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applicationsThe information provided by enunciation is that the elements of the set, have to respect the constraint x2=<3 We'll rewrite the constraint using the absolute value property: x2=<3...

applicationsTo compute the left side expression, we'll expand the squares, using the formulas: (a+b)^2 = a^2 + 2ab + b^2 (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc) (x+y)^2 = x^2 + 2xy + y^2 (x+y+1)^2 = x^2 +...

applicationsWe notice that the denominator of each fraction is the product of 2 consecutive numbers. We'll decompose the fraction inot partial fractions: 1/n(n+1) = A/n + B/(n+1) (1) We'll multiply the ratio...

applicationsThe intervals are sets of real numbers. Examples: The open interval (a;b) is the set of real numbers x that have the property: a < x < b All the real numbers located between a and b,...

applicationsWe'll recall the identities that helps us to complete the given squares: (a+b)^2 = a^2 + 2ab + b^2 (ab)^2 = a^2  2ab + b^2 We'll complete the first square x^214x+_ We'll identify a^2 = x^2 =>...

applicationsWe'll factor the numerator and denominator by x^n*y^n:x^n*y^n(x^3n  y^3n)/x^n*y^n(y^2n  x^2n)We'll reduce and we'll get:(x^3n  y^3n)/(y^2n  x^2n)We'll write the numerator using the formula of...

applicationsWe know that the values of x are positive in the 1st and the 4th quadrants.We also know that the values of y are negative in the 3rd and the 4th quadrants.We notice that the 4th quadrant...

applications6x1>6x We'll subtract 6b both sides: 6x  1  6 + x > 0 We'll combine like terms: 7x  7 > 0 We'll factorize by 7: 7(x1)>0 We'll divide by 7: x1>0 x>1 The range of possible...

applicationsThe value of discriminant is evaluated applying the formula: delta = b^2  4ac We'll identify a,b,c: 6y^2  2y  3 = 0 a = 6 b = 2 c = 3 We'll substitute the coefficients in the formula of...

applicationsGraphsFind the function y = mx + n if the points (1;2) and (1;1) are on the graph of the function.Any function is determined when it's coefficients are known. y = f(x) = mx + n Therefore, in order to determine the linear function, we'll have to calculate the coefficients m and n. Since the...

applicationsWe'll establish the conditions of existence of logarithms: (b+2)^2>0 The perfect square is always positive, for any value of b. log (b+2)^2 2log b = log 3 We'll shift 2log b to the right: log...

applicationsWe'll rewrite the given expression: 2*24/2(x+1) = 2(x+3) We'll reduce: 24/(x+1) = 2(x+3) We must multiply both sides by the denominator x+1. 24(x + 1)/ (x + 1) = 2(x+3)(x+1) We'll reduce and we'll...

applicationsLet the numbers be x and y: x  y = 28 x = 28 + y x^2 + y^2 = (28+y)^2 + y^2 We'll expand the square: 784 + 56y + y^2 + y^2 We'll combine the like terms and we'll rearrange the terms: 2y^2 + 56y +...

applicationsFirst, we'll build the determinant of the square matrix of the system: 1 1 1 det A = 2 1 2 1 4 m We'll evaluate the determinant: detA = 1*(1)*m + 2*4*1 + 1*1*(2) ...

applicationsWe'll calculate Sn (x) for x = 1. Sn (x) = 1 + 2*1 + 3*1 + .... + n*1 Since the expression above represents the sum of the first natural n terms, we'll replace it by the formula: Sn (1) = n(n+1)/2...

applicationsWe'll eliminate z from the equation (2) and (3). x + 2y = 10 (1) 9y – 3z =  6 (2) 6x – 5z = 2 (3) To eliminate z from (2) and (3), we'll multiply (2) by 5 and (3) by 3: 45y – 15z =  30...

applicationsWe'll put the equation of the lines in the general form: ax + by + c = 0 The first equation is: 7x2y17=0 The second equation is: 14x9y24=0 We'll form the system: 7x2y17=0 We'll add 17 both...

applicationsTo calculate the modulus of the complex number z, we'll rewrite the given complex number z. Since it is not allowed to keep a complex number to denominator, we'll multiply the ratio by the...

applicationsWe'll calculate the inverse function for the given functions. Let f(x) = y: y = x We'll multiply by 1 and we'll get: x = y f^1(x) = x So f(x) = f^1(x) for f^1(x) = x. We'll get the inverse...

applicationsThis is like having two equations. One is 2x  3 = 7 and the other is (2x  3) = 7 Which is the same as 2x + 3 = 7 So now we have to solve both equations. 2x  3 = 7 2x = 10 x = 5 And 2x + 3 =...

applicationsWe'll write the conditions: 1<(3x + 1)/2 < 1 We'll solve the simultaneous inequalities: 1<(3x + 1)/2 2<3x + 1 3<3x We'll divide by 3: x>1 We'll solve the other inequality:...

applicationsLet x and x+1 be the two consecutive numbers. We know, from enunciation that the sum of their squares is 61: x^2 + (x+1)^2 = 61 We'll expand the square using the formula: (a+b)^2 = a^2 + 2ab + b^2...

applications3x  2 >= 0 3x > =2 x>=2/3 10  x>=0 x=<10 The interval of admissible values for x is [2/3 ; 10]. Now, we'll solve the equation. We'll multiply both sides by the conjugate of the...

applicationsWe'll start from the given constraint X*Y=Y*X. We know that the product of two matrices is not commutative. Since X is not equal to Y, then, X*Y=Y*X if and only if Y = X^1 (Y is the inverse of the...

applicationsBY definition, the tangent line to a graph, at a given point, is the derivative of the given function: dy/dx = 1/x^2 The slope of the tangent line is: m = 1/3^2 m = 1/9 The equation of the line...

applicationsfof^1(x) = f(f^1(x)) = x, by definition Here y = x/4 + 3 The inverse function is found by expressing y in terms of x, x = (y  3)*4. Now interchange y and x. f^(x) = (x  3)*4 fof^1(x) = f((x ...

applicationsWe have to solve 3 log 4 (a2) = (9/2) for a 3 log(4) (a2) = (9/2) => log(4) (a2) = (9/6) => a  2 = 4^(3/2) => a  2 = 8 => a = 10 The solution for the equation is a = 10

applicationsAnother method to solve x^2 + 3x – 4 = 0 is by factorization. x^2 + 3x – 4 = 0 => x^2 + 4x  x  4 = 0 => x( x + 4)  1(x + 4) = 0 => (x  1)(x + 4) = 0 => x = 1 and x = 4 The...

applicationsWe have to simplify: x/ (1+1/x) + (1/x)/ (x + 1) x/ (1+1/x) + (1/x)/ (x + 1) => x/ ((x+1)/x) + (1/x)/ (x + 1) => x^2/(x + 1) + 1/x(x+1) => (x^3 + 1)/(x + 1)x => (x + 1)(x^2  x + 1)/(x...

applicationsThe equations to be solved are: x^2 + y^2 = 200  xy ...(1) x + y = 20  (xy)^(1/2) ...(2) (2) => (x + y)^2 = (20  (xy)^(1/2))^2 => x^2 + y^2 + 2xy = 400 + xy  40(xy)^(1/2) => x^2 + y^2...

applicationsWe have to solve: e^(3x4)  1/e^(x12)=0 e^(3x4)  1/e^(x12)=0 => e^(3x4)  e^((x12)) = 0 => e^(3x4) = e^((x12)) take the logarithm to the base e of both the sides => ln(e^(3x4))...

applicationsWe have to prove: (2x1)^2(x3)^2 = (3x4)(x+2) Take the left hand side (2x1)^2(x3)^2 Use the relation x^2  y^2 = (x  y)(x + y) => (2x  1  x + 3)(2x  1 + x  3) => (x + 2)(3x  4)...

applicationsWe have to find the square of (4+3i) (4+3i)^2 => (4+3i)(4+3i) => 4*4 + 4*3*i + 4*3*i + 3*i*3*i => 16 + 12i + 12i + 9i^2 i^2 = 1 => 16 + 24i  9 => 7 + 24i The required square is 7 +...

applicationsQuadraticThe quadratic equation has roots x=2/3 and x=4. Find one set of possible values for a,b,c.The form of the quadratic is ax^2 + bx + c = 0 We'll write the quadratic as a product of linear factors: ax^2 + bx + c = (x  2/3)(x+4) ax^2 + bx + c = x^2 + 4x  2x/3  8/3 Comparing both sides,...

applicationsThe sum of n terms of an arithmetic series is given as: Sn = 5n^2  11n Sn1 = 5(n  1)^2  11(n  1) => 5(n^2 + 1  2n)  11n + 11 => 5n^2 + 5  10n  11n + 11 => 5n^2  21n + 16 The term...

