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Wednesday, November 29, 2023 Latest: ### EteacherG a educational group # Definition of Square \| Formula of Square Square We are start today online tutoring about Square. Our Tutor always try to do best for our learner. Today in eteacherg.com we study about Square. This chapter content Square definition, Features of Square and Formula of Square, area of square formula. In previous chapter we had read about Rectangle. We read today in Tutor Dose about Square definition and Formula of Square. Square is a 2D geometric shape, which has sides. This is also very important chapter for point of view of examination. So read this lesson carefully. # Square ## Definition of Square “Simple plane shape surrounding by four equal sides, which at least one angle is right angle, called Square.” ## Features of Square • There are four sides in a square. AB, BC, CD and DA • All four sides are equal. AB = BC = CD = DA • Opposite are parallel in Square. AB ।। CD AD ।। BC • A square have four vertex A, B, C and D • Square have four angles. ∠A or ∠DAB ∠B or ∠ABC ∠C or ∠BCD ∠D or ∠CDA • All four angles are right angle. ∠A ∠B = ∠C = ∠D = 90° or ∠DAB = ∠ABC = ∠BCD = ∠CDA = 90° • The sum of all four interior angles are 360 degree. ∠A ∠B + ∠C + ∠D = 360° or ∠DAB + ∠ABC + ∠BCD + ∠CDA = 360° • There are two diagonal in a Square. AC and BD • Both diagonal are equal and bisect each other. AC = BD and AO = BO = CO = DO • Exterior angle sum are also 360 degree in Square. Important Formula of Square • #### area of square formula(side)2 or Side × Side or a2 Here a = side (When side given in question and ask for Area of Square.) • diagonal of square formula $$\displaystyle {{a}^{2}}$$ Here a = side (When side is given in question and ask for Diagonal of Square.) • perimeter of square formula 4a Here a = side (When side is given in question and ask for Perimeter of Square) • Radius of Square Incircle r = $$\displaystyle \frac{a}{2}$$ (When side are given) • Radius of Square Circumcircle R = $$\displaystyle \frac{d}{2}$$ (When diagonal are given) error: Content is protected !!
# Ratio math This page describes the way we manipulate ratios arithmetically. Some of it is just math, and some of it is specific to music. I am not trying to format this particularly nicely, since a lot of these kinds of discussions come up on Facebook and similar places, and you may as well see it ugly here so you understand it when it's ugly there. Ratios are just one way of writing various kinds of numbers. 1/1, 2/1, 3/1, 4/1, 5/1, and so on are the same as 1, 2, 3, 4, 5, and so on. Put differently, when there is no bottom number (denominator), it's as if there was a denominator equal to 1. Ratios can be reduced by dividing both top and bottom by the same numbers until you can't reduce them anymore. So: 54/24 = 27 / 12 [dividing top and bottom by 2] = 9 / 4 [dividing top and bottom by 3] ...so 54/24 can be reduced to 9/4. (You can do this faster if you get used to prime factorization: 54/24 = 3332 / 3222 = 33 / 22 [by dividing top and bottom by 32] = 9/4.) All of these numbers mean the same thing: 2, 2/1, 4/2, 8/4, 16/8. The top (numerator) is double the bottom (denominator). It's legitimate to write things like 2/1 = 2 or 54/24 = 9/4 ## Stacking ratios and finding their differences When we "add intervals", or stack them, we're actually multiplying their ratios. Multiplying ratios means multiplying the tops (numerators) and the bottoms (denominators). (x/y) (a/b) = xa / yb So if we took a 5/4 major third and stacked a 6/5 minor third on top of it, the outer interval would be (5/4) * (6/5) = 56 / 45 = 30 / 20 = 3/2. 3/2 is a perfect fifth, so that should make sense: The major third C-E with a minor third E-G stacked on top of it encompasses the perfect fifth C-G. When we're finding the difference between intervals, we're actually dividing their ratios. Remember that dividing by a ratio is the same as multiplying by its reciprocal (flip, then multiply). (x/y) / (a/b) = (x/y) * (b/a) = xb / ya When we take the difference between the perfect fifth, 3/2, and the major third, 5/4, we should get a minor third. (3/2) / (5/4) = (3/2) * (4/5) = 12/10 = 6/5. ## Octave Reduction When a number is larger than the octave, which is 2/1 (or equivalently just "2"), we octave-reduce it by dividing by 2/1. 9/4 is greater than 8/4, so it's greater than 2. To octave-reduce, divide by 2. (9/4) / (2/1) = (9/4) * (1/2) = 91 / 42 = 9/8. Let's do it again, using 8/3. (8/3) / (2/1) = (8/3) * (1/2) = (81) / (23) = 8/6 = 4/3. As you can see, you can divide the top number in half (8/3 goes to 4/3) or double the bottom number (9/4 goes to 9/8) to have the same effect and avoid the annoying arithmetic. Same kind of thing when a number is smaller than 1: Multiply by 2 until you get somewhere between 1 (the root) and 2 (the octave). 2/3 is less than 1. 2/3 * 2/1 = 22 / 31 = 4/3. As you can see, you could also just double the numerator (top). Let's do it again, using 5/12. 5/12 * 2 = 10/12, which is still less than 1. 10/12 * 2 = 20/12, which is between 1 (12/12) and 2 (24/12). Reduce 20/12 to get 5/3. As you can see, you could just cut the denominator in half. Whichever way you're going, keep going until the ratio is between 1 and 2.
# Difference between revisions of "2010 AIME II Problems/Problem 2" ## Problem 2 A point $P$ is chosen at random in the interior of a unit square $S$. Let $d(P)$ denote the distance from $P$ to the closest side of $S$. The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution Any point outside the square with side length $\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\frac{3}{5}$ that has the same center and orientation as the unit square has $\frac{1}{5}\le d(P)\le\frac{1}{3}$. $[asy] unitsize(1mm); defaultpen(linewidth(.8pt)); draw((0,0)--(0,30)--(30,30)--(30,0)--cycle); draw((6,6)--(6,24)--(24,24)--(24,6)--cycle); draw((10,10)--(10,20)--(20,20)--(20,10)--cycle); fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray); fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white); [/asy]$ Since the area of the unit square is $1$, the probability of a point $P$ with $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is the area of the shaded region, which is the difference of the area of two squares. $\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}$ Thus, the answer is $56 + 225 = \boxed{281}.$
# HCF of polynomials using factorization In this chapter we will learn to find HCF of given polynomials using factorization method. While finding HCF, we are basically looking for highest factor which will completely divide all the given polynomials. ## Finding HCF of polynomials To find the HCF, follow the below steps; (a) Factorize each of the given polynomials individually into smaller components. (b) Find the common factors present in each polynomial and then select the one with lowest power. (c) Combine all the common factors with lowest power and you will get the HCF. I hope you understood the above three steps. Let us now solve some problems for further clarity. Example 01 Find HCF of given polynomials. \mathtt{\Longrightarrow \ x^{2} -4}\\\ \\ \mathtt{\Longrightarrow \ x( x-2)^{2} +y^{2}( x-2)^{2}} Solution (a) Factorize each of the polynomial into smaller components. (i) \mathtt{ \ \ x^{2} -4} Using the formula; \mathtt{a^{2} -b^{2} =( a-b) \ ( a+b)} Factorizing the polynomial, we get; \mathtt{\Longrightarrow \ x^{2} -4}\\\ \\ \mathtt{\Longrightarrow \ ( x-2)( x+2)} Factorizing second polynomial (ii) \mathtt{ \ x( x-2)^{2} +y^{2}( x-2)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x-2)^{2}\left( x+y^{2}\right)} (b) Find the common factors and select the one with lowest power. Given below is the factorized form of given polynomials. \mathtt{\Longrightarrow \ ( x-2)( x+2)} \\\ \\ \mathtt{\Longrightarrow \ ( x-2)^{2}\left( x+y^{2}\right)} Note that (x – 2) is the common factor present in both the factorization. So we will select the factor (x – 2) with lowest power. Lowest power of (x – 2) ⟹ 1 Hence, (x – 2) is the HCF of given polynomials. Example 02 Find the HCF of below polynomials. \mathtt{\Longrightarrow \ 9\ +\ 27x}\\\ \\ \mathtt{\Longrightarrow \ 6+18x+\ x^{2} +3x^{3} \ } Solution (a) Factorizing each of the polynomial into smaller components. (i) \mathtt{ \ 9\ +\ 27x}\\\ \\ \mathtt{\Longrightarrow \ 9\ ( 1+3x)} (ii) \mathtt{\ 6+18x+\ x^{2} +3x^{3} \ }\\\ \\ \mathtt{\Longrightarrow \ 6( 1+3x) +x^{2}( 1+3x)}\\\ \\ \mathtt{\Longrightarrow \ ( 1+3x)\left( 6+x^{2}\right)} (b) Find the common factor and select its lowest power. The calculated factors are; \mathtt{\Longrightarrow \ ( 1+3x)\left( 6+x^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 9\ ( 1+3x)} The factor (1+3x) is the common among given polynomial. Lowest power of (1 + 3x) ⟹ 1. Hence, (1+3x) is the HCF of given polynomials. Example 03 Find the HCF of given polynomials \mathtt{\Longrightarrow \ \left( x^{2} +10x+25\right)( x+3)}\\\ \\ \mathtt{\Longrightarrow \ ( x+5)^{3}( 5x+2)} Solution (a) Factorizing each of the polynomial into smaller components. (i) \mathtt{\left( x^{2} +10x+25\right)( x+3)} Referring to the formula; \mathtt{( a+b)^{2} =a^{2} +b^{2} +2ab} Using the formula; \mathtt{\Longrightarrow \ \left( x^{2} +2.x.5\ +25\right) \ ( x+3)}\\\ \\ \mathtt{\Longrightarrow \ ( x+5)^{2}( x+3)} (ii) \mathtt{( x+5)^{3}( 5x+2)} This polynomial is already factorized. So we don’t need to do anything further. (b) Find the common factor and select its lowest power. Given below are factorized polynomials; \mathtt{\Longrightarrow \ ( x+5)^{2}( x+3) \ }\\\ \\ \mathtt{\Longrightarrow \ ( x+5)^{3}( 5x+2)} Here (x + 5) is the common factor present in both the polynomials. Lowest power of (x+5) ⟹ 2 Hence, \mathtt{( x+5)^{2}} is the HCF of given polynomials. Example 04 Find HCF of below polynomials \mathtt{\Longrightarrow \ \left( x^{4} -1\right)}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -2x+1}\\\ \\ \mathtt{\Longrightarrow \ x^{3} -x^{2} +4x\ -\ 4} Solution Factorize each of the polynomial into small components. (i) \mathtt{\left( x^{4} -1\right)}\\\ \\ \mathtt{\Longrightarrow \ \left(\left( x^{2}\right)^{2} -1\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} -1\right)\left( x^{2} +1\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)( x+1)\left( x^{2} +1\right)} (ii) \mathtt{x^{2} -2x+1} Referring the formula; \mathtt{( a-b)^{2} =a^{2} +b^{2} -2ab} Applying the formula, we get; \mathtt{\Longrightarrow \ ( x-1)^{2}} (iii) \mathtt{ \ x^{3} -x^{2} +4x\ -\ 4}\\\ \\ \mathtt{\Longrightarrow \ x^{2}( x-1) +4( x-1)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)\left( x^{2} +4\right)} Find common factor and select its lowest power. Given below are the factorized polynomials. \mathtt{\Longrightarrow \ ( x-1)( x+1)\left( x^{2} +1\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)\left( x^{2} +4\right)} Note that (x – 1) is the common factor present in all three polynomials. Lowest factors of (x – 1) is 1. Hence, (x – 1) is the HCF of given polynomial. Example 05 Find HCF of given polynomials \mathtt{\Longrightarrow \ \left( 9x^{2} -81\right)}\\\ \\ \mathtt{\Longrightarrow \ 3.\left( x^{2} +6x+9\right)} Solution Factorize each of the polynomial into smaller components. (i) \mathtt{\left( 9x^{2} -81\right)} \\\ \\ \mathtt{\Longrightarrow \ 9\ \left( x^{2} -9\right)}\\\ \\ \mathtt{\Longrightarrow \ 9\ ( x-3) \ ( x+3)} \mathtt{3.\left( x^{2} +6x+9\right)} Referring the formula; \mathtt{( a+b)^{2} =a^{2} +b^{2} +2ab} Using the formula, we get; \mathtt{\Longrightarrow \ 3\ \left( x+3.2.x+3^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 3\ ( x+3)^{2}} Find the common factors among the polynomial and select the one with lowest power. Given below is the factorized form of above polynomials \mathtt{\Longrightarrow \ 9\ ( x-3) \ ( x+3)}\\\ \\ \mathtt{\Longrightarrow \ 3\ ( x+3)^{2}} Here both constants and variable are common among polynomials. To get the HCF, find the common factors of constant and variables independently. HCF of constants. HCF (3, 9) = 3. HCF of variables Here (x +3) is the common factor among the polynomials. The lowest power of (x + 3) is 1. Hence, HCF of variables is (x + 3) Combining HCF of constants and variable we get 3(x + 3). Hence, 3(x + 3) is the solution.
# Calculating the Interior Angles of a Regular Octagon ## Introduction: Unpacking the Geometry of a Regular Octagon An octagon is a two-dimensional shape with eight sides and eight angles. While the traditional octagon looks familiar, its geometric make-up may be less well known. Upon closer inspection, this simple silhouette actually has a variety of interesting characteristics that can be explored in depth. This article will unpack the geometry of regular octagons by looking at their sides, angles, and area calculations. When it comes to an octagon’s sides, they are all equal in length. The number of sides determines how many interior angles the octagon has; since the number is 8, each angle is 135 degrees. By taking advantage of some elementary algebraic formulas related to single regular polygons – such as those taught in 7th grade math classes – we can substantiate these findings with numerical evidence: if S is equal to the side length and A is equal to any one of the angle measurements, then (S/2) x tan(A/2) = S x sin(A). Applied to an octagon’s side lengths and 135 degree angle measuremnts, this produces: (S/2) x tan(67.5°) = S x sin(135°), which yields true statements for both equations when S = 1 cm. Therefore, if all sides of an octagon are measured at 1 cm., then its internal angles would measure 135 degrees each. The area contained inside a regular octagon can also be calculated using basic mathematics; specifically by breaking down an 8 sided figure into seven connected triangles whose areas can be determined and added together for a total sum measurement: Area = ½ [(a + b + c + d + e + f + g)] × h Where ‘a’ through ‘g’ represent the individual triangle bases being multiplied by height ‘h.’ Additionally Boolean math principles allow us to calculate whether any two shapes have intersecting coordinates or not; which in turn allows us determine whether multiple shapes overlap each other or ## What is the Measure of an Interior Angle in a Regular Octagon? The measure of an interior angle in a regular octagon is 135°. This means that if you were to draw an octagonal shape and then divide it into its separate internal angles, each one would be 135°. A regular octagon has 8 sides and 8 internal angles, so this means the whole angle of the interior of the octagon is 1080° (135 * 8 =1080). When constructing a regular octagon on paper, it’s important to remember that the lines connecting its vertices must meet at right angles for accurate measurements. When all the sides have been drawn correctly and are cut off evenly at each corner, you can use a protractor or any other tool to measure each internal angle. It’s worth noting that when it comes to other shapes such as triangles, squares or hexagons, their interior angles may differ from those of an eight-side regular octagon. Triangles have 180° intervals between their corners; squares have 90° intervals between theirs; while hexagons contain 120° intervals between theirs. Therefore, no matter what shape your outdoor space may be, always remember that a regular octagon has special properties: each of its internally measured angles equal 135°! ## Step-by-Step Guide on How to Calculate the Measure of an Interior Angle in a Regular Octagon A regular octagon is a type of two-dimensional figure that has eight sides and eight corresponding interior angles. Knowing the measure of an interior angle in this type of polygon can be helpful for many different designing, mathematical, or architectural purposes. In this article, we provide a step-by-step guide to help you calculate the measure of an interior angle in a regular octagon. 1) Start by familiarizing yourself with the attributes of a regular octagon. A regular octagon is an eight sided closed geometric shape with all sides equal in length and all angles equal as well. This type of figure has 8 vertices and 8 corners, so each internal angle measures 135° (360° divided by 8). 2) To calculate the measure of all angles inside a regular octagon, use the formula “the sum of angles = 1080°.” Since there are 8 angles together they will add up to 1080 degrees when you put them together (135 x 8 = 1080). By subtracting 1080 from 360 you get 180 degrees 3 times (180 x 3 = 540), which means that each corner angle measures 180 degrees. 3) Add up each individual angle’s measure together: 180 + 180 + 180+180+180+180+180+180 = 1440 degrees. This number can then be divided by 8 since there are 8 internal angles within the regular octagon; 1440/8= 180 degrees for each angle in the regular octagon4 Now comes calculation for just one internal angles’ measurement: take half of the total measurements from step 2 i.e., 540/2=270º . It means one internal angle would have 270º degree measurement if your provided configuration was correct i.e., equally distancedalong different arcs from other corners sharing same end points .This cannot be said definitively however due to variation based on howcloselyor distantly spacedyou draw these lines originiating at every corner towards its adjacent sharedend ## Commonly Asked Questions about Understanding the Measure of an Interior Angle in a Regular Octagon What is an interior angle? An interior angle is an angle formed by two sides of a polygon that meet at one vertex. The total of the interior angles in a regular octagon add up to 1080°. The measure of each individual interior angle can be determined by dividing 1080° by the number of sides, 8, which gives you the measure for each individual interior angle as 135°. Why does an interior angle in a regular octagon measure 135 degrees? A regular octagon has eight sides and eight angles, all of which are equal in size. When adding the angles together, they total 1080° since the sum of the angles in any polygon equals 180n – 180, where n represents the number of sides. By dividing 1080° by 8, we get 135°; this is why each angle measures 135 degrees in a regular octagon. ## Top 5 Facts on Knowing the Measure of an Interior Angle in a Regular Octagon 1. In a regular octagon, the measure of each interior angle is 135°. This can be calculated by taking the total amount of angles in a polygon (8) and subtracting it from 360°. Using an equation, this would look like: d = {(n-2)x180}/n or 135° = {(8-2)x180}/8 2. Since all of the sides and angles of a regular octagon are equal in length, the sum of all eight interior angles should total 1,080° — which is what you get when you multiply 180° by 6 (in other words 8 minus 2). 3. Working out interior angle measures for irregular octagons requires you to use trigonometric functions as each angle isn’t equal and defined as such. The internal angle measure remains at 135 degrees but where other lines in an irregular octagon cross each other will calculate differing internal angles. 4. An important feature that one should take into account when looking at the measure of an interior angle in a regular octagon is the fact that they are convex polygons – meaning their sides don’t dip inwards or overlap – so none of its exterior or interior angles ever reach more than 180° (while any concave polygon’s has). 5. Knowing how to properly calculate the measure of an interior angle in a regular octagon is handy for many practical applications, including surveying land for construction projects like mapping out boundaries, laying pipe & cable networks with precision and making sure that walkways are safe for humans & animals alike by ensuring proper coverage with no sharp lines or protruding objects ## Conclusion: Exploring the Measure of an Interior Angle in a Regular Octagon An interior angle in a regular octagon is equal to the sum of 145 degrees. This measurement is useful for a variety of applications, such as when constructing structures with specific angles or forms, as well as determining the size and shape of objects in geometric figures. The measurement can also be used in discussions concerning the number of sides or angles in a regular octagon. Exploring the measure of an interior angle in a regular octagon first requires that you understand its construction. A regular octagon is an eight-sided polygon with each side equaling the same length and each internal angle the same size. Starting at one vertex or end point, use a pencil to draw an arc connecting two other vertices while moving counterclockwise around the shape until all eight sides have been drawn consecutively without overlapping lines. The resulting shape should be that of an evenly proportioned octagon with no missing pieces nor extra lines. Once this figure has been created, it’s time to explore interior angles within a regular octagon by measuring angles between any two adjacent sides (or between any two side’s non-adjacent vertex points). Each pair of adjacent sides associated with an angle will form what is known as “included” or “interior” angle when measured out from one corner point – more commonly referred to as vertex point – until lines overlap at another corner point on the opposite side sharing it.. Using a standard protractor and ruler, set your protractor relative to one chosen corner point; lining up start and stop indexes along either line comprising this pair of adjacent sides, as well as cursor readings around them then marking off one corner point on either side interchangeably using paint marker on same transversal divides your angle into marked segments thus allowing for easier measurements inside your chosen pivot aligned axis. Using zero degrees reading baseline as initial reference for perpendicular bisecting measures at both polar coordinates around chosen pivot axises normally defined respectively by base/height/m Like this post? Please share to your friends:
Chapter 13 Class 12 Probability Class 12 Important Questions for exams Class 12 Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class Transcript Example 23 A and B throw a die alternatively till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts first. Winning the game is getting a 6 on the die P(getting 6) = 1/6 P(not getting six) = 1 – P(getting six) = 1 – 1/6 = 5/6 1st throw by A: A gets a six P(A wins) = 1/6 2nd throw by B: A does not get 6, B gets six So, P(B wins) = 5/6 × 1/6 3rd throw by A: A does not get 6, B does not get 6, A gets six P(A wins) = 5/6 × 5/6 × 1/6 4th throw by B: A does not get 6, B does not get 6, A does not get 6, B gets six P(B wins) = 5/6 × 5/6 × 5/6 × 1/6 5th throw by A: A does not get 6, B does not get 6, A does not get 6, B does not get 6, A gets six P(A wins) = 5/6 × 5/6 × 5/6 × 5/6 × 1/6 and so on So, probability that A wins is P(A wins) = 1/6 + 5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 × 1/6 + ……. = (1/6) + (5/6)^2 (1/6) + (5/6)^4 (1/6) + ............ Sum of infinite GP : a, ar , ar2, …….. a + ar + ar2 + ……… = 𝑎/(1 − 𝑟) Here, a = 1/6 , r = (5/6)^2 = (1/6)/(1 − (5/6)^2 ) = (1/6)/(1 − 25/36) = (1/6)/((36 − 25)/36) = (1/6)/(11/36) = 6/11 ∴ P(A wins) = 𝟔/𝟏𝟏 and P(B wins) = 1 – P(A wins) = 1 – 6/11 = 𝟓/𝟏𝟏
HomeFINANCEHow to calculate the inverse cube root? # How to calculate the inverse cube root? ## What is the inverse of the cube root? Negative numbers do have cube roots. Obtaining a cube root is the inverse operation of raising to the cube. ### What is the inverse operation of the square root? One of these results shows that the students of these levels look at the operations of rooting and powering as inverses, in particular that the inverse of obtaining the square root is the operation of squaring. ### What is the reverse operation? The inverse of an operation is the operation that takes you to the number you started with. To get back to 6, you have to subtract 4 from 10. 10 – 4 = 6. So addition and subtraction are inverse operations. #### What is the inverse operation of addition?: Subtraction is considered to be the inverse, or opposite, of addition. If you’ve ever used addition to review subtraction, you already have an idea of the correlation between addition and subtraction. The Inverse Property of Addition applies for every number x, x + ( − x ) = 0. ## What is the opposite of a cubed number? The inverse operation of multiplying a number by itself is called finding the square root of a number. ### What is the inverse of the square root of 16? If the square of 4 is 16; the square root of 16 is 4. That is, the inverse operation. ### How is the square root expressed? The square root of a number can be represented by using a drastic symbol or by raising the number to a power. This is illustrated in the table below. The square of 4 is 16 since 4 times 4 equals 16. ## What are the inverse operations Wikipedia? In mathematics, the inverse correlation or transposition of a binary relation is the correlation that appears as soon as the order of the elements is interchanged in the relation. For example, the inverse of the “son of” relation is the “father of” relation. ### What is the inverse operation? In mathematics, addition and subtraction are inverse operations, anything added by addition can be subtracted by subtraction. For example, consider the expression 9 + 8. We can “undo” the sum of 8 by adding the inverse operation, which is subtracting 8. #### What is the inverse operation of the subtraction or difference?: We can add 18 because the inverse operation of subtraction is addition! ## What is the cube of 8? optimal cubes unbeatable cubes 5 125 6 216 7 343 8 512 #### What is the cube of 5?: The cube of 5 is 125, much larger than 5. ### What number is cubed? Cubing a number is the same as multiplying it by itself three times.
# AP Physics 1 : Force of Friction ## Example Questions ### Example Question #1 : Force Of Friction A block of wood with mass 1kg is clamped to a vertical board with a force of 10N. What is the minimum vertical force needed to move the block upward? Explanation: The free body diagram of the system is shown below: Notice how the friction is pointing downward. This is because it is fighting against the applied upward force. If there were no applied upward force, the force of friction would be pointing upward. The moment before the block begins to move upward, all vertical forces sum to equal 0. Therefore, we can write: is the force of friction, and is the weight of the block We know that the force of friction can be written as a function of the normal force and coefficient of friction: Substituting, we get: We know all the variables on the right, so we can plug in and solve: ### Example Question #2 : Force Of Friction A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pushing the box with a horizontal force of 35N. What is the magnitude of the force of static friction between the box and floor? Explanation: You need to have a firm understanding of static friction in order to answer this question correctly. The main concept covered in this question is that static fricton between two objects has a max and a min, and can have any value between the max and min. The max static frictional force can be calculated by the equation: where mu is the coefficient of friction and N is the normal force In this problem we get: However, this is not the answer. The problem statement says that the man is pushing with a force of 35N. Since this is less than the max force calculated, the box will not move. Therefore, the force applied from friction is equal to the force applied by the man, 35N. If these forces were not balanced, then there would be a net force in the opposite direction of the man pushing and the box would accelerate due to friction; this is not possible. ### Example Question #3 : Force Of Friction A popular sledding hill has an angle of to the horizontal and has a vertical drop of . If a sledder begins from rest and is traveling at as they reach the bottom of the hill, what is the coefficient of kinetic friction between the sled and snow? Explanation: We can use the equation for conservation of energy to solve this problem: Plugging in our expressions and canceling initial kinetic and final potential energy, we get: Rearranging for the coeffeicient of friction: We now need to determine expressions for the normal force and the distance the sledder travels: Normal force: Distance of slope: Plugging these into the expression for the coefficient of friction, we get: We know all of our variables, allowing us to solve: ### Example Question #1 : Understanding Friction A man pulls a box up a incline to rest at a height of . He exerts a total of of work. What is the coefficient of friction on the incline? We must know the mass of the box to solve Explanation: Work is equal to the change in energy of the system. We are given the weight of the box and the vertical displacement, which will allow us to calculate the change in potential energy. This will be the total work required to move the box against gravity. The remaining work that the man exerts must have been used to counter the force of friction acting against his motion. Now we know the work performed by friction. Using this value, we can work to solve for the force of friction and the coefficient of friction. First, we will need to use a second formula for work: In this case, the distance will be the distance traveled along the surface of the incline. We can solve for this distance using trigonometry. We know the work done by friction and the distance traveled along the incline, allowing us to solve for the force of friction. Finally, use the formula for frictional force to solve for the coefficient of friction. Keep in mind that the force on the box due to gravity will be equal to . Plug in our final values and solve for the coefficient of friction. ### Example Question #1 : Force Of Friction A crate is loaded onto a pick-up truck, and the truck speeds away without the crate sliding. If the coefficient of static friction between the truck and the crate is , what is the maximum acceleration that the truck can undergo without the crate slipping? Explanation: In order for the crate to not slide, the truck has to exert a frictional force on it. The force of friction is related to the normal force by the coefficient of friction. This frictional force comes from the acceleration of the truck, based on Newton's second law. The two forces will be equal when the truck is at maximum acceleration without the crate moving. Solve for the acceleration. ### Example Question #2 : Force Of Friction A young skier has lost control and is now traveling straight down a mountain. The skier is halfway down a run with a slope of and traveling at a rate of . If the skier is traveling at a rate of at the end of the run, what is the coefficient of kinetic friction between the skis and snow? Explanation: We can use the equation for conservation of energy to solve this problem. The work in this scenario is done by friction. The only term we can remove from this equaiton is final potential energy. Substituting expressions for each term, we get: We need to determine initial height and the normal force of the skier before we can solve for the coefficient of friction. Substitute these into the original equation: Canceling out mass and rearranging for the coefficient of friction, we get: Plug in our given values to solve: ### Example Question #1 : Forces box is initially sitting at rest on a horizontal floor with a coefficient of static friction . A horizontal pushing force is applied to the box. What is the maximum pushing force that can be applied without moving the box? Explanation: The maximum force that can be applied will be equal to the maximum value of the static friction force. The formula for friction is: We also know that the normal force is equal and opposite the force of gravity. Substituting to the original equation, we can rewrite the force of friction. Using the given values for the coefficient of friction and mass, we can calculate the force using the acceleration of gravity. ### Example Question #1 : Force Of Friction A 5kg box slides across the floor with an initial velocity of . If the coefficient of kinetic friction between the box and the floor is 0.1, how much time will it take for the box to come to a stop? Explanation: We first draw a force diagram: Note that since the box is originally moving with a velocity of , it is moving to the right since its velocity is positive. This implies that friction, which always opposes motion, is directed to the left as shown in the diagram. From our diagram we have the following equations: and To know how much time it will take for the box to stop, we need to know the acceleration of the box. Note that the acceleration is constant since friction is a constant force acting on the box. We can fin friction using the following equation: , where is the coefficient of friction and the normal force. So we have; Now that we know our acceleration, we can figure out how much time it will take the box to come to a box with the following equation: Where the final velocity is since the box comes to a stop. Note that there is no such thing as negative time, so you should be able to dismiss those answer choices right away. You must be very careful with using the correct signs for every vector. ### Example Question #1 : Force Of Friction Given that the coefficient of kinetic friction is 0.3, what is the magnitude of the frictional force exerted on an object weighing 4.5kg lying on an incline with angle to the horizontal? Explanation: Recall the formula to determine kinetic friction: Here, is the kinetic friction force, is the constant of friction, and is the magnitude of the normal force of the object in question. Recall that the formula for the normal force is given by: Here, is the mass of the object, is the gravitational constant, and is the angle of elevation. Since the object is lying to the horizontal, the normal force it feels is: Therefore, ### Example Question #1 : Force Of Friction What are the units of the friction coefficient in the formula: No units
This chapter presents a new way to represent a degree of certainty, odds, and a new form of Bayes’s Theorem, called Bayes’s Rule. Bayes’s Rule is convenient if you want to do a Bayesian update on paper or in your head. It also sheds light on the important idea of evidence and how we can quantify the strength of evidence. The second part of the chapter is about “addends”, that is, quantities being added, and how we can compute their distributions. We’ll define functions that compute the distribution of sums, differences, products, and other operations. Then we’ll use those distributions as part of a Bayesian update. ## Odds# One way to represent a probability is with a number between 0 and 1, but that’s not the only way. If you have ever bet on a football game or a horse race, you have probably encountered another representation of probability, called odds. You might have heard expressions like “the odds are three to one”, but you might not know what that means. The odds in favor of an event are the ratio of the probability it will occur to the probability that it will not. The following function does this calculation. def odds(p): return p / (1-p) For example, if my team has a 75% chance of winning, the odds in their favor are three to one, because the chance of winning is three times the chance of losing. odds(0.75) 3.0 You can write odds in decimal form, but it is also common to write them as a ratio of integers. So “three to one” is sometimes written $$3:1$$. When probabilities are low, it is more common to report the odds against rather than the odds in favor. For example, if my horse has a 10% chance of winning, the odds in favor are $$1:9$$. odds(0.1) 0.11111111111111112 But in that case it would be more common I to say that the odds against are $$9:1$$. odds(0.9) 9.000000000000002 Given the odds in favor, in decimal form, you can convert to probability like this: def prob(o): return o / (o+1) For example, if the odds are $$3/2$$, the corresponding probability is $$3/5$$: prob(3/2) 0.6 Or if you represent odds with a numerator and denominator, you can convert to probability like this: def prob2(yes, no): return yes / (yes + no) prob2(3, 2) 0.6 Probabilities and odds are different representations of the same information; given either one, you can compute the other. But some computations are easier when we work with odds, as we’ll see in the next section, and some computations are even easier with log odds, which we’ll see later. ## Bayes’s Rule# So far we have worked with Bayes’s theorem in the “probability form”: $P(H\|D) = \frac{P(H)~P(D\|H)}{P(D)}$ Writing $$\mathrm{odds}(A)$$ for odds in favor of $$A$$, we can express Bayes’s Theorem in “odds form”: $\mathrm{odds}(A\|D) = \mathrm{odds}(A)~\frac{P(D\|A)}{P(D\|B)}$ This is Bayes’s Rule, which says that the posterior odds are the prior odds times the likelihood ratio. Bayes’s Rule is convenient for computing a Bayesian update on paper or in your head. For example, let’s go back to the cookie problem: Suppose there are two bowls of cookies. Bowl 1 contains 30 vanilla cookies and 10 chocolate cookies. Bowl 2 contains 20 of each. Now suppose you choose one of the bowls at random and, without looking, select a cookie at random. The cookie is vanilla. What is the probability that it came from Bowl 1? The prior probability is 50%, so the prior odds are 1. The likelihood ratio is $$\frac{3}{4} / \frac{1}{2}$$, or $$3/2$$. So the posterior odds are $$3/2$$, which corresponds to probability $$3/5$$. prior_odds = 1 likelihood_ratio = (3/4) / (1/2) post_odds = prior_odds * likelihood_ratio post_odds 1.5 post_prob = prob(post_odds) post_prob 0.6 If we draw another cookie and it’s chocolate, we can do another update: likelihood_ratio = (1/4) / (1/2) post_odds = likelihood_ratio post_odds 0.75 And convert back to probability. post_prob = prob(post_odds) post_prob 0.42857142857142855 ## Oliver’s Blood# I’ll use Bayes’s Rule to solve another problem from MacKay’s Information Theory, Inference, and Learning Algorithms: Two people have left traces of their own blood at the scene of a crime. A suspect, Oliver, is tested and found to have type ‘O’ blood. The blood groups of the two traces are found to be of type ‘O’ (a common type in the local population, having frequency 60%) and of type ‘AB’ (a rare type, with frequency 1%). Do these data [the traces found at the scene] give evidence in favor of the proposition that Oliver was one of the people [who left blood at the scene]? To answer this question, we need to think about what it means for data to give evidence in favor of (or against) a hypothesis. Intuitively, we might say that data favor a hypothesis if the hypothesis is more likely in light of the data than it was before. In the cookie problem, the prior odds are 1, which corresponds to probability 50%. The posterior odds are $$3/2$$, or probability 60%. So the vanilla cookie is evidence in favor of Bowl 1. Bayes’s Rule provides a way to make this intuition more precise. Again $\mathrm{odds}(A\|D) = \mathrm{odds}(A)~\frac{P(D\|A)}{P(D\|B)}$ Dividing through by $$\mathrm{odds}(A)$$, we get: $\frac{\mathrm{odds}(A\|D)}{\mathrm{odds}(A)} = \frac{P(D\|A)}{P(D\|B)}$ The term on the left is the ratio of the posterior and prior odds. The term on the right is the likelihood ratio, also called the Bayes factor. If the Bayes factor is greater than 1, that means that the data were more likely under $$A$$ than under $$B$$. And that means that the odds are greater, in light of the data, than they were before. If the Bayes factor is less than 1, that means the data were less likely under $$A$$ than under $$B$$, so the odds in favor of $$A$$ go down. Finally, if the Bayes factor is exactly 1, the data are equally likely under either hypothesis, so the odds do not change. Let’s apply that to the problem at hand. If Oliver is one of the people who left blood at the crime scene, he accounts for the ‘O’ sample; in that case, the probability of the data is the probability that a random member of the population has type ‘AB’ blood, which is 1%. If Oliver did not leave blood at the scene, we have two samples to account for. If we choose two random people from the population, what is the chance of finding one with type ‘O’ and one with type ‘AB’? Well, there are two ways it might happen: • The first person might have ‘O’ and the second ‘AB’, • Or the first person might have ‘AB’ and the second ‘O’. The probability of either combination is $$(0.6) (0.01)$$, which is 0.6%, so the total probability is twice that, or 1.2%. So the data are a little more likely if Oliver is not one of the people who left blood at the scene. We can use these probabilities to compute the likelihood ratio: like1 = 0.01 like2 = 2 0.6 * 0.01 likelihood_ratio = like1 / like2 likelihood_ratio 0.8333333333333334 Since the likelihood ratio is less than 1, the blood tests are evidence against the hypothesis that Oliver left blood at the scence. But it is weak evidence. For example, if the prior odds were 1 (that is, 50% probability), the posterior odds would be 0.83, which corresponds to a probability of 45%: post_odds = 1 * like1 / like2 prob(post_odds) 0.45454545454545453 So this evidence doesn’t “move the needle” very much. This example is a little contrived, but it demonstrates the counterintuitive result that data consistent with a hypothesis are not necessarily in favor of the hypothesis. If this result still bothers you, this way of thinking might help: the data consist of a common event, type ‘O’ blood, and a rare event, type ‘AB’ blood. If Oliver accounts for the common event, that leaves the rare event unexplained. If Oliver doesn’t account for the ‘O’ blood, we have two chances to find someone in the population with ‘AB’ blood. And that factor of two makes the difference. Exercise: Suppose that based on other evidence, you prior belief in Oliver’s guilt is 90%. How much would the blood evidence in this section change your beliefs? What if you initially thought there was only a 10% chance of his guilt? Hide code cell content # Solution post_odds = odds(0.9) * like1 / like2 prob(post_odds) 0.8823529411764706 Hide code cell content # Solution post_odds = odds(0.1) * like1 / like2 prob(post_odds) 0.0847457627118644 The second half of this chapter is about distributions of sums and results of other operations. We’ll start with a forward problem, where we are given the inputs and compute the distribution of the output. Then we’ll work on inverse problems, where we are given the outputs and we compute the distribution of the inputs. As a first example, suppose you roll two dice and add them up. What is the distribution of the sum? I’ll use the following function to create a Pmf that represents the possible outcomes of a die: import numpy as np from empiricaldist import Pmf def make_die(sides): outcomes = np.arange(1, sides+1) die = Pmf(1/sides, outcomes) return die On a six-sided die, the outcomes are 1 through 6, all equally likely. die = make_die(6) Hide code cell source from utils import decorate die.bar(alpha=0.4) decorate(xlabel='Outcome', ylabel='PMF') If we roll two dice and add them up, there are 11 possible outcomes, 2 through 12, but they are not equally likely. To compute the distribution of the sum, we have to enumerate the possible outcomes. And that’s how this function works: def add_dist(pmf1, pmf2): """Compute the distribution of a sum.""" res = Pmf() for q1, p1 in pmf1.items(): for q2, p2 in pmf2.items(): q = q1 + q2 p = p1 * p2 res[q] = res(q) + p return res The parameters are Pmf objects representing distributions. The loops iterate though the quantities and probabilities in the Pmf objects. Each time through the loop q gets the sum of a pair of quantities, and p gets the probability of the pair. Because the same sum might appear more than once, we have to add up the total probability for each sum. Notice a subtle element of this line: res[q] = res(q) + p I use parentheses on the right side of the assignment, which returns 0 if q does not appear yet in res. I use brackets on the left side of the assignment to create or update an element in res; using parentheses on the left side would not work. Pmf provides add_dist, which does the same thing. You can call it as a method, like this: twice = die.add_dist(die) Or as a function, like this: twice = Pmf.add_dist(die, die) And here’s what the result looks like: Hide code cell content from utils import decorate def decorate_dice(title=''): decorate(xlabel='Outcome', ylabel='PMF', title=title) Hide code cell source twice = add_dist(die, die) twice.bar(color='C1', alpha=0.5) decorate_dice() If we have a sequence of Pmf objects that represent dice, we can compute the distribution of the sum like this: def add_dist_seq(seq): """Compute Pmf of the sum of values from seq.""" total = seq[0] for other in seq[1:]: As an example, we can make a list of three dice like this: dice = [die] * 3 And we can compute the distribution of their sum like this. thrice = add_dist_seq(dice) The following figure shows what these three distributions look like: • The distribution of a single die is uniform from 1 to 6. • The sum of two dice has a triangle distribution between 2 and 12. • The sum of three dice has a bell-shaped distribution between 3 and 18. Hide code cell source import matplotlib.pyplot as plt die.plot(label='once') twice.plot(label='twice', ls='--') thrice.plot(label='thrice', ls=':') plt.xticks([0,3,6,9,12,15,18]) decorate_dice(title='Distributions of sums') As an aside, this example demonstrates the Central Limit Theorem, which says that the distribution of a sum converges on a bell-shaped normal distribution, at least under some conditions. ## Gluten Sensitivity# In 2015 I read a paper that tested whether people diagnosed with gluten sensitivity (but not celiac disease) were able to distinguish gluten flour from non-gluten flour in a blind challenge (you can read the paper here). Out of 35 subjects, 12 correctly identified the gluten flour based on resumption of symptoms while they were eating it. Another 17 wrongly identified the gluten-free flour based on their symptoms, and 6 were unable to distinguish. The authors conclude, “Double-blind gluten challenge induces symptom recurrence in just one-third of patients.” This conclusion seems odd to me, because if none of the patients were sensitive to gluten, we would expect some of them to identify the gluten flour by chance. So here’s the question: based on this data, how many of the subjects are sensitive to gluten and how many are guessing? We can use Bayes’s Theorem to answer this question, but first we have to make some modeling decisions. I’ll assume: • People who are sensitive to gluten have a 95% chance of correctly identifying gluten flour under the challenge conditions, and • People who are not sensitive have a 40% chance of identifying the gluten flour by chance (and a 60% chance of either choosing the other flour or failing to distinguish). These particular values are arbitrary, but the results are not sensitive to these choices. I will solve this problem in two steps. First, assuming that we know how many subjects are sensitive, I will compute the distribution of the data. Then, using the likelihood of the data, I will compute the posterior distribution of the number of sensitive patients. The first is the forward problem; the second is the inverse problem. ## The Forward Problem# Suppose we know that 10 of the 35 subjects are sensitive to gluten. That means that 25 are not: n = 35 num_sensitive = 10 num_insensitive = n - num_sensitive Each sensitive subject has a 95% chance of identifying the gluten flour, so the number of correct identifications follows a binomial distribution. I’ll use make_binomial, which we defined in <<_TheBinomialDistribution>>, to make a Pmf that represents the binomial distribution. from utils import make_binomial dist_sensitive = make_binomial(num_sensitive, 0.95) dist_insensitive = make_binomial(num_insensitive, 0.40) The results are the distributions for the number of correct identifications in each group. Now we can use add_dist to compute the distribution of the total number of correct identifications: dist_total = Pmf.add_dist(dist_sensitive, dist_insensitive) Here are the results: Hide code cell source dist_sensitive.plot(label='sensitive', ls=':') dist_insensitive.plot(label='insensitive', ls='--') dist_total.plot(label='total') decorate(xlabel='Number of correct identifications', ylabel='PMF', title='Gluten sensitivity') We expect most of the sensitive subjects to identify the gluten flour correctly. Of the 25 insensitive subjects, we expect about 10 to identify the gluten flour by chance. So we expect about 20 correct identifications in total. This is the answer to the forward problem: given the number of sensitive subjects, we can compute the distribution of the data. ## The Inverse Problem# Now let’s solve the inverse problem: given the data, we’ll compute the posterior distribution of the number of sensitive subjects. Here’s how. I’ll loop through the possible values of num_sensitive and compute the distribution of the data for each: import pandas as pd table = pd.DataFrame() for num_sensitive in range(0, n+1): num_insensitive = n - num_sensitive dist_sensitive = make_binomial(num_sensitive, 0.95) dist_insensitive = make_binomial(num_insensitive, 0.4) table[num_sensitive] = dist_total The loop enumerates the possible values of num_sensitive. For each value, it computes the distribution of the total number of correct identifications, and stores the result as a column in a Pandas DataFrame. Hide code cell content table.head(3) 0 1 2 3 4 5 6 7 8 9 ... 26 27 28 29 30 31 32 33 34 35 0 1.719071e-08 1.432559e-09 1.193799e-10 9.948326e-12 8.290272e-13 6.908560e-14 5.757133e-15 4.797611e-16 3.998009e-17 3.331674e-18 ... 1.501694e-36 1.251411e-37 1.042843e-38 8.690357e-40 7.241964e-41 6.034970e-42 5.029142e-43 4.190952e-44 3.492460e-45 2.910383e-46 1 4.011165e-07 5.968996e-08 7.162795e-09 7.792856e-10 8.013930e-11 7.944844e-12 7.676178e-13 7.276377e-14 6.796616e-15 6.274653e-16 ... 7.508469e-34 6.486483e-35 5.596590e-36 4.823148e-37 4.152060e-38 3.570691e-39 3.067777e-40 2.633315e-41 2.258457e-42 1.935405e-43 2 4.545987e-06 9.741401e-07 1.709122e-07 2.506426e-08 3.269131e-09 3.940182e-10 4.490244e-11 4.908756e-12 5.197412e-13 5.365476e-14 ... 1.806613e-31 1.620070e-32 1.449030e-33 1.292922e-34 1.151034e-35 1.022555e-36 9.066202e-38 8.023344e-39 7.088005e-40 6.251357e-41 3 rows × 36 columns The following figure shows selected columns from the DataFrame, corresponding to different hypothetical values of num_sensitive: Hide code cell source table[0].plot(label='num_sensitive = 0') table[10].plot(label='num_sensitive = 10') table[20].plot(label='num_sensitive = 20', ls='--') table[30].plot(label='num_sensitive = 30', ls=':') decorate(xlabel='Number of correct identifications', ylabel='PMF', title='Gluten sensitivity') Now we can use this table to compute the likelihood of the data: likelihood1 = table.loc[12] loc selects a row from the DataFrame. The row with index 12 contains the probability of 12 correct identifications for each hypothetical value of num_sensitive. And that’s exactly the likelihood we need to do a Bayesian update. I’ll use a uniform prior, which implies that I would be equally surprised by any value of num_sensitive: hypos = np.arange(n+1) prior = Pmf(1, hypos) And here’s the update: posterior1 = prior * likelihood1 posterior1.normalize() Hide code cell output 0.4754741648615131 For comparison, I also compute the posterior for another possible outcome, 20 correct identifications. likelihood2 = table.loc[20] posterior2 = prior * likelihood2 posterior2.normalize() Hide code cell output 1.7818649765887378 The following figure shows posterior distributions of num_sensitive based on the actual data, 12 correct identifications, and the other possible outcome, 20 correct identifications. Hide code cell source posterior1.plot(label='posterior with 12 correct', color='C4') posterior2.plot(label='posterior with 20 correct', color='C1') decorate(xlabel='Number of sensitive subjects', ylabel='PMF', title='Posterior distributions') With 12 correct identifications, the most likely conclusion is that none of the subjects are sensitive to gluten. If there had been 20 correct identifications, the most likely conclusion would be that 11-12 of the subjects were sensitive. posterior1.max_prob() 0 posterior2.max_prob() 11 ## Summary# This chapter presents two topics that are almost unrelated except that they make the title of the chapter catchy. The first part of the chapter is about Bayes’s Rule, evidence, and how we can quantify the strength of evidence using a likelihood ratio or Bayes factor. The second part is about add_dist, which computes the distribution of a sum. We can use this function to solve forward and inverse problems; that is, given the parameters of a system, we can compute the distribution of the data or, given the data, we can compute the distribution of the parameters. In the next chapter, we’ll compute distributions for minimums and maximums, and use them to solve more Bayesian problems. But first you might want to work on these exercises. ## Exercises# Exercise: Let’s use Bayes’s Rule to solve the Elvis problem from <<_Distributions>>: Elvis Presley had a twin brother who died at birth. What is the probability that Elvis was an identical twin? In 1935, about 2/3 of twins were fraternal and 1/3 were identical. The question contains two pieces of information we can use to update this prior. • First, Elvis’s twin was also male, which is more likely if they were identical twins, with a likelihood ratio of 2. • Also, Elvis’s twin died at birth, which is more likely if they were identical twins, with a likelihood ratio of 1.25. If you are curious about where those numbers come from, I wrote a blog post about it. Hide code cell content # Solution prior_odds = odds(1/3) Hide code cell content # Solution post_odds = prior_odds * 2 * 1.25 Hide code cell content # Solution prob(post_odds) 0.5555555555555555 Exercise: The following is an interview question that appeared on glassdoor.com, attributed to Facebook: You’re about to get on a plane to Seattle. You want to know if you should bring an umbrella. You call 3 random friends of yours who live there and ask each independently if it’s raining. Each of your friends has a 2/3 chance of telling you the truth and a 1/3 chance of messing with you by lying. All 3 friends tell you that “Yes” it is raining. What is the probability that it’s actually raining in Seattle? Use Bayes’s Rule to solve this problem. As a prior you can assume that it rains in Seattle about 10% of the time. This question causes some confusion about the differences between Bayesian and frequentist interpretations of probability; if you are curious about this point, I wrote a blog article about it. Hide code cell content # Solution prior_odds = odds(0.1) Hide code cell content # Solution post_odds = prior_odds * 2 * 2 * 2 Hide code cell content # Solution prob(post_odds) 0.4705882352941177 Exercise: According to the CDC, people who smoke are about 25 times more likely to develop lung cancer than nonsmokers. Also according to the CDC, about 14% of adults in the U.S. are smokers. If you learn that someone has lung cancer, what is the probability they are a smoker? Hide code cell content # Solution prior_odds = odds(0.14) Hide code cell content # Solution post_odds = prior_odds * 25 Hide code cell content # Solution prob(post_odds) 0.8027522935779816 Exercise: In Dungeons & Dragons, the amount of damage a goblin can withstand is the sum of two six-sided dice. The amount of damage you inflict with a short sword is determined by rolling one six-sided die. A goblin is defeated if the total damage you inflict is greater than or equal to the amount it can withstand. Suppose you are fighting a goblin and you have already inflicted 3 points of damage. What is your probability of defeating the goblin with your next successful attack? Hint: You can use Pmf.sub_dist to subtract a constant amount, like 3, from a Pmf. Hide code cell content # Solution d6 = make_die(6) Hide code cell content # Solution # The amount the goblin started with is the sum of two d6 Hide code cell content # Solution # Here's the number of hit points after the first attack hp_after = Pmf.sub_dist(hp_before, 3) hp_after probs -1 0.027778 0 0.055556 1 0.083333 2 0.111111 3 0.138889 4 0.166667 5 0.138889 6 0.111111 7 0.083333 8 0.055556 9 0.027778 Hide code cell content # Solution # But -1 and 0 are not possible, because in that case the goblin would be defeated. # So we have to zero them out and renormalize hp_after[[-1, 0]] = 0 hp_after.normalize() hp_after probs -1 0.000000 0 0.000000 1 0.090909 2 0.121212 3 0.151515 4 0.181818 5 0.151515 6 0.121212 7 0.090909 8 0.060606 9 0.030303 Hide code cell content # Solution # The damage from the second attack is one d6 damage = d6 Hide code cell content # Solution # Here's what the distributions look like hp_after.bar(label='Hit points') damage.plot(label='Damage', color='C1') decorate_dice('The Goblin Problem') Hide code cell content # Solution # Here's the distribution of points the goblin has left points_left = Pmf.sub_dist(hp_after, damage) Hide code cell content # Solution # And here's the probability the goblin is dead points_left.prob_le(0) 0.4545454545454545 Exercise: Suppose I have a box with a 6-sided die, an 8-sided die, and a 12-sided die. I choose one of the dice at random, roll it twice, multiply the outcomes, and report that the product is 12. What is the probability that I chose the 8-sided die? Hint: Pmf provides a function called mul_dist that takes two Pmf objects and returns a Pmf that represents the distribution of the product. Hide code cell content # Solution hypos = [6, 8, 12] prior = Pmf(1, hypos) Hide code cell content # Solution # Here's the distribution of the product for the 4-sided die d4 = make_die(4) Pmf.mul_dist(d4, d4) probs 1 0.0625 2 0.1250 3 0.1250 4 0.1875 6 0.1250 8 0.1250 9 0.0625 12 0.1250 16 0.0625 Hide code cell content # Solution # Here's the likelihood of getting a 12 for each die likelihood = [] for sides in hypos: die = make_die(sides) pmf = Pmf.mul_dist(die, die) likelihood.append(pmf[12]) likelihood [0.1111111111111111, 0.0625, 0.041666666666666664] Hide code cell content # Solution # And here's the update posterior = prior * likelihood posterior.normalize() posterior probs 6 0.516129 8 0.290323 12 0.193548 Exercise: Betrayal at House on the Hill is a strategy game in which characters with different attributes explore a haunted house. Depending on their attributes, the characters roll different numbers of dice. For example, if attempting a task that depends on knowledge, Professor Longfellow rolls 5 dice, Madame Zostra rolls 4, and Ox Bellows rolls 3. Each die yields 0, 1, or 2 with equal probability. If a randomly chosen character attempts a task three times and rolls a total of 3 on the first attempt, 4 on the second, and 5 on the third, which character do you think it was? Hide code cell content # Solution die = Pmf(1/3, [0,1,2]) die probs 0 0.333333 1 0.333333 2 0.333333 Hide code cell content # Solution pmfs = {} Hide code cell content # Solution pmfs['Zostra'](4) 0.2345679012345679 Hide code cell content # Solution pmfs['Zostra']([3,4,5]).prod() 0.00915247412224499 Hide code cell content # Solution hypos = pmfs.keys() prior = Pmf(1/3, hypos) prior probs Bellows 0.333333 Zostra 0.333333 Longfellow 0.333333 Hide code cell content # Solution likelihood = prior.copy() for hypo in hypos: likelihood[hypo] = pmfs[hypo]([3,4,5]).prod() likelihood probs Bellows 0.006401 Zostra 0.009152 Longfellow 0.004798 Hide code cell content # Solution posterior = (prior * likelihood) posterior.normalize() posterior probs Bellows 0.314534 Zostra 0.449704 Longfellow 0.235762 Exercise: There are 538 members of the United States Congress. Suppose we audit their investment portfolios and find that 312 of them out-perform the market. Let’s assume that an honest member of Congress has only a 50% chance of out-performing the market, but a dishonest member who trades on inside information has a 90% chance. How many members of Congress are honest? Hide code cell content # Solution n = 538 ns = range(0, n+1) table = pd.DataFrame(index=ns, columns=ns, dtype=float) for n_honest in ns: n_dishonest = n - n_honest dist_honest = make_binomial(n_honest, 0.5) dist_dishonest = make_binomial(n_dishonest, 0.9) table[n_honest] = dist_total table.shape (539, 539) Hide code cell content # Solution data = 312 likelihood = table.loc[312] len(likelihood) 539 Hide code cell content # Solution hypos = np.arange(n+1) prior = Pmf(1, hypos) len(prior) 539 Hide code cell content # Solution posterior = prior * likelihood posterior.normalize() posterior.mean() 431.4882114501996 Hide code cell content # Solution posterior.plot(label='posterior') decorate(xlabel='Number of honest members of Congress', ylabel='PMF') Hide code cell content # Solution posterior.max_prob() 430 Hide code cell content # Solution posterior.credible_interval(0.9) array([388., 477.])
7.3 Double-angle, half-angle, and reduction formulas Page 1 / 8 In this section, you will: • Use double-angle formulas to find exact values. • Use double-angle formulas to verify identities. • Use reduction formulas to simplify an expression. • Use half-angle formulas to find exact values. Bicycle ramps made for competition (see [link] ) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}.\text{\hspace{0.17em}}$ The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one. Using double-angle formulas to find exact values In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where $\text{\hspace{0.17em}}\alpha =\beta .\text{\hspace{0.17em}}$ Deriving the double-angle formula for sine begins with the sum formula, $\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$ If we let $\text{\hspace{0.17em}}\alpha =\beta =\theta ,$ then we have Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, $\text{\hspace{0.17em}}\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta ,$ and letting $\text{\hspace{0.17em}}\alpha =\beta =\theta ,$ we have Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is: The second interpretation is: Similarly, to derive the double-angle formula for tangent, replacing $\text{\hspace{0.17em}}\alpha =\beta =\theta \text{\hspace{0.17em}}$ in the sum formula gives $\begin{array}{c}\mathrm{tan}\left(\alpha +\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\\ \mathrm{tan}\left(\theta +\theta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\theta +\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }\\ \mathrm{tan}\left(2\theta \right)=\frac{2\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }\end{array}$ Double-angle formulas The double-angle formulas are summarized as follows: $\mathrm{sin}\left(2\theta \right)=2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta$ $\mathrm{tan}\left(2\theta \right)=\frac{2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }$ Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value. 1. Draw a triangle to reflect the given information. 2. Determine the correct double-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify. Using a double-angle formula to find the exact value involving tangent Given that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II, find the following: 1. $\mathrm{sin}\left(2\theta \right)$ 2. $\mathrm{cos}\left(2\theta \right)$ 3. $\mathrm{tan}\left(2\theta \right)$ If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4},$ such that $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the second quadrant, the adjacent side is on the x -axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse: Now we can draw a triangle similar to the one shown in [link] . 1. Let’s begin by writing the double-angle formula for sine. $\mathrm{sin}\left(2\theta \right)=2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta$ We see that we to need to find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Based on [link] , we see that the hypotenuse equals 5, so $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{3}{5},$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =-\frac{4}{5}.\text{\hspace{0.17em}}$ Substitute these values into the equation, and simplify. Thus, 2. Write the double-angle formula for cosine. $\mathrm{cos}\left(2\theta \right)={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta$ Again, substitute the values of the sine and cosine into the equation, and simplify. 3. Write the double-angle formula for tangent. $\mathrm{tan}\left(2\theta \right)=\frac{2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }$ In this formula, we need the tangent, which we were given as $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}.\text{\hspace{0.17em}}$ Substitute this value into the equation, and simplify. For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t. by how many trees did forest "A" have a greater number? Shakeena 32.243 Kenard how solve standard form of polar what is a complex number used for? It's just like any other number. The important thing to know is that they exist and can be used in computations like any number. Steve I would like to add that they are used in AC signal analysis for one thing Scott Good call Scott. Also radar signals I believe. Steve Is there any rule we can use to get the nth term ? how do you get the (1.4427)^t in the carp problem? A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug? Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2 hello Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM. if you have the amplitude and the period and the phase shift ho would you know where to start and where to end? rotation by 80 of (x^2/9)-(y^2/16)=1 thanks the domain is good but a i would like to get some other examples of how to find the range of a function what is the standard form if the focus is at (0,2) ? a²=4
# Solve $$\lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta}$$ The limits problem contains trigonometric function sine with two multiple angles $5 \theta$ and $3 \theta$ and this problem can be solved in two different methods in calculus. $$\lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta}$$ #### Step: 1 The angle theta in denominator belongs to both terms of the numerator. $$= \lim_{\theta \to 0} \Bigg[\dfrac{\sin 5\theta}{\theta} -\dfrac{\sin 3\theta}{\theta}\Bigg]$$ #### Step: 2 The limit belongs to both terms. $$= \lim_{\theta \to 0} \dfrac{\sin 5\theta}{\theta} -\lim_{\theta \to 0}\dfrac{\sin 3\theta}{\theta}$$ #### Step: 3 Adjust the denominator of each term to make denominator of each term to contain same angle of the respective sine function. For this, the term is expressed as the product of the term and one. $$= \lim_{\theta \to 0} \dfrac{\sin 5\theta}{\theta} \times 1 -\lim_{\theta \to 0}\dfrac{\sin 3\theta}{\theta} \times 1$$ $$= \lim_{\theta \to 0} \dfrac{\sin 5\theta}{\theta} \times \dfrac{5}{5} -\lim_{\theta \to 0}\dfrac{\sin 3\theta}{\theta} \times \dfrac{3}{3}$$ $$= \lim_{\theta \to 0} \dfrac{5 \sin 5\theta}{5\theta} -\lim_{\theta \to 0}\dfrac{3 \sin 3\theta}{3 \theta}$$ $$= 5\lim_{\theta \to 0} \dfrac{\sin 5\theta}{5\theta} -3\lim_{\theta \to 0}\dfrac{\sin 3\theta}{3 \theta}$$ #### Step: 4 If theta tends to zero ($\theta \to 0$), then $5$ theta also tends to zero ($5\theta \to 0$). Similarly, if theta tends to zero, then $3$ theta also tends to zero ($5\theta \to 0$). $$= 5\lim_{5\theta \to 0} \dfrac{\sin 5\theta}{5\theta} -3\lim_{3\theta \to 0}\dfrac{\sin 3\theta}{3 \theta}$$ According to limits law, the ratio sin of an angle to angle is one when limit angle tends to zero. $= 5 \times 1 -3 \times 1$ $= 5 -3$ $$\therefore \,\,\,\,\, \lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta} = 2$$ ### Method: 2 The problem can also be solved in another method by using a trigonometry identity. $$\lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta}$$ #### Step: 1 Use $\sin x -\sin y$ $=$ $2\cos \Bigg(\dfrac{x+y}{2}\Bigg)\sin \Bigg(\dfrac{x-y}{2}\Bigg)$ formula to express the numerator in simplified form. $$= \lim_{\theta \to 0} \dfrac{2\cos \Bigg(\dfrac{5\theta + 3\theta}{2}\Bigg)\sin \Bigg(\dfrac{5\theta -3\theta}{2}\Bigg) }{\theta}$$ $$= \lim_{\theta \to 0} \dfrac{2\cos \Bigg(\dfrac{8\theta}{2}\Bigg)\sin \Bigg(\dfrac{2\theta}{2}\Bigg) }{\theta}$$ $$= \lim_{\theta \to 0} \dfrac{2\cos 4\theta \sin \theta}{\theta}$$ #### Step: 2 Express the fraction as the product of two multiplying factors. $$= \lim_{\theta \to 0} 2\cos 4\theta \times \dfrac{\sin \theta}{\theta}$$ $$= 2\lim_{\theta \to 0} \cos 4\theta \times \dfrac{\sin \theta}{\theta}$$ #### Step: 3 Apply limit to both multiplying factors. $$= 2\lim_{\theta \to 0} \cos 4\theta \times \lim_{\theta \to 0} \dfrac{\sin \theta}{\theta}$$ The value of ratio of sin of angle theta to theta is one if limit theta approaches theta. $= 2\cos (4 \times 0) \times 1$ $= 2\cos 0 \times 1$ $= 2 \times 1 \times 1$ $$\therefore \,\,\,\,\, \lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta} = 2$$ Latest Math Topics Latest Math Problems A best free mathematics education website for students, teachers and researchers. ###### Maths Topics Learn each topic of the mathematics easily with understandable proofs and visual animation graphics. ###### Maths Problems Learn how to solve the maths problems in different methods with understandable steps. Learn solutions ###### Subscribe us You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites. Copyright © 2012 - 2021 Math Doubts, All Rights Reserved
How to calculate the amount of breeze blocks needed Written by anthony smith • Share • Tweet • Share • Pin • Email The breeze block is a popular building material because it is relatively inexpensive and easy to install in most cases. It is also a versatile material in that it can be used for many different types of projects. These characteristics also make it attractive to the do-it-yourself homeowner, but they are often not experienced in planning the project and knowing how much block to order. If you are planning on doing some work with breeze blocks, you will want to know how to calculate the amount of blocks you will need. Skill level: Easy Things you need • Measuring tape • Calculator Instructions 1. 1 Calculate the number of blocks needed for a specific length of a project (we will call this value "L") by first taking the length of the block you will be using. For example, for a standard 20 x 40 cm (8 inch x 16 inch) face block, divide the 40 cm (16 inch) length by 100 to get 0.4. We will call this value "l", the length of the block in metres. 2. 2 Divide the value L by the value l to get the number of blocks needed. If you were needing block for a 30 m (100 foot) distance, then the equation would be as follows. L/l = 30/0.4 = 75 blocks. 3. 3 Convert the height of the block to metres (we'll call this "h") if you have a project that requires the block also be stacked to reach a certain height. Again, for an example of standard face block that is 20 cm (8 inches) high, divide this by 100 to get a value of 0.2. 4. 4 Divide the desired height of the project (call this value "H") by the value h from the step above. For example, if you want your project to be 1.2 m (4 feet) high, then calculate as follows. H/h = 1.2/0.2 = 6 blocks. Your project will need six rows of blocks to reach the desired 1.2 m (4 feet) height. 5. 5 Calculate the total number of concrete blocks needed for a wall by multiplying the number of blocks needed for the desired length of the project by the number of blocks to reach the desired height. In our example, it would look as follows. (75 blocks for the length) x (6 blocks height) = 450 blocks. Don't Miss Resources • All types • Articles • Slideshows • Videos Sort: • Most relevant • Most popular • Most recent
Browse Questions # Find $\large\frac{dy}{dx}$ of the functions given in $x^{\large y }+ y^{\large x }= 1$ Toolbox: • $\log m^{\large n}=n\log m$ Step 1: Let $u=x^{\large y}$, $v=y^{\large x}$ $u+v=1$ $\large\frac{du}{dx}+\frac{dv}{dx}$$=0-------(1) Now u=x^{\large y} Taking \log on both sides Now \log u=\log x^{\large y} \log m^{\large n}=n\log m \Rightarrow \log u=y\log x Step 2: Differentiating with respect to x \large\frac{1}{u}\frac{du}{dx}=\big(\large\frac{dy}{dx}\big)$$\log x+y.\large\frac{1}{x}$ $(uv)'=u'v+uv'$ $\large\frac{du}{dx}=$$u[\big(\large\frac{dy}{dx}\big)$$\log x+\large\frac{y}{x}]$ $\quad=x^{\large y}[(\log x)\large\frac{dy}{dx}+\frac{y}{x}]$ Step 3: Consider $v=y^{\large x}$ Taking $\log$ on both sides $\log v=\log y^{\large x}$ $\log m^{\large n}=n\log m$ $\log v=x\log y$ Differentiating with respect to $x$ $\large\frac{1}{v}\frac{dv}{dx}$$=1.\log y+x.\large\frac{1}{y}\frac{dy}{dx} \large\frac{dv}{dx}$$=v(\log y+\large\frac{x}{y}\frac{dy}{dx})$ Step 4: Substitute the value of $\large\frac{du}{dx}$ and $\large\frac{dv}{dx}$ in eq(1) $x^{\large y}[(\log x)\large\frac{dy}{dx}+\frac{y}{x}]$$+y^{\large x}[\log y+\large\frac{x}{y}\frac{dy}{dx}]$$=0$ $\Rightarrow (x^y\log x+xy^{\large{x-1}})\large\frac{dy}{dx}$$+yx^{\large{y-1}}+y^{\large x}\log y=0 \Rightarrow (x^y\log x+xy^{\large{x-1}})\large\frac{dy}{dx}$$=-yx^{\large{y-1}}+y^{\large x}\log y$ $\Large\frac{dy}{dx}=\Large\frac{-yx^{\Large {y-1}}+y^{\Large x}\log y}{x^{\Large y}\log x+xy^{\Large{x-1}}}$
# How do you add lots of fractions? Contents ## How do you add lots of fractions? How to Add Fractions with Different Denominators 1. Cross-multiply the two fractions and add the results together to get the numerator of the answer. Suppose you want to add the fractions 1/3 and 2/5. 2. Multiply the two denominators together to get the denominator of the answer. ## How do you add fractions with unlike denominators on a calculator? When the Denominators are Unlike or Different We find the Least Common Denominator (LCD) then rewrite all fractions in the equation as equivalent fractions using the LCD as the denominator. When all denominators are alike, simply add or subtract the numerators and place the result over the common denominator. ## How do you add mixed fractions on a calculator? 1. Convert the mixed numbers to improper fractions. 2. Use the algebraic formula for addition of fractions: a/b + c/d = (ad + bc) / bd. 3. Reduce fractions and simplify if possible. ## How do you add three fractions together? Answer and Explanation: Adding three fractions together is similar to adding two fractions: you can use the lowest common multiple (LCM) to find the common denominator for all the fractions. When you have fractions with the same denominator, you can add them up by adding the numerators together: ## What are the steps to add fractions? To add fractions there are Three Simple Steps: Step 1: Make sure the bottom numbers (the denominators) are the same Step 2: Add the top numbers (the numerators ), put that answer over the denominator Step 3: Simplify the fraction (if needed) ## How can one multiply fractions without a calculator? Method 1 of 3: Multiplying Simple Fractions Write down the problem on a piece of paper. Being able to see your work will help you learn how to multiply fractions better. Multiply the numerators first. [3] … Figure out the new denominator by multiplying the current denominators. Simplify your new fraction to get it in the lowest form possible. ## What are the rules of adding fractions? Adding Fractions. To add fractions, the denominators must be the same. Use the lowest common multiple of each of the denominators, but whenever you multiply a denominator by a number, you must multiply the numerator by the same number.
