<h2><u>Part A:</u></h2>
Let's denote no of seats in first row with r1 , second row with r2.....and so on.
r1=5
Since next row will have 10 additional row each time when we move to next row,
So,
r2=5+10=15
r3=15+10=25
<u>Using the terms r1,r2 and r3 , we can find explicit formula</u>
r1=5=5+0=5+0×10=5+(1-1)×10
r2=15=5+10=5+(2-1)×10
r3=25=5+20=5+(3-1)×10
<u>So for nth row,</u>
rn=5+(n-1)×10
Since 5=r1 and 10=common difference (d)
rn=r1+(n-1)d
Since 'a' is a convention term for 1st term,
<h3>
<u>⇒</u><u>rn=a+(n-1)d</u></h3>
which is an explicit formula to find no of seats in any given row.
<h2><u>Part B:</u></h2>
Using above explicit formula, we can calculate no of seats in 7th row,
r7=5+(7-1)×10
r7=5+(7-1)×10 =5+6×10
r7=5+(7-1)×10 =5+6×10 =65
which is the no of seats in 7th row.
Answer:
What do you need help with, it looks complete to me.
Step-by-step explanation:
Answer:
GE
Step-by-step explanation:
571,082,346,090 in word form would be Five hundred seventy one billion eighty two million three hundred forty six thousand ninety.
Answer:
The maximum revenue is $1,20,125 that occurs when the unit price is $155.
Step-by-step explanation:
The revenue function is given as:
where p is unit price in dollars.
First, we differentiate R(p) with respect to p, to get,
Equating the first derivative to zero, we get,
Again differentiation R(p), with respect to p, we get,
At p = 155
Thus by double derivative test, maxima occurs at p = 155 for R(p).
Thus, maximum revenue occurs when p = $155.
Maximum revenue
Thus, maximum revenue is $120125 that occurs when the unit price is $155.
