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scoundrel [369]
4 years ago
6

When completing the square on the equation c2 + 110 = 12, the resulting solution is: It

Mathematics
1 answer:
Delvig [45]4 years ago
6 0

Answer:

c=1 and c=-12

Step-by-step explanation:

The correct equation is:

c^2+11c=12

To solve by completing the square method.

Solution:

We have:

c^2+11c=12

In order to solve by completing the square method we will carry out the following operations to the given equation to get a perfect square binomial.

We can write as:

c^2+2.\frac{11}{2}c=12  

c^2+2.\frac{11}{2}c+(\frac{11}{2})^2-(\frac{11}{2})^2=12

(c+\frac{11}{2})^2-(\frac{11}{2})^2=12  [As  c^2+2.\frac{11}{2}c+(\frac{11}{2})^2=(c+\frac{11}{2})^2]

Adding both sides by (\frac{11}{2})^2

(c+\frac{11}{2})^2-(\frac{11}{2})^2+(\frac{11}{2})^2=12+(\frac{11}{2})^2

(c+\frac{11}{2})^2=12+\frac{121}{4}  

Taking LCD to add fraction.

(c+\frac{11}{2})^2=\frac{48}{4}+\frac{121}{4}

(c+\frac{11}{2})^2=\frac{169}{4}

Taking square root both sides.

\sqrt{(c+\frac{11}{2})^2}=\sqrt{\frac{169}{4}}

c+\frac{11}{2}=\pm\frac{13}{2}

Subtracting both sides by \frac{11}{2} :

c+\frac{11}{2}-\frac{11}{2}=\pm\frac{13}{2}-\frac{11}{2}

c=\pm\frac{13}{2}-\frac{11}{2}

So, we have:

c=\frac{13}{2}-\frac{11}{2} and c=-\frac{13}{2}-\frac{11}{2}

c=\frac{2}{2} and c=\frac{-24}{2}

c=1 and c=-12  (Answer)

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Answer:

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<u>Part A</u>

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 7 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.

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