working with fractions and decimals is an essential part of this process. Furthermore, the ability to convert between fractions and decimals helps individuals make comparisons, estimate values, and solve real-world problems involving measurements, proportions, and ratios. Developing these problem-solving skills equips individuals with a valuable toolset to tackle challenges in both academic and professional settings. In summary, understanding 1/8th as a decimal holds significant importance in everyday life, academic pursuits, professional careers, financial decision-making, and problem-solving. Proficiency in working with fractions and decimals, including 1/8th, empowers individuals to navigate various situations with confidence and accuracy. By grasping the concepts covered in this section, you will be equipped with essential skills that have practical applications in numerous aspects of life. ## Tips for Working with 1/8th as a Decimal ### Rounding Decimal Places When working with decimals, it is often necessary to round the decimal places to a specific number of digits. Rounding can help simplify calculations and make the results easier to understand. When rounding 1/8 as a decimal, it is important to consider the digit in the next decimal place. If the digit is 5 or greater, you round up to the next higher digit. If the digit is 4 or lower, you round down to the current digit. For example, when rounding 1/8, which is equal to 0.125, to two decimal places, we would round it to 0.13. ### Estimating with Decimals Estimating with decimals allows us to quickly approximate values without going through the process of calculating exact decimal equivalents. When working with 1/8 as a decimal, we can estimate its value by using a benchmark decimal, such as 0.1, which is equivalent to 1/10. Since 1/8 is slightly larger than 1/10, we can estimate it to be around 0.12. This estimation can be useful in situations where we need a quick approximation or when calculating mentally. ### Common Mistakes to Avoid When working with 1/8 as a decimal, there are a few common mistakes that people often make. One common mistake is forgetting to divide the numerator by the denominator when converting a fraction to a decimal. Another mistake is incorrectly rounding the decimal places, leading to inaccurate results. It is important to double-check calculations and pay attention to the details to avoid these mistakes. By being aware of these common errors, we can ensure the accuracy of our calculations and avoid unnecessary confusion. ### Practice and Review Suggestions To improve your understanding and proficiency in working with 1/8 as a decimal, it is essential to practice and review regularly. Here are some suggestions to help you strengthen your skills: 1. Solve practice problems: Find worksheets or online resources that provide exercises specifically focused on converting fractions to decimals and vice versa. Practice solving problems involving 1/8 as a decimal to reinforce your understanding. 2. Review decimal equivalents: Familiarize yourself with decimal equivalents of common fractions, including 1/8. By memorizing these decimal values, you can quickly and confidently work with fractions in decimal form. 3. Use real-world examples: Look for opportunities in your everyday life to apply your knowledge of 1/8 as a decimal. For instance, when dividing a pizza into equal slices, you can calculate the decimal value of each slice to understand its portion. 4. Seek feedback: Share your work with a teacher, tutor, or knowledgeable friend who can provide feedback and help identify any areas where you may need improvement. Constructive criticism can help you refine your skills and enhance your understanding of 1/8 as a decimal. By incorporating these practice and review suggestions into your learning routine, you can build a solid foundation and become confident in working with 1/8 as a decimal. Remember, practice makes perfect, so keep challenging yourself and seeking opportunities to apply your knowledge. Contact 3418 Emily Drive Charlotte, SC 28217 +1 803-820-9654
# How do you find the diagonals of a kite? Contents In order to solve this problem, first observe that the red diagonal line divides the kite into two triangles that each have side lengths of and. Notice, the hypotenuse of the interior triangle is the red diagonal. Therefore, use the Pythagorean theorem: , where the length of the red diagonal. ## How do you find the diagonals of a kite with the given area? If a kite has diagonals one and two, then its area can be found by calculating half of their product. In this question, the two diagonals of the kite are the lines and . And therefore, we have that the area is equal to one-half of the length of multiplied by the length of . ## How do you determine diagonals? You can find the diagonal of a rectangle if you have the width and the height. The diagonal equals the square root of the width squared plus the height squared. ## Are the diagonals of a kite 90 degrees? The intersection of the diagonals of a kite form 90 degree (right) angles. This means that they are perpendicular. The longer diagonal of a kite bisects the shorter one. This means that the longer diagonal cuts the shorter one in half. ## What is a formula for a kite? The area of a kite is half the product of the lengths of its diagonals. The formula to determine the area of a kite is: Area = 12×d1×d2 1 2 × d 1 × d 2 . Here d1 and d2 are long and short diagonals of a kite. ## Are diagonals of a kite equal? The two diagonals are not of the same length. The diagonals of a kite intersect each other at right angles. It can be observed that the longer diagonal bisects the shorter diagonal. A pair of diagonally opposite angles of a kite are said to be congruent. ## How many diagonals does a kite have? Every kite has two diagonals. ## What is the formula for finding diagonals in a polygon? According to the formula, number of diagonals = n (n-3)/ 2. So, 11-sided polygon will contain 11(11-3)/2 = 44 diagonals. Example 2: In a 20-sided polygon, one vertex does not send any diagonals. ## Are diagonals of a kite perpendicular? Proof: The diagonals of a kite are perpendicular. ## How do you find the area of a kite without diagonals? Kite area formula If you know two non-congruent side lengths and the size of the angle between those two sides, use the formula: area = a * b * sin(α) , where α is the angle between sides a and b . ## How do you measure a kite? What is the area of a kite equal to? The area of a kite measures the space inside the four sides. The most common way to find the area is by using the formula A = xy/2, where x and y are the lengths of the diagonals.
