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2 years ago

Help me pleaseeeeeeeeeeeeeee

Mathematics

1 answer:

Colt1911 [192]2 years ago

7 0

Answer:

WUJV

Step-by-step explanation:

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hope this helps.

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Please help What is the value of d?

PtichkaEL [24]

Rule: with any inscribed quadrilateral, the opposite angles are supplementary (they add to 180 degrees)

Based on that rule, we can say
d+100 = 180
d+100-100 = 180-100
d = 80

Answer: 80

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2 years ago

Gavin drank nine 8 ounce glasses of water today. How many quarts of water did he drink

dsp73

Answer:

2.25

Step-by-step explanation:

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2 years ago

Here is a triangle with the height on the side of the triangle. Use the formula: (1/2)*(Base)*(Height) to find the area of the t

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(If one square on the graph = one centimeter)

b = 10cm

b = 10cmh = 10cm

Area:

$= \frac{1}{2} \times b \times \: h$

$= \frac{1}{2 } \times 10 \times 10$

$= \frac{1}{2} \times 100$

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2 years ago

The cost C of manning household is partly constant and partly varies as the number of people, the cost is 70,000 and for 10 peop

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Step-by-step explanation:

god bless stay safe po

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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)$

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

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3 years ago