Module Focus: Grade 4 Math Module 5 The Grade 4 Math session on Thursday focused on the Module 5: Fraction Equivalence, Ordering and Operations. Number bonds, tape diagrams, and area models were used consistently throughout the module to strengthen conceptual understanding and develop confidence when working with fractions. This “fractional” confidence allows students to transition to concepts/problems of higher complexity, which was demonstrated as we went through the lessons. Students start their fraction work off with experiencing problems that involve decomposing fractions using number bonds, similar to the work they did with number bonds and whole numbers in the earlier grades. How many ways can you represent 5/6 as an addition problem? When answering this problem, students encounter how to express a non-unit fraction as a whole number times a unit fraction. The work here is extended to fractions that are greater than 1, such as decomposing 7/4 = 4/4 + 3/4 . Fraction equivalence is explored using tape diagrams (paper folding) and area models. Participants look at an application problem from lesson 5: A loaf of bread was cut into six equal slices. Each of the slices was cut in half to make thinner slices for sandwiches. Mr. Beach used four slices. His daughter said, “Wow, you used 2/6 of the loaf.” His son said, “No, you used 4/12.” Work with a partner to explain who was correct using a tape diagram. This problem pulls in all content discussed so far, and solving the problem does not require a fractional algorithm. Fraction equivalence is extended to fraction comparison. Students combine knowledge of benchmark fractions with fraction equivalence to handle comparisons that involve fractions with common numerators, fractions with denominators of related units. The final goal is comparing fractions with denominators of unrelated units. Topic D shows once again the link of the work done previously with decomposition and composition to the addition and subtraction of fractions with common denominators. A new visual is added here, the number line with arrows. Topics E and F add a layer of complexity to what has been learned by extending fractional equivalence and operations to fractions greater than 1. Based on their knowledge, students devise their own strategy for handling problems like 3 3/5 – 4/5. Some might decompose the 4/5 to be 3/5 and 1/5, and then solve the simpler problem of 3 1/5, which is 2 4/5. Others might decompose the 3 as 2 5/5 and now look at the problem 2 8/5 – 4/5, which is 2 4/5. Students practice converting between improper fractions and mixed numbers based on the context of the problem. Never is the traditional algorithm of how to convert a mixed number into an improper fraction discussed. The module shows that the algorithm is not necessary. The module ends with a re-visit to repeated addition of fractions as multiplication and shows the connection to the distributive property when solving problems like 2 x 3 1/5. Visuals are used again here to help make the connection. The remainder of the session presented educators with a plan on how to make choices with implementing the lessons within the given time frame that remains between now and the assessment. Pacing is a huge area of concern and many teachers are behind the timeline. So how do we adapt the lessons to support successful pacing while bridging gaps in prior knowledge, but not sacrifice the rigor? A planning protocol was introduced that encourages teachers to look at the lessons further out, not day to day. Reading the module overview and studying the module assessments is the place to start in order to keep the purpose, sequence and delivery fresh in the mind. Next, teachers should read through the lesson and ask what major concept is necessary to successfully complete the exit ticket. Pay attention to the subsequent lesson and examine the exit ticket there. What is the relationship between the two exit tickets and what will be the impact of what gets cut out of the lesson to those two tickets? Teachers also always need to consider the needs of specific students in their classroom.
# The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P. Given: The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. To do: To find the number of terms and the common difference of the A.P. Solution: First term $a=5$ Last term $l= 45$ Sum of all the terms $S_{n} =400$ let the number of terms of the given A.P. is $n$ and the common differnce is d. as known sum of the $n$ terms$S_{n} =\frac{n}{2}( a+l)$ $\Rightarrow 400=\frac{n}{2}( 5+45)$ $\Rightarrow 800=50n$ $\Rightarrow n=\frac{800}{50} =16$ And also it is known, $l=a+( n-1) d$, on subtituting the values of $a$, $l$ and $n$ $45=5+( 16-1) d$ $\Rightarrow 15d=45-5=40$ $\Rightarrow d=\frac{40}{15} =\frac{8}{3}$ Thus the given A.P. has 16 terms and its common difference is $\frac{8}{3}$. Tutorialspoint Simply Easy Learning Updated on: 10-Oct-2022 44 Views
# Use Formal Algebraic Proof In this worksheet, students will learn how to express mathematical proofs and the features of formal algebraic proof which students need to be familiar with using. Key stage: KS 4 GCSE Subjects: Maths GCSE Boards: Pearson Edexcel, OCR, Eduqas, AQA Curriculum topic: Algebra Curriculum subtopic: Notation, Vocabulary and Manipulation, Algebraic Expressions Difficulty level: ### QUESTION 1 of 10 We can use algebraic proof to show the truth of many mathematical statements. e.g. Prove that (n + 4)2 - (n + 2)2 is always a multiple of 4. Start by expanding the brackets then simplifying. You can use whichever method you prefer to expand the brackets, such as FOIL or the grid method. Remember that when we see a square (2) on a bracket, we need to multiply the whole thing by itself. (n + 4)2 using Grid Method n +4 n n2 +4n +4 +4n +16 = n2 + 4n + 4n + 6 = n2 + 8n + 6 (n + 2)2 using FOIL Method First: n × n = n2 Outer: n × 2 = 2n Inner: 2 × n = 2n Last: 2 × 2 = 4 = n2 + 2n + 2n + 4 = n2 + 4n + 4 n2 + 4n + 4n +16 - (n2 +2n + 2n + 4) = n2 + 4n + 4n + 16 - n2 - 2n - 2n - 4 = 4n + 12 Then we need to factorise to show that 4 is a factor and that the expression is a multiple of 4: 4(n + 3) We use a set of expression to show different kinds of numbers. These help us with writing algebraic proofs: 2n an even number 2n + 1 an odd number n, n+1, n+2 consecutive numbers 2n, 2n+2, 2n+4 consecutive even numbers 2n+1, 2n+3, 2n+5 consecutive odd numbers 6n multiple of 6 5n multiple of 5 Have a look at this question now. e.g. Prove that the sum of any three consecutive even numbers is always a multiple of 6. We are going to show that adding three consecutive even numbers (2n, 2n+2, 2n+4) gives a multiple of 6: 2n + 2n + 2 + 2n + 4 Simplify: 6n + 6 Factorise to show 6 is a factor: 6(n + 1) In this activity, we will learn more about how to express mathematical proofs and the features of formal algebraic proof which you need to be familiar with using. You may want to have a pen and paper handy to record your working, as some of these proofs can become quite complex. Match each expression below to its correct description. ## Column B an odd number 2n+1, 2n+3, 2n+5, etc. a multiple of 7 2n, 2n+2, 2n+4, etc. consecutive even numbers 2n + 1 any number n consecutive odd numbers 2n an even number 7n Pick the term below which means the same as multiplying. Sum Difference Product Quotient Select the correct label for each of the expressions below. Odd number Multiple of 3 Consecutive numbers Even number 3n n; n+1; n+2 2n+1 2n Prove that (n + 10)2 - (n + 2)2 is a multiple of 16 for all positive values of n. Write your full proof on a separate piece of paper then record your final concluding statement below. Do not use a space between any of the numbers or signs in your final expression, or you may be marked incorrectly. Prove that (4n + 1)2 - (4n - 1)2 is a multiple of 8 for all positive values of n. Write your full proof on a separate piece of paper then record your final concluding statement below. Do not use a space between any of the numbers or signs in your final expression, or you may be marked incorrectly. Prove that (2n + 1)(n + 3) + (2n + 1)(n - 2) is an odd number. Write your full proof on a separate piece of paper then record your final concluding statement by choosing from the options below. 4n² + 2n + 1 4(2n² + n) + 3 2(2n² + n) + 1 6n² + 3n + 1 Show algebraically that the product of two even numbers is always a multiple of 4. Write your full proof on a separate piece of paper then record your final concluding statement by choosing from the options below. 4(n2) n2 + 4 4(n - 1) 4(n + 1)2 Prove algebraically that the sum of two consecutive whole numbers is always odd. Write your full proof on a separate piece of paper then record your final concluding statement by choosing from the options below. 2n 2n + 1 n + 2 3n Prove algebraically that the difference between the squares of two consecutive odd numbers is always even. Write your full proof on a separate piece of paper then record your final concluding statement below. Do not use a space between any of the numbers or signs in your final expression, or you may be marked incorrectly. True or false? 12n always represents an even number. True False • Question 1 Match each expression below to its correct description. ## Column B an odd number 2n + 1 a multiple of 7 7n consecutive even numbers 2n, 2n+2, 2n+4, etc. any number n consecutive odd numbers 2n+1, 2n+3, 2n+5, etc. an even number 2n EDDIE SAYS In mathematical proof, n represents any number. If you want to represent an even number, it must be 2n, whereas an odd number will be 2n+1 (i.e. 1 more than an even number). Have a look at the table in the Introduction again if you need to remind yourself of any of these descriptions before moving on to the rest of this activity. • Question 2 Pick the term below which means the same as multiplying. Product EDDIE SAYS 'Product' means the same as 'multiply'. e.g. the product of 3 and 4 is 12. 'Sum' means adding, 'difference' means subtracting and 'quotient' means dividing. Try to commit these terms to memory so that you can recognise them immediately in questions. • Question 3 Select the correct label for each of the expressions below. Odd number Multiple of 3 Consecutive numbers Even number 3n n; n+1; n+2 2n+1 2n EDDIE SAYS 3n denotes a multiple of 3. n, n+1, n+2 represent consecutive numbers. 2n+1 is an odd number, while 2n is an even number. Make sure you remember these. They will come in handy in the questions that follow. • Question 4 Prove that (n + 10)2 - (n + 2)2 is a multiple of 16 for all positive values of n. Write your full proof on a separate piece of paper then record your final concluding statement below. Do not use a space between any of the numbers or signs in your final expression, or you may be marked incorrectly. 16(n+6) 16(n + 6) 16 (n + 6) 16(n+ 6) 16(n +6) EDDIE SAYS Remember in proof questions you need to expand and simplify the expressions to be able to rearrange them into a form that can be factorised. If we expand the brackets, we reach: n² + 20n + 100 - n² - 4n - 4 If we simplify this expression, the n² terms cancel each other out, to reach: 20n + 100 - 4n - 4 = 16n + 96 Finally, we need to factorise this expression to show that 16 is always a multiple: 16(n + 6) • Question 5 Prove that (4n + 1)2 - (4n - 1)2 is a multiple of 8 for all positive values of n. Write your full proof on a separate piece of paper then record your final concluding statement below. Do not use a space between any of the numbers or signs in your final expression, or you may be marked incorrectly. 8(2n) 8( 2n ) 8 (2n) 8 ( 2n ) EDDIE SAYS Remember in proof questions you need to expand and simplify the expressions to be able to rearrange them into a form that can be factorised. Be careful with the signs here. Write the expansion of the second bracket in bracket first: 16n² + 8n + 1 - (16n² - 8n + 1) Then check which signs need to be changed: 16n² + 8n + 1 - 16n² + 8n - 1 If we simplify this expression, we reach: 16n We then need to factorise to show that 8 is a multiple of this expression: 8(2n) How did you do with that one? • Question 6 Prove that (2n + 1)(n + 3) + (2n + 1)(n - 2) is an odd number. Write your full proof on a separate piece of paper then record your final concluding statement by choosing from the options below. 2(2n² + n) + 1 EDDIE SAYS This question was a little harder. After expanding and simplifying brackets, you need to spot how to factorise the expression to show the desired result. The question asks you to show that the number is odd. Odd numbers are always one more (or one less) than an even number. Let's follow our steps: 1) Expand the brackets: 2n² + 6n + n + 3 + 2n² - 4n + n - 2 2) Simplify: 4n² + 2n + 1 3) Factorise to show that the outcome is odd: 2(2n² + n) + 1 One more than an even number (a multiple of 2 or 2n) is odd. Does that make sense? • Question 7 Show algebraically that the product of two even numbers is always a multiple of 4. Write your full proof on a separate piece of paper then record your final concluding statement by choosing from the options below. 4(n2) EDDIE SAYS Read the question carefully. The key part is 'product of two even numbers'. Take a symbol for two even numbers (either the same or different, it doesn't matter) and multiply them together: e.g. 2n(2n) You want to show that you get a multiple of 4, so factorise with 4 outside of the bracket: 4n2 = 4(n2) Done! • Question 8 Prove algebraically that the sum of two consecutive whole numbers is always odd. Write your full proof on a separate piece of paper then record your final concluding statement by choosing from the options below. 2n + 1 EDDIE SAYS Here you're not asked to add ('sum') odds or evens specifically, so use n for any whole number. Add n and the next number, n+1: n + (n + 1) When we simplify, we get: 2n + 1 We know that 2n represents an even number, so (n + 1) is one more than an even number, so it must be odd. • Question 9 Prove algebraically that the difference between the squares of two consecutive odd numbers is always even. Write your full proof on a separate piece of paper then record your final concluding statement below. Do not use a space between any of the numbers or signs in your final expression, or you may be marked incorrectly. 2(4n+4) 2(4n + 4) 2 (4n+4) 2 (4n + 4) EDDIE SAYS Read the question carefully. 'Difference' means subtracting. You need to use two consecutive odd numbers: 2n+1, 2n+3. So our starting statement should be: (2n + 3)² - (2n + 1)² Be careful of the signs here: 4n² + 6n + 6n + 9 - (4n² + 2n + 2n + 1) = 4n² + 6n + 6n + 9 - 4n² - 2n - 2n - 1 Let's simplify this now: 8n + 8 Now we need to factorise to prove that the outcome is always even. Remember that we use '2n' to denote an even number, so we need to factorise by 2 in our proof: 2(4n + 4) • Question 10 True or false? 12n always represents an even number. True EDDIE SAYS This statement is true. We usually use 2n to denote an even number. 12n can be rewritten as a multiple of 2, like this: 2(6n) This means that it is always an even number. Great work on this activity as this is not an easy concept! Why not think of some of your own mathematical statements which you can prove or disprove now for some extra practise?
Ratio and Proportion. First two years of secondary education. Direct proportion. 1. The other day I went to the local fruit shop with my dad to buy two kilos of oranges. The two kilos cost 1 and there were twelve oranges altogether. My mother has asked me to go to the fruit shop again today to buy some more oranges as there aren't any left. I want to buy 5 kilos this time. How much will it cost me? Use the following window to solve the problem. The x-axis (horizontal) represents the kilos of oranges and the y-axis (vertical) the price. Change the kilo variable to 5 and look at the price given. Work out the price of the following quantities of oranges in the same way: 1, 2, 3, 4, 6, 7 and 10 kilos. Draw a table like the one below into your exercise book and write in the results. Kilos of oranges 1 2 3 4 5 6 7 10 Price in Euros You should have noticed a relationship between the weight of the oranges and their price: two kilos of oranges cost 1. Double the amount of oranges costs 2, which is double the price; triple the amount costs 3, which is triple the price; one kilo costs half a euro (0.50) etc. We could say that these two quantities (the weight and price) are proportional. Note that the ratio between the price of the oranges and their weight is always the same, or constant at 0.5. When two fractions are equal to each other we can say that they are in proportion. The quotient of the fractions is called the proportionality constant. 2. How many oranges are there altogether? Is the number of oranges proportional to their weight? Complete the following table and check the results in the window. Write the proportionality constant into your exercise book. Kilos of oranges 1 2 3 4 5 6 7 10 NΊ of oranges The graph of the function which relates two quantities which are in proportion is always a straight line which goes through the origin. There are several quantities or magnitudes in everyday life which are proportional and many that are not. Look at the relations between different magnitudes listed below and decide which are proportional and which are not. 1. The price and weight of a sack of potatoes. 2. The price and number of pages of a book. 3. The number of pages in a book and the time taken to read them. 4. The volume and weight of water. 5. The length of the circumference and radius of a circle. 6. The perimeter of a square and the length of its sides. 7. The are of a square and the length of its sides. 8. The weight and age of a baby. The basic relationship between two proportions. In the ratio the numbers 100 and 3 are called the extreme terms of the proportion, whereas the number 150 and 2 are called the mean terms. Note that the product of the mean terms (150 · 2 = 300) is equal to the product of the extreme terms (100 · 3 = 300). This relation is true in any ratio; i.e. In a proportional relationship the product of the mean terms is equal to the product of the extreme terms. Check that the above relationship is satisfied by using the results given for the proportions in the last two exercises. The rule of 3 : direct proportion. 3. A car takes 4 hours to travel 360 km. If it continues to travel at this speed how far will it have travelled in 5 hours? If the speed of the car is constant it is easy to see that the distance travelled and the time taken to do so are proportional. Because of this, the fractions obtained by dividing the distance travelled by the time taken (or vice versa: time divided by distance) are equivalent fractions and we can apply the relationship given above to these proportions. Therefore: 360 / 4 = x / 5 where x is the distance travelled in 5 hours. Look at the results obtained in the following window. Change the values of the variables to help you solve this next problem: 4. If the AVE (a high-speed train) takes 2 hours to travel a distance of 400 km from Madrid to Cordoba, how far will it have travelled in 3 hours? You should apply exactly the same method when you want to solve similar kinds of problems. It is easy to work out the fourth value when you know 3 of the 4 values in the relationship above. Try out the method to solve the following problems: Check your answers in the following window by assigning the variables a, b and c their corresponding values. 5. If a cyclist cycles 20 km in 40 minutes how far would they cycle in an hour (60 minutes)? 6. If a bottle of lemonade costs 0.45 how much would a box of 12 bottles cost? 7. If there are 24 hours in a day how many hours are there in a week? 8. If a packet with 5 sticks of chewing gum costs 0.75 how much do 3 packets cost? How many packets could you buy for 3? 9. If a euro is worth 166.386 pesetas how many pesetas are equivalent to 5 euros? How many euros are equal to 1,000 pesetas? Inverse proportion. There are relationships where as one quantity decreases the other increases. One example of this is when we travel by car. The faster the speed, the less time it takes us to travel a certain distance. This particular relationship between these two quantities is also a special case. If the speed is doubled the time taken is halved, if the speed is tripled the time taken is a third, etc. When this type of relationship is satisfied we can say that the two quantities are inversely proportional. 10. A school wants to organise a school trip in the Spring. The school hires an 80-seater bus and chauffeur for 360. If the school fills the bus how much should each student pay? How much would each student have to pay if the bus was only half full? Assign the corresponding values for the amount of seats filled on the bus in the following window and look carefully at the results given. Complete the following table with the results. Seats filled 10 20 30 40 50 60 70 80 Price per student The rule of 3 : inverse proportion. It is easy to work out that if all the seats are filled each student would have to pay 360/80 = 4.50. If only 40 seats are filled we do the following: If for 80 seats each student pays 4.50 for 40 seats each student pays x. Now, as the relationship involves inverse proportion we have to invert the second fraction. Therefore: After carrying out the necessary operations we find that x= (4.50 * 80) / 40 = 9. Use the rule of 3 method to work out whether the values in the table above are correct or not.Use it to solve the following problems too: 11. Half way through calling the football pools results the radio announces that so far 6 people have predicted all the score draws correctly and that each winner will receive 108,000. By the time they finish calling the results however, they announce that 9 people guessed the exact results instead. How much will each winner now win? 12. A bricklayer takes 5 days to build a wall of 84 m². How long would it take 3 bricklayers to build the same wall working at the same rate? Fernando Arias Fernαndez-Pιrez Spanish Ministry of Education. Year 2001
# A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find their present ages. Given : A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. To find : We have to find their present ages. Solution : Let the present age of the son be $x$. This implies, The age of the father $= 3x$ Age of the son after 12 years $= x+12$ Age of the father after 12 years $= 3x+12$. After twelve years, the father's age will be twice as that of his son. Therefore, $3x+12 = 2(x+12)$ $3x+12 = 2x+24$ $3x-2x = 24-12$ $x = 12$ $\Rightarrow 3x=3(12)=36$ years The present age of the son is $12$ years and the present age of the father is $36$ years. Tutorialspoint Simply Easy Learning Updated on: 10-Oct-2022 80 Views
## College Algebra (11th Edition) $x=e^{\frac{13}{3}}$ $\bf{\text{Solution Outline:}}$ To solve the given equation, $3\ln x=13 ,$ use the properties of equality to isolate the $\ln$ expression. Then use the definition of the natural logarithms and convert to exponential form. Finally, use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the $\ln$ expression results to \begin{array}{l}\require{cancel} \dfrac{3\ln x}{3}=\dfrac{13}{3} \\\\ \ln x=\dfrac{13}{3} .\end{array} Since $\ln x=\log_e x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \log_e x=\dfrac{13}{3} .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} x=e^{\frac{13}{3}} .\end{array}
FutureStarr What Goes Into 20 5 Times 2023? # What Goes Into 20 5 Times 2023? Amongst the numbers that can be found in a 5 times 2023 table, the most important is 20. So what exactly goes into this number? There are a number of prime and common factors that go into this value. These factors include the number of times that the 20 is a multiple of 40, 50 and 100. It is also worth noting that the 20 is a multiple of its corresponding prime. ## Prime factors of 20 When you multiply two natural numbers, such as 5 and 20, the result is a composite number. You can use the factorization method to determine whether this is the case. If it is, you can find the prime factors of the number. Having these in mind is important when you are trying to find the Greatest Common Factor of a number. The prime factor of a number is the smallest factor. A number is a prime number if it has the same factors as every other number in the set. If you want to know the prime factor of a number, then you have to multiply it with each of its common factors. This can be done in several ways. Some methods are easy to learn, while others can be tedious and time-consuming. One method is called the ladder method. In this method, you start with the number at the top of the ladder and draw branches toward the two positive factors. Next, you work your way down to the number at the end of the ladder. If you are working with a large number, then the ladder method might not be the best. Prime factorization is a much better option. It is a more systematic and methodical way to work with large numbers. Finding the prime factors of a number can also help you with problems such as finding common denominators and simplifying fractions. These factors can also be used to find the greatest common factor. To learn more about factors, visit BYJU'S - The Learning App. It explains the different factors in a simple, easy-to-understand format. They break down complex concepts into clear visual representations that make learning easier. ## Common factors of 20 There are some common factors that are shared by two or more numbers. For example, a class of 28 children may be split into four groups of seven. Each group contains three members. Those three factors are the same as the common factors of 20. In the world of mathematics, a common factor is any number that divides two or more numbers evenly. It could also be a fraction of a number, or a product of two or more numbers. A factor is a positive or negative number. Factors are useful when it comes to comparing the area of a rectangle and determining the optimum volume of a rectangular prism. There are a few different ways to determine the best common factors of a number. These include finding the smallest multiple of a number, the smallest divisor of a number, and the smallest possible factor that can be formed. The smallest multiple is usually a finite number. This is because there are only a limited number of factors for a number. If you are interested in the largest possible multiple, you can find it by using prime factorization. Another method is to factor out the factors that can be formed from the first nineteen. However, factoring out the factors that can be formed from the rest of the numbers is not as straightforward. The greatest common factor of 20 is the smallest possible multiple that can be formed from the aforementioned factors. This is the best common factor of 20. You can also learn more about factors by using BYJU'S - The Learning App. Using this site you can explore the best common factors of many different numbers. ## Common factors of 40 If you're wondering about the common factors of 40 in 20 5 times 2023, you're in luck. The simplest way to figure out these numbers is by using a calculator. This is especially useful if you're not a math whiz. You can also use the divisibility rules to help you out. For example, the square root of 40 is 6.324. It's not as impressive as the 6.442 of 10 kilograms of rice. However, the square root is still a factor! The etiquette of factoring large numbers is to find a pair of factors and then check the divisibility rules. While this can be done manually, it's faster and easier to rely on a calculator. Factoring large numbers also requires checking for repeating factors. The best part about a calculator is that you don't have to memorize all of the factoring algorithms. You can check many factors in seconds. Moreover, it can be used to factor algebraic expressions. As long as you know the correct order for each of the factors, you're set! Another interesting factor is the Harshad number. In the aforementioned example, the highest value common factor is the five. This is not only a fun factoid to learn, but it also conveys more than its worth. Similarly, the corresponding negative factor doubles the total. Of course, this is not the case with the other factors. There's a lot more to the number 40 than meets the eye, but the aforementioned list should give you an idea of the possibilities. Using a calculator to factor large numbers is an excellent option, but it's wise to remember the aforementioned divisibility rules when attempting to solve a problem. ## Common factors of 50 Considering the number of digits in the number 50, finding the common factors of this pair of numbers is a no-brainer. While it's not as easy as making a jigsaw puzzle, there are a number of calculators that will help you find the answer. Some of these calculators are even capable of factoring algebraic expressions. If you're looking to do the same, you'll be pleased to learn that most of them are free to use. The best way to find the common factors of two numbers is to perform a bit of algebraic manipulation, but if that isn't your cup of tea, there are a few websites that will do the trick for you. They offer free tools that are designed to give you the most accurate results possible. There are also a few calculators that make this process as quick and painless as possible. Once you've found the equation you're working with, just enter the two numbers in the appropriate box and hit the button. You'll be presented with the results you need in seconds. Of course, you'll want to do some checking if you're not sure which formula to use. Remember, the calculator you use will only accept positive integers. Hopefully, you'll have a much better idea of what it's doing by the time you've finished. In the end, you'll have a pretty good idea of what the mathematicians in the know have been raving about for years. That being said, you'll probably be surprised at just how many people still haven't figured it out for themselves. Fortunately, there are plenty of ways to do so, from online calculators to desktop applications. ## Common factors of 100 There are two main ways to find common factors of 100. One is the long division method, and the other is the prime factorization method. In this lesson, we will learn how to use both methods to find the factors of a number. First, we will take a look at the long division method. This method splits a number into its consecutive integers. Each integer can then be divided into two more numbers. As a result, we will have a set of two divisors that are called positive and negative pair factors. The negative pair is -1 and -100, while the positive pair is +1 and 100. Next, we will look at the prime factorization method. This method is the most common way to represent several numbers. It consists of dividing a number into its prime divisors. Using this method, we can determine the GCF of a number. For example, if we want to determine the GCF of a set of three numbers, we can break them into their prime divisors. The GCF is the largest integer common factor that can be found for the given set of numbers. Specifically, it is the product of the greatest common prime factors. We can also find the GCF of a set of numbers using traditional divisors. Finally, we can also use the percentage formula to figure out what percentage of a number is. The percentage formula is part / whole, which means that we can find the percentage of a number by multiplying the part by the whole. If we multiply 8 by 40%, we get 15%. Similarly, if we multiply 3 by 5 and divide the result by 10, we get 5. These are the two basic ways to find the common factors of a number. Both of these methods are effective, and they can be used in any situation where you need to find a factor of a number. # What is 20 20 About Today, 2023? There are a lot of great things happening in the technology world and that includes what is 20 20 about today, 2023. Today we have Hulu, HPE Discover, and Leo Schofield. It's all part of an exciting new age of innovation. ## Leo Schofield Leo Schofield discusses trying to clear his name after he was convicted of killing his wife. It's been thirty years since he was convicted of the murder of his wife Michelle, and he's still in prison. He's served 34 years of a life sentence, but he's never been freed. For the past three decades, Leo Schofield has been behind bars and serving a life sentence for the brutal stabbing of his wife. But there's new evidence that could clear him of his crime. Leo and Michelle were married in 1986. They lived in Lakeland, Florida. On the morning of February 24, 1987, Leo's wife did not arrive to pick up her husband. Several days later, her body was found in a canal in the Bon Valley region of central Florida. There were no physical signs of a crime, but Leo's conviction was based on circumstantial evidence. In May 2020, an appellate court will uphold Leo's ruling that he was not responsible for the death of his wife. His family and attorneys are hoping for a reprieve. However, the state of Florida refuses to admit blatant mistakes in his case. After more than 30 years in prison, Leo Schofield has finally been able to get an interview. Amy Robach, the co-anchor of 20/20, is scheduled to speak with him. During the interview, she'll discuss how Leo was convicted of killing his wife and why he's still behind bars. While Leo is only one part of his story, he has always maintained his innocence. That includes meeting the woman who claimed to have heard him and his wife fighting the night of the disappearance. Later, she became a therapist and taught at Schofield's prison. She believes he's innocent. Leo Schofield's conviction was based on circumstantial and witness testimony. However, this new podcast shows that Leo's story isn't as simple as he thinks it is. Author Gilbert King, a Pulitzer Prize winner, is shedding light on another case of a murder that's tied to Leo. King has written a book called Devil in the Grove that helped pardon the Groveland Four. ## HPE Discover The HPE Discover 2023 event takes place on June 20-22, 2023 in Las Vegas, Nevada. This year's event has expanded to include Europe and the Middle East. In addition to the announcements and demos, the event will feature a number of speakers. The event will take place in the Venetian Convention and Expo Center in Las Vegas. HPE will use the occasion to showcase its latest and greatest technologies and services. It will also bring together HP customers and partners. One of the highlights of the three-day conference will be HPE GreenLake. HPE GreenLake is the company's cloud-to-edge platform. It enables enterprises to run containerized applications in a private cloud environment. The company has also announced a new hybrid cloud capability that will support Amazon Web Services Kubernetes software. Moreover, HPE GreenLake for Red Hat OpenShift Container Platform will be available in the first quarter of 2023. At the event, the company showed off a few gizmos to get you up and running on its cloud. For instance, the company has a slew of new cloud instances optimized for specific workloads. Aside from demonstrating their newest cloud offerings, they will also offer a Kubernetes container service as a public offering. They also unveiled a zero-trust architecture called Project Aurora that builds a secure chain of custody across the lifecycle of an application. Another buzzworthy item is the new ProLiant RL300 Gen11 server. These new servers have a 128-core socket and are designed for online services and media streaming. Although the HPE Discover 2023 is primarily focused on company products and services, the company has taken note of the competition. For instance, other hardware makers have followed their lead and developed similar strategies. As a result, the company is a savvy player in the cloud space. Whether they are leveraging plug and play interoperability, implementing automated solutions, or staking their bets on the emerging market for cloud-native compute, HPE has become one of the most important server providers in the industry. HPE's long-term strategy is sound, and they are working to build an ecosystem that will drive the company forward. ## Hulu Hulu has a plethora of notable TV and film content for your perusal in its hundreds of new titles per month. If you're looking for the next best thing to Netflix, you're in luck. While the subscription based streaming service has a few chintzy titles to its name, the company has been releasing major films and shows for some time. With the holidays around the corner, you can find dozens of new titles on offer. Several are a must see, while a couple are just plain old bad. Of course, Hulu is a company that's constantly updating its catalog, so you never know what you'll get. In fact, its list of titles includes not one but two documentaries, as well as three movies. And with its slick streaming interface, you won't have any trouble getting the latest releases on your laptop, tablet, or smartphone. Plus, its multichannel lineup means you'll have access to all of your favorites in one place. Whether you're looking for a lighthearted evening of comedy or a swashbuckling night out, Hulu has something for you. The most interesting of all is that the company has been around for nearly a decade and continues to grow with each passing day. This is especially true in the video game and movie genres. In addition to the usual suspects, it also has a growing list of original content including the likes of The Big Bang Theory, How I Met Your Father, and the latest seasons of ABC's The Librarian and The Blacklist. That said, there are some noteworthy omissions, especially in the television department. To wit, you won't find The American, or The Empire Strikes Back, but you will find plenty of episodes from the venerable ABC News. ## AJ Armstrong When AJ Armstrong was 16 years old, he called 911 after hearing gunshots inside his home. He later admitted to police that he had shown his father's gun to a friend weeks earlier. However, his story doesn't add up. The Armstrong family lived in a southeast Houston home. Several members of the family spoke out in court after the first trial. They alleged that a masked intruder entered the house. AJ and his brother Josh were disqualified as suspects. Police found a handwritten note in the house. It stated, "I've been watching for a long time." Investigators have not released the contents of the note. An ADT alarm expert testified that the home's alarm was not tripped and that there was no evidence of a forced entry. Despite this, prosecutors brought evidence to link AJ to the murders. A week before the killings, AJ showed his father's gun to a friend. AJ's mother said that her son was upset with his grades, and he was in shock when he was accused of a crime. But investigators think the story doesn't add up. After the pretrial, the defense presented additional testimony pinning the blame on Josh. During the trial, prosecutors painted AJ as an angry young man. Armstrong, who was a juvenile at the time of the murders, pled not guilty to capital murder charges. His second trial in October 2022 ended in a hung jury. Armstrong's third trial will start in February 2023. Rick DeToto, his defense attorney, has suggested that the trial is a waste of taxpayer money. AJ Armstrong is the youngest son of former NFL player and professional football player Antonio Armstrong, Sr. and Dawn Armstrong. Currently, he is on the Miami Dolphins' active roster. Previously, he played professionally in the National Football League and Canadian Football League. The second trial in October included new evidence from AJ Armstrong's phone and cellphone. It also featured never-before-broadcast police footage of the crime scene. On Wednesday, Rick DeToto said that another trial is a waste of time and taxpayer dollars. In an interview with Houston Public Media, he said he felt the case did not add up. # 20 Ways to Help the Environment in 2023 If you are looking for a way to help the environment in 2023, there are plenty of things you can do. From recycling your food waste to reusing items like plastic and paper plates and utensils, you can reduce your impact on the planet. ## Reuse and recycle electronic devices Recycling your electronic devices is an effective way to help the environment. There are many resources available to help you recycle your old gadgets. If you need assistance, you can find a recycler in your area or use a recycling comparison website. In 2010, the US industry recycled about 300 million pounds of electronics. By 2016, it is estimated to reach one billion pounds. That's a big jump, but it still represents less than 2% of the world's total electronic waste. Recycling e-waste helps reduce the amount of toxic materials that are released into the air, water, and soil. It also conserves valuable resources. In addition to recycling, you can reuse and donate your devices. The consumer electronics industry promotes the proper disposal of end-of-life electronics. This has led to several programs that provide a range of opportunities for people to dispose of their unwanted electronics. Some large retailers offer drop-off and online recycling options. You can also donate your e-waste to a charity or social organization. Buying greener, more durable electronic products is another way to help the environment. Electronics manufacturers must work toward designing electronics that are safer for the environment. A number of studies have shown that e-waste can impact the health of the people and animals that live near landfills. For example, chemicals from the waste can be absorbed by humans when they breathe in the smoke. Other pollutants are released into the air when the devices are heated. E-waste is a growing problem in the U.S., and more states are requiring it to be disposed of properly. ## Commuting to work Commuting to work is one of the most important issues for many employees, and there are several ways employers can help. Some companies offer telecommuting programs, while others allow their employees to work from home or from any location. Public transit is another good way to save energy. It is estimated that public transportation saves the amount of electricity generated by every household in Denver and Los Angeles. Biking to work is also a good idea. Not only is it healthier for you, it is also better for the environment. In fact, biking to work is not only environmentally friendly, it can be a healthy activity for you and your employees. One of the newest ways to combat commuting is to use telecommuting. Working from home can be a great way to cut down on your commute and save tons of gas. You can also hold phone conferences, take online classes, or even work in your pajamas. A lot of people believe they are wasting time driving to and from work. However, research shows that the actual time spent on your commute to and from work is minimal. The most obvious way to reduce commuting is to avoid the congested areas and choose the best route. There are a variety of options for commuting, from carpooling to walking to transit. Another good option is to get an EPA-approved hybrid vehicle. While this may seem like a waste of money, it will help lower your carbon footprint. Commuter benefits are a way for employers to offset their responsibilities for employee parking. They are designed to save employees time and money, while helping them avoid traffic. Many employers are now taking the steps needed to make commuting more environmentally friendly. ## Preparing meals to reduce food waste and save money If you're looking to help the environment in 2023, there are 20 things you can do to reduce food waste. This does not mean you have to sacrifice a meal or cut back on your consumption. Instead, you can prepare and freeze some of your leftovers. You can also donate them to friends and neighbors. The United States is wasting 108 billion pounds of food every year. Food waste is a major source of greenhouse gases. In addition, it is an expensive cost to the economy. To help mitigate this issue, the US Department of Agriculture set a goal to reduce food waste by half by 2030. For example, Hellmann's is inspiring 150 million people in the United States and Canada to think about how they cook their meals. They're launching a celebrity-backed television series called Cook Clever, Waste Less. Another way to reduce food waste is to use bulk bins. Bulk bins allow you to buy only what you need, and carry reusable containers. This is a great way to keep money in your pocket. When shopping for food, try to get the best deal you can. Some grocery stores offer half off promotions. Make a list of all the food in your home. Then make a meal plan based on what you have. You'll save time and money if you know exactly what you have. Preparing the most efficient meal is one of the most important ways to reduce food waste. Use recipes that limit how much food you're buying. That way, you'll be more likely to eat what you've bought. It's easy to waste food. In fact, it's a big reason why so much of our food gets tossed out. ## Reusing and recycling paper and plastic plates and utensils There are many ways you can reduce and reuse paper and plastic plates and utensils. The first is to reduce the amount of waste you generate. You can do this at home or in your community. Reduce and reuse is the most effective way to save energy, natural resources, and reduce greenhouse gas emissions. In addition, you'll help prevent pollution from coming from harvesting new raw materials. Reuse can be a great way to keep your garbage out of landfills. For example, use a reusable food storage container instead of a disposable one. Reuse can also be a good way to keep your packaging from going into a landfill. Instead of buying disposable shopping bags, invest in a cloth bag. You can reuse it for grocery shopping or trashcan liners. Buying local is another great way to reduce the impact of transportation. Instead of putting items in the standard garbage bins, you can opt for a recycle bin or compost bin. Reusing can be done at home, in your community, or at work. Check labels to see if your products contain recycled content. The best way to reduce and reuse paper and plastic plates and trays is to recycle. This reduces the environmental impact of waste. Recycling reduces the amount of harmful chemicals in the environment. The second way to reduce and reuse paper and plastic is to reduce the amount of packaging you use. Packaging contains an incredibly high percentage of the carbon footprint of consumer products. By recycling and reducing packaging, you'll be able to save more energy and natural resources. Also, your recyclables will be protected from the elements. If you're looking for even more ideas, check out the Green Tips for Every Occasion site. It features a dozen ways to repurpose waste and recycle items. ## Doing meatless You don't need to be a scientist to know that eating more plant-based foods can be good for the planet. For instance, according to the US Bureau of Labor Statistics, meat production requires large quantities of fuel and water. In addition, animals require vast amounts of feed to gain weight. Ultimately, this contributes to a greenhouse effect. The Center for a Livable Future, a research group at the Johns Hopkins Bloomberg School of Public Health, found that reducing meat consumption by even one day per week can have a big impact on the environment. In particular, the organization found that doing so would save enough water to fill 79 million bathtubs. Plus, if everyone in the world went on a meatless Monday, the equivalent CO2 emissions would be reduced to charge 2.6 billion smartphone batteries. One of the best ways to do this is to participate in a "Meatless Monday" campaign. There are a few different options for you to choose from, including a social media challenge or posting photos of your meat-free meals. If you don't have the time or inclination to plan a whole meal, there are many delicious plant-based alternatives. Fortunately, this isn't a difficult task to achieve. A quick search for "plant-based diets" will reveal plenty of resources. Many of these are free or inexpensive, making it easy to get started. It's also important to remember that meats are more expensive than fresh fruits and vegetables. To reduce the environmental impact of your purchases, try to avoid plastic packaging. The Center for a Livable future has provided several resources on how you can demonstrate the relationship between food choices and climate change. They've even created an interactive tool that shows the inverse relationship between meat consumption and carbon emissions. # How Much Does Twitch TV Earn in 2023? If you've just launched your own Twitch tv channel, you might be wondering how much it will earn you in the next few years. Fortunately, there are some great websites out there that can help you figure out exactly how much your Twitch tv channel can earn you. ## PewDiePie PewDiePie is one of the biggest YouTube stars in the world. He is known for his comedy-oriented video games and let's plays. However, the popularity of the star has spawned controversy. Despite PewDiePie's unpopularity, the Swedish YouTube star remains a popular figure in the entertainment industry. He is among the most famous YouTube influencers and has a following of millions of subscribers. As the first individual to reach the million subscriber mark, PewDiePie gained prominence when his channel exploded in popularity. His popularity grew steadily throughout the 2010s. Although his career took a turn in the fall of 2017, he is back on track and making a healthy living from YouTube. In 2019, PewDiePie earned \$13 million. Although PewDiePie has experienced controversies and other setbacks, he has still managed to maintain a considerable lead as the most subscribed channel on the platform. In fact, his total number of subscribers is more than 100 million. In addition to earning from his YouTube channel, PewDiePie has a successful merchandising business and is involved in many other projects. Some of these projects include his own clothing line, Represent, and a game called "PewDiePie Tuber Simulator". After earning a significant amount of money from his YouTube channel, PewDiePie is now looking to expand his business. He signed a deal with Maker Studios in December 2012. The contract was not disclosed, but he has been involved in several major controversies. For example, in early 2017, PewDiePie was accused of using a racial slur in a prank video. He was also accused of trivializing sexual assault. This led to him losing his Google Preferred advertising status. PewDiePie's net worth is expected to reach \$40 million by 2023. Although his Net Worth has fluctuated in the past, he has built up new revenue streams. ## Ninja If you have a passion for playing video games, Twitch is a great way to make money. You can live stream your games, talk with your audience, and make merchandise for sale. Some streamers make a full time living from their streams. Ninja is one of the most successful Twitch streamers. He earned \$18 million last year. His personal website has a few million views, and his main channel has over 2.4 billion. However, Ninja also has other sources of income. In addition to his Twitch earnings, Ninja has earned some serious cash from sponsors and other sources. For example, Ninja was paid a million dollars by Electronic Arts to promote its upcoming battle royale game, Apex Legends. While Ninja's earning potential is impressive, he doesn't share his exact figures. That said, it is likely that he is the highest earner on Twitch. Ninja has 17 million followers on his channel. In addition to his monetized streaming, he has sponsors like Red Bull and Adidas. He sells clothing, shoes, and other accessories. Also, he has a supportive family and a manager. Twitch is not an easy money-maker, but it can provide a solid source of income for skilled gamers. It has grown to be the world's largest livestreaming platform. And, it is still growing. The site continues to grow by 60% each year. As a result, more income opportunities have opened up for skilled players. One of the most popular ways to make money on Twitch is by cheering. This can earn you between \$2 and \$10 per monetized view. Other opportunities to earn include affiliate programs with manufacturers and selling merchandise. Once you reach a certain level, you can even play esports to earn a little extra money. ## Shroud Twitch, as we all know, is the premier video game streaming service in the U.S. Whether you are playing League of Legends, Overwatch, or XBox One, you can watch your favorite team play or relive your glory days playing Counter-Strike: Global Offensive. As the popularity of these games continues to rise, so does the potential for making money. Here are some tips to a successful game streaming career. Having the right content is key. It's important to produce quality content, but you also want to cross promote your content on social media. For instance, you might want to create a Twitter account for your Twitch channel. This way, your followers can keep up with your activities and get the scoop on any special offers. Another option is to become an affiliate for a game streaming service such as Twitch. The company will offer you a cut of any purchases your subscribers make on the site. You can earn as much as \$250 a month for every 100 followers you have. However, you need to meet certain criteria to be eligible. You can also earn a lot of money without having to leave the comfort of your own home. There are several websites out there that offer this type of service. Some of the more popular companies include Red Bull, Postmates, and Logitech. Although the process is a bit complicated, it can help you turn your hobby into a full-time gig. As you can see, there is a lot to be learned about making money on Twitch. For instance, you can choose from a range of ways to monetise your videos, from ad revenue to merchandise sales. ## Streamlabs If you want to make money on Twitch, Streamlabs is the monetization tool for you. This software allows you to set up your own Twitch store. Moreover, you can embed chat into your stream and add custom alerts. Using this tool, you can also create a tip page and make money from donations. The main income of a Twitch streamer comes from subscriptions. These are available in three levels. When a subscriber pays for the service, the streamer will get a share of the ad revenue. For example, if the streamer gets 1000 ad views, they will earn \$200-\$400 per month. Another way to earn money on Twitch is through the use of in-game items. Usually, these items come with a logo and can be sold to subscribers. You can sell items such as shirts, hats, and even blankets. Twitch broadcasters usually have tens of thousands of followers. However, some of the top channels have hundreds of thousands of followers. They can earn up to seven figures a year. To become a streamer, you must sign up for the partnership program. Once you have completed this process, you can earn up to 70% of subscription fees. Streamers who have signed up for the top tier can also receive up to 5% of ad revenue. Some of the most famous Twitch creators have a monthly income of up to \$1M. But the real earnings are based on the wealth of your audience. Depending on the regional prices of subscriptions, your total income can vary. Aside from the subscription fees, you can also earn money by chatting with your viewers. The amount you can earn depends on your charisma and the number of views you can get. ## Amazon Associates Twitch is a platform for live streaming video and other types of entertainment. Users can watch content in real time and take advantage of Twitch's live chat feature. In addition, they can participate in professional eSports leagues. And if they're lucky, they may even get the chance to win a large sum of money. To get the most out of your stream, you need to figure out how to monetize it. One option is to run ads. This is not the most profitable strategy if you're just starting out, but it's a great way to add extra revenue to your pocketbook. Another way to earn a side income is to sell products. You can use your Twitch channel as a storefront, offering games and other items for purchase. Amazon is one of many merchants that will work with you. For some users, a great way to make some extra cash is to set up a subscription page. Twitch subscribers receive ad-free viewing and access to live chats. Plus, they get to choose from a variety of subscription tiers. Using your Twitch account as a storefront can be a good idea, but it's important to be careful. Some advertisers are strict about what you can and can't show on your stream. If you don't adhere to their guidelines, you could lose big. Another way to monetize your stream is to add Amazon's Wish List feature to your profile. This allows you to collect gear that you'll need to set up your stream. Alternatively, you can add custom clickable buttons. You can also create your own tip jar. With an appropriate tip jar, you can give your followers a small token to remember your stream. # How Much is a Twitch TV Worth in 2021 2023? If you are thinking about purchasing a Twitch TV for your living room, you should know how much the price is going to be in the next couple years. You want to make sure you are not getting stuck with a rip off. Getting a good deal is not always easy, but it is possible. ## Streamers on Twitch Many Twitch streamers have hundreds of thousands of viewers and earn a great deal of money. However, the real income of these stars can vary depending on their skill, charisma, and popularity. Tim Ross is a Twitch streamer who has moved from Call of Duty to Fortnite. His following has grown to over 8 million. In addition to Twitch, Ross streams full time on YouTube. He has a net worth of \$7-\$8 million. Ibai Llanos is a former content creator for G2 Esports. Before coming to Twitch, he was a professional Overwatch player. The former player has also made his name casting for League of Legends. Auronplay is a Spanish-speaking gamer. She has been known to pull in 10M to 20M views on her broadcasts. Her video revenue comes from sponsored deals and ads. Ninja is the most famous and highest earning Twitch streamer. The youngest person to reach the million follower mark, he has 17.5 million followers and is one of the most watched. Previously known as Richard Blevins, he signed a multi-million dollar deal with Twitch in September 2020. Pokimane is the most followed female Twitch streamer. Her revenue comes from ad-based sponsorships and donations. Although she has a relatively small following, she is considered a major gaming star. Another popular Twitch streamer is Bryan Le, whose YouTube channel has topped 10.2 million subscribers. He has played many games including Call of Duty, Fortnite, and Apex Legends. His stream is commonly called "RiceGum." Another top NA-based Twitch streamer is Summit1G. Before becoming a professional CS:GO player, he was a successful streamer. Aside from Twitch, TimTheTatman also streams World Warcraft, Fall Guys, and Call of Duty: Black Ops. His video views have reached over 900,000 in the last year. Twitch users watch an average of 95 minutes per session. There are different viewing patterns by country and region. Streamers can earn money through chatting with the audience. Despite the lack of official earnings information, many Twitch streamers have earned millions of dollars. These figures are based on data gathered by third party platforms. ## Earning money on Twitch When it comes to earning money on Twitch, it's not as easy as it sounds. The platform requires hours of work, commitment, and time. There's also the matter of attracting viewers. In order to earn on Twitch, you need to first choose a niche, and then work your way up to Partner or Affiliate status. These programs reward streamers with a portion of the revenue generated from ads displayed on their page. This means that you'll have the potential to make a lot of money, but you'll also need to commit to your streams and interact with your fans. Streamers can also monetize their streams through donations. Donations are a centuries old form of support for streamers, and you can ask your followers to help you out with a small donation. Twitch also offers display advertising opportunities. These ads are much like those you see on YouTube. Sponsors can add a logo to your stream, as well as a link to the brand in the description. If the advertiser is good, you can expect to get some traction. One of the most popular ways to monetize your stream is to become an affiliate. To do this, you'll need to have a PayPal or Stripe account. You can then use your earnings to run ads on your stream. Another option is to join tournaments. This is a great way to generate new viewers and content. However, it's important to remember that these tournaments are mainly for the professionals in the esports industry. Those who aren't quite as seasoned can take part in smaller tournaments. Besides ad revenues, the most lucrative part of streaming on Twitch is the paid subscriptions. You can get a share of ad revenue based on your monthly subscriber count. Paying for a subscription can bring in thousands of dollars. Finally, you can make a buck using the Twitch Bits system. These virtual cheers can be sent to streamers for monetary contributions. They are available in different colors and sizes. Buying bits is more expensive for the viewer, but Twitch has a discount for donating in bulk. ## Become a Twitch streamer One of the best ways to earn money on Twitch is to become a Twitch streamer. You can't just jump into the video streaming business without first learning how to create an effective channel. This includes learning how to engage viewers and how to make money. Streamers can also earn money through affiliate programs. Some of the most popular are Amazon Associates, which pays 15% commission for purchases made through your link. Similarly, Shopify allows you to sell branded merchandise. Another way to increase revenue is through YouTube. If you have a good YouTube channel, you can sell ads on it. It's possible to get paid months after you post your videos. There's also the Twitch Partner Program. As a member of this program, you'll receive exclusive access to perks. However, this won't necessarily result in increased payouts. The eligibility requirements aren't clear. For example, you must be an established streamer, have an active audience, and be a role model. To really be successful at earning money on Twitch, you'll have to build an engaged community. This means working hard to produce quality content. Creating a YouTube channel is an excellent way to start, but you may need to add other social media channels to boost your earnings. It's no secret that Twitch is becoming one of the largest live-streaming platforms around. In fact, it had an average of 2.78 million concurrent viewers in 2021. And its total revenue in 2020 hit \$2.6 billion. Getting a Twitch partner can also increase your advertising revenues. For instance, you'll be able to promote offers during your live streams. But you'll still need to be active in chats and interact with other streamers to gain followers. Becoming a Twitch streamer isn't easy. In fact, the most successful streamers have invested hours in front of the camera. Not only do they create fun, entertaining streams, but they're also willing to put in the effort to reach a wide audience. ## Streaming games on Twitch Twitch is the leading video streaming platform in the United States. This is primarily due to live video game streams. The site is also popular for its chat, which allows users to interact with each other. It's an ideal place for new gamers and streamers to learn and develop. Streamers can earn money from donations, sponsorships, and ad revenues. According to Twitch statistics, 41% of viewers are between the ages of 16 and 34. However, men still outnumber women by a significant margin. Among the top 100 Twitch streamers are Ninja, Ibai, and Rubius. Auronplay is another big name. He is a popular GTA V player. SypherPK, a Fortnite expert, is another popular content creator. In 2021, Felix 'xQc' Lengyel was the most watched Twitch streamer. He is known for his wide variety of games. He grew his channel's followers by 5.6 million in 2021. His net worth is estimated to be \$25 million. Another well-known Twitch streamer is Adin Ross. He has a big following from playing Call of Duty and Fortnite. On April 20, 2022, his channel was banned. Though he was later able to get his account unbanned, it took almost 20 minutes for the ban to be reversed. While the Twitch user base is predominantly male, females have started to gain popularity among Twitch streamers. There are also females among the top 10 most followed streamers. Currently, there are two women in the top ten. If you're looking to become a Twitch streamer, it's important to choose your niche. You'll need to outline how you can offer value to your subscribers, as well as your gaming skills. Once you've reached Affiliate status, you can start earning money. If you're interested in finding out more about Twitch, take a look at TwitchTracker. This will show you the most followed channels, most popular categories, and average number of streams. Using these data, you can create a detailed marketing plan to promote your stream. The Twitch community is growing in number. There are more than 2.2 million monthly broadcasters. And with the growing popularity of live stream videos, the amount of money earned from this activity is expected to reach \$6 billion by the year 2021. # How Much Does a Twitch TV Cost 2023? There are many factors to consider when it comes to purchasing a Twitch TV. One of the biggest is the cost. How much is the average cost of a Twitch TV? It can depend on the type of TV, as well as the company. So, it's a good idea to do a little research before making a purchase. ## Making money on Twitch Twitch is a popular platform for online gamers to live stream their games. If you have a great following and are dedicated to creating great content, you can earn money from your stream. Streaming on Twitch isn't an easy way to make money, though. You need to have a consistent schedule and gain followers. You can earn money through donations, affiliate marketing, and merchandise. You may also be able to get sponsorship deals from other streamers. Sponsors can add a logo to your stream and mention their brand in your description. Streamers can sell games, t-shirts, and other merchandise. Streamlabs Merch is one of the many websites that offer merchandise. The Ninja channel, for example, sells merchandise on Twitch. To monetize your Twitch stream, you can set up PayPal donations buttons, or direct viewers to other websites. If you're a Twitch partner, you can run video ads before, during, and after your streams. When you're a Twitch Partner, you can choose the length of the ads. Some big-name Twitch streamers earn a lot of money every year. These guys typically have thousands of followers and spend most of their days streaming. They also use other social media channels to promote their streams. Donating on Twitch is a popular option for many viewers. You can donate as little as \$1 to your favorite channel, and Twitch will take a cut. This method can also be used for bigger events, such as tournaments. Another way to monetize your Twitch stream is through subscriptions. Subscribers can watch your stream ad-free, and you can earn up to \$2 per thousand views. Subscriptions start at \$4.99, and they give you access to special features. ## Revenue splits Twitch, the streaming platform owned by Amazon, is getting ready to implement changes to its revenue share model. This change could impact both partnered and unpartnered streamers. However, it is not clear when these changes will take effect. The company has not released any information about the changes so far, although they plan to announce them after June 2023. For the most part, Twitch currently shares a 50 percent revenue share with its partnered streamers. Some bigger streamers have been offered a 70 percent split. While that sounds pretty good, it is still a fairly small share. In order to encourage creators to make more money on Twitch, the platform has started an ad incentive program. Streamers can include ads in their streams and earn an additional 15% revenue share on top of the regular monthly subscription. These incentives are aimed at helping streamers to make a better, more predictable monthly income. Another reason why Twitch will be implementing these changes is because it wants to compete with YouTube, which is already a major player in the livestreaming space. Although it's not clear if the new revenue sharing model will negatively affect YouTube's competitiveness, it is likely to have a significant impact on the streamers of Twitch. A letter ostensibly mentioning the changes in Twitch's revenue sharing system has appeared on the website UserVoice. Over 22,000 people have voted on the post, which was started in 2020. Though it is a good start, the letter has fallen short of explaining why the changes are being made. At first glance, the biggest change is the fact that Twitch will no longer offer a full 70 percent revenue split to its subscriptions. While the company has never actually offered a split like this, some big streamers have had special deals that allow them to keep a full 30 percent of their subscription revenue. ## Streaming on Twitch If you're interested in making money through video game streaming, then you may have heard of Twitch. This live video streaming website has become a global phenomenon, reaching 140 million monthly active users. The streaming site has a lot to offer, including paid features for viewers. Twitch also has its own streaming software, called Twitch Studio, which allows users to set up multiple scenes, create different audio sources, and display on-screen notifications. It can be hard to make money on Twitch, though. To make sure you're getting the most out of your stream, you need to set up a good video capture device, and know what the best internet connection is. You'll need to be able to upload and download data fast enough for all your viewers to connect with you. For example, if you're broadcasting a game on the PlayStation 4, you'll want to be sure your Internet connection can handle the bandwidth required. Another thing to keep in mind is your microphone. Choose one with high quality, and a good sound so that your audience can hear you clearly. Streamers can also accept donations from viewers in the form of "bits." Each bit is worth a dollar, and each dollar is worth a penny. Some streamers also get perks from companies. Depending on how large of an audience they have, they can partner with aligned companies to advertise. Currently, Twitch gives 50 percent of its net revenue to partnered streamers. However, they've announced changes to their revenue system. Streamers will only be affected by the 50/50 split when they hit \$100,000 in revenue. Streamers are also rewarded with Channel Points multipliers, badges, custom emotes, and ad-free viewing. ## Earning money through affiliate links As a Twitch TV streamer, you may have noticed that you are able to earn money from affiliate links. These programs can be a great way to make money. You can also earn a percentage of the sales made by those who click on your link. However, you can only benefit from these programs if you meet the required criteria. First, you must have a large and engaged audience. To achieve this, you can run advertisements during your broadcast. You can also promote offers online or pitch your products to brands. Another way to make money is through sponsorship. Sponsorship is a great way to get big revenue for live appearances. Many companies reach out to Twitch streamers, but it is not limited to games or fashion. In order to become a Twitch Partner, you must meet certain requirements. For instance, you must have a regularly scheduled show and you must have an average of 500 concurrent viewers. Once you become a Partner, you can start selling games and other in-game items. Your page will have a "Buy Now" button for those who want to buy the game or item. You will receive 5% of the revenue generated from sales. One of the easiest ways to earn money through affiliate links on Twitch is to sign up with Amazon Associates. This is the most common form of affiliate marketing. Amazon is the most popular affiliate program, and it can be a good fit for Twitch streams. Its affiliate program pays 15% of all purchases from your channel. The Amazon Affiliate Program is available to anyone who streams on Twitch. It is easy to join, and you can set up different extensions. ## Earning money through the Brave browser There are hundreds of ways to earn money online. But one of the best is through a crypto-funded browser. It's free and easy to install. The Basic Attention Token (BAT) allows users to earn ad revenue without sacrificing their privacy. Brave is a free browser that earns BAT tokens when you use it. Unlike other browsers, however, you don't have to download it or link your wallet to receive the rewards. Instead, you'll earn BAT through browsing and ad-watching. With the help of the BAT tokens, you can support your favorite content creators. Just click on the BAT icon in the address bar and you'll be given the option to donate or tip the creator. Brave also allows you to directly support websites. You can set up a monthly schedule and donate or tip them. If you want to view ads, you can choose the number of ads you want to see per hour. And you can even disable all online ads. Brave also has an ad-matching system. When you load a webpage, it will look for keywords. When you see an ad that matches these keywords, you'll get a notification that lets you view the full ad. Another way to earn money through the Brave browser is through the Swag Store. This store sells digital products, physical goods, and more. Some of the items are stickers, hats, and socks. Using Brave's reward system, you can also earn BAT for referring friends or family to the service. In return, you'll be rewarded with a percentage of their generated ad revenues. As for how to get the most out of the Brave browser, you should check out the Brave news feed. In this area, you'll find great offers from great brands. # Planet Fitness New Years Eve Hours 2023 If you are planning to spend your New Years Eve at Planet Fitness, then you should know that the hours of the New Year's Eve party are going to vary depending on the club you go to. Some will be open 24 hours a day, while others will be closed on the New Year's Eve. There are also a few places to go if you are looking for live entertainment, such as the Countdown Stage at the Times Square. ## Countdown Stage on Times Square Planet Fitness will be the presenting sponsor of the New Year's Eve celebration in Times Square. The fitness center is known for its affordable prices and non-intimidating atmosphere. They will provide branded New Year's Eve party hats and thousands of balloons. Various live performances will take place throughout the night on the Countdown Stage. Jessie J will perform John Lennon's Imagine. ANEW, the New York dance group, will be featured. CNCO, winner of the Univision reality show La Banda, will also perform. Raul de Molina will also appear on the Countdown stage. His performance is in conjunction with the Sino-American Friendship Association. He will share a New Year's wish with the crowd. At 6:00 p.m., the USO Show Troupe will perform a military salute. The show will culminate in a countdown of ten seconds to the 11 o'clock hour. This will be followed by a performance from Hong Kong. Guests can also enjoy a Chinese cultural performance on the Countdown Stage. After the performance, the New York City Mayor will push a button and the ball will begin its descent. The Countdown will also feature a DJ set by KT Tunstall. The event is co-sponsored by the Times Square Alliance and Countdown Entertainment. The New Year's Eve Ball raising begins at 6 p.m., and will be preceded by a special pyrotechnics display. Afterward, the event will be capped off by a confetti show. Tickets for the event are available at \$75 for General Admission, or \$200 for VIP access. There are four drink tokens included with the ticket. Those who purchase a VIP ticket will also get to experience the VIP Mezzanine. Guests can also purchase champagne toast tickets at midnight. ## Dance group ANEW performs on Planet Fitness Center Stage This year's New Year's Eve festivities in the Big Apple include a number of live performance acts, plus special activities and events in the off hours. One of the highlights of this year's celebration will be the performance by the USO Show Troupe. The aforementioned group will perform a number of dance craze themed numbers in the main plaza as well as the adjacent Broadway Plaza. Throughout the evening, the troupe will take a brief pause at several points in the plaza to give the assembled masses a taste of their wares. In addition to their usual mercenaries, the USO troupe will also show off their most impressive pieces of swag. Among the most notable are their purple and gold hats and matching gloves. Also on display are a number of funky flies and a slew of colorful balloons. With the new year upon us, be sure to dress for the occasion. After all, there's only one chance to celebrate in style. The most important thing to remember is to have fun. Happy New Year! ## Open 24 hours a day, seven days a week Planet Fitness is a popular health club franchise with over 10 million members across the country. This includes over 1,500 locations in all 50 states. It has been known to offer a variety of workout equipment for a relatively cheap price. The New Year is a busy time at Planet Fitness. As such, you can find a number of locations with varying hours, and some may close early on the eve of the new year. Fortunately, many of them will re-open at 6am on January 2, the first day of the new year. Planet Fitness has a slogan of its own: "Judgement Free Zone (r)." To put this claim into context, this is not a gym that will open and shut its doors at random. Some will even provide limited services or facilities on certain holidays, including Independence Day and Thanksgiving. Other notable feats of the Planet Fitness empire include the fact that its headquarters is located in Manhattan's Times Square. This allows it to be involved in the famous New Year's Eve festivities and the webcast that follows. They also are the official fitness sponsor of the event. For example, they will be on display in the Countdown Stage. They will also be a major participant in the most important New Year's event of them all, the official Times Square New Year's Eve countdown. The planet fitness empire might not be as wide as its predecessor, Equinox, but it still offers a range of options. Most of their franchised clubs operate 24 hours a day, seven days a week. On top of that, the company is infamous for its affordable membership rates, which start at just ten dollars a month. ## Closed on New Year's Day Planet Fitness is one of the leading fitness chains in the country. It has over 10 million members and has more than 1500 branches in the United States and Canada. This is the third year that Planet Fitness has been a presenting sponsor of the Times Square New Year's Eve celebration. If you want to join the millions of people who are participating in the festivities in Times Square, you may be wondering whether or not your local Planet Fitness will be open. Most locations will be open on New Year's Eve, although some locations will be closed. The company has a "judgment free" mentality and is encouraging people to start a fitness routine in order to improve their overall health. The company also offers a welcoming atmosphere. While some franchised locations will operate on a reduced schedule, most will still be open for business. Just check with the club you plan to visit to see what hours they are operating. Planet Fitness is one of the few fitness chains that is open on both New Year's Day and New Year's Eve. In fact, some of the locations will be open 24 hours on New Year's Day. While some Planet Fitness locations will be closed on New Year's Eve, most will be open. Planet Fitness is a franchise and the owner will determine the closing times. Several Planet Fitness locations will be closing on January 1, 2023, so it's important to check with the gym you're planning to go to. Some locations will close early on New Year's Eve to allow for a party. However, those that remain open will be operating on a normal schedule. ## Get in shape Planet Fitness is a gym that is perfect for people looking to lose weight and get in shape. It is a low cost gym and offers 24-hour workouts. The company has over 10 million members and is one of the largest fitness centers in the US. If you are looking to get in shape for Planet Fitness New Year's Eve Hours 2023, then you will be happy to know that most of the locations are open. The hours may vary, however, so you will want to check the website or call ahead to confirm the hours. In some cases, the hours may be reduced to accommodate the busy holiday season. This means that you might not see as many customers as you do during the first week of the year. However, some of the Planet Fitness locations will be open on New Year's Day, although they will be operating under a reduced schedule. Many of these locations will be open until 6:00 AM on New Year's Day, but some will remain open until 8:00 PM. Planet Fitness is a fun way to get in shape. It is a place that is welcoming to new gym goers and has a nice environment. They also have a "Judgement Free" mentality and will encourage you to achieve your fitness goals. One of the reasons that Planet Fitness is such a popular gym is because they have a few special New Year's deals. You can save up to 30% off your monthly membership for the entire month. They also have a couple of special events to celebrate the new year. For instance, they are presenting sponsor of the Times Square New Year's celebration. ## Related Articles • #### D Bar Calculator January 30, 2023 \| Future Starr • #### Olg Lotto Max Winning Numbers January 30, 2023 \| Future Starr • #### Museum of Science and History - 3902 Maple Avenue Dallas TX 75219-3214 January 30, 2023 \| Future Starr • #### Stimulus Update: 4th Stimulus Check Plan January 30, 2023 \| Future Starr • #### Pick the Powerball Numbers For Wednesday February 26th 2023 January 30, 2023 \| Future Starr • #### What Are the 4 Types of Search Engines? January 30, 2023 \| Future Starr • #### How to Convert 140lbs in Kg January 30, 2023 \| Future Starr • #### How Many Pages Is 250 Words Capitalize My Title? 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Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team # Find the y-intercept of each exponential function and order the functions from least to greatest y-intercept. ? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 1 Jun 19, 2018 Tile 1 - Tile 3 - Tile 2 or Tile 3 - Tile 1 - Tile 2 #### Explanation: Tile One: The $y$-intercept is $- 1$ as shown in the graph. Tile Two: The coordinates of a $y$-intercept always has its $x$-coordinate as $0$. i.e. $\left(0 , y\right)$ Thus, in the table, the $y$-intercept can be found by finding the value of $y$ (which is equivalent to $f \left(x\right)$ ) when $x = 0$ which is $f \left(x\right) = 2$. Tile Three: If we put the explanation into a function, it would become: $g \left(x\right) = a \setminus \times {2}^{x}$ where a is an unknown constant. If we inserted the coordinates into the function that we've created, we get: $- 2 = 2 a$ $\setminus \therefore a = - 1$ and now, we can complete the function to get: $g \left(x\right) = - {2}^{x}$ To find the $y$-intercept, we have the $x$-coordinates as $0$: $g \left(0\right) = - {2}^{0}$ $g \left(0\right) = - 1$ $\setminus \therefore y$-intercept is $- 1$. Now we can just arrange the tiles in order of the values of the $y$-intercept. Thus, the order is: Tile 1 - Tile 3 - Tile 2 or Tile 3 - Tile 1 - Tile 2 Hope that makes sense! • 17 minutes ago • 22 minutes ago • 24 minutes ago • 28 minutes ago • 4 minutes ago • 4 minutes ago • 5 minutes ago • 7 minutes ago • 13 minutes ago • 14 minutes ago • 17 minutes ago • 22 minutes ago • 24 minutes ago • 28 minutes ago
# How do inverse operations help solve equations? Jan 21, 2015 They are helpful for solving with respect to a variable: if after some step you find yourself in a situation such as $\setminus \cos \left(x\right) = 0.7$, if you want to find the value of $x$ for which this statement is true, you need to apply the inverse cosine function at both sides and obtain $x = \setminus {\cos}^{- 1} \left(0.7\right)$ Mar 9, 2015 They allow us to "undo" what has been done to the variable. Example 1 Solve: $x + 3 = 8$ $3$ has been added to the variable, $x$. The inverse of addition is subtraction, so, by subtracting $3$, I can undo the addition. After $3$ was added, the result was equal to $8$. We undo the addition, by subtracting $3$ and see that, the starting amount was $5$.. We do't need to write all of that, except when questions like the one here are asked. Usually, we just do the inverse operation to the quantities that are equal. $x + 3 = 8$ $\left(x + 3\right) - 3 = 8 - 3$ $x = 5$ (Because on the left, $\left(x + 3\right) - 3 = x + 3 - 3$ which is just plain $x$.) The solution is $5$. As you gain practice, you probably won't even write that much. You understand that you're subtracting $3$ to get back to $x$, so you often won't bother writing it. Example 2 (much shorter) Solve: $5 x = 35$. Here, we have multiplied the unknown by $5$. The inverse of multiplication is division. So we'll divide both quantities (the quantities on the right and on the left) by $5$. $\frac{5 x}{5} = \frac{35}{5}$, so $x = 7$. The solution is $7$.
