Matura (maj 2011, poziom podstawowy, próbna, OKE Łomża)

$-2$-2

$\displaystyle - \frac{1}{2}$\displaystyle - \frac{1}{2}

$\displaystyle \frac{1}{2}$\displaystyle \frac{1}{2}

$2$2

trzy rozwiązania: $x=-4$x=-4, $x=-1$x=-1, $x=1$x=1

trzy rozwiązania: $x=-1$x=-1, $x=1$x=1, $x=4$x=4

dwa rozwiązania: $x=-4$x=-4, $x=1$x=1

dwa rozwiązania: $x=-4$x=-4, $x=0$x=0

$25p$25p

$p+25$p+25

$2p$2p

$p+2$p+2

$0$0

$2\sqrt{3}$2\sqrt{3}

$4\sqrt{3}$4\sqrt{3}

$8$8

$\left(-\infty,2\right)\cup\left(2,+\infty\right)$\left(-\infty,2\right)\cup\left(2,+\infty\right)

$\displaystyle \left(-\infty, \frac{ 3}{2} \right)\cup\left( \frac{3}{2} ,+\infty\right)$\displaystyle \left(-\infty, \frac{ 3}{2} \right)\cup\left( \frac{3}{2} ,+\infty\right)

$\left(-\infty,1\right)\cup\left(1,5\right)\cup\left(5,+\infty\right)$\left(-\infty,1\right)\cup\left(1,5\right)\cup\left(5,+\infty\right)

$\left(-\infty,-2\right)\cup\left(-2,+\infty\right)$\left(-\infty,-2\right)\cup\left(-2,+\infty\right)

$f(x)=-4x+2$f(x)=-4x+2

$f(x)=-4x-2$f(x)=-4x-2

$\displaystyle f(x)= \frac{1}{4}x-2$\displaystyle f(x)= \frac{1}{4}x-2

$\displaystyle f(x)=- \frac{1}{4}x-2$\displaystyle f(x)=- \frac{1}{4}x-2

$a_7=3$a_7=3

$\displaystyle a_7=3 \frac{1}{5}$\displaystyle a_7=3 \frac{1}{5}

$\displaystyle a_7=3 \frac{2}{5}$\displaystyle a_7=3 \frac{2}{5}

$a_7=4$a_7=4

$\displaystyle \frac{4}{3}$\displaystyle \frac{4}{3}

$\displaystyle \frac{3}{4}$\displaystyle \frac{3}{4}

$\displaystyle \frac{3}{5}$\displaystyle \frac{3}{5}

$\displaystyle \frac{4}{5}$\displaystyle \frac{4}{5}

$9{,}6\ \text{cm}$9{,}6\ \text{cm}

$24\ \text{cm}$24\ \text{cm}

$60\ \text{cm}$60\ \text{cm}

$150\ \text{cm}$150\ \text{cm}

$y=-2x+3$y=-2x+3

$y=2x-1$y=2x-1

$y=2x+5$y=2x+5

$y=-x+1$y=-x+1

$20\%$20\%

$25\%$25\%

$50\%$50\%

$80\%$80\%

$\left(-\infty,2\right)$\left(-\infty,2\right)

$\left(-\infty,3\right]$\left(-\infty,3\right]

$\left(-\infty,3\right)$\left(-\infty,3\right)

$\left(-\infty,2\right]$\left(-\infty,2\right]

$\left(x+4\right)^2+\left(y-5\right)^2=16$\left(x+4\right)^2+\left(y-5\right)^2=16

$\left(x-4\right)^2+\left(y+5\right)^2=25$\left(x-4\right)^2+\left(y+5\right)^2=25

$\left(x+4\right)^2+\left(y-5\right)^2=25$\left(x+4\right)^2+\left(y-5\right)^2=25

$\left(x-4\right)^2+\left(y+5\right)^2=16$\left(x-4\right)^2+\left(y+5\right)^2=16