32 is the correct answer I believe if using PEMDAS method
Answer:
D
Step-by-step explanation:
Since in a pass code, the placement of the digits is
important, therefore this means that to solve for the total number of
possibilities we have to make use of the principle of Permutation. The formula
for calculating the total number of possibilities using Permutation is given
as:
P = n! / (n – r)!
where,
n = is the total amount of numbers to choose from = 20
r = is the total number of digits needed in the passcode =
4
Therefore solving for the total possibilities P:
P = 20! / (20 – 4)!
P = 20! / 16!
P = 116,280
<span>Hence there are a total of 116,280 possibilities of pass
codes.</span>
Answer:
- <u><em>A dilation by a scale factor of 4 and then a reflection across the x-axis </em></u>
Explanation:
<u>1. Vertices of triangle FGH:</u>
- F: (-2,1)
- G: (-3,3)
- H: (0,1)
<u>2. Vertices of triangle F'G'H':</u>
- F': (-8,-4)
- G': (-12,-12)
- H': (0, -4)
<u>3. Solution:</u>
Look at the coordinates of the point H and H': to transform (0,1) to (0,-4) you can muliply each coordinate by 4 and then change the y-coordinate from 4 to -4. That is<em> a dilation by a scale factor of 4 and a reflection across the x-axis.</em> This is the proof:
- Rule for a dilation by a scale factor of 4: (x,y) → 4(x,y)
(0,1) → 4(0,1) = (0,4)
- Rule for a reflection across the x-axis:{ (x,y) → (x, -y)
(0,4) → (0,-4)
Verfiy the transformations of the other vertices with the same rule:
- Dilation by a scale factor of 4: multiply each coordinate by 4
4(-2,1) → (-8,4)
4(-3,3) → (-12,12)
- Relfection across the x-axis: keep the x-coordinate and negate the y-coordinate
(-8,4) → (-8,-4) ⇒ F'
(-12,12) → (-12,-12) ⇒ G'
Therefore, the three points follow the rules for <em>a dilation by a scale factor of 4 and then a reflection across the x-axis.</em>
Answer:
maybe
Step-by-step explanation:
Dora is apparently assuming the dimensions are integers. In that case she is correct.
If the dimensions are unconstrained, the perimeter will be largest when a pair of opposite sides will be the smallest measure allowed.
For some perimeter P and side length x, the area is ...
A = x(P/2 -x)
Conversely, the perimeter for a given area is ...
P = 2(A/x +x)
This gets very large when x gets very small, so Dora is correct in saying that the side lengths that are as small as they can be will result in the largest perimeter. We have no way of telling if her assumption of integer side lengths is appropriate. If it is not, her statement makes no sense.