Question

19.4. Найдите значение аргумента x для функции: 1 1) y=ax + х+8, если у 1; ; 3 ; 0,3; 8; 30; 2) y = 0,01x - 2,5, если у = 2,5; 0,01; 25 3) у = + 25% x, если у = 0,5; 10. % 1; 8​

Answer 1

Ответ:

$1)\ \ y=\dfrac{1}{3}\, x+8\\\\x=\dfrac{1}{3}:\ \ y(\dfrac{1}{3})=\dfrac{1}{3}\cdot \dfrac{1}{3}+8=\dfrac{1}{9}+8=8\dfrac{1}{9}\\\\x=0,3:\ \ y(0,3)=\dfrac{0,3}{3}+8=\dfrac{3}{30}+8=8\dfrac{1}{10}=8,1\\\\x=8:\ y(8)=\dfrac{1}{3}\cdot 8+8=\dfrac{32}{3}=10\dfrac{2}{3} \\\\x=30:\ y(30)=\dfrac{1}{3}\cdot 30+8=10+8=18$

$2)\ \ y=0,01x-2,5\\\\x=2,5:\ \ y(2,5)=0,01\cdot 2,5-2,5=0,025-2,5=-2,475\\\\x=0,01:\ \ y(0,01)=0,01\cdot 0,01-2,5=0,0001-2,5=-2,4999\\\\x=\dfrac{1}{25}:\ \ y(\dfrac{1}{25})=0,01\cdot \dfrac{1}{25}-2,5=0,0004-2,5=-2,4996\\\\\\3)\ \ y=\dfrac{1}{8}+25\%\, x\\\\x=\dfrac{1}{4}:\ \ y(\dfrac{1}{4})=\dfrac{1}{8}+0,25\cdot \dfrac{1}{4}=\dfrac{1}{8}+\dfrac{1}{4}\cdot \dfrac{1}{4}=\dfrac{1}{8}+\dfrac{1}{16}=\dfrac{3}{16}$

$x=0,5=\dfrac{1}{2}:\ \ y(\dfrac{1}{2})=\dfrac{1}{8}+0,25\cdot \dfrac{1}{2}=\dfrac{1}{8}+\dfrac{1}{4}\cdot \dfrac{1}{2}=\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{2}{8}=\dfrac{1}{4}=0,25$

$x=10:\ \ y(10)=\dfrac{1}{8}+0,25\cdot 10=\dfrac{1}{8}+2,5=\dfrac{1}{8}+\dfrac{5}{2}=\dfrac{21}{8}=2\dfrac{5}{8}=2,625$