Show that sin( + 180°) + 2 cos( − 360°) − sin( − 180°) = 2x

Mathematics

1 answer:

den301095 [7]3 years ago

8 0

Answer:

Example 1: Change sin 80° cos 130° + cos 80° sin 130° into a trigonometric function in ... Example 2: Verify that cos (180° − x) = − cos x ... Example 7: Verify that sin (360° − x) = − sin x.

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Attendance at last year's show was 16,000 people. This year 16,800 people attended the show. What was the percent change in atte

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The percent change in attendance, relative to last year's show is a 5% increase.

<h3>What is a percentage? </h3>

A Percentage can be described as a fraction out of an amount that is usually expressed as a number out of hundred.

<h3>What is the percentage change in attendance? </h3>

The value of the percentage change in attendance can be determined by subtracting the number of people in this year's show from the number of people at last year's show. The result would be divided by the number of people at last year's show and multiplied by 100.

[(16,800 - 16,000)/16,000] x 100 = 5%

To learn more about percentages, please check: brainly.com/question/25764815

3 0

2 years ago

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)$