Help I have a Placement test and I am so lost!!!!!!!!!

P

Princezz3286

Junior Member
Joined
Nov 12, 2005
Messages
66
can somebody please show me how this is done????????

Problem 1:
(x+y)^2 + (x- 3y)^2

a) 2x^2 + 8xy 10y^2
b) 2x^2 + 10y^2
c) 2x^2 + 7y
d) 2x^2 -8y^2
e) 2x^2 - 4xy + 10y^2

Problem 2:
If f(x)= x-5/2 and g(x) = x^2-2x then g(f(-1))= ????

a)1
b)3
c)15
d)9
e)-3

Problem 3:
8x/3x-6 * x^2-4/2x+4 = ????

I need how to set up and solve these operations...... If you want to show me the set up I can do it and send you back the answer........ so I can practice!!! :p
Thank You Very Much!!!!!!!!
 
Princezz3286 said:
can somebody please show me how this is done????????

Problem 1:
(x+y)^2 + (x- 3y)^2

Expand and simplify.

The first part:
(x + y)^2 = x^2 + 2xy + y^2, right?

a) 2x^2 + 8xy 10y^2
b) 2x^2 + 10y^2
c) 2x^2 + 7y
d) 2x^2 -8y^2
e) 2x^2 - 4xy + 10y^2

Problem 2:
If f(x)= x-5/2 and g(x) = x^2-2x then g(f(-1))= ????

Can you tell me what f(-1) = ?

a)1
b)3
c)15
d)9
e)-3
Click to expand...
 
Your problem is you don't understand function notation.
If f(x) = x-5/2 then
f(y) = y-5/2
You replace the x with whatever is in the F(?) ()s. In this case it is f(-1). When you do the replacement you get
(-1)-5/2
 
right, so we have f(x)=x-5/2 and g(x)=x^2-2x then g(f(-1))

f(x)= -1-5/2 which is -6/2 so further reduced=-3

then that being said, g(x)= x^2-2x, -3^2=9 + 6 = 15 right?
 
Hmmm. -1 -5/2 = -2/2 - 5/2 = -7/2
The rest is much better with that change. Maybe you do understand. :evil:
 
This one I have under control now............ it is the other one that i dont understand, 8x/3x-6 * x^2-4/2x+4
 
8x/3x-6 * x^2-4/2x+4 =
8x/(3(x-2))*(x-2)(x+2)/(2(x+2)) =
8x/(3(x-2))*(x-2)/2 =
8x/6
 
my problem is i don't know weather to cross multiply? multiply just the top and just the bottom or what?
 
You would cross multiply only across = signs. Yhis is justa multiplication prpblem just like
(1/4)*(4/5) = 4/20 = 1/5
What we did was multiply top*top/(bottom*bottom) with top and bottom cancelation first.
(1/4)*(4/5) = (1/1)(1/5) = 1/5
 
ok, i am good on that now, but how would i simplify the numbers?
so now i am working with what you were saying before. 8x/3(x-2) * (x-2)(x+2)/(2(x+2)) how do i make the (x+ or - 2) go away?
 
8x/3x-6 * x^2-4/2x+4 =
First we factor
8x/(3(x-2))*(x-2)(x+2)/(2(x+2)) =
The (x+2) in the numerator cancels the (x+2) in the denominator.
8x/(3(x-2))*(x-2)/2 =
The (x-2) in the right numerator cancels the (x-2) in the left denominator.
8x/(2*3) = 8x/6 = 4x/3
 
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