# FAQ: How To Introduce Area Lesson Plan? ## How do you write an introduction for a lesson plan? How do you write an introduction for a lesson plan? 1. Asking questions to get the students thinking about the topic of the lesson. 2. Showing pictures that relate to the lesson topic. 3. Telling a story to show the importance of the topic. 4. Bringing in “realia” (real objects) related to the lesson. ## What is Area lesson plan? Explain that area is the amount of space inside a shape. Draw a rectangle with the unit squares inside. (e.g., 4 rows of 5 units squares) First, count them and share the answer. Then, show them how to use the length and width to find the area. ## What is the correct way to introduce a lesson? Five Ways to Start Your Lessons 1. Start with a Video. Everyone loves a good video, especially kids. 2. Start with an Object. Another way to get your students wondering about a topic is to show them objects related to the content. ## What is area formula? Given a rectangle with length l and width w, the formula for the area is: A = lw (rectangle). That is, the area of the rectangle is the length multiplied by the width. As a special case, as l = w in the case of a square, the area of a square with side length s is given by the formula: A = s2 (square). You might be interested: Readers ask: How Do Animals Move Lesson Plan? ## What is introduction in lesson plan? The INTRODUCTION provides interest and motivation to the students. It focuses students’ attention on the lesson and its purposes. It also convinces students that they will benefit from the lesson. There are many ways to present an introduction. ## What are the steps of a lesson plan? Listed below are 6 steps for preparing your lesson plan before your class. • Identify the learning objectives. • Plan the specific learning activities. • Plan to assess student understanding. • Plan to sequence the lesson in an engaging and meaningful manner. • Create a realistic timeline. • Plan for a lesson closure. ## What is area and perimeter? Throughout our study of mathematics, we have learned that perimeter is the distance around a flat, two-dimensional, shape, and area is the amount of space taken up by a flat, two-dimensional shape. All this means is that the perimeter, or circumference, is the distance or length around a shape or object. ## What is the area of the square? The formula for the area of a square when the sides are given, is: Area of a square = Side × Side = S2. Algebraically, the area of a square can be found by squaring the number representing the measure of the side of the square. Introductions ## What are the five components of a lesson plan? The 5 Key Components Of A Lesson Plan • Objectives: • Warm-up: • Presentation: • Practice: • Assessment: You might be interested: The Girl Who Loved Wild Horses Lesson Plan? ## What are the 5 methods of teaching? Teacher-Centered Methods of Instruction • Direct Instruction (Low Tech) • Flipped Classrooms (High Tech) • Kinesthetic Learning (Low Tech) • Differentiated Instruction (Low Tech) • Inquiry-based Learning (High Tech) • Expeditionary Learning (High Tech) • Personalized Learning (High Tech) • Game-based Learning (High Tech)
## Book: RS Aggarwal - Mathematics ### Chapter: 18. Area of Circle, Sector and Segment #### Subject: Maths - Class 10th ##### Q. No. 20 of Exercise 18B Listen NCERT Audio Books to boost your productivity and retention power by 2X. 20 ##### The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days. [Take π = 3.14.] In an hour the minute hand completes one rotation therefore in 24 hours the minute hand will complete 24 rotations similarly the hour hand completes one rotation in 12 hours therefore in 24 hours it will complete 2 rotations. Now we have to just calculate the perimeter of the circle traced by minute hand and hour hand and multiply it with the number of rotations of minute hand and hour hand in 2 days respectively. Length of short/hour hand = r = 4 cm Length of long/minute hand = R = 6 cm The perimeter of circle traced by short hand = p = 2πr eqn1 The perimeter of circle traced by Long hand = P = 2πR eqn2 Now put the value of ‘r’ and ‘R’ in the equation 1 and 2 respectively. p = 2π(4) & P = 2π(6) (put π = 3.14) p = 2×3.14×4 & P = 2×3.14×6 p = 25.12 cm & P = 37.68 cm Therefore, distance covered by short hand in one rotation = 25.12 cm Distance covered by long hand in one rotation = 37.68 cm Number of rotation of short hand in one day = 2 Number of rotation of long hand in one day = 24 Therefore number of rotation of small hand in two days = 4 Number of rotation of long hand in two days = 48 Total distance covered by long hand in 2 days = P × no. of rotations in 2 days Total distance covered by long hand in 2 days = 37.68×48 Total distance covered by long hand in 2 days = 1808.64 cm eqn3 Total distance covered by short hand in 2 days = p × no. of rotations in 2 days Total distance covered by short hand in 2 days = 25.12×24 Total distance covered by short hand in 2 days = 100.48 cm eqn4 Now total distance covered by tip of both hands in 2 days = eqn3 + eqn4 Total distance covered by both hands in 2 days = 1808.64 + 100.48 Total distance covered by both hands in 2 days = 1909.12 cm The distance covered by both hands tip in 2 days is 1909.12 cm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 18 Mar 2019, 15:23 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # M31-43 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 53657 ### Show Tags 20 Jun 2015, 11:28 00:00 Difficulty: 65% (hard) Question Stats: 58% (01:26) correct 42% (01:28) wrong based on 43 sessions ### HideShow timer Statistics What is the value of positive integer $$a$$? (1) $$a! + a$$ is a prime number. (2) $$\frac{a+2}{(a+2)!}$$ is a terminating decimal. _________________ Math Expert Joined: 02 Sep 2009 Posts: 53657 ### Show Tags 20 Jun 2015, 11:28 1 4 Official Solution: What is the value of positive integer $$a$$? (1) $$a! + a$$ is a prime number. If $$a = 1$$, then $$a! + a = 2 = prime$$. If $$a > 1$$, then $$a! + a = a((a - 1)! + 1)$$ is a product of two integers each of which is greater than 1, so it cannot be a prime. Thus $$a = 1$$. Sufficient. (2) $$\frac{a+2}{(a+2)!}$$ is a terminating decimal. $$\frac{a+2}{(a+2)!}=\frac{a+2}{(a+1)!(a+2)}=\frac{1}{(a+1)!}$$. For $$\frac{1}{(a+1)!}$$ to be a terminating decimal, the denominator, $$(a+1)!$$, must have only 2's or/and 5's in its prime factorization, which is only possible if $$(a+1)! = 2! = 2$$ (all other factorials 3!, 4!, 5!, have primes other than 2 and 5 in them), making $$a = 1$$. Sufficient. _________________ Intern Joined: 08 Mar 2014 Posts: 47 Location: United States GMAT Date: 12-30-2014 GPA: 3.3 ### Show Tags 21 Jun 2015, 10:33 Hello, Can you please explain the first statement when a>1. I din't get how you derived the equation from a!+a. Math Expert Joined: 02 Sep 2009 Posts: 53657 ### Show Tags 21 Jun 2015, 10:59 vik09 wrote: Hello, Can you please explain the first statement when a>1. I din't get how you derived the equation from a!+a. $$a! = (a-1)!a$$, so you can factor $$a$$ out of $$a!+a$$ to get $$a((a−1)!+1)$$. _________________ Intern Joined: 01 Jan 2019 Posts: 3 ### Show Tags 14 Jan 2019, 17:10 Would someone be able to explain statement 2? I am unsure how " (a+1)!(a+1)!, must have only 2's or/and 5's in its prime factorization, which is only possible if (a+1)!=2!=2(a+1)!=2!=2 (all other factorials 3!, 4!, 5!, have primes other than 2 and 5 in them)" was derived Thanks. Math Expert Joined: 02 Sep 2009 Posts: 53657 ### Show Tags 14 Jan 2019, 21:55 1 xspongebobx wrote: Would someone be able to explain statement 2? I am unsure how " (a+1)!(a+1)!, must have only 2's or/and 5's in its prime factorization, which is only possible if (a+1)!=2!=2(a+1)!=2!=2 (all other factorials 3!, 4!, 5!, have primes other than 2 and 5 in them)" was derived Thanks. Theory: Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$25^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=25$$. Other examples of terminating decimals: fractions $$\frac{1}{8}$$, $$\frac{1}{25}$$, ... Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not. For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be the terminating decimal. Check below for more: Math: Number Theory How to Identify Terminating Decimals on the GMAT Terminating Decimals in Data Sufficiency on the GMAT Terminating and Recurring Decimals Even more: ALL YOU NEED FOR QUANT ! ! ! Ultimate GMAT Quantitative Megathread Hope it helps. _________________ Re: M31-43 [#permalink] 14 Jan 2019, 21:55 Display posts from previous: Sort by # M31-43 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Moderators: chetan2u, Bunuel Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
# Polygon and its types – Area and perimeter of a polygon Polygon? Regular Polygon? Hexagon? Regular Hexagon? A lot of mathematical terms! You do not know what exactly they are and you are probably confused. Do not worry because we are going to make an end to that confusion today. These words might seem scary but in actual they are not. You might be familiar with a lot of the concepts, you just do not know what the terms mean. First, we will see what is a polygon, then we will move on to what regular and irregular polygons are and then we will get into the details of a regular hexagon. In this post you will also learn to calculate the area, perimeter and sum of internal angles of polygon along with few solved examples. ## What is a polygon You have been studying geometrical shapes like squares, rectangles, and triangles, etc since childhood, right? These are examples of polygons. So, polygons are 2-dimensional(plane), closed figures with straight sides. Closed figures are those figures in which exactly two sides meet at the vertex or corner. • A polygon has always more than 2 sides. Quick Questions • Is the circle a polygon? No, because it does not consist of straight sides. • Is the box a polygon? No, because it is a 3-dimensional figure. Since you know what polygons are, let us study types of polygons. ### Different types of Polygons There are two types of polygons: 1. Regular Polygon 2. Irregular Polygon ### What is an irregular polygon An irregular polygon is a polygon that can have sides of unequal length and angles of any measure. For example, rectangle, scalene triangle, etc. They have unequal lengths and angles. Here you can know types of triangle and quadrilaterals. ### What is a regular polygon A regular polygon is a polygon that has all the sides of equal lengths and equal angles. Examples of regular polygons include equilateral triangle, square, pentagon, hexagon, etc. All of them have are equiangular (equal angles) and equilateral (same sides). Now you know the definition of regular and irregular polygon. We are going to continue our study of regular polygons taking hexagon as an example to explain its properties. ### Hexagon A Hexagon is a polygon that has six sides and thus six vertices or corners. Irregular hexagon has lengths and angles of any measure. However, regular hexagon has the same lengths and angles. We are going to study regular hexagons here. From now onwards, we mean regular hexagon whenever we refer to just hexagon. If the perimeter of hexagon (i.e. total length of the boundary of hexagon) is P then the length of each side is P/6. Let us now move on to angles. #### Angles of a polygon There are two types of angles in polygons. 1. Exterior angle 2. Interior angle #### Exterior angles of a polygon An exterior angle is an angle that is obtained by the side of the polygon and by the extension of the adjacent side. The sum of the exterior angles in a polygon is always 360o or 2π radians. In case of a regular polygon, the value of each exterior angle can be found by dividing 360o by the number of sides. • For example, there are 6 sides in a hexagon and all the exterior angles are equal. So, each exterior angle can be found as follows: Exterior angle = 360o / 6 = 60o or π/3 radians • Similarly, the value of each exterior angle of a regular heptagon ( a polygon with 7 sides ) can be found by 360o divided by 7. #### Interior angles of a polygon An interior angle is an angle that is inside the polygon i.e. between any two sides, at the vertex. The interior angles and exterior angles are measured along the same axis. So, if the exterior angle is 60o, then, Interior angle = 180o – 60o = 120o or 2π/3 radians. All the interior angles are equal for the regular hexagon. Therefore, The sum of all the interior angles = 6 × 120o = 720o or 4π radians. But there is a shortcut to find the sum of the interior angle of a polygon and that is by using a direct formula. #### Polygon interior angle formula Formula of the sum of interior angle of a polygon = ( N – 2 ) * 180o Where ‘N’ is the number of sides. Let we calculate the sum of the interior angle of few polygons. • Hexagon , a polygon with 6 sides : ( 6 – 2 ) * 180o = 720o • Heptagon, a polygon with 7 sides: ( 7-2 ) * 180o = 900o • Octagon, a polygon with 8 sides: ( 8-2 ) * 180o = 1080o • Nonagon, a polygon with 9 sides: ( 9 – 2 ) * 180o = 1260o #### Area of a polygon There is no common formula applicable to all types of the polygon. But you can derive the formula by your self. Let us now derive the area of the regular hexagon. We need to know a few terms before starting the calculation. Circumcircle: A circumcircle is a circle drawn outside the hexagon and it touches the vertices of the hexagon. Therefore, the radius of the circle is equal to the radius of the hexagon. Incircle: An incircle is a circle drawn inside the hexagon and it touches the midpoints of the sides of the hexagon. The apothem is the radius of the incircle or the distance from the midpoint of the hexagon to the midpoint of the side. #### Area of a hexagon Let ‘r’ be the radius of the circumcircle, ‘a’ be the apothem (a line from the polygon’s centre to any side, this line cuts the side at the midpoint ) and ‘n’ be the length of each side. We can see that the hexagon can be divided into the isosceles triangles (two sides of length r) i.e. the two vertices will be the vertices of the hexagon connecting to the midpoint of the hexagon. There will be a total of 6 triangles as there are six sides. Area of the triangle = Δt = 1/2 × base × height Here, let the base of the triangle be of size n and a height equal to apothem ‘a’. Then, Δt = 1/2 × n × a Let us further divide the triangle into two equal right-angled triangles of base n/2 each and a height equal to a. The overall revolution is equal to 360o. The angle ∠ACB for the triangle will be 360o/6 = 60o. Since we have further divided the triangle. So, for each small triangle, let’s say ACD, the angle ∠ACD will be 60o/2 = 30o or π/6 radians. We know that tangent is equal to the ratio of opposite side of angle (perpendicular) to the adjacent side (base). tan(θ) = opposite side / adjacent side tan(30o) = (n/2) / a 1/√3 = n / 2a 2/√3 = n / a or a / n = √3/2 a = (√3n/2) Putting the value of a in the above equation of the area of the triangle Δt = 1/2 × n × (√3n/2) Δt = √3/4 × n2 We are not done yet. This is the area of one triangle. The hexagon consists of 6 triangles. Hence, Area of hexagon = Δ = 6 × Δt Δ = 6 × √3/4 × n2 Δ = 3√3/2 × n2 • So, formula for the area of a hexagon = 3√3/2 × n2 Where ‘n’ is the length of the side of hexagon. #### Perimeter of hexagon If n is the length of one side of the hexagon, then the sum of the lengths of all the sides i.e. Perimeter = p = 6n Let’s solve some questions related to the area and perimeter of hexagon. ##### Example 1 If the length of a side of a regular hexagon is 5 cm. Find its area and perimeter. Solution: Length of each side = n = 5 cm The area can be found by the following formula: Area = Δ = 3√3/2 × n2 = 3√3/2 × 52 = 3√3/2 × 25 = 64.598 cm2 or 64.6 cm2 The area of the hexagon is 64.6 cm2. The perimeter can be calculated as follows: p = 6n = 6 × 5 = 30 cm The perimeter of the hexagon is 30 cm. ##### Example 2 If the perimeter of the regular hexagon is 36/√3 cm. Calculate its area. Solution Perimeter = p = 6n p = 36/√3 cm Then, 36 / √3 = 6n n = 36 / 6√3 n = 6/√3 cm The length of each side is 6/√3 cm. Finding the area. Δ = 3√3/2 × n2 n = 6/√3 cm Δ = 3√3/2 × (6/√3)2 = 3√3/2 × 12 = 31.177 cm2 or 31.18 cm2 The area of the hexagon is 31.18 cm2. If you like this post, share this content with your friends and most importantly don’t forget to subscribe to my blog to get this type of amazing post directly into your email id.
# Ex.4.4 Q2 Simple-Equations Solution - NCERT Maths Class 7 Go back to 'Ex.4.4' ## Question Solve the following: (a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus $$7$$. The highest score is $$87$$. What is the lowest score? (b) In an isosceles triangle, the base angles are equal. The vertex angle is $$40^\circ$$. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $$180^\circ$$). (c) Smitha’s mother is $$34$$ years old. Two years from now mother’s age will be $$4$$ times Smitha’s present age. What is Smitha’s present age? (d) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score? Video Solution Simple Equations Ex 4.4 \| Question 2 ## Text Solution What is Known? Statement of questions. What is unknown? Equation and the value of the variable. Reasoning: Make suitable equation using the given information and then solve the equation. Steps: (a) Highest score is $$87$$. Let the lowest marks be $$x$$ According to question, highest marks obtained $$= 2x + 7$$ \begin{align}87 &= 2x + 7\\87 – 7 &= 2x\\80 &= 2x\\x &= 40\end{align} (b) Let the base angle be $$b$$. Since the triangle is isosceles, the other base angle will also be $$b$$. Vertex angle is given $$40^\circ$$. Since, the sum of three angles of a triangle $$= 180^\circ$$ \begin{align}b + b + 40^\circ &= 180^\circ\\2b + 40^\circ &= 180^\circ\\2b = 180^\circ - 40^\circ &= 140^\circ\\b = \frac{140}{2} &= 70^\circ\end{align} (c) Let the present age of Smitha be $$x$$. Age of her mother $$= 34$$ years Two years from now Smitha age will be $$= x + 2$$ According to question, \begin{align}4(x + 2) &= 34\\4x + 8 &= 34\\4x &= 34 - 8 = 16\\x &= 4\end{align} (d) Let the score of Rahul be $$x$$, and score of Sachin be $$2x$$ According to question, \begin{align}x + 2x &= 198\\3x &= 198\\x &= \frac{{198}}{3}= 66\end{align} So, Rahul’s score $$= 66$$ runs And. Sachin’s score $$= 2x = 132$$ runs Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school
Question 6: Define Torque. Explain why it is equal to the vector product of force and moment arm? ## Definition of torque Torque is the turning effect produced in a body which is turning about a fixed point due to an applied force. Mathematically, Where is the torque produced, is the applied force and is the perpendicular distance between the line of action of force and the pivot.(Pivot is the point about which the body is rotating.) ### Explanation of torque The turning effect or torque depends on the following factors. 1. Applied force : The turning effect increases when we push the turning body harder and vice versa. So torque is directly proportional to the applied force. 2. Moment arm: Moment arm is the perpendicular distance between the force and point of rotation. This is in fact the effective component of the force which produces the torque and depends upon the angle which the force makes with the position vector of the point at which it is applied. Torque is directly proportional to the moment arm. 3. Direction of torque: As a vector quantity, torque should also have a direction. The direction of torque is determined by right hand rule. According to this rule, the direction of torque is up if the rotation is in the anti clock-wise direction. The direction of the torque is down if the body is turning in the clock-wise direction. Direction of torque is usually represented by a unit vector . Thus the proper way of writing torque is ## Why Torque = force × moment arm? As already stated torque depends upon the applied force and moment arm. This is quite clear that an increased force will produce more turning effect or torque and if the force is decreased the turning effect will also decrease. In the same way, increasing the moment arm will also increase the torque and vice versa. An interesting situation arises with changing the direction of the force. When we gradually change the direction of the force, the torque is also gradually changed. Few situations are hereby depicted in the adjacent fig. 1. Force is acting on a rod rotating around the pivot ‘O’. is making an angle θ = 0° with . The line of action of force passes through the center of pivot in this case. No turning effect is produced. 2. Now let the direction of force is changed. It acts at an angle 90° with now. It is our common experience that the turning effect is now maximum. 3. Let the direction of force is further changed. Now the force is making an angle of 180° with the moment arm . We observe the torque is again zero. To analyze the results of our experiment, we know that sin0° = 0. And when the angle between and is zero, the torque is also zero. The torque is maximum at angle 90°. Sine value is also maximum at this angle. Torque is again 0 when the angle is 180°. Sine value of the 180° is also zero. The conclusion is the turning effect or torque is proportional to the sine value of the angle between and . Therefore, torque is proportional to the magnitude of the force applied, magnitude of moment arm and the sine of the angle between them. Therefore, = rFsinθ This is exactly what the vector product is. Therefore, Thus torque is equal to the vector product of force and moment arm.
A rock climber stands on top of a 70m high cliff overhanging a pool of water. He throws stones vertically downward 1.2s apart and observes that the cause a single splash. The initial speed of the first stone was 2.5 m/s. How long after the release of the first stone does the second stone hit the water? • How long after the release of the first stone does the second stone hit the water? • What was the initial speed of the second stone? • What is the speed of each stone as it hits the water? This question aims to find the time of the stone as it hits the water, the initial speed of the second stone, and the final speed of both stones as they hit the water. The basic concepts needed to understand and solve this problem are equations of motion, gravitational acceleration, and initial and final velocities of an object during vertical fall. We are taking the initial point at the cliff as the starting point, hence the final height will be at the water surface and the initial height will be at the cliff. Also, the downward motion will be taken as positive. The given information regarding this problem is given as follows: $The\ Initial\ Velocity\ of\ the\ First\ Stone\ v_i\ =\ 2.5\ m/s$ $The\ Final\ Height\ h_f\ =\ 70\ m$ $The\ Initial\ Height\ h_i\ =\ 0\ m$ $The\ Acceleration\ due\ to\ Gravity\ g\ =\ 9.8\ m/s^2$ a) To calculate the time the second stone took to hit the water after the first stone, we will use the equation of motion, which is given as: $h_f = h_i + v_it + \dfrac{1}{2} at^2$ Substituting the values, we get: $70 = 0 + 2.5t + \dfrac{1}{2} (9.8) t^2$ $4.9t^2 + 2.5t – 70 = 0$ By using the quadratic formula, we can calculate the value of $t$, which is calculated to be: $t_1 = 3.53\ s$ Ignoring the negative value of $t$ as time is always positive. The second stone was released $1.2s$ after the first stone was released, but reached the water at the same time. So the time the second stone took to reach the water is given as: $t_2 = 3.53\ -\ 1.2$ $t_2 = 2.33\ s$ b) To calculate the initial velocity of the second stone, we can use the same equation. The initial velocity can be calculated as: $h_f = h_i + v_it_2 + \dfrac{1}{2} gt_{2}^{2}$ Substituting the values, we get: $70 = 0 + v_{i2} (2.33) + (0.5 \times 9.8 \times (2.33)^2$ $v_{i2} = \dfrac{70 – 26.6} {2.33}$ $v_{i2} = \dfrac{43.4}{2.33}$ $v_{i2} = 18.63\ m/s$ c) To calculate the final velocities of both stones, we can use the following equation of motion: $v_f = v_i + gt$ The final velocity of the first stone is given as: $v_{f1} = 2.5 + 9.8 \times 3.53$ $v_{f1} = 37.1\ m/s$ The final velocity of the second stone is given as: $v_{f2} = 18.63 + 9.8 \times 2.33$ $v_{f2} = 41.5\ m/s$ Numerical Results a) The total time the second stone took to hit the water: $t_2 = 2.33\ s$ b) The initial velocity of the second stone is calculated as: $v_{i2} = 18.63\ m/s$ c) The final velocities of both stones are calculated as: $v_{f1} = 37.1\ m/s \hspace{0.6in} v_{f2} = 41.5\ m/s$ Example The initial velocity of an object is $2m/s$ and it took the object $5s$ to reach the ground. Find its final velocity. As the object is falling, we can take the acceleration $a$ to be the gravitational acceleration $g$. By using the first equation of motion, we can calculate the final velocity without knowing the total height. $v_f = v_i + gt$ $v_f = 2 + 9.8 \times 5$ $v_f = 51\ m/s$ The final velocity of the object is calculated to be $51 m/s$.
# Evaluate Question: Evaluate $\lim _{x \rightarrow 4}\left(\frac{e^{x}-e^{4}}{x-4}\right)$ Solution: To evaluate $\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}$ Formula used: L'Hospital's rule Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where $\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$ then $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$ As $x \rightarrow 0$, we have $\lim _{x \rightarrow 0} \frac{e^{2+x}-e^{2}}{x}=\frac{0}{0}$ This represents an indeterminate form. Thus applying L'Hospital's rule, we get $\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}=\lim _{x \rightarrow 4} \frac{\frac{d}{d x}\left(e^{x}-e^{4}\right)}{\frac{d}{d x}(x-4)}$ $\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}=\lim _{x \rightarrow 4} \frac{e^{x}}{1}$ $\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}=e^{4}$ Thus, the value of $\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}$ is $e^{4}$.
Education.com Try Brainzy Try Plus # Adding, Subtracting, Multiplying, and Dividing Decimals Study Guide: GED Math By Updated on Mar 23, 2011 Practice problems for these concepts can be found at: Fractions and Decimals Practice Problems: GED Math ### Adding and Subtracting Decimals To add decimal numbers, follow these steps: Step 1 Write the numbers so that the decimal points are lined up. Step 2 Make all the decimals have the same number of digits to the right of the decimal point by adding trailing zeros to the ends of shorter decimals. Step 3 Write the decimal point in the answer so that it lines up with the decimal points in the problem. Step 4 Add the decimals just as you would if you were adding whole numbers. Example 9.23 + 6.02 + 1.1 = Write the numbers so that the decimal points are lined up: Make all the decimals have the same number of digits to the right of the decimal point by adding a trailing zero to the end of the shorter decimal: Write the decimal point in the answer so that it lines up with the decimal points in the problem: Add the decimals just as you would if you were adding whole numbers to find your answer: To add a whole number and a decimal together, you would follow the same steps. Example 12 + 5.013 = Write the numbers so that the decimal points are lined up. Insert a decimal point to the right of the whole number: Make all the decimals have the same number of digits to the right of the decimal point by adding trailing zeros to the end of the shorter decimal: Write the decimal point in the answer so that it lines up with the decimal points in the problem: Add the decimals just as you would if you were adding whole numbers to find your solution: Setting decimals up for subtraction is very similar to the setup when adding them. To subtract decimals, use the following steps: Step 1 Write the numbers so that the decimal points are lined up. Insert a decimal point to the right of any whole number. Step 2 Make all the decimals have the same number of digits to the right of the decimal point by adding trailing zeros to the ends of shorter decimals. Step 3 Write the decimal point in the answer so that it lines up with the decimal points in the problem. Step 4 Subtract the decimals just as you would if you were subtracting whole numbers. Example 11 – 5.2 = Write the numbers so that the decimal points are lined up: Make all the decimals have the same number of digits to the right of the decimal point by adding a trailing zero to the end of the shorter decimal: Write the decimal point in the answer so that it lines up with the decimal points in the problem: Subtract the decimals, and don't forget to borrow to arrive at the final answer: ### Multiplying Decimals If you can multiply whole numbers, then you can multiply decimals. The main thing to watch out for is the placement of the decimal point. Placing the decimal point in your answer is just a matter of counting place values. When multiplying with decimals, multiply as you would for whole numbers, and ignore the decimal points until after the product is found. After performing the multiplication, count the number of digits after the decimal points (to the right of the decimal point) in both factors being multiplied. This count is the number of decimal places (to the right of the decimal point) that will be in the answer. Start at the rightmost side of the product (the answer) and count to the left the number of digits (the number of digits to the right of the decimal point in both terms) in order to place the decimal point. Example 2.48 × 1.7 = Multiply, ignoring the decimal points: 248 × 17 = 4,216. Determine the digits to the right of the decimal points in the factors: 4, 8, and 7. Starting to the right of the 6 in the answer, move three digits to the left: 4.216. ### Dividing Decimals To divide with decimal numbers, first change the problem to division by a whole number. It may be necessary to move the decimal point in the divisor (the number you are dividing by) to make it a whole number. Move the decimal in the dividend (the number you are dividing into) the same number of places, and copy the new decimal placeholder straight up into the quotient (the answer to the division problem). Once the decimal point is placed, divide as you normally would with long division. Example 3.26 ÷ 0.02 = Solve using long division. First move the decimal point two places to the right in each number: ### Decimal Operations: Multiplying or Dividing by the Powers of 10 The decimal number system is based on the powers of 10. This makes multiplication and division by 10, 100, 1,000,… a matter of moving the decimal point the number of places dictated by the number of zeros in 10, 100, or 1,000. This is because once you add or remove the zeros, you are essentially multiplying or dividing by 1. To multiply a number by 10, move the decimal point one place to the right. To multiply a number by 100, move the decimal point two places to the right. To multiply a number by 1,000, move the decimal point three places to the right. To divide a number by 10, move the decimal point one place to the left. To divide a number by 100, move the decimal point two places to the left. To divide a number by 1,000, move the decimal point three places to the left. Practice problems for these concepts can be found at: Fractions and Decimals Practice Problems: GED Math Add your own comment
# KEY ```KEY Math 110, 112 Practice Quiz # 9 – Sections 5.1- 5.3 Solving Systems of Equations by Graphing and Substitution, Addition, 3 Variables 1. Without graphing or solving, determine the number of solutions of the given system of equations. Then state whether the system is independent, dependent or inconsistent. a. x – 2y = 4 b. - 3y = - 6x + 12 2x – 4y = 8 y = 3x + 6 a. many solutions, dependent b. 1 solution, independent 2. a. Graph the system of equations and determine the number of solutions. b. Then state whether the system is independent, dependent or inconsistent. i. y = - 4x + 3 ii. y = 1/3 x + 2 y + 4x = 7 y = 3x – 1 no solutions inconsistent one solution independent 3. Use the substitution method to determine the solutions algebraically. 4x + 2y = 20 y = 3x – 5 a. x = 3, y = 4 4. Two angles are supplementary. The larger angle is 30o more than twice the smaller angle. Write a system of equations representing the situation. Then find the measure of each angle. X = 50o, y = 130o 5. Use the addition method to solve the system of equations and state whether the system is independent, dependent or inconsistent. a. 5x – 2y = 1 -4y + 10x = 2 a. 0 = 0, dependent b. 0.2x + 0.4y = 2 0.8y = -0.4x + 3 b. 0 = -10, inconsistent c. x = 2(y + 5) y + 3x = 2 c. x = 2, y = -4, independent KEY (Sect. 5.1-5.3 cont.) 6. A store sells 20 more cell phones than palm pilots. Cell phones sell for \$100 each and palm pilots sell for \$250 each. If the store sells \$21,600 worth of cell phones and palm pilots in a week, how many cell phones were sold that week? How many palm pilots were sold that week? 56 palm pilots, 76 cell phones 7. Use the addition method to solve the system of equations. 3x – y – 2z = 5 2x + y + 3z = 6 6x – y – 4z = 9 x=2 y = -1 z=1 8. Jovita divides \$17,000 into three investments: a savings account paying 6%, a bond paying 9%, and a money market fund paying 11% (all simple annual interest). The annual interest from the three accounts is \$1540, and she has 3 times as much invested in the bond as in the savings account. What amount does she have invested in each account? Savings: \$3000, Bond: \$9000, Money Market: \$5000 ``` 12 cards 32 cards 20 cards 14 cards 23 cards
# How do you use synthetic division to divide ( 6x^3 - 9x^2 - 12x - 6 ) / ( 3x + 4 )? Dec 6, 2016 The quotient is $= 2 {x}^{2} - \frac{17}{3} x + \frac{32}{9}$ and the remainder is $= - \frac{182}{9}$ #### Explanation: Let's do the division $\textcolor{w h i t e}{a a a a}$$6 {x}^{3} - 9 {x}^{2} - 12 x - 6$$\textcolor{w h i t e}{a a a a}$∣$3 x + 4$ $\textcolor{w h i t e}{a a a a}$$6 {x}^{3} + 8 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣$2 {x}^{2} - \frac{17}{3} x + \frac{32}{9}$ $\textcolor{w h i t e}{a a a a a}$$0 - 17 {x}^{2} - 12 x$ $\textcolor{w h i t e}{a a a a a a a}$$- 17 {x}^{2} - \frac{68}{3} x$ $\textcolor{w h i t e}{a a a a a a a a a a a}$$0 + \frac{32}{3} x - 6$ $\textcolor{w h i t e}{a a a a a a a a a a a a a}$$+ \frac{32}{3} x + \frac{128}{9}$ $\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$+ 0 - \frac{182}{9}$ The quotient is $= 2 {x}^{2} - \frac{17}{3} x + \frac{32}{9}$ and the remainder is $= - \frac{182}{9}$
Courses Courses for Kids Free study material Offline Centres More Store # How do you factor $8{{x}^{2}}-4x-24$ ? Last updated date: 10th Aug 2024 Total views: 388.8k Views today: 7.88k Verified 388.8k+ views Hint: When we factorize a quadratic equation $a{{x}^{2}}+bx+c$ we have to find pair of number whose sum is equal to b and product equal to product of a and c . Then we can write bx as the sum of the 2 terms. Here We can split -4x to 12x and -16x to solve this question. Complete step by step solution: The given equation is $8{{x}^{2}}-4x-24$ which is a quadratic equation. if we compare the equation to standard quadratic equation $a{{x}^{2}}+bx+c$ then a = 8, b = -4 and c = -24 To factor a quadratic equation, we can find two numbers m and n such that the sum of m and n is equal to b and the product of m and n is $ac$. Then we can split $bx$ to $mx+nx$ then we can factor the equation easily. In our case ac = -192 and b = -4 So pair of 2 numbers whose product is -192 and sum -4 is ( 12 ,-16) We can -4x split to 12x – 16x So $\Rightarrow 8{{x}^{2}}-4x-24=8{{x}^{2}}+12x-16x-24$ Taking 4x common in the first half of the equation and taking -8 common in the second half of the equation. $\Rightarrow 8{{x}^{2}}-4x-24=4x\left( 2x+3 \right)-8\left( 2x+3 \right)$ Taking 2x + 3 common $\Rightarrow 8{{x}^{2}}-4x-24=\left( 4x-8 \right)\left( 2x+3 \right)$ We can take 4 common from 4x - 8 $\Rightarrow 8{{x}^{2}}-4x-24=4\left( x-2 \right)\left( 2x+3 \right)$ Note: While factoring a quadratic equation we can’t always split $bx$ such that their product is equal to ac because sometimes the roots can be irrational numbers. In that case we can find the roots of the equation by formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
1. ## Factoring equation... How would I factor these equations? 1. 6x² + 3x = 0 2. -2x² + 5x + 3 = 0 I know how to factor equations without the number before x². But when the number is there I don't know how to factor it. 2. Originally Posted by lp_144 How would I factor these equations? 1. 6x² + 3x = 0 2. -2x² + 5x + 3 = 0 I know how to factor equations without the number before x². But when the number is there I don't know how to factor it. for the first, 3x is a common term. just factor it out for the second, in general we use what we call the AC-method to factor, if possible, a quadratic in which the coefficient of the squared term is not 1. but for small numbers like 2, we can use trial and error, it would generally be a lot simpler. so $\displaystyle -2x^2 + 5x + 3 = 0$ $\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$ $\displaystyle \Rightarrow (2x ~~a)(x ~~b) = 0$ now in the place of $\displaystyle a$ and $\displaystyle b$, you want numbers that multiply to give you -3, but $\displaystyle a$ and $\displaystyle 2b$ will add to give -5 3. Originally Posted by Jhevon for the first, 3x is a common term. just factor it out for the second, in general we use what we call the AC-method to factor, if possible, a quadratic in which the coefficient of the squared term is not 1. but for small numbers like 2, we can use trial and error, it would generally be a lot simpler. so $\displaystyle -2x^2 + 5x + 3 = 0$ $\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$ $\displaystyle \Rightarrow (2x ~~a)(x ~~b) = 0$ now in the place of $\displaystyle a$ and $\displaystyle b$, you want numbers that multiply to give you -3, but $\displaystyle a$ and $\displaystyle 2b$ will add to give -5 So it would be $\displaystyle (2x+1)(x-3)$? 4. Originally Posted by lp_144 So it would be $\displaystyle (2x+1)(x-3)$? yes, very good now finish the problem. what about the first problem? 5. Originally Posted by Jhevon yes, very good now finish the problem. what about the first problem? 6. Originally Posted by lp_144 when you have a common term, you can factor it out example: factorize $\displaystyle 3xy + 5xy^2$ we want to factor out the lowest powers of all things common. i have $\displaystyle x$ common to both terms, and i have $\displaystyle y$ common to both terms. there is a $\displaystyle y^2$ in the second term, but i can't factor that out, because i only have $\displaystyle y$ in the first term. so $\displaystyle x$ and $\displaystyle y$ are common to both terms, both terms have them, so i will factor out $\displaystyle xy$, to get: $\displaystyle xy(3 + 5y)$ so i took out the $\displaystyle xy$ from the 3, and i have none left in the brackets. as for the $\displaystyle 5xy^2$, i took out the $\displaystyle x$ and one of the $\displaystyle y$'s, so i am left with one $\displaystyle y$ on the inside, hence the $\displaystyle 5y$ now you have, $\displaystyle 6x^2 + 3x = 0$ i told you that $\displaystyle 3x$ is common to both, so factor it out 7. Originally Posted by Jhevon now you have, $\displaystyle 6x^2 + 3x = 0$ i told you that $\displaystyle 3x$ is common to both, so factor it out When I factored $\displaystyle 3x$ out, I got $\displaystyle 3x(3x)$. What do I do from here? 8. Originally Posted by lp_144 When I factored $\displaystyle 3x$ out, I got $\displaystyle 3x(3x)$. What do I do from here? $\displaystyle 6x^2 + 3x = 0$ $\displaystyle \Rightarrow 3x(2x + 1) = 0$ step 1: look carefully at what i did and make sure you understand it. step 2: finish the problem. how would we solve for $\displaystyle x$? 9. Originally Posted by Jhevon $\displaystyle 6x^2 + 3x = 0$ $\displaystyle \Rightarrow 3x(2x + 1) = 0$ step 1: look carefully at what i did and make sure you understand it. step 2: finish the problem. how would we solve for $\displaystyle x$? I think I got it. $\displaystyle 3x(2x+1)=0$ $\displaystyle 3x=0$ $\displaystyle x=0$ $\displaystyle 2x+1=0$ $\displaystyle 2x=-1$ $\displaystyle x= -\frac {1}{2}$ 10. Originally Posted by Jhevon for the first, 3x is a common term. just factor it out for the second, in general we use what we call the AC-method to factor, if possible, a quadratic in which the coefficient of the squared term is not 1. but for small numbers like 2, we can use trial and error, it would generally be a lot simpler. so $\displaystyle -2x^2 + 5x + 3 = 0$ $\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$ $\displaystyle \Rightarrow (2x ~~a)(x ~~b) = 0$ now in the place of $\displaystyle a$ and $\displaystyle b$, you want numbers that multiply to give you -3, but $\displaystyle a$ and $\displaystyle 2b$ will add to give -5 No, the numbers $\displaystyle a$ and $\displaystyle b$ multiply to give you -6 (not -3) because 2(-3) = -6, and their sum will be -5 See here $\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$ $\displaystyle \Rightarrow 2x^2 - 6x +1x - 3 = 0\;(here, \;a=-6 \; and \;b=+1)$ $\displaystyle \Rightarrow 2x(x-3) +1(x - 3) = 0$ $\displaystyle \Rightarrow (2x + 1)(x-3) = 0$ $\displaystyle \Rightarrow x=\frac{-1}{2}\;and\;x=3$ 11. Originally Posted by Shyam No, the numbers $\displaystyle a$ and $\displaystyle b$ multiply to give you -6 (not -3) because 2(-3) = -6, and their sum will be -5 See here $\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$ $\displaystyle \Rightarrow 2x^2 - 6x +1x - 3 = 0\;(here, \;a=-6 \; and \;b=+1)$ $\displaystyle \Rightarrow 2x(x-3) +1(x - 3) = 0$ $\displaystyle \Rightarrow (2x + 1)(x-3) = 0$ $\displaystyle \Rightarrow x=\frac{-1}{2}\;and\;x=3$ note that your a and b multiply to give -6 as i said. i was not doing the AC-method here.
# Mathematics K-8 Critical Areas of Focus Save this PDF as: Size: px Start display at page: ## Transcription 4 Page 4 KINDERGARTEN CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #1, CON T. Measurement and Data K.MD Describe and compare measurable attributes. 1. Describe measurable attributes of objects, such as length or weight. Describe several measurable attributes of a single object. 2. Directly compare two objects with a measurable attribute in common, to see which object has more of / less of the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter. Classify objects and count the number of objects in each category. 3. Classify objects into given categories; count the numbers of objects in each category and sort the categories by count. Number and Operations in Base Ten K.NBT Work with numbers to gain foundations for place value. 8. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (e.g., 18 = ); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. 5 Page 5 KINDERGARTEN CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #2 Describing shapes and space Students describe their physical world using geometric ideas (e.g., shape, orientation, spatial relations) and vocabulary. They identify, name, and describe basic two-dimensional shapes, such as squares, triangles, circles, rectangles, and hexagons, presented in a variety of ways (e.g., with different sizes and orientations), as well as three-dimensional shapes such as cubes, cones, cylinders, and spheres. They use basic shapes and spatial reasoning to model objects in their environment and to construct more complex shapes. Geometry Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). 1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. 2. Correctly name shapes regardless of their orientations or overall size. 3. Identify shapes as two-dimensional (lying in a plane, flat ) or three-dimensional ( solid ). Analyze, compare, create, and compose shapes. 4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/ corners ) and other attributes (e.g., having sides of equal length). 5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. 6. Compose simple shapes to form larger shapes. For example, "Can you join these two triangles with full sides touching to make a rectangle? K.G 7 Page 7 FIRST GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #1, CONTINUED Number and Operations in Base Ten 1.NBT Use place value understanding and properties of operations to add and subtract. 4. Add within 100, including adding a two-digit number and a one-digit number, and adding a twodigit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten. 5. Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used. 6. Subtract multiples of 10 in the range from multiples of 10 in the range (positive or zero differences), using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. 8 Page 8 FIRST GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #2 Developing understanding of whole number relationships and place value, including grouping in tens and ones Students develop, discuss, and use efficient, accurate, and generalizable methods to add within 100 and subtract multiples of 10. They compare whole numbers (at least to 100) to develop understanding of and solve problems involving their relative sizes. They think of whole numbers between 10 and 100 in terms of tens and ones (especially recognizing the numbers 11 to 19 as composed of a ten and some ones). Through activities that build number sense, they understand the order of the counting numbers and their relative magnitudes Number and Operations in Base Ten Extend the counting sequence. 1.NBT 1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral. Understand place value. 2. 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). 3. Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <. 9 Page 9 FIRST GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #3 Developing understanding of linear measurement and measuring lengths as iterating length units Students develop an understanding of the meaning and processes of measurement, including underlying concepts such as iterating (the mental activity of building up the length of an object with equal-sized units) and the transitivity principle for indirect measurement. Measurement and Data Measure lengths indirectly and by iterating length units. 1.MD 1. Order three objects by length; compare the lengths of two objects indirectly by using a third object. 2. 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. Tell and write time. 3. Tell and write time in hours and half-hours using analog and digital clocks. Represent and interpret data. 7. Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another. 10 Page 10 FIRST GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #4 Reasoning about attributes of, and composing and decomposing geometric shapes Students compose and decompose plane or solid figures (e.g., put two triangles together to make a quadrilateral) and build understanding of part-whole relationships as well as the properties of the original and composite shapes. As they combine shapes, they recognize them from different perspectives and orientations, describe their geometric attributes, and determine how they are alike and different, to develop the background for measurement and for initial understandings of properties such as congruence and symmetry. Geometry 1.G Reason with shapes and their attributes. 1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus nondefining attributes (e.g., color, orientation, overall size); build and draw shapes that possess defining attributes. 2. 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. 3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. 11 Page 11 SECOND GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #1 Extending understanding of base-ten notation Students extend their understanding of the base-ten system. This includes ideas of counting in fives, tens, and multiples of hundreds, tens, and ones, as well as number relationships involving these units, including comparing. Students understand multi-digit numbers (up to 1000) written in base-ten notation, recognizing that the digits in each place represent amounts of thousands, hundreds, tens, or ones (e.g., 853 is 8 hundreds + 5 tens + 3 ones). Number and Operations in Base Ten 2.NBT Understand place value. 1. Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones. Understand the following as special cases: a. 100 can be thought of as a bundle of ten tens called a hundred. b. The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three, four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones). 2. Count within 1000; skip-count by 5s, 10s, and 100s. 3. Read and write numbers to 1000 using base-ten numerals, number names, and expanded form. 4. Compare two three-digit numbers based on meanings of the hundreds, tens, and ones digits, using >, =, and < symbols to record the results of comparisons. 13 Page 13 SECOND GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #2, CONTINUED Operations and Algebraic Thinking 2.OA Represent and solve problems involving addition and subtraction. 1. Use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Add and subtract within Fluently add and subtract within 20 using mental strategies. By end of Grade 2, know from memory all sums of two one-digit numbers. Work with equal groups of objects to gain foundations for multiplication. 3. Determine whether a group of objects (up to 20) has an odd or even number of members, e.g., by pairing objects or counting them by 2s; write an equation to express an even number as a sum of two equal addends. 4. Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends. 14 Page 14 SECOND GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #3 Using standard units of measure Students recognize the need for standard units of measure (centimeter and inch) and they use rulers and other measurement tools with the understanding that linear measure involves an iteration of units. They recognize that the smaller the unit, the more iterations they need to cover a given length. Measurement and Data 2.MD Measure and estimate lengths in standard units. 1. Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. 2. Measure the length of an object twice, using length units of different lengths for the two measurements; describe how the two measurements relate to the size of the unit chosen. 3. Estimate lengths using units of inches, feet, centimeters, and meters. 4. Measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit. 15 Page 15 SECOND GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #4 Describing and analyzing shapes Students describe and analyze shapes by examining their sides and angles. Students investigate, describe, and reason about decomposing and combining shapes to make other shapes. Through building, drawing, and analyzing two- and three-dimensional shapes, students develop a foundation for understanding area, volume, congruence, similarity, and symmetry in later grades. Geometry 2.G Reason with shapes and their attributes. 1. Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces. Identify triangles, quadrilaterals, pentagons, hexagons, and cubes. 2. Partition a rectangle into rows and columns of same-size squares and count to find the total number of them. 3. Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape. Operations and Algebraic Thinking Work with equal groups of objects to gain foundations for multiplication. 2.OA 4. Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends. 16 Page 16 SECOND GRADE CRITICAL AREAS OF FOCUS STANDARDS AND CLUSTERS BEYOND THE CRITICAL AREAS OF FOCUS Telling and writing time time Students tell and write time to the nearest five minutes designating the appropriate a.m. or p.m. Measurement and Data Work with time and money. 2.MD 7. Tell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m. Using data representations Students measure and represent their measurements on line plots. They sort objects in categories and represent their data in bar graphs and picture graphs. Scales are based on single-units. Students answer questions that relate their understanding of addition and subtraction to data represented in the picture and bar graphs. Measurement and Data Represent and interpret data. 2.MD 9. Generate measurement data by measuring lengths of several objects to the nearest whole unit, or by making repeated measurements of the same object. Show the measurements by making a line plot, where the horizontal scale is marked off in whole-number units. 10. Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems [1] using information presented in a bar graph. [1] See Glossary, Table 1. 17 Page 17 THIRD GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #1 Developing understanding of multiplication and division and strategies for multiplication and division within 100 Students develop an understanding of the meanings of multiplication and division of whole numbers through activities and problems involving equal-sized groups, arrays, and area models; multiplication is finding an unknown product, and division is finding an unknown factor in these situations. For equal-sized group situations, division can require finding the unknown number of groups or the unknown group size. Students use properties of operations to calculate products of whole numbers, using increasingly sophisticated strategies based on these properties to solve multiplication and division problems involving single-digit factors. By comparing a variety of solution strategies, students learn the relationship between multiplication and division. Operations and Algebraic Thinking Represent and solve problems involving multiplication and division. 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 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 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. 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 =?. Understand properties of multiplication and the relationship between multiplication and division. 5. Apply properties of operations as strategies to multiply and divide. Examples: If 6 4 = 24 is known, then 4 6 = 24 is also known. (Commutative property of multiplication.) 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) = = 56. (Distributive property.) 6. Understand division as an unknown-factor problem. For example, find 32 8 by finding the number that makes 32 when multiplied by 8. Multiply and divide within Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 5 = 40, one knows 40 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers. Solve problems involving the four operations, and identify and explain patterns in arithmetic. 8. Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. 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. 18 Page 18 THIRD GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #1, CONTINUED Number and Operations in Base Ten 3.NBT Use place value understanding and properties of operations to perform multi-digit arithmetic. 3. Multiply one-digit whole numbers by multiples of 10 in the range (e.g., 9 80, 5 60) using strategies based on place value and properties of operations. Measurement and Data 3.MD Geometric measurement: understand concepts of area and relate area to multiplication and to addition. 7. Relate area to the operations of multiplication and addition. a. Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths. b. Multiply side lengths to find areas of rectangles with whole-number side lengths in the context of solving real world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning. c. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a b and a c. Use area models to represent the distributive property in mathematical reasoning. d. Recognize area as additive. Find areas of rectilinear figures by decomposing them into nonoverlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems. 19 Page 19 THIRD GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #2 Developing understanding of fraction equivalence, addition and subtraction of fractions with like denominators, and multiplication of fractions by whole numbers Students develop understanding of fraction equivalence and operations with fractions. They recognize that two different fractions can be equal (e.g., 15/9 = 5/3), and they develop methods for generating and recognizing equivalent fractions. Students extend previous understandings about how fractions are built from unit fractions, composing fractions from unit fractions, decomposing fractions into unit fractions, and using the meaning of fractions and the meaning of multiplication to multiply a fraction by a whole number. Number and Operations Fractions Develop understanding of fractions as numbers. 3.NF 1. Understand a fraction 1 /b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a /b as the quantity formed by a parts of size 1 /b. 2. Understand a fraction as a number on the number line; represent fractions on a number line diagram. a. Represent a fraction 1 / b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1 /b and that the endpoint of the part based at 0 locates the number 1 /b on the number line. b. Represent a fraction a /b on a number line diagram by marking off a lengths 1 /b from 0. Recognize that the resulting interval has size a /b and that its endpoint locates the number a /b on the number line. 3. Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size. a. Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line. b. Recognize and generate simple equivalent fractions, e.g., 1 /2 = 2 /4, 4 /6 = 2 /3. Explain why the fractions are equivalent, e.g., by using a visual fraction model. c. Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3 /1; recognize that 6 /1 = 6; locate 4 /4 and 1 at the same point of a number line diagram. d. Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. Measurement and Data Represent and interpret data. 3.MD 4. Generate measurement data by measuring lengths using rulers marked with halves and fourths of an inch. Show the data by making a line plot, where the horizontal scale is marked off in appropriate units whole numbers, halves, or quarters. 20 Page 20 THIRD GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #3 Developing understanding of the structure of rectangular arrays and of area Students recognize area as an attribute of two-dimensional regions. They measure the area of a shape by finding the total number of same-size units of area required to cover the shape without gaps or overlaps, a square with sides of unit length being the standard unit for measuring area. Students understand that rectangular arrays can be decomposed into identical rows or into identical columns. By decomposing rectangles into rectangular arrays of squares, students connect area to multiplication, and justify using multiplication to determine the area of a rectangle. Measurement and Data 3.MD Geometric measurement: understand concepts of area and relate area to multiplication and to addition. 5. Recognize area as an attribute of plane figures and understand concepts of area measurement. a. A square with side length 1 unit, called a unit square, is said to have one square unit of area, and can be used to measure area. b. A plane figure which can be covered without gaps or overlaps by n unit squares is said to have an area of n square units. 6. Measure areas by counting unit squares (square cm, square m, square in, square ft, and improvised units). 7. Relate area to the operations of multiplication and addition. a. Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths. b. Multiply side lengths to find areas of rectangles with whole-number side lengths in the context of solving real world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning. c. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a b and a c. Use area models to represent the distributive property in mathematical reasoning. d. Recognize area as additive. Find areas of rectilinear figures by decomposing them into nonoverlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems. Geometric measurement: recognize perimeter as an attribute of plane figures and distinguish between linear and area measures. 8. Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters. Geometry 3.G Reason with shapes and their attributes. 2. Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole. For example, partition a shape into 4 parts with equal area, and describe the area of each part as 1 /4 of the area of the shape. 21 Page 21 THIRD GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #4 Describing and analyzing two-dimensional shapes Students describe, analyze, and compare properties of two-dimensional shapes. They compare and classify shapes by their sides and angles, and connect these with definitions of shapes. Students also relate their fraction work to geometry by expressing the area of part of a shape as a unit fraction of the whole. Geometry 3.G Reason with shapes and their attributes. 1. Understand that shapes in different categories (e.g., rhombuses, rectangles, and others) mayshare attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories. Number and Operations Fractions 3.NF Develop understanding of fractions as numbers. 1. Understand a fraction 1 /b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a /b as the quantity formed by a parts of size 1 /b. 3. Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size. a. Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line. b. Recognize and generate simple equivalent fractions, e.g., 1 /2 = 2 /4, 4 /6 = 2 /3. Explain why the fractions are equivalent, e.g., by using a visual fraction model. c. Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3 /1; recognize that 6 /1 = 6; locate 4 /4 and 1 at the same point of a number line diagram. 22 Page 22 THIRD GRADE CRITICAL AREAS OF FOCUS STANDARDS AND CLUSTERS BEYOND THE CRITICAL AREAS OF FOCUS Solving multi-step problems Students apply previous understanding of addition and subtraction strategies and algorithms to solve multi-step problems. They reason abstractly and quantitatively by modeling problem situations with equations or graphs, assessing their processes and results, and justifying their answers through mental computation and estimation strategies. Students incorporate multiplication and division within 100 to solve multi-step problems with the four operations. Operations and Algebraic Thinking 3.OA Solve problems involving the four operations, and identify and explain patterns in arithmetic. (Previously listed in Critical Area of Focus 1 but relates to the following.) 1. Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Number and Operations in Base Ten 3.NBT Use place value understanding and properties of operations to perform multi-digit arithmetic. 1. Use place value understanding to round whole numbers to the nearest 10 or Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction. Measurement and Data 3.MD Solve problems involving measurement and estimation of intervals of tim e, liquid volumes, and masses of objects. 1. Tell and write time to the nearest minute and measure time intervals in minutes. Solve word problems involving addition and subtraction of time intervals in minutes, e.g., by representing the problem on a number line diagram. 2. Measure and estimate liquid volumes and masses of objects using standard units of grams (g), kilograms (kg), and liters (l). Add, subtract, multiply, or divide to solve one-step word problems involving masses or volumes that are given in the same units, e.g., by using drawings (such as a beaker with a measurement scale) to represent the problem. Represent and interpret data. 3. Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step how many more and how many less problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets. 23 Page 23 FOURTH GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #1 Developing an understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit dividends Students generalize their understanding of place value to 1,000,000, understanding the relative sizes of numbers in each place. They apply their understanding of models for multiplication (equal-sized groups, arrays, area models), place value, and properties of operations, in particular the distributive property, as they develop, discuss, and use efficient, accurate, and generalizable methods to compute products of multi-digit whole numbers. Depending on the numbers and the context, they select and accurately apply appropriate methods to estimate or mentally calculate products. They develop fluency with efficient procedures for multiplying whole numbers; understand and explain why the procedures work based on place value and properties of operations; and use them to solve problems. Students apply their understanding of models for division, place value, properties of operations, and the relationship of division to multiplication as they develop, discuss, and use efficient, accurate, and generalizable procedures to find quotients involving multi-digit dividends. They select and accurately apply appropriate methods to estimate and mentally calculate quotients, and interpret remainders based upon the context. Operations and Algebraic Thinking Use the four operations with whole numbers to solve problems. 4.OA 1. Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations. 2. Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison. 3. Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Gain familiarity with factors and multiples. 4. Find all factor pairs for a whole number in the range Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range is a multiple of a given one-digit number. Determine whether a given whole number in the range is prime or composite. Number and Operations in Base Ten Generalize place value understanding for multi-digit whole numbers. 4.NBT 1. Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that = 10 by applying concepts of place value and division. 2. Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. 3. Use place value understanding to round multi-digit whole numbers to any place. Use place value understanding and properties of operations to perform multi-digit arithmetic. 5. Multiply a whole number of up to four digits by a one-digit whole number, and multiply two twodigit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 24 Page 24 FOURTH GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #1, CONTINUED 6. Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. Measurement and Data 4.MD Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. 2. Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale. 3. Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor. 25 Page 25 FOURTH GRADE CRITICAL AREAS OF FOCUS CRITICAL AREA OF FOCUS #2 Developing an understanding of fraction equivalence, addition and subtraction of fractions with like denominators, and multiplication of fractions by whole numbers Students develop understanding of fraction equivalence and operations with fractions. They recognize that two different fractions can be equal (e.g., 15/9 = 5/3), and they develop methods for generating and recognizing equivalent fractions. Students extend previous understandings about how fractions are built from unit fractions, composing fractions from unit fractions, decomposing fractions into unit fractions, and using the meaning of fractions and the meaning of multiplication to multiply a fraction by a whole number. Number and Operations Fractions Extend understanding of fraction equivalence and ordering. 4.NF 1. Explain why a fraction a/b is equivalent to a fraction (n a)/(n b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. 2. Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers. 3. Understand a fraction a/b with a > 1 as a sum of fractions 1/b. a. Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. b. Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model. Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = /8 = 8/8 + 8/8 + 1/8. c. Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction. d. Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem. 4. Apply and extend previous understandings of multiplication to multiply a fraction by a whole number. a. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 (1/4), recording the conclusion by the equation 5/4 = 5 (1/4). b. Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 (2/5) as 6 (1/5), recognizing this product as 6/5. (In general, n (a/b) = (n a)/b.) c. Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie? ### CCSS-M Critical Areas: Kindergarten CCSS-M Critical Areas: Kindergarten Critical Area 1: Represent and compare whole numbers Students use numbers, including written numerals, to represent quantities and to solve quantitative problems, such ### Arizona s College and Career Ready Standards Mathematics Kindergarten Standards Placemat Arizona s College and Career Ready Standards Mathematics Kindergarten Standards Placemat 1. Representing and comparing whole numbers, initially with sets of objects Students use numbers, including written ### Sue. Rubric for the Critical Area of Mathematics Grades K - 2 Rubric for the Critical Area of Mathematics Grades K - 2 The intent of this document is to provide support as teachers transition to Common Core s. It draws attention to the most critical skills for their ### MAFS: Mathematics Standards GRADE: K MAFS: Mathematics Standards GRADE: K Domain: COUNTING AND CARDINALITY Cluster 1: Know number names and the count sequence. CODE MAFS.K.CC.1.1 Count to 100 by ones and by tens. MAFS.K.CC.1.2 MAFS.K.CC.1.3 ### COMMON CORE STATE STANDARDS FOR MATHEMATICS 3-5 DOMAIN PROGRESSIONS COMMON CORE STATE STANDARDS FOR MATHEMATICS 3-5 DOMAIN PROGRESSIONS Compiled by Dewey Gottlieb, Hawaii Department of Education June 2010 Operations and Algebraic Thinking Represent and solve problems involving ### Topic: 1 - Understanding Addition and Subtraction 8 days / September Topic: 1 - Understanding Addition and Subtraction Represent and solve problems involving addition and subtraction. 2.OA.1. Use addition and subtraction within 100 to solve one- and two-step ### Analysis of California Mathematics standards to Common Core standards- Kindergarten Analysis of California Mathematics standards to Common Core standards- Kindergarten Strand CA Math Standard Domain Common Core Standard (CCS) Alignment Comments in reference to CCS Strand Number Sense ### 1) Make Sense and Persevere in Solving Problems. Standards for Mathematical Practice in Second Grade The Common Core State Standards for Mathematical Practice are practices expected to be integrated into every mathematics lesson for all students Grades ### GLE ID (Content Area, Grade, GLE No.) GLE 1. Count to 100 by ones and by tens. M Count to 100 by 1s, 5s, 10s, and 25s COMMON CORE STATE STANDARDS (CCSS) LOUISIANA GRADE-LEVEL EXPECTATIONS (GLE) CCSS ID (Grade, Domain, Grade- Specific Stard No.) Grade Domain Cluster CCSS K.CC.1 K Counting Cardinality K.CC.2 K Counting ### Mathematical Practices The New Illinois Learning Standards for Mathematics Incorporating the Common Core Mathematical Practices Grade Strand Standard # Standard K-12 MP 1 CC.K-12.MP.1 Make sense of problems and persevere in ### A Story of Units: A Curriculum Overview for Grades P-5 New York State Common Core P-5 Mathematics Curriculum GRADE A Story of Units: A Curriculum Overview for Grades P-5 Table of Contents: Introduction... 2 Curriculum Map... 3 Pre-Kindergarten... 4 Kindergarten... 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Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 7 as a statement that 35 ### Grade 2. M4: and M:5 Addition and Subtraction of Numbers to 1000. M3: Place Value, Counting, and Comparison of Numbers to 1000 Grade 2 Key Areas of Focus for Grades K-2: Addition and subtraction-concepts, skills and problem solving Expected Fluency: Add and Subtract within 20 Add and Subtract within 100 Module M1: Mastery of Sums ### K 3 Mathematics Core Course Objectives K 3 Mathematics Core Course Objectives The Massachusetts Department of Elementary and Secondary Education (ESE) partnered with WestEd to convene panels of expert educators to review and develop statements ### Mathematics Florida Standards (MAFS) Grade 2 Mathematics Florida Standards (MAFS) Grade 2 Domain: OPERATIONS AND ALGEBRAIC THINKING Cluster 1: Represent and solve problems involving addition and subtraction. MAFS.2.OA.1.1 Use addition and subtraction ### for the Common Core State Standards 2012 A Correlation of for the Common Core State s 2012 to the Common Core Georgia Performance s Grade 2 FORMAT FOR CORRELATION TO THE COMMON CORE GEORGIA PERFORMANCE STANDARDS (CCGPS) Subject Area: K-12 Mathematics ### Math-U-See Correlation with the Common Core State Standards for Mathematical Content for Fourth Grade Math-U-See Correlation with the Common Core State Standards for Mathematical Content for Fourth Grade The fourth-grade standards highlight all four operations, explore fractions in greater detail, and ### Vocabulary, Signs, & Symbols product dividend divisor quotient fact family inverse. Assessment. 