Question

two postive numbers are in the ratio of 7:11 if the sum of there square is equal to 680 find the smaller of two numbers​

Answer 1

Step-by-step explanation:

$let \: the \: numbers \: are \: 7x \: and \: 11x \: where \: x > 0 \\ by \: the \: question \: we \: get \: {(7x)}^{2} + {(11x)}^{2} = 680 \\ 49 {x}^{2} + 121 {x}^{2} = 680 \\ 170 {x }^{2} = 680 \\ {x}^{2} = \frac{680}{170} \\ {x}^{2} = 4 \\ x = \sqrt{4} \\ x = 2 \\ so \: the \: numbers \: are \\ 7x = 7 \times 2 = 14 \: and \:11x = 11 \times 2 = 22$