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anyanavicka [17]
3 years ago
5

What is the area of 7 1/3 mm x 4 1/2 mm

Mathematics
1 answer:
sasho [114]3 years ago
5 0

 

\displaystyle\bf\\7\frac{1}{3}~mm\times4\frac{1}{2}~mm=\\\\\\=\frac{7\times3+1}{3}\times\frac{4\times2+1}{2}=\\\\\\=\frac{21+1}{3}\times\frac{8+1}{2}=\\\\\\=\frac{22}{3}\times\frac{9}{2}=\\\\\\=\frac{22\times9}{3\times2}=\\\\\\=\frac{11\times3}{1\times1}=11\times3=\boxed{\bf33~mm^2}

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Answer:

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If without replacement:

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8 Balls:

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If with replacement:

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If without replacement:

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3 years ago
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Hello from MrBillDoesMath!

Answer:

See below


Discussion:

Triangles BAD and BCD are congruent by the SAS postulate.

          * Side AD = DC because BD is the perpendicular bisector  of AC.

          * Angle ADB and CDB are both right angles, and hence equal, because

            BD is the perpendicular bisector of AC)


Corresponding parts of congruent triangles are equal so angle BAD = angle BCD,  which was to be proven.


Regards,  

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Answer:

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Step-by-step explanation:

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