What grade are u in. because i can help u if u tell me bout it
Answer:
three hundred and eight five twenty sevenths
Answer:
Population of fish in 2010 = 1,646 fish (Approx.)
Step-by-step explanation:
Given:
Population of fish in 2002 = 2100 fish
Decrease rate = 3% per year
Find:
Population of fish in 2010
Computation:
Number of year = 2010 - 2002
Number of year = 8 year
A = P[1+r]ⁿ
Population of fish in 2010 = Population of fish in 2002[1+(-3%)]ⁿ
Population of fish in 2010 = 2100[1-0.03]⁸
Population of fish in 2010 = 2100[0.97]⁸
Population of fish in 2010 = 2100[0.7837]
Population of fish in 2010 = 1645.77
Population of fish in 2010 = 1,646 fish (Approx.)
Answer:
Step-by-step explanation:
![(\sqrt{3} +4)(1+\sqrt{3})\\\\= \sqrt{3}(1+\sqrt{3} )+4(1+\sqrt{3})\\\\= \sqrt{3} + (\sqrt{3} )^2 + 4 + 4\sqrt{3} \\\\= \sqrt{3} + 3+4+4\sqrt{3} \\\\= 7 + \sqrt{3} + 4\sqrt{3} \\\\Take \ \sqrt{3} \ common\\\\= 7 + \sqrt{3} (1+4)\\\\= 7 + \sqrt{3}(5)\\\\= 7 + 5\sqrt{3} \\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7B3%7D%20%2B4%29%281%2B%5Csqrt%7B3%7D%29%5C%5C%5C%5C%3D%20%5Csqrt%7B3%7D%281%2B%5Csqrt%7B3%7D%20%29%2B4%281%2B%5Csqrt%7B3%7D%29%5C%5C%5C%5C%3D%20%5Csqrt%7B3%7D%20%2B%20%28%5Csqrt%7B3%7D%20%29%5E2%20%2B%204%20%2B%204%5Csqrt%7B3%7D%20%5C%5C%5C%5C%3D%20%5Csqrt%7B3%7D%20%2B%203%2B4%2B4%5Csqrt%7B3%7D%20%20%5C%5C%5C%5C%3D%207%20%2B%20%5Csqrt%7B3%7D%20%20%2B%204%5Csqrt%7B3%7D%20%5C%5C%5C%5CTake%20%5C%20%5Csqrt%7B3%7D%20%5C%20common%5C%5C%5C%5C%3D%207%20%2B%20%5Csqrt%7B3%7D%20%281%2B4%29%5C%5C%5C%5C%3D%207%20%2B%20%5Csqrt%7B3%7D%285%29%5C%5C%5C%5C%3D%207%20%2B%205%5Csqrt%7B3%7D%20%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
