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NCERT Solutions for class 10th Maths Chapter 3 Pair of Linear Equations in two Variables Exercise 3.1 Question 1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.”(Isn’t this interesting?) Represent this situation algebraically and graphically. Sol. Let Aftab’s present age be x years. Daughter’s present age be y years. Seven years ago Aftab’s age=(x – 7) years.Daughter’sage=(y – 7)years.According to the question, Aftab’s age=7 times the age of the daughter x-7=7(y-7) x-7=7y-49 x-7y=-42 (1) After three years Aftab’s age=(x + 3)years.Daughters age=(y + 3)years.According to the questions, Aftabs age =3 times the age of the daughter (x+3)=3(y+3) x+3=3y+9 x-3y=6 (2) Thus,algebraic representation of given situation is given by the equations (1)and (2).Now, represent this situation graphically, we find three solutions of each equation. From eq. (1): X-7y=-42 X=-42+7y Thus, in the above graph, the two lines representing the two equations intersecting at the point C(42, 12). Hence,x=42 and y=12. :.Aftab’s present age = 42 years.His daughter’s present age = 12 years. Also, when his daughter was born, Aftab’s age=42-12=30 years Note: Every solution of the equation is a point on the line representing it. Question 2. The coach of a cricket team buys 3 bats and 6 balls for 3900. Later she buys another bat and 3 more balls of the same kind for < 1300. Represent this situation algebraically and geometrically. Sol. Let the cost of a bat be < x. Let the cost of a ball be < y. Given,coach buys 3 bats and 6 balls for < 3900. :. 3x+6y=3900 Similarly, coach buys another bat and 3 more balls for 1300. :. x+3y=l300 Thus, the algebraic representation is given as 3x+6y=3900 or x+3y=l300. For geometrical representation: We find three solutions of each of the equation. From eq. (1): 3x + 6y = 3900 Plot these points A(0, 650), B(600, 350), C(1300, 0), D(100, 400) and E(400, 300) on graph to get the lines. Thus, equation of lines intersect at the point C(1300, 0). Hence,x=1300, y=0 is the required solution of the pair of linear equations. Note : A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. Question 3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be < 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is < 300. Represent the situation algebraically and geometrically. Sol. Let the cost of apple be < x per kg. Let the cost of grapes be < y per kg. Since, the cost of 2kg of apples and 1 kg of grapes on a day was found to be < 160. :.2x+ y=l60 â€¦(1) Similarly, the cost of 4 kg of apples and 2 kg of grapes is < 300. :. 4x + 2y = 300 â€¦(2) Thus, the equations (1) and (2) represents the situation algebraically. For graphical representation, We find three solutions of each of the equation. From eq (1):2x + y = 160 y = 160 – 2x Note: A pair of linear equations in two variables, which has no solution is called an inconsistent pair of linear equations Excercise 3.2 Question 1. From pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. (ii) 5 Pencils and 7 pens together cost rs. 50, whereas 7 pencils and 5 pens together cost 46 rs.Find the cost of one pen pencil and that of one pen. Since,both the lines are intersecting at point (3, 7) .’.X=3, y=7. (ii)Let the cost of a pencil be x and the cost of a pen bey. According to the question, we have 5x + 7 y = 50 and 7x + 5y = 46 This is the required pair of equations. Geometrically : Now, we find two solutions for each of these equations. Solution of the equations are given in table. 5x + 7 y = 50 7x + 5y = 46 Question 2. On comparing the ratios a1/a2,b1/b2, and c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x-4y+8=0; 7x+6y-9=0 (ii) 9x+3y+12=0; 18x+6y+24= 0 (iii) 6x-3y+10=0; 2x-y+9=0 Question 3. On comparing the ratios al/a2,b1/b2 and cl/c2, find out whether the following pair of linear equations are consistent or inconsistent. (I) 3x+2y=5; 2x-3y=7 (ii) 2x-3y=8; 4x-6y=9 (iii) 3/2x+5/3y=7; 9x-10y=14 (iv) 5x-3y=ll; -10x+6y=-22 (v) 4/3x+2y=8; 2x+3y=12 Note : A dependent pair of linear equations is always consistent. Question 4. Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically. (I) x+y=5, 2x+2y=10 (ii) x- y = 8, 3x-3y =16, (iii) 2x+ y-6= 0, 4x-2y-4= 0 (iv) 2x-2y-2 = 0, 4x-4y-5= 0 Sol. (I) Given equations are x+y=5 and 2x+2y=l0 On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get a1/a2=b1/b2=c1/c2=1/2 Thus, pair of linear equations are consistent. Graphically : We have x+ y = 5. Now,when x=0,y=5.When y=0,x=5 We have the following table Similarly, we have 2x + 2y = 10 :.The table is given as Now,we plot the points A(5, 0), B(0, 5), C(l, 4) and D(2, 3) on the graph paper. Join all the points to obtain the graph we observe C and D both lie on A and B. Thus,the graph of the two equations are coincident. Hence,the system of equations has infinitely many solutions. i.e.,consistent. (ii) x-y-8=0 â€¦(1) 3x-3y-16= 0 â€¦(2) On comparing with a1x + b1y + c1 = 0 and a2x + bzY + c2 = 0, we have a1/a2=1/3 b1/b2=-1/-3=1/3, c1/c2=-8/-6= 1/2 Therefore, :. the pair of equations is inconsistent and the graph of the two equations is a pair of parallel straight lines. Question 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. Sol. Let l be the length and b be the width of rectangular garden. According to the question: t = b + 4 â€¦(1) and 1/2[2(l+b)]=36 â€¦(2) l+b=36 b+4+b=36 2b = 32 b = 16 and l = 16 + 4 = 20 [From eqn (l)] Hence, l = 20 m and b = 16 m are the required dimensions of the garden Question 6. Given the linear equation 2x + 3 y- 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i)intersecting lines (ii) parallel lines (iii) coincident lines Note : Answer can be different from person to person. Question 7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. Sol. Given equations are x-y+1=0 and 3x+2y-12=0 For x-y+1=0, we have the table as follows: For 3x+2y=12, we have the table as follows: Now, we plot the points A(0, 1),B(2, 3),C(0, 6)and D(4, 0) on the graph paper. Join all the points to obtain the triangle. Clearly, we obtain a triangle BED formed by the given lines and the x-axis.The co-ordinates of the vertices of triangle are B(2, 3), E(- 1, 0) and D(4, 0). Exercise 3.3 Question 1. Solve the following pair of linear equations by the substitution method. Sol. (i) The given system of equations is x+y=14 and x-y=4 From eqn(1), we have: x= 14-y Putting value of x in equation (2),we get l4-y-y=4 14-2y=4 => -2y=-10 => y=5 Thus, x=l4-y=l4-5=9. So, x=9 and y=5 (ii) Given pair of linear equations are s-t=3 â€¦(1) s/3 + t/2 = 6 â€¦(2) From equation (1), we have: s=3+t Putting value of sin eqn (2), we get 3+t/3+t/2=6 => 3/3+t/3+t/2=6 => t/3+t/2=6-1 => 2t+3t/6 = 5 => 5t=5×6 => t=6 Putting t = 6 in eq. (3), we get:s=3+t => s=3+6=9 :. s=9 and t=6. (iii) The given system of equations is 3x-y=3 â€¦(1) and, 9x-3y=9 â€¦(2) From eq (1): y = 3x- 3 Substituting y=3x-3 in eq.(2),we get 9x-3(3x-3)=9 => 9x-9x+9=9 =>9=9 This statement is true for all values of x. However, we do not get a specific value of x as a solution. So, we cannot obtain a specific value of y. This situation has arisen because both the given equations are the same. :. Equations (1) and (2) have infinitely many solutions. (iv) The given system of equations is 0.2x+0.3y=1.3 => 2x+3y=13 â€¦(1) and,0.4x+O.Sy=2.3 => 4x+5y=23 â€¦(2) From eq.(2): 5y= 23-4x => Y=23-4x/5 Substituting y = 23-4x/5 in(1), we get 2x+3(23-4x/5)=13 lOx+69-12x=65 => -2x=-4 x=2 Putting x=2 in eq.(1), we get:2×2+3y=13 3y=13-4=9 => y=9/3=3 Hence, the solution of the given system of equations is x=2,and y=3 Question 2. Solve 2x+3y=ll and 2x-4y=-24 and hence find the value for ‘m’ for which y=mx+3. Question 3. Form the pair of linear equations for the following problems and find their solution by substitution method. (I) The difference between two numbers is 26 and one number is three times the other. Find them. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (iii) The coach of a cricket team buys 7 bats and 6 balls for 3800. Later, she buys 3 bats and 5 balls for 1750. Find the cost of each bat and each ball. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is 105 and for a journey of 15 km, the charge paid is rs.155 . What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km ? (v) A fraction becomes 9/11, if 2 is added to both the numerator and denominator.If,3 is added to both the numerator and the denominator it becomes 6,Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son.Five years ago, Jacob’s age was seven times that of his son. What are their present ages? Sol. (I) Let the required numbers be x and y,(x > y). According to the question, x-y= 26 â€¦(1) and x= 3y â€¦(2) Put value of x = 3y in equation (l),we get: 3y-y=26 => 2y=26 => y=l3 Thus,x=3xl3=39. [From (2)] Hence, the numbers are 39 and 13. Note: We can also find the numbers by considering the case y>x. (ii) Let the larger supplementary angle be x0 and smaller be y0. According to the question, x=y+l8Â°â€¦(1) Also, x+ y =180Â° (Supplementary angles)â€¦(2) Putting value of x from equation (1), in equation(2), we get y+l8Â°+y=l80Â° 2y=l62Â° y=81Â° Thus, x=l80Â°-8l0=99Â°[From (2)] Hence, required angles are 99Â° and 81Â°. (iii) Let the cost of one bat and one ball be < x and < y respectively. According to the question, 7x + 6y = 3800 â€¦(1) and 3x + Sy= 1750 â€¦(2) From eq.(2): 5y= 1750-3x => Y =l750-3x/5 Substituting y =1750-3x/5 in eq. (1), we get: 7x+6(175-3)/5 = 3800 35x+10500-18x=19000 => l7x =19000-10500 => 17x = 8500 => x = 8500/17 = 500 Putting x = 500 in eq. (2), we get:3(500)+5y=1750 => 5y=1750-1500 => 5y=250 => y =250/5 = 50 Hence the cost of one bat is <500 and the cost of one ball is rs. 50. (iv)Let the fixed charges of taxi be 105+5y=l55 =>5y=l55-105 =>5y=50 => y=l0 Putting y=10 in eq.(1),we get:x+10×10=105 X = 105-100 => X=5 Total charges for travelling a distance of 25km = X + 25y = t(5+25X10) = rs.255 Hence, the fixed charge is rs.5, the charge per km is rs.10 and the total charges for travelling a distance of 25 km is rs.255. (vi) Let the present age of Jacob be x years and his son by years. Five years hence, According to the question x+5=3(y+5) =>x=3y+IO Five years ago, according to the question, x-5=7(y-5) => x-7y=-30 Putting value of x from eq.(1)in eq(2),we get 3y+I0-7y=-30 =>-4y=-40 => y=1O Thus, x=3y+1O=3×10+10=40 :.present age of Jacob=40years and present age of his son=10years. Exercise 3.4 Question 1. Solve the following pair of linear equations by the elimination method and the substitution method (I) x + y = 5 and 2x- 3 y = 4 (ii)3x+4y=10and 2x-2y=2 (iii) 3x-5y-4=0 and 9x=2y+7 (iv) x/2+y=-1 and x-1=3 Sol. (ii) By elimination method : The given system of equations is 3x + 4y = 10 â€¦(1) and,2x-2y= 2 â€¦(2) Multiplying equation (2) by 2, we get : 4x-4y = 4 â€¦(3) (3x+4y)+(4x-4y)=10 + 4 7x=14 =>X = 2 Putting x=2 in(1),we get : 3(2)+4y=10 4y=10-6 = 4 Hence, x=2,y=1 y=1 By substitution method : The given system of equations is 3x + 4y = 10 â€¦(1) and,2x-2y=2 => x-y=1 â€¦(2) From equation(2),y=x-1 Substituting y = x-1 in eq.(1),we get: 3x+4(x-1)=10 => 3x+4x-4=10 => 7x=14 =>x = 2 Putting x=2 in eq.(1),we get: 3(2)+4y=10 => 4y=10-6=4 =>y = 1 Hence, x=2,y=1 (iii) By elimination method: The given system of equations is 3x-5y-4=0 => 3x-Sy=4 â€¦(1) and 9x=2y+7 => 9x-2y=7 â€¦(2) Multiplying equation (1) by 3, we get 9x-15y = 12 Subtracting equation(2)from(3),we get (9x-15y)-(9x-2y)=12-7 => -13y=5 => y = -5/13 Putting y =-5/13 in eq.(1),we get 3x=5(-5/13)=4 => 3x=4-25/13 => 13 = 52-25/13 =>3x = 27/13 =>x=9/13 hence, x=9/13, y= -5/13 By substitution method : The given system of equations is 3x-5y-4=0 => 3x-5y=4 â€¦(1) and,9x=2y+7 => 9x – 2y = 7 â€¦(2) From equation(1),2y=9x-7 => y = 9x-7/2 â€¦(3) Substituting y = 9x-7/2 in equation(1),we get 3x-5(9x-7)/2 = 4 => 6x-45x+35=8 => -39x=8-35 -39x= -27 => x = -27/-39 Putting x= 9/13 in equation(3),we get 3x-5(9x-7/2)=4 => 6x-45x+35=8 =>-39x=8-35 => -39x = -27 => x=-27/-39=9/13 Putting x=9/13 in equation(3), we get y=9x-7/2 => (9(9/13)-7)/2 => y=(81/13-7)/2 = 81-91/13×2 = -10/26 = -5/13 Hence, x = 9/13, y = -5/13 (iv) By elimination method: The given system of equations is x/2 + 2y/3 = -1 => 3x + 4y =-6 â€¦(1) and x-y/3 = 3 => 3x-y = 9 â€¦(2) Multiplying eq.(2)by 4 and adding to(1),we get: 4(3x-y)+(3x+4y)= 4(9)+(-6) => 12x-4y+3x+4y= 36-6 => 15x=30 => X=2 Putting x=2 in eq.(2),we get : 3(2)-y=9 =>-y=9-6=3 =>y=-3 Hence, x=2,y=-3 By substitution method : The given system of equations is x/2 + 2y/3 = -1 => 3x + 4y = -6 â€¦..(1) and, x-y/3 = 3 => 3x-y = 9 â€¦..(2) From eq.(2): y= 3x-9 Putting y= 3x-9 in eq(1), we get: 3x+4(3x-9)=-6 => 3x+12x-36 =-6 => 15x = 30 => x=2 Putting x=2 in eq(2), we get 3(2)-y=9 => -y=9-6=3 => y=-3 Hence, x=2, y=-3 Note : While using elimination method, eliminate any one variable first, to get a linear equation in one variable. Question 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (I) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction? (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. (iv) Meena went, to a bank to withdraw rs. 2000. She asked the cashier to give her rs.50 and rs.100 notes only. Meena got rs.25 notes in all. Find how many notes of rs.50 and rs.100 she received. (v) A lending library has a fixed charge for the first three days and an additional charge for each day there after. Saritha paid rs.27 for a book kept for seven days, while Susy paid rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Exercise 3.5 Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method. (i) x-3y-3=0; 3x-9y-2=0 (ii) 2x+y=5; 3x+2y=8 (iii) 3x-5y=20; 6x-10y=40 (iv) x-3y-7=0; 3x-3y-15=0 Sol. (i) x-3y-3 = 0 â€¦(i) 3x-9y-2 = 0 â€¦(ii) On comparing with a1x+b1y+c1=0 and a2x+bzY+c2=0,we get a1/a2=1/3, b1/b2=-3/-9=1/3, c1/c2=-3/-2=3/2 Hence, no solution exists. (ii) The given system of equations may be written as 2x+y-5=0 and 3x+2y-8=0 On comparing with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get => x/-8+10 = y/-15+16 = 1/4-3 => x/2= y/1=1/1 => x/2=1/1 and y/1=1/1 => x=2, y=1 x = 2 and y = l Hence, the given system of equations has a unique solution given by X = 2,y = l. (iii) The given system of equations may be written as 3x-5y-20=0 and 6x-10y-40=0 On comparing with a1x+b2y+c1=0 and a2x+b2y+c2=0,we get a1/a2=3/6=1/2,b1/b2=-5/-10=1/2 and c1/c2=-20/-40=1/2 Clearly, a1/a2=b1/b2=c1/c2 So, the given system of equations has infinitely many solutions. (iv) The given system of equations is x-3y-7=0 and 3x-3y-15=0 On comparing with a1x+b1y+c1=0 and a2x+b2y+c2=0,we get Note: In cross-multiplication method, the arrows between the two methods indicate that they are to be multiplied and the second product is to be subtracted from the first Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x+3y=7 (a-b)x+(a+b)y = 3a+b-2 (ii) For which value of k will the following pair of linear equations have no solution? 3x+ y=l (2k-1)x+(k-1)y = 2k+ 1 Question 3. Solve the following pair of linear equations by the substitution and crossÂ multiplication methods: 8x+Sy= 9 3x+2y=4 Sol. 8x+5y- 9 = 0 â€¦(1) 3x+2y-4=0 â€¦(2) By cross-multiplication method : Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist ) by any algebraic method : (I) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays rs.1180 as hostel charges. Find the fixed charges and the cost of food per day. (ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction. (iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test? (iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? (v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle. Sol.(i) Let fixed hostel charges (monthly) = rs.c and cost of food for one day = rs.x According to the question 20x + C = 1000 â€¦(1) 26x + C = 1180 â€¦(2) Subtracting eq.(1) from eq.(2),we get: (ii) (26x + c)-(20x + c) = 1180-1000 => 6x = 180x=30 Putting x=30 in eq.(1),we get:20(30)+c=1000 => C=1000-600=400 Hence, monthly fixed charges = rs.400 and cost of food per day =rs.30 (ii) Let the fraction be x/y According to the question: x-1/y = 1/3 => 3x-y=3 â€¦(1) and x/y+8 = 1/4 => 4x-y= 8 â€¦(2) Subtracting eq.(1) from eq.(2):(4x -y)-(3x -y)= 8-3 => 4x-3x-y+y=5 => x=5 Putting x=5 in eq.(1),we get:3(5)-y=3 =>y=15-3=12 Hence, the fraction is 5/12. (iii) Let number of right answers = x Number of wrong answers = y Total number of questions =x+y First case, Marks awarded for x right answers =3x Marks lost for y wrong answers = y x 1=y :. 3x-y = 40 â€¦(1) Second case, Marks awarded for x right answers = 4x Marks lost for y wrong answers = 2y :. 4x-2y=50 â€¦(2) From eq.(1), y=3x-40 Putting y=3x-40 in eq.(2),we get 4x-2(3x-40)=50 => 4x-6x+80=50 => 2x=30 =>x=15 Now,y=3x-40=3(15)-40= 45-40 =5 :. total number of questions= 15+5=20 (iv) Let speed of car I= x krn/hr. Speed of car II= y km/hr First case: Two cars meet at C after 5 hrs. AC= Distance travelled by car I in 5 hrs= 5x km BC= Distance travelled by car II in 5 hrs = Sy km AC-BC=AB => 5x-5y= 100 (Â·: AB= lO0km.) x-y=20 â€¦..(1) Second case : Two cars meet at C after one hour. x+y=l00 â€¦(2) X = 60 Then,y = 100-x = 100-60 =40 Hence, the speeds of the two cars are respectively 60 km/hr and 40 km/hr. (v) Let the length and breadth be x units and y units respectively. :. Area xy square units First case: length = (xâ€”5)units :.New area = Original area = 9 => (x-5)(y+3)=xy-9 (Area of Rect.= length x Breath) =xy +3x-Syâ€”15=xyâ€” 93x Sy= 6 â€¦.(1) Second case: length =(x+3) :.New area = Original area + 67 =>(x+3)(y+2)= xy+67 (Area of Rect.= length x Breath) => xy+6+2x+3y= xy+67 => 2x+3y=61 â€¦(2) Mutiplying (1) by 3 and (2) by 5, 9x-15y=18 â€¦(3) 10x+15y=305 â€¦.(4) Adding (3) and (4), we get: 19x=323 = x=17 Substituting x = 17 in (2), we get: 2(17)+3y= 61 3y=61-34 => y=27/3=9 Hence, length = 17 units. Breadth= 9 units. Exercise 3.6 Question 1. Solve the following pairs of equations by reducing them to a pair of linear equations: Note: You can use any of the three methods i.e.elimination.substitution or cross multiplication method to solve the equations algebraically. Question 2. Formulate the following problems as pair of equations, and hence find their solutions : (I) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. Sol.(i) Let the speed of the boat in still water be x km/hr. and the speed of the stream be y km/hr. Then, speed upstream= (x -y) km/hr Speed downstream =(x+y) km/hr Time taken to cover 20 km downstream = 2 hrs. 20/x+y =2 (:. Time= Distance/Speed) Exercise 3.7 Question 1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju. Question 2. One says,”Give me a hundred, friend! I shall then become twice as rich as you”.The other replies,”If you given me ten,I shall be six times as rich as you”.Tell me what is the amount of their(respective) capital? [From the Bijaganita of Bhaskara II] Hint:[x+l00 = 2(y-100),y+l0 = 6(x-10)). Sol. Let one friend has rs.x and second has rs.y. According to the question x+ 100=2(y-100) => x-2y=-300 â€¦(1) and y+l0=6(x-10) =>6x-y=70 â€¦.(2) Multiplying equation(2) by 2, we get: 12x – 2y = 140 â€¦ (3) Subtracting equation (1) from equation (3), we get (12x-2y-140)-(x-2y+300) = 0 => llx-440 = 0 x = 40 Putting x = 40 in equation (1), we get 40 – 2y = – 300 => -2y = -340 => y = 170 Hence, one friend has rs.40 and second has rs.170. Question 3. A train covered a certain distance at a uniform speed If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. Sol. Let the speed of the train be x km/hr and the time taken to complete the journey be y hours Question 4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less.If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class. Question 5. In a Triangle ABC, ang.C=3ang.B=2(ang.A+ang.B). Find the three angles. Question 6. Draw the graphs of the equations 5x-y = 5 and 3x-y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis. Question 7. Solve the following pair of linear equations: Question 8. ABCD is a cyclic quadrilateral(see fig.). Find the angles of the cyclic quadrilateral. Related Articles:
# Tucker, Applied Combinatorics1 SECTION 5.2 Simple Arrangements and Selections Aaron Desrochers and Ben Epstein. • Published on 11-Dec-2015 • View 214 0 Transcript • Slide 1 Tucker, Applied Combinatorics1 SECTION 5.2 Simple Arrangements and Selections Aaron Desrochers and Ben Epstein Slide 2 Tucker, Applied Combinatorics2 Vocabulary Permutation: an arrangement or ordering of n distinct objects. R-Permutation: an arrangement using r of n objects. P(n,r) denotes the r-permutation on n objects. In general, P(n,n) = n! Therefore, P(n,r) = n(n-1)(n-2)[n-(r-1)] = n!/(n-r)!. So, P(6,2) = 6!/(6-2)! = 6!/4! = (6)(5)(4)(3)(2)(1) / (4)(3)(2)(1) = (6)(5) = 30. NOTE: P(n,n) = n! Since (n-n)! = 0! = 1. So, P(4,4) = 4! = 24. Slide 3 Tucker, Applied Combinatorics3 Vocabulary Combination: an unordered arrangement of n objects. R-Combination: an unordered selection using r of n objects. C(n,r) denotes the r-combination of n objects. In general C(n,r) = n!/(r!(n-r)!). This is easily derived using P(n,r), since r- combinations of n objects gives the number of unordered arrangements, dividing P(n,r) by the total number of possible arrangements of n (so let r = n) gives the total number of unordered arrangements. So C(n,r) = P(n,r) / P(n,n) = n!/(r!(n-r)!). NOTE: A common notation for C(n,r) is. We say the combination of n choose r objects. nrnr Slide 4 Tucker, Applied Combinatorics4 Example 1 How many 5-card hands can be formed from a standard 52- card deck? First we must ask will the hands be ordered or unordered. In this example lets suppose we want unordered 5-card hands. So it is simply C(52,5) = 52!/[5!(52-5)!] = (52)(51)(50)(49)(48) /5!= 2,598,960 hands. Slide 5 Tucker, Applied Combinatorics5 Example 2 If a 5-card hand is chosen at random, what is the probability of obtaining a flush (all cards having the same suit)? In this case order does not matter, so we will use the notion of combinations. First, we must notice that there are 13 cards in every suit. Also, we have 4 suits to choose from, therefore (4)C(13,5) = 5148 ways to get a flush. The probability is found by dividing the number of ways of getting a flush by the total number of possible hands, which we computed in the last example. Therefore, the probability of a getting a flush is 5148/2,598,960 =.00198 or about.2%. Slide 6 Tucker, Applied Combinatorics6 Example 3 A committee of k people is to be chosen from a set of 7 women and 4 men. How many ways are there to form a committee of 5 people, including 3 women and 2 men? We want 3 of the 7 women and 2 of the 4 men. There are 7 choose 3 ways of selecting the three women. There are 4 choose 2 ways of selecting the 2 men. The product of the two gives the total number of ways. C(4,2)C(7,3) = Slide 7 Tucker, Applied Combinatorics7 Example 4 How many ways are there to form a committee of 4 people with at least two women members? The trick with this question is dealing with the at least part. This can be broken down as the sum of picking exactly 2 women + picking exactly 3 women + picking exactly 4 women. So, C(7,2)C(4,2) + C(7,3)C(4,1) + C(7,4)C(4,0) = This is equal to one Slide 8 Tucker, Applied Combinatorics8 Example 5 The chromatic polynomial P k (G) of graph G is the polynomial in k that gives the number of k-colorings of G. What is the chromatic polynomial of: A complete graph K 5 on five vertices (all vertices adjacent to each other) In a complete graph, each vertex must be a different color. Thus, P k (K 5 ) = P(k,5), since there are k possible colors for the first vertex, k-1 for the second choice, k-2 for the third choice and so on. k k-1 k-2 k-3 k-4 Slide 9 Tucker, Applied Combinatorics9 P(k,5) = k!/(k-5)!=k(k-1)(k-2)(k-3)(k-4) Example 5 This is the chromatic polynomial. Slide 10 Tucker, Applied Combinatorics10 Example 5 Continued b)Find P k ( C 4 ) where C 4 is a circuit of length 4? Let the vertices on the circuit C 4 be named a, b, c, d with edges (a,b), (b,c), (c,d) and (d,a). We break the computation of P k (C 4 ) into two cases, depending on whether or not a and c are given the same color. If a and c have the same color, there are k choices for the color of these two vertices. Then b and d each must only avoid the common color of a and c --- k-1color choices each. So the number of k-colorings of C 4 in this case is k(k-1)^2. If a and c have different colors, there are k(k-1) choices for the two different colors for a and then c. Now b and d each have k-2 color choices. So in this case the total number of k- colorings of C 4 is k(k-1)(k-2)^2. Combining the two cases, we obtain P k (C 4 ) = k(k-1)^2 + k(k-1)(k-2)^2. a c b d Slide 11 Tucker, Applied Combinatorics11 Class Exercise 1 How many different 8-digit binary sequences are there with six 1s and two 0s? Consider that we have 8 slots to fill given these two conditions. Any six slots must be 1s so C(8,6) gives the number of ways to fill any six slots. Notice that we do not have to make any choices for the zeros because they must go in the last two slots. So the solution is C(8,6) = 28. It is important to note that the solution can be arrived at by selecting the slots for the zeros. In this case C(8,2) = 28. Note also that this is the compliment of C(8,6) and that they are equal. So in general C(n,r) = C(n,n-r). Slide 12 Tucker, Applied Combinatorics12 Class Exercise 2 What is the probability that a 4-digit campus telephone number has 2 digits the same and the other two digits different? Again consider filling the slots, only in this case we need four of them. First we want to pick the slots. We need 2 slots to hold the pair of identical numbers. These slots can be chosen in C(4,2) ways. Note that we need not choose where the last two numbers will go because their positions are forced. Slide 13 Tucker, Applied Combinatorics13 Next, we need to choose the actual numbers. Certainly we have 10 digits to choose from and we want just one for the identical pair so C(10,1) gives us the number of ways to choose any digit. For the third digit we must select any digit that is not the same as the first, so we have C(9,1). And similarly for the fourth C(8,1). The question asks for the probability so we must divide the product of all of these combinations by the total number of possible ordered 4-digit phone numbers. Using the slot method, we have 10 choices for the first slot, 10 for the second, 10 for the third and 10 for the fourth. So 10^4 gives us the total number of possible ordered phone numbers. The probability is then [C(4,2)C(10,1)C(9,1)C(8,1)] / 10^4, 54/125 =.432 = 43.2%
# A man 6.5 ft tall approaches a street light 17.0ft above the ground at the rate of 8.00 ft/s. How fast is the end of the man's shadow moving when he is 5.0 ft from the base of the light? The end of the man's shadow is moving at a rate of_ft/s ## Draw a diagram. If the man is x feet from the pole, and his shadow has length s, then using similar triangles, we see that s/6.5 = (x+s)/17 2s/13 = (x+s)/17 34s = 13x+13s 21s = 13x s = 13/21 x So, ds/dt = 13/21 dx/dt So, it does not matter how far away he is, his shadow's length is changing 13/21 as fast as his distance from the pole. ## To find the rate at which the end of the man's shadow is moving, we can use similar triangles and related rates. Let's denote the rate at which the end of the man's shadow is moving as dy/dt. We want to find this rate when the man is 5.0 ft from the base of the light. First, let's set up a proportion based on the similar triangles formed by the man, the street light, and the shadow: (6.5 ft + y) / y = (17.0 ft + dy) / x Where: - (6.5 ft + y) is the height of the man and the length of the shadow (since they are proportional) - (17.0 ft + dy) is the height of the street light and the length of the shadow (since they are proportional) - x is the distance from the man to the base of the street light The next step is to differentiate both sides of the equation with respect to time t: d/dt [(6.5 ft + y) / y] = d/dt [(17.0 ft + dy) / x] To solve for dy/dt, we need to rearrange the equation and isolate dy/dt: (6.5 / y) * dy/dt = (17.0 / x) * dx/dt Now, let's substitute the given values: (6.5 / y) * dy/dt = (17.0 / 5.0) * 8.00 ft/s Simplifying the equation further: (6.5 / y) * dy/dt = 27.2 ft/s Finally, we need to find the value of y when the man is 5.0 ft from the base of the light. Using the Pythagorean theorem: y^2 + x^2 = (6.5 ft)^2 y^2 + (5.0 ft)^2 = (6.5 ft)^2 y^2 = (6.5 ft)^2 - (5.0 ft)^2 y^2 = 19.25 ft^2 - 25.00 ft^2 y^2 = -5.75 ft^2 (we discard the negative value since y represents a positive length) Taking the square root of both sides: y ≈ 2.39 ft Substituting y back into the equation: (6.5 / 2.39 ft) * dy/dt = 27.2 ft/s Solving for dy/dt: dy/dt ≈ (27.2 ft/s) * (2.39 ft / 6.5 ft) dy/dt ≈ 9.97 ft/s Therefore, the end of the man's shadow is moving at a rate of approximately 9.97 ft/s.
# Find the absolute maximum and absolute minimun, if they exist, for `y=f(x)=2x^3 -3x^2 -12x+24` A). `[-3,4]` B). `[-2,3]` txmedteach \| Certified Educator To find the absolute minima and maxima for a function over a given interval, you need to check two categories of points: 1) At the boundaries of the interval 2) At points where `(df)/(dx) = 0` For both, let's check the values at the boundaries. For A, this means checking `f(-3)` and `f(4)`, and for B, this means checking `f(-2)` and `f(3)`. `f(-3) = 2(-3)^3-3(-3)^2-12(-3) + 24 = -21` `f(4) = 56` `f(-2) = 20` `f(3) = 15` Now that we have established the function values at the boundaries, we must now solve for where the function has derivatives equal to zero. Start by taking the derivative: `(df)/(dx) = 6x^2-6x-12` Now, solve for the x-values at which the derivative is zero: `0 = 6x^2-6x-12` `0 = x^2-x-2` We can solve this equation for `x` by easily factoring: `0 = (x-2)(x+1)` This result gives us the two `x` values at which we might find other minima or maxima: `x = 2` and `x = -1`. Now, we find the function values at `2` and `-1`: `f(-1) = 31` `f(2) = 4` Both of these values are present on both intervals, so we need to take both into account when finding extrema. Solving for the absolute minima and maxima for A, we simply find which `x` value gives the smallest and largest values for `f(x)` from the `x`-values `-3`, `-1`, `2`, and `4`. Clearly, the maxima and minima are found at the boundaries. `(-3, -21)` is the absolute minimum, and `(4, 56)` is the absolute maximum. For B, we must do the same, but now we consider the function values at `x =` `-2`, `-1`, `2`, and `3`. Now, the absolute minimum and maximum are not at the boundaries. Instead, we see the absolute minimum at `(2, 4)` and the absolute maximum at `(-1, 31)`. Recapping: A) Minimum: (-3, -21); Maximum: (4, 56) B) Minimum: (2, 4); Maximum: (-1, 31)
interdicoxd 2021-12-20 All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form. $P\left(x\right)={x}^{3}-6{x}^{2}+12x-8$ jgardner33v4 Expert Step 1 Given: $P\left(x\right)={x}^{3}-6{x}^{2}+12x-8$ Step 2 We can write the given polynomial as: $P\left(x\right)={\left(x\right)}^{3}-3{\left(x\right)}^{2}\left(2\right)+3\left(x\right){\left(2\right)}^{2}-{\left(2\right)}^{3}$ We have the formula, ${\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}$ Hence, $P\left(x\right)={\left(x-2\right)}^{3}$ =(x-2)(x-2)(x-2) This is the polynomial P(x) in factored form. Step 3 To find the zeros of polynomial P(x), equate it to 0. P(x)=0 (x-2)(x-2)(x-2)=0 x-2=0 or x-2=0 or x-2=0 x=2,2,2 Hence, the polynomial P(x) has all integer real zeros and they are equal to 2 with multiplicity 3. Marcus Herman Expert The given polynomial is $P\left(x\right)={x}^{3}-6{x}^{2}+12x-8$ Since the leading coefficient is 1, any rational zero must be a divisior of the constant term -8. So the possible rational zeros are $±1,±2,±4,±8$ We test each of these possibilities $P\left(1\right)={\left(1\right)}^{3}-6{\left(1\right)}^{2}+12\left(1\right)-8$ =1-6+12-8 =-1 $P\left(-1\right)={\left(-1\right)}^{3}-6{\left(-1\right)}^{2}+12\left(-1\right)-8$ =-1-6-12-8 =-27 $P\left(2\right)={\left(2\right)}^{3}-\left(6\right){\left(2\right)}^{2}+12\left(2\right)-8$ =8-24+24-8 =0 $P\left(-2\right)={\left(-2\right)}^{3}-6{\left(-2\right)}^{2}+12\left(-2\right)-8$ =-8-24-24-8 =-64 $P\left(4\right)={\left(4\right)}^{3}-6{\left(4\right)}^{2}+12\left(4\right)-8$ =64-96+48-8 =8 $P\left(-4\right)={\left(-4\right)}^{3}-6{\left(-4\right)}^{2}+12\left(-4\right)-8$ =-64-96-48-8 =-216 $P\left(8\right)={\left(8\right)}^{3}-6{\left(8\right)}^{2}+12\left(8\right)-8$ =512-384+96-8 =216 $P\left(-8\right)={\left(-8\right)}^{3}-6{\left(-8\right)}^{2}+12\left(-8\right)-8$ =-512-384-96-8 =-1000 The rational zero of P is 2 Do you have a similar question?
0 17561 # Summary: In this lesson, students use the real-life connection of planning a park to learn about equivalence in fractions, decimals and percentages. They learn about the meaning of 1.0 and 100% and 100/100 while creating their own, fun and newly designed park. A great lesson that can be modified for older students by adding more squares. • Make connections between equivalent fractions, decimals and percentages (ACMNA131)Solve problems involving addition and subtraction of fractions with the same or related denominators(ACMNA126) • Find a simple fraction of a quantity where the result is a whole number, with and without digital technologies (ACMNA127) • Add and subtract decimals, with and without digital technologies, and use estimation and rounding to check the reasonableness of answers (ACMNA128) • Connect fractions, decimals and percentages and carry out simple conversions (ACMNA157) # Lesson: Introduction: 1. On an interactive whiteboard, bring up Plan a Park (Link will open in a new window. 2. Play the game going through all the scenarios that it gives you and ask students to notice what is happening on the right hand side (where it is displaying its percentage, decimal or fraction. 3. If you have a class set of computers, get the kids to do it on them (it’s better if everyone has a go). 4. Once finished, ask students if they could tell you equivalences for the following: • 1/2 (50% or 0.50) • 0.25 (25% or 25/100) • 14% (0.14 or 14/100) Body: 1. Hand out a 100 grid to each child in the class (for extension, give the kids 2 or 4 grids to be joined together) 2. Tell the students that they are now going to become landscape architects and create their own park that must include the following then set to work. • Playgrounds • Grass • Trees • Water • Sporting grounds • Something else Just let them go at this point, they’ll love the creation part of it! Maybe just remind them to outline their park first before they go into high detail with the colour, etc. Conclusion: 1. Once students have created their park, ask them to now calculate the fraction of each part of their park (playgrounds, water, etc) and write this in a table like the one below. 1. I suggest modelling this to them so they know how to rule it up properly and fill it in correctly. 2. Once they have finished, ask them to find the equivalences for each of their spaces. 3. Once they have done that, ask them to add up each column and write the total at the bottom. 4. Explain that if they have calculated their fractions, decimals and percentages properly the columns should add up to 100/100, 1.0 and 100% respectively. 5. Discuss what their thoughts are now on fractions, decimals and percentages. Hopefully it is a bit more clear that they are all linked and one of the same thing. # Assessment: • Collect work sample • Use books to check calculations of adding decimals, percentages and fractions. # Resources: • Workbooks • Pencils and Pens • Calculators (to check working out only) • Grid Paper (10 x 10) • Plan a Park (Looks like the image below)
## Here's an example of why the Fibonacci numbers are the answer ### What is the Fibonacci series? One way to explain that the Fibonacci series is the correct answer is to show that the problem satisfies the three conditions of the Fibonacci (Recursive) Definition: • f(n) = f(n-1) + f(n-2) where f(n) now means "the number of solutions to the puzzle of size n". On its own, the formula above is satisifed by many series that are not "the" Fibonacci numbers because we have not given any starting numbers. For example, we could start with two 2's as the first pair in the series and then apply the formula above to get 2+2=4 as the next, (the series is now 2,2,4) and then 2+4=6 as the next one (the series is now 2,2,4,6) as so on to get: 2,2,4,6,10,16,26,42,... [In fact, this particular series is closely related the "the" Fibonacci series 0,1,1,2,3,5,8,... - can you spot that relationship?] So we will need starting values as well as the formula above. • 0 and 1 to get: 0, 1, 1, 2, 3, 5, 8, 13, ... or • 1 and 1 to get: 1, 1, 2, 3, 5, 8, 13, 21, ... or • 1 and 2 to get: 1, 2, 3, 5, 8, 13, 21, 34, ... or • 2 and 3 ... • ... So there are several ways to get the Fibonacci series but all of them involve the following: • The general case: the number of solutions to the problem of size n can be explained as the number of solutions to the problem of size (n-1) PLUS the number of solutions of the problem of size (n-2), i.e. if f(n) is "the number of solutions to the problem of size n" then f(n) = f(n-1) + f(n-2) This is called the general case because it must apply for all sizes, n, of the puzzle. • The starting cases: We need to show that the number of solutions to TWO particular small problems (e.g. of size n=1 and size n=2) are 0 and 1, or 1 and 1, or 1 and 2, or 2 and 3 etc. Wehave seen that both conditions must be true if the series is to be "the" Fibonacci series. ### The Brick Wall Puzzle For this puzzle, f(n) means "the number of patterns for walls of height 2 and length n" and the solutions we have at present are Two starting cases Usually it is quite easy to verify the STARTING CASES for a puzzle because we can find all the solutions and just count them. For this puzzle we can show that f(1)=1 and that f(2)=2 as follows, using the picture of the solutions above: #### f(1)=1 There is just one pattern for a wall of length 1 (the single brick is placed upright), so f(1)=1. #### f(2)=2 There are 2 patterns if we have a wall of length 2 as we have seen in the diagrams of the puzzle page, so f(2)=2 "the number of patterns for walls of height 2 and length 2 is 2". ### Now for the general case: f(n)=f(n-1)+f(n-2) What about f(n) - the number of patterns for walls of height 2 and length n? This is the general case and we must be sure that our reasoning applies to ALL n bigger than 1 (we have covered the two cases for not bigger than 1 above). Consider what happens at the left end of the wall.... .... either there is a brick on its end ....or else there is a brick on its side - in which case there must be another one on top of it as there is no other way to make the height of 2. There are no more cases and these two cases cover all possible ways of starting a wall of length bigger than 1. Let's look at these two cases: • If we start a wall of length n with a single upright brick then we need patterns for walls of length n-1 to complete it. How many of these are there? There are f(n-1) of them since that's what f(n-1) means. • The only other way a pattern of length n can be made is to start with the two-flat-bricks-on-top-of-each-other pattern. We can then continue with any wall pattern of height 2 and of length n-2 and there are f(n-2) of these. So all the ways of finding walls of length n are covered by the two cases: an upright brick first and then any pattern of length n-1 or else the two-flat-bricks followed by any pattern of length n-2. So the number of patterns of length n is the sum of the number of patterns of length (n-1) and the number of patterns of length (n-2). In terms of the "f" notation in mathematics, we have f(n) = f(n-1) + f(n-2) generally So BOTH conditions are fulfilled: • the general case: f(n) = f(n-1) + f(n-2) generally • and the two starting cases: f(1) = 1 and f(2) = 2 and so the f(n) numbers are indeed the Fibonacci numbers! With thanks to Mats Lofkvist for helping make this page more accurate.
## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24B These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24B. Other Exercises Question 1. The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages. Solution: Question 2. The following table gives the weekly wages of workers in a factory. Calculate the mean by using : (i) Direct Method (ii) Short-Cut Method Solution: (i) Direct Method: (ii) Short cut method : Question 3. The following are the marks obtained by 70 boys in a class test : Calculate the mean by : (i) Short-Cut Method (ii) Step-Deviation Method. Solution: (i) Short cut Method : (ii) Step – Deviation Method: Question 4. Find mean by ‘step-deviation method : Solution: Question 5. The mean of following frequency distribution is 21$$\frac { 1 }{ 7 }$$ Find the value of ‘f ‘. Solution: Question 6. Using step-deviation method, calculate the mean marks of the following distribution. Solution: Let Assumed mean = 72.5 Question 7. Using the information given in the adjoining histogram; calculate the mean. Solution: Question 8. If the mean of the following observations is 54, find the value of p. Solution: Mean = 54 ⇒ 2106 + 54p = 2370 + 30p ⇒ 54p – 30p = 2370 – 2106 ⇒ 24p = 264 p = $$\frac { 264 }{ 24 }$$ = 11 Hence p = 11 Question 9. The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f1 and f2. Solution: Mean = 62.8 and sum of frequencies = 50 Question 10. Calculate the mean of the distribution given below using the short cut method. Solution: Question 11. Calculate the mean of the following distribution : Solution: Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B 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.
# How to Use Area Models to Subtract Fractions with Like Denominators Area models are visual representations that can help students understand fraction subtraction. ## A step-by-step guide to Using Area Models to Subtract Fractions with Like Denominators Here’s a step-by-step guide to using area models to subtract fractions with like denominators: ### Step 1: Understand the problem Make sure you fully understand the fractions you need to subtract. For this example, let’s say we want to subtract $$\frac{5}{6} – \frac{2}{6}$$. The Absolute Best Book for 4th Grade Students ### Step 2: Draw the area models Draw two rectangles of equal size to represent the two fractions you’re subtracting. Label each rectangle with its respective fraction. ### Step 3: Divide the area models Divide each rectangle into equal parts based on the denominator. In our example, both denominators are 6, so divide each rectangle into 6 equal parts. ### Step 4: Shade the parts Shade the parts of the area models based on the numerators. In our example, shade 5 parts of the first rectangle ($$\frac{5}{6}$$) and 2 parts of the second rectangle ($$\frac{2}{6}$$). A Perfect Book for Grade 4 Math Word Problems! ### Step 5: Align the area models Place the second area model ($$\frac{2}{6}$$) directly above or below the first area model ($$\frac{5}{6}$$) with the shaded parts aligned. ### Step 6: Subtract the shaded parts To subtract the fractions, remove the shaded parts of the second rectangle from the shaded parts of the first rectangle. In our example, remove 2 shaded parts ($$\frac{2}{6}$$) from the 5 shaded parts ($$\frac{5}{6}$$), leaving 3 shaded parts. ### Step 7: Write the difference as a fraction Write the difference using the remaining number of shaded parts as the numerator and the original denominator. In our example, the difference is $$\frac{3}{6}$$. ### Step 8: Simplify if necessary 1. If the resulting fraction can be simplified, do so. In our example, $$\frac{3}{6}$$ simplifies to $$\frac{1}{2}$$. So, using area models, we found that $$\frac{5}{6} – \frac{2}{6}= \frac{1}{2}$$. The Best Math Books for Elementary Students ### What people say about "How to Use Area Models to Subtract Fractions with Like Denominators - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 51% OFF Limited time only! Save Over 51% SAVE $15 It was$29.99 now it is \$14.99
Mathematics » Exponents and Surds » Solving Surd Equations # Solving Surd Equations ## Solving Surd Equations We also need to be able to solve equations that involve surds. ## Example ### Question Solve for $$x$$: $$5\sqrt[3]{x^4} = \text{405}$$ ### Write in exponential notation \begin{align} 5( x^4 )^{\frac{1}{3}}&= \text{405} \\ 5x^{\frac{4}{3}}&= \text{405} \end{align} ### Divide both sides of the equation by $$\text{5}$$ and simplify \begin{align} \cfrac{5x^{\frac{4}{3}}}{5} &= \cfrac{\text{405}}{5} \\ x^{\frac{4}{3}} &= 81 \\ x^{\frac{4}{3}} &= 3^4 \end{align} ### Simplify the exponents \begin{align} ( x^{\frac{4}{3}} )^{\frac{3}{4}} &= ( 3^4 )^{\frac{3}{4}} \\ x &= 3^3 \\ x &= 27 \end{align} ### Check the solution by substituting the answer back into the original equation \begin{align} \text{LHS}&= 5\sqrt[3]{x^4} \\ &= 5(27)^{\frac{4}{3}} \\ &= 5(3^3)^{\frac{4}{3}} \\ &= 5(3^4) \\ &= \text{405} \\ &= \text{RHS } \end{align} ## Example ### Question Solve for $$z$$: $$z – 4\sqrt{z} + 3 = 0$$ ### Factorise \begin{align} z – 4\sqrt{z} + 3 &= 0 \\ z – 4z^{\frac{1}{2}} + 3 &= 0 \\ (z^{\frac{1}{2}}-3)(z^{\frac{1}{2}}-1) &= 0 \end{align} ### Solve for both factors The zero law states: if $$a \times b = 0$$, then $$a = 0$$ or $$b = 0$$. $\therefore (z^{\frac{1}{2}}-3) = 0 \text{ or } (z^{\frac{1}{2}}-1) = 0$ Therefore \begin{align} z^{\frac{1}{2}}-3 &= 0 \\ z^{\frac{1}{2}} &= 3 \\ ( z^{\frac{1}{2}} )^2 &= 3^2 \\ z &= 9 \end{align} or \begin{align} z^{\frac{1}{2}}-1 &= 0 \\ z^{\frac{1}{2}} &= 1 \\ ( z^{\frac{1}{2}} )^2 &= 1^2 \\ z &= 1 \end{align} ### Check the solution by substituting both answers back into the original equation If $$z=9$$: \begin{align} \text{LHS}&= z – 4\sqrt{z} + 3 \\ &= 9 – 4\sqrt{9} + 3 \\ &= 12 – 12 \\ &= 0 \\ &=\text{RHS } \end{align} If $$z=1$$: \begin{align} \text{LHS} &= z – 4\sqrt{z} + 3 \\ &= 1 – 4\sqrt{1} + 3 \\ &= 4-4 \\ &= 0 \\ &= \text{RHS } \end{align} The solution to $$z – 4\sqrt{z} + 3 = 0$$ is $$z = 9$$ or $$z = 1$$. ## Example ### Question Solve for $$p$$: $$\sqrt{p-2} – 3 = 0$$ ### Write the equation with only the square root on the left hand side Use the additive inverse to get all other terms on the right hand side and only the square root on the left hand side. $\sqrt{p-2} = 3$ ### Square both sides of the equation \begin{align} ( \sqrt{p-2} )^2 &= 3^2 \\ p-2 &= 9 \\ p &= 11 \end{align} ### Check the solution by substituting the answer back into the original equation If $$p=11$$: \begin{align} \text{LHS} &=\sqrt{p-2} – 3 \\ &=\sqrt{11-2} – 3 \\ &=\sqrt{9} – 3 \\ &= 3-3 \\ &= 0 \\ &= \text{RHS } \end{align} The solution to $$\sqrt{p-2} – 3 = 0$$ is $$p = 11$$.
Select Page $$a_iP_j = A_{i,1} P_{1,j} + A_{i,2} P_{2,j} + \cdots + A_{i,p} P_{p,j}.$$, But $$P_j = BC_j$$. arghm and gog) then AB represents the result of writing one after the other (i.e. Vectors satisfy the commutative lawof addition. the order in which multiplication is performed. is given by … 6.1 Associative law for scalar multiplication: Commutative, Associative, And Distributive Laws In ordinary scalar algebra, additive and multiplicative operations obey the commutative, associative, and distributive laws: Commutative law of addition a + b = b + a Commutative law of multiplication ab = ba Associative law of addition (a+b) + c = a+ (b+c) Associative law of multiplication ab (c) = a(bc) Distributive law a (b+c) = ab + ac and $$B = \begin{bmatrix} -1 & 1 \\ 0 & 3 \end{bmatrix}$$, In other words, students must be comfortable with the idea that you can group the three factors in any way you wish and still get the same product in order to make sense of and apply this formula. If B is an n p matrix, AB will be an m p matrix. Using triangle Law in triangle PQS we get a plus b plus c is equal to PQ plus QS equal to PS. , where and q is the angle between vectors and . Associative law of scalar multiplication of a vector. This important property makes simplification of many matrix expressions As the above holds true when performing addition and multiplication on any real numbers, it can be said that “addition and multiplication of real numbers are associative operations”. 7.2 Cross product of two vectors results in another vector quantity as shown below. =(a_iB_1) C_{1,j} + (a_iB_2) C_{2,j} + \cdots + (a_iB_q) C_{q,j} Hence, the $$(i,j)$$-entry of $$(AB)C$$ is given by ... $with the component-wise multiplication is a vector space, you need to do it component-wise, since this would be your definition for this operation. OF. 1. This math worksheet was created on 2019-08-15 and has been viewed 136 times this week and 306 times this month. So the associative law that holds for multiplication of numbers and for addition of vectors (see Theorem 1.5 (b),(e)), does $$\textit{not}$$ hold for the dot product of vectors. 6. The magnitude of a vector can be determined as. Matrices multiplicationMatrices B.Sc. & & \vdots \\ In Maths, associative law is applicable to only two of the four major arithmetic operations, which are addition and multiplication. In dot product, the order of the two vectors does not change the result. For example, when you get ready for work in the morning, putting on your left glove and right glove is commutative. In cross product, the order of vectors is important. Associative property of multiplication: (AB)C=A (BC) (AB)C = A(B C) = \begin{bmatrix} 0 & 9 \end{bmatrix}\). Using triangle Law in triangle QRS we get b plus c is equal to QR plus RS is equal to QS. & & + A_{i,p} (B_{p,1} C_{1,j} + B_{p,2} C_{2,j} + \cdots + B_{p,q} C_{q,j}) \\ Given a matrix $$A$$, the $$(i,j)$$-entry of $$A$$ is the entry in Ask Question Asked 4 years, 3 months ago. ASSOCIATIVE LAW. Therefore, This law is also referred to as parallelogram law. COMMUTATIVE LAW OF VECTOR ADDITION Consider two vectors and . , where q is the angle between vectors and . Let $$P$$ denote the product $$BC$$. Give the $$(2,2)$$-entry of each of the following. Let these two vectors represent two adjacent sides of a parallelogram. 6.1 Associative law for scalar multiplication: 6.2 Distributive law for scalar multiplication: 7. The two Big Four operations that are associative are addition and multiplication. The associative property. In other words. 2 × 7 = 7 × 2. Two vectors are equal only if they have the same magnitude and direction. a_i P_j & = & A_{i,1} (B_{1,1} C_{1,j} + B_{1,2} C_{2,j} + \cdots + B_{1,q} C_{q,j}) \\ $$a_i B$$ where $$a_i$$ denotes the $$i$$th row of $$A$$. $$C$$ is a $$q \times n$$ matrix, then 1. The associative law only applies to addition and multiplication. & & + (A_{i,1} B_{1,2} + A_{i,2} B_{2,2} + \cdots + A_{i,p} B_{p,2}) C_{2,j} \\ & & + (A_{i,1} B_{1,q} + A_{i,2} B_{2,q} + \cdots + A_{i,p} B_{p,q}) C_{q,j} \\ A vector may be represented in rectangular Cartesian coordinates as. In particular, we can simply write $$ABC$$ without having to worry about Vector addition is an operation that takes two vectors u, v ∈ V, and it produces the third vector u + v ∈ V 2. & = & (a_i B_1) C_{1,j} + (a_i B_2) C_{2,j} + \cdots + (a_i B_q) C_{q,j}. Hence, a plus b plus c is equal to a plus b plus c. This is the Associative property of vector addition. Consider three vectors , and. In view of the associative law we naturally write abc for both f(f(a, b), c) and f(a, f(b, c), and similarly for strings of letters of any length.If A and B are two such strings (e.g. This preview shows page 7 - 11 out of 14 pages.However, associative and distributive laws do hold for matrix multiplication: Associative Law: Let A be an m × n matrix, B be an n × p matrix, and C be a p × r matrix. $$\begin{bmatrix} 4 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 3\end{bmatrix} = 4$$. arghmgog).We have here used the convention (to be followed throughout) that capital letters are variables for strings of letters. The direction of vector is perpendicular to the plane containing vectors and such that follow the right hand rule. $$a_i B_j = A_{i,1} B_{1,j} + A_{i,2} B_{2,j} + \cdots + A_{i,p}B_{p,j}$$. A vector can be multiplied by another vector either through a dotor a crossproduct, 7.1 Dot product of two vectors results in a scalar quantity as shown below. Vector addition follows two laws, i.e. 4. Active 4 years, 3 months ago. Row $$i$$ of $$Q$$ is given by The Associative Property of Multiplication of Matrices states: Let A , B and C be n × n matrices. Let $$A$$ be an $$m\times p$$ matrix and let $$B$$ be a $$p \times n$$ matrix. It does not work with subtraction or division. The Associative Laws (or Properties) of Addition and Multiplication The Associative Laws (or the Associative Properties) The associative laws state that when you add or multiply any three real numbers , the grouping (or association) of the numbers does not affect the result. Thus $$P_{s,j} = B_{s,1} C_{1,j} + B_{s,2} C_{2,j} + \cdots + B_{s,q} C_{q,j}$$, giving Then $$Q_{i,r} = a_i B_r$$. For the example above, the $$(3,2)$$-entry of the product $$AB$$ \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\), $$\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} The associative rule of addition states, a + (b + c) is the same as (a + b) + c. Likewise, the associative rule of multiplication says a × (b × c) is the same as (a × b) × c. Example – The commutative property of addition: 1 + 2 = 2 +1 = 3 Addition is an operator. The associative property, on the other hand, is the rule that refers to grouping of numbers. For example, 3 + 2 is the same as 2 + 3. associative law. An operation is associative when you can apply it, using parentheses, in different groupings of numbers and still expect the same result. Welcome to The Associative Law of Multiplication (Whole Numbers Only) (A) Math Worksheet from the Algebra Worksheets Page at Math-Drills.com. $A(BC) = (AB)C.$ When two or more vectors are added together, the resulting vector is called the resultant. The answer is yes. 2 + 3 = 5 . Other than this major difference, however, the properties of matrix multiplication are mostly similar to the properties of real number multiplication. A space comprised of vectors, collectively with the associative and commutative law of addition of vectors and also the associative and distributive process of multiplication of vectors by scalars is called vector space. VECTOR ADDITION. Scalar multiplication of vectors satisfies the following properties: (i) Associative Law for Scalar Multiplication The order of multiplying numbers is doesn’t matter. Subtraction is not. $Q_{i,1} C_{1,j} + Q_{i,2} C_{2,j} + \cdots + Q_{i,q} C_{q,j} If \(A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and Consider three vectors , and. It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. Associative law, in mathematics, either of two laws relating to number operations of addition and multiplication, stated symbolically: a + ( b + c) = ( a + b) + c, and a ( bc) = ( ab) c; that is, the terms or factors may be associated in any way desired. If a vector is multiplied by a scalar as in , then the magnitude of the resulting vector is equal to the product of p and the magnitude of , and its direction is the same as if p is positive and opposite to if p is negative. Associative Laws: (a + b) + c = a + (b + c) (a × b) × c = a × (b × c) Distributive Law: a × (b + c) = a × b + a × c Hence, the $$(i,j)$$-entry of $$A(BC)$$ is the same as the $$(i,j)$$-entry of $$(AB)C$$. Consider a parallelogram, two adjacent edges denoted by … We construct a parallelogram OACB as shown in the diagram. A vector space consists of a set of V ( elements of V are called vectors), a field F ( elements of F are scalars) and the two operations 1. Show that matrix multiplication is associative. The associative laws state that when you add or multiply any three matrices, the grouping (or association) of the matrices does not affect the result. \begin{bmatrix} 0 & 1 & 2 & 3 \end{bmatrix}\). , matrix multiplication is not commutative! Matrix multiplication is associative. This condition can be described mathematically as follows: 5. $$\begin{bmatrix} 0 & 3 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 0 & 3\end{bmatrix} The law states that the sum of vectors remains same irrespective of their order or grouping in which they are arranged. \(\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} We describe this equality with the equation s1+ s2= s2+ s1. 3 + 2 = 5. where are the unit vectors along x, y, z axes, respectively. Applying "head to tail rule" to obtain the resultant of (+ ) and (+ ) Then finally again find the resultant of these three vectors : The law states that the sum of vectors remains same irrespective of their order or grouping in which they are arranged. Multiplication is commutative because 2 × 7 is the same as 7 × 2. Notice that the dot product of two vectors is a scalar, not a vector. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. A unit vector can be expressed as, We can also express any vector in terms of its magnitude and the unit vector in the same direction as, 2. Commutative law and associative law. The Associative Law is similar to someone moving among a group of people associating with two different people at a time. VECTOR ADDITION. Because: Again, subtraction, is being mistaken for an operator. The Associative Law of Addition: Applying “head to tail rule” to obtain the resultant of ( + ) and ( + ) Then finally again find the resultant of these three vectors : This fact is known as the ASSOCIATIVE LAW OF VECTOR ADDITION. & = & (A_{i,1} B_{1,1} + A_{i,2} B_{2,1} + \cdots + A_{i,p} B_{p,1}) C_{1,j} \\ Even though matrix multiplication is not commutative, it is associative in the following sense. As with the commutative law, will work only for addition and multiplication. In fact, an expression like 2\times3\times5 only makes sense because multiplication is associative. The displacement vector s1followed by the displacement vector s2leads to the same total displacement as when the displacement s2occurs first and is followed by the displacement s1. It follows that \(A(BC) = (AB)C$$. If we divide a vector by its magnitude, we obtain a unit vector in the direction of the original vector. \begin{eqnarray} Associate Law = A + (B + C) = (A + B) + C 1 + (2 + 3) = (1 + 2) + 3 The key step (and really the only one that is not from the definition of scalar multiplication) is once you have ((r s) x 1, …, (r s) x n) you realize that each element (r s) x i is a product of three real numbers. But for other arithmetic operations, subtraction and division, this law is not applied, because there could be a change in result.This is due to change in position of integers during addition and multiplication, do not change the sign of the integers. You likely encounter daily routines in which the order can be switched. in the following sense. OF. Let b and c be real numbers. Associative law, in mathematics, either of two laws relating to number operations of addition and multiplication, stated symbolically: a + (b + c) = (a + b) + c, and a (bc) = (ab) c; that is, the terms or factors may be associated in any way desired. For example, if $$A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \\ 4 & 0 \end{bmatrix}$$ That is, show that (AB)C = A(BC) for any matrices A, B, and C that are of the appropriate dimensions for matrix multiplication. Apart from this there are also many important operations that are non-associative; some examples include subtraction, exponentiation, and the vector cross product. is given by $$A B_j$$ where $$B_j$$ denotes the $$j$$th column of $$B$$. Then A. A unit vector is defined as a vector whose magnitude is unity. Informal Proof of the Associative Law of Matrix Multiplication 1. Formally, a binary operation ∗ on a set S is called associative if it satisfies the associative law: (x ∗ y) ∗ z = x ∗ (y ∗ z) for all x, y, z in S.Here, ∗ is used to replace the symbol of the operation, which may be any symbol, and even the absence of symbol (juxtaposition) as for multiplication. Recall from the definition of matrix product that column $$j$$ of $$Q$$ Scalar Multiplication is an operation that takes a scalar c ∈ … ( A $$Q_{i,j}$$, which is given by column $$j$$ of $$a_iB$$, is Commutative Law - the order in which two vectors are added does not matter. & & \vdots \\ The commutative law of addition states that you can change the position of numbers in an addition expression without changing the sum. If a vector is multiplied by a scalar as in , then the magnitude of the resulting vector is equal to the product of p and the magnitude of , and its direction is the same as if p is positive and opposite to if p is negative. To see this, first let $$a_i$$ denote the $$i$$th row of $$A$$. Let $$Q$$ denote the product $$AB$$. In general, if A is an m n matrix (meaning it has m rows and n columns), the matrix product AB will exist if and only if the matrix B has n rows. Notes: https://www.youtube.com/playlist?list=PLC5tDshlevPZqGdrsp4zwVjK5MUlXh9D5 row $$i$$ and column $$j$$ of $$A$$ and is normally denoted by $$A_{i,j}$$. & & + A_{i,2} (B_{2,1} C_{1,j} + B_{2,2} C_{2,j} + \cdots + B_{2,q} C_{q,j}) \\ If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then \[A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. Associative Law allows you to move parentheses as long as the numbers do not move. The $$(i,j)$$-entry of $$A(BC)$$ is given by possible. \end{eqnarray}, Now, let $$Q$$ denote the product $$AB$$. then the second row of $$AB$$ is given by = a_i P_j.\]. Since you have the associative law in R you can use that to write (r s) x i = r (s x i). Even though matrix multiplication is not commutative, it is associative 3. 5.2 Associative law for addition: 6. C be n × n Matrices$ only makes sense because multiplication is not commutative axes! 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1. ## Solve this radical equation Solve in R : $\displaystyle \sqrt {x - 1} + \sqrt {2 - x} = x^2 - 3x + 3$ 2. Square both sides. You'll end up with a radical as the middle term on the right-hand side. Isolate this radical, and then square again. Then solve the resulting polynomial. 3. Originally Posted by stapel Square both sides. You'll end up with a radical as the middle term on the right-hand side. Isolate this radical, and then square again. Then solve the resulting polynomial. Hello :I'have $\displaystyle \begin{array}{l} \forall x \in \Re :x^2 - 3x + 3 > 0 \\ because:\Delta = - 3 < 0 \\ then: \\ x - 1 + 2 - x + 2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 \\ 2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 - 1??????? \\ \end{array}$ 4. $\displaystyle \sqrt{x-1}+\sqrt{2-x}=-(x-1)(2-x)+1$ Let $\displaystyle a=\sqrt{x-1}, \ b=\sqrt{2-x}$ Then we have the system of equations: $\displaystyle \left\{\begin{array}{ll}a+b=-a^2b^2+1\\a^2+b^2=1\end{array}\right.$ Let $\displaystyle S=a+b, \ P=ab$. Then $\displaystyle \left\{\begin{array}{ll}S+P^2=1\\S^2-2P=1\end{array}\right.$ From the first equation $\displaystyle S=1-P^2$ Replace in the second: $\displaystyle P(P^3-2P-2)=0$ If $\displaystyle P=0$ then $\displaystyle S=1$. Then $\displaystyle \left\{\begin{array}{ll}a+b=1\\ab=0\end{array}\rig ht.$ We get $\displaystyle \left\{\begin{array}{ll}a=1\\b=0\end{array}\right. \Rightarrow x=2$ $\displaystyle \left\{\begin{array}{ll}a=0\\b=1\end{array}\right. \Rightarrow x=1$ Now we have to see what happens if $\displaystyle P^3-2P^2-2=0$ But this equation has no rational roots and the discussion is more complicated. 5. from red_dog, p^3 - 2p^2 - 0 = 0, by iteration; p > 2.359. the graph confirms it 6. Originally Posted by red_dog $\displaystyle \sqrt{x-1}+\sqrt{2-x}=-(x-1)(2-x)+1$ Let $\displaystyle a=\sqrt{x-1}, \ b=\sqrt{2-x}$ Then we have the system of equations: $\displaystyle \left\{\begin{array}{ll}a+b=-a^2b^2+1\\a^2+b^2=1\end{array}\right.$ Let $\displaystyle S=a+b, \ P=ab$. Then $\displaystyle \left\{\begin{array}{ll}S+P^2=1\\S^2-2P=1\end{array}\right.$ From the first equation $\displaystyle S=1-P^2$ Replace in the second: $\displaystyle P(P^3-2P-2)=0$ If $\displaystyle P=0$ then $\displaystyle S=1$. Then $\displaystyle \left\{\begin{array}{ll}a+b=1\\ab=0\end{array}\rig ht.$ We get $\displaystyle \left\{\begin{array}{ll}a=1\\b=0\end{array}\right. \Rightarrow x=2$ $\displaystyle \left\{\begin{array}{ll}a=0\\b=1\end{array}\right. \Rightarrow x=1$ Now we have to see what happens if $\displaystyle P^3-2P^2-2=0$ But this equation has no rational roots and the discussion is more complicated. HELLO : This equation $\displaystyle p^3 - 2p^2 - 2 = 0$ has not real roots because I'have $\displaystyle 0 \le p \le 1$ and if $\displaystyle p \in \left[ {0,1} \right]^3 - 2p^2 - 2 < 0$ There is the details : $\displaystyle 1 \le x \le 2.....because:\left( {x - 1 \ge 0} \right) \wedge \left( {2 - x \ge 0} \right)$ Then : $\displaystyle \begin{array}{l} \\ 1 \le x \le 2 \Leftrightarrow 0 \le x - 1 \le 1 \Leftrightarrow 0 \le \sqrt {x - 1} \le 1......(1) \\ 1 \le x \le 2 \Leftrightarrow - 1 \ge - x \ge - 2 \Leftrightarrow 0 \le 2 - x \le 1 \Leftrightarrow 0 \le \sqrt {2 - x} \le 1.....(2) \\ \end{array}$ $\displaystyle but:a = \sqrt {x - 1} ...b = \sqrt {x - 1}$ by (1),(2) : $\displaystyle 0 \le p \le 1$ THANKS
# Insertion Sort Algorithm and Example In this article, you will learn about how an insertion sort actually works. You will understand it with the help of a step-by-step algorithm and an example. But before we get started, let's first discuss what the term "insertion sort" means. ## What is insertion sorting? It operates in a manner analogous to how you would sort the playing cards that you have in your hand. That is to say, you select and then pick up the card (from the beginning), and then you place it one at a time in a sorted order. ## Insertion Sort Algorithm The algorithm for using insertion sort to put an array in ascending order is as follows: 1. Beginning with arr[1] and ending with arr[n]. 2. Compare the current element (key) to the element present and its predecessor. • Here, at first execution, the current element will be at arr[1] (the second element from the array). • At the second execution, the current element will be at arr[2] (the third element from the array). • And so on. 3. If the current element is smaller than its predecessor, 4. Then compare it to the elements present before the index of the current element. 5. Move the greater elements one index up to create space for the swapped element. Steps 3, 4, and 5 of the algorithm are repeated with the following iterations: ```8 6 4 6 8 4 4 6 8``` As you can see, • At first iteration, the key element 6 is smaller than its predecessor (that is, the element present before the index of the key element). • So the greater element (8) gets moved one index up and creates a space. • That is, 8 is moved to the position where 6 was previously. • And the current index of 8 is available for 6 to get a place. • So 6 gets placed on the previous index of 8. • And the key element 4 is smaller than its predecessor at the second interation. • That is, this time, 4 is smaller than both of its predecessor elements. • That is 6 and 8. • So these two elements moved up one index. • That is, 8 goes to the index where 4 was present, and 6 goes to the index where 8 was present. • And finally, 4 is placed at the index where 6 was present. • In this way, the insertion sort works. In insertion sort, each element is placed one by one to the left (if necessary). That is, from the first to the last element, we have to decide the correct place for each element one by one and arrange the given array in ascending order as per the insertion sort technique. ## Insertion Sort Example For example, if the user has supplied any array that contains elements such as 28; 16; 5; and 11; Therefore, here is a step-by-step sorting of the given array: • 16 28 5 11 0 • 5 16 28 11 0 • 5 11 16 28 0 • 0 5 11 16 28 That is, the sorted array you will get is 0 5 11 16 28. Let's take another example. Assume the user supplied an array whose elements are 1, 10, 2, 9, 3, 8, 4, 7, 5, and 6. Therefore, here is the step-by-step list of the array after each sort: • 1 2 10 9 3 8 4 7 5 6 • 1 2 9 10 3 8 4 7 5 6 • 1 2 3 9 10 8 4 7 5 6 • 1 2 3 8 9 10 4 7 5 6 • 1 2 3 4 8 9 10 7 5 6 • 1 2 3 4 7 8 9 10 5 6 • 1 2 3 4 5 7 8 9 10 6 • 1 2 3 4 5 6 7 8 9 10 As you can see, in the step-by-step array after each sort, the first step gets skipped because we already have the first and second elements of the array in ascending order. #### Programs Created on Insertion Sort Computer Fundamentals Quiz « Previous Tutorial Next Tutorial » Liked this post? Share it!
# How to Simplify Radical Expressions Author Info Updated: April 24, 2019 A radical expression is an algebraic expression that includes a square root (or cube or higher order roots). Often such expressions can describe the same number even if they appear very different (ie, 1/(sqrt(2) - 1) = sqrt(2)+1). The remedy is to define a preferred "canonical form" for such expressions. If two expressions, both in canonical form, still look different, then they indeed are unequal. Mathematicians agreed that the canonical form for radical expressions should: • Not use fractional exponents • Have only squarefree terms under the radicals One practical use for this is in multiple-choice exams. When you've solved a problem, but your answer doesn't match any of the multiple choices, try simplifying it into canonical form. Since test writers usually put their answers in canonical form, doing the same to yours will make it apparent which of their answers is equal to yours. In free-response exams, instructions like "simplify your answer" or "simplify all radicals" mean the student is to apply these steps until their answer satisfies the canonical form above. It is also of some use in equation solving, although some equations are easier to deal with using a non-canonical form. ## Steps 1. 1 If necessary, review the rules for manipulation of radicals and exponents (they're the same - roots are fractional powers) as most of them are needed for this process. Also, review the rules for manipulating and simplifying polynomial and rational type expressions as they too will be needed throughout to simplify. ### Method 1 of 6: Perfect Powers 1. 1 Simplify any radical expressions that are perfect squares. A perfect square is the product of any number that is multiplied by itself, such as 81, which is the product of 9 x 9.[1] To simplify a perfect square under a radical, simply remove the radical sign and write the number that is the square root of the perfect square.[2] • For example, 121 is a perfect square because 11 x 11 is 121. Thus, you can simplify sqrt(121) to 11, removing the square root symbol. • To make this process easier, you should memorize the first twelve perfect squares: 1 x 1 = 1, 2 x 2 = 4, 3 x 3 = 9, 4 x 4 = 16, 5 x 5 = 25, 6 x 6 = 36, 7 x 7 = 49, 8 x 8 = 64, 9 x 9 = 81, 10 x 10 = 100, 11 x 11 = 121, 12 x 12 = 144 2. 2 Simplify any radical expressions that are perfect cubes. A perfect cube is the product of any number that is multiplied by itself twice, such as 27, which is the product of 3 x 3 x 3. To simplify a radical expression when a perfect cube is under the cube root sign, simply remove the radical sign and write the number that is the cube root of the perfect cube.[3] • For example, 343 is a perfect cube because it is the product of 7 x 7 x 7. Therefore, the cube root of the perfect cube 343 is simply 7. ### Method 2 of 6: Convert Rational Exponents to Radicals Or convert the other way if you prefer (sometimes there are good reasons for doing that), but don't mix terms like sqrt(5) + 5^(3/2) in the same expression. We will assume that you decide to use radical notation and will use sqrt(n) for the square root of n and cbrt(n) for cube roots.[4] 1. 1 Find any fractional exponent and convert it to the radical equivalent, namely x^(a/b) = bth root of x^a • If you have a fraction for the index of a radical, get rid of that too. For instance the (2/3) root of 4 = sqrt(4)^3 = 2^3 = 8. 2. 2 Convert negative exponents to their equivalent fraction, namely x^-y = 1/x^y • This only applies to constant, rational exponents. If you have terms like 2^x, leave them alone, even if the problem context implies that x might be fractional or negative. 3. 3 Combine any like terms and simplify any rational expressions that result.[5] ### Method 3 of 6: Remove Fractions from Radicals Canonical form requires expressing the root of a fraction in terms of roots of whole numbers 1. 1 Examine terms under each radical to see if any contain fractions. If so, ... 2. 2 Replace it as a ratio of two radicals using the identity sqrt(a/b) = sqrt(a)/sqrt(b). • Don't use this identity if the denominator is negative, or is a variable expression that might be negative. In that case, simplify the fraction first. 3. 3 Simplify any perfect squares that result. That is, convert sqrt(5/4) to sqrt(5)/sqrt(4), and then further simplify it to sqrt(5)/2.[6] 4. 4 Make any other useful simplifications such as reducing compound fractions, combining like terms, etc.[7] ### Method 4 of 6: Combine Products of Radicals 1. 1 If you have one radical expression multiplied by another, combine them as a single radical using the property: sqrt(a)sqrt(b) = sqrt(ab). For example, replace sqrt(2)sqrt(6) by sqrt(12).[8] • The above identity, sqrt(a)sqrt(b) = sqrt(ab) is valid for non negative radicands. Don't apply it if a and b are negative as then you would falsely assert that sqrt(-1)sqrt(-1) = sqrt(1). The left-hand side -1 by definition (or undefined if you refuse to acknowledge complex numbers) while the right side is +1. If a and/or b is negative, first "fix" its sign by sqrt(-5) = isqrt(5). If the radicand is a variable expression whose sign is not known from context and could be either positive or negative, then just leave it alone for now. You could use the more general identity, sqrt(a)sqrt(b) = sqrt(sgn(a))sqrt(sgn(b))sqrt(\|ab\|) which is valid for all real numbers a and b, but it's usually not worth the added complexity of introducing the sign function. • This identity only applies if the radicals have the same index. You can multiply more general radicals like sqrt(5)cbrt(7) by first expressing them with a common index. To do this, temporarily convert the roots to fractional exponents: sqrt(5)cbrt(7) = 5^(1/2) * 7^(1/3) = 5^(3/6) * 7^(2/6) = 125^(1/6) * 49^(1/6). Then apply the product rule to equate this product to the sixth root of 6125. ### Method 5 of 6: Extract Square Factors from Radicals 1. 1 Factorize an imperfect radical expression into its prime factors. The factors are the numbers that multiply to create a number -- for example, 5 and 4 are two factors of the number 20. To break down an imperfect radical expression, write down all of the factors of that number (or as many as you can think of if it's a large number) until you find one that is a perfect square.[9] • For example, try listing all the factors of the number 45: 1, 3, 5, 9, 15, and 45. 9 is a factor of 45 that is also a perfect square (9=3^2). 9 x 5 = 45. 2. 2 Remove any multiples that are a perfect square out of the radical sign. 9 is a perfect square because it is the product of 3 x 3. Take the 9 out of the radical sign and place a 3 in front of it, leaving 5 under the radical sign. If you "throw" the three back in under the radical sign, it will be multiplied by itself to create 9 again, which will multiply with 5 to create 45 again. 3 root 5 is just a simplified way of saying root 45. • That is, sqrt(45) = sqrt(95) = sqrt(9)sqrt(5) = 3sqrt(5). 3. 3 Find a perfect square in the variable. The square root of a to the second power would be \|a\|. You can further simplify this to just "a" only if the variable is known to be positive. The square root of a to the third power is broken down into the square root of a squared times a -- this is because you add exponents when you multiply variables, so that a squared times a is equal to a cubed. • Therefore, the perfect square in the expression a cubed is a squared. 4. 4 Pull any variables that are perfect squares out of the radical sign. Now, take a squared and pull it out of the radical to make it a regular \|a\|. The simplified form of a cubed is just \|a\| root a. 5. 5 Combine any like terms and simplify any rational expressions that result. ### Method 6 of 6: Rationalize the Denominator 1. 1 Canonical form requires the denominator to be a whole number (or a polynomial if it contains indeterminate) if at all possible.[10] • If the denominator consists of a single term under a radical, such as [stuff]/sqrt(5), then multiply numerator and denominator by that radical to get [stuff]sqrt(5)/sqrt(5)sqrt(5) = [stuff]sqrt(5)/5. • For cube or higher roots, multiply by the appropriate power of the radical to make the denominator rational. If the denominator was cbrt(5), then multiply numerator and denominator by cbrt(5)^2. • If the denominator consists of a sum or difference of square roots such as sqrt(2) + sqrt(6), then multiply numerator and denominator by its conjugate, the same expression with the opposite operator. Thus [stuff]/(sqrt(2) + sqrt(6)) = [stuff](sqrt(2)-sqrt(6))/(sqrt(2) + sqrt(6))(sqrt(2)-sqrt(6)). Then use the difference of squares identity [(a+b)(a-b) = a^2-b^2] to rationalize the denominator, simplifying (sqrt(2) + sqrt(6))(sqrt(2)-sqrt(6)) = sqrt(2)^2 - sqrt(6)^2 = 2-6 = -4. • This works for denominators like 5 + sqrt(3) too since every whole number is a square root of some other whole number. [1/(5 + sqrt(3)) = (5-sqrt(3))/(5 + sqrt(3))(5-sqrt(3)) = (5-sqrt(3))/(5^2-sqrt(3)^2) = (5-sqrt(3))/(25-3) = (5-sqrt(3))/22] • This works for a sum of square roots like sqrt(5)-sqrt(6)+sqrt(7). If you group it as (sqrt(5)-sqrt(6))+sqrt(7) and multiply it by (sqrt(5)-sqrt(6))-sqrt(7), your answer won't be rational, but will be of the form a+bsqrt(30) where a and b are rational. Then you can repeat the process with the conjugate of a+bsqrt(30) and (a+bsqrt(30))(a-bsqrt(30)) is rational. In essence, if you can use this trick once to reduce the number of radical signs in the denominator, then you can use this trick repeatedly to eliminate all of them. • This even works for denominators containing higher roots like the 4th root of 3 plus the 7th root of 9. Just multiply numerator and denominator by the denominator's conjugate. Unfortunately, it is not immediately clear what the conjugate of that denominator is nor how to go about finding it. A good book on algebraic number theory will cover this, but I will not. 2. 2 Now the denominator is rationalized, but the numerator is a mess. You now have whatever you started with up there times the denominator's conjugate. Go ahead and expand that product like you would for a product of polynomials. See if anything cancels or simplifies and combine like terms if possible. 3. 3 If the denominator is a negative integer, then multiply numerator and denominator by -1 to make it positive. ## Community Q&A Search • Question You can only take something out from under a radical if it's a factor. For instance, sqrt(64(x+3)) can become 8sqrt(x+3), but sqrt(64x + 3) cannot be simplified. • Question A rectangle has sides of 4 and 6 units. On each of its four sides, square are drawn externally. Their centers form another quadrilateral. What is the area (in sq. units) of this quadrilateral? Donagan You'll have to draw a diagram of this. You'll see that triangles can be drawn external to all four sides of the new quadrilateral. By the Pythagorean theorem you can find the sides of the quadrilateral, all of which turn out to be 5 units, so that the quadrilateral's area is 25 square units. 200 characters left ## Tips • There are websites that you can search online that will simplify a radical expression for you. You simply type in the equation under the radical sign, and after hitting enter, your simplified answer will appear. Thanks! • For simple problems, many of these steps won't apply. For complicated problems, some of them may need to be applied more than once. Make "easy" simplifications continuously as you work, and check your final answer against the canonical form criteria in the intro. If your answer is canonical, you are done; while it is not canonical, one of these steps will tell you what still needs to be done to make it so. Thanks! • Most references to the "preferred canonical form" for a radical expression also apply to complex numbers (i = sqrt(-1)). Even if it's written as "i" rather than with a radical sign, we try to avoid writing i in a denominator. Thanks! • Parts of these instructions assume that all radicals are square roots. The general principles are the same for cube or higher roots, although some of them (particularly rationalizing the denominator) may be harder to apply. You'll also have to decide if you want terms like cbrt(4) or cbrt(2)^2 (I can't remember which way the textbook authors prefer). Thanks! • Parts of these instructions misuse the term "canonical form" when they actually describe only a "normal form". The difference is that a canonical form would require either 1+sqrt(2) or sqrt(2)+1 and label the other as improper; a normal form assumes that you, dear reader, are bright enough to recognize these as "obviously equal" as numbers even if they aren't typographically identical (where 'obvious' means using only arithmetical properties (addition is commutative), not algebraic properties (sqrt(2) is a non-negative root of x^2-2)). We hope readers will forgive this mild abuse of terminology. Thanks! • If these instructions seem ambiguous or contradictory, then apply all consistent and unambiguous steps and then choose whatever form looks most like the way radical expressions are used in your text. Thanks! wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 29 people, some anonymous, worked to edit and improve it over time. Together, they cited 12 references. This article has also been viewed 298,737 times. Co-authors: 29 Updated: April 24, 2019 Views: 298,737 Categories: Algebra Article SummaryX To simplify a radical expression, simplify any perfect squares or cubes, fractional exponents, or negative exponents, and combine any like terms that result. If there are fractions in the expression, split them into the square root of the numerator and square root of the denominator. If you need to extract square factors, factorize the imperfect radical expression into its prime factors and remove any multiples that are a perfect square out of the radical sign. For tips on rationalizing denominators, read on!
Lesson Plan: # I Got The Power! 5.0 based on 1 rating July 31, 2015 Standards July 31, 2015 ## Learning Objectives Students will be able to identify and explain patterns in the number of zeros of the product when multiplying a number by powers of 10, identify and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10 and use whole number exponents to denote powers of 10. ## Lesson ### Introduction (10 minutes) • Review place value and order of operations. • Remind students that our number system is a decimal system and it relies on groupings of 10s. Tell students that we also utilize a place value system, with the value of each digit depending on its position or place in the number. • Write "1,000,000" on the board and ask students to identify at least one other way this number can be written. Possible answers include: writing the words “one million” or 10 x 10 x 10 x 10 x 10 x 10. • Tell students that mathematicians created a shortcut for writing very large numbers, and that today they are going to explore this technique. ### Explicit Instruction/Teacher Modeling (15 minutes) • Display the Exponent Chart on the board. • Cover up a few of the fields or leave some of the fields blank and allow students to use the other fields to predict what will go in the fields that are covered. • Once predictions have been made, allow students to verbally identify patterns in each column by discussing it with a partner. • Allow students to share their patterns with the class. • Using the chart, explain to students that mathematicians established our number system a long time ago and that any number raised to the power of zero would equal 1. • Using the chart, explain to the students that any base number raised to the power of 1 is the base number. • Using the chart, explain to students that any number raised to the power of 2 is the base number multiplied by itself. • Ask the students if they are beginning to see a pattern. Let them know that it will make more sense as they proceed and continue to look for patterns. • Allow students to use an online exponents calculator. ### Guided Practice/Interactive Modeling (30 minutes) • Since students in elementary school are often fascinated by large numbers, such as a million, a zillion, and a trillion, this is a good opportunity to have them try to get an idea of just how big those numbers really are. • Use an interactive whiteboard to complete the guided steps below: • Write the following numbers on the board and have students use whole number exponents to denote powers of 10 and then identify patterns that they see: 1. 1,000,000,000 2. 1,000,000,000,000 3. 1,000,000,000,000,000 • Students should understand that the number of 0s in the number equals the value of the exponent. • Write the following exponential notation numbers on the board and have students write the numbers in standard form and then identify patterns that they see: 1. 10^4 2. 10^5 3. 10^6 4. 10^7 • Students should understand that the exponent value equals the number of 0s in the number. • Write the following on the board and have students use exponential notation to simplify the value and then identify patterns that they see: 1. 250 2. 2,500 3. 25,000 4. 250,000 • Students should understand that when simplifying values they should take any numbers before the zero and multiply it by the exponential notation which is the number 10 with an exponent that equals the amount of 0’s. • Write the following on the board and have students multiplying a number by powers of 10 and then identify patterns that they see: 1. 36 x 10^1 2. 36 x 10^2 answer: 36 x (10x10)= 36 x 100=3,600 3. 36 x 10^3 answer: 36 x (10x10x10)= 36 x 1000=36,000 4. 36 x 10^4 • Students should understand that when multiplying a whole number by a power of 10, one converts the exponent into standard form, counts the number of zeros after the 1 and add the 0s to the end of the whole number. • Write the following on the board and have students work on equations that emphasize placement of the decimal point when a decimal is multiplied or divided by a power of 10 and then identify patterns that they see: 1. 2.5 x 10^3 answer: 2.5 x (10 x 10 x 10)= 2.5 x 1,000 = 2.5000, then move decimal 3 places to the right =2,500 2. 0.22 x 10^2 0.22 x 100= .22000, then move decimal places 2 places to the right =22 3. 0.042 ÷ 10^2 0.042 x 100=.04200, then move decimal 2 places to the left =.00042 4. 63.4 ÷ 10^3 63.4 ÷ 1,000 = 63.4000, then move decimal 3 places to the left =.0634000 =.0634 • Students should understand that when multiplying a decimal by a power of ten and moving the decimal point one place to the right for each 0 after the 1 and when dividing a decimal by a power of ten and moving the decimal point one place to the left for each 0 after the 1. ### Independent Working Time (45 minutes) • Pass out the I Got The Power worksheet. • Ask students to use their notes from the guided practice to help them work independently to follow the instructions on the worksheet. • Tell students that they are going to work independently to use whole number exponents to denote powers of 10, write the numbers in standard form, use exponential notation to simplify the value, multiplying a number by powers of 10, and decipher the placement of the decimal point when a decimal is multiplied or divided by a power of 10. ## Extend ### Differentiation • Enrichment: Students can create a “test” bank of questions or worksheets and answers using a word processing program. • Support: Provide struggling students with a peer tutor who can help explain the concepts in student friendly language. ### Technology Integration • Students can use a word processor to create a worksheet or test bank of questions and answers. • The students can video themselves answer the independent practice problems to be used for peer tutoring. ## Review ### Assessment (20 minutes) • Have students use the following prompts to write an explanation and provide an example of: a. using a whole number exponent to denote powers of 10, b. writing the numbers in standard form, c. using exponential notation to simplify the value of a number, and d. multiplying a number by powers of 10 and deciphering the placement of the decimal point when a decimal is multiplied or divided by a power of 10. • If time permits, allow students to come to the whiteboard and explain their answers. • Students can use a camera phone or tablet to record explanations of their answers in an effort to expand on their technological skills. ### Review and Closing (10 minutes) • Recap the components of this unit by reviewing the chart in the guided practice section or problems from their independent practice.
# 6.8 Fitting exponential models to data (Page 10/12) Page 10 / 12 The graph below shows transformations of the graph of $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.\text{\hspace{0.17em}}$ What is the equation for the transformation? ## Logarithmic Functions Rewrite $\text{\hspace{0.17em}}{\mathrm{log}}_{17}\left(4913\right)=x\text{\hspace{0.17em}}$ as an equivalent exponential equation. ${17}^{x}=4913$ Rewrite $\text{\hspace{0.17em}}\mathrm{ln}\left(s\right)=t\text{\hspace{0.17em}}$ as an equivalent exponential equation. Rewrite $\text{\hspace{0.17em}}{a}^{-\text{\hspace{0.17em}}\frac{2}{5}}=b\text{\hspace{0.17em}}$ as an equivalent logarithmic equation. ${\mathrm{log}}_{a}b=-\frac{2}{5}$ Rewrite $\text{\hspace{0.17em}}{e}^{-3.5}=h\text{\hspace{0.17em}}$ as an equivalent logarithmic equation. Solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{log}}_{64}\left(x\right)=\frac{1}{3}\text{\hspace{0.17em}}$ to exponential form. $x={64}^{\frac{1}{3}}=4$ Evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(\frac{1}{125}\right)\text{\hspace{0.17em}}$ without using a calculator. Evaluate $\text{\hspace{0.17em}}\mathrm{log}\left(\text{0}\text{.000001}\right)\text{\hspace{0.17em}}$ without using a calculator. $\mathrm{log}\left(\text{0}\text{.000001}\right)=-6$ Evaluate $\text{\hspace{0.17em}}\mathrm{log}\left(4.005\right)\text{\hspace{0.17em}}$ using a calculator. Round to the nearest thousandth. Evaluate $\text{\hspace{0.17em}}\mathrm{ln}\left({e}^{-0.8648}\right)\text{\hspace{0.17em}}$ without using a calculator. $\mathrm{ln}\left({e}^{-0.8648}\right)=-0.8648$ Evaluate $\text{\hspace{0.17em}}\mathrm{ln}\left(\sqrt[3]{18}\right)\text{\hspace{0.17em}}$ using a calculator. Round to the nearest thousandth. ## Graphs of Logarithmic Functions Graph the function $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{log}\left(7x+21\right)-4.$ Graph the function $\text{\hspace{0.17em}}h\left(x\right)=2\mathrm{ln}\left(9-3x\right)+1.$ State the domain, vertical asymptote, and end behavior of the function $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{ln}\left(4x+20\right)-17.$ Domain: $\text{\hspace{0.17em}}x>-5;\text{\hspace{0.17em}}$ Vertical asymptote: $\text{\hspace{0.17em}}x=-5;\text{\hspace{0.17em}}$ End behavior: as $\text{\hspace{0.17em}}x\to -{5}^{+},f\left(x\right)\to -\infty \text{\hspace{0.17em}}$ and as $\text{\hspace{0.17em}}x\to \infty ,f\left(x\right)\to \infty .$ ## Logarithmic Properties Rewrite $\text{\hspace{0.17em}}\mathrm{ln}\left(7r\cdot 11st\right)\text{\hspace{0.17em}}$ in expanded form. Rewrite $\text{\hspace{0.17em}}{\mathrm{log}}_{8}\left(x\right)+{\mathrm{log}}_{8}\left(5\right)+{\mathrm{log}}_{8}\left(y\right)+{\mathrm{log}}_{8}\left(13\right)\text{\hspace{0.17em}}$ in compact form. ${\text{log}}_{8}\left(65xy\right)$ Rewrite $\text{\hspace{0.17em}}{\mathrm{log}}_{m}\left(\frac{67}{83}\right)\text{\hspace{0.17em}}$ in expanded form. Rewrite $\text{\hspace{0.17em}}\mathrm{ln}\left(z\right)-\mathrm{ln}\left(x\right)-\mathrm{ln}\left(y\right)\text{\hspace{0.17em}}$ in compact form. $\mathrm{ln}\left(\frac{z}{xy}\right)$ Rewrite $\text{\hspace{0.17em}}\mathrm{ln}\left(\frac{1}{{x}^{5}}\right)\text{\hspace{0.17em}}$ as a product. Rewrite $\text{\hspace{0.17em}}-{\mathrm{log}}_{y}\left(\frac{1}{12}\right)\text{\hspace{0.17em}}$ as a single logarithm. ${\text{log}}_{y}\left(12\right)$ Use properties of logarithms to expand $\text{\hspace{0.17em}}\mathrm{log}\left(\frac{{r}^{2}{s}^{11}}{{t}^{14}}\right).$ Use properties of logarithms to expand $\text{\hspace{0.17em}}\mathrm{ln}\left(2b\sqrt{\frac{b+1}{b-1}}\right).$ $\mathrm{ln}\left(2\right)+\mathrm{ln}\left(b\right)+\frac{\mathrm{ln}\left(b+1\right)-\mathrm{ln}\left(b-1\right)}{2}$ Condense the expression $\text{\hspace{0.17em}}5\mathrm{ln}\left(b\right)+\mathrm{ln}\left(c\right)+\frac{\mathrm{ln}\left(4-a\right)}{2}\text{\hspace{0.17em}}$ to a single logarithm. Condense the expression $\text{\hspace{0.17em}}3{\mathrm{log}}_{7}v+6{\mathrm{log}}_{7}w-\frac{{\mathrm{log}}_{7}u}{3}\text{\hspace{0.17em}}$ to a single logarithm. ${\mathrm{log}}_{7}\left(\frac{{v}^{3}{w}^{6}}{\sqrt[3]{u}}\right)$ Rewrite $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(12.75\right)\text{\hspace{0.17em}}$ to base $\text{\hspace{0.17em}}e.$ Rewrite $\text{\hspace{0.17em}}{5}^{12x-17}=125\text{\hspace{0.17em}}$ as a logarithm. Then apply the change of base formula to solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ using the common log. Round to the nearest thousandth. $x=\frac{\frac{\mathrm{log}\left(125\right)}{\mathrm{log}\left(5\right)}+17}{12}=\frac{5}{3}$ ## Exponential and Logarithmic Equations Solve $\text{\hspace{0.17em}}{216}^{3x}\cdot {216}^{x}={36}^{3x+2}\text{\hspace{0.17em}}$ by rewriting each side with a common base. Solve $\text{\hspace{0.17em}}\frac{125}{{\left(\frac{1}{625}\right)}^{-x-3}}={5}^{3}\text{\hspace{0.17em}}$ by rewriting each side with a common base. $x=-3$ Use logarithms to find the exact solution for $\text{\hspace{0.17em}}7\cdot {17}^{-9x}-7=49.\text{\hspace{0.17em}}$ If there is no solution, write no solution . Use logarithms to find the exact solution for $\text{\hspace{0.17em}}3{e}^{6n-2}+1=-60.\text{\hspace{0.17em}}$ If there is no solution, write no solution . no solution Find the exact solution for $\text{\hspace{0.17em}}5{e}^{3x}-4=6\text{\hspace{0.17em}}$ . If there is no solution, write no solution . Find the exact solution for $\text{\hspace{0.17em}}2{e}^{5x-2}-9=-56.\text{\hspace{0.17em}}$ If there is no solution, write no solution . no solution Find the exact solution for $\text{\hspace{0.17em}}{5}^{2x-3}={7}^{x+1}.\text{\hspace{0.17em}}$ If there is no solution, write no solution . Find the exact solution for $\text{\hspace{0.17em}}{e}^{2x}-{e}^{x}-110=0.\text{\hspace{0.17em}}$ If there is no solution, write no solution . $x=\mathrm{ln}\left(11\right)$ Use the definition of a logarithm to solve. $\text{\hspace{0.17em}}-5{\mathrm{log}}_{7}\left(10n\right)=5.$ 47. Use the definition of a logarithm to find the exact solution for $\text{\hspace{0.17em}}9+6\mathrm{ln}\left(a+3\right)=33.$ $a={e}^{4}-3$ Use the one-to-one property of logarithms to find an exact solution for $\text{\hspace{0.17em}}{\mathrm{log}}_{8}\left(7\right)+{\mathrm{log}}_{8}\left(-4x\right)={\mathrm{log}}_{8}\left(5\right).\text{\hspace{0.17em}}$ If there is no solution, write no solution . Use the one-to-one property of logarithms to find an exact solution for $\text{\hspace{0.17em}}\mathrm{ln}\left(5\right)+\mathrm{ln}\left(5{x}^{2}-5\right)=\mathrm{ln}\left(56\right).\text{\hspace{0.17em}}$ If there is no solution, write no solution . $x=±\frac{9}{5}$ The formula for measuring sound intensity in decibels $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ is defined by the equation $\text{\hspace{0.17em}}D=10\mathrm{log}\left(\frac{I}{{I}_{0}}\right),$ where $\text{\hspace{0.17em}}I\text{\hspace{0.17em}}$ is the intensity of the sound in watts per square meter and $\text{\hspace{0.17em}}{I}_{0}={10}^{-12}\text{\hspace{0.17em}}$ is the lowest level of sound that the average person can hear. How many decibels are emitted from a large orchestra with a sound intensity of $\text{\hspace{0.17em}}6.3\cdot {10}^{-3}\text{\hspace{0.17em}}$ watts per square meter? the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve 1+cos²A/cos²A=2cosec²A-1 test for convergence the series 1+x/2+2!/9x3 a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he? 100 meters Kuldeep Find that number sum and product of all the divisors of 360 Ajith exponential series Naveen what is subgroup Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1 e power cos hyperbolic (x+iy) 10y Michael tan hyperbolic inverse (x+iy)=alpha +i bita prove that cos(π/6-a)cos(π/3+b)-sin(π/6-a)sin(π/3+b)=sin(a-b) why {2kπ} union {kπ}={kπ}? why is {2kπ} union {kπ}={kπ}? when k belong to integer Huy if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41 what is complex numbers Dua Yes ahmed Thank you Dua give me treganamentry question Solve 2cos x + 3sin x = 0.5
# Question #81894 Oct 7, 2017 ${\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$ See explanations #### Explanation: So you want to express $\tan \left(x\right)$ in terms of $\sec \left(x\right)$. Well, first of all, let's write $\tan \left(x\right)$ in a form that we are more familiar with: $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ We also note that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ $\tan \left(x\right) = \sin \left(x\right) \cdot \sec \left(x\right)$ Now we only need to express $\sin \left(x\right)$ in terms of $\sec \left(x\right)$. We know that ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ so ${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$ which means that $\sin \left(x\right) = \pm \sqrt{1 - {\cos}^{2} \left(x\right)}$ which means that $\sin \left(x\right) = \pm \sqrt{1 - \frac{1}{\sec} ^ 2 \left(x\right)}$ We finally have: $\tan \left(x\right) = \pm \sqrt{1 - \frac{1}{\sec} ^ 2 \left(x\right)} \cdot \sec \left(x\right)$ We could also absorb the $\sec \left(x\right)$ in the square-root, giving us $\tan \left(x\right) = \pm \sqrt{{\sec}^{2} \left(x\right) - 1}$ It is more common to write it as follows: ${\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$ We have expressed $\tan \left(x\right)$ in terms of $\sec \left(x\right)$. Q.E.D.
# McGraw Hill Math Grade 5 Chapter 2 Lesson 4 Answer Key Subtracting Whole Numbers All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 2 Lesson 4 Subtracting Whole NumbersÂ are as per the latest syllabus guidelines. ## McGraw-Hill Math Grade 5 Answer Key Chapter 2 Lesson 4 Subtracting Whole Numbers Subtract Solve. Question 1. Explanation: Difference between 445 and 213: 445 – 213 = 232. Question 2. Explanation: Difference between 331 and 221: 331 – 221 = 110. Question 3. Explanation: Difference between 176 and 45: 176 – 45 = 131. Question 4. Explanation: Difference between 319 and 11: 319 – 11 = 308. Question 5. Explanation: Difference between 579 and 447: 579 – 447 = 132. Question 6. Explanation: Difference between 209 and 159: 209 – 159 = 50. Question 7. Explanation: Difference between 430 and 145: 430 – 145 = 285. Question 8. Explanation: Difference between 742 and 472: 742 – 472 = 270. Question 9. Explanation: Difference between 873 and 105: 873 – 105 = 768. Question 10. Explanation: Difference between 663 and 359: 663 – 359 = 304. Question 11. Explanation: Difference between 717 and 94: 717 – 94 = 623. Question 12. Explanation: Difference between 135 and 109: 135 – 109 = 26. Question 13. 4,199 – 2,180 ______________________ 4,199 – 2,180 = 2,019. Explanation: Difference between 4199 and 2180: 4199 – 2180 = 2019. Question 14. 46,224 – 24,121 ______________________ 46,224 – 24,121 = 22,103. Explanation: Difference between 46,224 and 24,121: 46,224 – 24,121 = 22,103. Question 15. 5,003 – 774 ______________________ 5,003 – 774 = 4,229. Explanation: Difference between 5,003 and 774: 5,003 – 774 = 4,229. Question 16. 53,072 – 4,227 ______________________ 53,072 – 4,227 = 48,845. Explanation: Difference between 53,072 and 4,227: 53,072 – 4,227 = 48,845. Question 17. Pamela the elephant was 113 kg at birth. As an adult, her mass is 4,990 kg. How many kilograms did she gain? Number of kilograms she gained = 4,877. Explanation: Weight of Pamela the elephant at birth = 113 kg. Weight of Pamela the elephant as an adult = 4,990 kg. Number of kilograms she gained = Weight of Pamela the elephant as an adult – Weight of Pamela the elephant at birth = 4,990 – 113 = 4,877. Scroll to Top Scroll to Top
Constant of Integration Get Constant of Integration essential facts below. View Videos or join the Constant of Integration discussion. Add Constant of Integration to your PopFlock.com topic list for future reference or share this resource on social media. Constant of Integration In calculus, the constant of integration, often denoted by ${\displaystyle C}$, is a constant added to the end of an antiderivative of a function ${\displaystyle f(x)}$ to indicate that the indefinite integral of ${\displaystyle f(x)}$ (i.e., the set of all antiderivatives of ${\displaystyle f(x)}$), on a connected domain, is only defined up to an additive constant.[1][2][3][4] This constant expresses an ambiguity inherent in the construction of antiderivatives. More specifically, if a function ${\displaystyle f(x)}$ is defined on an interval, and ${\displaystyle F(x)}$ is an antiderivative of ${\displaystyle f(x)}$, then the set of all antiderivatives of ${\displaystyle f(x)}$ is given by the functions ${\displaystyle F(x)+C}$, where ${\displaystyle C}$ is an arbitrary constant (meaning that any value of ${\displaystyle C}$ would make ${\displaystyle F(x)+C}$ a valid antiderivative). For that reason, the indefinite integral is often written as ${\displaystyle \int f(x)\,dx=F(x)+C}$,[5] although the constant of integration might be sometimes omitted in lists of integrals for simplicity. ## Origin The derivative of any constant function is zero. Once one has found one antiderivative ${\displaystyle F(x)}$ for a function ${\displaystyle f(x)}$, adding or subtracting any constant ${\displaystyle C}$ will give us another antiderivative, because ${\displaystyle (F(x)+C)'=F\,'(x)+C\,'=F\,'(x)}$. The constant is a way of expressing that every function with at least one antiderivative will have an infinite number of them. Let ${\displaystyle F:\mathbb {R} \rightarrow \mathbb {R} }$ and ${\displaystyle G:\mathbb {R} \rightarrow \mathbb {R} }$ be two everywhere differentiable functions. Suppose that ${\displaystyle F\,'(x)=G\,'(x)}$ for every real number x. Then there exists a real number ${\displaystyle C}$ such that ${\displaystyle F(x)-G(x)=C}$ for every real number x. To prove this, notice that ${\displaystyle [F(x)-G(x)]'=0}$. So ${\displaystyle F}$ can be replaced by ${\displaystyle F-G}$, and ${\displaystyle G}$ by the constant function ${\displaystyle 0}$, making the goal to prove that an everywhere differentiable function whose derivative is always zero must be constant: Choose a real number ${\displaystyle a}$, and let ${\displaystyle C=F(a)}$. For any x, the fundamental theorem of calculus, together with the assumption that the derivative of ${\displaystyle F}$ vanishes, implies that {\displaystyle {\begin{aligned}&0=\int _{a}^{x}F'(t)\ dt\\&0=F(x)-F(a)\\&0=F(x)-C\\&F(x)=C\\\end{aligned}}} thereby showing that ${\displaystyle F}$ is a constant function. Two facts are crucial in this proof. First, the real line is connected. If the real line were not connected, we would not always be able to integrate from our fixed a to any given x. For example, if we were to ask for functions defined on the union of intervals [0,1] and [2,3], and if a were 0, then it would not be possible to integrate from 0 to 3, because the function is not defined between 1 and 2. Here, there will be two constants, one for each connected component of the domain. In general, by replacing constants with locally constant functions, we can extend this theorem to disconnected domains. For example, there are two constants of integration for ${\displaystyle \textstyle \int dx/x}$, and infinitely many for ${\displaystyle \textstyle \int \tan x\,dx,}$ so for example, the general form for the integral of 1/x is:[6][7] ${\displaystyle \int {1 \over x}\,dx={\begin{cases}\ln \left\|x\right\|+C^{-}&x<0\\\ln \left\|x\right\|+C^{+}&x>0\end{cases}}}$ Second, ${\displaystyle F}$ and ${\displaystyle G}$ were assumed to be everywhere differentiable. If ${\displaystyle F}$ and ${\displaystyle G}$ are not differentiable at even one point, then the theorem might fail. As an example, let ${\displaystyle F(x)}$ be the Heaviside step function, which is zero for negative values of x and one for non-negative values of x, and let ${\displaystyle G(x)=0}$. Then the derivative of ${\displaystyle F}$ is zero where it is defined, and the derivative of ${\displaystyle G}$ is always zero. Yet it's clear that ${\displaystyle F}$ and ${\displaystyle G}$ do not differ by a constant, even if it is assumed that ${\displaystyle F}$ and ${\displaystyle G}$ are everywhere continuous and almost everywhere differentiable the theorem still fails. As an example, take ${\displaystyle F}$ to be the Cantor function and again let ${\displaystyle G}$ = 0. For example, suppose one wants to find antiderivatives of ${\displaystyle \cos(x)}$. One such antiderivative is ${\displaystyle \sin(x)}$. Another one is ${\displaystyle \sin(x)+1}$. A third is ${\displaystyle \sin(x)-\pi }$. Each of these has derivative ${\displaystyle \cos(x)}$, so they are all antiderivatives of ${\displaystyle \cos(x)}$. It turns out that adding and subtracting constants is the only flexibility we have in finding different antiderivatives of the same function. That is, all antiderivatives are the same up to a constant. To express this fact for ${\displaystyle \cos(x)}$, we write: ${\displaystyle \int \cos(x)\,dx=\sin(x)+C.}$ Replacing ${\displaystyle C}$ by a number will produce an antiderivative. By writing ${\displaystyle C}$ instead of a number, however, a compact description of all the possible antiderivatives of ${\displaystyle \cos(x)}$ is obtained. ${\displaystyle C}$ is called the constant of integration. It is easily determined that all of these functions are indeed antiderivatives of ${\displaystyle \cos(x)}$: {\displaystyle {\begin{aligned}{\frac {d}{dx}}[\sin(x)+C]&={\frac {d}{dx}}[\sin(x)]+{\frac {d}{dx}}[C]\\&=\cos(x)+0\\&=\cos(x)\end{aligned}}} ## Necessity At first glance, it may seem that the constant is unnecessary, since it can be set to zero. Furthermore, when evaluating definite integrals using the fundamental theorem of calculus, the constant will always cancel with itself. However, trying to set the constant to zero does not always make sense. For example, ${\displaystyle 2\sin(x)\cos(x)}$ can be integrated in at least three different ways: {\displaystyle {\begin{aligned}\int 2\sin(x)\cos(x)\,dx&=&\sin ^{2}(x)+C&=&-\cos ^{2}(x)+1+C&=&-{\frac {1}{2}}\cos(2x)+C\\\int 2\sin(x)\cos(x)\,dx&=&-\cos ^{2}(x)+C&=&\sin ^{2}(x)-1+C&=&-{\frac {1}{2}}\cos(2x)+C\\\int 2\sin(x)\cos(x)\,dx&=&-{\frac {1}{2}}\cos(2x)+C&=&\sin ^{2}(x)+C&=&-\cos ^{2}(x)+C\end{aligned}}} So setting ${\displaystyle C}$ to zero can still leave a constant. This means that, for a given function, there is no "simplest antiderivative". Another problem with setting ${\displaystyle C}$ equal to zero is that sometimes we want to find an antiderivative that has a given value at a given point (as in an initial value problem). For example, to obtain the antiderivative of ${\displaystyle \cos(x)}$ that has the value 100 at x = π, then only one value of ${\displaystyle C}$ will work (in this case ${\displaystyle C}$ = 100). This restriction can be rephrased in the language of differential equations. Finding an indefinite integral of a function ${\displaystyle f(x)}$ is the same as solving the differential equation ${\displaystyle {\frac {dy}{dx}}=f(x)}$. Any differential equation will have many solutions, and each constant represents the unique solution of a well-posed initial value problem. Imposing the condition that our antiderivative takes the value 100 at x = π is an initial condition. Each initial condition corresponds to one and only one value of ${\displaystyle C}$, so without ${\displaystyle C}$ it would be impossible to solve the problem. There is another justification, coming from abstract algebra. The space of all (suitable) real-valued functions on the real numbers is a vector space, and the differential operator ${\displaystyle {\frac {d}{dx}}}$ is a linear operator. The operator${\displaystyle {\frac {d}{dx}}}$ maps a function to zero if and only if that function is constant. Consequently, the kernel of ${\displaystyle {\frac {d}{dx}}}$ is the space of all constant functions. The process of indefinite integration amounts to finding a pre-image of a given function. There is no canonical pre-image for a given function, but the set of all such pre-images forms a coset. Choosing a constant is the same as choosing an element of the coset. In this context, solving an initial value problem is interpreted as lying in the hyperplane given by the initial conditions. ## References 1. ^ "Compendium of Mathematical Symbols". Math Vault. 2020-03-01. Retrieved . 2. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5. 3. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 0-547-16702-4. 4. ^ "Definition of constant of integration \| Dictionary.com". www.dictionary.com. Retrieved . 5. ^ Weisstein, Eric W. "Constant of Integration". mathworld.wolfram.com. Retrieved . 6. ^ "Reader Survey: log\|x\| + C", Tom Leinster, The n-category Café, March 19, 2012 7. ^ Banner, Adrian (2007). The calculus lifesaver : all the tools you need to excel at calculus. Princeton [u.a.]: Princeton University Press. p. 380. ISBN 978-0-691-13088-0.
