Answer:
We readily separate the variables and integrate:
∫dP/P=∫(k+bcos2
t)dt
ln P=kt+(b/2)*sin2t+ln C
Clearly C = Po, so we find that P(t) = Poexp(kt + (b/2)* sin 2t). The 271- curve with the typical numerical values P_o = 100, k = 0.03, and b = 0.06. It oscillates about the curve which represents natural growth with P_o and k = 0.03. We see that the two agree at the end of each full year.
note:
find the attached graph
Answer:
Yes
Step-by-step explanation:
This relation is a function. Functions must have unique x-values for every y-value, this relation meets those requirements. Also, since this is a graph you can use the vertical line test. To pass the vertical line test you must be able to draw a vertical line on the graph and have it intersect with the function in no more than one place.
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Similar polygons vary directly with each other. They have the same scale.
Something I notice about the polygons is that both of the sides are decreased by 10.
40-30= 10 and 28-18=10
Try to find the scale.
40/28= 1.42857....
30/18= 1.666666777
I would say these are not similar polygons. They do not share the same scale.
I hope this helps!
~kaikers
Answer:
9
Step-by-step explanation:
