Answer:
Mr. Dairo can produce 20580 pieces of Papaya rosette in two weeks.
Step-by-step explanation:
Papaya rosette pieces on first day = 5890
Papaya rosette Pieces on second day = 7020
Papaya rosette Pieces of third day = 8150
Let

We can check if these numbers are part of a sequence.
In order to check, common difference will be found first.

It can be observed that the common difference is same. When the common difference is same, the sequence is said to be an arithmetic sequence.
The formula for arithmetic sequence is given by:

Putting the values

The formula for nth term can be used to find any term. As we have to find the number of papaya rosette pieces after two weeks which means that we need to find the number of pieces on 14th day.
So,

Hence,
Mr. Dairo can produce 20580 pieces of Papaya rosette in two weeks.
Answer:
(a)96.77%
(b)3.23%
Step-by-step explanation:
Starting with the Michaelis-Menten equation which is used to model biochemical reactions:
Dividing both sides by 
![\dfrac{v}{V_{max}}=\dfrac{[S]}{K_M + [S]}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B%5BS%5D%7D%7BK_M%20%2B%20%5BS%5D%7D)
Where:
maximum rate achieved by the system
=The Michaelis constant
Substrate concentration
(a) When ![[S]=30K_M](https://tex.z-dn.net/?f=%5BS%5D%3D30K_M)
![\dfrac{v}{V_{max}}=\dfrac{[S]}{K_M + [S]}\\\dfrac{v}{V_{max}}=\dfrac{30K_M}{K_M + 30K_M}\\\dfrac{v}{V_{max}}=\dfrac{30}{1 + 30}\\\dfrac{v}{V_{max}}=\dfrac{30}{31}\\$Expressed as a percentage\\\dfrac{v}{V_{max}}=\dfrac{30}{31}X100=96.77\%](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B%5BS%5D%7D%7BK_M%20%2B%20%5BS%5D%7D%5C%5C%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B30K_M%7D%7BK_M%20%2B%2030K_M%7D%5C%5C%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B30%7D%7B1%20%2B%2030%7D%5C%5C%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B30%7D%7B31%7D%5C%5C%24Expressed%20as%20a%20percentage%5C%5C%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B30%7D%7B31%7DX100%3D96.77%5C%25)
(b)When ![K_M=30[S]](https://tex.z-dn.net/?f=K_M%3D30%5BS%5D)
![\dfrac{v}{V_{max}}=\dfrac{[S]}{K_M + [S]}\\\dfrac{v}{V_{max}}=\dfrac{[S]}{30[S] + [S]}\\\\=\dfrac{1[S]}{30[S] + 1[S]}\\=\dfrac{1}{30 + 1}\\\dfrac{v}{V_{max}}=\dfrac{1}{31}\\$Expressed as a percentage\\\dfrac{v}{V_{max}}=\dfrac{1}{31}X100=3.23\%](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B%5BS%5D%7D%7BK_M%20%2B%20%5BS%5D%7D%5C%5C%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B%5BS%5D%7D%7B30%5BS%5D%20%2B%20%5BS%5D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%5BS%5D%7D%7B30%5BS%5D%20%2B%201%5BS%5D%7D%5C%5C%3D%5Cdfrac%7B1%7D%7B30%20%2B%201%7D%5C%5C%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B1%7D%7B31%7D%5C%5C%24Expressed%20as%20a%20percentage%5C%5C%5Cdfrac%7Bv%7D%7BV_%7Bmax%7D%7D%3D%5Cdfrac%7B1%7D%7B31%7DX100%3D3.23%5C%25)
Taking into account the repeated ones it would be 7! what is 7*6*5*4*3*2*1
X = # of perfomances
y = total amount of money
Income: y = 7500x
Expense: y = 5900x + 88000
Substitution:
7500x = 5900x + 88000
1600x = 88000
x = 55
y = 7500(55) = $412,500
55 performances are needed to break-even.
14= 3k-4
_ less than _ = -
3* k simplified= 3k