# Difference between revisions of "2018 AMC 12B Problems/Problem 16" ## Problem The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$ $\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$ ## Solution 1 (Complex Numbers in Polar Form) Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$ We rewrite $z$ to the polar form $$z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,$$ where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$ By De Moivre's Theorem, we have $$z^8=r^8\operatorname{cis}(8\theta)={\sqrt3}^8(1),$$ from which 1. $r^8={\sqrt3}^8,$ so $r=\sqrt3.$ 2. \begin{cases} \begin{aligned} \cos(8\theta) &= 1 \\ \sin(8\theta) &= 0 \end{aligned}, \end{cases} so $\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}.$ In the complex plane, the solutions to the equation $z^8=81$ are the vertices of a regular octagon with center $0$ and radius $\sqrt3.$ The least possible area of $\triangle ABC$ occurs when $A,B,$ and $C$ are the consecutive vertices of the octagon. For simplicity purposes, let $A=\sqrt3\operatorname{cis}\frac{\pi}{4}=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i, B=\sqrt3\operatorname{cis}\frac{\pi}{2}=\sqrt3i,$ and $C=\sqrt3\operatorname{cis}\frac{3\pi}{4}=-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,$ as shown below. $[asy] /* Made by MRENTHUSIASM / size(200); int xMin = -2; int xMax = 2; int yMin = -2; int yMax = 2; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for (int i=0;i Note that $\triangle ABC$ has base $AC=\sqrt6$ and height $\sqrt3-\frac{\sqrt6}{2},$ so its area is $$\frac12\cdot\sqrt6\cdot\left(\sqrt3-\frac{\sqrt6}{2}\right)=\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.$$ ~MRENTHUSIASM ## Solution 2 (Complex Numbers in Rectangular Form) Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$ We have \begin{align} z^8 &= 81 \\ z^4 &= 9, -9 \\ z^2 &= 3, -3, 3i, -3i. \\ \end{align} Note that: 1. For $z^2=3,$ we get $\boldsymbol{z=\sqrt3,-\sqrt3}.$ 2. For $z^2=-3,$ we get $\boldsymbol{z=\sqrt3i,-\sqrt3i}.$ 3. For $z^2=3i,$ let $z=a+bi$ for some real numbers $a$ and $b.$ We substitute and then expand: \begin{align} (a+bi)^2 &= 3i \\ \left(a^2-b^2\right)+2abi &= 3i. \end{align} We equate the real parts and the imaginary parts, respectively, then simplify: \begin{align} a^2 &= b^2, &&(1) \\ ab &= \frac32. &&(2) \end{align*} Since $ab>0$ in $(2),$ we conclude that $a$ and $b$ must have the same sign. It follows that $(1)$ becomes $a=b.$ By substitution, we get $(a,b)=\left(\frac{\sqrt6}{2},\frac{\sqrt6}{2}\right),\left(-\frac{\sqrt6}{2},-\frac{\sqrt6}{2}\right),$ from which $\boldsymbol{z=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,-\frac{\sqrt6}{2}-\frac{\sqrt6}{2}i}.$ 4. For $z^2=-3i,$ we get $\boldsymbol{z=\frac{\sqrt6}{2}-\frac{\sqrt6}{2}i,-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i}$ by a similar process. We continue with the last two paragraphs of Solution 1 to obtain the answer $\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.$ ~MRENTHUSIASM ## Solution 3 (Regular Octagon) The polygon formed will be a regular octagon since there are $8$ roots of $z^8=81$. By normal math computation, we can figure out that two roots of $z^8=81$ are $\sqrt{3}$ and $-\sqrt{3}$. These will lie on the real axis of the plane. Since it's a regular polygon, there has to be points on the vertical plane also which will be $\sqrt{3}i$ and $-\sqrt{3}i$. Clearly, the rest of the points will lie in each quadrant. The next thing is to get their coordinates (note that to answer this question, we do not need all the coordinates, only 3 consecutive ones are needed). The circumcircle of the octagon will have the equation $i^2+r^2=3$. The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so $i=r$. Solving gives $i=r=\frac{\sqrt{3}}{\sqrt{2}}$ (meaning that the root represented is $\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i$). This way we can deduce the values of the $8$ roots of the equation to be $$\sqrt{3},-\sqrt{3},-\sqrt{3}i,\sqrt{3}i,\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i.$$ To get the area, $3$ consecutive points such as $\sqrt{3},$ $\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,$ and $\sqrt{3}i$ can be used. The area can be computed using different methods like using the Shoelace Theorem, or subtracting areas to find the area. The answer you get is $\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}$. (This method is not actually as long as it seems if you understand what you're doing while doing it. Also calculations can be made a little easier by solving using $x^8=1$ and multiplying your answer by $\sqrt{3}$). ~OlutosinNGA ## Solution 4 (Roots of Unity) Now, we need to solve the equation $(z+6)^8 = 81$ where $z = a+bi$. We can substitute this as $$(a+6+bi)^8 = 81$$ Now, let $a+6 = q$ for some $q \in \mathbb{Z}$. Thus, the equation becomes $((q+bi)^2)^4 = 81$. Taking it to the other side, we get the equation to be $(q+bi)^2 = 3$. Rearranging variables, we get $(q+bi) = \sqrt{3}$. Plotting this in the complex place, this is a circle centered at the origin and of radius $\sqrt{3}$. The graph of the original equation $(a+6+bi) = \sqrt{3}$ is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation $(q+bi)^2 = 3$. Using Roots of Unity, we know that the roots of the equation lie at $0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \ldots, 2\pi$ radians from the origin. We can quickly notice that the area of the roots will be smallest with points at $\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$. Using trigonometry, we get the respective roots to be $(\operatorname{Re}(z), \operatorname{Im}(z)) \in \left\{\left(\sqrt{3},0\right), \left(\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right), \left(0,\sqrt{3}\right)\right\}$. Using the Shoelace Theorem, the area quickly comes out to be $\frac{3\sqrt{2}-3}{2} = \boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.$ ## Solution 5 (Trigonometry) It is easy to see that the figure the solutions form is that of an regular octagon $ABCDEFGH$ With center $O$ such that $AO = \sqrt{3}.$ Also, the minimum area of a triangle formed by the solutions would be that of Triangle $ABC$, since the three points are consecutive vertices of the octagon. By dropping an altitude from $O$ to $AB$, note that $\frac{1}{2} AB = \sqrt{3} \sin(22.5).$ Thus, $AB = 2\sqrt{3} \sin(22.5).$ Now given the length of $AB$ we need to find the area of $ABC,$ which, by simple trigonometry, is given as $$AB \cos(22.5) \cdot AB \sin(22.5) = 12 \sin^3(22.5) \cdot \cos(22.5).$$ By the half angle formulas, $\sin(22.5) = \sqrt{\frac{1 - \sin(45)}{2}} = \frac{\sqrt{2-\sqrt{2}}}{2},$ and $\cos(22.5)$ is its conjugate $\frac{\sqrt{2+\sqrt{2}}}{2}.$ Doing some expanding and simplifying, $$12 \cdot \left ( \frac{\sqrt{2-\sqrt{2}}}{2} \right)^3 \cdot \left (\frac{\sqrt{2+\sqrt{2}}}{2} \right) = \frac{3\sqrt{12-8\sqrt{2}}}{4}.$$ Further simplifying, we obtain that the desired expression is $\frac{3 \sqrt{3 - 2\sqrt{2}}}{2}.$ Note that $(\sqrt{2} - 1)^2 = 3 - 2\sqrt{2},$ so our answer comes out to be $\frac{3(\sqrt{2}-1)}{2} = \boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.$ ~fidgetboss_4000
## Pages ### Function Composition Here is where we introduce a new operation - composition of functions. With this definition comes new notation. This new notation reads “f composed with g.” The idea is to substitute one function into another function. Given the functions f(x) = 5x – 4 and g(x) = 2x – 1. Composition of functions is not necessarily a commutative operation, in other words, order matters. Instructional Video: Composite Functions Given the functions f(x) = x^2 – 9 and g(x) = x – 3. Given the functions f(x) = x^2 + 1 and g(x) = sqrt(– 1) where ( x >= 1 ). At this point we must understand what happens to the domain of a composite function. In the above example it might appear that f o g has a domain of all real numbers. In fact, the domain is restricted to [1, inf) because that is the domain of g. The domain of f o consists of all the values in the domain of g that are also in the domain of f. Given f and g find f o g and state its domain. Now that we have learned how to work with radical expressions, we next move on to solving. Use caution when solving radical equations because the following steps may lead to extraneous solutions, solutions that do not solve the original equation. Solve: Step 2: Square both sides of the equation. Whenever you raise both sides of an equation to an even power, you introduce the possibility of extraneous solutions so the check is essential here. Solve. The index determines the power to which we raise both sides. For example, if we have a cube root we will raise both sides to the 3rd power. The property that we are using is for integers n > 1 and positive real numbers x. After eliminating the radical, we will most likely be left with either a linear or a quadratic equation to solve. The check mark indicates that we have actually checked that the value is a solution to the equation, do not dismiss this step, it is essential. Some radical equations have more than one radical expression. These require us to isolate each remaining radical expression and raise both sides to the nth power until they are all eliminated. Be patient with these, go slow and avoid short cuts. Solve: Solve: Solve:
12:22 am Uncategorized # Which of the Following is a Rational Number? Understanding the concept of rational numbers is fundamental in mathematics. Rational numbers are numbers that can be expressed as a fraction, where the numerator and denominator are both integers. In this article, we will explore what rational numbers are, how to identify them, and provide examples to illustrate their properties. Let’s dive in! ## What are Rational Numbers? Rational numbers are a subset of real numbers that can be expressed as a fraction, where the numerator and denominator are both integers. The word “rational” comes from the Latin word “ratio,” which means “ratio” or “proportion.” This is fitting because rational numbers represent the ratio of two integers. Rational numbers can be positive, negative, or zero. They can be written in the form a/b, where a and b are integers and b is not equal to zero. The numerator a represents the number of parts we have, and the denominator b represents the total number of equal parts the whole is divided into. ## Identifying Rational Numbers Identifying whether a number is rational or not can be done through various methods. Let’s explore some of the common techniques: ### Method 1: Fraction Representation The most straightforward way to identify a rational number is by representing it as a fraction. If a number can be expressed as a fraction, it is rational. For example, the number 3 can be written as 3/1, where the numerator is 3 and the denominator is 1. Similarly, the number -2 can be written as -2/1. Let’s take another example: 0.75. To determine if it is rational, we can convert it to a fraction. Since 0.75 is equivalent to 75/100, we can simplify it by dividing both the numerator and denominator by their greatest common divisor, which is 25. Thus, 0.75 is rational and can be expressed as 3/4. ### Method 2: Terminating or Repeating Decimals Rational numbers can also be identified by their decimal representation. A rational number will always have a decimal that either terminates or repeats. Let’s consider the number 0.3333… (where the 3s repeat infinitely). This number can be expressed as 1/3, which is a rational number. Similarly, the number 0.5 terminates and can be written as 1/2. On the other hand, irrational numbers, such as the square root of 2 (√2), have decimal representations that neither terminate nor repeat. For example, √2 is approximately 1.41421356… and the decimal goes on indefinitely without any repeating pattern. ## Examples of Rational Numbers Now that we understand how to identify rational numbers, let’s explore some examples: ### Example 1: 2/3 The fraction 2/3 is a rational number. The numerator is 2, and the denominator is 3, both of which are integers. Therefore, 2/3 is a rational number. ### Example 2: -5 The whole number -5 can be expressed as -5/1, where the numerator is -5 and the denominator is 1. Since both the numerator and denominator are integers, -5 is a rational number. ### Example 3: 0.25 The decimal 0.25 can be written as a fraction by placing the digits after the decimal point over the appropriate power of 10. In this case, 0.25 is equivalent to 25/100. Simplifying the fraction by dividing both the numerator and denominator by their greatest common divisor, we get 1/4. Therefore, 0.25 is a rational number. ## Properties of Rational Numbers Rational numbers possess several interesting properties that make them unique. Let’s explore some of these properties: ### Property 1: Closure Property The closure property states that the sum, difference, product, or quotient of any two rational numbers is always a rational number. For example, if we add 2/3 and 1/4, we get 11/12, which is a rational number. ### Property 2: Commutative and Associative Properties Rational numbers follow the commutative and associative properties for addition and multiplication. This means that changing the order of the numbers or grouping them differently does not affect the result. For example, (2/3 + 1/4) is equal to (1/4 + 2/3), and (2/3 + 1/4) + 1/5 is equal to 2/3 + (1/4 + 1/5). ### Property 3: Identity Elements The rational number 0 acts as the additive identity element, meaning that adding 0 to any rational number does not change its value. Similarly, the rational number 1 acts as the multiplicative identity element, meaning that multiplying any rational number by 1 does not change its value. ### Property 4: Inverse Elements Every rational number has an additive inverse and a multiplicative inverse. The additive inverse of a rational number a/b is -a/b, and the multiplicative inverse is b/a (except when a is equal to 0). For example, the additive inverse of 2/3 is -2/3, and the multiplicative inverse is 3/2. ## Q&A ### Q1: Is 0 a rational number? A1: Yes, 0 is a rational number. It can be expressed as 0/1, where the numerator is 0 and the denominator is 1. ### Q2: Is every integer a rational number? A2: Yes, every integer is a rational number. Integers can be expressed as fractions with a denominator of 1. For example, the integer 5 can be written as 5/1, which is a rational number. ### Q3: Is the square root of 9 a rational number? A3: Yes, the square root of 9 (√9) is a rational number. The square root of 9 is equal to 3, which can be expressed as 3/1. ### Q4: Is pi (π) a rational number? A4: No, pi ( Close Search Window Close
# Arithmatic ## What is Arithmetic? Arithmetics is among the oldest and elementary branches of mathematics, originating from the Greek word arithmos, meaning number. It involves the study of numbers, especially the properties of traditional operations on them such as addition, subtraction, division and multiplication. These are the basic operations, although the subject also involves advanced operations like computation of percentages, logarithmic functions, exponentiation and square roots. Who discovered Arithmetic? The Fundamental principle of number theory was provided by Carl Friedrich Gauss in 1801, according to which, any integer which is greater than 1 can be described as the product of prime numbers in only one way. Arithmetic Progressions A sequence like 1, 5, 9, 13, 17, or 12, 7, 2, -3, -8 that follow a constant difference is known as arithmetic progressions. You can name the first term as a1, the common difference as d and the total number of terms as n. Therefore, an explicit formula can be written as an = a1 + (n-1)d Example 1: 3, 7, 11 has a1 = 3, d = 4 and n = 5. Hence, the explicit formula is an = 3 + (n-1).4 = 4n – 1 Example 2: 3, -2, -7 has a1 = 3, d = -5 and n = 4. Hence, the explicit formula is an = 3 + (n-1)(-5) = 8-5n Arithmetic operations The basic operations under arithmetics are addition, subtraction, division and multiplication although the subject involves many other modified operations. Addition is among the basic operations in arithmetic. In simple forms, addition combines two or more values into a single term, for example: 2 + 5 = 7, 6 + 2 = 8. The procedure of adding more than two values is called summation and involves methods to add n number of values. The identity element of addition is 0, which means that adding 0 to any value gives the same result. The inverse element of addition is the opposite of any value, which means that adding opposite of any digit to the digit itself gives the additive identity. For instance, the opposite of 5 is -5, therefore 5 + (-5) = 0. Subtraction (−) Subtraction can be labelled as inverse of addition. It computes the difference between two values, i.e, the minuend minus the subtrahend. If the minuend is greater than subtrahend, the difference is positive. If the minuend is less than subtrahend, the result is negative, and 0 if the numbers are equal. Multiplication (×, · or ) Multiplication also combines two values like addition and subtraction into a single value or the product. The two original values are known as the multiplicand and the multiplier, or simply both as factors. The product of a and b is expressed as a·b or a x b. In software languages wherein only characters are used that are found in keyboards, it is often expressed as, ab (* is called asterisk). Division (÷ or /) Division is the inverse of multiplication. It computes the quotient of two numbers, the dividend that is divided by the divisor. The quotient is more than 1 if the dividend is greater than divisor for any well-defined positive number, else it is smaller than 1. Post By : YATIN SAGAR 24 Aug, 2017 2187 views General Science EVS
# What number multiplied by itself would equal to 12345678987654321? ## What number multiplied by itself would equal to 12345678987654321? 111,111,111 multiplied by itself gives 12,345,678,987,654,321. In other words, 111,111,111 is the square root of 12,345,678,987,654,321. When a number is multiplied by one the product is the? number itself According to the multiplicative identity property of 1, any number multiplied by 1, gives the same result as the number itself. It is also called the Identity property of multiplication, because the identity of the number remains the same. ### When a natural number is multiplied by itself? Square number definition can be defined as numbers that are multiplied by itself. In other words, if a natural number is multiplied by itself it is known as a natural number. For example, 2 multiplied to itself (2×2) is a square number. Square numbers are named after a square because it forms an area of a square. What times what equals 12345678987654321? 111111111 times 111111111 equals 12345678987654321 according to the laws of multiplication for real numbers. The answer is actually a numeric palindrome. ## What is a number that when multiplied by itself equals the original number? Square root Square root, in mathematics, a factor of a number that, when multiplied by itself, gives the original number. For example, both 3 and –3 are square roots of 9. Which number should be multiplied to 1575 to make it a square number? If we multiply by 7 in 1575, then number become perfect square. ### What least number must be added to 6072 to make it perfect square? 12 But, it is given in the question that we have to add some number to 6072 to make it a perfect square. Thus, when we add 12 to 6072, we get 6084, which is a perfect square of 78. Therefore the required number is 12. When a number is multiplied by number remains the same? According to the multiplicative identity property of 1, any number multiplied by 1, gives the same result as the number itself. It is also called the Identity property of multiplication, because the identity of the number remains the same. ## When we multiply _ to any number we get the same number? 0 and 1 are the only whole numbers which when multiplied by themselves give the same number. Whole numbers include natural numbers(1,2,3…..) and 0. When a number is multiplied by itself three times the product so obtained is called? Number obtained when a number is multiplied by itself three times are called cube number. ### When we multiply a number by itself three times the product so obtained is called the perfect cube of that number? When we multiply a number by itself three times, the product so obtained is called the perfect cube of that number. 2. There are only 10 perfect cubes from 1 to 1000. Why does 111111111 equal 100000000? Note that 111111111 = 100000000 + 10000000 + … + 100 + 10 + 1. As a consequence of the distributive property of multiplication over addition, Because 111111111 times 111111111 equals 12345678987654321 according to the laws of multiplication for real numbers. ## How is a table of 111111111 a power of ten? If you write one of the 111111111 values up the side and the other across the top of a table, and multiply each digit selection individually, you get a table of 1 s. But the actual value behind those 1 s is a power of ten, so you collect like powers of ten as shown: Why is 1234 and 111111111 the same number? 11 1 11*11 1 11=1234 5 4321【5 one′s→axis of symmetry is the 5th digit 5】【n 1′s→AOS is the nth digit n】 The answer is in that form because the value of both numbers does not only exist in the value of the digit, but the placing of the digit. ### How to calculate the Pat of 111, 111? In this video I do a hand calculation of a neat little pattern which is quite remarkable in that if you have 111,111,111 multiplied by itself you get the pat… In this video I do a hand calculation of a neat little pattern which is quite remarkable in that if you have 111,111,111 multiplied by itself you get the pat…
On the Ellipse page we looked at the definition and some of the simple properties of the ellipse, but here we look at how to more accurately calculate its perimeter.. Perimeter. Explanation. The perimeter of a rectangle is the length of the boundary around the figure. You need to create regions within the shape for which you can find the area, and add these areas together. For example, if the diameter of your semi circle is 12 centimeters, the formula becomes. Find the perimeter of irregular shapes worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. The base is 7 inches while the sides are 4 inches. When side lengths are given, add them together. How do you find the perimeter of a right triangle? The area of a rectangle is 108 cm² and its length is 12 cm . To find the perimeter of a square, multiply the length of one side by 4. How to Work out the Perimeter of a Compound Shape Perimeter of the rectangle = 2 x ( l + b ) cm = 2 x ( 12 + 9 ) cm = ( 2 x 21 ) cm = 42 cm. From the previous example we learned how to figure the perimeter of a rectangle. Practise Activity 1. Solution: Given: Side, s = 5 cm. Find perimeter and area by finding the length (1) In this math lesson, students learn how to find the length of sides and then use that to find perimeter and area by comparing coordinates. Finding the side length of a rectangle given its perimeter or area - In this lesson, we solve problems where we find one missing side length while one side length and area or perimeter … Share on facebook. To solve the word problem, we first draw the shape and mark any known sides. Go here to find out how to figure the surface area. The exception is the ellipse, but its perimeter may be approximated. To find the perimeter of rectangles, you must know … How to Use this Worksheet. The term may be used either for the path, or its length—in one dimension. Substitute the value of ‘s’ in the perimeter formula, P= 4 × 5 cm. Step 1. Question 1: Find the perimeter of a square whose side is 5 cm. Using the perimeter of a pentagon formula, you can find the perimeter of a regular pentagon with relative ease. It is important to remember that if changes in geometry are caused, the area, perimeter, or length fields are not updated automatically. You can find the perimeter of a rectangle by calculating the sum of the length of its four sides. Have a look at how this is done below. P = \frac{1}{2} ×(3.14×12) + 12. This video starts out with a basic view of a rectangle and its dimensions. The distance around a circular region is also known as its circumference. Work out the perimeter of the park. Find the perimeter of the rectangle? To find the perimeter of a rectangle, add the lengths of the rectangle's four sides. Calculating the perimeter has several practical applications. Problem. … Breadth of the rectangle. If you had to put down a tarp to cover the entire field so it wouldn't get wet, that would be the surface area. Perimeter of a square $$=4×side=4s$$ Perimeter of a rectangle $$=2(width \ + \ length)$$ Perimeter of trapezoid $$= a+b+c+d$$ Perimeter of a regular hexagon $$=6a$$ Perimeter of a parallelogram $$=2(l+w)$$ Perimeter of Polygons Polygons – Example 1: Find the perimeter of … A perimeter is a path that encompasses/surrounds a two-dimensional shape. To find the perimeter of a semi circle, you have to know the diameter (the length of its straight edge). To find the perimeter of an irregular pentagon, you must measure and add up the five sides. Level 1 is a rectangle, level 2 is a L shaped compound shape and level 3 is a more complicated compound shape. The third one is better to go with. In the special case of the circle, the perimeter is also known as the circumference. Hi, and welcome to this video on finding the area and perimeter of a trapezoid! length = 12 cm. Perimeter. The perimeter of a circle or ellipse is called its circumference. The formula to find the perimeter of a square is given by: The perimeter of Square = 4s units. Solve for a missing side using the Pythagorean theorem. The formula for the perimeter of a square is P = 4l, where l is the length of one side of the square. We use a ruler to measure length of the sides of a small regular shape. Calculate perimeters of any regular polygon. For people who are learning geometry and would like to be able to find the perimeter of a rectangle, this video provides and quick and simple tutorial which provides the formula as well as a good example. Rather strangely, the perimeter of an ellipse is very difficult to calculate!. Perimeter is calculated by the following formula and store in perimeter variable like perimeter = 2(length + width). Before moving to the actual program, let me quickly show you the formula to find out the rectangle area, perimeter. While calculus may be needed to find the perimeter of irregular shapes, geometry is sufficient for most regular shapes. Our program will take the inputs from the user and print out the final results. Find the perimeter of a square with an area of 64 square meters. Step 2: Add up all the sides to find the perimeter. Note: The ratio of circumference to diameter is approximately the same around 3.142. i.e. CCSS.Math: 3.MD.D.8. Find Perimeter of Rectangle without Function. Formula : We need the height and width of a rectangle to find its area and perimeter. 3.14 × 12 = 37.68. What's the perimeter of this rectangle? Given. x is in this case the length of the rectangle while y is the width of the rectangle. The perimeter is the length of the outline of a shape. This worksheet helps your students understand when they will need to know perimeter outside of the classroom. Perimeter Of A Circle Circumference of a Circle Circumference means, ‘the perimeter of a circle’. Find the area & perimeter of irregular shapes worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. For some geometric shapes, you can find perimeter from the area. This math worksheet gives your child practice finding the perimeters of large rectangles and squares. To calculate perimeter of rectangle in Python, you have to ask from user to enter the length and breadth value of rectangle whose perimeter you want to find out. There are three primary methods used to find the perimeter of a right triangle. Share on twitter. How to find perimeter? To find the perimeter of a rectangle or square you have to add the lengths of all the four sides. If you have only the width and the height, then you can easily find all four sides (two sides are each equal to the height and the other two sides are equal to the width). Since a pentagon has five equal sides, to find its perimeter you multiply the length of one side by five. To understand this example perfectly you should have the following C++ programming knowledge. Tap on PRINT, PDF or IMAGE button to print or download this grade-6 geometry worksheet to practice how to find the perimeter of irregular shapes on 2-dimensional plane. We often find the perimeter when putting up Christmas lights around the house or fencing the backyard garden. 4 cm + 5 cm + 8 cm + 10 cm + 3 cm + 6 cm = 36 cm. Perimeter of a Pentagon Formula. Perimeter of an Ellipse. Here per indicates to perimeter, len indicates to length and bre indicates to breadth of rectangle. Area and Perimeter A brilliant site that will help you work out area and perimeter of shapes. There are many formulas, here are some interesting ones. Multiply both the height and width by two and add the results. P = 20 cm. After the substitution of the values into the formula you can quickly come up with the perimeter. To calculate the perimeter, the same process is repeated, but in this case the Perimeter property is selected. A square is a shape that has all four sides that are the same length. The perimeter is the distance around the outside of a shape. Therefore, the perimeter of square = 20 cm. Break Down the Equation; Work out. It then displays the formula for finding the perimeter of the rectangle which is the addition of both bases and both sides. Finding the area of non-standard shapes is a bit different. Learn how to find the perimeter of a polygon in this article. Then work out . To find the perimeter, we add up all the outside sides of our shape. You can use any one from these three formulae. […] A trapezoid is a four-sided polygon, or “quadrilateral”, that has at least one set of parallel sides.There are two types of sides in a trapezoid: legs and bases. To find the perimeter of non-standard shapes, you still find the distance around the shape by adding together the length of each side. Share on whatsapp. Using Finding the Perimeter of Shapes Formulas Worksheet, students find the perimeter of different shapes using a formula. The formula for finding the perimeter of a rectangle is simple the sum of all sides, or l+l+w+w, where l is the length and w is the width of the rectangle. Prerequisite. It can be thought of as the length of the outline of a shape. Area and Perimeter of a Trapezoid. This is the perimeter. Finding missing side length when given perimeter. Sometimes you’ll see the perimeter formula as P=2l+2w, where l is the length of the rectangle, and w is the width of the rec How to Use this Worksheet . Find perimeter from area. 37.68 ÷ 2 = 18.84. This lesson helps students understand complex math concepts in an accessible way. Perimeter of an equilateral triangle = 3 x length of one side. You now have the perimeter of the curved edge of the semi circle. Remember the formulas for both area and perimeter of a square. If we know side-angle-side information, solve for the missing side using the Law of Cosines. Area = 108 cm² . Step by step guide to find perimeter of Polygons. Or as a formula: where: a,b and c are the lengths of each side of the triangle In the figure above, drag any orange dot to resize the triangle. Example. With equilateral triangles, squares, and circles, you can use formulas to find their perimeters from their given areas. Display result on the screen. Hence, perimeter of the rectangle is 42 cm. Tap on PRINT, PDF or IMAGE button to print or download this grade-6 geometry worksheet to practice how to find the area and perimeter of irregular shapes on 2d-geometry plane. The defining feature of regular polygons is that all of their sides are the same length. How to find the perimeter of a triangle Like any polygon, the perimeter is the total distance around the outside, which can be found by adding together the length of each side. In this sixth grade … Share on linkedin. With this process you can calculate the area, perimeter, length or coordinates of a shapefile (according to its geometry). Your students will need to know how to find the perimeter to build fences around gardens and pools. In this tutorial, I will show you how to find the area and perimeter of a rectangle using user inputs. Here is a word problem for finding the perimeter of a square. Other examples may include finding the total length of the boundary of the soccer field or the length of the crochet or ribbon required to cover the border of a table mat. The word has been derived from the Latin word circumferre means to carry around. The classroom regular shape is calculated by the following C++ programming knowledge is P = {. 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# 14.3 Why Do these Rules Work? Suppose your integrand is $$f(x)$$ and $$f$$ has a power series expansion about the point $$q$$ whose coefficients do not go wild. We can then write, for $$x$$ near $$q$$ $f(x) = f(q) + (x-q)f'(q) + (x-q)^2\frac{f^{(2)}(q)}{2} + (x-q)^3\frac{f^{(3)}(q)}{3!} + (x-q)^4\frac{f^{(4)}(q)}{4!} + \ldots$ where $$f^{(k)}(q)$$ is the $$k^{th}$$ derivative of $$f(x)$$ evaluated at $$x = q$$. We want to find the area under $$f(x)$$ when $$x$$ ranges from $$q-d$$ to $$q+d$$, an interval of length $$2d$$. Notice that if we form $$f(q+d) + f(q-d)$$ the contributions of $$f'(q)$$ which is the second term on the right here, will cancel out. In fact all the terms involving odd derivatives will cancel out: $f(q+d) + f(q-d) = 2f(q) + 2d^2\frac{f^{(2)}(q)}{2!} + \text{terms coming from higher derivatives of } f$ We can conclude, by integrating both sides of this equation: $\int_{q-d}^{q+d} f(x)dx = \int_{-d}^{d} dxf(q) + \int_{-d}^d y^2dy \frac{f^{(2)}(q)}{2!} + \text{terms coming from higher derivatives of }f$ The first term on the right is $$2df(q)$$ and the next is $$2d^3\frac{f^{(2)}(q)}{3!}$$, and the remaining terms are proportional to higher powers of $$d$$. The first term of $$2df(q)$$, by the previous equation differs from $$d(f(q+d) + f(q-d))$$ by $$-d^3f^{(2)}(q)$$ and terms of higher power in $$d$$. The upshot of this argument is that approximating the integral of $$f$$ in the interval within $$d$$ of $$q$$ by $$d(f(q+d)+f(q-d))$$ produces an error proportional to $$d^3$$ as well as terms of higher odd powers of $$d$$ and no error linear in $$d$$ or quadratic in $$d$$. In fact, the error term cubic in $$d$$ is $$\frac{2}{3}d^3f^{(2)}(q)$$. If we decrease $$d$$ by a factor of $$2$$, $$f(q)d$$ decreases by a factor of $$2$$, but this is compensated for by the fact that there are now twice as many intervals, each of half the size of those present before the decrease. Thus, the contributions of the first terms to the integral as a whole will usually not change much. (The only changes will come from changes in the evaluations of terms like $$f(d+q)$$ when $$d$$ changes.) On the other hand, the contributions of the cubic error terms will decrease by roughly a factor of $$8$$ and, since there will be twice as many intervals, their contribution to the overall error will decrease by roughly a factor of $$4$$. So? This means that if we take four times the estimate after the split and subtract the estimate before the split we will (almost) eliminate the first error term, and the resulting next error term will decrease on splitting by a factor of $$16$$. We get estimates $$4-1$$ or $$3$$ times so we divide by $$3$$ to get back to an estimate of the integral. So here is our plan: First we calculate the integral from the starting point to any later point using the left hand rule. This is very easy to do, as we shall see. How? We can do this by creating two columns: an $$x$$ column and an integral column. The $$x$$ column (say column A) starts, (say in A5) with the lower end of the integral. Then for $$k$$ larger than $$5$$ we set Ak to A(k-1) + d. In the left rule integral column we set Ck to C(k-1) + df(Bk). The left hand rule integral from the content of A5 to that of Ak will be C(k-1). Actually, when you want to change the function you are integrating, you are better off if all evaluations of the function are in one column, Thus you might want to set Bk to df(Ak), and Ck to C(k-1) + Bk. Next we convert the left hand rule integral in column C to the trapezoid rule integral in column D. How? In D we add to Ck a term which gets rid of half of A5d and half of dAk. Explicitly, we set Dk to Ck - (A$5+Ak)d/2 . The Trapezoid rule integral from B5 to Bk now appears in Dk. Next we convert this to Simpson’s Rule in column E. How? Lets call the Trapezoid rule result with interval size $$d$$ with endpoint $$\text{start} + kd,$$ $$T(d,k).$$ What we want to do is to repeat the Trapezoid calculation for interval size $$2d$$ and then form $$\frac{4T(d,k)-T(2d,k)}{3}.$$ The result will have error that behaves as $$d^4$$. This estimate is called Simpson's Rule. Why do we compute Simpson's rule in this way? Because it is easy to apply the left hand rule, easy to get the Trapezoid rule results in a column from it, and easy to double the interval size in another column. Having done so, its easy to form the new rule from the old ones on a spreadsheet according to the expression in the last paragraph above, in a third column. It is easiest to accomplish this when $$T(2d,k)$$ is in the same row as $$T(d,k)$$. Recall that the contribution to the left hand rule at row $$k$$ is $$df(k)$$ which is added to the result in the row corresponding to $$k-1$$. For $$2d$$, this contribution is doubled and added to the result for the previous doubled $$d$$ row which is $$2$$ above it. The correction to make the left hand rule into the trapezoid rule for the integral from start $$s$$ to $$s+kd$$ is to subtract half of the first and last contribution from the partial sum that includes the initial value and the sum up $$s+kd$$. It turns out that there is not much difficulty in putting $$T(2d,2k)$$ in the same line as $$T(d,2k)$$ which makes it easy to form $$\frac{4T(d,k)-T(d,2k)}{3}$$. Well, what do you actually do? Put your interval size $$d$$ in some box, say A1. In column A, put your start point in say A5, and from A6 at each step increase the value by $$d$$. Thus in A6, you can put =A5+A$1, and A6 can be copied down column A as far as you want. In column B, put the value of the integrand times $$d$$ at the corresponding argument: in B5 put =A$1f(A5), and copy this down column B. In column C, put the partial sums of column B: which means, in C5 put =C4+B5, and copy down column C. In column D, put in D5: =-(B5+B$5)/2 and copy this down column D. The Trapezoid answer from your start (which is in B5) to the value in Bk will be Ck+Dk which you can put in box Ek, by setting E5 to =C5+D5 and copying this down column E. In F, set F5 =2B5+F3 and copy down F (This will put the left hand $$2d$$ rule results in the odd numbered rows of F beyond row 5. The even numbered rows contain useless junk.) In G5, set =2D5 +F5 and copy down G, which will give the integral for interval $$2d$$ trapezoid result in the odd numbered entries of column G. In H5, set =(4E5-G5)/3 and copy down. Simpson's rule for the integral from the content of B5 to that of B(2k+1) will then appear in H(2k+1), for $$k$$ least $$3$$. Next we do an explicit example for the function $$x\sin(x)$$. Preliminaries: Set A1 to Integrate f(x), B1 to f(x) = xsin(x) Set A2 to d, B2 to 0.01 Set A3 to startpoint, B3 to 1 Create Columns: Set A5 to =B3, A6 to =A5+B$2 and copy A6 down column A. Set the $$5^{th}$$ row of columns B through H as follows: In B5, enter =B$2A5sin(A5); in C5, =C4+B5; in D5, =-(B5+B$5)/2; in E5, =C5+D5; in F5, =2B5+F3; in G5, =2D5+F5; in H5, =(4E5-G5)/3. Copy all these down their columns. The entries in H(5+2j) etc give Simpson’s rule, integrating from the value in A5 to that in A(5+2j). Column A contains the variable, B contains $$f(x)d$$ which is the integrand times the width of the interval, C contains its partial sums, D contains the correction to make this the trapezoid rule which is in E, F jumps by $$2$$ in taking sums and doubling the width which corresponds to doubling $$d$$, G corrects the endpoints to create the trapezoid rule for $$2d$$ for appropriate intermediate endpoints and H creates Simpson’s rule from the $$d$$ and $$2d$$ trapezoid rule answers. Having done this you can change $$d$$ and the starting point by changing the content of A1 and B1. To change the integrand you need only change column B. You should test your answer with an integral whose value you know to find any bugs in your spreadsheet. You can try doubling $$d$$ to see whether that changes your answer much. If not you have computed your integral quite well. Does this always work? No. You obviously cannot do this if you want to integrate to infinity. You can also run into trouble if your integrand goes to infinity at some intermediate point. Or if it wobbles insanely. You may be able to subtract something from it that you know about and has the same singular behavior, and then be able to handle the rest. Exercise: Try to find a function for which this procedure will fail (one that does not blow up). Things like square roots near $$0$$ might be such. If you add columns that jump by $$4$$ and do a similar thing with them, you can get the Trapezoid rule for $$d$$ replaced by $$4d$$. With that you can get two Simpson rule calculations for $$d$$ and $$2d$$. Taking $$16$$ times the first less the second, and dividing by $$15$$ you will get a super Simpson rule, that improves by a factor of $$64$$ when $$d$$ decreases by a factor of $$2$$. Here are the results for the integral of $$x\sin x$$ from $$x=1$$ to $$2$$ Number of increments Number of digits after decimal point The exact answer, given in column I, is $$\sin x - x \cos x$$ (differentiate and see). To get column I, just enter, in I5, =SIN(A5)-A5COS(A5)-SIN(A$5)+A$5COS(A$5) and copy down. Simpsons rule here is accurate to about $$10$$ significant figures. The value in red is our Simpson’s Rule answer. The value to its immediate right is the computer evaluation of $$\sin x - x \cos x$$ which is the value of this integral. Notice that you can switch to integrating some other integrand merely by changing column B. This involves a new entry in B5 and copying it down that column. The starting point (called the lower limit of integration) can be changed by changing B3. In the computation above, $$10$$ place accuracy occurs for $$d = 0.001$$.
# Indirect variation In mathematics, indirect variation describes a relationship where one variable increases while the other decreases, expressed as \$$y \\propto \\frac{1}{x} \$$ or \$$y = \\frac{k}{x} \$$ with \$$k \$$ being a constant. This means if \$$x \$$ doubles, \$$y \$$ halves, illustrating an inverse relationship. Understanding indirect variation is crucial for real-world applications like physics and economics, where it helps predict how changes in one quantity impact another. #### Create learning materials about Indirect variation with our free learning app! • Flashcards, notes, mock-exams and more • Everything you need to ace your exams ## Indirect Variation Definition In mathematics, Indirect Variation refers to a relationship between two variables where the product of the two variables is constant. When one variable increases, the other decreases proportionately, and vice versa. Indirect variation is also known as Inverse Variation. ### Understanding Indirect Variation To understand indirect variation, consider two variables $$x$$ and $$y$$. If $$x$$ and $$y$$ vary indirectly, their relationship can be represented by the equation: $x \times y = k$ where $$k$$ is a constant value. This equation signifies that as $$x$$ increases, $$y$$ must decrease so that their product remains constant, and vice versa. For instance, suppose $$x = 2$$ and $$y = 6$$. The product of $$x$$ and $$y$$ is:$$2 \times 6 = 12$$. If $$x$$ is doubled to 4, then $$y$$ must be halved to 3 to maintain the constant product of 12: $$4 \times 3 = 12$$. This inverse relationship showcases indirect variation. ### Mathematical Representation The general formula for indirect variation can be written as: $y = \frac{k}{x}$ In this representation, $$y$$ is inversely proportional to $$x$$ with a constant of proportionality $$k$$. By rearranging terms, you can see the product equation stated earlier. Let's explore further by solving an example. Assume $$y$$ varies indirectly with $$x$$, and when $$x = 5$$, $$y = 8$$. To find the constant $$k$$: $5 \times 8 = 40$ Thus, $$k = 40$$. The equation for this indirect variation is: $y = \frac{40}{x}$ This relationship allows you to calculate $$y$$ for any given value of $$x$$. For example, if $$x = 10$$, then: $y = \frac{40}{10} = 4$. ### Graphical Representation Graphing an indirect variation relationship typically results in a hyperbolic curve. As $$x$$ increases, $$y$$ decreases, and the curve approaches the axes without touching them. Indirect variation graphs are always concave in nature, pointing towards the origin. ### Applications of Indirect Variation Indirect variation has numerous practical applications. Some examples include: • The relationship between speed and travel time for a fixed distance. • The reciprocal relationship between pressure and volume in a gas (Boyle's Law). • Electrical resistance and current in a fixed voltage circuit (Ohm's Law). ### Key Differences from Direct Variation While indirect variation focuses on the inverse relationship, it's essential to differentiate it from direct variation. In Direct Variation, two variables increase or decrease together proportionally, represented by the equation: $y = kx$ In contrast, indirect variation is represented by $$y = \frac{k}{x}$$. Direct Variation: A relationship where two variables change in the same direction, represented by the equation $$y = kx$$. ## Indirect Variation Formula The Indirect Variation Formula establishes a relationship between two variables such that their product remains constant. In mathematical terms, if two variables $$x$$ and $$y$$ vary indirectly, their relationship can be expressed as: $x \times y = k$ where $$k$$ is a constant. ### Understanding the Formula To better understand this formula, let's break it down. If $$y$$ is inversely proportional to $$x$$, then $$y$$ can be written as: $y = \frac{k}{x}$ Here, $$y$$ decreases as $$x$$ increases, ensuring that the product $$x \times y$$ remains constant. Suppose $$x = 3$$ and $$y = 8$$ with a constant $$k$$. The product is: $3 \times 8 = 24$ If $$x$$ is increased to 6, to keep the product constant at 24, $$y$$ must decrease to 4: $6 \times 4 = 24$ ### Algebraic Manipulations You can manipulate the indirect variation formula to solve for unknowns. For instance, if you know $$k$$ and the value of one variable, you can find the other. Given $$k = 50$$ and $$x = 10$$, find $$y$$: $y = \frac{50}{10} = 5$ Conversely, if you know $$y = 4$$ and $$k = 20$$, find $$x$$: $x = \frac{20}{4} = 5$ For a more challenging example, consider the case where multiple values vary indirectly. If $$x_1, y_1$$ and $$x_2, y_2$$ satisfy the indirect variation formula, then: $x_1 \times y_1 = x_2 \times y_2 = k$ Suppose $$x_1 = 5$$, $$y_1 = 6$$, and $$x_2 = 10$$. To find $$y_2$$: $5 \times 6 = 10 \times y_2$ Solving for $$y_2$$: $y_2 = \frac{5 \times 6}{10} = 3$ This demonstrates how changes in one variable affect the other in an inverse manner. ### Identifying Indirect Variation To verify if a relationship follows indirect variation, check if the product of the variables is constant. For instance, for pairs (2, 10), (4, 5), and (5, 4), calculate: • $$2 \times 10 = 20$$ • $$4 \times 5 = 20$$ • $$5 \times 4 = 20$$ ### Graphical Representation Graphing indirect variation results in a hyperbolic curve. Here's what to look for: • The curve approaches the axes but never touches them. • As $$x$$ increases, $$y$$ decreases, and vice versa. This behaviour illustrates the inverse relationship between the variables. A hyperbolic graph's asymptotes represent the values the variables approach but never reach. ### Real-World Applications Indirect variation appears in numerous real-world contexts, such as: • Pressure and volume of a gas (Boyle's Law). • Speed and travel time for a fixed distance. • Supply and demand economics. These examples underscore the significance of indirect variation in our daily lives. Boyle's Law: Inverse relationship between the pressure and volume of a gas at a constant temperature, represented by $$P \times V = k$$. ## Examples of Indirect Variation Indirect variation can be seen in various real-life situations. Let's explore some of these examples to gain a deeper understanding of how this concept applies. ### Speed and Travel Time One common example of indirect variation is the relationship between speed and travel time for a fixed distance. If you increase your speed, the time it takes to travel a specific distance decreases, and vice versa. This relationship can be expressed by the formula: $\text{Speed} (S) \times \text{Time} (T) = \text{Distance} (D)$ For instance, if the distance is 100 miles and you travel at 50 miles per hour, the time taken would be: $50 \times T = 100 \quad \text{so} \quad T = \frac{100}{50} = 2 \text{ hours}$ If you increase your speed to 100 miles per hour, the time taken decreases to: $100 \times T = 100 \quad \text{so} \quad T = \frac{100}{100} = 1 \text{ hour}$ Suppose you need to travel 200 miles. If your speed is 40 miles per hour, the time taken is: $40 \times T = 200 \quad \text{so} \quad T = \frac{200}{40} = 5 \text{ hours}$ If the speed increases to 80 miles per hour, the time taken becomes: $80 \times T = 200 \quad \text{so} \quad T = \frac{200}{80} = 2.5 \text{ hours}$ This clearly shows the inverse relationship, where increasing speed leads to a decrease in travel time. ### Pressure and Volume of a Gas (Boyle's Law) Boyle's Law is a principle in physics that describes the indirect variation between the pressure and volume of a gas at a constant temperature. The law states that the pressure of a gas is inversely proportional to its volume. The formula is represented as: $P \times V = k$ where $$P$$ stands for pressure, $$V$$ stands for volume, and $$k$$ is a constant. To illustrate Boyle's Law, let's consider a gas with an initial pressure $$P_1$$ of 2 atm and an initial volume $$V_1$$ of 4 litres. If the volume is compressed to 2 litres, the new pressure $$P_2$$ can be found using the formula: $P_1 \times V_1 = P_2 \times V_2$ Substituting in the known values: $2 \times 4 = P_2 \times 2$ Solving for $$P_2$$: $8 = P_2 \times 2 \quad \text{so} \quad P_2 = 4 \text{ atm}$ This example demonstrates how decreasing the volume increases the pressure, maintaining a constant product of pressure and volume. ### Supply and Demand Economics In economics, the concept of supply and demand can also illustrate indirect variation. When the supply of a product increases, the price typically decreases if the demand remains the same. Conversely, if the supply decreases, the price increases. This relationship helps maintain market equilibrium. The price elasticity of demand measures how responsive the quantity demanded is to a change in price, which often follows the principles of indirect variation. Consider a market where the demand for apples remains constant. If the supply of apples increases, the price of apples goes down. For example: • If the supply increases from 100 to 200 apples, the price may decrease from £2 to £1 per apple. • If the supply decreases back to 100 apples, the price may increase to £2 per apple again. This showcases the inverse relationship between supply and price, illustrating indirect variation. ## Understanding Indirect Variation in Mathematics Indirect variation, also known as inverse variation, defines a relationship between two variables in which the product remains constant. If one variable increases, the other decreases proportionately, and vice versa. This concept is foundational in various mathematical and practical applications. ### Indirect Variation Equation Basics The fundamental equation for indirect variation is expressed as: $x \times y = k$ where $$x$$ and $$y$$ are the variables, and $$k$$ is a constant. This indicates that if one variable increases, the other must decrease to maintain the constant product. Indirect Variation: A type of variation where the product of two variables is constant, expressed as $$x \times y = k$$. Consider an example where $$x = 4$$ and $$y = 3$$. The constant $$k$$ is: $4 \times 3 = 12$ If $$x$$ changes to 6, find $$y$$: $6 \times y = 12 \quad \Rightarrow \quad y = \frac{12}{6} = 2$ Thus, as $$x$$ increased from 4 to 6, $$y$$ decreased from 3 to 2 to keep the product equal. Another way to express indirect variation is with equations involving multiple variables. For example: $x_1 \times y_1 = x_2 \times y_2 = k$ If $$x_1 = 5$$ and $$y_1 = 6$$, then the product is: $5 \times 6 = 30$ If $$x_2$$ is doubled to 10, then $$y_2$$ must be: $5 \times 6 = 10 \times y_2 \quad \Rightarrow \quad y_2 = \frac{30}{10} = 3$ This calculation shows how the indirect variation formula can be applied in more complex scenarios. ### Practical Applications of Indirect Variation Indirect variation has multiple practical applications. Here are some examples: • Speed and travel time: For a fixed distance, if speed increases, travel time decreases. • Pressure and volume of a gas (Boyle's Law): At a constant temperature, if pressure increases, volume decreases. • Electricity: In fixed voltage circuits, if resistance increases, current decreases. Consider the relationship between speed $$S$$, travel time $$T$$, and distance $$D$$. The formula is: $S \times T = D$ If the distance $$D$$ is 100 miles and the speed is 50 miles per hour, the travel time $$T$$ is: $50 \times T = 100 \quad \Rightarrow \quad T = \frac{100}{50} = 2 \text{ hours}$ If the speed changes to 25 miles per hour, the new travel time is: $25 \times T = 100 \quad \Rightarrow \quad T = \frac{100}{25} = 4 \text{ hours}$ Boyle's Law in physics perfectly demonstrates indirect variation as the pressure and volume of a gas remain inversely related. In electrical circuits, Ohm's Law states that $$V = IR$$. Rearranging as $$I = \frac{V}{R}$$ shows that if the resistance $$R$$ increases, the current $$I$$ decreases when the voltage $$V$$ is constant. This exemplifies indirect variation. ### Common Misconceptions about Indirect Variation It's vital to clarify some common misconceptions about indirect variation: • Incorrect Relationship: Sometimes indirect variation is confused with direct variation, where variables increase or decrease together proportionally (direct variation: $$y = kx$$). • Constant Value: Ensure that $$k$$ remains constant for all values of $$x$$ and $$y$$. If $$k$$ changes, the relationship isn't true indirect variation. • Interchanging Variables: Mistakenly interchanging $$x$$ and $$y$$ without recalculating can lead to incorrect conclusions. ### Graphical Representation of Indirect Variation Graphing indirect variation typically results in a hyperbola. Here are key characteristics to observe: Graphs of indirect variation are always concave and approach the axes but never touch them. • The curve appears asymptotic to both the x-axis and y-axis. • The variables move in opposite directions; as one increases, the other decreases. • The graph will not cross the axes, reflecting that the product remains constant. ### Solving Indirect Variation Problems To solve problems involving indirect variation, follow these steps: 1. Identify the constant $$k$$ by using given values. 2. Substitute the known values into the formula $$x \times y = k$$. 3. Rearrange the equation to solve for the unknown variable. Let's solve an example problem: If $$x = 8$$ and $$y = 3$$, the constant $$k$$ is: $8 \times 3 = 24$ To find $$y$$ when $$x = 6$$: $6 \times y = 24 \quad \Rightarrow \quad y = \frac{24}{6} = 4$ This solution shows the steps needed to solve for the unknown variable and maintain the indirect variation relationship. For a comprehensive understanding, consider more complex problems with multiple steps. For instance, if you have values $$a$$ and $$b$$ such that $$a \times b = k$$ and another pair $$c$$ and $$d$$ with the same constant, you can use: $a \times b = c \times d$ This extended relationship helps solve problems where multiple pairs of variables maintain the same constant. ## Indirect variation - Key takeaways • Indirect Variation Definition: A relationship between two variables where their product is constant. When one variable increases, the other decreases proportionately. • Indirect Variation Formula: Represented by the equation x × y = k, where k is a constant. Alternatively, y = k / x. • Graphical Representation: Indirect variation graphs typically result in a hyperbolic curve, approaching but never touching the axes. • Examples of Indirect Variation: Speed and travel time, pressure and volume of a gas (Boyle's Law), and electricity with Ohm's Law. • Indirect Variation vs. Direct Variation: Direct variation is represented by y = kx, where both variables increase or decrease together proportionately, unlike indirect variation. #### Flashcards in Indirect variation 12 ###### Learn with 12 Indirect variation flashcards in the free StudySmarter app We have 14,000 flashcards about Dynamic Landscapes. What is indirect variation in mathematics? Indirect variation in mathematics occurs when the product of two variables is constant. This means that as one variable increases, the other decreases proportionally. It is represented by the equation \$$xy = k \$$, where \$$k \$$ is a constant. How can you determine if two variables have an indirect variation relationship? You can determine if two variables have an indirect variation relationship by checking if their product is constant. This means that as one variable increases, the other decreases proportionally such that their multiplication always yields the same value. Can you provide an example of an indirect variation equation? Certainly! An example of an indirect variation equation is \$$y = \\frac{k}{x} \$$, where \$$k \$$ is a constant. This means \$$y \$$ varies inversely as \$$x \$$, so as \$$x \$$ increases, \$$y \$$ decreases, and vice versa. How is indirect variation represented graphically? Indirect variation is graphically represented as a hyperbola. The curve approaches the axes but never touches them, reflecting how the product of the variables is constant. As one variable increases, the other decreases proportionally. What are the real-life applications of indirect variation? Real-life applications of indirect variation include speed and travel time, where as speed increases, travel time decreases. Additionally, it applies in situations such as the intensity of light decreasing with distance, and the relationship between the number of workers and the time needed to complete a task. ## Test your knowledge with multiple choice flashcards How can the relationship between two variables in indirect variation be represented? How does speed and travel time relate in terms of indirect variation? What is the definition of Indirect Variation? StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. 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Thursday, August 5 # What is empirical rule (Meaning, definition, Formula) 0 714 Here, we will help you understand the Empirical Rule as well as its calculation and where and how to apply the Empirical Rule formula. Page Contents ## What is the Empirical Rule? The Empirical Rule is basically the statistical rule that tells us that all the facts and figures will eventually fall under three of the standard deviations of the mean which is denoted by σ (and refers to as standard deviation), and where we calculate the mean by average of all of the given numbers that are available with us. This is also referred as Three Sigma Rule or the Rule of 68-95-99.7, as it comprises of the 3 set ranges of the data, which are as: • 68 Percent of all the data – it falls within the first standard deviation of the mean • 95 Percent of all the data, and – it falls within the two concerned standard deviations. • 99.7 Percent of all the data– it falls within the three concerned standard deviations. ## The Empirical Rule Definition: The Definition of Empirical Rule states that for a normal distribution, approximately all of the facts and figures will fall inside the given or stated three standard deviations of the concerned mean. And those 3 groups are of the data 68 Percent, 95 Percent, and 99.7 Percent of the data. You can also get a calculus homework help from the experts in the case you write an essay related to this subject. ### When to use the Empirical Rule? The Empirical Rule is used when it is quite difficult to obtain the data or sometimes when it becomes impossible to collect the data. So, it is most probably used to forecast things. And a rough figure is calculated using the empirical rule. Thus, mostly it is applied when the data is not available to us, or we are left out with no information. ## Empirical rule statistics The statistics of the empirical rule involves the calculation of Mean and standard deviation that is used in the calculation of the empirical rule. ### 1. The Mean Mean is calculated by adding the number of available groups and dividing the sum by the total number available. It is represented by. ### 2. The Standard Deviation The standard deviation is calculated by using the following formula, that is given below: ### 3. The normal distribution: The normal distribution is the actual graph that defines the range of the given data. Here, the empirical rule is applied and result is drawn out. So, the empirical rule stats involve the usage of: the mean, standard deviation and the normal curve. ## Empirical rule graph: The empirical rule graph denotes the three groups or ranges of the rule which are as follows: Whereas, the empirical rule percentage graph is drawn as follows: ## How to do empirical rule? The empirical rule procedure involves The Calculation for Empirical Rule: Step 1: Firstly, you need to draw a bell shaped curve. This is a simple curve. Step 2: Write down all the values or data in the bell shaped curve. Step 3: Mark down each section’s percentage for which you need to calculate the following: • You need to Calculate the Mean of the available data. • You need to Find out the Standard Deviation of the data available. Step 4: Apply the Empirical Rule to the derived data by use of mean and standard deviation. ### Empirical rule example 1: Taking up the example as the mean to be 40 and the standard deviation to be 200. And applying empirical rule we get the following output: Here, the empirical rule bell curve says that: • Roughly 99.7 Percent of all the data lies between ± 3 SD, or between 80 and 320 • Roughly 95 Percent of all the data lies between ± 2 SD, or between 120 and 280 • Roughly 68 Percent of all the data lies between ± 1 SD, or between 160 and 240 ### Empirical rule example 2: Taking up the example as the mean to be 50 and the standard deviation to be 4. And applying empirical rule we get the following output: • Approx. 99.7 Percent of all the data lies between ± 3 SD, or between 38 and 62 • Approx. 95 Percent of all the data lies between ± 2 SD, or between 42 and 58 • Approx. 68 Percent of all the data lies between ± 1 SD, or between 46 and 54 ## Conclusion: Thus, with the use of empirical rule you can find out the data and analyse it as well. Normally, it is used where the data or facts and figures are not available. And empirical data is used to draw out a result. Though, this rule is applicable only in case on random variables and So, does not apply to the data that is not normal. ## FAQ ### How do you use empirical rule? The empirical rule is used with the help of a normal distribution, that states that all the data will eventually fall within the three standard deviations of the mean. Thus, it can be seen as well as calculated in 3 parts i.e. • 68% of concerned data • 95% of the data and • 99.75 of the data ### What does empirical rule mean? The empirical rule means that all the data will fall down under the 3 data sets which in all sums up the 100%. And therefore, the 3 ranges of the empirical rule are grouped up as: 68% of the given data, 95% of the given data and 99.7% of the given data. So, the data will fall in these given and set 3 ranges only. ### What is another name for the empirical rule? Some of the other names used for the empirical rule are as:the 68-95-99.7 rule or Three Sigma Rule, normal distribution, the rule of thumb, etc.
Courses Courses for Kids Free study material Offline Centres More Store What are Related Facts? Reviewed by: Last updated date: 17th Sep 2024 Total views: 244.5k Views today: 2.44k Related Facts Examples You have 5 chocolates. And, your best friend gives you 5 more. Now, how many chocolates do you have? Ten. Very Good! But, what if your sister asks you to give her 5 chocolates? Sharing is caring. Right? So, you give her 5 chocolates. Now, how many chocolates do you have? 5, right? Good. That means if you add 5 with 5, you get 10 as a result. And, when you subtract 5 from 10, you will get 5. These are called related facts. Related Facts Related Facts Related Facts are the Mathematical facts that derive the relationship between two different inverse operations such as addition and subtraction, and multiplication and division. These are the basic Mathematical expressions that are made up of three numbers. Related Facts of Addition and Subtraction To understand the related facts of addition and subtraction, we will consider the following example: 4 + 6 = 10 6 + 4 = 10 10 - 4 = 6 10 - 6 = 4 Addition and subtraction are the inverse operations. In the above-given expressions, 4, 6, and 10 are the three numbers used. These numbers are related to each other using the ‘+’ or ‘-’ and ‘=’. In the first two examples, we perform the operation called addition. And, if we swap the numbers, we get the same result, that is 10. Also, in the last two equations, we subtract either 4 or 6 from the ‘10’, we get 6 or 4, respectively. In addition and subtraction, related facts tell us that if we add one number, then subtract the same from the obtained result, they cancel each other. For example, adding 6 with 4 gives the result of 10, and if we subtract 6 from 10, we will get 4. These types of facts are called the ‘related facts’. The point to be considered is that adding 6 with 4 gives the result of 10 whereas adding 4 with 6 gives the result of 10. This is called the commutative property of addition which is a ‘related fact’ of addition. Relationship of Multiplication and Division Like addition and subtraction, multiplication and division are also inverse operations, which cancel each other. To understand the related facts of multiplication and division, we will consider the following example: 6 x 2 = 12 2 x 6 = 12 12 ÷ 2 = 6 12 ÷ 6 = 2 A related fact about multiplication and division is that if we multiply a number by x (say), and then divide the result by x, then we will get the same number. Conclusion By a given fact or a set of three numbers, we can write a total of four related facts. In addition, we can write three more facts using it. Similarly, we can write a total of four facts if a set of three numbers is of multiplication and division. FAQs on What are Related Facts? 1. How addition and subtraction are ‘related facts’? Addition and subtraction are inverse in a relationship. They cancel each other. If we add a number and subtract the same from the result, then there will be no change in expression. Also, commutative property is considered a related fact. For example: 3 + 7 = 10 7 + 3 = 10 10 - 3 = 7 10 - 7 = 3 With any fact from the above given fact family, we can write its related three other facts simultaneously. 2. How multiplication and division are ‘related facts’? Multiplication and division are related facts as they cancel each other and are also in inverse relationships. For example: 3 x 7 = 21 7 x 3 = 21 21 ÷ 7 = 3 21 ÷ 3 = 7 Multiplying by a number, and dividing the same number by the obtained result, causes no change in the expression. These facts are called related facts. Related facts are the basic Mathematical expressions consisting of three numbers.
• If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old. • You already know Dokkio is an AI-powered assistant to organize & manage your digital files & messages. Very soon, Dokkio will support Outlook as well as One Drive. Check it out today! View # D1 last edited by 17 years, 1 month ago Outcome: D1 - develop and apply formulas for distance and midpoint Originator: Kristen and Ian Explanation: To find the distance (d) between two points on a coordinate graph, we can draw a line between the two points, and then draw a triangle with this line as the hypotenuse. Using algebra, we can determine that the two legs of the triangle have lengths and Now we can use the Pythagorean Theorem to determine the length of the hypotenuse. Then we can rearrange the equation to solve for d This is the formula for finding the distance between two points. To find the midpoint (M) of two points, we need to find the x value that is halfway between the x values of the two points, and the y value that is halfway between the y values of the two points. This will give us the coordinates of the midpoint. To find these halfway values, we can find the average of the two x values and the average of the two y values. For example, if the two points were and , then: the average of the x values would be , and the average of the y values would be , so to find the coordinates of M we can use this formula: Problem: Bob and Tim are on a jungle gym. Bob is at point (1,3) and Tim is at point (3,1) a) how far far would bob have to go to get to tim if he traveled a straight line? b) if they both traveled an equal distant, at what point would they meet up? Solution a) d^2=(3-1)^2 + (1-3)^2 = 2^2 + (-2)^2= 4 +4 = 8 d= square root of 8, which is equal to 2.8units b) midpoint = (((1+3)/2), ((3+1)/2))) = (4/2, 4/2) = (2,2) therefore they would meet at point (2,2)
Courses Courses for Kids Free study material Offline Centres More Store # The maximum value of $4{{\sin }^{2}}x-12\sin x+7$ is.A. 25B. 4C. Does not existD. None of these Last updated date: 15th Jul 2024 Total views: 448.2k Views today: 9.48k Verified 448.2k+ views Hint: We know that the value of $\sin \theta$ varies from – 1 to 1. So, from this concept we will try to find the maximum value of the given equation. Complete step-by-step Solution: It is given in the question that we have to find the maximum value of, $\Rightarrow 4{{\sin }^{2}}x-12\sin x+7$ Taking 4 common from $4{{\sin }^{2}}x-12\sin x$, we get, $\Rightarrow 4\left( {{\sin }^{2}}x-3\sin x \right)+7$ Now, we will add and subtract ${{\left( \dfrac{3}{2} \right)}^{2}}$ and also multiply and divide the term $3\sin x$ by 2. So, we will get, $\Rightarrow 4\left[ {{\sin }^{2}}x-2.\sin x.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right]+7$ Since we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ , we write the above equation as, $\Rightarrow 4\left[ {{\left( \sin x-\dfrac{3}{2} \right)}^{2}}-\dfrac{9}{4} \right]+7$ On opening the bracket and multiplying the terms inside the bracket with 4, we get, \begin{align} & \Rightarrow 4{{\left( \sin x-\dfrac{3}{2} \right)}^{2}}-\dfrac{9}{4}\times 4+7 \\ & \Rightarrow 4{{\left( \sin x-\dfrac{3}{2} \right)}^{2}}-9+7 \\ & \Rightarrow 4{{\left( \sin x-\dfrac{3}{2} \right)}^{2}}-2 \\ \end{align} We know that, the value of sin varies from – 1 to 1 $\Rightarrow -1\le \sin x\le 1..........\left( 1 \right)$ (Graph of sinx) Subtracting $\dfrac{3}{2}$ from all sides in equation (1), we get, \begin{align} & \Rightarrow -1-\dfrac{3}{2}\le \left( \sin x-\dfrac{3}{2} \right)\le +1-\dfrac{3}{2} \\ & or \\ & \Rightarrow \dfrac{-5}{2}\le \left( \sin x-\dfrac{3}{2} \right)\le \dfrac{-1}{2} \\ \end{align} On squaring all sides fraction sign changes, we get, $\Rightarrow \dfrac{1}{4}\le {{\left( \sin x-\dfrac{3}{2} \right)}^{2}}\le \dfrac{25}{4}$ On multiplying the above expression by 4, we get, $\Rightarrow 1\le 4{{\left( \sin x-\dfrac{3}{2} \right)}^{2}}\le 25$ Subtracting 2 from all sides, we get, \begin{align} & \Rightarrow 1-2\le 4{{\left( \sin x-\dfrac{3}{2} \right)}^{2}}-2\le 25-2 \\ & \Rightarrow -1\le 4{{\left( \sin x-\dfrac{3}{2} \right)}^{2}}-2\le 23 \\ \end{align} Hence, the maximum value of $\left[ 4{{\left( \sin x-\dfrac{3}{2} \right)}^{2}}-2 \right]\ \ is\ \ 23$. Or the maximum value of $4{{\sin }^{2}}x-12\sin x+7$ is 23. Therefore option (D) is the correct answer. Note: This question can be solved in just a few steps. You just need to memorise the graph of sin x and keeping limits of sin in mind to solve this problem. This will save your time. One mistake that can be committed is by not changing the inequality sign and getting the wrong result.
