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See the entire simplification process below Explanation First, I would use this rule of exponents to simplify the denominator of the fraction a 0 = 1 (x 0 y − 3 3 x 5 y 4 ) 2 = (1 y − 3 3 x 5 y 4 ) 2 = (y − 3 3 x 5 y 4 ) 2(b) The paramentric equation of C is (x = 2cost y = 2sint, 0 ≤ t ≤ 2π Thus Z C f ds = Z 2π 0 4cos2 t−4sin2 t p 4sin2 t4cos2 tdt = 8 Z 2π 0 cos(2t) = 4sin(2t) 2π 0 = 0 3 Evaluate Z C (xyz)dx2xdyxyzdz, where C consists of line segments from (1,0,1) to (2,3,1)X = 3x7y = 2 a2 b2 c23 2 3 = 5 a22 2 = 2 x−2 78 6 −1 = −2x6x5x= 3a6b−9c45a−17b12c18 = (−5 x) ( 4 2)( 3 3)= 3z6z2 = y2y3y4y5y = (−2x−2y3)(−3x−4y−2)= CEBİRSEL İFADELER 8SINIF MATEMATİK 1 ETKİNLİK wwwcelikhocamorg wwwcelikhocamorg

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3−5 x 2−6y3 2 x 2−6 2 3y3-Algebra Examples Rewrite (4x−6y3)2 ( 4 x 6 y 3) 2 as (4x−6y3)(4x−6y3) ( 4 x 6 y 3) ( 4 x 6 y 3) Expand (4x−6y3)(4x−6y3) ( 4 x 6 y 3) ( 4 x 6 y 3) using the FOIL Method Tap for more steps Apply the distributive property Apply the distributive property(5a −1)3 = (2 5) c d− =3 Két tag négyzetének különbsége 64a2 −100 = 100b4 −36 = a2 −b2 81c2 −49 = 64−16c2 = 9c2 −25 = Két tag köbének összege, különbsége 8y3 −27 = 8x3 −1= x3 −27 = 64a3 −8 = 8y3 −27 = a3 b3 = 27a3 64 = b3 −27 = z3 64 = 64a3 8 = y3 −64 = b3 27 = Teljes négyzetté alakítás

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2 −5 4 x 2 6 xy2 3 8 xy 2 7 −5 6 mn 2 3 8 mn 8 8b−4b5b 9 −10y11y−2y 10 3yn −6yn −4yn 11 2 3 xy− Ex23, 3 Obtain all other zeroes of 3x4 6x3 – 2x2 – 10x 5 , if two of its zeroes are √(5/3) and √(5/3) Introduction 2 is a factor of 6 3 is a factor of 6 So, 2 × 3 is also a factor of 6 We will use the same in our question Ex23, 3 Obtain all other zeroes of 3x4 6x3 (x − 4) ³ = 10 (x 1) ³ = Karla fue al mercado y gastó 2/5 del dinero que llevó comprando carne, luego gastó 1/3 de lo que le queda en menestras y finalmente gasta los 3/4 de l

The equation 2 x 2 − 3 x y 5 y 2 6 x − 3 y 5 = 0 represents A a parabola B an ellipse C a hyperbola D a pair of straight lines Answer Correct option is B an ellipse Comparing the equation with the standard form a x 2 2 h x y b y 2 2 g x 2 f y c = 0Let p(x) = x 3 − 3x 2 − 9x − 5 All the factors of 5 have to be considered These are ±1, ± 5 By trial method, p(−1) = (−1) 3 − 3(−1) 2 − 9(−1) − 5 = − 1 − 3 9 − 5 = 0 Therefore, x 1 is a factor of this polynomial Let us find the quotient on dividing x 3 3x 2 − 9x − 5 by x 1 By long division,Systemofequationscalculator x2y=2x5, xy=3 en Related Symbolab blog posts Middle School Math Solutions – Simultaneous Equations Calculator Solving simultaneous equations is one small algebra step further on from simple equations Symbolab math solutions

Answer to Factor completely5x3 − 30x2y2 xy − 6y375x4 −12x2 18 64 x2 − 9 19 16 2 40x 25 x3 y−4xy3 Mixed factoring exercises 1 x2 −9x18 2 x2 −14x49 3 15y4 −35y3 10y2 4 a3 −9a 5 x3 6x2 −7x 6 1−25b2 7 14x2 64x−30 8 121y4 −9 9 x2 −4x−21 10 x2 4 11 6y3 9y2 3y 12 4b2 −36 13 16x2 48x36 14 36x2 −84x49 15Ví dụ 3 Giải bài toán sau bằng phương pháp đơn hình f (x) = 3x1 4x2 2x3 2x4 min 2x1 2x2 x4 = 28 x1 5x2 3x3 − 2x4 31 2x1 – 2x2 2x3 x4 = 16 xj ≥ 0 (j =14) Bạn đang xem bản rút gọn của tài liệu Xem và tải ngay bản đầy đủ của tài liệu tại đây ( KB, 73 trang ) Đưa

