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daser333 [38]
3 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]3 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]3 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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The temperature at 6 pm was 0f. At 10 pm the temperature was -11.2f. Write an expression that you can use to find the average ch
Mariana [72]

Answer:

Average change in temperature is - 2.8 Fahrenheit per hour.

Step-by-step explanation:

Average change in temperature is the change in temperature divided by the time taken for the change.

Change in temperature = (-11.2) F- 0 F

                                       = -11.2 F

Time taken for the change in temperature = 10 pm - 6 pm

                                                                       = 4 hours

Average change in temperature = Change in temperature / time taken for the change

                                                      = \frac{-11.2}{4}

                                                      = - 2.8 Fahrenheit per hour

3 0
3 years ago
Write the equation of the line in fully simplified slope-intercept form. ​
Anettt [7]

Answer: y= -3/2 -5

explanation:

The slope is 3/-2, and it slopes down. making it a negative slope/fraction

And the point is on -5 on the Y axis.

Making the answer y= -3/2-5

3 0
3 years ago
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Jade is going out for pizza with her friend.For 60 dollars how many pizzas can she buy if each pizza costs 12 dollars
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Answer:

jade can by 5 pizzas

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2 years ago
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What is t'(x) when t(x) = In<br> x2<br> 2x2+3x+2<br> ?
Korvikt [17]

Step-by-step explanation:

2-×^2

t(×)=3×

(s o t )(x)=st(×)=s(t(x))=2-(3×)^2=2-9x^2

(s o t)(-7) =2-9(-7)^2=2-9(49)= 2-441= -439

7 0
3 years ago
Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
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