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3 years ago

What’s the answer???

Mathematics

1 answer:

Harrizon [31]3 years ago

5 0

first

second

and fifth are true

Basically the square root of -1 is i. From that you should be able to deduce the answers.

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0.6w = 0.48 how u do this

vovangra [49]

A variable next to any number means to multiply, so we would do the inverse operation -- We would divide 0.48 by 0.6, which is 0.8. Therefore, W=0.48. To check your work, multiply 0.6 by 0.8, and what do you know? It's 0.48! :)

3 0

2 years ago

Read 2 more answers

the width of a rectangle is 3 inches, and the area is 88 square inches. What are the length and width of the rectangle

inessss [21]

Answer:

length is 29.3, width is 3

Step-by-step explanation:

A= L×W

and if A = 88 and W is 3

88= L×3

Divide

88/3

legth =29.3

8 0

2 years ago

What does 10.44 x 2.5 equal?

SSSSS [86.1K]

It is equal to 26 if you didn’t understand you can ask more questions and I’ll help

7 0

2 years ago

Read 2 more answers

Help will give branliest

Daniel [21]

Answer:

They make heart shapes! hearts look like <3

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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.

7 0

2 years ago