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What comes next in this sequence 10 30 15 16 48 24 25?
75x3, /2, +1, x3...10 x3=30 /2=15 +1=16 x3=48 /2=24 +1=25 x3=75
X cubed plus x equals -2?
x3 + x = -2 => x3 + x + 2 = 0 => x3 + x2 - x2 - x + 2x - 2 = 0 => x2(x+1) - x(x+1) + 2(x+1) = 0 => (x+1)*(x2-x+2) = 0 Setting the first bracket equal to zero gives the only real solution, which is x = -1 The second bracket gives the complex roots, x = ½*[1 +or- i*sqrt(7)]
What is the solution set for -x3-2x2-x-2?
There can be no solution set because there is no equation or inequality, only an expression.
What is x cubed times 1 half times the squared root of x cubed?
x3*(1/2)*sqrt(x3) = 1/2*x4.5 or 1/2*sqrt(x9)
How do you factorise x cubed take one?
If it's (x3 -1) that you want to factorize, then find the solutions to (x3 -1) = 0.So if P(x) is a polynomial of x, and x=a is a solution for P(x) = 0, then (x - a) is a factor of P(x).So x = 1 solves (x3 -1) = 0, so (x - 1) is a factor. Use long division (x3 -1)/(x-1) = x2 + x + 1. Use the quadratic formula to find the roots of this: -1/2 ± i*sqrt(3)/2, which is complex. So the factorization is:(x3 -1) = (x - 1)( x2 + x + 1)Multiply the polynomials together to check that your answer is correct.
What is the solution to x cube -4 equals 2?
X3 - 4 = 2 X3 = 6 X = 61/3 X = about 1.817
What comes next in this sequence 10 30 15 16 48 24 25?
75x3, /2, +1, x3...10 x3=30 /2=15 +1=16 x3=48 /2=24 +1=25 x3=75
X cubed plus x equals -2?
x3 + x = -2 => x3 + x + 2 = 0 => x3 + x2 - x2 - x + 2x - 2 = 0 => x2(x+1) - x(x+1) + 2(x+1) = 0 => (x+1)*(x2-x+2) = 0 Setting the first bracket equal to zero gives the only real solution, which is x = -1 The second bracket gives the complex roots, x = ½*[1 +or- i*sqrt(7)]
What is the solution set for -x3-2x2-x-2?
There can be no solution set because there is no equation or inequality, only an expression.
What is x cubed times 1 half times the squared root of x cubed?
x3*(1/2)*sqrt(x3) = 1/2*x4.5 or 1/2*sqrt(x9)
Factor the polynomial x3-2x2 plus x-2?
x3 - 2x2 + x - 2 =(x - 2)(x2 + 1)
What is 1 plus 2 X3?
7
How do you factorise x cubed take one?
If it's (x3 -1) that you want to factorize, then find the solutions to (x3 -1) = 0.So if P(x) is a polynomial of x, and x=a is a solution for P(x) = 0, then (x - a) is a factor of P(x).So x = 1 solves (x3 -1) = 0, so (x - 1) is a factor. Use long division (x3 -1)/(x-1) = x2 + x + 1. Use the quadratic formula to find the roots of this: -1/2 ± i*sqrt(3)/2, which is complex. So the factorization is:(x3 -1) = (x - 1)( x2 + x + 1)Multiply the polynomials together to check that your answer is correct.
Can you multiply 3 consecutive numbers and get 2015?
This is the same as asking is there an integer solution to x(x+1)(x+2)=2015.x3 -3x2 +2x=2015So we must solve x3 -3x2 +2x-2015=0 for integer solutions.The solution is x=13.657 which is not an integer. That is the only real solution.So the answer is no.
How do you factor x3 plus 1?
x3 + 1 = x3 + x2 - x2 - x + x + 1 = x2(x + 1) - x(x + 1) +1(x + 1) = (x + 1)(x2 - x + 1)
What number is next in the following sequence 10 30 15 16 48 24?
10 30 15 16 48 24 25 x3 1/2 +1 x3 1/2 +1
What is the inverse of x3 plus 1?
The inverse of a number is 1 divided by that number. So the inverse of x3 + 1 is 1/(x3 + 1).
