They are called relation symbols
they tell how something relates to something else
greater than tells ou that the the result must be more than, but not eqal to that numberd so example, we have to have something that flys higher than that bridge (100ft) but it cannot fly 100 or else it will crash
less than tells you that you must be less than, but not euqal to. example
this thing must be less than 7 feet tall to fit under the doorway, it cannot be 7 feet or else it wont fit
equal to is an exact amount, nothing more or less
we have to have a screw that is this big, nothing more or less
less and greater than is nice since if you found a solution and you wanted to notate it, example
if the naswer is any number less than 10, it would be nicer to write x<10 instead of listing all numbers less than 10 since they could get very close example 9.999999999 forever
greater is the same, any number bigger than 30, you woul dhave to include 30.0000000001 and smaller which is hard, but with the x>30, it is easier
the equals just shows equality, this is equal to this, period
First solve the triangle:
75° + 75° + x = 180°
150° + x = 180°
x = 180° - 150°
Therefore x = 30°
Then, use the geometrical property of vertically opposite angles.
Therefore. x° = 30°
Answer:
7% percent off of $33.00 would be $30.69 cents
Hope helps
Answer:
In option a, w < 10
in option b, w = 10 and 2(20) + 2(10) = 40 + 20 = 60 < 150
in option c, w > 10 but 2(60) + 2(20) = 120 + 40 = 160 > 150
In option d, w > 10 but 2(55) + 2(30) = 110 + 60 = 170 > 150
Therefore the corect answer is option b.
Step-by-step explanation:
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.
