You’re answer is D. Y=3/2x+5/2
Answer:
60 times will they ring together at the same second in one hour excluding the one at the end.
Step-by-step explanation:
Given : Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively.
To find : How many times will they ring together at the same second in one hour excluding the one at the end?
Solution :
First we find the LCM of 3, 6, 10, 12 and 15.
2 | 3 6 10 12 15
2 | 3 3 5 6 15
3 | 3 3 5 3 15
5 | 1 1 5 1 5
| 1 1 1 1 1


So, the bells will ring together after every 60 seconds i.e. 1 minutes.
i.e. in 1 minute they rand together 1 time.
We know, 1 hour = 60 minutes
So, in 60 minute they rang together 60 times.
Therefore, 60 times will they ring together at the same second in one hour excluding the one at the end.
<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.
Part A : y=$0.75m+$4.50
Part B : $8.25