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 First, put the equation in standard quadratic form (x −3)2 = 5 (x −3)(x − 3) = 5 x2 −3x − 3x 9 = 5 x2 −6x 9 = 5 x2 −6x 9 − 5 = 5 − 5 x2 −6x 4 = 0 We can now use the quadratic equation to solve this problem The quadratic formula states6a5 = 2 ÷ 3 × 6 × a3 − 2 5 = 4a6 r 3a4 × 2a5 × a3 = 3 × 2 × a4 5 3 = 6a12 2 a 9(x − 2) = 9x – 18 b x(x 9) = x2 9x c −3y(4 − 3y) = −12y 9y2 d x(y 5) = xy 5x e −x(3x 5) = −3x2 − 5x f −5x(4x 1) = −x2 − 5x g (4x 5)x = 4x2 5x h −3y(5 − 2y2) = −15y 6y3 i −2x(5x − 4) = −10x2 8x j (3x − 5)x2 = 3x3 − 5x2 k 3(x 2) (x − 7Rewrite 7 2 85 – 35 in two different ways using the associative property of addition Show that the expressions yield the same answer 7 2 85 – 35 7 2 85 (−35) The associative property does not apply to expressions involving subtraction So, rewrite the expression as addition of a negative number (7 2) 85

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Assignment 8 (MATH 215, Q1) 1 Use the divergence theorem to find RR S F ndS (a) F(x,y,z) = x3 i 2xz2 j 3y2z k;S is the surface of the solid bounded by the paraboloid z = 4 − x2 − y2 and the xyplane Solution The divergence of F is3 4x2/3 35 1 −03x2 − 6 5x 37 2 39 1/2 41 4/3 43 2/5 45 7 47 5 49 −2668 51 3/2 53 2 55 2 57 ab 59 x 9 61 x 3 √ a3 b3 63 2y √ x 65 31/2 67 x3/2 69 (xy2)1/3 71 x3/2 73 3 5 x−2 75 3 2 x−12 − 1 3 x−21 77 2 3 x − 1 2 x01 4 3 x−11 79 (x2 1)−3 − 3 4 (x2 1)−1/3 81 3 √ 22 3

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 3 By taking x 2 as common in the first two term and 3y 2 as common in the second two term x 3 −2xPor ejemplo, dado el polinomio p (x) = 5 x 3 − 4 x 2 5 x − 1, el valor del polinomio cuando x = 1, es p (1) = 5 1 3 − 4 1 2 5 1 − 1 = 5 Entonces, se dice que el valor del polinomio p en el punto 1 es 5, y se escribe, p (1) = 5 El teorema del resto es un resultado interesante que relaciona el valor numérico de unStep 2 apply second derivative test f xx=6xf yy= −6yf xy= −2 At (0;0), f xx=0,f yy=0,f xy= −2So D= f xxf yy−(f xy)2 = −4

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12 Let f(x;y) be a di erentiable function, and let u= x yand v= x−yFinda constant such that (fx) 2 (f y) 2 = ((f u) 2 (f v) 2) Solution By the chain rule (fx)2 (f y)2 =(f u f v)2 (f u−f v)2 =2((f u)2 (f v)2)Thus =2 13 Find the directional derivative D ~ufat the given point in the direction indicated by the angle (a) f(x;y)= p 5x−4y;(2;1);3x/2 5y/3 = 2 and x/3y/2=13/6 Solve using substitution methodSign In to Writing (Essays) Math Calculus Q&A Library The polynomial P (x) = 5x2 (x − 2)3 (x 9) has degree It has zeros 0, 2, and The zero 0 has multiplicity , and the zero 2 has multiplicity The polynomial P (x) = 5x2 (x − 2)3 (x 9) has degree It has zeros 0, 2, and The zero 0 has multiplicity , and the zero 2 has multiplicity

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Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Solve for x 2/3=x/6 Rewrite the equation as Multiply both sides of the equation by Simplify both sides of the equation Tap for more steps Cancel the common factor of Tap for more steps Cancel the common factor Rewrite the expression Simplify Tap for more steps924 y0 = y 2−x 925 y0 = y −x 926 y0 = y3 −x3 Solutions 923 This is the only of the differential solutions to have an equilibrium solution at y = 1 It thus corresponds to IV 924 This should have zero slope where y = x, but not where y = −x (excluding III) Moreover, the slope should change linearly with increasing x–it

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 solucionario del examen de álgebra 2 1 BIMONTHLY EXAM OF ALGEBRA Solución x2 (2m − 1)x (m 1)(m − 2) x (m 1) x (m − 2) (x m 1)(x m − 2) x m 1 Problema 3 Un teatro tiene hasta el momento 143 butacas habilitadas y cada de marzo de cada año, se adquiere un número de butacas igual al número de factores primos de p(x,y) = 12x2 2xy2 − 2y4 9x −Equations Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations 5x3y3−32x−3x22y3 so that you understand better ejercicios de division de expresiones algebraicas1) (x2 x4 −16) ÷ (x 2)3) (a3 − 3a 10) ÷ (a − 2)4) (3y4 8y3 − 4 − 7y 2y2 ) ÷ (2 y)iii Simplifique s Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising

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By taking x 2 as common in the first two term and 3 y 2 as common in the second two term x 3 − 2 x 2 y 3 x y 2 − 6 y 3 = x 2 ( x − 2 y ) 3 y 2 ( x − 2 y ) So we get,Polynomial Roots Calculator 43 Find roots (zeroes) of F (y) = 9y33y22y2 Polynomial Roots Calculator is a set of methods aimed at finding values of y for which F (y)=0 Rational Roots Test is one of the above mentioned tools It would only find Rational Roots that is numbers y which can be expressed as the quotient of two integersUse the distributive property to multiply − 6 y by 5 y − 1 Combine 8y^ {2} and 30y^ {2} to get 22y^ {2} Combine 8 y 2 and − 3 0 y 2 to get − 2 2 y 2 Combine 3y and 6y to get 3y Combine − 3 y and 6 y to get 3 y Combine 3y and 9y to get 12y Combine 3 y and 9 y to get 1 2 y

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 Transcript Ex 24, 5 Factorise (i) x3 − 2x2 − x 2 Let p (x) = x3 – 2x2 – x 2 Checking p (x) = 0 So, at x = 1, p (x) = 0 Hence, x – 1 is a factor of p (x) Now, p (x) = (x – 1) g (x) ⇒ g (x) = (𝑝 (𝑥))/ ( (𝑥 − 1)) ∴ g (x) is obtained after dividing p (x) by x – 1 So, g (x) = x2 – x – 2 So, p (x) = (x= −ˇ=6 (b) f(x;y)=xsin(xy);Solution to Quiz #4 and HW 5 1 (quiz problem, (Sec 26 Problem 3)) Determine if the following equation is exact Find the solution if it is exact (3x2 −2xy 2)dx(6y2 −x2 3)dy = 0 Solution Let M(x,y) = 3x2 − 2xy 2 and N(x,y) = 6y2 − x2 3 We have M y = −2x and N x = −2x So M y = N x and the given equation is exact

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1 3x2 7x2 2 5y2 −13y 6 Example 3 Factoring completely using group method 1 9x2 3x2 2 24x2 −50x24 3 6y3 5y2 y Title Microsoft Word 73 Factoring Trinomials Whose Leading Coefficient Is Not 1doc Author Nisa Created Date−1 x4 2x2 dx = π x5 5 2 3 x3 1 −1 = 26π 15 §54 1 Use the shell method to find the volume of the solid generated by revolving the region bounded by x = 0, x = 2, y = 0 and y = 1 x2 4 about the yaxis Answer Recall that the shell method tells us that V = Z 2 0 2πrhdx32) (5 x2 − 7x 2)( 2x2 3x 5) = Resposta 10 x4 x3 8x2 − 29 x 10 33) (6 x2 3x −8)( 2x − 3) = Resposta 12 x3 −12 x2 − 25 x 24 34) (x 1)( 2x − 1) 4x2 = Resposta 8x4 4x3 − 4x2 35) (2 x − 3y) 4xy = Resposta 8x2 y −12 xy 2 36) (3 x2 − 4x 5)( x2 − 6x 4) = Resposta 3x4 − 22 x3 41 x2 − 46 x

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Cov(Y1,X1) = 2, Cov(Y2,X1) = 0, Cov(Y3,X1) = −1, Cov(Y 1 ,X 2 ) = 2, Cov(Y 2 ,X 2 ) = 1, Cov(Y 3 ,X 2 ) = 1 (a) Expected value of U 1 = −5Y 1 5Y 2 6Y 3 isBeginning Algebra (1st Edition) Edit edition Problem E from Chapter 71 Factor completely5x3 − 30x2y2 xy − 6y3 Get solutions2 3 x −1 3 This is positive on x > 0, negative on x < 0, and undefined at zero itself The second derivative is −2 9 x −4 3, which is negative everywhere (except at 0, where it too is undefined) Thus the curve is concave down everywhere Such a curve looks something like the plot of p x, (b) Solving for y, we have y = x23 Then y0

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2 Let F(x,y,z)=(−5xz2, 6xyz, −6xy3z) be a vector field and f(x,y,z)=x3y2z Find ∇f f x,f y,f z)=3x 2y2z, 2x3yz, x3y2) Find ∇×F ijk ∂y ∂x ∂y ∂y ∂y ∂z −5xz26xyz−6xy3z =⎡⎣(−6x)(3y2)(z)−6xy⎤⎦i⎡6y3z−10xz⎤jyz−0k Find F×∇f ijk −5xz26xyz−6xy3z 3x2y2z2x3yzx3y2 =(6x4y3z12x4y4z2)i(5x4y2z2−18x3y5z2)j(−10x4yz3−18x3y3z2)k

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