Is the pond a rectangle? Can you post a pic?
Problem 16
<h3>Answer: i</h3>
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Work Shown:
The exponent 41 divided by 4 leads to
41/4 = 10 remainder 1
The "remainder 1" means that
i^(41) = i^1 = i
The reason why I divided by 4 is because the pattern shown below
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
repeats itself over and over. So this is a block of four items repeated forever.
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Problem 18
<h3>Answer: 1</h3>
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Work Shown:
Divide 3136 over 4 to get
3136/4 = 784 remainder 0
Therefore,
i^3136 = i^0 = 1
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Problem 20
<h3>Answer: i</h3>
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Work Shown:
Combine i^6*i^7 into i^13. We add the exponents here
Now divide by 4 to find the remainder
13/4 = 3 remainder 1
So, i^13 = i^1 = i
999946
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10,000,000
Answer:
cos(logx)*(1/x)
Step-by-step explanation:
(sin(logx))'=cos(logx)*logx'=cos(logx)*(1/x)
Answer:
The second choice: graph begins in the second quadrant near the axis and increases slowly while crossing the ordered pair 3, 1. The graph then begins to increase quickly throughout the first quadrant.
Explanation:
1) As per the set of choices, the function is:

2) Therefore it is an exponential function with these characteristics:
- Since, the bases is greater than 1 (2), the function is growing in all the domain.
- The domain is all real numbers (- ∞, ∞).
- To know where the function starts, calculate the limit of f(x) as x trends to negative infinity:

That means that the range is y > 0, and so the graph starts in the second quadrant.
- You can find the y-intersection making x = 0, which is 2⁰ ⁻ ³ = 2 ⁻³ = 1/8 = 0.25. So, the graph cross the y axis at y = 0.25.
- That tells you, that the function increases slowly, at least, until that point (0, 0.25).
- The other bullet point is when x = 3: 2 ³ ⁻ ³ = 2⁰ = 1. Therefore, the graph passes through the point (3, 1).
- From that point, the function starts to increase rapidly (since it is exponential).
- Those are the characteristics given by the second choice: graph begins in the second quadrant near the x-axis and increases slowly while crossing the ordered pair 3, 1. The graph then begins to increase quickly throughout the first quadrant.
You surely will find useful to watch the graph that I have attached.