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2 years ago

PLEASE HELP DUE TONIGHT

Mathematics

1 answer:

lukranit [14]2 years ago

6 0

The height & speed are mathematically given as

t = 0.2 s

u = 4 m/s

<h3>What is the height & speed?</h3>

Generally, the equation for height is mathematically given as

h = 0.2gt^2

Therefore

0.2 = 0.5 * 9.8 * $t^2$

0.2 = 4.9 * $t^2$

t = 0.2 s

In conclusion, the initial speed is

s = ut

Therefore

u = 0.8 / 0.2

u = 4 m/s

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1 year ago

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3 years ago

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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3 years ago