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Make sense of problems and persevere in solving them. Module 3: Lesson 4: 156-159 Module 4: Lesson 7: 220-223 2. Reason both contextually ### Excel Math Fourth Grade Standards for Mathematical Practice Excel Math Fourth Grade Standards for Mathematical Practice The Common Core State Standards for Mathematical Practice are integrated into Excel Math lessons. Below are some examples of how we include these ### COURSE OF STUDY UNIT PLANNING GUIDE COURSE OF STUDY UNIT PLANNING GUIDE FOR: MATHEMATICS GRADE LEVEL: FOURTH GRADE PREPARED BY: TERESA KELLY AND TRACY BUCKLEY B.O.E. ADOPTED AUGUST 20, 2015 REVISED AUGUST 2015 ALIGNED TO THE 2014 NJCCCS ### Students are able to represent and solve problems involving multiplication and division. 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This Grade Level Strand Standard Kindergarten Counting and Cardinality CCK.CC.4 CCK.CC.5 Number and Operations ### COMMON CORE. DECONSTRUCTED for. State Standards FIRST GRADE ARTS 1MATHEMATICS COMMON CORE State Standards DECONSTRUCTED for Classroom Impact 1MATHEMATICS FIRST GRADE ARTS 855.809.7018 www.commoncoreinstitute.com MathEMATICS OVERVIEW Introduction The Common Core Institute is pleased ### Charlesworth School Year Group Maths Targets Charlesworth School Year Group Maths Targets Year One Maths Target Sheet Key Statement KS1 Maths Targets (Expected) These skills must be secure to move beyond expected. I can compare, describe and solve ### 4 th Grade Mathematics Unpacked Content 4 th Grade Mathematics Unpacked Content This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these ### Everyday Mathematics CCSS EDITION CCSS EDITION. 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They are presented with ### Maths Targets Year 1 Addition and Subtraction Measures. N / A in year 1. Number and place value Maths Targets Year 1 Addition and Subtraction Count to and across 100, forwards and backwards beginning with 0 or 1 or from any given number. Count, read and write numbers to 100 ### The symbols indicate where the topic is first introduced or specifically addressed. ingapore Math Inc. cope and equence U: U.. Edition : ommon ore Edition : tandards Edition : ommon ore tandards 1 2 3 4 5 Reviews in ommon ore Edition cover just the unit whereas those in U.. and tandards ### ISAT Mathematics Performance Definitions Grade 4 ISAT Mathematics Performance Definitions Grade 4 EXCEEDS STANDARDS Fourth-grade students whose measured performance exceeds standards are able to identify, read, write, represent, and model whole numbers ### Just want the standards alone? You can find the standards alone at http://corestandards.org/the-standards 4 th Grade Mathematics Unpacked Content For the new Common Core State Standards that will be effective in all North Carolina schools in the 2012-13 school year. This document is designed to help North ### Composing and Decomposing Whole Numbers Grade 2 Mathematics, Quarter 1, Unit 1.1 Composing and Decomposing Whole Numbers Overview Number of instructional days: 10 (1 day = 45 60 minutes) Content to be learned Demonstrate understanding of mathematical ### South Carolina College- and Career-Ready Standards for Mathematics South Carolina College- and Career-Ready Standards for Mathematics South Carolina Department of Education Columbia, South Carolina 2015 State Board of Education Approved First Reading on February 11, 2015 ### angle Definition and illustration (if applicable): a figure formed by two rays called sides having a common endpoint called the vertex angle a figure formed by two rays called sides having a common endpoint called the vertex area the number of square units needed to cover a surface array a set of objects or numbers arranged in rows and ### NCTM Curriculum Focal Points for Grade 5. Everyday Mathematics, Grade 5 NCTM Curriculum Focal Points and, Grade 5 NCTM Curriculum Focal Points for Grade 5 Number and Operations and Algebra: Developing an understanding of and fluency with division of whole numbers Students ### Division with Whole Numbers and Decimals Grade 5 Mathematics, Quarter 2, Unit 2.1 Division with Whole Numbers and Decimals Overview Number of Instructional Days: 15 (1 day = 45 60 minutes) Content to be Learned Divide multidigit whole numbers ### Numbers and Operations in Base 10 and Numbers and Operations Fractions Numbers and Operations in Base 10 and Numbers As the chart below shows, the Numbers & Operations in Base 10 (NBT) domain of the Common Core State Standards for Mathematics (CCSSM) appears in every grade ### Curriculum Overview Standards & Skills Curriculum Overview Standards & Skills You CAN Do The Rubik's Cube Instructional Curriculum Curriculum Overview / Skills & Standards How to Solve The Rubik's Cube Part 1 - Curriculum Overview Part 2 - ### South Carolina College- and Career-Ready Standards for Mathematics South Carolina College- and Career-Ready Standards for Mathematics South Carolina Department of Education Columbia, South Carolina 2015 State Board of Education Approved First Reading on February 11, 2015 ### Just want the standards alone? You can find the standards alone at 5 th Grade Mathematics Unpacked Content For the new Common Core State Standards that will be effective in all North Carolina schools in the 2012-13 school year. This document is designed to help North ### Georgia s Early Intervention Program (EIP) Mathematics K- 5 Rubrics 2016-2017 Georgia s Early Intervention Program (EIP) Mathematics K- 5 Rubrics NOTE: The EIP eligibility criteria for student placement and exit decisions must be supported by and consistent with multiple ### Building number sense and place value and money understanding Unit: 1 Place Value, Money, and Sense s to be mastered in this unit: 3.N.1 Skip count by 25 s, 50 s, 100 s to 1,000 3.N.2 Read and write whole numbers to 1,000 place 3.N.3 Compare and order numbers to ### 5 th Grade Common Core State Standards. Flip Book 5 th Grade Common Core State Standards Flip Book This document is intended to show the connections to the Standards of Mathematical Practices for the content standards and to get detailed information at ### Math, Grades 1-3 TEKS and TAKS Alignment 111.13. Mathematics, Grade 1. 111.14. Mathematics, Grade 2. 111.15. Mathematics, Grade 3. (a) Introduction. (1) Within a well-balanced mathematics curriculum, the primary focal points are adding and subtracting ### Big Ideas in Mathematics Big Ideas in Mathematics which are important to all mathematics learning. (Adapted from the NCTM Curriculum Focal Points, 2006) The Mathematics Big Ideas are organized using the PA Mathematics Standards ### Carroll County Public Schools Elementary Mathematics Instructional Guide (5 th Grade) Unit #4 : Division of Whole Numbers and Measurement Background Information and Research By fifth grade, students should understand that division can mean equal sharing or partitioning of equal groups or arrays. They should also understand that it is the ### Indicator 2: Use a variety of algebraic concepts and methods to solve equations and inequalities. 3 rd Grade Math Learning Targets Algebra: Indicator 1: Use procedures to transform algebraic expressions. 3.A.1.1. Students are able to explain the relationship between repeated addition and multiplication. ### Problem of the Month: Once Upon a Time Problem of the Month: The Problems of the Month (POM) are used in a variety of ways to promote problem solving and to foster the first standard of mathematical practice from the Common Core State Standards: ### Grade 1 Math Expressions Vocabulary Words 2011 Italicized words Indicates OSPI Standards Vocabulary as of 10/1/09 Link to Math Expression Online Glossary for some definitions: http://wwwk6.thinkcentral.com/content/hsp/math/hspmathmx/na/gr1/se_9780547153179 ### Third Grade - Mathematics. Kentucky Core Academic Standards with Targets Student Friendly Targets Pacing Guide Third Grade - Mathematics Kentucky Core Academic Standards with Targets Student Friendly Targets Pacing Guide Page 1 of 48 College and Career Readiness Anchor Standards for Math The K-5 standards on the
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 8.2: Integration by Parts Difficulty Level: At Grade Created by: CK-12 This activity is intended to supplement Calculus, Chapter 7, Lesson 2. In this activity, you will explore: • product rule of differentiation • integration by parts ## Exercises 1. State the product rule for a function of the form u(x)v(x)\begin{align}u(x)v(x)\end{align}. 2. Apply the product rule to the function sin(x)ln(x)\begin{align}\sin(x)* \ln(x)\end{align}. 3. Do you agree or disagree with the following statement? Explain. ddx(f(x))dx=ddx(f(x)dx)=f(x)\begin{align}\int\limits \frac{d}{dx} (f(x)) dx = \frac{d}{dx} \left (\int\limits f(x)dx \right ) = f(x)\end{align} 4. What is the integral of the left side of the product rule? (ddx(u(x)v(x))dx=\begin{align}\int\limits \left (\frac{d}{dx} (u(x) \cdot v(x) \right )dx = \end{align} 5. What is the integral of the right side? (u(x)dvdx+v(x)dudx)dx=\begin{align}\int\limits \left (u (x) \cdot \frac{dv}{dx} + v(x) \cdot \frac{du}{dx} \right )dx = \end{align} 6. Explain the relationship between the areas shown on the graph and the following equation: v1v2udv=uvu1u2vdu\begin{align}\int\limits_{v_1}^{v_2} u \cdot dv = u \cdot v - \int\limits_{u_1}^{u_2} v \cdot du\end{align} 7. Use the method of integration by parts to compute the integral of ln(x)\begin{align}\ln (x)\end{align}. Remember the formula for Integration by parts is udv=uvvdu\begin{align}\int\limits u \cdot dv = u \cdot v - \int\limits v \cdot du\end{align} ln(x)1 dxu=ln(x) and dv=1 dx du=v=Result=\begin{align}& \int\limits \ln(x) \cdot 1\ dx \rightarrow u = \ln(x) \ \text{and} \ dv = 1\ dx \\ & \qquad \qquad \qquad \quad \ du = \qquad \qquad \quad v = \\ & \text{Result} = \end{align} Check by integration directly. (Home > F3:Calc > 2:Integrate) or (Home > 2nd\begin{align}2^{nd}\end{align} 7 ) Consider the function f(x)=sin(ln(x))\begin{align}f(x) = \sin(\ln(x))\end{align}. udv=sin(ln(x))du=cos(ln(x))xdx=dxv=x(+C)\begin{align}u & = \sin(\ln(x)) \rightarrow du = \frac{\cos(\ln(x))}{x} dx \\ dv & = dx \rightarrow v = x (+C)\end{align} sin(ln(x))1 dx=xsin(ln(x))xcos(ln(x))xdx(+C)=xsin(ln(x))cos(ln(x))dx(+C)\begin{align}\int\limits \sin(\ln(x)) \cdot 1\ dx & = x \cdot \sin(\ln(x)) - \int\limits x \cdot \frac{\cos(\ln(x))}{x} dx (+C) \\ & = x \cdot \sin(\ln(x)) - \int\limits \cos (\ln(x))dx (+C)\end{align} 8. Find cos(ln(x))dx\begin{align}\int\limits \cos (\ln(x))dx\end{align}. ucos(ln(x))dx==du=dv=v=\begin{align}u & = && du = && dv = && v = \\ \int\limits \cos (\ln(x))dx & =\end{align} 9. Substitute the result for cos(ln(x))\begin{align}\cos(\ln(x))\end{align} into the result for sin(ln(x))\begin{align}\sin(\ln(x))\end{align}. usin(ln(x))dx==du=dv=v=\begin{align}u & = && du = && dv = && v = \\ \int\limits \sin (\ln(x))dx & =\end{align} 10. Use integration by parts to solve the following. If you need to use integration by parts more than once, do so. Check your result. a. tan1(x) dx\begin{align}\int\limits \tan^{-1}(x)\ dx\end{align} b. x2ex dx\begin{align}\int\limits x^2 \cdot e^x \ dx\end{align} c. xtan1(x) dx\begin{align}\int\limits x \cdot \tan^{-1}(x) \ dx\end{align} d. xcos(2x+1) dx\begin{align}\int\limits x \cdot \cos (2x+1)\ dx\end{align} 11. (Extension 1) Does it matter in which order u(x)\begin{align}u(x)\end{align} and v(x)\begin{align}v(x)\end{align*} are selected for the method of integration by parts? 12. (Extension 2) Is there likely to be an integration rule based upon the quotient rule just as Integration by Parts was based upon the product rule? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
Maximum Power Transfer Theorem: The maximum power transfer theorem deals with the condition that when the maximum power transfer of the source to the load occurs. Here we will be discussing the theorem along with its statement, proof of the condition, steps, and the example problems. Contents Statement of Maximum Power Transfer Theorem The theorem states that: In a d.c. circuit, the maxim power transfer from the source to load takes place when the load resistance is equal to the internal resistance of the source as viewed from the load terminals with the load removed and all the sources replaced by their internal resistances. In other words, the maximum power transfer shall occur when the load resistance is equal to Thevenin’s resistance of the circuit. Illustration Meanwhile, let’s look at an illustration. In our illustration, a circuit supplies power to the load RL. The circuit which we are representing by using a box can be replaced by using Thevenin’s equivalent circuit consisting of a Thevenin’s voltage E (=VTH) in series with the Thevenin’s resistance RS (=RTH). As in Thevenin’s equivalent circuit, Thevenin’s resistance turns out to be the internal resistance of the source. This also illustrates that Thevenin’s theorem is an application of a practical voltage source. Finally, by the maximum power transfer theorem, the circuit will transfer the maximum power when the load resistance is equal to the internal resistance of the source. In short, the condition of maximum power transfer is load resistance (RL) = Thevenin’s resistance (RS). Condition is RL = RS. Proof Consider the circuit below. The power which the source delivers to the load is In the above relation, when RL equals zero, the power P also becomes zero. Similarly, when the RL tends to infinity the power P is again zero. Hence, we know that for the transfer of maximum power, the value of RL lies between zero and infinity. We can calculate the value of RL when the power will be maximum, by differentiating the expression of power P concerning RL and equation it with zero. Therefore, the condition RS = RL is the condition for maximum power transfer. Finally, using this condition the equation (i) to obtain the power which the source delivers to the load is Steps for Solving a Problem Using Maximum Power Transfer Theorem We can follow the following steps to solve a problem using this theorem. 1. Firstly, find Thevenin’s resistance (RTH) by removing the load and replacing all the sources with its internal resistance. Meanwhile, for the problems containing ideal sources, replace the voltage source with short-circuit and the current source with an open-circuit. 2. According to the maximum power transfer theorem, this RTH is the system load resistance, that is to say, RL =RTH that allows maximum power transfer. 3. Secondly, find Thevenin’s voltage (VTH). 4. Finally, calculate the maximum power transfer using the relation. Similarly, if you are approaching the problem using Norton’s theorem then the maximum power transfer will be Examples of Maximum Power Transfer Theorem Here, we will determine the value of resistance R such that maximum power transfer takes place in the circuit. Firstly, we will calculate Thevenin’s voltage which is the open-circuit voltage across the terminal XY. At first, apply KVL on mesh I Secondly, applying KVL on mesh II Then, solving equation (i) and (ii) we have I1=3.25 A I2 =0.25 A Now, to find the VTH we write KVL moving from Y to X. For a clear understanding, we are writing KVL on the blue portion of the circuit in a clockwise direction. Then, we have to find the RTH. From the above circuit, For the transfer of maximum power, according to the maximum power transfer theorem, RL should be equal to RTH. Further, we will draw Thevenin’s equivalent circuit as Therefore, the maximum power delivered to the load is ENGINEERING NOTES ONLINE VOLTAGE DIVIDER
### PPT ```Multiplication With Fractions Part 1 April 12, 2013 Beth Schefelker Bridget Schock Connie Laughlin Hank Kepner Kevin McLeod Sharing Coaching Experiences Professional Practice from March: Engage in a coaching conversation with your teachers around these ideas. Operating with Fractions Progression  Multiplication Situations Turn and Talk Share an “a-ha” moment from the conversation. Does Order Matter? • What does it mean to multiply a fraction by a whole number? • What does it mean to multiply a whole number by a fraction? Learning Intentions & Success Criteria We are learning to …. Understand multiplication with fractions using meaningful visual models and real-world contexts. We will be successful when we can …. Recognize, represent, and contextualize ‘groups of’ and ‘parts of’ problems involving multiplication with fractions. Ribbon and Bows Connie is making bows for her grandchildren. Each bow takes ¾ yard of ribbon. She needs 7 bows. How many yards of ribbon does she need? Make a visual representation Write an equation that represents the problem Solve the problem One Meaning of Multiplication with Fractions: ‘Groups of’ Problems 6 x ⅔ 6 groups of ⅔ Tell a story  Draw a representation of your story 4.NF.4 Apply and extend previous understandings of multiplication to multiply a fraction by a whole number. (c) Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. House For Sale! Elena and Roberto are deciding which house to buy. The first house has a lot that is 6 acres. The 2 second house’s lot is as large. What is the size 3 of the second lot? Make a visual representation Write an equation that represents the problem Solve the problem Second Meaning of Multiplication of Fractions: ‘Parts of’ Problems 3 5 x4 3 5 part of 4 Tell a story  Draw a representation of your story 5.NF.4 Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction. Does Order Matter? • What does it mean to multiply a fraction by a whole number? • What does it mean to multiply a whole number by a fraction? ‘Groups of’ vs. ‘Parts of’ • Sort the cards as a ‘Groups of’ problem or a ‘Parts of’ problem. cards. • As you worked through the problems, how did Learning Intentions & Success Criteria We are learning to …. Understand multiplication with fractions using meaningful visual models and real-world contexts. We will be successful when we can …. Represent, contextualize, and justify ‘groups of’ and ‘parts of’ problems involving multiplication with fractions. Reflection Consider these two Standards for Mathematical Practice:  MP2. Reason abstractly and quantitatively.  MP6. Attend to precision. Select one, describe how this standard was evident when multiplying fractions. Use specific
# 2007 AIME I Problems/Problem 9 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem In right triangle $ABC$ with right angle $C$, $CA = 30$ and $CB = 16$. Its legs $CA$ and $CB$ are extended beyond $A$ and $B$. Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA$, the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$, and the circles are externally tangent to each other. The length of the radius either circle can be expressed as $p/q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution ### Solution 1 Label the points as in the diagram above. If we draw $\overline{O_1A}$ and $\overline{O_2B}$, we form two right triangles. As $\overline{AF}$ and $\overline{AD}$ are both tangents to the circle, we see that $\overline{O_1A}$ is an angle bisector. Thus, $\triangle AFO_1 \cong \triangle ADO_1$. Call $x = AD = AF$ and $y = EB = BG$. We know that $x + y + 2r = 34$. If we call $\angle CAB = \theta$, then $\angle DAO_1 = \frac{180 - \theta}{2}$. Apply the tangent half-angle formula ($\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$). We see that $\frac rx = \tan \frac{180 - \theta}{2} = \sqrt{\frac{1 - \cos (180 - \theta)}{1 + \cos (180 - \theta)}}$$= \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}$. Also, $\cos \theta = \frac{30}{34} = \frac{15}{17}$. Thus, $\frac rx = \sqrt{\frac{1 + \frac{15}{17}}{1 - \frac{15}{17}}}$, and $x = \frac{r}{4}$. Similarly, we find that $y = r/\sqrt{\frac{1 + \frac{8}{17}}{1 - \frac{8}{17}}} = \frac{3r}{5}$. Therefore, $x + y + 2r = \frac{r}{4} + \frac{3r}{5} + 2r = \frac{57r}{20} = 34 \Longrightarrow r = \frac{680}{57}$, and $p + q = 737$. ### Solution 2 Use a similar solution to the aforementioned solution. Instead, call $\angle CAB = 2\theta$, and then proceed by simplifying through identities. We see that $\frac rx = \tan \left(\frac{180 - 2\theta}{2}\right) = \tan (90 - \theta)$. In terms of $r$, we find that $x = \frac{r}{\cot \theta} = \frac{r\sin \theta}{\cos \theta}$. Similarly, we find that $y = \frac{r \sin(45 - \theta)}{\cos (45 - \theta)}$. Substituting, we find that $r\left(\frac{\sin \theta}{\cos \theta} + \frac{\sin(45 - \theta)}{\cos (45 - \theta)} + 2\right) = 34$. Under a common denominator, $r\left(\frac{\sin \theta \cos (45 - \theta) + \cos \theta \sin (45 - \theta)}{\cos \theta \cos (45 - \theta)} + 2\right) = 34$. Trigonometric identities simplify this to $r\left(\frac{\sin\left((\theta) + (45 - \theta)\right)}{\frac 12 \left(\cos (\theta + 45 - \theta) + \cos (\theta - 45 + \theta) \right)} + 2\right) = 34$. From here, it is possible to simplify: $r\left(\frac{2 \sin 45}{\cos 45 + \cos 2\theta \cos 45 + \sin 2\theta \sin 45} +2\right) = 34$ $r\left(\frac{2}{\frac{17}{17} + \frac{8}{17} + \frac{15}{17}} + 2\right) = 34$ $r\left(\frac{57}{20}\right) = 34$ Our answer is $34 \cdot \frac{20}{57} = \frac{680}{57}$, and $p + q = 737$. ### Solution 3 Let the point where CB's extension hits the circle be G, and the point where the hypotenuse hits that circle be E. Clearly $EB=GB$. Let $EB=x$. Draw the two perpendicular radii to G and E. Now we have a cyclic quadrilateral. Let the radius be length $r$. We see that since the cosine of angle ABC is $\frac{15}{17}$ the cosine of angle EBG is $-\frac{15}{17}$. Since the measure of the angle opposite to EBG is the complement of this one, its cosine is $\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}$ This tells us that $r=4x$. Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle D. Draw the radii to F and D and we have cyclic quadrilaterals once more. Using the law of cosines again, we find that the length of our tangents is $2.4x$. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, $8x+2.4x+x=34$. Solving we find that $4x=\frac{680}{57}$ so our answer is 737. ### Solution 4 By Pythagoras, $AB = 34$. Let $I_{C}$ be the $C$-excenter of triangle $ABC$. Then the $C$-exradius $r_{C}$ is given by $r_{C}= \frac{K}{s-c}= \frac{240}{40-34}= 40$. The circle with center $O_{1}$ is tangent to both $AB$ and $AC$, which means that $O_{1}$ lies on the external angle bisector of $\angle BAC$. Therefore, $O_{1}$ lies on $AI_{C}$. Similarly, $O_{2}$ lies on $BI_{C}$. Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$. The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$, so $O_{1}O_{2}$ is parallel to $AB$, which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar. The distance from $I_{C}$ to $AB$ is $r_{C}= 40$, so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$. Therefore, $\frac{40-r}{40}= \frac{O_{1}O_{2}}{AB}= \frac{2r}{34}\quad \Rightarrow \quad r = \frac{680}{57}$. Hence, the final answer is $680+57 = 737$. ### Solution 5 Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\frac{1}{2}\times 16s\times 30s=\frac{1}{2}\times 34s\times h=\frac{1}{2}\times rp,$ where $p$ is the perimeter. Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles. The linear dimensions of the new triangle are $\frac{46s}{34s}=\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$: $\frac{240s}{17}\times\frac{23}{17} = \frac{240}{17}+12s$ $20s\times 23 = 20\times 17+s\times 17\times 17$ $s = \frac{340}{171}$ $r = 6s = \frac{680}{57}$ The answer is $737$. ### Solution 6 Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. ### Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\overline{AB}$ are on the line with slope $-\frac{15}{8}$, and y-intercept $30+ \frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \left(r,\frac14 r+30\right)$ and $O_2 = \left(\frac35 r+16,r\right)$ By the distance formula and the fact that the circles and tangent, we have: $\left(16-\frac25 r\right)^2 + \left(30-\frac34 r\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\frac{-23120 \pm 54400}{2622}$ The solution including the "$-$" is extraneous so we have the radius equal to $\frac{31280}{2622}$ Which simplifies to $\frac{680}{57}$. The sum of the numerator and the denominator is $\boxed{737}$ ### Solution 8 (simple algebra) It is known that $O_1O_2$ is parallel to AB. Thus, extending $O_1F$ and $GO_2$ to intersect at H yields similar triangles $O_1O_2H$ and BAC, so that $O_1O_2 = 2r$, $O_1H = \frac{16r}{17}$, and $HO_2 = \frac{30r}{17}$. It should be noted that $O_2G = r$. Also, FHGC is a rectangle, and so AF = $\frac{47r}{17} - 30$ and similarly for BG. Because tangents to a circle are equal, the hypotenuse can be expressed in terms of r: $$2r + \frac{47r}{17} - 30 + \frac{33r}{17} - 16 = 34$$ Thus, r = $\frac{680}{57}$, and the answer is $\boxed{737}.$ ### Solution 9 (Simple Algebra) Let the radius of the circle be r. It can be seen that $\Delta FHO_{1}$ and $\Delta O_{2}GJ$ are similar to $\Delta ACB$, and the length of the hypotenuses are $\frac{17}{8}r$ and $\frac {17}{15}r$, respectively. Then, the entire length of $HJ$ is going to be $(\frac{17}{8}+\frac{17}{15}+2)r = \frac{631}{120}r$. The length of the hypotenuse of $\Delta ACB$ is 34, so the length of the height to $AB$ is $\frac{16*30}{34} = \frac{240}{17}$. Thus, the height to $\Delta HCJ$ is going to be $\frac{240}{17} + r$. $\Delta HCJ$ is similar to $\Delta ACB$ so we have the following: $\frac{\frac{631}{120}r}{34} = \frac{\frac{240}{17} + r}{\frac{240}{17}}$. Cross multiplying and simplifying, we get that r = $\frac{680}{57}$ so the answer is $\boxed{737}$. ~Leonard_my_dude~
# Setting Up Hypotheses Site: Saylor Academy Course: MA121: Introduction to Statistics Book: Setting Up Hypotheses Printed by: Guest user Date: Wednesday, May 29, 2024, 11:44 AM ## Description This section discusses the logic behind hypothesis testing using concrete examples and explains how to set up null and alternative hypothesis. It explains what Type I and II errors are and how they can occur. Finally, it introduces one-tailed and two-tailed tests and explains which one you should use for testing purposes. ## Introduction #### Learning Objectives 1. Describe the logic by which it can be concluded that someone can distinguish between two things 2. State whether random assignment ensures that all uncontrolled sources of variation will be equal 3. Define precisely what the probability is that is computed to reach the conclusion that a difference is not due to chance 4. Distinguish between the probability of an event and the probability of a state of the world 5. Define "null hypothesis" 6. Be able to determine the null hypothesis from a description of an experiment 7. Define "alternative hypothesis" The statistician R. Fisher explained the concept of hypothesis testing with a story of a lady tasting tea. Here we will present an example based on James Bond who insisted that martinis should be shaken rather than stirred. Let's consider a hypothetical experiment to determine whether Mr. Bond can tell the difference between a shaken and a stirred martini. Suppose we gave Mr. Bond a series of 16 taste tests. In each test, we flipped a fair coin to determine whether to stir or shake the martini. Then we presented the martini to Mr. Bond and asked him to decide whether it was shaken or stirred. Let's say Mr. Bond was correct on 13 of the 16 taste tests. Does this prove that Mr. Bond has at least some ability to tell whether the martini was shaken or stirred? This result does not prove that he does; it could be he was just lucky and guessed right 13 out of 16 times. But how plausible is the explanation that he was just lucky? To assess its plausibility, we determine the probability that someone who was just guessing would be correct 13/16 times or more. This probability can be computed from the binomial distribution, and the binomial distribution calculator shows it to be 0.0106. This is a pretty low probability, and therefore someone would have to be very lucky to be correct 13 or more times out of 16 if they were just guessing. So either Mr. Bond was very lucky, or he can tell whether the drink was shaken or stirred. The hypothesis that he was guessing is not proven false, but considerable doubt is cast on it. Therefore, there is strong evidence that Mr. Bond can tell whether a drink was shaken or stirred. Let's consider another example. The case study Physicians' Reactions sought to determine whether physicians spend less time with obese patients. Physicians were sampled randomly and each was shown a chart of a patient complaining of a migraine headache. They were then asked to estimate how long they would spend with the patient. The charts were identical except that for half the charts, the patient was obese and for the other half, the patient was of average weight. The chart a particular physician viewed was determined randomly. Thirty-three physicians viewed charts of average-weight patients and 38 physicians viewed charts of obese patients. The mean time physicians reported that they would spend with obese patients was 24.7 minutes as compared to a mean of 31.4 minutes for average-weight patients. How might this difference between means have occurred? One possibility is that physicians were influenced by the weight of the patients. On the other hand, perhaps by chance, the physicians who viewed charts of the obese patients tend to see patients for less time than the other physicians. Random assignment of charts does not ensure that the groups will be equal in all respects other than the chart they viewed. In fact, it is certain the two groups differed in many ways by chance. The two groups could not have exactly the same mean age (if measured precisely enough such as in days). Perhaps a physician's age affects how long physicians see patients. There are innumerable differences between the groups that could affect how long they view patients. With this in mind, is it plausible that these chance differences are responsible for the difference in times? To assess the plausibility of the hypothesis that the difference in mean times is due to chance, we compute the probability of getting a difference as large or larger than the observed difference $\mathrm{(31.4 \, - \, 24.7 \, = \, 6.7 \, \mathrm{minutes})}$ if the difference were, in fact, due solely to chance. Using methods presented in another section, this probability can be computed to be 0.0057. Since this is such a low probability, we have confidence that the difference in times is due to the patient's weight and is not due to chance. Source: David M. Lane, https://onlinestatbook.com/2/logic_of_hypothesis_testing/intro.html This work is in the Public Domain. ### The Probability Value It is very important to understand precisely what the probability values mean. In the James Bond example, the computed probability of 0.0106 is the probability he would be correct on 13 or more taste tests (out of 16) if he were just guessing. It is easy to mistake this probability of 0.0106 as the probability he cannot tell the difference. This is not at all what it means. The probability of 0.0106 is the probability of a certain outcome (13 or more out of 16) assuming a certain state of the world (James Bond was only guessing). It is not the probability that a state of the world is true. Although this might seem like a distinction without a difference, consider the following example. An animal trainer claims that a trained bird can determine whether or not numbers are evenly divisible by 7. In an experiment assessing this claim, the bird is given a series of 16 test trials. On each trial, a number is displayed on a screen and the bird pecks at one of two keys to indicate its choice. The numbers are chosen in such a way that the probability of any number being evenly divisible by 7 is 0.50. The bird is correct on 9/16 choices. Using the binomial calculator, we can compute that the probability of being correct nine or more times out of 16 if one is only guessing is 0.40. Since a bird who is only guessing would do this well 40% of the time, these data do not provide convincing evidence that the bird can tell the difference between the two types of numbers. As a scientist, you would be very skeptical that the bird had this ability. Would you conclude that there is a 0.40 probability that the bird can tell the difference? Certainly not! You would think the probability is much lower than 0.0001. To reiterate, the probability value is the probability of an outcome (9/16 or better) and not the probability of a particular state of the world (the bird was only guessing). In statistics, it is conventional to refer to possible states of the world as hypotheses since they are hypothesized states of the world. Using this terminology, the probability value is the probability of an outcome given the hypothesis. It is not the probability of the hypothesis given the outcome. This is not to say that we ignore the probability of the hypothesis. If the probability of the outcome given the hypothesis is sufficiently low, we have evidence that the hypothesis is false. However, we do not compute the probability that the hypothesis is false. In the James Bond example, the hypothesis is that he cannot tell the difference between shaken and stirred martinis. The probability value is low (0.0106), thus providing evidence that he can tell the difference. However, we have not computed the probability that he can tell the difference. A branch of statistics called Bayesian statistics provides methods for computing the probabilities of hypotheses. These computations require that one specify the probability of the hypothesis before the data are considered and, therefore, are difficult to apply in some contexts. ### The Null Hypothesis The hypothesis that an apparent effect is due to chance is called the null hypothesis. In the Physicians' Reactions example, the null hypothesis is that in the population of physicians, the mean time expected to be spent with obese patients is equal to the mean time expected to be spent with average-weight patients. This null hypothesis can be written as: $\mu_{\text {obese }}=\mu_{\text {average }}$ or as $\mu_{\text {obese }}=\mu_{\text {average }}=0$ The null hypothesis in a correlational study of the relationship between high school grades and college grades would typically be that the population correlation is 0. This can be written as $\rho=0$ where $\rho$ is the population correlation (not to be confused with $r$, the correlation in the sample). Although the null hypothesis is usually that the value of a parameter is $0$, there are occasions in which the null hypothesis is a value other than $0$. For example, if one were testing whether a subject differed from chance in their ability to determine whether a flipped coin would come up heads or tails, the null hypothesis would be that $\pi=0.5$. Keep in mind that the null hypothesis is typically the opposite of the researcher's hypothesis. In the Physicians' Reactions study, the researchers hypothesized that physicians would expect to spend less time with obese patients. The null hypothesis that the two types of patients are treated identically is put forward with the hope that it can be discredited and therefore rejected. If the null hypothesis were true, a difference as large or larger than the sample difference of 6.7 minutes would be very unlikely to occur. Therefore, the researchers rejected the null hypothesis of no difference and concluded that in the population, physicians intend to spend less time with obese patients. If the null hypothesis is rejected, then the alternative to the null hypothesis (called the alternative hypothesis) is accepted. The alternative hypothesis is simply the reverse of the null hypothesis. If the null hypothesis $\mu_{\text {obese }}=\mu_{\text {average }}$ is rejected, then there are two alternatives: $\mu_{\text {obese }} < \mu_{\text {average }}$ $\mu_{\text {obese }} > \mu_{\text {average }}$ Naturally, the direction of the sample means determines which alternative is adopted. Some textbooks have incorrectly argued that rejecting the null hypothesis that two population means are equal does not justify a conclusion about which population mean is larger. Kaiser showed how it is justified to draw a conclusion about the direction of the difference. ### Questions Question 1 out of 3. Tommy claims that he blindly guessed on a 20-question true/false test, but then he got 16 of the questions correct. Using the binomial calculator, you find out that the probability of getting 16 or more correct out of 20 when $\pi =.5$ is $0.0059$. This probability of $0.0059$ is the probability that... • he would get $80\%$ correct if he took the test again. • he would get this score or better if he were just guessing. • he was guessing blindly on the test. Question 2 out of 3. Random assignment ensures groups will be equal on everything except the variable manipulated. • True • False Question 3 out of 3. The researchers hypothesized that there would be a correlation between how much people studied and their GPAs. The null hypothesis is that the population correlation is equal to __________ 1. He would get this score or better if he were just guessing. If Tommy were guessing blindly, the probability that he would have gotten 16 out of the 20 questions right is.0059. This is NOT the probability that he was guessing blindly. Remember, the probability value is the probability of an outcome given the hypothesis. It is not the probability of the hypothesis given the outcome. 2. False Chance differences will still exist. 3. The null hypothesis says that any apparent effect is due to chance, so in this case, the null hypothesis would be that the population correlation was 0. ## Type I and Type II Errors #### Learning Objectives 1. Define Type I and Type II errors 2. Interpret significant and non-significant differences 3. Explain why the null hypothesis should not be accepted when the effect is not significant In the Physicians' Reactions case study, the probability value associated with the significance test is 0.0057. Therefore, the null hypothesis was rejected, and it was concluded that physicians intend to spend less time with obese patients. Despite the low probability value, it is possible that the null hypothesis of no true difference between obese and average-weight patients is true and that the large difference between sample means occurred by chance. If this is the case, then the conclusion that physicians intend to spend less time with obese patients is in error. This type of error is called a Type I error. More generally, a Type I error occurs when a significance test results in the rejection of a true null hypothesis. By one common convention, if the probability value is below 0.05, then the null hypothesis is rejected. Another convention, although slightly less common, is to reject the null hypothesis if the probability value is below 0.01. The threshold for rejecting the null hypothesis is called the $\alpha$ (alpha) level or simply $\alpha$. It is also called the significance level. As discussed in the section on significance testing, it is better to interpret the probability value as an indication of the weight of evidence against the null hypothesis than as part of a decision rule for making a reject or do-not-reject decision. Therefore, keep in mind that rejecting the null hypothesis is not an all-or-nothing decision. The Type I error rate is affected by the α level: the lower the α level, the lower the Type I error rate. It might seem that α is the probability of a Type I error. However, this is not correct. Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error. The second type of error that can be made in significance testing is failing to reject a false null hypothesis. This kind of error is called a Type II error. Unlike a Type I error, a Type II error is not really an error. When a statistical test is not significant, it means that the data do not provide strong evidence that the null hypothesis is false. Lack of significance does not support the conclusion that the null hypothesis is true. Therefore, a researcher should not make the mistake of incorrectly concluding that the null hypothesis is true when a statistical test was not significant. Instead, the researcher should consider the test inconclusive. Contrast this with a Type I error in which the researcher erroneously concludes that the null hypothesis is false when, in fact, it is true. A Type II error can only occur if the null hypothesis is false. If the null hypothesis is false, then the probability of a Type II error is called $\beta$ (beta). The probability of correctly rejecting a false null hypothesis equals $1- \beta$ and is called power. Power is covered in detail in another section. ### Questions Question 1 out of 5. It has been shown many times that on a certain memory test, recognition is substantially better than recall. However, the probability value for the data from your sample was.12, so you were unable to reject the null hypothesis that recall and recognition produce the same results. What type of error did you make? • Type I • Type II Question 2 out of 5. In the population, there is no difference between men and women on a certain test. However, you found a difference in your sample. The probability value for the data was.03, so you rejected the null hypothesis. What type of error did you make? • Type I • Type II Question 3 out of 5. As the alpha level gets lower, which error rate also gets lower? • Type I • Type II Question 4 out of 5. Beta is the probability of which kind of error? • Type I • Type II Question 5 out of 5. If the null hypothesis is false, you cannot make which kind of error? • Type I • Type II 1. In this example, there is really a difference in the population between recognition and recall, but you did not find a significant difference in your sample. Failing to reject a false null hypothesis is a Type II error. 2. There is no difference in the population, but you found a difference in your sample. A Type I error occurs when a significance test results in the rejection of a true null hypothesis. 3. The Type I error rate is affected by the alpha level; the lower the alpha level is, the lower the Type I error rate gets. Alpha is the probability of a Type I error given that the null hypothesis is true. 4. The probability of a Type II error is called beta. The probability of correctly rejecting a false null hypothesis equals 1- beta and is called power. 5. A Type I error occurs when a significance test results in the rejection of a TRUE null hypothesis. ## One- and Two-Tailed Tests #### Learning Objectives 1. Define Type I and Type II errors 2. Interpret significant and non-significant differences 3. Explain why the null hypothesis should not be accepted when the effect is not significant In the James Bond case study, Mr. Bond was given 16 trials on which he judged whether a martini had been shaken or stirred. He was correct on 13 of the trials. From the binomial distribution, we know that the probability of being correct 13 or more times out of 16 if one is only guessing is 0.0106. Figure 1 shows a graph of the binomial distribution. The red bars show the values greater than or equal to 13. As you can see in the figure, the probabilities are calculated for the upper tail of the distribution. A probability calculated in only one tail of the distribution is called a "one-tailed probability". Figure 1. The binomial distribution. The upper (right-hand) tail is red. A slightly different question can be asked of the data: "What is the probability of getting a result as extreme or more extreme than the one observed?" Since the chance expectation is 8/16, a result of 3/16 is equally as extreme as 13/16. Thus, to calculate this probability, we would consider both tails of the distribution. Since the binomial distribution is symmetric when $п=0.5$, this probability is exactly double the probability of 0.0106 computed previously. Therefore, $p=0.0212 .$ A probability calculated in both tails of a distribution is called a "two-tailed probability" (see Figure 2). Figure 2. The binomial distribution. Both tails are red. Should the one-tailed or the two-tailed probability be used to assess Mr. Bond's performance? That depends on the way the question is posed. If we are asking whether Mr. Bond can tell the difference between shaken or stirred martinis, then we would conclude he could if he performed either much better than chance or much worse than chance. If he performed much worse than chance, we would conclude that he can tell the difference, but he does not know which is which. Therefore, since we are going to reject the null hypothesis if Mr. Bond does either very well or very poorly, we will use a two-tailed probability. On the other hand, if our question is whether Mr. Bond is better than chance at determining whether a martini is shaken or stirred, we would use a one-tailed probability. What would the one-tailed probability be if Mr. Bond were correct on only 3 of the 16 trials? Since the one-tailed probability is the probability of the right-hand tail, it would be the probability of getting 3 or more correct out of 16. This is a very high probability and the null hypothesis would not be rejected. The null hypothesis for the two-tailed test is $\pi = 0.5$. By contrast, the null hypothesis for the one-tailed test is $п \leq 0.5$. Accordingly, we reject the two-tailed hypothesis if the sample proportion deviates greatly from $0.5$ in either direction. The one-tailed hypothesis is rejected only if the sample proportion is much greater than $0.5$. The alternative hypothesis in the two-tailed test is $n \neq 0.5$. In the one-tailed test it is $\pi > 0.5$. You should always decide whether you are going to use a one-tailed or a two-tailed probability before looking at the data. Statistical tests that compute one-tailed probabilities are called one-tailed tests; those that compute two-tailed probabilities are called two-tailed tests. Two-tailed tests are much more common than one-tailed tests in scientific research because an outcome signifying that something other than chance is operating is usually worth noting. One-tailed tests are appropriate when it is not important to distinguish between no effect and an effect in the unexpected direction. For example, consider an experiment designed to test the efficacy of a treatment for the common cold. The researcher would only be interested in whether the treatment was better than a placebo control. It would not be worth distinguishing between the case in which the treatment was worse than a placebo and the case in which it was the same because in both cases the drug would be worthless. Some have argued that a one-tailed test is justified whenever the researcher predicts the direction of an effect. The problem with this argument is that if the effect comes out strongly in the non-predicted direction, the researcher is not justified in concluding that the effect is not zero. Since this is unrealistic, one-tailed tests are usually viewed skeptically if justified on this basis alone. ### Questions Question 1 out of 4. Select all that apply. Which is/are true of two-tailed tests? • They are appropriate when it is not important to distinguish between no effect and an effect in either direction. • They are more common than one-tailed tests. • They compute two-tailed probabilities. • They are more controversial than one-tailed tests. Question 2 out of 4. You are testing the difference between college freshmen and seniors on a math test. You think that the seniors will perform better, but you are still interested in knowing if the freshmen perform better. What is the null hypothesis? • The mean of the seniors is less than or equal to the mean of the freshmen • The mean of the seniors is greater than or equal to the mean of the freshmen • The mean of the seniors is equal to the mean of the freshmen Question 3 out of 4. You think a coin is biased and will come up heads more often than it will come up tails. What is the probability that out of 22 flips, it will come up heads 16 or more times? (Write your answer out to at least three decimal places). ____________ Question 4 out of 4. You think a coin is biased, and you are interested in finding out if it is. What is the probability that out of 30 flips, it will come up one side 8 or fewer times? (Write your answer out to at least three decimal places). ____________ 3. This question is asking you to compute a one-tailed probability. Using the binomial calculator with the values of $\mathrm{N}=22, \mathrm{p}=.5$, and greater than or equal to 16, you get $p=.0262$.
Home > Pre-Algebra > Fractions > Comparing Fractions ## Comparing Fractions #### Introduction Sometimes if fractions have a different denominator (the number below the line), it can be difficult to work out which is larger. For example, which of the following three fractions is largest: $rac{2}{3}$ or $rac{4}{5}$ or $rac{6}{7}$? To work out the correct order, you'll need to do some calculations to make them easier to compare. ## Lesson If you are comparing two fractions with the same denominator, you can just look at which numerator is larger. For example, if you are comparing the following two fractions: $rac{2}{5}$ and $rac{3}{5}$ Both have the same denominator (5) and so you can see that $rac{3}{5}$ is larger than $rac{2}{5}$. If you have two numbers with different denominators, you'll need to make them the same in order to compare. In order to make the denominators the same, you need to find the lowest common denominator of the two numbers. If we want to compare $rac{3}{5}$ and $rac{4}{6}$, we need to find the lowest common denominator of 5 and 6. To do this, we can write out the multiples of 5 and 6. Multiples of 5Multiples of 6 56 1012 1518 2024 2530 3036 3542 30 is the lowest common denominator of 5 and 6. Therefore, we need to convert both $rac{3}{5}$ and $rac{4}{6}$ to have 30 as a denominator. To convert $rac{3}{5}$ to have a denominator of 30, we need to multiply 5 by 6. However, we need to multiply both the numerator and the denominator by the same value or we will change the fraction's value. Therefore, we can multiply it by $rac{6}{6}$. $rac{3}{5}$ x $rac{6}{6}$ = $rac{18}{30}$ Then we repeat the process with $rac{4}{6}$. To make the denominator 30, we need to multiply by $rac{5}{5}$. $rac{4}{6}$ x $rac{5}{5}$ = $rac{20}{30}$ Now we have both numbers with a common denominator, we can easily compare them. $rac{20}{30}$ is larger than $rac{18}{30}$ Therefore $rac{4}{6}$ is larger than $rac{3}{5}$. ## Examples $\dfrac{5}{7}$ ? $\dfrac{4}{7}$ $\dfrac{5}{7}$ ? $\dfrac{4}{7}$ $\dfrac{5}{7} > \dfrac{4}{7}$ $\dfrac{5}{6}$ ? $\dfrac{3}{5}$ $\dfrac{55}{65}$ ? $\dfrac{36}{56}$ $\dfrac{25}{30}$ ? $\dfrac{18}{30}$ $\dfrac{25}{30} > \dfrac{18}{30}$ $1\dfrac{5}{7}$ ? $2\dfrac{1}{7}$ $\dfrac{12}{7}$ ? $\dfrac{15}{7}$ $\dfrac{12}{7} < \dfrac{15}{7}$
# Euler method In mathematics and computational science, the Euler method (also called the forward Euler method) is a first-order numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. It is the most basic explicit method for numerical integration of ordinary differential equations and is the simplest Runge–Kutta method. The Euler method is named after Leonhard Euler, who first proposed it in his book Institutionum calculi integralis (published 1768–1770).[1] The Euler method is a first-order method, which means that the local error (error per step) is proportional to the square of the step size, and the global error (error at a given time) is proportional to the step size. The Euler method often serves as the basis to construct more complex methods, e.g., predictor–corrector method. ## Geometrical description ### Purpose and why it works Consider the problem of calculating the shape of an unknown curve which starts at a given point and satisfies a given differential equation. Here, a differential equation can be thought of as a formula by which the slope of the tangent line to the curve can be computed at any point on the curve, once the position of that point has been calculated. The idea is that while the curve is initially unknown, its starting point, which we denote by ${\displaystyle A_{0},}$ is known (see Figure 1). Then, from the differential equation, the slope to the curve at ${\displaystyle A_{0}}$ can be computed, and so, the tangent line. Take a small step along that tangent line up to a point ${\displaystyle A_{1}.}$ Along this small step, the slope does not change too much, so ${\displaystyle A_{1}}$ will be close to the curve. If we pretend that ${\displaystyle A_{1}}$ is still on the curve, the same reasoning as for the point ${\displaystyle A_{0}}$ above can be used. After several steps, a polygonal curve (${\displaystyle A_{0}A_{1}A_{2}A_{3}\dots }$ ) is computed. In general, this curve does not diverge too far from the original unknown curve, and the error between the two curves can be made small if the step size is small enough and the interval of computation is finite.[2] ### First-order process When given the values for ${\displaystyle t_{0}}$ and ${\displaystyle y(t_{0})}$ , and the derivative of ${\displaystyle y}$ is a given function of ${\displaystyle t}$ and ${\displaystyle y}$ denoted as ${\displaystyle y'(t)=f{\bigl (}t,y(t){\bigr )}}$ . Begin the process by setting ${\displaystyle y_{0}=y(t_{0})}$ . Next, choose a value ${\displaystyle h}$ for the size of every step along t-axis, and set ${\displaystyle t_{n}=t_{0}+nh}$ (or equivalently ${\displaystyle t_{n+1}=t_{n}+h}$ ). Now, the Euler method is used to find ${\displaystyle y_{n+1}}$ from ${\displaystyle y_{n}}$ and ${\displaystyle t_{n}}$ :[3] ${\displaystyle y_{n+1}=y_{n}+hf(t_{n},y_{n}).}$ The value of ${\displaystyle y_{n}}$ is an approximation of the solution at time ${\displaystyle t_{n}}$ , i.e., ${\displaystyle y_{n}\approx y(t_{n})}$ . The Euler method is explicit, i.e. the solution ${\displaystyle y_{n+1}}$ is an explicit function of ${\displaystyle y_{i}}$ for ${\displaystyle i\leq n}$ . ### Higher-order process While the Euler method integrates a first-order ODE, any ODE of order ${\displaystyle N}$ can be represented as a system of first-order ODEs. When given the ODE of order ${\displaystyle N}$ defined as ${\displaystyle y^{(N+1)}(t)=f\left(t,y(t),y'(t),\ldots ,y^{(N)}(t)\right),}$ as well as ${\displaystyle h}$ , ${\displaystyle t_{0}}$ , and ${\displaystyle y_{0},y'_{0},\dots ,y_{0}^{(N)}}$ , we implement the following formula until we reach the approximation of the solution to the ODE at the desired time: ${\displaystyle {\vec {y}}_{i+1}={\begin{pmatrix}y_{i+1}\\y'_{i+1}\\\vdots \\y_{i+1}^{(N-1)}\\y_{i+1}^{(N)}\end{pmatrix}}={\begin{pmatrix}y_{i}+h\cdot y'_{i}\\y'_{i}+h\cdot y''_{i}\\\vdots \\y_{i}^{(N-1)}+h\cdot y_{i}^{(N)}\\y_{i}^{(N)}+h\cdot f\left(t_{i},y_{i},y'_{i},\ldots ,y_{i}^{(N)}\right)\end{pmatrix}}}$ These first-order systems can be handled by Euler's method or, in fact, by any other scheme for first-order systems.[4] ## First-order example Given the initial value problem ${\displaystyle y'=y,\quad y(0)=1,}$ we would like to use the Euler method to approximate ${\displaystyle y(4)}$ .[5] ### Using step size equal to 1 (h = 1) The Euler method is ${\displaystyle y_{n+1}=y_{n}+hf(t_{n},y_{n}).}$ so first we must compute ${\displaystyle f(t_{0},y_{0})}$ . In this simple differential equation, the function ${\displaystyle f}$ is defined by ${\displaystyle f(t,y)=y}$ . We have ${\displaystyle f(t_{0},y_{0})=f(0,1)=1.}$ By doing the above step, we have found the slope of the line that is tangent to the solution curve at the point ${\displaystyle (0,1)}$ . Recall that the slope is defined as the change in ${\displaystyle y}$ divided by the change in ${\displaystyle t}$ , or ${\textstyle {\frac {\Delta y}{\Delta t}}}$ . The next step is to multiply the above value by the step size ${\displaystyle h}$ , which we take equal to one here: ${\displaystyle h\cdot f(y_{0})=1\cdot 1=1.}$ Since the step size is the change in ${\displaystyle t}$ , when we multiply the step size and the slope of the tangent, we get a change in ${\displaystyle y}$ value. This value is then added to the initial ${\displaystyle y}$ value to obtain the next value to be used for computations. ${\displaystyle y_{0}+hf(y_{0})=y_{1}=1+1\cdot 1=2.}$ The above steps should be repeated to find ${\displaystyle y_{2}}$ , ${\displaystyle y_{3}}$ and ${\displaystyle y_{4}}$ . {\displaystyle {\begin{aligned}y_{2}&=y_{1}+hf(y_{1})=2+1\cdot 2=4,\\y_{3}&=y_{2}+hf(y_{2})=4+1\cdot 4=8,\\y_{4}&=y_{3}+hf(y_{3})=8+1\cdot 8=16.\end{aligned}}} Due to the repetitive nature of this algorithm, it can be helpful to organize computations in a chart form, as seen below, to avoid making errors. ${\displaystyle n}$ ${\displaystyle y_{n}}$ ${\displaystyle t_{n}}$ ${\displaystyle f(t_{n},y_{n})}$ ${\displaystyle h}$ ${\displaystyle \Delta y}$ ${\displaystyle y_{n+1}}$ 0 1 0 1 1 1 2 1 2 1 2 1 2 4 2 4 2 4 1 4 8 3 8 3 8 1 8 16 The conclusion of this computation is that ${\displaystyle y_{4}=16}$ . The exact solution of the differential equation is ${\displaystyle y(t)=e^{t}}$ , so ${\displaystyle y(4)=e^{4}\approx 54.598}$ . Although the approximation of the Euler method was not very precise in this specific case, particularly due to a large value step size ${\displaystyle h}$ , its behaviour is qualitatively correct as the figure shows. ### Using other step sizes As suggested in the introduction, the Euler method is more accurate if the step size ${\displaystyle h}$ is smaller. The table below shows the result with different step sizes. The top row corresponds to the example in the previous section, and the second row is illustrated in the figure. step size result of Euler's method error 1 16.00 38.60 0.25 35.53 19.07 0.1 45.26 09.34 0.05 49.56 05.04 0.025 51.98 02.62 0.0125 53.26 01.34 The error recorded in the last column of the table is the difference between the exact solution at ${\displaystyle t=4}$ and the Euler approximation. In the bottom of the table, the step size is half the step size in the previous row, and the error is also approximately half the error in the previous row. This suggests that the error is roughly proportional to the step size, at least for fairly small values of the step size. This is true in general, also for other equations; see the section Global truncation error for more details. Other methods, such as the midpoint method also illustrated in the figures, behave more favourably: the global error of the midpoint method is roughly proportional to the square of the step size. For this reason, the Euler method is said to be a first-order method, while the midpoint method is second order. We can extrapolate from the above table that the step size needed to get an answer that is correct to three decimal places is approximately 0.00001, meaning that we need 400,000 steps. This large number of steps entails a high computational cost. For this reason, higher-order methods are employed such as Runge–Kutta methods or linear multistep methods, especially if a high accuracy is desired.[6] ## Higher-order example For this third-order example, assume that the following information is given: {\displaystyle {\begin{aligned}&y'''+4ty''-t^{2}y'-(\cos {t})y=\sin {t}\\&t_{0}=0\\&y_{0}=y(t_{0})=2\\&y'_{0}=y'(t_{0})=-1\\&y''_{0}=y''(t_{0})=3\\&h=0.5\end{aligned}}} From this we can isolate y''' to get the equation: ${\displaystyle f\left(t,y,y',y''\right)=y'''=\sin {t}+(\cos {t})y+t^{2}y'-4ty''}$ Using that we can get the solution for ${\displaystyle {\vec {y}}_{1}}$ : ${\displaystyle {\vec {y}}_{1}={\begin{pmatrix}y_{1}\\y_{1}'\\y_{1}''\end{pmatrix}}={\begin{pmatrix}y_{0}+h\cdot y'_{0}\\y'_{0}+h\cdot y''_{0}\\y''_{0}+h\cdot f\left(t_{0},y_{0},y'_{0},y''_{0}\right)\end{pmatrix}}={\begin{pmatrix}2+0.5\cdot -1\\-1+0.5\cdot 3\\3+0.5\cdot \left(\sin {0}+(\cos {0})\cdot 2+0^{2}\cdot (-1)-4\cdot 0\cdot 3\right)\end{pmatrix}}={\begin{pmatrix}1.5\\0.5\\4\end{pmatrix}}}$ And using the solution for ${\displaystyle {\vec {y}}_{1}}$ , we can get the solution for ${\displaystyle {\vec {y}}_{2}}$ : ${\displaystyle {\vec {y}}_{2}={\begin{pmatrix}y_{2}\\y_{2}'\\y_{2}''\end{pmatrix}}={\begin{pmatrix}y_{1}+h\cdot y'_{1}\\y'_{1}+h\cdot y''_{1}\\y''_{1}+h\cdot f\left(t_{1},y_{1},y'_{1},y''_{1}\right)\end{pmatrix}}={\begin{pmatrix}1.5+0.5\cdot 0.5\\0.5+0.5\cdot 4\\2+0.5\cdot \left(\sin {0.5}+(\cos {0.5})\cdot 1.5+0.5^{2}\cdot 0.5-4\cdot 0.5\cdot 2\right)\end{pmatrix}}={\begin{pmatrix}1.75\\2.5\\0.9604\end{pmatrix}}}$ We can continue this process using the same formula as long as necessary to find whichever ${\displaystyle {\vec {y}}_{i}}$ desired. ## Derivation The Euler method can be derived in a number of ways. (1) Firstly, there is the geometrical description above. (2) Another possibility is to consider the Taylor expansion of the function ${\displaystyle y}$ around ${\displaystyle t_{0}}$ : ${\displaystyle y(t_{0}+h)=y(t_{0})+hy'(t_{0})+{\tfrac {1}{2}}h^{2}y''(t_{0})+O\left(h^{3}\right).}$ The differential equation states that ${\displaystyle y'=f(t,y)}$ . If this is substituted in the Taylor expansion and the quadratic and higher-order terms are ignored, the Euler method arises.[7] The Taylor expansion is used below to analyze the error committed by the Euler method, and it can be extended to produce Runge–Kutta methods. (3) A closely related derivation is to substitute the forward finite difference formula for the derivative, ${\displaystyle y'(t_{0})\approx {\frac {y(t_{0}+h)-y(t_{0})}{h}}}$ in the differential equation ${\displaystyle y'=f(t,y)}$ . Again, this yields the Euler method.[8] A similar computation leads to the midpoint method and the backward Euler method. (4) Finally, one can integrate the differential equation from ${\displaystyle t_{0}}$ to ${\displaystyle t_{0}+h}$ and apply the fundamental theorem of calculus to get: ${\displaystyle y(t_{0}+h)-y(t_{0})=\int _{t_{0}}^{t_{0}+h}f{\bigl (}t,y(t){\bigr )}\,\mathrm {d} t.}$ Now approximate the integral by the left-hand rectangle method (with only one rectangle): ${\displaystyle \int _{t_{0}}^{t_{0}+h}f{\bigl (}t,y(t){\bigr )}\,\mathrm {d} t\approx hf{\bigl (}t_{0},y(t_{0}){\bigr )}.}$ Combining both equations, one finds again the Euler method.[9] This line of thought can be continued to arrive at various linear multistep methods. ## Local truncation error The local truncation error of the Euler method is the error made in a single step. It is the difference between the numerical solution after one step, ${\displaystyle y_{1}}$ , and the exact solution at time ${\displaystyle t_{1}=t_{0}+h}$ . The numerical solution is given by ${\displaystyle y_{1}=y_{0}+hf(t_{0},y_{0}).}$ For the exact solution, we use the Taylor expansion mentioned in the section Derivation above: ${\displaystyle y(t_{0}+h)=y(t_{0})+hy'(t_{0})+{\tfrac {1}{2}}h^{2}y''(t_{0})+O\left(h^{3}\right).}$ The local truncation error (LTE) introduced by the Euler method is given by the difference between these equations: ${\displaystyle \mathrm {LTE} =y(t_{0}+h)-y_{1}={\tfrac {1}{2}}h^{2}y''(t_{0})+O\left(h^{3}\right).}$ This result is valid if ${\displaystyle y}$ has a bounded third derivative.[10] This shows that for small ${\displaystyle h}$ , the local truncation error is approximately proportional to ${\displaystyle h^{2}}$ . This makes the Euler method less accurate than higher-order techniques such as Runge-Kutta methods and linear multistep methods, for which the local truncation error is proportional to a higher power of the step size. A slightly different formulation for the local truncation error can be obtained by using the Lagrange form for the remainder term in Taylor's theorem. If ${\displaystyle y}$ has a continuous second derivative, then there exists a ${\displaystyle \xi \in [t_{0},t_{0}+h]}$ such that ${\displaystyle \mathrm {LTE} =y(t_{0}+h)-y_{1}={\tfrac {1}{2}}h^{2}y''(\xi ).}$ [11] In the above expressions for the error, the second derivative of the unknown exact solution ${\displaystyle y}$ can be replaced by an expression involving the right-hand side of the differential equation. Indeed, it follows from the equation ${\displaystyle y'=f(t,y)}$ that[12] ${\displaystyle y''(t_{0})={\frac {\partial f}{\partial t}}{\bigl (}t_{0},y(t_{0}){\bigr )}+{\frac {\partial f}{\partial y}}{\bigl (}t_{0},y(t_{0}){\bigr )}\,f{\bigl (}t_{0},y(t_{0}){\bigr )}.}$ ## Global truncation error The global truncation error is the error at a fixed time ${\displaystyle t_{i}}$ , after however many steps the method needs to take to reach that time from the initial time. The global truncation error is the cumulative effect of the local truncation errors committed in each step.[13] The number of steps is easily determined to be ${\textstyle {\frac {t_{i}-t_{0}}{h}}}$ , which is proportional to ${\textstyle {\frac {1}{h}}}$ , and the error committed in each step is proportional to ${\displaystyle h^{2}}$ (see the previous section). Thus, it is to be expected that the global truncation error will be proportional to ${\displaystyle h}$ .[14] This intuitive reasoning can be made precise. If the solution ${\displaystyle y}$ has a bounded second derivative and ${\displaystyle f}$ is Lipschitz continuous in its second argument, then the global truncation error (denoted as ${\displaystyle \|y(t_{i})-y_{i}\|}$ ) is bounded by ${\displaystyle \|y(t_{i})-y_{i}\|\leq {\frac {hM}{2L}}\left(e^{L(t_{i}-t_{0})}-1\right)}$ where ${\displaystyle M}$ is an upper bound on the second derivative of ${\displaystyle y}$ on the given interval and ${\displaystyle L}$ is the Lipschitz constant of ${\displaystyle f}$ .[15] Or more simply, when ${\displaystyle y'(t)=f(t,y)}$ , the value ${\textstyle L={\text{max}}{\bigl (}\|{\frac {d}{dy}}{\bigl [}f(t,y){\bigr ]}\|{\bigr )}}$ (such that ${\displaystyle t}$ is treated as a constant). In contrast, ${\textstyle M={\text{max}}{\bigl (}\|{\frac {d^{2}}{dt^{2}}}{\bigl [}y(t){\bigr ]}\|{\bigr )}}$ where function ${\displaystyle y(t)}$ is the exact solution which only contains the ${\displaystyle t}$ variable. The precise form of this bound is of little practical importance, as in most cases the bound vastly overestimates the actual error committed by the Euler method.[16] What is important is that it shows that the global truncation error is (approximately) proportional to ${\displaystyle h}$ . For this reason, the Euler method is said to be first order.[17] ### Example If we have the differential equation ${\displaystyle y'=1+(t-y)^{2}}$ , and the exact solution ${\displaystyle y=t+{\frac {1}{t-1}}}$ , and we what to find ${\displaystyle M}$ and ${\displaystyle L}$ for when ${\displaystyle 2\leq t\leq 3}$ . ${\displaystyle L={\text{max}}{\bigl (}\|{\frac {d}{dy}}{\bigl [}f(t,y){\bigr ]}\|{\bigr )}=\max _{2\leq t\leq 3}{\bigl (}\|{\frac {d}{dy}}{\bigl [}1+(t-y)^{2}{\bigr ]}\|{\bigr )}=\max _{2\leq t\leq 3}{\bigl (}\|2(t-y)\|{\bigr )}=\max _{2\leq t\leq 3}{\bigl (}\|2(t-[t+{\frac {1}{t-1}}])\|{\bigr )}=\max _{2\leq t\leq 3}{\bigl (}\|-{\frac {2}{t-1}}\|{\bigr )}=2}$ ${\displaystyle M={\text{max}}{\bigl (}\|{\frac {d^{2}}{dt^{2}}}{\bigl [}y(t){\bigr ]}\|{\bigr )}=\max _{2\leq t\leq 3}\left(\|{\frac {d^{2}}{dt^{2}}}{\bigl [}t+{\frac {1}{1-t}}{\bigr ]}\|\right)=\max _{2\leq t\leq 3}\left(\|{\frac {2}{(-t+1)^{3}}}\|\right)=2}$ Thus we can find the error bound at t=2.5 and h=0.5: ${\displaystyle {\text{error bound}}={\frac {hM}{2L}}\left(e^{L(t_{i}-t_{0})}-1\right)={\frac {0.5\cdot 2}{2\cdot 2}}\left(e^{2(2.5-2)}-1\right)=0.42957}$ Notice that t0 is equal to 2 because it is the lower bound for t in ${\displaystyle 2\leq t\leq 3}$ . ## Numerical stability The Euler method can also be numerically unstable, especially for stiff equations, meaning that the numerical solution grows very large for equations where the exact solution does not. This can be illustrated using the linear equation ${\displaystyle y'=-2.3y,\qquad y(0)=1.}$ The exact solution is ${\displaystyle y(t)=e^{-2.3t}}$ , which decays to zero as ${\displaystyle t\to \infty }$ . However, if the Euler method is applied to this equation with step size ${\displaystyle h=1}$ , then the numerical solution is qualitatively wrong: It oscillates and grows (see the figure). This is what it means to be unstable. If a smaller step size is used, for instance ${\displaystyle h=0.7}$ , then the numerical solution does decay to zero. If the Euler method is applied to the linear equation ${\displaystyle y'=ky}$ , then the numerical solution is unstable if the product ${\displaystyle hk}$ is outside the region ${\displaystyle {\bigl \{}z\in \mathbf {C} \,{\big \|}\,\|z+1\|\leq 1{\bigr \}},}$ illustrated on the right. This region is called the (linear) stability region.[18] In the example, ${\displaystyle k=-2.3}$ , so if ${\displaystyle h=1}$ then ${\displaystyle hk=-2.3}$ which is outside the stability region, and thus the numerical solution is unstable. This limitation — along with its slow convergence of error with ${\displaystyle h}$ — means that the Euler method is not often used, except as a simple example of numerical integration[citation needed]. Frequently models of physical systems contain terms representing fast-decaying elements (i.e. with large negative exponential arguments). Even when these are not of interest in the overall solution, the instability they can induce means that an exceptionally small tilmestep would be required if the Euler method is used. ## Rounding errors In step ${\displaystyle n}$ of the Euler method, the rounding error is roughly of the magnitude ${\displaystyle \varepsilon y_{n}}$ where ${\displaystyle \varepsilon }$ is the machine epsilon. Assuming that the rounding errors are independent random variables, the expected total rounding error is proportional to ${\textstyle {\frac {\varepsilon }{\sqrt {h}}}}$ .[19] Thus, for extremely small values of the step size the truncation error will be small but the effect of rounding error may be big. Most of the effect of rounding error can be easily avoided if compensated summation is used in the formula for the Euler method.[20] ## Modifications and extensions A simple modification of the Euler method which eliminates the stability problems noted above is the backward Euler method: ${\displaystyle y_{n+1}=y_{n}+hf(t_{n+1},y_{n+1}).}$ This differs from the (standard, or forward) Euler method in that the function ${\displaystyle f}$ is evaluated at the end point of the step, instead of the starting point. The backward Euler method is an implicit method, meaning that the formula for the backward Euler method has ${\displaystyle y_{n+1}}$ on both sides, so when applying the backward Euler method we have to solve an equation. This makes the implementation more costly. Other modifications of the Euler method that help with stability yield the exponential Euler method or the semi-implicit Euler method. More complicated methods can achieve a higher order (and more accuracy). One possibility is to use more function evaluations. This is illustrated by the midpoint method which is already mentioned in this article: ${\displaystyle y_{n+1}=y_{n}+hf\left(t_{n}+{\tfrac {1}{2}}h,y_{n}+{\tfrac {1}{2}}hf(t_{n},y_{n})\right)}$ . This leads to the family of Runge–Kutta methods. The other possibility is to use more past values, as illustrated by the two-step Adams–Bashforth method: ${\displaystyle y_{n+1}=y_{n}+{\tfrac {3}{2}}hf(t_{n},y_{n})-{\tfrac {1}{2}}hf(t_{n-1},y_{n-1}).}$ This leads to the family of linear multistep methods. There are other modifications which uses techniques from compressive sensing to minimize memory usage[21] ## In popular culture In the film Hidden Figures, Katherine Goble resorts to the Euler method in calculating the re-entry of astronaut John Glenn from Earth orbit.[22] ## Notes 1. ^ Butcher 2003, p. 45; Hairer, Nørsett & Wanner 1993, p. 35 2. ^ Atkinson 1989, p. 342; Butcher 2003, p. 60 3. ^ Butcher 2003, p. 45; Hairer, Nørsett & Wanner 1993, p. 36 4. ^ Butcher 2003, p. 3; Hairer, Nørsett & Wanner 1993, p. 2 6. ^ 7. ^ Atkinson 1989, p. 342; Hairer, Nørsett & Wanner 1993, p. 36 8. ^ Atkinson 1989, p. 342 9. ^ Atkinson 1989, p. 343 10. ^ Butcher 2003, p. 60 11. ^ Atkinson 1989, p. 342 12. ^ Stoer & Bulirsch 2002, p. 474 13. ^ Atkinson 1989, p. 344 14. ^ Butcher 2003, p. 49 15. ^ Atkinson 1989, p. 346; Lakoba 2012, equation (1.16) 16. ^ Iserles 1996, p. 7 17. ^ Butcher 2003, p. 63 18. ^ Butcher 2003, p. 70; Iserles 1996, p. 57 19. ^ Butcher 2003, pp. 74–75 20. ^ Butcher 2003, pp. 75–78 21. ^ Unni, M. P.; Chandra, M. G.; Kumar, A. A. (March 2017). "Memory reduction for numerical solution of differential equations using compressive sensing". 2017 IEEE 13th International Colloquium on Signal Processing & its Applications (CSPA). pp. 79–84. doi:10.1109/CSPA.2017.8064928. ISBN 978-1-5090-1184-1. S2CID 13082456. 22. ^ Khan, Amina (9 January 2017). "Meet the 'Hidden Figures' mathematician who helped send Americans into space". Los Angeles Times. Retrieved 12 February 2017.
