Answer:
d
Step-by-step explanation:
(3a^2 - 3b)(2a^2 - b)
3a^2(2a^2 - b) - 3b(2a^2 - b) =
6a^4 - 3a^2b - 6a^2b + 3b^2 =
6a^4 -9a^2b + 3b^2 <===
3/25 is your answer
12/100-->
6/50-->
=3/25
Step-by-step explanation:
Answer: The third one is the correct answer
Step-by-step explanation: I hope this helps
Answer:
<h3><u>Option 1</u></h3>
Earn $50 every month.
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000

This is a <u>linear function</u>.
<h3><u>Option 2</u></h3>
Earn 3% interest each month.
(Assuming the interest earned each month is <u>compounding interest</u>.)
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000

This is an <u>exponential function</u>.
<h3><u>Table of values</u></h3>
<u />

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.
However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the <u>point of intersection</u>.
From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.
<h3><u>Conclusion</u></h3>
If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.