Question Number 7902 by tawakalitu last updated on 24/Sep/16
$${The}\:{diameter}\:{of}\:{the}\:{wheel}\:{of}\:{a}\:{car}\:{is}\:\mathrm{36}\:{cm}.\: \\ $$$${how}\:{many}\:{revolutions}\:{correct}\:{to}\:\mathrm{3}\:{significant} \\ $$$${figure}\:,\:{will}\:{it}\:{make}\:{to}\:{cover}\:{a}\:{distance}\:{of}\:\mathrm{1}.\mathrm{05}{km} \\ $$$${take}\:\pi\:=\:\frac{\mathrm{22}}{\mathrm{7}} \\ $$
Answered by sandy_suhendra last updated on 24/Sep/16
$$\mathrm{1}.\mathrm{05}\:{km}\:=\:\mathrm{1}.\mathrm{05}×\mathrm{10}^{\mathrm{5}} \:{cm} \\ $$$$\frac{\mathrm{1}.\mathrm{05}×\mathrm{10}^{\mathrm{5}} }{\frac{\mathrm{22}}{\mathrm{7}}×\mathrm{36}}\:=\:\mathrm{928},\mathrm{030}\:=\:\mathrm{928}\:{revolutions} \\ $$
Commented by tawakalitu last updated on 24/Sep/16
$${i}\:{really}\:{appreciate}. \\ $$