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2 years ago

For each explicitly-defined sequence below, write the first few terms. Then use the terms to write a recursive equation.

Mathematics

1 answer:

julia-pushkina [17]2 years ago

6 0

Answer:

Step-by-step explanation:

Problem A

t(1) = 2(1) + 5

t(2) = 2*2 + 5 = 9

t(3) = 2*3 + 5 = 11

t(4) = 2*4 + 5 = 13

So this is the explicit result. Now try it recursively.

t_3 = t_2 + 2

t_3 = 9 + 2

t_3 = 11 which is just what it should do.

t_n = t_(n - 1) + 2

Problem B

t(1) = 3 * 1/2

t(1) = 3/2

t(2) = 3*(1/2)^2

t(2) = 3 * 1/4

t(2) = 3/4

t(3) = 3*(1/2)^3

t(3) = 3 * 1/8

t(3) = 3/8

t(4) = 3 (1/2)^4

t(4) = 3 (1/16)

t(4) = 3/16

So in general

t_n = t_n-1 * 1/2

For example t(5)

t_5 = t_4 * 1/2

t_5 = 3 /16 * 1/2 = 3/32

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2 years ago

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)

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

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2 years ago