d. both a relation and a function:
Given:
Mark records his science scores in each monthly assessment over a period of 5 months. In the first assessment he scores 76%. In the second assessment he scores 73%. After that, his scores keep increasing by 2% in every assessment.
x represents the number of assessments since he starts recording and y represents the scores in each assessment.
In order for a relation to be a function the association has to be unambiguous that means that for a given input only one output can exist.If an input can have two or more outputs then you cannot determine which is the correct output for that input.
In the given situation:
x is the input that is number of assessments since mark starts recording the scores so there is only one assessment no repeating.so there is only one output.
Hence the relation is a function.
Learn more about the function here:
brainly.com/question/5975436
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The solution depends on the value of
![k](https://tex.z-dn.net/?f=k)
. To make things simple, assume
![k>0](https://tex.z-dn.net/?f=k%3E0)
. The homogeneous part of the equation is
![\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D-16ky%3D0)
and has characteristic equation
![r^2-16k=0\implies r=\pm4\sqrt k](https://tex.z-dn.net/?f=r%5E2-16k%3D0%5Cimplies%20r%3D%5Cpm4%5Csqrt%20k)
which admits the characteristic solution
![y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}](https://tex.z-dn.net/?f=y_c%3DC_1e%5E%7B-4%5Csqrt%20kx%7D%2BC_2e%5E%7B4%5Csqrt%20kx%7D)
.
For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be
![y_p=ae^{4x}+be^x](https://tex.z-dn.net/?f=y_p%3Dae%5E%7B4x%7D%2Bbe%5Ex)
. Then
![\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y_p%7D%7B%5Cmathrm%20dx%5E2%7D%3D16ae%5E%7B4x%7D%2Bbe%5Ex)
So you have
![16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x](https://tex.z-dn.net/?f=16ae%5E%7B4x%7D%2Bbe%5Ex-16k%28ae%5E%7B4x%7D%2Bbe%5Ex%29%3D9.6e%5E%7B4x%7D%2B30e%5Ex)
![(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x](https://tex.z-dn.net/?f=%2816a-16ka%29e%5E%7B4x%7D%2B%28b-16kb%29e%5Ex%3D9.6e%5E%7B4x%7D%2B30e%5Ex)
This means
![16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}](https://tex.z-dn.net/?f=16a%281-k%29%3D9.6%5Cimplies%20a%3D%5Cdfrac3%7B5%281-k%29%7D)
![b(1-16k)=30\implies b=\dfrac{30}{1-16k}](https://tex.z-dn.net/?f=b%281-16k%29%3D30%5Cimplies%20b%3D%5Cdfrac%7B30%7D%7B1-16k%7D)
and so the general solution would be
The greatest common factor of 645 and 570 would be <u>35</u>.
Answer:
each tube costs $1.5 so put it the point in between 1 and 2
thank you
Step-by-step explanation: