Help it’s for my final most of my grade
1 answer:
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Answer:
5). x = 15
7). JL = 78
Step-by-step explanation:
We can solve these problems with the help of midsegment theorem.
Midsegment theorem:
"Segment joining midpoints of two sides of a triangle is parallel and half the measure of third side of the triangle"
Question 5
8x - 23 =
8x - 23 = 5x + 22
8x - 5x = 22 + 23
3x = 45
x = 15
Therefore, x = 15 is the answer.
Question 7
Since, MN is the midsegment,
MN =
5x - 16 =
5x - 16 = 2x + 17
5x - 2x = 17 + 16
3x = 33
x = 11
Therefore, Length of JL = 4(11) + 34
= 44 + 34
= 78 units
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
