Find the length of the missing side. Leave your answer in simplest radical form.
1 answer:
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Answer:
c(p) = {0.2 p ⇒ p < 1000
0.24 p - 40 ⇒ 1000 ≤ p ≤ 2000
0.29 p - 140 ⇒ p > 2000}
Step-by-step explanation:
* Lets explain how to solve the problem
- The profit is represented by p
1. If the profit is under $1,000, the commission rate is 20%
∵ The profit is p < 1000
∵ 20% of p = = 0.2 p
∵ c(p) is the function of the commission
∴ c(p) = 0.2 p when p < 1000
2. If the profit is at least $1,000 and less than or equal to $2,000, the
commission rate is 20% of the first $1,000 and 24% of the remainder
of the profit
- At least means greater than or equal
∵ The profit 1000 ≤ p ≤ 2000
- The commission is divided into 20% of first $1000 and 24% of
the reminder
∵ 20% of 1000 = = 200
∵ The remainder of the profit = p - 1000
∵ 24% of the remainder profit =
= 0.24(p - 1000) = 0.24 p - 240
∴ The total commission = 200 + 0.24 p - 240
∴ The total commission = 0.24 p - 40
∴ c(p) = 0.24 p - 40 when 1000 ≤ p ≤ 2000
3. If the profit is above $2,000, the rate is 20% of the first $1,000
of profit, 24% of the next $1,000 of profit, and 29% of the amount
of profit over $2,000
∵ The profit p > 2000
- The commission is divided into 20% of first $1000 and 24% of the
next $1,000 of profit, and 29% of the amount of profit over $2,000
∵ 20% of 1000 = = 200
∵ 24% of 1000 = = 240
- The amount of profit over $2,000 = p - 2000
∵ 29% of the amount of profit over $2,000 =
= 0.29(p - 2000)
= 0.29 p - 580
∴ The total commission = 200 + 240 + 0.29 p - 580
∴ The total commission = 0.29 p - 140
∴ c(p) = 0.29 p - 140 when p > 2000
* The commission function is:
c(p) = {0.2 p ⇒ p < 1000
0.24 p - 40 ⇒ 1000 ≤ p ≤ 2000
0.29 p - 140 ⇒ p > 2000}
Answer:
The value of the computer is given by .
Step-by-step explanation:
A computer sells for $900.
If the price of the computer loses 30% of its value per year, then it is compounded every year.
Now, the price of the computer t years after it is sold will be
........... (1)
Therefore, the value of the computer is given by the above equation (1). (Answer)