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yulyashka [42]
2 years ago
5

A box is 50 cm long. Which of these is the closet to the length of this box in feet ? (1 inch = 2.54cm)

Mathematics
2 answers:
maksim [4K]2 years ago
7 0
19 feet is the closest without going over 50cm
Nitella [24]2 years ago
4 0

Answer:

1ft 7.7in/1.64 ft

Step-by-step explanation:

50/2.54=19.69in

12in=1ft

so 1ft 7.7in

19.69/12=1.64

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6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the
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Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

a_1 the horizontal distance between the first observer and the ballonist

a_2 the horizontal distance between the second observer and the ballonist

\alpha _1 and \alpha _2 the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

S=a*h (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

S=S_1+S_2 (equation 2)

S_1=a_1*h

But we can write S_1 in terms of \alpha _1, like this:

\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}

And for S_2 will be the same:

S_2=\frac{h^{2} }{\tan(\alpha _2)}

Replacing in the equation 2:

S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})

And replacing in the equation 1:

h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}

So, we can replace all the known data in the last equation:

h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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