Chris sees 6 fish.
Mike sees 3 more fish than Chris.
6 + 3 = 9
So on your bar, you have Chris's number of fish he saw, 6, and you have 3 being added to that on Mike's bar.
Answer:
Therefore,

Step-by-step explanation:
Given:
![A=\left[\begin{array}{ccc}3&6&9\\2&4&8\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%269%5C%5C2%264%268%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To Find:
a₂₁ = ?
Solution:
Let,
![A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26a_%7B12%7D%26a_%7B13%7D%5C%5Ca_%7B21%7D%26a_%7B22%7D%26a_%7B23%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
We require ' a₂₁ ' i.e Second Row First Column Element
So on Comparing we get
∴ 
Therefore,

Answer:
2x^2+6x+3
Step-by-step explanation:
answer : 2x^2+6x+3
-7 is the missing exponent
<span>B(n) = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
where B is the balance after n payments are made, i is the monthly interest rate, P is the monthly payment and A is the initial amount of loan.
We require B(n) = 0...i.e. balance of 0 after n months.
so, 0 = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
Then, with some algebraic juggling we get:
n = -[log(1 - (Ai/P)]/log(1 + i)
Now, payment is at the beginning of the month, so A = $754.43 - $150 => $604.43
Also, i = (13.6/100)/12 => 0.136/12 per month
i.e. n = -[log(1 - (604.43)(0.136/12)/150)]/log(1 + 0.136/12)
so, n = 4.15 months...i.e. 4 payments + remainder
b) Now we have A = $754.43 - $300 = $454.43 so,
n = -[log(1 - (454.43)(0.136/12)/300)]/log(1 + 0.136/12)
so, n = 1.54 months...i.e. 1 payment + remainder
</span>