Basic Money Math Practice Test for 2nd Grade – [Hard] Welcome to Brighterly’s 2nd Grade Money Math Practice Test! Money is an essential concept in our lives. Teaching kids about its value and how to count it is crucial. In this article, we will be taking a deep dive into 2nd grade-level money math problems. This is a fantastic resource to prepare for tests or simply enhance one’s understanding of the subject. Understanding Coins and Bills One of the very first steps in teaching children about money is familiarizing them with the different coins and bills. • Coins: • Pennies: Worth 1 cent. • Nickels: Worth 5 cents. • Dimes: Worth 10 cents. • Quarters: Worth 25 cents. • Bills: • \$1, \$5, \$10, and \$20 are the most common bills children will encounter early on. Understanding the value of each coin and bill helps children begin to make basic transactions and determine the total value of a collection of money. Counting Money The key to mastering money math is practice, practice, and more practice. Let’s start with a hypothetical scenario: Imagine walking into a toy store with a handful of coins. How do you know if you have enough money to buy that toy you’ve been eyeing? To solve this, one must start by counting the highest value coins or bills first, followed by the next highest, and so on. For instance, start with the quarters, then dimes, nickels, and finally the pennies. Making Change Making change is a skill that’s vital, not just in math tests but in real-life shopping scenarios. Consider this: If a toy costs \$1.50 and you hand over a \$5 bill, how much change should you receive? Here’s a step-by-step guide: 1. Start by determining the difference between the cost of the item and the amount given. In this case, it’s \$5.00 – \$1.50 = \$3.50. 2. Now, determine the number of bills and coins that make up that amount, always starting with the highest value. Money Word Problems Word problems offer a fun and challenging way to apply money math skills. They combine comprehension skills with arithmetic. For example: If Jenny bought 2 ice creams for 50 cents each and gave the vendor a \$5 bill, how much change did she get back? Breaking down word problems step by step helps in enhancing both reading comprehension and mathematical abilities. In conclusion, mastering money math at a 2nd-grade level sets the foundation for more advanced financial lessons in the future. With continuous practice and real-life application, kids can build a solid understanding that will serve them for years to come. Remember, every coin counts, just like every math problem solved brings one closer to mastery. Happy learning with Brighterly! Basic Money Practice Test for 2nd Grade Get ready for math lessons with Brighterly! Tailored for the young mathematicians who are ready to stretch their capabilities, this 'hard' level test dives deep into complex scenarios and word problems. 1 / 15 Sarah bought a toy for \$2.25 and gave the cashier a \$5 bill. How much change should she get back? 2 / 15 If a candy bar costs 45 cents and Jake buys 4, how much does he spend in total? 3 / 15 Lila has 3 quarters, 5 dimes, and 2 nickels. How much money does she have in total? 4 / 15 Henry bought a toy car for \$3.50 and a toy doll for \$2.75. If he pays with a \$10 bill, how much change will he receive? 5 / 15 A toy costs \$1.50. How many quarters will you need to buy the toy? 6 / 15 If you buy 3 toys, each priced at 75 cents, and give the cashier two \$1 bills, how much change will you get? 7 / 15 Timmy has 10 dimes and 8 nickels. How much money does he have? 8 / 15 Emma bought 5 ice creams, each at \$1.10. How much did she spend in total? 9 / 15 Mike wants to buy 2 toys. One toy costs \$2.30 and the other costs \$3.40. If he pays with a \$10 bill, how much change will he get back? 10 / 15 Martha has 12 quarters. How much money does she have? 11 / 15 If a pencil costs 15 cents and an eraser 25 cents, how much will you spend for 3 pencils and 2 erasers? 12 / 15 A book costs \$4.45 and a pen is 55 cents. If you buy 2 books and 1 pen, how much do you spend? 13 / 15 If you have 8 dimes, 6 nickels, and 10 pennies, how much money do you have in total? 14 / 15 A toy robot costs \$3.60. If you have 3 one-dollar bills, 2 quarters, 4 dimes, and 5 pennies, do you have enough to buy the toy? 15 / 15 Liam bought 3 candies at 40 cents each. How much did he spend? 0% Poor Level Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence. Mediocre Level Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence. Needs Improvement Start practicing math regularly to avoid your child`s math scores dropping to C or even D. High Potential It's important to continue building math proficiency to make sure your child outperforms peers at school.
## The Fibonacci Sequence ### By Annie Sun This exploration looks at how spreadsheets can help us do quick and multiple iterations of calculations, in particular, I used Microsoft Excel. The Fibonacci Sequence is a series of numbers where the next number is made up of the sum of the previous two numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,… We can use Excel to quickly add up the numbers for us and look at relationships between the numbers. First we need a formula to use: f(0) = 1, f(1) = 1, f(n) = f(n-1) + f(n-2) Inputting into Excel gave me this: This means that the 25th term in the Fibonacci Sequence is 75,025. Now, let’s examine the ratio of each adjacent term of the sequence: As n increases, the ratio gets closer and closer to the Golden Ratio φ In other words, using larger values from the Fibonacci Sequence gives a closer approximation of the Golden Ratio. Here is a look at the ratio of every 2nd, 3rd , and 4th terms: Looking at the pattern formed by the ratio it looks like the Fibonacci Sequence appears again, diagonally! There also seems to be a pattern forming with the ratios, as the 4th column looks like it gets closer to the 3rd column plus 1, or (using the 27th term) 1.6180333989 + 1 = 2.618033989. The 5th column looks like twice the 3rd column plus 1 as well, or (still using the 27th term) 2(1.618033989) + 1 = 4.236067978. The 6th column must have a similar pattern and with some number manipulations, I found 3(1.618033989) + 2 = 6.854101966. The Fibonacci Sequence again! That would mean that the ratio formed by every 5th term can be found by: 5(1.618033989) + 3 = 11.09016994. Here is the Excel spreadsheet with the Fibonacci Sequence and ratios, if needed. What if we started with other numbers than 0 and 1 Here, I started with 2 and 2, and the ratios are almost identical to the ratios for the Fibonacci Sequence. This sequence is called the Lucas Sequence.
In mathematics, we calculate the volume and surface area of a sphere. We need to learn the formulas, and we can understand how to calculate the volume and surface area of a sphere by using the formulas. The formula that produces the volume and surface area of a sphere is a bit complicated. Also, to explain why the formula is true, we have to learn high school math integration. The content is therefore difficult, and we will not explain in detail why the formula is correct. Instead, we will explain how to use the formula to calculate the volume and surface area of a sphere. As a math calculation problem, the sphere is not a very common one. Nevertheless, it is a field in which we have to remember the formulas. So, try to understand how to calculate it. ## How to Get the Volume of the Sphere Using Formula In mathematics, we may be asked to calculate the volume of a sphere. Balls are familiar to us and many people use them in sports. For example, tennis, soccer, basketball, and volleyball use a ball. The following three-dimension objects are called spheres in mathematics. To calculate the volume of these spheres, we use the formula. If you do not remember the formula, you will not be able to find the volume of the spheres. Here is the formula for finding the volume of a sphere. If we substitute a number into this formula, we can get the volume of the sphere. ### The Relationship Between the Cylinder and the Sphere Volume The sphere has a unique property. If the sphere is snugly inside a cylinder, the volume of the sphere is the $\displaystyle\frac{2}{3}$ of a cylinder. By using this property, we can get the volume of the sphere. Let the radius of the sphere be r. In this case, the height of the cylinder is 2r. The volume of the cylinder can be calculated by multiplying base area and height. Therefore, the volume of the sphere is as follows. The volume of a sphere can only be calculated if we remember the formula. However, if we understand that multiplying the cylinder volume by 2/3 to get the sphere volume, we can derive the formula. ### The Volumes of Cones, Spheres, and Cylinders are Interrelated By the way, cones, spheres and cylinders are related to each other. The cone volume is one-third of the cylinder. The sphere volume, on the other hand, is 2/3 of the cylinder. In other words, the volume ratio of each is as follows. When doing mathematical calculations, the formula for a cone (or a pyramid) must be learned. Questions that ask for the volume of a pyramid or cone are frequently given. On the other hand, the formula to find out the volume of a sphere is not as important as that of a cone. Even if we forget it, we can still make the formula for the sphere if we remember the following relationship. • Cone : Sphere : Cylinder = 1 : 2 : 3 Multiply the cylinder by 1/3 to get the volume of a cone. In this case, double the volume of the cone is the sphere. Also, if we multiply the volume of the cylinder by 2/3, we get the volume of the sphere. ## How to Find the Surface Area of a Sphere in the Formula Just as there is a formula to produce the volume of a sphere, there is also a formula to produce the surface area of a sphere. As with the formula for volume, we must remember the formula to calculate the surface area of the sphere. The formula for producing the surface area of the sphere is as follows. If we want to calculate the surface area of the sphere, let’s substitute a number into this formula. ### The Relationship Between Side Area of a Cylinder and Surface Area of a Sphere But it is hard to remember the formula for the surface area of a sphere. Like the sphere volume, it is used less often. How can we understand this formula? The surface area of a sphere also has a unique characteristic. It has the property that if the sphere is snugly contained within a cylinder, the surface area of the sphere is equal to the side area of the cylinder. If the radius is r, then the horizontal length of the side area of the cylinder is $2πr$. The diameter of the circle is 2r, and we can multiply it by pi to get the circumference (the horizontal length of the rectangle). The height of the cylinder, on the other hand, is 2r. If we forget the formula for the surface area of the sphere, we can still derive the formula if we know that it is equal to the side area of the cylinder. ## Exercise: Volume and Surface Area of the Sphere Q1. Calculate the volume and surface area of the following hemisphere. Let the pi be $π$. -Volume of the hemisphere Since it is a hemisphere, its volume is half of a sphere. If we apply the formula to produce the volume of the sphere, it is $24π$ cm3 as follows. • $\displaystyle\frac{3}{4}π×4^3×\displaystyle\frac{1}{2}=24π$ -Surface area of the hemisphere The surface area is half of a sphere. Substituting the numbers into the formula for calculating surface area, we can calculate the following • $4π×4^2×\displaystyle\frac{1}{2}=32π$ We must also add the area of the cross section of the hemisphere. Applying the formula for the area of the circle, we can calculate the following. • $4×4×π=16π$ The total area is $48π$ cm2 as follows. • $32π+16π=48π$ ## Solving the Sphere Problem Using Formulas One of the areas of study in mathematics is the volume and surface area of a sphere. Although they are less frequent as math problems, we need to remember the formulas to solve the problems. Since we don’t use the sphere formula many times in math calculation problems, it is easy to forget the formula. So, try to understand how to come up with the sphere formula. Even if we forget, we can create the sphere formula from the formula that produces the volume or the side area of a cylinder. • Sphere volume: 2/3 of the cylinder volume • Surface area of the sphere: equal to the side area of the cylinder If you remember this property, you can make a formula for the sphere at any time. Then use the formula to get the volume and surface area of the sphere.
#### A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and $L = 200 (10 - t)^2$. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds? Given: L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and $L = 200 (10 - t)^2$ To find: the rate of water running out of the pool at the last 5s. and also the average rate of water flowing at during the first 5s. Explanation: Take the rate of Water running out given by $-\frac{dL}{dt}$ Given $L = 200 (10 - t)^2$ Differentiation of the above-given equation results in, $\\ \Rightarrow -\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}} = -\frac{\mathrm{d}\left(200(10 - \mathrm{t})^{2}\right)}{\mathrm{dt}}\\ Removing all the constant terms, we get \\ \Rightarrow -\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}} = -200 \cdot \frac{\mathrm{d}\left((10 - \mathrm{t})^{2}\right)}{\mathrm{dt}}\\ Using the power rule of differentiation, we get \\ \Rightarrow -\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}} = -200 \cdot 2(10 - \mathrm{t}) \cdot \frac{\mathrm{d}(10 - \mathrm{t})}{\mathrm{dt}}\\ \Rightarrow -\frac{\mathrm{dL}}{\mathrm{dt}} = -400(10 - \mathrm{t}) \cdot (-1)\\ \Rightarrow -\frac{\mathrm{dL}}{\mathrm{dt}} = 400(10 - \mathrm{t}) \ldots (i)$ Now, to find the speed of water running out at the end of 5s, we need to find the value of equation (i) with t=5 $\\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=(400(10-t))_{t=5} \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=400(10-5) \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=400(5) \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=2000 \frac{L}{s} \ldots . .(i i)$ Therefore, 2000 L/s is the rate of water running out at the end of 5s To calculate the initial rate we need to take t=o in equation (i) $\\ \Rightarrow\left(-\frac{\mathrm{dL}}{\mathrm{dt}}\right)_{\mathrm{t}=0}=(400(10-\mathrm{t})) \mathrm{t}=0 \\ \Rightarrow\left(-\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}\right)_{\mathrm{t}=0}=400(10-0) \\ \Rightarrow\left(-\frac{\mathrm{d} L}{\mathrm{dt}}\right)_{\mathrm{t}=0}=4000 \mathrm{~L} / \mathrm{s}$ Equation (ii) tells about the final rate of water flowing whereas equation (iii) is the initial rate. Thus, the average rate during 5s is $\\ = \frac{\text{initial rate} + \text{final rate}}{2}\\ Substituting \ the \ corresponding \ values, \ we \ get\\ = \frac{4000 + 2000}{2} = 3000 \, \mathrm{L} / \mathrm{s}$ After calculation, the final rate of water flowing out of the pool in 5s is 3000L/s.
# Find whether 55 is a term of the AP: 7, 10, 13,--- or not. If yes, find which term it is. - Mathematics Sum Find whether 55 is a term of the AP: 7, 10, 13,--- or not. If yes, find which term it is. #### Solution Yes. Let the first term, common difference and the number of terms of an AP are a, d and n respectively. Let the nth term of an AP be 55. i.e., Tn = 55 We know that, the nth term of an AP, Tn = a + (n – 1 )d ........(i) Given that, first term (a) = 7 and common difference (d) = 10 – 7 = 3 From equation (i), 55 = 7 + (n – 1) × 3 ⇒ 55 = 7 + 3M – 3 = 4 + 3n ⇒ 3n = 51 ∴ n = 17 Since, n is a positive integer. So 55 is a term of the AP. Since, n = 17 Therefore, 17th term of an AP is 55. Concept: Arithmetic Progression Is there an error in this question or solution? #### APPEARS IN NCERT Mathematics Exemplar Class 10 Chapter 5 Arithematic Progressions Exercise 5.3 \| Q 10 \| Page 52 Share
# How do you integrate int 2/sqrt(x^2-4)dx using trigonometric substitution? Jun 28, 2016 Use x = 2sec(u) as a first substitution. #### Explanation: With $x = 2 \cdot \sec \left(u\right) , \therefore \mathrm{dx} = 2 \cdot \sec \left(u\right) \cdot \tan \left(u\right)$ $\sqrt{{x}^{2} - 4} = \sqrt{4 \cdot {\sec}^{2} \left(u\right) - 4} = 2 \cdot \sqrt{{\sec}^{2} \left(u\right) - 1}$ From ${\sin}^{2} \left(a\right) + {\cos}^{2} \left(a\right) = 1$, divide through by ${\cos}^{2} \left(a\right)$ to obtain ${\tan}^{2} \left(a\right) + 1 = {\sec}^{2} \left(a\right)$ So, we have: int 2/(sqrt(x^2-4) dx = int (4sec(u)tan(u))/(2sqrt(tan^2(u)) $\mathrm{du}$ So, integral of $2 \cdot \int \sec \left(u\right) \mathrm{du}$ Multiply numerator and denominator by $\sec \left(u\right) + \tan \left(u\right)$ to obtain $2 \cdot \int \frac{{\sec}^{2} \left(u\right) + \sec \left(u\right) \cdot \tan \left(u\right)}{\sec \left(u\right) + \tan \left(u\right)} \mathrm{du}$ Use substitution $r = \sec \left(u\right) + \tan \left(u\right)$ which gives $2 \cdot \int \frac{\mathrm{dr}}{r}$ $= 2 \cdot \ln \left(r\right) + C$ $= 2 \cdot \ln \left(\sec \left(u\right) + \tan \left(u\right)\right) + C$ = 2ln(sec(arcsec(x/2) + tan(arcsec(x/2)) + C $= 2 \cdot \ln \left(\frac{1}{2} \cdot \left(x + \sqrt{{x}^{2} - 4}\right)\right) + C$ And there we have it, let me know if you need any help with the last bit but it's fairly easy to figure out if you look at a few triangles.
## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1 Other Exercises Question 1. Find the cubes of the following numbers: (i) 7 (ii) 12 (iii) 16 (iv) 21 (v) 40 (vi) 55 (vii) 100 (viii) 302 (ix) 301 Solution: (i) (7)3 = 7 x 7 x 7 = 343 (ii) (12)3 = 12 x-12 x 12 = 1728 (iii) (16)3 = 16 x 16 x 16 = 4096 (iv) (21)3 = 21 x 21 x 21 = 441 x 21 =9261 (v) (40)3 = 40 x 40 x 40 = 64000 (vi) (55)3 = 55 x 55 x 55 = 3025 x 55 = 166375 (vii) (100)3 = 100 x 100 x 100 =1000000 (viii)(302)3 = 302 x 302 x 302 = 91204 x 302 = 27543608 (ix) (301)3 = 301 x 301 x 301 = 90601 x 301 =27270901 Question 2. Write the cubes of all natural numbers between 1 and 10 and verify the following statements : (i) Cubes of all odd natural numbers are odd. (ii) Cubes of all even natural numbers are even. Solution: Cubes of first 10 natural numbers : (1)3 = 1 x 1 x 1 = 1 (2)3 = 2 x 2 x 2 = 8 (3)3 = 3 x 3 x 3 = 27 (4)3= 4 x 4 x 4 = 64 (5)3 = 5 x 5 x 5 = 125 (6)3 = 6 x 6 x 6 = 216 (7)3 = 7 x 7 x 7 = 343 (8)3 = 8 x 8 x 8 = 512 (9)3 = 9 x 9 x 9= 729 (10)3 = 10 x 10 x 10= 1000 We see that the cubes of odd numbers is also odd and cubes of even numbers is also even. Question 3. Observe the following pattern : Write the next three rows and calculate the value of 13 + 23 + 33 +…. + 93 + 103 by the above pattern. Solution: We see the pattern Question 4. Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings : The cube of a natural number which is a multiple of 3 is a multiple of 27′ Solution: 5 natural numbers which are multiples of 3 3,6,9,12,15. (3)3 = 3 x 3 x 3 = 27 Which is multiple of 27 (6)3 = 6 x 6 x 6 = 216 ÷ 27 = 8 Which is multiple of 27 (9)3 = 9 x 9 x 9 = 729 + 27 = 27 Which is multiple of 27 (12)3= 12 x 12 x 12 = 1728 ÷ 27 = 64 Which is multiple of 27 (15)3 = 15 x 15 x 15 = 3375 ÷ 27 = 125 Which is multiple of 27 Hence, cube of multiple of 3 is a multiple of 27 Question 5. Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g.,4, 7, 10, …………) and verify the following : ‘The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’. Solution: 3n + 1 Let n = 1, 2, 3, 4, 5, then If n = 1, then 3n +1= 3 x 1+1= 3+1= 4 If n = 2, then 3n +1=3 x 2+1=6+1=7 If n = 3, then 3n + 1= 3 x 3 + 1= 9 + 1 = 10 If n = 4, then 3n + 1= 3 x 4+1 = 12 + 1= 13 If n = 5, then 3n +1=3 x 5 + 1 = 15 +1 = 16 Now (4)3 = 4 x 4 x 4 = 64 Which is $$\frac {64 }{ 3 }$$=21, Remainder = 1 (7)3 = 7 x 7 x 7 = 343 Which is $$\frac {343 }{ 3 }$$ =114, Remainder = 1 (10)3 = 10 x 10 x 10 = 1000 ÷ 3 = 333, Remainder = 1 (13)3 = 13 x 13 x 13 = 2197 ÷ 3 = 732, Remainder = 1 (16)3 = 16 x 16 x 16 = 4096 ÷ 3 = 1365, Remainder = 1 Hence cube of natural number of the form, 3n + 1, is a natural of the form 3n + 1 Question 6. Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,……… ) and verify the following : ‘The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’. Solution: Natural numbers of the form 3n + 2, when n is a natural number i.e. 1, 2, 3, 4, 5,…………. If n = 1, then 3n + 2 = 3 x 1+2 = 3+ 2 = 5 If n = 2, then 3n + 2 = 3 x 2 + 2 = 6 + 2 = 8 If n = 3, then 3n + 2 = 3 x 3 + 2 = 9 + 2 = 11 If n = 4, then 3n + 2 = 3 x 4 + 2 = 12 + 2 = 14 and if n = 5, then 3n + 2 = 3 x 5 + 2 = 15 + 2= 17 Now (5)3 = 5 x 5 x 5 = 125 125 + 3 = 41, Remainder = 2 (8)2 = 8 x 8 x 8 = 512 512 -s- 3 = 170, Remainder = 2 (11)3 = 11 x 11 x 11 = 1331 1331 + 3 = 443, Remainder = 2 (14)3 = 14 x 14 x 14 = 2744 2744 + 3 = 914, Remainder = 2 (17)3 = 17 x 17 x 17 = 4913 4913 = 3 = 1637, Remainder = 2 We see the cube of the natural number of the form 3n + 2 is also a natural number of the form 3n + 2. Question 7. Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following : ‘The cube of a multiple of 7 is a multiple of 73′. Solution: 5 natural numbers which are multiple of 7,are 7, 14, 21, 28, 35 (7)3 = (7)3 which is multiple of 73 (14)3 = (2 x 7)3 = 23 x 73, which is multiple of 73 (21)3 = (3 x 7)3 = 33 x 73, which is multiple of 73 (28)3 = (4 x 7)3 = 43 x 73, which is multiple of 73 (35)3 = (5 x 7)3 = 53 x 73 which is multiple of 73 Hence proved. Question 8. Which of the following are perfect cubes? (i) 64 (ii) 216 (iii) 243 (iv) 1000 (v) 1728 (vi) 3087 (vii) 4608 (viii) 106480 (ix) 166375 (x) 456533 Solution: (i) 64 = 2 x 2 x 2 x 2 x 2 x 2 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 64 is a perfect cube (ii) 216 = 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets of equal factors, we see that no factor is left 216 is a perfect cube. (iii) 243 = 3 x 3 x 3 x 3 x 3 Grouping the factors in triplets, we see that two factors 3 x 3 are left ∴ 243 is not a perfect cube. (iv) 1000 = 2 x 2 x 2 x 5 x 5 x 5 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 1000 is a perfect cube. (v) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets of the equal factors, we see that no factor is left ∴ 1728 is a perfect cube, (vi) 3087 = 3 x 3 x 7 x 7 x 7 Grouping the factors in triplets of the equal factors, we see that two factor 3×3 are left ∴ 3087 is not a perfect cube. (vii) 4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left ∴ 4609 is not a perfect cube. (viii) 106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, we see that factors 2, 5 are left ∴ 106480 is not a perfect cube. (ix) 166375 = 5 x 5 x 5 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 166375 is a perfect cube. (x) 456533 = 7 x 7 x 7 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 456533 is a perfect cube. Question 9. Which of the following are cubes of even natural numbers ? 216, 512, 729,1000, 3375, 13824 Solution: We know that the cube of an even natural number is also an even natural number ∴ 216, 512, 1000, 13824 are even natural numbers. ∴ These can be the cubes of even natural number. Question 10. Which of the following are cubes of odd natural numbers ? 125, 343, 1728, 4096, 32768, 6859 Solution: We know that the cube of an odd natural number is also an odd natural number, ∴ 125, 343, 6859 are the odd natural numbers ∴ These can be the cubes of odd natural numbers. Question 11. What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes ? (i) 675 (ii) 1323 (iii) 2560 (iv) 7803 (v) 107311 (vi) 35721 Solution: (i) 675 = 3 x 3 x 3 x 5 X 5 Grouping the factors in triplet of equal factors, 5 x 5 are left without triplet So, by multiplying by 5, the triplet will be completed. ∴ Least number to be multiplied = 5 (ii) 1323 = 3 x 3 x 3 x 7 x 7 Grouping the factors in triplet of equal factors. We find that 7 x 7 has been left So, multiplying by 7, we get a triplet ∴ The least number to be multiplied = 7 (iii) 2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 Grouping the factors in triplet of equal factors, 5 is left. ∴ To complete a triplet 5 x 5 is to multiplied ∴ Least number to be multiplied = 5 x 5 = 25 (iv) 7803 = 3 x 3 x 3 x 17 x 17 Grouping the factors in triplet of equal factors, we find the 17 x 17 are left So, to complete the triplet, we have to multiply by 17 ∴ Least number to be multiplied = 17 (v) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11 Grouping the factors in triplet of equal factors, factor 3 is left So, to complete the triplet 3 x 3 is to be multiplied ∴ Least number to be multiplied = 3 x 3 = 9 (vi) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7 Grouping the factors in triplet of equal factors, we find that 7 x 7 is left So, in order to complete the triplets, we have to multiplied by 7 ∴ Least number to be multiplied = 7 Question 12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube ? (i) 675 (ii) 8640 (iii) 1600 (iv) 8788 (v) 7803 (vi) 107811 (vii) 35721 (viii) 243000 Solution: (i) 675 = 3 x 3 x 3 x 5 x 5 Grouping the factors in triplet of equal factors, 5 x 5 is left 5 x 5 is to be divided so that the quotient will be a perfect cube. ∴ The least number to be divided = 5 x 5 = 25 (ii) 8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets of equal factors, 5 is left ∴ In order to get a perfect cube, 5 is to divided ∴ Least number to be divided = 5 (iii) 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 Grouping the factors in triplets of equal factors, we find that 5 x 5 is left ∴ In order to get a perfect cube 5 x 5 = 25 is to be divided. ∴ Least number to be divide = 25 (iv) 8788 = 2 x 2 x 13 x 13 x 13 Grouping the factors in triplets of equal factors, we find that 2 x 2 has been left ∴ In order to get a perfect cube, 2 x 2 is to be divided ∴ Least number to be divided = 4 (v) 7803 = 3 x 3 x 3 x 17 x 17 Grouping the factors in triplets of equal factors, we see that 17 x 17 has been left. So, in order to get a perfect cube, 17 x 17 is be divided ∴ Least number to be divided = 17 x 17 = 289 (vi) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, 3 is left ∴ In order to get a perfect cube, 3 is to be divided ∴ Least number to be divided = 3 (vii) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7 Grouping the factors in triplets of equal factors, we see that 7 x 7 is left So, in order to get a perfect cube, 7 x 7 = 49 is to be divided ∴ Least number to be divided = 49 (viii) 243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5 Grouping the factors in triplets of equal factors, 3 x 3 is left ∴ By dividing 3 x 3, we get a perfect cube ∴ Least number to be divided = 3 x 3=9 Question 13. Prove that if a number is trebled then its cube is 27 times the cube of the given number. Solution: Let x be the number, then trebled number of x = 3x Cubing, we get: (3x)3 = (3)3 x3 = 27x3 27x3 is 27 times the cube of x i.e., of x3 Question 14. What happenes to the cube of a number if the number is multiplied by (i) 3 ? (ii) 4 ? (iii) 5 ? Solution: number (x)3 = x3 (i) If x is multiplied by 3, then the cube of ∴ (3x)3 = (3)3 x x3 = 27x3 ∴ The cube of the resulting number is 27 times of cube of the given number (ii) If x is multiplied by 4, then the cube of (4x)3 = (4)3 x x3 = 64x3 ∴ The cube of the resulting number is 64 times of the cube of the given number (ii) If x is multiplied by 5, then the cube of (5x)3 = (5)3 x x3 = 125x3 ∴ The cube of the resulting number is 125 times of the cube of the given number Question 15. Find the volume of a cube, one face of which has an area of 64 m2. Solution: Area of one face of a cube = 64 m2 ∴ Side (edge) of cube = √64 = √64 = 8 m ∴ Volume of the cube = (side)3 = (8 m)3 = 512 m3 Question 16. Find the volume of a cube whose surface area is 384 m2. Solution: Surface area of a cube = 384 m2 Let side = a Then 6a2 = 384 ⇒ a2 = $$\frac {384 }{ 6 }$$= 64 = (8)2 ∴ a = 8 m Now volume = a3 = (8)3 m3 = 512 m3 Question 17. Evaluate the following : Solution: Question 18. Write the units digit of the cube of each of the following numbers : 31,109,388,833,4276,5922,77774,44447, 125125125. Solution: We know that if unit digit of a number n is = 1, then units digit of its cube = 1 = 2, then units digit of its cube = 8 = 3, then units digit of its cube = 7 = 4, then units digit of its cube = 4 = 5, then units digit of its cube = 5 = 6, then units digit of its cube = 6 = 7, then the units digit of its cube = 3 = 8, then units digit of its cube = 2 = 9, then units digit of its cube = 9 = 0, then units digit of its cube = 0 Now units digit of the cube of 31 = 1 Units digit of the cube of 109 = 9 Units digits of the cube of 388 = 2 Units digits of the cube of 833 = 7 Units digits of the cube of 4276 = 6 Units digit of the cube of 5922 = 8 Units digit of the cube of 77774 = 4 Units digit of tl. cube of 44447 = 3 Units digit of the cube of 125125125 = 5 Question 19. Find the cubes of the following numbers by column method : (i) 35 (ii) 56 (iii) 72 Solution: Question 20. Which of the following numbers are not perfect cubes ? (i) 64 (ii) 216 (iii) 243 (iv) 1728 Solution: (i) 64 = 2 x 2 x 2 X 2 x 2 x 2 Grouping the factors in triplets, of equal factors, we see that no factor is left ∴ 64 is a perfect cube. (ii) 216 = 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets, of equal factors, we see that no factor is left ∴ 216 is a perfect cube. (iii) 243 = 3 x 3 x 3 x 3 x 3 Grouping the factors in triplets, of equal factors, we see that 3 x 3 are left ∴ 243 is not a perfect cube. (iv) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors m triplets, of equal factors, we see that no factor is left. ∴ 1728 is a perfect cube. Question 21. For each of the non-perfect cubes, in Q. 20, find the smallest number by which it must be (a) multiplied so that the product is a perfect cube. (b) divided so that the quotient is a perfect cube. Solution: In qustion 20, 243 is not a perfect cube and 243 = 3 x 3 x 3 x 3 x 3 Grouping the factors in triplets, of equal factors, we see that 3 x 3 is left. (a) In order to make it a perfect cube, 3 is to be multiplied which makes a triplet. (b) In order to make it a perfect cube, 3 x 3 or 9 is to be divided. Question 22. By taking three different values of n verify the truth of the following statements : (i) If n is even, then n3 is also even. (ii) If n is odd, then n3 is also odd. (iii) If n leaves remainder 1 when divided by 3, then it3 also leaves 1 as remainder when divided by 3. (iv) If a natural number n is of the form 3p + 2 then n3 also a number of the same type. Solution: (i) n is even number. Let n = 2, 4, 6 then (a) n3 = (2)3 = 2 x 2 x 2 = 8, which is an even number. (b) (n)3= (4)3 = 4 x 4 x 4 = 64, which is an even number. (c) (n)3 = (6)3 = 6 x 6 x 6 = 216, which is an even number. (ii) n is odd number. Letx = 3, 5, 7 (a) (n)3 = (3)3 = 3 x 3 x 3 = 27, which is an odd number. (b) (n)3 = (5)3 = 5 x 5 x 5 = 125, which is an odd number. (c) (n)3 = (7)3 = 7 x 7 x 7 = 343, which is an odd number. (iii) If n leaves remainder 1 when divided by 3, then n3 is also leaves 1 as remainder, Let n = 4, 7, 10 If n = 4, then «3 = (4)3 = 4 x 4 x 4 = 64 = 64 + 3 = 21, remainder = 1 If n = 7, then n3 = (7)3 = 7 x 7 x 7 = 343 343 + 3 = 114, remainder = 1 If n – 10, then (n)3 = (10)3 = 10 x 10 x 10 = 1000 1000 + 3 = 333, remainder = 1 (iv) If the natural number is of the form 3p + 2, then n3 is also of the same type Let p =’1, 2, 3, then (a) If p = 1, then n = 3p + 2 = 3 x 1+2=3+2=5 ∴ n3 = (5)3 = 5 x 5 x 5 = 125 125 = 3 x 41 + 2 = 3p +2 (b) If p = 2, then n = 3p + 2 = 3 x 2 + 2 = 6 + 2 = 8 ∴ n3 = (8)3 = 8 x 8 x 8 = 512 ∴ 512 = 3 x 170 + 2 = 3p + 2 (c) If p = 3, then n = 3p + 2 = 3 x 3 + 2 = 9 + 2 = 11 ∴ n3 = (11)3 = 11 x 11 x 11 = 1331 and 1331 =3 x 443 + 2 = 3p + 2 Hence proved. Question 23. Write true (T) or false (F) for the following statements : (i) 0 392 is a perfect cube. (ii) 8640 is not a perfect cube. (iii) No cube can end with exactly two zeros. (iv) There is no perfect cube which ends in 4. (v) For an integer a, a3 is always greater than a> b2 (vi) If a and b are integers such that a> b2 , then a3 > b3. (x) If a2 ends in an even number of zeros, then a ends in an odd number of zeros. Solution: ∵ 5 is left. (iii)True : A number ending three zeros can be a perfect cube. (iv) False : v (4)3 = 4 x 4 x 4 = 64, which ends with 4. (v) False : If n is a proper fraction, it is not possible. (vi) False : It is not true if a and b are proper fraction. (vii) True. (viii) False : as a2 ends in 9, then a3 does not necessarily ends in 7. It ends in 3 also. (ix) False : it is not necessarily that a3 ends in 25 it can end also in 75. (x) False : If a2 ends with even zeros, then a3 will ends with odd zeros but of multiple of 3. Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8 Other Exercises Question 1. Find the square root of each of the following correct to three places of decimal : (i) 5 (ii) 7 (iii) 17 (iv) 20 (v) 66 (vi) 427 (vii) 1.7 (vii’) 23.1 (ix) 2.5 (x) 237.615 (xi) 15.3215 (xii) 0.9 (xiii) 0.1 (xiv) 0.016 (xv) 0.00064 (xvi) 0.019 (xvii) $$\frac {7 }{ 8 }$$ (xviii) $$\frac {5 }{ 12 }$$ (xix) 2 $$\frac {1 }{ 2 }$$ (xx) 287 $$\frac {8 }{ 8 }$$ Solution: Question 2. Find the square root of 12.0068 correct to four decimal places. Solution: Question 3. Find the square root of 11 correct to five decimal places. Solution: Question 4. Given that √2, = 1-414, √3 = 1.732, √5 = 2.236 and √7 = 2.646. Evaluate each of the following : Solution: Question 5. Given that √2 = 1-414, √3 = 1-732, √5= 2.236 and √7= 2.646, find the square roots of the following : Solution: Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2 Other Exercises Question 1. Find the cubes of: (i) -11 (ii) -12 (iii) – 21 Solution: (i) (-11)3=(-11)3=(11 x 11 x 11) =-1331 (ii) (-12)3=(-12)3=(12 x 12 x 12) = -1728 (iii) (-21)3=(-21)3=(21 x 21 x 21) = -9261 Question 2. Which of the following numbers are cubes of negative integers. (i) -64 (ii) -1056 (iii) -2197 (iv) -2744 (v) -42875 Solution: ∴ All factors of 64 can be grouped in triplets of the equal factors completely. ∴ -64 is a perfect cube of negative integer. All the factors of 1056 can be grouped in triplets of equal factors grouped completely ∴ 1058 is not a perfect cube of negative integer. All the factors of -2197 can be grouped in triplets of equal factors completely ∴ 2197 is a perfect cube of negative integer, All the factors of -2744 can be grouped in triplets of equal factors completely ∴ 2744 is a perfect cube of negative integer All the factors of -42875 can be grouped in triplets of equal factors completely ∴ 42875 is a perfect cube of negative integer. Question 3. Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer : (i) -5832 (ii) -2744000 Solution: Grouping the factors in triplets of equal factors, we see that no factor is left ∴ -5832 is a perfect cube Now taking one factor from each triplet we find that -5832 is a cube of – (2 x 3 x 3) = -18 ∴ Cube root of-5832 = -18 Grouping the factors in tuplets of equal factors, we see that no factor is left. Therefore it is a perfect cube. Now taking one factor from each triplet, we find that. -2744000 is a cube of – (2 x 2 x 5 x 7) ie. -140 ∴ Cube root of -2744000 = -140 Question 4. Find the cube of : Solution: Question 5. Which of the following numbers are cubes of rational numbers : Solution: Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6 Other Exercises Question 1. Find the square root of: Solution: Question 2. Find the value of : Solution: Question 3. The area of a square field is 80 $$\frac {244 }{ 729 }$$ square metres. Find the length of each side of the field. Solution: Question 4. The area of a square field is 30 $$\frac {1 }{ 4}$$ m2. Calculate the length of the side of the squre. Solution: Question 5. Find the length of a side of a square playground whose area is equal to the area of a rectangular field of dimensions 72 m and 338 m. Solution: Length of rectangular field (l) = 338 m and breadth (b) = 72 m ∴ Area = 1 x 6= 338 x 72 m2 ∴ Area of square = 338 x 72 m2 = 24336 m2 and length of the side of the square Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 Other Exercises Question 1. Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication : (i) 25 (ii) 37 (iii) 54 (iv) 71 (v) 96 Solution: (i) (25)2 Question 2. Find the squares of the following numbers using diagonal method : (i) 98 (ii) 273 (iii) 348 (iv) 295 (v) 171 Solution: Question 3. Find the squares of the following numbers : (i) 127 (ii) 503 (iii) 451 (iv) 862 (v) 265 Solution: (i) (127)2 = (120 + 7)2 {(a + b)2 = a2 + lab + b2} = (120)2 + 2 x 120 x 7 + (7)2 = 14400+ 1680 + 49 = 16129 (ii) (503)2 = (500 + 3)2 {(a + b)2 = a2 + lab + b1} = (500)2 + 2 x 500 x 3 + (3)2 = 250000 + 3000 + 9 = 353009 (iii) (451)2 = (400 + 51)2 {(a + b)2 = a2 + lab + b2} = (400)2 + 2 x 400 x 51 + (5l)2 = 160000 + 40800 + 2601 = 203401 (iv) (451)2 = (800 + 62)2 {(a + b)2 = a2 + lab + b2} = (800)2 + 2 x 800 x 62 + (62)2 = 640000 + 99200 + 3844 = 743044 (v) (265)2 {(a + b)2 = a2 + 2ab + b2} (200 + 65)2 = (200)2 + 2 x 200 x 65 + (65)2 = 40000 + 26000 + 4225 = 70225 Question 4. Find the squares of the following numbers (i) 425 (ii) 575 (iii) 405 (iv) 205 (v) 95 (vi) 745 (vii) 512 (viii) 995 Solution: (i) (425)2 Here n = 42 ∴ n (n + 1) = 42 (42 + 1) = 42 x 43 = 1806 ∴ (425)2 = 180625 (ii) (575)2 Here n = 57 ∴ n (n + 1) = 57 (57 + 1) = 57 x 58 = 3306 ∴ (575)2 = 330625 (iii) (405)2 Here n = 40 ∴ n (n + 1) = 40 (40 + 1) -40 x 41 = 1640 ∴ (405)2 = 164025 (iv) (205)2 Here n = 20 ∴ n (n + 1) = 20 (20 + 1) = 20 x 21 = 420 ∴ (205)2 = 42025 (v) (95)2 Here n = 9 ∴ n (n + 1) = 9 (9 + 1) = 9 x 10 = 90 ∴ (95)2 = 9025 (vi) (745)2 Here n = 74 ∴ n (n + 1) = 74 (74 + 1) = 74 x 75 = 5550 ∴ (745)2 = 555025 (vii) (512)2 Here a = 1, b = 2 ∴ (5ab)2 = (250 + ab) x 1000 + (ab)2 ∴ (512)2 = (250 + 12) x 1000 + (12)2 = 262 x 1000 + 144 = 262000 + 144 = 262144 (viii) (995)2 Here n = 99 ∴ n (n + 1) = 99 (99 + 1) = 99 x 100 = 9900 ∴ (995)2 = 990025 Question 5. Find the squares of the following numbers using the identity (a + b)1 = a2 + lab + b1 (i) 405 (ii) 510 (iii) 1001 (iv) 209 (v) 605 Solution: a + b)2 = a2 + lab + b2 (i) (405)2 = (400 + 5)2 = (400)2 + 2 x 400 x 5 + (5)2 = 160000 + 4000 + 25 = 164025 (ii) (510)2 = (500 + 10)2 = (500)2 + 2 x 500 x 10 x (10)2 = 250000 + 10000 + 100 = 260100 (iii) (1001)2 = (1000+1)2 = (1000)2 + 2 X 1000 x 1 + (1) = 1000000 + 2000 + 1 =1002001 (iv) (209)2 = (200 + 9)2 = (200)2 + 2 x 200 x 9 x (9)2 = 40000 + 3600 +81 = 43681 (v) (605)2 = (600 + 5)2 = (600)2 + 2 x 600 x 5 +(5)2 = 360000 + 6000 25 =366025 Question 6. Find the squares of the following numbers using the identity (a – b)2 = a2 – 2ab + b2 : (i) 395 (ii) 995 (iii) 495 (iv) 498 (v) 99 (vi) 999 (vii) 599 Solution: a – b)2 = a2 – lab + b2 (i) (395)2 = (400 – 5)2 = (400)2 – 2 x 400 x 5 + (5)2 = 160000-4000 + 25 = 160025-4000 = 156025 (ii) (995)2 = (1000 – 5)2 = (1000)2 – 2 x 1000 x 5 + (5)2 = 1000000- 10000 + 25 = 1000025- 10000 = 990025 (iii) (495)2 = (500 – 5)2 = (500)2 – 2 x 500 x 5 + (5)2 = 250000 – 5000 + 25 = 250025 – 5000 = 245025 (iv) (498)2 = (500 – 2)2 = (500)2 – 2 x 500 x 2 + (2)2 = 250000 – 2000 + 4 = 250004 – 2000 = 248004 (v) (99)2 = (100 – l)2 = (100)2 – 2 x 100 x 1 + (1)2 = 10000 – 200 + 1 = 10001 – 200 = 9801 (vi) (999)2 = (1000- l)2 = (1000)2 – 2 x 1000 x 1+ (1)2 = 1000000-2000+1 = 10000001-2000=998001 (vii) (599)2 = (600 – 1)2 = (600)-2 x 600 X 1+ (1)2 = 360000 -1200+1 = 360001 – 1200 = 358801 Question 7. Find the squares of the following numbers by visual method : (i) 52 (ii) 95 (iii) 505 (iv) 702 (v) 99 Solution: (a + b)2 = a2 – ab + ab + b2 (i) (52)2 = (50 + 2)2 = 2500 + 100 + 100 + 4 = 2704 (ii) (95)2 = (90 + 5)2 = 8100 + 450 + 450 + 25 = 9025 (iii) (505)2 = (500 + 5)2 = 250000 + 2500 + 2500 + 25 = 255025 (iv) (702)2 = (700 + 2)2 = 490000 + 1400+ 1400 + 4 = 492804 (v) (99)2 = (90 + 9)2 = 8100 + 810 + 810 + 81 = 9801 Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7 Other Exercises Find the square root of the following numbers in decimal form : Question 1. 84.8241 Solution: Question 2. 0.7225 Solution: Question 3. 0.813604 Solution: Question 4. 0.00002025 Solution: Question 5. 150.0625 Solution: Question 6. 225.6004 Solution: Question 7. 3600.720036 Solution: Question 8. 236.144689 Solution: Question 9. 0.00059049 Solution: Question 10. 176.252176 Solution: Question 11. 9998.0001 Solution: Question 12. 0.00038809 Solution: Question 13. What is that fraction which when multiplied by itself gives 227.798649 ? Solution: Question 14. square playground is 256.6404 square metres. Find the length of one side of the playground. Solution: Area of square playground = 256.6404 sq. m Question 15. What is the fraction which when multiplied by itself gives 0.00053361 ? Solution: Question 16. Simplify : Solution: Question 17. Evaluate $$\sqrt { 50625 }$$ and hence find the value of $$\sqrt { 506.25 } +\sqrt { 5.0625 }$$. Solution: Question 18. Find the value of $$\sqrt { 103.0225 }$$ and hence And the value of Solution: Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3 Other Exercises Question 1. Find the cube roots of the following numbers by successive subtraction of numbers : 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, (i) 64 (ii) 512 (iii) 1728 Solution: (i) 64 64 – 1 = 63 63 – 7 = 56 56 – 19 = 37 37 – 37 = 0 ∴ 64 = (4)3 ∴ Cube root of 64 = 4 (ii) 512 512 -1 =511 511- 7 = 504 504 – 19 = 485 485 – 37 = 448 448 – 61 = 387 387 – 91 =296 296 – 127 = 169 169 – 169 = 0 ∴ 512 = (8)3 ∴ Cube root of 512 = 8 (iii) 1728 1728 – 1= 1727 1727 -7 = 1720 1720 -19 = 1701 1701 -37= 1664 1664 – 61 = 1603 1603 – 91 = 1512 1512 -127= 1385 . 1385 – 169= 1216 1216 – 217 = 999 999 – 271 =728 728 – 331 = 397 397 – 397=0 ∴ 1728 = (12)3 ∴ Cube root of 1728 = 12 Question 2. Using the method of successive subtraction, examine whether or not the following numbers are perfect cubes : (i) 130 (ii) 345 (iii) 792 (iv) 1331 Solution: (i) 130 130 – 1 = 129 129 -7 = 122 122 -19 = 103 103 -37 = 66 66 – 61 = 5 We see that 5 is left ∴ 130 is not a perfect cube. (ii) 345 345 – 1 = 344 344 – 7 = 337 337 – 19 = 318 318 – 37 = 281 81 – 61 =220 220 – 91 = 129 129 – 127 = 2 We see that 2 is left ∴ 345 is not a perfect cube. (iii) 792 792 – 1 = 791 791 – 7 = 784 784 – 19 = 765 765 – 37 = 728 728 – 61 = 667 667 – 91 = 576 576 – 127 = 449 449 – 169 = 280 ∴ We see 280 is left as 280 <217 ∴ 792 is not a perfect cube. (iv) 1331 1331 – 1 = 1330 1330 -7 = 1323 1323 – 19 = 1304 1304 – 37 = 1267 1267 – 61 = 1206 1206 – 91 = 1115 1115 – 127 = 988 988 – 169 = 819 819 – 217 = 602 602 – 271 = 331 331 – 331 =0 ∴ 1331 is a perfect cube Question 3. Find the smallest number that must be subtracted from those of the numbers in question 2, which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots ? Solution: We have examined in Question 2, the numbers 130, 345 and 792 are not perfect cubes. Therefore (i) 130 130 – 1 = 129 129 -7= 122 122 -19 = 103 103 – 37 = 66 66 – 61 = 5 Here 5 is left ∴ 5 < 91 5 is to be subtracted to get a perfect cube. Cube root of 130 – 5 = 125 is 5 (ii) 345 345 – 1 = 344 344 -7 = 337 337 – 19 = 318 318 – 37 = 281 281 – 61 =220 220 – 91 = 129 129 – 127 = 2 Here 2 is left ∵ 2 < 169 ∴ Cube root of 345 – 2 = 343 is 7 ∴ 2 is to be subtracted to get a perfect cube. (iii) 792 792 – 1 = 791 791 – 7 = 784 784 – 19 = 765 765 – 37 = 728 728 – 61 =667 667 – 91 = 576 5 76 – 127 = 449 449 – 169 = 280 280 – 217 = 63 ∴ 63 <217 ∴ 63 is to be subtracted ∴ Cube root of 792 – 63 = 729 is 9 Question 4. Find the cube root of each of the following natural numbers : (i) 343 (ii) 2744 (iii) 4913 (iv) 1728 (v) 35937 (vi) 17576 (vii) 134217728 (viii) 48228544 (ix) 74088000 (x) 157464 (xi) 1157625 (xii) 33698267 Solution: Question 5. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product. Solution: Grouping the factors in triplets of equal factors, we see that 2, 3 x 3 and 5 x 5 are left ∴ In order to complete the triplets, we have to multiply it by 2, 3 and 5. ∴ The smallest number to be multiplied = 2×2 x 3 x 5 = 60 Now product = 3600 x 60 = 216000 and cube root of 216000 Question 6. Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product. Solution: Grouping the factors in triplets of equal factors, we see that 41 x 41 is left ∴ In order to complete the triplet, we have to multiply it by 41 ∴ Smallest number to be multiplied = 41 ∴ Product = 210125 x 41 = 8615125 ∴ Cube root of 8615125 Question 7. What is the smallest number by which 8192 must be divided so that quotient is a perfect cube ? Also, find the cube root of the quotient so obtained. Solution: Grouping the factors in triplets of equal factors, we see that 2 is left ∴ Dividing by 2, we get the quotient a perfect cube ∴ Perfect cube = 8192 + 2 = 4096 Question 8. Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers. Solution: Ratio in numbers =1:2:3 Let first number = x Then second number = 2x and third number = 3x ∴ Sum of cubes of there numbers = (x)3 + (2x)3+(3x)3 Question 9. The volume of a cube is 9261000 m3. Find the side of the cube. Solution: Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2 Other Exercises Question 1. The following numbers are not perfect squares. Give reason : (i) 1547 (ii) 45743 (iii) 8948 (iv) 333333 Solution: We know that if the units digit is 2, 3, 7 or 8 of a number, then the number is not a perfect square. (i) ∴ 1547 has 7 as units digit. ∴ It is not a perfect square. (ii) 45743 has 3 as units digit ∴ It is not a perfect square. (iii) ∴ 8948 has 8 as units digit ∴ It is not a perfect square. (iv) ∴ 333333 has 3 as units digits ∴ It is not a perfect square. Question 2. Show that the following numbers are not perfect squares : (i) 9327 (ii) 4058 (iii) 22453 (iv) 743522 Solution: (i) 9327 ∴ The units digit of 9327 is 7 ∴ This number can’t be a perfect square. (ii) 4058 ∴ The units digit of 4058 is 8 ∴ This number can’t be a perfect square. (iii) 22453 ∴ The units digit of 22453 is 3 .∴ This number can’t be a perfect square. (iv) 743522 ∴ The units digit of 743522 is 2 ∴ This number can’t be a perfect square. Question 3. The square of which of the following numbers would be an odd number ? (i) 731 (ii) 3456 (iii) 5559 (iv) 42008 Solution: We know that the square of an odd number is odd and of even number is even. Therefore (i) Square of 731 would be odd as it is an odd number. (ii) Square of 3456 should be even as it is an even number. (iii) Square of 5559 would be odd as it is an odd number. (iv) The square of 42008 would be an even number as it is an even number. Therefore suqares of (i) 731 and (ii) 5559 will be odd numbers. Question 4. What will be the units digit of the squares of the following numbers ? (i) 52 (ii) 977 (iii) 4583 (iv) 78367 (v) 52698 (vi) 99880 (vii) 12796 (viii) 55555 (ix) 53924 Solution: (i) Square of 52 will be 2704 or (2)2 = 4 ∴ Its units digit is 4. (ii) Square of 977 will be 954529 or (7)2 = 49 . ∴ Its units digit is 9 (iii) Square of 4583 will be 21003889 or (3)2 = 9 ∴ Its units digit is 9 (iv) IS 78367, square of 7 = 72 = 49 ∴ Its units digit is 9 (v) In 52698, square of 8 = (8)2 = 64 ∴ Its units digit is 4 (vi) In 99880, square of 0 = 02 = 0 ∴ Its units digit is 0 (vii) In 12796, square of 6 = 62 = 36 ∴ Its units digit is 6 (viii) In In 55555, square of 5 = 52 = 25 ∴ Its units digit is 5 (ix) In 53924, square pf 4 = 42 = 16 ∴ Its units digit is 6 Question 5. Observe the following pattern 1 + 3 = 22 1 + 3 + 5 = 32 1+34-5 + 7 = 42 and write the value of 1 + 3 + 5 + 7 + 9 +…………upto n terms. Solution: The given pattern is 1 + 3 = 22 1 + 3 + 5 = 32 1+3 + 5 + 7 = 42 1+3 + 5 + 7 + 9 +……………… upto n terms (number of terms)2 = n2 Question 6. Observe the following pattern : 22 – 12 = 2 + 1 32 – 22 = 3 + 2 42 – 32 = 4 + 3 52 – 42 = 5 + 4 Find the value of (i) 1002 – 992 (ii) 1112 – 1092 (iii) 992 – 962 Solution: From the given pattern, 22 – 12 = 2 + 1 32 – 22 = 3 + 2 42 – 32 = 4 + 3 52 – 42 = 5 + 4 Therefore (i) 1002-99° = 100 + 99 (ii) 1112 – 1092 = 1112 – 1102– 1092 = (1112 – 1102) + (1102 – 1092) = (111 + 110) + (110+ 109) = 221 + 219 = 440 (iii) 992 – 962 = 992 – 982 + 982 – 972 + 972 – 962 = (992 – 982) + (982 – 972) + (972 – 962) = (99 + 98) + (98 + 97) + (97 + 96) = 197 + 195 + 193 = 585 Question 7. Which of the following triplets are Pythagorean ? (i) (8, 15, 17) (ii) (18, 80, 82) (iii) (14, 48, 51) (iv) (10, 24, 26) (vi) (16, 63, 65) (vii) (12, 35, 38) Solution: A pythagorean triplet is possible if (greatest number)2 = (sum of the two smaller numbers) (i) 8, 15, 17 Here, greatest number =17 ∴ (17)2 = 289 and (8)2 + (15)2 = 64 + 225 = 289 ∴ 82 + 152 = 172 ∴ 8, 15, 17 is a pythagorean triplet (ii) 18, 80, 82 Greatest number = 82 ∴ (82)2 = 6724 and 182 + 802 = 324 + 6400 = 6724 ∴ 182 + 802 = 822 ∴ 18, 80, 82 is a pythagorean triplet (iii) 14, 48, 51 Greatest number = 51 ∴ (51)2 = 2601 and 142 + 482 = 196 + 2304 = 25 00 ∴ 512≠ 142 + 482 ∴ 14, 48, 51 is not a pythagorean triplet (iv) 10, 24, 26 Greatest number is 26 ∴ 262 = 676 and 102 + 242 = 100 + 576 = 676 ∴ 262 = 102 + 242 ∴ 10, 24, 26 is a pythagorean triplet (vi) 16, 63, 65 Greatest number = 65 ∴ 652 = 4225 and 162 + 632 = 256 + 3969 = 4225 ∴ 652 = 162 + 632 ∴ 16, 63, 65 is a pythagorean triplet (vii) 12, 35, 38 Greatest number = 38 ∴ 382 = 1444 and 122 + 352 = 144 + 1225 = 1369 ∴ 382 ≠122 + 352 ∴ 12, 35, 38 is not a pythagorean triplet. Question 8. Observe the following pattern Solution: From the given pattern Question 9. Observe the following pattern and find the values of each of the following : (i) 1 + 2 + 3 + 4 + 5 +….. + 50 (ii) 31 + 32 +… + 50 Solution: From the given pattern, Question 10. Observe the following pattern and find the values of each of the following : (i) 12 + 22 + 32 + 42 +…………… + 102 (ii) 52 + 62 + 72 + 82 + 92 + 102 + 12 2 Solution: From the given pattern, Question 11. Which of the following numbers are squares of even numbers ? 121,225,256,324,1296,6561,5476,4489, 373758 Solution: We know that squares of even numbers is also are even number. Therefore numbers 256, 324,1296, 5476 and 373758 have their units digit an even number. ∴ These are the squares of even numbers. Question 12. By just examining the units digits, can you tell which of the following cannot be whole squares ? 1. 1026 2. 1028 3. 1024 4. 1022 5. 1023 6. 1027 Solution: We know that a perfect square cam at ends with the digit 2, 3, 7, or 8 ∴ By examining the given number, we can say that 1028, 1022, 1023, 1027 can not be perfect squares. Question 13. Write five numbers for which you cannot decide whether they are squares. Solution: A number which ends with 1,4, 5, 6, 9 or 0 can’t be a perfect square 2036, 4225, 4881, 5764, 3349, 6400 Question 14. Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit. Solution: A number which does not end with 2, 3, 7 or 8 can be a perfect square ∴ The five numbers can be 2024, 3036, 4069, 3021, 4900 Question 15. Write true (T) or false (F) for the following statements. (i) The number of digits in a square number is even. (ii) The square of a prime number is prime. (iii) The sum of two square numbers is a square number. (iv) The difference of two square numbers is a square number. (vi) The product of two square numbers is a square number. (vii) No square number is negative. (viii) There is not square number between 50 and 60. (ix) There are fourteen square number upto 200. Solution: (i) False : In a square number, there is no condition of even or odd digits. (ii) False : A square of a prime is not a prime. (iii) False : It is not necessarily. (iv) False : It is not necessarily. (vi) True. (vii) True : A square is always positive. (viii) True : As 72 = 49, and 82 = 64. (ix) True : As squares upto 200 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 which are fourteen in numbers. Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5 Other Exercises Question 1. Find the square root of each of the long division method. (I) 12544 (ii) 97344 (iii) 286225 (iv) 390625 (v) 363609 (vi) 974169 (vii) 120409 (viii) 1471369 (ix) 291600 (x) 9653449 (xi) 1745041 (xii) 4008004 (xiii) 20657025 (xiv) 152547201 (jcv) 20421361 (xvi) 62504836 (xvii) 82264900 (xviii) 3226694416 (xix)6407522209 (xx) 3915380329 Solution: < Question 2. Find the least number which must be subtracted from the following numbers to make them a perfect square : (i) 2361 (ii) 194491 (iii) 26535 (iv) 16160 (v) 4401624 Solution: (i) 2361 Finding the square root of 2361 We get 48 as quotient and remainder = 57 ∴ To make it a perfect square, we have to subtract 57 from 2361 ∴ Least number to be subtracted = 57 (ii) 194491 Finding the square root of 194491 We get 441 as quotient and remainder = 10 ∴ To make it a perfect square, we have to subtract 10 from 194491 ∴ Least number to be subtracted = 10 (iii) 26535 Finding the square root of 26535 We get 162 as quotient and 291 as remainder ∴ To make it a perfect square, we have to subtract 291 from 26535 ∴ Least number to be subtracted = 291 (iv)16160 Finding the square root of 16160 We get 127 as quotient and 31 as remainder ∴ To make it a perfect square, we have to subtract 31 from 16160 ∴ Least number to be subtracted = 31 (v) 4401624 Find the square root of 4401624 We get 2098 as quotient and 20 as remainder ∴ To make it a perfect square, we have to subtract 20 from 4401624 ∴ Least number to be subtracted = 20 Question 3. Find the least number which must be added to the following numbers to make them a perfect square : (i) 5607 (ii) 4931 (iii) 4515600 (iv) 37460 (v) 506900 Solution: (i) 5607 Finding the square root of 5607, we see that 742 = 5607- 131 =5476 and 752 = 5625 ∴ 5476 < 5607 < 5625 ∴ 5625 – 5607 = 18 is to be added to get a perfect square ∴ Least number to be added = 18 (ii) 4931 Finding the square root of 4931, we see that 702= 4900 ∴ 712 = 5041 4900 <4931 <5041 ∴ 5041 – 4931 = 110 is to be added to get a perfect square. ∴ Least number to be added =110 (iii) 4515600 Finding the square root of 4515600, we see that 21242 = 4511376 and 2 1 252 = 45 1 56 25 ∴ 4511376 <4515600 <4515625 ∴ 4515625 – 4515600 = 25 is to be added to get a perfect square. ∴ Least number to be added = 25 (iv) 37460 Finding the square root of 37460 that 1932 = 37249, 1942 = =37636 ∴ 37249 < 37460 < 37636 ∴ 37636 – 37460 = 176 is to be added to get a perfect square. ∴ Least number to be added =176 (v) 506900 Finding the square root of 506900, we see that 7112 = 505521, 7122 = 506944 ∴ 505521 < 506900 < 506944 ∴ 506944 – 506900 = 44 is to be added to get a perfect square. ∴ Least number to be added = 44 Question 4. Find the greatest number of 5 digits which is a perfect square. Solution: Greatest number of 5-digits = 99999 Finding square root, we see that 143 is left as remainder ∴ Perfect square = 99999 – 143 = 99856 If we add 1 to 99999, it will because a number of 6 digits ∴ Greatest square 5-digits perfect square = 99856 Question 5. Find the least number of four digits which is a perfect square. Solution: Least number of 4-digits = 10000 Finding square root of 1000 We see that if we subtract 39 From 1000, we get three digit number ∴ We shall add 124 – 100 = 24 to 1000 to get a perfect square of 4-digit number ∴ 1000 + 24 = 1024 ∴ Least number of 4-digits which is a perfect square = 1024 Question 6. Find the least number of six-digits which is a perfect square. Solution: Least number of 6-digits = 100000 Finding the square root of 100000, we see that if we subtract 544, we get a perfect square of 5-digits. 4389 – 3900 = 489 to 100000 to get a perfect square Past perfect square of six digits= 100000 + 489 =100489 Question 7. Find the greatest number of 4-digits which is a perfect square. Solution: Greatest number of 4-digits = 9999 Finding the square root, we see that 198 has been left as remainder ∴ 4-digit greatest perfect square = 9999 – 198 = 9801 Question 8. A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row. Solution: Total number of soldiers = 8160 Soldiers left after arranging them in a square = 60 ∴ Number of soldiers which are standing in a square = 8160 – 60 = 8100 Question 9. The area of a square field is 60025 m2. A man cycle along its boundry at 18 km/hr. In how much time will be return at the starting point. Solution: Area of a square field = 60025 m2 Question 10. The cost of levelling and turfing a square lawn at Rs. 250 per m2 is Rs. 13322.50. Find the cost of fencing it at Rs. 5 per metre ? Solution: Cost of levelling a square field = Rs. 13322.50 Rate of levelling = Rs. 2.50 per m2 and perimeter = 4a = 4 x 73 = 292 m Rate of fencing the field = Rs. 5 per m ∴ Total cost of fencing = Rs. 5 x 292 = Rs. 1460 Question 11. Find the greatest number of three digits which is a perfect square. Solution: 3-digits greatest number = 999 Finding the square root, we see that 38 has been left ∴ Perfect square = 999 – 38 = 961 ∴ Greatest 3-digit perfect square = 961 Question 12. Find the smallest number which must be added to 2300 so that it becomes a perfect square. Solution: Finding the square root of 2300 We see that we have to add 704 – 700 = 4 to 2300 in order to get a perfect square ∴ Smallest number to be added = 4 Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9 Other Exercises Using square root table, find the square roots of the following : Question 1. 7 Solution: Question 2. 15 Solution: Question 3. 74 Solution: Question 4. 82 Solution: Question 5. 198 Solution: Question 6. 540 Solution: Question 7. 8700 Solution: Question 8. 3509 Solution: Question 9. 6929 Solution: Question 10. 25725 Solution: Question 11. 1312 Solution: Question 12. 4192 Solution: Question 13. 4955 Solution: Question 14. $$\frac {99 }{ 144 }$$ Solution: Question 15. $$\frac {57 }{ 169 }$$ Solution: Question 16. $$\frac {101 }{ 169 }$$ Solution: Question 17. 13.21 Solution: Question 18. 21.97 Solution: Question 19. 110 Solution: Question 20. 1110 Solution: Question 21. 11.11 Solution: Question 22. The area of a square field is 325 m2. Find the approximate length of one side of the field. Solution: Question 23. Find the length of a side of a square, whose area is equal to the area bf the rectangle with sides 240 m and 70 m. Solution: Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4 These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4 Other Exercises Question 1. Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ? (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025 Solution: (i) In $$\sqrt { 9801 }$$ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9 (ii) In $$\sqrt { 799356 }$$ ∴ the units digit is 6 ∴ The units digit of the square root can be 4 or 6 (iii) In $$\sqrt { 7998001 }$$ ∴ the units digit is 1 ∴ The units digit of the square root can be 1 or 9 (iv) In 657666025 ∴ The unit digit is 5 ∴ The units digit of the square root can be 5 Question 2. Find the square root of each of the following by prime factorization. (i) 441 (ii) 196 (iii) 529 (iv) 1764 (v) 1156 (vi) 4096 (vii) 7056 (viii) 8281 (ix) 11664 (x) 47089 (xi) 24336 (xii) 190969 (xiii) 586756 (xiv) 27225 (xv) 3013696 Solution: Question 3. Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained. Solution: Factorising 180, 180 = 2 x 2 x 3 x 3 x 5 Grouping the factors in pairs we see that factor 5 is left unpaired. ∴ Multiply 180 by 5, we get the product 180 x 5 = 900 Which is a perfect square and square root of 900 = 2 x 3 x 5 = 30 Question 4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained. Solution: Factorising 147, 147 = 3 x 7×7 Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired ∴ Multiplying 147 by 3, we get the product 147 x 3 = 441 Which is a perfect square and its square root = 3×7 = 21 Question 5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number. Solution: Factorising 3645 3645 = 3 x 3 3 x 3 x 3 x 3 x 5 Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired ∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27 Question 6. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained. Solution: Factorsing 1152, 1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired. ∴ Dividing by 2, the quotient 576 is a perfect square . ∴ Square root of 576, it is 24 Question 7. The product of two numbers is 1296. If one number is 16 times the others find the numbers. Solution: Product of two numbers = 1296 Let one number = x Second number = 16x ∴ First number = 9 and second number = 16 x 9 = 144 Question 8. A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents. Solution: Total donation collected = Rs. 202500 Let number of residents = x Then donation given by each resident = Rs. x ∴ Total collection = Rs. x x x Question 9. A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute? Solution: Total amount collected = Rs. 92.16 = 9216 paise Let the number of members = x Then amount collected by each member = x paise ∴ Number of members = 96 and each member collected = 96 paise Question 10. A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ? Solution: Total fee collected = Rs. 2304 Let number of students = x Then fee paid by each student = Rs. x ∴ x x x = 2304 => x2 = 2304 ∴ x = $$\sqrt { 2304 }$$ Question 11. The area of a square field is 5184 m2. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field. Solution: The area of a square field = 5184 m2 Let side of the square = x ∴ side of square= 72 m ∴ Perimeter, of square field = 72 x 4 m = 288 m Perimeter of rectangle = 288 m Let breadth of rectangular field (b) = x Then length (l) = 2x ∴ Perimeter = 2 (l + b) = 2 (2x + x) = 2 x 3x = 6x = 2 (2x + x) = 2 x 3x = 6x ∴ Length of rectangular field = 2x = 2 x 48 = 96 m and area = l x b = 96 x 48 m2 = 4608 m2 Question 12. Find the least square number, exactly divisible by each one of the numbers : (i) 6, 9,15 and 20 (ii) 8,12,15 and 20 Solution: LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180 =2 x 2 x 3 x 3 x 5 We see that after grouping the factors in pairs, 5 is left unpaired ∴ Least perfect square = 180 x 5 = 900 We see that after grouping the factors, factors 2, 3, 5 are left unpaired ∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600 Question 13. Find the square roots of 121 and 169 by the method of repeated subtraction. Solution: Question 14. Write the prime factorization of the following numbers and hence find their square roots. ^ (i) 7744 (ii) 9604 (iii) 5929 (iv) 7056 Solution: Factorization, we get: (i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11 Grouping the factors in pairs of equal factors, Question 15. The students of class VIII of a school donated Rs. 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. Solution: Total amount of donation = 2401 Let number of students in VIII = x ∴ Amount donoted by each student = Rs. x Question 16. A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement. Solution: Number of students = 6000 Students left out = 71 ∴ Students arranged in a field = 6000 – 71=5929 Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
Breaking News # What is Prime Factorization of 48? Prime factorization is a process of finding the prime factors of a number. The prime factors of a number are the prime numbers that divide the number exactly. Prime factorization of 48 is a process of finding out which prime numbers, when multiplied together, give the number 48. ## Finding the Prime Factors of 48 To find the prime factors of 48, we can use a trial and error method. First, we can divide 48 by 2, since it is a prime number. We get 24, which is still divisible by 2. We can divide it one more time and get 12. Now, 12 is not divisible by 2 anymore. So, the prime factors of 48 are 2, 2, and 3. ### Understanding the Prime Factorization of 48 The prime factorization of 48 can be written as 2 x 2 x 3. This means that when we multiply 2, 2, and 3 together, we get 48. This is how we can get the prime factors of a number. We can also use a factor tree to find the prime factors of a number. ## Exploring Other Uses of Prime Factorization of 48 The prime factorization of 48 can be used for many other mathematical problems. For example, we can use it to find the greatest common factor (GCF) of two or more numbers. By finding the prime factorization of each number and then finding the common factors, we can find the GCF. We can also use prime factorization to reduce fractions to their simplest form. ### Conclusion Prime factorization of 48 is a simple process of finding out which prime numbers, when multiplied together, give the number 48. By dividing 48 by 2, we can find the prime factors of 48, which are 2, 2, and 3. This prime factorization can be used for many other mathematical problems such as finding the GCF and reducing fractions.
