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Oliga [24]
3 years ago
6

A vehicle travels at a steady rate of 35 miles per hour. A student will graph the total distance, D, as a function of the time i

n hours, h, showing the distance for 1 to 3 hours. If the student counts by 10's, what should be the largest value on the D-axis of his or her graph so that the scale is the most appropriate for the given information?

Mathematics
2 answers:
adelina 88 [10]3 years ago
7 0

Solution:

Speed of Vehicle which travels at steady rate = 35 miles per hour

As we know , Distance = Speed × Time

For , Time = 1 Hours

Distance traveled = 35 miles per hour × 1 Hour = 35 miles

Time =  2 Hours

Distance traveled = 35 miles per hour × 2 Hour = 70 miles

Time = 3 Hours

Distance traveled = 35 miles per hour × 3 Hour = 105 miles

So points which are represented as ⇒ (Distance , Time) is (1,35),(2,70), (3,105).

We can find the equation of line using two point formula.

\frac{y-35}{x-1}=\frac{70-35}{2-1}\\\\y-35=35(x-1) \\\\y - 35 = 35 x -35 \\\\ y = 35 x

For , T = 10's = \frac{10}{60} = \frac{1}{6} Hour

y = 35 \times \frac{1}{6}= 6 miles (approximately)

1 unit = 6 miles (on D axis, i.e Y axis.)

To locate 105 on D axis , Largest value on D axis= \frac{105}{6}=17.5 Units

The largest value on the D-axis of his or her graph so that the scale is the most appropriate for the given information is 17.5 i.e Y axis.


scZoUnD [109]3 years ago
6 0
110 is the answer i got it wrong on my test and that was it.
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A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
bonufazy [111]

Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve ...

A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50

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p = 300

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q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

___

The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.

6 0
3 years ago
Read 2 more answers
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