What is the fat percentage in one serving of a whole grain
1 answer:
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Answer:
denominators are zero for x=1
there is no solution
Step-by-step explanation:
We suspect your equation is supposed to be ...
10/(5x -5) +1/5 = 2/(x -1) . . . . . parentheses are required on denominators
The denominators will be zero for x=1.
This version of the equation has no solution. It simplifies to ...
2/(x-1) + 1/5 = 2/(x -1)
1/5 = 0 . . . . . subtract 2/(x-1) from both sides
There is no value of x that will make this equation true.
_____
<em>Alternative interpretation</em>
The way your equation is written, it must be interpreted according to the order of operations to be ...
(10/5)x -5 +1/5 = (2/x) -1 . . . . x=0 makes the denominator zero
2x -3.8 = 2/x
2x^2 -3.8x = 2 . . . . multiply by x
2(x^2 -1.9x +.95^2) = 2 +2(0.95^2)
(x -0.95)^2 = 1.9025
x = 0.95 ± √1.9025 ≈ {-0.4293, 2.3293}
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
Then the ODE becomes
which has the characteristic equation
and in terms of
From the given initial conditions, we find
so the particular solution to the IVP is