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galben [10]
3 years ago
12

How to you get 28 from the numbers 3, 6, 2, and 8, while only using each number once?

Mathematics
2 answers:
Serhud [2]3 years ago
5 0

Answer:

9 1/3 or 9.33, 28/6 or 4 4/6, 14 and 28/8 or 3 4/8

Step-by-step explanation:

( 1 ) <em>The number 28 divided by 3 gives </em><em>9 1/3 </em><em>or </em><em>9.33.</em>

<em>( 2 ) Using a calculator, if you typed in 28 divided by 6, you'd get 4.6667. You could also express </em><em>28/6</em><em> as a mixed fraction:</em><em> 4 4/6.</em>

<em>( 3 ) 28 divided by 2 is </em><em>14.</em>

<em>( 4 ) Using a calculator, if you typed in 28 divided by 8, you'd get </em><em>3.5.</em><em> You could also express </em><em>28/8 </em><em>as a mixed fraction: </em><em>3 4/8.</em>

valentina_108 [34]3 years ago
3 0

Answer:

3x6+2+8=28

Step-by-step explanation:

3x6=18

18 + 2 = 20

20 + 8 = 28

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"Find an integer, x, such that 5, 10, and x represent the lengths of the sides of an acute triangle.
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1^o\\5\leq10\leq x\\\\\left\{\begin{array}{ccc}5+10 \ \textgreater \  x\\5^2+10^2 \ \textgreater \  x^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x^2 \ \textless \  125\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x \ \textless \  \sqrt{125}\approx11.1\end{array}\right\\\\\boxed{x=11}\\\\2^o\\5\leq x\leq10\\\\\left\{\begin{array}{ccc}5+x \ \textgreater \  10\\ 5^2+x^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x^2 \ \textgreater \  75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x \ \textgreater \  \sqrt{75}\approx8.7\end{array}\right\\\boxed{x=9}

3^o\\x\leq5\leq10\\\\\left\{\begin{array}{ccc}x +5 \ \textgreater \  10\\ x^2+5^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x\in\O}\\\\Answer:\boxed{x=9\ or\ x=11}\to your\ answer:\boxed{\boxed{x=11}}
8 0
4 years ago
Read 2 more answers
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