Answer:
Step-by-step explanation:
The nth term of an AP is expressed as;
Tn = a+(n-1)d
a is the first term
n is the number of terms
d is the common difference
Sixth term
T6 = a+(6-1)d
T6 = a+5d
Third term;
T3 = a+(3-1)d
T3 = a+2d
If the sixth term of an AP is twice the 3rd term, then
a+5d = 2(a+2d)
a+5d = 2a+4d
Given a = 3
3+5d = 2(3)+4d
5d-4d = 6-3
d = 3
Common difference is 3
For the 10th term
T10 = a+9d
T10 = 3+9(3)
T10 = 3+27
T10 = 30
Hence the tenth term is 30
3) 3.sec² Ф = 4.tan² Ф
but sec² Ф = 1/cos² Ф
3/cos² Ф = 4.tan² Ф
3/4 = (tan² Ф).(cos² Ф) ; tan² Ф = sin² Ф/cos² Ф
3/4 = (sin² Ф)x(cos² Ф)/(cos² Ф)
sin² Ф = 3/4
a) sin Ф = +(√3)/2 and b) sin Ф = - (√3)/2
sin Ф = + - (√3)/2 → Ф =π/3 + kπ/3
Answer:
539
Step-by-step explanation:
Answer:
the answer is (23.47,4.38)
Step-by-step explanation: