Part A)
If f(x) - 3 is the new equation, it means there is a vertical translation of f(x) down 3 units. The y-intercept will decrease by 3 units. Areas of increasing on the function may be lessened as the function is being translated down 3 units. The areas of decrease will increase because the function is being translated down. End behaviour will not change from a translation as long as the function is continuous at each end, (not a finite function with end points). The evenness or oddness of f(x) will not change either.
Part B:
The y-intercept will be flipped horizontally about the x-axis and multiplied by 2. This will mean that if the y-intercept was positive, it will now be negative and vice versa. The increasing and decreasing regions of the graph will be flipped, so anywhere f(x) was positive will now be negative and vice versa. They will also be double what they were before because all values are multiplied by 2. The end behaviour will switch. If f(x) was from Quad1->Quad3 for example, it will now be Quad2->Quad4 because of the flip at the x-axis. The evenness and oddness of the function will not change seeing as the degree of f(x) is not affected.
Answer: option D is the correct answer.
Step-by-step explanation:
The given sequence is a geometric sequence because the consecutive terms differ by a common ratio.
The formula for determining the nth term of a geometric progression is expressed as
an = a1r^(n - 1)
Where
a1 represents the first term of the sequence.
r represents the common ratio.
n represents the number of terms.
From the information given,
a1 = 36
r = 12/36 = 4/12 = 1/3
Therefore, the formula for the nth term of the sequence is
an = 36 × 1/3^(n - 1)
an = 36 × 3^-1(n - 1)
an = 36 × 3^(-n + 1)
an = 36 × 3^(1 - n)
Answer:
answer is choice 4 because in te given angle A and angle B are congrunt and if they are acute the degree measure is 0-90
We use P = i•e^rt for exponential population growth, where P = end population, i = initial population, r = rate, and t = time
P = 2•i = 2•15 = 30, so 30 = 15 [e^(r•1)],
or 30/15 = 2 = e^(r)
ln 2 = ln (e^r)
.693 = r•(ln e), ln e = 1, so r = .693
Now that we have our doubling rate of .693, we can use that r and our t as the 12th hour is t=11, because there are 11 more hours at the end of that first hour
So our initial population is again 15, and P = i•e^rt
P = 15•e^(.693×11) = 15•e^(7.624)
P = 15•2046.94 = 30,704