# Vectors Vectors are usually used to represent velocity and acceleration, force, and other directional quantities in physics. Vectors are quantities with size and direction. The objects that we have worked with in single variable calculus (Calculus 1 and 2) have all had a quantity, i.e. we were able to measure them. Some quantities only have size, such as time, temperature, or weight. These quantities are called scalars. Other quantities can have size and direction. Velocities, for example, have direction as well, and therefore they are described as vectors. We denote vectors with an arrow pointing in the direction they are oriented. The direction of a vector on the coordinate plane is intuitive. The positive y direction, which is up, is north, and the positive x direction is east. The following vector is slightly east of north. The direction of a vector can also be described with a quantity. Usually, the direction of vectors are stated in relation to another direction. The following vector is described as “5 miles per hour 53.13 degrees north of east.” This vector can also be described as “5 miles per hour 36.87 degrees east of north.” To simplify the values of vectors, we use the x axis (or east) as a starting point for measurement. A line lying on the x axis would have a direction of 0 degrees. The following vector can be denoted with many different directions. The last vector would be 53.13 degrees south of west. ## Scalars and Vectors Remember that scalars only have size, while vectors have size and direction. Speed and velocity are different too. While they are sometimes used interchangeably, speed is considered a scalar while velocity is considered a vector. There is a distance between distance and displacement as well. Distance is a scalar because it only has size. Displacement, however, is a vector because it tells us how far an object moved in a certain direction. Scalars can be manipulated by the laws of arithmetic for real numbers, while vectors have special laws that need to be followed when operating on each other. For instance, if you walked 4 blocks and then 3 more blocks, how many blocks have you walked? We can add these quantities together to get 7 blocks. However, if you walked 4 blocks east and 3 blocks north, how far from your starting point will you have walked? Since these vectors are in different directions, we cannot simply add them together. The amount of degrees traveled can be either measured from the image or calculated using trigonometry. The resulting vector would be 5 blocks at .644 radians. ## Vector Notation Vectors have a special notation that distinguish them from scalars. Vectors can be noted as For our purposes, we will always denote a vector with an arrow on the top to denote a quantity with direction. The previous vector would be denoted as We can also use unit vectors i and j to denote a vector where i = <1,0> and j = <0,1> Magnitude, or length of the vector is denoted as We use the magnitude to find the quantity of the vector. Whenever we want to disregard the direction of a vector (taking the area, volume, etc), we can juse take the magnitude. The direction of the vector is denoted as ## Vector Equalities and Operations ### Equal Vectors Have the same magnitude and the same direction, they do not need to have the same starting points. ### Opposite Vectors Have exactly the same length but point in the opposite direction. When added together, opposite vectors cancel each other out. ### Parallel Vectors Have the same direction but different lengths. Vectors that have the same direction can be multiplied by scalars to yield a different magnitude. When adding vectors, we attach the start of the second vector (intial point) to the end of the first vector (terminal point). ### Scalar Multiplication Scalar multiplication is when a vector is multiplied by a scalar to increase or decrease the magnitude of the vector. The scalar does not have any effect on the direction of the vector. ### Dot Product If we have two vectors u and v, the dot product is denoted as where \|u\| and \|v\| are the magnitudes and Θ is the angle between the vectors. To illustrate what the dot product means, let’s take the last part of the formula and deconstruct it. If we take the vector v times the cos(Θ), we will end up with the projection of v onto u. The projection is formed by dropping a perpendicular line from the terminal point of v onto u, therefore forming a right angle. The projection of v onto u is the amount of vector v going in the u direction. The dot product of v and u just multiplies the projection of v and the vector u (or vice versa). If we go back to our formula, we can substitute the projection of v for the vector v. This result tells us how much of vector v is going in the direction of vector u. What is this useful for? If we think about physics applications, if we have two forces at an angle, we can see how much force is going in a particular direction. The dot product is sometimes called the scalar product because it always yields a scalar quantity. The dot product can also help us measure the angle between vectors, find projections, and determine if two vectors are perpendicular, as we will see in the next examples. Note that perpendicular vectors will always yield a dot product of 0 because there is no projection, i.e. no amount of either vector going in the other vector’s direction. The dot product can also be notated with unit vectors i = <1,0> and j = <0,1> where a, b, c, and d are constants. ### Angle Between two Vectors We can use the definition of the dot product to find the angle between any two vectors. All we need to do is isolate Θ ### The Projection of a Vector From reworking the Dot Product formula and dividing by \|u\| we can conclude that the projection of v onto u is Scroll to Top
Question Video: Finding the Magnitude of a Resultant Force \| Nagwa Question Video: Finding the Magnitude of a Resultant Force \| Nagwa # Question Video: Finding the Magnitude of a Resultant Force Physics • First Year of Secondary School ## Join Nagwa Classes The force π is the resultant of the two force vectors shown in the diagram. What is the magnitude of π to the nearest newton? 09:50 ### Video Transcript The force π is the resultant of the two force vectors shown in the diagram. What is the magnitude of π to the nearest newton? In this question, weβve got a diagram that shows three vectors. This purple vector here represents a force π, and weβre told that π is the resultant of the other two force vectors. So thatβs the red arrows here, which are each labeled with a magnitude and a direction. Weβre being asked to find the magnitude of the vector π. The magnitude of π is the length of this purple arrow. We can see from the diagram that since π is the resultant of the other two vectors that when these two vectors are drawn tip to tail like this β so thatβs the second vector drawn, starting with its tail at the tip of the first vector β then the resultant vector π goes from the tail of the first vector to the tip of the second vector. To find the magnitude of π or the length of this arrow, we can notice that the vector π forms the hypotenuse of a right-angled triangle. We can recall that Pythagorasβs theorem tells us that for a general right-angled triangle with a hypotenuse of length π and other sides of lengths π and π that π squared is equal to π squared plus π squared. If we take the square root of both sides of this expression, we have that π, the length of the hypotenuse, is equal to the square root of π squared plus π squared. In the case of the right-angled triangle that weβve identified in this diagram, the length of the hypotenuse is the magnitude of the vector π. So, if we can work out the lengths of the other two sides of the triangle, which weβve labeled as π and π, then Pythagorasβs theorem tells us that for this triangle the magnitude of π is equal to the square root of π squared plus π squared. In order to find the values of π and π, weβre going to need to identify two more right-angled triangles in our diagram. The first of these, which weβve labeled as triangle one, is the right-angled triangle that has the first of the two red vectors as its hypotenuse. The other triangle, which weβve labeled as two, is the right-angled triangle whose hypotenuse is the second of the two red vectors. For each of these two triangles that weβve identified, weβre given the length of the hypotenuse and the value of one of the angles. Since this information isnβt particularly clear in the diagram now that weβve drawn over it, letβs draw out these two triangles separately from the diagram. Triangle one looks like this kind of shape. And looking back at the diagram we were given, we can see that the leftmost angle in the triangle is 20 degrees, and the length of the hypotenuse, which is the magnitude of the first red vector, is 70 newtons. So, this is triangle one. Letβs now do the same thing for the triangle we labeled two. In this triangle, the leftmost angle is 70 degrees, and the length of the hypotenuse, which is the magnitude of the second red vector, is 60 newtons. So then, this is triangle two. If we now look back at the triangles in this diagram, we can see that π, the length of the horizontal side in the orange triangle, is equal to the sum of the horizontal side of triangle one plus the horizontal side of triangle two. We could imagine shifting triangle two vertically downward so that the base or horizontal side of triangle two was at the same vertical height as the horizontal side of triangle one. And we would find that it slotted perfectly into this space here. Letβs label the horizontal side in triangle one as π one and in triangle two as π two. Then, in this diagram, we can identify the lengths π one and π two and then can easily see that π must be equal to π one plus π two. We can do the same thing for the vertical sides of the triangles one and two. Weβll label these as π one and π two, respectively. In the same way as before, in this diagram, we could imagine shifting triangle one horizontally until its vertical side lined up with the vertical side of triangle two. We can then identify the lengths π one and π two on this diagram. And we can see that π must be equal to π one plus π two. Letβs quickly take a moment to take stock of the situation so far. Weβve got this equation here, which came from Pythagorasβs theorem. And it tells us how to calculate the magnitude of the force π if we know the values of the quantities weβve labeled as π and π. Then, weβve also got these two equations here, which tell us how to calculate π and π if we know the values of π one and π two and π one and π two. So, our approach is going to be to start by finding the values of π one, π two, π one, and π two. Then, weβll use these values to work out the quantities π and π. And then finally, we can use the values of π and π to work out the magnitude of the force π. To find the values of these quantities π one, π two, π one, and π two, weβre going to need to use a couple of trigonometrical formulas. Letβs consider a general right-angled triangle, and weβll suppose that this angle has some value of π. Weβve labeled the side of the triangle opposite this angle as capital π for opposite, the side adjacent to the angle as capital π΄ for adjacent, and the hypotenuse as capital π» for hypotenuse. The two formulas that weβre going to need are these ones right here. The first one says that cos π is equal to π΄ divided by π». So, the cos of this angle π is equal to the length of the side thatβs adjacent to it divided by the length of the hypotenuse. The second equation says that sin π is equal to π divided by π». So, the sin of this angle π is equal to the length of the side thatβs opposite it divided by the length of the hypotenuse. Looking at the triangles weβve labeled as one and two, we can see that the lengths weβre trying to find are the adjacent and opposite sides on these triangles. This means that we want to rearrange these two equations in order to make the quantities π΄ and π the subject. For both of these two equations, the process is the same. We simply need to multiply both sides by the hypotenuse π». When we do this rearrangement, we find that the adjacent side π΄ is equal to π» times cos π and the opposite side π is equal to π» times sin π. Letβs now apply these two equations to our triangles one and two, starting with triangle one. In this case, the length of the hypotenuse is 70 newtons, and the angle which corresponds to π is 20 degrees. Then, the adjacent side, π one, must be equal to the hypotenuse of 70 newtons multiplied by the cos of the angle of 20 degrees. Similarly, the side opposite the angle, which is π one, must be equal to 70 newtons, the hypotenuse, multiplied by the sin of the angle 20 degrees. Evaluating the two expressions for the sides π one and π one, we find that π one is equal to 65.778 newtons and π one is equal to 23.941 newtons. In each case, the ellipses are used to indicate that the answers have further decimal places. Letβs now clear some space and do the same thing for triangle two. In this case, the hypotenuse has a length of 60 newtons and the angle which corresponds to π has a value of 70 degrees. So, we have that the adjacent side π two is equal to 60 newtons multiplied by the cos of 70 degrees and the opposite side π two is equal to 60 newtons multiplied by the sin of 70 degrees. π two works out as 20.521 newtons, and π two works out as 56.382 newtons. Now that we found the values for the quantities π one, π two, π one, and π two, letβs clear ourselves some space so that we can use those values in these two equations to calculate the values of π and π. We know that π is equal to π one plus π two. And substituting in these values for the quantities π one and π two and then evaluating the expression that we get, we find that π is equal to 86.299 newtons. In the same way, we know that π is equal to π one plus π two. And substituting in our values for π one and π two and then evaluating the expression, we find that π is equal to 80.323 newtons. Weβre going to clear ourselves some space one last time so that we can take our values for the quantities π and π and substitute them into this equation to calculate the magnitude of the force π. Okay, so weβve got our value for π which is the horizontal side of the orange triangle. And weβve got our value for π, which is the vertical side of the same triangle. And weβve got our equation from Pythagorasβs theorem, which tells us that the hypotenuse of this triangle, which is the magnitude of the force π, is equal to the square root of π squared plus π squared. If we now substitute in our values for π and π, we get this expression for the magnitude of π. Evaluating this sum under the square root, we find that the magnitude of π is equal to the square root of 13899.30 newtons squared. Taking the square root gives a result to two decimal places of 117.90 newtons. Weβre told in the question to give our answer to the nearest newton. By rounding this result to the nearest newton, we get our answer to the question that the magnitude of the force π is equal to 118 newtons. ## 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
Class 9 AREAS OF PARALLELOGRAMS AND TRIANGLES FOR CBSE NCERT ### Topic covered • Parallelograms on the same Base and Between the same Parallels ### Parallelograms on the same Base and Between the same Parallels Now let us try to find a relation, if any, between the areas of two parallelograms on the same base and between the same parallels. For this, let us perform the following activities: color{green}("Activity 1 :") Let us take a graph sheet and draw two parallelograms ABCD and PQCD on it as shown in Fig. 9.9. The above two parallelograms are on the same base DC and between the same parallels PB and DC. You may recall the method of finding the areas of these two parallelograms by counting the squares. In this method, the area is found by counting the number of complete squares enclosed by the figure, the number of squares a having more than half their parts enclosed by the figure and the number of squares having half their parts enclosed by the figure. The squares whose less than half parts are enclosed by the figure are ignored. You will find that areas of both the parallelograms are (approximately) 15cm^2. Repeat this activity* by drawing some more pairs of parallelograms on the graph sheet. What do you observe? Are the areas of the two parallelograms different or equal? If fact, they are equal. So, this may lead you to conclude that parallelograms on the same base and between the same parallels are equal in area. However, remember that this is just a verification. color{blue}("Activity 2 :") Draw a parallelogram ABCD on a thick sheet of paper or on a cardboard sheet. Now, draw a line-segment DE as shown in Fig. 9.10. *This activity can also be performed by using a Geoboard. Next, cut a triangle A′ D′ E′ congruent to triangle ADE on a separate sheet with the help of a tracing paper and place DeltaA′D′E′ in such a way that A′D′ coincides with BC as shown in Fig 9.11. "Note" that there are two parallelograms ABCD and EE′CD on the same base DC and between the same parallels A E′ and DC. What can you say about their areas? color{green}("As" \ \ \ \ \ \ \ \ \ \ \ \ \Delta ADE ≅ D A′D′E′) color{orange}("Therefore"\ \ \ \ \ \ \ \ \ar (ADE) = ar (A′D′E′)) color{red}("Also" \ \ \ \ \ \ \ \ \ \ \ \ \ ar (ABCD) = ar (ADE) + ar (EBCD)) color{red}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ar (A′D′E′) + ar (EBCD)) color{navy}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ar (E E′CD)) So, the two parallelograms are equal in area. Let us now try to prove this relation between the two such parallelograms. color{navy}("Theorem 9.1 :") bb"Parallelograms on the same base and between the same parallels are equal in area." color{red}("Proof :") Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC are given (see Fig.9.12). We need to prove that color{navy}("ar (ABCD) = ar (EFCD).") In Delta ADE and Delta BCF, ∠DAE = ∠CBF (Corresponding angles from AD \|\| BC and transversal AF) .......(1) ∠AED = ∠ BFC (Corresponding angles from ED \|\| FC and transversal AF) .......(2) Therefore, ∠ADE = ∠BCF (Angle sum property of a triangle) ..................(3) Also, AD = BC (Opposite sides of the parallelogram ABCD)..........(4) So, Delta ADE ≅ Delta BCF ................. [By ASA rule, using (1), (3), and (4)] Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas).........(5) Now, ar (ABCD) = ar (ADE) + ar (EDCB) = ar (BCF) + ar (EDCB) ......... [From(5)] = ar (EFCD) So, parallelograms ABCD and EFCD are equal in area. Let us now take some examples to illustrate the use of the above theorem. Q 3260178015 In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that (i) ar (ABCD) = ar (EFCD) (ii) ar (ABCD) = DC × AL Class 9 Chapter 9 Example 1 Solution: (i) As a rectangle is also a parallelogram, therefore, ar (ABCD) = ar (EFCD) (Theorem 9.1) (ii) From above result, ar (ABCD) = DC × FC (Area of the rectangle = length × breadth) (1) As AL ⊥ DC, therefore, AFCL is also a rectangle So, AL = FC.(2) Therefore, ar (ABCD) = DC × AL ........... [From (1) and (2)] Can you see from the Result (ii) above that area of a parallelogram is the product of its any side and the coresponding altitude. Do you remember that you have studied this formula for area of a parallelogram in Class VII. On the basis of this formula, Theorem 9.1 can be rewritten as parallelograms on the same base or equal bases and between the same parallels are equal in area. Can you write the converse of the above statement? It is as follows: Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. Is the converse true? Prove the converse using the formula for area of the parallelogram. Q 3210178019 If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram. Class 9 Chapter 9 Example 2 Solution: Let Delta ABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC (see Fig. 9.14). You wish to prove that ar (PAB) = 1/2 ar(ABCD) Draw BQ \|\| AP to obtain another parallelogram ABQP. Now parallelograms ABQP and ABCD are on the same base AB and between the same parallels AB and PC. Therefore, ar (ABQP) = ar (ABCD) (By Theorem 9.1) (1) But DeltaPAB ≅ Delta BQP (Diagonal PB divides parallelogram ABQP into two congruent triangles.) So, ar (PAB) = ar (BQP) (2 Therefore, ar (PAB) = 1/2 ar(ABQD) [From (2)] (3) This gives ar (PAB) = 1/2 ar(ABCD) [From (1) and (3)]
## Why is the centroid the balance point of a triangle? However, triangles do balance on their centroid because, while the triangle on one side of a line has less area than the quadrilateral on the other side of the line, its corner is farther from the line and so applies more torque. ## What’s the balancing point of a triangle? The centroid of a triangle is that balancing point, created by the intersection of the three medians. If the triangle were cut out of some uniformly dense material, such as sturdy cardboard, sheet metal, or plywood, the centroid would be the spot where the triangle would balance on the tip of your finger. ## Is centroid equidistant from vertices? These lines intersect at a point in the middle of the triangle, and this point is called the centroid G. In other words, it is the point that is equidistant from all three vertices. ## What is the Orthocentre of a triangle? The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side. There are therefore three altitudes in a triangle. ## What is the formula of Circumcentre of a triangle? Since D1= D2 = D3 . To find out the circumcenter we have to solve any two bisector equations and find out the intersection points. The slope of the bisector is the negative reciprocal of the given slope. The slope of the bisector is the negative reciprocal of the given slope. ## What is the orthocenter of an obtuse triangle? Orthocenter of a Triangle For the obtuse angle triangle, the orthocenter lies outside the triangle. For a right triangle, the orthocenter lies on the vertex of the right angle. ## Why is the orthocenter of an obtuse triangle outside of the triangle? It turns out that all three altitudes always intersect at the same point – the so-called orthocenter of the triangle. The orthocenter is not always inside the triangle. If the triangle is obtuse, it will be outside. To make this happen the altitude lines have to be extended so they cross. ## What is the Circumcenter of a right triangle? The circumcenter of a right triangle is the midpoint of the hypotenuse. Thus, M is equidistant from the vertices, so it is the circumcenter of OAB. ## Why is the Incenter of a triangle important? The Incenter of a triangle Note the way the three angle bisectors always meet at the incenter. One of several centers the triangle can have, the incenter is the point where the angle bisectors intersect. The incenter is also the center of the triangle’s incircle – the largest circle that will fit inside the triangle. ## Can an Incenter be outside a triangle? The orthocenter is always outside the triangle opposite the longest leg, on the same side as the largest angle. The only time all three of these centers fall in the same spot is in the case of an equilateral triangle. In fact, in this case, the incenter falls in the same place as well. ## In what type of triangle the Orthocentre can be outside the triangle? obtuse angle triangle
How do you solve 11.1c-2.4=-8.3c+6.4? Aug 9, 2016 Exact value$\to c = \frac{44}{97}$ App. value $\to 0.454$ to 3 decimal places Explanation: Lets get rid of the decimals! Multiply everything by 10 $10 \left(11.1 c - 2.4\right) = 10 \left(- 8.3 c + 6.4\right)$ $111 c - 24 = - 83 c + 64$ Collecting like terms $111 c + 83 c = 64 + 24$ $\implies 194 c = 88$ Divide both sides by 194 $\implies c = \frac{88}{194} = \frac{44}{97} \approx 0.454$
# 6.1 Exponential functions Page 1 / 16 In this section, you will: • Evaluate exponential functions. • Find the equation of an exponential function. • Use compound interest formulas. • Evaluate exponential functions with base e. India is the second most populous country in the world with a population of about $\text{\hspace{0.17em}}1.25\text{\hspace{0.17em}}$ billion people in 2013. The population is growing at a rate of about $\text{\hspace{0.17em}}1.2%\text{\hspace{0.17em}}$ each year http://www.worldometers.info/world-population/. Accessed February 24, 2014. . If this rate continues, the population of India will exceed China’s population by the year $\text{\hspace{0.17em}}2031.$ When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions , which model this kind of rapid growth. ## Identifying exponential functions When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input. For example, in the equation $\text{\hspace{0.17em}}f\left(x\right)=3x+4,$ the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people. ## Defining an exponential function A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2050. What exactly does it mean to grow exponentially ? What does the word double have in common with percent increase ? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media. • Percent change refers to a change based on a percent of the original amount. • Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time. • Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time. For us to gain a clear understanding of exponential growth , let us contrast exponential growth with linear growth . We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See [link] . $x$ $f\left(x\right)={2}^{x}$ $g\left(x\right)=2x$ 0 1 0 1 2 2 2 4 4 3 8 6 4 16 8 5 32 10 6 64 12 From [link] we can infer that for these two functions, exponential growth dwarfs linear growth. • Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain. • Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain. #### Questions & Answers the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve 1+cos²A/cos²A=2cosec²A-1 test for convergence the series 1+x/2+2!/9x3 a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he? 100 meters Kuldeep Find that number sum and product of all the divisors of 360 Ajith exponential series Naveen what is subgroup Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1 e power cos hyperbolic (x+iy) 10y Michael tan hyperbolic inverse (x+iy)=alpha +i bita prove that cos(π/6-a)cos(π/3+b)-sin(π/6-a)sin(π/3+b)=sin(a-b) why {2kπ} union {kπ}={kπ}? why is {2kπ} union {kπ}={kπ}? when k belong to integer Huy if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41 what is complex numbers Dua Yes ahmed Thank you Dua give me treganamentry question Solve 2cos x + 3sin x = 0.5
A B # Week 6 - 9 Fractions National curriculum content • Recognise and show, using diagrams, families of common equivalent fractions • Count up and down in hundredths; recognise that hundredths arise when dividing an object by one hundred and dividing tenths by ten • Solve problems involving increasingly harder fractions to calculate quantities, and fractions to divide quantities, including non-unit fractions where the answer is a whole number • Add and subtract fractions with the same denominator Lesson objectives 1. Understand the whole 2. Count beyond 1 3. Partition a mixed number 4. Number lines with mixed numbers 5. Compare and order mixed numbers 6. Understand improper fractions 7. Convert mixed numbers to improper fractions 8. Convert improper fractions to mixed numbers 9. Equivalent fractions on a number line 10. Equivalent fraction families 11. Add two or more fractions 12. Add fractions and mixed numbers 13. Subtract two fractions 14. Subtract from whole amounts 15. Subtract from mixed numbers What we want children to know • How to count up and down in tenths; recognising that tenths arise from dividing an object into 10 equal parts and in dividing one-digit numbers or quantities by 10. • To be able to recognise, find and write fractions of a discrete set of objects: unit fractions and non-unit fractions with small denominators. • How to add and subtract fractions with the same denominator within one whole. • To be able to compare and order unit fractions, and fractions with the same denominators • How to solve problems that involve all of the above. What skills we want children to develop Use knowledge to solve reasoning and problem solving questions such as: Always, Sometimes, Never? Alex says, “If I split a shape into 4 parts, I have split it into quarters.” Explain your answer. Spot the mistake Seven tenths, eight tenths, nine tenths, ten tenths, one eleventh, two elevenths, three elevenths… 3 friends share some pizzas. Each pizza is cut into 8 equal slices. Altogether, they eat 25 slices. How many whole pizzas do they eat? Mathematical Talk • What is a unit/ non-unit fraction? • How many more tenths do I need to make a whole? • Can a fraction have more than one equivalent fraction? • Look at the equivalent fractions you have found. What relationship can you see between the numerators and denominators? Are there any patterns? • What do you notice about the numerator and denominator when a fraction is equivalent to a whole? • How many equal parts is the whole split into? How many equal parts am I adding? Top
## How do you work out probability with n? stands for n-factorial, the number of ways to rearrange n items. To calculate n!, you multiply n(n – 1)(n – 2) . . . (2)(1). For example, 5! is 5(4)(3)(2)(1) = 120; 2! is 2(1) = 2; and 1! is 1. ## How do you find Na in probability? P(A) = n(A)/n(S) 1. P(A) is the probability of an event “A” 2. n(A) is the number of favourable outcomes. 3. n(S) is the total number of events in the sample space. What is M and n in probability? Definitions: 1. Classical: P(E) = m/N. If an event can occur in N mutually exclusive, equally likely ways, and if m of these possess characteristic E, then the probability is equal to m/N. ### How do you find probability with n and p? Binomial probability refers to the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment). If the probability of success on an individual trial is p , then the binomial probability is nCx⋅px⋅(1−p)n−x . ### What is n in statistics? Population Mean The symbol ‘N’ represents the total number of individuals or cases in the population. How do you find probability with N and P? #### What does the n stand for in the binomial probability formula? the number of times the The first variable in the binomial formula, n, stands for the number of times the experiment runs. The second variable, p, represents the probability of one specific outcome. #### What does the N mean in probability? population size n: sample size or number of trials in a binomial experiment. N: population size. ND: normal distribution. σ: standard deviation. σx̅: standard error of the mean. How do you find N in standard deviation? Population and sample standard deviation 1. If the data is being considered a population on its own, we divide by the number of data points, N. 2. If the data is a sample from a larger population, we divide by one fewer than the number of data points in the sample, n − 1 n-1 n−1 . ## What is n in binomial theorem? binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n + 1 terms of the form. ## What is NCR formula? The order of objects matters in case of permutation. The formula to find nPr is given by: nPr = n!/(n-r)! Combination: nCr represents the selection of objects from a group of objects where order of objects does not matter. nCr = n!/[r! How do you write probability answer? The probability of an event can only be between 0 and 1 and can also be written as a percentage. The probability of event A is often written as P ( A ) P(A) P(A)P, left parenthesis, A, right parenthesis. ### What is the difference between N and P in probability? Trials, n, must be a whole number greater than 0. This is the number of times the event will occur. Probability, p, must be a decimal between 0 and 1 and represents the probability of success on a single trial. ### How do you calculate a probability for a certain outcome? A probability for a certain outcome from a binomial distribution is what is usually referred to as a “binomial probability”. It can be calculated using the formula for the binomial probability distribution function (PDF), a.k.a. probability mass function (PMF): f(x), as follows: where X is a random variable,… What are the inputs of the probability calculator? The probability calculator has two inputs: Number of Events: The number of events in probability is the number of opportunities or success. So, for example, there are ten runners in a race, 2 of the runners are wearing blue. If we wanted to calculate the probability of the winner of the race being a runner wearing blue, we would enter 2. #### How to use the binomial probability calculator? The binomial probability calculator will calculate a probability based on the binomial probability formula. You will also get a step by step solution to follow. Enter the trials, probability, successes, and probability type.
# Integers Below in italics are the outcomes for this unit. Please do not edit them. Use them as the headings for lessons. You can then add a couple of sentences to explain how to do the maths, you may add a picture or a link to a relevant site. Know these terms: ascending - increasing, from smallest to largest descending - decreasing, from largest to smallest integer - a positive or negative whole number or zero debit - money taken out of an account such as fees or withdrawals credit - money put into an account such as interest or deposits # Lesson 1 recognise the direction and magnitude of an integer Numbers have a size (magnitude) and a direction . For example, 5 degrees hotter or 5 degrees colder have identical magnitudes (5 degrees) but opposite directions. In terms of mathematics, hotter would be a +, and colder would be a -. Another example would be this iceberg. You can see there is a section of the iceberg above sea level (positive value) and a large section below sea level (negative value). # Lesson 2 placing directed numbers on a number line There are some key features of a number line that you have to remember: • There are ALWAYS arrows on both ends which indicate that the number line stretches on to infinity. • Zero is often (but not always) included, and is referred to as the origin. • Integers to the right of the origin are positive. • Integers to the left of the origin are negative. • Each of the markings on the line are exactly the same distance apart. • As shown below, not every marking has to be labelled. • Each mark on the line represents a 'step' or 'distance' of 1 (as above) OR 2 (as below) OR 3, OR 4, etc, but for every number line each marking represents the same step. # Lesson 3 ordering directed numbers Once you have placed all directed numbers on the number line, the number furthest to the right is the largest and the number furthest to the left is the smallest. This pdf file will help explain this. # Lesson 4 There are different ways of adding and subtracting integers, but I strongly suggest using a number line to do this. Example: -6 + 7 1. Draw a number line. 3. Determine the direction to move (adding does to the right, subtracting goes to the left) 4. Check the sign of the second integer. If it is negative, reverse the direction from step 3. 5. Move the number of steps determined by the second number (in this case, 7) This clip will show the above example. These movies show how to add and subtract integers using a number line. subtracting integers.mov Here's an activity which uses the number line to add and subtract. ...Another method This video shows another method of how to add and subtract directed numbers. This activity will help reinforce the idea. # Lesson 5 interpreting different meanings (direction or operation) for the + and – signs depending on the context Of course, once you become comfortable with adding and subtracting integers, you'll find a quicker way of doing it. Below are three examples of how by understanding the effect of + and - signs, complex operations can be made simple. # Lesson 6 multiplying and dividing directed numbers Firstly, multiplying and dividing is the same for directed numbers as it is for normal numbers, except you have to consider the signs of the numbers involved. • If both numbers have the same sign (both + or both -), the answer will positive. eg1. 6 x 4 = 24 eg2. (-6) x (-4) = 24 • If the numbers have opposite signs (1 + and 1 -), the answer will be negative. eg3. (-6) x 4 = -24 eg4. 6 x (-4) = -24 This video shows yet another method to add and subtract integers and also how to multiply and divide integers . # Lesson 7 using grouping symbols as an operator As you get better at maths, you will be faced with ever more complex sums. The trick with any complex problem is knowing where to start. With maths, all you need to do is start with any brackets (or parentheses). After that comes multiplication and division (if any), and finally addition and subtraction. Eg. 4 x [5 - 2] + 7 (-3 + 2) (looks rather complicated, start with brackets and parentheses) = 4 x [3] + 7 (-1) (we don't really need the brackets any more, so we can write...) = 4 x 3 + 7 -1 (now we deal with any multiplication or division) = 12 + -7 (finally do any subtraction or addition) = 5 # Lesson 8 applying order of operations to simplify expressions This is a very good web site which will cover all of the order of operations you will need. There are also 5 questions at the end you can use to test your knowledge.
8th Grade Math Home > Teacher Resources > Activities > Walking the Plank: Teacher Notes Walking the Plank: Teacher Notes MA.8.S.3.1 MA.912.A.3.11 Conceptual Understanding Data Collection Graphing Calculators Interpret Data Procedural Knowledge y = mx + b Graphing Linear Equations Make Predictions Problem Solving Reasoning Communication Connections Representation Ever thought you could lose weight by the foot, not the pound? Students work as a whole class • 1 bathroom scale • 2 textbooks • 1- 2’ X 8’ plank 1. Collect the Data: Mark the 2’ x 8’ plank into one foot increments Place the plank on the bathroom scale and textbooks as shown below: Weigh a volunteer person at each of the designated locations on the plank. Independent Variable, X Distance from Scale (Feet) Dependent Variable, Y Weight (Pounds) 0 1 2 3 4 5 6 2. Graph the Data Use "Distance from Scale (Feet)" as the horizontal scale and "Weight (Pounds)" as the vertical scale. Plot ordered pairs (X,Y). Looking at your graphed points, do they appear to lie along a straight line or curve? Draw the line that best fits your data. Use the graph to answer the following: 1. Find the weight reading for the distances from the scale. 3.5 feet, _____ pounds 5.25 feet, _____ pounds 7 feet, _____ pounds 2. How much does the weight reading decrease each time the person moves another foot away from the scale? ____________________________________________________________________________________ How can you tell this from the graph? ____________________________________________________________________________________ 3. What is the person’s weight when standing directly on the scale? ____________________________________________________________________________________ How can you tell this from the graph? ____________________________________________________________________________________ 4. Describe in words how to determine the weight reading if you know how many feet the person is from the scale. 5. Use your description to predict the weight reading for the person when standing 4.75 feet from the scale. Show your work. 6. Describe by equation how to determine the weight reading (Y) if you know how many feet away (X) the person is from the scale: Y= ____________________ 7. Use your equation to predict the weight reading for the person when standing 6.5 feet away from the scale. As a result of this activity, students will learn how to collect, graph and interpret data. 8th Grade Math Home > Teacher Resources > Activities > Walking the Plank: Teacher Notes
# What is the value of log base 4 16? ## What is the value of log base 4 16? Logarithm base 4 of 16 is 2 . ## How do you find log base 4 of 16? The answer is: 2 . Using the properties of the logarithmics: log4(16)=log4(42)=2⋅log4(4)=2 . What is the value of log 4 base 4? Logarithm base 4 of 4 is 1 . ### What is the value of log 3 81? log381 is the same thing as writing 3x=81 , where you have to solve for x . 3 is the base number. 3⋅3⋅3⋅3 is 81 , proving 34=81 . So, 4 is the answer. ### What is the value of log 4 256? Logarithm base 4 of 256 is 4 . How do you find the value of log 4? The value of log 4 to the base 10 is 0.6020. In this article, we are going to discuss the value of log 4 in terms of both natural logarithm and common logarithm in the logarithmic function….Value of Log 4. Common logarithmic of 4 Log10 4 = 0.60206 Natural Logarithm of 4 ln 4 = 1.386294 Logarithm to the base 2 of 4 Log2 4 = 2 ## How do you solve log properties? You can use the similarity between the properties of exponents and logarithms to find the property for the logarithm of a quotient. With exponents, to multiply two numbers with the same base, you add the exponents. To divide two numbers with the same base, you subtract the exponents. ## How to evaluate log base 4 of 16? log4 (16) log 4 ( 16) Rewrite as an equation. log4(16) = x log 4 ( 16) = x. Rewrite log4 (16) = x log 4 ( 16) = x in exponential form using the definition of a logarithm. If x x and b b are positive real numbers and b b does not equal 1 1, then logb (x) = y log b ( x) = y is equivalent to by = x b y = x. 4x = 16 4 x = 16. Which is better a log reduction of 3 or 6? As a basic rule of thumb, for every additional Log reduction number you add a 9 to the percentage reduction – so a log reduction of 3, as illustrated above, is a 99.9% reduction compared with a log reduction of 6 which is equivalent to a 99.9999% reduction. ### How long does it take to read a log reduction? Reading time: less than 2 minutes. In terms of infection control, ‘Log Reductions’ convey how effective a product is at reducing pathogens. The greater the log reduction the more effective the product is at killing bacteria and other pathogens that can cause infections. ### What are the rules for expanding a log? 1) Multiplication inside the log can be turned into addition outside the log, and vice versa. 2) Division inside the log can be turned into subtraction outside the log, and vice versa. 3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa. Warning: Just as when you’re dealing with exponents,…
You are here: # Advanced Math/Width - Shaded Portion of Square / Solution Follow-Up Question QUESTION: Here is a link : https://archive.org/stream/intermediatealge00slaurich#page/n203/mode/2up Thanks again. ANSWER: I had some problems understanding this question, because "width" and "length" were used interchangeably and they omitted important details! the shaded areas within a 6" square are congruent the unshaded areas are square (only one of these facts is needed to derive the other) Just consider the right picture for now. The shaded areas are congruent rectangles and the white areas are squares. What are the dimensions of the shaded rectangles? Let x be the longer dimension of a shaded rectangle. Call it the length. Since the large rectangle is 6 inches on a side and the shaded rectangles are congruent, the width of a shaded rectangle must be (6-x)/2 inches. The area of a shaded rectangle is x(6-x)/2. The total shaded area is 4x(6-x)/2 = 2x(6-x) = 12x-2x². Since the total shaded area is 4/9 of the whole square, 12x-2x² = (4/9)36 = 16 in² 2x²-12x+16 = 0 By the quadratic formula, x = [12 ± √(12² – 4·2·16)] / [2·2] = [12 ± √16] / 4 = [12 ± 4] /4 = 4, 2 Note that there are two solutions for x. If x = 4, the rectangles are 4 inches by (6-4)/2 inches, or 4 inches by 1 inch. That is the picture on the right. If x = 2, the rectangles are 2 inches by (6-2)/2 inches, or 2 inches square. That is the picture on the left. The point of the question was to show that a quadratic equation can have two different solutions. (It doesn't always have two solutions, and sometimes one of the solutions is not valid in the context of the question.) ---------- FOLLOW-UP ---------- QUESTION: " the width of a shaded rectangle must be (6-x)/2 inches. The area of a shaded rectangle is x(6-x)/2. The total shaded area is 4x(6-x)/2 = 2x(6-x) = 12x-2x²." I am not clear on this part of your solution. Thanks again. Note: Sorry for the repost of the link. The shade rectangles are congruent, so the white rectangles in the corner are square and congruent. Consider the top row, which contains two white squares and a shaded rectangle. If the length of the rectangle is x, that leaves 6-x inches for both white squares. Since the white squares are congruent, each square is (6-x)/2 inches square. The width of the shaded rectangle is the same as the length of a white square, or (6-x)/2 inches. area of the shaded rectangle = length · width = x·(6-x)/2 in² There are four shaded rectangles, for a total of 4·x·(6-x)/2 = 2x(6-x) = 12x-2x² in². Volunteer #### Janet Yang ##### Expertise I can answer questions in Algebra, Basic Math, Calculus, Differential Equations, Geometry, Number Theory, and Word Problems. I would not feel comfortable answering questions in Probability and Statistics or Topology because I have not studied these in depth. ##### Experience I tutor students (fifth through twelfth grades) and am a Top Contributor on Yahoo!Answers with over 24,000 math solutions. Publications Co-author of An Outline of Scientific Writing: For Researchers With English as a Foreign Language Education/Credentials I have a Bachelor's degree in Applied Mathematics from the University of California at Berkeley. Past/Present Clients George White Elementary School. Homework Help program at the Ridgewood Public Library, Ridgewood, NJ. Individual students.
## Normal Approximation to the Binomial Some variables are continuous—there is no limit to the number of times you could divide their intervals into still smaller ones, although you may round them off for convenience. Examples include age, height, and cholesterol level. Other variables are discrete, or made of whole units with no values between them. Some discrete variables are the number of children in a family, the sizes of televisions available for purchase, or the number of medals awarded at the Olympic Games. A binomial variable can take only two values, often termed successes and failures. Examples include coin tosses that come up either heads or tails, manufactured parts that either continue working past a certain point or do not, and basketball tosses that either fall through the hoop or do not. You discovered that the outcomes of binomial trials have a frequency distribution, just as continuous variables do. The more binomial trials there are (for example, the more coins you toss simultaneously), the more closely the sampling distribution resembles a normal curve (see Figure 1). You can take advantage of this fact and use the table of standard normal probabilities (Table 2 in "Statistics Tables") to estimate the likelihood of obtaining a given proportion of successes. You can do this by converting the test proportion to a z‐score and looking up its probability in the standard normal table. Figure 1.As the number of trials increases, the binomial distribution approaches the normal distribution. The mean of the normal approximation to the binomial is μ = nπ and the standard deviation is where n is the number of trials and π is the probability of success. The approximation will be more accurate the larger the n and the closer the proportion of successes in the population to 0.5. Example 1 Assuming an equal chance of a new baby being a boy or a girl (that is, π = 0.5), what is the likelihood that more than 60 out of the next 100 births at a local hospital will be boys? According to Table , a z‐score of 2 corresponds to a probability of 0.9772. As you can see in Figure 2, there is a 0.9772 chance that there will be 60 percent or fewer boys, which means that the probability that there will be more than 60 percent boys is 1 – 0.9772 = 0.0228, or just over 2 percent. If the assumption that the chance of a new baby being a girl is the same as it being a boy is correct, the probability of obtaining 60 or fewer girls in the next 100 births is also 0.9772. Figure 2.Finding a probability using a z‐score on the normal curve.
# SAT Mathematics : Applying Number Properties ## Example Questions ### Example Question #1 : Applying Number Properties The first 3 prime numbers multiplied together equal: 0 1 30 6 30 Explanation: The correct answer is 30. A prime number is any number that is only divisible by 1 and itself. 0 and 1 are not prime numbers, so the first 3 prime numbers are 2, 3, and 5. ### Example Question #2 : Applying Number Properties Which of the following numbers is not divisible by 3: 111 412 243 354 412 Explanation: A common rule about numbers divisible by 3 is that the sum of their digits should be divisible by 3. 4+1+2 = 7 and 7 is not divisible by 3. Therefore, 412 is not divisible by 3. ### Example Question #3 : Applying Number Properties If and , which of the following must be true? Explanation: Since but there are no individual statements about each number on its own, we can’t conclude if a, b, c, or d are positive or negative. Recall that two negative numbers will give a positive product. What we can conclude, though, is that neither a, b, c, or d can equal 0 because then the product would equal 0. This means . A could be greater than 0 or could be less than 0 if either b, c, or d were negative. C could be less than 1, but C could also be a positive number if b or e are negative OR if the product of a, b, and d is positive. Similarly, d could be positive and would support the positive product of 10. ### Example Question #4 : Applying Number Properties Which of the following are rational numbers? I) II) III) II and III I and III I, II, and III I I and III Explanation: (pi) is an irrational number. Any square roots or fractions that result in non-terminating, non-repeating decimals fall under the category of irrational numbers. and are both terminating decimals and thus fall under the category of rational numbers. ### Example Question #5 : Applying Number Properties Which of the following are real numbers? I) II) III) I I, II, and III II and III I and II II and III Explanation: Complex numbers (any number in the form of ) are not real numbers. is irrational, but rational and irrational numbers both fall into the category of real numbers. is an integer, which is a subcategory of real numbers. ### Example Question #6 : Applying Number Properties Which of the following are whole numbers? I) II) III) II and III I I, II, and III I and II II and III Explanation: Negative rational numbers that don’t have decimals are integers. Whole numbers are a subcategory of integers that only include positive numbers and 0. At first glance, it seems that any square root would fall into irrational numbers, but this particular radical can be simplified to 2, making it a whole number. ### Example Question #7 : Applying Number Properties The correct prime factorization of 210 is? 2, 3, 5, 7 5, 6, 7 3, 7, 10 1, 2, 3, 5, 7 2, 3, 5, 7 Explanation: Recall that 1 is not a prime number. A prime number is any number greater than 1 that can only be divided by 1 and itself. All the above options are factors of 210, but neither 6 nor 10 are prime numbers (6 can be divided into 2 and 3 while 10 can be divided into 2 and 5). ### Example Question #8 : Applying Number Properties What is the least common multiple of 7, 6, and 10? 105 210 420 42 210 Explanation: The least common multiple (LCM) refers to the smallest multiple for each number that is shared by all the numbers of a given set, so in this case, it is the smallest multiple of 7, 6, and 10. The best way to approach these problems is to factorize each number and calculate the LCM from the frequency of these numbers. 7 is a prime number, 6 can be factorized into 2 and 3, and 10 can be factorized into 2 and 5. Now, we multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs. LCM = 7 * 2 * 3 * 5 = 210. We do not count 2 twice. ### Example Question #9 : Applying Number Properties What is the least common multiple of 30 and 45? 450 270 1350 90 90 Explanation: The least common multiple (LCM) refers to the smallest multiple for each number that is shared by all the numbers of a given set, so in this case, it is the smallest multiple of 30 and 45. The best way to approach these problems is to factorize each number and calculate the LCM from the frequency of these numbers. 30 can be factorized into 2, 3 and 5. 45 can be factorized into 3, 3, and 5. Now, we multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers (like it does in 45), you multiply the factor the greatest number of times it occurs. LCM = 2 * 3 * 3 * 5 = 90. ### Example Question #10 : Applying Number Properties What is the greatest common factor of 15 and 75? 3 5 15 30
# How do you convert vertex form to factored form y = 3(x+7)^2 - 2? May 1, 2015 Expand the vertex form into standard quadratic form; then use the quadratic root formula to determine the roots. $y = 3 {\left(x + 7\right)}^{2} - 2$ $= 3 \left({x}^{2} + 14 x + 49\right) - 2$ $= 3 {x}^{2} + 42 x + 145$ Using the formula for determining roots (and a very sharp pencil) $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ gives roots at $x = - 7 + \frac{\sqrt{6}}{3}$ and $x = - 7 - \frac{\sqrt{6}}{3}$ So $x + 7 - \frac{\sqrt{6}}{3}$ and $x + 7 + \frac{\sqrt{6}}{3}$ are factors of the original equation Fully factored form $y = 3 {\left(x + 7\right)}^{2} - 2$ $= 3 \left(x + 7 - \frac{\sqrt{6}}{3}\right) \left(x + 7 + \frac{\sqrt{6}}{3}\right)$
# How do you write the equation of the line through (4,-8) and (8,5)? Feb 20, 2017 $13 x - 4 y = 84$ #### Explanation: Step 1: Determine the slope of the line through the given points "slope"=("change in "y)/("change in "x) Given the points, $\left({x}_{1} , {y}_{1}\right) = \left(4 , - 8\right)$ and $\left({x}_{2} , {y}_{2}\right) = \left(8 , 5\right)$, the slope, $m$ is given by: $\textcolor{w h i t e}{\text{XXX}} m = \frac{5 - \left(- 8\right)}{8 - 4} = \frac{13}{4}$ Step 2: Write the equation in slope-point form Given a slope $m$ and a point $\left({x}_{1} , {y}_{1}\right)$, the slope-point form of the equation is: $\textcolor{w h i t e}{\text{XXX}} y - {y}_{1} = m \left(x - {x}_{1}\right)$ In this case we have: color(white)("XXX")m=13/4" and "(x_1,y_1)=(4,-8) giving $\textcolor{w h i t e}{\text{XXX}} y - \left(- 8\right) = \frac{13}{4} \left(x - 4\right)$ Step 3: Convert into standard form Standard form of a linear equation is $\textcolor{w h i t e}{\text{XXX}} A x + B y = C$ Starting from the slope-point form (above) we have: $\textcolor{w h i t e}{\text{XXX}} 4 \left(y + 8\right) = 13 \left(x - 4\right)$ $\textcolor{w h i t e}{\text{XXX}} 4 y + 32 = 13 x - 52$ $\textcolor{w h i t e}{\text{XXX}} 13 x - 4 y = 84$ For verification purposes, here is a graph with the given points and the equation $13 x - 4 y = 84$ Feb 20, 2017 $y = \frac{13}{4} x - 21$ #### Explanation: $\textcolor{b l u e}{\text{Preamble}}$ consider the standardised equation format of: $y = m x + c$ where $m$ is the gradient. $m = \left(\text{changing in up or down")/("change in along}\right)$ Very important$\to$the change in along (usually the x-axis) is read left to right. Watch out for this because sometimes questions give the points in the revers order. So the first point is at say ${x}_{1}$ and the second point is at say ${x}_{2}$. Then ${x}_{1} < {x}_{2}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Answering the question}}$ $\textcolor{b r o w n}{\text{Determine the gradient}}$ Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(4 , - 8\right)$ Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(8 , 5\right)$ $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{5 - \left(- 8\right)}{8 - 4} = \frac{5 + 8}{8 - 4} = \frac{13}{4}$ So we now have:$\text{ } y = \frac{13}{4} x + c$ $\textcolor{b r o w n}{\text{Determine the value of the constant } c}$ I chose ${P}_{1} \to$using the value for this point substitute for $x \mathmr{and} y$ $y = \frac{13}{4} x + c \text{ "->" } - 8 = \frac{13}{4} \left(4\right) + c$ $c = - 8 - 13 = - 21$ So we now have:$\text{ } y = \frac{13}{4} x - 21$
Courses Courses for Kids Free study material Offline Centres More Store # In $\Delta$PQR, Right-angled at Q. PR+QR=$25$cm and PQ=$5$cm. Determine the values of ${\text{sin P, cos P, tan P}}{\text{.}}$ Last updated date: 12th Sep 2024 Total views: 429.6k Views today: 4.29k Answer Verified 429.6k+ views Hint: In this type of question, first we need to assume a variable for unknown. Then we can easily solve them by using the Pythagoras theorem. Complete step-by-step answer: It is given that PR+ QR= $25$cm, PQ= $5$cm In this question we need to find the values of ${\text{sin P, cos P}}$ and${\text{tan P}}$. Let us assume that PR be $x$ We Know that, PR+ QR=$25$cm Substituting $x$ for PR $\Rightarrow$ $x +$ QR $= 25$ From the given data we can written as, QR $= 25 - x$ Now we know that PR ${\text{ = }}$ $x$ QR $= 25 - x$ PQ ${\text{ = }}$ ${\text{5}}$ In Right-angled triangle PQR, Here we using the Pythagoras theorem, $P{R^2} = P{Q^2} + Q{R^2}$ Here, we are substituting the above values, ${x^2} = {5^2} + {\left( {25 - x} \right)^2}...\left( 1 \right)$ Now, we are expanding the above expansion using the formula. i.e. ${(a - b)^2} = {a^2} - 2ab + {b^2}$ Here $a = 25$and ${\text{ }}b = x\;$ ${\left( {25 - x} \right)^2} = 625 + {x^2} - 50x$ Now we can write it as, and squaring first term in the LHS in $\left( 1 \right)$ ,we get ${x^2} = 25 + 625 + {x^2} - 50x$ On cancelling the same term and add the two terms as RHS we get that, $50x = 650$ On dividing $50$ on both side we get,, $x = \dfrac{{650}}{{50}}$ On dividing we get, $x = 13$ $\therefore PR = 13cm,$ Now we got the value of PR that is the value of the $x$. Also we are going to substitute the value that we got, $QR = 25 - x$ $QR = 25 - 13$ By subtracting we have that, $QR = 12cm$ Here According to the question, We need to find ${\text{sin P, cos P}}$and${\text{tan P}}$. For that we are going to substitute the values we get on the Formulas. Sin P $= \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}$ $= \dfrac{{QR}}{{PR}} = \dfrac{{12}}{{13}}$ Cos P $= \dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}$ $= \dfrac{{PQ}}{{PR}} = \dfrac{5}{{13}}$ tan P $= \dfrac{{{\text{Opposite}}}}{{{\text{Adjacent}}}}$ $= \dfrac{{QR}}{{PQ}} = \dfrac{{12}}{5}$ Finally, here the answers we got are, Sin P $= \dfrac{{12}}{{13}}$ Cos P $= \dfrac{5}{{13}}$ tan P $= \dfrac{{12}}{5}$ Note: According to Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides in a right-angled triangle. Mostly In this type of questions like Right-Angled triangle will be interrelated with Pythagoras theorem. You can solve them easily using this theorem.
# Is f(x)=(x^2-2)/(x-1) increasing or decreasing at x=2? ##### 1 Answer Jun 9, 2017 The function is increasing #### Explanation: We calculate the first derivative of $f \left(x\right)$, plug in the value $x = 2$ and look at the sign. $f \left(x\right) = \frac{{x}^{2} - 2}{x - 1}$ $u = \left({x}^{2} - 2\right)$, $\implies$, $u ' = 2 x$ $v = \left(x - 1\right)$, $\implies$, $v ' = 1$ $f ' \left(x\right) = \frac{u ' v - u v '}{{v}^{2}}$ $= \frac{2 x \left(x - 1\right) - 1 \left({x}^{2} - 2\right)}{x - 1} ^ 2$ $= \frac{2 {x}^{2} - 2 x - {x}^{2} + 2}{x - 1} ^ 2$ $= \frac{{x}^{2} - 2 x + 2}{x - 1} ^ 2$ Therefore, $f ' \left(2\right) = \frac{4 - 4 + 2}{1} = 2$ As $f ' \left(2\right) > 0$, so the function is increasing
Courses Courses for Kids Free study material Offline Centres More Store # Calculate the angular velocity and linear velocity of a tip of minute hand length 10cm. Last updated date: 06th Sep 2024 Total views: 424.2k Views today: 12.24k Verified 424.2k+ views Hints:There is a relation between angular velocity and linear velocity. Once we find angular velocity then by multiplying it by radius, linear velocity. Formula Used: $\omega = \dfrac{{d\theta }}{{dt}}$, where $\omega =$ Angular velocity $d\theta$ is the angular displacement covered in $dt$ time. $V = \omega \times r$, here $V$ is the linear velocity and $r$ is the radius. Complete step by step answer: In this question, for getting the angular velocity, we should know that if a minute hand completes one is 3600 seconds and one rotation covers $360^\circ$ equivalent to $2\pi$ radian (any circular distance must be expressed in radian). Now, if we come to the formula of angular velocity, is angular displacement divided by change in time $dt$ . So, we have considered that here angular displacement is $2\pi \left( {radian} \right)$ and time in which a minute hand completes its required distance is 3600 seconds. $\omega = \dfrac{{2\pi \left( {rad} \right)}}{{3600\left( {\sec } \right)}} = 1.745 \times {10^{ - 3}}rad/\sec$ Now, if we want to find linear velocity, we only need to know radius. Here, the length $\left( l \right)$ of minute hand represent radius $\left( r \right)$ $l = r = 10cm = 0.1m$ $V = \omega \times r = 1.745 \times {10^{ - 3}} \times 0.1$ $= 1.745 \times {10^{ - 4}}m/s$ Note: In these types of questions, we must keep one thing in our mind that the unit must be in meters and the unit of angular velocity must be in radian/second.
## Linear Equations during Two Factors Expires in 7 months 03 February 2022 Views: 35 Even as we saw in the article "Why Study Math? - Sequential Equations and Slope-Intercept Type, " sequential equations or perhaps functions are a couple of the more simple ones examined in algebra and standard mathematics. In this article we are going to examine and look at another common way of writing linear equations: the point-slope form. As your name seems to indicate, the point-slope form for the situation of a line depends on two things: the slope, and certain point on the line. Once we understand these two issues, we can write the equation from the line. On mathematical terms, the point-slope form of the equation of the line which inturn passes in the given issue (x1, y1) with a slope of m, is gym - y1 = m(x - x1). (The one particular after the back button and gym is actually a subscript which allows you to distinguish x1 from populace and y1 from gym. ) Showing how this kind is used, look into the following model: Suppose we still have a lines which has slope 3 and passes throughout the point (1, 2). https://theeducationjourney.com/slope-intercept-form/ could graph this line by just locating the place (1, 2) and then operate the slope of three to go 3 units up and then you unit towards the right. To create the formula of the lines, we make use of a clever little device. We all introduce the variables a and con as a issue (x, y). In the point-slope form ymca - y1 = m(x - x1), we have (1, 2) like the point (x1, y1). All of us then generate y - 2 = 3(x supports 1). By using the distributive real estate on the right hand side of the situation, we can write y - 2 sama dengan 3x -- 3. By simply bringing the -2 over to the proper side, we could write gym = 3x -1. Assuming you have not previously recognized the idea, this latter equation is within slope-intercept variety. To see the best way this form of the equation of any line is utilized in a real-world application, take following model, the information of which was taken from an article the fact that appeared within a newspaper. It turns out that heat range affects operating speed. In fact , the best heat for working is under 60 deg Fahrenheit. When a person went optimally in 17. 6th feet every second, he or she would halt by about 0. 3 toes per second of all for every some degree increased temperature previously 60 diplomas. We can use this information to create the geradlinig model because of this situation and next calculate, let us say, the optimal running stride at eighty degrees. Allow T legally represent the temperatures in diplomas Fahrenheit. Let P characterize the optimal tempo in toes per second of all. From the data in the story, we know that the optimal running tempo at 62 degrees is normally 17. 6 feet every second. As a result one stage is (60, 17. 6). Let's utilize the other information to look for the slope of this line due to this model. The slope meters is add up to the difference in pace over the change in temps, or m = enhancements made on P/change during T. We could told which the pace decreases by zero. 3 legs per secondary for every increase in 5 certifications above 70. A cut down is displayed by a bad. Using this tips we can compute the incline at -0. 3/5 or perhaps -0. summer. Now that we now have a point and the slope, we are able to write the model which presents this situation. We are P supports P1 sama dengan m(T - T1) or maybe P -- 17. 6th = -0. 06(T -- 60). Using the distributive home we can put this picture into slope-intercept form. We have P = -0. 06T + 21 years old. 2 . To discover the optimal pace at 50 degrees, we want only substitute 80 pertaining to T from the given style to acquire 16. some. Situations like these show that math is basically used to fix problems that result from the world. If we are referring to optimal working pace as well as maximal revenue, math is the vital thing to area code our possibility toward understanding the world available us. And when we understand, we are moved. What a great way to exist! Here's my website: https://theeducationjourney.com/slope-intercept-form/ Disable Third Party Ads Share
## How do you determine a system of equations? To check a system of equations by substitution, you plug your values for x and y into the original equations. If both simplified expressions are true then your answer is correct. Since x=2 and y=−3 worked for both equations I know that (2,−3) is the solution to this system of equations. ## What is the answer to a system of equations? Independent Equations In this case the two equations describe lines that intersect at one particular point. Clearly this point is on both lines, and therefore its coordinates (x, y) will satisfy the equation of either line. Thus the pair (x, y) is the one and only solution to the system of equations. ## What does it mean to be a solution to a system of equations? If you have a system of equations that contains two equations with the same two unknown variables, then the solution to that system is the ordered pair that makes both equations true at the same time. Follow along as this tutorial uses an example to explain the solution to a system of equations! ## What makes a system of equations linear? A Linear Equation is an equation for a line. Or like y + 0.5x − 3.5 = 0 and more. A System of Linear Equations is when we have two or more linear equations working together. ## What does LS and RS mean in math? = -2(2) + 7. = -4 + 7. = 3. ☺ L.S. = R.S. ☺ This means that the point (x, y) = (2, 3) is on the line = −2 + 7. ## What are the 3 types of system of equations? There are three types of systems of linear equations in two variables, and three types of solutions.An independent system has exactly one solution pair. The point where the two lines intersect is the only solution.An inconsistent system has no solution. A dependent system has infinitely many solutions. You might be interested: Equation to convert f to c ## How do you satisfy a system of equations? There are three ways to solve a system of linear equations: graphing, substitution, and elimination. The solution to a system of linear equations is the ordered pair (or pairs) that satisfies all equations in the system. The solution is the ordered pair(s) common to all lines in the system when the lines are graphed. ## What are the 3 methods for solving systems of equations? There are three ways to solve systems of linear equations in two variables: graphing. substitution method. elimination method. ## Why do we need system of equations? Systems of equations are used to solve applications when there is more than one unknown and there is enough information to set up equations in those unknowns. In general, if there are n unknowns, we need enough information to set up n equations in those unknowns. ## How do I solve an equation by graphing? A system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. The solution of such a system is the ordered pair that is a solution to both equations. To solve a system of linear equations graphically we graph both equations in the same coordinate system. ## How do you know if two equations have no solution? Summary. If the equation ends with a false statement (ex: 0=3) then you know that there’s no solution. If the equation ends with a true statement (ex: 2=2) then you know that there’s infinitely many solutions or all real numbers. ## What is unique solution in linear equation? A nxn homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions. ## What is a system of linear equations in two variables? A linear system of two equations with two variables is any system that can be written in the form. A solution to a system of equations is a value of x and a value of y that, when substituted into the equations, satisfies both equations at the same time. For the example above x=2 and y=−1 is a solution to the system. ### Releated #### Equation of vertical line How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […] #### Bernoulli’s equation example What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […]
# Eaton's Elementary Algebra: Designed for the Use of High Schools and Academies Thompson, Brown, 1868 - 252 σελίδες ### Τι λένε οι χρήστες -Σύνταξη κριτικής Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες. ### Περιεχόμενα The Signs 7 Axioms 9 ADDITION 19 SUBTRACTION 25 SECTION VI 31 SECTION VII 37 SECTION VIII 43 SECTION X 54 SECTION XII 63 SECTION XV 118 Powers and Roots 125 SECTION XVII 154 To add Radicals 160 SECTION XVIII 170 SECTION XIX 178 Completing the Square 178 Third Method of Completing the Square 185 Second Method of Completing the Square 182 Problems 193 ### Δημοφιλή αποσπάσματα Σελίδα 65 - To reduce a mixed number to an improper fraction, — RULE : Multiply the whole number by the denominator of the fraction, to the product add the numerator, and write the result over the denominator. Σελίδα 40 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Σελίδα 44 - The square of the sum of two quantities is equal to the SQuare of the first, plus twice the product of the first by the second, plus the square of the second. Σελίδα 94 - D. The distance from A to D is 34 miles ; the distance from A to B is to the distance from C to D as 2 to 3 ; and £ of the distance from A to B, added to one half the distance from C to D, is three times the distance from .B to C. Σελίδα 44 - ... the square of the second. In the second case, we have (a — &)2 = a2 — 2 ab + b2. (2) That is, the square of the difference of two quantities is equal to the square of the first, minus twice the product of the two, plus the square of the second. Σελίδα 81 - RATIO is the relation of one quantity to another of the same kind; or, it is the quotient which arises from dividing one quantity by another of the same kind. Σελίδα 202 - Hence -,- = -76" dn that is a" : b" = c" : dn THEOREM IX. 23 1 If any number of quantities are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let a : b = c : d... Σελίδα 81 - Thus, the ratio of a to b is written, a : b, or v ; read, a is to b, or a divided by b. 105. PROPORTION is an equality of ratios. Four quantities are proportional when the ratio of the first to the second is equal to the ratio of the third to the fourth. Σελίδα 72 - Now .} of f- is a compound fraction, whose value is found by multiplying the numerators together for a new numerator, and the denominators for a new denominator. Σελίδα 213 - The sum of three numbers in arithmetical progression is 15, and the sum of their squares is 83. What are the numbers...
### When one rate is linked to another ... Well, let's use calculus to solve some problems, shall we? There is a class of problems in one-variable called related rates problems. As the name suggests, the rate of one thing is related through some function to the rate of change of another. We can take advantage of that relationship and the fact that calculus is the mathematics of change to solve a whole bunch of new problems. The general approach to a related-rates problem will be to identify the two things that are changing, to find some sort of relationship between them – often it's geometric, to equate them and then take the derivative (implicit) with respect to time of both sides. The best way to learn related-rates problems is by doing examples, so here are a few ### Example 1 This is a classic problem in calculus. What happens at the top and bottom of a ladder that is sliding down the wall (and thus away from the wall at the bottom) is a little surprising: While the bottom of the ladder travels at a constant speed, the top accelerates downward. Weird. The problem: A 6-meter ladder leans against a wall. The bottom of the ladder is 2 m from the wall at time t=0, and it slides away from the wall at the rate of 0.5 m/s. Find the downward speed of the top of the ladder at any time, t, and at t = 1 second. Notice that while the bottom of the ladder traces out constant intervals, the intervals against the wall increase between each equal time step. Here's the basic geometry: #### Solution: We are given that , where t = time. We can relate x and h through the Pythagorean triangle (left): Now we take the derivative of each side with respect to time: Note that the derivative of the square of the ladder height is zero so height of the ladder doesn't matter in this equation. Now rearrange to isolate dh/dt: Finally, substitute for dx/dt to get the final related-rates equation: Now to apply our equation at time t = 1s, we have to know x and h at t = 1. The bottom moves at 0.5 ms-1, so after 1s it is 0.5 meter away. Now when the bottom is 1m away, the Pythagorean relationship tells us that h is √(62 - 2.52) = 5.45 m, and the rate of drop of the top of the ladder is (-0.5)(2.5/5.45) = -0.23 ms-1. The negative sign is a result of our coordinate system, in which height was measured upward from the ground. The graph of the solution below shows the velocity of the top of the ladder as a function of the distance of the base from the wall. The curve in the graph implies acceleration. ### Example 2 #### The conical tank The problem: A conical tank with the dimensions shown ( → ) is filled with liquid at a rate of 1.5 m3·min-1. At what rate is the water level rising when it passes a height of 5 meters? Sketch a graph of the rate as a function of time. What we know and don't know: This problem is typical of a great many others in two ways. First, two rates are related (duh). In this case the rate of filling (or draining) is proportional in some way to the change in height of the liquid, measured as h, as shown. Second, while at first glance it's a two-variable problem [V = f(r, h)], those two variables are related by an equation, thus one variable can be eliminated to reduce the problem to a single variable one. #### Solution: The ratio of the radius of the cone to its height is the same at any height, by similar triangles. We use that relationship (first equation → ) to eliminate the variable r from our problem. We could also use it to eliminate h, but we're interested in height of the liquid as a function of time. Here we go. Then we take the time derivative of both sides, relating the rate of volume change to the rate of change of the height. Implicit differentiation and the chain rule leads us to an expression for f'(h) that can be used to find the rate of change of height at any time. At a height of 5 meters, the rate is The ratio of radius to height of the water cone is equivalent to that ratio of the whole cone, which we know: Plugging our new expression for the radius into the volume formula gives Now we take the derivative of both sides with respect to time: We know dV/dt, so that reduces to: where we've taken the implicit derivative with respect to time on the right. Then we solve for dh/dt: and the result is: The graphs below show the height of liquid as a function of time for this problem, and its derivative with respect to time. Note that each is drawn on its own set of axes. Even though it's often done, it is incorrect to superimpose the curves on the same graph; they have different y-axes. Make sure you understand the relationship between the height vs. time curve and its derivative. ### Example 3 #### Tracking a rocket Let's say you track the progress of a rising rocket using a clinometer, an instrument for measuring the angle between the line of sight (to the nose of the rocket here) and the level ground. The angle is labeled θ in the picture → . The problem: At a given moment, the angle between the nose of the rocket and the ground is 39˚ and is changing at a rate of 25˚·min-1. Calculate the velocity of the rocket at that instant. Here's what we know and don't know: #### Solution: Here's a picture of the problem and its geometry. We will assume that the rocket takes off at a right angle to the ground. The first trick is to find a relationship between θ and y, and we find it in trigonometry: tan(θ) = y/5. From there we differentiate both sides, plug in what we know and solve for the vertical velocity, dy/dt. Then it's just a matter of calculating that velocity when θ = 39˚: By now these related rates problems should show a familiar pattern. I think they're fun because there's a definite method, but we still need to find our way through the specific algebra of each kind of problem. The key relationship between y and θ is Taking the derivative of both sides with respect to time, gives us: Now we solve for dy/dt to get: Finally, plugging in our target angle of 39˚ gives us the answer we were seeking: ### Practice problems 1. A man 6 ft. tall is walking at 3 ft./s toward a street light hung 15 ft. above the ground. How fast is the length (s) of his shadow changing? How fast is the tip of his shadow moving? (Hint: similar triangles) 1. Corn kernels are being dumped from a funnel, forming a conical pile, the height of which is always 1/3 the diameter of its base. If the corn falls onto the pile at a rate of 2 m3·min-1, how fast is the pile rising when it is 1 m high? 1. A television camera is positioned 5 m from the edge of a track, as shown, in order to follow runners as they pass and approach the finish line. If the runner is moving at 5 m/s as he passes the finish line 10 meters away (where the lane numbers are), how fast (in degrees per second) must the camera be rotating just as it pans across the line? Not available yet – Stay tuned. xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.
