1.) Find the domain of the function:
h(y) = (y^3-8) (y+2)^-3

2.) Simplify:
m^2 = m

3.) Simplify:
7x^2/9y [4x]/[15y^2]∙[35/6xy]^-1

4.) Simplify:

5.) Simplify:

6.) Simplify:

7.) Simplify:

1.) When h(y) = (y - 8)/(y+2) is h(y) = (y-2)(y+2y+4)/(y+2).
The domain is the variable possibilities, and that is all y but -2.

2.) To simplify m = m, divide by m { noting that m can't be 0 } and get m = 1.

3.) It is not clear what [35/6xy]^-1 would be.
If it is really [35/(6xy)]^-1, then it is 6xy/35.
As written, in math, division and multiplication are both done from left to right,
so this would be 35/6 times xy, and to the -1 would gives 6/(35xy).

Once this has been determined, in proper mathematics, it would end up being times the preceding fraction, so it would be times (4x)/(15y), but perhaps it is in the denominator.

4.) Convert all fractions to 20ths, so multiply the 1st fraction by 2/2.
This gives 14/20 + 11/20 - 9/20 = 16/20.  That reduces to 4/5.

5.) Convert all fractions to 15ths.  To do this, multiply the 1st by 5/5 and the 2nd by 3/3.
This gives 10/15 - 9/15 + 4/15 =5/15 = 1/3.

6.) Taking x - 2x gives -x.
Taking 3/5 + 1/10 ... multiply 3/5 by 2/2, so we have 6/10 + 1/10 = 7/10.
This makes the answer be -x + 7/10.

7.) This looks like it should really be 1/x - (1/y)/(y/x) - x/y.
Multiplying the 2nd terms by (y/x)/(y/x) gives (y/(xy))/((xy)/(xy)).
The fraction on the bottom disappears, giving y/(xy).
Since y/y is 1, this leaves 1/x for the middle term.

Rewriting what's given gives 1/x - 1/x - x/y, and this is just -x/y.


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Scott A Wilson


Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.


I have solved story problems, linear equations, parabolic equations. I have also solved some 3rd order equations and equations with multiple variables.

Documents at Boeing in assistance on the manufacturiing floor.

MS at math OSU in mathematics at OSU, 1986. BS at OSU in mathematical sciences (math, statistics, computer science), 1984.

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Both my BS and MS degrees were given with honors.

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Students in a wide variety of areas since the 80's; over 1,000 of them have been in algebra.

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