Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates

throughout the year.

The dog's weight W(t) (in kg) as a function of time t (in days) over the course of a year can be modeled by

a sinusoidal expression of the form a. cos(b. t) + d.

Att = 0, the start of the year, he is at his maximum weight of 9.1 kg. One-quarter of the year later, when

t = 91.25, he is at his average weight of 8.2 kg.

Find W(t).

t should be in radians.

W(t) =

W(t) = 0.9cos(2πt/366) + 8.2  

Step-by-step explanation:

W(t) = a cos(bt) + d

1. Calculate the phase shift, b

At t= 0, the dog is at maximum weight, so the cosine function is also at a maximum.

The cosine function is not shifted, so b = 1.

W(t) = a cos t + d

2. Calculate d

The dog's average weight is 8.2 kg, so the mid-line d = 8.2.

W(t) = a cos t + 8.2  

3. Calculate a

The dog's maximum weight is 9.1 kg.

The deviation from the average (the amplitude, a) is 9.1 kg - 8.2 kg = 0.9 kg.

W(t) = 0.9cos t  + 8.2

3. Calculate t

The period p = 2π/b = 2π/1 = 2π

From t = 0 to t = 91.25 da is one-quarter of a period, so

p = 4 × 91.25 da = 365 da = 2π rad

The conversion factor is 1 da =2π/365 rad

The function with t in radians is

W(t) = 0.9cos(2πt/365) + 8.2

The figure below shows the graph of the function.

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40

step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.