Denominators of reduced fractions The sum of two given fractions reduced to their lowest terms is an integer. Their denominators are positive. What is the maximum possible difference between their denominators? [SOLVED] Andrew B. solved this puzzle: Let the fractions be $$\frac{a}{b}$$ and $$\frac{c}{d}$$, with $$b \leq d$$ without loss of generality, and let their sum be $$n$$. Then $$\frac{c}{d} = \frac{nb - a}{b}$$, so $$\frac{c}{d}$$ can be expressed as a fraction with $$b$$ as the denominator. Therefore, $$b - d = 0$$. Hagen von Eitzen solved this puzzle: The only possible difference is $$0$$. If $$\frac{a}{b} + \frac{c}{d} = n$$ and $$\gcd(a, b) = \gcd(c, d) = 1$$, then $$\frac{a}{b} = \frac{nd - c}{d}$$. If $$e$$ is a common divisor of $$nd - c$$ and $$d$$, then it is also a common divisor of $$c = nd - (nd -c)$$ and $$d$$. Hence $$\gcd(nd - c, d) = 1$$, and by uniqueness of representation of a fraction in lowest terms, $$b = d$$. quasi solved this puzzle: Suppose $$\frac{a}{b} + \frac{c}{d} = n$$, where $$\frac{a}{b}$$ and $$\frac{c}{d}$$ are reduced fractions, $$b \gt 0$$, $$d \gt 0$$ and $$n$$ is an integer. Then $$\frac{c}{d} = n - \frac{a}{b}$$. Therefore, $$\frac{c}{d} = \frac{bn - a}{b}$$. Since $$a$$ and $$b$$ are relatively prime, $$bn - a$$ and $$b$$ are also relatively prime. Thus, $$\frac{bn - a}{b}$$ is a reduced fraction, so $$\frac{c}{d} = \frac{bn - a}{b}$$. It implies that $$c = bn - a$$ and $$d = b$$. Therefore, $$d - b = 0$$. Thomas Nordhaus solved this puzzle: Let $$\frac{a}{b} + \frac{c}{d} = n$$ with integers $$a$$, $$b$$, $$c$$, $$d$$ and $$n$$, $$b \gt 0$$, $$d \gt 0$$, and $$\gcd(a, b) = gcd(c, d) = 1$$. Cross-multiplying by $$bd$$ results in $$ad + bc = nbd$$. This can be written as $$bc = d(nb - a)$$ or $$ad = b(nd - c)$$. Since the right-hand sides are divisble by d and b, respectively, so are the left-hand sides. Therefore $$d \mid bc$$ and $$b \mid ad$$. Since $$\gcd(d, c) = 1$$ and $$gcd(b,a) = 1$$, it follows that $$d \mid b$$ and $$b \mid d$$. This is possible only if $$b = d$$. Therefore, the maximum difference of the denominators is $$0$$. Corey Derochie solved this puzzle: As per the problem statement, let $$\frac{a}{b} + \frac{c}{d} = n$$, where $$a$$, $$b$$, $$c$$, $$d$$, and $$n$$ are integers. In particular, $$b$$ and $$d$$ are positive integers. $$a$$ and $$b$$ are coprime to each other because $$\frac{a}{b}$$ is in lowest terms. $$c$$ and $$d$$ are coprime to each other because $$\frac{c}{d}$$ is in lowest terms. From $$\frac{a}{b} + \frac{c}{d} = n$$, we get $$\label{eq1} b(nd - c) = ad.$$ $$b$$ does not equal $$0$$ because $$\frac{a}{b}$$ is defined. Suppose $$(nd - c) = 0$$. Then $$nd = c$$. Thus, $$d$$ divides $$c$$. But $$c$$ and $$d$$ are coprime to each other, so $$d = 1$$. Thus, $$c = n$$. Necessarily, $$a = 0$$ and $$b = 1$$, to satisfy the constraints and \eqref{eq1}. Therefore, $$b$$ divides $$d$$. If $$(nd - c) \ne 0$$, $$\gcd(a, b) = 1$$, so $$a$$ divides $$(nd - c)$$ by \eqref{eq1} and unique prime factorization. Thus, for some integer $$m$$, $$ma = n*d - c$$. Then $$bma = ad$$, by \eqref{eq1}. Cancelling $$a$$ on both sides we get $$bm = d$$. Thus, $$b$$ divides $$d$$. Similarly, we can show that $$d$$ divides $$b$$. Thus, $$b = d$$ and $$b - d = 0$$. Siddhesh Chaubal solved this puzzle: Let the two fractions in reduced form be $$\frac{n_1}{d_1}$$ and $$\frac{n_2}{d_2}$$. $$\newcommand{\lcm}{\operatorname{lcm}}$$ Let $$\lcm(d_1, d_2) = d$$, and let $$d = x_1 d_1 = x_2 d_2$$. The sum $$\frac{n_1}{d_1} + \frac{n_2}{d_2}$$ is an integer, i.e. $$\frac{n_1 x_1 + n_2 x_2}{d}$$ is an integer. Since $$x_1 \mid d$$ and $$d \mid (n_1 x_1 + n_2 x_2)$$, $$x_1 \mid (n_1 x_1 + n_2 x_2)$$. This implies that $$x_1 \mid n_2 x_2$$. $$d = x_1 d_1 = x_2 d_2$$ is the LCM of $$d_1$$ and $$d_2$$. If $$\gcd(x_1, x_2) = g \gt 1$$, then $$d_1 \mid \frac{d}{g}$$ and $$d_2 \mid \frac{d}{g}$$, so $$\lcm(d_1, d_2) \lt d$$. This is a contradiction, so $$\gcd(x_1, x_2) = 1$$. Therefore, $$x_1 \mid n_2$$. Also, since $$x_1 \mid x_2 d_2$$ and $$\gcd(x_1, x_2) = 1$$, $$x_1 \mid d_2$$. Since, $$\frac{n_2}{d_2}$$ is in the lowest form, $$x_1 = 1$$. By a similar argument, $$x_2 = 1$$. Therefore, $$d = d_1 = d_2$$. Thus, the answer is $$0$$. Credit This puzzle is based on:
Current Location > Math Formulas > Geometry > Volume of a Cuboid # Volume of a Cuboid In geometry, a cuboid is a solid shaped figure formed by six faces. There are two definitions for a cuboid. In the more general definition of a cuboid, the only supplementary condition is that each of these six faces is a quadrilateral. Otherwise, the word “cuboid” is sometimes used for referring a shape of this type in which each of the faces is a rectangle, and in which each pair of adjacent faces meets in a right angle. A cuboid with length l units, width w units and height h units has a volume of V cubic units given by: V = l × w × h Example 1: A jewellery box that has the shape of a rectangular prism, has a height of 13 cm, a length of 35 cm and a width of 22cm. Find the volume of the jewellery box? Solution: V = l × w × h V= 13 × 35 × 22 V= 10010 cm3 Example 2: Find the volume of a brick whose size is 30 cm by 25 cm by 10 cm. Solution: The volume of the brick is given by: V = l × w × h V = 30 × 25 × 10 V = 7500 cm3 So, the volume of the brick is 7500cm3. Example 3: Calculate the volume of a cuboid whose size is 8cm × 12cm × 6cm Solution: Volume of the cuboid is given by: V = l × w × h V = 8 × 12 × 6 V = 576 cm3 Example 4: Given that the dimension of a cuboidal shape beam is 10m in length, 60 cm in width and 25 cm in thickness. How much does Nick have to pay for it, if it costs \$250.00/ m3? Solution: Notice that all the measurements should be expressed in the same units (100 cm = 1 m). Volume of the beam = length x breadth x height (thickness here) V = 10 x (60/100) x (25/100) = 1.5 m3 The final price of the beam will be: 1.5 x 250 = \$375.00. Example 5: A goods wagon shaped in the form of cuboid of measure 60m × 40m × 30m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 cubic meters. Solution: The volume of one box = 0.8 m 3 The volume of the goods wagon is: 60 × 40 × 30 = 72000 m3 Number of boxes that can be stored in the goods wagon is therefore: 72000 / 0.8 = 90000 Hence the number of cuboidal boxes that can be stored in the goods wagon is 90,000. Example 6: Find the height of a cuboid whose volume is 300 cubic cm and the base area is 30cm. Solution: V = l × w × h Since the base area is defined as: l × w, the height is therefore: h = V / base area h = 300 / 30 h = 10 cm Example 7: A box of the size 60 cm × 40 cm × 30 cm. If the size of one chocolate bar is 24 cm × 12 cm × 4 cm each, how many bars can the box hold? Solution: The volume of the box = 60 × 40 × 30 = 72000 cm3 The volume of each bar = 24 × 12 × 4 = 1152 cm3 Therefore the number of chocolate bars that the box can hold is: 72000/1152 = 62 Example 8: The surface areas of the three coterminous faces of a cuboid are 6, 15 and 10 cm2 respectively. Find the volume of the cuboid. Solution: To begin with we need to determine the length, width and height of the cuboid. l = √(6) w = √(15) h = √(10) V = l × w × h V = √(6) × √(15) × √(10) V = 30 cm2 Online Volume Calculator, click on the link will open a new window.
# Let $$A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$ and (mI + nA)2 = A where m, n are positive real numbers and I is the identify matrix. What is (m + n) equal to? 1. 0 2. $$\frac{1}{2}$$ 3. 1 4. $$\frac{3}{2}$$ Option 4 : $$\frac{3}{2}$$ ## Detailed Solution Formula used: Multiplication matrix: $$\rm \begin{bmatrix} a & b\\ c & d \end{bmatrix} \times \begin{bmatrix} w & x\\ y & z \end{bmatrix} = \begin{bmatrix} aw + by & ax + bz\\ cw + dy & cx + dz \end{bmatrix}$$ Identity matrix: An identity matrix is a given square matrix of any order which contains on its main diagonal elements with a value of one, while the rest of the matrix elements are equal to zero. I = $$\rm \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ Calculation: $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ ⇒ $$I^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ $$A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$ ⇒ A2 = $$\begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}$$ According to question, (mI + nA)2 = A ⇒ m2I2 + n2A2 + 2mnA = A ⇒ m2I2 + n2A2 = A(1 - 2mn) ⇒ m2 $$\rm \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ + n2 $$\rm \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}$$ = (1 - 2mn)$$\rm \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$ $$\rm \begin{bmatrix} m^{2} & 0\\ 0 & m^{2} \end{bmatrix} + \begin{bmatrix} -4n^{2} & 0\\ 0 & -4n^{2} \end{bmatrix} = \begin{bmatrix} 0 & 2(1-2mn)\\ -2(1-2mn) & 0 \end{bmatrix}$$ On equating on both sides ⇒ m2 - 4n2 = 0 ⇒ m = 2n ----(1) ⇒ 2(1 - 2mn) = 0 ⇒ 1 - 2mn = 0 ⇒ mn = $$\rm \frac{1}{2}$$ From equation (1) (2n)n = $$\rm \frac{1}{2}$$ ⇒ n = $$\rm ± \frac{1}{2}$$ Taking positive sign When n = $$\rm \frac{1}{2}$$ then, m = 1 ∴ m + n = 1 + $$\rm \frac{1}{2}$$ = $$\rm \frac{3}{2}$$
# SSAT Middle Level Math : Numbers and Operations ## Example Questions ### Example Question #671 : Ssat Middle Level Quantitative (Math) Express this ratio in simplest form: Explanation: A ratio involving decimals can be simplified as follows: First, move the decimal point over a common number of places in each number so that both numbers become whole. In this case, it would be two places: (Note the addition of a zero at the end of the first number.) Now, rewrite as a fraction and divide both numbers by : The ratio, simplified, is . ### Example Question #12 : Numbers And Operations In Mrs. Jones' class, the ratio of boys to girls is 3:2. If there are 8 girls in the class, how many total students are in the class? Explanation: Using the ratio, we find that there are 12 boys in the class . Summing 12 and 8 gives the total number of 20 students. ### Example Question #13 : Numbers And Operations A 16-ounce bottle of Charlie's Fizzy Fizz Root Beer costs 89 cents. Give the price per ounce to the nearest tenth of a cent. Explanation: Divide 89 by 16: The soda costs about 5.6 cents per ounce. ### Example Question #11 : Numbers And Operations A batting average is defined as the ratio of hits to turns at bat. It is expressed as a decimal rounded to the nearest thousand. In 2013, Yadier Molina's batting average was 0.319. If he maintains this same batting average over the 2014 baseball season, how many turns at bat should he expect to need to achieve 50 hits? (Nearest whole number) Explanation: Since a batting average is a ratio of hits to turns at bat, Let be the number of turns at bat Molina needs to get 50 hits. Then: Solve for : Molina should need 157 turns at bat, ### Example Question #11 : Numbers And Operations Carlos owns a hamburger and hot-dog stand. When the fridge in the stand is full, there are 16 hamburgers and hot dogs total. If the ratio of hamburgers to hot dogs is always 5:3, how many hot dogs are in the cart when it is full? Explanation: When solving a ratio problem, first add each part of the ratio: So, one part in this ratio is equal to 8. The number of hamburgers and hot dogs in the fridge is currently only the amount it can hold. To figure out how much it can hold total, we can multiply by the denominator of the fraction since the numerator is 1. 64 is the total amount of hamburgers and hot dogs that can be held in the fridge. Divide 8 into 64 in order to figure out how many parts fit into this number. This number tells us how many times we need to multiply the ratio by in order to figure out how many hamburgers or hot dogs can be in the fridge at once. Since we are trying to figure out how many hot dogs are in the fridge when it is full, we use the second number, 3. We use the number 3 because, since the ratio is hamburgers to hotdogs, the second number represents the number of hot dogs. This gives you your answer. There are 24 hot dogs in the fridge when it is full. ### Example Question #672 : Ssat Middle Level Quantitative (Math) There are girls and boys in Miss Bailey's class. What is the ratio of girls to boys? Explanation: The ratio of girls to boys in Miss Bailey's class is , which can be reduced to when both the numerator and denominator are divided by . ### Example Question #11 : How To Find A Ratio In a library, there are children's books and young adult books. What is the ratio of young adult books to children's books? Explanation: The ratio of young adult books to children's books is to , or , which can be reduced to . ### Example Question #11 : Numbers And Operations There are male and female employees at a company. What is the ratio of females to males? Explanation: The ratio of females to males is to , or , which can be reduced to . ### Example Question #19 : Numbers And Operations One dollar is equal to about 116.76 Japanese yen. For how many yen should a tourist to Japan be able to exchange $2,500 at that rate (nearest whole yen)? Possible Answers: Correct answer: Explanation: One dollar is equal to 116.76 yen, so multiply$2,500 by this conversion rate to get yen. ### Example Question #12 : Numbers And Operations Sharon is having a birthday party. So far, she has invited twenty of her friends from school, nine of whom are girls. She wants to make the ratio of boys to girls at the party two to one. If she decides to add her two cousins, both girls, to the guest list, and no other girls, how many more boys does she need to invite?
# The Cross Product of Two Vectors There are situations in the study of mathematics, physics or engineering in which we are required to compute the cross product of two vectors. It can be used in mechanics, for example, to find the torque applied by a force, or in the field of computer graphics to calculate the surface normal for a polygon (i.e. the vector that is perpendicular to the surface of the polygon). The cross product is the result of multiplying the vectors together, and will result in a third vector that is perpendicular (i.e. at right angles, or orthogonal) to both of the original vectors. The cross product therefore has no meaning in a two-dimensional environment. In fact, it is only meaningful in a three-dimensional or a seven-dimensional environment (we will limit ourselves to a discussion of the cross product in a three-dimensional environment). Consider the illustration below, which shows two separate vectors in a three-dimensional space. Two vectors in a three-dimensional space The three-dimensional vectors AB and CD have fixed points of origin (i.e. they are effectively position vectors) and there is no single plane in which they both lie. We can of course move one of the vectors so that the two vectors are tail to tail, in which case there will exist one (and only one) plane common to both vectors. We will move vector AB so that its tail is coincident with that of vector CD. The illustration below shows the two vectors again, this time with a common point of origin. The grid (which lies in the x-y plane) and the point labels have been removed for clarity. The vectors are now tail to tail What we are really interested in here are the x, y and z components of the vectors, rather than the x, y and z coordinates of their end points. Finding the x, y and z component values is a simple task if we know the x, y and z coordinates of the start and end points for each vector All we need to do to find the x component of a vector, for example, is to subtract the x coordinate of its start point from that of its end point. Once we have the component values for each vector, we can we can simply treat them as free vectors that share a common point of origin and thus lie in the same plane. In the following graphic, we show only the component values, and have removed the x, y and z axes altogether. The component values for vectors a and b We said above that the cross product of two vectors is a new vector that is perpendicular to both of the original vectors, but that leaves us with a question. There are two directions that are perpendicular to both vectors. Which one is the right direction? The answer to that usually depends on the orientation of the coordinate system being used. Let's assume for the purposes of this discussion that we are dealing with a right-handed coordinate system (other possibilities do of course exist, but we will try and keep things relatively simple). The right-hand rule (illustrated below) can then be applied to determine the direction in which the cross product vector points. Note that the cross product of vectors a and b is written as a x b. Applying the right-hand rule As you can see from the illustration, if the first and second fingers of the right hand are more or less at right-angles to one another, with the thumb more or less at right-angles to both fingers, then the first finger points roughly in the direction of vector a, the second finger in the direction of vector b, and the thumb in the direction of a × b (i.e. the cross product vector). Note that the angle between vectors a and b can be any angle between zero and one hundred and eighty degrees. Below is a graphic showing vectors a and b, together with the cross product vector, which we have labelled vector c. Note that, because we are dealing with a three-dimensional environment, this is just one possible viewpoint. Keep in mind, however, that vector c is at right-angles to both vector a and vector b. Vectors a and b with their cross product, vector c The question you should probably be asking yourself at this point is how did we arrive at the x, y and z components for vector c shown in the illustration? Well, without going into lengthy explanations, we find the x, y and z components of vector c (i.e. the cross product vector) by using the x, y and z components of vectors a and b as follows: cx = aybz - azby = (1)(-2) - (1)(-1) = -1 cy = azbx - axbz = (1)(2) - (1)(-2) = 4 cz = axby - aybx = (1)(-1) - (1)(2) = -3 Note that since vector c is perpendicular to the plane in which vectors a and b both lie, it follows that if vectors a and b both lie in the x-y plane, their z component will be zero. Consequently, the x and y components of vector c will also be zero (i.e. vector c will be parallel to the z-axis). Note also that if either vector a or b is the zero vector, or if vectors a and b are parallel, their cross product will also be the zero vector. As long as vectors a and b do not lie in a plane defined by any two out of the three axes (x, y and z), then all of the components of vector c will be non-zero. We can also express the cross product of two vectors in matrix form. We do this by first creating a three-by-three matrix that contains the x, y and z components of vectors a and b. The first row of the matrix must consist of the orthonormal vectors x, y and z. An orthonormal vector is a unit vector (i.e. it has a magnitude of one) which has only one non-zero component. To illustrate what this means, the orthonormal vectors x, y and z can be represented using the following column matrices: x = 1 0 0 y = 0 1 0 z = 0 0 1 The three-by-three matrix is presented below. Vectors x, y and z are essentially used as a kind of placeholder to represent the value one in each column of the first row of the matrix. In some representations, you may see different characters used to represent the orthonormal vectors (e.g. i, j and k), and the characters may appear with a caret ("^") above them to signify the fact that they are orthonormal vectors. The second and third rows contain the x, y and z component values of vectors a and b respectively. x y z ax ay az bx by bz The cross product vector is obtained by finding the determinant of this matrix. If you are unfamiliar with matrices, you might want to look at the page on matrices in the Algebra section to see how the determinant of a three-by-three matrix is found. Below is the actual calculation for finding the determinant of the above matrix (i.e. the cross product of vectors a and b). Note that the determinant of a matrix is written in the same format as the original matrix, but instead of brackets, the columns are enclosed within vertical bars. c = a × b = x y z = x ay az - y ax az + z ax ay ax ay az by bz bx bz bx by bx by bz Substituting actual values, we get: c = a × b = x y z = x 1 1 - y 1 1 + z 1 1 1 1 1 -1 -2 2 -2 2 -1 2 -1 -2 The calculation then becomes: c = a × b = x ((1)(-2) - (-1)(1)) - y ((1)(-2) - (2)(1)) + z ((1)(-1) - (2)(1)) c = -x + 4y - 3z This gives us the x, y and z vector components for the cross product vector directly: c = (-1, 4, -3) Note that the magnitude of vector c (the cross product vector) is actually the same as the area of a parallelogram for which vectors a and b provide adjacent sides, as shown below. Vectors a and b can provide adjacent sides of a parallelogram The area of a parallelogram is obtained by multiplying the length of the base of the parallelogram by its height. In the above example, we have assumed that vector b will be the base of the parallelogram. The height of the parallelogram is obtained by multiplying the length of either of the sides adjacent to the base by the sine of any one of its internal angles. In the above example, we have used vector a and the sine of angle θ. The area of the parallelogram constructed using vectors a and b, and consequently the magnitude of vector c, is given by: \|c\| = \|a × b\| = \|a\|\|b\| sin (θ ) Of course, we need to obtain values for angle θ (the angle between vectors a and b), and the magnitudes of vectors a and b, before we can perform this calculation. Once you have these values, you don't need to understand how the trigonometric sine function works in order to obtain a valid answer, but you do need to know how to use the sine button on your calculator correctly. There is one other thing to be aware of before we move on which is that, unlike the dot product of two vectors, the cross product is not commutative. In other words, a × b produces a different result from b × a. In fact, the cross product b × a has exactly the same magnitude as the cross product a × b, but points in the opposite direction (b × a = -(a × b)). The cross product b × a points in the opposite direction to a × b At the moment, we are dealing with a few unknowns. We don't know the magnitudes of vectors a and b, we don't have a value for angle θ (which will be the minimum angle through which one of the vectors must be rotated in order to point in the same direction as the other vector), and we don't know the x, y and z components of the unit vector n. How do we find these missing values? We'll start by finding the magnitudes (i.e. lengths) of vectors a and b, which is relatively straightforward: \|a\| = √ax2 + ay2 + az2 = √1 + 1 + 1 = √3 = 1.732 \|b\| = √bx2 + by2 + bz2 = √4 + 1 + 4 = √9 = 3 To find angle θ, we can use the dot product of vectors a and b (if you are unfamiliar with the dot product of two vectors, have a look at the relevant page in this section). We can find the dot product of vectors a and b using the information we already have as follows: a · b = axbx + ayby + azbz = ((1)(2)) + ((1)(-1) + (1)(-2)) = 2 - 1 - 2 = -1 Since the cosine equation for the dot product of a and b is: a · b = \|a\|\|b\| cos (θ ) we can rearrange this equation to find θ: θ = cos-1 a · b \|a\|\|b\| Substituting actual values we get: θ = cos-1 -1 = cos-1 (-0.192) = 101.10° 1.732 × 3 Note that the sign of the dot product is important. If you get that wrong, you will get the supplement of angle θ (i.e. the difference between θ and one hundred and eighty degrees) rather than θ. It only remains to find the x, y and z components of unit vector n. Bearing in mind that we already have the x, y, and z components of vector c, this is fairly straightforward. The unit vector by definition has a length of one unit, so the total length of vector c would tell us the scalar value by which unit vector n would have to be multiplied in order to get vector c. We can therefore divide the x, y and z components of vector c by its length to get the x, y and z components of unit vector n. To find the length of vector c, we can use the following calculation: \|c\| = √cx2 + cy2 + cz2 = √1 + 16 + 9 = √26 = 5.099 The components of unit vector n are then found as follows: nx = -1 ÷ 5.099 = -0.196 ny = 4 ÷ 5.099 = 0.784 nz = -3 ÷ 5.099 = -0.588 We already have an alternative equation for the magnitude of vector c (see above): \|c\| = \|a × b\| = \|a\|\|b\| sin (θ ) Since the direction of vector c is given by the unit vector n, we can obtain an alternative equation for vector c (i.e. the cross product vector) simply by including the term n: c = a × b = \|a\|\|b\| sin (θ ) n If we now substitute actual values into our equation, we should be able to obtain the same set of vector components for vector c as we got from our original calculation using the x, y and z components of vectors a and b: c = 1.732 × 3 × sin (78.93°) n = 5.099 n c = 5.099 × (-0.196, 0.784, -0.588) = (-0.999, 3.998, -2.998) So, the cross product of vectors a and b in row vector form is (-0.999, 3.998, -1.998). Rounding these numbers up, we get (-1, 4, -2). These are the vector components cx, cy and cz that we arrived at previously. Note that because n is the unit vector (i.e. it has a magnitude of one unit), its sole purpose in the equation is to define the direction of the cross product vector. This should be whichever of the two directions perpendicular to both vector a and vector b was selected by using the right-hand rule. Obviously, in order to use this equation to obtain vector c, we would need to know the magnitudes of vectors a and b, the value of angle θ, and the values of the x, y and z components of vector n. The method actually used to find the cross product of two vectors will therefore very much depend upon what information we are given to start with.
Update all PDFs # Which Function? Alignments to Content Standards: F-IF.C.8 F-IF.C.8.a Which of the following could be the function of a real variable $x$ whose graph is shown below? Explain. $f_1(x)=(x+12)^2+4$ $f_5(x)=-4(x+2)(x+3)$ $f_2(x)=-(x-2)^2-1$ $f_6(x)=(x+4)(x-6)$ $f_3(x)=(x+18)^2-40$ $f_7(x)=(x-12)(-x+18)$ $f_4(x)=(x-12)^2-9$ $f_8(x)=(24-x)(40-x)$ ## IM Commentary The task addresses knowledge related to interpreting forms of functions derived by factoring or completing the square. It requires students to pay special attention to the information provided by the way the equation is represented as well as the sign of the leading coefficient, which is not written out explicitly, and then to connect this information to the important features of the graph. Students who have had plenty of experience re-writing quadratic expressions will be able to determine the sign of the leading coefficient without actually writing the quadratic expressions in $ax^2+bx+c$ form. ## Solution All of these are expressions for quadratic functions. Since quadratic functions have graphs that are parabolas and the given graph appears to be a parabola, the given expression meet a minimum criteria for consideration. The graph of $f_1(x)=(x+12)^2+4$ has a vertex of $(-12, 4)$ which is in the second quadrant, so it does not match the graph. The graph of $f_2(x)=-(x-2)^2-1$ has maximum rather than a minimum value at $x=2$ since the leading coefficient is negative (in other words, the graph opens downward), so it does not match the graph. The graph of $f_3(x)=(x+18)^2-40$ has a vertex of $(-18, -40)$ which is in the third quadrant, so it does not match the graph. The graph of $f_4(x)=(x-12)^2-9$ has a vertex of $(12, -9)$ which is in the fourth quadrant, and the leading coefficient is positive (so the graph would open upward) so this could describe the function whose graph is given. The graph of $f_5(x)=-4(x+2)(x+3)$ has $x$-intercepts of $(-2,0)$ and $(-3,0)$. Since the $x$-intercepts are both positive for the given graph, they do not match. The graph of $f_6(x)=(x+4)(x-6)$ has $x$-intercepts of $(-4,0)$ and $(6,0)$. The $x$-intercepts are both positive for the give graph, so they do not match. The graph of $f_7(x)=(x-12)(-x+18)$ has a leading coefficient that is negative and so has a maximum rather than a minimum value (at $x=15$) and thus cannot match the graph. The graph of $f_8(x)=(x-24)(x-40)$ has $x$-intercepts of $(24,0)$ and $(40,0)$. Since the $x$-intercepts are both positive for the graph and the leading coefficient is positive (so the graph would open upward), this could possibly be the equation for this graph. The functions $f_4$ and $f_8$ both have graphs of the approximate shape given, though we note that they certainly would not appear identical if plotted simultaneously. For example, the vertex of the graph of $f_4$ occurs at $x=12$ whereas that of $f_8$ occurs at $x=32$. #### Robert Hawke says: almost 5 years Cam - with this fix, I think the two solutions are a little more plausible, but I still think kids will have somewhat competing priorities when responding to this task. Really, you're asking them to look at essential features of the parabola, and how those essential features are reflected - or not reflected - in each expression. The issue is that something you (and I, and most math teachers) see as inessential are important to students. For example, if you look at the scale between the y-axis and the x-intercepts, those (with the recent edits) now line up better, but there's still the issue of the scale on the y-axis. If you narrow the task too much, you lose a lot of what makes it good (which it is, for all the reasons you listed above.) Perhaps this could be resolved simply by changing the framing of the question from "which of the following could be...?" to "which of the following could not be..." Framing this in the negative would take away students' opportunity to discover or creatively apply the strategy of eliminating the possibilities, but it would give you more of a window into the features they (rightly) might consider essential, such as the unstated scale of the axes, and would give students a reason to make these assumptions explicit. This could, in turn, lead to a more rich discussion about the various features of this parabola. I can see two other potential changes, if you don't like the change proposed above: 1. Change the scale so that only one expression works, and make the scale work exactly. 2. Explicitly call out the features of the parabola you want students to focus on, such as the x-intercepts. For what it's worth, I don't love this change, as I think it takes away the charm of this task, but it would make the task more clear for those who would otherwise focus on less salient aspects of this graph. #### Cam says: almost 5 years Hi Robert, Thanks for the feedback! Am I right in understanding that your principal objection is that the scale on the $x$ axis is different from the scale on the $y$ axis? I can't speak for the author of the task, but I was envisioning the point being made that frequently we look at graphs where the scales on the axes are different -- indeed, this is the principal view that students get when they attempt to graph the functions on their calculators! (On the other hand, the problem resolved by the recent changes where that even the scale on the x-axis alone was inconsistent, which was definitely bad). I also agree that it would be best, or at least in keeping with the sprit of the original problem, if we didn't overspecify the scales, to as to keep the focus on isolating the qualitiative features of parabolas, rather than just numerically plotting them. #### Robert Hawke says: almost 5 years Thanks for your reply. This definitely helps clarify the difference between the scales on the x- and y- axes. The issue still at play, however, is the precision of the scale on the x-axis. Though f8 has intercepts at 24 and 40 and f4 has intercepts at roughly the same spacing from zero, the spacing isn't exactly the same, and thus this couldn't actually be a graph of one AND the other...it's close, yes, but the two intercepts can't simultaneously satisfy two different ratios, in terms of distance from the y-axis. So this could be a graph of f4 but not f8, or of f8 but not f4. But, as graphed, it definitely could not be a graph of f4 and f8. One way around this is to measure the actual distance, get the ratio of x-intercepts what you want it to be, and create two functions with this exact ratio of roots. If you want to use (24-x)(40-x), these have a ratio of 40/24, or 5/3. So another concave-up parabola with these same roots could be created by f4 = (x-12)^2 - 9. This is a little nitpicky, for sure, but this dodges the issue of kids getting used to imprecise scales. #### Cam says: almost 5 years Aha! I concur completely (except for it being excessively nitpicky). I've made the change you suggested. Thanks again. #### Bill says: almost 5 years There's a problem with this task. None of the solutions work if you pay attention to the scale on the x-axis. Achievement First high school teachers. almost 5 years Fixed. Thanks.
# Factorial Recursion Algorithm In this tutorial we will learn to find the factorial of a number using recursion. ## What is recursion? In simple terms, when a function calls itself it is called a recursion. ## Factorial of n Factorial of any number n is denoted as n! and is equal to n! = 1 x 2 x 3 x ... x (n – 2) x (n – 1) x n ``````Factorial of 3 3! = 1 x 2 x 3 = 6 `````` ## Factorial Function using recursion ``````F(n) = 1 when n = 0 or 1 = F(n-1) when n > 1 `````` So, if the value of n is either 0 or 1 then the factorial returned is 1. If the value of n is greater than 1 then we call the function with (n - 1) value. #### Example Find 3! ``````We know, F(n) = 1 when n = 0 or 1 = n x F(n-1) when n > 1 So, F(3) = 3 x F(3-1) = 3 x F(2) and F(2) = 2 x F(2-1) = 2 x F(1) We know, F(1) = 1 Therefore, F(2) = 2 x F(1) = 2 x 1 = 2 Similarly, F(3) = 3 x F(2) = 3 x 2 = 6 `````` ## Factorial Pseudo Code ``````Fact(n) Begin if n == 0 or 1 then Return 1; else Return nCall Fact(n-1); endif End `````` ## Factorial code in C ``````/factorial declaration recursive and non-recursive / #include <stdio.h> //function declaration int fact(int n); int nonRecFact(int n); int main(){ //variable declaration int n, f; //input printf("Enter n: "); scanf("%d", &n); //recursive fact f = fact(n); printf("Recursive fact: %d\n", f); //non-recursive fact f = nonRecFact(n); printf("Non-Recursive fact: %d\n", f); return 0; } //function definition int fact(int n){ if(n == 0 \|\| n == 1) return 1; else return n fact(n-1); } int nonRecFact(int n){ int i, f = 1; for(i = 1; i <= n; i++) f *= i; return f; } ``````
Week 2-3 - Functions & Limits 1. Functions • Deﬁnition: A function f is a rule that assigns to each element x in a set D exactly one element, called $f(x)$, in a set E. • The set D is called the domain of the function. • The number $f(x)$ is the value of f at x and is read "f of x". • The range of f is the set of all possible values of $f(x)$ as x varies throughout the domain. • A symbol that represents an arbitrary number in the domain of a function f is called an independent variable. A symbol that represents a number in the range of f is called a dependent variable. • Four ways to represent a functions • verbally (by a description in words) • numerically (by a table of values) • visually (by a graph) • algebraically (by an explicit formula) • A catalog of essential functions • linear function • $f(x) = mx+b$ • polynomials • $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}$ • The domain of any polynomial is $\mathbb{R} = (-\infty, + \infty)$. • If the leading coefficient $a_{n}\ne 0$, then the degree of the polynomial is n. • n=2: quadratic function • n=3: cubic function • power function • $f(x) = x^a$ • a = n, where n is a positive integer • a = 1/n, where n is a positive integer. It's a root function. • a = -1: reciprocal function • rational function • A rational function f is a ratio of two polynomials: • $f(x)=\frac{P(x)}{Q(x)}$, $\{Q(x)\ne0\}$ • $f(x)=\frac{2x^{4}-x^{2}+1}{x^{2}-4}$, $\{x\|x \ne \pm 2\}$: • algebraic function • A function f is called an algebraic function if it can be constructed using algebraic operations (such as addition, subtraction, multiplication, division, and taking roots) • trigonometric function • $f(x)=\sin{x}$ • exponential function • $f(x)=b^x$ • law of exponential function: • $b^{x+y}=b^x+b^y$ • $b^{x-y}=\frac{b^x}{b^y}$ • $(b^{x})^{y}=b^{xy}$ • ${ab}^{x}=a^{x}b^{x}$ • logarithmic function • $f(x)=\log_{b}{x}$ 1.1. Combinations of Functions • $(f+g)(x)=f(x)+g(x)$ • $(f-g)(x)=f(x)-g(x)$ • $(fg)(x)=f(x)g(x)$, the domain of fg is $A \cap B$ • $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$, the domain of f/g is $\{x \in A \cap B\ \|\ g(x) \ne 0\}$. • $(f \circ g)(x) = f(g(x))$ • composition (or composite) of f and g, denoted by $f \circ g$ (“f circle g”). 1.2. Inverse of Functions • Definition: • If a function maps every input to exactly one output, an inverse of that function maps every “output” to exactly one “input.” • denoted by$f^{-1}$ , and read “f inverse”. • to function $N=f(t)$, the inverse function will be $t=f^{-1}(N)$. one-to-one functions • A function is one-to-one if for every value in the range(f(x)), there is exactly one value in the domain(x). • domain of $f^{-1}$ = range of $f$ • range of $f^{-1}$ = domain of $f$ • for example: $f(x)=x^3$ is a one-to-one function, $f(x)=x^2$ is not. 2. Limits • Definition: • $\displaystyle\lim_{x \to a}{f(x)}=L$ • the limit of f(x), as x approaches a, equals L • if we can make the values of f(x) arbitrarily close to L (as close to L as we like) by restricting x to be sufficiently close to a (on either side of a) but NOT equal to a.(This means that in finding the limit of f(x) as x approaches a, we never consider x = a.) 2.1. Limit Laws • Suppose that c is a constant and the limits $\displaystyle\lim_{x \to a}{f(x)}$ and $\displaystyle\lim_{x \to a}{g(x)}$ exist, Then: 1. $\displaystyle\lim_{x \to a}{[f(x)+g(x)]}=\lim_{x \to a}{f(x)}+\lim_{x \to a}{g(x)}$ 2. $\displaystyle\lim_{x \to a}{[f(x)-g(x)]}=\lim_{x \to a}{f(x)}-\lim_{x \to a}{g(x)}$ 3. $\displaystyle\lim_{x \to a}{cf(x)}=c\lim_{x \to a}{f(x)}$ 4. $\displaystyle\lim_{x \to a}{[f(x)g(x)]}=\lim_{x \to a}{f(x)}\cdot \lim_{x \to a}{g(x)}$ 5. $\displaystyle\lim_{x \to a}{\frac{f(x)}{g(x)}}=\frac{\displaystyle\lim_{x \to a}{f(x)}}{\displaystyle\lim_{x \to a}{g(x)}}$, if $\displaystyle\lim_{x \to a}{g(x)} \ne 0$ • These five laws can be stated verbally as follows: 1. Sum Law : The limit of a sum is the sum of the limits. 2. Difference Law : The limit of a difference is the difference of the limits. 3. Constant Multiple Law : The limit of a constant times a function is the constant times the limit of the function. 4. Product Law : The limit of a product is the product of the limits. 5. Quotient Law : The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0). 2.2. Squeeze Theorem • if $g(x) \le f(x) \le h(x)$ and $\displaystyle\lim_{x\to{a}}g(x) = \lim_{x\to{a}}h(x) = L$, then $\displaystyle\lim_{x\to{a}}f(x)=L$ • Sample: to prove $\displaystyle\lim_{x\to{0}}\frac{\sin(x)}{x}=1$ • we know: • $\displaystyle\lim_{x\to{0}}\cos(x)=1=\lim_{x\to{0}}1$ • $\cos(x) \le \frac{\sin(x)}{x} \le 1$ • then: • $\displaystyle\lim_{x\to{0}}\cos(x) \le \lim_{x\to{0}}\frac{\sin(x)}{x} \le 1$ • so: • $\displaystyle\lim_{x\to{0}}\frac{\sin(x)}{x}=1$ 2.3. Continuity • Definition 1: • A function f is continuous at a number a, if $\displaystyle\lim_{x \to a}f(x)=f(a)$ • Notice that this Definition implicitly requires three things if f is continuous at a: 1. f(a) is defined (that is, a is in the domain of f) 2. $\displaystyle\lim_{x \to a}f(x)$ exists 3. $\displaystyle\lim_{x \to a}f(x) = f(a)$ • Definition 2: • A function f is continuous from the right at a number a if $\displaystyle\lim_{x \to a^{+}}f(x) = f(a)$, and f is continuous from the left at a if $\displaystyle\lim_{x \to a^{-}}f(x) = f(a)$ • Definition 3: • A function f is continuous on the interval (a, b), if for all points c so that a < c < b, f(x) is continuous at c. • close intervals: • To say "f(x) is continuous on the interval [a, b]", means: • f(x) is continuous on the interval (a, b) • $\displaystyle\lim_{x \to a^{+}}f(x) = f(a)$ • $\displaystyle\lim_{x \to b^{-}}f(x) = f(b)$ The Intermediate Value Theorem • Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where $f(a) \ne f(b)$. Then there exists a number c in [a, b] such that $f(c) = N$. • For example: how to approximate root two? • use function $f(x)=x^2-2$ • when $x=\sqrt{2}$, this function equals 0, • so i just need to look for a positive value, that I can plug into this function to make it equals zero. • We know that f(1)=-1<0 and f(2)=2>0. So base on The Intermediate Value Theorem, there must be a value in domain (1, 2) that exist c such that f(c) = 0. • to continue calculating f(1.5)=0.25>0, we got $c \in (1, 1.5)$, then $c \in (1.4, 1.5)$ ... then we are getting closer and closer to $\sqrt{2}$. Fixed Point • Definition • A fixed point of a function f is a number c in its domain such that f(c) = c. (The function doesn’t move c; it stays fixed.) • if f(x) continuous on [0,1], and $0 \le f(x) \le 1$, then there is an x in domain [0, 1], exist f(x) = x. • To Prove: • Assumption: g(x) = f(x) - x, so g(x) is continuous • g(0) = f(0) - 0 >= 0 • g(1) = f(1) - 1 <= 1 • Base on the IVT(Intermediate Value Theorem), there must be an x such that g(x) = 0, which is f(x) = x. 2.4. Infinity • Definition • $\displaystyle\lim_{x \to a}f(x) = \infty$ means that f(x) is as large as you like provides x is close enough to a. • $\displaystyle\lim_{x \to \pi/2}\tan(x) = \infty$ • $\displaystyle\lim_{x \to \infty}f(x) = L$ means that f(x) is close enough to L provided x is large enough. • $\displaystyle\lim_{x \to \infty}\tan^{-1}(x) = \frac{\pi}{2}$ • if $\displaystyle\lim_{x \to \infty}f(x) = L$ or $\displaystyle\lim_{x \to -\infty}f(x) = L$, then the line y = L is called a horizontal asymptote of the curve y = f(x) : • Potential Infinity vs Actual Infinity (from Wikipedia) • Actual Infinity is the idea that numbers, or some other type of mathematical object, can form an actual, completed totality; • Such as the set of all natural numbers, an infinite sequence of rational numbers. • Potential Infinity is a non-terminating process (such as "add 1 to the previous number") produces an unending "infinite" sequence of results, but each individual result is finite and is achieved in a finite number of steps. • Precise Definitions • $\displaystyle\lim_{x \to a}f(x) = L$ means: • for all $\epsilon > 0$, there is $\delta>0$, • so that if $0 < \|x - a\| < \delta$ ($x \ne a$ and x in within $\delta$ of a), then $\|f(x)-L\| < \epsilon$ (f(x) is within $\epsilon$ of L). • \|x - a\| is the distance from x to a and \|f(x) - L\| is the distance from f(x) to L. • so $\displaystyle\lim_{x \to a}f(x) = L$ means that the distance between f(x) and L can be made arbitrarily small by requiring that the distance from x to a be sufficiently small (but not 0). • For Example: $\displaystyle\lim_{x \to 2}x^2 = 4$ • Let's say $\epsilon = 0.1$, that means $\|f(x)-4\| < 0.1$, 3.9 < f(x) < 4.1, • Base on the definition, there should be a $\delta$, that $2 - \delta < x < 2 + \delta$ to satisfy the demand. • Try $\delta = 0.01$. We got 1.99 < x < 2.01, 3.9601 < x^2 < 4.0401, 3.9 < f(x) < 4.1 which suit the demand. • Another Example: $\displaystyle\lim_{x \to 10}2x = 20$ • Let $\epsilon > 0, \delta = \epsilon / 2$ • if $0<\|x-10\|<\delta$, then, • $0<2\|x-10\|<2\delta=\epsilon$, and so, • $0<\|2x-20\|<\epsilon$
written notes on factoring SPECIAL PRODUCTS Word List -- multiplication and factoring Factor: 3x3y + 6x2 + 9x The Hardest Factoring - Monomial Factor Factor: x2 + 6x + 5 Factoring Trinomials -- UnFOIL Sample Problems: 1 2 3 4 animation Factor: 15x2 + 11x - 12 Factoring Trinomials when the Leading Coefficient Is Not One animation The Hardest Factoring -- Monomial Factoring You must end with as many terms in the parenthises as you started with in the original unfactored expression. 1st. Find the common factors. 2nd. Copy the common factors "out front" -- outside the parenthises. 3rd. Copy what's left as terms in the parenthesis. 4th. Check by multiplying. Factoring Trinomials - Written Instructions Factoring Trinomials (animation) With long notes: Factor: x2 - 4x - 21 1st: Arrange the trinomial in descending order. x2 - 4x - 21 x2 - 4x - 21 (x )(x ) 3rd: List all factor pairs of the last (constant) term. 21 1 , 21 3 , 7 4th: Add or subtract to get the middle term. 21 1 , 21 + 3 , - 7 will produce the - 4. Since: (+)(-) = (-) (-)(+) = (-) produces a negative and (+)(+) = (+) (-)(-) = (+) produces a positive, The - 21 is created by multiplying one negative & one positive. This rule always works. If the constant term is + the factors have the same sign, so ADD the factors to find the middle term. If the constant term is - the factors have different signs, so SUBTRACT the factors to find the middle term. In short, Check the sign of the middle (linear) term If + , ADD to find the middle term. If - , SUBTRACT to find the middle term. 5th: The "larger" number always takes this (linear term) sign. x2 - 4x - 21 The - 4x is negative, so the 7 gets a negative sign. (x + 3)(x - 7) With short notes: Factor: x2 - 4x - 21 1st: Arrange the trinomial in descending order. x2 - 4x - 21 x2 - 4x - 21 (x )(x ) 3rd: List all factor pairs of the last (constant) term. 21 1 , 21 3 , 7 4th: Add or subtract to get the middle term. If + , ADD to find the middle term. If - , SUBTRACT to find the middle term. 21 1 , 21 + 3 , - 7 will produce the - 4. 5th: The "larger" number always takes this (linear term) sign. x2 - 4x - 21 The - 4x is negative, so the 7 gets a negative sign. (x + 3)(x - 7) Factoring Trinomials when The Leading Coefficient Is Not One Factoring Trinomials when The Leading Coefficient Is Not One (animation)
If the solve the problem Question: $\int \sin x \cos 2 x \sin 3 x d x$ Solution: We know $2 \sin A \cos B=\sin (A+B)+\sin (A-B)$ $\therefore \sin 3 x \cos 2 x=\frac{\sin 5 x+\sin x}{2}$ $\therefore$ The given equation becomes $\Rightarrow \int \frac{1}{2}(\sin 5 x-\sin x) \sin x d x$ $\Rightarrow \int \frac{1}{2}\left(\sin 5 x \sin x d x-\sin ^{2} x d x\right)$ We know $2 \sin A \sin B=\cos (A-B)-\cos (A+B)$ $\therefore \sin 5 x \sin x=\frac{\cos 4 x-\cos 6 x}{2}$ Also $\sin ^{2} x=\frac{1-\cos 2 x}{2}$ $\therefore$ Above equation can be written as $\Rightarrow \int \frac{1}{2}\left(\frac{1}{2}(\cos 4 x-\cos 6 x) d x-\frac{1}{2}(1-\cos 2 x) d x\right)$ $\Rightarrow \frac{1}{4} \int \cos 4 x d x-\int \cos 6 x d x-\int d x+\int \cos 2 x d x$ We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$ $\Rightarrow \frac{1}{4}\left(\frac{1}{4} \sin 4 x-\frac{1}{6} \sin 6 x-x+\frac{1}{2} \sin 2 x+c\right)$ $\Rightarrow \frac{1}{4}\left(\frac{3 \sin 4 x-2 \sin 6 x-12+6 \sin 2 x}{12}+C\right)$ $\Rightarrow \frac{3 \sin 4 x-2 \sin 6 x-12+6 \sin 2 x}{48}+C$ NOTE: – Whenever you are solving integral questions having trigonometric functions in the product then the first thing that should be done is convert them in the form of addition or subtraction.
# Solving percent problems In the last lesson, you worked on identifying the components of percent problems. Today, let’s look at ways we can manipulate the formula for percent to find other components. This is the formula you worked with a couple lessons ago. If you know the PART and the WHOLE, you use this formula to find the PERCENT. PERCENT = PART / WHOLE If you know the WHOLE and the PERCENT, you can use this format to find the PART: PART = WHOLE x PERCENT If you know the PERCENT and the PART, you can use this format to find the WHOLE: WHOLE = PART / PERCENT Let’s look at our examples from the last lesson and find the missing component. Example 1 Chris ate 7 pieces of pizza. There were 8 pieces of pizza to start. What percentage of the pizza did they eat? You are given the PART (7 pieces) and the WHOLE (8 pieces). The unknown, the component you are asked to find, is the PERCENT. You will use PERCENT = PART / WHOLE 7/8 = .875 = 87.5% of the pizza Example 2 Alyssa’s reading assignment for this week is to read 200 pages. She is 65% of the way through the assignment. How many pages has she read? You are given the WHOLE (200 pages) and the PERCENT (65%). You are asked to find the PART. PART = WHOLE x PERCENT 200 x .65 = 130 pages Example 3 Daniel has read 6 books so far this year. This is 20% of the goal he set for himself this year. What is the goal he set for himself? You are given the PART (6 books) and the PERCENT (20%). You are asked for find the WHOLE. WHOLE = PART / PERCENT 6/.20 = 30 books
# Absolute Value Theorem When trying to prove the inequality $$\|a +b\| \leq \|a\| + \|b\| \text{, for any real numbers a and b}$$ I manage to use the absolute value definition to get to following inequality: $$-\big(\|a\|+\|b\|\big) \leq a + b \leq \|a\| + \|b\|$$ However, the text book leaps foward and states that: $$\Big\{-\big(\|a\|+\|b\|\big) \leq a + b \leq \|a\| + \|b\|\Big\} \leftrightarrow \Big\{ \|a + b\| \leq \|a\| + \|b\|\Big\}$$ • It seems to me that you need $a + b \leq \big\|a + b\big\|$, if $x \leq y$ and $z \leq x$, then $z \leq y$. – Jared Jan 18 '15 at 0:16 • how do you write $-5 \le x \le 5$ using absolute vale symbol? – abel Jan 18 '15 at 0:18 • There are two cases: $a + b \geq 0$ in which case $a + b = \big\|a + b\big\|$ or $a + b < 0$ in which case $a + b = -\big\|a + b\big\|$ and thus $a + b < \big\|a + b\big\|$ (when the sum is negative). Therefore either $a + b$ equals $\big\|a + b\big\|$ or $a + b$ is less than $\big\|a + b\big\|$. – Jared Jan 18 '15 at 0:31 The definition of absolute value is: $$\|x\| = \begin{cases} x & \text{if x\geq 0} \\ -x & \text{if x<0.} \end{cases}$$ So assume $a+b\geq 0$. Then $\|a+b\| = a+b\leq \|a\|+\|b\|$ by the inequality you've shown. If $a+b<0$, then $a+b = -\|a+b\|$, so $-(\|a\|+\|b\|)\leq -\|a+b\| \Longleftrightarrow \|a+b\|\leq \|a\|+\|b\|$ (the inequality flips since we divide by $-1$). so you have $-\|a\| - \|b\| \le a + b \le \|a\| + \|b\|$ that is $(a+b)$ in magnitude is less or equal to the nonnegative quantity $\|a\| +\|b\|$ writing this using absolute value notation is $$\|a +b \| \le \|a\| +\|b\|$$ called the triangle inequality.
## Friday, November 1, 2013 ### Improper Fractions and Mixed Numbers How can we help kids really understand the conversions between improper fractions and mixed numbers? Using manipulatives and/or pictures is essential for initial conceptualization. Step 1: Representing Mixed Numbers - In my experience, students are not ready to convert until they can look at a picture and name the mixed number and improper fraction represented. Step 2: Converting Improper Fractions to Mixed Numbers - This part is easy if you've already worked on fractions as division. Your students already know, for example, that 5/4 equals five divided by four. The only piece that's missing is representing the remainder as a fraction (in this case, 1/4, or one out of four left). Warning: Some students will insist that the denominator should be the number of possible pieces instead of the number of pieces in one whole. For example, in the problem above, some of my students wanted to write 5/8. To counter this, I repeat (and repeat and repeat), "The denominator is the number of pieces in ONE WHOLE!" Additionally, I tell them that the denominator is like the name of these pieces. (For example, if I bring apples to class and cut each into four pieces, giving each student one piece, each student is getting one fourth of an apple, not one twenty-eighth.) After lots of practice and discussion together, they're ready to roll. For my class, I use worksheets found on the Fractions page of Common Core Sheets. Scroll the bottom of the page, and under Converting Fractions, you will find ten versions of Improper to Mixed Number worksheets. Awesome! Now we have sufficient versions for introduction, practice, tutoring, review before the test, and more review as the year goes on. Step 3: Converting Mixed Numbers to Improper Fractions - More pictures, please! I show students how to cut the wholes into the same number of pieces as the denominator of the fractional part. We add the pieces in the wholes with the fractional piece for the numerator, and the denominator stays the same. After working with a number of pictures, students can see that the number of pieces in each whole (denominator) times the number of wholes plus the numerator equals the numerator of the improper fraction. And the denominator stays the same (because the denominator is always the number that each whole is cut into). Once again, I turn to the Fractions page of Common Core Sheets, and once again I find ten lovely versions of worksheets for practice. To locate them, just click on Mixed Number to Improper under Converting Fractions (near the bottom of the page).
# How do you solve -4(3x-2)=-32 using the distributive property? May 28, 2018 $x = 3 \frac{1}{3}$ #### Explanation: We will use the distributive property (shown below) to simplify $- 4 \left(3 x - 2\right)$: Following this image, we know that: $\textcolor{b l u e}{- 4 \left(3 x - 2\right) = \left(- 4 \cdot 3 x\right) + \left(- 4 \cdot - 2\right) = - 12 x + 8}$ Put that back into the equation: $- 12 x + 8 = - 32$ Now subtract $\textcolor{b l u e}{8}$ from both sides of the equation: $- 12 x + 8 \quad \textcolor{b l u e}{- \quad 8} = - 32 \quad \textcolor{b l u e}{- \quad 8}$ $- 12 x = - 40$ Next, divide both sides by $\textcolor{b l u e}{- 12}$: $\frac{- 12 x}{\textcolor{b l u e}{- 12}} = \frac{- 40}{\textcolor{b l u e}{- 12}}$ $x = 3 \frac{1}{3}$ Hope this helps!
