Answer:
12870ways
Step-by-step explanation:
Combination has to do with selection
Total members in a tennis club = 15
number of men = 8
number of women = 7
Number of team consisting of women will be expressed as 15C7
15C7 = 15!/(15-7)!7!
15C7 = 15!/8!7!
15C7 = 15*14*13*12*11*10*9*8!/8!7!
15C7 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2
15C7 = 15*14*13*12*11/56
15C7 = 6,435 ways
Number of team consisting of men will be expressed as 15C8
15C8 = 15!/8!7!
15C8 = 15*14*13*12*11*10*9*8!/8!7!
15C8 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2
15C8 = 6,435 ways
Adding both
Total ways = 6,435 ways + 6,435 ways
Total ways = 12870ways
Hence the required number of ways is 12870ways
Because both positive and negative values have<span> a positive </span>absolute value.
F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
Answer:
Let's define the variables:
A = price of one adult ticket.
S = price of one student ticket.
We know that:
"On the first day of ticket sales the school sold 1 adult ticket and 6 student tickets for a total of $69."
1*A + 6*S = $69
"The school took in $150 on the second day by selling 7 adult tickets and student tickets"
7*A + 7*S = $150
Then we have a system of equations:
A + 6*S = $69
7*A + 7*S = $150.
To solve this, we should start by isolating one variable in one of the equations, let's isolate A in the first equation:
A = $69 - 6*S
Now let's replace this in the other equation:
7*($69 - 6*S) + 7*S = $150
Now we can solve this for S.
$483 - 42*S + 7*S = $150
$483 - 35*S = $150
$483 - $150 = 35*S
$333 = 35*S
$333/35 = S
$9.51 = S
That we could round to $9.50
That is the price of one student ticket.
