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How do you simplify x2-9 over x2-6x plus 9?
First you distribute: x2-9 = (x-3)(x+3) and that is you numerator. Distribute x2-6x+9. Find two #'s that mult. to make 9 and add to make 6. 3x3=9 and 3+3=6. now you replace the 6x with -3x and -3x so rewrite the problem as x2-3x-3x+9 then you distribute x(x-3)-3(x-3) so x2-6x+9 becomes (x-3)(x-3) and that is your denominator. Tharefore: (x-3)(x+3) over (x-3)(x-3) then cross out like terms and your answer is (x+3) over (x-3)
What is another way to write x2 over y3?
x^2/y^3 = x^2*y^(-3)
What is 4 over 5 plus 2 over 3?
The answer is x2/36+(82+y3n) over 17 squared
This circle is centered at the origin and the length of its radius is 3 over 2 What is the equation of the circle?
Since the circle is centered at the origin, the equation of the circle is x2 + y2 = r2. So we have: x2 + y2 = (3/2)2 x2 + y2 = 9/4
What is the answer to 9-x2?
7
x2+x-6 over x2-x-2 (fraction) multiply by x2+11x+24 over x2-x-2 (fraction) divide by x+3 over x2+3x+2.multiply and divide as indicated. please help?
5
Integrate 2exp-x2 from 3 to 7?
See http://www.wolframalpha.com/input/?i=integrate+2exp(-x2)+over+3+to+7
What is the simplest form of 3x over x3?
3/x2
What is equal to the rational expression when x 5 or negative 3 x2 plus x-6 over x2-2x-15?
When x = -3
How do you simplify x2-9 over x2-6x plus 9?
First you distribute: x2-9 = (x-3)(x+3) and that is you numerator. Distribute x2-6x+9. Find two #'s that mult. to make 9 and add to make 6. 3x3=9 and 3+3=6. now you replace the 6x with -3x and -3x so rewrite the problem as x2-3x-3x+9 then you distribute x(x-3)-3(x-3) so x2-6x+9 becomes (x-3)(x-3) and that is your denominator. Tharefore: (x-3)(x+3) over (x-3)(x-3) then cross out like terms and your answer is (x+3) over (x-3)
What is 2 over x plus 1 over x squared?
2/x + 1/x2 = (2x+1)/x2
What is another way to write x2 over y3?
x^2/y^3 = x^2*y^(-3)
What is 3x-3 multiplied by x over 6 multiplied by x2-x?
(3x - 3)(x/6)(x2 - x)= (3)(x - 1)(x/6)(x)(x - 1) = (1/2)(x2)(x -1)2 = (1/2)(x2)(x2 - 2x + 1) = (1/2)x4 - x3 + (1/2)x2
What is the equivalent fraction for 3 over 3?
1 Actually, once the numerator and denominator are both multiplied by the same number, our options are limitless as to what the equivalent fraction for 3 over 3 can be. 3 x2 = 6 3 x5 = 15 - - or - --- etc. 3 x2 = 6 3 x5 = 15
X6 plus 3x4-x2-3 equals 0?
x6 + 3x4 - x2 - 3 = 0(x6 + 3x4) - (x2 + 3) = 0x4(x2 + 3) - (x2 + 3) = 0(x2 + 3)(x4 - 1) = 0(x2 + 3)[(x2)2 - 12] = 0(x2 + 3)(x2 + 1)(x2 - 1) = 0(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0x2 + 3 = 0x2 = -3x = ±√-3 = ±i√3 ≈ ±1.7ix2 + 1 = 0x2 = -1x = ±√-1 = ±i√1 ≈ ±ix + 1 = 0x = -1x - 1 = 0x = 1The solutions are x = ±1, ±i, ±1.7i.
What is a grouping factor for polynomial x3 ax 3a 3x2?
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
What do you do when you are dividing exponents and there are 3 number as a numerator and only 1 as the denominator?
Subtract them. example: x3 --- x1 Subtract the 1 from the 3 and you get x2 over 1 or just x2