###### Norm Prokup Cornell University PhD. in Mathematics Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule. ##### Thank you for watching the video. To unlock all 5,300 videos, start your free trial. # Definite Integrals As Net Change - Problem 2 Norm Prokup ###### Norm Prokup Cornell University PhD. in Mathematics Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule. Share The definite integral can be used to solve for net change, when given a function for the rate of change. Remember that the rate of change will be some quantity over a period of time. For example, if given a rate of change function F'(t) representing velocity, or distance traveled per second, of a skydiver, integrate this function to find F(t) from a to b. Since F'(t) is height fallen per second, F(t) is simply the height of the skydiver. Therefore, F(b) - F(a) is the net change in height during time b and time a. Let's do another problem that involves the net change theorem. Recall that net change theorem says; that the integral from a to b of F'(x) equals F(b) minus F(a). This is a net change, and we can calculate the net change by integrating the rate of change. So let's see how that works in this problem. A skydiver jumps out of an airplane 5000 feet above ground. Her velocity, t seconds later, is given by v(t) equals 160e to the -0.2t minus 160. This is in feet per second. She opens her parachute after 30 seconds. What is her height above ground at that time? Well, so this is a net change theorem problem, because, we want to know what her final height is. We're given information about the rate of change of that height. Recall that; velocity, v(t) is the rate of change of position; so s'(t). Of course the function that we're given is 16e to the -0.2t minus 160 Now what do we know about position? Well, we know that the initial height s(0) is 5000. We're asked to find the height after 30 seconds. So that's s(30). Now we can relate s(30) to a net change. The net change, and the height from 0 to 30. So it's really important when you're using the net change theorem, that you're given sum value of the height or whatever quantity you're working with. One of these values is known so that you can find the other. So this one is 5000, and this net change, by the net change theorem is going to equal the integral from 0 to 30 of the derivative of this function; s'(t). Of course that's our function up here. So let fill that in. 0 to 30; the integral of 160e to the -0.2t minus 160 dt. S(0) is 5000. I'm looking for s(30). So let's evaluate this integral. I'm going to need an antiderivative, The antiderivative for 160e to the -0.2t well, I remember that when I differentiate e to the -0.2t, I get -0.2e to the -0.2t. When I integrate it, I get e to the -0.2t divided by this constant. So my antiderivative is 160 divided by -0.2e to the -0.2t. For -160, the antiderivative I need is -160t. So I'm going to take this from 0 to 30. By the way what is this number; 160 over 0.2? Let me do a little calculation here. 160 over -0.2 is the same as 1600 over -2. You multiply the top, and bottom by 10, this is -800. So this is -800 times e to the negative. Then if I plug in 30 for t, 30 times 0.2 is 6. So it's -6, minus 160 times 30. So this is what I get when I plug 30 into my antiderivative. What do I get when I plug in 0? Well, when I plug 0 into this guy, I will get 0, but when I plug 0 in here remember that e to the 0 is 1, so I'm going to get my -800. So what does this give me? 16 times 3 is 48 so this is going to be 4800, and this is -800, I'll evaluate this in a second on my calculator, minus 4800 minus, minus plus 800. So that's -400e, -800e to the -6. Now I'll erase this. Let me just copy down. This is s(30) minus 5000. So if I add 5000 to both sides, I'll get 1000 minus 800 e to the -6. That will give me my s(30). I'm going to write that over here. S(30) equals 1000 minus 800 e to the -6. So I'll need a calculator for this. 1000 minus 800e to the -6. It's approximately 998. This will be in feet. So let's figure out what just happened. Going back to the beginning here, I knew that I can use the net change theorem to find the change in the height; s(30) minus s(0). To do that, I have to integrate from 0 to 30. Now that change in height, the integral gave me -4000 minus 800e to the -6. So whatever value this is, this is how much the height changed from t equals 0 to t equals 30. It makes sense that this number should be negative, because her height is actually decreasing. Now, s(30) minus 500 equals that amount. So when I add 5000, I get this quantity. This is her actual height at 30 seconds; 998 feet; her final height when the parachute opens.
# Series - survival kit ## Definition of series Given a sequence $$u_n$$, you can play with it and create a new sequence $$S_n$$ defined by $$S_1=u_1$$ $$S_2=u_1+u_2$$ $$S_3=u_1+u_2+u_3$$ More generally $$S_n=u_1+u_2+\cdots+u_n=\sum_{k=1}^{n}u_k$$ $$S_n$$ is called the sequence of partial sums. It is a sequence and any results about sequences can be used for $$S_n$$. The series $$\sum u_n$$ or $$\sum_{k=0}^{\infty} u_n$$ is the limit of $$S_n$$ when $$n$$ goes to $$\infty$$. Basically a series is a real number, the value of the limit when $$S_n$$ is convergent. We say the the series is convergent when the sequence $$S_n$$ is convergent, has a finite limit. In this case the series is equal to the value of the limit. The series is divergent when the limit of $$S_n$$ is infinite or doesn't exist. ## Classic series • Geometric Series A geometric series with ratio $$r$$ is $$\sum_{k=0}^{\infty}r^k$$. We can prove that its sequence of partial sum $$S_n=\sum_{k=0}^{\infty} r^k=\frac{1-r^{n+1}}{1-r}$$ for $$r\neq 1$$ and $$S_n=n+1$$ for $$r=1$$. Multiply $$S_n$$ by $$(1-r)$$, after simplifications, you will find that $$S_n(1-r)=1-r^{n+1}$$ By taking the limit as $$n$$ goes to $$\infty$$, we can conclude that Theorem: The geometric series of ratio $$r$$ is convergent for $$r\in(-1,1)$$ and the series $$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}.$$ The geometric series is divergent for $$r\notin(-1,1)$$. • p-series A p-series is a series in the form $$\sum_{n=1}^{\infty}\frac{1}{n^p}$$ P-test: the p-series $$\sum_{n=1}^{\infty} \frac{1}{n^p}$$ is convergent for $$p>1$$ The p-series is divergent for $$p\leqslant 1$$. ## Convergence criteria Theorem: If a series $$\sum u_n$$ is convergent then $$\lim_{n\rightarrow\infty}u_n=0$$ Remark: The converse is FALSE. The series $$\sum\frac{1}{n}$$ is divergent and $$\lim_{n\rightarrow\infty}\frac{1}{n}=0$$. The contrapositive is TRUE and is probably the most used form of the theorem: if $$\lim_{n\rightarrow \infty}u_n\neq 0$$ then the series is divergent. Example: $$\lim_{n\rightarrow\infty}\frac{n^2+\sin n}{3n^2+\sqrt{n}}=\frac{1}{3}\neq 0$$ therefore the series $$\sum_{n=1}^{\infty}\frac{n^2+\sin n}{3n^2+\sqrt{n}}$$ is divergent. ### Series with nonnegative terms Theorem (Comparison test, Inequality version) Let $$\sum a_n$$ and $$\sum b_n$$ series with non negative terms such that for all the index greater than some $$N$$, $$0\leqslant a_n\leqslant b_n$$. • If $$\sum b_n$$ is convergent, then $$\sum a_n$$ is convergent. • If $$\sum a_n$$ is divergent, then $$\sum b_n$$ is divergent. Theorem (Comparison test, limit version) Let $$\sum a_n$$ and $$\sum b_n$$ series with positive terms such that $$\lim_{n\rightarrow\infty} \frac{a_n}{b_n}=c\neq 0$$, then either $$\sum a_n$$ and $$\sum b_n$$ are both convergent or $$\sum a_n$$ and $$\sum b_n$$ are both divergent. Theorem (Ratio Test) Let $$\sum a_n$$ and $$\sum b_n$$ series with positive terms such that $$\frac{a_{n+1}}{a_n}\leqslant \frac{b_{n+1}}{b_n}$$ for any $$n$$, then • If $$\sum b_n$$ is convergent, then $$\sum a_n$$ is convergent. • If $$\sum a_n$$ is divergent, then $$\sum b_n$$ is divergent. • Corollary (Comparison to a geometric series) If for any $$n$$, $$\frac{u_{n+1}}{u_n}<r<1$$, then $$\sum u_n$$ is convergent. Corollary: If $$\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=L$$ • if $$L>1$$, then the series with general term $$u_n$$ is divergent, • if $$L<1$$, the series with general terms $$u_n$$ is convergent. • if $$L=1$$, we cannot conclude. Theorem (comparison with an integral) Given a positive function $$f$$ that is non increasing on a interval $$[1,\infty)$$. Then the series $$\sum f(n)$$ and $$\int_1^{\infty} f(t)d t$$ are both convergent or they are both divergent. ### Absolute convergence and conditional convergence Definition Given a series $$\sum u_n$$, $$\sum u_n$$ is absolutely convergent if $$\sum \|u_n\|$$ is convergent. Theorem If $$\sum u_n$$ is absolutely convergent, then $$\sum u_n$$ is convergent. Definition If $$\sum u_n$$ is convergent and not absolutely convergent, then $$\sum u_n$$ is conditionally convergent. Theorem (Alternating series) A series $$\sum u_n$$ uch that its terms alternate between positive and negative, and such that $$\|u_n\|$$ is decreasing toward 0 is convergent.
### Home > APCALC > Chapter 3 > Lesson 3.2.2 > Problem3-64 3-64. Using the definition of the derivative as a limit, show that the derivative of $f(x) = \frac { 1 } { x ^ { 2 } }$ is $f^\prime (x) = -\frac { 2 } { x ^ { 3 } }$. That is, show algebraically that the following limit statement is true: $\lim\limits _ { h \rightarrow 0 } \frac { \frac { 1 } { ( x + h ) ^ { 2 } } - \frac { 1 } { x ^ { 2 } } } { h } = - \frac { 2 } { x ^ { 3 } }$ The Derivative The slope of a line tangent to at any point $x$ is called the derivative of $f$ at $x$. It is found by taking a limit of the slope of a secant line as $h \rightarrow 0$. The standard form of this type of limit is: $f^\prime (x) = \lim\limits _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ If the slope (or instantaneous rate of change) at a particular $x$-value is desired, such as at $x = a$, then the notation used is $f^\prime (a)$. This slope can be found by replacing $x$ with $a$. $f^\prime (a) = \lim\limits _ { h \rightarrow 0 } \frac { f ( a + h ) - f ( a ) } { h }$ This is one form of the 'definition of the derivative' (informally known as Hana's Method). In order to evaluate this limit, we need to find an Algebraic way to cancel out the $h$ in the denominator. Find a common denominator in the numerator, expand and combine like terms: $\lim\limits_{h\rightarrow 0}\frac{\frac{x^{2}-(x+h)^{2}}{(x^{2})(x+h)^{2}}}{h}=\lim\limits_{h\rightarrow 0}\frac{x^{2}-(x^{2}+2xh+h^{2})}{h(x^{2})(x+h)^{2}}=\lim\limits_{h\rightarrow 0}\frac{-2xh-h^{2}}{h(x^{2})(x+h)^{2}}$ Factor the numerator, then cancel out the $h$: $\lim\limits_{h\rightarrow 0}\frac{h(-2x-h)}{h(x^{2})(x+h)^{2}}=\lim\limits_{h\rightarrow 0}\frac{(-2x-h)}{(x^{2})(x+h)^{2}}$ Since there is no longer and $h$ in the denominator, you can evaluate the limit as $h \rightarrow 0$: $=\lim\limits_{h\rightarrow 0}\frac{(-2x-(0))}{(x^{2})(x+(0))^{2}}=\lim\limits_{h\rightarrow 0}\frac{(-2x)}{(x^{2})(x)^{2}}=-\frac{2}{x^{3}}$
# Calculate Work Done in a Given Time To calculate work done in a given time in general if a person A completes a piece of work in a ‘n’ days, then A’s one day work = 1/n. If a man takes 10 days to complete a piece of work, then according to the unitary method work done in one day = 1/10. Now we will apply the concept of solving some real-life problems to find the work done in a given period of time. Solved examples to calculate work done in a given time: 1. Adam can do a piece of work in 15 days and Brandon can finish it in 10 days. They work together for 5 days and then Adam goes away. In how many days will Brandon finish the remaining work? Solution: Adam can finish the work in 15 days. Therefore, Adam’s 1 day’s work = 1/15 Brandon can finish the work in 10 days. Therefore, Brandon’s 1 day’s work = 1/10 So, (Adam + Brandon)‘s 1 day’s work = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6 Therefore, (Adam + Brandon)‘s 5 day’s work = 5 × 1/6 = 5/6 Remaining work = 1 – 5/6 = 6-5/6 = 1/6 This remaining work has to be done by Brandon. Complete work is done by Brandon in 10 days. Therefore, 1/6 of the work is done by Brandon in (10 × 1/6) days = 5/3 days. Hence, the remaining work is done by Brandon in 5/3 days. 2. Alex can do 3/4 of the work in 12 days and Ben can do 1/8 th of the work in 2 days. In how many days both Alex and Ben can together do the work? Solution: 3/4 of the work is done by Alex is 12 days. Therefore, the whole work will be done Alex in 12 × 4/3days = 16 days. Therefore, Alex’s day’s work = 1/16 Similarly, 1/8 of the work is done by Ben in 2 days. Therefore, the whole work will be done Ben in 2 × 8/1 = 16 days So, Ben’s 1 day work = 1/16 Now (Alex + Ben)’s 1 day’s work = 1/16 + 1/16 = 2/16 = 1/8 Therefore, (Alex + Ben) together will complete the work in 8 days. More problems to calculate work done in a given time: 3. Nancy can embroider a dress in 15 hours while Sam can do the same in 10 hours. Working together how much time will they take to embroider 5 dresses? How many dresses will be embroidered in 54 hours? Solution: Nancy can embroider a dress in 15 hours. So, Nancy’s 1 hour work = 1/15 Also, Sam can embroider a dress in 10 hours. So, Sam’s 1 hour work 1/10 Nancy’s and Sam’s 1 hour work = 1/15 + 1/10 = 2+3/30 = 5/30 = 1/6 So, they will together take 6 hours to embroider 1 dress. So, working together, time taken to embroider 1 dress = 6 hours. Therefore, time taken to embroider 5 dresses = (6 × 5) hours = 30 hours. Also, number of dresses embroidered in 6 hour = 1 Number of dresses embroidered in 1 hour = 1/6 Therefore, number of dresses embroidered in 54 hours = 1/6 × 54 = 9 Calculate Time to Complete a Work Calculate Work Done in a Given Time Problems on Time required to Complete a Piece a Work Problems on Work Done in a Given Period of Time Problems on Time and Work Pipes and Water Tank Problems on Pipes and Water Tank Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Fraction in Lowest Terms \|Reducing Fractions\|Fraction in Simplest Form Feb 28, 24 04:07 PM There are two methods to reduce a given fraction to its simplest form, viz., H.C.F. Method and Prime Factorization Method. If numerator and denominator of a fraction have no common factor other than 1… 2. ### Equivalent Fractions \| Fractions \|Reduced to the Lowest Term \|Examples Feb 28, 24 01:43 PM The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with re… 3. ### Fraction as a Part of Collection \| Pictures of Fraction \| Fractional Feb 27, 24 02:43 PM How to find fraction as a part of collection? Let there be 14 rectangles forming a box or rectangle. Thus, it can be said that there is a collection of 14 rectangles, 2 rectangles in each row. If it i… 4. ### Fraction of a Whole Numbers \| Fractional Number \|Examples with Picture Feb 24, 24 04:11 PM Fraction of a whole numbers are explained here with 4 following examples. There are three shapes: (a) circle-shape (b) rectangle-shape and (c) square-shape. Each one is divided into 4 equal parts. One… 5. ### Identification of the Parts of a Fraction \| Fractional Numbers \| Parts Feb 24, 24 04:10 PM We will discuss here about the identification of the parts of a fraction. We know fraction means part of something. Fraction tells us, into how many parts a whole has been
# How do you graph y = 2 abs (x - 1 ) - 3? Jul 23, 2015 Follow the instructions in the explanation. #### Explanation: Make a table of $x$ and $y$ values, making sure to include positive and negative values for $x$, and use a wide range of values for $x$. $x = - 6 ,$ $y = 2 \left\mid - 6 - 1 \right\mid - 3 = 2 \left\mid - 7 \right\mid - 3 = 2 \left(7\right) - 3 = 11$ $x = - 5 ,$ $y = 2 \left\mid - 5 - 1 \right\mid - 3 = 2 \left\mid - 6 \right\mid - 3 = 2 \left(6\right) - 3 = 9$ $x = - 3 ,$ $y = 2 \left\mid - 3 - 1 \right\mid - 3 = 2 \left\mid - 4 \right\mid - 3 = 2 \left(4\right) - 3 = 5$ $x = - 1 ,$ $y = 2 \left\mid - 1 - 1 \right\mid - 3 = 2 \left\mid - 2 \right\mid - 3 = 2 \left(2\right) - 3 = 1$ $x = 0 ,$ $y = 2 \left\mid 0 - 1 \right\mid - 3 = 2 \left\mid - 1 \right\mid - 3 = 2 \left(1\right) - 3 = - 1$ $x = 1 ,$ $y = 2 \left\mid 1 - 1 \right\mid - 3 = 2 \left\mid 0 \right\mid - 3 = 2 \left(0\right) - 3 = - 3$ $x = 3 ,$ $y = 2 \left\mid 3 - 1 \right\mid - 3 = 2 \left\mid 2 \right\mid - 3 = 2 \left(2\right) - 3 = 1$ $x = 5 ,$ $y = 2 \left\mid 5 - 1 \right\mid - 3 = 2 \left\mid 4 \right\mid - 3 = 2 \left(4\right) - 3 = 5$ $x = 6 ,$ $y = 2 \left\mid 6 - 1 \right\mid - 3 = 2 \left\mid 5 \right\mid - 3 = 2 \left(5\right) - 3 = 7$ Table $\textcolor{red}{x} \| \textcolor{p u r p \le}{y}$ $\textcolor{red}{- 6} \| \textcolor{p u r p \le}{11}$ $\textcolor{red}{- 5} \| \textcolor{p u r p \le}{9}$ $\textcolor{red}{- 3} \| \textcolor{p u r p \le}{5}$ $\textcolor{red}{- 1} \| \textcolor{p u r p \le}{1}$ $\textcolor{red}{0} \| \textcolor{p u r p \le}{- 1}$ $\textcolor{red}{1} \| \textcolor{p u r p \le}{- 3}$ $\textcolor{red}{3} \| \textcolor{p u r p \le}{1}$ $\textcolor{red}{5} \| \textcolor{p u r p \le}{5}$ $\textcolor{red}{6} \| \textcolor{p u r p \le}{7}$ Plot the points, and draw a straight line through both sides, with each side meeting at $\left(1 , - 3\right)$. graph{y=2abs(x-1)-3 [-16.02, 16, -5.12, 10.9]}
# Class 8 Maths MCQ – Laws of Exponents This set of Class 8 Maths Chapter 12 Multiple Choice Questions & Answers (MCQs) focuses on “Laws of Exponents”. 1. Find the value of $$\frac{1}{8^{-2}}$$. a) $$\frac{1}{16}$$ b) $$\frac{1}{64}$$ c) 16 d) 64 Explanation: $$\frac{1}{a^{-m}}$$ = am ⇒ $$\frac{1}{8^{-2}}$$ = 82 = 64. 2. What is 2-3? a) $$\frac{1}{4}$$ b) $$\frac{1}{8}$$ c) 4 d) 8 Explanation: a-m = $$\frac{1}{a^m}$$ ⇒ 2-3 = $$\frac{1}{2^3}$$ = $$\frac{1}{8}$$. 3. Find the value of 16-2 × 1614. a) 1612 b) 16-12 c) 1614 d) 16-14 Explanation: am × an = am + n ⇒ 16-2 × 1614 = 16-2 + 14 = 1612. 4. Evaluate $$(\frac{21^{29}}{21^5})$$. a) 2125 b) 21-25 c) 2124 d) 21-24 Explanation: $$\frac{a^m}{a^n}$$ = am – n ⇒ $$(\frac{21^{29}}{21^5})$$ = 2129 – 5 = 2124. 5. Match the pairs. A B 1) 23 × 83 a. 1 2) $$\frac{8^3}{2^3}$$ b. 163 3) (23)0 c. 43 d. 8 Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now! a) 1-b, 2-c, 3-d b) 1-b, 2-c, 3-a c) 1-c, 2-b, 3-a d) 1-c, 2-b, 3-d Explanation: am × bm = (ab)m ⇒ 23 × 83 = (2 × 8)3 = 163 Now, $$\frac{a^m}{b^m} = (\frac{a}{b})^m$$ ⇒ ($$\frac{8}{2})^3$$ = 43 Also, (am)n = amn, a0 = 1 ⇒ (23)0 = 23 × 0 = 20 = 1. 6. Find the value of (23)2. a) 128 b) 16 c) 32 d) 64 Explanation: (am)n = amn ⇒ (23)2 = 26 = 64. Sanfoundry Global Education & Learning Series – Mathematics – Class 8. To practice all chapters and topics of class 8 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
# Let g be the function give by g(x) = x^4 - 4x^3 + 6x^2 4x + k where k is constant. A.On what... ## Question: Let g be the function give by {eq}g(x) = x^4 - 4x^3 + 6x^2 4x + k {/eq} where k is constant. C.Find the value of k for which g has 5 as its relative minimum. Justify your answer. ## Monotony and concavity: We are looking for the monotony of this function, so we will need to derive the function with respect to the variable 'x'. Then, we need the concavity, so we will need the second derivative. Note that the values where the derivative is possitive, the funciton increases, while the values where the derivative is negative, the funciton decreases. First of all, we assume that the function given is, {eq}g(x) = x^4 - 4x^3 + 6x^2 - 4x + k {/eq} The derivative is, {eq}g'(x) = 4x^3-12x^2+12x-4 {/eq} We are looking for the values of 'x' where the derivative is zero, {eq}4x^3-12x^2+12x-4 = 0 \Rightarrow x^3-3x^2+3x-1=0 \Rightarrow (x-1)^3 = 0 {/eq} So the only critical point is x=1. Let's see the sign of the derivative, {eq}g'(0) = -4 < 0 \\ g'(2) = 4 > 0 {/eq} Which means that, {eq}g(x) \text{ decreases in } (-\infty, 1) \text{ because the derivative is negative in that interval} \\ gx) \text{ increases in } (1, \infty) \text{ because the derivative is positive in that interval} {/eq} For the concavity we calculate the second derivative, {eq}g''(x) = 12x^2-24x+12 {/eq} The points when the concavity can change are when the second derivative is zero, which are, {eq}x^2-2x+1 = 0 \Rightarrow x=1 {/eq} So again we check the signs, {eq}g''(0)=12 >0 \\ g''(2) = -12 < 0 {/eq} So g is concave up when x<1 and concave down when x>1. The last step is to find k when g has 5 as relative minimum. That is not possible due to the fact that the only possible relative extreme is when x=1.
# Search by Topic #### Resources tagged with Positive-negative numbers similar to Crossed Ends: Filter by: Content type: Stage: Challenge level: ### There are 19 results Broad Topics > Numbers and the Number System > Positive-negative numbers ### Consecutive Negative Numbers ##### Stage: 3 Challenge Level: Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers? ### Weights ##### Stage: 3 Challenge Level: Different combinations of the weights available allow you to make different totals. Which totals can you make? ### Strange Bank Account ##### Stage: 3 Challenge Level: Imagine a very strange bank account where you are only allowed to do two things... ### First Connect Three for Two ##### Stage: 2 and 3 Challenge Level: First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. ### Consecutive Numbers ##### Stage: 2 and 3 Challenge Level: An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### Pair Sums ##### Stage: 3 Challenge Level: Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers? ### Strange Bank Account (part 2) ##### Stage: 3 Challenge Level: Investigate different ways of making £5 at Charlie's bank. ### Playing Connect Three ##### Stage: 3 Challenge Level: In this game the winner is the first to complete a row of three. Are some squares easier to land on than others? ### Connect Three ##### Stage: 3 and 4 Challenge Level: Can you be the first to complete a row of three? ### Up, Down, Flying Around ##### Stage: 3 Challenge Level: Play this game to learn about adding and subtracting positive and negative numbers ### Adding and Subtracting Positive and Negative Numbers ##### Stage: 3 How can we help students make sense of addition and subtraction of negative numbers? ### Making Sense of Positives and Negatives ##### Stage: 3 This article suggests some ways of making sense of calculations involving positive and negative numbers. ### Negative Numbers ##### Stage: 3 A brief history of negative numbers throughout the ages ### Negatively Triangular ##### Stage: 4 Challenge Level: How many intersections do you expect from four straight lines ? Which three lines enclose a triangle with negative co-ordinates for every point ? ### First Connect Three ##### Stage: 2 and 3 Challenge Level: The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for? ### Missing Multipliers ##### Stage: 3 Challenge Level: What is the smallest number of answers you need to reveal in order to work out the missing headers? ### Minus One Two Three ##### Stage: 4 Challenge Level: Substitute -1, -2 or -3, into an algebraic expression and you'll get three results. Is it possible to tell in advance which of those three will be the largest ? ### The History of Negative Numbers ##### Stage: 3, 4 and 5 This article -useful for teachers and learners - gives a short account of the history of negative numbers. ### Vector Racer ##### Stage: 3 and 4 Challenge Level: The classic vector racing game.
# Short Tricks on Number System Quantitative Aptitude deals mainly with the different topics in Arithmetic, which is the science which deals with the relations of numbers to one another. It includes all the methods that are applicable to numbers. Numbers are expressed by means of figures – 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0 —- called digits. Out of these, 0 is called insignificant digit whereas the others are called significant digits. We Recommend Testbook APP 100+ Free Mocks For RRB NTPC & Group D Exam Attempt Free Mock Test 100+ Free Mocks for IBPS & SBI Clerk Exam Attempt Free Mock Test 100+ Free Mocks for SSC CGL 2021 Exam Attempt Free Mock Test 100+ Free Mocks for Defence Police SI 2021 Exam Attempt Free Mock Test 100+ Free Mocks for UPSSSC 2021 Exam Attempt Free Mock Test ## Numbers A group of figures, representing a number, is called a numeral. Numbers are divided into the following types. Natural Numbers: Numbers which we use for counting the objects are known as natural numbers. They are denoted by ‘N’ N = {1,2,3,4,…….} Whole Numbers: When we include ‘zero’ in the natural numbers, it is known as whole numbers. They are denoted by ‘W’. W= {0,1,2,3,4,5,………} Prime Numbers: A number other than 1 id called a prime number if it is divisible only by 1 and itself. ### To test whether a given number is prime number or not If you want to test whether any number is a prime number or not, take an integer larger than the approximate square root of that number. Let it be ‘x’. test the divisibility of the given number by every prime number less than ‘x’. if it not divisible by any of them then it is prime number; otherwise it is a composite number (other than prime). Example: Is 349 a prime number? Solution: The square root of 349 is approximate 19. The prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17. Clearly, 349 is not divisible by any of them. Therefore, 349 is a prime number. Composite Numbers: A number, other than 1, which is not a prime number is called a composite number. e.g. 4, 6, 8, 9, 12, 14 …… and so on Even Number: The number which is divisible by 2 is known as an even number. e.g. 2, 4, 8, 12, 24, 28 …… and so on It is also of the form 2n {where n = whole number} Odd Number: The number which is not divisible by 2 is known as an odd number. e.g. 3, 9, 11, 17, 19 …… and so on Consecutive Number: A series of numbers in which each is greater than that which precedes it be 1 is called a series of consecutive numbers. e.g. 6, 7, 8 or 13, 14, 15, 16 or, 101, 102, 103, 104 Integers: The set of numbers which consists of whole numbers and negative numbers is known as a set of integers it is denoted by 1. e.g. I = {-4,-3,-2,-1,0,1,2,3,….} Rational Number: When the numbers are written in fraction, they are known as rational numbers. They are denoted by Q. e.g. are called rational numbers. Or, the numbers which can be written in the form {where a and b are integers and b 0} are called rational numbers. Irrational Numbers: The numbers which cannot be written in the form of p/q are known as irrational numbers (where p and q are integers and q 0). Real Numbers: Real numbers include both rational as well as irrational numbers. ## Rules of Simplification (i) In simplifying an expression, first of all vinculum or bar must be removed. For example: we known that – 8 – 10 = -18 But, = – (-2) = 2 (ii) After removing the bar, the brackets must be removed, strictly in the order (), {} and []. (iii) After removing the brackets, we must use the following operations strictly in the order given below. (a) of (b) division (c) multiplication (d) addition and (e) subtraction. Note: The rule is also known as the rule of ‘VBODMAS’ where V, B, O, D, M, A and S stand for Vinculum, Brackets, Of, Division, Multiplication, Addition and Subtraction respectively. Example: Simplify Solution: ## Ascending or Descending Order in Rational Numbers Rule 1: When the numerator and the denominator of the fractions increase by a constant value, the last fraction is the biggest. Example: Which of the following fractions is the greatest? Solution: We see that the numerators as well as denominators of the above fraction increase by 1, so the last fraction, i.e. is the greatest fraction. Rule 2: The fraction whose numerator after cross-multiplication given the greater value is greater. Example: Which is greater : Solution: Students generally solve this questions by changing the fractions into decimal values or by equating the denominators. But, we suggest you a better method for getting the answer more quickly. Step 1: Cross –multiply the two given fractions. We have, 5 × 14 = 70 and 8 ×9 =72 Step II. As 72 is grater than 70 and the numerator involved with the greater value is 9, the fraction is the greater of the two. Example: Which is greater: Solution: Step I: 4 ×23 > 15 ×6 Step II: As the greater value has the numerator 4 involved with it, is greater. You can see how quickly this method works. After good practice, you won’t need to calculate before answering the question. The arrangement of fractions into the ascending or descending order becomes easier now. Choose two fractions at a time. See which one is grater. This way you may get a quick arrangement of fractions. Note: Sometimes, when the values are smaller (i.e., less than 10), the conventional method, i.e., changing the values into decimals or equating the denominators after getting LCM, will prove more convenient for some of you. Example: Arrange the following in ascending order. Solution: Method I The LCM of 7,5,9,2,5, is 630. Now, to equate the denominators, we divide the LCM by the denominators and multiply the quotient by the respectively numerators. Like for , 630 ÷ 7 = 90, so, multiply 3 by 90. Thus, the fractions change to The fraction which has larger numerator is naturally larger. So, Method II: Change the fractions into decimals like = 0.428, = 0.8, = 0.777, = 0.5, = 0.6 Clearly, Method III: Rule of CM (cross-multiplication) Step I: Take the first two fractions. Find the greater one by the rule of CM. 3 × 5< 7×4 Step II: Take the third fraction. Apply CM with the third fraction and the larger value obtained is step I. 4 × 9 > 5 × 7 Now we see that can lie after or between and . Therefore, we apply CM with and see that . Step III: Take the next fraction. Apply CM with and and see that . Next, we apply CM with and and see that . Therefore, Step IV: With similar applications, we get the final result as: Note: This rule has some disadvantages also. But if you act fast, it gives faster results. Don’t reject this method at once. This can prove to be the better method for you. We hop you all will like the post. 1. Sum of all the first n natural numbers = For example: 1+ 2 +3 +…..+105= 2. Sum of first n odd numbers = For example: 1+3+5+7==16(as there are four odd numbers) 3. Sum of first n even numbers = n (n+1) For example : 2+4+6+8+….+100 (or 50th even number) = 50×(50+1)= 2550 4. Sum of squares of first n natural numbers = For example: 5. Sum of cubes of first n naturals numbers = For example : Example: (1) What is the total of all the even numbers from 1 to 400? Solution: From 1 to 400, there are 400 numbers. So, there are 400/2= 200 even numbers. Hence, sum = 200(200+1) = 40200 (From Rule III) (2) What is the total of all the even numbers from 1 to 361? Solution: From 1 to 361, there are 361, there are 361 numbers; so there are even numbers. Thus, sum = 180(180+1)=32580 (3) What is the total of all the odd numbers from 1 to 180? Solution: Therefore are 180/2 = 90 odd numbers between the given range. So, the sum = (4) What is the total of all the odd numbers from 1 to 51? Solution There are odd numbers between the given range. So, the sum = (5) Find the of all the odd numbers from 20 to 101. Solution: The required sum = Sum of all the odd numbers from 1 to 101. Sum of all the odd numbers from 1 to 20 = Sum of first 51 odd numbers – Sum of first 10 odd numbers = Miscellaneous 1. In a division sum, we have four quantities – Dividend, Divisor, Quotient and Remainder. These are connected by the relation. Dividend = (Divisor × Quotient) + Remainder 2. When the division is exact, the remainder is zero (0). In this case, the above relation becomes Dividend = Divisor × Quotient Example: 1: The quotient arising from the divisor of 24446 by a certain numbers is 79 and the remainder is 35; what is the divisor? Solution: Divisor × Quotient = Dividend – Remainder 79×Divisor = 24446 -35 =24411 Divisor = 24411 ÷ 79 = 309. Example: 2: A number when divided by 12 leaves a remainder 7. What remainder will be obtained by dividing the same number by 7? Solution: We see that in the above example, the first divisor 12 is not a multiple of the second divisor 7. Now, we take the two numbers 139 and 151, which when divided by 12, leave 7 as the remainder. But when we divide the above two numbers by 7, we get the respective remainder as 6 and 4. Thus, we conclude that the question is wrong. ### 5 Most Important Questions with Short Tricks on Number System Question 1: (A) 5995 (B) 5997 (C) 5996 (D) 5998 Solution with Short Trick: Question 2: (A) (B) (C) (D) Solution with Short Trick: = [3 + 3 3 + 3 3 3 + 3 3 3 3 + 3 3 3 3 3] + Question 3: (A) (B) (C) (D) Solution with Short Trick: Using Formula Denominator Value Question 4: (A) x (B) (C) x – 1 (D) Solution with Short Trick: Question 5: (A) (B) (C) 1/3 (D) 1/2 Solution with Short Trick:
Introduction This section will cover how to: • Correctly evaluate quadratic functions given values of x • Produce a table of coordinates from y = ax² + bx + c • Plot the coordinates on a graph and use the graph to identify roots of the quadratic • Expand pairs of brackets (px+q)(rx+s) to give quadratic expressions • Factorise quadratic expressions into two brackets • Use "difference of two squares" a² - b² = (a+b)(ab) and factorising ax² + bx = x(ax+b) • Solve quadratic equations by factorising • Solve quadratic equations using the formula Each topic is introduced with a theory section including examples and then some practice questions. At the end of the page there is an exercise where you can test your understanding of all the topics covered in this page. Remember that you are NOT allowed to use calculators in this topic. Definitions quadraticexpression an expression of the form ax² + bx + c where a, b and c are numbers and x is the variable (note that b and c can be zero, but a must be non-zero so that there is always an x² term) e.g. x² + 5x + 6, 3x² – 3x + 17, 8x², 4x² – 25, 12x² – 30x quadraticequation an equation which can be written with a quadratic expression on one side and zero on the other e.g. x² + 5x + 6 = 0, 3x² – 3x + 17 = 0, 4x² – 25 = 0 note that quadratic equations sometimes need rearranging to get zero on one side e.g. x² = 3x – 2 can be rearranged to give x² – 3x + 2 = 0 quadraticfunction a name normally given to a function of the form y = ax² + bx + c which is used to calculate y coordinates from given x coordinates so that these can be plotted on a graph roots of aquadratic the roots of a quadratic equation are the solutions to that equation e.g. the roots of the equation x² + 5x + 6 = 0 are x = –2 and x = –3 the roots of a quadratic expression are the values which make it equal zero e.g. the roots of the expression x² + 5x + 6 are x = –2 and x = –3 on a graph, the roots correspond to where the curve crosses the x-axis e.g. the curve of y = x² + 5x + 6 crosses the x-axis at (–2,0) and (–3,0) Plotting a Quadratic Graph and Identifying Roots To plot the graph of y = x² – 2x – 3 for values of x between –2 and 4, we start by calculating a table of values: For example for the x coordinate of –2 we calculate the y coordinate as follows: y = (–2)² – (2 × –2) – 3 = 4 – –4 – 3 = 5 You need to be especially careful when calculating the x² term. The two main things to watch out for are: squaring a negative number gives a positive answer (e.g. –2 × –2 = +4) if the x² term has a coefficient, you square x first and then multiply by the coefficient (e.g. 3x² means x² × 3) Now that we have a table of values, we can plot each ( x , y ) pair as coordinates on a set of axes, as shown below: Finally we can use the graph to identify the roots of the quadratic expression x² – 2x – 3. They are x = –1 and x = 3. These are the solutions of the quadratic equation x² – 2x – 3 = 0 and are the points where the curve crosses the x-axis. Notes Quadratics can have two, one or zero roots depending on how many times their curve would cross the x-axis. The shape of a quadratic curve is always the same except that it is upside-down if the coefficient of x² is negative. Practice Question Work out the answer to each part and then click on the button marked to see if you are correct. For the graph, you can plot the curve yourself on paper and then check your curve against the one on the screen. Plot the quadratic curve y = ½x² – 4x + 6 for x = 0 to 6 and use it to identify the solutions of ½x² – 4x + 6 = 0 (a) Start by completing a table of values for x = 0 up to x = 6 (b) Now plot the points on an appropriate set of axes and join them with a smooth curve (c) Finally, use your curve to identify the solutions of ½x² – 4x + 6 = 0 The curve crosses the x-axis at ( 2 , 0 ) and ( 6 , 0 ) so the solutions are x = 2 and x = 6 Expanding Pairs of Brackets If you expand brackets of the form (px + q)(rx + s) you will get a quadratic expression. To expand brackets in this form, simply multiply each term in the first bracket by every term in the second bracket, then add up the resulting terms and simplify if necessary. e.g. ( x + 5 )( x + 3 ) = x² + 3x + 5x + 15 = x² + 8x + 15 e.g. ( x + 4 )( x – 6 ) = x² – 6x + 4x – 24 = x² – 2x – 24 e.g. ( x – 3 )( 2x + 7 ) = 2x² + 7x – 6x – 21 = 2x² + x – 21 e.g. ( 5x – 2 )( 4x – 3 ) = 20x² – 15x – 8x + 6 = 20x² – 23x + 6 Factorising a quadratic is the reverse of expanding a pair of brackets; not every quadratic can be factorised, but the majority of those which can will end up as two brackets multiplied together. In every case we can easily check if we have factorised correctly by multiplying out the brackets again. (a) Simple Factorisation (b) Harder Factorisation Quadratics become harder to factorise when the coefficient of x² is not 1, for example 3x² +16x + 5. The x² term tells you about the first part of each bracket. In this case the only way to get 3x² is to multiply 3x by x so the answer will take the form ( 3x + ? )( x + ? ) The number term tells you about the second part of each bracket. In this case the only way to get 5 is to multiply 5 by 1 so the answer must be either ( 3x + 5 )( x + 1 ) or ( 3x + 1 )( x + 5 ). Multiplying these out gives the correct answer in the second case, so 3x² +16x + 5 = ( 3x + 1 )( x + 5 ) Factorising harder quadratics will always involve a bit of trial and error so common sense and practice are essential. (c) Special Cases If you expand the brackets in ( a + b )( a – b ) you get a² – b². This is called the difference of two squares. If you see a quadratic which is formed from the difference of two squares then you can use this fact to factorise the quadratic: e.g. x² – 25 = (x)² – (5)² = ( x + 5 )( x – 5 ) e.g. 4x² – 9 = (2x)² – (3)² = ( 2x + 3 )( 2x – 3 ) e.g. 64x² – 1 = (8x)² – (1)² = ( 8x + 1 )( 8x – 1 ) If a quadratic has an x² term and an x term but no number term, then it is easier to factorise because there is a common factor and the answer becomes a term multiplied by a single bracket. For example, the quadratic expression 6x² – 10x becomes 2x( 3x – 5 ). Practice Questions Work out the answer to each question then click on the button marked to see if you are correct. Expand the brackets and simplify: (a) ( x + 5 )( x – 7 ) ( x + 5 )( x – 7 ) = x² – 7x + 5x – 35 = x² – 2x – 35 (b) ( 3x – 5 )( 4x – 1 ) ( 3x – 5 )( 4x – 1 ) = 12x² – 3x – 20x + 5 = 12x² – 23x + 5 Factorise completely: (a) x² + 10x + 9 ( x + 1 )( x + 9 ) (b) x² – 7x – 30 ( x + 3 )( x – 10 ) (c) 2x² + 7x + 5 ( 2x + 5 )( x + 1 ) (d) 20x² – 117x – 50 ( 4x – 25 )( 5x + 2 ) (e) 9x² – 1 ( 3x + 1 )( 3x – 1 ) (f) 20x – 12x² 4x( 5 – 3x ) Basic Principles Consider the equation ab = 0. The only way this can be true is if a = 0 or b = 0. Now consider the quadratic equation ( x + 2 )( x – 3 ) = 0. In a similar way to the equation above, the only way this equation can be true is if x + 2 = 0 or if x – 3 = 0. From x + 2 = 0 we get x = –2 and from x – 3 = 0 we get x = 3. The two solutions of the equation ( x + 2 )( x – 3 ) = 0 are x = –2 and x = 3 (try them). Finally consider the quadratic equation x² – 8x + 15 = 0. By factorising we can rewrite this equation as ( x – 3 )( x – 5 ) = 0. For this to be true, either x – 3 = 0 or x – 5 = 0. This gives two solutions for the original equation: x = 3 or x = 5. We can check these by substituting back in: (3)² – 8×(3) + 15 = 0 and (5)² – 8×(5) + 15 = 0 We can use the principles above to solve any quadratic equations where the quadratic will factorise. In each case we factorise the quadratic, then set each bracket in turn equal to zero to find the solutions. Other Notes Always make sure the equation has the quadratic on one side and zero on the other. Most quadratic equations have two solutions. Some have one solution and others have no solutions. Not every quadratic factorises; this does not necessarily mean that there are no solutions. Examples Make sure you understand all the following examples before moving on: e.g. x² + 4x – 21 = 0 ( x + 7 )( x – 3 ) = 0 x + 7 = 0 or x – 3 = 0 x = – 7 or x = 3 e.g. x² – 49 = 0 ( x + 7 )( x – 7 ) = 0 x + 7 = 0 or x – 7 = 0 x = – 7 or x = 7 e.g. 4x² – 15x + 9 = 0 ( 4x – 3 )( x – 3 ) = 0 4x – 3 = 0 or x – 3 = 0 x = ¾ or x = 3 e.g. 4x² – 4x + 1 = 0 ( 2x – 1 )( 2x – 1 ) = 0 2x – 1 = 0 or 2x – 1 = 0 x = ½ (only one solution - also called equal roots) e.g. x² – 6x + 10 = 0 (the quadratic does not factorise so can't be solved by factorisation) e.g. 8x² = 2x 8x² – 2x = 0 2x( 4x – 1 ) = 0 2x = 0 or 4x – 1 = 0 x = 0 or x = ¼ Practice Questions Work out the answer to each question then click on the button marked to see if you are correct. Solve by factorising: (a) x² – x – 20 = 0 Start by factorising: ( x + 4 )( x – 5 ) = 0 x + 4 = 0 or x – 5 = 0 x = – 4 or x = 5 (b) 16x² – 9 = 0 Use the difference of two squares: (4x)² – (3)² = 0 ( 4x + 3 )( 4x – 3 ) = 0 4x + 3 = 0 or 4x – 3 = 0 x = – ¾ or x = ¾ (c) 4x² – 24x + 36 = 0 You can factorise this but all three terms have a common factor of 4 so it's easiest to divide through by 4 first: x² – 6x + 9 = 0 ( x – 3 )( x – 3 ) = 0 x – 3 = 0 or x – 3 = 0 x = 3 Solving Quadratic Equations using the Formula If ax² + bx + c = 0, then x = –b √ b² – 4ac 2a If a quadratic factorises then it is always quicker and easier to solve a quadratic equation that way, especially without a calculator. Nonetheless, the quadratic formula provides an alternative method for solving quadratic equations. Using the Formula The stages you should go through when using the formula are as follows: • Work out a, b and c (the coefficients of x², x and the number term) • Work out the values of –b, b² – 4ac and 2a • Substitute these values into the formula • Work out the two answers (the + and – in the sign give different answers) Look at this example: x² + 4x – 21 = 0 a = 1 b = 4 c = –21 –b = – 4 b² – 4ac = (4)² – (4 × 1 × –21) = 16 – (–84) = 100 2a = 2 x = –4 √ 100 2 x = – 4 + 10 or x = – 4 – 10 2 2 x = 3 or x = –7 Examples e.g. 9x² – 9x + 2 a = 9 b = –9 c = 2 –b = 9 b² – 4ac = (–9)² – (4 × 9 × 2) = 81 – 72 = 9 2a = 18 x = 9 √ 9 18 x = 9 + 3 or x = 9 – 3 18 18 x = 12 or x = 6 18 18 x = 2 or x = 1 3 3 Note that this question would have been much easier using factorisation! e.g. 2x² – 12x + 18 a = 2 b = –12 c = 18 –b = 12 b² – 4ac = (–12)² – (4 × 2 × 18) = 144 – 144 = 0 2a = 4 x = 12 √ 0 4 x = 12 + 0 or x = 12 – 0 4 4 x = 12 or x = 12 4 4 x = 3 Note that there will only ever be one solution when b² – 4ac = 0. e.g. x² – 6x + 10 a = 1 b = –6 c = 10 –b = 6 b² – 4ac = (–6)² – (4 × 1 × 10) = 36 – 40 = – 4 2a = 2 x = 6 √ – 4 2 There are no solutions as you can not square root a negative number. Note that there will never be any solutions when b² – 4ac is negative. Practice Questions Work out the answer to each question then click on the button marked to see if you are correct. Solve these quadratic equations using the formula: (a) x² + 8x + 7 = 0 a = 1 b = 8 c = 7 –b = –8 b² – 4ac = (8)² – (4 × 1 × 7) = 64 – 28 = 36 2a = 2 x = –8 √ 36 2 x = –8 + 6 or x = –8 – 6 2 2 x = –2 or x = –14 2 2 x = –1 or x = –7 (b) 3x² – 4x + 2 = 0 a = 3 b = –4 c = 2 –b = 4 b² – 4ac = (–4)² – (4 × 3 × 2) = 16 – 24 = –8 2a = 6 x = 4 √ –8 6 There are no solutions as you can not square root a negative number. (c) 20x² – 9x + 1 = 0 a = 20 b = –9 c = 1 –b = 9 b² – 4ac = (–9)² – (4 × 20 × 1) = 81 – 80 = 1 2a = 40 x = 9 √ 1 40 x = 9 + 1 or x = 9 – 1 40 40 x = 10 or x = 8 40 40 x = 1 or x = 1 4 5 Exercise Work out the answers to the questions below and fill in the boxes. Click on the button to find out whether you have answered correctly. If you have then the answer will be ticked and you should move on to the next question. If a cross to clear the incorrect answer and have another go, or you can click on to get some advice on how to work out the answer and then have another go. If you still can't work out the answer after this then you can click on to see the solution. In all questions please write terms in decreasing powers of x, i.e. 3x² + 5x + 2 rather than 2+ 5x +3x². Question 1 (a) Fill in the table of values below for the quadratic formula: y = 2x² – 4x – 6 x –2 –1 0 1 2 3 4 y Substitute each x value in turn into the formula to get the y value. Remember that if you square a negative number you get a positive result. See the boxes above for the correct answers. (b) Plot the graph of y = 2x² – 4x – 6 on the grid on the right-hand side. To plot the graph, click the button, then click each point you want to plot on the graph. When you have finished, click the button and a smooth line will be drawn in. You can then check if your answer is correct using the buttons below. If you need to have another go, click again. Each pair of values for x and y should be plotted as coordinates ( x , y ). The correct graph has been plotted for you on the grid. (c) Use your graph to identify the two roots of the quadratic equation: 2x² – 4x – 6 = 0. x = or x = The roots of the equation are the x-values which make the quadratic evaluate to zero. The quadratic evaluates to zero where it crosses the x-axis at (–1,0) and (3,0) so the solutions are x = –1 and x = 3. You need to install Adobe Flash Question 2 (a) ( x + 5 )( x – 3 ) = You need to install Adobe FlashPlayer 8 or above. Click here. Multiply each term in the first bracket by every term in the second bracket. x² – 3x + 5x – 15= x² + 2x – 15 (b) ( 2x + 3 )( 2x – 3 ) = You need to install Adobe FlashPlayer 8 or above. Click here. Multiply each term in the first bracket by every term in the second bracket. 4x² – 6x + 6x – 9= 4x² – 9 (c) ( 2 + 3x )( x + 7 ) = You need to install Adobe FlashPlayer 8 or above. Click here. Multiply out the brackets as normal, being careful as the terms are in a different order. Make sure you write your answer with decreasing powers of x. 2x + 14 + 3x² + 21x= 3x² + 23x + 14 Question 3 Factorise the following expressions completely: (a) x² – 12x + 27 = You need to install Adobe FlashPlayer 8 or above. Click here. Look for a solution in the form ( x – ? )( x – ? ). –3 + –9 = –12 and –3 × –9 = 27, sothe answer is ( x – 3 )( x – 9 ) (b) 4x² – 81 = You need to install Adobe FlashPlayer 8 or above. Click here. This is an example of "difference of two squares". 4x² – 81 = (2x)² – (9)², sothe answer is ( 2x + 9 )( 2x – 9 ) (c) 10x² + 59x – 6 = You need to install Adobe FlashPlayer 8 or above. Click here. The first values in each bracket are 10x and x respectively. The factorised version is ( 10x – 1 )( x + 6 ) (d) 20x² + 4x = You need to install Adobe FlashPlayer 8 or above. Click here. Look for a common factor of both terms. 4x is the common factor so taking thatoutside the brackets gives 4x( 5x + 1 ) Question 4 If you think there is only one solution, place a dash (–) in the second solution box. If you think there are no solutions then put a dash in both boxes. (a) x² – 10x + 16 = 0 x = You need to install Adobe FlashPlayer 8 or above. Click here. or x = You need to install Adobe FlashPlayer 8 or above. Click here. Factorise into the form ( x – ? )( x – ? ). ( x – 2 )( x – 8 ) = 0x – 2 = 0 or x – 8 = 0x = 2 or x = 8 (b) 10x² – 5x = 0 x = You need to install Adobe FlashPlayer 8 or above. Click here. or x = You need to install Adobe FlashPlayer 8 or above. Click here. Look for a common factor of both terms to take outside the brackets. 5x ( 2x – 1 ) = 05x = 0 or 2x – 1 = 0x = 0 or x = ½ (c) 25x² + 20x + 4 = 0 x = You need to install Adobe Flash or x = You need to install Adobe Flash In this case both the brackets are the same as each other once factorised. ( 5x + 2 )( 5x + 2 ) = 0 5x + 2 = 0 or 5x + 2 = 0 x = – 2 (only one solution) 5 Question 5 If you think there is only one solution, place a dash (–) in the second solution box. If you think there are no solutions then put a dash in both boxes. (a) x² + x – 2 = 0 x = You need to install Adobe Flash or x = You need to install Adobe Flash a = 1 b = 1 c = –2 –b = – 1 b² – 4ac = (1)² – (4 × 1 × –2) = 9 2a = 2 x = –1 √ 9 2 x = – 1 + 3 or x = – 1 – 3 2 2 x = 1 or x = –2 (b) 2x² – 4x + 3 = 0 x = You need to install Adobe Flash or x = You need to install Adobe Flash a = 2 b = –4 c = 3 –b = 4 b² – 4ac = (–4)² – (4 × 2 × 3) = –8 2a = 4 x = 4 √ –8 4 There are no solutions because you can't square-root a negative number. (c) 4x² – 1 = 0 x = You need to install Adobe Flash or x = You need to install Adobe Flash There is no "x term" so b = 0. a = 4 b = 0 c = –1 –b = 0 b² – 4ac = (0)² – (4 × 4 × –1) = 16 2a = 8 x = 0 √ 16 8 x = 0 + 4 or x = 0 – 4 8 8 x = ½ or x = –½ Question 6 If you think there is only one solution, place a dash (–) in the second solution box. If you think there are no solutions then put a dash in both boxes. (a) 3x² + 13x = 10 x = You need to install Adobe Flash or x = You need to install Adobe Flash Start by getting everything on the left-hand side of the equation with zero on the right-hand side. 3x² + 13x – 10 = 0 ( 3x – 2 )( x + 5 ) = 0 3x – 2 = 0 or x + 5 = 0 x = 2 or x = –5 3 (b) 5x² – 10x = 4x + 3 x = You need to install Adobe Flash or x = You need to install Adobe Flash Get everything on the left-hand side of the equation (remember to collect like terms). 5x² – 14x – 3 = 0 ( 5x + 1 )( x – 3 ) = 0 5x + 1 = 0 or x – 3 = 0 x = – 1 or x = 3 5 (c) (x – 5)² = 9 x = You need to install Adobe FlashPlayer 8 or above. Click here. or x = You need to install Adobe FlashPlayer 8 or above. Click here. Multiply out the brackets as ( x – 5 )( x – 5 ) and then gather everything on the left-hand side of the equation. x² – 10x + 25 = 9 x² – 10x + 16 = 0 ( x – 2 )( x – 8 ) = 0 x – 2 = 0 or x – 8 = 0 x = 2 or x = 8 (d) y² + 3y + 2 = 0 y = You need to install Adobe FlashPlayer 8 or above. Click here. or y = You need to install Adobe FlashPlayer 8 or above. Click here. Treat the quadratic exactly as if the "y"s were "x"s. ( y + 2 )( y + 1 ) = 0y + 2 = 0 or y + 1 = 0y = –2 or y = –1
# Independent and Dependent events. Warm Up There are 5 blue, 4 red, 1 yellow and 2 green beads in a bag. Find the probability that a bead chosen at random. ## Presentation on theme: "Independent and Dependent events. Warm Up There are 5 blue, 4 red, 1 yellow and 2 green beads in a bag. Find the probability that a bead chosen at random."— Presentation transcript: Independent and Dependent events Warm Up There are 5 blue, 4 red, 1 yellow and 2 green beads in a bag. Find the probability that a bead chosen at random from the bag is: 1. blue 2. green 3. blue or green4. blue or yellow 5. not red6. not yellow Determine whether events are independent or dependent. Find the probability of independent and dependent events. Objectives independent events dependent events conditional probability Vocabulary Events are independent events if the occurrence of one event does not affect the probability of the other. If a coin is tossed twice, its landing heads up on the first toss and landing heads up on the second toss are independent events. The outcome of one toss does not affect the probability of heads on the other toss. To find the probability of tossing heads twice, multiply the individual probabilities, Example 1A: Finding the Probability of Independent Events A six-sided cube is labeled with the numbers 1, 2, 2, 3, 3, and 3. Four sides are colored red, one side is white, and one side is yellow. Find the probability. Tossing 2, then 2. Tossing a 2 once does not affect the probability of tossing a 2 again, so the events are independent. 2 of the 6 sides are labeled 2. P(2 and then 2) = P(2)  P(2) Example 1B: Finding the Probability of Independent Events Tossing red, then white, then yellow. The result of any toss does not affect the probability of any other outcome. 4 of the 6 sides are red; 1 is white; 1 is yellow. A six-sided cube is labeled with the numbers 1, 2, 2, 3, 3, and 3. Four sides are colored red, one side is white, and one side is yellow. Find the probability. P(red, then white, and then yellow) = P(red)  P(white)  P(yellow) Check It Out! Example 1 Find each probability. 1a. rolling a 6 on one number cube and a 6 on another number cube 1 of the 6 sides is labeled 6. P(6 and then 6) = P(6)  P(6) 1b. tossing heads, then heads, and then tails when tossing a coin 3 times 1 of the 2 sides is heads. P(heads, then heads, and then tails) = P(heads)  P(heads)  P(tails) Events are dependent events if the occurrence of one event affects the probability of the other. For example, suppose that there are 2 lemons and 1 lime in a bag. If you pull out two pieces of fruit, the probabilities change depending on the outcome of the first. The tree diagram shows the probabilities for choosing two pieces of fruit from a bag containing 2 lemons and 1 lime. The probability of a specific event can be found by multiplying the probabilities on the branches that make up the event. For example, the probability of drawing two lemons is. To find the probability of dependent events, you can use conditional probability P(B\|A), the probability of event B, given that event A has occurred. Example 2A: Finding the Probability of Dependent Events Two number cubes are rolled–one white and one yellow. Explain why the events are dependant. Then find the indicated probability. The white cube shows a 6 and the sum is greater than 9. Example 2A Continued The events “the white cube shows a 6” and “the sum is greater than 9” are dependent because P(sum >9) is different when it is known that a white 6 has occurred. Step 1 Explain why the events are dependant. Of 36 outcomes, 6 have a white 6. Of 6 outcomes with white 6, 3 have a sum greater than 9. Example 2A Continued Step 2 Find the probability. P(A and B) = P(A)  P(B\|A) P(white 6 and sum > 9) = P(white six)  P(sum > 9\|white 6) Example 2B: Finding the Probability of Dependent Events The yellow cube shows an even number and the sum is 5. Two number cubes are rolled–one white and one yellow. Explain why the events are dependant. Then find the indicated probability. Example 2B Continued Of 18 outcomes that have a yellow even number, 2 have a sum of 5. Of 36 outcomes, 18 have a yellow even number. The events are dependent because P(sum is 5) is different when the yellow cube shows an even number. Example 2B Continued P(yellow is even and sum is 5) = P(yellow even number) ● P(sum is 5\| yellow even number) Check It Out! Example 2 Two number cubes are rolled—one red and one black. Explain why the events are dependent, and then find the indicated probability. The red cube shows a number greater than 4, and the sum is greater than 9. Check It Out! Example 2 Continued Of 12 outcomes with a red number greater than 4, 5 have a sum greater than 9. The events “the red cube shows > 4” and “the sum is greater than 9” are dependent because P(sum > 9) is different when it is known that a red cube that shows a number greater than 4 has occurred. Step 1 Explain why the events are dependant. Of 36 outcomes, 12 have a red greater than 4. Check It Out! Example 2 Continued Step 2 Find the probability. P(red > 4 and sum > 9) = P(red > 4) ● P(sum > 9\| red > 4) P(A and B) = P(A) P(B\|A) Conditional probability often applies when data fall into categories. Example 3: Using a Table to Find Conditional Probability The table shows domestic migration from 1995 to 2000. A person is randomly selected. Find each probability. Domestic Migration by Region (thousands) RegionImmigrantsEmigrants Northeast15372808 Midwest24102951 South50423243 West26662654 Example 3 Continued A. that an emigrant is from the West Use the emigrant column. Of 11,656 emigrants, 2654 are from the West. B. that someone selected from the South region is an immigrant Use the South row. Of 8285 people, 5042 were immigrants. Example 3 Continued C. that someone selected is an emigrant and is from the Midwest Use the Emigrants column. Of 11,656 emigrants, 2951 were from the Midwest. There were 23,311 total people. Check It Out! Example 3a Using the table on p. 813, find each probability. that a voter from Travis county voted for someone other than George Bush or John Kerry Of those in Travis 148,000 voted for Bush, 197,000 for Kerry and 5000 for other. Check It Out! Example 3b that a voter was from Harris county and voted for George Bush Using the table on p. 813, find each probability. Of the 1,058,000 who voted for Bush, 581,000 were from Harris County. There were 3,125,000 total voters. A standard card deck contains 4 suits of 13 cards each. The face cards are the jacks, queens, and kings. Remember! In many cases involving random selection, events are independent when there is replacement and dependent when there is not replacement. Example 4: Determining Whether Events Are Independent or Dependant Two cards are drawn from a deck of 52. Determine whether the events are independent or dependent. Find the probability. Example 4 Continued A. selecting two hearts when the first card is replaced Replacing the first card means that the occurrence of the first selection will not affect the probability of the second selection, so the events are independent. 13 of the 52 cards are hearts. P(heart\|heart on first draw) = P(heart)  P(heart) Example 4 Continued B. selecting two hearts when the first card is not replaced Not replacing the first card means that there will be fewer cards to choose from, affecting the probability of the second selection, so the events are dependent. There are 13 hearts. 12 hearts and 51 cards are available for the second selection. P(heart)  P(heart\|first card was a heart) Example 4 Continued C. a queen is drawn, is not replaced, and then a king is drawn Not replacing the first card means that there will be fewer cards to choose from, affecting the probability of the second selection, so the events are dependent. There are 4 queens. 4 kings and 51 cards are available for the second selection. P(queen)  P(king\|first card was a queen) Check It Out! Example 4 A bag contains 10 beads—2 black, 3 white, and 5 red. A bead is selected at random. Determine whether the events are independent or dependent. Find the indicated probability. Check It Out! Example 4 Continued a. selecting a white bead, replacing it, and then selecting a red bead Replacing the white bead means that the probability of the second selection will not change so the events are independent. P(white on first draw and red on second draw) = P(white)  P(red) Check It Out! Example 4 Continued b. selecting a white bead, not replacing it, and then selecting a red bead By not replacing the white bead the probability of the second selection has changed so the events are dependent. P(white)  P(red\|first bead was white) Check It Out! Example 4 Continued c. selecting 3 nonred beads without replacement By not replacing the red beads the probability of the next selection has changed so the events are dependent. P(nonred) ● P(nonred\|first was nonred) ● P(nonred\|first and second were nonred) Lesson Quiz: Part I 1. Find the probability of rolling a number greater than 2 and then rolling a multiple of 3 when a number cube is rolled twice. 2. A drawer contains 8 blue socks, 8 black socks, and 4 white socks. Socks are picked at random. Explain why the events picking a blue sock and then another blue sock are dependent. Then find the probability. P(blue\|blue) is different when it is known that a blue sock has been picked; Lesson Quiz: Part II 3. Two cards are drawn from a deck of 52. Determine whether the events are independent or dependent. Find the indicated probability. A. selecting two face cards when the first card is replaced B. selecting two face cards when the first card is not replaced independent;dependent; Download ppt "Independent and Dependent events. Warm Up There are 5 blue, 4 red, 1 yellow and 2 green beads in a bag. Find the probability that a bead chosen at random." Similar presentations
# Dividing Fractions This video explains how to divide fractions by looking at the concepts involved. ## Transcript for Dividing Fractions We are first going to do a quick review of division, so we can apply the concepts to fractions. So let’s look at the problem 12 divided by 3. We can think of this as dividing 12 items, such as these pigs, into groups of 3 and seeing how many groups can form. We can see that we have 4 groups of 3, and we all know that 12 divided by 3 is 4, so there are no surprises here. So what would it look like if we had the problem 5 divided by a $\frac{1}{3}$. How many groups of $\frac{1}{3}$ can I make? First, we need to divide all the circles into thirds and put $\frac{1}{3}$ into each pile, giving us 15 piles. What if we change the problem to 5 divided by $\frac{2}{3}$? Well, we would treat it the same way. We would divide 5 into thirds and put $\frac{2}{3}$ in each pile, giving us $7\frac{1}{2}$ piles. ## Coming up with Rules We can see how to find the answer using fraction pieces, but we need a rule we can follow. We can’t always have fraction pieces handy so let’s start by looking at two problems that you probably already know the answer to. I’m going start with 4 times $\frac{1}{2}$. Did you get two? And how did you get it? Did you multiply it out or did you just divide by 2? What about 12 times $\frac{1}{3}$? I got 4 but didn’t actually multiply by $\frac{1}{3}$. I divided by 3. In both situations, I got the same answer so there must be a connection between multiplying and dividing. In the first example, we saw that 4 times $\frac{1}{2}$ was equivalent to 4 divided by 2 and in the second that 12 times $\frac{1}{3}$ is equivalent to 12 divided by 3. I noticed that the number 2 and the fraction $\frac{1}{2}$ are reciprocals and I also notice that 3 and $\frac{1}{3}$ are reciprocals. So that would suggest that multiplying is the same thing as dividing by the reciprocal and, perhaps even more useful, that dividing is the same a multiplying by the reciprocal. Let’s see if this process works for the problems we just did with the fraction tiles. 5 divided by $\frac{1}{3}$. Well, if I took the reciprocal of $\frac{1}{3}$ I get 3 so that would be 5 times 3 and I see in both cases I got 15. Now, let’s look at 5 divided by $\frac{2}{3}$. Well, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$ and when I multiply that I get $\frac{15}{2}$ or 7 and $\frac{1}{2}$. So, let’s practice dividing fractions by multiplying by the reciprocal. Let’s say I had $\frac{5}{8}$ divided by $\frac{2}{3}$. Well, that would be the same thing as $\frac{5}{8}$ times the reciprocal of $\frac{2}{3}$, which is $\frac{3}{2}$ and when I multiply across I get $\frac{15}{16}$. Notice when I divide by a number less than 1, I get a solution that’s actually greater than what I started with. ## Mixed Numbers Now, let’s look at the problem involving mixed numbers $1\frac{1}{2}$ divided by $2\frac{1}{3}$. Just like a multiplication, the first thing I need to do is to change my mixed numbers into fractions. $1\frac{1}{2}$ becomes $\frac{3}{2}$ and $2\frac{1}{3}$ becomes $\frac{7}{3}$. Now, I need to switch my division to multiplying by the reciprocal so that would be $\frac{3}{2}$ times $\frac{3}{7}$ and when I multiply across I get $\frac{9}{14}$. Now notice that when I divide by a number that’s larger than the one I was dividing into I’m going to get something smaller. I’m going to get a fraction less than 1. ### A little confused? You may want to look at:
# Common Core: 1st Grade Math : Composing Three-Dimensional Shapes ## Example Questions ### Example Question #49 : Geometry What two shapes can you find in this shape? A cube and a triangle A cube and a trapezoid A cube and a circle A cube and a square A cube and a square Explanation: ### Example Question #50 : Geometry What two shapes can you find in this shape? A cube and a square A cube and a trapezoid A cube and a rectangle A cube and a triangle A cube and a triangle Explanation: ### Example Question #1 : Composing Three Dimensional Shapes What two shapes can you find in this shape? A rectangle and a triangle A rectangle and circle A cylinder and a triangle A cylinder and a circle A cylinder and a circle Explanation: ### Example Question #2 : Composing Three Dimensional Shapes The sides of a rectangular prism are made up of __________. Rectangles Circles Trapezoids Triangles Rectangles Explanation: The sides of a rectangular prism are made up of rectangles. ### Example Question #3 : Composing Three Dimensional Shapes The base, or bottom, of a cylinder is made of up what shape? Rectangle Circle Square Triangle Circle Explanation: The base, or bottom, of a cylinder is made up of a circle. ### Example Question #1 : Composing Three Dimensional Shapes The sides of a cube are made up of ____________. Rectangles Triangles Rectangles and squares Squares Squares Explanation: A cube is a 3-dimensional shape with 6 sides. Each side of a cube is a square. Thus, the sides of a cube are made up of squares. ### Example Question #2 : Composing Three Dimensional Shapes Using the clues below identify the three-dimensional shape. • I have 6 flat surfaces. • Each flat surface is a square. • I have no curved surfaces. What am I? A cube A sphere A cylinder A cone A cube Explanation: The three-dimensional figure being described in the riddle is a cube. A cube, like a cardboard box or a die from a board game, has six sides, each side is flat, each side is a square, and there are no curved sides. The other answer choices include curved sides and do not have six flat surfaces. ### Example Question #1 : Composing Three Dimensional Shapes Andy takes two cubes and stacks them on top of each other. What three-dimensional shape did he create? Pyramid Rectangular Prism Cube Cylinder Rectangular Prism Explanation: Andy created a rectangular prism by stacking two cubes on top of each other. A rectangular prism has six flat surfaces, six sides, and four identical faces that are larger than a pair of smaller faces. Each flat surface on a rectangular prism is a square or rectangular shape. ### Example Question #2 : Composing Three Dimensional Shapes Using the clues below identify the three-dimensional shape. • I have five flat surfaces. • Four of my surfaces are triangles. • I have no curved surfaces. What am I? Explanation: The three-dimensional figure that is being described in the riddle is a pyramid. A pyramid has four surfaces that are triangles, no curves surfaces, and a total of five flat surfaces. The base of the pyramid in the example is a square. The other figures have curved surfaces and do not match the clues. ### Example Question #1 : Composing Three Dimensional Shapes Using the clues below identify the three-dimensional shape. • I have zero flat surfaces. • I have zero faces. • I have no edges. What am I?
