30.2 is correct according to PEMDAS
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
Given:
A space shuttle can travel at 28968 km/h.
To find:
The distance it can cover in 8 hrs.
Solution:
We have,
Speed of space shuttle = 28968 km/h.
So,
Distance covered by the space shuttle in 1 hour = 28968 km/h.
Distance covered by the space shuttle in 8 hour = (28968 × 8) km
= 231744 km
Therefore, the distance it can cover in 8 hrs is 231744 km.
Answer: 8x + 40
Step-by-step explanation:
To distribute, we multiply 8 by both of the values inside the parentheses;
Given:
8(x + 5)
Distribute:
(8 * x) + (8 * 5)
Multiply and simplify:
8x + 40
<u>Answer:</u>
The variable that has the highest power is considered to be the degree of polynomials in an algebraic equation.
A column:
1) 
.
The degree is 3.
2)
.
The degree is 2.
3) 
The degree is 1.
B column:
1) 
The degree is 2.
2) 
The degree is 2.
3) 
The degree is 3.
A×B columns:
While Multiplying two terms in a equation, if the variables are same then multiply the constant value and sum the exponent value.
1) 
.
=
The degree is 5.
2) 
.
=
.
The degree is 4.
3) 
.
The degree is 4.
