find the constant of variation k.?
363034
m = kn\nk = m/n\n\nk = 144/8\nk = 18\n\nANS : 18
859781
It says y varies inversely as the square of x.  This means that you can write an equation for it:\n\ny = k/x^2\n\nwhere k is the proportionality constant.\n\nYou're told that y = 7.2 when x = 10, so solve the equation for k:\n\nk = yx^2\n\nand plug in:\n\nk = 7.2(10^2) = 720.\n\nSo your equation is\n\ny = 720/x^2\n\nNow plug in:\n\n20 = 720/x^2\n\nx^2 = 720/20 = 36\n\nx = 6 (positive or negative 6, actually, but take the answer you think most reasonable).
1149287
pV = n R T => if pressure is constant => p = n R T / V = const. => p / n R = T/ V = const.\n=> T / V = const.\nT1 = 333 K\nT2 = 533 K\nV1 = 0.2 m3\n\nT1 / V1 = T2 / V2 =>\nV2 = V1 x T2 / V2 = 0.2 x 533 / 333 = 0.32 m^3\nanswer D is correct
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u
963577
4x^2 is the key.\n\nYou get 1 x^2 by multiplying 1 by x^2, so kx times x must equal 4-1=3.\n\nIn the same way for the second:\n\nYou get -1 x^2 by multiplying -1 by x, so 2*k*x^2 = -5-(-1)x^2\n\n2*k*x^2 = -4x^2\nk=-4/2\nk=-2.\n\nThe second is similar.  Basically, look at the x^2 coefficient and look at how many you have through the terms NOT including k, then subtract the number of other ones from the total number of x^2.  kx must be multiplied by x to get x^2, so take into account the other x coefficient.\n\ni.e. (2x)(kx) = 2kx^2
400015
By newron's ii law\nMa=F-f\nf=10-10*0.6\nf=4 N\nf= -4 i (vector notation)
622930
Inverse variation  y = k/x therefore k = yx\nk is a constant so both cases equal the same k so set them equal to each other...\n-2(8) = 2/3 x\n-16 = 2/3 x\n-48 = 2x\n-24 = x\n\ntosmith
735121
y varies kx\n30 = .... wait a minute this question is impossible to do lol, u need to know two sets of values ( y, x).. either you said the question wrongly or the person who made it is dumb
744496
A row of AB is of the form RB where R is a row of A. Now, if\nR1, R2, ...Rk are linearly dependent, there are constants c1,..ck\nwhere\nc1*R1+c2*R2+..ck*Rk=0.\nNow apply B on the right to get your result.
69686
I think the answer is x=1/2.\n\nIf x and y vary inversely, then xy = k, which is some fixed number. Taking the first two values, (3)(14)=k, so k=42.\n\nIf y=84, and k=42, then (84*x)=42, or x = 0.5
226307
= 24 k^2   +  6 k^3 +  18k^2  - 54 k^3\n= 42k^2 - 48 k^3\n= 6k^2 (7 - 8k)\n\nif 6k^2 (7 - 8k)=0\n7 - 8k=0\n8k=7\nk=7/8
529324
Give me your email id. I will send a small chapter containing discussions on "Variation" as an attachment. Contact me at psbhowmick@yahoo.com
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
1154468
make a common denominator\n\na/b - c/d = a*d/b*d - c*b/d*b = (a*d-c*b)/b*d
1137927
I think the best you will be able to do is get a recurrence relation for the coefficients. Even that isn't going to be so easy. For example, it is clear that if we write\nt_k (x)=sum p_nk x^n,\nthen\na_n=sum p_nk  c_k.\n\nNotice that even a_0 will depend on infinitely many c_k (since t_k(0)=p_0k  is non-zero for all even k).\n\nAdded:\nIt looks like\na_0 = sum (-1)^k c_{2k}.\nThere is a recurrence relation among the p_nk which you might be able to exploit to get formulas for the other a_k. I'll think about it a bit more.\n\nAdded:\nOK. In an e-mail you suggested doing the integrals for the expnasions of x^alpha. Here's a suggestion:\nYou want to evaluate \nint_-1^1 x^n t_k(x)/sqrt(1-x^2) dx.\nif you let x=cos(A), then t_k(x)=cos(kA), so the integral transforms into\nint_-pi/2^pi/2 cos^n(A) cos(kA) dA.\nIf you now write this in terms of complex exponentials and expand by the binomial theorem, several terms will zero out. I'm not sure if this does what you want, but it makes some progress, I think.
735022
There will be a term proportional to distance added to a constant delay term, so the form will be t=kD + c. Solving for k and c:\nt1=.152=100k + c\nt2=.229=170k + c\nt2-t1=.077=70k\nThen k=.0011\nt1=.152=.11 + c --> c=.042\nt2=.229=.187 + c --> c=.042\nResult is\nt=.0011*D + .042
364905
Use Charles' Law\n\nLet V initial = V\nLet T initial = T\n\nLet V final = V'\nLet T final = T'\n\n\nV/T = V'/T'\n\nSolve for V' you might have to convert from mL to L though.