CAN SOMEONE PLEASE ADD ALL OF THIS FOR ME.?

Answer: B. same distance

There is the same distance between the point on the preimage to the center of rotation and that of the distance from the corresponding point on the image to the center of rotation. There is change in the distance of the two.

8 0

2 years ago

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A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position

pantera1 [17]

Answer:

233.48s

3.84 min

Step-by-step explanation:

In order to solve this problem, we can start by drawing what the situation looks like. See attached picture.

We can model this situation by making use of a trigonometric function. Trigonometric functions have the following shape:

$y=A cos(\omega t+\phi)+C$

where:

A= amplitude =-20m because the model starts at the lowest point of the trajectory.

f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.

t= time

$\omega=$ angular speed.

in this case:

$\omega=\frac{2\pi}{T}$

where T is the period, in this case 6 min or

$6min(\frac{60s}{1min})=360s$

so:

$\omega=\frac{2\pi}{360}$

$\omega = \frac{\pi}{160}$

and

$\phi$ = phase angle

C= vertical shift

in this case our vertical shift will be:

2m+20m=22m

in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:

$y=-20 cos(\frac{\pi}{180}t)+22$

Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:

$-20 cos(\frac{\pi}{180}t)+22>13$

so we can solve this inequality, we can start by turning it into an equation we can solve for t:

$-20 cos(\frac{\pi}{180}t)+22=13$

and solve it:

$-20 cos(\frac{\pi}{180}t)=13-22$

$-20 cos(\frac{\pi}{180}t)=-9$

$cos(\frac{\pi}{180}t)=\frac{9}{20}$

and we can take the inverse of cos to get:

$\frac{\pi}{180}t=cos^{-1}(\frac{9}{20})$

which yields two possible answers: (see attached picture)

so

$\frac{\pi}{180}t=1.104$ or $\frac{\pi}{180}t=5.179$

so we can solve the two equations. Let's start with the first one:

$\frac{\pi}{180}t=1.104$

$t =1.104(\frac{180}{\pi})$

t=63.25s

and the second one:

$\frac{\pi}{180}t=5.179$

$t=5.179(\frac{180}{\pi})$

t=296.73s

so now we can build our possible intervals we can use to test the inequality:

[0, 63.25] for a test value of 1

[63.25,296.73] for a test value of 70

[296.73, 360] for a test value of 300

let's test the first interval:

[0, 63.25] for a test value of 1

$-20 cos(\frac{\pi}{180}(1))+22>13$

2>13 this is false

let's now test the second interval:

[63.25,296.73] for a test value of 70

$-20 cos(\frac{\pi}{180}(70))+22>13$

15.16>13 this is true

and finally the third interval:

[296.73, 360] for a test value of 300

$-20 cos(\frac{\pi}{180}(300))+22>13$

12>13 this is false.

We only got one true outcome which belonged to the second interval:

[63.25,296.73]

so the total time spent above a height of 13m will be:

196.73-63.25=233.48s

which is the same as:

$233.48(\frac{1min}{60s})=3.84 min$

see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.

3 0

2 years ago