702 Вправа 702Алгебра 8 клас Бевз 2025 · § 14 · стор. 137

702. Користуючись калькулятором, обчисли з точністю до сотих: а) √18 · √50; б) √40 + √90; в) π + √18; г) 3√30 − 2/3; ґ) 2/3 + 2√3; д) π · √10 − 2/7.

а) ${\displaystyle \sqrt{18}\approx 4,2426}$, ${\displaystyle \sqrt{50}\approx 7,0711}$;

${\displaystyle 4,2426\cdot 7,0711\approx 30,0000\approx 30,00}$;

б) ${\displaystyle \sqrt{40}\approx 6,3246}$, ${\displaystyle \sqrt{90}\approx 9,4868}$;

${\displaystyle 6,3246+9,4868=}$${\displaystyle 15,8114\approx 15,81}$;

в) ${\displaystyle \pi \approx 3,1416}$, ${\displaystyle \sqrt{18}\approx 4,2426}$;

${\displaystyle 3,1416+4,2426=}$${\displaystyle 7,3842\approx 7,38}$;

г) ${\displaystyle \sqrt{30}\approx 5,47723}$, ${\displaystyle \frac{2}{3}\approx 0,66667}$;

${\displaystyle 3\cdot 5,47723-0,66667=}$${\displaystyle 16,43169-0,66667=}$${\displaystyle 15,76502\approx 15,77}$;

ґ) ${\displaystyle \sqrt{3}\approx 1,7321}$, ${\displaystyle \frac{2}{3}\approx 0,6667}$;

${\displaystyle 0,6667+2\cdot 1,7321=}$${\displaystyle 0,6667+3,4642=}$${\displaystyle 4,1309\approx 4,13}$;

д) ${\displaystyle \sqrt{10}\approx 3,1623}$, ${\displaystyle \frac{2}{7}\approx 0,2857}$;

${\displaystyle 3,1416\cdot 3,1623-0,2857\approx 9,9347-0,2857=}$${\displaystyle 9,6490\approx 9,65}$.