Will Fox Pick Up Roseanne now That ABC Has Cancelled It?

I think the discussion will happen after some polling and a temp gauge on Barr's apology and the cloud of scandal is lifted.But I doubt it will make this upcoming season. If anything, it might be next. But if there is a lack of good programming choices, I am certain FOX would love to have the show in their back pocket.

1. Prove that $a^ab^bc^cge (abc)^(abc)/3$ [duplicate]

Use the fact that $log $ is monotone increasing function:$$(a-b)(log(a)-log(b))ge0 implies left(fracab

ight)^a-b ge 1$$$$(c-a)(log(c)-log(a))ge0 implies left(fracbc

ight)^b-c ge 1$$$$(b-c)(log(b)-log(c))ge0 implies left(fracca

ight)^c-a ge 1$$Multiply the three inequality yields your desired form !

2. why is abc family called abc family?

Its called ABC Family that means its for the family not necessarily that every show on the channel has to be good for little kids to watch. All of the shows you mentioned are good shows for teenagers, and I think most of the shows on it have some good messages there just not super obvious

3. How much longer is NASCAR going to stick with ABC?

Not sure. I have no prob with ABC. Great audio, video and commenting. FOX once did not show the start of a great race this year and it was not the first time. I get over it after the 3 seconds it takes to change the channel on the remote.

4. What channel, date, and time does Greek air on ABC?

I am not sure of the exact timeslot, but I believe it airs around 8:00p.m on ABC Family, not the regular ABC channel

5. Is the ABC diet safer of you're overweight?

If you go on this ABC diet...you are going to mess up your metabolism--which is what essentially controls your weight loss. You are going to slow your metabolism down, and then from that point on, it is going to be even harder for you to lose weight..you may even gain it. Do not be an idiot, and do it the right way.

6. Help! How to prove this? abc are positive numbers..?

divide both sides by abc 1/c1/a1/b > 1 This is because a b c are all positive, and less than one. 1/anything less than one > 1 Add three of those up and you get something more than one.

7. What are your ABC Diet results?

i had anorexia for 3 years, (11-14) i've done part of the ABC diet but never have finished it. i lost weight. i've droped to 83lbs lowest. so you will lose weight pretty fast, but if you binge then you will gain it back faster. (oh and i was 107lbs before anorexia so that was like a 24lbs loss?) but just adding i do not condone it and i've passed out before.

8. What channel is ABC on Direct TV?

Ok so for Denver and direct tv it's actually 311

9. Prove that the area of $triangle DEF$ is twice the area of $triangle ABC$

It is very interesting to solve this problem by only using our greek friends Pythagoras and Menelaus, together with some simple algebra.Without loss of generality assume that $P$ lies on the half-plane determined by $BC$ and not containing $A$. Note that there is no need to use the circle, once we fix $angle BPC = 120$. In the figure above I added point $G$ given by the intersection of line $FD$ with $AC$. Suppose also $overlineAB=1$.Once $overlineBF = x$ is chosen, the entire Figure is defined. We aim therefore at showing that, independently of $x$, $$[DEF] = 2[ABC].$$Below the triangle in question has been isolated, to better help you in the demontrations. Recall that $overlineBC = 1$ and $overlineBF = x$. Use Pythagorean Theorem on $triangle CKF$ in oder to determine $$overlineCF = sqrtx^2x1.$$ From the similarity $$triangle BPC sim triangle BCF,$$ show that $$overlineCFcdotoverlineCP = 1,$$ so that, in the end, you have $$ begincases overlineCF = sqrtx^2x1, overlineCP = frac1sqrtx^2x1, overlineFP = fracx(x1)sqrtx^2x1. endcases $$MT on $triangle ACF$ with the line $BE$, gives $$fracoverlineCEoverlineAE = frac1x1.$$ Together with $overlineAE-overlineCE = 1$ this leads to $$overlineCE = frac1x$$ and $$overlineAE = fracx1x.$$Similarly, MT on $triangle BFC$ and line $AP$, with the known fact that $overlineBD overlineCD = 1$ yields $$overlineCD = frac1x1$$ and $$overlineDB = fracxx1.$$ MT on $triangle ABC$ and line $FG$, with $overlineAG overlineCG = 1$, gives you $$ overlineAG = fracx1x2$$ and $$ overlineCG = frac1x2.$$ Finally, MT on $triangle CGF$ with line $AP$ yields $$ fracoverlineGDoverlineDF = frac1x(x2).$$ Now we only need to compute areas of triangles with fixed altitude and a given ratio between bases.Firstly we have $$[ACF] = [ABC](1x).$$Then $$[AFE] = [ACF] fracoverlineAEoverlineAC,$$ that is $$[AFE] = [ABC]frac(x1)^2x.$$We also have $$[GFE] = [AFE]fracoverlineGEoverlineAE,$$ yielding $$ [GFE] =2[ABC]frac(x1)^2x(x2).$$Finally observe that $$frac[DFE][GDE] = fracoverlineGDoverlineDF = frac1x(x2).$$Thus we have the system of equations $$ begincases frac[DFE][GDE] = frac1x(x2) [DFE] [GDE] =2[ABC]frac(x1)^2x(x2). endcases $$leading, once solved, to the desired result, i.e. $$boxed[DFE] = 2[ABC] $$$blacksquare$