Given J(1, 1), K(3, 1), L(3, -4), and M(1, -4) and that J'(-1, 5), K'(1, 5), L'(1, 0), and M'(-1, 0). What is the rule that tran
anastassius [24]
(x; y) -> (x - 2; y + 4)
J(1; 1) ⇒ J'(1 - 2; 1 + 4) = (-1; 5)
K(3; 1) ⇒ K'(3 - 2; 1 + 4) = (1; 5)
L(3;-4) ⇒ L'(3 - 2; -4 + 4) = (1; 0)
M(1;-4) ⇒ M'(1 - 2;-4 + 4) = (-1; 0)
<h3>
Answer:</h3>
C (see attached)
<h3>
Step-by-step explanation:</h3>
The linear portion of the graph is defined for x ≥ 1, with the point x=1 included. Only selections A and C do that.
The quadratic portion of the graph is defined for x < 1. Only selection C does that. (Selection A is doubly-defined for x > 1, so is not a function. It is undefined for x < 1.)
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Step-by-step explanation:
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Answer:
51.1 ≈ r
Step-by-step explanation:
"A bicycle wheel travels 321 in for each revolution"- means that 321 is the circumference of the wheel that is a circle.
C = 2· π· r
What is the radius of the wheel ?
321 = 2· π· r, divide both sides of the equation by 2 π
(321/2·π) = r , use calculator, solve and round the answer
51.1 ≈ r
