Answer:
(a)123 km/hr
(b)39 degrees
Step-by-step explanation:
Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.
Distance = Speed X Time
Therefore: PQ =50km/hr X 2 hr =100 km
It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.
Distance, QA=50km/hr X 2.5 hr =125 km
Using alternate angles in the diagram:
(a)First, we calculate the distance traveled, PA by plane Y.
Using Cosine rule
SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am
Time taken =1.5 hour
Therefore:
Average Speed of Y
(b)Flight Direction of Y
Using Law of Sines
![\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)](https://tex.z-dn.net/?f=%5Cdfrac%7Bp%7D%7B%5Csin%20P%7D%20%3D%5Cdfrac%7Bq%7D%7B%5Csin%20Q%7D%5C%5C%5Cdfrac%7B125%7D%7B%5Csin%20P%7D%20%3D%5Cdfrac%7B184.87%7D%7B%5Csin%20110%7D%5C%5C123%20%5Ctimes%20%5Csin%20P%3D125%20%5Ctimes%20%5Csin%20110%5C%5C%5Csin%20P%3D%28125%20%5Ctimes%20%5Csin%20110%29%20%5Cdiv%20184.87%5C%5CP%3D%5Carcsin%20%5B%28125%20%5Ctimes%20%5Csin%20110%29%20%5Cdiv%20184.87%5D%5C%5CP%3D39%5E%5Ccirc%20%24%20%28to%20the%20nearest%20degree%29)
The direction of flight Y to the nearest degree is 39 degrees.
Answer:
Step-by-step explanation:
<u>Arithmetic Sequences</u>
The arithmetic sequences are identified because any term n is obtained by adding or subtracting a fixed number to the previous term. That number is called the common difference.
The equation to calculate the nth term of an arithmetic sequence is:
Where
an = nth term
a1 = first term
r = common difference
n = number of the term
We are given the first terms of a sequence:
-12, -28, -44,...
Find the common difference by subtracting consecutive terms:
r = -28 - (-12) = -16
r = -44 - (-28) = -16
The first term is a1 = -12. Now we calculate the term n=61:
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