8x^2-5x-2
method
open bracket
6x^2-2x-3+x^2-3x+1
collect like terms
6x^2+2x^2-3-3x+1
8x^2-3-3x+1
again collect like terms
8x^2-2x-3x-3+1
8x^2-5x-3+1. (-×-=+ addition but keeping sign minus
8x-5x-2. (-×+=-)
Withdrawals must be subtracted from the beginning balance since to withdraw means to take out, so withdrawing money would lower the balance.
Deposits are added to the balance, so they must be added in the equation.
The answer is <span>c) Beginning Balance - Withdrawals + Deposits = Ending Balance</span>
Answer:
Maximum revenue = $8000
The price that will guarantee the maximum revenue is $40
Step-by-step explanation:
Given that:
Price of product = $35
Total sale of items = 225
For every dollar increase in the price, the number of items sold will decrease by 5.
The total cost of item sold = 225 ×35
The total cost of item sold = 7875
If c should be the dollar unit in price increment;
Therefore; the cost function is : ![[35+c(1)][225-5(c)]](https://tex.z-dn.net/?f=%5B35%2Bc%281%29%5D%5B225-5%28c%29%5D)
For maximum revenue;

![\dfrac{d}{dc}[[35+c(1)][225-5(c)]]=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdc%7D%5B%5B35%2Bc%281%29%5D%5B225-5%28c%29%5D%5D%3D0)
0+225-35× 5 -10c = 0
225 - 175 =10c
50 = 10c
c = 50/10
c = 5
Maximum revenue = ![[35+c(1)][225-5(c)]](https://tex.z-dn.net/?f=%5B35%2Bc%281%29%5D%5B225-5%28c%29%5D)
Maximum revenue = ![[35+5(1)][225-5(5)]](https://tex.z-dn.net/?f=%5B35%2B5%281%29%5D%5B225-5%285%29%5D)
Maximum revenue = (35 + 5)(225-25)
Maximum revenue = (40 )(200)
Maximum revenue = $8000
The price that will guarantee the maximum revenue is :
=(35 +c)
= 35 + 5
= $40
Use the distributive property.
3*2a - 3*b = 6a - 3b.