sec ^6 A−tan ^6A=(1+tan ^2A−tan ^2 A)(1+tan ^4
A+2tan ^2 A+tan ^2 A+tan ^4 A+tan ^4 A)
sec ^6 A−tan ^6 A=1+3tan ^2 A+3tan ^4A
L.H.S. = R.H.S.
Hence proved
Step-by-step explanation:
♡´･ᴗ･`♡
Answer:
(2n+1)(n+2)
Step-by-step explanation:
Use basic factor pairs, then figure out the two 2s in the factor pairs add with the one to be five. Then just make the answer above.
x - first number
x + 18 - second number
x + (x + 18) = 36
x + x + 18 = 36
2x + 18 = 36 | - 18
2x = 18 | : 2
x = 9
First number: x = 9
Second number: x + 18 = 9 + 18 = 27
Those numbers are 9 and 27.
Answer:
133/143
Step-by-step explanation:
Let S be the sample space
Let E be the event of selecting three committee partners with at least one junior partner.
Partners in the law firm include:
Senior partners = 6
Junior partners = 7
Total partners = 13
n(S) = number of ways of selecting 3 partners from 13 = 13C3
n(S) = 13C3 = 13!/(10!3!) = (13x12x11)/(3x2x1) = 286
To get n(E) i.e least 1 junior partner in the selected committee, we may have:
(2 senior and 1 junior) or ( 1 senior and 2 junior) or (3 junior).
Therefore, the required number of way is given below:
= (6C2 x 7C1) + (6C1 x 7C2) + 7C3
= [(6x5)/2 x 7] + [6 x (7x6)/2] + [(7x6x5)/(3x2)]
= 105 + 126 + 35
n(E) = 266
Therefore, the probability P(E) that at least one of the junior partners is on the committee is given below:
P(E) = n(E) /n(S)
P(E) = 266/286
P(E) = 133/143
Answer:
x= 105
y= 40
Step-by-step explanation:
3y=4y-40 because they are alternate interior
Solve for y.
y=40
4y-40=x+15 because they are vertical angles.
Substitute 40 in for y.
4(40)-40=x+15
Solve.
x=105
