c=-15=0. The new equation is (2m^2-m-15) Original equation: 2m^2-m-15=0 Step 1. (2m^2-6m+5=15=0) To start we separate the fraction. Then solve both sides of the original equation separately Step 2. 2m(m-3)+5(m-3)=0 Move both factors to the left of the 0 a) M-3=0 Step 3. 2m+5=0 Divide each term in A by “2” Solve: a) m=-5/2 For the second problem the quadratic formula will be used to solve the equation. Since a=2y^2 b=3y
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of elevation of the sun when the shadow of a pole ‘h’ meters high is 3 h m. long is… A) 30° B) 45° C) 60° D) 90° 2. The perimeter of a triangle with vertices (0‚4)‚ (0‚0) and (3‚0) is … A) 5 B) 7 C) 11 D) 12 2 3. The roots of a quadratic equation 2 x -kx+k=0 are equal. The value of ‘k’ is … A) 0 only B) 4 C) 8 only D) 0 and 8 4. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is … A) 3:4 B) 4:3 C) 9:16 D) 16:9 5. If the common difference of
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Al-Khwarizmi: The Father of Algebra Muhammed Ibn Musa al-Khwarizmi‚ was a mathematical pioneer‚ and is considered by many to be the greatest mathematician of the Islamic world‚ as well as the founder algebra. His book entitled Kitâb al-Mukhtasar fî Hisâb al-Jabr wa ’l-Muqâbala‚ which means “The Compendious Book on Calculation by Completion and Balancing‚” established algebra as an independent discipline. While his arithmetic work‚ possibly entitled Kitāb al-Jamʿ wa-l-tafrīq bi-ḥisāb al-Hind
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MIDTERM NOTES Solve Linear Equations 0. Cancel all denominators by multiplying every term by the LCD. 1. Simplify LHS and RHS. 2. Eliminate variable term on RHS. 3. Eliminate variable term on LHS. 4. Eliminate the coefficient of the variable. Solve Rational Equations 1. Find LCD. 2. Cancel all denominators by multiplying every term by the LCD. 3. Solve. 4. Omit those solutions that make LCD=0. Complex Numbers: a + bi Powers of i: i0 = 1‚ i1 = i‚ i2 = −1‚ i3 = −i in = ir where r is the
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their generators. Technology Used: Technology that had been used is shown below 1) Autograph (Version 3.3) Graphing Display Calculator TI-84 Plus Texas Instruments 2) Defining terms:i Quadratic‚ cubic‚ quartic functions are members of the family of polynomials. A quadratic function is a function of the form constants and A cubic function is a function of the form are constants and A quartic function is a function of the form where are constants and Complex numbers
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Abstract Algebra 1 of 29 SOLVABILITY BY RADICALS - Linear Equation Clearly the root of the linear equation ax + b = 0 (1) is given in terms of the coefficients a and b by x = −b/a as long as a = 0. Nicomedes Alonso III Abstract Algebra 2 of 29 SOLVABILITY BY RADICALS - Quadratic Equation We know that the roots of the quadratic equation‚ ax 2 + bx + c = 0 (2) are given by the well-known quadratic formula x= −b ± b2 − 4ac ‚ 2a a=0 in terms of
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Contents MODULE 2 1 Scatter graphs 1.1 Scatter graphs and relationships 1.2 Lines of best fit and correlation 1.3 Using lines of best fit Chapter summary Chapter review questions 1 1 5 6 10 10 4 Processing‚ representing and interpreting data 4.1 Frequency polygons 4.2 Cumulative frequency 4.3 Box plots 4.4 Comparing distributions 4.5 Frequency density and histograms Chapter summary Chapter review questions 51 51 56 64 65 68 73 73 2 Collecting and recording data 14 2.1 Introduction
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b b 16. a0 = 1 where a ∈ R‚ a = 0 1 1 17. a−n = n ‚ an = −n a a √ 18. ap/q = q ap 19. If am = an and a = ±1‚ a = 0 then m = n 20. If an = bn where n = 0‚ then a = ±b √ √ √ √ 21. If x‚ y are quadratic surds and if a + x = y‚ then a = 0 and x = y √ √ √ √ 22. If x‚ y are quadratic surds and if a + x = b + y then a = b and x = y 23. If a‚ m‚ n are positive real numbers and a = 1‚ then loga mn = loga m+loga n m 24. If a‚ m‚ n are positive real numbers‚ a = 1‚ then loga = loga m −
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produce a box with the maximum volume. The concept used to solve the problem To calculate the volume of the cube‚ length x breadth x height was utilized. This gave a cubic equation. This equation was then differentiated which gave a quadratic formula. dydx was then equated = 0. The quadratic was then solved using the quadratic formula ( x=-b±b2-4ac2a) to obtain two values of (x). These values were then substituted into the second differential (d2ydx2). If the value substituted produced a negative
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radius. Rewriting the equation with one variable would result in a polynomial equation that you could solve to find the radius. 3. Rewrite the formula using the variable x for the radius. Substitute the value of the volume found in step 2 for V and express the height of the object in terms of x plus or minus a constant. For example‚ if the height measurement is 4 inches longer than the radius‚ then the expression for the height will be (x + 4). 4. Simplify the equation and write it in standard
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