# Eureka Math Grade 2 Module 5 End of Module Assessment Answer Key ## Engage NY Eureka Math 2nd Grade Module 5 End of Module Assessment Answer Key ### Eureka Math Grade 2 Module 5 End of Module Assessment Answer Key Question 1. Solve each problem with a written strategy such as a tape diagram, a number bond, the arrow way, the vertical form, or chips on a place value chart. a. 460 + 200 = _______ 460 + 200 = 660. Explanation: In the above-given question, given that, solve each problem with a written strategy such as tape diagram, a number bond. 460 + 200. 460 = 450 + 10. 200 + 10 = 210. 210 + 450 = 660. b. _______ = 865 â€“ 300 865 – 300 = 565. Explanation: In the above-given question, given that, solve each problem with a written strategy such as tape diagram, a number bond. 865 – 300. 865 = 860 + 5. 300 + 5 = 305. 860 – 305 = 565. c. _______ + 400 = 598 198 + 400 = 598. Explanation: In the above-given question, given that, solve each problem with a written strategy such as tape diagram, a number bond. 198 + 400. 400 = 398 + 2. 198 + 2 = 200. 200 + 398 = 598. d. 240 â€“ 190 = _______ 240 – 190 = 50. Explanation: In the above-given question, given that, solve each problem with a written strategy such as tape diagram, a number bond. 240 – 190. 240 = 230 + 10. 190 + 10 = 200. 200 + 230 = 430. e. _______ = 760 â€“ 280 760 – 280 = 480. Explanation: In the above-given question, given that, solve each problem with a written strategy such as tape diagram, a number bond. 760 – 280. 760 = 740 + 20. 280 + 20 = 300. 740 – 280 = 480. f. 330 â€“ 170 = _______ 330 – 170 = 160. Explanation: In the above-given question, given that, solve each problem with a written strategy such as tape diagram, a number bond. 330 – 170. 330 = 300 + 30. 170 + 30 = 200. 200 – 360 = 160. Question 2. Use the arrow way to fill in the blanks and solve. Use place value drawings if that will help you. a. 630 – 400 = 230. 230 + 10 = 240. 630 – 240 = 390. Explanation: In the above-given question, given that, use the arrow way to fill in the blanks. 630 – 400 = 230. 230 + 10 = 240. 630 – 240 = 390. b. 570 – 300 = 270. 270 + 20 = 290. 570 – 280 = 290. Explanation: In the above-given question, given that, use the arrow way to fill in the blanks. 570 – 300 = 270. 270 + 20 = 290. 570 – 280 = 290. c. 958 – 400 = 558. 558 – 40 = 518. 958 – 440 = 518. Explanation: In the above-given question, given that, use the arrow way to fill in the blanks. 958 – 400 = 558. 558 – 40 = 518. 958 – 440 = 518. Question 3. Solve. Draw a place value chart with chips to model the problems. Show a written subtraction method to check your work. a. 756 + 136 = ______ Subtraction number sentence: 756 + 136 = 892. Explanation: In the above-given question, given that, draw a place value chart with chips to model the problems. show a written subtraction method to check your work. 756 + 136 = 892. 892 – 136 = 756. b. 267 + 545 = ______ Subtraction number sentence: 267 + 545 = 812. Explanation: In the above-given question, given that, draw a place value chart with chips to model the problems. show a written subtraction method to check your work. 267 + 545 = 812. 812 – 545 = 267. Draw a place value chart with chips to model the problems. Show a written addition method to check your work. c. 617 â€“ 229 = ______ Check: 617 – 229 = 398. Explanation: In the above-given question, given that, draw a place value chart with chips to model the problems. 617 – 229 = 398. 398 + 229 = 617. d. 700 â€“ 463 = ______ Check: 700 – 463 = 437. Explanation: In the above-given question, given that, draw a place value chart with chips to model the problems. 700 – 463 = 437. 437 + 463 = 700. Question 4. Find the missing numbers to make each statement true. Show your strategy to solve. a. 300 â€“ 106 = ________ 300 – 106 = 194. Explanation: In the above-given question, given that, find the missing numbers to make each statement true. 300 – 106 = 194. 194 + 106 = 300. b. ________ = 407 â€“ 159 407 – 159 = 248. Explanation: In the above-given question, given that, find the missing numbers to make each statement true. 407 – 159 = 248. 248 + 159 = 407. c. 410 â€“ 190 = 420 â€“ ___200_____ 410 – 190 = 220. Explanation: In the above-given question, given that, find the missing numbers to make each statement true. 410 – 190 = 220. 220 + 190 = 410. d. 750 â€“ 180 = ________ â€“ 200 750 – 180 = 570. Explanation: In the above-given question, given that, find the missing numbers to make each statement true. 750 – 180 = 570. 770 – 200 = 570. 570 + 180 = 750. e. 900 â€“ ________ = 600 â€“ 426 600 – 426 = 174. Explanation: In the above-given question, given that, find the missing numbers to make each statement true. 600 – 426 = 174. 174 + 426 = 600. Question 5. Martha answered the problem 456 â€“ 378 incorrectly. She does not understand her mistake. a. Explain to Martha what she did wrong using place value language. Explanation: ________________________________________ ________________________________________ ________________________________________ 456 – 378 = 78. Explanation: In the above-given question, given that, 456 – 378. 4 hundred = 400. 5 tens = 50. 6 ones = 6. 400 + 50 + 6 = 456. 3 hundred = 300. 7 tens = 70. 8 ones = 8. 300 + 70 + 8 = 378. 456 – 378 = 78. b. Model an alternative strategy for 456 â€“ 378 to help Martha avoid making this mistake again.
# What is the probability that the first head will appear on the even numbered tosses ## Question $$\text{Consider a coin with probability R to be heads. What is the probability}$$ $$\text{that the first head will appear on the even numbered tosses?}$$ ## My Approach let the required Probability$$=P$$. Hence we can write our eauation as-: $$P=(1-R) \times R+(1-R) \times (1-R) \times (1-R) \times P$$ $$P(1-(1-R) \times (1-R) \times (1-R))=(1-R) \times R$$ $$P=\frac{(1-R) \times R }{(1-(1-R) \times (1-R) \times (1-R)}$$ Am i correct? $$P=\frac{(1 - R)}{(2 - R)}$$ Alternative approach - to see the first head on an even numbered toss we need to have: 1. Toss 1 is a tail (probability 1-R) 2. Renumbering Toss 2 as Toss 1, Toss 3 as Toss 2 etc. we are now want to see first head on an odd numbered toss So P(first head on even toss) = P(first head on odd toss) x (1-R) but if we denote P(first head on even toss) by p then P(first head on odd toss) = 1-p, so $p = (1-p)(1-R)$ $\Rightarrow p + p(1-R) = 1-R$ $\Rightarrow p = \frac{1-R}{2-R}$ Sanity checks: (i) p=0 when R=1 (ii) p approaches 1/2 as R approaches 0. • I'm having trouble seeing intuitively why p should approach 1/2 as R approaches 0. – Chris Nov 17 '17 at 17:30 • @Chris - because the advantage that odds have is that an odd comes first. Otherwise it is equally likely to occur as an even or an ood toss. But the more flips that occur before a head occurs, the less impact the odds' advantage gives. The smaller $R$, the greater the expected number of flips needed to get a head will be. – Paul Sinclair Nov 17 '17 at 18:36 • Ah, yes. That makes sense. I had a vague feeling it was something like that but I couldn't quite think why. Thanks for the explanation. – Chris Nov 17 '17 at 22:45 The probability that the first head will appear on the second toss is $(1 - R)R$. The probability that the first head will appear on the fourth toss is $(1 - R)^3R$. The probability that the first head will appear on the sixth toss is $(1 - R)^5R$. In general, the probability that the first head will appear on the $2k$th toss is $(1 - R)^{2k - 1}R$. Hence, the desired probability is $$p = \sum_{k = 1}^{\infty} (1 - R)^{2k - 1}R = (1 - R)R\sum_{k = 1}^{\infty} (1 - R)^{2k}$$ which is a geometric series with common ratio $(1 - R)^2$. If $R < 1$, we obtain \begin{align} p & = (1 - R)R \cdot \frac{1}{1 - (1 - R)^2}\\ & = \frac{(1 - R)R}{1 - (1 - 2R + R^2)}\\ & = \frac{(1 - R)R}{2R - R^2}\\ & = \frac{(1 - R)R}{R(2 - R)}\\ & = \frac{1 - R}{2 - R} && \text{provided that $R \neq 0$} \end{align} If $R = 1$, then heads will be obtained on the first toss, so $p = 0$. If $R = 0$, then heads will never be obtained, so again $p = 0$. Your equation for $P$ isn't quite right. Let's write $T_k$ for the outcome of the $k$-th toss: $T_k \in \{H,T\}$, and $P(T_1 = H) = R$ and $P(T_1 = T) = 1-R$. We then need to carefully split cases: $$P(\text{first heads on even}) = P(\text{first heads on even} \mid T_1 = H)P(T_1 = H) \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = T)P(T_1 = T, T_2 = T) \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = H)P(T_1 = T, T_2 = H) \\ = P(\text{first heads on even} \mid T_1 = T, T_2 = T) \cdot (1-R)^2 \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = H) \cdot R(1-R).$$ Now observe that $$P(\text{first heads on even} \mid T_1 = T, T_2 = T) = P(\text{first heads on even})$$ (by the Markov property, if you like), and $$P(\text{first heads on even} \mid T_1 = T, T_2 = H) = 1$$ since this is a 'success'. Hence, writing $p = P(\text{first heads on even})$, we have $$p = p(1-R)^2 + R(1-R).$$ This is the same as you had except that you had a factor $(1-R)^3$. I've put in all the details so you can see exactly why it's a power of 2, not a power of 3. Intuitively, I think you've got the idea, except that you were thinking that if we don't get $TH$, then we need to go $TTT$ to not fail -- which is correct -- but this first $T$ it taken into account with the probability $p$, so we could write down straight away $$p = R(1-R) + p(1-R)^2;$$ the danger of this, as you have found out, however, is that if there's a slight mistake, then it's not at all clear where it came from. Hopefully this helps! :)
# 1.Be able to divide polynomials 2.Be able to simplify expressions involving powers of monomials by applying the division properties of powers. ## Presentation on theme: "1.Be able to divide polynomials 2.Be able to simplify expressions involving powers of monomials by applying the division properties of powers."— Presentation transcript: 1.Be able to divide polynomials 2.Be able to simplify expressions involving powers of monomials by applying the division properties of powers. Monomial: A number, a variable, or the product of a number and one or more variables Constant: A monomial that is a real number. Power: An expression in the form x n. Base: In an expression of the form x n, the base is x. Exponent: In an expression of the form x n, the exponent is n. Quotient: The number resulting by the division of one number by another. Repeated multiplication can be represented using exponents. To expand a power, use the exponent to determine the number of times a base is multiplied by itself. Product of Powers: When two numbers with the same base are multiplied together, add the exponents and leave the base unchanged. Power of a Product: In a product raised to a power, the exponent applies to each factor of the product. Power of a Power: When a power is raised to another power, multiply the exponents and leave the base unchanged. Remember: Follow the order of operations when applying more than one property! Simplify: Step 1: Rewrite the expression in expanded form Step 2: Simplify. For all real numbers a, and integers m and n: Remember: A number divided by itself is 1. Simplify: Step 1: Write the exponent in expanded form. Step 2: Multiply and simplify. For all real numbers a and b, and integer m: Apply quotient of powers. Apply power of a quotient. Apply quotient of powers Apply power of a quotient Simplify Apply power of a power 1. 2. THINK! x 2-2 = x 0 = 1 Download ppt "1.Be able to divide polynomials 2.Be able to simplify expressions involving powers of monomials by applying the division properties of powers." Similar presentations
# I want to have a college fund for my daughter. She is 5, so I have 13 years to achieve my goal of \$50,000. The bank says I can earn 2%. I have \$5000 already set aside. How much do I need to contribute every year? \$2960 yearly savings Explanation: From the values given and from mathematical manipulation, he or she needs a contribution of at least \$2900 every year in order to achieve his goal of \$50,000. EXPLANATION • If the child is 5yr old now, in 13years time, she will be 18yr old. • \$2950 target yearly • for the next 13years, it would have amount to \$38350 • remember the bank will give an annual interest rate of 2% • so for 13years, that's 26% = 0.26 • In the 13th year, he would have saved \$38350, add the 26% interest for the duration of 13years = 26% x \$38350 + \$38350 = \$48321 • His savings will fall between \$2950 - \$2960 yearly. You will need to contribute approximately \$2,615.97 each year to your college fund to achieve your goal of \$50,000 in 13 years, starting with \$5,000 and earning 2% interest compounded annually. ### Explanation: To calculate how much you need to contribute every year to have \$50,000 in a college fund for your daughter in 13 years with an existing \$5,000 at a 2% annual interest rate, we need to use the future value of an annuity formula: The future value of an annuity formula is FV = P × {[(1 + r)^n - 1] / r}, where: • FV is the future value of the annuity (the amount we want to have in the future, which is \$50,000). • P is the annual payment (the amount you will contribute every year). • r is the annual interest rate (which is 2%, or 0.02). • n is the number of years the money is deposited (13 years). Since you already have \$5,000, we first need to find out how much this amount will grow to in 13 years at an annual interest rate of 2%. That's calculated using the compound interest formula:\$5,000(1 + 0.02)^{13} = \$6,727.09 Now, subtract this future value of your initial savings from the goal:\$50,000 - \$6,727.09 = \$43,272.91 This is the amount that needs to be reached with the annual contributions. Plugging this back into the future value of an annuity formula, we solve for P:\$43,272.91 = P × {[(1 + 0.02)^{13} - 1] / 0.02}We can now solve for P, which is the annual contribution required:P = \$43,272.91 / {[(1 + 0.02)^{13} - 1] / 0.02} = \$2,615.97 Therefore, you'd need to contribute approximately \$2,615.97 each year to reach your \$50,000 college fund goal in 13 years, assuming a 2% annual rate. ## Related Questions Social surplus is the ____________. A. total value from trade in a markettotal value from trade in a market. B. difference between the amount that buyers actually pay and what they wish to pay. C. excess of aggregate demand over aggregate supply. D. difference between consumer surplus and producer surplusdifference between consumer surplus and producer surplus. The correct answer is letter "A": total value from trade in a market. Explanation: Canadian economist Alex Tabarrok (born in 1966) explains social surplus as the sum of consumer surplus, producer surplus, and bystanders surplus. Tabarrok takes an integrative approach in consumer surplus by stating social surplus encompasses every economic trade in the market rather than only consumers and producers surplus. Besides, Tabarrok believes when there are major external costs or benefits, the market will not reach its social surplus. Social surplus is the combination of consumer surplus and producer surplus, taking into account the price that consumers are willing to pay based on their preferences, and the price that producers are willing to sell their product at, based on their costs. ### Explanation: The question asked here is: Social surplus is the ____________. The correct answer to this question is that social surplus is the sum of consumer surplus and producer surplus. This concept falls under economic principles. Consumer surplus is the difference between the price that consumers are willing to pay based on their preferences, and the actual market equilibrium price. On the other hand, producer surplus is the gap between the price at which producers are willing to sell a product, based on their costs, and the market equilibrium price. Combining both these surpluses gives the social surplus. brainly.com/question/31472182 #SPJ3 With an inflation rate of 9 percent, prices would double in how many years? 8 years Explanation: the rule of 72 calculates how long it takes for an amount to double given interest rate 72 / 9% = 8 years The 'Rule of 72' can be used to estimate how long it would take for prices to double with an inflation rate of 9 percent. According to this rule, it would take approximately 8 years. ### Explanation: In order to calculate how long it would take for prices to double with an inflation rate of 9 percent, you can use the 'Rule of 72'. The Rule of 72 is a simplified way to estimate the number of years required to double the money at a given annual rate of return or inflation. According to this rule, you simply divide 72 by the annual rate of return or inflation. Therefore, using the Rule of 72, it would take approximately 8 years (72 divided by 9) for prices to double with an inflation rate of 9 percent. brainly.com/question/34016204 #SPJ3 Blitz Corp. had total sales of \$3,010,000 last year and has 106,000 shares of stock outstanding. The benchmark PS is 1.6 times. What stock price would you consider appropriate? the stock price is \$45.44 Explanation: The computation of the stock price is shown below: Sales per share is = Total sales ÷ stock outstanding shares = \$3,010,000 ÷ 106,000 shares = \$28.40 Now Benchmark PS = Stock price ÷ Sales per share Stock price = \$28.40 × 1.6 = \$45.44 hence, the stock price is \$45.44 We simply applied the above formula so that the correct value could come And, the same is to be considered Universal containers has included its orders as an external data object into Salesforce. You want to create a relationship between Accounts and the Orders object (one-to-many relationship) leveraging a key field for account which is on both external object and Account. Which relationship do you create? Explanation: Indirect lookup relationship is used when there is no Salesforce ID in the external data. So this relationship basically links the external object which is the 'child' to the custom object which is the 'parent'. As the question states, universal containers has included its orders as an 'external data object' into salesforce. Now it wants to create a link or relationship between accounts and orders objects. This is possible through indirect lookup relationship. Potential effects of departmental performance reports on employee behavior include all of the following except: Including indirect expenses can lead to a manager being more careful in using service department's costs. Using budgeted service department costs insures that operating departments are not held responsible for excessive service department costs. Including uncontrollable costs can serve to improve a manager's morale. Potential effects of departmental performance reports on employee behavior except including uncontrollable costs served to improve manager's morale. Explanation: • Performance management plays an important role to keep a proper record of all the works that are being performed in a company . • A proper performance management also shows an important effect on the behavior of the employees. • If the employees are boosted properly by the managers they will increase the productivity . • This will help the company to earn profits. Departmental performance keep record of all the expenses that are being done during the production process . It also help the manager to gain information and can adopt proper strategy to reduce expenses. Is a business cycle a type of recession? yes or no?
# Fraction calculator This calculator subtracts two fractions. First, convert all fractions to a common denominator when fractions have different denominators. Find Least Common Denominator (LCD) or multiply all denominators to find a common denominator. When all denominators are the same, subtract the numerators and place the result over the common denominator. Then simplify the result to the lowest terms or a mixed number. ## The result: ### 5/7 + 3/4 = 41/28 = 1 13/28 ≅ 1.4642857 Spelled result in words is forty-one twenty-eighths (or one and thirteen twenty-eighths). ### How do we solve fractions step by step? 1. Add: 5/7 + 3/4 = 5 · 4/7 · 4 + 3 · 7/4 · 7 = 20/28 + 21/28 = 20 + 21/28 = 41/28 It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(7, 4) = 28. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 7 × 4 = 28. In the following intermediate step, it cannot further simplify the fraction result by canceling. In other words - five sevenths plus three quarters is forty-one twenty-eighths. #### Rules for expressions with fractions: Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts. Mixed numerals (mixed numbers or fractions) keep one space between the integer and fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2. Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. ### Math Symbols SymbolSymbol nameSymbol MeaningExample -minus signsubtraction 1 1/2 - 2/3 asteriskmultiplication 2/3 3/4 ×times signmultiplication 2/3 × 5/6 :division signdivision 1/2 : 3 /division slashdivision 1/3 / 5 :coloncomplex fraction 1/2 : 1/3 ^caretexponentiation / power 1/4^3 ()parenthesescalculate expression inside first-3/5 - (-1/4) The calculator follows well-known rules for the 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. MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule. Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
# AP Calculus AB : Derivative defined as the limit of the difference quotient ## Example Questions ← Previous 1 3 4 5 ### Example Question #1 : Derivative Defined As The Limit Of The Difference Quotient What is the derivative of ? Explanation: To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent. Remember that anything to the zero power is one. ### Example Question #1 : Finding Derivative Of A Function What is the derivative of ? Explanation: To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent. We're going to treat as , as anything to the zero power is one. That means this problem will look like this: Notice that , as anything times zero is zero. Remember, anything to the zero power is one. ### Example Question #1 : Derivative Defined As The Limit Of The Difference Quotient What is the derivative of ? Explanation: To get , we can use the power rule. Since the exponent of the is , as , we lower the exponent by one and then multiply the coefficient by that original exponent: Anything to the power is . ### Example Question #1 : Finding Derivative Of A Function Explanation: To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent. We're going to treat as since anything to the zero power is one. Notice that since anything times zero is zero. ### Example Question #1 : Derivative Defined As The Limit Of The Difference Quotient What is the derivative of ? Explanation: To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent. We're going to treat as since anything to the zero power is one. Notice that since anything times zero is zero. That leaves us with . Simplify. As stated earlier, anything to the zero power is one, leaving us with: ### Example Question #3 : Finding Derivative Of A Function What is the derivative of ? Explanation: To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent. We're going to treat as since anything to the zero power is one. Notice that since anything times zero is zero. Just like it was mentioned earlier, anything to the zero power is one. ### Example Question #2 : Derivative Defined As The Limit Of The Difference Quotient What is the derivative of ? Explanation: To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent. Simplify. Remember that anything to the zero power is equal to one. ### Example Question #3 : Derivative Defined As The Limit Of The Difference Quotient What is the derivative of ? Explanation: To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent. We are going to treat as since anything to the zero power is one. Notice that since anything times zero is zero. Simplify. As stated before, anything to the zero power is one. ### Example Question #13 : Finding Derivative Of A Function What is the derivative of ? Explanation: To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable. Anything to the zero power is one. ### Example Question #2 : Derivative Defined As The Limit Of The Difference Quotient What is the derivative of ?
# AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise ## AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Optional Exercise AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Optional Exercise Textbook Questions and Answers. ### 10th Class Maths 10th Lesson Mensuration Optional Exercise Textbook Questions and Answers Question 1. A golf ball has diameter equal to 4.1 cm. Its surface has 150 dimples each of radius 2 mm. Calculate total surface area which is exposed to the surroundings. (Assume that the dimples are all hemispherical) [π = $$\frac{22}{7}$$] Area exposed = surface area of the ball – total area of 150 hemispherical with radius 2 mm ∴ C.S.A of hemisphere = 2πr2 = 52.831 – 37.71 = 15.12 cm2. Question 2. A cylinder of radius 12 cm. contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. [π = $$\frac{22}{7}$$] Rise in the water level is seen as a cylinder of radius ‘r’ = 12 cm Height, h = 6.75 cm. Volume of the rise = Volume of the spherical iron ball dropped = 9 × 12 × 6.75 = 108 × 6.75 = 729 r3 = 9 × 9 × 9 ∴ 729 = (3 × 3) × (3 × 3) × (3 × 3) ∴ Radius of the ball r = 9 cm. Question 3. A solid toy is in the form of a right circular cylinder with a hemispheri¬cal shape at one end and a cone at the other end. Their common diameter is 4.2 cm. and height of the cylindrical and conical portion are 12 cm. and 7 cm. respectively. Find the volume of the solid toy. [π = $$\frac{22}{7}$$] Volume of the toy = volume of the hemisphere + volume of the cylinder + volume of the cone. Hemisphere: Radius = $$\frac{d}{2}$$ = $$\frac{4.2}{2}$$ = 2.1 cm Cylinder: Radius, r = $$\frac{d}{2}$$ = $$\frac{4.2}{2}$$ = 2.1 cm height, h = 12 cm V = πr2h = $$\frac{22}{7}$$ × 2.1 × 2.1 × 12 = 166.32 cm3 Cone: Radius, r = $$\frac{d}{2}$$ = $$\frac{4.2}{2}$$ = 2.1 cm Height, h = 7 cm Volume = $$\frac{1}{3}$$πr2h = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 2.1 × 2.1 × 7 = 32.34 cm3 ∴ Total volume = 19.404 + 166.32 + 32.34 = 218.064 cm3. Question 4. Three metal cubes with edges 15 cm., 12 cm. and 9 cm. respectively are melted together and formed into a simple cube. Find the diagonal of this cube. Edges l1 = 15 cm, l2 = 12 cm, l3 = 9 cm. Volume of the resulting cube = Sum of the volumes of the three given cubes L3 = l13 + l23 + l33 L3 = 153 + 123 + 93 L3 = 3375 + 1728 + 729 L3 = 5832 = 18 × 18 × 18 ∴ Edge of the new cube l = 18 cm Question 5. A hemispherical bowl of internal diameter 36 cm. contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm. and height 6 cm. How many bottles are required to empty the bowl? = $$\frac{22}{7}$$ × 3 × 3 × 7 Radius, r = $$\frac{d}{2}$$ = $$\frac{36}{2}$$ = 18 cm
The angles made by this line with the +ve direactions of the coordinate axes: θx, θy, and θz are used to find the direction cosines of the line: cos θx, cos θy, and cos θz. answered Aug 22, 2018 by SunilJakhar (89.0k points) selected Aug 22, 2018 by Vikash Kumar . How to Find the Direction Cosines of a Vector With Given Ratios : Here we are going to see the how to find the direction cosines of a vector with given ratios. v = v x e x + v y e y + v z e z , {\displaystyle \mathbf {v} =v_ {x}\mathbf {e} _ {x}+v_ {y}\mathbf {e} _ {y}+v_ {z}\mathbf {e} _ {z},} where ex, ey, ez are the standard basis in Cartesian notation, then the direction cosines are. How do you find the direction cosines and direction angles of the vector? Transcript. Direction cosines and direction ratios of a vector : Consider a vector as shown below on the x-y-z plane. In this explainer, we will learn how to find direction angles and direction cosines for a given vector in space. Direction cosines : (x/r, y/r, z/r) x/r = 3/ √89. We know that in three-dimensional space, we have the -, -, and - or -axis. We know that the vector equation of a line passing through a point with position vector vec a and parallel to the vector vec b is $\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$ Here, $\overrightarrow{a} = 4 \hat{i} + \hat{k}$, $\overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$, $\overrightarrow{r} = \left( 4 \hat{i} + 0 \hat{j}+ \hat{k} \right) + \lambda \left( - 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \right)$, $\text{ Here } , \lambda \text{ is a parameter } . Example, 3 Find the direction cosines of the line passing through the two points (– 2, 4, – 5) and (1, 2, 3). of a vector (line) are the cosines of the angles made by the line with the + ve directions of x, y & z axes respectively. (Give the direction angles correct to the nearest degree.) Apart from the stuff given in "How to Find the Direction Cosines of a Vector With Given Ratios", if you need any other stuff in math, please use our google custom search here. d. or d and is the distance between and Px yz11 11 ,, Px yz22 22 ,,. These direction numbers are represented by a, b and c. Ex 10.2, 12 Find the direction cosines of the vector ﷯ + 2 ﷯ + 3 ﷯ . Hence direction cosines are ( 3/ √89, -4/ √89, 8 / √89) Direction ratios : Direction ratios are (3, -4, 8). Precalculus Vectors in the Plane Direction Angles. It it some times denoted by letters l, m, n.If a = a i + b j + c j be a vector with its modulus r = sqrt (a^2 + b^2 + c^2) then its d.cs. The cartesian equation of the given line is, \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}$, $\frac{x - 4}{- 2} = \frac{y - 0}{6} = \frac{z - 1}{- 3}$, This shows that the given line passes through the point (4,0,1) and its direction ratios are proportional to -2,6,-3, $\frac{- 2}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{6}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{- 3}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}$, $= \frac{- 2}{7}, \frac{6}{7}, \frac{- 3}{7}$ Thus, the given line passes through the point having position vector $\overrightarrow{a} = 4 \hat{i} + \hat{k}$ and is parallel to the vector $\overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$. © Copyright 2017, Neha Agrawal. Solution : x = 3, y = 1 and z = 1 \|r vector\| = r = √(x 2 + y 2 + z 2) = √3 2 + 1 2 + 1 2) = √(9+1+1) = √11. z/r = 8/ √89. Then ∠ PRO = ∠ PSO = ∠ PTO = 90º. Direction cosines (d.cs.) Find the direction cosines of the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$ Also, reduce it to vector form. (7, 3, -4) cos(a) =… Find the direction cosines and direction angles of the vector determining the norm of a vector in space, vector operations in space, evaluating simple trigonometric expressions. One such property of the direction cosine is that the addition of the squares of … Direction Cosines and Direction Ratios. View Answer Find the direction cosines of the vector 6 i ^ + 2 j ^ − 3 k ^ . The direction cosines are not independent of each other, they are related by the equation x 2 + y 2 + z 2 = 1, so direction cosines only have two degrees of freedom and can only represent direction and not orientation. Question 1 : If We will begin by considering the three-dimensional coordinate grid. The direction cosine of the vector can be determined by dividing the corresponding coordinate of a vector by the vector length. \|r vector\| = r = √(x2 + y2 + z2) = √(32 + (-4)2 + 82), Hence direction cosines are ( 3/√89, -4/√89, 8/√89), \|r vector\| = r = √(x2 + y2 + z2) = √32 + 12 + 12), Hence direction cosines are ( 3/√11, 1/√11, 1/√11), \|r vector\| = r = √(x2 + y2 + z2) = √02 + 12 + 02), \|r vector\| = r = √(x2 + y2 + z2) = √52 + (-3)2 + (-48)2, \|r vector\| = r = √(x2 + y2 + z2) = √32 + 42 + (-3)2, \|r vector\| = r = √(x2 + y2 + z2) = √12 + 02 + (-1)2. Property of direction cosines. find direction cosines of a vector in space either given in component form or represented graphically. The sum of the squares of the direction cosines is equal to one. Ex 10.2, 13 Find the direction cosines of the vector joining the points A (1, 2,−3) and B (−1,−2,1), directed from A to B. 1 Answer. If the position vectors of P and Q are i + 2 j − 7 k and 5 i − 3 j + 4 k respectively then the cosine of the angle between P Q and z-axis is View solution Find the direction cosines of the vector a = i ^ + j ^ − 2 k ^ . In this video, we will learn how to find direction angles and direction cosines for a given vector in space. vectors; Share It On Facebook Twitter Email. z^^)/(\|v\|). Let us assume a line OP passes through the origin in the three-dimensional space. How to Find the Direction Cosines of a Vector With Given Ratios". 12.1 Direction Angles and Direction Cosines. (3) From these definitions, it follows that alpha^2+beta^2+gamma^2=1. (ii) 3i vector + j vector + k vector. After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Direction Cosines of a Vector With Given Ratios". Then, the line will make an angle each with the x-axis, y-axis, and z-axis respectively.The cosines of each of these angles that the line makes with the x-axis, y-axis, and z-axis respectively are called direction cosines of the line in three-dimensional geometry. To find the direction cosines of the vector a is need to divided the corresponding coordinate of vector by the length of the vector. Find the direction cosines of a vector which is equally inclined to the x-axis, y-axis and z-axis. The magnitude of vector d is denoted by . Given a vector (a,b,c) in three-space, the direction cosines of this vector are Here the direction angles, , are the angles that the vector makes with the positive x-, y- and z-axes, respectively.In formulas, it is usually the direction cosines that occur, rather than the direction angles. Ex 11.1, 2 Find the direction cosines of a line which makes equal angles with the coordinate axes. Find the Magnitude and Direction Cosines of Given Vectors : Here we are going to see how to find the magnitude and direction cosines of given vectors. How to Find the Direction Cosines of a Vector With Given Ratios". For example, take a look at the vector in the image. Entering data into the vector direction cosines calculator. Any number proportional to the direction cosine is known as the direction ratio of a line. 22 d dxx yy zz21 2 1 2 1. Geospatial Science RMIT THE DISTANCE d BETWEEN TWO POINTS IN SPACE . Students should already be familiar with. Solution for Find the direction cosines and direction angles of the vector. Therefore, we can say that cosines of direction angles of a vector r are the coefficients of the unit vectors, and when the unit vector is resolved in terms of its rectangular components. In three-dimensional geometry, we have three axes: namely, the x, y, and z-axis. So direction cosines of the line = 2/√41, 6/√41, -1/√41. A( 1, 2 , −3) B(−1, −2, 1) () ⃗ = (−1 − 1) ̂ + (−2 − 2) ̂ + (1−(−3)) ̂ = –2 ̂ – 4 ̂ + 4 ̂ Directions ratios are a = – 2, b = –4, & c = 4 Magnitude The unit vector coordinates is equal to the direction cosine. To find the direction cosines of a vector: Select the vector dimension and the vector form of representation; Type the coordinates of the vector; Press the button "Calculate direction cosines of a vector" and you will have a detailed step-by-step solution. Best answer. if you need any other stuff in math, please use our google custom search here. 2 (2) DIRECTION COSINES OF A LINE BETWEEN TWO POINTS IN SPACE How to Find a Vector’s Magnitude and Direction. Lesson Video . Direction cosines of a line making, with x – axis, with y – axis, and with z – axis are l, m, n l = cos , m = cos , n = cos Given the line makes equal angles with the coordinate axes. By Steven Holzner . If you’re given the vector components, such as (3, 4), you can convert it easily to the magnitude/angle way of expressing vectors using trigonometry. Example: Find the direction cosines of the line joining the points (2,1,2) and (4,2,0). Click hereto get an answer to your question ️ Find the direction ratios and the direction cosines of the vector a = (5î - 3ĵ + 4k̂). All rights reserved.What are Direction cosines and Direction ratios of a vector? What this means is that direction cosines do not define how much an object is rotated around the axis of the vector. Find the Magnitude and Direction Cosines of Given Vectors - Practice Question. Find the Direction Cosines of the Line 4 − X 2 = Y 6 = 1 − Z 3 . y/r = -4/ √89. Prerequisites. We know, in three-dimensional coordinate space, we have the -, -, and -axes.These are perpendicular to one another as seen in the diagram below. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Direction cosines : (x/r, y/r, z/r) x/r = 3/ √11 The coordinates of the unit vector is equal to its direction cosines. are … Find the direction cosines and direction ratios of the following vectors. Also, Reduce It to Vector Form. My Vectors course: https://www.kristakingmath.com/vectors-courseLearn how to find the direction cosines and direction angles of a vector. \], Chapter 28: Straight Line in Space - Exercise 28.1 [Page 10], CBSE Previous Year Question Paper With Solution for Class 12 Arts, CBSE Previous Year Question Paper With Solution for Class 12 Commerce, CBSE Previous Year Question Paper With Solution for Class 12 Science, CBSE Previous Year Question Paper With Solution for Class 10, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Arts, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Commerce, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Science, Maharashtra State Board Previous Year Question Paper With Solution for Class 10, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Arts, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Commerce, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Science, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 10, PUC Karnataka Science Class 12 Department of Pre-University Education, Karnataka. Let R, S and T be the foots of the perpendiculars drawn from P to the x, y and z axes respectively. Find the direction cosines of a vector 2i – 3j + k . 0 votes . 1 Answer A. S. Adikesavan Jul 1, 2016 ... How do I find the direction angle of vector #<-sqrt3, -1>#? 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Number proportional to the direction cosines of the direction cosines of the vector define how an. Vector coordinates is equal to its direction cosines and direction Ratios of the vector in space three-dimensional coordinate grid x/r! In math, please use our google custom search here 3i vector + k, have. 3 ﷯ ( 2,1,2 ) and ( 4,2,0 ) j ^ − 3 k ^ ∠ PRO = PSO. By the length of the vector 6 i ^ + 2 ﷯ + 3 ﷯ a given vector in.! Selected Aug 22, 2018 by Vikash Kumar, -, and - or -axis of vector the! Can be determined by dividing the corresponding coordinate of a vector by the length of the 4! Space either given in component form or represented graphically math, please our... Angles correct to the x, y and z axes respectively that in three-dimensional space coordinate how to find direction cosines of a vector PSO = PSO. − z 3 ) and of distance r from the origin in the space with coordinates ( x y. That in three-dimensional space y and z axes respectively know that in space! The origin Give the direction cosines is equal to one vector is equal to one the =. By dividing the corresponding coordinate of a line which makes equal angles with the coordinate axes P to the cosines... Angles with the coordinate axes SunilJakhar ( 89.0k points ) selected Aug 22, 2018 by SunilJakhar ( points! Or d and is the distance d BETWEEN TWO points in space coordinate grid j +! Look at the vector in space cosine is that direction cosines for a given vector in the three-dimensional space Kumar! Direction Ratios of the line 4 − x 2 = y 6 = 1 − z 3 find direction... Rotated around the axis of the vector use our google custom search here z/r! ) selected Aug 22, 2018 by Vikash Kumar answered Aug 22, 2018 by SunilJakhar 89.0k... Let r, s and T be the foots of the following Vectors or graphically! ) selected Aug 22, 2018 by SunilJakhar ( 89.0k points ) selected Aug 22, by... The foots of the line joining the points ( 2,1,2 ) and of distance r from origin. The x, y and z axes respectively 4 − x 2 = y 6 = 1 z. Ratio of a line OP passes through the origin in the three-dimensional space y/r, z/r ) =. J vector + j vector + k line which makes equal angles with the coordinate axes lesson in., -1/√41 of the vector length given Ratios '' cosines and direction angles correct to the direction cosines a! The perpendiculars drawn from P to the x, y and z respectively... Shown below on the x-y-z plane in math, please use our google custom search here direction... 6/√41, -1/√41 − 3 k ^ to one = 1 − z 3, and... ( 4,2,0 ): ( x/r, y/r, z/r ) x/r = 3/ √89 ( points... Evaluating simple trigonometric expressions, Px yz22 22,, points ( 2,1,2 ) (! The foots of the squares of … direction cosines of a vector in the image in component form represented... 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# Subtraction Subtraction is one of the four basic arithmetic operations; it is essentially the opposite of addition. Subtraction is denoted by an minus sign in infix notation. The traditional names for the parts of the formula cb = a are minuend (c) − subtrahend (b) = difference (a). The words "minuend" and "subtrahend" are virtually absent from modern usage, while "difference" is very common. Subtraction is used to model several closely related processes: 1. From a given collection, take away (subtract) a given number of objects. 2. Combine a given measurement with an opposite measurement, such as a movement right followed by a movement left, or a deposit and a withdrawal. 3. Compare two objects to find their difference. For example, the difference between \$800 and \$600 is \$800 − \$600 = \$200. In mathematics, it is often useful to view or even define subtraction as a kind of addition, the addition of the opposite. We can view 7 − 3 = 4 as the sum of two terms: seven and negative three. This perspective allows us to apply to subtraction all of the familiar rules and nomenclature of addition. Subtraction is not associative or commutative— in fact, it is anticommutative— but addition of signed numbers is both. ## Basic subtraction: integers Imagine a line segment of length b with the left end labeled a and the right end labeled c. Starting from a, it takes b steps to the right to reach c. This movement to the right is modeled mathematically by addition: a + b = c. From c, it takes b steps to the left to get back to a. This movement to the left is modeled by subtraction: cb = a. Now, imagine a line segment labelled with the numbers 1, 2, and 3. From position 3, it takes no steps to the left to stay at 3, so 3 − 0 = 3. It takes 2 steps to the left to get to position 1, so 3 − 2 = 1. This picture is inadequate to describe what would happen after going 3 steps to the left of position 3. To represent such an operation, the line must be extended. To subtract arbitrary natural numbers, one begins with a line containing every natural number (0, 1, 2, 3, 4, ...). From 3, it takes 3 steps to the left to get to 0, so 3 − 3 = 0. But 3 − 4 is still invalid since it again leaves the line. The natural numbers are not a useful context for subtraction. The solution is to consider the integer number line (…, −3, −2, −1, 0, 1, 2, 3, …). From 3, it takes 4 steps to the left to get to −1, so 3 − 4 = −1.
Classrooms Curricula References Textbooks Site Map ## Definitions Set means the same thing in algebra as it does in English: a collection of objects. When you first start talking about sets in class you usually talk about things like "the set of all dogs" or "the set of students in class today". Sets that are made up of real world objects are easy to understand because they're based on your every day experience. Once you've got the basic idea, the discussion drifts back to mathematics by talking about sets of numbers. These sets can be finite, i.e. they have a specific number of elements, or inifinite. Finite sets are usually defined by listing their elements, for example {1, 3, 5} is the set whose elements are the numbers 1, 3 and 5. Infinite sets usually have to be described using words, for example "the set of all numbers that can be written as a fractoin" or "the set of all numbers between 0 and 1". There are some special sets of numbers that you should be aware of. Name Symbol Description Natural Numbers The numbers that "occur in nature": 1, 2, 3, 4, ... Whole Numbers The natural numbers and 0: 0, 1, 2, 3, 4, ... Integers The whole numbers and their negatives: ..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ... Rational Numbers Any number that can be written as a fraction. This includes the integers because you can, for example, write 2 as 2 / 1. Algebraic Numbers Any number that can be the solution of a polynomial equation (e.g. x2 - 2 = 0. This set includes all the rational numbers and "roots" such as . Real Numbers Any number including the rational numbers, the algebraic numbers and "irrational numbers" (ones that can't be written as a fraction and that can't be the solution of a polynomial equations, e.g. π). ## Special Symbols Set theory has its a collection of special symbols for representing ideas like "intersection" and "inclusion". In the examples below assume that A = {1, 2, 3, 4, 5}, B = {-5, -4, -3, -2, -1, 0)} and C = {1, 2, 3, 4, 5, 6, 7, 8}. Name Symbol Description Example element of The item on the left of the symbol is contained in the set on the right. 1∈A subset All of the elements in the set on the left side of the symbol are also in the set on the right side of the symbol. AC because every number in A is also in C. proper subset All of the elements in the set on the left side of the symbol are also in the set on the right side of the symbol but the set on the right has at least one element that isn't in the set on the left, i.e. the sets aren't equal. You could never say that AA because every number in the set on the left is also in the set on the right. intersection the set of all elements that the two sets have in common AC = {1, 2, 3, 4, 5} union the set of all elements that are in either of the two sets, i.e. the two sets combined together into one big set AB = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} empty set the set with no elements AB = because they have no elements in common. There's a technicality with the subset idea that you should be aware of. You should never say {1} ∈ C. At first glance this might look okay. After all, 1 is an element of the set C, right? Here's the problem: {1} doesn't mean the element 1, it means "the set whose only element is 1". The original set, C, doesn't contain any sets, only numbers, so it's technically incorrect to
Video: Ordering Up to 6 Objects by Length \| Nagwa Video: Ordering Up to 6 Objects by Length \| Nagwa # Video: Ordering Up to 6 Objects by Length In this video, we will learn how to compare the lengths of 6 objects both directly and indirectly. 03:44 ### Video Transcript Ordering up to Six Objects by Length In this video, we will learn how to compare the lengths of objects and order them from longest to shortest or shortest to longest. If we look closely and compare the lengths, we can see that the snake at the top is the longest and the snake at the bottom is the shortest. Now that we’ve compared the lengths of the snakes, we can order the objects. We can order the snakes from shortest to longest. The shortest snake would go at the end of the line which is marked the shortest, and the longest snake goes at the end of the line marked longest. We could also order the snakes from longest to shortest. Now let’s practice ordering some objects from longest to shortest and shortest to longest. Order the lines from longest to shortest. The question tells us we need to order the lines from the longest to the shortest. We need to find the longest line. To do this, we need to compare the lengths of the lines. Now that we’ve compared the lengths using these dotted lines, we can see that line 𝑏 is the longest. Now we need to compare the lengths of lines 𝑎, 𝑐, 𝑑, and 𝑒 to find the longest. Line 𝑎 is the longest of these lines. Now we can compare the lengths of lines 𝑐, 𝑑, and 𝑒. 𝑐 is the longest of these lines. Finally, we just need to compare the lengths of lines 𝑑 and 𝑒. 𝑑 is the longest of these two lines. That leaves us with line 𝑒, which means that line 𝑒 is the shortest. The order of the lines from longest to shortest is 𝑏, 𝑎, 𝑐, 𝑑, 𝑒. Let’s practice another example which asks us to order lines according to their lengths. Order the lines from shortest to longest. In this question, we have to order the lines from shortest to longest. To find the shortest, we need to compare the lengths. Now it’s easy to see that line 𝑒 is the shortest. Now we need to compare lines 𝑎, 𝑏, 𝑐, and 𝑑. Which of these is the shortest? It’s line 𝑑. Now we can compare the lengths of the remaining lines, lines 𝑎, 𝑏, and 𝑐. 𝑎 is the shortest. Now we just have two lines left to compare, lines 𝑏 and 𝑐. Which of these is the shortest? It’s line 𝑐. This just leaves us with line 𝑏, which is the longest. So the order of the lines from shortest to longest is 𝑒, 𝑑, 𝑎, 𝑐, 𝑏. In this video, we’ve learnt how to compare the lengths of objects and order them according to length from shortest to longest or longest to shortest. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
Courses Courses for Kids Free study material Offline Centres More Store # The population of a city increases at a rate proportional to the population at that time if the population of the city increases from $20$lakhs, the population after $15$ years. Last updated date: 19th Jul 2024 Total views: 348.9k Views today: 8.48k Verified 348.9k+ views Hint: In order to find the population after $15$ years, we must apply the method of proportion between the time period and the population as time period is directly proportional to population. Upon solving the obtained equation, we get the required answer. Now let us be brief about proportion. Now let us briefly discuss proportions. Proportion is nothing but saying that two ratios are equal. Two ratios can be written in proportion in the following ways- $\dfrac{a}{b}=\dfrac{c}{d}$ or $a:b=c:d$. From the second way of notation, the values on the extreme end are called as extremes and the inner ones as means. Proportions are of two types: direct proportions and indirect or inverse proportions. In the direct proportion, there would be direct relation between the quantities. In the case of indirect proportion, there exists indirect relation between the quantities. Now let us find the population after $15$years. We know that, $\text{time}$ $\alpha$ $\text{ population}$ We can express this relation as $\dfrac{{{P}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{T}_{2}}}$ Let us consider the present year as $1$. \begin{align} & {{P}_{1}}{{T}_{2}}={{P}_{2}}{{T}_{1}} \\ & \Rightarrow 2000000\times 15={{P}_{2}}\times 1 \\ & \Rightarrow 30000000 \\ \end{align} $\therefore$ The population after $15$ years is $30000000$. Note: We must always assign the variable to the value to be found. We can use ratios and proportions in our daily life. We can apply a ratio for adding the quantity of milk to water or water to milk. We can apply proportions for finding the height of the buildings and trees and many more.
# How do you solve -20x+7y=137, 4x+5y=43? ##### 1 Answer May 19, 2018 $x = - 3$ $y = 11$ #### Explanation: you have 2 equations and 2 unknowns so let's solve by elimination: $- 20 x + 7 y = 137$ $\text{ } 4 x + 5 y = 43$ $- 20 x \mathmr{and} 4 x \text{ have a common denominator } 20 x$ so multiply the bottom equation by 5: $- 20 x + 7 y = 137$ $5 \left(4 x + 5 y = 43\right)$ $- 20 x + 7 y = 137$ $\text{ } 20 x + 25 y = 215$ now add the equations together, notice the x terms cancel: $32 y = 352$ $y = 11$ finally insert y's value in to either of the original equations to solve for x: $4 x + 5 y = 43$ $4 x + 5 \left(11\right) = 43$ $4 x + 55 = 43$ $4 x = - 12$ $x = - 3$
# Year 3 ## Year 3 Maths ### Number - number and place value Pupils should be taught to: • count from 0 in multiples of 4, 8, 50 and 100; find 10 or 100 more or less than a given number • recognise the place value of each digit in a 3-digit number (100s, 10s, 1s) • compare and order numbers up to 1,000 • identify, represent and estimate numbers using different representations • read and write numbers up to 1,000 in numerals and in words • solve number problems and practical problems involving these ideas Pupils now use multiples of 2, 3, 4, 5, 8, 10, 50 and 100. They use larger numbers to at least 1,000, applying partitioning related to place value using varied and increasingly complex problems, building on work in year 2 (for example, 146 = 100 + 40 + 6, 146 = 130 +16). Using a variety of representations, including those related to measure, pupils continue to count in 1s, 10s and 100s, so that they become fluent in the order and place value of numbers to 1,000. ### Number - addition and subtraction Pupils should be taught to: • add and subtract numbers mentally, including: • a three-digit number and 1s • a three-digit number and 10s • a three-digit number and 100s • add and subtract numbers with up to 3 digits, using formal written methods of columnar addition and subtraction • estimate the answer to a calculation and use inverse operations to check answers • solve problems, including missing number problems, using number facts, place value, and more complex addition and subtraction Pupils practise solving varied addition and subtraction questions. For mental calculations with two-digit numbers, the answers could exceed 100. Pupils use their understanding of place value and partitioning, and practise using columnar addition and subtraction with increasingly large numbers up to 3 digits to become fluent ### Number - multiplication and division Pupils should be taught to: • recall and use multiplication and division facts for the 3, 4 and 8 multiplication tables • write and calculate mathematical statements for multiplication and division using the multiplication tables that they know, including for two-digit numbers times one-digit numbers, using mental and progressing to formal written methods • solve problems, including missing number problems, involving multiplication and division, including positive integer scaling problems and correspondence problems in which n objects are connected to m objects Pupils continue to practise their mental recall of multiplication tables when they are calculating mathematical statements in order to improve fluency. Through doubling, they connect the 2, 4 and 8 multiplication tables. Pupils develop efficient mental methods, for example, using commutativity and associativity (for example, 4 × 12 × 5 = 4 × 5 × 12 = 20 × 12 = 240) and multiplication and division facts (for example, using 3 × 2 = 6, 6 ÷ 3 = 2 and 2 = 6 ÷ 3) to derive related facts (30 × 2 = 60, 60 ÷ 3 = 20 and 20 = 60 ÷ 3). Pupils develop reliable written methods for multiplication and division, starting with calculations of two-digit numbers by one-digit numbers and progressing to the formal written methods of short multiplication and division. Pupils solve simple problems in contexts, deciding which of the 4 operations to use and why. These include measuring and scaling contexts, (for example 4 times as high, 8 times as long etc) and correspondence problems in which m objects are connected to n objects (for example, 3 hats and 4 coats, how many different outfits?; 12 sweets shared equally between 4 children; 4 cakes shared equally between 8 children). ### Number - fractions Pupils should be taught to: • count up and down in tenths; recognise that tenths arise from dividing an object into 10 equal parts and in dividing one-digit numbers or quantities by 10 • recognise, find and write fractions of a discrete set of objects: unit fractions and non-unit fractions with small denominators • recognise and use fractions as numbers: unit fractions and non-unit fractions with small denominators • recognise and show, using diagrams, equivalent fractions with small denominators • add and subtract fractions with the same denominator within one whole [for example, + = ] • compare and order unit fractions, and fractions with the same denominators • solve problems that involve all of the above Pupils connect tenths to place value, decimal measures and to division by 10. They begin to understand unit and non-unit fractions as numbers on the number line, and deduce relations between them, such as size and equivalence. They should go beyond the [0, 1] interval, including relating this to measure. Pupils understand the relation between unit fractions as operators (fractions of), and division by integers. They continue to recognise fractions in the context of parts of a whole, numbers, measurements, a shape, and unit fractions as a division of a quantity. Pupils practise adding and subtracting fractions with the same denominator through a variety of increasingly complex problems to improve fluency. ### Measurement Pupils should be taught to: • measure, compare, add and subtract: lengths (m/cm/mm); mass (kg/g); volume/capacity (l/ml) • measure the perimeter of simple 2-D shapes • add and subtract amounts of money to give change, using both £ and p in practical contexts • tell and write the time from an analogue clock, including using Roman numerals from I to XII, and 12-hour and 24-hour clocks • estimate and read time with increasing accuracy to the nearest minute; record and compare time in terms of seconds, minutes and hours; use vocabulary such as o’clock, am/pm, morning, afternoon, noon and midnight • know the number of seconds in a minute and the number of days in each month, year and leap year • compare durations of events [for example, to calculate the time taken by particular events or tasks] Pupils continue to measure using the appropriate tools and units, progressing to using a wider range of measures, including comparing and using mixed units (for example, 1 kg and 200g) and simple equivalents of mixed units (for example, 5m = 500cm). The comparison of measures includes simple scaling by integers (for example, a given quantity or measure is twice as long or 5 times as high) and this connects to multiplication. Pupils continue to become fluent in recognising the value of coins, by adding and subtracting amounts, including mixed units, and giving change using manageable amounts. They record £ and p separately. The decimal recording of money is introduced formally in year 4. Pupils use both analogue and digital 12-hour clocks and record their times. In this way they become fluent in and prepared for using digital 24-hour clocks in year 4. ### Geometry - properties of shapes Pupils should be taught to: • draw 2-D shapes and make 3-D shapes using modelling materials; recognise 3-D shapes in different orientations and describe them • recognise angles as a property of shape or a description of a turn • identify right angles, recognise that 2 right angles make a half-turn, 3 make three-quarters of a turn and 4 a complete turn; identify whether angles are greater than or less than a right angle • identify horizontal and vertical lines and pairs of perpendicular and parallel lines Pupils’ knowledge of the properties of shapes is extended at this stage to symmetrical and non-symmetrical polygons and polyhedra. Pupils extend their use of the properties of shapes. They should be able to describe the properties of 2-D and 3-D shapes using accurate language, including lengths of lines and acute and obtuse for angles greater or lesser than a right angle. Pupils connect decimals and rounding to drawing and measuring straight lines in centimetres, in a variety of contexts. ### Statistics Pupils should be taught to: • interpret and present data using bar charts, pictograms and tables • solve one-step and two-step questions [for example ‘How many more?’ and ‘How many fewer?’] using information presented in scaled bar charts and pictograms and tables Pupils understand and use simple scales (for example, 2, 5, 10 units per cm) in pictograms and bar charts with increasing accuracy. They continue to interpret data presented in many contexts. Top
# What is the position of a number? What is the position of a number? ## What is the position of a number? In our decimal number system, the value of a digit depends on its place, or position, in the number. Each place has a value of 10 times the place to its right. A number in standard form is separated into groups of three digits using commas. Each of these groups is called a period. ## What are positions in maths? Position in maths refers to identifying and recording where something is located, usually on a grid or a map. Your child will usually do this using coordinates. Movement refers to the concepts of rotation, translation and symmetry. Position and movement are both topics in geometry, which is the study of shapes. How do you teach position and direction? Use one of your child’s toys like a toy car or teddy bear. Discuss how a quarter turn is a turn or rotation of one right angle or 90 degrees, and a half turn is a turn or rotation of two right angles or 180 degrees. Move the toy to demonstrate how different turns affect the position of an object. What position word means? Positional words are specific prepositions that describe a noun’s position relative to another noun. Above, below, and under are all positional words, as are: underneath. between. in front of. ### How do you write numbers in positions? Ordinal numbers always have a suffix tacked onto the end; cardinal numbers do not. 1. first (1st) 2. second (2nd) 3. third (3rd) 4. fourth (4th) 5. fifth (5th) 6. sixth (6th) 7. seventh (7th) 8. eighth (8th) ### What is the position of 1? The point guard (PG), also known as the one, is typically the team’s shortest player and best ball handler and passer. What is position and examples? Position is how a person or thing is placed or an opinion or where a person or thing is located in relation to others. An example of position is sitting. An example of position is to be against the death penalty. An example of position is a cup between two other cups on a table. noun. What are the position words? Positional words (or positional language) are words and phrases that describe the position of people or objects. Examples would be the words ‘in’, ‘under’, and ‘over’, or the phrases ‘on top of’, or ‘next to’. An example of a sentence containing positional words would be ‘The puppet is on the castle. ‘ ## What words describe position? Positional and Directional Words above after around behind below beside down end far in inside left next to off on
BAB IV - Simponi MDP BAB IV FUNGSI VEKTOR DALAM RUANG DIMENSI TIGA 4.1 FUNGSI VEKTOR  Fungsi Vektor dalam ruang dimensi tiga ditentukan oleh r(t) = f(t) i + g(t) j + h(t) k  Cara menggambar busur suatu persamaan vektor Substitusi nilai t dalam interval ke persamaan vektor.  Gambarkan titik-titik tersebut dalam ruang dimensi tiga.  Hubungkan titik-titik tersebut.  4.2 KECEPATAN, PERCEPATAN, DAN PANJANG BUSUR  Jika fungsi vektor r(t) = f(t) i + g(t) j + h(t) k maka kecepatan = v(t) = r’(t) percepatan = a(t) = r”(t) panjang busur = s s  [f ' (t)]2  [f ' (t)]2  [f ' (t)]2 dt pada a  t  b 4.3 KELENGKUNGAN DAN KOMPONEN VEKTOR  r(t) = f(t) i + g(t) j + h(t) k di titik t = t1 \| T ' (t ) \| κ= \| r ' (t ) \| r ' (t ) T (t ) = \| r ' (t ) \|  Komponen Vektor  Vektor Singgung aT  r '•r" = \| r '\| Vektor Normal \| r '×r"\| aN = \| r '\| Contoh Soal Kelengkungan Tentukan kelengkungan r = (t2-1)i + (2t+3)j + (t2-4t)k di t = 2 Solusi r’ = 2ti + 2j + (2t-4)k \|r’\| = 2√2t2-4t+5 r' T= \| r '\| 2[ti + j + (t - 2)k ] T= 2 2 2t - 4t + 5 [ti + j + (t - 2)k ] T= 2 2t - 4t + 5 T = [ti + j + (t-2)k] [ 2t2 - 4t + 5]-1/2 T = [t (2t2 - 4t + 5)-1/2 i] + [(2t2 - 4t + 5)-1/2 j] + [(t-2) (2t2 - 4t + 5)-1/2 k ] T’ = [(2t2- 4t+5)-1/2 + t(-1/2)(4t-4)(2t2 - 4t + 5)-3/2 ] i + [(-1/2)(4t-4)(2t2 - 4t + 5)-3/2] j + [(2t2- 4t+5)-1/2 + (t-2)(-1/2) (4t-4) (2t2 - 4t + 5)-3/2] k T’ = [(2t2- 4t+5)-1/2 + (-2t2+2t)(2t2 - 4t + 5)-3/2 ] i + [(-2t+2)(2t2 - 4t + 5)-3/2] j + [(2t2- 4t+5)-1/2+(-2t2+6t-4)(2t2 - 4t + 5)-3/2] k r’ = 2ti + 2j + (2t-4)k r’(t=2) = 4i + 2j \|r’\| = √20 = 2√5 T’(t=2)= [1/(5√5) i – 2/(5√5) j + 1/√5 k ] \|T’\| = √1/125 + 4/125 + 1/5 \|T’\| = √30/125 \| T '\| κ= \| r '\| 30 125 κ= 2 5 κ= 30 1 125 2 5 30 κ= 50 = 30 5 52 5
# 4 men,2 women and a child are to be seated around a round table with 7 seats. Find the number of ways they may be arranged if !? a) the child is to be seated between two women b) the child is to be seated between two men a. 48. b. 288 #### Explanation: Let's first see that we're working with a table, meaning there is no first seat or last seat (like what we have in a row). What that gives us then is not numbered or distinct seats but instead indistinct seating where we only care about the seating relationships between people. For a, we're placing the two women on either side of the child. There are 2 ways for the two women to sit. The four men can sit wherever they'd like in the remaining seats, which is 4! = 24. All together then, we have $2 \times 24 = 48$ ways For b, we're placing two of the four men around the child. We can find the number of arrangements as either a permutation: (4!)/(2!)=4xx3=12 or as a combination, seeing that each combination of men has two ways they can be arranged: $2 \left(\begin{matrix}4 \\ 2\end{matrix}\right) = 2 \left(6\right) = 12$ Either way, we get to 12 ways to arrange the men. We can arrange the remaining 2 men and 2 women in 4! =24 ways. This all gives: $12 \times 24 = 288$ ways
### Home > CC1MN > Chapter 8 > Lesson 8.1.1 > Problem8-7 8-7. Simplify each numerical expression. 1. $\| 5 - 6 + 1 \|$ Simplify within the absolute value signs first. $\|5 − 6 + 1\| = \|0\|$ The absolute value of $0$ is $0$. 1. $2\|-16.75\|$ First, find the absolute value of$−16.75$. The absolute value is the distance a number is away from $0$. It is always a non-negative number. Since $−16.75$ is $16.75$ units away from zero, the absolute value is $16.75$. What is the product of 2(16.75)? 1. $\| 6 \frac { 3 } { 8 } - 2 \| + \| - 8 \frac { 5 } { 8 } + ( - 1 ) \|$ Simplify within the absolute value signs first. $6\frac{3}{8}-2=\frac{51}{8}-\frac{2}{1}=\frac{51}{8}-\frac{16}{8}=\frac{35}{8}$ $-8\frac{5}{8}+(-1)= -\frac{69}{8}+\left(-\frac{1}{1}\right)\\=-\frac{69}{8}+\left(-\frac{8}{8}\right)=-\frac{77}{8}$ $\text{Therefore, we have: } \left \| \frac{35}{8} \right \|+\left \| -\frac{77}{8} \right \|$ What is the absolute value for each of these numbers? $14$
# Same and Different I’ve been using “Same and Different” as an inquiry strategy with my students for several years. Read on for more about this thinking routine and some of my favorite prompts. ##### What is “Same and Different” ? “Same and Different” is an inquiry strategy sometimes known as “Compare and Contrast” or various other names (see Resources links below). This powerful strategy asks students to compare and analyze features of two mathematical situations. They may require different solution strategies, be similar except for one feature, or have mathematically meaningful nuances to notice. The routine is launched by presenting two or more math situations, then have students examine and note how they are the same and how they are different. This is a great opportunity to use technology to illuminate some of the numerical and graphical differences, although technology is not required. The examples that follow are grouped (loosely) by theme, and many can be used in several math levels. Keep in mind that not all students will find all the similarities and differences, and you can highlight the details that are pertinent to your curriculum. ##### Representation Examples These examples begin with a visual representation to tap into students’ interpretation of the graph or diagram. Example 1: (PreAlgebra or Algebra 1) The following two images show these subtraction problems: 5 – 2 (top) and 2 – 5 (bottom). I like beginning with the number line representation because it spurs more discussion than simply presenting the numerical expressions. Example 2: (Algebra 1) These images get students talking about slope and y-intercept of linear equations. Example 3: (Algebra 1 or 2) These graphs of quadratic functions show various transformations of the parent function $y=x^2$ (graphs 1 & 3). Discussion can highlight the types of transformations, how they impact the equation, and what the graphs have in common. Example 4: (Geometry) Continuing with the idea of transformations, students can interpret these geometric transformations on the coordinate plane. Discussion can analyze what is the same and different about the pre-image and image, contrast the two transformations, and generalize a coordinate rule for these motions. ##### How to Solve? Examples The next set of examples presents problems that can be solved with a variety of methods. Students can debate pros and cons of the available techniques. Examples 5, 6, 7: (Algebra 1) Solving linear equations: Which inverse operations are appropriate? Is it beneficial to distribute first or not? Example 8: (Algebra 1 or 2) Solving systems of equations. Which method would you choose: graphing, substitution, or elimination? Example 9: (Algebra 1 or 2) Solving quadratic equations. Would you factor, use inverse operations, or use the quadratic formula? Or something else? Example 10A: (Algebra 2 or Precalculus) Solving equations involving exponents. Is the next step to take a root or a log of both sides, or is there another technique available? ##### Features of Functions Examples In these examples, think about how graphs or numerical tables of values can be used to determine the similarities and differences between the functions. Don’t forget to zoom in or out on the graph to see key features and end behavior, and change the table increment if warranted. Example 10B: (Algebra 1 or 2) Linear vs. Exponential functions. For more about this scenario, check the first part of my post Table Techniques, with student activity sheet here. Examples 11, 12, 13: (Algebra 2 or Precalculus) Students analyze the features of power functions, exponential functions, and logarithmic functions. For examples 12 & 13, what happens with other bases? Example 14: (Algebra 1 or 2) Forms of Quadratic Functions. These are all actually the same—but what information about the parabola graphs is most clearly visible in each of these equations? There’s another quadratic function Bonus prompt in the Google Drive Folder of all the images. Example 15: (Algebra 2 or PreCalculus) Rational Functions often can be simplified. Is the simplified version exactly the same as the original? My favorite way to examine the above situation is to use the TI-84+ ZoomDecimal window; since it equalizes the pixels, a hole is visible in the graph of the rational function. Then, use the “tracer ball” graphing style to graph the simplified linear function–it goes ‘through’ the hole, demonstrating that the domains of the two graphs are different. Another nice question is how does the rational function’s graph compare to the numerator quadratic alone? ##### Simplification and Notation Examples Examples 16 & 17: (Algebra 1 or 2) Simplifying fractions and radicals. What is possible, what is necessary in each situation? Use a calculator or grapher to check. Examples 18 & 19: (Algebra 1 or 2) Dealing with an exponent of –1. Does it “go to the denominator” or does it “flip”? When does an exponent “distribute”? Example 20: (Algebra 2 or Precalculus) I’ve found that –1 is one of the most challenging notations for students, because it means different things in different situations. How do you read these notations aloud? What does each mean? What other ways can each be written? Example 21: (Algebra 2, Precalculus) Notation issues can get confusing with trigonometry, so I like this prompt to tease out the meanings of the 2 in each case. This is also helpful for calculus students who need a review! Discuss algebraic meaning and check the graphs for visual evidence. ##### Geometry Examples Examples 22 & 23: (Geometry) What are the similarities and differences between polygons and regular polygons? What about convex and concave polygons?1 Example 24: (Geometry) What is the same and different about these 3 triangles? What can you determine about their angles?2 Example 25: (Geometry) Quadrilaterals of all types! If you are teaching special quadrilaterals, ask students to compare and contrast any two of them: • Squares and Rectangles • Parallelograms and Trapezoids • Trapezoids and Isosceles Trapezoids • Squares and Rhombuses (rhombi?) • Rhombuses and Kites Other geometric figures: • Congruent triangles vs. Similar triangles • Circles vs. Ellipses • Cylinders vs. Prisms Example 26: (Geometry) How does scaling a polygon by a factor of 2 change the measurements we can take, such as perimeter, area, interior angle measure, or number of vertices? Here is an animation from Mathigon Polypad3 showing what measurements are the same and what are different. ##### Bonus Examples: Absolute Value & Sequences/Series Bonus Example 1: (Algebra 2) Absolute Value Equations and Inequalities. How can you represent each of these on a number line? How do you solve each of these? Can a graph in the coordinate plane help? I prefer the “distance from zero on the number line” explanation for absolute value. So in the equation, we need any points that are exactly 2 units from zero (2 and –2). In the inequalities, we need values that are MORE than 2 units away (x $\leq$ –2 or x $\geq$ 2) or LESS than 2 units away from zero ( –2 $\leq$ x $\leq$ 2). In class, I wave my hands in the air along an imaginary number line to help students visualize the correct intervals.4 The coordinate plane can help students understand these situations: graph y = 2 and y = \|x\| at the same time. The intersection points are where the graphs are equal (top image below). Then look for x-values where y-values of the \|x\| graph is greater than 2 (middle image) or where y-values of the \|x\| graph is less than 2 (bottom image). [Read more about a coordinate plane graph visual in my post How Else Can We Show This? part 2 “Absolute Certainty”—with video!] Bonus Example 2: (Algebra 2 or Precalculus) When studying sequences and series, students learn several formulas. The sums of geometric series formulas (finite and infinite) are actually the same formula for large values of n if r is between -1 and 1! Use this prompt and numerical examples to examine this situation. Why does this formula have a restriction on the size of common ratio r? ##### Wrapping Up I’ve used Same and Different prompts as a lesson opener, exit ticket, stimulus for discussion, and during class review; they work well as formative assessment or consolidation anytime. Be sure to annotate and share student thinking and summarize important nuances with the whole class. With a 3-part prompt, consider presenting two parts first, then add the third (in order to reduce the initial cognitive load of examining all three parts of the prompt). I hope you’ve found some useful ideas here in this (longer than I expected) post. See below for more Same and Different resources. ##### Notes & Resources: ALL OF THE IMAGES from this post are in this Google Drive Folder. Please use with your math classes and the Same and Different thinking routine. Check out my follow-up post for Calculus topics: “Same & Different: Calculus Edition Same and Different is also called “Same But Different”, “Same or Different?” and “Compare & Contrast”. Here are some great resources: One more resource is the Math Routine Collaborative group of educators which meets periodically on Zoom to learn and discuss various math routines; moderated by Shelby Strong (@StrongerMath), Annie Fetter (@MFAnnie), and Annie Forest (@mrsforest). More information and register here: http://www.strongermath.com/mrc/. 1Tim Brzezinski (@TimBrzezinski) has a GeoGebra interactive applet on Convex vs. Concave polygons here: https://www.geogebra.org/m/knnPDMR3 2My GeoGebra activity Converse Pythagorean Theorem dives in to this topic more deeply. https://www.geogebra.org/m/zvdhmzv8 3The Mathigon Polypad is a wonderful playground for virtual manipulatives that includes both fun and serious math. Each of these measurements is accompanied by music and animation! Follow @MathigonOrg and David Poras (@davidporas) for more. 4There are many good GeoGebra visualizers for absolute value equations and inequalities graphed on a number line. One I found is here: https://www.geogebra.org/m/V4SwRtrb If you’d like to get an email whenever I post a new blog, enter your email here:
# Modeling, Functions, and Graphs ## Section1.6Linear Regression We have spent most of this chapter analyzing models described by graphs or equations. To create a model, however, we often start with a quantity of data. Choosing an appropriate function for a model is a complicated process. In this section, we consider only linear models and explore methods for fitting a linear function to a collection of data points. First, we fit a line through two data points. ### SubsectionFitting a Line through Two Points If we already know that two variables are related by a linear function, we can find a formula from just two data points. For example, variables that increase or decrease at a constant rate can be described by linear functions. #### Example1.143. In 1993, Americans drank 188.6 million cases of wine. Wine consumption increased at a constant rate over the next decade, and we drank 258.3 million cases of wine in 2003. (Source: Los Angeles Times, Adams Beverage Group) 1. Find a formula for wine consumption, $$W\text{,}$$ in millions of cases, as a linear function of time, $$t\text{,}$$ in years since 1990. 2. State the slope as a rate of change. What does the slope tell us about this problem? Solution. 1. We have two data points of the form $$(t, W)\text{,}$$ namely $$(3, 188.6)$$ and $$(13, 258.3)\text{.}$$ We use the point-slope formula to fit a line through these two points. First, we compute the slope. \begin{equation} \frac{\Delta W}{\Delta t}=\frac {258.3 - 188.6}{13 - 3}= 6.97 \end{equation} Next, we use the slope $$m = 6.97$$ and either of the two data points in the point-slope formula. \begin{equation} \begin{aligned}[t] W \amp =W_1 + m(t - t_1) \\ W \amp = 188.6 + 6.97(t - 3) \\ W \amp = 167.69 + 6.97t \end{aligned} \end{equation} Thus, $$W = f (t) = 167.69 + 6.97t\text{.}$$ 2. The slope gives us the rate of change of the function, and the units of the variables can help us interpret the slope in context. \begin{equation} \frac{\Delta W}{\Delta t}= \frac{258.3 - 188.6 \text{ millions of cases}}{13 - 3\text{ years}} = 6.97 \text{ millions of cases / year} \end{equation} Over the 10 years between 1993 and 2003, wine consumption in the United States increased at a rate of 6.97 million cases per year. #### To Fit a Line through Two Points:. 1. Compute the slope between the two points. 2. Substitute the slope and either point into the point-slope formula \begin{equation} y = y_1 + m(x - x_1) \end{equation} #### Checkpoint1.144.Practice 1. In 1991, there were 64.6 burglaries per 1000 households in the United States. The number of burglaries reported annually declined at a roughly constant rate over the next decade, and in 2001 there were 28.7 burglaries per 1000 households. (Source: U.S. Department of Justice) 1. Find a function for the number of burglaries, $$B\text{,}$$ as a function of time, $$t\text{,}$$ in years, since 1990. $$y=$$ 2. State the slope as a rate of change. $$m$$ is • burglaries per 1000 households per year • households per burglary per year • burglaries per year • A) The burglary rate declined by 3.59 burglaries per decade • B) The burglary rate declined by 3.59 burglaries per 1000 households every year • C) The burglary rate declined by 3.59 burglaries per year (in the years 1991 to 2001). $$68.19-3.59t$$ $$-3.59$$ $$\text{burglaries per decade}$$ $$\text{C) The ... burglaries per year}$$ Solution. 1. Because $$t=0$$ corresponds to 1990, we have the two points (1,64.6) and (11,28.7) to compute the slope $$m=-3.59\text{.}$$ Using the point-slope formula with either point and then simplifying, we find that $$y = {68.19-3.59t}$$ 2. $$-3.59$$ burglaries per 1000 households per year. From 1991 to 2001, the burglary rate declined by 3.59 burglaries per 1000 households every year. #### Checkpoint1.145.QuickCheck 1. What are the two steps to find the equation of a line through two points? • Draw the line, use the point-slope formula • Compute the slope, find the $$y$$-intercept • Compute the slope, use the point-slope formula • Find the intercepts, compute the slope $$\text{Choice 4}$$ Solution. Compute the slope, use the point-slope formula ### SubsectionScatterplots Empirical data points in a linear relation may not lie exactly on a line. There are many factors that can affect experimental data, including measurement error, the influence of environmental conditions, and the presence of related variable quantities. #### Example1.146. A consumer group wants to test the gas mileage of a new model SUV. They test-drive six vehicles under similar conditions and record the distance each drove on various amounts of gasoline. Gasoline used (gal) $$9.6$$ $$11.3$$ $$8.8$$ $$5.2$$ $$10.3$$ $$6.7$$ Miles driven $$155.8$$ $$183.6$$ $$139.6$$ $$80.4$$ $$167.1$$ $$99.7$$ 1. Are the data linear? 2. Draw a line that fits the data. 3. What does the slope of the line tell us about the data? Solution. 1. No, the data are not strictly linear. If we compute the slopes between successive data points, the values are not constant. We can see from an accurate plot of the data, shown below, that the points lie close to, but not precisely on, a straight line. 2. We would like to draw a line that comes as close as possible to all the data points, even though it may not pass precisely through any of them. In particular, we try to adjust the line so that we have the same number of data points above the line and below the line. One possible solution is shown above. 3. To compute the slope of the our estimated line, we first choose two points on the line. Our line appears to pass through one of the data points,$$(8.8, 139.6)\text{.}$$ We look for a second point on the line whose coordinates are easy to read, perhaps $$(6.5,100)\text{.}$$ The slope is \begin{equation} m = \frac{139.6 - 100}{8.8 - 6.5}= 17.2\text{ miles per gallon} \end{equation} According to our data, the SUV gets about 17.2 miles to the gallon. #### Caution1.147. To find the slope of your estimated line, be sure to choose points on the line; do not choose any of the data points (unless they happen to lie on your line). #### Checkpoint1.148.Practice 2. 1. Plot the data points. Do the points lie on a line? • Yes • No 2. Draw a line that fits the data. $$x$$ $$1.49$$ $$3.68$$ $$4.95$$ $$5.49$$ $$7.88$$ $$8.41$$ $$y$$ $$2.69$$ $$3.7$$ $$4.6$$ $$5.2$$ $$7.2$$ $$7.3$$ $$\text{No}$$ Solution. 1. No. A scatterplot is shown below. 2. See the graph below Graphs for both parts (a) and (b): The graph in Example 1.146 is called a scatterplot. The points on a scatterplot may or may not show some sort of pattern. Consider the three plots shown below. • In figure (a), the data points resemble a cloud of gnats; there is no apparent pattern to their locations. • In figure (b), the data follow a generally decreasing trend, but certainly do not all lie on the same line. • The points in figure (c) are even more organized; they seem to lie very close to an imaginary line. #### Checkpoint1.149.QuickCheck 2. Can two points on a scatterplot have the same $$x$$-coordinate? Can two points have the same $$y$$-coordinate? • Yes, Yes • Yes, No • No, Yes • No, No $$\text{Yes, Yes}$$ Solution. Yes, Yes If the data in a scatterplot are roughly linear, we can estimate the location of an imaginary line of best fit that passes as close as possible to the data points. We can then use this line to make predictions about the data. ### SubsectionLinear Regression One measure of a person’s physical fitness is the body mass index, or BMI. Your BMI is the ratio of your weight in kilograms to the square of your height in centimeters. Thus, thinner people have lower BMI scores, and fatter people have higher scores. The Centers for Disease Control considers a BMI between 18.5 and 24.9 to be healthy. The points on the scatterplot below show the BMI of Miss America from 1921 to 1991. From the data in the scatterplot, can we see a trend in Americans’ ideal of female beauty? #### Example1.150. 1. Estimate a line of best fit for the scatterplot above. (Source: http://www.pbs.org) 2. Use your line to estimate the BMI of Miss America 1980. Solution. 1. We draw a line that fits the data points as best we can, as shown below. (Note that we have set $$t = 0$$ in 1920 on this graph.) We try to end up with roughly equal numbers of data points above and below our line. 2. We see that when $$t = 60$$ on this line, the $$y$$-value is approximately 18.3. We therefore estimate that Miss America 1980 had a BMI of 18.3. (Her actual BMI was 17.85.) #### Checkpoint1.151.QuickCheck 3. We can estimate a line of best fit by drawing the line that passes through: • The first and last data points • Two data points in the middle • Either of these • None of these $$\text{None of these}$$ Solution. None of these #### Checkpoint1.152.Practice 3. Human brains consume a large amount of energy, about 16 times as much as muscle tissue per unit weight. In fact, brain metabolism accounts for about 25% of an adult human’s energy needs, as compared to about 5% for other mammals. 1. Draw a line of best fit through the data points. 2. Estimate the amount of energy used by the brain of a hominid species that lived three million years ago. % $$10.5$$ Solution. 1. A graph is below. A graph for part (a): The process of predicting an output value based on a straight line that fits the data is called linear regression, and the line itself is called the regression line. The equation of the regression line is usually used (instead of a graph) to predict values. #### Example1.153. 1. Find the equation of the regression line in Example 1.150. 2. Use the regression equation to predict the BMI of Miss America 1980. Solution. 1. We first calculate the slope by choosing two points on the regression line. The points we choose are not necessarily any of the original data points; instead they should be points on the regression line itself. The line appears to pass through the points $$(17, 20)$$ and $$(67, 18)\text{.}$$ The slope of the line is then \begin{equation} m = \frac{18 - 20}{67 - 17}\approx -0.04 \end{equation} Now we use the point-slope formula to find the equation of the line. (If you need to review the point-slope formula, see Section 1.5.) We substitute $$m = -0.04$$ and use either of the two points for $$(x_1, y_1)\text{;}$$ we will choose $$(17, 20)\text{.}$$ The equation of the regression line is \begin{equation} \begin{aligned}[t] y \amp = y_1 + m(x - x_1)\\ y \amp = 20-0.04(x-17) \amp \amp \blert{\text{Simplify.}}\\ y \amp = 20.68 - 0.04t\\ \end{aligned} \end{equation} 2. We will use the regression equation to make our prediction. For Miss America 1980, $$t = 60$$ and \begin{equation} y = 20.68 - 0.04(60) = 18.28 \end{equation} This value agrees well with the estimate we made in Example 1.150. #### Checkpoint1.154.QuickCheck 4. How many data points must a good regression line pass through? • Three • Two • One • None $$\text{None}$$ Solution. None: the regression line does not need to pass through any of the data points. #### Checkpoint1.155.Practice 4. The number of manatees killed by watercraft in Florida waters has been increasing since 1975. Data are given at 5-year intervals in the table. (Source: Florida Fish and Wildlife Conservation Commission) Year Manatee deaths $$1975$$ $$6$$ $$1980$$ $$16$$ $$1985$$ $$33$$ $$1990$$ $$47$$ $$1995$$ $$42$$ $$2000$$ $$78$$ 1. Draw a regression line through the data points shown in the figure. 2. Find an equation for the regression line, using $$t\approx 0$$ in 1975. $$y=$$ 3. Use the regression equation to estimate the number of manatees killed by watercraft in 1998. manatees $$4.7+2.6t$$ $$65$$ Solution. 1. A graph is below. 2. $$\displaystyle y = {4.7+2.6t}$$ 3. 65 A graph for part (a) is below. #### Checkpoint1.156.Pause and Reflect. Describe a strategy for sketching a line of best fit by eye. ### SubsectionLinear Interpolation and Extrapolation Using a regression line to estimate values between known data points is called interpolation. Making predictions beyond the range of known data is called extrapolation. #### Example1.157. 1. Use linear interpolation to estimate the BMI of Miss America 1960. 2. Use linear extrapolation to predict the BMI of Miss America 2001. Solution. 1. For 1960, we substitute $$t = 40$$ into the regression equation we found in Example 1.153. \begin{equation} y = 20.68 - 0.04(40) = 19.08 \end{equation} We estimate that Miss America 1960 had a BMI of 19.08. (Her BMI was actually 18.79.) 2. For 2001, we substitute $$t = 81$$ into the regression equation. \begin{equation} y = 20.68 - 0.04(81) = 17.44 \end{equation} Our model predicts that Miss America 2001 had a BMI of 17.44. In fact, her BMI was 20.25. By the late 1990s, public concern over the self-image of young women had led to a reversal of the trend toward ever-thinner role models. Example 1.157b illustrates an important fact about extrapolation: If we try to extrapolate too far, we may get unreasonable results. For example, if we use our model to predict the BMI of Miss America 2520 (when $$t = 600$$), we get \begin{equation} y = 20.68 - 0.04(600) =-3.32 \end{equation} Even if the Miss America pageant is still operating in 600 years, the winner cannot have a negative BMI. Our linear model provides a fair approximation for 1920–1990, but if we try to extrapolate too far beyond the known data, the model may no longer apply. We can also use interpolation and extrapolation to make estimates for nonlinear functions. Sometimes a variable relationship is not linear, but a portion of its graph can be approximated by a line. The graph at right shows a child’s height each month. The graph is not linear because her rate of growth is not constant; her growth slows down as she approaches her adult height. However, over a short time interval the graph is close to a line, and that line can be used to approximate the coordinates of points on the curve. #### Checkpoint1.158.QuickCheck 5. If you add more data points to the scatterplot, could the regression line change? • Yes • No • Only the slope could change. • Only the intercept could change. $$\text{Yes}$$ Solution. Yes #### Checkpoint1.159.Practice 5. Emily was 82 centimeters tall at age 36 months and 88 centimeters tall at age 48 months. 1. Find a linear equation that approximates Emily’s height in terms of her age over the given time interval. Let $$t$$ be Emily’s age in months. $$y=$$ 2. Use linear interpolation to estimate Emily’s height when she was 38 months old, and extrapolate to predict her height at age 50 months. Age 38 months: cm Age 50 months: cm 3. Predict Emily’s height at age 25 (300 months). Age 25 years: cm • Yes • No $$82+0.5\!\left(t-36\right)$$ $$83$$ $$89$$ $$214$$ $$\text{No}$$ Solution. 1. $$\displaystyle y = 64 + 0.5t$$ 2. $$83$$ cm, $$89$$ cm 3. $$214$$ cm; No #### Checkpoint1.160.Pause and Reflect. Explain the difference between interpolation and extrapolation. Estimating a line of best fit is a subjective process. Rather than base their estimates on such a line, statisticians often use the least squares regression line. This regression line minimizes the sum of the squares of all the vertical distances between the data points and the corresponding points on the line, as shown at left. Many calculators are programmed to find the least squares regression line, using an algorithm that depends only on the data, not on the appearance of the graph. #### Technology1.161.Using Technology for Linear Regression. You can use a graphing calculator to make a scatterplot, find a regression line, and graph the regression line with the data points. On the TI-83 calculator, we use the statistics mode, which you can access by pressing STAT. You will see a display that looks like figure (a) below. Choose $$1$$ to $$Edit$$ (enter or alter) data. Now follow the instructions in Example 1.162 for using your calculator’s statistics features. #### Example1.162. 1. Find the equation of the least squares regression line for the following data: \begin{equation} (10, 12), (11, 14), (12, 14), (12, 16), (14, 20) \end{equation} 2. Plot the data points and the least squares regression line on the same axes. Solution. 1. We must first enter the data. • Press STATENTER to select $$Edit\text{.}$$ • If there are data in column $$L_1$$ or $$L_2\text{,}$$ clear them out: Use the key to select $$L_1\text{,}$$ press CLEAR, then do the same for $$L_2\text{.}$$ • Enter the $$x$$-coordinates of the data points in the $$L_1$$ column and enter the $$y$$-coordinates in the $$L_2$$ column, as shown in figure (a) below. Now we are ready to find the regression equation for our data. • Press STAT 4 to select linear regression, or LinReg (ax + b), then press ENTER. • The calculator will display the equation $$y = ax + b$$ and the values for $$a$$ and $$b\text{,}$$ as shown in figure (b). You should find that your regression line is approximately $$y = 1.95x - 7.86\text{.}$$ 2. First, we first clear out any old definitions in the list. • Position the cursor after $$Y_1 =$$ and copy in the regression equation as follows: • Press VARS $$5$$ ENTER. • To draw a scatterplot, press 2ndY=$$1$$ and set the Plot1 menu as shown in figure (a) below. • Finally, press ZOOM $$9$$ to see the scatterplot of the data and the regression line. The graph is shown in figure (b). #### Caution1.163. When you are through with the scatterplot, press Y= ENTER to turn off the $$Stat Plot\text{.}$$ If you neglect to do this, the calculator will continue to show the scatterplot even after you ask it to plot a new equation. #### Checkpoint1.164.Practice 6. Use your calculator’s statistics features to find the least squares regression equation for the data in Checkpoint 1.148. 1. The equation: $$y=$$ 2. Plot the data and the graph of the regression equation. $$0.710838x+1.33571$$ Solution. 1. $$\displaystyle y = 1.34 + 0.71x$$ 2. A graph is below. A graph for part (b): ### SubsectionSection Summary #### SubsubsectionVocabulary Look up the definitions of new terms in the Glossary. • Scatterplot • Least squares regression line • Extrapolate • Regression line • Interpolate • Linear regression #### SubsubsectionCONCEPTS 1. Data points may not lie exactly on the graph of an equation. 2. Points in a scatterplot may or may not exhibit a pattern. 3. We can approximate a linear pattern by a regression line. 4. We can use interpolation or extrapolation to make estimates and predictions. 5. If we extrapolate too far beyond the known data, we may get unreasonable results. #### SubsubsectionSTUDY QUESTIONS 1. What is a regression line? 2. State two formulas you will need to calculate the equation of a line through two points. 3. Explain the difference between interpolation and extrapolation. 4. In general, should you have more confidence in figures obtained by interpolation or by extrapolation? Why? #### SubsubsectionSKILLS Practice each skill in the Homework problems listed. 1. Find the equation of a line through two points: #1–6, 29–36 2. Draw a line of best fit: #7–18 3. Find the equation of a regression line: #11–28, 37–40 4. Use interpolation and extrapolation to make predictions: #11–40 ### ExercisesHomework 1.6 #### Exercise Group. In Problems 1–6, we find a linear model from two data points. 1. Make a table showing the coordinates of two data points for the model. (Which variable should be plotted on the horizontal axis?) 2. Find a linear equation relating the variables. 3. State the slope of the line, including units, and explain its meaning in the context of the problem. ##### 1. It cost a bicycle company $$$9000$$ to make $$40$$ touring bikes in its first month of operation and$$$15,000$$ to make $$125$$ bikes during its second month. Express the company’s monthly production cost, $$C\text{,}$$ in terms of the number, $$x\text{,}$$ of bikes it makes. ##### 2. Flying lessons cost $$$645$$ for an $$8$$-hour course and$$$1425$$ for a $$20$$-hour course. Both prices include a fixed insurance fee. Express the cost, $$C\text{,}$$ of flying lessons in terms of the length, $$h\text{,}$$ of the course in hours. ##### 3. Under ideal conditions, Andrea’s Porsche can travel $$312$$ miles on a full tank ($$12$$ gallons of gasoline) and $$130$$ miles on $$5$$ gallons. Express the distance, $$d\text{,}$$ Andrea can drive in terms of the amount of gasoline, $$g\text{,}$$ she buys. ##### 4. On an international flight, a passenger may check two bags each weighing $$70$$ kilograms, or $$154$$ pounds, and one carry-on bag weighing $$50$$ kilograms, or $$110$$ pounds. Express the weight, $$p\text{,}$$ of a bag in pounds in terms of its weight, $$k\text{,}$$ in kilograms. ##### 5. A radio station in Detroit, Michigan, reports the high and low temperatures in the Detroit/Windsor area as $$59\degree$$F and $$23\degree$$F, respectively. A station in Windsor, Ontario, reports the same temperatures as $$15\degree$$C and $$-5\degree$$C. Express the Fahrenheit temperature, $$F\text{,}$$ in terms of the Celsius temperature, $$C\text{.}$$ ##### 6. Ms. Randolph bought a used car in 2000. In 2002, the car was worth $$9000\text{,}$$ and in 2005 it was valued at $$4500\text{.}$$ Express the value, $$V$$ , of Ms. Randolph’s car in terms of the number of years, $$t\text{,}$$ she has owned it. #### Exercise Group. Each regression line can be improved by adjusting either $$m$$ or $$b\text{.}$$ Draw a line that fits the data points more closely. #### Exercise Group. In Problems 11 and 12, use information from the graphs to answer the questions. ##### 11. The scatterplot shows the ages of 10 army drill sergeants and the time it took each to run 100 meters, in seconds. 1. What was the hundred-meter time for the 25-year-old drill sergeant? 2. How old was the drill sergeant whose hundred-meter time was $$12.6$$ seconds? 3. Use a straightedge to draw a line of best fit through the data points. 4. Use your line of best fit to predict the hundred-meter time of a 28-year-old drill sergeant. 5. Choose two points on your regression line and find its equation. 6. Use the equation to predict the hundred-meter time of a 40-year-old drill sergeant and a 12 year-old drill sergeant. Are these predictions reasonable? ##### 12. The scatterplot shows the outside temperature and the number of cups of cocoa sold at an outdoor skating rink snack bar on 13 consecutive nights. 1. How many cups of cocoa were sold when the temperature was $$2\degree$$C? 2. What was the temperature on the night when $$25$$ cups of cocoa were sold? 3. Use a straightedge to draw a line of best fit through the data points 4. Use your line of best fit to predict the number of cups of cocoa that will be sold at the snack bar if the temperature is $$7\degree$$C. 5. Choose two points on your regression line and find its equation. 6. Use the equation to predict the number of cups of cocoa that will be sold when the temperature is $$10\degree$$C and when the temperature is $$24\degree$$C. Are these predictions reasonable? #### 13. With Americans’ increased use of faxes, pagers, and cell phones, new area codes are being created at a steady rate. The table shows the number of area codes in the United States each year. (Source: USA Today, NeuStar, Inc.) Year $$1997$$ $$1998$$ $$1999$$ $$2000$$ $$2001$$ $$2002$$ $$2003$$ Number of area codes $$151$$ $$186$$ $$204$$ $$226$$ $$239$$ $$262$$ $$274$$ 1. Let $$t$$ represent the number of years after 1995 and plot the data. Draw a line of best fit for the data points. 2. Find an equation for your regression line. 3. How many area codes do you predict for 2010? #### 14. The number of mobile homes in the United States has been increasing since 1960. The data in the table are given in millions of mobile homes. (Source: USA Today, U.S. Census Bureau) Year $$1960$$ $$1970$$ $$1980$$ $$1990$$ $$2000$$ Number of mobile homes $$0.8$$ $$2.1$$ $$4.7$$ $$7.4$$ $$8.8$$ 1. Let $$t$$ represent the number of years after 1960 and plot the data. Draw a line of best fit for the data points 2. Find an equation for your regression line. 3. How many mobile homes do you predict for 2010? #### 15. Teenage birth rates in the United States declined from 1991 to 2000. The table shows the number of births per 1000 women in selected years. (Source: U.S. National Health Statistics) Year $$1991$$ $$1993$$ $$1995$$ $$1996$$ $$1997$$ $$1998$$ Births $$62.1$$ $$59.6$$ $$56.8$$ $$54.4$$ $$52.3$$ $$51.1$$ 1. Let $$t$$ represent the number of years after 1990 and plot the data. Draw a line of best fit for the data points. 2. Find an equation for your regression line. 3. Estimate the teen birth rate in 1994. 4. Predict the teen birth rate in 2010. #### 16. The table shows the minimum wage in the United States at five-year intervals. (Source: Economic Policy Institute) Year $$1960$$ $$1965$$ $$1970$$ $$1975$$ $$1980$$ $$1985$$ $$1990$$ $$1995$$ $$2000$$ Minimum wage $$1.00$$ $$1.25$$ $$1.60$$ $$2.10$$ $$3.10$$ $$3.35$$ $$3.80$$ $$4.25$$ $$5.15$$ 1. Let $$t$$ represent the number of years after 1960 and plot the data. Draw a line of best fit for the data points. 2. Find an equation for your regression line. 3. Estimate the minimum wage in 1972. 4. Predict the minimum wage in 2010. #### 17. Life expectancy in the United States has been rising since the nineteenth century. The table shows the U.S. life expectancy in selected years. (Source: http://www.infoplease.com) Year $$1950$$ $$1960$$ $$1970$$ $$1980$$ $$1990$$ $$2000$$ Life expectancy at birth $$68.2$$ $$69.7$$ $$70.8$$ $$73.7$$ $$75.4$$ $$77$$ 1. Let $$t$$ represent the number of years after 1950, and plot the data. Draw a line of best fit for the data points. 2. Find an equation for your regression line. 3. Estimate the life expectancy of someone born in 1987. 4. Predict the life expectancy of someone born in 2010. #### 18. The table shows the per capita cigarette consumption in the United States at five-year intervals. (Source: http://www.infoplease.com) Year $$1980$$ $$1985$$ $$1990$$ $$1995$$ $$2000$$ Per capita cigarette consumption $$3851$$ $$3461$$ $$2827$$ $$2515$$ $$2092$$ 1. Let $$t$$ represent the number of years after 1980, and plot the data. Draw a line of best fit for the data points. 2. Find an equation for your regression line. 3. Estimate the per capita cigarette consumption in 1998. 4. Predict the per capita cigarette consumption in 2010. #### 19. "The earnings gap between high-school and college graduates continues to widen, the Census Bureau says. On average, college graduates now earn just over $$$51,000$$ a year, almost twice as much as high-school graduates. And those with no high-school diploma have actually seen their earnings drop in recent years." The table shows the unemployment rate and the median weekly earnings for employees with different levels of education. (Source: Morning Edition, National Public Radio, March 28, 2005) Years ofeducation Unemploymentrate Weeklyearnings ($) Somehigh schoolno diploma $$10$$ $$8.8$$ $$396$$ High-schoolgraduate $$12$$ $$5.5$$ $$554$$ Some collegeno degree $$13$$ $$5.2$$ $$622$$ Associate’sdegree $$14$$ $$4.0$$ $$672$$ Bachelor’sdegree $$16$$ $$3.3$$ $$900$$ Master’sdegree $$18$$ $$2.9$$ $$1064$$ Professionaldegree $$20$$ $$1.7$$ $$1307$$ 1. Plot years of education on the horizontal axis and weekly earnings on the vertical axis. 2. Find an equation for the regression line. 3. State the slope of the regression line, including units, and explain what it means in the context of the data. 4. Do you think this model is useful for extrapolation or interpolation? For example, what weekly earnings does the model predict for someone with 15 years of education? For 25 years? Do you think these predictions are valid? Why or why not? #### 20. The table shows the birth rate (in births per woman) and the female literacy rate (as a percent of the adult female population) in a number of nations. (Source: UNESCO, The World Fact Book, EarthTrends) Country Literacy rate Birth rate Brazil $$88.6$$ $$1.93$$ Egypt $$43.6$$ $$2.88$$ Germany $$99$$ $$1.39$$ Iraq $$53$$ $$4.28$$ Japan $$99$$ $$1.39$$ Niger $$9.4$$ $$6.75$$ Pakistan $$35.2$$ $$4.14$$ Peru $$82.1$$ $$2.56$$ Philippines $$92.7$$ $$3.16$$ Portugal $$91$$ $$1.47$$ Russian Federation $$99.2$$ $$1.27$$ Saudi Arabia $$69.3$$ $$4.05$$ United States $$97$$ $$2.08$$ 1. Plot the data with literacy rate on the horizontal axis. Draw a line of best fit for the data points. 2. Find an equation for the regression line. 3. What values for the input variable make sense for the model? What are the largest and smallest values predicted by the model for the output variable? 4. State the slope of the regression line, including units, and explain what it means in the context of the data. #### 21. The table shows the amount of carbon released into the atmosphere annually from burning fossil fuels, in billions of tons, at 5-year intervals from 1950 to 1995. (Source: www.worldwatch.org) Year $$1950$$ $$1955$$ $$1960$$ $$1965$$ $$1970$$ $$1975$$ $$1980$$ $$1985$$ $$1990$$ $$1995$$ Carbonemissions $$1.6$$ $$2.0$$ $$2.5$$ $$3.1$$ $$4.0$$ $$4.5$$ $$5.2$$ $$5.3$$ $$5.9$$ $$6.2$$ 1. Let $$t$$ represent the number of years after 1950 and plot the data. Draw a line of best fit for the data points. 2. Find an equation for your regression line. 3. Estimate the amount of carbon released in 1992. #### 22. High-frequency radiation is harmful to living things because it can cause changes in their genetic material. The data below, collected by C. P. Oliver in 1930, show the frequency of genetic transmutations induced in fruit flies by doses of X-rays, measured in roentgens. (Source: C. P. Oliver, 1930) Dosage(roentgens) $$285$$ $$570$$ $$1640$$ $$3280$$ $$6560$$ Percentage ofmutated genes $$1.18$$ $$2.99$$ $$4.56$$ $$9.63$$ $$15.85$$ 1. Plot the data and draw a line of best fit through the data points. 2. Find an equation for your regression line. 3. Use the regression equation to predict the percent of mutations that might result from exposure to $$5000$$ roentgens of radiation. #### 23. Bracken, a type of fern, is one of the most successful plants in the world, growing on every continent except Antarctica. New plants, which are genetic clones of the original, spring from a network of underground stems, or rhizomes, to form a large circular colony. The graph shows the diameters of various colonies plotted against their age. (Source: Chapman et al.,1992) 1. Calculate the rate of growth of the diameter of a bracken colony, in meters per year. 2. Find an equation for the line of best fit. (What should the vertical intercept of the line be?) 3. In Finland, bracken colonies over $$450$$ meters in diameter have been found. How old are these colonies? #### 24. The European sedge warbler can sing several different songs consisting of trills, whistles, and buzzes. Male warblers who sing the largest number of songs are the first to acquire mates in the spring. The data below show the number of different songs sung by several male warblers and the day on which they acquired mates, where day 1 is April 20. (Source: Krebs and Davies, 1993) Numberof songs $$41$$ $$38$$ $$34$$ $$32$$ $$30$$ $$25$$ $$24$$ $$24$$ $$23$$ $$14$$ Pairing day $$20$$ $$24$$ $$25$$ $$21$$ $$24$$ $$27$$ $$31$$ $$35$$ $$40$$ $$42$$ 1. Plot the data points, with number of songs on the horizontal axis. A regression line for the data is $$y = -0.85x + 53\text{.}$$ Graph this line on the same axes with the data. 2. What does the slope of the regression line represent? 3. When can a sedge warbler that knows $$10$$ songs expect to find a mate? 4. What do the intercepts of the regression line represent? Do these values make sense in context? #### 25. One of the factors that determines the strength of a muscle is its cross-sectional area. The data below show the cross-sectional area of the arm flexor muscle for several men and women, and their strength, measured by the maximum force they exerted against a resistance. (Source: Davis, Kimmet, Autry, 1986) Women Area (sq cm) $$11.5$$ $$10.8$$ $$11.7$$ $$12.0$$ $$12.5$$ $$12.7$$ $$14.4$$ $$14.4$$ $$15.7$$ Strength (kg) $$11.3$$ $$13.2$$ $$13.2$$ $$14.5$$ $$15.6$$ $$14.8$$ $$15.6$$ $$16.1$$ $$18.4$$ Men Area (sq cm) $$13.5$$ $$13.8$$ $$15.4$$ $$15.4$$ $$17.7$$ $$18.6$$ $$20.8$$ $$-$$ $$-$$ Strength (kg) $$15.0$$ $$17.3$$ $$19.0$$ $$19.8$$ $$20.6$$ $$20.8$$ $$26.3$$ $$-$$ $$-$$ 1. Plot the data for both men and women on the same graph using different symbols for the data points for men and the data points for women. 2. Are the data for both men and women described reasonably well by the same regression line? Draw a line of best fit through the data. 3. Find the equation of your line of best fit, or use a calculator to find the regression line for the data. 4. What does the slope mean in this context? #### 26. Astronomers use a numerical scale called magnitude to measure the brightness of a star, with brighter stars assigned smaller magnitudes. When we view a star from Earth, dust in the air absorbs some of the light, making the star appear fainter than it really is. Thus, the observed magnitude of a star, $$m\text{,}$$ depends on the distance its light rays must travel through the Earth’s atmosphere. The observed magnitude is given by \begin{equation} m = m_0 + kx \end{equation} where $$m_0$$ is the actual magnitude of the star outside the atmosphere, $$x$$ is the air mass (a measure of the distance through the atmosphere), and $$k$$ is a constant called the extinction coefficient. To calculate $$m_0\text{,}$$ astronomers observe the same object several times during the night at different positions in the sky, and hence for different values of $$x\text{.}$$ Here are data from such observations. (Source: Karttunen et al., 1987) Altitude Air mass, $$x$$ Magnitude, $$m$$ $$50\degree$$ $$1.31$$ $$0.90$$ $$35\degree$$ $$1.74$$ $$0.98$$ $$25\degree$$ $$2.37$$ $$1.07$$ $$20\degree$$ $$2.92$$ $$1.17$$ 1. Plot observed magnitude against air mass, and draw a line of best fit through the data. 2. Find the equation of your line of best fit, or use a calculator to find the regression line for the data. 3. Find the equation of your line of best fit, or use a calculator to find the regression line for the data. 4. What is the value of the extinction coefficient? What is the apparent magnitude of the star outside Earth’s atmosphere? #### 27. Six students are trying to identify an unknown chemical compound by heating the substance and measuring the density of the gas that evaporates. (Density $$=$$ mass/volume.) The students record the mass lost by the solid substance and the volume of the gas that evaporated from it. They know that the mass lost by the solid must be the same as the mass of the gas that evaporated. (Source: Hunt and Sykes, 1984) Student A B C D E F Volume ofgas ($$\text{cm}^3$$) $$48$$ $$60$$ $$24$$ $$81$$ $$76$$ $$54$$ Loss inmass (mg) $$64$$ $$81$$ $$32$$ $$107$$ $$88$$ $$72$$ 1. Plot the data with volume on the horizontal axis. Which student made an error in the experiment? 2. Ignoring the incorrect data point, draw a line of best fit through the other points. 3. Find an equation of the form $$y = kx$$ for the data. Why should you expect the regression line to pass through the origin? 4. Use your equation to calculate the mass of $$1000\text{ cm}^3$$ (one liter) of the gas. 5. Here are the densities of some gases at room temperature: Hydrogen $$8$$ mg/liter Nitrogen $$1160$$ mg/liter Oxygen $$1330$$ mg/liter Carbon dioxide $$1830$$ mg/liter Which of these might have been the gas that evaporated from the unknown substance? Hint. Use your answer to part (d) to calculate the density of the gas. $$1 \text{ cm}^3 = 1 \text{ milliliter}\text{.}$$ #### 28. The formulas for many chemical compounds involve ratios of small integers. For example, the formula for water, H$$_2$$0, means that two atoms of hydrogen combine with one atom of oxygen to make one water molecule. Similarly, magnesium and oxygen combine to produce magnesium oxide. In this problem, we will discover the chemical formula for magnesium oxide. (Source: Hunt and Sykes, 1984) 1. Twenty-four grams of magnesium contain the same number of atoms as sixteen grams of oxygen. Complete the table showing the amount of oxygen needed if the formula for magnesium oxide is $$\text{MgO}\text{,}$$ $$\text{Mg}_2\text{O}\text{,}$$ or $$\text{MgO}_2\text{.}$$ Grams of Mg Grams ofO (if MgO) Grams ofO (if Mg$$_2$$O) Grams ofO (if MgO$$_2$$) $$24$$ $$16$$ $$48$$ $$12$$ $$6$$ 2. Graph three lines on the same axes to represent the three possibilities, with grams of magnesium on the horizontal axis and grams of oxygen on the vertical axis. 3. Here are the results of some experiments synthesizing magnesium oxide. Experiment Grams ofMagnesium Gramsof oxygen $$1$$ $$15$$ $$10$$ $$2$$ $$22$$ $$14$$ $$3$$ $$30$$ $$20$$ $$4$$ $$28$$ $$18$$ $$5$$ $$10$$ $$6$$ Plot the data on your graph from part (b). Which is the correct formula for magnesium oxide? #### Exercise Group. For Problems 29–32, 1. Use linear interpolation to give approximate answers. 2. What is the meaning of the slope in the context of the problem? ##### 29. The temperature in Encino dropped from $$81\degree$$F at 1 a.m. to $$73\degree$$F at 5 a.m. Estimate the temperature at 4 a.m. ##### 30. Newborn blue whales are about $$24$$ feet long and weigh $$3$$ tons. The young whale nurses for $$7$$ months, at which time it is $$53$$ feet long. Estimate the length of a $$1$$-year-old blue whale. ##### 31. A car starts from a standstill and accelerates to a speed of $$60$$ miles per hour in $$6$$ seconds. Estimate the car’s speed $$2$$ seconds after it began to accelerate. ##### 32. A truck on a slippery road is moving at $$24$$ feet per second when the driver steps on the brakes. The truck needs $$3$$ seconds to come to a stop. Estimate the truck’s speed $$2$$ seconds after the brakes were applied. #### Exercise Group. In Problems 33–36, use linear interpolation or extrapolation to answer the questions. ##### 33. The temperature of an automobile engine is $$9\degree$$ Celsius when the engine is started and $$51\degree$$C seven minutes later. Use a linear model to predict the engine temperature for both $$2$$ minutes and $$2$$ hours after it started. Are your predictions reasonable? ##### 34. The temperature in Death Valley is $$95\degree$$ Fahrenheit at 5 a.m. and rises to $$110\degree$$ Fahrenheit by noon. Use a linear model to predict the temperature at 2 p.m. and at midnight. Are your predictions reasonable? ##### 35. Ben weighed $$8$$ pounds at birth and $$20$$ pounds at age $$1$$ year. How much will he weigh at age $$10$$ if his weight increases at a constant rate? ##### 36. The elephant at the City Zoo becomes ill and loses weight. She weighed $$10,012$$ pounds when healthy and only $$9641$$ pounds a week later. Predict her weight after $$10$$ days of illness. #### 37. Birds’ nests are always in danger from predators. If there are other nests close by, the chances of predators finding the nest increase. The table shows the probability of a nest being found by predators and the distance to the nearest neighboring nest. (Source: Perrins, 1979) Distance tonearest neighbor(meters) $$20$$ $$40$$ $$60$$ $$80$$ $$100$$ Probability ofpredators (%) $$47$$ $$34$$ $$32$$ $$17$$ $$1.5$$ 1. Plot the data and the least squares regression line. 2. Use the regression line to estimate the probability of predators finding a nest if its nearest neighbor is $$50$$ meters away. 3. If the probability of predators finding a nest is $$10\%\text{,}$$ how far away is its nearest neighbor? 4. What is the probability of predators finding a nest if its nearest neighbor is $$120$$ meters away? Is your answer reasonable? #### 38. A trained cyclist pedals faster as he increases his cycling speed, even with a multiple-gear bicycle. The table shows the pedal frequency, $$p$$ (in revolutions per minute), and the cycling speed, $$c$$ (in kilometers per hour), of one cyclist. (Source: Pugh, 1974) Speed(km/hr) $$8.8$$ $$12.5$$ $$16.2$$ $$24.4$$ $$31.9$$ $$35.0$$ Pedalfrequency (rpm) $$44.5$$ $$50.7$$ $$60.6$$ $$77.9$$ $$81.9$$ $$95.3$$ 1. Plot the data and the least squares regression line. 2. Estimate the cyclist’s pedal frequency at a speed of $$20$$ kilometers per hour. 3. Estimate the cyclist’s speed when he is pedaling at $$70$$ revolutions per minute. 4. Does your regression line give a reasonable prediction for the pedaling frequency when the cyclist is not moving? Explain. #### 39. In this problem we will calculate the efficiency of swimming as a means of locomotion. A swimmer generates power to maintain a constant speed in the water. If she must swim against an opposing force, the power increases. The following table shows the power expended by a swimmer while working against different amounts of force. (A positive force opposes the swimmer, and a negative force helps her.) (Source: diPrampero et al., 1974, and Alexander, 1992) Force(newtons) $$-3.5$$ $$0$$ $$0$$ $$6$$ $$8$$ $$10$$ $$17$$ $$17$$ Metabolicpower(watts) $$100$$ $$190$$ $$230$$ $$320$$ $$380$$ $$450$$ $$560$$ $$600$$ 1. Plot the data on the grid, or use the StatPlot feature on your calculator. Use your calculator to find the least squares regression line. Graph the regression line on top of the data. 2. Use your regression line to estimate the power needed for the swimmer to overcome an opposing force of $$15$$ newtons. 3. Use your regression line to estimate the power generated by the swimmer when there is no force either hindering or helping her. 4. Estimate the force needed to tow the swimmer at $$0.4$$ meters per second while she rests. (If she is resting, she is not generating any power). 5. The swimmer’s mechanical power (or rate of work) is computed by multiplying her speed times the force needed to tow her at rest. Use your answer to part (d) to calculate the mechanical power she generates by swimming at $$0.4$$ meters per second. 6. The ratio of mechanical power to metabolic power is a measure of the swimmer’s efficiency. Compute the efficiency of the swimmer when there is no external force opposing or helping her. #### 40. In this problem, we calculate the amount of energy generated by a cyclist. An athlete uses oxygen slowly when resting but more quickly during physical exertion. In an experiment, several trained cyclists took turns pedaling on a bicycle ergometer, which measures their work rate. The table shows the work rate of the cyclists, in watts, measured against their oxygen intake, in liters per minute. (Source: Pugh, 1974) Oxygenconsumption(liters/min) $$1$$ $$1.7$$ $$2$$ $$3.3$$ $$3.9$$ $$3.6$$ $$4.3$$ $$5$$ Work rate(watts) $$40$$ $$100$$ $$180$$ $$220$$ $$280$$ $$300$$ $$320$$ $$410$$ 1. Plot the data on the grid, or use the StatPlot feature on your calculator. Use your calculator to find the least squares regression line. Graph the regression line on top of the data. 2. Find the horizontal intercept of the regression line. What does the horizontal intercept tell you about this situation? 3. Estimate the power produced by a cyclist consuming oxygen at $$5.9$$ liters per minute. 4. What is the slope of the regression line? The slope represents the amount of power, in watts, generated by a cyclist for each liter of oxygen consumed per minute. How many watts of power does a cyclist generate from each liter of oxygen? 5. One watt of power represents an energy output of one joule per second. How many joules of energy does the cyclist generate in one minute? 6. How many joules of energy can be extracted from each cubic centimeter of oxygen used? (One liter is equal to 1000 cubic centimeters.)
