This problem can be solved through simple arithmetic
progression
Let
a1 = the first term of the sequence
a(n) = the nth term of the sequence
n = number of terms
d = common difference
Sn = sum of all terms
given
a1 = 12
a2 = 16
n = 10
d = 16 -12 = 4
@n = 10
a(n) = a1 + (n-1)d
a(10) = 12 + (9)4
a(10) = 48 seats
Sn = (n/2) * (a1 + a(10))
Sn = 5* (12 + 48)
Sn = 300 seats
Therefore the total number of seats is 300.
Is it 22m...look at the other side shouldn't it be the same (sorry if I didn't help :))))))))
I gotchu my guy let me know what you need help with I’d gladly revive those free points
Sec theta = 1/cos theta = 1/0.5 = 2
Option D is the correct answer.
<h3>
Answer:</h3>
6 hours
<h3>
Step-by-step explanation:</h3>
The two hoses together take 1/3 the time (4/12 = 1/3), so the two hoses together are equivalent to 3 of the first hose.
That is, the second hose is equivalent to 2 of the first hose. Two of the first hose could fill the vat in half the time one of them can, so 6 hours.
The second hose alone can fill the vat in 6 hours.
_____
The first hose's rate of doing work is ...
... (1 vat)/(12 hours) = (1/12) vat/hour
If h is the second hose's rate of doing work, then working together their rate is ...
... (1/12 vat/hour) + h = (1/4 vat/hour)
... h = (1/4 - 1/12) vat/hour = (3/12 -1/12) vat/hour = 2/12 vat/hour
... h = 1/6 vat/hour
so will take 6 hours to fill 1 vat.
