sec ( t ) cos ( t ) = (1/cos (t)) *cos(t) = 1
Answer:
Ai. Common ratio = 2/3
Aii. First term = 54
B. Sum of the first five terms = 422/3
Step-by-step explanation:
From the question given above, the following data were obtained:
3rd term (T3) = 24
6Th term (T6) = 64/9
First term (a) =?
Common ratio (r) =?
Sum of the first five terms (S5) =?
Ai. Determination of the common ratio (r).
T3 = ar²
T3 = 24
24 = ar²....... (1)
T6 = ar⁵
T6 = 64/9
64/9 = ar⁵......... (2)
The equation are:
24 = ar²....... (1)
64/9 = ar⁵......... (2)
Divide equation 2 by equation 1.
64/9 ÷ 24 = ar⁵ / ar²
64/9 × 1/24 = r³
8/27 = r³
Take the cube root of both side
r = 3√(8/27)
r = 2/3
Thus, the common ratio is 2/3
Aii. Determination of the first term (a).
T3 = ar²
3rd term (T3) = 24
Common ratio (r) = 2/3
First term (a) =?
24 = a(2/3)²
24 = 4a/9
Cross multiply
24 × 9 = 4a
216 = 4a
Divide both side by 4
a = 216/4
a = 54
Thus, the first term (a) is 54
B. Determination of the sum of the first five terms.
Common ratio (r) = 2/3
First term (a) = 54
Number of term (n) = 5
Sum of first five terms (S5) =?
Sn = a[1 –rⁿ] / 1 – r
S5 = 54[1 – (⅔)⁵] / 1 – ⅔
S5 = 54 [1 – 32/243] / ⅓
S5 = 54 (211/243) × 3
S5 = 54 × 211/81
S5 = 6 × 211/9
S5 = 2 × 211/3
S5 = 422/3
Thus, the sum of the first five terms is 422/3
5 tens= 50
23 ones= 23
Add them together and you get 73. But multiply them you get 1150.
Here's the general formula for bacteria growth/decay problems
Af = Ai (e^kt)
where:
Af = Final amount
Ai = Initial amount
k = growth rate constant
<span>t = time
But there's another formula for a doubling problem.
</span>kt = ln(2)
So, Colby (1)
k1A = ln(2) / t
k1A = ln(2) / 2 = 0.34657 per hour.
So, Jaquan (2)
k2A = ln(2) / t
<span>k2A = ln(2) /3 = 0.23105 per hour.
</span>
We need to use the rate of Colby and Jaquan in order to get the final amount in 1 day or 24 hours.
Af1 = 50(e^0.34657(24))
Af1 = 204,800
Af2 = 204,800 = Ai2(e^0.23105(24))
<span>Af2 = 800</span>
Answer:
b. 66
Step-by-step explanation:
s2 Could refer to the variance of the 66 employees' salaries
