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3 years ago

Help me pleaseeee pleaseee

Mathematics

1 answer:

murzikaleks [220]3 years ago

4 0

All your doing is placing the x numbers in the x spot than solving and just filling out the chart with the remaining number

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Solve the problem. Type the number in the box that correctly completes the sentence.

motikmotik

Answer:

5 miles

Step-by-step explanation:

Think of this like a triangle. From the bottom of the tower, to the top of the tower, to the point 3 miles away, and back to the bottom of the tower.

So we already have 2 side lengths. The height of the tower, 3 miles, and the base, 4 miles. In order to find the 3rd length, the distance from the top of the tower to the point 4 miles away from the bottom, we need to apply the formula A squared + B squared = C squared.

We have A and B, (3 and 4) and we need C.

A squared (3 squared) is 9

B squared (4 squared) is 16

so 9 + 16 = C squared

9 + 16 = 25

C squared = 25

square root of 25 is 5

C = 5

The distance from the top of the tower to the point 4 miles away is 5.

3 0

3 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.

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