# Standards-Aligned Resources for Your Classroom Manage curriculum, personalize instruction, and detect early warnings with Kiddom ### Kiddom Supports All Standards, Including 3.OA.9 Common Core Mathematics Standards ### 3.OA.9 Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends. ## View Standards Related to3.OA.9 ### 3.OA.1 Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7. View 3.OA.1 ### 3.OA.2 Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8. View 3.OA.2 ### 3.OA.3 Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.1 View 3.OA.3 ### 3.OA.4 Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 × ? = 48, 5 = ? ÷ 3, 6 × 6 = ?. View 3.OA.4 ### 3.OA.5 Apply properties of operations as strategies to multiply and divide.2 Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.) View 3.OA.5 View All Common Core Mathematics Standards ### Looks like we don't have content for 3.OA.9 yet We've noted your interest and will work on it. In the mean time, you can search 70,000 other pieces of standards-aligned content in Kiddom’s K-12 library. Search Kiddom's Library # Save Time With Kiddom Only the Kiddom platform bridges the curriculum, instruction, and assessment cycle to save teachers time and improve student outcomes regardless of the curriculum used. Free for Teachers! Quickly Find Resources Select from 70,000+ standards-aligned videos, quizzes, lessons, and more. Share With Students Assign lessons and assessments to a single student (or group of students). Measure Mastery See how students are performing against individual standards or skills. Free for Teachers! "I can see where my class and any student is at any moment in their educational journey. This way I can take action to assist them to work towards mastery." #### Mr. Albrecht High School Teacher
# Factors of 980: Prime Factorization, Methods, and Examples The factors of 980 are the numbers that make 980 completely divisible upon them. They give a zero remainder. The given number’s factors can be positive and negative, provided that the given number is achieved upon multiplication of two-factor integers. ### Factors of 980 Here are the factors of number 980. Factors of 980: 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 49, 70, 98, 140, 196, 245, 490, 980 ### Negative Factors of 980 The negative factors of 980 are similar to its positive aspects, just with a negative sign. Negative Factors of 980: -1, -2, -4, -5, -7, -10, -14, -20, -28, -35, -49, -70, -98, -140, -196, -245, -490, and -980 ### Prime Factorization of 980 The prime factorization of 980 is the way of expressing its prime factors in the product form. Prime Factorization: $2^{2}$ x 5 x $7^{2}$ In this article, we will learn about the factors of 980 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree. ## What Are the Factors of 980? The factors of 980 are 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 49, 70, 98, 140, 196, 245, 490, and 980. These numbers are the factors as they do not leave any remainder when divided by 980. The factors of 980 are classified as prime numbers and composite numbers. The prime factors of the number 980 can be determined using the prime factorization technique. ## How To Find the Factors of 980? You can find the factors of 980 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero. To find the factors of 980, create a list containing the numbers that are exactly divisible by 980 with zero remainders. One important thing to note is that 1 and 980 are the 980’s factors as every natural number has 1 and the number itself as its factor. 1 is also called the universal factor of every number. The factors of 980 are determined as follows: $\dfrac{980}{1} = 980$ $\dfrac{980}{2} = 490$ $\dfrac{980}{4} = 245$ $\dfrac{980}{5} = 196$ $\dfrac{980}{7} = 140$ $\dfrac{980}{10} = 98$ $\dfrac{980}{14} = 70$ $\dfrac{980}{20} = 49$ $\dfrac{980}{28} = 35$ Therefore, 1, 2, 4, 5, 7, 10, 14, 20, and 28, and their whole number quotients 35, 49, 70, 98, 140, 196, 245, 490, and 980 are the factors of 980. ### Total Number of Factors of 980 For 980, there are 18 positive factors and 18 negative ones. So in total, there are 36 factors of 980. To find the total number of factors of the given number, follow the procedure mentioned below: 1. Find the factorization of the given number. 