Tips Menjawab Soal Latihan TPA Numerik Aljabar Part 6

Posted by arman on 04.03

$11.\;(20,11)\;x\;{(0,001)}^{2}\;x\;n\;=\;2011,\;adalah\;...$

$a.\;{10}^4$

$b.\;{10}^5$

$c.\;{10}^6$

$d.\;{10}^7$

$e.\;{10}^8$

$12.\;6\frac{2}{5}\;x\;\frac{2}{7}\;x\;4\frac{1}{3}\;=$

$a.\;9\frac{97}{105}$

$b.\;8\frac{97}{105}$

$c.\;7\frac{97}{105}$

$d.\;6\frac{97}{105}$

$e.\;6\frac{57}{105}$

$13.(\frac{8}{7}:\frac{2}{3}):\frac{1}{2}\;=\;...$

$a.\;\frac{7}{26}$

$b.\;\frac{24}{7}$

$c.\;\frac{26}{7}$

$d.\;\frac{27}{4}$

$e.\;\frac{27}{26}$

$14.7,50\;:\;(\frac{4}{10}\;+\;16%)\;x\;8,27\;=\;$<\p>

$a.110,760$

$b.110,850$

$c.110,970$

$d.110,125$

$e.110,105$

$15.\frac{{17}^2\;-\;{13}^2}{5\;x\;24}\;-\;16,67%\;-\;0,67\;=\;...$

$a.0,1633$

$b.\frac{1}{3}$

$c.0,333$

$d.\frac{1}{4}$

$e.\frac{1}{2}$




Nama Anda
New Johny WussUpdated: 04.03

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