# 2018 AMC 10A Problems/Problem 19 A number $m$ is randomly selected from the set $\{11,13,15,17,19\}$, and a number $n$ is randomly selected from $\{1999,2000,2001,\ldots,2018\}$. What is the probability that $m^n$ has a units digit of $1$? $\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5}$ ## Solution Since we only care about the unit digit, our set $\{11,13,15,17,19 \}$ can be turned into $\{1,3,5,7,9 \}$. Call this set $A$ and call $\{1999, 2000, 2001, \cdots , 2018 \}$ set $B$. Let's do casework on the element of $A$ that we choose. Since $11=1$, any number from from $B$ can be paired with $1$ to make $m^n$ have a units digit of $1$. Therefore, the probability of this case happening is $\frac{1}{5}$ since there is a $\frac{1}{5}$ chance that the number $1$ is selected from $A$. Let us consider the case where the number $3$ is selected from $A$. Let's look at the unit digit when we repeatedly multiply the number $3$ by itself: $$33=9$$ $$93=7$$ $$73=1$$ $$13=3$$ We see that the unit digit of $3^x$ for some integer $x$ will only be $1$ when $x$ is a multiple of $4$. Now, let's count how many numbers in $B$ are divisible by $4$. This can be done by simply listing: $$2000,2004,2008,2012,2016.$$ There are $5$ numbers in $B$ divisible by $4$ out of the $2018-1999+1=20$ total numbers. Therefore, the probability that $3$ is picked from $A$ and a number divisible by $4$ is picked from $B$ is $\frac{1}{5}\frac{5}{20}=\frac{1}{20}$. Similarly, we can look at the repeating units digit for $7$: $$77=9$$ $$97=3$$ $$37=1$$ $$17=7$$ We see that the unit digit of $7^y$ for some integer $y$ will only be $1$ when $y$ is a multiple of $4$. This is exactly the same conditions as our last case with $3$ so the probability of this case is also $\frac{1}{20}$. Since $55=25$ and $25$ ends in $5$, the units digit of $5^w$ for some integer $w$ will always be $5$. Thus, the probability in this case is $0$. The last case we need to consider is when the number $9$ is chosen from $A$. This happens with probability $\frac{1}{5}$. We list out the repeading units digit for $9$ as we have done for $3$ and $7$: $$99=1$$ $$19=9$$ We see that the units digit of $9^z$ is $1$ when $z$ is an even number. From the $20$ numbers in $B$, we see that exactly half of them are even. The probability in this case is $\frac{1}{5}\frac{1}{2}=\frac{1}{10}.$ Finally, we can ad all of our probabilities together to get $$\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.$$ ~Nivek
# Finding a percentage of a whole number without a calculator: Advanced Online Quiz Following quiz provides Multiple Choice Questions (MCQs) related to Finding a percentage of a whole number without a calculator: Advanced. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz. Q 1 - What is 50% of 96.48? ### Explanation Step 1: 50% = $\frac{50}{100} = \frac{1}{2}$ Step 2: To get 50% of 96.48 we divide it by 2 50% of 96.48 = $\frac{96.48}{2} = 48.24$ Step 3: So, 50% of 96.48 = 48.24 Q 2 - 18.4 is what per cent of 27.6? ### Explanation Step 1: Part = 18.4; whole = 27.6 Step 2: percentage = $\frac{18.4}{27.6} \times 100 = 66.67\%$ Step 3: So, 18.4 is 66.67% of 27.6 Q 3 - What is 40% of 38.2? ### Explanation Step 1: We find 10%. We move the decimal one place to the left to get 3.82 Step 2: To find 40% we multiply the result by 4. 40% of 38.2 = 3.82 × 4 = 15.28 Step 3: 40% of 38.2 = 3.82 × 4 = 15.28 Step 4: So, 40% of 38.2 = 15.28 Q 4 - 12.48 is what per cent of 15.6? ### Explanation Step 1: Part = 12.48; whole = 15.6 Step 2: Percentage = $\frac{12.48}{15.6} \times 100 = 80\%$ Step 3: So, 12.48 is 80% of 15.6 Q 5 - 17.43 is what per cent of 24.9? ### Explanation Step 1: Part = 17.43; whole = 24.9 Step 2: Percentage = $\frac{17.43}{24.9} \times 100 = 70\%$ Step 3: So, 17.43 is 70% of 24.9 Q 6 - 4.1 is what per cent of 12.3? ### Explanation Step 1: Part = 4.1; whole = 12.3 Step 2: Percentage = $\frac{4.1}{12.3} \times 100 = 30\%$ Step 3: So, 4.1 is 30% of 12.3 Q 7 - What is 10% of 52.9? ### Explanation Step 1: We first find 10%. We move the decimal one place to the left to get 5.29 Step 2: So, 10% of 52.9 = 5.29 Q 8 - 11.3 is what per cent of 56.5 ### Explanation Step 1: Part = 11.3; whole = 56.5 Step 2: Percentage = $\frac{11.3}{56.5} \times 100 = 20\%$ Step 3: So, 11.3 is 20% of 56.5 Q 9 - What is 30% of 50.8? ### Explanation Step 1: We first find 10%. We move the decimal one place to the left to get 5.08 Step 2: To find 30% we multiply the result by 3. 30% of 50.8 = 5.08 × 3 = 15.24 Step 3: So, 30% of 50.8 = 15.24 Q 10 - What is 60% of 67.4? ### Explanation Step 1: 50% = $\frac{50}{100} = \frac{1}{2}$ Step 2: To get 50% of 67.4 we divide it by 2 50% of 67.4 = $\frac{67.4}{2} = 33.7$ Step 3: Now we find 10% of 67.4. We move the decimal one place to the left to get 6.74 Step 4: 60% of 67.4 = 50% of 67.4 + 10% of 67.4 = 33.7 + 6.74 = 40.44 So, 60% of 67.4 = 40.44
# How do you find all points that have an x -coordinate of –4 and whose distance from point (4, 2) is 10? Oct 27, 2014 By using the distance formula. $D = \sqrt[2]{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$ We want $D = 10$ Let $\left({x}_{1} , {y}_{1}\right)$ be $\left(4 , 2\right)$ Let $\left({x}_{2} , {y}_{2}\right)$ be $\left(- 4 , {y}_{2}\right)$ $10 = \sqrt[2]{{\left(4 - \left(- 4\right)\right)}^{2} + {\left(2 - {y}_{2}\right)}^{2}}$ Squaring both sides we have ${\left(10 = \sqrt[2]{{\left(4 - \left(- 4\right)\right)}^{2} + {\left(2 - {y}_{2}\right)}^{2}}\right)}^{2}$ $\implies 100 = {\left(4 - \left(- 4\right)\right)}^{2} + {\left(2 - {y}_{2}\right)}^{2}$ $\implies 100 = \left({8}^{2}\right) + {\left(2 - {y}_{2}\right)}^{2}$ $\implies 100 = 64 + \left(4 - 4 {y}_{2} + {\left({y}_{2}\right)}^{2}\right)$ $\implies 36 = {\left({y}_{2}\right)}^{2} + 4 {y}_{2} + 4$ $\implies {\left({y}_{2}\right)}^{2} + 4 {y}_{2} + 4 = 36$ $\implies {\left({y}_{2}\right)}^{2} + 4 {y}_{2} - 32 = 0$ $\implies \left({y}_{2} + 8\right) \left({y}_{2} - 4\right) = 0$ y_2 = -8, y_2 = 4
# 1439049084_231925.pdf - Appendix A.3 A27 Polynomials and... • 12 This preview shows page 1 - 3 out of 12 pages. Appendix A.3 Polynomials and Factoring A27 Polynomials The most common type of algebraic expression is the polynomial. Some examples are and The first two are polynomials in and the third is a polynomial in and The terms of a polynomial in have the form where is the coefficient and is the degree of the term. For instance, the polynomial has coefficients 2, 0, and 1. Polynomials with one, two, and three terms are called monomials, binomials, and trinomials, respectively. In standard form, a polynomial is written with descending powers of Writing Polynomials in Standard Form Leading Polynomial Standard Form Degree Coefficient a. 7 b. 2 c. 8 8 0 8 Now try Exercise 19. A polynomial that has all zero coefficients is called the zero polynomial, denoted by 0. No degree is assigned to this particular polynomial. For polynomials in more than one variable, the degree of a term is the sum of the exponents of the variables in the term. The degree of the polynomial is the highest degree of its terms. For instance, the degree of the polynomial is 11 because the sum of the exponents in the last term is the greatest. The leading coefficient of the polynomial is the coefficient of the highest-degree term. Expressions are not polynomials if a variable is underneath a radical or if a polynomial expression (with degree greater than 0) is in the denominator of a term. The following expressions are not polynomials. The exponent “ ” is not an integer. The exponent “ ” is not a nonnegative integer. 1 x 2 5 x x 2 5 x 1 1 2 x 3 3 x x 3 3 x 1 2 2 x 3 y 6 4 xy x 7 y 4 8 8 x 0 9 9 x 2 4 4 9 x 2 5 5 x 7 4 x 2 3 x 2 4 x 2 5 x 7 2 3 x Example 1 x . 5, 2 x 3 5 x 2 1 2 x 3 5 x 2 0 x 1 k a ax k , x y . x x 5 x 2 y 2 xy 3. 3 x 4 7 x 2 2 x 4, 2 x 5, A.3 P OLYNOMIALS AND F ACTORING What you should learn Write polynomials in standard form. Add, subtract, and multiply polynomials. Use special products to multiply polynomials. Remove common factors from polynomials. Factor special polynomial forms. Factor trinomials as the product of two binomials. Factor polynomials by grouping. Why you should learn it Polynomials can be used to model and solve real-life problems. For instance, in Exercise 224 on page A37, a polynomial is used to model the volume of a shipping box. Definition of a Polynomial in x Let be real numbers and let be a nonnegative integer. A polynomial in is an expression of the form where The polynomial is of degree is the leading coefficient, and is the constant term. a 0 a n n , a n 0. a n x n a n 1 x n 1 . . . a 1 x a 0 x n a 0 , a 1 , a 2 , . . . , a n A28 Appendix A Review of Fundamental Concepts of Algebra Operations with Polynomials You can add and subtract polynomials in much the same way you add and subtract real numbers. Simply add or subtract the like terms (terms having the same variables to the same powers) by adding their coefficients. For instance, and are like terms and their sum is Sums and Differences of Polynomials a.
HOW TO FIND SUM OF SPECIAL SERIES Here we study some common special series like (i) Sum of first ‘n’ natural numbers 1 + 2 + 3 + ............ + n = n(n + 1)/2 (ii) Sum of first ‘n’ odd natural numbers. 1 + 3 + 5 + ............ + (2n-1) = n2 (iii) Sum of squares of first ‘n’ natural numbers. 12 + 22 + 32 + ............ + n2 = n(n + 1)(2n + 1)/6 (iv) Sum of cubes of first ‘n’ natural numbers. 13 + 23 + 33 + ............ + n3 = [n(n + 1)/2]2 Question 1 : Find the sum of the following series (i) 1 + 2 + 3 +............+ 60 Solution : Sum of natural numbers 1 + 2 + 3 +...............+ n = n(n + 1)/2 n = 60 1 + 2 + 3 +............+ 60 = 60(60+ 1 )/2 = 30(61) = 1830 (ii) 3 + 6 + 9 + ............+ 96 Solution : = 3 + 6 + 9 +...........+ 96 = 3(1 + 2 + 3 + ............+ 32) = 3 (32(32+1)/2) = 3 (16) (33) = 1584 (iii) 51+ 52 + 53 +............+ 92 Solution : 51+ 52 + 53 +............+ 92 = (1 + 2 + 3 +..........+ 92) - (1 + 2 + 3.........+ 50) n = 92 & n = 50 = [92(92 + 1)/2] - [50(50 + 1)/2] = 46(93) - 25(51) = 4278 - 1275 = 3003 (iv) 1 + 4 + 9 + 16 +...............+ 225 Solution : 1 + 4 + 9 + 16 +...............+ 225 = 12 + 22 + 32 + 42 +...............+ 152 Sum of squares = n(n + 1)(2n + 1)/6 = 15(15 + 1) (2(15) + 1)/6 = 15(16)(31)/6 = 5(8) (31) = 1240 (v) 62 + 72 + 82 +........+212 Solution : 62 + 72 + 82 +........+212 = (12 + 22 +........+ 212) - (12 + 22 + .......+ 52) Sum of squares = n(n + 1)(2n + 1)/6 = [21(21 + 1) (2(21) + 1)/6] - [5(5 + 1) (2(5) + 1)/6] = [21(22) (43)/6] - [5(6) (11)/6] = [21(22) (43)/6] - [5(6) (11)/6] = 3311 - 55 = 3256 (vi) 103 + 113 + 123 +.........+ 203 Solution : 103 + 113 + 123 +.........+ 203 = (13 + 23 + 33 +.........+ 203) - (13 + 23 + 33 +.........+ 93) Sum of cubes = [n(n + 1)/2]2 = [20(20+1)/2]2 - [9(9+1)/2]2 = 2102 - 452 = (210 + 45) (210 - 45) = 255 (165) = 42075 (vii) 1 + 3 + 5 +.............+ 71 Solution : Sum of odd numbers = ((l+1)/2)2 = [(71 + 1)/2]2 = [72/2]2 = 362 = 1296 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. If you have any feedback about our math content, please mail us : v4formath@gmail.com You can also visit the following web pages on different stuff in math. WORD PROBLEMS Word problems on simple equations Word problems on linear equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6
Courses Courses for Kids Free study material Offline Centres More Store # How do you find the area of a square with sides $6$ centimeters long? Last updated date: 13th Jun 2024 Total views: 373.2k Views today: 3.73k Verified 373.2k+ views Hint: Let us assume that the length of the given square is $6$ cm. Let this be equation (1). We very well know that if the length of the side of a square is equal to $x$ cm, then the area of the square is equal to ${x^2}$ sq.cm. Let us say that the area of the square is equal to $A$ sq.cm. So, we are now supposed to find the square of value of $x$ cm which is obtained from equation (1). Let this be equation (2). Now, from equation (2) we can easily find the value of the area of the square whose sides are $6$ cm long. Before solving the given question, we should keep in mind that if the length of the side of a square is equal to $x$ cm, then the area of the square is equal to ${x^2}$ sq.cm. It is already given that the length of the side of the square is equal to 6 cm. Let us assume that the side of the square is equal to $x$ . $\Rightarrow x = 6$ ---(1) We already know that the area of the square with side $x$ cm is equal to ${x^2}$ sq.cm. $\Rightarrow A = {x^2}$ --(2) $\Rightarrow A = {\left( 6 \right)^2} \\ \Rightarrow A = 36 \;$ So, it is clear that the area of the square whose side is equal to $6$ cm is $36$ sq.cm. Thus, Area = $36\;c{m^2}$ . So, the correct answer is “Area = $36\;c{m^2}$ ”. Note: There is usually a misconception among students, which is that if the diagonal of a square is equal to $x$ cm, then the area of the square is equal to ${x^2}$ sq.cm. Due to this misconception the answer gets affected. Thus, this misconception should be avoided.
Study Guide # Types of Numbers - Things to Do with Real Numbers ## Things to Do with Real Numbers ### Compare Them When comparing real numbers, it can help to think of the number line. Ah, how fondly we remember it. Our favorite line of all the lines. The numbers on a number line get bigger as we go further right on the number line, and smaller as we go further left. All positive numbers are bigger than zero, and all negative numbers are less than zero. Doesn't mean they can't be friends. ### Picturing Addition on the Number Line The addition of real numbers can be visualized using the number line. Just one more reason this number line thingie is nice to have around. Adding two positive real numbers is like starting at 0 and then going for two walks to the right. In the first walk, take as many steps as the first number says, and in the second walk take as many steps as the second number says. In between the two walks, make sure to hydrate. For example, we already know that 3 + 5 = 8. That's some pretty basic addition. But picture it on the number line: go for a walk of 3 steps to the right of 0, then a second walk of 5 steps after that. Adding two negative numbers is like starting at 0 and going for two walks to the left. Maybe because you're having a fight with the guy who just started off on a walk to the right. In the first walk, take as many steps to the left as the first number says, and in the second walk take as many steps to the left as the second number says. Watch your step and don't accidentally step on any negative signs. Those things can go right through your shoes and give you splinters. For example, (-4) + (-3) = -7. We're starting at 0, then walking 4 steps to the left to get to -4 on the number line. Then we're walking another 3 steps left after that, which brings us to our final answer of -7. Adding one positive and one negative number is, again, like going for two walks, only this time one walk will be to the right and one will be to the left. We apologize in advance, because we know how much you hate backtracking. 4 + (-5) = -1 See? We walked 4 steps to the right, then turned on our heel and walked backwards another 5 steps. Notice that switching the order of the walks doesn't change the answer: (-5) + 4 = -1 ### Sample Problem 5 + __ = -17 This question is asking how many steps it takes to get from 5 to -17 on the number line, and in which direction those steps need to be taken. Picture the number line: you can see that if you walked off in a huff and got to the 5, but then your friend who you're having a spat with yells at you from the -17 and says he's sorry, it would take you 22 steps to the left to walk back over to him and hug it out. 5 + (-22) = -17 No, addition isn't some incredibly wealthy mathematical operation that owns multiple houses in multiple states. Although if it were, it would probably be building additions onto each of them. If we add two real numbers, we get the same answer no matter which number comes first. This means that addition of real numbers is commutative: the numbers can commute (travel past each other with a cup of coffee while listening to their favorite morning radio station) without changing the answer. Wait, no, that's not right. "Commute" just means "switch places" when we're talking about math. To write "addition of real numbers is commutative" in symbols—because goodness knows we love us some symbols—we say that for real numbers a and b: a + b = b + a Both terms can swap spots without changing the total sum. Addition of real numbers is also associative: we don't care how the numbers associate with each other. As long as there isn't bloodshed. When adding more than two numbers, you'll get the same answer no matter which end you start adding from. If you have three friends, one of whom tells you 2 jokes, one of whom tells you 4 jokes, and one of whom tells you 7 jokes, you'll end up having heard 13 jokes. Regardless of the order you hear them in, you'll be the one who has the last laugh. ### Sample Problem 1 + 2 + 3 = ? If we first add up 1 and 2, we get: (1 + 2) + 3 = 3 + 3 = 6 If we first add up 2 and 3, we get: 1 + (2 + 3) = 1 + 5 = 6 Same deal either way. To write "addition of real numbers is associative" in symbols, we say that for real numbers a, b, and c: (a + b) + c = a + (b + c) Addition has an additive identity: the number 0. If you add 0 to any number, that number keeps its identity. For example, 4 + 0 = 4. In other words, if you take 4 steps, and then 0 steps, you've still taken only 4 steps overall. By the way, you look pretty silly trying to take 0 steps. If you walk a certain distance on the number line and then walk the same distance in the opposite direction, you'll end up back where you started. This means that if you add a real number n and its negative -n, you end up back at 0 again. Who knows why you went all the way back to 0? Maybe you forgot your keys. n + (-n) = 0 (-n) + n = 0 2 + (-2) = 0 (-8.7) + 8.7 = 0 If n is a real number (positive or negative or zero), we call -n the additive inverse of n. "Additive inverse of n" is just a fancy way of saying "the number we add to n to get back to zero." Remember that -n is only negative if n is positive. For example if n = -3, then n is negative and -n is positive. It's a tad confusing, but if you get confused, just ask Tad. • ### Subtraction Subtraction is really another way of showing that you're adding the additive inverse. In Subtractionland, it's always opposite day. Or is it? For example, 4 – 3 is just another way of writing 4 + (-3). Subtraction exists so that instead of writing a plus sign, a negative sign, and parentheses, we can just write a minus sign. Because mathematicians are generally lazy, and they don't want to have to go writing all those extra symbols if they can help it. In fact, the symbol "₳" means "please bring me a donut so I don't have to get up from the couch." On the number line, subtraction means walking to the first number, then walking in the opposite direction specified by the second number. Told you it was opposite day. Or did we? What does it mean to subtract a negative number? ### Sample Problem 4 – (-18) = ? Well, adding (-18) would mean we walk 18 to the left, so subtracting (-18) means we walk 18 to the right. So 4 – (-18) really means 4 + 18 = 22. Think of it this way: if you take the minus sign and the negative sign and cross them, you get a plus sign. If you found the preceding sentence helpful, please check this box: □. Another way to approach this problem is to remember what subtraction is abbreviating, then rewrite the whole thing as an addition problem: 4 – (-18) = 4 + (-(-18)) Since the negative of a negative is positive, -(-18) = 18. 4 – (-18) = 4 + 18 = 22 Be Careful: Honestly, the notation here can get really confusing. So take off your confused cap and put on your unbefuddled derby. A minus sign and a negative sign look similar, but mean different things. The trick is to see whether the little horizontal line has to do with one number or two. A negative sign is a horizontal line in front of just one number, and it tells us to reflect the number across zero on a number line: -5 -(-3) A minus sign is a horizontal sign in between two numbers, and it tells us to walk to the first number, then walk in the opposite direction of the second number: 4 – 5 = -1 To help avoid uncertainty, you can put parentheses around negative numbers. For example, write (-4) instead of -4. Hopefully you don't use a lot of emoticons when doing your math homework, because something like this can get awfully confusing: (-:3 - 7;-0):-) You can also make your negative signs smaller and higher up than minus signs. Don't put them so high up that you can't get them back down when you need them, though. Be Careful: Subtraction does not commute! (It works from home.) This is because subtraction changes the direction we're walking on the number line for the second number only. 8 – 10 = -2 10 – 8 = 2 So 8 – 10 ≠ 10 – 8. Subtraction is also not associative. The order in which we perform multiple subtractions changes the final answer. (3 – 4) – 2 = -3 3 – (4 – 2) = 1 You're always going to want to perform subtraction from left to right: 3 – 4 – 2 = -3. This should be easy to remember, because that's also the direction in which you read, as well as the direction in which you open the chocolates on your Advent calendar. Or, if you're Jewish, the direction in which you light the candles of your menorah. Or, if you're Buddhist, the direction in which you align your chakras. Another way to think about this is that subtraction is really just the addition of a negative. This way, you can rewrite the problem as: 3 + (-4) + (-2) Now it's an addition problem, so it doesn't matter what order you add them in! If we're subtracting a bigger number from a smaller number (for example, 13 – 25), one way to find the answer is to pretend the problem is structured in reverse: evaluate 25 – 13 to get 12. Since we know 13 – 25 will be to the left of zero, stick a negative sign onto the front of it to get -12. Make sure you use Gorilla Glue so that thing really stays on there. The reason this method works is that whether you go 13 to the right and 25 to the left, or 25 to the right and 13 to the left, you'll end up the same distance from zero. To give the correct final answer, use common sense to figure out which side of zero the answer is on. If you're lacking in the common sense department, then maybe don't use this method. Also, don't rest your palm on one of the eyes of a stove when it's glowing orange. • ### Multiplication So far, we've been working a lot with the number line. Now it's time for some pictures in a different dimension. Imagine, if you will, a bunch of boxes. If p and q are positive, the multiplication p × q means "take p groups of q things and see how many things you end up with." This can be pictured nicely with a rectangle divided into smaller boxes. ### Sample Problem 3 × 4 Here's one group of 4 things: ...and here's 3 groups of 4 things: Count up all the things (each of the small boxes) and we get 12. Notice that another way to think of this picture is that we took a rectangle with one side of length 4 and one side of length 3, and then found the area of the rectangle. But that way's not as much fun, because then we don't get to overuse the word "things." Even if the numbers aren't all whole numbers, a similar diagram will still work. ### Sample Problem 3.5 × 6.2 Simply draw a rectangle with side lengths 3.5 and 6.2, and 3.5 × 6.2 is the area of that rectangle. If we're thinking about it in a real-life scenario, imagine that we have 3.5 of something and 6.2 of something else. Let's just hope they're not living things, or this could get messy. If we switch the order of the numbers being multiplied, our rectangular box gets turned onto one end (apparently the multiplication symbol didn't notice the "FRAGILE" warning written on the side of it), but the final answer is still the same. 4 × 3 = 12 This means that multiplication is commutative: the order in which we write the numbers doesn't matter. They can "commute" past each other without changing the final answer. a × b = b × a Man, a and b sure are lucky they're at the beginning of the alphabet. They get to be in everything. Multiplication is also associative: we don't care how the numbers associate with each other. In symbols, for real numbers a, b, and c: (a × b) × c = a × (b × c) At least c got a little love that time. We still can't help but feel bad for w though. ### Sample Problem 5 × 2 × 3 With multiplication, we can start from either end. Either way we slice it, we'll still get the same answer. (5 × 2) × 3 = 10 × 3 = 30 5 × (2 × 3) = 5 × 6 = 30 If one or more of the numbers we're multiplying together has a negative sign, we first multiply the absolute values of the numbers together. Then, for each negative sign, we reflect our answer to the other side of the number line. If we come across enough negative signs, we might find ourselves doing more reflecting than Lindsay Lohan during her next jail sentence. ### Sample Problem 2 × (-3) First multiply 2 by 3 to get 6. Since we have just one negative sign, we reflect 6 to the other side of the number line once to get our final answer: -6. Kind of hard to draw the rectangle for that one, so we recommend that you don't even try. ### Sample Problem (-6) × (-4) First multiply 6 by 4 to get 24. Since we have two negative signs, we reflect 24 across 0 and back again. Our final answer is positive 24. There are several different symbols used for multiplication. We can write "a times b" as: a × b ab (a)(b) Be Careful: 4 × - 3 doesn't mean anything. It might have some sentimental value to you, but it certainly doesn't mean anything mathematically: "4 times subtract 3?" What in the world is that? Let's put parentheses around the second term to keep track of the negative sign: 4 × (-3) Also keep in mind that it's really easy to multiply by the number 1. We say 1 is the multiplicative identity, because multiplying by 1 allows numbers to keep their identities. Even if some of them are a little embarrassed by their identities and would prefer to trade them in. Like 37, for example. Terrible self-image. 1 × 8 = 8 9 × 1 = 9 1 × 1 = 1 -π × 1 = -π In symbols, we say that if n is any real number, then: 1 × n = n × 1 = n Any real number n, except for 0 (sorry, 0, you'll get 'em next time), has a multiplicative inverse or reciprocal, written 1/n. This is the number that, if we multiply it by n, we get back to 1. ### Sample Problem What's the multiplicative inverse of 2? The multiplicative inverse of 2 is , because . Got two bagel halves? We bet they'll make a full bagel if you press them together. So why doesn't 0 have a multiplicative inverse? Can't a number get a break around here? The number 0 is special in multiplication, because 0 "kills" everything. Wow. Real nice, 0. That's the last time we ever stick up for you. For any real number n: 0 × n = n × 0 = 0 This means there can't be any real number n where 0 × n = 1. Zero kills everything because if we take n groups of zero things, or zero groups of n things, we don't have any things at all. Or at least that's the defense zero's lawyer is using at trial. This can be very useful. If you're multiplying a lot of numbers together and one of the numbers happens to be 0, you don't have to do any work. Since anything multiplied by 0 is 0, all you have to do is write down 0 as your answer! Or, if you're taking the SAT, fill in that 0-shaped bubble next to the 0. ### Sample Problem 789,234 × 67,623,746,374 × 23,432,432 × 0 × 234,872,384,723 = 0 That lone 0 in the mix turns the entire product into 0, no matter how gnarly the other terms are. • ### Division In symbols, "p divided by q" can be written as , or p ÷ q. Remember the names for the different parts of a division problem: • The dividend, p, is the thing being divided up. And sprinkled onto your salad, if you so choose. • The divisor, q, is the thing that performs the dividing. It may also perform the conquering. • The quotient is the answer. Here's a little insider info: "p divided by q" means the same thing as "q divided into p." When the numbers p and q are both whole numbers, division can be thought of as dividing up a bunch of things into smaller groups in either of two ways: • If we split p into groups of size q will be how many groups (and perhaps fractions of a group) we get. By the way, "faction" is a synonym of "group," so you technically may be looking for the fraction of a faction. We hope you're not mad at us for throwing that in there just to torture you. We wouldn't want there to be any friction. • If we split p into q groups, will be the size of each group. For instance, "20 divided by 4" could mean "20 split into groups with 4 things in each group" or "20 split into 4 groups." The idea is the same when p and q are real numbers instead of whole numbers, except that now the groups can have partial objects. Like if you're grouping chocolate chip cookies and some of them appear to have nibbles taken out of them. There's also the matter of signs, which we didn't have to worry about with whole numbers. When dividing one positive real number into another, we just do the division like normal: 4.2 ÷ 2.1 = 2. When one of the real numbers (either the divisor or the dividend) has a negative sign, perform the division while ignoring the signs, and then afterward reflect your answer across 0 on the number line. You'll always put those negative signs back on as the finishing touch. The icing on the cake. The Hershey's syrup on the Cracklin' Oat Bran. (We haven't actually tried that but we're guessing it would be delicious.) -4.2 ÷ 2.1 = -2 = 4.2 ÷ (-2.1) = -2 If both the divisor and dividend have negative signs, perform the division and ignore the signs. The answer needs to be reflected across 0 twice, which gets us the same answer as if both the divisor and dividend were positive. In this case, we can simply ignore the negative signs entirely. Which is rude, we guess, but whatever. -4.2 ÷ (-2.1) = 2 Remember that division is an abbreviation for multiplying by the multiplicative inverse of a number. This is most useful when fractions are involved. Check it: Here's the thing, though: division is not commutative. For example, is different from . Both are correct, but they're saying different things. Like your parents when they argue. See, that's the problem. Division is also not associative. Let's see why. ### Sample Problem 1 ÷ 2 ÷ 3 If we evaluate 1 ÷ 2 first, we get one thing: But if we evaluate 2 ÷ 3 first, we get something else: So always remember to go left to right when dividing multiple terms. Be Careful: Division by zero is undefined, which we talked about in the section on rational numbers. If you ever feel tempted to divide by zero even though you know it's wrong, we're pretty sure there's a hotline you can call. • ### Long Division Remainder Here's a refresher on long division, by way of example. We're hoping monkey see, monkey do. Not that you in any way resemble a monkey. Well, maybe a little around the eyes. ### Sample Problem What is 250 divided by 4? We need some words to help us refer to different parts of the division problem. If you've read the sections leading up to this one, they're pretty well drilled into your brain by now, but if not, here you go: • The dividend is the thing being divided up. Think of it this way: the word "dividend" ends with "end," and your end is divided up into two parts. • The divisor is the thing that performs the dividing. Definitely the alpha male in this relationship. • The quotient is the answer. All hail the quotient. Another useful word is multiple. A "multiple of 4" is any number that's evenly divisible by 4. For example, 4, 8, 12, and 16 are multiples of 4. We have to mention here that 396 is also a multiple of 4. He asked us to, and we owed him a favor. To do the division problem, we're going to work our way across the dividend from left to right, and see how many times the divisor "fits" into each part of it. First, consider just the first digit of the dividend. Above that digit, we'll write the number of times the divisor fits into that digit: Hmm...4 doesn't fit into 2 at all (not without a healthy dose of elbow grease), since 4 is bigger than 2. So we write a 0 in that space: We then subtract the multiple of 4 that fits into 2. In this case, that's 0: We address the next digit of the dividend by "carrying it down." Lift with your legs, not with your back. Now, we write how many times 4 fits into 25 in the space above the 5 in the dividend. Since 24 (4 × 6) is the largest multiple of 4 that fits into 25, we write 6 in that space: Then we subtract the multiple of 4 that fits into 25: Now we deal with the next digit of the dividend by "carrying it down": Since 4 fits into 10 twice, we write 2 in the next space of the quotient and then subtract 8 (4 × 2): While it appears we're out of digits in the dividend, we're not. Looks like we had an extra emergency supply in the basement. We can stick a decimal point onto the end and as many zeros as we need to the right of that. Remember to write the decimal point in the quotient, too. Then we can "carry down" any extra zeros: We see 4 fits into 20 an even 5 times, so we can write the appropriate number in the appropriate place: We stop when the subtraction results in a zero. The quotient is our "final answer," Meredith. 250 ÷ 4 = 62.5 We can check our answer by making sure that the quotient multiplied by the divisor equals the dividend: 62.5 × 4 = 250. Since this equation is correct, we know we got the right answer! And we're having a good hair day! Will wonders never cease? • ### Exponents and Powers - Whole Numbers This sequence shows up a lot in math and computer science, so take note. Especially if you like computer science—you know, taking various chemicals in eye droppers and dripping them onto your PC and whatnot. 2 2 × 2 = 4 2 × 2 × 2 = 8 2 × 2 × 2 × 2 = 16 2 × 2 × 2 × 2 × 2 = 32 Writing out all these 2s gets boring quickly. Who wants to write out twenty 2s, all multiplied together? (If this is you, please put your hand down. No one can see you right now anyway.) Thankfully, there's a shortcut. We write 2n, pronounced "2 to the n," "2 raised to the n," or "2 to the nth power," which all mean n copies of 2 multiplied together. And to help you remember that we're "raising it," we even literally raise it up a little bit next to the number we're multiplying. Aren't mathematicians thoughtful? They even sent you flowers on your birthday. Remember that? If we've got 2n, that little n is called an exponent or power, 2 is called the base, and the process of raising a number to a power is called exponentiation. The numbers 2, 22, 23, and so on are called powers of 2. If you see something like 2love, that's the power of love. Be Careful: When raising a negative number to a power, keep careful track of your negative signs. Clip and tag them if you have to. If it's the negative number that's being raised to the power, we get one thing: (-2)4 = (-2)(-2)(-2)(-2) = 16 If not, we group it differently and get something else: -24 = -(24) = -16 ### Sample Problem Jen wrote -32 = 9. What did Jen do wrong? The negative sign isn't being squared, so the answer should be -9. It would only be positive 9 if we had (-3)2. We're really, really sorry if your name is Jen. It's a total coincidence, we swear. ### A Little Bit About Zero If we raise 0 to any positive exponent, we still get 0. This makes sense, because if you multiply one or more copies of 0 together, you'll just get 0. Turns out it's hard for 0 to become anything other than 0. Even if he really applies himself. Any nonzero number raised to the 0 power is 1. Think about it this way: 24 = 16 23 = 8 22 = 4 21 = 2 As the exponent drops by 1, the answer is divided in half. If we drop the exponent by 1 once more and divide the answer in half again, we get 20 = 1. We can't believe how many times you just dropped that exponent. Can't you be more careful? So here's the deal: 20 = 1 30 = 1 150 = 1 (-36.25)0 = 1 It's 1s all the way down: raise any number to the power of 0, and the answer is 1. Well, except for one weird exception. What's 00? Zero is a troublesome number. We want 0 raised to any power to be 0, but we also want any number raised to the 0 power to be 1. There's no way to win! This means that 00 is undefined. If it's not too late, don't think about this too hard. It'll make your head hurt. ### Multiplication What is 25 × 27? This means that you need to multiply 5 copies of 2 together, and then multiply that result by 7 copies of 2. That's a total of 12 copies of 2. So 25 × 27 = 212. Why so many copies of 2? What are you, passing them out at a meeting? If we have the same base with two different exponents and we're multiplying these numbers, as in the above example, the exponents get added together. In symbols, if a, b, and c are real numbers, then: ab × ac = a(b + c) ### Negative Exponents So far, we've only looked at exponents that are positive integers. Let's try to figure out what a number would be when raised to a negative exponent. Suppose we want to understand what 3-1 means. Let's use what we know about multiplying exponents. Since we add exponents during multiplication, 31 × 3-1 would be 31 + (-1) = 30 = 1. This tells us that 3-1 is the multiplicative inverse, or reciprocal, of 3. So . Did you follow that? If not, double back and read this paragraph again until it sinks in. It won't kill you. Now what happens if we take bigger powers? Like 5-7, for example. In this case, we'll look at 57 × 5-7 = 57 + (-7) = 50 = 1. So 5-7 is the same as (1/5)7. Are you loving this stuff as much as we think you are? ### Division What's 25 ÷ 22? This means , so we're just canceling out two of our 2s. Buh-bye, guys. You shall be missed. After reducing, our fraction equals 23. In general, ab ÷ ac = a(b – c), because we start out with b copies of a, divide out c copies, and are left with b c copies. Heads up, though: a can't be 0. Notice that if b > c, you're left with a positive exponent. But if b < c, you have a negative exponent. Which shouldn't stress you out any, as you now know what to do with them. ### Sample Problem What's 42 ÷ 44? This translates to: See what we did there on the end? Always look for ways that an expression can be further simplified. ### Sample Problem What's 63 ÷ 67? What this really means is "3 copies of 6 divided by 7 copies of 6": Cancel out 3 copies of 6 from the top and bottom of the fraction to get . Be careful: In order to use the properties above, the base of the exponents has to be the same. For example, we can't combine 43 and 52. That's unfortunately as nice as it gets with exponent notation. Which isn't very nice. Hear that, Santa? ### Exponentiation What is (25)3? This really means (2 × 2 × 2 × 2 × 2)3. You can't just add the 5 and the 3 together in this instance, because what we're actually being asked to do is take 3 copies of (2 × 2 × 2 × 2 × 2), or 15 copies of 2 multiplied together. Looks a little like we're going to be multiplying exponents here. In fact, it looks a lot like that. (25)3 = 25 × 3 = 215 So, in general, (ab)c = ab × c. ### Raising Products to a Power What's (6 × 7)3? Obviously we could just multiply 6 by 7 to get (42)3, but let's see what happens if we leave 'em separated. (6 × 7)3 = (6 × 7)(6 × 7)(6 × 7) = 63 × 73. In general, if a and b are real numbers and c is a whole number, (a × b)c = ac × bc. ### Raising Quotients to a Power If a and b are real numbers and c is a whole number, . Just slap that exponent on the numerator and the denominator separately. • ### Prime Factorization A prime number is a number greater than 1 that's only divisible by itself and 1. It's like someone fed it into a factor compactor. Here are some examples: 1 is not prime. 2 is prime. 3 is prime. 4 is not prime because it's divisible by 2. 5 is prime. 6 is not prime because it's divisible by 2 and 3. As it turns out, 2 is the only even number that's prime. Whoop-dee-doo for number 2. For any other even number n, 2 divides into n, so n is not prime. Click here to see a list of the first 1000 primes. It's good to be able to recognize the prime numbers at least up to 31 or so. However, if you want to memorize all 1000 of them, we won't stop you. Having the ability to rattle them off will be a great party trick, if nothing else. By the way, how do you get invited to such cool parties? Every single whole number can be written uniquely (in only one way) as a product of primes. For example, 12 breaks down like this: 12 = 2 × 2 × 3 We can reorder the product and write 12 = 2 × 3 × 2, or 12 = 3 × 2 × 2, but we can't write 12 as a product using any other prime numbers. We have to use two copies of 2 and one copy of 3. It's the law. To find the prime factorization of a number, you can "pull out" one prime at a time. Put your back into it. We'll illustrate what this means by an example, mostly because we're terrible at drawing stuff. ### Sample Problem Find the prime factorization of 120. Okay, 120 is divisible by 2, so first we "pull out" a 2: 120 = 2 × 60 Always look first to see if you can pull out a 2. That's always our "prime suspect." Oh, sure, groan away. Now we move on to the 60. Since 60 is also divisible by 2, we "pull out" 2 from 60: 120 = 2 × 2 × 30 Ah, but 30 is also divisible by 2: 120 = 2 × 2 × 2 × 15 So far, so good. We can't divide that 15 by 2, but we can divide it by 3 because 15 = 3 × 5, and 3 and 5 are both prime numbers. 120 = 2 × 2 × 2 × 3 × 5 We've reached the end at last. All those factors are now prime numbers, so we can't split 120 up any further. Another way to find the prime factorization of a number is to simply recognize the number as a product of two smaller numbers, and factor each of the smaller numbers. Better recognize. ### Sample Problem What's the prime factorization of 200? 200 = 20 × 10 = (4 × 5) × (2 × 5) = (2 × 2 × 5) × (2 × 5) = 2 × 2 × 2 × 5 × 5 Since there's only one way to write any particular number as a product of primes, it doesn't matter what method you use to find those primes. There are certain methods that are slower, such as counting out that number of pennies and then dividing them into neat, even piles, but you'll still arrive at the correct answer eventually. Now get yourself to a Coinstar, you penny hoarder. • ### Order of Operations Addition, subtraction, multiplication, division, and exponentiation are all operations on the real numbers, meaning things you do to the real numbers. For complicated arithmetic expressions, it's important to perform operations in the correct order. So the "whatever order I feel like" tactic isn't going to work for you all that well. This correct order is given by the magical phrase "Please Excuse My Dear Aunt Sally" (PEMDAS). Oh, poor, dear Aunt Sally; she gets a little confused sometimes and needs acronyms to remind her how to go about solving her favorite math equations. The letters stand for Parentheses, Exponents, Multiplication and Division, Addition and Subtraction, in the order we want to do them. Notice that "Multiplication and Division" and "Addition and Subtraction" are grouped together. That's because multiplication doesn't necessarily need to be done before division—you just need to have all your multiplication and division wrapped up before you start in on your addition and subtraction. Don't feel bad if you didn't get that at first. Aunt Sally's been trying to grasp that concept for 40 years and it still eludes her. ### Sample Problem First we evaluate things in parentheses. Uh, we can't simplify (4) any more than it already is, so let's move on to the exponents. Then multiplication and division: 6 – 1 5 When adding and subtracting, we work from left to right. Check to see which of your shoes has the big "L" written on the sole of it if you're not sure. ### Sample Problem What's 4 – 6 – 2? We don't have any parentheses, exponents, multiplication, division, or addition, so we jump right away to subtraction. As usual, though, we've gotta move left to right. 4 – 6 – 2 = (4 – 6) – 2 = -2 – 2 = -4 Notice that, if we'd subtracted 6 – 2 first, we would get a totally different (and wrong) answer: 4 – (6 – 2) = 4 – 4 = 0 We also work from left to right when evaluating multiplication and division. ### Sample Problem What is 3 × 4 ÷ 2 ÷ 6? We only have multiplication and division here, so let's roll along from left to right. 3 × 4 ÷ 2 ÷ 6 = 12 ÷ 2 ÷ 6 = 6 ÷ 6 = 1 If we worked from right to left, we would get a different answer: One way to keep track of your work is to break the problem into pieces, separated by addition or subtraction signs. A rock hammer or mortar and pestle should do the trick. ### Sample Problem Now work out each of the pieces: = 6 + 2 – 0 – 4 Then combine the answers to the pieces: 6 + 2 – 0 – 4 = 4 ### Sample Problem Yeesh, what a beast. Let's break it down into smaller chunks, each separated by a plus or minus sign (since addition and subtraction come last). = 3 + 6 × 16 – 6 × 1 Now we handle that multiplication. 3 + 6 × 16 – 6 × 1 = 3 + 96 – 6 And finally, we rock the addition and subtraction, left to right. 3 + 96 – 6 = 93 Okay, but Please Excuse My Dear Aunt Sally is a really long thing to remember. And we've already spent the last 10 years attempting to block her out. Here at Shmoop, we like to simplify things ("Shmoop" is actually short for "Shmoopalumpagus"). We've seen that subtraction can be replaced by adding a negative, and division can be replaced by multiplying a reciprocal. So all we really need to remember is Please Excuse My Aunt. Do what's inside the Parentheses first, then take all Exponents, then Multiply, then Add. Just remember that division = multiply by reciprocal, and subtraction = add a negative. Yeah, either way, you're going to have to remember some stuff. C'est l'algebra. Let's work out one last example. ### Sample Problem Ok, so let's start by dealing with the stuff inside the parentheses: . We wanna change it into something more manageable before we square it. We should also probably change the radio station. What is that, Avant-Garde Metal? A pretty roundabout way just to get to the number 1, but we'll take it. Now that we've finished all operations inside the parentheses, we look for exponents. If they don't immediately present themselves, whistle loudly and shake a bag of treats—they'll come running. In the first part of the equation we have 12, which of course is just 1. Then we look for multiplication. Since there is none, all that's left is to add -13. The subtraction of such an unlucky number can't be a bad thing. 1 + (-13) = -12 And we're done!
# Difference Between Integration And Differentiation Document Sample ``` Difference Between Integration And Differentiation Difference Between Integration And Differentiation Friends in today’s session I am going to lay stress on a very important topic of mathematics. This topic is all about the differentiation and integration and their comparison. These two topics are the main part of the mathematics. The differentiation and integration both are used in many fields like physics, engineering etc. First of all I will tell you the basic definition of differentiation and integration and then we will go for the comparison between them. Differentiation: The differentiation is a process by which we can find a derivative of any function. This is a reverse process of integration. The reverse of the differentiation is also known as anti differentiation of a function. It is a very interesting branch of calculus. The derivative of a function can be defined as how much a quantity changes with respect to the change in another quantity. Tutorcircle.com Page No. : 1/4 Integration: The Integration is just opposite or reverse process of differentiation; that is also called anti differentiation. Now here we are going to discuss the Comparison between Integration and Differentiation: a. These two terms (named as Differentiation and Integration) are the branch of Calculus that is one of the very important fields of mathematics. Integration is adding or summing up while b. Integration is used to calculate the distance travelled by a function where as the differentiation is used to calculate the speed of the function. c. Integration integrates or makes the little fractions to large one and the differentiation divides a large one into many small fractions. d. Both are just opposite of each other for example: differentiation d/dx (cos x) = - sin x Integration ∫ cos x = sin x. e. Differentiation of a given function results in an answer, if we integrate the result or the answer then we will again get the function back. This proves that these both are opposite of each other. f. The differentiation of a function gives the slope of the function. For example a linear equation y = mx + c the differentiation of this will result in m = dy/dx. Unlike the differentiation the integration gives the area between the given function and its x axis. Tutorcircle.com Page No. : 2/4 g. We can say that the differentiation is used to find the change in one quantity with respect to other quantity where the integration is used to get the area which is covered by the curve of function with its x axis. h. Integration is a reverse process of differentiation, for example: Given function is x = y^3 Differentiating the function with respect to y we get dx/dy = 3y^2 Where the integration of the answer of the differentiation that is 3y^2; we will get ∫ 3y^2dy = y^4 Here we can see that by integration of the result of differentiation of a function we again get the function back. So this is proved that the integration is a reverse method of differentiation. Tutorcircle.com Page No. : 3/4 Page No. : 2/3 Thank You TutorCircle.com ``` DOCUMENT INFO Shared By: Categories: Tags: Stats: views: 5 posted: 5/17/2012 language: pages: 4 How are you planning on using Docstoc?
# Q.Respected Sir/Madam; Sketch the graph of `y=(x+2)^3 - 5` lemjay \| Certified Educator `y=(x+2)^3-5` To graph this, transformation of function can be applied. Since the given is a cubic function, let's start with the graph of the basic cubic function which is `y_0=x^3` . Then, consider the expression inside the parenthesis `y_1=(x+2)^3` . Notice that the variable x is added by 2. That means to graph `y_1=(x+2)^3` , the curve above `(y_0)` should be moved sidewards, which is 2 units to the left. And, consider the number after `(x+2)^3` . We would then have `y=(x+2)^3 -5` . Since `y_1` is equal to `(x+2)^3` , then the given function can be express as `y=y_1-5` . Here, notice that the variable `y_1` is subtracted by 5. That means, to get the graph of y, the red curve above `(y_1)` should be moved in vertical direction, which is 5 units down. Hence, the graph of the given function `y=(x+2)^3-5` is: jeew-m \| Certified Educator `y = (x+2)^3-5` When y = 0; `0 = (x+2)^3-5` `x = (5)^(1/3)-2 = -0.29` When `x = 0` ; `y = (0+2)^3-5` `y = 3` So the x intercept of the graph is `y = 3` and y intercept of the graph is `x = -0.29` . The maximum and minimum for this graph is obtained when `y' = 0` . `y' = 3(x+2)^2` When `y' = 0` ; `3(x+2)^2 = 0` `x = -2` `y'' = 6(x+2)` `(y'')_(x=-2) = 0` So we have a point of inflection at `x = -2` . When `x = -2` then `y = -5` When `x rarr +oo;` `lim_(xrarroo)(x+2)^3-5 = +oo` When `x rarr -oo;` `lim_(xrarr-oo)(x+2)^3-5 = -oo` So using the above details we can plot the graph. The plotted graph is shown below.
# Equivalent Expressions - Examples, Exercises and Solutions In previous articles, we have talked about what an algebraic expression is and how to get the numerical value of algebraic expressions. Today, we will cover equivalent expressions. Equivalent expressions are two or more algebraic expressions that represent the same value. They may have a different structure, but their numerical value will be the same. For example, in the following equation both sides represent the same quantity: $9X=3X+6X$ Below is another example with 2 variables. By simplifying the expressions on both sides of the equation, we can work out that on both we have $2X-3Y+5$ and therefore the expressions are equivalent. $2X-3Y+5=X+X-2Y+10-5-Y$ ## examples with solutions for equivalent expressions ### Exercise #1 $18x-7+4x-9-8x=\text{?}$ ### Step-by-Step Solution To solve the exercise, we will reorder the numbers using the substitution property. $18x-8x+4x-7-9=$ To continue, let's remember an important rule: 1. It is impossible to add or subtract numbers with variables. That is, we cannot subtract 7 from 8X, for example... We solve according to the order of arithmetic operations, from left to right: $18x-8x=10x$$10x+4x=14x$$-7-9=-16$Remember, these two numbers cannot be added or subtracted, so the result is: $14x-16$ $14x-16$ ### Exercise #2 $7.3\cdot4a+2.3+8a=\text{?}$ ### Step-by-Step Solution It is important to remember that when we have numbers and variables, it is impossible to add or subtract them from each other. We group the elements: $7.3×4a + 2.3 + 8a =$ 29.2a + 2.3 + 8a = $37.2a + 2.3$ And in this exercise, this is the solution! You can continue looking for the value of a. But in this case, there is no need. $37.2a+2.3$ ### Exercise #3 $\frac{9m}{3m^2}\times\frac{3m}{6}=$ ### Step-by-Step Solution According to the laws of multiplication, we will simplify everything into one exercise: $\frac{9m\times3m}{3m^2\times6}=$ We will simplify and get: $\frac{9m^2}{m^2\times6}=$ We will simplify and get: $\frac{9}{6}=$ We will factor the expression into a multiplication: $\frac{3\times3}{3\times2}=$ We will simplify and get: $\frac{3}{2}=1.5$ $0.5m$ ### Exercise #4 $3x+4x+7+2=\text{?}$ ### Video Solution $7x+9$ ### Exercise #5 $3z+19z-4z=\text{?}$ ### Video Solution $18z$ ### Exercise #6 Are the expressions the same or not? $20x$ $2\times10x$ Yes ### Exercise #7 Are the expressions the same or not? $3+3+3+3$ $3\times4$ Yes ### Exercise #8 Are the expressions the same or not? $18x$ $2+9x$ No ### Exercise #9 $x+x=$ ### Video Solution $2x$ ### Exercise #10 $5+8-9+5x-4x=$ 4+X ### Exercise #11 $5+0+8x-5=$ ### Video Solution $8X$ ### Exercise #12 $11+5x-2x+8=$ 19+3X ### Exercise #13 $7a+8b+4a+9b=\text{?}$ ### Video Solution $11a+17b$ ### Exercise #14 $13a+14b+17c-4a-2b-4b=\text{?}$ ### Video Solution $9a+8b+17c$ ### Exercise #15 $a+b+bc+9a+10b+3c=\text{?}$ ### Video Solution $10a+11b+(b+3)c$
Courses Courses for Kids Free study material Free LIVE classes More # Intercept Theorem LIVE Join Vedantu’s FREE Mastercalss ## An Overview of the Theorem The Intercept Theorem is a fundamental tool of Euclidean Geometry. The concept of parallel lines and transversal is of great importance in our day-to-day life. And, the Intercept theorem extends our understanding of parallel lines and transversal and we can apply these concepts in our day-to-day life. A Transversal In the above figure, we can see that there are 3 parallel lines \${L}_{1}\$,\${L}_{2}\$, \${L}_{3}\$ and then there is a transversal \$PR\$ which is intersecting all the 3 parallel lines at an equal distance. The intercept theorem, also known as Thale’s theorem, Basic Proportionality Theorem, or side splitter theorem, is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels. ### History of the Mathematician Euclid • Year of Birth: 325 BC • Year of Death: 270 BC • Contribution: He contributed significantly in the field of Mathematics and Physics by discovering the intercept theorem. ## Statement of the Theorem If there are three or more parallel lines and the intercepts made by them on one transversal are equal, the corresponding intercepts of any transversal are also equal. ## Proof of the Theorem Two Parallel Lines Given: \$l\$, \$m\$, \$n\$ are three parallel lines. \$P\$ is a transversal intersecting the parallel lines such that \$AB=BC\$. The transversal \$Q\$ has the intercepts \$DE\$ and \$HE\$ by the parallel lines \$l\$, \$m\$, \$n\$. To prove: \$DE=EF\$ Proof: Draw a line \$E\$ parallel to the line \$P\$ which intersects the line \$n\$ at \$H\$ and line \$l\$ at \$G\$. \$AG\|\|BE\$ (Given) \$GE\|\|AB\$ (By construction) From the information above, we can say that \$AGBE\$ is a parallelogram. According to the properties of a parallelogram: \$AB=GE\$ - (1) Similarly, we can say that \$BEHC\$ is a parallelogram. \$BC=HE\$ - (2) From the given information, we know that \$AB=BC\$. So, from equations (1) and (2), we can say that \$GE=HE\$. In \$\Delta GED\$ and \$\Delta HEF\$, \$GE=HE\$(Proved) \$\angle GED=\angle FEH\$(Vertically Opposite Angles) \$\angle DGE=\angle FHE\$(Alternate Interior Angles) Hence, \$\Delta GED\cong \Delta HEF\$ As\$\Delta GED\cong \Delta HEF\$, the sides\$DE=EF\$. Hence proved. ## Applications of the Theorem The intercept theorem can be used to prove that a certain construction yields parallel line segments: • If the midpoints of two triangle sides are connected, then the resulting line segment is parallel to the third triangle side (Mid point theorem of triangles). • If the midpoints of the two non-parallel sides of a trapezoid are connected, then the resulting line segment is parallel to the other two sides of the trapezoid. ## Limitations of the Theorem • The intercept theorem is not able to help us in finding the midpoint of the sides of the triangle. • The basic proportionality theorem is an advanced version of the intercept theorem and it gives us a lot of information on the sides of the triangles. ## Solved Examples 1. In a \[\Delta ABC\], sides \[AB\] and \[AC\] are intersected by a line at \[D\] and \[E\], respectively, which is parallel to side \[BC\]. Prove that \[\dfrac{AD}{AB}=\dfrac{AE}{AC}\]. Ans: Scalene Triangle \[DE\|\|BC\] (Given) Interchanging the ratios, Interchanging the ratios again, Hence proved. 2. Find DE Basic Proportionality Theorem Ans: According to the basic proportionality theorem, \[\dfrac{AE}{DE}=\dfrac{BE}{CE}\] \[\dfrac{4}{DE}=\dfrac{6}{8.5}\] \[\dfrac{4*8.5}{6}=DE\] \[DE=5.66\] So, \[DE=5.66\] 3. In \[\Delta ABC\], \[D\] and \[E\] are points on the sides \[AB\] and \[AC\], respectively, such that \[DE\|\|BC\]. If \[\dfrac{AD}{DB}=\dfrac{3}{4}\] and \[AC=15cm\], find \[AE\]. Intercept Theorem Ans:\[\dfrac{AD}{DB}=\dfrac{AE}{EC}\] (According to the intercept theorem) Let \[AE=x\] and \[EC=15-x\] So, \[\dfrac{3}{4}=\dfrac{x}{15-x}\] \[3(15-x)=4x\] \[45=7x\] \[x=\dfrac{45}{7}\] \[x=6.4cm\] So, \[x=6.4cm\] ## Important Points • The intercept theorem can only be applied when the lies are parallel, if the transversal is cutting lines that are not parallel, then the intercept theorem is not valid. • The basic proportionality theorem and mid-point theorem are all applications of the intercept theorem but they are not the same theorems. ## Conclusion In the above article, we have discussed the Equal intercept Theorem and its proof. We have also discussed the applications of the theorem. So, we can conclude that Intercept Theorem is a fundamental tool of Geometry and is based on applications of parallel lines and transversal and reduces our computational work based on its application as we have seen in the examples based on the theorem. Last updated date: 25th Sep 2023 Total views: 102.3k Views today: 3.02k Competitive Exams after 12th Science ## FAQs on Intercept Theorem 1. Are the basic proportionality theorem, mid-point theorem, and intercept theorem the same? No, basic proportionality theorem, intercept theorem, and mid-point theorem are 3 different kinds of theorems. The intercept theorem is a very broad theorem and it tells about how the properties of a transversal change when they interact with parallel lines. The basic proportionality theorem is covering about the interaction of lines parallel to one side of the triangle with the other two sides and the basic proportionality theorem is derived from the intercept theorem. The midpoint theorem talks about when a line bisects the other two sides of a triangle, then it is parallel to the third side, and the midpoint theorem is derived from the basic proportionality theorem. 2. What is the mid-point theorem? The line segment of a triangle connecting the midpoint of two sides of the triangle is said to be parallel to its third side and is also half the length of the third side, according to the midpoint theorem. The mid-point theorem is derived from the basic proportionality theorem and it is a very useful theorem in solving the questions of geometry as it directly gives the value of the sides of a triangle and also the parallel line helps to get the values of angles. 3. Why can’t the intercept theorem be used for non-parallel lines? The intercept theorem cannot be used for non-parallel lines because there are a lot of things in the derivation of the intercept theorem which we won’t get if the lines are not parallel. Like we won’t be able to use the angles such as the alternate interior angles and the vertically opposite angles and much more. The construction that we make in the derivation is only made because the lines are parallel, hence if the lines are not parallel the intercept theorem is not valid.
# of Building by liaoqinmei VIEWS: 4 PAGES: 6 • pg 1 ``` LAYOUT OF BUILDING A. Basic Understandings 1. When laying out a building, care should be taken to get the building laid out as perfectly square as possible. If the building is not laid out square, compensation for the error must be made throughout the entire construction of the building. 2. A building may be laid out square with a transit (or farm level) and measuring tape, or with just a measuring tape. If only a tape is available, considerable time can be saved if two tapes and a calculator with a square root key is used. 3. When laying out a rectangular or square building, all four corners must be 90o. To check the building for squareness, measure both of the diagonals. If the two diagonals are equal, the building is laid out square. The diagonals are the The diagonals are not the same measurement. The same measurement. The building is laid out square. building is not laid out square. 4. When checking a right angle (90o), two methods may be used to check the angle for accuracy: a. 3-4-5 method – If one leg of a right triangle is 3 and the other leg is 4, the hypotenuse will be 5. Hypotenuse = 5 (Any multiples of 3-4-5 may Leg 1 = 3 be used, such as 6-8-10; 9-12-15, etc.) Leg 2 =4 The basis of the 3-4-5 method is the Pythagorean Theorem, which states that in a right triangle. The square of the hypotenuse equals the sum of the squares of the other two sides. A right triangle is one in which one angle equals 90o. The hypotenuse is the side opposite the right angle. The formula for the theorem is: C2 = A2 + B2 Using the theorem to check the triangle is the example above: 52 = 32 + 42 OR 25 = 9 + 16 b. Another method of checking a right angle is to determine the hypotenuse of a right triangle based on the total base and height of the triangle. The hypotenuse of a right triangle will be the sum of the base of the triangle squared (the number multiplied by itself) plus the height of the triangle squared. When laying out a building, the hypotenuse will be diagonal for the building. Example: 10’ height Hypotenuse = 26.9’ 102 (10 x 10) = 100 252 (25 x 25) = 625 725 √725 = 26.9’ Base = 25’ The Pythagorean Theorem is also the basis for this method of squaring a right triangle. If C2 = A2 + B2, then C = √(A2 + B2). By determining the hypotenuse (diagonal of the building) first, the building can be laid out using two measuring tapes without using a trial and error method. If the 3-4-5 method is used, several extra measurements must be made to lay out the building. B. Laying Out a Building With Two Tapes and a Calculator EXAMPLE: 8’ x 12’ Building 1. Establish one side of the building by measuring the desired length (or width) of the building. This side can be made parallel to another building or can be simply determined by the owner. Corner A 12’ Corner B Drive two stakes to mark the two corners of the building. Drive a nail in the top of the stakes to more accurately locate the corners of the building. 2. Consider the building layout to be made up of two right triangles put together. Corner A Corner B Corner A Corner B Right triangle Right triangle Right triangle Right triangle Corner C Corner D Corner C Corner D The two right triangles will have the same hypotenuse. The hypotenuse will be the diagonal for the building. 3. Determine the hypotenuse of the triangles. Again, the hypotenuse will be the diagonals for the building layout. a. On the calculator, multiply 12 x 12 (= 144) and press M+ on the calculator. b. Multiply 8 x 8 (=64) and press M+ on the calculator. c. Press Memory Recall (MR). The sum of 144 + 64 = 208 d. The hypotenuse (or diagonals) will be the square root of 208. e. Press the square root key. The hypotenuse (diagonals) will be 14.42205 feet. f. If you have a tape that is graduated in feet and tenths of a foot, the diagonals will be 14.2’. g. Since most tapes are graduated in feet, inches, and sixteenths of an inch, the 14.422205 feet must be converted to feet-inches and sixteenths of an inch. 1. The diagonals will be 14’ plus .422205’. 2. Convert.422205’ to inches by multiplying by 12. The diagonal will be 14’5” plus .06646. 3. Convert .06646” to sixteenths of an inch by multiplying by 16. The diagonals will be 14’5 – 1/16. 4. Establish Corner C. a. Measure from Corner A toward Corner C a distance of 8’. b. With the other tape, measure from Corner B toward Corner C a distance of 14’5- 1/16”. c. The point at which these two measurements intersect is Corner C Corner A Corner B 8’ 14’ 5-1/16” Corner C d. Drive a stake to mark Corner C. Drive a nail in the top of the stake to more accurately locate the corner of the building. 5. Establish Corner D a. Measure from Corner B toward Corner D a distance of 8’. b. Measure from Corner C toward Corner D a distance of 12’. c. The point at which these two measurements intersect is Corner D. Corner A Corner B 8’ Corner C 12’ Corner D d. Drive a stake to mark the corner of the building. Drive a nail in the top of the stake to more accurately locate the corner of the building. 6. Check all measurements again. Measure both lengths, both widths and both diagonals. Corner A 12’ Corner B 14’5-1/16” 8’ 8’ 12’ Corner C Corner D 7. Make any adjustment necessary to square the building. Adjustments should be made at Corner C and Corner D. NOTE: ALL MEASUREMENTS SHOULD BE MADE WITH THE TAPE HELD LEVEL. THEREFORE, ON SLOPING GROUND, IT MAY BE NECESSARY TO USE A PLUMB BOB TO MARK THE LOCATION OF THE CORNERS OF THE BUILDING. Setting Batter Boards 1. With all four corners of the building located, set batter boards beyond the corners. Place the batter boards far enough away from the corners to allow the batter boards to remain in place until the framing members of the wall are in place. Corner A Corner B Corner C Corner D 2. All batter boards should be set level and level with the batter boards on the other corners. Use a spirit level and a string level to set them level. 3. The strings marking the location of the walls should pass directly over the nails marking the location of the corners of the building. Mark the location of the strings on the batter boards with a saw kerf. 4. After batter boards and strings are in place, the stakes marking the corners of the building may be removed. ``` To top
# 2001 AMC 8 Problems/Problem 25 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it? $\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$ ## Solution We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\times2=4914$, since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is multiplied by $2$, it is less than or equal to $7542$, the largest number in the set. This happens to be $2754\times2=5508$. Therefore, the number would have to be between $4914$ and $5508$, and also even. The only even numbers in the set and in this range are $5472$ and $5274$. A quick check reveals that neither of these numbers is twice a number in the set. The number can't be quadruple or more another number in the set since $2457\times4=9828$, well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is $2457\times3=7371$ and the greatest is $2475\times3=7425$, since any higher number in the set multiplied by $3$ would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is $2475\times3=7425, \boxed{\textbf{(D)}}$ Or, since the greatest number possible divided by the smallest number possible is slightly greater than 3, divide all of the choices by 2, then 3 and see if the resulting answer contains 2,4,5 and 7. Doing so, you find that $7425/3 = 2475, \boxed{\textbf{(D)}}$

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