Answer:
common ratio(r) = 10
place of the term = n
a = 3
now
sequence of 1 term = ar^(n-1)
= 3×10^(1-1)
= 3× 10^0
= 3×1
= 3
sequence of 2 term = ar^(n-1)
= 3×10¹
= 30
and so on
A. I would suggest including <em>all</em> units in your answer. Both answers are correct but the unit is also important.
average rate of change:
($750 - $350) / (2014 - 2010) = $400 / (4 years) = $100 per year
initial value:
$350
B. Take 2010 to be year 0. Then for year <em>x</em> = 0, the fee is <em>y</em> = $350.
If the fee increases at a rate of $100 per year, this means that when <em>x</em> = 1, <em>y</em> = $350 + $100 = $450. So the graph of the function that models the dog-walking fee is a line that passes through the points (0, 350) and (1, 450), adn the slope of this line is
(450 - 350) / (1 - 0) = 100/1 = 100
which is the same as the average rate of change.
Using the point-slope formula for a line, the equation you want is then
<em>y</em> - 350 = 100 (<em>x</em> - 0)
<em>y</em> - 350 = 100<em>x</em>
<em>y</em> = 100<em>x</em> + 350
Answer: 3.75 + - 7
Step-by-step explanation:
1.5x - 3 + 1.5x - 3 + 0.75x - 1
1.5x + 1.5x + 0.75x - 3 + - 3 + - 1
3.75x + - 7
Answer:
70 cars per year.
Step-by-step explanation:
Given:
In 1984 a locally-owned car company sold 2,743 cars.
In 2003, the car sales rose to 4,069.
Question asked:
What was the average rate of change for the total number of cars sold?
Solution:
In 2003 number of cars sold = 4069
In 1984 number of cars sold = 2743
Change in the number of car sold = 4069 - 2743
= 1326
Now, time interval between 2003 and 1984 = 19
Thus, the average rate of change for the total number of cars sold is 70 cars per year.
Answer:
D
Step-by-step explanation:
