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Shalnov [3]
3 years ago
5

What are the roots of 4x2 + 35x − 9 = 0?

Mathematics
1 answer:
ElenaW [278]3 years ago
4 0
X= - 1/4 = -0.250 OR x=9
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PLEASE HELP! WILL MARK BRAINLIEST!
makkiz [27]

Answer:

The maximum height a rider will experience is 55 feet.

Step-by-step explanation:

Let's start writing the function that defines the path of a seat on the new Ferris wheel. This function will depend of the variable ''t'' which is time.

X(t)=(x,y)

In which X(t) are the coordinates of the seat (the x - coordinate and the y - coordinate) that depend from time.

''x'' and ''y'' are functions that depend from the variable ''t''.

For this exercise :

X(t)=[-25sin(\frac{\pi}{30}t);-25cos(\frac{\pi}{30}t)+30]

In order to find the maximum height a rider will experience we will study the behaviour of the y - component from the function X(t).

The function to study is y(t)=-25cos(\frac{\pi}{30}t)+30

To find its maximum, we will derivate this function and equalize it to 0. Doing this, we will find the ''critical points'' from the function.

⇒ y(t)=-25cos(\frac{\pi}{30}t)+30  ⇒

y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t)

Now we equalize y'(t) to 0 ⇒

y'(t)=0 ⇒ \frac{5}{6}\pi sin(\frac{\pi}{30}t)=0

In this case it is easier to look for the values of ''t'' that verify :

sin(\frac{\pi}{30}t)=0

Now we need to find the values of ''t''. We know that :

sin(0)=0\\\\sin(\pi)=0\\sin(-\pi)=0

Therefore we can write the following equivalent equations :

\frac{\pi}{30}t=0 (I)

\frac{\pi}{30}t=\pi (II)

\frac{\pi}{30}t=-\pi (III)

From (I) we obtain t_{1}=0

From (II) we obtain t_{2}=30

And finally from (III) we obtain t_{3}=-30

We found the three critical points of y(t). To see if they are either maximum or minimum we will use the second derivative test. Let's calculate the second derivate of y(t) :

y'(t)=\frac{5}{6}\pi sin(\frac{\pi}{30}t) ⇒

y''(t)=\frac{\pi ^{2}}{36}cos(\frac{\pi }{30}t)

Now given that we have an arbitrary critical point ''t_{n}'' ⇒

If y''(t_{n})>0  then we will have a minimun at t_{n}

If y''(t_{n}) then we will have a maximum at t_{n}

Using the second derivative test with t_{1},t_{2} and t_{3} ⇒

y''(t_{1})=y''(0)=\frac{\pi ^{2}}{36} >0 ⇒ We have a minimum for t_{1}=0

y''(t_{2})=y''(30)=\frac{-\pi^{2}}{36} ⇒ We have a maximum for t_{2}=30

y''(t_{3})=y''(-30)=\frac{-\pi^{2}}{36} ⇒ We have a maximum for t_{3}=-30

The last step for this exercise will be to find the values of the maximums.

We can do this by replacing in the equation of y(t) the critical points t_{2} and t_{3} ⇒

y(t_{2})=y(30)=55

y(t_{3})=y(-30)=55

We found out that the maximum height a rider will experience is 55 feet.

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3 years ago
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