Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates throughout
1 answer:
Based on the weight and the model that is given, it should be noted that W(t) in radians will be W(t) = 0.9cos(2πt/366) + 8.2.
<h3>How to calculate the radian.</h3>
From the information, W(t) = a cos(bt) + d. Firstly, 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.
Then 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. Then calculate a, the dog's maximum weight is 9.1 kg. The deviation from the average is 9.1 kg - 8.2 kg = 0.9 kg. W(t) = 0.9cost + 8.2
Lastly, calculate t. The period p = 2π/b = 2π/1 = 2π. The conversion factor is 1 da =2π/365 rad. Therefore, the function with t in radians is W(t) = 0.9cos(2πt/365) + 8.2.
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Answer:
(3 + 4x)(9 - 12x + 16x²) = 0
(4m - 1)(16m² + 4m + 1) = 0
Step-by-step explanation:
Here we have to solve the sum of two cubes which is
Now, the equation is
⇒ 3³ + (4x)³ = 0
⇒ (3 + 4x)[3² - 3(4x) + (4x)²] = 0
⇒ (3 + 4x)(9 - 12x + 16x²) = 0
So, (3 + 4x) = 0 or (9 - 12x + 16x²) = 0
Therefore, from the above two relation we can solve for x.
One root will be and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.
Again, we have to solve the difference of two cubes which is
Now, the equation is
⇒ (4m)³ - 1³ = 0
⇒ (4m - 1)[(4m)² + 4m(1) + 1²] = 0
⇒ (4m - 1)(16m² + 4m + 1) = 0
So, (4m - 1) = 0 or (16m² + 4m + 1) = 0
Therefore, from the above two relation we can solve for m.
One root will be and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation. (Answer)