The answer A is the unit rate:)
We have the equation:

We know two points and we will use them to calculate the parameters a and b.
The point (0,3) will let us know a, as b^0=1.

Now, we use the point (2, 108/25) to calcualte b:
![\begin{gathered} y=3\cdot b^x \\ \frac{108}{25}=3\cdot b^2 \\ 3\cdot b^2=\frac{108}{25} \\ b^2=\frac{108}{25\cdot3}=\frac{108}{3}\cdot\frac{1}{25}=\frac{36}{25} \\ b=\sqrt[]{\frac{36}{25}} \\ b=\frac{\sqrt[]{36}}{\sqrt[]{25}} \\ b=\frac{6}{5} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D3%5Ccdot%20b%5Ex%20%5C%5C%20%5Cfrac%7B108%7D%7B25%7D%3D3%5Ccdot%20b%5E2%20%5C%5C%203%5Ccdot%20b%5E2%3D%5Cfrac%7B108%7D%7B25%7D%20%5C%5C%20b%5E2%3D%5Cfrac%7B108%7D%7B25%5Ccdot3%7D%3D%5Cfrac%7B108%7D%7B3%7D%5Ccdot%5Cfrac%7B1%7D%7B25%7D%3D%5Cfrac%7B36%7D%7B25%7D%20%5C%5C%20b%3D%5Csqrt%5B%5D%7B%5Cfrac%7B36%7D%7B25%7D%7D%20%5C%5C%20b%3D%5Cfrac%7B%5Csqrt%5B%5D%7B36%7D%7D%7B%5Csqrt%5B%5D%7B25%7D%7D%20%5C%5C%20b%3D%5Cfrac%7B6%7D%7B5%7D%20%5Cend%7Bgathered%7D)
Then, we can write the equation as:
Let
x-------> the number of dinner
y-------> the number of lunch
we know that
-------> equation A
------> equation B
Substitute equation B in equation A
![8[y]+5y \leq 42](https://tex.z-dn.net/?f=8%5By%5D%2B5y%20%5Cleq%2042)



so
the greatest number of lunch is 

Hence
the greatest number of dinner is 
therefore
the greatest number of meals is

<u>the answer is</u>

Answer: 5+3 times 4+2 is 48
Step-by-step explanation:
Circle formula
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center
and r=radius
to find the radius
we are given one of the points and the center
distnace from them is the radius
distance formula
D=

points (-3,2) and (1,5)
D=

D=

D=

D=

D=5
center is -3,2
r=5
input
(x-(-3))^2+(y-2)^2=5^2
(x+3)^2+(y-2)^2=25 is equation
radius =5
input -7 for x and solve for y
(-7+3)^2+(y-2)^2=25
(-4)^2+(y-2)^2=25
16+(y-2)^2=25
minus 16
(y-2)^2=9
sqqrt
y-2=+/-3
add 2
y=2+/-3
y=5 or -1
the point (-7,5) and (7,-1) lie on this circle
radius=5 units
the points (-7,5) and (-7,1) lie on this circle