Answer:
24 terms
Step-by-step explanation:
The sum of an arithmetic sequence is the average of the first and last terms, multiplied by the number of terms. The last term is given by ...
an = a1 + (n-1)d
We have a sequence with first term a1 = 2 and common difference d = 2. So the last term is ...
an = 2+ 2(n -1) = 2n
Then the average of first and last terms times the number of terms is ...
Sn = 600 = n(2 + 2n)/2 = n(n+1) . . . . . . close to n²
We can solve the quadratic in n, or we can estimate the value of n as the integer just below the square root of 600.
√600 ≈ 24.5
so we believe n = 24.
_____
<em>Check</em>
S24 = 24·25 = 600 . . . . . . as required.
Answer:
y=5x
Step-by-step explanation:
Because if you look at it 5x any of them would equal the answer
1. x = 15120/12 = 1260 (monthly salary)
2. 1200 x 0.05 = 60 (raise)
Check: 1200+60=1260
This question is not complete
Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°