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48
12
8

{:[148=2(z+12)],[188=z^(2)+62 z],[0=z^(2)+12 z-10],[(z+3.2)(z-15.2)],[2=-3.2quad z=15.2]:}
Find the diameter of 
o.O. A line that appears to be tangent is tangent. If your answer is not a whole number, round to the nearest tenth.
10.
11.
12.
12.5

0=36

0=8.

4848\newline1212\newline88\newline148=2(z+12)188=z2+62z0=z2+12z10(z+3.2)(z15.2)2=3.2z=15.2 \begin{array}{l} 148=2(z+12) \\ 188=z^{2}+62 z \\ 0=z^{2}+12 z-10 \\ (z+3.2)(z-15.2) \\ 2=-3.2 \quad z=15.2 \end{array} \newlineFind the diameter of O \odot O . A line that appears to be tangent is tangent. If your answer is not a whole number, round to the nearest tenth.\newline1010.\newline1111.\newline1212.\newline1212.55\newline0=36 0=36 \newline0=8 0=8 .

Full solution

Q. 4848\newline1212\newline88\newline148=2(z+12)188=z2+62z0=z2+12z10(z+3.2)(z15.2)2=3.2z=15.2 \begin{array}{l} 148=2(z+12) \\ 188=z^{2}+62 z \\ 0=z^{2}+12 z-10 \\ (z+3.2)(z-15.2) \\ 2=-3.2 \quad z=15.2 \end{array} \newlineFind the diameter of O \odot O . A line that appears to be tangent is tangent. If your answer is not a whole number, round to the nearest tenth.\newline1010.\newline1111.\newline1212.\newline1212.55\newline0=36 0=36 \newline0=8 0=8 .
  1. Question Prompt: question_prompt: Find the diameter of a circle where a line that appears to be tangent is tangent.
  2. Step 11: Step 11: Solve the first equation 148=2(z+12)148 = 2(z + 12).
  3. Step 22: Step 22: Plug the value of zz into the second equation 188=z2+62z188 = z^2 + 62z.

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