Answer:
sorry.......dud.....
Step-by-step explanation:
hope some one will help you........
<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.
3, 6, 9, 12, 15
I did what you asked
Answer:
12:9
Step-by-step explanation:
I think the answer is C-No solutions
