Please help will fan and medal!

- love_to_love_you

Please help will fan and medal!

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- love_to_love_you

h of the following functions represents a geometric sequence? Why?
a. The function f(x) = x^3 represents a geometric sequence because each term is cubed to make it greater.
b. The function f(x) = 4x represents a geometric sequence because each term
is the next higher multiple of 4.
c. The function f(x) = 4x- 4 represents a geometric sequence because each term is 4 more than the previous term.
d. The function f(x) = 4^x represents a geometric sequence because each term is 4 times as great as the previous term.

- anonymous

soo what do u know about the sequence

- anonymous

@jamesr lol she's off line

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## More answers

- anonymous

ok less work 4 me

- anonymous

@jamesr lol she's back

- love_to_love_you

Sorry my computer died

- love_to_love_you

Is it a?

- SolomonZelman

ok, i am back had to unload some packages....

- SolomonZelman

An exponental function represents a geometric sequence.
An exponential function is an form of:
\(y=b^x\) or \(y=a(b)^x\)

- love_to_love_you

so it is A?

- anonymous

-100 is the y intercept and if you change it 25 it would go up
i believe it is a

- SolomonZelman

it is not A.

- SolomonZelman

An exponental function represents a geometric sequence.
An exponential function is an form of:
y=b\(^x\) or y=a(b)\(^x\)
----------------------------------------------------
why is this so that exponential function represents a geometric sequencet?
A geometric sequence is a sequence that follows a pattern of multiplying.
In other words you multiply the first term \(a_1\) by common ratio \(\rm r\) to
get the the second term \(a_2\).
Then you multiply the second term \(a_2\) by common ratio \(\rm r\) to get to the
third term \(a_3\). (AND SO ON......)
For example: \(3,~~6,~~12,~~24,~~48~...\)
You can see that the sequence starts from 3, so \(a_1=3\)
And that you multiply times 2, so \(\rm r\)=2.
Now!
We can represent the same thing exponentially.... how (read this while I am typing)

- SolomonZelman

So, exponential function
y=a(b)\(^x\)
(we will refer to the same example)
we can write y=3(2)\(^{x-1}\)
and it will be equivalent.
when x=1 you get: f(1)=3(2)\(^{1-1}\) = 3(2)\(^{0}\) =3 (and that is \(a_1\))
when x=2 you get: f(2)=3(2)\(^{2-1}\) = 3(2)\(^{1}\) =3•2=6 (and that is \(a_2\))
when x=3 you get: f(3)=3(3)\(^{3-1}\) = 3(2)\(^{2}\) =3•4=12 (and that is \(a_3\))
when x=4 you get: f(3)=3(3)\(^{4-1}\) = 3(2)\(^{3}\) =3•8=24 (and that is \(a_4\))
and so forth

- love_to_love_you

So it is definitely not a. Could it be B or D?

- SolomonZelman

do you know what arithemtic sequence is?

- love_to_love_you

I'm leaning more towards b.

- SolomonZelman

do you know what an arithmetic sequence is?

- love_to_love_you

It's a sequence where the difference is constant

- SolomonZelman

yes, where you add or subtract the same number.

- anonymous

i have a feelin she guessin lol

- SolomonZelman

And this (arithmetic sequence) is represented by linea function (in a form of y=mx+b)

- SolomonZelman

linear*

- love_to_love_you

You are not being any help Jamesr so if you do not have anything helpful to say, please leave.

- SolomonZelman

So, option B
y=4x
when x=1, you get 4•1=4
when x=2, you get 4•2=8
when x=3, you get 4•3=12
when x=4, you get 4•4=16
when x=5, you get 4•5=25
so this function (for natural number x values) would represent an arithemtic sequence with the first term 4 and a common difference of 4.

- SolomonZelman

And any other function in a form of y=mx+b is an arithmetic sequence (roughly speaking, of course)

- SolomonZelman

So tell me now, is B the correct option or not?

- love_to_love_you

Not

- SolomonZelman

yes, B is not the answer.

- love_to_love_you

And I'm guessing D isn't either.

- SolomonZelman

why not?

- love_to_love_you

Sorry *C and because its a form of mx+b

- SolomonZelman

you need a function that represents a geometric sequence (not arithmetic sequence).

- love_to_love_you

So d?

- SolomonZelman

yes

- love_to_love_you

Alrighty. Thanks!

- anonymous

im sorry i was gonna help but @SolomonZelman beat me to it

- anonymous

LOL @jamesr leave if you don't have nothing good ta say

- anonymous

im sorry i am im just an donkey term some times

- anonymous

LOL u said donkey

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