Answer:
910 students
Step-by-step explanation:
<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.
by the double angle identity for sine. Move everything to one side and factor out the cosine term.
Now the zero product property tells us that there are two cases where this is true,
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of
, so
where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval
for
and
. More generally, if you think of
as a point on the unit circle, this occurs whenever
also completes a full revolution about the origin. This means for any integer
, the general solution in this case would be
and
.
Answer:
x=8
Step-by-step explanation:
Those angles are supplementary so:
140+5x=180
5x=40
x=8
