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sineoko [7]
3 years ago
10

Solve. -1/8c=2 a. 4 b. –16 c. –4 d. 16

Mathematics
1 answer:
kotykmax [81]3 years ago
4 0
\frac{-1}{8c} =2
This requires that  c ≠ 0 , meaning c≠0
\frac{-1}{8c} =2
Multiply by x to eliminate var in denominator
\frac{ \not c*(-1)}{ 8 \not c} =c*2
\frac{-1}{8} =c*2
Common denominator  is 8 
\frac{\not 8(-1)}{\not 8} =8c*2
-1=8c*2
-1=16c
x=-16
The answer is: b. -16
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 <span>number times itself equals 196 
</span>
= n * n = 196

14 * 14 = 196 


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Yes, the total cost of Lacey lesson is a split function of hours scheduled

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\huge \boxed{\mathfrak{Question} \downarrow}

  • Expand & simplify ⇨ ( \sqrt{10}  -  \sqrt{2} ) ^{2}. Give your answer in the form b - c \:  \sqrt{5} where b & c are integers.

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{10}-\sqrt{2}\right)^{2}.

\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}

The square of \sqrt{10} is 10.

10-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}

Multiply \sqrt{2} and \sqrt{2} to get 2.

10-2\times 2\sqrt{5}+\left(\sqrt{2}\right)^{2}

Multiply -2 and 2 to get -4.

10-4\sqrt{5}+\left(\sqrt{2}\right)^{2}

The square of \sqrt{2} is 2.

10-4\sqrt{5}+2

Add 10 and 2 to get 12.

\boxed{ \boxed{\bf\:12-4\sqrt{5} }}

  • Here, b & c are integers where \boxed{ \sf \: b = 12 \: and \: c = 4}
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so... whatever "b" and "m" are, their sum for that week was 18
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whatever "b" and "m" are
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solve for "b", to see how many hours she spent babysitting
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