There are 96 people consisting of 48 married couples in a room. Assuming that no husband or wife shakes thr others hand but everyone else shakes hands exactly once, how many handshakes are there?

4512 handshakes

Step-by-step explanation:

We are told in the question, that there are

96 people in the room = 48 couples

Assuming that no husband or wife shakes each others hand but everyone else shakes hands exactly once

This means

48 couples will shake 94 people because they are excluded

This is calculated as

48 × 94 = 4512 handshakes

k = 1

P(t) = (t + 20)²

Step-by-step explanation:

P' = k√P = k P⁰•⁵

To solve for k,

P' = 10 rabbits/month

P = 100 rabbits

10 = k √100

10 = 10k

k = 1

To solve for P(t)

P' = dP/dt

(dP/dt) = kP⁰•⁵

dP/P⁰•⁵ = k dt

P⁻⁰•⁵ dP = k dt

∫ P⁻⁰•⁵ dP = ∫ k dt

2P⁰•⁵ = kt + c

At t = 0, P = 100

2(100)⁰•⁵ = 0 + c

2 × 10 = c

c = 20

2P⁰•⁵ = kt + c

2P⁰•⁵ = kt + 20

Recall, k = 1

2P⁰•⁵ = t + 20

P⁰•⁵ = (t + 20)

P(t) = (t + 20)²