The scale factor of dilation from AB to EF is 0.25
<h3>How to determine the scale factor of dilation?</h3>
The given parameters are:
- EFJL is dilation of ABIK
- AB has a length of 12
- EF has a length of 3
From the above parameters, side lengths AB and EF are corresponding sides
This means that:
The scale factor of dilation (k) is
k = AB/EF
So, we have
k = 3/12
Evaluate
k = 0.25
Hence, the scale factor of dilation from AB to EF is 0.25
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Answer:
12
Step-by-step explanation:
I'm not entirely sure abt this but here goes nothing
75.36 divided by 3.14 = 24
24 = diameter
radius = 12
Answer:
The answer is below
Step-by-step explanation:
An equation shows the relationship between two or more variables. An equation is a statement that shows the equality between expressions. An equation with infinitely many solution is when all numbers are solutions, that is there is no one solution. Example is: x + 3 = x + 3.
When each side of an equation has been simplified, equations that have the same coefficients and the same constants on each side have infinitely many solutions
9514 1404 393
Answer:
q = 40
Step-by-step explanation:
When the quadratic has roots p and r, it can be factored as ...
(x -p)(x -r) = x² -(p+r)x +pr
So, the sum of the roots is 14, and their difference is 6. This lets us find the roots from ...
p + r = 14
p - r = 6
2p = 20 . . . add the two equations
p = 10
r = 14 -p = 4
The value of interest is then ...
q = pr = (10)(4)
q = 40
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The graph shows the roots to be 4 and 10, as we found.
Answer:
if repetition is allowed,
if repetition is not allowed.
Step-by-step explanation:
For the first case, we have a choice of 26 letters <em>each step of the way. </em>For each of the 26 letters we can pick for the first slot, we can pick 26 for the second, and for each of <em>those</em> 26, we can pick between 26 again for our third slot, and well, you get the idea. Each step, we're multiplying the number of possible passwords by 26, so for a four-letter password, that comes out to 26 × 26 × 26 × 26 =
possible passwords.
If repetition is <em>not </em>allowed, we're slowly going to deplete our supply of letters. We still get 26 to choose from for the first letter, but once we've picked it, we only have 25 for the second. Once we pick the second, we only have 24 for the third, and so on for the fourth. This gives us instead a pretty generous choice of 26 × 25 × 24 × 23 passwords.