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Alexus [3.1K]
3 years ago
5

Which statement is correct?

Mathematics
2 answers:
Lana71 [14]3 years ago
5 0

Answer:

A line that falls from right to left has negative slope.

Step-by-step explanation:

Due to the fact that negative slope looks like this

zloy xaker [14]3 years ago
5 0

Answer:

A line that rises from left to right has positive slope.

Step-by-step explanation:

A horizontal line just goes from left to right but not up or down so y does not change at all. Change in y/change in x makes slope. Since y does not change, it has 0 slope which isn’t no slope.

A vertical line moves up and down but not left to right so there is no change in x. You cannot divide by 0 if you follow the equation y/0. So slope is undefined.

A line that rises from left to right goes in a diagonal direction is like a 2d hill. It is positive because change in y would be positive and change in x would be also positive. Positive/positive is positive making the slope positive.

A line that falls from right to left literally gives the same line so that means it is also positive slope, not negative.

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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]

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(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.

7 0
3 years ago
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