Let the bushels of wheat is b and weight of the wheat is w.
We can say that more the bushels of wheat more will be the weight of the wheat.
Hence, the quantities vary directly.
Therefore, we have
, where k is the constant of variation.
Now, we have been given that 5 bushels of wheat weigh 136 kg. Thus, we have
Thus, the constant of variation is
Now, we have been given 3.5 bushels of wheat. Hence, we have
Therefore, 3.5 bushels of wheat weigh 95.2 kg
Answer:
a) y = 0, y = 4 and y = 5
b) y ⊂ (- ∞, 0) ∪ (0, 4) ∪ (5, ∞)
c) y ⊂ (4,5)
Step-by-step explanation:
Data provided in the question:
function y(t) satisfies the differential equation:
= y⁴ − 9y³ + 20y²
Now,
a) For constant solution
= 0
or
y⁴ − 9y³ + 20y² = 0
or
y² (y² - 9y + 20 ) = 0
or
y²(y² -4y - 5y + 20) = 0
or
y²( y(y - 4) -5(y - 4)) = 0
or
y²(y - 4)(y - 5) = 0
therefore, solutions are
y = 0, y = 4 and y = 5
b) for y increasing
> 0
or
y²(y - 4)(y - 5) > 0
or
y²
y ⊂ (- ∞, 0) ∪ (0, 4) ∪ (5, ∞)
c) for y decreasing
< 0
or
y²(y - 4)(y - 5) > 0
or
y²
y ⊂ (4,5)