Question

help

Answer 1

@Vocaloid

Answer 2

@Vocaloid

Answer 3

Please post your question

Answer 4

i plotted yesterday but they all look the same

Answer 5

?

Answer 6

that's what i got but 12 is the period i know that

Answer 7

judging by this graph - the period is 12 and the amplitude is about 4, so we can eliminate C and D

let's try and think about the horizontal shift

Answer 8

im thinking b

Answer 9

a looked correct when i graphed it

Answer 10

hm - if we change the function to cos this is what we get

Answer 11

notice how it starts decreasing from 0 instead of increasing like the actual function

Answer 12

okay could you help with more this topic is kind of confusing

Answer 13

@Vocaloid

Answer 14

@sillybilly123

Answer 15

is it a

Answer 16

it's a negative cosine function

with some shifting up the y - axis

Answer 17

@sillybilly123

Answer 18

@sillybilly123

Answer 19

so this looks like a negative sine function



you just gotta roll with it

Answer 20

period is: trough to trough; or peak to peak. which is about 10

amplitude is sorta 47 - (-4)


seems to have a period of about 10

so you just roll from there

Answer 21

well that is still all choices that follow that

Answer 22

@sillybilly123

Answer 23

not really

it is in the form \(y = - A \sin (kx) + y_o\)

and \(y_o \approx 26\)

So

\(y - 26 = - A \sin (kx) \)

And then you can add in an amplitude:

\(y - 26 = - (44 - 4) \sin (kx) \)

for k ,you have \(k = \dfrac{2 \pi}{\lambda}\)

And: \(\lambda \approx 10\), which was the approximation mentioned above


etc