Arumor spreads through a school. let y(t)y(t) be the fraction of the population that has heard the rumor at time tt and assume that the rate at which the rumor spreads is proportional to the product of the fraction yy of the population that has heard the rumor and the fraction 1−y1−y that has not yet heard the rumor.

Step-by-step explanation:

Rate the rumor spread so proportional to y∙(1 - y).

So the rate of change of y equals this rate. Therefore  

dy/dt = k∙y∙(1 - y)

That differential is a separable first order equation. Separation of variables yields:

(1/(y∙(1 - y)) dy = k dt

Hence,

∫ (1/(y∙(1 - y)) dy = ∫ k dt

Partial fraction expansion for the left hand side:

1/(y∙(1 - y) = (A/y) + (B/(1 - y))

Residue method gives

A = 1/(1 - 0) = 1

B = 1/1 = 1

∫ (1/(y∙(1 - y)) dy = ∫ k dt

<=>

∫ (1/y) + (1/(1 - y)) dy = ∫ k dt

<=>

ln(y) - ln(1 - y) = k∙t + C

(C is the constant of integration)

Since ln(a) - ln(b) = ln(a/b) you can rewrite  

ln( y/(1 - y) )= k∙t + C

To evaluate C apply the initial value

y(t=0) = 10% = 0.1

=>

ln( 0.1/(1 - 0.1) )= k∙0 + C

=>

C = ln(0.1/0.9) = ln(1/9)

So  

ln( y/(1 - y) ) = k∙t + ln(1/9)

To evaluate k, solve this equation for k and apply the given value

y = 0.4 at t = 2 days

ln( y/(1 - y) ) = k∙t + ln(1/9)

<=>

k = ( ln( y/(1 - y) ) - ln(1/9) ) / t

= ( ln( (y/(1 - y) ) / (1/9) ) / k∙t

= ( ln( 9∙y/(1 - y) ) / t

= ( ln( 9∙0.4/(1 - 0.4) ) / 2 d

= (1/2)∙ln( 6 ) d⁻¹

≈ 0.8959 d⁻¹

go on the tube you might find some information there

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation: