Hey! Are you good at math? I really need some help over here...?

Question

How can I change volume to centimetres? Can anyone please....... help me with the work? Or I am going to be busted.

Answer 1

Centimeters is distance, not volume. Did you mean cubic meters or millileters ? Maybe this can help anyway - 1 cubic foot = 1 cubic meter times(X) .03 . I think I need a little more info. What are you trying to convert into what?

Answer 2

look online for a site that converts things. theres plenty! search something like "volume to centimeters converter or calculator"

Answer 3

>> one million) (one million / 15) < (one million / two) (three / four) (five / 6) ... (ninety nine / one hundred) < (one million / 10) If you multiply it out (a one-liner in Perl), you get one million/15 = zero.0666... < zero.079589... < zero.one million = one million/10 Symbolically, the denominator is two^50 * 50! and the numerator is one hundred! / (denominator), so total the product is P = one hundred! / (two^one hundred * 50! * 50!) = "one hundred Choose 50" / two^one hundred So P is the possibility of tossing a reasonable coin one hundred occasions and getting precisely 50 heads. You can approximate it with both Sitrling's approximation for the factorial or with the average distribution approximation to the Binomial(N = one hundred, p = q = one million/two) distribution. The latter could be Prob(forty nine.five <= X <= 50.five) whilst X is almost always disbursed with imply Np = 50 and variance Npq = 25 (and so normal deviation five). Thus we wish the possibility that a average is inside +/- zero.five/five = +/- zero.one million normal devi ations from the imply, that's approximately zero.079656, among one million/15 and one million/10. The former, Stirling's approximation, says one million = lim n -> pos. inf. n! / (sqrt(two pi n) (n/e)^n) So P is approx. sqrt(two pi one hundred) (one hundred/e)^one hundred / (two^one hundred * (sqrt(two pi 50) (50/e)^50)^two) = 10 sqrt(two pi) (one hundred/e)^one hundred / (sqrt(two pi) sqrt(two pi) sqrt(50) sqrt(50) two^one hundred (50/e)^one hundred) = (10 / 50) (one million / sqrt(two pi) (one hundred/e)^one hundred / ((one hundred/e)^one hundred) = one million / (five sqrt(two pi)) two pi is 6.28... is a little bit over 6.25 = two.five^two, so sqrt(two pi) is a little bit over two.five, and five sqrt(two pi) is a little bit over 12.five. So 10 < five sqrt(two pi) < 15, for this reason one million/15 < (one million / (five sqrt(two pi))) approx= P < one million/10. >> two) Consider a 25 x 25 rectangular grid in every of whose 625 squares is located both 'a+one million' or 'a-one million'. Suppose a(i) denotes the made from all of the numbers within the (i)th row and b(j) denotes the made from all of the numbers within the (j)th column. Prove that a(one million) + b(one million) + a(two) + b(two) + ... + a(25) + b(25) isn't identical to zero. I consider you mentioned this improper. If you utilize 'a+one million' and 'a-one million' as though a have been a variable, then the certain sum is a polynomial of measure at so much 25 in a, and so there will probably be values of a which make the sum zero (25 problematic root, no longer always certain; a minimum of one truly root on the grounds that the coefficients are truly and the measure is bizarre). For instance, permit a = one million, and assign adequate a-one million labels in order that every row and every column comprises a minimum of one a-one million = zero. Instead, I consider they supposed to mention you assign one of the most values { -one million, one million } to every of the 625 squares. That is, "assign a '+one million' or a '-one million'." Anyway, with that interpretation, each row product and each column product is one million or -one million. Start with all one million's. Each row product a(i) and every column product b(j) is one million. Flip one million's to -one million's one by one and watch what occurs to the sum of the a(i) + b(j). At every turn, one of the most a(i) alterations from -one million to one million or vice versa, and the sum a(one million) + ... + a(25) both is going up two or down two. Likewise precisely one of the most b(j) alterations from one million to -one million or vice versa, inflicting the sum b(one million) + ... + b(25) to both pass up two or pass down two. So the total sum alterations via both of + two + two = four + two - two = zero - two + two = zero - two - two = -four As every one million is flipped to a minus one million, the sum mod four is left unchanged. Since the sum with all one million's was once 50 = two mod four, the sum for each configuration of one million's and -one million's is two mod four. It is in no way zero. >> three) If p and p^two + two are primes, end up that p^three + two may be high. P can not be two on the grounds that then p^two + two = four + two = 6 isn't high. p can also be three on the grounds that then p^two + two = nine + two = eleven and p^three + two = 27 + two = 29 are all primes. Suppose p is a primary larger than three. p isn't divisible via three, so both p = one million mod three or p = two mod three. In both case, p^two + two = one million + two = three = zero mod three and so three divides p^two + two and so p^two + two isn't high. So the one time each p and p^two + two are each high is whilst p = three, wherein case p^three + two may be a primary. Dan