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  • What is the antiderivative of # 1 (x^5)#? - Socratic
    "antiderivative "-> -1 (4x^4)+C Apply the revers of differentiation '
  • How do you solve (x - 5) (x - 8) = (x + 1) (x - Socratic
    x=11 Multiply both sides by (x-5)(x-8) This will allow you to cancel common factors and get rid of the denominators
  • What does (1+2 x-15 x^2) (1+4 x-5 x^2) simplify to? - Socratic
    #(1+2 x-15 x^2) (1+4 x-5 x^2)# It is easier to tackle if it is written like this instead: #color(red)((1 1+2 x-15 x^2)) div color(blue)((1 1+4 x-5 x^2))#
  • How do you evaluate #cos^-1(cos((pi) 10))#? - Socratic
    The answer is pi 10 By definition, cos^(-1)(x) is an angle phi that satisfies two requirements: (1) cos phi = x (the main part of a definition) (2) 0<=phi<=pi (to assure uniqueness of the result) In our case x=cos(pi 10), so the first requirement is cos(phi)=cos(pi 10) and the obvious candidate for phi is pi 10
  • How do you solve abs(5 - 1 x) lt; 2? - Socratic
    First interpret the modulus to get: -2 < 5 - 1 x < 2 Then apply arithmetic operations to arrive at: 1 7 < x < 1 3 abs(5-1 x) < 2 is equivalent to -2 < 5-1 x < 2 Next subtract 5 from all parts of this inequality to get: -7 < -1 x < -3 Next multiply all parts by -1 and reverse the inequality signs to get: 7 > 1 x > 3 Since this forces 1 x to be positive, that means x > 0 too Multiply all parts
  • Question #03617 - Socratic
    How do I use matrices to find the solution of the system of equations #y=−2x−4# and #y+4=−2x
  • How do you write a polynomial in standard form given zeros 1,5,3-i?
    x^4-12x^3+51x^2-70x+50=0 Complex roots occur in conjugate pairs So, as 3-i is a root, 3+i is also a root The roots are 1, 5, 3-i and 3+i The biquadratic having these as roots is in the form x^4-S_1 x^3 + S_2 x^2-S_3x+S_4=0, where S_1=sum(roots) = 12, S_2=sum(products of the roots, taken two at a time) = 51, S_3=sum(product of the roots, taken three at a time) = 70 and S_4=product of all
  • How do you solve the following system?: -x -3y =5, -3x +4y = -7 - Socratic
    How do you solve #12y-3x=-1# and #x-4y=1# using the substitution method? Which method do you use to solve the system of equations #y=1 4x-14# and #y=19 8x+7#? What are the 2 numbers if the sum is 70 and they differ by 11?
  • How to find solution set in RR^3 of x-pi=5? - Socratic
    1) (x,y,z)=(57 7, 0, 0) 2) (x, y, z) = ((-3-sqrt2) 4, (sqrt2-5) 4, (5+3sqrt2) 4) 3) (x, y, z) = (7,2,0) 1) x-pi=5 x=5+22 7 = 57 7 Therefore, (x,y,z)=(57 7, 0, 0) 2) Solving using substitution method: 2x-y+z=1 (i) x-2y+z=3 (ii) y=sqrt2-z (iii) Solving (i) with substitution method we get: x=1 2 + sqrt2 2 - z Putting value of x and y into (ii) we get: z= (5+3sqrt2) 4 Putting value of z into (ii





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