Question

help? @jhonyy9

Answer 1

a) divide by 7
b) take the arccosine of both sids
c) multiply both sides by \(\frac{3}{\pi}\)

Answer 2

Okay so the formula is h=7cos(pi/3 t) what am I dividing by 7?

Answer 3

both sides

Answer 4

you are solving for \(t\) right? get rid of that seven out front first by writing \[\frac{h}{7}=\cos(\frac{\pi}{3}t)\]

Answer 5

Yes!

Answer 6

you are not "taking away" the seven, you are dividing both sides by seven

Answer 7

@Michele_Laino

Answer 8

first step:
I divide both sides by \(7\), so I get:
\[\frac{h}{7} = \frac{7}{7}\cos \left( {\frac{\pi }{3}t} \right)\]
please simplify

Answer 9

what is \(7/7=...?\)

Answer 10

\[\frac{h}{7} = \cos \left( {\frac{\pi }{3}t} \right)\]
is it right?

Answer 11

7/7=1

Answer 12

ok!
now in order to simplify further, I have to apply the inverse function of cosine, to both sides, so I get:
\[{\cos ^{ - 1}}\left( {\frac{h}{7}} \right) = {\cos ^{ - 1}}\left\{ {\cos \left( {\frac{\pi }{3}t} \right)} \right\} = \frac{\pi }{3}t\]

Answer 13

Okay, yeah I know what inverse is.

Answer 14

finally I multiply both sides of last equation by \(3/\pi\), and I get:
\[\left\{ {{{\cos }^{ - 1}}\left( {\frac{h}{7}} \right)} \right\}\left( {\frac{3}{\pi }} \right) = \left( {\frac{\pi }{3}} \right)\left( {\frac{3}{\pi }} \right)t\]

Answer 15

and after a simplification, I write this:
\[\left\{ {{{\cos }^{ - 1}}\left( {\frac{h}{7}} \right)} \right\}\left( {\frac{3}{\pi }} \right) = t\]

Answer 16

or:
\[t = \left\{ {{{\cos }^{ - 1}}\left( {\frac{h}{7}} \right)} \right\}\left( {\frac{3}{\pi }} \right)\]

Answer 17

Now how would we apply part b?

Answer 18

we have to solve this equation:
\[t = \left\{ {{{\cos }^{ - 1}}\left( {\frac{1}{7}} \right)} \right\}\left( {\frac{3}{\pi }} \right) = ...\]
namely I have substituted \(h=1\) cm
please solve for \(t\)

Answer 19

we have to solve the computation, using the radians, namely you have to set your calculator on "rad" mode

Answer 20

I still got the same answers when I changed it...? I dont get it.

Answer 21

If I use "rad" mode, I get:
\(t=1.36\) approximated value

Answer 22

Yeah I know.

Answer 23

if you change from "deg" to "rad" also the answer changes

Answer 24

next, you have to repeat such computation, using \(h=3\) and \(h=5\) cm

Answer 25

for example, if \(h=3\) cm, I get:
\[{t_0} = \left\{ {{{\cos }^{ - 1}}\left( {\frac{3}{7}} \right)} \right\}\left( {\frac{3}{\pi }} \right) = ...\]

Answer 26

please wait a moment

Answer 27

one second

Answer 28

So I can put those decimals with approx behind it?!

Answer 29

at \(t=0\) the weight is at \(h=7\) cm, since if I substitute \(t=0\), I get:
\[h = 7\cos \left( {\frac{\pi }{3} \cdot 0} \right) = 7 \cdot 1 = 7\]