Answer:
KL = 3
Step-by-step explanation:
It appears we are to assume that K lies between J and L, so we have ...
JK +KL = JL
(2x+3) +(x) = (4x)
3x +3 = 4x
3 = x . . . . . . . subtract 3x
KL = 3
(x-1/2)² + (y-3/2)² = 4/9
For any circle in the standard form (x-h)² + (y-k)² = r², the center is at (h, k) and the radius is r.
In the case of our equation here. the center is at (1/2, 3/2) with a radius of 2/3.
I REALIZED I DID IT WRONG SO UH IGNORE WHAT I SAID IM SORRY D:
Part (a)
Plug in y = 0 and solve for x. Use the zero product property
y = x(x+3)(x-2)
0 = x(x+3)(x-2)
x(x+3)(x-2) = 0
x = 0 or x+3 = 0 or x-2 = 0 .... zero product property
x = 0 or x = -3 or x = 2
The three roots or zeros are x = 0 or x = -3 or x = 2
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Part (b)
The roots of the graph are: x = -2, x = 1, x = 3. Each root is where the graph crosses or touches the horizontal x axis.
Note how x = 0 is found in part (a), but not found here. This is one example where the graphs don't match. Another would be x = -3 is in part (a), but not here.
So that's why the graph does <u>not</u> match with the function in part (a)