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3 years ago

I'm confused pls i need help

Mathematics

2 answers:

Scrat [10]3 years ago

6 0

You have to multiply the 12.4 and 9.3 and then the 12.4 and 15.5 and then whatever to answers you get split them in half then add them together and there’s your answer

yan [13]3 years ago

5 0

Answer:

I am pretty sure it is 249.86 square feet.

Step-by-step explanation:

Find the area of the rectangle:

$15.5*12.4=192.20$ (base x height)

Find the area of the triangle:

$9.3*12.4=115.32\\115.32/2=57.66$ (base x height divided by 2)

Add the 2 areas together:

$192.20+57.66=249.86$

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ANSWER

Option D is correct

EXPLANATION

We resolve the forces into component forms.

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Also the second is resolved to obtain,

$F_2= 280 \cos(135) i + 280 \sin(135) j.$

$F_2= - 140 \sqrt{2} i + 140 \sqrt{2} j$

To find the resultant vector, V, We add the corresponding components of the two forces to get:

$V=F_1+F_2$

$\implies \: V = (200 \sqrt{3} - 140 \sqrt{2} )i + (200 + 140 \sqrt{2})j$

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3 years ago

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$