1/3 cubed = ???? what
2 answers:
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Answer:
2 solutions
Step-by-step explanation:
I like to use a graphing calculator to find solutions for equations like these. The two solutions are ...
- x = 1
- x = 4
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To solve this algebraically, it is convenient to subtract 2x-7 from both sides of the equation:
3x(x -4) +5 -x -(2x -7) = 0
3x^2 -12x +5 -x -2x +7 = 0 . . . . . eliminate parentheses
3x^2 -15x +12 = 0 . . . . . . . . . . . . collect terms
3(x -1)(x -4) = 0 . . . . . . . . . . . . . . . factor
The values of x that make these factors zero are x=1 and x=4. These are the solutions to the equation. There are two solutions.
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<em>Alternate method</em>
Once you get to the quadratic form, you can find the number of solutions without actually finding the solutions. The discriminant is ...
d = b^2 -4ac . . . . where a, b, c are the coefficients in the form ax^2+bx+c
d = (-15)^2 -4(3)(12) = 225 -144 = 81
This positive value means the equation has 2 real solutions.
the first step is to deal with the negative exponents, which causes numbers to go to the other side of the fraction
8x^{-9}+(-3x^{-3}y^{-2})(\frac{5}{x^6y^{-2}})
goes to
now we can simplify the multiplied terms
Now that we have common bases, we can add the 2 fractions together to get