slashragnarok

Yes the answer indeed should be 1D and 2D. 1. A drunken person relies on a lamp post for support. Assistance could have been a similar word but the first blank seals it. Notice that Housman assails or attacks his rival. Invective in fact means highly critical or abusive language. 2. Woolf actually ridicules or mocks traditional concepts of truth by saying that truth is not something that one can touch and store on a shelf.

I think Cuento is right.

The radius of the cylinder is obviously 3*sqrt(2) and since the height of the cylinder is 1/sqrt(2) times the length of radius, the height is 3 units. Now the square that can be inscribed in the cross sectional area of the cylinder has a diagonal length of 2 times this radius i.e 6*sqrt(2). So the length of a side of this square is 6. And so the square can accommodate 6x6 i.e 36 squares of dimensions 1 unit by 1 unit. So each layer of cubes has a total of 36 cubes. And there are 3 such layers of cubes since the height of the cylinder is 3 units for a total of 108 cubes. And again there will be some space left around the edges of this square and on each of the four sides of the square we can further fit 6 squares, 2 in each layer upto the height of 3 units. This is because after packing the cylinder in the fashion stated above, we still have space left around the edges of the square. So answer is 108+24=132. Answer B. Sorry for the bad explanation.

1. except 42 the rest can be written as products of 3's and 2's. 42 has a 7 as factor which cannot make it a factor of 2^n.3^k. 2^n.3^k doesn't have a 7 in it. Ever. 2. We can write 72.42 as 72+0.42 i.e. (3*24)+(42/100) = 3*(24+(14/100)). so we can clearly see that k and n are 3 and 14 respectively. 3. Since the binding just fits around the perimeter of the rug, we can say that the perimeter of the rug is 30. So 2(l+b)=30 or l+b=15 and area is 54 so l*b=54 i.e. l=54/b. So from the previous equation 54/b+b=15 or b^2-15b+54=0. Solving this quadratic equation we get 9 and 6 as solutions, which are the length and breadth respectively. So 10 is definitely greater than the longer side. Best of luck.

Look at this page here GRE General Test: Getting Your Scores If you scroll down a bit, you'll see that the scores are sent to you after 10-15 days.

Call me stupid but I still don't get my head around how technological advances can make installation of some feature obligatory. Maybe technology makes it easy, affordable, less cumbersome, whatever. But obligatory? How?

For me, it's plain and simple repetition. Drill the words into your head. And every time you read something and come across a word that less than familiar, search it up on a dictionary.

I think it's A. If the product of any number of integers is to be odd, then all the integers forming the product must necessarily be odd as only one even integer among the integers is required to make the product even. So here there have to be 10 odd integers. And so number of negative integers have to be less than 10 as 10 negative integers make the product positive.

I think it should be 2 because querulous means having a complaining attitude. It's not surprising if person likely to complain, greets you cordially. But a insouciant person is indifferent, so any warmth of behavior on her part is surprising.

In addition to the grammatical corrections made, I would like to say that this analysis of the argument is pretty good. Nice examples, good structure. Overall a great argument essay. Few suggestions I would like to make are that I have seen people with 3 or more paragraphs take home the highest grades and the conclusion needs to sum up your claims and it needs to be a bit more elaborate. The essay is too short in my opinion. I would award it 4.5. And sorry for the late post.

For consecutive number sum I think the answer can be made to tally if you don't consider 2x+2, 2x+4 etc but rather x, x+2, x+4, etc. then the sum comes to be 5x+20. So t=5x+20, or x=(t-20)/5 and the greatest integer i.e x+8 can be written as (t-20)/5+8=(t-20+40)/5=(t+20)/5. Answer then is E I think.

The sum of the angles in a triangle is always constant and the value is a constant for any triangle. So if the angle subtended at the center increases the other two angles must decrease to compensate for the increase. The biggest side does correspond to the biggest angle but that biggest angle is opposite the the biggest side. So in the problem x is part of the biggest side, not opposite to it.

Judy Adam problem: You wrote: "Now 2x x 6 years 2x-6 x-6 ago 5(2x-6) = x-6 10x-30 = x-6 9x=24 x = 24/9 2(24/9) =48/9 ....so what's the problem?" The problem is in bold. Judy was 6 times as old as Adam. So it will be 2x-6=5(x-6) Let the 5 consecutive even integers be 2x-4, 2x-2, 2x, 2x+2 and 2x+4. So by the problem statement, t=10x. Therefore 2x=t/10 and so 2x+4=(t/10)+4=(t+40)/10. None of the answers match I think. Because let the numbers be 2,4,6,8,10. The sum is 30 which is denoted by t. Now in terms of t, the greatest integer 10 would be t/3. So the expression varies from case to case. No general term exists in my opinion. 3/5 th take Spanish. Remaining is 2/5th. 1/4th of 2/5th German. All others take French. So 3/4th of 2/5th take French. i.e. 3/10th. So the answer is 30%. Let the sales in 1994 be x. In 1995 there was a decrease by 80%. So in 1995 the sales was 0.2x. In 1996 again, the sales went up to the sales in 1994 i.e. x. Therefore increase in sales in 1995-1996= x-0.2x=0.8x. Expressed as a percentage of sales in 1995 this is (0.8x/0.2x)*100=400%. Page 307, example 18: Let the side of the big square be a. Area=a^2. Then length of AC=sqrt2*a. So length of EG=0.5*sqrt2*a=sqrt2*a/2 Now EG^2=EH^2+HG^2. Pythagoras theorem. And EH=HG. So 2(EH^2)=(a^2)/2 or EH^2=a^2/4, EH=a/2. So length of side of smaller square is a/2. Area=(a^2)/4. So shaded region's area is (a^2)-(a^2)/4=3(a^2)/4. So answer is 3:4.

someone authoritative demands acceptance, something conspicuous demands attention I think.