One column of numbers consists of 61, 24, and 47. when the digits of the numbers are added together, the result is 6 + 1 + 2 + 4 + 4 + 7 = 24, and when the digits of 24 are then added together, the end result is 2 + 4 = 6. if the same process is performed on the numbers in a second column, what can be concluded?

The question is incomplete because it must content a list of choices to select the right one.

Any way, a conclusion that you can make, and that is a common one for this kind of questions, is about whether the sum of the numbers of the second column may or not be the same sum of the numbers of the first column.

The condition for the two sums be the same is that when the digits of the second column are added together the result be the same obtained for the sum of the digits of the first column. In this case that is 6.

So, the possible answer is:

If the end result from the second column is not 6, then the sum of the numbers in the first column is not equal to the sum of the numbers in the second column.

we know that

the volume of the oblique pyramid with a square base is equal to

V=\frac{1}{3}BhV=

3

1

Bh

where

B is the area of the base

h is the height of the pyramid

In this problem

The area of the base is

B=EF^{2}\ units^{2}B=EF

2

units

2

h=AB\ unitsh=AB units

The two formulas to calculate h are

Applying the Pythagoras Theorem

h^{2}=AC^{2}-CB^{2}h

2

=AC

2

−CB

2

> h=\sqrt{AC^{2}-CB^{2}}h=

AC

2

−CB

2

h^{2}=AD^{2}-DB^{2}h

2

=AD

2

−DB

2

> h=\sqrt{AD^{2}-DB^{2}}h=

AD

2

−DB

2

therefore

the answer is

AB and EF

AC, CB and EF

AD, DB and EF