Im positive its false because they dont add up
Answer:
a) 30 kangaroos in 2030
b) decreasing 8% per year
c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo
Step-by-step explanation:
We assume your equation is supposed to be ...
P(t) = 76(0.92^t)
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a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30
In the year 2030, the population of kangaroos in the province is modeled to be 30.
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b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.
The population is decreasing by 8% each year.
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c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.
P(100) = 75(0.92^100) = 76(0.0002392)
P(100) ≈ 0.0182, about 1/55th of a kangaroo
A square with the same perimeter as a rectangle will have larger area.
However, 42 is not divisible by 4, so we can do 10 by 11.
Length = 11, Width = 10, Area = 110
42 / 4 = 10.5
Length and Width = 10.5, Area = 110.25
The answer is 56.5 you take pie times the radius squared
Volume of the prism is given by:
Volume=length×width×height
but
V=<span>18x^3+5x^2-2x
V=x(18x^2+5x-2)
V=x(9x-2)(2x+1)
Hence the possible expression for the dimensions of the shipping box is x by (9x-2) by (2x+1)</span>
