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3 years ago

Jason can travel 24 3/4 miles in 1/2 hour. What is his average speed in miles per hour?

Mathematics

2 answers:

Sunny_sXe [5.5K]3 years ago

4 0

49.5 mph
just multiply the miles by 2 because it is only half an hour

almond37 [142]3 years ago

4 0

Answer:

24 3/4 miles is the distance

1/2 hour - is time

V=S/t

24\frac{3}{4}

/ 0.5 = \frac{(24*4+3)2}{4}

(24∗4+3)2

=49.5 miles per hour

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Vinvika [58]

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C. -7x⁸ + 5x⁷ - 11x + 2

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4 years ago

It is estimated that t-weeks into a semester, the average amount of sleep a college math student gets per day S(t) decreases at

goldenfox [79]

Answer:

s(15)= $\frac{-3}{2} e^{15^{2} } +9.6$

Step-by-step explanation:

Before to getting started ,we have to consider the function which represents sleep decline rate S'(t) = $-3te^{t^{2} } \\$

Since this function is interpreted in this case as a rate of change, we can deduce this funcion is the first derivative of the function of average amount of sleep (S(t)).

In order to find S(t), we will integrate S'(t)

$\int\limits^ {} -3t3e^{t^{2} } \, dx \\$

This case corresponds to intergration by substitution

Substituting u= $x^{2} \\$

Computting the differential of u

$\frac{du}{dt} =2tdt\\$

Now, let's go back to our integral and write it in terms of u, using both definition of u and its defferential.

$\int\limits^{}-3te^{u} \, \frac{1}{2t}du \\\\\int\limits^{}\frac{-3}{2} e^{u}du\\$

The integral of a exponential function is itself. So the result is:

$\frac{-3}{2} e^{u} +c$

Now substitute u back in:

$\frac{-3}{2} e^{t^{2} } +c$

This answer correspond to function of average amount of sleep.

We know that when the semester begins (t=0) , math students sleep an average of 8.1 hours per day so we will use this info in order to find the value of c.

$s(0)=\frac{-3}{2} e^{0^{2} } +c=8.1\\\\\frac{-3}{2} +c=8.1$