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den301095 [7]
3 years ago
9

Could someone explain to me how to answer this question.

Mathematics
2 answers:
lyudmila [28]3 years ago
6 0
1/6(9/9)+8/8
1/6+1
7/6

Paul [167]3 years ago
5 0
1/6 (\sqrt{81} / 3^2) + 2^3 / |-8|

Let's simplify some things first:
The square root of 81 is 9.
3^2 = 9
2^3 = 8
|-8| = 8

1/6 (9/9) + 8/8

1/6(1) + 1

1/6 + 1 = 7/6

The answer is 7/6, or 1 and 1/6
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An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then
Zigmanuir [339]

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:

\angle Q=110^\circ

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)

(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)

The direction of flight Y to the nearest degree is 39 degrees.

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Explain why -4 is greater than -10 even though 4 is less than 10
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