Shadow that is 24 meters long. How tall is the building?

Mathematics

2 answers:

anyanavicka [17]3 years ago

8 0

Answer:

Answer

Find out the what is the height of the building.

To prove

Let us assume that the height of the tree be x .

Let us assume that the shadow of the tree be y .

x = ky

Where k is the constant of proportionality.

As given

A 21 meter tall tree near a building casts a shadow that is 24 meters long.

21 = k × 24

k = 0.875

Let us assume that the height of the building be u .

As given

The building casts a shadow that is 56 meters long.

put values in the x = ky

u = 0.875 × 56

u = 49 cm

Therefore the height of the building is 49 cm .

<h3>i used another problems answer to help you answer yours.</h3>

vfiekz [6]3 years ago

4 0

Answer: The building is around 18 meters tall which is choice C.

Step-by-step explanation:

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Step-by-step explanation:

This is a combinatorial problem of balls and sticks. In order to represent a way of distributing n identical chairs to k classrooms we can align n balls and k-1 sticks. The first classroom will receive as many chairs as the amount of balls before the first stick. The second one will receive as many chairs as the amount of balls between the first and the second stick, the third classroom will receive the amount between the second and third stick and so on (if 2 sticks are one next to the other, then the respective classroom receives 0 chairs).

The total amount of ways to distribute n chairs to k classrooms as a result, is the total amount of ways to put k-1 sticks and n balls in a line. This can be represented by picking k-1 places for the sticks from n+k-1 places available; thus the cardinality will be the combinatorial number of n+k-1 with k-1, ${n+k-1 \choose k-1}$ .

For the 2 largest classrooms we distribute n = 50 chairs. Here k = 2, thus the total amount of ways to distribute them is ${50+2-1 \choose 2-1} = 51$ .

For the 3 remaining classrooms (k=3) we need to distribute the remaining 50 chairs, here we have ${50+3-1 \choose 3-1} = {52 \choose 2} = 1326$ ways of making the distribution.