1/25
(5^2)^-3 x 5^4= 0.04 which is 1/25 in fraction form
We know that
Part a) <span>Find the fifth term of the arithmetic sequence in which t1 = 3 and tn = tn-1 + 4
t1=3
t2=t1+4----> 3+4-----> 7
t3=t2+4-----> 7+4----> 11
t4=t3+4-----> 11+4---> 15
t5=t4+4-----> 15+4---> 19
the answer Part a) is 19
Part b) </span><span>Find the tenth term of the arithmetic sequence in which t1 = 2 and t4 = -10
we know that
tn=t1+(n-1)*d-----> d=[tn-t1]/(n-1)
t1=2
t4=-10
n=4
find the value of d
d=[-10-2]/(4-1)-----> d=-12/3----> d=-4
find the </span>tenth term (t10)
t10=t1+(10-1)*(-4)----> t10=2+9*(-4)----> t10=-34
the answer Part b) is -34
Part c) <span>Find the fifth term of the geometric sequence in which t1 = 3 and tn = 2tn-1
t1=3
t2=2*t1----> 2*3----> 6
t3=2*t2----> 2*6----> 12
t4=2*t3-----> 2*12---> 24
t2=2*t4----> 2*24----> 48
the answer Part c) is 48</span>
Answer:
59
Step-by-step explanation:
ur welcome
Answer:
the correct answer is option D
Answer:
A) -sqrt(3)/3
Step-by-step explanation:
If you know your unit circle (attached), 11pi/6 has the value of (sqrt(3)/2,-1/2), where the coords are (cosine, sine).
Tangent is sine/cosine:
(-1/2)/(sqrt(3)/2)
We can convert this to multiplication by miltiplying the reciprocal of sqrt(3)/2:
(-1/2)*(2/sqrt(3))
Multiplied out to:
-2/(2sqrt(3))
We have to rationalize this by multiplying the numerator and denominator by sqrt(3):
-2(sqrt(3))/2sqrt(3)(sqrt(3)
two of the same square roots multiplied together equals the number without the sqrt:
-2(sqrt(3))/2*3
we can factor out 2 from the numerator and denominator:
-sqrt(3)/3
Therefore, the answer is A
