I don't know if I understand this question that well, but if I am right, the first digit of the question you are asking is 5, and it is in the hundreds place. If this is not what you mean, then can you put more description in this question please?
Answer:
- 33 1/3 liters of 30%
- 16 2/3 liters of 45%
Step-by-step explanation:
Let x represent the liters of 45% solution needed. Then the amount of HCl in the mix is ...
0.45x +0.30(50 -x) = 0.35(50)
0.15x = 0.05(50) . . . . . simplify, subtract 0.30(50)
x = (0.05/0.15)(50) = 50/3 = 16 2/3 . . . liters of 45% HCl
33 1/3 liters of 30% and 16 2/3 liters of 45% HCl are needed.
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<em>Comment on the solution</em>
You may notice that the general solution to a mixture problem of this sort is that the fraction of the mix that is the highest contributor is ...
(mix % - low %) / (high % - low %) = (.35 -.30) / (.45 -.30) = .05/.15 = 1/3
Substitute
, so that
![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%5D%3D-%5Cdfrac1%7Bx%5E2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%2B%5Cdfrac1x%5Cleft%28%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D%5Cright%29%3D%5Cdfrac1%7Bx%5E2%7D%5Cleft%28%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%29)
Then the ODE becomes
which has the characteristic equation
with roots at
. This means the characteristic solution for
is
and in terms of
, this is
From the given initial conditions, we find
so the particular solution to the IVP is
Answer:
it A
Step-by-step explanation:
