The populations of termites and spiders in a certain house are growing exponentially. the house contains 120 termites the day you move in. after four days, the house contains 210 termites. three days after moving in, there are two times as many termites as spiders. eight days after moving in, there were four times as many termites as spiders. how long (in days) does it take the population of spiders to triple? (round your answer to one decimal place.)

in order to triple the inicial population of spiders, will take 50395 days

Step-by-step explanation:

we can define the termite population function as T(t) and the one for spiders as S(t) , where t represents time measured in days

since both have and exponencial growth

T(t)= a*e^(b*t)

S(t)= c*e^(d*t)

1) when the day the person moves in , t=0 and T(0)= 120 termites

T(0) = a*e^(b*0) = a = 120

2) after 4 days , t=4 and  the house contains T(4) = 210 termites

T(4)= 120*e^(b*4) = 210 → 4*b = ln (210/120) → b = (1/4)* ln(210/120)= 0.14

therefore

T(t) = 120*e^(0.14*t)

3) 3 days after moving in , t=3, there were T(3) = 120*e^(0.14*3)=182.63≈ 182 termites . The number of spiders is half of the number of termites → S(3) = T(3) * 1 spider/ 2 termites  =91.31 spiders ≈ 91 spiders

4) after 8 days of moving in , t=8, there were T(8) = 120*e^(0.14*8)=367.78≈ 368 termites . The number of spiders is 0.25 times the number of termites → S(8) = T(8) * 1 spider/ 4 termites =91.94 spiders   ≈ 92 spiders

from

S(t)= c*e^(d*t) → d = ln [S (tb)/S (ta) ] / (tb-ta)

therefore d = ln [ S(8)/S(3) ] / (8 - 3 ) = 2.18*10^-5

in order to triple the initial population

S(t3) = 3 *S(0) = 3*[c*e^(d*0)] = 3*c

S(t3) = c*e^(d*t3) = 3* c → t3 = ln(3) / d = 50395 days

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: