5 would be chur answer i hope this helps chu
Answer:
<h3>Answer c</h3>
Step-by-step explanation:
♥ thus y’/(yf-y) = k, where yf=6000;
<span>♠ hence ln(yf-y) =-kx +C; </span>
<span>at x=0 y(0)=y0=1000, hence C= ln(yf-y0); </span>
<span>and we get: </span>
<span>ln((yf-y)/(yf-y0) =-kx; or; </span>
<span>yf-y =(yf-y0)*exp(-kx); at last </span>
<span>♦ y = 6000- 5100*exp(-0.0015*x);
</span>
<span><span><span>dy</span><span>dt</span></span>∝y=ky</span>separate the variables<span><span><span>dy</span>y</span>=kdt</span>integrate both sides:<span>lny=kt+C</span><span>y(t)=C<span>e<span>kt</span></span></span>by plugging in t=0 we find that C the original population<span>y(0)=C<span>e0</span>=C⟹C=<span>y0</span></span>so we get<span><span>y=<span>y0</span><span>e<span>kt</span></span></span></span>
We have the next inequality:
This represents that y can take all real numbers over 4 (not including the number four).
Hence, we can graph this like this:
Where the limit with the number four is represented with the line ----------