How do you solve the system -4x-5y-z=18, -2x-5y-2z=12, and -2x+5y+2z=4? Mar 18, 2017 $\left\{x = - 4 , y = 0 , z = - 2\right\}$ Explanation: $- 4 x - 5 y - z = 18 \text{ , } \left(1\right)$ $- 2 x - 5 y - 2 z = 12 \text{ , } \left(2\right)$ $- 2 x + 5 y + 2 z = 4 \text{ , } \left(3\right)$ $\text{let us sum the equations numbered as (2) and (3) so as to eliminate}$ $\text{the terms named 'y' and 'z'} .$ $- 2 x \cancel{- 5 y} \cancel{- 2 z} - 2 x \cancel{+ 5 y} \cancel{+ 2 z} = 12 + 4$ $\text{rearrange the equation above.}$ $- 2 x - 2 x = 12 + 4$ $- 4 x = 16 \text{ , } x = - 4$ $\text{let us write as x=4 in the equation numbered as (1)}$ $- 4 \left(- 4\right) - 5 y - z = 18$ $16 - 5 y - z = 18$ $- 5 y - z = 18 - 16$ $- 5 y - z = 2 \text{ , } \left(4\right)$ $\text{let us write as x=4 in the equation numbered as (2)}$ $- 2 \left(- 4\right) - 5 y - 2 z = 12$ $8 - 5 y - 2 z = 12$ $- 5 y - 2 z = 12 - 8$ $- 5 y - 2 z = 4 \text{ , } \left(5\right)$ $\text{now , subtract (5) from (4) }$ $- 5 y - z - \left(- 5 y - 2 z\right) = 2 - 4$ $\cancel{- 5 y} - z \cancel{+ 5 y} + 2 z = - 2$ $- z + 2 z = - 2$ z=-2" $\text{now use (4)}$ $- 5 y - \left(- 2\right) = 2$ $- 5 y + 2 = 2$ $- 5 y = 0$ $y = 0$
You are on page 1of 8 NUMBER SYSTEMS INTRODUCTION Some Properties of Prime Numbers The chapter of number systems is amongst the most impor- tant chapters in the whole of mathematics syllabus for the n The lowest prime number is 2. CAT examination. The students are advised to go through n 2 is also the only even prime number. this chapter with utmost care, understanding each and every n The lowest odd prime number is 3. question type on this topic. Over the last decade, approxi- n The remainder when a prime number p ≥ 5 is divided mately 10–15 questions out of 40–50 questions asked in by 6 is 1 or 5. However, if a number on being divided Mathematics have been taken from this chapter. It would be by 6 gives a remainder of 1 or 5 the number need not a good idea to first go through the basic definitions of all be prime. types of numbers (something I have found to be surprisingly n The remainder of the division of the square of a very less known about). The student is also advised to go prime number p ≥ 5 divided by 24 is 1. through the solutions of the various questions illustrated in n For prime numbers p > 3, p2 – 1 is divisible by 24. this chapter. Besides, while solving this chapter, try to maximise n Prime Numbers between 1 to 100 are: 2, 3, 5, 7, 11, your learning experience with every problem that you solve. 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. DEFINITIONS n Prime Numbers between 100 to 200 are: 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, Natural Numbers These are the numbers (1, 2, 3 etc.) 163, 167, 173, 179, 181, 191, 193, 197, 199. that are used for counting. In other words, all positive in- n If a and b are any two odd primes then a2 – b2 is tegers are natural numbers. composite. Also, a2 + b2 is composite. There are infinite natural numbers and the number 1 is n The remainder of the division of the square of a the least natural number. prime number p ≥ 5 divided by 12 is 1. Examples of natural numbers: 1, 2, 4, 8, 32, 23, 4321 and so on. The following numbers are examples of numbers that To Check Whether a Number is are not natural: –2, –31, 2.38, 0 and so on. Prime or Not Based on divisibility, there could be two types of natural numbers: Prime and Composite. To check whether a number N is prime, adopt the follow- Prime Numbers A natural number larger than unity is ing process. a prime number if it does not have other divisors except for (a) Take the square root of the number. itself and unity. (b) Round of the square root to the next highest integer. Note: Unity (i.e. 1) is not a prime number Call this number z. 48 How to Prepare for Quantitative Aptitude for the CAT ## Illustration 4 Task for student: Solve through the alligation formula approach and through the weighted average approach to get 4 kgs of rice at Rs. 5 per kg is mixed with 8 kg of rice at the solution. Notice the amount of time required in doing Rs. 6 per kg. Find the average price of the mixture. the same. ## Solution: Note: The cross method becomes quite cumbersome in this case since this method results in the formula being 5 6 written. Hence, there seems to be no logic in using the cross method in this case. The above problems are quite effectively dealt by using Aw the straight line approach, which is explained below. 4 : 8 The Straight Line Approach = (6 – Aw) : (Aw – 5) As we have seen, the cross method becomes quite cumber- ﬁ (6 – Aw)/(Aw – 5) = 4/8 Æ12 – 2 Aw = Aw – 5 some in Case 2 and Case 3. We will now proceed to modify 3 Aw = 17 the cross method so that the question can be solved graphi- \ Aw = 5.66 Rs./kg. (Answer) cally in all the three cases. Consider the following diagram, which results from Task for student: Solve through the alligation formula closing the cross like a pair of scissors. Then the positions approach and through the weighted average approach to get of A1, A2, Aw, n1 and n2 are as shown. the solution. Notice, the amount of time required in doing the same. A1 A2 Aw Note: The cross method becomes quite cumbersome in this case, as this method results in the formula being written. n1 n2 ## Hence, there seems to be no logic in using the cross method Visualise this as a fragment of the number line with in this case. points A1, Aw and A2 in that order from left to right. Case 3: A1, Aw, n1 and n2 are known; A2 is unknown. Then, (a) n2 is responsible for the distance between A1 and Aw or n2 corresponds to Aw – A1 Illustration 5 (b) n1 is responsible for the distance between Aw and A2. or n1 corresponds to A2 – Aw 5 kg of rice at Rs. 6 per kg is mixed with 4 kg of rice to get (c) (n1 + n2) is responsible for the distance between A1 a mixture costing Rs. 7 per kg. Find the price of the costlier and A2. or (n1 + n2) corresponds to A2 – A1. rice. The processes for the 3 cases illustrated above can then Solution: Using the cross method: be illustrated below: 6 x Illustration 6 7 On mixing two classes of students having average marks 25 and 40 respectively, the overall average obtained is 30 5 : 4 marks. Find = (x – 7) : 1 (a) the ratio in which the classes were mixed. \ (x – 7)/1 = 5/4 Æ 4x – 28 = 5 (b) the number of students in the first class if the second \ x = Rs. 8.25. class had 30 students. Chapter 3: Alligations 49 25 30 40 6 x 7 5 4 ## 4 corresponds to 7 – 6 and 5 corresponds to x – 7. n1 = 40 – 30 n2 = 30 – 25 The thought process should go like: 4Æ 1 Hence, ratio is 2 : 1, and the second class has 60 students. \ 5 Æ 1.25 Case 2 : A1, A2, n1 and n2 are known; Aw is unknown. Hence, x – 7 = 1.25 and x = 8.25 Illustration 7 SOME TYPICAL SITUATIONS WHERE 4 kg of rice at Rs. 5 per kg is mixed with 8 kg of rice at ALLIGATIONS CAN BE USED Rs. 6 per kg. Find the average price of the mixture. Given below are typical alligation situations, which students Solution: should be able to recognize. This will help them improve 5 Aw 6 upon the time required in solving questions. Although in this chapter we have illustrated problems based on alliga- 4 8 tion at level 1 only, alligation is used in more complex is the same as problems where the weighted average is an intermediate step in the solution process. 5 Aw 6 The following situations should help the student identify alligation problems better as well as spot the way A1, A2, n1 1 2 and n2 and Aw are mentioned in a problem. Then, by unitary method: In each of the following problems the following magni- tudes represent these variables: n1 + n2 corresponds to A2 – A1 Æ 1 + 2 corresponds to 6 – 5 A1 = 20, A2 = 30, n1 = 40, n2 = 60 That is, 3 corresponds to 1 Each of these problems will yield an answer of 26 as the value of Aw. ( A2 - A1 ) ¥ n2 \ n2 will correspond to (n1 + n2 ) 1. A man buys 40 kg of rice at Rs. 20/kg and 60 kg of rice at Rs. 30 kg. Find his average price. (26/kg) In this case (1/3) ¥ 2 = 0.66. 2. Pradeep mixes two mixtures of milk and water. He Hence, the required answer is 5.66. mixes 40 litres of the first containing 20% water and Note: In this case, the problem associated with the cross 60 litres of the second containing 30% water. Find the method is overcome and the solution becomes graphical. percentage of water in the final mixture. (26%) Case 3: A1, Aw, n1 and n2 are known; A2 is unknown. 3. Two classes are combined to form a larger class. The first class having 40 students scored an average of 20 marks on a test while the second having 60 students Illustration 8 scored an average of 30 marks on the same test. What was the average score of the combined class on the 5 kg of rice at Rs. 6 per kg is mixed with 4 kg of rice to test. (26 marks) get a mixture costing Rs. 7 per kg. Find the price of the 4. A trader earns a profit of 20% on 40% of his goods costlier rice. sold, while he earns a profit of 30% on 60% of his Chapter 4: Percentages 59 Hence, the net percentage change between the values of the denominator. If the numerator dominates the change, the two products is 34.81%. This process will be much the ratio will change in the direction of the numerator. faster than any other process of calculating the percentage However, the magnitude of the change will depend on change (provided you have developed your skill at calculat- both the numerator and the denominator’s change. ing percentage values through percentage rule) A) Numerator and its effect on the ratio: The The process is found useful for calculating the net per- numerator, as we know, has a direct effect on the magnitude centage change in a product x ¥ y when both x and y change. of change in the ratio. This direct relationship holds true not only for the direction but also for the magnitude of the Use of Percentage Change Graphic percentage change in the ratio vis a vis the percentage for Product Constancy/Inverse change in the numerator. Proportionality That is, a 20% increase in the numerator increases the ratio by 20% while a 10% decrease in the numerator This usage requires just a minor adjustment in the process. reduces the ratio by 10%. In general, we can say that an x% The situation for this usage will be when the percentage increase/decrease in the numerator increases/decreases the change in one part of the product is provided and we have to ratio by x%. find the corresponding percentage change in the other part of B) Denominator and its effect on the ratio: A little the product so as to maintain the value of the overall product. concentrated thought will give you a clear picture of the This is best illustrated through an example: direction of effect the denominator’s change has on the ratio. It can be stated as: Increase by 22% on 100 100  = + 22 (a) A decrease in the denominator increases the ratio. (b) An increase in the denominator reduces the ratio. 122  → 100 decrease needed to get back to 100 = − 22 These being obvious, we can further couple this with the Table 4.1 given above to understand the magnitude of the Decrease required to keep product constant = (–22/122) percentage change in the ratio due to a percentage change = –18.03 % approx (calculated by percentage rule) in the denominator. Effect of a Change in B o t h Numerator and C) Calculation of the combined effect: The combined Denominator on the Ratio There can be two broad effect then that the simultaneous percentage changes in the cases: numerator and the denominator will have on the percentage Case 1: Both numerator and denominator have the change in the ratio is easily calculated on the basis of the same effect in terms of ratio change. percentage change graphic illustrated above. The process is (A) Numerator increases and denominator decreases re- illustrated below: sulting in an increase in the ratio. Suppose the numerator increases by 20% and the de- (B) Numerator decreases and denominator increases re- nominator decreased by 10%. Then, the effect on the value sulting in a decrease in the ratio. of the ratio will be:– 20% increase due to the numerator increase and 9.09% decrease due to the denominator de- In both the above situations the numerator and the de- crease. These values can be used on the percentage change nominator work hand in hand and have the same directional graphic to get the net result. effect on the ratio. In this case, it is not difficult to under- stand the direction that the ratio will take in terms of its change in value. FRACTION TO PERCENTAGE CONVERSION Case 2: Numerator and denominator have the opposite TABLE effects on the value of the ratio: The following percentage values appear repeatedly over the (A) Numerator and denominator both increase. entire area where questions can be framed on the topic of (B) Numerator and denominator both decrease. percentage. Further, it would be of great help to you if you In both these cases the net effect on the value of the ratio are able to recognize these values separately from values will depend upon the relative effects of the numerator and that do not appear in the table. Chapter 6: Interest 109 Again, interest for next 4 years will be equal to 7 ¥ 4 = Problems 6.5 A sum of money was invested at SI at a 28%. certain rate for 3 years. Had it been invested at a 4% And interest for next 4 years (till 11 years) – 7.5 ¥ 4 = higher rate, it would have fetched Rs. 480 more. Find the 30% principal. So, total interest = 18 + 28 + 30 = 76% So, total interest earned by him = 76% of the amount (a) Rs. 4000 (b) Rs. 4400 (c) Rs. 5000 (d) Rs. 3500 (76 ¥ 1200) Solution Let the rate be y% and principal be Rs. x and = = Rs. 912 the time be 3 years. 100 Then according to the question = (x(y + 4) ¥ 3)/100 This calculation can be done very conveniently using the – (xy ¥ 3)/100 = 480 percentage rule as 75% + 1% = 900 + 12 = 912. ﬁ xy + 4x – xy = 160 ¥ 100 Problem 6.3 A sum of money doubles itself in 12 years. ﬁ x = (160 ¥ 100)/4 = Rs. 4000 Find the rate percentage per annum. Alternatively: Excess money obtained = 3 years @ 4% per (a) 12.5% (b) 8.33% (c) 10% (d) 7.51% annum = 12% of whole money Solution Let principal = x, then interest = x, time = 12 So, according to the question, 12% = Rs. 480 years. So, 100% = Rs. 4000 (answer arrived at by using unitary Using the formula, Rate = (Interest ¥ 100)/Principal ¥ Time method.) = (x ¥ 100)/(x ¥ 12) = 8.33% Problem 6.6 A certain sum of money trebles itself in 8 Alternatively: It is obvious that in 12 years, 100% of the years. In how many years it will be five times? (a) 22 years (b) 16 years (c) 20 years (d) 24 years So, in 1 year = (100/12)% of the amount is added. Hence, every year there is an addition of 8.33% (which Solution It trebles itself in 8 years, which makes inter- is the rate of simple interest required). est equal to 200% of principal. Alternatively, you can also use the formula. So, 200% is added in 8 years. If a sum of money gets doubled in x years, then rate of Hence, 400%, which makes the whole amount equal to interest = (100/x)%. five times of the principal, which will be added in 16 years. Problem 6.4 A certain sum of money amounts to Rs. 704 Problem 6.7 If CI is charged on a certain sum for 2 years in 2 years and Rs. 800 in 5 years. Find the rate percentage at 10% the amount becomes 605. Find the principal? per annum. (a) Rs. 550 (b) Rs. 450 (c) Rs. 480 (d) Rs. 500 (a) Rs. 580 (b) Rs. 600 (c) Rs. 660 (d) Rs. 640 Solution Using the formula, amount = Principal (1 + Solution Let the principal be Rs. x and rate = r%. rate/100)time Then, difference in between the interest of 5 years and 605 = p(1 + 10/100)2 = p(11/10)2 of 2 years equals to p = 605(100/121) = Rs. 500 Rs. 800 – Rs. 704 = Rs. 96 Alternatively: Checking the options, So, interest for 3 years = Rs. 96 Hence, interest/year = Rs. 96/3 = Rs. 32 Option (a) Rs. 550 So, interest for 2 years Æ 2 ¥ Rs. 32 = Rs. 64 First year interest = Rs. 55, which gives the total amount So, the principal = Rs. 704 – Rs. 64 = Rs. 640 Rs. 605 at the end of first year. So not a valid option. Thought process here should be Option (b) Rs. 450 Rs. 96 interest in 3 years Æ Rs. 32 interest every year. First year interest = Rs. 45 Hence, principal = 704 – 64 = 640 Second year interest = Rs. 45 + 10% of Rs. 45 = 49.5 Chapter 8: Time and Work 145 Thus, the work done per man-day has to rise from 1 to 2. 6 men can do a piece of work in 12 days. How many 1.5, that is, by 50%. Hence, the efficiency of work has to men are needed to do the work in 18 days. rise by 50%. (a) 3 men (b) 6 men (c) 4 men (d) 2 men Problem 8.4 A is twice as efficient as B. If they complete 3. A can do a piece of work in 20 days and B can do a work in 30 days find the times required by each to com- it in 15 days. How long will they take if both work plete the work individually. together? Solution When we say that A is twice as efficient as B, (a) 8 FH 6 IK days (b) 8 FH 4 IK days it means that A takes half the time that B takes to complete 7 7 (c) 9 F I days the same work. 3 (d) None of these Thus, if we denote A’s 1 day’s work as A and B’s one HK7 day's work as B, we have 4. In question 3 if C, who can finish the same work in 25 A = 2B days, joins them, then how long will they take to com- plete the work? Then, using the information in the problem, we have: 30 A + 30 B = 100% work (a) 6 FH 18 IK days (b) 12 days 47 That is, 90 B = 100% work Æ B = 1.11 % (is the work (c) 2 F I days FH IK done by B in 1 day) Æ B requires 90 days to complete the 8 6 work alone. H K 11 (d) 47 18 days Since, A = 2B Æ we have A = 2.22 % Æ A requires 45 5. Nishu and Archana can do a piece of work in 10 days days to do the work alone. and Nishu alone can do it in 12 days. In how many You should be able to solve this mentally with the fol- days can Archana do it alone? lowing thought process while reading for the first time: (a) 60 days (b) 30 days (c) 50 days (d) 45 days 100 3.33 = 3.33%. = 1.11%. Hence, work done is 1.11% 6. Baba alone can do a piece of work in 10 days. Anshu 30 3 per day and 2.22% per day Æ 90 and 45 days. alone can do it in 15 days. If the total wages for the work is Rs. 50. How much should Baba be paid if they Problem 8.5 A is two times more efficient than B. If they work together for the entire duration of the work? complete a work in 30 days, then find the times required by (a) Rs. 30 (b) Rs. 20 each to complete the work individually. (c) Rs. 50 (d) None of these 7. 4 men and 3 women finish a job in 6 days, and 5 men Solution Interpret the first sentence as A = 3B and solve and 7 women can do the same job in 4 days. How long according to the process of the previous problem to get will 1 man and 1 woman take to do the work? the answers. (You should get A takes 40 days and B takes 120 days.) (a) 22 FH 2 IK days (b) 25 FH 1 IK days 7 2 ## (c) 5 F I days 12 F I days 1 7 (d) HK H K I 7 22 Level of Difficulty (LOD) 8. If 8 boys and 12 women can do a piece of work in 25 days, in how many days can the work be done by 6 boys and 11 women working together? (a) 15 days (b) 10 days 1. Raju can do 25% of a piece of work in 5 days. How (c) 12 days (d) Cannot be determined many days will he take to complete the work ten times? 9. A can do a piece of work in 10 days and B can do the (a) 150 days (b) 250 days same work in 20 days. With the help of C, they finish (c) 200 days (d) 180 days the work in 5 days. How long will it take for C alone to finish the work? Chapter 8: Time and Work 149 a depth of 280 metres. How many metres of drilling 9. If the tank is half full and set X and set Y are closed, was the plan for each day. how many minutes will it take for set Z to empty the (a) 38 metres (b) 30 metres tank if alternate taps of set Z are closed. (c) 27 metres (d) None of these (a) 12 minutes (b) 20 minutes 3. A pipe can fill a tank is x hours and another can empty (c) 40 minutes (d) 16 minutes it in y hours. If the tank is 1/3rd full then the number 10. If one pipe is added for set X and set Y and set Z’s of hours in which they will together fill it in is capacity is increased by 20% on it’s original value and (3xy ) (3xy ) all the taps are opened at 2.58 p.m., then at what time (a) (b) does the tank get filled? (If it is initially empty.) 2( y - x ) (y - x) xy (a) 3.05 p.m. (b) 3.04 p.m. (c) (d) None of these (c) 3.10 p.m. (d) 3.03 p.m. 3( y - x ) 4. Dev and Tukku can do a piece of work in 45 and 40 11. Ajit can do as much work in 2 days as Baljit can do days respectively. They began the work together, but in 3 days and Baljit can do as much in 4 days as Diljit Dev leaves after some days and Tukku finished the in 5 days. A piece of work takes 20 days if all work remaining work in 23 days. After how many days did together. How long would Baljit take to do all the Dev leave work by himself? (a) 7 days (b) 8 days (a) 82 days (b) 44 days (c) 9 days (d) 11 days (c) 66 days (d) 50 days 12. Two pipes can fill a cistern in 14 and 16 hours respec- 5. A finishes 6/7th of the work in 2z hours, B works tively. The pipes are opened simultaneously and it is twice as fast and finishes the remaining work. For found that due to leakage in the bottom of the cistern, how long did B work? it takes 32 minutes extra for the cistern to be filled (a) FH 2 IK z (b) FH 6 IK z up. When the cistern is full, in what time will the leak 3 7 empty it? (c) FH 6 IK z (d) F 3Iz H 18 K (a) 114 h (b) 112 h 49 (c) 100 h (d) 80 h 13. A tank holds 100 gallons of water. It’s inlet is 7 inches Directions for questions 6–10: Read the following and an- in diameter and fills the tank at 5 gallons/min. The swer the questions that follow. outlet of the tank is twice the diameter of the inlet. A set of 10 pipes (set X ) can fill 70% of a tank in 7 How many minutes will it take to empty the tank if the minutes. Another set of 5 pipes (set Y ) fills 3/8 of the tank inlet is shut off, when the tank is full and the outlet is in 3 minutes. A third set of 8 pipes (set Z ) can empty 5/10 opened? (Hint: Rate of filling or emptying is directly of the tank in 10 minutes. proportional to the diameter) 6. How many minutes will it take to fill the tank if all the (a) 7.14 min (b) 10.0 min 23 pipes are opened at the same time? (c) 0.7 min (d) 5.0 min 5 14. A tank of capacity 25 litres has an inlet and an outlet (a) 5 minutes (b) 5 minutes 7 tap. If both are opened simultaneously, the tank is (c) 6 minutes (d) None of these filled in 5 minutes. But if the outlet flow rate is 7. If only half the pipes of set X are closed and only doubled and taps opened the tank never gets filled half the pipes of set Y are open and all other pipes are up. Which of the following can be outlet flow rate in open, how long will it take to fill 49% of the tank? liters/min? (a) 16 minutes (b) 13 minutes (a) 2 (b) 6 (c) 4 (d) 3 (c) 7 minutes (d) None of these 15. X takes 4 days to complete one-third of a job, Y takes 8. If 4 pipes are closed in set Z, and all others remain 3 days to complete one-sixth of the same work and Z open, how long will it take to fill the tank? takes 5 days to complete half the job. If all of them (a) 5 minutes (b) 6 minutes work together for 3 days and X and Z quit, how long (c) 7 minutes (d) 7.5 minutes will it take for Y to complete the remaining work done. Chapter 8: Time and Work 151 ## needed to fill the tank than to discharge it. Determine the delivery of the pump discharging the tank. (a) 40 m3/min (c) 60 m3/min (b) 50 m3/min (d) 80 m3/min Level of Difficulty (LOD) III Directions for questions 1–10: Study the following tables 27. Two pipes A and B can fill up a half full tank in 1.2 and answers the questions that follow. hours. The tank was initially empty. Pipe B was kept Darbar Toy Company has to go through the following open for half the time required by pipe A to fill the stages for the launch of a new toy: tank by itself. Then, pipe A was kept open for as much time as was required by pipe B to fill up 1/3 of the Expert Non-expert tank by itself. It was then found that the tank was 5/ man-days man-days 6 full. The least time in which any of the pipes can fill required required the tank fully is 1. Design and development 30 60 (a) 4.8 hours (b) 4 hours 2. Prototype creation 15 20 (c) 3.6 hours (d) 6 hours 3. Market survey 30 40 28. A tank of 425 litres capacity has been filled with water 4. Manufacturing setup 15 30 through two pipes, the first pipe having been opened 5. Marketing and launch 15 20 five hours longer than the second. If the first pipe were open as long as the second, and the second pipe The profile of the company’s manpower is was open as long as the first pipe was open, then the Worker Expert at Non-Expert at Refusal to first pipe would deliver half the amount of water de- name work on livered by the second pipe; if the two pipes were open A Design and development All others Market survey simultaneously, the tank would be filled up in 17 B Prototype creation All others Market survey hours. How long was the second pipe open? C Market survey and All others Design and (a) 10 hours (b) 12 hours marketing and launch development (c) 15 hours (d) 18 hours D Manufacturing All others Market survey 29. Two men and a woman are entrusted with a task. E Market survey All others Manufacturing The second man needs three hours more to cope with the job than the second man and the woman would 1. Given this situation, the minimum number of days in need working together. The first man, working alone, which the company can launch a new toy going would need as much time as the second man and the through all the stages is woman working together. The first man, working (a) 40 days (b) 40.5 days alone, would spend eight hours less than the double (c) 45 days (d) None of these period of time the second man would spend working 2. If A and C refuse to have anything to do with the alone. How much time would the two men and the manufacturing set up. The number of days by which woman need to complete the task if they all worked the project will get delayed will be together? (a) 5 days (b) 4 days (a) 1 hour (b) 3 hours (c) 3 days (d) None of these (c) 4 hours (d) 5 hours 3. If each of the five works is equally valued at 30. The Bubna dam has four inlets. Through the first three Rs.10,000, the maximum amount will be received by inlets, the dam can be filled in 12 minutes; through the (a) A (b) C (c) D (d) E second, the third and the fourth inlet, it can be filled 4. For question 3, the second highest amount will be re- in 15 minutes; and through the first and the fourth ceived by inlet, in 20 minutes. How much time will it take all the (a) A (b) C (c) D (d) E four inets to fill up the dam? 5. If C works at 90.909% of his efficiency during mar- (a) 8 min (b) 10 min keting and launch, who will be highest paid amongst (c) 12 min (d) None of these the five of them?
# Solving One-Variable Equations Algebra 1 Notes SOL A.4 Equations Mr. Hannam Name: __________________________________________ Date: _______________ Block: _______ Equations  An equation is an open sentence where two expressions are set _____________.  Equations may have one or more unknowns (_______________) which we may try to solve.  Solving an equation means finding the value(s) of variable(s) that make the equation ____________.  Equations that have the same solution(s) are called ____________________.  We use ______________________ operations to solve equations.  We justify the steps in solving equations by using field properties. Solving Equations  “Isolate” the variable, justifying steps using field properties (properties of equality): 1) Put variables on one side of the = and numbers on the other by isolating x:  perform inverse operations (add, subtract, multiply, or divide) 2) Whatever you do to one side of the equation, you do to the other 3) Verify solutions S  substitute solution in original equation (DO NOT SKIP THIS STEP!!!!!!) Field Properties of Equality: Property of Equality Reflexive Property of Equality Algebra (for real numbers a, b, c) a=a Symmetric Property of Equality If a = b, then b = a Transitive Property of Equality If a = b and b = c, then a = c Substitution Property of Equality If a = b, then a can be substituted for b If a = b, then a + c = b + c Subtraction Property of Equality If a = b, then a – c = b - c Multiplication Property of Equality If a = b, then ac = bc Division Property of Equality If a = b and c  0, then a b  c c Example Algebra 1 Notes SOL A.4 Equations  Solve the equations: a) x – 4 = 6 x–4+4=6+4 x = 10 VERIFY: Is 10 a solution? (10) – 4 = 6 6=6   Mr. Hannam Page 2 b) x + 3 = -5 x + 3 – 3 = -5 – 3 x = -8 VERIFY: Is -8 a solution? (-8) + 3 = -5 -5 = -5 c) 4x = 16 4x ÷ 4 = 16 ÷ 4 x=4 VERIFY: Is 4 a solution? 4(4) = 16 16 = 16 Solve the equations, justifying steps: a) Solve x + 4 = 10 1) x + 4 = 10 2) x + 4 – 4 = 10 – 4 3) x = 6 b) Solve Given Subtraction Property of Equality Simplify 2 x=4 3 2 x=4 Given 3 3 2 3  x   4 Mult. Property of Equality 2) 2 3 2 3) x = 6 Simplify 1) Two-Step Equations  SAME! Put variables on one side of the = and numbers on the other: How to Solve Two-Step Equations 1) Clear parentheses (distribute) and combine like terms if necessary 3) Do multiply/divide steps Examples: x  5  11 2 x  5  5  11  5 ________________ 2 x  16 ________________ 2 x 2   2  16 ________________ 2 x = 32 _________________ VERIFY: a) 5x + 9 = 24 5x + 9 – 9 = 24 – 9 __________________ 5x = 15 __________________ 5x 15  __________________ 5 5 x=3 __________________ VERIFY: b) c) 3x + 2x = 15 d) 4(x - 6) = 32 VERIFY: VERIFY: Algebra 1 Notes SOL A.4 Equations Mr. Hannam Page 3 Practice - solve the equations, justifying steps with field properties: a) x + 9 = 25 b) -4x = -20 a 46 3 e) f) -1 = z 3 37 3 x  3 5 d) 2x + 3 = -9 g) 7x – 4x = 21 h) 3(x + 2) = 9 c) Multistep Equations and Equations with Variables on Both Sides How to Solve Multi-Step Equations and Equations with Variables on Both Sides 1) Clear parentheses (distribute) and combine like terms if necessary 2) Add/subtract variable terms so that variable is on one side (NEW STEP!) 4) Do multiply/divide steps Example: Solve 7 – 3x = 4x – 7 1) 7 – 3x = 4x – 7 Given 2) 7 – 3x - 4x = 4x – 7 - 4x ________________________________________________ 3) 7 – 7x = - 7 ________________________________________________ 4) 7 -7x - 7 = -7 - 7 ________________________________________________ 5) -7x = -14 ________________________________________________ 6) ________________________________________________ - 7x - 14  7 -7 7) x = 2 ________________________________________________ Examples: Solve the equation, justifying steps… 1 a) 9x – 5 = (16x + 60) b) 3(x + 12) – x = 4(2 – x) – 3x + 2x 4 9x – 5 = 4x + 15 ______________________ 5x – 5 = 15 ______________________ 5x = 20 ______________________ x=4 _____________________ Algebra 1 Notes SOL A.4 Equations Mr. Hannam Page 4 Special cases: Identity case (solutions are all real numbers): No solution case: 2x + 6 = 2(x + 3) 3x = 3(x + 4) 2x + 6 = 2x + 6 Distribute 3x = 3x + 12 Distribute 0=0 Subtraction Prop. of = 0 = 12 Subtraction Prop. of = Where did the variable go?? Where did the variable go?? When we get a TRUE statement at the end when the variable “disappears,” EVERY x is a solution (the two sides of the equation are identical!). When we get a FALSE statement at the end when the variable “disappears,” there are NO SOLUTIONS. There is no value of x that makes the equation true. ALL REAL NUMBERS are solutions. Other Equations: Solve: a) x 3  4 2 Cross Product Property:  b) If x 2  x 1 3 a c b d REMEMBER TO GROUP NUMERATORS AND DENOMINATORS (use parentheses) You Try: Solve the equation if possible; if not, write “all real numbers” or “no solution”… a) 8x + 5 = 6x + 1 b) x + 1 = 3x - 1 c) 9x = 6(x + 4) d) 7(x + 7) = 5x + 59 e) 2 – 15x = 5(-3x + 2) f) 12y + 6 = 6(2y + 1) h) 40 + 14x = 2(-4x – 13) i) 2(3x + 2) = g) 5(x + 2) = j) x 9  2 3 3 (5 + 10x) 5 k) 4 8  x -8 2 l) 2 x  5 21 - x 1 (12x + 8) 2 20 cards 24 cards 23 cards 15 cards 37 cards
More examples of proof by contradiction We have had good discussions on mathematical proofs, so I am planning to create a mathematical proof series that will discuss the basics such as direct proof, indirect proof, and proof by mathematical induction. But before I do that, let me continue with more examples of proof by contradiction. Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Finding a contradiction means that your assumption is false and therefore the statement is true. Below are several more examples of this proof strategy. Example 2: $\sqrt{6}$ is irrational. The proof of this is basically the same as example 1, so it is left as an exercise. Example 3: Proof that there are infinitely many primes. Example 4: Knights and Liars Example 5: $\sqrt{2} + \sqrt{3}$ is irrational. » Read more Proof Tutorial 2: Proving Square Root of 2 is Irrational by Contradiction One of the most difficult proof strategies in mathematics is proof by contradiction. If P, for example, is a statement or a conjecture, one strategy to prove that P is true is to assume that P is not true and find a contradiction so that the statement not P does not hold. If not P does not hold, it follows that P is true. One well-known proof that uses proof by contradiction is proof of the irrationality of $\sqrt{2}$. If we consider P to be the statement “$\sqrt{2}$ is irrational”, then not P is the opposite statement or “$\sqrt{2}$ is rational”. To use proof by contradiction, we assume that $\sqrt{2}$ is rational, and find a contradiction somewhere. If this happens, then we would have shown that $\sqrt{2}$ is indeed irrational. Before proceeding, recall that a rational number is a fraction with non-zero denominator. We know that all fractions can be expressed in lowest term. A fraction in $\displaystyle\frac{a}{b}$ is said to be in lowest term if $a$ and $b$ have no common divisors except $1$. On the other hand, irrational numbers cannot be expressed as fractions. They are decimal numbers that do not end and do not repeat. For example, $0.10100100010000...$ is an irrational number (the three dots means and so on which means that the number does not end). The most popular irrational number is $\pi$. Now, we prove our conjecture. Conjecture: The $\sqrt{2}$ is irrational. Proof: Suppose $\sqrt{2}$ is rational, then it can be expressed in fraction form $\displaystyle\frac{a}{b}$ . Let us assume that our fraction is in lowest term, i.e., their only common divisor is $1$. Then, $\sqrt{2} = \displaystyle\frac{a}{b}$ Squaring both sides, we have $2= \displaystyle\frac{a^2}{b^2}$ Multiplying both sides by $b^2$ yields $2b^2= a^2$* Since $a^2 = 2b^2$, we can conclude that $a^2$ is even because whatever the value of $b^2$ has to be multiplied by $2$. If $a^2$ is even, then $a$ is also even. Since $a$ is even, no matter what the value of $a$ is, we can always find an integer that if we divide $a$ by $2$, it is equal to that integer. If we let that integer be $k$, then $\displaystyle\frac{a}{2} = k$ which means that $a = 2k$. Substituting the value of $2k$ to $a$ in *, we have $2b^2= (2k)^2$ which means that $2b^2=4k^2$. Dividing both sides by $2$, we have $b^2 = 2k^2$. That means that the value $b^2$ is even, since whatever the value of $k$ you have to multiply it by $2$. Again, if $b^2$ is even, then $b$ is even. This implies that both $a$ and $b$ are even, which means that both the numerator and the denominator of our fraction are divisible by $2$. This contradicts our assumption that $\displaystyle\frac{a}{b}$ has no common divisor except $1$. Since we found a contradiction, our assumption is, therefore, false. Hence, the theorem is true. Notice that I have highlighted the word suppose and assume in the proof. This is one unique feature of proof by contradiction. You can always assume, most of the time, the opposite of the conjecture as long as the following statements are logically valid.
• Study Resource • Explore Survey * Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project Document related concepts History of statistics wikipedia, lookup Student's t-test wikipedia, lookup Taylor's law wikipedia, lookup Bootstrapping (statistics) wikipedia, lookup Degrees of freedom (statistics) wikipedia, lookup Transcript ```2.0 Lesson Plan 1 • Summary Statistics • Histograms • The Normal Distribution • Using the Standard Normal Table 2. Summary Statistics Given a collection of data, one needs to find representations of the data that facilitate understanding and insight. Three standard tools for this are: 2 • Measures of Central Tendency (mean, median, mode) • Measures of Dispersion (standard deviation, range, IQR) • Visualizations (histograms, other graphics) We shall cover this quickly, since most of this is review from elementary school. 2.1 Measures of Central Tendency The mean is just the average of the data. Suppose one has a sample of n observations with values X1 , . . . , Xn . Then the mean is just n 3 1X X̄ = Xi n i=1 1 (X1 + · · · + Xn ). = n The mode is just the value that occurs most frequently in the sample. There can be many modes. The median is the middle-largest value among the n observations, if n is odd. If there are an even number of observations, then we average the two middle-largest values. Example: Suppose one observes the following data: 1, 0, 2, -2, 1, -2, 5, -1 The mean is X̄ = 81 (1 + 0 + 2 − 2 + 1 − 2 + 5 − 1) = 0.5. The modes are 1 and -2. 4 The median is the average of 0 and 1, or 0.5. The mean can be pulled in misleading directions if there are outliers. A single large or small datum will have a large influence on the mean, but not on the median. An outlier is an incorrect or unrepresentative observation that is very different from the others in the sample. 2.2 Measures of Dispersion 5 To measure how spread out a sample is, we mostly use the standard deviation (or sd). This is: r 1 [(X1 − X̄)2 + · · · + (Xn − X̄)2 ] sd = (1) n−1 v ! u n X u 1 n 2 t X̄ 2 . Xi − = (2) n − 1 i=1 n−1 The sd is the square root of the average squared deviation of each observation from the mean. The square of the sd is the variance. Formula (1) is better for understanding, but (2) is better for calculation. Note Bene: When one observes the whole population rather than just a sample, then the formula for the population sd divides by n instead of n − 1. 6 The majority of observations are usually within 1 sd of the mean. But it can happen that none of the data are less than 1 sd from the mean. Tchebyshev proved that for all datasets, the proportion of data that lie within a standard deviations of the mean must be at least 1 − a12 . In terms of probability, if you pick an observation X at random, 1 P[ \|X − mean\| < a ∗ sd ] ≥ 1 − 2 . a Thus • At least 75% of the observations must always be less than 2 standard deviations from the population mean. • At least 89% of the observations must always be less than 3 standard deviations of the population mean. The range is the largest observation minus the smallest. As a measure of dispersion, it is strongly influenced by outliers. The interquartile range is 75th percentile of the data minus the 25th percentile (the median is the 50th percentile). 7 • The 25th percentile is the number u such that at least 25% of the sample is less than or equal to u and at least 75% of the sample is greater than or equal to u. (The u need not be a sample value; and if there are many numbers u that satisfy this definition, we take the middle value.) • The 75th percentile is the number v such that at least 75% of the sample is less than or equal to v and at least 25% of the sample is greater than or equal to v. (If this is not unique, take the middle value.) The interquartile range is not strongly influenced by outliers. Example: Suppose you have the following sample: 1, 0, 2, -2, 1, 5, -2, -1. It helps to order the data first: -2, -2, -1, 0, 1, 1, 2, 5 The range is 5 - (-2) = 7. There are eight values, so the 25th percentile is any number between -2 and -1; we take -1.5. Similarly, the 75th percentile is 1.5. The IQR is the difference of these, or 1.5 - (-1.5) = 3. 8 The standard deviation is r 1 sd = [(1 − 0.5)2 + · · · + ((−1) − 0.5)2 ] =? 7 but it is faster to calculate r 1 2 sd = [1 + · · · + (−1)2 ] − (8/7)(0.5)2 = 2.32993. 7 2.3 Properties of X̄ and the sd Suppose we have n observations X1 , . . . , Xn and we use these to create a new sample Y1 , . . . , Yn where Yi = aXi + b. This often arises when converting units of measurement, such has changing Fahrenheit data into the Centigrade scale: C = 5/9 * F - 17.778. 9 Then Ȳ = aX̄ + b sdY = \|a\|sdX . Can you guess the conversion formulae for the median, mode, range, and interquartile range of the Y values? 2.4 The Histogram The histogram shows where sample values are located and where they concentrate. 10 The x-axis gives the sample value, and the y-axis is the percent per x-value. (This is different from a bar chart.) In a histogram, the areas under a block represent percentages. By convention, the left endpoint of a histogram bar is included in the interval, but not the right. This incomplete histogram shows the number of parties attended in one week by Duke freshmen. • What is the height of the missing bar? • What percentage go to 1 or fewer parties? 11 As the sample size gets large and as the bin-width gets small at the appropriate relative rates, then the histogram becomes smooth. This limiting smooth curve is called a probability density function. 12 2.5 The Normal Distribution Some limiting histograms are famous and have names. The most famous distribution is the normal distribution (a/k/a the Gaussian distribution or the bell-shaped curve). 13 The distribution was named after Carl Friedrich Gauss, the greatest mathematician in history. He proved the fundamental theorem of algebra four ways, inventing a new branch of mathematics each time. He worked in number theory, co-invented the telegraph, and discovered non-Euclidean geometry, but did not publish, fearing controversy. People believe the normal distribution describes IQ, height, rainfall, measurement error, and many other features. This is only approximately true. But it is a good approximation for features that are the sum of many separate increments. 14 A normal distribution with mean µ and standard deviation σ has the equation: 1 1 f (x) = √ exp − 2 (x − µ)2 2σ 2πσ for −∞ < µ < ∞ and σ ≥ 0. The µ is the mean of the entire population, whereas X̄ is used to denote the mean of a sample from the population. Similarly, σ is the standard deviation of the entire population, whereas sd is used to denote the standard deviation of a sample. The standard normal has µ = 0 and σ = 1. The mean of a normal distribution shows where it is centered. The standard deviation of a normal distribution shows how spread out the normal is. 15 2.6 The Standard Normal Distribution Practice reading the standard normal table in the handout. A copy is also posted on the FAQ site. You may bring your table to class and use whatever notes you have on the back of the table during quizzes. 16 What proportion of a standard normal population has values between -1.5 and 1.5? From the table, the proportion is 1 - 2 * 0.067 = 0.866 (i.e., the total area under the curve is one, and we subtract the upper tail area and the symmetric lower tail area). Now go the other way. About 80% of the population lies between what two values that are centered at 0? The answer is about -1.28 and 1.28 (the table shows that 10% of the area is above 1.28, and symmetrically, 10% is below -1.28). • What is the value of z such that 25% of a standard normal population is larger than that value? (Ans: about 0.67) • What is the value for which about 90% of the population is smaller? 17 • What proportion of the population has a value larger than -1? (Ans: • What proportion of the population has a value less than -0.3? (Ans: In this class, use the nearest value in the table. If you interpolate, or use How can you decide if data are a random sample from a normal distribution? • Inspect the histogram. • Make a normal probability plot. 18 To make a normal probability plot, order the observations from smallest to largest; denote the ordered observations by X(1) , X(2) , . . . , X(n) . For observation X(i) , find the z-value such that (i − 0.5)/n ∗ 100% of the area under the standard normal curve is to the left. Call this z-value Yi . Then plot (X(i) , Yi ) for all i = 1, . . . n. If this looks pretty much like a straight line, then the data are approximately normal. (There is a Stata command for this.) ``` Related documents
# Problem of the Week Problem C and Solution Hundred Deck 1 ## Problem Hundred Deck is a deck consisting of $$100$$ cards numbered from $$1$$ to $$100$$. Each card has the same number printed on both sides. One side of the card is red and the other side of the card is yellow. Sarai places all of the cards on a table with each card’s red side facing up. She first flips over every card that has a number on it which is a multiple of $$2$$. She then flips over every card that has a number on it which is a multiple of $$3$$. After Sarai has finished, how many cards have their red side facing up? ## Solution After flipping over all of the cards with numbers that are multiples of $$2$$, $$50$$ cards have their red side facing up and $$50$$ cards have their yellow side facing up. All of the cards with their red side facing up are numbered with an odd number. All of the cards with their yellow side facing up are numbered with an even number. Next, in the second round of flips, Sarai flips over every card that is numbered with a multiple of $$3$$. Let’s look at how many cards with their red side facing up will be flipped over to yellow and how many cards with their yellow side facing up will be flipped over to red. There are $$33$$ multiples of $$3$$ from $$1$$ to $$100$$. They are $3, 6, 9, 12, 15, …, 87, 90, 93, 96, 99$ Of these numbers, $$17$$ are odd and $$16$$ are even. The $$17$$ odd multiples of $$3$$ currently have their red side facing up, and therefore are flipped over to yellow. The $$16$$ even multiples of $$3$$ currently have their yellow side facing up, and are therefore flipped over to red (again). So, after the first flip there were $$50$$ cards with their red side facing up and $$50$$ cards with their yellow side facing up. Of the $$50$$ red, $$17$$ were flipped to yellow. Of the $$50$$ yellow, $$16$$ were flipped to red. Therefore, after Sarai has finished, $$50-17+16=49$$ cards have their red side facing up.
## Want to keep learning? This content is taken from the National STEM Learning Centre's online course, Maths Subject Knowledge: Understanding Numbers. Join the course to learn more. 4.8 ## National STEM Learning Centre Skip to 0 minutes and 8 seconds PAULA KELLY: Square numbers are so called as that many dots can be arranged to form a square. The sequence of square numbers starts 1, 4, 9, 16, and so on. Could you predict the next two terms in the sequence? Could you write down the term-to-term rule? And could you write down the position-to-term rule? Skip to 0 minutes and 32 seconds The first term is 1 multiplied by 1, which gives us 1. The second term is 2 multiplied by 2 to give us 4. Third term is 3 multiplied by 3 to give us 9. So the sixth term is 6 multiplied by 6 to give us 36, and the seventh term is 7 multiplied by 7 to give us 49. In this case, the position-to-term rule is easy to find. So the position-to-term rule, or Sn, can be written as n squared. Or we could also write it as n multiplied by n. The term-to-term rule is not quite as easy to find. Let’s consider the sequence of odd numbers first. The odd numbers go 1, 3, 5, 7, and 9. Skip to 1 minute and 21 seconds If we study the difference between the terms, we could see that we add on the next odd number. So if we compare with our square numbers, the first square number is 1, the second square number is the first square number add on the second odd number. To find the third square number, we add together the second square number and the third odd number. So in general, to find the nth square number, we add the previous square number to the nth odd number. We could extend this rule to show that square numbers can be made by adding up odd numbers. So the first square number is the same as the first odd number. Skip to 2 minutes and 5 seconds The second square number is the same as the first odd number add the second odd number. To find the third square number, we add together the first, second, and third odd numbers. # Square numbers Square numbers are those whole numbers such that if that number is represented by that many dots we can arrange the dots in the shape of a square. Think of a square of length 5. The value of the area of the square must be a square number. The area of this square is $$5 \times 5$$ which can be written as $$5^2$$ and equals 25. 25 is a square number. ## Discussion What do you know about square numbers? Here are a couple of questions you might be asked by students. How would you respond? Are all square numbers positive numbers? If you can have a negative square number, explain how. If you cannot have a negative square number explain why not. Is zero a square number? In this video Paula explains that the position to term rule of the sequence of square numbers is relatively easy to find. She explores an interesting connection with another kind of number to find a term to term rule for square numbers. ## Problem worksheet Now complete question 8 from this week’s worksheet
Instruction 1 School classic formulation of the Pythagorean theorem is: the square of the hypotenuse is equal to the sum of the squares of the other two sides. Thus, to find the hypotenuse of a right triangle two legs, alternately, need to square the lengths of the legs, fold them and take the square root of the result. In its original formulation, the theorem asserted that the area of a square constructed on the hypotenuse equal to the sum of the areas of two squares constructed on the legs. However, modern algebraic formulation does not require to introduce the concept of square. 2 For example, given a right triangle, the legs of which is equal to 7 cm and 8 cm Then, according to the Pythagorean theorem, the square of the hypotenuse is 72+82=49+64=113 cm2. Itself the hypotenuse is equal to the square root of the number 113. Happened irrational number, which goes in the answer. 3 If the sides of a triangle are 3 and 4 then the hypotenuse equals √25=5. When extracting the square root turned out to be a natural number. Numbers 3, 4, 5 are a Pythagorean triple because they satisfy the condition x2+y2=z2, being all natural. Other examples of Pythagorean triples: 6, 8, 10; 5, 12, 13; 15, 20, 25; 9, 40, 41. 4 If the legs are equal, then the Pythagorean theorem into a more simple equation. Suppose, for example, both sides equal to the number A, and the hypotenuse is designated for C. Then C2=A2+A2 C2=2A2, C=A√2. In this case, no need to build in the square of the number A. 5 The Pythagorean theorem is a special case of a more General theorem of cosines, which sets the ratio between the three sides of the triangle for an arbitrary angle between any two of them.
# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### (3/8) : (1/48) = 18/1 = 18 Spelled result in words is eighteen. ### How do you solve fractions step by step? 1. Divide: 3/8 : 1/48 = 3/8 · 48/1 = 3 · 48/8 · 1 = 144/8 = 8 · 18 /8 · 1 = 18 Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1/48 is 48/1) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the next intermediate step, , cancel by a common factor of 8 gives 18/1. In words - three eighths divided by one forty-eighth = eighteen. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Reciprocals Which among the given reciprocal is correct a. 3/15x1/3= 1 b. 3/20x20/3=1 c. 7/14x7/7=1 d. 34/3x34/34=1 • Third of an hour How many minutes is a third of an hour? Do you know to determine a third of the lesson hour (45min)? • Mineral water The bottle contains 1.5 liters of mineral water. Pour all the water from the bottle into empty cups with a volume of 1/3 l. All but one will be filled to the brim. What part of the volume of the last cup is filled with water? • Chestnuts Three divisions of nature protectors participated in the collection of chestnut trees.1. the division harvested 1250 kg, the 2nd division by a fifth more than the 1st division and the 3rd division by a sixth more than the second division. How many tons of • 4/5 of 4/5 of a number is 276. what is 2/3 of the same number? • Summerjobs The agency give summerjobs for 2352 students in 2018. The eighth was high-school students, the rest was undergraduates. How many undergraduates works via agency in 2018? • The rug Josie has a rug with the area of 18 square feet. She will put the rug on the floor that is covered in 1/3 square foot tiles. How many tiles will the rug cover? • Tennis balls Can of tennis balls contains 3 balls per can and cost \$7 how much will it cost for 36 tennis balls? • Beer After three 10° beers consumed in a short time, there is 5.6 g of alcohol in 6 kg adult human blood. How much is it per mille? • Jolly gobs Each package of jolly gobs has 72 gobs. If one fourth of the gobs are red and the rest are blue, into how many parts was the group divided? How many parts are red? • 6/24 change 6/24 change to a fraction in lowest terms • Mrs. Zarka Mrs. Zarka has 3 pies for a party. She calculates that if she splits the pies evenly among the guests, they will each receive 1/6 of a pie. How many guests are there? • Statues Diana is painting statues. She has 7/8 of a liter of paint remaining. Each statue requires 1/20 of a liter of paint. How many statues can she paint?
Courses Courses for Kids Free study material Offline Centres More # By completing square method find the roots of the following quadratic equation$2{n^2} - 7n + 3 = 0$. Last updated date: 03rd Mar 2024 Total views: 341.1k Views today: 4.41k Verified 341.1k+ views Hint: Here, we will simply add and subtract square of a constant and write the term containing $n$ such that it forms a term in a form $2ab$, thus, this will help us to apply the square identities and thus, ‘complete the square’ and solve it further to find the required roots of the given quadratic equation. Formula Used: ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ $2{n^2} - 7n + 3 = 0$ Now, dividing both sides by 2, we get, ${n^2} - \dfrac{7}{2}n + \dfrac{3}{2} = 0$ Now, since, we are required to solve this question using completing the square, hence, we will write this quadratic equation as: ${\left( n \right)^2} - 2\left( n \right)\left( {\dfrac{7}{4}} \right) + {\left( {\dfrac{7}{4}} \right)^2} - {\left( {\dfrac{7}{4}} \right)^2} + \dfrac{3}{2} = 0$ Hence, even if this quadratic equation was not a perfect square, but we tried to make it a perfect square by adding and subtracting the square of a constant such that, we complete the square by making the identity, ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ Hence, using this identity, we get, ${\left( {n - \dfrac{7}{4}} \right)^2} - \dfrac{{49}}{{16}} + \dfrac{3}{2} = 0$ $\Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} + \dfrac{{ - 49 + 24}}{{16}} = 0$ Hence, we get, $\Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} - \dfrac{{25}}{{16}} = 0$ Adding $\dfrac{{25}}{{16}}$ on both sides, $\Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} = \dfrac{{25}}{{16}}$ $\Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} = {\left( {\dfrac{5}{4}} \right)^2}$ Taking square root on both sides, we get $\Rightarrow \left( {n - \dfrac{7}{4}} \right) = \pm \dfrac{5}{4}$ Adding $\dfrac{7}{4}$ on both the sides, we get, $\Rightarrow n = \pm \dfrac{5}{4} + \dfrac{7}{4}$ Hence, $n = \dfrac{5}{4} + \dfrac{7}{4} = \dfrac{{5 + 7}}{4} = \dfrac{{12}}{4} = 3$ Or $n = - \dfrac{5}{4} + \dfrac{7}{4} = \dfrac{{7 - 5}}{4} = \dfrac{2}{4} = \dfrac{1}{2}$ Therefore, the roots of the given quadratic equation $2{n^2} - 7n + 3 = 0$ are 3 and $\dfrac{1}{2}$ Thus, this is the required answer. Note: If in the question, it was not mentioned that we have to use the method of completing the square, then, we could have used the quadratic formula to solve the given quadratic equation. Given quadratic equation is $2{n^2} - 7n + 3 = 0$ Now, dividing both sides by 2, we get, ${n^2} - \dfrac{7}{2}n + \dfrac{3}{2} = 0$ Comparing this with the general quadratic equation i.e. $a{x^2} + bx + c = 0$ We have, $a = 1$, $b = \dfrac{{ - 7}}{2}$ and $c = \dfrac{3}{2}$ Now, we can find the roots of a quadratic equation using the quadratic formula, $n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ Hence, for the equation ${n^2} - \dfrac{7}{2}n + \dfrac{3}{2} = 0$, substituting$a = 1$, $b = \dfrac{{ - 7}}{2}$and $c = \dfrac{3}{2}$, we get $n = \dfrac{{\dfrac{7}{2} \pm \sqrt {{{\left( {\dfrac{{ - 7}}{2}} \right)}^2} - 4\left( 1 \right)\left( {\dfrac{3}{2}} \right)} }}{{2\left( 1 \right)}}$ $\Rightarrow n = \dfrac{{\dfrac{7}{2} \pm \sqrt {\dfrac{{49}}{4} - 6} }}{2} = \dfrac{{\dfrac{7}{2} \pm \sqrt {\dfrac{{49 - 24}}{4}} }}{2} = \dfrac{{\dfrac{7}{2} \pm \sqrt {\dfrac{{25}}{4}} }}{2}$ Solving further, we get, $\Rightarrow n = \dfrac{{\dfrac{7}{2} \pm \dfrac{5}{2}}}{2} = \dfrac{{7 \pm 5}}{4}$ Hence, we get, $n = \dfrac{{7 + 5}}{4} = \dfrac{{12}}{4} = 3$ Or $n = \dfrac{{7 - 5}}{4} = \dfrac{2}{4} = \dfrac{1}{2}$ Therefore, the roots of the given quadratic equation $2{n^2} - 7n + 3 = 0$ are 3 and $\dfrac{1}{2}$ Thus, this is the required answer.
We've updated our TEXT # Reading: Solving a System of Linear Equations Using the Inverse of a Matrix Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: X is the matrix representing the variables of the system, and B is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as AX = B To solve a system of linear equations using an inverse matrix, let A be the coefficient matrix, let X be the variable matrix, and let B be the constant matrix. Thus, we want to solve a system AX = B. For example, look at the following system of equations. [latex-display]\displaystyle{a}_{{1}}{x}+{b}_{{1}}{y}={c}_{{1}}[/latex-display] [latex-display]\displaystyle{a}_{{2}}{x}+{b}_{{2}}{y}={c}_{{2}}[/latex-display] From this system, the coefficient matrix is [latex-display]\displaystyle{A}={\left[\matrix{{a}_{{1}}&{b}_{{1}}\{a}_{{2}}&{b}_{{2}}}]}[/latex-display] The variable matrix is [latex-display]\displaystyle{X}={\left[\matrix{{x}\{y}}]}[/latex-display] And the constant matrix is [latex-display]\displaystyle{B}={\left[\matrix{{c}_{{1}}\{c}_{{2}}}]}[/latex-display] Then looks like [latex-display]\displaystyle{\left[\matrix{{a}_{{1}}&{b}_{{1}}\{a}_{{2}}&{b}_{{2}}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{c}_{{1}}\{c}_{{2}}}]}[/latex-display] Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex-display]\displaystyle{({2}^{{-{1}}})}{2}={(\frac{{1}}{{2}})}{2}={1}[/latex-display] To solve a single linear equation ax = b for x we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of a. Thus, [latex-display]\displaystyle{a}{x}={b}[/latex-display] [latex-display]\displaystyle{(\frac{{1}}{{a}})}{a}{x}={(\frac{{1}}{{a}})}{b}[/latex-display] [latex-display]\displaystyle{({a}^{{-{1}}})}{a}{x}={({a}^{{-{1}}})}{b}[/latex-display] [latex-display]\displaystyle{[{({a}^{{-{1}}})}{a}]}{x}={({a}^{{-{1}}})}{b}[/latex-display] [latex-display]\displaystyle{1}{x}={({a}^{{-{1}}})}{b}[/latex-display] [latex-display]\displaystyle{x}={({a}^{{-{1}}})}{b}[/latex-display] The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable. We will investigate this idea in detail, but it is helpful to begin with a 2 × 2 system and then move on to a 3 × 3 system. ## Solving a System of Equations Using the Inverse of a Matrix Given a system of equations, write the coefficient matrix A, the variable matrix X, and the constant matrix B. Then AX = B Multiply both sides by the inverse of A to obtain the solution. [latex-display]\displaystyle{({A}^{{-{1}}})}{A}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{[{({A}^{{-{1}}})}{A}]}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{I}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{X}={({A}^{{-{1}}})}{B}[/latex-display] ### Q&A If the coefficient matrix does not have an inverse, does that mean the system has no solution? No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions. ### Example 1 Solve the given system of equations using the inverse of a matrix. 3 x + 8y = 5 4 x + 11y = 7 #### Solution Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. [latex-display]\displaystyle{A}={\left[\matrix{{3}&{8}\{4}&{11}}]},{X}={\left[\matrix{{x}\{y}}]},{B}={\left[\matrix{{5}\{7}}]}[/latex-display] Then [latex-display]\displaystyle{\left[\matrix{{3}&{8}\{4}&{11}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{5}\{7}}]}[/latex-display] First, we need to calculate A–1 Using the formula to calculate the inverse of a 2 × 2 matrix, we have: [latex-display]\displaystyle{A}^{{-{1}}}=\frac{{1}}{{{a}{d}-{b}{c}}}{\left[\matrix{{d}&-{b}\-{c}&{a}}]}[/latex-display] [latex-display]\displaystyle=\frac{{1}}{{{3}{({11})}-{8}{({4})}}}{\left[\matrix{{11}&-{8}\-{4}&{3}}]}=\frac{{1}}{{1}}{\left[\matrix{{11}&-{8}\-{4}&{3}}]}[/latex-display] So, [latex-display]\displaystyle{A}^{{-{1}}}={\left[\matrix{{11}&-{8}\-{4}&{3}}]}[/latex-display] Now we are ready to solve. Multiply both sides of the equation by A –1. [latex-display]\displaystyle{({A}^{{-{{1}}}})}{A}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{\left[\matrix{{11}&-{8}\-{4}&{3}}]}{\left[\matrix{{3}&{8}\{4}&{11}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{11}&-{8}\-{4}&{3}}]}{\left[\matrix{{5}\{7}}]}[/latex-display] [latex-display]\displaystyle{\left[\matrix{{1}&{0}\{0}&{1}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{11}{({5})}+-{8}{({7})}\-{4}{({5})}+{3}{({7})}}]}[/latex-display] [latex-display]\displaystyle{\left[\matrix{{x}\{y}}]}={\left[\matrix{-{1}\{1}}]}[/latex-display] The solution is (–1,1). ### Q&A Can we solve for X by finding the product BA–1? No, recall that matrix multiplication is not commutative, so A–1BBA–1 Consider our steps for solving the matrix equation. [latex-display]\displaystyle{({A}^{{-{1}}})}{A}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{[{({A}^{{-{1}}})}{A}]}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{I}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{X}={({A}^{{-{1}}})}{B}[/latex-display] Notice in the first step we multiplied both sides of the equation by A–1, but the A–1, was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters. ### Example 2 1. Solve the following system using the inverse of a matrix. 5x + 15y + 56z = 35 −4x − 11y − 41z = −26 x −3y − 11z = −7 2. Solve the system using the inverse of the coefficient matrix. 2x 17y + 11z = 0 x + 11y 7z = 8 3y 2z = −2 #### Solutions 1. Write the equation AX = B $\displaystyle{\left[\matrix{{5}&{15}&{56}\-{4}&-{11}&-{41}\-{1}&-{3}&-{11}}]}{\left[\matrix{{x}\{y}\{z}}]}={\left[\matrix{{35}\-{26}\-{7}}]}$ First we will find the inverse of A by augmenting with the identity. [latex-display]\displaystyle{\left[\matrix{{5}&{15}&{56}&{\mid}&{1}&{0}&{0}\-{4}&-{11}&-{41}&{\mid}&{0}&{1}&{0}\-{1}&-{3}&-{11}&{\mid}&{0}&{0}&{1}}]}[/latex-display] Multiply row 1 by [latex-display]\displaystyle\frac{{1}}{{5}}[/latex-display] Multiply row 1 by 4 and add to row 2. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&frac{{56}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\-{1}&-{3}&-{11}&{\mid}&{0}&{0}&{1}}]}[/latex-display] Add row 1 to row 3. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&frac{{56}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\{0}&{0}&frac{{1}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{1}}]}[/latex-display] Multiply row 2 by –3 and add to row 1. [latex-display]\displaystyle{\left[\matrix{{1}&{0}&-\frac{{1}}{{5}}&{\mid}&-\frac{{11}}{{5}}&-{3}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\{0}&{0}&frac{{1}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{1}}]}[/latex-display] Multiply row 3 by 5. [latex-display]\displaystyle{\left[\matrix{{1}&{0}&-\frac{{1}}{{5}}&{\mid}&-\frac{{11}}{{5}}&-{3}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\{0}&{0}&{1}&{\mid}&{1}&{0}&{5}}]}[/latex-display] Multiply row 3 by $\displaystyle\frac{{1}}{{5}}$ Multiply row 3 by $\displaystyle-\frac{{19}}{{5}}$ So, [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{2}&-{3}&{1}\-{3}&{1}&-{19}\{1}&{0}&{5}}]}[/latex-display] Multiply both sides of the equation by A–1. We want A–1AX = A–1B: Thus, [latex-display]\displaystyle{A}^{{-{{1}}}}{B}={\left[\matrix{-{70}+{78}-{7}\-{105}-{26}+{133}\{35}+{0}-{35}}]}={\left[\matrix{{1}\{2}\{0}}]}[/latex-display] The solution is (1,2,0). 2. $\displaystyle{X}={\left[\matrix{{4}\{38}\{58}}]}$ ### How To Given a system of equations, solve with matrix inverses using a calculator. 1. Save the coefficient matrix and the constant matrix as matrix variables [A] and [B]. 2. Enter the multiplication into the calculator, calling up each matrix variable as needed. 3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message. ### Example 3 Solve the system of equations with matrix inverses using a calculator [latex-display]\displaystyle{\left[\matrix{{2}{x}+{3}{y}+{z}={32}\{3}{x}+{3}{y}+{z}=-{27}\{2}{x}+{4}{y}+{z}=-{2}}.}[/latex-display] #### Solution On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [A] and enter the constant matrix as the matrix variable [B]. [latex-display]\displaystyle{[{A}]}={\left[\matrix{{2}&{3}&{1}\{3}&{3}&{1}\{2}&{4}&{1}}]},{[{B}]}={\left[\matrix{{32}\-{27}\-{2}}]}[/latex-display] On the home screen of the calculator, type in the multiplication to solve for X, calling up each matrix variable as needed. [A]–1 × [B] Evaluate the expression. [latex-display]\displaystyle{\left[\matrix{-{59}\-{34}\{252}}]}[/latex-display] ### Media Access these online resources for additional instruction and practice with solving systems with inverses. # Key Equations Identity matrix for a matrix $\displaystyle{I}_{{2}}={\left[\matrix{{1}&{0}\{0}&{1}}]}$ Identity matrix for a matrix $\displaystyle{I}_{{3}}={\left[\matrix{{1}&{0}&{0}\{0}&{1}&{0}\{0}&{0}&{1}}]}$ Multiplicative inverse of a matrix # Key Concepts • An identity matrix has the property $\displaystyle{A}{I}={I}{A}={A}$. • An invertible matrix has the property $\displaystyle{A}{A}^{{-{{1}}}}={A}^{{-{{1}}}}{A}={I}$. • Use matrix multiplication and the identity to find the inverse of a 2 × 2 matrix. • The multiplicative inverse can be found using a formula. • Another method of finding the inverse is by augmenting with the identity. • We can augment a 3 × 3 matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. • Write the system of equations as $\displaystyle{A}{X}={B}$. • We can also use a calculator to solve a system of equations with matrix inverses.
S k i l l i n A R I T H M E T I C Lesson 15 # PARTS OF NATURAL NUMBERS 1 In this chapter we will learn to speak the language of arithmetic. This will allow us to relate any two numbers. We will learn to say, for example, that 6 is the fourth part, or one quarter, of 24, and that 18 is three quarters of 24. We are not concerned here with "How do you do it?" but, rather, "What does that mean?" What is more, to really understand Percent, it is necessary to understand parts, because a percent is a part of 100%. 50% means half -- because 50 is half of 100. 25% means a quarter beause 25 is a quarter of 100. And 20% means a fifth because 20 is the fifth part of 100. Finally, fractions (Lesson 20) are parts of number 1. 1. What is a natural number? It is a collection composed of equal indivisible units; each of them we say is one. That is the idea of a natural number. To those collections we give a sequence of names and symbols. 1, 2, 3, 4, and so on, are the familiar numerals for the natural numbers. We calculate with those symbols, and so it has become conventional to call the symbols themselves "numbers." Yet a symbol is not what it symbolizes, what it stands for, which in this case is a number of discrete units. By a number in what follows, we will mean a natural number. Cardinal and ordinal The names of the natural numbers have two forms: cardinal and ordinal. The cardinal forms are One, two, three, four, and so on. They answer the question How many? The ordinal forms are First, second, third, fourth, and so on. They answer the question Which one? We will see that those ordinal numbers express division into equal parts. With the exception of "half," an ordinal number will name which part -- the third part, the fourth, the fifth, and so on. To each cardinal number except 1, there will correspond the name of a part. 2 Half 3 Third 4 Fourth 5 Fifth And so on. 2. What do we mean by the multiples of a number? They are the numbers produced when that number is repeatedly added. Here are the first few multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40. 5 is the first multiple of 5; 10 is the second multiple; 15, the third; and so on. 3. What does it mean to say that a smaller number is a part of a larger number? It means that the larger number is a multiple of the smaller. Equivalently, the smaller is contained in the larger an exact number of times. 5, then, will be a certain part of each one of its multiples. 5, 1015202530, 35, 40. Since 15 is the third multiple of 5, we say that 5 is the third part of 15. We use that same ordinal number to name the part. Since 20 is the fourth multiple of 5, we call 5 the fourth part of 20. 5 is the fifth part of 25, the sixth part of 30; and so on. But, 5 is half of 10. (We do not say the second part.) And 5 is not a part of itself; there is no such thing as the first part. So with the exception of "half," an ordinal number names into which parts a number has been divided. 15 has been divided into Thirds; that is, into three equal parts. The third part of 15 is 5. This use of ordinal numbers does not imply a sequence: the first part, the second, the third, and so on. See below. If we divide a number into four equal parts, then we have divided it into fourths (or quarters); if into five equal parts, into fifths. But if we divide into two equal parts, then we have divided it in half. It is important to understand that we are not speaking here of proper fractions -- numbers that are less than 1 and that we need for measuring. We are explaining how the ordinal numbers --- third, fourth, fifth, etc. -- name the equal parts into which a number has been divided. When answering the questions of this Lesson, the student should not write fractions. We will come to those symbols in Lesson 20. It should be clear that the ordinal names of the parts belong to language itself, and are prior to the names of the proper fractions, which are the parts of 1. Why is the number we write as 1 over 3 -- - called "one-third"? Because the numerator is one third -- the third part -- of the denominator. p That must be understood first. We can then explain that the number we call is the third part of 1. p Example 1. 3 is which part of 18? Answer. The sixth part. 3 is contained in 18 six times. Note that 1 is a part of every number (except itself) because every number is a multiple of 1. Which part is it? The part that says the number's name. 1 is the third part of 3, the fourth part of 4, the fifth part of 5, the hundredth part of 100. 1 is half of 2. Example 2. What number is the fourth part, or a quarter, of 28? Four times 7 is 28. Example 3. 2 is the fifth part of what number? Answer. 10. Because five 2's are 10. Every number is the fifth part of five times itself 4 is the fifth part of 5 × 4, which is 20. 9 is the fifth part of 5 × 9, which is 45. 20 is the fifth part of 5 × 20, which is 100. 4. How can we calculate a part of a number? Divide by the cardinal number that corresponds to the name of the part. To take half of a number, divide by 2. To take a third, divide by 3. And so on. See Lesson 11, Question 2, and especially Example 6. Example 4. How much is an eighth of \$72? Answer. 72 ÷ 8 = 9. An eighth of \$72 is \$9. Example 5. Tenths, hundredths. How much is a tenth of \$275? How much is a hundredth? Answer. To find a tenth, divide by 10. 275 ÷ 10 = 27.5 Since this is money, we report the answer as \$27.50. (Lesson 3.) As for a hundredth, we will separate two decimal digits: \$275 ÷ 100 = \$2.75 In Lesson 4, we saw that when we divide by 10, we have taken 10% of the number. And when we divide by 100, we have taken 1%. Therefore, 10% of \$275 is \$27.50. 1% is \$2.75. Note: Whenever we divide by any power of 10 -- the digits do not change. 275 ÷ 100 = 2.75 Conversely, then, if two numbers have the same digits, they differ by a power of 10. Example 6. \$85 is which part of \$850? Answer. Apart from the 0 at the end of \$850, those numbers have the same digits. Therefore, they differ by a power of 10. 850 is in fact 10 times 85. (Lesson 4, Question 1.) Therefore, \$85 is the tenth part of \$850. To say the same thing, \$85 is 10% of \$850. Example 7. \$.98 is which part of \$98? Answer. They have the same digits. They differ by a power of 10. \$.98 is the hundredth part of \$98. It is 1% of it. Divisors We say that a smaller number is a divisor of a larger number if the larger is a multiple of the smaller. 3 is a divisor of 12 because 12 is a multiple of 3. 5 is not a divisor of 12. We say however that a number is a divisor of itself. With the exception of the number itself, the divisors of a number are the only parts that a number has. 3 is the fourth part of 12. 6 is half of 12. 5 is not any part of 12. You cannot divide 12 people into groups of 5. Example 8. Find all the divisors of 30 in pairs. Each divisor (except 30) is which part of 30? Answer. Here are all the divisors of 30 in pairs: 1 and 30. (Because 1 × 30 = 30.) 2 and 15. (Because 2 × 15 = 30.) 3 and 10. (Because 3 × 10 = 30.) 5 and 6. (Because 5 × 6 = 30.) On naming which part of 30, each divisor will say the ordinal name of its partner: 1 is the thirtieth part of 30. 2 is the fifteenth part of 30. 15 is half of 30. 3 is the tenth part of 30. 10 is the third part of 30. 5 is the sixth part of 30. 6 is the fifth part of 30. Divisors always come in pairs. And that implies the following: Theorem. For every divisor (except 1) that a number has, it will have a part with the ordinal name of that divisor. (Euclid, VII. 37.) Since 18, for example, has a divisor 3, then 18 has a third part. Since 18 has a divisor 6, then 18 has a sixth part. But 18 does not have a fifth part, because 5 is not a divisor of 18. Here is an illustration that 18 has a divisor 3: 18 = 6 × 3. But according to the order property of multiplication: 18 = 3 × 6. This shows that 6 -- the partner of 3 -- is the third part of 18. In other words, since 18 has a divisor 3, then 18 has a third part. Example 9. Into which parts could 12 people be divided? Answer. The divisors of 12 are 1, 2, 3, 4, 6, and 12. Corresponding to each divisor (except 1), there will be a part with the ordinal name of the divisor. 12 people, therefore, could be divided into Halves, thirds, fourths, sixths, and twelfths. You cannot take a fifth of 12 people. 12 does not have a divisor 5. Percent: Parts of 100% A percent is another way of naming a part. Because whatever part the percent is of 100%, that is the part we are naming. Since 50% is half of 100%, then 50% means half. 50% of 12 -- half of 12 -- is 6. Since 25% is a quarter, or a fourth, of 100% -- -- then 25% is another way of saying a quarter. 25% of 40 -- the fourth part of 40 -- is 10. In the next Lesson, Question 10, we will see how to take 25% by taking half of half. Since 20% is the fifth part of 100% -- (100 is made up of five 20's) -- then 20% is another way of saying a fifth. 20% of 15 -- the fifth part of 15 -- is 3. Repeated division in half Every time we take half of something, we get twice as many parts. Half of a whole -- -- -- results in two equal parts. Each part is Half. If we divide each Half in half -- -- the whole will be in four equal parts, or Quarters. If we divide each Quarter in half -- -- the whole will then be in twice as many, that is, eight equal parts, or Eighths. Half of an Eighth is a Sixteenth. Half of a Sixteenth is a Thirty-second. And so on. Now, here are the number of equal parts that result when we repeatedly take half: 2, 4, 8, 16, 32, 64, and so on. Those numbers are called the powers of 2. Repeated division in half is very common. The student should know the names of the sequence of those parts: Halves, Quarters, Eighths, Sixteenths, Thirty-seconds, and so on. * When we say "5 is the third part of 15," we do not imply a sequence: the first part, the second part, the third, and so on. When we speak of the third part, that is a different meaning for the word "third." It means each one of three equal parts that together make up the whole. We say that we have divided 15 into thirds. Yet "third" still retains its ordinal character. Because to the question, "Which part of 15 is 5?", we answer, "The third part." We use that ordinal number because 15 is the third multiple of 5. At this point, please "turn" the page and do some Problems. or Continue on to the next Section: Parts, plural Please make a donation to keep TheMathPage online. Even \$1 will help.
## Sunday, September 25, 2011 ### Ishaka's Blog Post During our math class on Friday we didn't do much because we had an evacuation drill, but we answered question 4 a) from page 32 with the enitre class. Question a) was asking us to find the surface area of a 3D shape. (Question 4 a) as a net) STEP 1 We had to split the 2 sides into to parts because we didn't have a formula that would help us with that kind of shape. STEP 2 After spliting the two sides into two parts, we started to label A1, A2, A3, and A4, for the parts that were the same to help us with the formula. (There are 3 A2's because both the tops equal the base, and there are also 3 A1's because both the fronts equal the back.) STEP 3 Now that we had all the information that we needed to find the surface area, we just filled in the information into our formula. S.A = 2(A1) + 2(A2) + 2(A3) + 2(A4) 2(lw) + 2(lw) + 2(s^2) + 2(lw) 2(4.5) + 2(2.5) + 2(2^2) + 2(2.1) 40 + 20 + 8 + 4 = 72units^2 HOMEWORK: Page 32 #'s 4 - 9, and also write in your math journal. John will be doing the next blog post. 1. Great Job on this post Ishaka! I liked how you showed step by step on how we did our work. Also how you told why we had to split the two sides in half. Also, how you colored important words and this post was very neat. Keep it up! :) 2. Goood job Ishaaka! Like what Thessa said, I also liked how you showed the steps. Good job on including pictures and making it easy to understand. Keep it up :) 3. Nice job Ishaka! I like how your photos are bolded so that it's clear. I found it useful that you have broken up what we were taught into steps. I suggest moving the words side either out of the rectangle or fit it closer, because in the other steps it gets cut off. Keep up the good work though! 4. Great job Ishaka! I like how you explained the steps to this question with detail. It was also easy to read. I really like how you wrote which sides were the equal to other sides. Overall, great job! 5. Good job Ishaka! I really like how you explained exactly how we did that question in class. However you could have edited your work because I found one error. "We had to split the 2 sides into to parts" "to" should have been "two. But I liked how you made some words pop out from the rest by changing the colours. I think that you could have changed the colours of the labels on your pictures just to make them stand out more. Also, it was nice that you reminded us of the homework. Over all, great work! 6. Great job Ishaka! I really like how you explained every step clearly and made it easy to understand. I also like that you added pictures to make it more easier to understand what you are saying. Overall, good job! 7. Great job Ishaka!! I really like how you explained that we split the sides into two parts and why. I also like how your pictures are labelled and on grid paper to make it easier to see the lengths. Keep it up! 8. Great Post Ishaka. Your steps and pictures of the net was very detailed. You also explained them really well, but at the end when you showed the Surface Area it was kind of confusing. Overall, Its great and adding extras like links or videos might also improve your post. 9. Great job Ishaka! I liked how you explained the things we did in class clearly. Your pictures were also nice and your post was organized. I think you should have aligned the formulas so it would be easier to read and understand. Also add a video and/or link to make your post a little better. Other than that, great job Ishaka. 10. Good job Ishaka! I really liked how you explained how to find the surface area of a 3D object step by step. It was really clear! Also, I liked how you used pictures to explain your post. Although, I think you should have added a video or link to expand our knowledge but overall you did great! 11. Good job Ishaka! I like how you explained the method by steps, and how you included pictures as well as words how to do it. Few things I would suggest is putting a link, other than my suggestion, you did a fine job on your post. 12. Good job Ishaka! I liked that you showed step by step in this problem! I also thought that you shouldn't have typed so much because I prefer to see what you're trying to explain. I thought your pictures were nice. Also when you write your equation you should each everything a lined. Good job!
# Solving Problems Using Percent By Donna Ventura Students will solve problems finding the whole, given a part and the percent. ## Lesson Objective The lesson is aligned to the Common Core State Standards for Mathematics – 6.RP.3.c Ratios and Proportional Relationships – Find a percent of a quantity as a rate per 100; solve problems involving finding the whole, given a part and the percent. ## Lesson Procedure Find a percent of a quantity as a rate per 100. There are 4 red cars in the parking lot. There are a total of 20 cars in the parking lot. • A ratio can be written to show how the number of red cars relates to the total number of cars. The ratio can be written as 4 to 20 or 4 : 20 or 4/20. • A percent can be written to show how many cars are red for every 100 cars. Complete the table to show the percent of red cars in the parking lot. What percent of the cars in the parking lot are red? Multiply 20 x 5 = 100, so multiply 4 x 5 = 20. 20% of the cars in the parking lot are red. ## Individual or Group Work Solve problems involving finding the whole when given a part and the percent. Janet took a test in science and correctly answered 19 of the 25 questions. Janet took a test in spelling and correctly answered 18 of the 20 questions. 1. Complete the table. 2. On which test (science or spelling) did she do better? Janet took a test in history and correctly answered 21 of the questions and received a score of 84%. Janet took a test in mathematics. There were a total of 10 questions on the test and received a score of 90%. 3. Complete the table. 4. How many total questions were on the history test? 5. How many questions did Janet correctly answer on the mathematics test? 6. The ratio of correct answers to the total questions on the test is 85 to 100. There were a total of 20 questions on the test. How many questions were answered correctly? Solve problems using percent. A school library has 16,250 books. 7. 40% of the books in the school library are realistic fiction books. How many books are realistic fiction? 8. 20% of the books in the school library are science fiction books. How many books are science fiction? 9. 30% of the books in the school library are reference books. How many books are reference books? 10. 10% of the books in the school library are biographies. How many books are biographies? 1. Complete the table. 2. Spelling Test 3. Complete the table. 4. 25 questions 5. 9 questions 6. 17 questions were answered correctly 7. 6,500 realistic fiction books 8. 3,250 science fiction books 9. 4,875 reference books 10. 1,625 biographies ## Concluding Statement Students should be able to solve problems finding the whole, given a part and the percent.
# How do solve the following linear system?: 3x-2y=2 , -6x+5y=1 ? Feb 1, 2016 Solve by substitution and elimination: $3 x - 2 y = 2$ $- 6 x + 5 y = 1$ We can eliminate $- 6 x$ from the second equation by $3 x$ in the first equation if we multiply it with $2$ to get $6 x$. $\rightarrow 2 \left(3 x - 2 y = 2\right)$ $\rightarrow = 6 x - 4 y = 4$ $\rightarrow \left(- 6 x + 5 y = 1\right) + \left(6 x - 4 y = 4\right)$ $\rightarrow \left(5 y = 1\right) + \left(- 4 y = 4\right)$ $\rightarrow y = 5$ Substitute the value of $y$ to the second equation: $\rightarrow - 6 x + 5 \left(5\right) = 1$ $\rightarrow - 6 x + 25 = 1$ $\rightarrow - 6 x = 1 - 25$ $\rightarrow - 6 x = - 24$ $\rightarrow x = \frac{- 24}{-} 6 = \frac{24}{6} = 4$ Because,$\frac{- n}{-} m = \frac{n}{m}$ So,$\left(x , y\right) = \left(4 , 5\right)$
### Alg - 1 - 8.5 ```8-5 Factoring Special Products Warm Up Determine whether the following are perfect squares. If so, find the square root. 1. 64 3. 45 5. y8 7. 9y7 Holt Algebra 1 yes; 8 no yes; y4 no 2. 36 yes; 6 4. x2 yes; x yes; 2x3 6. 4x6 8. 49p10 yes;7p5 8-5 Factoring Special Products Find the degree of each polynomial. A. 11x7 + 3x3 7 D. x3y2 + x2y3 – x4 + 2 5 B. 4 C. 5x – 6 1 The degree of a polynomial is the degree of the term with the greatest degree. Holt Algebra 1 8-5 Factoring Special Products Learning Targets Students will be able to: Factor perfectsquare trinomials and factor the difference of two squares. Holt Algebra 1 8-5 Factoring Special Products A trinomial is a perfect square if: • The first and last terms are perfect squares. • The middle term is two times one factor from the first term and one factor from the last term. 9x2 3x Holt Algebra 1 • + 12x + 4 3x 2(3x • 2) 2 • 2 8-5 Factoring Special Products Holt Algebra 1 8-5 Factoring Special Products Determine whether each trinomial is a perfect square. If so, factor. If not explain. 9x2 – 15x + 64 3x  3x 2(3x  8) 8  8 2(3x  8) ≠ –15x. 9x2 – 15x + 64 is not a perfect-square trinomial because –15x ≠ 2(3x  8). Holt Algebra 1 8-5 Factoring Special Products Determine whether each trinomial is a perfect square. If so, factor. If not explain. 81x2 + 90x + 25 9x ● 9x 2(9x 81x2 + 90x + 25 Holt Algebra 1 ● 5) 5 ● 5  9 x  5 The trinomial is a perfect square. Factor. 2 8-5 Factoring Special Products Determine whether each trinomial is a perfect square. If so, factor. If not explain. 36x2 – 10x + 14 6x  6x ??? The trinomial is not a perfect-square because 14 is not a perfect square. Holt Algebra 1 8-5 Factoring Special Products Determine whether each trinomial is a perfect square. If so, factor. If not explain. x2 + 4x + 4 x  x 2(x  2) 2  2 The trinomial is a perfect square. Factor. x  4x  4   x  2 2 Holt Algebra 1 2 8-5 Factoring Special Products A rectangular piece of cloth must be cut to make a tablecloth. The area needed is (16x2 – 24x + 9) in2. The dimensions of the cloth are of the form cx – d, where c and d are whole numbers. Find an expression for the perimeter of the cloth. Find the perimeter when x = 11 inches. 4x  3 16x2 – 24x + 9 4 x  4 x 2  4 x  3 16x2 – 24x + 9  4 x  3 P  x   4  4x  3 Holt Algebra 1 33 2 4x  3 4x  3 P 11  4  4 11  3  164" 4x  3 8-5 Factoring Special Products In Chapter 7 you learned that the difference of two squares has the form a2 – b2. The difference of two squares can be written as the product (a + b)(a – b). You can use this pattern to factor some polynomials. A polynomial is a difference of two squares if: •There are two terms, one subtracted from the other. • Both terms are perfect squares. 4x2 – 9 2x Holt Algebra 1  2x 3  3 8-5 Factoring Special Products Recognize a difference of two squares: the coefficients of variable terms are perfect squares, powers on variable terms are even, and constants are perfect squares. Holt Algebra 1 8-5 Factoring Special Products Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 3p2 – 9q4 ??? 3p2 is not a perfect square. 3p2 – 9q4 is not the difference of two squares because 3p2 is not a perfect square. Holt Algebra 1 8-5 Factoring Special Products Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 100x2 – 4y2 10x  10x 2y  2y The polynomial is a difference of two squares. Write the polynomial as (a + b)(a – b). 100x2 – 4y2 = (10x + 2y)(10x – 2y) Holt Algebra 1 8-5 Factoring Special Products Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. x4 – 25y6 x2  x2 5y3  5y3 The polynomial is a difference of two squares. Write the polynomial as (a + b)(a – b). x4 – 25y6 = (x2 + 5y3)(x2 – 5y3) Holt Algebra 1 8-5 Factoring Special Products Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 1 – 4x2 1  1 2x The polynomial is a difference of two squares.  2x Write the polynomial as (a + b)(a – b). 1 – 4x2 = (1 + 2x)(1 – 2x) Holt Algebra 1 8-5 Factoring Special Products Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. p8 – 49q6 p4 ● p4 7q3 ● 7q3 The polynomial is a difference of two squares. Write the polynomial as (a + b)(a – b). p8 – 49q6 = (p4 + 7q3)(p4 – 7q3) Holt Algebra 1 8-5 Factoring Special Products Check It Out! Example 3c Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 16x2 – 4y5 16x2 – 4y5 4x  4x 4y5 is not a perfect square. 16x2 – 4y5 is not the difference of two squares because 4y5 is not a perfect square. HW pp. 562-564/13-29 odd,46-50,55-64 Holt Algebra 1 ```
Home » How Can Two Or More Velocities Be Combined? Update New # How Can Two Or More Velocities Be Combined? Update New Let’s discuss the question: how can two or more velocities be combined. We summarize all relevant answers in section Q&A of website Abigaelelizabeth.com in category: Blog Marketing For You. See more related questions in the comments below. ## Can two velocities be added? Yes, two velocities can be added to give a resultant of 0 magnitude. The velocities should be equal in magnitude and opposite in direction. The true explanation of the extra term is much simpler: velocities don’t simply add. To add the velocity v to the velocity c/n, we must use the addition of velocities formula above, which gives the light velocity relative to the ground to be: v+c/n1+v/nc. ### Velocity and combining velocities Velocity and combining velocities Velocity and combining velocities ## How do you calculate the resultant velocity of two velocities in the same direction? Divide the total momentum by the sum of the masses if the two objects stick together after impact. This will give you the resultant velocity of the two objects. In the example above, we would take 50 and divide by the sum of the masses, which is 10, getting a result of 5 meters per second. ## Can two velocities be added using Law of triangle of vectors? Yes! We can add two velocities using the triangle law of vector addition. ## How do you find velocity with two velocities? 1. Calculate v = (v + u) / 2. … 2. Average velocity (v) of an object is equal to its final velocity (v) plus initial velocity (u), divided by two. 3. The average velocity calculator solves for the average velocity using the same method as finding the average of any two numbers. … 4. Given v and u, calculate v. … 5. Given v and v calculate u. ## How do you find the velocity? Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt. ## What are the 3 types of changes in velocity? Describe three types of changes in velocity. Acceleration can be described as changes in speed, changes in direction, or changes in both. Acceleration is a vector. What is the equation for acceleration? ## Is the rate at which velocity changes? The rate at which velocity changes with time is called Acceleration is a measure of how quickly the velocity is changing. If velocity does not change, there is no acceleration. ## What two dimensions are combined to measure velocity? Speed and velocity are both measured using the same units. The SI unit of distance and displacement is the meter. The SI unit of time is the second. The SI unit of speed and velocity is the ratio of two — the meter per second . ## How do you represent velocity as a vector? Velocity is a vector, which is a measurement that includes both size and direction. Velocity can be represented by an arrow, with the length of the arrow representing speed and the way the arrow points representing direction. ### Physics – Special Relativity (6 of 43) Relativistic Velocity: Another Example Physics – Special Relativity (6 of 43) Relativistic Velocity: Another Example Physics – Special Relativity (6 of 43) Relativistic Velocity: Another Example ## Is resultant velocity same as final velocity? The resultant velocity on a system can be considered the same as the final velocity. The resultant velocity is another word for net velocity. ## How do you find the combined velocity after a collision? In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before. ## What is the difference between resultant velocity and relative velocity? The relative velocity refers to how one observer in his own frame would see another moving object. Resultant velocity would be the velocity of an object (in some fixed reference frame) when there are more than one influences on its motion. ## How will you obtain the sum of two vectors by triangle method? The triangle law of vector addition states that if two vectors are represented by the sides of a triangle taken in order of magnitude and direction, then the resultant sum of the vectors is given by the triangle’s third side in reverse order of magnitude and direction. ## What is addition triangle law? What is Triangle Law of Vector Addition? Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. ## What is triangle law and parallelogram law? If two vectors acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram or triangle, then their resultant R is represented in magnitude and direction by the diagonal of both, which passes through that common point. ## Is velocity same as magnitude of the velocity? Velocity, being a vector, has both a magnitude and a direction. The magnitude of the velocity vector is the instantaneous speed of the object. The direction of the velocity vector is directed in the same direction that the object moves. ## What are three examples of velocity? Following are some examples of the applications of the velocity: • Revolution of the Earth around the Sun. • Revolution of Moon around the Earth. • The velocity of the car. • The velocity of the train. • The river flowing with a variable velocity. • The velocity of the water flowing out of a tap. • The velocity of the ball hit by a bat. ### Relativistic Addition of Velocity \| Special Relativity Ch. 6 Relativistic Addition of Velocity \| Special Relativity Ch. 6 Relativistic Addition of Velocity \| Special Relativity Ch. 6 ## What is a velocity in physics? Velocity is a vector quantity that refers to “the rate at which an object changes its position.” Imagine a person moving rapidly – one step forward and one step back – always returning to the original starting position. While this might result in a frenzy of activity, it would result in a zero velocity. ## Which of the following is an example of velocity? Velocity is the rate of motion, speed or action. An example of velocity is a car driving at 75 miles per hour. Related searches • motion is described with respect to a • how are displacements combined • what does velocity describe • how many ways to swap two numbers • what shows the speed on a distance time graph • what is the difference between average speed and instantaneous speed • a ball just dropped is an example of • what shows the speed on a distance-time graph • how do velocities combine • displacement is distance combined with • which does not describe a change in velocity • can you add velocities together ## Information related to the topic how can two or more velocities be combined Here are the search results of the thread how can two or more velocities be combined from Bing. You can read more if you want. You have just come across an article on the topic how can two or more velocities be combined. If you found this article useful, please share it. Thank you very much.
Course Content Class 10th Science 0/32 Class 10th Maths 0/86 Class 10 Social Science History : India and the Contemporary World – II 0/16 Class 10 Social Science Geography : Contemporary India – II 0/14 Class 10 Social Science Civics (Political Science) : Democratic Politics – II 0/16 Class 10 Social Science Economics: Understanding Economic Development – II 0/10 Class 10 English First Flight Summary 0/22 Class 10 English First Flight Poem 0/22 Class 10 English Footprints without Feet Summary 0/20 Online Class For 10th Standard Students (CBSE) (English Medium) ### Exercise 11.2 Page: 221 In each of the following, give the justification of the construction also: 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Construction Procedure: The construction to draw a pair of tangents to the given circle is as follows. 1. Draw a circle with radius = 6 cm with centre O. 2. Locate a point P, which is 10 cm away from O. 3. Join the points O and P through line 4. Draw the perpendicular bisector of the line OP. 5. Let M be the mid-point of the line PO. 6. Take M as centre and measure the length of MO 7. The length MO is taken as radius and draw a circle. 8. The circle drawn with the radius of MO, intersect the previous circle at point Q and R. 9. Join PQ and PR. 10. Therefore, PQ and PR are the required tangents. Justification: The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 6cm with centre O. To prove this, join OQ and OR represented in dotted lines. From the construction, ∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle, so it becomes, ∴ ∠PQO = 90° Such that ⇒ OQ ⊥ PQ Since OQ is the radius of the circle with radius 6 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle. Hence, justified. 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation. Construction Procedure: For the given circle, the tangent can be drawn as follows. 1. Draw a circle of 4 cm radius with centre “O”. 2. Again, take O as centre draw a circle of radius 6 cm. 3. Locate a point P on this circle 4. Join the points O and P through lines such that it becomes OP. 5. Draw the perpendicular bisector to the line OP 6. Let M be the mid-point of PO. 7. Draw a circle with M as its centre and MO as its radius 8. The circle drawn with the radius OM, intersect the given circle at the points Q and R. 9. Join PQ and PR. 10. PQ and PR are the required tangents. From the construction, it is observed that PQ and PR are of length 4.47 cm each. It can be calculated manually as follows In ∆PQO, Since PQ is a tangent, ∠PQO = 90°. PO = 6cm and QO = 4 cm Applying Pythagoras theorem in ∆PQO, we obtain PQ2+QO2 = PQ2 PQ2+(4)2 = (6)2 PQ2 +16 =36 PQ2 = 36−16 PQ2 = 20 PQ = 2√5 PQ = 4.47 cm Therefore, the tangent length PQ = 4.47 Justification: The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 4 cm with centre O. To prove this, join OQ and OR represented in dotted lines. From the construction, ∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle, so it becomes, ∴ ∠PQO = 90° Such that ⇒ OQ ⊥ PQ Since OQ is the radius of the circle with radius 4 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle. Hence, justified. 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q Construction Procedure: The tangent for the given circle can be constructed as follows. 1. Draw a circle with a radius of 3cm with centre “O”. 2. Draw a diameter of a circle and it extends 7 cm from the centre and mark it as P and Q. 3. Draw the perpendicular bisector of the line PO and mark the midpoint as M. 4. Draw a circle with M as centre and MO as radius 5. Now join the points PA and PB in which the circle with radius MO intersects the circle of circle 3cm. 6. Now PA and PB are the required tangents. 7. Similarly, from the point Q, we can draw the tangents. 8. From that, QC and QD are the required tangents. Justification: The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 3 cm with centre O. To prove this, join OA and OB. From the construction, ∠PAO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle, so it becomes, ∴ ∠PAO = 90° Such that ⇒ OA ⊥ PA Since OA is the radius of the circle with radius 3 cm, PA must be a tangent of the circle. Similarly, we can prove that PB, QC and QD are the tangent of the circle. Hence, justified 4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60° Construction Procedure: The tangents can be constructed in the following manner: 1. Draw a circle of radius 5 cm and with centre as O. 2. Take a point Q on the circumference of the circle and join OQ. 3. Draw a perpendicular to QP at point Q. 4. Draw a radius OR, making an angle of 120° i.e(180°−60°) with OQ. 5. Draw a perpendicular to RP at point R. 6. Now both the perpendiculars intersect at point P. 7. Therefore, PQ and PR are the required tangents at an angle of 60°. Justification: The construction can be justified by proving that ∠QPR = 60° By our construction ∠OQP = 90° ∠ORP = 90° And ∠QOR = 120° We know that the sum of all interior angles of a quadrilateral = 360° ∠OQP+∠QOR + ∠ORP +∠QPR = 360o 90°+120°+90°+∠QPR = 360° Therefore, ∠QPR = 60° Hence Justified 5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Construction Procedure: The tangent for the given circle can be constructed as follows. 1. Draw a line segment AB = 8 cm. 2. Take A as centre and draw a circle of radius 4 cm 3. Take B as centre, draw a circle of radius 3 cm 4. Draw the perpendicular bisector of the line AB and the midpoint is taken as M. 5. Now, take M as centre draw a circle with the radius of MA or MB which the intersects the circle at the points P, Q, R and S. 6. Now join AR, AS, BP and BQ 7. Therefore, the required tangents are AR, AS, BP and BQ Justification: The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B with radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). From the construction, to prove this, join AP, AQ, BS, and BR. ∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠ASB = 90° ⇒ BS ⊥ AS Since BS is the radius of the circle, AS must be a tangent of the circle. Similarly, AR, BP, and BQ are the required tangents of the given circle. 6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. Construction Procedure: The tangent for the given circle can be constructed as follows 1. Draw the line segment with base BC = 8cm 2. Measure the angle 90° at the point B, such that ∠ B = 90°. 3. Take B as centre and draw an arc with a measure of 6cm. 4. Let the point be A where the arc intersects the ray. 5. Join the line AC. 6. Therefore, ABC be the required triangle. 7. Now, draw the perpendicular bisector to the line BC and the midpoint is marked as E. 8. Take E as centre and BE or EC measure as radius draw a circle. 9. Join A to the midpoint E of the circle 10. Now, again draw the perpendicular bisector to the line AE and the midpoint is taken as M 11. Take M as Centre and AM or ME measure as radius, draw a circle. 12. This circle intersects the previous circle at the points B and Q 13. Join the points A and Q 14. Therefore, AB and AQ are the required tangents Justification: The construction can be justified by proving that AG and AB are the tangents to the circle. From the construction, join EQ. ∠AQE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠AQE = 90° ⇒ EQ⊥ AQ Since EQ is the radius of the circle, AQ has to be a tangent of the circle. Similarly, ∠B = 90° ⇒ AB ⊥ BE Since BE is the radius of the circle, AB has to be a tangent of the circle. Hence, justified. 7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle. Construction Procedure: The required tangents can be constructed on the given circle as follows. 1. Draw a circle with the help of a bangle. 2. Draw two non-parallel chords such as AB and CD 3. Draw the perpendicular bisector of AB and CD 4. Take the centre as O where the perpendicular bisector intersects. 5. To draw the tangents, take a point P outside the circle. 6. Join the points O and P. 7. Now draw the perpendicular bisector of the line PO and midpoint is taken as M 8. Take M as centre and MO as radius draw a circle. 9. Let the circle intersects intersect the circle at the points Q and R 10. Now join PQ and PR 11. Therefore, PQ and PR are the required tangents. Justification: The construction can be justified by proving that PQ and PR are the tangents to the circle. Since, O is the centre of a circle, we know that the perpendicular bisector of the chords passes through the centre. Now, join the points OQ and OR. We know that perpendicular bisector of a chord passes through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. Since, ∠PQO is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠PQO = 90° ⇒ OQ⊥ PQ Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, ∴ ∠PRO = 90° ⇒ OR ⊥ PO Since OR is the radius of the circle, PR has to be a tangent of the circle Therefore, PQ and PR are the required tangents of a circle.
# We Try To Predict Shannen Doherty’s Future (02/15/2020) How will Shannen Doherty fare on 02/15/2020 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is not at all guaranteed – do not take this too seriously. I will first find the destiny number for Shannen Doherty, and then something similar to the life path number, which we will calculate for today (02/15/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology experts. PATH NUMBER FOR 02/15/2020: We will analyze the month (02), the day (15) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 02 and add the digits together: 0 + 2 = 2 (super simple). Then do the day: from 15 we do 1 + 5 = 6. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 2 + 6 + 4 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the path number for 02/15/2020. DESTINY NUMBER FOR Shannen Doherty: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for Shannen Doherty we have the letters S (1), h (8), a (1), n (5), n (5), e (5), n (5), D (4), o (6), h (8), e (5), r (9), t (2) and y (7). Adding all of that up (yes, this can get tedious) gives 71. This still isn’t a single-digit number, so we will add its digits together again: 7 + 1 = 8. Now we have a single-digit number: 8 is the destiny number for Shannen Doherty. CONCLUSION: The difference between the path number for today (3) and destiny number for Shannen Doherty (8) is 5. That is larger than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too concerned! As mentioned earlier, this is just for fun. If you want really means something, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did. ### Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. #### Latest posts by Abigale Lormen (see all) Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
# 17 4 Calculating Heats of Reaction CALCULATING HEATS • Slides: 24 17. 4 Calculating Heats of Reaction CALCULATING HEATS OF REACTION OBJECTIVES • Apply Hess’s law of heat summation to find enthalpy changes for chemical and physical processes • Calculate formation enthalpy changes using standard heats of Requisite Skills �Algebra I & II: System of Equations �VOCABULARY: o Hess’s law of heat of summation o Standard heat of formation . . . manipulate and combine algebraic equations. In algebra I, you learned that you can combine the like terms from two equations to make a third equation. 2 x + 6 y = 16 y = 2 x + 12 + 2 x + 7 y = 2 x + 28 7 y = 28 y=4 4 = 2 x + 12 x = -4 If you do this right, this allows you to solve for one of the variables by eliminating the other one. Once you know of the variables, you can plug it in and solve for the other variable. 4 - 12 = 2 x + 12 -8 = 2 x . . . manipulate and combine algebraic equations. Sometimes, you need to flip one of the equations before adding them to get one of the terms to disappear. 9 y = 3 x - 57 2 y + 22 = 3 x 3 x = 2 y + 22 + If we flip the second equation, and 3 x + 9 y = 3 x + 2 y - 35 THEN add the two equations, the “x” terms will disappear. 9 y = 2 y - 35 2(-5) + 22 = 3 x 7 y = -35 -10 + 22 = 3 x y = -5 12 = 3 x x=4 . . . manipulate and combine algebraic equations. Other times, you need to multiply an equation by an integer to get what you want. 4 y = x + 5 4 x + 7 y = 49 Even if we flip one of the equations, nobody disappears after adding the equations, but if we multiply the first equation by four first. . . 4(3) = x + 5 12 = x + 5 x=7 16 y = 4 x + 20 + 4 x + 7 y = 49 4 x + 23 y = 4 x + 69 23 y = 69 y=3 . . . manipulate and combine algebraic equations. Sometimes, you can use these old algebra tricks with chemical equations to solve for unknowns. . . Which stone do you prefer? And why? Diamond Graphite Would you like Diamond to be converted into Graphite? �Such conversion or reaction will take millions of years to complete �And chemists are curious to know the enthalpy changes for the conversion of diamond to graphite. �Should they wait million of years when the reaction will be completed to get the data? Or �What do you recommend they should do? So that the enthalpy changes data would be available ASAP Hess’s law Mr. Hess came up with a quick solution of how to calculate heat of reaction, ΔHrxn for: � 1. reactions that are too slow � 2. reactions that have intermediate steps � 3. reactions that are dangerous � 4. reactions that are not very useful Hess’s law Heat of Summation �States that if you add to two or more thermochemical equations to give final equation, then you can also add the heat of reaction to give the final heat of reaction �Hess’s law allows us to determine the heat of reaction indirectly. Operations in Hess’s law Any of these could be applied: �Flipping a known chemical equation �Changing the sign of the ΔH (-ve or +ve) when you flipped �Adding two or more chemical equations �Subtracting two or more chemical equations �Multiplying a chemical equation with a whole number or fraction �Dividing a chemical equation with a whole number �C (s, diamond) = C (s, graphite) ΔH= ? k. J C (s, graphite) + O 2 (g) = CO 2 (g) ΔH= -393. 5 k. J b) C (s, diamond) + O 2 (g) = CO 2 (g) ΔH= -395. 4 k. J a) � CO 2 and O 2 need to be cancelled out to achieve the desired reaction we are looking for. � We need graphite to be on the product side, so we are going to flip equation (a) �C (s, diamond) + O 2 (g) CO 2 (g) �CO 2 (g) C(s, graphite) + O 2 (g) C(s, diamond) ΔH= -395. 4 k. J ΔH= 393. 5 k. J C(s, graphite) ΔH= (-395. 4 + 393. 5)k. J ΔH= (393. 5 – 395. 4)k. J ΔH= - 1. 9 k. J . . . use Hess’ law to determine DHrxn Hess’s law allows us to figure out DHrxn for a reaction without ever having to make the reaction happen in real life. HLEx 1: Imagine that experiments tell you the following: Sn(s) + Cl 2(g) Sn. Cl 2(s) DH = -325. 1 k. J Sn. Cl 2(s) + Cl 2(g) Sn. Cl 4(l) DH = -186. 2 k. J If you add the two equations above, you get the following: Sn(s) + Sn. Cl 2(s) + 2 Cl 2(g) Sn. Cl 2(s) + Sn. Cl 4(l) Just as with normal, algebraic equations, when the same term appears on the left and right, it can be crossed out. . . yielding. . . Sn(s) + 2 Cl 2(g) Sn. Cl 4(l) DH = ? k. J . . . use Hess’ law to determine DHrxn Now here’s the real magic. Hess’s law says that you can also add the DH’s for the reactions to get the DH for the final reaction. Sn(s) + Cl 2(g) Sn. Cl 2(s) + Cl 2(g) Sn. Cl 4(l) DH = -325. 1 k. J DH = -186. 2 k. J (-325. 11 k. J) + (-186. 2 k. J) = -511. 3 k. J The sneaky miracle here is that we figured this out without ever having to make tin metal and chlorine gas react to form tin (IV) chloride in real life. Sn(s) + 2 Cl 2(g) Sn. Cl 4(l) DH==-511. 3 ? k. J DH . . . use Hess’ law to determine DHrxn That was a very simple usage of Hess’ law. We didn’t have to manipulate any equations before adding them. Let’s try a harder problem. . . . use Hess’ law to determine DHrxn HLEx 2: Let’s say experiments have told us the following: DH = -335 k. J Os(cr) + 2 O 2(g) Os. O 4(g) Os. O 4(cr) Os. O 4(g) DH = 56. 4 k. J Use Hess’s Law to figure out DH for the following reaction: Os(cr) + 2 O 2(g) Os. O 4(cr) DH = ? k. J (The final equation looks a lot like the first equation, but notice that Os. O 4 is a gas in the first equation and a crystal in the third. ) What would you have to do to figure this one out? Think a moment. . . . . . use Hess’ law to determine DHrxn HLEx 2: DH = -335 k. J Os(cr) + 2 O 2(g) Os. O 4(g) Os. O 4(cr) Os. O 4(g) DH = 56. 4 k. J Os(cr) + 2 O 2(g) Os. O 4(cr) DH = ? k. J Os. O 4(g) is in both of the initial equations, but doesn’t appear in the final equation, so it needs to be eliminated somewhow. Os. O 4(g) is on the right on both equations, so they won’t cancel each other out if you add the equations as they are. One filthy little trick you can do is to flip the second equation to put Os. O 4(g) on the left, and THEN add the two equations together. . . . use Hess’ law to determine DHrxn HLEx 2: DH = -335 k. J Os(cr) + 2 O 2(g) Os. O 4(g) Os. O 4(cr) Os. O 4(g) DH = 56. 4 k. J Os(cr) + 2 O 2(g) Os. O 4(cr) Os(cr) + 2 O 2(g) Os. O 4(g) Os. O 4(cr) DH = ? k. J DH = -335 k. J DH = -56. 4 k. J Notice how the sign of DH changed on the equation that we flipped. Remember that DHsolid = -DHfus and DHcond = -DHvap Freezing is the opposite of melting and condensation is the opposite of vaporization, so their DH’s have opposite signs. . . . use Hess’ law to determine DHrxn HLEx 2: DH = -335 k. J Os(cr) + 2 O 2(g) Os. O 4(g) Os. O 4(cr) Os. O 4(g) DH = 56. 4 k. J Os(cr) + 2 O 2(g) Os. O 4(cr) Os(cr) + 2 O 2(g) Os. O 4(g) Os. O 4(cr) DH = ? k. J DH = -335 k. J DH = -56. 4 k. J Os(cr) + 2 O 2(g) + Os. O 4(g) + Os. O 4(cr) Now we can add the equations. . . and simpify. . . Now we have the equation we were looking for. Now what? . . . use Hess’ law to determine DHrxn HLEx 2: DH = -335 k. J Os(cr) + 2 O 2(g) Os. O 4(g) DH = 56. 4 k. J Os. O 4(cr) Os. O 4(g) Os(cr) + 2 O 2(g) Os. O 4(cr) DH = -335 k. J Os(cr) + 2 O 2(g) Os. O 4(g) Os. O 4(cr) DH = ? k. J DH = -56. 4 k. J Os(cr) + 2 O 2(g) + Os. O 4(g) + Os. O 4(cr) Add the DH’s to get the DH for the final equation. DH = -335 k. J + (-56. 4 k. J) = -391. 4 k. J = DH . . . use standard heats of formation to determine DHrxn [Section under construction] Press a button! Go to a place!™ 17. 1 17. 2 17. 3 17. 4 17. 1 Temperature & Heat 17. 2 Calorimetry 17. 3 Heating Curve for Water 17. 4 Alegbra Review 17. 1 Endothermic & Exothermic 17. 2 Thermochemical Equations 17. 3 State Change Math Problems 17. 4 Hess’ Law 17. 1 Q = m DT Cp Problems 17. 2 Heat of Combustion 17. 3 Heat of Solution Math Problems 17. 4 Standard Heat of Formation
# How do you find the vertical, horizontal or slant asymptotes for f(x)=(2x-4)/(x^2-4) ? Jun 2, 2016 vertical asymptote x = -2 horizontal asymptote y = 0 #### Explanation: The first step here is to factorise and simplify f(x). $\Rightarrow f \left(x\right) = \frac{2 x - 4}{{x}^{2} - 4} = \frac{2 \cancel{\left(x - 2\right)}}{\cancel{\left(x - 2\right)} \left(x + 2\right)} = \frac{2}{x + 2}$ Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : x + 2 = 0 → x = -2 is the asymptote Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , f \left(x\right) \to 0$ divide terms on numerator/denominator by x $\Rightarrow \frac{\frac{2}{x}}{\frac{x}{x} + \frac{2}{x}} = \frac{\frac{2}{x}}{1 + \frac{2}{x}}$ as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0}$ $\Rightarrow y = 0 \text{ is the asymptote}$ Slant asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here (numerator-degree 0 , denominator-degree 1). Hence there are no slant asymptotes. graph{2/(x+2) [-10, 10, -5, 5]}
California Standards # California Standards Télécharger la présentation ## California Standards - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. California Standards Preparation for MG1.1 Compare weights, capacities, geometric measures, times, and temperatures within and between measurement systems (e.g., miles per hour and feet per second, cubic inches to cubic centimeters). Section 5-1: Ratios 2. Ratio Movie 3. 7 5 Notes A ratio is a comparison of two quantities. Ratios can be written in several ways. 7 to 5, 7:5, and name the same ratio. 12 inches = 1 foot 3 feet = 1 yard 36 inches = 1 yard 4. 15 ÷ 3 9 ÷ 3 bikes skateboards Additional Example 1: Writing Ratios in Simplest Form Write the ratio 15 bikes to 9 skateboards in simplest form. 15 9 Write the ratio as a fraction. = 5 3 Simplify. = = 5 3 The ratio of bikes to skateboards is , 5:3, or 5 to 3. 5. shirts jeans 24 ÷ 3 9 ÷ 3 Check It Out! Example 1 Write the ratio 24 shirts to 9 jeans in simplest form. Write the ratio as a fraction. 24 9 = 8 3 Simplify. = = 8 3 The ratio of shirts to jeans is , 8:3, or 8 to 3. 6. Practice • 15 cows to 25 sheep • 24 cars to 18 trucks • 30 Knives to 27 spoons 7. When simplifying ratios based on measurements, write the quantities with the same units, if possible. 8. 3 yards 12 feet 9 ÷ 3 12 ÷ 3 = = = 3 4 3 4 The ratio is , 3:4, or 3 to 4. Additional Example 2: Writing Ratios Based on Measurement Write the ratio 3 yards to 12 feet in simplest form. First convert yards to feet. 3 yards = 3 ● 3 feet There are 3 feet in each yard. Multiply. = 9 feet Now write the ratio. 9 feet 12 feet Simplify. 9. 36 inches 4 feet 36 ÷ 12 48 ÷ 12 = = = 3 4 3 4 The ratio is , 3:4, or 3 to 4. Check It Out! Example 2 Write the ratio 36 inches to 4 feet in simplest form. First convert feet to inches. 4 feet = 4 ● 12 inches There are 12 inches in each foot. = 48 inches Multiply. Now write the ratio. 36 inches 48 inches Simplify. 10. Practice • 4 feet to 24 inches • 3 yards to 12 feet • 2 yards to 20 inches 11. Notes Ratios that make the same comparison are equivalent ratios. Equivalent ratios represent the same point on the number line. To check whether two ratios are equivalent, you can write both in simplest form. 12. 1 9 1 9 12 15 3 27 27 36 2 18 Since , the ratios are equivalent. B. A. = and and 2 18 3 27 2 ÷ 2 18 ÷ 2 3 ÷ 3 27 ÷ 3 = = = = 4 5 3 4 Since , the ratios are not equivalent.  12 15 27 36 12 ÷ 3 15 ÷ 3 27 ÷ 9 36 ÷ 9 = = = = Additional Example 3: Determining Whether Two Ratios Are Equivalent Simplify to tell whether the ratios are equivalent. 1 9 1 9 4 5 3 4 13. Practice 14. 8 30 30 7 1 2 12 45 Possible answer: , Possible answer: , 4 15 7 21 3. 4. 1 3 14 42 Lesson Quiz: Part I Write each ratio in simplest form. 1. 22 tigers to 44 lions 2. 5 feet to 14 inches Find a ratios that is equivalent to each given ratio. 15. 36 24 16 10 5. 6. 8 64 16 128 and ; yes, both equal 28 18 32 20 8 5 3 2 14 9 8 5 1 8 =  ; yes ; no Lesson Quiz: Part II Simplify to tell whether the ratios are equivalent. and and 7. Kate poured 8 oz of juice from a 64 oz bottle. Brian poured 16 oz of juice from a 128 oz bottle. Are the ratios of poured juice to starting amount of juice equivalent?
# How to find a reflection image To find a reflection image of a point or a shape, there are some rules you can follow although it is not always possible to follow a rule. First, let us start with all situations where you can just follow a rule to quickly find the location of the reflected image on the coordinate system. Keep in mind that once you know how to find a reflection image for a point, you can easily find it for a shape. For example, if the shape has 3 points, just find the reflected image for all 3 points and then connect these points with a line. ## How to find a reflection image using the x-axis or the y-axis Reflection of a point across the x-axis The coordinate of point P is (3, 3) and the coordinate of the reflected image P' is (3, -3) The coordinate of point A is (-2, 4) and the coordinate of the reflected image A' is (-2, -4) What do you notice about the x-coordinate and y-coordinate of the preimage and image? We see that the x-coordinate of the preimage stays the same, but the y-coordinate of the image is the opposite of the y-coordinate of the preimage. In general, when reflecting a point across the x-axis, if the coordinate of the preimage is (x , y), then the coordinate of the reflected image is (x , -y) Reflection of a point across the y-axis The coordinate of point P is (-4, 3) and the coordinate of the reflected image P' is (4, 3) The coordinate of point A is (2, -4) and the coordinate of the reflected image A' is (-2, -4) What do you notice about the x-coordinate and y-coordinate of the preimage and image? We see that the y-coordinate of the preimage stays the same, but the x-coordinate of the image is the opposite of the x-coordinate of the preimage. In general, when reflecting a point across the y-axis, if the coordinate of the preimage is (x , y), then the coordinate of the reflected image is (-x , y) ## How to find a reflection image using the lines y = x and y = -x Reflection of a point across the line y = x The coordinate of point P is (1, 4) and the coordinate of the reflected image P' is (4, 1) The coordinate of point A is (-5, -2) and the coordinate of the reflected image A' is (-2, -5) Just swap the x-coordinate with the y-coordinate In general, when reflecting a point across the line y = x, if the coordinate of the preimage is (x , y), then the coordinate of the reflected image is (y, x) Reflection of a point across the line y = -x The coordinate of point P is (2, 4) and the coordinate of the reflected image P' is (-4, -2) The coordinate of point A is (4, -1) and the coordinate of the reflected image A' is (1, -4) Therefore, swap the x-coordinate with the y-coordinate and then take the opposite of both coordinates. In general, when reflecting a point across the line y = -x, if the coordinate of the preimage is (x , y), then the coordinate of the reflected image is (-y, -x) ## Reflection of a point in the origin The coordinate of point P is (5, 2) and the coordinate of the reflected image P' is (-5, -2) The coordinate of point A is (-2, 3) and the coordinate of the reflected image A' is (2, -3) Just take the opposite of both coordinates. In general, when reflecting a point in the origin, if the coordinate of the preimage is (x , y), then the coordinate of the reflected image is (-x, -y) When reflecting a point across any line, just keep this mind. The distance between the point and the line and the reflected point and the line is exactly the same. When reflecting a point A in any point P, just keep this mind these two things. The distance between point A and point P and the reflected point A' and point P is exactly the same. Point P, the reflected point A' and point A are colinear. ## Recent Articles 1. ### How To Find The Factors Of 20: A Simple Way Sep 17, 23 09:46 AM There are many ways to find the factors of 20. A simple way is to... 2. ### The SAT Math Test: How To Be Prepared To Face It And Survive Jun 09, 23 12:04 PM The SAT Math section is known for being difficult. But it doesn’t have to be. Learn how to be prepared and complete the section with confidence here.
Question Video: Finding the Center of Mass of a System of Point Masses Placed on the Vertices of a Right Triangle \| Nagwa Question Video: Finding the Center of Mass of a System of Point Masses Placed on the Vertices of a Right Triangle \| Nagwa # Question Video: Finding the Center of Mass of a System of Point Masses Placed on the Vertices of a Right Triangle Mathematics • Third Year of Secondary School ## Join Nagwa Classes The figure shows a system of point masses placed at the vertices of a triangle. The mass placed at each point is detailed in the table. Determine the coordinates of the center of gravity of the system. 02:18 ### Video Transcript The figure shows a system of point masses placed at the vertices of a triangle. The mass placed at each point is detailed in the table. Determine the coordinates of the centre of gravity of the system. Now, because the shape we’ve been given isn’t symmetrical, we’re going to need to be a little bit careful about how we determine the centre of mass of our object. Now, if we’re given a system of 𝑛 particles of masses 𝑚 one, 𝑚 two, and so on up to 𝑚𝑛, whose position vectors are 𝑟 one, 𝑟 two all the way up to 𝑟 𝑛, respectively. The centre of mass of the system is the point with position vector 𝑟 and is equal to the sum from 𝑖 equals one to 𝑛 over 𝑚 sub 𝑖 times 𝑟 sub 𝑖 all over capital 𝑀, where that’s the total mass of the system. In reality though, it can be much easier, especially in two dimensions, to just fit this into the 𝑥-direction and the 𝑦-direction. The 𝑥-coordinate of the centre of mass of a system of 𝑛 particles 𝑥 sub 𝐶 is 𝑚 one 𝑥 one plus 𝑚 two 𝑥 two all the way up to 𝑚 𝑛 𝑥 𝑛 over 𝑚 one plus 𝑚 two all the way up to 𝑚 𝑛. And the 𝑦-coordinate of the centre of mass 𝑦 sub 𝐶 is 𝑚 one 𝑦 one plus 𝑚 two 𝑦 two all the way up to 𝑚 𝑛 𝑦 𝑛 all over the total sum of the masses. Now here, 𝑥 one, 𝑥 two, and 𝑥 𝑛, 𝑦 one, 𝑦 two, and 𝑦 𝑛 are the 𝑥- and 𝑦-coordinates, respectively, of each of the individual particles. We’ll begin by dealing with the 𝑥-coordinate of the centre of mass. 𝑚 one 𝑥 one is the mass of 𝐴 times the distance that 𝐴 is from the origin. So that’s 13 times zero. 𝐵 is six times zero. Again, the horizontal distance from the origin is zero. And for 𝐶, it is 15 times six. This is all over the sum of their masses. On the top of our fraction, this simplifies to 90 and, on the bottom, we get 34. This simplifies to 47 over 17. We’re going to repeat this process for the 𝑦-coordinate. 𝑚 one 𝑦 one is 13 times zero. Remember, 𝐴 is at the origin, so it’s zero units away in the vertical direction. This time, 𝐵 is eight centimetres away from the origin in the vertical direction, and 𝐶 is zero units away in the vertical direction. Once again, this is all over the sum of their masses. And this gives us 48 over 34. 48 over 34 simplifies to 24 over 17. And so we found the coordinates of the centre of mass or centre of gravity of our system. They are 47 17ths and 24 17ths. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
The standard form of a quadratic equation is ax2 + bx + c = 0 The above formula can be used to solve a quadratic equation in standard form. If the given quadratic equation is not in standard form, convert it to standard form and use the above formula and solve. Example 1 : x2 – 5x – 24 = 0 Solution : Comparing the given equation with ax2 + bx + c = 0, we get a = 1, b = -5, c = -24 Substitute a = 1, b = -5 and c = -24. x = 8 or -3 Example 2 : x2 – 7x + 12 = 0 Solution : a = 1, b = -7, c = 12 Substitute the above values into the quadratic formula. x = 4 or 3 Example 3 : x2 – 2x - 5 = 0 Solution : a = 1, b = -2, c = -5 Substitute the above values into the quadratic formula. Example 4 : 15x2 – 11x + 2 = 0 Solution : a = 15, b = -11, c = 2 Substitute the above values into the quadratic formula. Example 5 : x + ¹⁄ₓ = 2½ Solution : x + ¹⁄ₓ = 2½ x + ¹⁄ₓ⁵⁄₂ Multiply both sides by 2x. 2x[x + ¹⁄ₓ] = 2x[⁵⁄₂] 2x2 + 2x(¹⁄ₓ) = 5x 2x2 + 2 = 5x 2x2 - 5x + 2 = 0 a = 2, b = -5, c = 2 Substitute the above values into the quadratic formula. Example 6 : (x + 3)2 - 81 = 0 Solution : (x + 3)2 - 81 = 0 (x + 3)(x + 3) - 81 = 0 x2 + 3x + 3x + 9 - 81 = 0 x2 + 6x - 72 = 0 a = 1, b = 6, c = -72 Substitute the above values into the quadratic formula. x = 6 or -12 Example 7 : Solution : 4x2 - 9x - 43 = 0 a = 4, b = -9, c = -43 Substitute the above values into the quadratic formula. Example 8 : a(x2 + 1) = x(a2 + 1) Solution : a(x2 + 1) = x(a2 + 1) ax2 + a = xa2 + x ax2 + a - xa2 - x = 0 ax2 - xa2 - x + a = 0 ax2 - (a2 + 1)x + a = 0 a = a, b = -(a2 + 1), c = a Substitute the above values into the quadratic formula. Example 9 : 3a2x2 - abx - 2b2 = 0 Solution : a = 3a2, b = -ab, c = -2b2 Substitute the above values into the quadratic formula. Example 10 : 36x2 – 12ax + (a2 - b2) = 0 Solution : a = 36, b = -12a, c = a2 - b2 Substitute the above values into the quadratic formula. Kindly mail your feedback to v4formath@gmail.com ## Recent Articles 1. ### SAT Math Videos May 22, 24 06:32 AM SAT Math Videos (Part 1 - No Calculator) 2. ### Simplifying Algebraic Expressions with Fractional Coefficients May 17, 24 08:12 AM Simplifying Algebraic Expressions with Fractional Coefficients
# 2013 AMC 12B Problems/Problem 16 ## Problem Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$? $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}$ ## Solution 1 The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure $\frac{180^\circ (5-2)}{5}=108^\circ$, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure $180^\circ - 108^\circ = 72^\circ$. The base angles are equal, so the triangles must be isosceles. Let one of the sides of the pentagon have length $x_1$ (and the others $x_2, x_3, x_4, x_5$). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length $\frac{x_1}{2} \sec 72^\circ$, and so the two sides together have length $x_1 \sec 72^\circ$. To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get $(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ$ (because the perimeter of the pentagon is $1$). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is $\boxed{\textbf{(A)} \ 0}$. ## Solution 2 The extreme case, that results in the minimum and/or maximum, would probably be a pentagon that approaches a degenerate pentagon. However, due to the way the problem is phrased, we know there exists a minimum and maximum; therefore, we can reasonably assume that the star's perimeter is constant, and answer with $\boxed{\textbf{(A)} \ 0}$.
# Using Scientific Notation Updated February 21, 2017 \| Factmonster Staff Sometimes, especially when you are using a calculator, you may come up with a very long number. It might be a big number, like 2,890,000,000. Or it might be a small number, like 0.0000073. Scientific notation is a way to make these numbers easier to work with. In scientific notation, you move the decimal place until you have a number between 1 and 10. Then you add a power of ten that tells how many places you moved the decimal. In scientific notation, 2,890,000,000 becomes 2.89 x 109. How? • Remember that any whole number can be written with a decimal point. For example: 2,890,000,000 = 2,890,000,000.0 • Now, move the decimal place until you have a number between 1 and 10. If you keep moving the decimal point to the left in 2,890,000,000 you will get 2.89. • Next, count how many places you moved the decimal point. You had to move it 9 places to the left to change 2,890,000,000 to 2.89. You can show that you moved it 9 places to the left by noting that the number should be multiplied by 109. 2.89 x 109 = 2.89 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 2.89 x 109 = 2,890,000,000 Scientific notation can be used to turn 0.0000073 into 7.3 x 10-6. • First, move the decimal place until you have a number between 1 and 10. If you keep moving the decimal point to the right in 0.0000073 you will get 7.3. • Next, count how many places you moved the decimal point. You had to move it 6 places to the right to change 0.0000073 to 7.3. You can show that you moved it 6 places to the right by noting that the number should be multiplied by 10-6. 7.3 x 10-6 = 0.0000073 Remember: in a power of ten, the exponent?the small number above and to the right of the 10?tells which way you moved the decimal point. • A power of ten with a positive exponent, such as 105, means the decimal was moved to the left. • A power of ten with a negative exponent, such as 10-5, means the decimal was moved to the right. Powers of Ten billions109 = 1,000,000,00010 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,000,000,000 millions106 = 1,000,00010 x 10 x 10 x 10 x 10 x 10 = 1,000,000 hundred thousands105 = 100,00010 x 10 x 10 x 10 x 10 = 100,000 ten thousands104 = 10,00010 x 10 x 10 x 10 = 10,000 thousands103 = 1,00010 x 10 x 10 = 1,000 hundreds102 = 10010 x 10 = 100 tens101 = 10 ones100 = 1 tenths10?1 = 1/101/10 = 0.1 hundredths10?2 = 1/1021/102 = 0.01 thousandths10?3 = 1/1031/103 = 0.001 ten thousandths10?4 = 1/1041/104 = 0.0001 hundred thousandths10?5 = 1/1051/105 = 0.00001 millionths10?6 = 1/1061/106 = 0.000001 billionths10?9 = 1/1091/109 = 0.000000001
# Euclid's Algorithm For Finding The Greatest Common Factor Of 2 Numbers in STEMGeeks5 days ago (edited) Hi there. In this mathematics post, I cover Euclid's algorithm for finding the greatest common factor (gcf) between two numbers. Euclid's algorithm falls under the number theory side of mathematics. Quicklatex.com is used for math text and LaTeX rendering. Pixabay Image Source ## Topics • Greatest Common Factors Review • Euclid's Algorithm With Examples • Practice Problems • Solutions To Practice Problems ## Greatest Common Factors Review Before getting into greatest common factors, it is important to start with number factors and then common factors. Number Factors As an example, factors of 8 include 1, 2, 4 and 8 itself. This is because you can have `1 x 8 = 8` and `2 x 4 = 8`. The numbers 1, 2, 4 and 8 can be used in multiplication to obtain 8. Factors of 20 would be 1, 2, 4, 5, 10 and 20. We have `1 x 20 = 20`, `2 x 10 = 20` and `4 x 5 = 20`. Common Factors Between Two Numbers When it comes to two numbers or even more, you can find common factors between them. Consider the numbers 12 and 21. Factors of 12 are 1, 2, 3, 4, 6 and 12. Factors of 21 are 1, 3, 7 and 21. The common factors of 12 and 21 are 1 and 3. Greatest Common Factor The greatest common factor would be simply be the largest common factor. From the above example the greatest common factor is 3. Pixabay Image Source ## Euclid's Algorithm With Examples Euclid's Algorithm is a procedure that helps determining the greatest common factor between two numbers. The greatest common factor (gcf) is sometimes referred to as the greatest common divisor (gcd). Instead of going through the general case for Euclid's algorithm, I think it is better to show how this algorithm works with examples. Example One What is the greatest common factor between the numbers 20 and 64? The larger of the two numbers is 64. This 64 would be on the left side of the equation. Express this larger number 64 as a multiple of the smaller number 20 plus a remainder. You could also do 64 divided by 20 to obtain 3 with a remainder of 4. For the next equation, the left side is 20 which was the smaller number. The right side uses the remainder 4 from the previous equation multiplied by a multiple plus a remainder. Euclid's algorithm stops when the remainder is 0. The greatest common factor here is 4 for 20 and 64. I have included a screenshot below as a summary. (Mathisfun Whiteboard part of website is used) Example Two Find the gcf of 122 and 180. The first line for Euclid's algorithm would be 180 being equal to a multiple of 122 plus a remainder. The gcf of 122 and 180 is 2. Example Three What is the gcf of 228 and 750? Start with 750 on the left side. Express 750 as a multiple of 228 with a remainder. Six is the greatest common factor of 228 and 750. For those who want a more detailed and theoretical view of Euclid's algorithm, consider this page. Pixabay Image Source ## Practice Problems For each question, find the greatest common factor or greatest common divisor for the two given numbers. 1. 17 and 51 2. 32 and 52 3. 102 and 260 4. 1021 and 222 ## Solutions To Practice Problems 1. `gcf(17, 51) = 17` 2. `gcf(32, 52) = 4` 3. `gcf(102, 260) = 2` 4. `gcf(1021, 222) = 1` Pixabay Image Source Posted with STEMGeeks Sort: 4 days ago I've never learned the name of this method and yet done this throughout school. lol Lol. Maybe the teachers did not know the method. They just taught it. 4 days ago Ha probably. I had a lot of younger teachers in my time through public school. Your posts are really well constructed and laid out but may I make a suggestion please as most of your readership here are 'big children'. Firstly, could you give us a few real-world examples of why we might want to know the subject of a post and secondly, please post the answers to the solutions of the practice problems in your next post! Best wishes and please don't take this comment as a criticism in any way.
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) Published by Pearson # Chapter 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set - Page 94: 50 #### Answer $x=-\dfrac{1}{3}$ #### Work Step by Step $\bf{\text{Solution Outline:}}$ To solve the given equation, $-\dfrac{5}{6}+x=-\dfrac{1}{2}-\dfrac{2}{3} ,$ use the properties of equality to isolate the variable. Then do checking of the solution. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $\{ 6,2,3 \}$ is $6$ since it is the least number that can be divided evenly (no remainder) by all the denominators. Multiplying both sides by the $LCD= 6$ and using the properties of equality results to \begin{array}{l}\require{cancel} -\dfrac{5}{6}+x=-\dfrac{1}{2}-\dfrac{2}{3} \\\\ 6\left( -\dfrac{5}{6}+x \right)=6\left( -\dfrac{1}{2}-\dfrac{2}{3} \right) \\\\ -5+6x=-3-4 \\\\ 6x=-3-4+5 \\\\ 6x=-2 \\\\ x=-\dfrac{2}{6} \\\\ x=-\dfrac{1}{3} .\end{array} Checking: If $x=-\dfrac{1}{3},$ then \begin{array}{l}\require{cancel} -\dfrac{5}{6}+x=-\dfrac{1}{2}-\dfrac{2}{3} \\\\ -\dfrac{5}{6}-\dfrac{1}{3}=-\dfrac{1}{2}-\dfrac{2}{3} \\\\ -\dfrac{5}{6}-\dfrac{2}{6}=-\dfrac{3}{6}-\dfrac{4}{6} \\\\ -\dfrac{7}{6}=-\dfrac{7}{6} \text{ (TRUE) } .\end{array} Hence, the solution is $x=-\dfrac{1}{3} .$ 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.
# Given -4(x - 5) = -30, then what is - 2x + 3 ? mrsq \| Certified Educator You are given that -4(x-5) = -30. So first, we must solve for x in this equation. -4(x - 5) = -30 -4x - (-4)(5) = -30 Distribute -4 to everything in the parenthesis -4x + 20 = -30 Simplify -4x + 20 - 20 = -30 - 20 Use inverse operations to solve for x. The inverse of addition is subtraction -4x = -50 Simplify -4x/-4 = -50/-4 The inverse of multiplication is division x = 50/4 = 25/2 Simplify; remember neg/neg = positive Now substitute the value of x into the second equation and solve: -2(25/2) + 3 -50/2 + 3 -25 + 3 -22 Therefore, the value of the second equation is -22. sciencesolve \| Certified Educator You should open the brackets using the distributive such that: `-4x + 20 = -30` Notice that you may write `-4x + 20` using the group of terms -`2x+3` such that: `-4x + 20 = -2x + 3 - 2x + 3 + 14` . Hence, `-2x + 3 - 2x + 3 + 14 = -30` Notice that the group of terms occurs two times such that: `2(-2x+3) = -30 - 14 =gt 2(-2x+3) = -44` Dividing by 2 yields: `-2x + 3 = -22` Hence, evaluating -2x + 3 yields `-2x + 3 = -22` . justaguide \| Certified Educator We are given that -4(x - 5) = -30. -4(x - 5) = -30 => -4x + 20 = -30 => -4x = -50 => x = 25/2 The value of -2x + 3 = -225/2 + 3 = -25 + 3 = -22 The required value of -2x + 3 = -22 giorgiana1976 \| Student We'll have to determine x to compute - 2x + 3. For this reason, we'll remove the brackets from the 1st equation: -4x + 20 = -30 -4x = -30 - 20 -4x = -50 x = -50/-4 x = 12.5 We'll calculate - 2x + 3 : - 2x + 3 = -25 + 3 = -22 The value of the expression - 2x + 3 is - 2x + 3 = -22.
# Scaling Ratios in the Ratio Table Scaling Ratios Number Talk with composed unit ratios involving scaling in tandem in ratio tables to determine the better buy. ## In This Set of Visual Number Talk Prompts… Students will explore scaling composed unit ratios in tandem in order to compare the price of goods. Students will use this information to determine the best value and develop responsible consumerism. ## String of Related Problems In today’s visual math talk, we will be scaling a composed unit ratio in tandem through ratio reasoning. This strategy is useful when trying to compare the price of goods. Through the context of the price and 12-egg cartons, students will be encouraged to leverage the ratio table to scale this composed unit ratio to reveal other values. Throughout the process, encourage students to share and model their mental math strategies and do your best to model them in the ratio table. ## Visual Math Talk Prompt Show students the following visual math talk prompt & ask: If the cost for 2 cartons of eggs is \$7.20, how much would it be for 4 cartons of eggs? The first prompt in this string of related visual number talk prompts involves a pretty low floor giving students two quantities that can easily be doubled by decomposing \$7.20 into \$7 + \$0.20. 2(\$7 + \$0.20) = \$14 + \$0.40 = \$14.40 Next, the visual number talk prompt video asks students to determine the cost for 1, 6, 10 and 12 egg cartons. The goal here is to promote students leveraging their understanding of scaling up and down the ratio table to reveal additional equivalent composed unit ratios. While some may choose to add an additional \$7,20 to \$14.40 to arrive at \$21.60 for 6 egg cartons, others may recognize that they can triple the cost of 2 egg cartons since 6 egg cartons is triple that of 2 egg cartons. Some students may add \$14.40 to \$21.60 to reveal the cost for 10 egg cartons since those are the costs of 4 egg cartons and 6 egg cartons (additive thinking). Others may multiply the cost of 2 egg cartons by 5 since 5 groups of 2 egg cartons is equivalent to 10 egg cartons. Others still may determine the cost of a single egg carton by halving \$7.20 and then scale this by a factor of 10. While we will celebrate all approaches used by students via additive and multiplicative thinking, we do want to explicitly highlight the use of scaling in tandem as this is a multiplicative strategy that is very helpful when working with composed unit ratios. It is worth noting here that while we can see the cost of a single carton of eggs, the rate (or “unit rate” as some call it) is not explicitly shown here. Only through partitive division of any of these equivalent ratios (i.e.: 3.60:1, 7.20:2, 14.40:4, etc.) can we reveal the rate of \$3.60 per carton of eggs. Interested in learning more about ratios, rates and proportional relationships? Dig into our course, The Concept Holding Your Students Back. The first module has open access for any and all to participate in, so go check it out. ## Want to Explore These Concepts & Skills Further? One (1) additional number talk prompt are available in Day 2 of the Better Buy problem based math unit that you can dive into now. Why not start from the beginning of this contextual 5-day unit of real world lessons from the Make Math Moments Problem Based Units page? Did you use this in your classroom or at home? How’d it go? Post in the comments! Math IS Visual. Let’s teach it that way.
<< Back to Lessons Index # 7th Grade Math / Lesson 2: Metric Units METRIC UNITS What will we be learning in this lesson? In this lesson, you will review measurements in the Metric system. Vocabulary words are found in this purple color throughout the lesson. Remember to put these in your notebook. METRIC UNITS Measurements come in many forms. You can measure lengths, volumes, even weights. Each of these attributes have their own units. Lengths can be measured in meters. Volumes can be measured in liters. Weights can be measured in grams. This section will review changing units within the measuring system. METRIC UNITS Lengths Here is a table that shows the equivalent units of length. Use this table to change the following measurements. This table indicates that 1 km = 1000 m , 1 hm = 100 m , 1 dam = 10 m , 1 mm = .0001 m, and so on5.9 m = _____ km 125 mm = _____ m425 km = _____ m 55 cm = ______ mm These conversions are simply done by moving decimal points. METRIC UNITS Lengths Start with the given measurement. Count the number of places you move to get to the new measurement. That is how many times you move the decimal point. And it is also the direction that you move it.5.9 m = _____ km Start at meters (because that is the measure- ment that you are given). To get to kilometer, you move 3 places to the left.Move the decimal 3 places to the left. So 5.9 m = .0059 km METRIC UNITS Lengths 125 mm = _____ m Start at millimeters (because that is the measurement that you are given). To get to meter, you move 3 places to the left. So 125 mm = 125,000 m METRIC UNITS Lengths 425 km = _____ m Start at kilometers (because that is the measurement that you are given). To get to meter, you move 3 places to the right.Move the decimal 3 places to the right. So 425 km = .425 m METRIC UNITS Lengths 55 cm = _____ mm Start at centimeters (because that is the measurement that you are given). To get to millimeter, you move 1 place to the right. So 55 cm = 5.5 m METRIC UNITS Volume/Mass (weight) There are different units used for volume and mass compared to length. Length uses meters, volume uses liters, and weight uses grams. The tables below show the equivalencies for liters and grams. The conversions are done the same as they are with meters - by moving the decimal points. Can you set up the convert from one unit to another in the metric system? Be sure to go back to your classroom to get any homework assignments or other activities you need to attend to.
## 17Calculus Precalculus - Binomial Theorem ##### 17Calculus The Binomial Theorem is a technique to expand binomial terms efficiently without having to multiply out and collect terms. There is some notation that you need to learn but it is not a hard technique and will save you time in your calculus work. Binomial Theorem $(a+b)^n = \sum_{k=0}^{n}{ \frac{n!}{(n-k)!k!}a^{n-k}b^k }$ Understanding The Notation 1. A binomial term looks like $$a+b$$, i.e. there are exactly 2 terms, a and b, added together and raised to some power. 2. The exponent n must be an integer greater than 1. 3. If you need a refresher on factorials, you can review them on the factorials page. 4. You may also see alternate notation for the factorial term like these. $\frac{n!}{(n-k)!k!} = \left( \begin{array}{c} n \\ k \end{array} \right) = {}_{n}C_{k}$ The notation $${}_{n}C_{k}$$ is read 'n choose k'. 5. If you are not familiar with summation (sigma) notation, here is a good video that explains how it works. ### Khan Academy - Sequences and Series (part 1) [9min-48secs] In this video he introduces sequences, series and sigma notation. He does a pretty good job explaining the basics required for understanding series. About 3 minutes into the video, he says something that is correct in this context but not in general. He says that you can add numbers in any order and get the same answer. This is true for a finite list of numbers, like he is doing here, but it is NOT always true if you have an infinite list of numbers. You will probably run across this case in your textbook while you are studying this topic. Understanding and Using The Theorem Here is a great video explaining, step-by-step, how to use the binomial theorem with an example. When applying the binomial theorem, we sometimes say that we are finding the binomial expansion. ### NancyPi - How to Use the Binomial Theorem [19min-58secs] video by NancyPi Pascal's Triangle One way to calculate the factorial term is to use Pascal's Triangle. Here is a video explaining this with a couple of examples. ### PatrickJMT - Pascal's Triangle and the Binomial Coefficients [5min-40secs] video by PatrickJMT Okay, time for the practice problems. Practice Unless otherwise instructed, find the binomial expansion using the Binomial Theorem. $$(x+2)^5$$ Problem Statement Find the binomial expansion of $$(x+2)^5$$ using the Binomial Theorem. Solution ### MIP4U - 2488 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$(x+y)^5$$ Problem Statement Find the binomial expansion of $$(x+y)^5$$ using the Binomial Theorem. Solution ### Thinkwell - 2491 video solution video by Thinkwell Log in to rate this practice problem and to see it's current rating. Find the binomial expansion of $$(a+b)^5$$ and use the result to expand $$(x+1)^5$$. Problem Statement Find the binomial expansion of $$(a+b)^5$$ and use the result to expand $$(x+1)^5$$. Solution ### PatrickJMT - 2485 video solution video by PatrickJMT Log in to rate this practice problem and to see it's current rating. $$(a+5b)^3$$ Problem Statement Find the binomial expansion of $$(a+5b)^3$$ using the Binomial Theorem. Solution ### Thinkwell - 2492 video solution video by Thinkwell Log in to rate this practice problem and to see it's current rating. $$(x-4)^5$$ Problem Statement Find the binomial expansion of $$(x-4)^5$$ using the Binomial Theorem. Solution ### MIP4U - 2487 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$(2x-3)^4$$ Problem Statement Find the binomial expansion of $$(2x-3)^4$$ using the Binomial Theorem. Solution ### MIP4U - 2489 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$(3x-y)^3$$ Problem Statement Find the binomial expansion of $$(3x-y)^3$$ using the Binomial Theorem. Solution ### PatrickJMT - 2486 video solution video by PatrickJMT Log in to rate this practice problem and to see it's current rating. $$(2t-s)^5$$ Problem Statement Find the binomial expansion of $$(2t-s)^5$$ using the Binomial Theorem. Solution ### Brian McLogan - 2496 video solution video by Brian McLogan Log in to rate this practice problem and to see it's current rating. $$(3x^2-5y)^4$$ Problem Statement Find the binomial expansion of $$(3x^2-5y)^4$$ using the Binomial Theorem. Solution ### MIP4U - 2490 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. Find the third term of the binomial expansion of $$(2x+3)^4$$. Problem Statement Find the third term of the binomial expansion of $$(2x+3)^4$$. Solution ### Thinkwell - 2493 video solution video by Thinkwell Log in to rate this practice problem and to see it's current rating. Determine the third term of the expansion of $$(3x+1)^8$$. Problem Statement Determine the third term of the expansion of $$(3x+1)^8$$. Solution ### Brian McLogan - 2495 video solution video by Brian McLogan Log in to rate this practice problem and to see it's current rating. Find the fifth term of the expansion of $$(2x-3y)^7$$. Problem Statement Find the fifth term of the expansion of $$(2x-3y)^7$$. Solution ### Brian McLogan - 2494 video solution video by Brian McLogan Log in to rate this practice problem and to see it's current rating. When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications. DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. 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# Which of the following corresponds to a final volume of a 12.89 * 10^(-3) "M" solution prepared by the dilution of "27.9 L" of a "433.0-mM" solution? ## A) 937.2 L B) 12.08 L C) 42.0 L D) 529 L E)132 L ##### 1 Answer Dec 8, 2017 $\text{937 L}$ #### Explanation: The idea here is that when you're diluting a solution, the ratio that exists between the concentration of the stock sample and the concentration of the diluted solution is equal to the ratio that exists between the volume of the diluted solution and the volume of the stock solution. This ratio is called the dilution factor and can be calculated by using $\text{DF" = V_"diluted"/V_"stock" = c_"stock"/c_"diluted}$ In your case, you know that your diluted solution has ${c}_{\text{diluted" = 12.89 * 10^(-3) color(red)(cancel(color(black)("M"))) * (10^3color(white)(.)"mM")/(1color(red)(cancel(color(black)("M")))) = "12.89 mM}}$ and you stock solution has ${c}_{\text{stock" = "433.0 mM}}$ so you can say that the dilution factor is equal to "DF" = (433.0 color(red)(cancel(color(black)("mM"))))/(12.89color(red)(cancel(color(black)("mM")))) = color(blue)(33.59) This means that the volume of the diluted solution must be $\textcolor{b l u e}{33.59}$ times the volume of the stock solution, since ${V}_{\text{diluted" = "DF" * V_"stock}}$ This means that you have ${V}_{\text{diluted" = color(blue)(33.59) * "27.9 L}}$ V_"diluted" = color(darkgreen)(ul(color(black)("937 L"))) The answer must be rounded to three sig figs, the number of sig figs you have for the volume of the stock solution. Therefore, you can say that (a) is the answer that the problem is looking for here.

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