~For this problem, you need to divide.
~The easy way that I think might help you is if you change the fraction into a decimal.
Now divide $8.47 by 0.875
The price per pound of the coffee was $9.68
Answer:
It would be the side opposite the smallest angle.
Side DE (which is opposite angle F) is the shortest.
Step-by-step explanation:
Using derivatives, it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval:
(-15,-10).
<h3>What is the slope of the tangent line to a function f(x) at point x = x0?</h3>
It is given by the derivative at x = x0, that is:
.
In this problem, the function is:
Hence the derivative is:
For a slope of -1, we have that:
0.4x + 5 = -1
0.4x = -6
x = -15.
For a slope of 1, we have that:
0.4x + 5 = 1.
0.4x = -4
x = -10
Hence the interval is:
(-15,-10).
More can be learned about derivatives and tangent lines at brainly.com/question/8174665
#SPJ1
I need question for the "factor" problem
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
