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Elan Coil [88]
3 years ago
9

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

Mathematics
1 answer:
My name is Ann [436]3 years ago
4 0

Answer:

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.

 

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It is handy to know your times tables.

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Step-by-step explanation:

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(a) The Leading coefficient test for polynomial functions
The test is as follows.  The sign of the leading coefficient (LC, coefficient of the highest power of the variable) together with the degree (odd or even)  of the polynomial function tells us the end behaviour of the function according to the following table:

LC/degree     Even       Odd
Positive        +inf/+inf    -inf/+inf   
Negative       -inf/-inf     +inf/-inf

It can be easily memorized by comparing with a line or a parabola
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y=x^2 (positive leading coefficient, even degree)  goes from +inf to +inf.
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For the given polynomial function
f(x) = x^2(x+2) = x^3+x^2
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(b) x-intercepts
Since f(x)=x^2(x+2), the x-intercepts are x=0 (multiplicity 2) and x=-2.
=>
From the left, graph is increasing and crosses x=-2 once from negative to positive, after which graph is decreasing.  At x=0, because of the root with multiplicity 2, graph touches the origin (0,0) and turns back towards +inf.

(c) the y-intercept
from the polynomial f(x)=x^2(x+2)=x^3+x^2 = x^3+x^2+0,
we see that the y-intercept =0 because the "initial value" or the constant term is zero.

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