Question

Someone plis help me with algebro 3 homeowrrok :((((((

Answer 1

Josie's pay
`a_n` from company A in the
`n`-th year is given recursively by

`\begin{cases} a_1 = 42000 \\ a_(n+1) = a_n + 4000 & \text{for } n\ge1 \end{cases}`

while her pay
`b_n` from company B is

`\begin{cases} b_1 = 42000 \\ b_(n+1) = 1.09 b_n & \text{for } n \ge1 \end{cases}`

From these recurrences, we have

`a_2 - a_1 = 4000`

`a_3 - a_2 = 4000`

`a_4 - a_3 = 4000`

and so on - this is just confirming that the pay from company A increases at flat rate of $4000 each year.

Similarly,

`b_2 = 1.09 b_1 = 45780 \implies b_2 - b_1 = 3780`

`b_3 = 1.09 b_2 \approx 49900 \implies b_3 - b_2 \approx 4120`

`b_4 = 1.09 b_3 \approx 54391 \implies b_4 - b_3 \approx 4491`

and we see here that
`b_n` increases at a larger rate year. In only 2 years of working at company B, Josie would be making more money, and the amount by which her salary increases each year is also increasing.

Assuming Josie hopes to make more money, invest in her future, etc, she should take the offer from company B.

Answer 2

Explanation:

Company A:

The recursive formula is, aₙ=aₙ₋₁+4,000, where a₁=42,000

Company B:

The recursive formula is, aₙ₊₁=(1+0.09) x aₙ, where a₁= 42,000

I'm sure she should go for company B, in the long run, she would make much more money. This is proven because company a is increasing using an arithmetic sequence; based on this all they're doing is adding 4,000. But company B is increasing at a percentage of 9 after a while that can really amount to a lot.

Hope this helps, if you are still confused don't hesitate to ask :))