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Snowcat [4.5K]
3 years ago
5

What is the difference in height between an albatross flying at 100 m above the surface of the ocean and a shark swimming 30 m b

elow the surface?
Mathematics
1 answer:
Lelechka [254]3 years ago
6 0

Answer:  130 meters

Step-by-step explanation:

We have to let the "surface of the ocean" be at 0 height.

So,

anything above surface has POSITIVE height

and anything below surface has NEGATIVE height

The albatross is 100 m above, so we can say its height is:

+100 meters

The shark is 30 m below, so we can say its height is:

-30 meters

The difference in height is thus:

100 - (-30) = 100 + 30 = 130 meters

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In ΔUVW, w = 44 cm, u = 83 cm and ∠V=141°. Find the area of ΔUVW, to the nearest square centimeter.
Tema [17]

Answer:

Area  \approx 1149\ cm^2

Step-by-step explanation:

<u>Given that:</u>

ΔUVW,

Side w = 44 cm, (It is the side opposite to \angle W)

Side u = 83 cm (It is the side opposite to \angle U)

and ∠V=141°

Please refer to the attached image with labeling of the triangle with the dimensions given.

Area of a triangle with two sides given and angle between the two sides can be formulated as:

A = \dfrac{1}{2}\times a\times b\times sinC

Where a and b are the two sides and

\angle C is the angle between the sides a and b

Here we have a = w = 44cm

b = u = 44cm

and ∠C= ∠V=141

Putting the values to find the area:

A = \dfrac{1}{2}\times 44\times 83\times sin141\\\Rightarrow A = \dfrac{1}{2} \times 3652 \times sin141\\\Rightarrow A =1826 \times 0.629\\\Rightarrow A  \approx 1149\ cm^2

So, the <em>area </em>of given triangle to the nearest square centimetre is:

Area  \approx 1149\ cm^2

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