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Answer: 2
Explanation:
I hope this helped!
<!> Brainliest is appreciated! <!>
- Zack Slocum
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I believe the answer is 2 :) .
Answer:
It is a many-to-one relation
Step-by-step explanation:
Given
See attachment for relation
Required
What type of function is it?
The relation can be represented as:
![\left[\begin{array}{c}y\\ \\10\\11\\4\\10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dy%5C%5C%20%5C%5C10%5C%5C11%5C%5C4%5C%5C10%5Cend%7Barray%7D%5Cright%5D)
Where
and 
Notice that the range has an occurrence of 10 (twice)
i.e.
and 
In function and relations, when two different values in the domain point to the same value in the range implies that, <em>the relation is many to one.</em>
15% becomes 0.15
multiply it by 8700
0.15x8700=1305
the second duckling is wandering by 2.6 units distance than the first duckling .
<u>Step-by-step explanation:</u>
Here we have , Two ducklings wander away from the nest while their mother is away. The first duckling's displacement (distance and direction) from the nest is (12,5) The second duckling's displacement is (13,-8) . We need to find How much farther did the second duckling wander than the first duckling. Let's find out:
Let a = (12,5) and b =(13,-8)
The distance each duckling wandered is the magnitude of its displacement vector. Therefore, the expression Distance second duck wandered is given by :
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Therefore , the second duckling is wandering by 2.6 units distance than the first duckling .