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Lisa [10]
3 years ago
14

What is the square root of 683

Mathematics
2 answers:
Mashutka [201]3 years ago
4 0

Answer:

The square root of six hundred and eighty-three √683 = 26.13426869074



Amiraneli [1.4K]3 years ago
3 0

Answer:

26.1342686907

Step-by-step explanation:


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Which expressions are equal to the expression –16.8 – 4.7? Choose all answers that are correct.
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Use\ commutative\ property:a+b=b+a\\\\-16.8-4.7=-16.8+(-4.7)\\\\therefore:\\\\a.\ -4.7-16.8=-4.7+(-16.8)=-16.8+(-4.7)=-16.8-4.7\ CORRECT\\\\b.\ -4.7+16.8\ NOT\\\\c.\ -16.8+(-4.7)=-16.8-4.7\ CORRECT\\\\d.\ 16.8-(-14.7)=16.8+14.7\ NOT

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A bag contains 6 one-dollar coins and 4 two-dollar coins. Three coins are taken at random without replacement.
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1/(x-1)x+1/x(x+1)+...+1/(x+9)(x+10)=11/12
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Notice that

\dfrac1{(x-1)x} = \dfrac1{x-1} - \dfrac1x \\\\ \dfrac1{x(x+1)} = \dfrac1x - \dfrac1{x+1} \\\\ \vdots \\\\ \dfrac1{(x+9)(x+10)} = \dfrac1{x+9} - \dfrac1{x+10}

That is, the <em>n</em>-th term (where <em>n</em> = -1, 0, 1, …, 9) in the sum on the left side has a partial fraction decomposition of

\dfrac{1}{(x+n)(x+n+1)} = \dfrac1{x+n} - \dfrac1{x+n+1}

and in the sum, some adjacent terms will cancel and leave you with

\dfrac1{(x-1)x} + \dfrac1{x(x+1)} + \cdots + \dfrac1{(x+9)(x+10)} = \dfrac1{x-1} - \dfrac1{x+10} = \dfrac{11}{12}

Now solve for <em>x</em>, bearing in mind that we cannot have <em>x</em> = 0, -1, -2, …, -10 :

\dfrac1{x-1} - \dfrac1{x+10} = \dfrac{11}{12}

Combine the fractions on the left side:

\dfrac{(x+10)-(x-1)}{(x-1)(x+10)} = \dfrac{11}{(x-1)(x+10)} = \dfrac{11}{12}

Then we must have

(x-1)(x+10) = x^2 + 9x - 10 = 12 \implies x^2+9x-22 = (x+11)(x-2) = 0

so that either <em>x</em> = 2 or <em>x</em> = -11.

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