Matura (sierpień 2012, poziom podstawowy, poprawka)

$10\%$10\%

$11\%$11\%

$20\%$20\%

$21\%$21\%

$3^{-4}$3^{-4}

$3^{-9}$3^{-9}

$9^{-1}$9^{-1}

$9^{-9}$9^{-9}

$0$0

$1$1

$2$2

$3$3

$-14$-14

$22$22

$-14-12\sqrt{2}$-14-12\sqrt{2}

$22-12\sqrt{2}$22-12\sqrt{2}

$m=3$m=3

$m=1$m=1

$m=-2$m=-2

$m=-4$m=-4

$x=-8$x=-8

$x=-4$x=-4

$x=4$x=4

$x=8$x=8

$\left(-6,0\right)$\left(-6,0\right)

$\left(0,6\right)$\left(0,6\right)

$\left(-\infty,-6\right)\cup\left(0,+\infty\right)$\left(-\infty,-6\right)\cup\left(0,+\infty\right)

$\left(-\infty,0\right)\cup\left(6,+\infty\right)$\left(-\infty,0\right)\cup\left(6,+\infty\right)

$\left(x^3+1\right)\left(x^2-2\right)$\left(x^3+1\right)\left(x^2-2\right)

$\left(x^3-1\right)\left(x^3+2\right)$\left(x^3-1\right)\left(x^3+2\right)

$\left(x^2+2\right)\left(x^4-1\right)$\left(x^2+2\right)\left(x^4-1\right)

$\left(x^4-2\right)\left(x+1\right)$\left(x^4-2\right)\left(x+1\right)

dokładnie jedno rozwiązanie

dokładnie dwa rozwiązania

dokładnie trzy rozwiązania

dokładnie cztery rozwiązania

$\displaystyle a_3= \frac{1}{2}$\displaystyle a_3= \frac{1}{2}

$\displaystyle a_3= -\frac{1}{2}$\displaystyle a_3= -\frac{1}{2}

$\displaystyle a_3= \frac{3}{8}$\displaystyle a_3= \frac{3}{8}

$\displaystyle a_3= -\frac{3}{8}$\displaystyle a_3= -\frac{3}{8}

$a_4=-18$a_4=-18

$a_4=0$a_4=0

$a_4=4{,}5$a_4=4{,}5

$a_4=144$a_4=144

$\displaystyle \frac{7}{6}$\displaystyle \frac{7}{6}

$\displaystyle \frac{7 \cdot 13}{120}$\displaystyle \frac{7 \cdot 13}{120}

$\displaystyle \frac{7}{\sqrt{120}}$\displaystyle \frac{7}{\sqrt{120}}

$\displaystyle \frac{7}{13\sqrt{120}}$\displaystyle \frac{7}{13\sqrt{120}}

$\displaystyle \cos\alpha= \frac{9}{11}$\displaystyle \cos\alpha= \frac{9}{11}

$\displaystyle \sin\alpha= \frac{9}{11}$\displaystyle \sin\alpha= \frac{9}{11}

$\displaystyle \sin\alpha= \frac{11}{2\sqrt{10}}$\displaystyle \sin\alpha= \frac{11}{2\sqrt{10}}

$\displaystyle \cos\alpha= \frac{2\sqrt{10}}{11}$\displaystyle \cos\alpha= \frac{2\sqrt{10}}{11}

$8$8

$4\sqrt{10}$4\sqrt{10}

$2\sqrt{58}$2\sqrt{58}

$10$10

$65^\circ$65^\circ

$100^\circ$100^\circ

$115^\circ$115^\circ

$130^\circ$130^\circ

$36$36

$18$18

$12$12

$6$6

$y=3x$y=3x

$y=-3x$y=-3x

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

$\displaystyle y= \frac{1}{3}x+2$\displaystyle y= \frac{1}{3}x+2

$29$29

$40$40

$58$58

$74$74

$\left(-4,-6\right)$\left(-4,-6\right)

$\left(4,6\right)$\left(4,6\right)

$\left(4,-6\right)$\left(4,-6\right)

$\left(-4,6\right)$\left(-4,6\right)

$512$512

$384$384

$96$96

$16$16

$\displaystyle \frac{\sqrt{3}}{6}\pi a^3$\displaystyle \frac{\sqrt{3}}{6}\pi a^3

$\displaystyle \frac{\sqrt{3}}{8}\pi a^3$\displaystyle \frac{\sqrt{3}}{8}\pi a^3

$\displaystyle \frac{\sqrt{3}}{12}\pi a^3$\displaystyle \frac{\sqrt{3}}{12}\pi a^3

$\displaystyle \frac{\sqrt{3}}{24}\pi a^3$\displaystyle \frac{\sqrt{3}}{24}\pi a^3

$3400\ \text{zł}$3400\ \text{zł}

$3500\ \text{zł}$3500\ \text{zł}

$6000\ \text{zł}$6000\ \text{zł}

$7000\ \text{zł}$7000\ \text{zł}

$\displaystyle p< \frac{1}{5}$\displaystyle p< \frac{1}{5}

$\displaystyle p= \frac{1}{5}$\displaystyle p= \frac{1}{5}

$\displaystyle p= \frac{1}{4}$\displaystyle p= \frac{1}{4}

$\displaystyle p> \frac{1}{4}$\displaystyle p> \frac{1}{4}