<h3>2
Answers: </h3><h3>
Choice B) Shift down</h3><h3>
Choice C) Shift right</h3>
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Explanation:
Start with the point (-2,-4). Let's try to move it to (2, -9)
To do so, we need to shift down and to the right (in either order).
Specifically we shift 4 units to the right to go from x=-2 to x=4
Note how x=-2 moves to x+4 = -2+4 = 2
Also, we shift 5 units down. We have y = -4 turn into y = -9 as shown below
y ---> y-5 = -4-5 = -9
So the translation rule is 
Let's see what happens when we apply the translation rule to (2,4)

which is the other endpoint of the blue segment. This shows that the translation rule works for (2,4) to move to (6,-1)
We do not apply any dilations. The red and blue segments are the same length because translations preserve distance. All we're effectively doing is moving the red segment to land on the blue segment. This means no vertical or horizontal stretches are done. The same can be said about compressions as well.
Maybe try using Photomath it works for me. Good luck
Answer:
P(Coin landing heads and blue marble is randomly selected) = 3/20.
Step-by-step explanation:
In this question, two events simultaneously take place. First, all the probabilities have to be identified. It is mentioned that the coin is a fair coin, therefore the probabilities of all the outcomes associated with the coin tossing will be equal. Therefore:
P(Coin landing heads) = 1/2.
P(Coin landing tails) = 1/2.
There are a total of 3 blue + 4 green + 3 red = 10 marbles in the bag. Therefore:
P(Selected marble is blue) = 3/10.
P(Selected marble is green) = 4/10.
P(Selected marble is red) = 3/10.
Assuming that both the events are independent, the probabilities of both the events can safely be multiplied. Therefore:
P(Coin landing heads and blue marble is randomly selected) = 1/2 * 3*10 = 3/20.
Therefore, the answer is 3/20!!!