2. Demonstrate the prime factorization of the number in the form of exponent form. 3. Add 1 to each of the exponents of the prime factor. 4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number. By following this procedure, the total number of factors of 980 is given as: Factorization of 980 is 1 x $2^{2}$ x 5 x $7^{2}$. The exponent of 1 and 5 is 1 and that of 2 and 7 is 2. Adding 1 to each and multiplying them together results in 36. Therefore, the total number of factors of 980 is 36. 18 are positive, and 18 factors are negative. ### Important Notes Here are some essential points that must be considered while finding the factors of any given number: • The factor of any given number must be a whole number. • The factors of the number cannot be in the form of decimals or fractions. • Factors can be positive as well as negative. • Negative factors are the additive inverse of the positive factors of a given number. • The factor of a number cannot be greater than that number. • Every even number has 2 as its prime factor, the smallest prime factor. ## Factors of 980 by Prime Factorization The number 980 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors. Before finding the factors of 980 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves. To start the prime factorization of 980, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor. Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 980 can be expressed as: 980 = $2^{2}$ x 5 x $7^{2}$ ## Factors of 980 in Pairs The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given. For 980, the factor pairs can be found as: 1 x 980 = 980 2 x 490 = 980 4 x 245 = 980 5 x 196 = 980 7 x 140 = 980 10 x 98 = 980 14 x 70 = 980 20 x 49 = 980 28 x 35 = 980 The possible factor pairs of 980 are given as (1, 980), (2, 490), (4, 245), (5, 196), (7, 140), (10, 98), (14, 70), (20, 49), and (28, 35). All these numbers in pairs, when multiplied, give 980 as the product. The negative factor pairs of 980 are given as: -1 x -980 = 980 -2 x -490 = 980 -4 x -245 = 980 -5 x -196 = 980 -7 x -140 = 980 -10 x -98 = 980 -14 x -70 = 980 -20 x -49 = 980 -28 x -35 = 980 It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -2, -4, -5, -7, -10, -14, -20, -28, -35, -49, -70, -98, -140, -196, -245, -490, and -980 are called negative factors of 980. The list of all the factors of 980, including positive as well as negative numbers, is given below. Factor list of 980: 1, -1, 2, -2, 4, -4, 5, -5, 7, -7, 10, -10, 14, -14, 20, -20, 28, -28, 35, -35, 49, -49, 70, -70, 98, -98, 140, -140, 196, -196, 245, -245, -490, 490, 980, and -980 ## Factors of 980 Solved Examples To better understand the concept of factors, let’s solve some examples. ### Example 1 How many factors of 980 are there? ### Solution The total number of Factors of 980 is 18. Factors of 980 are 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 49, 70, 98, 140, 196, 245, 490, and 980. ### Example 2 Find the factors of 980 using prime factorization. ### Solution The prime factorization of 980 is given as: 980 $\div$ 2 = 490 490 $\div$ 2 = 245 245 $\div$ 5 = 49 49 $\div$ 7 = 7 7 $\div$ 7 = 1 So the prime factorization of 980 can be written as: $2^{2}$ x 5 x $7^{2}$ = 980
## Binomial Expansion and Binomial Series ### Binomial Expansion and Binomial Series In algebra, we all have learnt the following basic algebraic expansion: $$\hspace{3em} {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$$. We can keep multiplying the expression $${ \small (a + b) }$$ by itself to find the expression for higher index value. For example: $$\\[12pt] {(a + b)}^{3} \ = \ {(a + b)}^{2} \times (a + b)$$ $$\\[12pt] \hspace{1.5em} \ = \ ({a}^{2} + 2ab + {b}^{2}) \times (a + b)$$ $$\\[12pt] \hspace{1.5em} \ = \ {a}^{3} + {a}^{2}b + 2{a}^2b + 2a{b}^2 + a{b}^2 + {b}^{3}$$ $$\hspace{1.5em} \ = \ {a}^{3} + 3{a}^{2}b + 3a{b}^2 + {b}^{3}$$ Of course this is a really tedious way for a very large index/power/exponent number. Instead of doing it manually, we could use a formula called the Binomial Theorem which is shown below: $${(a + b)}^{n} \ = \ \displaystyle \sum_{k=0}^{n} \binom{n}{k} {a}^{n \ – \ k} \ {b}^{k}$$ with $$\displaystyle \binom{n}{k} \ = \ \frac{n!