The product of two consecutive odd numbers is 19043. Which is the smaller number? [#869]

137

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137

X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28

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X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28

125

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125

100 * 100% + 200 * 200% = $100\times \frac{100}{100} + 200 \times \frac{200}{100}$ = $\frac{10000}{100} + \frac{40000}{100}$ = 100 + 400 = 500

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100 * 100% + 200 * 200% = $100\times \frac{100}{100} + 200 \times \frac{200}{100}$ = $\frac{10000}{100} + \frac{40000}{100}$ = 100 + 400 = 500

The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

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The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

= $\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}$ = $\sqrt{\mathrm{0.02*0.02}} + \sqrt{\mathrm{1.2*1.2}} + \sqrt{\mathrm{1.3*1.3}} + \sqrt{\mathrm{0.03*0.03}}$ = 0.2 + 1.2 + 1.3 + 0.03 = 2.73

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= $\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}$ = $\sqrt{\mathrm{0.02*0.02}} + \sqrt{\mathrm{1.2*1.2}} + \sqrt{\mathrm{1.3*1.3}} + \sqrt{\mathrm{0.03*0.03}}$ = 0.2 + 1.2 + 1.3 + 0.03 = 2.73

376 -> 367 629 -> 269 921 -> 129 * 397 -> 379

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376 -> 367 629 -> 269 921 -> 129 * 397 -> 379

10 m

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10 m

42 - (4*3) = 16 - 12 = 4

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4^{2 - (4*3) = 16 - 12 = 4}

Area = πr² π1² : π3² = π1 : π9 = 1:9

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Area = πr² π1² : π3² = π1 : π9 = 1:9

720

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720