For the answer to the question above,
<span>V(n) = a * b^n, where V(n) shows the value of boat after n years.
V(0) = 3500
V(2) = 2000
n = 0
V(0) = a * b^0 = 3500
a = 3500
V(2) = a * b^2
2000 = 3500 * b^2
b = sqrt (2000/3500)
b ≈ 0.76
V(n) = 3500 * 0.76^n
We can check it for n = 1 which is close to 2500 in the graph:
V(1) = 3500 * (0.76)^1
V(1) = 2660
And in the graph we have V(3) ≈ 1500,
V(n) = 3500 * (0.76)^3 ≈ 1536
Now n = 9.5
V(9.5) = 3500 * (0.76)^(9.5)
V(9.5) ≈ 258</span>
50 Miles at an average rate
<span>Let be the length, and be the width.
If the length is 1.6 times the width, then
If the sum of the length and with (in feet) is 130, then
</span><span>Substituting in the equation above, the expression for , we get
--> --> -->
Ten, substituting the value found for in we find
-->
The two equations we set up at the start form a system of linear equations:
</span>
Answer:
14 ÷ 9 = 1.55555555556
9 ÷ 14 = 0.64285714285
14 ÷ 5 = 2.8
5 ÷ 14 = 0.35714285714
Step-by-step explanation:
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Answer:
3
Step-by-step explanation:
hope this helps with the problem
