Answer:
(x -11) * t + 1067
Step-by-step explanation:
We know that in her tutoring job she earns x amount of money and how waitress earns $ 11 / hour.
They give us the combined number of hours that he works with all the jobs, which is 97, but from here we can say that in the time that he does NOT work as a tutor it is actually 97 - t.
Total earned this month = x * t + 11 * (97 -t)
Total earned this month = x * t + 1067 - 11 * t = (x -11) * t + 1067
Answer with Step-by-step explanation:
We are given that a function f(x) is continuous on (
).
1.f'(-1)=0 and f''(-1)=-7
We have to find information about f.
When f'(-1)=0 and f''(-1)=-7 < 0
Then, function is maximum at x=-1.
Therefore, at x=-1, f has local maximum.
Answer:a)at x=-1 ,f has local maximum.
2.) if f'(4)=0 and f''(4)=0
We know that when f''(x)=0 then test fails then the function has not maximum or minimum.
Therefore, at x=4 , f has not a maximum or minimum.
Answer:c) at x=4, f has not a maximum or minimum.
Answer:
Step-by-step explanation:
Let P be the population of the community
So the population of a community increase at a rate proportional to the number of people present at a time
That is
![\frac{dp}{dt} \propto p\\\\\frac{dp}{dt} =kp\\\\ [k \texttt {is constant}]\\\\\frac{dp}{dt} -kp =0](https://tex.z-dn.net/?f=%5Cfrac%7Bdp%7D%7Bdt%7D%20%5Cpropto%20p%5C%5C%5C%5C%5Cfrac%7Bdp%7D%7Bdt%7D%20%3Dkp%5C%5C%5C%5C%20%5Bk%20%5Ctexttt%20%7Bis%20constant%7D%5D%5C%5C%5C%5C%5Cfrac%7Bdp%7D%7Bdt%7D%20-kp%20%3D0)
Solve this equation we get
![p(t)=p_0e^{kt}---(1)](https://tex.z-dn.net/?f=p%28t%29%3Dp_0e%5E%7Bkt%7D---%281%29)
where p is the present population
p₀ is the initial population
If the initial population as doubled in 5 years
that is time t = 5 years
We get
![2p_o=p_oe^{5k}\\\\e^{5k}=2](https://tex.z-dn.net/?f=2p_o%3Dp_oe%5E%7B5k%7D%5C%5C%5C%5Ce%5E%7B5k%7D%3D2)
Apply In on both side to get
![Ine^{5k}=In2\\\\5k=In2\\\\k=\frac{In2}{5} \\\\\therefore k=\frac{In2}{5}](https://tex.z-dn.net/?f=Ine%5E%7B5k%7D%3DIn2%5C%5C%5C%5C5k%3DIn2%5C%5C%5C%5Ck%3D%5Cfrac%7BIn2%7D%7B5%7D%20%5C%5C%5C%5C%5Ctherefore%20k%3D%5Cfrac%7BIn2%7D%7B5%7D)
Substitute
in
to get
![\large \boxed {p(t)=p_oe^{\frac{In2}{5}t }}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%20%7Bp%28t%29%3Dp_oe%5E%7B%5Cfrac%7BIn2%7D%7B5%7Dt%20%7D%7D)
Given that population of a community is 9000 at 3 years
substitute t = 3 in ![{p(t)=p_oe^{\frac{In2}{5}t }}](https://tex.z-dn.net/?f=%7Bp%28t%29%3Dp_oe%5E%7B%5Cfrac%7BIn2%7D%7B5%7Dt%20%7D%7D)
![p(3)=p_oe^{3 (\frac{In2}{5}) }\\\\9000=p_oe^{3 (\frac{In2}{5}) }\\\\p_o=\frac{9000}{e^{3(\frac{In2}{5} )}} \\\\=5937.8](https://tex.z-dn.net/?f=p%283%29%3Dp_oe%5E%7B3%20%28%5Cfrac%7BIn2%7D%7B5%7D%29%20%7D%5C%5C%5C%5C9000%3Dp_oe%5E%7B3%20%28%5Cfrac%7BIn2%7D%7B5%7D%29%20%7D%5C%5C%5C%5Cp_o%3D%5Cfrac%7B9000%7D%7Be%5E%7B3%28%5Cfrac%7BIn2%7D%7B5%7D%20%29%7D%7D%20%5C%5C%5C%5C%3D5937.8)
<h3>Therefore, the initial population is 5937.8</h3>
7 dimes+7 pennies
or 77 pennies
or 7 dimes+1 nickle+2 pennies
<u>3 different ways!!!</u>
DE and EF are segments of the line DF, so they will add up to DF. This can be represented with this equation:
![(4x + 10) + (2x - 1) = 9x - 15](https://tex.z-dn.net/?f=%20%284x%20%2B%2010%29%20%2B%20%282x%20-%201%29%20%3D%209x%20-%2015%20)
Combine like terms:
![4x + 2x = 6x](https://tex.z-dn.net/?f=%204x%20%2B%202x%20%3D%206x%20)
![10 + (-1) = 9](https://tex.z-dn.net/?f=%2010%20%2B%20%28-1%29%20%3D%209%20)
![6x + 9 = 9x - 15](https://tex.z-dn.net/?f=%206x%20%2B%209%20%3D%209x%20-%2015%20)
Add 15 to both sides:
![6x + 24 = 9x](https://tex.z-dn.net/?f=%206x%20%2B%2024%20%3D%209x%20)
Subtract 6x from both sides:
![24 = 3x](https://tex.z-dn.net/?f=%2024%20%3D%203x%20)
Divide both sides by 3 to get x by itself:
![\boxed{x = 8}](https://tex.z-dn.net/?f=%20%5Cboxed%7Bx%20%3D%208%7D%20)
The value of x will be 8.
Plug this value of x into the expression for DF:
![9(8) - 15 = 72 - 15 = \boxed{57}](https://tex.z-dn.net/?f=%209%288%29%20-%2015%20%3D%2072%20-%2015%20%3D%20%5Cboxed%7B57%7D%20)
DF will equal 57.