Matemática, perguntado por mriajuwlia, 8 meses atrás


alguém poderia me ajudar a fazer essas contas?

a) V12 - 316 + V8 - 24
b) 3V8 + 4/18 - 3V50 + 32
c) 125+ 227-20 + 3/12
d) 320 + 32 -245 + 50
e 2 + 32​

Soluções para a tarefa

Respondido por osextordq
1

a)V\cdot \:12-316+V\cdot \:8-24\\\\=12V+8V-316-24\\\\=20V-316-24\\\\=20V-340\\\\

b) 3V\cdot \:8+\frac{4}{18}-3V\cdot \:50+32\\\\=24V+\frac{4}{18}-3\cdot \:50V+32\\\\=24V+\frac{2}{9}-3\cdot \:50V+32\\\\=24V+\frac{2}{9}-150V+32\\\\=24V-150V+\frac{2}{9}+32\\\\=-126V+\frac{2}{9}+32\\\\=\frac{32\cdot \:9}{9}+\frac{2}{9}\\\\=\frac{32\cdot \:9+2}{9}\\\\=-126V+\frac{290}{9}

c) 125+227-20+3/12\\\\=125+227-20+\frac{3}{12}\\\\=\frac{1329}{4}\\\\\frac{1329}{4}=332\frac{1}{4}\\\\=332\frac{1}{4}

d) 320+32-245+50\\\\320+32=352\\\\=352-245+50\\\\352-245=107\\\\=107+50\\\\107+50=157\\\\=157

e) 2+32\\\\2+32=34\\\\=34

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