Yes
remember you can do anything to an equaiton as long as you do it to boh sides
ad try to get unknown to one side
9n-10=4n+10
minus 4n both sides
9n-4n-10=4n-4n+10
5n-10=0+10
add 10
5n+10-10=0+10+10
5n+0=20
5n=20
divide both sies by 5
5n/5=20/5
5/5n=4
1n=4
n=4
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: 1
Step-by-step explanation:
1. 2 4/16 + 3 2/3 = 2 12/48 + 3 32/48 = 71/12 reduced is 5 11/12
( lcm of 16 and 3 is 48)
2. 6 11/12 - 5 11/12 = 1
Please give me a brainliest i did all my hard work. The answer 1 is 100% right.
Answer:
1/4
Step-by-step explanation:
that is the answer
I found the constant which was -3
a = 1/4
b=-1