Question Number 191191 by mathlove last updated on 20/Apr/23
Answered by mehdee42 last updated on 20/Apr/23
$$\mathrm{14} \\ $$
Answered by mr W last updated on 20/Apr/23
$${x}+{y}=\mathrm{19}\:\:\:\:\:\:\:\:\:×\mathrm{4} \\ $$$${x}+\mathrm{4}{y}=\mathrm{34} \\ $$$$\mathrm{3}{x}=\mathrm{4}×\mathrm{19}−\mathrm{34} \\ $$$$\Rightarrow{x}=\frac{\mathrm{4}×\mathrm{19}−\mathrm{34}}{\mathrm{3}}=\mathrm{14} \\ $$
Commented by mathlove last updated on 21/Apr/23
$${tnks} \\ $$
Answered by anr0h3 last updated on 21/Apr/23
$$ \\ $$$${x}:\:\mathrm{1}\:{glass}\:{heigth} \\ $$$${r}:\:{the}\:{part}\:{exceeds}\:{one}\:{glass}\:{when}\:{putting}\:{another}\:{one}\:{inside},\:{and}\:{r}\:{uniform} \\ $$$$ \\ $$$$\mathrm{19}−{r}={x} \\ $$$$\mathrm{34}−\mathrm{4}{r}={x} \\ $$$$\mathrm{19}−{r}=\mathrm{34}−\mathrm{4}{r} \\ $$$$\mathrm{4}{r}−{r}=\mathrm{34}−\mathrm{19} \\ $$$$\mathrm{3}{r}=\mathrm{15} \\ $$$${r}=\frac{\mathrm{15}}{\mathrm{3}}=\mathrm{5} \\ $$$${r}=\mathrm{5}\rightarrow\mathrm{19}−{r}={x}\:{and}\:{x}=\mathrm{14} \\ $$$${answer}:\:{one}\:{glass}\:{heigth}\:{is}\:\mathrm{14}{cm} \\ $$$$ \\ $$$$ \\ $$
Commented by mathlove last updated on 21/Apr/23
$${thanks} \\ $$