Answer:
Step-by-step explanation:
Exponential growth
Answer:
k=0
Step-by-step explanation:
R(x)=kx-2x²
R(1/4)=k(1/4)-2(1/4)²=k/4-1/8
R(-1/4)=k(-1/4)-2(-1/4)²=-k/4-1/8
R(1/4)=R(-1/4)
k/4-1/8=-k/4-1/8
k/4+k/4=1/8-1/8
2k/4=0
2k=0
k=0
Lets write equation of a function:
y = kx + n
Direct variation in simple is equation of a line which has n=0 or in other words which y to x ratio is k.
First option gets 7=7 but it isnt direct variation because n is not equal to 0
third option is indeed correct. once we implement coordinates (2,7) we get 7=7
Answer is
y = 7/2x
Answer:
The answers are given below.
Step-by-step explanation:
The computation is shown below:
1.a.
Profit Margin = Net Income ÷ Sales × 100
= $374 ÷ $6,900 ×100
= 5.4%
1-b:
Average Assets = (Beginning Assets + Ending Assets) ÷ 2
= ($3,200 + $3,600) ÷ 2
= $3,400
Now
Return on Assets = Net Income ÷ Average Assets
= $374 ÷ $3,400
= 11%
1-c
Average Equity = ($700 + $700 + $320 + $270) ÷ 2
= $995
Now
Return on Equity = Net Income ÷ Average Equity *100
= $374 ÷ $995
= 37.59%
2:
Dividends Paid = Beginning Retained Earnings + Net Income – Ending Retained Earnings
= $270 + $374 - $320
= $324
Answer:
Step-by-step explanation:
2005 AMC 8 Problems/Problem 20
Problem
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$
Solution
Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.
See Also
2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
