For this case, the first thing we must do is define variables.
We have then:
t: number of hours
F (t): total charge
We write the function that models the problem:
Where,
b: represents an initial fee.
We must find the value of b.
For this, we use the following data:
Her total fee for a 4-hour job, for instance, is $ 32.
We have then:
From here, we clear the value of b:
Then, the function that models the problem is:
Answer:
the function's formula is:

Answer:
47 pounds i think \/'_'\/
Step-by-step explanation:

Answer:
12.0 tablet computers/month
Step-by-step explanation:
The average price of the tablet 25 months from now will be:

Next, we determine the rate at which the quantity demanded changes with respect to time.
Using Chain Rule (and a calculator)

![\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdp%7D%3D%20%5Cdfrac%7Bd%7D%7Bdp%7D%5Cleft%5B%7B%20%5Cdfrac%20%7B%20100%20%7D%20%7B%209%20%7D%20%5Csqrt%20%7B%20810%2C000%20-%20p%20%5E%20%7B%202%20%7D%20%7D%20%7D%5Cright%5D%20%3D-%5Cdfrac%7B100%7D%7B9%7Dp%28810%2C000-p%5E2%29%5E%7B-1%2F2%7D)
![\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdp%7D%7Bdt%7D%3D%5Cdfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%5Cdfrac%20%7B%20400%20%7D%20%7B%201%20%2B%20%5Cdfrac%20%7B%201%20%7D%20%7B%208%20%7D%20%5Csqrt%20%7B%20t%20%7D%20%7D%20%2B%20200%20%5Cright%5D%3D-25%5Cleft%5B1%20%2B%20%5Cdfrac%20%7B%201%20%7D%20%7B%208%20%7D%20%5Csqrt%20%7B%20t%20%7D%20%5Cright%5D%5E%7B-2%7Dt%5E%7B-1%2F2%7D)
Therefore:
![\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D%20%5Cleft%5B-%5Cdfrac%7B100%7D%7B9%7Dp%28810%2C000-p%5E2%29%5E%7B-1%2F2%7D%5Cright%5D%5Cleft%5B-25%5Cleft%5B1%20%2B%20%5Cdfrac%20%7B%201%20%7D%20%7B%208%20%7D%20%5Csqrt%20%7B%20t%20%7D%20%5Cright%5D%5E%7B-2%7Dt%5E%7B-1%2F2%7D%5Cright%5D)
Recall that at t=25, 
Therefore:
![\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%2825%29%3D%20%5Cleft%5B-%5Cdfrac%7B100%7D%7B9%7D%5Ctimes%20446.15%28810%2C000-446.15%5E2%29%5E%7B-1%2F2%7D%5Cright%5D%5Cleft%5B-25%5Cleft%5B1%20%2B%20%5Cdfrac%20%7B%201%20%7D%20%7B%208%20%7D%20%5Csqrt%20%7B25%7D%20%5Cright%5D%5E%7B-2%7D25%5E%7B-1%2F2%7D%5Cright%5D%5C%5C%3D12.009)
The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.
Step-by-step explanation:
1. hundred thousand
2.hundred thousand
Answer:
0
Step-by-step explanation:
We have the fraction
Step 1. Use LCM of the fraction, (2m+3)(2m-3), to simplify the fraction:
![\frac{(2m-3)(2m)-(2m+3)(2m)}{(2m+3)(2m-3)} =\frac{2m^2-6m-[2m^2+6m]}{(2m+3)(2m-3)} =\frac{2m^2-6m-2m^2-6m}{(2m+3)(2m-3)} =\frac{-12m}{(2m+3)(2m-3)}](https://tex.z-dn.net/?f=%5Cfrac%7B%282m-3%29%282m%29-%282m%2B3%29%282m%29%7D%7B%282m%2B3%29%282m-3%29%7D%20%3D%5Cfrac%7B2m%5E2-6m-%5B2m%5E2%2B6m%5D%7D%7B%282m%2B3%29%282m-3%29%7D%20%3D%5Cfrac%7B2m%5E2-6m-2m%5E2-6m%7D%7B%282m%2B3%29%282m-3%29%7D%20%3D%5Cfrac%7B-12m%7D%7B%282m%2B3%29%282m-3%29%7D)
Step 2. Equate the resulting fraction to zero and solve for
:

![-12m=0[(2m+3)(2m-3)]](https://tex.z-dn.net/?f=-12m%3D0%5B%282m%2B3%29%282m-3%29%5D)



Step 3. Replace the value in the original equation and check if it holds:


Since
,


Since the only solution of the equation holds, the equation bellow doesn't have any extraneous solution