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.

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation:

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.