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
Answer: 
Step-by-step explanation:
First, we divide the composite figure into two distinct geometric shapes. This would be a triangle and a square. To find the triangle we use the formula:
A = 1/2bh
To find the base of the triangle we can add:
9 + 9 + 18 = 36
Now we plug in and solve:
A = 1/2bh
A = 1/2(36)(27)
= 1/2(972)
= 486
To find the square, we can use length x width:
18×18 = 324
Now, we add the sums together:
486 + 324 = 810
Answer:
The multiple zero is 
.
The multiplicity is 2
Step-by-step explanation:
The given function is
To find the zero of this function, we equate it to zero.
We factor to obtain;
Observe that; the expression within the parenthesis is a perfect square.
This implies that;
This implies that;
Either 
or 
has a multiple zero which is
and its multplicity is 2.
Answer:
x = -4.604
Step-by-step explanation:
Here, we want to get the value of x
We start by dividing through by 5
e^(2x + 11) = 30/5
e^(2x + 11) = 6
Writing this in natural logarithm form, we have;
ln 6 = 2x + 11
2x + 11 = 1.792
2x = 1.792-11
2x = -9.208
x = -9.208/2
x = -4.604
POH is the -log [ OH- ]. Using this equation, simply find the -log of the OH- concentration to find the pOH:
pOH = - log ( 0.000086 ) = 4.07