}{k!(n-k)!}$$ $$\hspace{4.4em} = \ {\large \frac{n (n \ – \ 1) (n \ – \ 2) … (n \ – \ k \ + \ 1)}{k(k \ – \ 1)(k \ – \ 2) \ … \ 1} }$$ for $$k \geq 1$$ and $$\displaystyle \binom{n}{0} \ = \ 1$$ $$\\[10pt]$$ Example: $$\\[12pt] {(a + b)}^{10}$$ $$\\[15pt] = \binom{10}{0} {a}^{10}{b}^{0} + \binom{10}{1} {a}^{9}{b}^{1}+…+ \binom{10}{10} {a}^{0}{b}^{10}$$ $$\\[15pt] = {a}^{10} + 10 {a}^{9}b+ 45 {a}^{8}{b}^{2} + … + 10{a}{b}^{9} + {b}^{10}$$ $$\\[10pt]$$ Note that, $$\\[20pt]\quad \ {\displaystyle \binom{10}{0} } \ = \ 1$$ $$\\[20pt]\quad \ {\displaystyle \binom{10}{1} } \ = \ {\large \frac{ 10 }{ 1 } } \ = \ 10$$ $$\\[20pt]\quad \ {\displaystyle \binom{10}{2} } \ = \ {\large \frac{ 10 \ \times \ 9 }{ 2 \ \times \ 1 } } \ = \ 45$$ $$\\[20pt]\quad \ {\displaystyle \binom{10}{3} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 }{ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 120$$ $$\\[20pt]\quad \ {\displaystyle \binom{10}{4} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 \ \times \ 7 }{4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 210$$ $$\\[20pt]\quad \ {\displaystyle \binom{10}{5} } \ = \ {\large \frac{10 \ \times \ 9 \ \times \ 8 \ \times \ 7 \ \times \ 6 }{5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 252$$ , etc. $$\binom{n}{k}$$ is read as “n choose k” or sometimes referred to as the binomial coefficients. Notice that this binomial expansion has a finite number of terms with the k values take the non-negative numbers from 0, 1, 2, … , n. Then the next question would be: Can we still use the binomial theorem for the expansion with negative number or fractional number for the index value? Thankfully, Sir Isaac Newton has shown that the binomial theorem can be generalized to take in any numbers for the index value including the negative and fractional numbers as long as it is within a convergence rule. Using this result, we have the Binomial Series which can be expressed as follows: $$\\[25pt]{(1 + x)}^{n} = \displaystyle \sum_{k=0}^{\infty} \binom{n}{k} {x}^{k}$$ $$\\[10pt]{\small = \ 1 + nx + \frac{(n)(n-1)}{2!}{x}^{2} + \frac{(n)(n-1)(n-2)}{3!}{x}^{3} + … }$$ with $${\small k }$$ is any real number and $${\small -1 \lt x \lt 1 }$$. $$\\[10pt]$$ Example: $$\\[12pt] { {\large \frac{1}{1 + x}} \ = \ (1 + x)}^{-1}$$ $$\\[15pt]{\small = \ 1 + (-1)x + \frac{(-1)(-2)}{2!}{x}^{2} + \frac{(-1)(-2)(-3)}{3!}{x}^{3} + … }$$ $$\\[15pt]{\small = \ 1 \ – \ x \ + \ {x}^{2} \ – \ {x}^{3} \ + \ … \ – \ … }$$ $$\\[15pt]$$ The expansion is valid when $${\small -1 \lt x \lt 1 }$$. $$\\[10pt]$$ Another example: $$\\[12pt] { \frac{1}{\sqrt{(1 + x)}} \ = \ (1 + x)}^{\frac{1}{2}}$$ $$\\[15pt]{\small = \ 1 + \frac{1}{2}x + \frac{(\frac{1}{2})(- \frac{1}{2})}{2!}{x}^{2} + \frac{(\frac{1}{2})(- \frac{1}{2})( – \frac{3}{2})}{3!}{x}^{3} + … }$$ $$\\[15pt]{ = \ 1 + \frac{x}{2} – \frac{{x}^{2}}{8} + \frac{{x}^{3}}{16} \ – \ … + … }$$ $$\\[15pt]$$ The expansion is valid when $${\small -1 \lt x \lt 1 }$$. Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) . $$\\[1pt]$$ EXAMPLE: $${\small 1.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }$$ $$\\[1pt]$$ Express $${\small f(x) }$$ in partial fractions and hence obtain the expansion of $${\small f(x) }$$ in ascending powers of x, up to and including the term in $${\small {x}^{2}}$$. $$\\[1pt]$$ $${\small 2.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }$$ $$\\[1pt]$$ Express $${\small f(x) }$$ in partial fractions and hence obtain the expansion of $${\small f(x) }$$ in ascending powers of x, up to and including the term in $${\small {x}^{3}}$$. $$\\[1pt]$$ $${\small 3.