Q. solve fo r R in the equation
(R/R+3)+(4/R-2)=(20/R^2+R-6)
show work please im doing this for extra credit and i am clueless on what to do... thanks : )
A. By factoring, we have:
r/(r + 3) + 4/(r - 2) = 20/((r - 2)(r + 3))
Multiply both sides by (r + 3)(r - 2) to get:
[r/(r + 3) + 4/(r - 2) = 20/((r - 2)(r + 3))](r + 3)(r - 2)
r(r - 2) + 4(r + 3) = 20
r² - 2r + 4r + 12 = 20
r² + 2r + 12 = 30
Then,
r² + 2r + 12 - 30 = 0
r² + 2r - 18 = 0
By ar² + br + c, let:
a = 1
b = 2
c = -18
Next, by quadratic formula:
r = (-b ± â(b² - 4ac))/2a
==> (-2 ± â((2)² - 4*1*-18))/2
==> (-2 ± â(4 + 72))/2
==> (-2 ± â76)/2
==> (-2 ± â(4*19))/2
==> (-2 ± 2â19)/2
==> 2(-1 ± â19)/2
==> -1 ± â19
I hope this helps!
Isolate r?! pls help with math?
Q. solve equation for r >< if possible pls solve all
s=7r-33
s=16-40
s=5(3/r-5)
s= 3(r-2)/5(r+2)
thanx so much! yea its suppose to be 16r-40 x] sorry
A. okay, here goes:
s=7r-33
r=(s+33)/7
s=16-40
there is no r, so you cannot solve for it, if it is supposed to say s=16r-40
16r=-40-s
r=(-40-s)/16
s=5(3/r-5)
s=15/r-5
sr-5s=15
r=(15+5s)/s
s= 3(r-2)/5(r+2)
s=3r-6/(5r+2)
5rs+2s=3r-6
5rs-3r=-6-2s
r(5s-3)=-6-2s
r=(-6-2s)/(5s-3)
Find the limit as r approaches 0?
Q. ((a^r + b^r)/2)^(1/r)
A. lim ((a^r + b^r)/2)^(1/r) (as r --> 0)
LN lim = lim LN
lim LN ((a^r + b^r)/2)^(1/r) =
= lim (1/r)LN ((a^r + b^r)/2) (as r --> 0) is a indetermination form 0/0
L'Hopital rule
lim ((a^r LN(a) + b^r LN(b))/(a^r + b^r)) (as r --> 0)
= LN(a)/2 + LN(b)/2 = LN â(ab)
as
LN lim ((a^r + b^r)/2)^(1/r) = LN â(ab)
lim ((a^r + b^r)/2)^(1/r) = â(ab)
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Title : how do you solve for R?
Description : Q. solve fo r R in the equation (R/R+3)+(4/R-2)=(20/R^2+R-6) show work please im doing this for extra credit and i am clueless on what to ...