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)
__
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.
__
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.
__
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
Answer:
15 games
Step-by-step explanation:
12X3=36 5X3=15
-7 + 10 = 3
=
10 + (-7)
Is the same as
10 - 8
…
Not sure if this was that helpful.. I hope it did.
Answer: 72 u^2
<h3>
Explanation:</h3>
What we know:
- Both triangles are identical
- Both rectangles are different
- There are values in units^2 given
- There are right angles
How to solve:
We need to find the area of at least one of the triangles and double it. Then, we need to find the areas of both rectangles. Finally, we need to add these areas to find the total area. The final area will be represented in units squared (u^2)
<h2>
Process:</h2>
Triangles
Set up equation A = 1/2(bh)
Substitute A = 1/2(4*3)
Simplify A = 1/2(12)
Solve A = 6
Double *2
A = 12 u^2
Rectangles
Set up equation A = lh
Substitute A = (14)(3)
Simplify A = 42 u^2
Set up equation A = lh
Substitute A = [14-(4+4)](3)
Simplify A = (14-8)(3)
Simplify A = (6)(3)
Multiply A = 18 u^2
Total Area
Set up equation A = R1+R2+T
Substitute A = 42 + 18 + 12
Simplify A = 60 + 12
Solve A = 72 u^2
<h3>
Answer: 72 u^2</h3>
