Step-by-step explanation:
Whenever we put a negative number inside a modulus function it will give us the positive output. For example , |-3| = 3 , |-6|=6, |5|= 5 ,etc.
So a modulus function i.e. |x| is always greater than zero ( positive ) when x is any number except 0 and it is equal to zero when the value of x is 0.
So |x| can't be less than -4 as |x| is always positive . So the statement is false.
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:
<span>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.</span>
First, we have to find how many small cubes formed by the large cube.
To find the number of small cubes, we should calculate the volume of each large cube and small cube. Equalize the volume units to dm³
Dimension of large cube
s = 1 m
s = 10 dm
Volume of large cube
v = s³
v = 10³
v = 1,000 dm³
Dimension of small cube
s = 1 dm
Volume of small cube
v = s³
v = 1³
v = 1 dm³
Second, calculate the number of small cubes formed
n = volume of large cube / volume of small cube
n = 1,000 dm³ / 1 dm³
n = 1,000
There are 1,000 small cubes.
Third, calculate the height of the structure.
The structure is formed by 1,000 cubes. Each of them is 1 dm high.
The height of the structure is
h = 1 dm × 1,000
h = 1,000 dm
h = 100 m
The height of the structure is 1,000 dm or 100 m
Answer:
Step-by-step explanation:
For a 30-60-90 triangle the sides always have the same relationship
Short leg = a
Long leg = a√3
Hypotenuse = 2a
BC is the short leg of ∆ABC
Given BC = 2
BC = a
Therefor
a = 2
AB = 2a = 4
AC = a√3 = 2√3
For ∆ACD
As above AC = 2√3
Since AC is the hypotenuse of ∆ACD
2a = 2√3
a = √3
CD = a = √3
AD = a√3 = 3
For ∆BCD
As above
BC = 2
CD = √3
Since BC is the hypotenuse of ∆BCD
2a = 2
a = 1
DB = a = 1
