Question

The number of people in a town of 10,000 who have heard a rumor started by a small group of people is given by the following function:



a. How many people were in the group that started the rumor?





b. How many people have heard the rumor after 1 day? After 5 days?





c. How long will it be until 1,000 of the people in the town have heard the rumor?

WILL GIVE MEDAL!!! Need help asap though!

Answer 1

Where is the "following function"?..

Answer 2

N(t) = 10,000/ 5 + 1245e ^ -0.97t

Answer 3

that is the function :)

Answer 4

Okay.
Let's start with A.
when t = 0 .

Answer 5

evaluate it for when t = 0
you get.
n(0) = 10000/(5+1245(e^-0.97(0) = 10000/1250
= 8
so 8 people.

Answer 6

All you have to do is substitute for the value of t , and evaluate the functions. For b.

Answer 7

ok

Answer 8

Now, for C.
You want to know when 1000 people have heard the rumor..so substitute N with 1000 and solve for t.

Answer 9

1000 = 10000/(5+1245e^-.97t)
1/10 = 1/(5+1245e^-.97t)
5 + 1245e^-.97t = 10
1245e^.97t = 5
divide both sides..
e^-.97t =.0040
take the natural log of both sides.
-.97t = ln .004
simplify.
t = 5.7

Answer 10

about 5 days
ok so 5.7 weeks?

Answer 11

and for b I get 2471 for 1 day and 4359.79 for b

Answer 12

t = 5.7 so about 6 days.

Answer 13

after 1 day 21 people will hear the rumor,
after 5 days bout 678 people will hear the rumor...

Answer 14

for 5 days day I get 678
and and 21 for 1 day

Answer 15

right sorry I made an error in my calculator

Answer 16

yeah. good job :)

Answer 17

thanks can you help me on one more problem?

Answer 18

Solve each equation for x: 5 = log3(x^2 + 18) and -2 = log(3x+5)

Answer 19

5 = log 3(x^2+18)
10^-5 = 3(x^2+18)
does that help?

Answer 20

-2 = log (3x+5)
10^-2 = 3x + 5
just solve for x now :)

Answer 21

ok I think I have the answers could you post both for me to double check?

Answer 22

I don't have the first one done, too lazy to do a quadratic..
but for the second one x = -1.663

Answer 23

ok I got that one too :)

Answer 24

can u answer this last one?

Answer 25

e^x e^(x+1) = 1

Answer 26

\[e^x e^{x+1}=1\]
not sure how to do this one sorry.

Answer 27

but im pretty sure it involves using the natural log >.<

Answer 28

ok thanks