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daser333 [38]
2 years ago
10

Throughout any given month, the maximum and minimum ocean tides follow a periodic pattern. Last year, at a certain location on t

he California coast, researchers recorded the height of low tide, with respect to sea level, each day during the month of July. The lowest low tide was first measured on July 11, at -1.4 feet. The highest low tide was first measured on July 4, at 1.8 feet. The average low tide for the month of July was measured to be 0.2 feet.
Why do the oceans tide follow a periodic pattern?
Mathematics
2 answers:
Dahasolnce [82]2 years ago
4 0

Idk if you need it but here are my answers for part C and D (the questions after this one on plato)

C: To model this function I would choose cosine because it's graph fits better with the cycle of the moon. It would start at the highest low tide, when the moon is new, and then the graph would go down steadily back to a full moon, where the lowest low tide is when when the moon is full (strongest gravitational pull).

D: The amplitude of the function would have to be around 1.3 because the function is like a line of best fit, and it would have to be close to the points (4, 1.8) and (11, -1.4), but not include them since they are in a way maximums and minimums and the graph has to fit along with the low and high low tides of the other days.

aksik [14]2 years ago
3 0
The flow of of ocean tides are affected by the gravitational pull of the moon primarily and the sun. We have periods where the moon is closer with the earth's surface as it orbits around us, this is also described by the phases of the moon we observed. The closer the moon orbits around us are the periods where ocean tide is high. This is a cycle and follows a pattern.
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weqwewe [10]

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Step-by-step explanation:

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What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0
Tom [10]

Answer:

x=\frac{-7-\sqrt{5}}{2} or x=\frac{-7+ \sqrt{5}}{2}

Step-by-step explanation:

The given equation is

(2x+3)^2+8(2x+3)+11=0

Let us treat this as a quadratic equation in (2x+3).

where a=1,b=8,c=11

The solution is given by the quadratic formula;

(2x+3)=\frac{-b\pm \sqrt{b^2-4ac} }{2a}

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(2x+3)=\frac{-8\pm \sqrt{8^2-4(1)(11)} }{2(1)}

(2x+3)=\frac{-8\pm \sqrt{64-44} }{2}

(2x+3)=\frac{-8\pm \sqrt{20} }{2}

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(2x+3)=-4\pm \sqrt{5}

(2x+3)=-4-\sqrt{5} or (2x+3)=-4+ \sqrt{5}

2x=-3-4-\sqrt{5} or 2x=-3-4+ \sqrt{5}

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x=\frac{-7-\sqrt{5}}{2} or x=\frac{-7+ \sqrt{5}}{2}

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