\enspace}$$ Find the value of $${\small {(2 \ + \ x)}^{6} \ – \ {(2 \ – \ x)}^{6} }$$ in ascending powers of x, up to and including the term in $${\small {x}^{3}}$$. Hence, find the value of $${\small {(1.99)}^{6} \ – \ {(2.01)}^{6} }$$. $$\\[1pt]$$ $${\small 4.\enspace}$$ Find the last four terms in the expansion in ascending powers of x of $$(2x – {x}^{2})^{10}$$. $$\\[1pt]$$ $${\small 5.\enspace}$$ Find the value of $${\small \frac{a}{b} }$$ in $${\small {(a + bx)}^{12}}$$ given that the coefficient of $${\small {x}^{2} }$$ is 11 times the coefficient of $${\small x }$$. $$\\[1pt]$$ $${\small 6.\enspace}$$ Find the value of a, b and n in the following expansion: $$\\[1pt]$$ $${\small {(a + bx)}^{n} \ = \ … + {\large\frac{20}{3}}{x}^{2} + 20 {x}^{3} + … }$$. $$\\[1pt]$$ given that $${\small {\large \frac{a}{b}} \ = \ {\large \frac{4}{9}} }$$. $$\\[1pt]$$ $${\small 7.\enspace}$$ Consider the binomial expansion of $${\small {\large \frac{1}{\sqrt{4-x}} } }$$. $$\\[1pt]$$ $${\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}$$ Write down the first 4 terms. $${\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}$$ State the interval of convergence for the complete expression. $${\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}$$ Use the expansion to estimate $${\small {\large \frac{1}{\sqrt{3.6}} } }$$. Check your answer by direct calculation. $$\\[1pt]$$ $${\small 8.\hspace{0.4em}(\textrm{i}).\hspace{0.5em}}$$ Find the first 3 terms in the expansion of $${\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} }$$ in descending powers of x. $$\\[1pt]$$ $${\small\hspace{1em}(\textrm{ii}).\hspace{0.3em}}$$ Hence find the coefficient of $${\small {x}^{4}}$$ in the expansion of $${\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} {( {\large \frac{1}{{x}^{2}}} \ + \ 1 )}^{2} }$$. $$\\[1pt]$$ $${\small 9.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 2(a) and (b) $$\\[1pt]$$ $${\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}$$ Expand $${(2 − 3x)}^{−2}$$ in ascending powers of $$x$$, up to and including the term in $$x^2$$, simplifying the coefficients. $$\\[1pt]$$ $${\small\hspace{1em}(\textrm{b}).\hspace{0.5em}}$$ State the set of values of $$x$$ for which the expansion is valid. $$\\[1pt]$$ $${\small 10.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 1 $$\\[1pt]$$ Expand $${(1 + 3x)}^{\frac{2}{3}}$$ in ascending powers of $$x$$, up to and including the term in $$x^3$$, simplifying the coefficients. $$\\[1pt]$$ $${\small 11.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 9(a) and (b) $$\\[1pt]$$ Let $${\small f(x) \ = \ {\large \frac{ 14 \ – \ 3x \ + \ 2x^{2} }{(2 \ + \ x)( 3 \ + \ x^{2} ) }} }$$. $$\\[1pt]$$ $${\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}$$ Express $${\small f(x) }$$ in partial fractions. $$\\[1pt]$$ $${\small \hspace{1em}(\textrm{b}).\hspace{0.5em}}$$ Hence obtain the expansion of $$f(x)$$ in ascending powers of $$x$$, up to and including the term in $$x^2$$. $$\\[1pt]$$ $${\small 12.\enspace}$$ 9709/13/O/N/21 – Paper 13 Nov 2021 Pure Maths 1 No 2 $$\\[1pt]$$ $${\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}$$ Find the first three terms, in ascending powers of $$x$$, in the expansion of $${(1 \ + \ ax)}^{6}$$. $$\\[1pt]$$ $${\small \hspace{1em}(\textrm{b}).\hspace{0.5em}}$$ Given that the coefficient of $$x^2$$ in the expansion of $$(1 \ – \ 3x){(1 \ + \ ax)}^{6}$$ is $$−3$$, find the possible values of the constant $$a$$. $$\\[1pt]$$ $${\small 13.\enspace}$$ 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 4 $$\\[1pt]$$ The coefficient of $$x$$ in the expansion of $$\ {(4x \ + \ {\large\frac{10}{x}} )}^{3} \$$ is $$p$$. The coefficient of $${\large\frac{1}{x} }$$ in the expansion of $$\ {(2x \ + \ {\large\frac{k}{{x}^{2}} })}^{5} \$$ is $$q$$. $$\\[1pt]$$ Given that $$\ p \ = \ 6q$$, find the possible values of $$k$$. $$\\[1pt]$$ $${\small 14.