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April 1, 2015

April 1, 2015

Posted by **SC** on Thursday, November 5, 2009 at 8:17pm.

What I done so far.

TERMITES

y(not)=100

t=4

y=200

200=100b^4

b=2^(1/4)

T(t)=100(1.189207)^t

for the spiders.S(t).. i tried plugging t=3 and t=8and divided by two and four to the T(x) equation.... and for two numbers 84.089 and 99.99..... but I think Im doing it wrong

or is it

T(t) = 2(S(t))?

- Math -
**TeacherJim**, Thursday, November 5, 2009 at 8:47pmI agree with your first part: b=2^(1/4) = 1.189207.

Now, after 3 days, according to the question, we must have 84.089 spiders - wonder what .089 of a spider looks like? - and after 8 days, we have 100.

If S0 is the number of spiders on day zero, and s is the multiplication rate of spiders, then

S0 * s^3 = 84.089

S0 * s^8 = 100

From which we can quickly get s^5.

And then we get s.

And from there, there's not much left to do, I think.

- Math -
**SC**, Thursday, November 5, 2009 at 8:57pmI made a mistake. after eight days , its 199.99 spiders.

I still don't understand how would we get s?

- Math -
**TeacherJim**, Thursday, November 5, 2009 at 9:07pmYou didn't make a mistake:

"Eight days after moving in, there were four times as many termites as spiders"

But 100 * (2^.25)^8 = 400 (termites)

So if there were 400 termites, there were 100 spiders.

So how do we get s?

S0 * s^3 = 84.089

S0 * s^8 = 100

(S0 * s^8) / (S0 * s^3)

= s^5

= 100 / 84.089

s = (100 / 84.089)^(1/5)

- Math -
**SC**, Thursday, November 5, 2009 at 9:10pmOhh all i needed to do was divide! heh thanks a lot!

- Math -
**TeacherJim**, Thursday, November 5, 2009 at 9:13pmYou're welcome!

- Math -
**nissreen**, Thursday, January 10, 2013 at 7:43pmyou guys never gave an answer..

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