To adeebnafees: The diameter of a car's back wheels are one and a half times the diameters of the front wheels. When the car travels one metre, the back wheels go around 6 times. How many times do the front wheels go around when the car travels one meter?
AdeebNafees: Your physical chains are nothing to the physical might of the bed.
AdeebNafees: Cya!
AdeebNafees: So circumference of back wheel is 1/6  (reply)
AdeebNafees: If the diameter is one and a half times as great, so is the circumference  (reply)
AdeebNafees: So the circumference of the front wheel is 1.5/6  (reply)
AdeebNafees: Oh no, wait, the front sheels are 1.5 times smaller  (reply)
AdeebNafees: Some 0.111 meters I think  (reply)
AdeebNafees: So you divide 1 by 1.111  (reply)
AdeebNafees: Some 9 times I think  (reply)
To AdeebNafees: yay
To adeebnafees: I cut my finger :(
AdeebNafees: *puts bandage on*  (reply)
To AdeebNafees: d'aww
To adeebnafees: Let R be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of R ?
AdeebNafees: Ooh, interesting.  (reply)
AdeebNafees: The remaining three integers can occupy either the space below 2, [5, 7, 8, 10, 11, 12, 13] or numbers above 14.  (reply)
AdeebNafees: If all of them are below 2, 3 will be the new median value.  (reply)
AdeebNafees: If all of them are above 14, 9 will be the median value.  (reply)
AdeebNafees: 2 and 14 are thus never the correct median value.  (reply)
AdeebNafees: This implies that the range of possible values will be somewhere from 3 to 9.  (reply)
AdeebNafees: There are only 7 possible values: 3, 4, 5, 6, 7, 8, 9  (reply)
AdeebNafees: However, all are not possible.  (reply)
AdeebNafees: 3 and 9 have been demonstrated to be possible. 4 is not possible as it cannot be the central value for any distribution of integers.  (reply)
AdeebNafees: 5 is possible as we get 7 values at that time and can place two other values behond 2 and 14. 7, again, is possible since there are 9 values in all.  (reply)
AdeebNafees: 8 is possible for a similar reasoning.  (reply)
AdeebNafees: So the possible values, I'd hazard, are 3, 5, 6, 7, 8  (reply)
AdeebNafees: This is actually, come to think of it, a rearrangement puzzle.  (reply)
AdeebNafees: "How many numbers can you have which will have 4 numbers on either side provided the current numbers"  (reply)
AdeebNafees: For 2 and 14, we already have 5 numbers on each side.   (reply)
AdeebNafees: Wait a second. 4 is possible.  (reply)
AdeebNafees: 0, 1, 2, 3, 4, 6, 9, 13, 14  (reply)
AdeebNafees: So the correct range of values is from 3 to 9, inclusive.  (reply)
To AdeebNafees: good
To AdeebNafees: The first and seventh terms of a sequence are both 8. Each term of the sequence is the sum of the two previous terms. What is the fifth term?
AdeebNafees: Taking x as second term. 8+x is the 3rd term. 8+2x is the 4th. 16 + 3x is the 5th. 24 + 5x is the 6th. 40 + 8x is the seventh. But 40 + 8x = 8. So 8x = -32. Therefore x = -4.  (reply)
AdeebNafees: Therefore 5th term = 16 - 12 = 4  (reply)
To AdeebNafees: icas
AdeebNafees: Wut is that Malay?  (reply)
To AdeebNafees: what is 3+4
AdeebNafees: Depends on the base of the number system you are using.  (reply)
AdeebNafees: In base 7, that's 10/  (reply)
To AdeebNafees: A stone weighs 2.5N. When it is fully submerged in a liquid, its apparent weight is 2.2N. Calculate the density of liquid if its volume displaced by the stone is 25cm^3.
AdeebNafees: The liquid is exerting an upward force of 0.3N, which is the weight of the displaced liquid (25cm^3). But weight is related to mass by good old g.  (reply)
AdeebNafees: So we have the volume of the displaced liquid AND its mass.  (reply)
AdeebNafees: Some 1.2k kg/m^3 I think  (reply)
AdeebNafees: Or 1.2g/cm^3, if you are into CGS  (reply)
AdeebNafees: I would guess the liquid to be glycerol, which has a density very close to 1.2g/cm^3  (reply)
AdeebNafees: (courtesty Wikip)  (reply)
To AdeebNafees: If y(x) = u(x) v(x) w(x) with y, w, u, v as functions of x, prove that y' = u(x) v(x) w'(x) + u(x) v'(x) w(x) + u'(x) v(x) w(x)
AdeebNafees: I just need to prove the basic product rule, which automagically proves the above.  (reply)
AdeebNafees: And seriously, I have proved the product rule like thousands of times. It involves a rearrangement midway.  (reply)
AdeebNafees: Oh, and you need to add a term midway.  (reply)
AdeebNafees: *add-subtract  (reply)
To AdeebNafees: How many pineapples does it take to feed an elephant?
AdeebNafees: To be more specific, d/dx(f(x)g(x)) = lim [h->0] f(x+h)g(x+h)-f(x)g(x)/h  (reply)
AdeebNafees: It depends on the time of the day and the mood of the elefun.  (reply)
KADleon: Swag  (reply)
To adeebnafees: what did you eat
To lethalmutiny: Evening. Lunch?
LethalMutiny: no  (reply)