Question

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19.3. Найдите значение функции: 1) y=x + 8, если х = 1782; 1101; ; 0,3; 2) y = 0,01x - 2,5, если х = 25; 250; 2,5; 3) y=A + 25% x, если х = 40; 100; 40; 100; . 3 1 1 ; 8. 8​

Answer 1

Ответ:

$1)\ \ y=\dfrac{1}{3}\, x+8\\\\x=1782:\ y(1782)=\dfrac{1}{3}\cdot 1782+8=594+8=602\\\\x=1101:\ y(1101)=\dfrac{1}{3}\cdot 1101+8=367+8=375\\\\x=\dfrac{2}{3}:\ \ y(\dfrac{2}{3})=\dfrac{1}{3}\cdot \dfrac{2}{3}+8=\dfrac{2}{9}+8=8\dfrac{2}{9}\\\\x=0,3:\ \ y(0,3)=\dfrac{0,3}{3}+8=\dfrac{3}{30}+8=8\dfrac{1}{10}=8,1$

$2)\ \ y=0,01x-2,5\\\\x=25:\ \ y(25)=0,01\cdot 25-2,5=0,25-2,5=-2,25\\\\x=250:\ \ y(250)=0,01\cdot 250-2,5=2,5-2,5=0\\\\x=2,5:\ \ y(2,5)=0,01\cdot 2,5-2,5=0,025-2,5=-2,475\\\\\\3)\ \ y=\dfrac{1}{8}+25\%\, x\\\\x=40:\ \ y(40)=\dfrac{1}{8}+0,25\cdot 40=\dfrac{1}{8}+10=10\dfrac{1}{8}\\\\x=100:\ \ y(100)=\dfrac{1}{8}+0,25\cdot 100=\dfrac{1}{8}+25=25\dfrac{1}{8}$

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