\enspace}$$ 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 1 $$\\[1pt]$$ The coefficient of $$x^2$$ in the expansion of $$\ (4 \ + \ ax){(1 \ + \ {\large\frac{x}{2}})}^{6} \$$ is $$3$$. $$\\[1pt]$$ Find the value of the constant $$a$$. $$\\[1pt]$$ PRACTICE MORE WITH THESE QUESTIONS BELOW! $${\small 1.\enspace}$$ Expand $$\frac{1}{\sqrt[3]{(1 \ + \ 6x)}}$$ in ascending powers of x up to and including the term in $${\small x^3 }$$, simplifying the coefficients. $${\small 2. \enspace}$$ Expand $$\frac{4}{\sqrt{(4 \ – \ 3x)}}$$ in ascending powers of x up to and including the term in $${\small x^2 }$$, simplifying the coefficients. $${\small 3. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{3{x}^{2} \ + \ x \ + \ 6}{(x \ + \ 2)({x}^{2} \ + \ 4) }} }$$ (i). Express $${\small f(x) }$$ in partial fractions. (ii). Hence obtain the expansion of $${\small f(x) }$$ in ascending powers of x, up to and including the term in $${\small {x}^{2}}$$. $${\small 4. \enspace}$$ Express $${\small {(3 \ + \ 4x)}^{\frac{1}{2}} }$$ as a series of descending powers of x up to and including the third non-zero coefficient. State the set of values of x for which the series expansion is valid. $${\small 5. \enspace}$$ (i). Given that the first three terms in the expansion of $${\small {(1 \ + \ ax)}^{b}}$$ in ascending powers of x are $${\small 1 \ + \ x \ + \ \frac{3}{2}{x}^{2}}$$, where a and b are constants, find the values of a and b. (ii). Hence, with the values of a and b found in (i), expand $$\frac{ {(1 \ + \ ax)}^{b} }{1 \ – \ x}$$ as a series in ascending powers of x up to and including the term in $${\small x^2 }$$. $${\small 6. \enspace}$$ Expand $${\small {(1-2p)}^{8} }$$ up to and including the term in $${\small p^3 }$$. Hence, find the first four terms in the expansion in ascending powers of x of $${\small {(1 \ – \ 4x \ + \ \frac{2}{x})}^{8} }$$. $${\small 7. \enspace}$$ Find the coefficient of $${\small {x}^{2}}$$ in the expansion of $${\small {(1 \ + \ 3x)}^{2}{(1 \ – \ 3x)}^{10} }$$. $${\small 8. \enspace}$$ Find the coefficient of $${ \small x }$$ in the expansion of $${\small {(3x \ – \ \frac{2}{x})}^{9} }$$. $${\small 9. \enspace}$$ Find the value of a and b in the following expansion: $${\small \hspace{1.5em} {(a + bx)}^{9} \ = \ … + 672{x}^{3} + 252 {x}^{4} + … }$$ $${\small 10.\enspace}$$ Find the value of a and b in the expansion of $${\small {(1 \ + \ ax)}^{5}{(1 \ – \ bx)}^{7} }$$ if the coefficients of $${ \small x }$$ and $${\small {x}^{2}}$$ are 3 and -9 respectively. As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
# CAT 2017 Important Topics in Logical Reasoning: Venn Diagrams Updated : Sep 9, 2017, 10:00 By : N Shiva Guru Venn Diagrams Venn diagram is the pictorial representation of the set and also the operation involved in the sets. We often use circles to represent the sets and overlapping of the circles to represent the common elements in two or more sets. The universal set U is represented by interior of a rectangle and its subsets are represented by interior of circles within the rectangle. Any values that belong to more than one set will be placed in the sections where the circles overlap. Below are some examples of Venn-diagrams. U = universal set A = subset of U A’ = complement of A The representation of A U B in Venn diagram is The representation of A ∩ B in Venn diagram is Difference of set A from B i.e. A – B Only A or Only B Formula: n (A U B) = n(A)+n(B)- n(A∩B) The Venn diagram for 3 sets Formula: n(AUBUC)= n(A)+n(B)+n(C)- n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C) Concept of Maxima and Minima Let's have a look at the Venn diagram of two sets again: In the beginning, the number of elements in all the areas is 0, as shown above Let’s insert 1 element in A∩B now, we can see that just by adding 1 to A∩B the number of elements in both A and B increased by 1. The total number of elements in all areas combined is 1 only (0+1+0) but if we add the number of