applicationsTangent line equation.Given x = 5cost, y =3sint, what is the equation of the tangent line if t=pi/4.It is given that x = 5*cos t and y = 3*sin t. We have to find the equation of the tangent if t = pi/4 When t = pi/4 , x = 5*(1/sqrt 2) and y = 3/sqrt 2 dx/dt = 5*sin t and dy/dt = 3*cos t dy/dx =...

applicationsIt is given that log(72) 48 = a and log(6) 24 = b a = log(72) 48 = log(6) 48/ log(6) 72 => log(6) (6*8)/log(6) 6*12 => [1 + log(6) 8]/[1 + log(6) 12] => [1 + 3*log(6) 2]/[2 + log(6) 2] b =...

applicationsTo find the derivative dy/dx of y= x^25xy+3y^2 implicit differentiation can be used and differentiation carried out in the following way: y= x^25xy+3y^2 dy/dx = 2x  5*x*(dy/dx)  5y +...

applicationsWe'll write the equation of the ellipse: (x/a)^2 + (y/b)^2 = 1 To put the given equation in this form, we'll divide by 64 both sides: 4x^2 + 16y^2 = 64 4x^2/64 + 16y^2/64 = 64/64 We'll simplify and...

applicationsWe have to find x and y given x(1+2i)+y(2i)=4+3i x(1+2i)+y(2i)=4+3i => x + 2xi + 2y  yi = 4 + 3i equate the real and imaginary coefficients x + 2y = 4 and 2x  y = 3 x = 4  2y substitute in...

applicationsAccording to the fundamental theorem of calculus: Int f(x)dx = F(b)  F(a) for x = a to x = b In this case a = 1 and b = 1 We recall the property of additivity of Integrals: Int (x^3x)dx = Int...

applicationsWe have to prove that (cospi/4+i*sinpi/4)^2008 is real. cos (pi/4) = 1/sqrt 2 sin (pi/4) = 1/sqrt 2 (cos pi/4 + i*sin pi/4)^2008 => [1/sqrt 2 + i*(1/sqrt 2)]^2008 => [1/sqrt 2 + i*(1/sqrt...

applicationsApplying combinatorics we'll get: C(x,2) = x!/2!*(x2)! But x! = (x2)!*(x1)*x 2! = 1*2 = 2 C(x,2) =(x2)!*(x1)*x/2!*(x2)! We'll simplify and we'll get: C(x,2) =(x1)*x/2 A(x,2) = x!/(x2)!...

applicationsWe have to find the tangent to the curve y = x^3  7x^2 + 14x  8 at the point where x=1 At x = 1, y = 1  7 + 14  8 = 0. The slope of the tangent to the curve at x = 1 is the value of y' at x =...

applicationsWe have to solve x + y = 5 ...(1) x^2 + y^2 = 13 ...(2) From (1) x = 5  y Substituting in (2) (5  y)^2 + y^2 = 13 => 25 + y^2  10y + y^2 = 13 => 2y^2  10y + 12 = 0 => y^2  5y + 6 = 0...

applicationsSince the given logarithm does exist, then the base and the argument of logarithm respect the conditions of existence. We'll impose the constraints of existence of logarithms: 1) x+1 > 0 2) x +...

applicationsThe equation given is lg(x+1)  lg 9 = 1  lg x I assume the base of the logarithm is 10. Use the property that lg a  lg b = lg a/b and 1 = lg 10 lg(x+1)  lg 9 = 1  lg x => lg(x+1)  lg 9 =...

applicationsThe consecutive terms of a geometric sequence have a common ratio. We have the sequence given by 3 ,a ,b , 24 24/b = b/a = a/3 => 24a = b^2 or a = b^2/24 substitute in b/a = a/3 => 24b/b^2 =...

applicationsFor the triangle ABC AB = 6, B=pi/4, C=pi/6. As the angles of a triangle have a sum of pi. A = 7*pi/12 Use the property sin A/a = sin B/b = sin C/c c = 6, C = pi/6, B = pi/4 and A = 7*pi/12 =>...

applicationsBy definition, if x = i is cancelling out the polynomial, that means that x = i is the root of the polynomial. f(i) = 0 We'll replace x by i and we'll have: f(i) = 4*i^3  12*i^2 + m*i + n We'll...