If LM = 6, what is the perimeter of △PKQ?
2 answers:
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Answer:
The original function was transformed by a a horizontal shift to the right in 1 unit, and also a vertical shift upwards of 5 units.
Step-by-step explanation:
Recall the four very important rules regarding translations (shifts) of the graph of functions:
1) In order to shift the graph of a function vertically c units upwards, we must transform f (x) by adding c to it.
2) In order to shift the graph of a function vertically c units downwards, we must transform f (x) by subtracting c from it.
3) In order to shift the graph of a function horizontally c units to the right, we must transform the variable x by subtracting c from x.
4) In order to shift the graph of a function horizontally c units to the left, we must transform the variable x by adding c to x.
We notice that in our case, The original function has been transformed by "subtracting 1 unit from x", and by adding 5 units to the full function. Therefore we are in the presence of a horizontal shift to the right in 1 unit (as explained in rule 3), and also a vertical shift upwards of 5 units (as explained in rule 1).
Answer:
B) T = 2πr/v
Step-by-step explanation:
To solve the given equation for T, multiply it by T/v.
The given equation is where h is the height, in feet, of a ball and t is the time, in seconds.
<u>Part a: The height of the ball when t = 2 seconds:</u>
The height of the ball above the ground 2 seconds after it is released can be determined by substituting t= 2 in the equation , we get;
Simplifying the terms, we get;
Thus, the height of the ball after 2 seconds is 100 feet.
<u>Part b: The height of the ball when t = 4 seconds:</u>
The height of the ball above the ground 4 seconds after it is released can be determined by substituting t = 4 in the equation , we get;
Simplifying the terms, we get;
Thus, the height of the ball after 4 seconds is 68 feet.