Answer: We know that an "exponential growth" is that where a function start growing slowly and after start increasing more quickly.
a "logarithmic growth" is the opposite of this, at first you will see a quick increase, that after some time gets slower.
But in this case we are searching for decays: The decays are the opposite of the growths, an "logarithmic decay" starts slow and gets faster latter, and an "exponential decay" decreases very quickly at first, and latter more slowly.
Then, the model for the temperature of the hot tea over the time is a "exponential decay", this is a function of the form T(x) = T₀
, where T₀ is the initial temperature, t is the time, and k is a constant number.
Answer:
Answer D
Step-by-step explanation:
The formula is 
. We have our r (radius) and h (height), so plugging it all in would give us A = (3.14)(5 + sqrt(12^2)+(5^2). After computing this, you would get answer D, 282.6.
Answer:
Kindly check explanation
Step-by-step explanation:
Given the data:
Education : 11 11 8 13 17 11 11 11 19 13 15 9 15 15 11
Internet use 10 5 0 14 24 0 15 12 20 10 5 8 12 15 0
Labeled scatter plot of Education and Internet Use is attached in the picture below.
Yes there appears to be a linear relationship between the two variables (Education and Internet Use) as the data points appears to have an upward trend depicted by the linear trend line in the graph.
The correlation Coefficient value which is measures the degree of linear relationship between education and internet use is 0.7048
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
5n + (-6) = -2
add 6 to -2
5n = -2 + 6
5n = 4
divide both sides by 5
n = 4/5
