Answer:
a) 1/2; reduction
b) 5/4; enlargement
Step-by-step explanation:
In each case, the scale factor is CP'/CP. When it is more than 1, the dilation is an enlargement.
Even before you run the numbers, you can tell if it is an enlargement or not. If the dilated figure is larger, P is closer to C than is P'. If P' is closer to C, then it is a reduction.
a) CP'/CP = (4-2)/4 = 2/4 = 1/2 . . . . a reduction
__
b) CP'/CP = 25/20 = 5/4 . . . . an enlargement
Answer:
A and B is the correct answer
Step-by-step explanation:
<span>A glide reflection is the composition of a reflection and a translation, where the line of reflection, m, is parallel to the directional vector line, v, of the translation. Example: A glide reflection is commutative. Reversing the direction of the composition will not affect the outcome.
I hope this helps you!</span>
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