Answer:
Step-by-step explanation:
a. Draw a direction field for the given differential equation
b. Based on the inspection of the direction field, describe ow solutions behave for large t.
The solution appear oscillatory
All solutions seems to converge to the function y0(t)=4
All solutions seems to converge to the function y0(t)=0
All solutions seems to seems to eventually have negative slopes a and hence decrease without bound
All solutions seems to seems to eventually have positive slopes a and hence increase without bound
C
As t-infinity
All solutions seems to seems to eventually have positive slopes a and hence decrease without bound
All solutions seems to converge to the function y0(t)=0
All solutions seems to seems to eventually have negative slopes a and hence decrease without bound
All solutions seems to converge to the function y0(t)=4
The solution are oscillatory
Using the identity:
, we get:
There are two solutions to this equation:
1) Since the period of cosine is 2π, so 0 + 2π = 2π will also be a solution to the given equation
2) Therefore, there are 3 solutions to the given trigonometric equation.
x^2 - 4
------------------------------
x + 7 | x^3 + 7x^2 - 4x - 28
x^3 + 7x^2
------------------
0 - 4x - 28
-4x - 28
-------------
0
Now we know that
x^3 + 7x^2 - 4x - 28 = (x^2 - 4)(x + 7)
x^2 - 4 can be factored since it is the difference of two squares.
x^2 - 4 = (x + 2)(x - 2)
The complete factoring is:
x^3 + 7x^2 - 4x - 28 = (x +2)(x - 2)(x + 7)
Answer:
∝ = 60.50°
Step-by-step explanation:
∡VAM = ∝
Answer:
15, 6.4, 13
Step-by-step explanation:
That is the most simple way I could answer it for you.
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