elements in A and B, the addition will come up 2. Take 3 sets Venn diagram Here we have added 1 element to intersection of A and B but not C . we can see that A and B both increase by 1 and therefore we get a surplus of 1 element. Now in this above figure we have added 1 element to intersection of all the three sets (A and B and C). We can see that A, B and C all increase by 1element each and therefore, we get a surplus of 2 elements. Now let’s see how this is related to maxima and minima with below example Question 1: According to the survey, at least 70% of people like TOI, at least 75% like HT and at least 80% TheHindu What is the minimum percentage of people who like all three? Solution: Percentage of people who like TOI+HT+TheHindu = 70% +75% +80% => 225% => surplus of 125% Now this surplus can be accommodated by adding elements to either intersection of only two sets or to intersection of only three sets. As the intersection of only two sets can accommodate only a surplus of 100%, the surplus of 25% will still be left. This surplus of 25% can be accommodated by adding elements to intersection of three sets. For that we have to take 25% out of the intersection of only two sets and add it to intersection of three sets. Therefore, the minimum percentage of people who like all three =25%. Alternatively, Let the elements added to intersection of only two sets and intersection of three sets be x and y, respectively. These elements will have to cover the surplus. • x+2y=125% where x+y <=100%. for minimum value of y, we need value of x. • x= 75%, y=25%. Question 2: If in a Survey, organized by any N.G.O on the cold drinks after effects found that 80% of the total people like Coca - Cola and 70% like Limca. What can be the minimum and maximum number of people who drink both the drinks? Solution: Here n(L ∪ C) ≤ 100% Using the formula, n(L) + n(C) – n(L ∩ C) = n(L ∪ C) ≤ 100% 80 + 70 – X ≤ 100 Solving we get X ≥ 50%, Also X ≤ 70% Minimum value = 50% and Maximum value = 70%. Question 3: In a certain city only two newspapers A & B are published. It is known that 25% of the city population reads A & 20% read B while 8% read both A & B. It is also known that 30% of those Solution: Let A & B denote sets of people who read paper A & paper B respectively and in all there are 100 people, then n(A) = 25, n(B) = 20, n (A ∩ B) = 8. Hence the people who read paper A only i.e. n(A – B) = n (A) – n(A ∩ B) = 25 – 8 = 17. And the people who read paper B only i.e. n(B – A) = n (B) – n (A ∩ B) = 20 – 8 = 12. Percentage of people reading an advertisement = [(30% of 17) + (40% of + 12) + (50% of 8)] % = 13.9%. How to score 100 Percentile in CAT 2017 All the very Best, Sep 9CAT & MBA Posted by: Pursuing Unconventionality: Forever Member since Jul 2016
# Procedure for finding absolute extrema for a continuous function on a closed bounded interval Suppose $f$ is a continuous function defined on a closed bounded interval of the form $[a,b]$ with $a < b$. The extreme value theorem tells us that $f$ attains its absolute maximum value and absolute minimum value. Our goal is to determine the following four things: 1. The absolute maximum value of $f$ on $[a,b]$. 2. All points in $[a,b]$ at which this absolute maximum value is attained. 3. The absolute minimum value of $f$ on $[a,b]$. 4. All points in $[a,b]$ at which this absolute minimum value is attained. ## Description of the procedure ### Procedure via determination of local extrema We know the following: For a function whose domain is an interval, any point of absolute maximum must be either a point of local maximum or a point of endpoint maximum. Similarly, any point of absolute minimum must be either a point of local minimum or a point of endpoint minimum. Therefore, the following procedure can be used to compute the absolute maximum and minimum values and all the points where these values are attained. Step no. Step summary Step details 1 Find points of local maximum, local minimum, endpoint