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I have a few problems which are really confusing me, if someone can do one of these out for me, i'd appriciate it.
Code:y^2-9 8
------- X -----
4 y+3
s^2 + 2s - 8 s+5
---------------- X -----------
s^2 +3s -10 s^2-16
5m 25m^2
------- / ------------
m+1 m^2 + 2m +1
thanks,
Dimava
The first one
-1- (y^2-9)/4 * 8/(y+3)
-2- ((y^2-9)*8)/(4*(y+3))
-3- ((y+3)*(y-3)*8)/(4*(y+3)) remember the 3. Binom?
-4- ((y+3)*(y-3)*2)/(1*(y+3))
-5- ((y+3)*(y-3)*2)/(y+3)
-6- (y-3)*2)
-7- 2y-6
The third comes to
1/5* (1+1/m^2)
The second comes to:
Code:1
---
s-4
I gotQuote:
Originally posted by opus
The third comes to
1/5* (1+1/m^2)
For the third.Code:m + 1
-----
5m
Your solution has the m's squared. Otherwise we got the same solution.
a few more problems...
please bear with me i know im posting more than a couple, but i've done a few hundred problems on my own already...
Code:6
x - ------
x+2
_______________
8
x + -------
x+5
Code:What are the excluded values of y in the expression
y^2 + y - 2 y^2-5y+5
------------------ * -------------------
y^2 - 2y - 15 y^2 + 5y +6
Code:Use long division to find the quotient
(t^2 +4t -6) / (t+5)
Code:An airplane can fly at a rate of 525 mi/h in still air. It can fly 2800 miles with the wind in the same time that it can fly 2450 miles against the wind. Find the rate of the wind.
If somebody can help me out with a few of these i'd really appricate it. And if possible I would like to see how you got the answer so I can learn from it.Code:I attached this problem because it uses subscripts
thanks,
Dimava
OK, I see my error, thanksQuote:
Originally posted by DiGiTaIErRoR
I got
For the third.Code:m + 1
-----
5m
Your solution has the m's squared. Otherwise we got the same solution.
For sure I have to do the airplane question:
speed =525 mph
Distance against wind 2450 m
distance with wind 2800 m
2800m/(525mph+x mph)=2450m/(525mph-x mph)
solve that and you get a wind of 35 mph
on the second
excluded are all values that will leed to a division thru 0.
first one
y^2 -2y-15
same as (y-5)*(y+5)
which would be 0 if y=5 or y=3
second one
y^2+5y+6 will only by = if y=-2
So excluded are -2 , 3 and 5
thanks, wouldn't it be 5, -2, -3 though?
You're correct, sorry
it must read:
first one
y^2 -2y-15
same as (y-5)*(y+3 )
which would be 0 if y=5 or y=-3
on the last one
1/Rt=1/R1 + (1/(R2 +R3)) | -(1/R1)
1/Rt -1/R1=1/(R2 + R3) | * (R2 + R3)
(1/Rt -1/R1)*(R2 + R3)=1 |/(1/Rt -1/R1)
(R2 + R3)=1/(1/Rt -1/R1) | -R3
R2= 1/(1/Rt -1/R1) -R3
R2= 1/(1/2 - 1/5) - 3
R2=1/3
ok sounds good thanks
now if i could get the 1st and 3rd, it would be perfect
The first puh, er, ?????
The third, what is a Long division?
on the third I only get
(t^2+4t-6)/(t+5)
(t+5)*(t-1)-5/(t+5)
[(t+5)*(t-1)/(t+5)] -5/(t+5)
(t-1) - 5/(t+5)