The minimum cost is achieved when 150 trees are produced giving a minimum cost of $27.5
Polynomial is an expression that involves the <em>operations of addition, subtraction, multiplication of variables.</em>
Let C represent the cost for buying and caring for n trees. Given that:
C = 0.001n² - 0.3n + 50.
The minimum cost is at dC/dn = 0, hence:
dC/dn = 0.002n - 0.3
0.002n - 0.3 = 0
0.002n = 0.3
n = 150
C(150) = 0.001(150)² - 0.3(150) + 50 = 27.5
The minimum cost is achieved when 150 trees are produced giving a minimum cost of $27.5
Find out more at: brainly.com/question/25836610
Answer:
54
Step-by-step explanation:
First you take two and multiply it by three, which gives you 6. Then you multiply it by three and get 18. Then you multiply and get 54.
Answer:
No
Step-by-step explanation:
How to solve :
(a-4)/(a^2+8a+12)*(a^2-2a-48)/(a^2+4a-32)
Factor:
a^2+8a+12 = (a+6)(a+2)
a^2-2a-48 = (a-8)(a+6)
a^2+4a-32 = (a+8)(a-4)
simplify a+6, and a-4
you are left with (a-8)/(a+2)(a+8) = (a-8)/(a^2+10x+16)
Answer: 90% decrease
Step-by-step explanation: First determine whether the number is increasing or decreasing. Since it changes from 2,000 to 200, it's going down so it's decreasing.
Now to find the percent decrease, we divide the
amount of change by the original number.
The <em>amount of change</em> is the difference between the two numbers
which in this case is 2,000 - 200 and the original number is 2,000.
.
2,000 - 200 is 1,800.
So we're left with 1,800/2,000 and dividing
2,000 into 1,800 gives us 0.9.
Remember however that our problem is
asking for a <em>percent</em>, not a decimal.
So we need to write 0.9 as a percent and we can do
that by multiplying it by 100 and adding the percent sign.
So 0.9 becomes 90%.
So when a number changes from 2,000 to 200,
it has decreased by 90%.
So first (3 1/2) equals to 3+1/2 which is 7/2, same for(2 1/3), 2+1/3=7/3. It's asking for speed in laps per minute, so like how many labs did she swim per minute, which is labs over time——(7/2)/(7/3)=(7/2)*(3/7)=3/2. So she swam 3/2 labs per minute.