maximum, endpoint minimum First, compute all the points of local maximum for $f$and alongside, compute all the points of local minimum $f$. Along with this, also determine, for each of the endpoints $a,b$, whether the function has a one-sided local maximum (i.e., endpoint maximum) or one-sided local minimum (i.e., endpoint minimum) at that endpoint. Use the techniques for finding local extrema to do this. 2 Evaluate function at all points and compare Evaluate the function at all the points of local maximum plus endpoint maximum. Compare these values. The largest of the values is the absolute maximum value. All the points where that largest value is attained are the points of absolute maximum. Evaluate the function at all the points of local minimum plus endpoint minimum. Compare these values. The smallest of the values is the absolute minimum value. All the points where that smallest value is attained are the points of absolute minimum. The main problem with this procedure is that in Step (1), we may spend a lot of effort trying to determine the nature of local extremum using the first derivative test or second derivative test for a large number of critical points which ultimately will not turn out to be points of absolute extremum. ### Procedure via determination of critical points The idea here is to combine the observation that absolute extremum implies local extremum or endpoint extremum, along with the observation that point of local extremum implies critical point. Step no. Step summary Step details 1 Find all the critical points Compute all the critical points of $f$ by computing the derivative $f'$, then determining the points in $(a,b)$ where this is either zero or undefined. 2 Evaluate function at critical points, endpoints, and compare Evaluate the function $f$ at all critical points found in Step (1), and at both endpoints $a,b$. Among all these values, the largest is the absolute maximum value, and all the points (among the critical points and endpoints) where this largest value is attained are the points of absolute maximum. Similarly, the smallest among these is the absolute minimum value, and all the points (among the critical points and endpoint) where this smallest value is attained are the points of absolute minimum. ### Comparison of the two procedures The two procedures are fairly similar. There are the following key differences: • Advantage of procedure via determination of critical points: We do not need to determine for each critical point whether it is a point of local maximum, local minimum, or neither. This saves on some effort that might have gone into using the first derivative test or second derivative test. • Disadvantage of procedure via determination of critical points: We could potentially need to evaluate the function on a larger set of points, because there may well be many more critical points than points of local extremum. This disadvantage becomes even sharper if the goal is to only find the absolute maximum). In that case, if we first filter down to the points of local maximum and endpoint maximum, then we need only evaluate the function at those points. A similar remark applies if we need to only find the absolute minimum. Comparing the advantage and disadvantage, we see that the main trade-off is between the ease of evaluating and comparing function values and the ease of evaluating the signs of the first derivative on intervals (if using the first derivative test) or the second derivative at critical points (if using the second derivative test). If the function is easy to evaluate and compare between points, it makes sense to simply determine the critical points and evaluate and compare values between the critical points and endpoints. If, on the other hand, the function is hard to evaluate and compare, but its derivatives are easier to study, it makes sense to use the appropriate derivative tests to find local extrema and then compare.

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