Wykres funkcji f(x) = (x - 2) 2 + 1 otrzymamy, przesuwając parabolę y = x 2 o:

2 jednostki w prawo i 1 jednostkę w górę

2 jednostki w prawo i 1 jednostkę w dół

2 jednostki w lewo i 1 jednostkę w dół

2 jednostki w lewo i 1 jednostkę w górę

13 4 + 3 = \frac{13}{4+\sqrt{3}}= 4 + 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> 1 3 =

4 − 3 4-\sqrt{3} 4 − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

13 + 4 3 13+4\sqrt{3} 1 3 + 4 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

13 ( 4 − 3 ) 13\left(4-\sqrt{3}\right) 1 3 ( 4 − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> )

4 + 3 4+\sqrt{3} 4 + 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Kąt α \alpha α jest ostry i t g α = 2 3 tg\alpha=\frac{2}{3} t g α = 3 2 . Wówczas sin ⁡ α = \sin\alpha= sin α =

2 13 13 \frac{2\sqrt{13}}{13} 1 3 2 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 13 26 \frac{3\sqrt{13}}{26} 2 6 3 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

13 13 \frac{\sqrt{13}}{13} 1 3 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 13 13 \frac{3\sqrt{13}}{13} 1 3 3 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Punkty B(-1;4), i C(5;2) są sąsiednimi wierzchołkami kwadratu ABCD. Pole kwadratu jest równe:

40 \sqrt{40} 4 0 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

L i c z b a 4 − 1 ⋅ 2 1 4 ⋅ 1 6 1 3 Liczba\ \sqrt{4^{-1}}\cdot2^{\frac{1}{4}}\cdot16^{\frac{1}{3}} L i c z b a 4 − 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ⋅ 2 4 1 ⋅ 1 6 3 1 jest równa

2 7 12 2^{\frac{7}{12}} 2 1 2 7

2 25 12 2^{\frac{25}{12}} 2 1 2 2 5

2 − 1 9 2^{\frac{-1}{9}} 2 9 − 1

Zadania maturalne

Authored by Karolina Faszczewska

Mathematics

1st - 5th Grade

Used 84+ times

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MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Ciąg (1,x,16) jest ciągiem geometrycznym dla:

x=4

x=8

x=4 lub x=8

x=4 lub x=-4

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Ile jest liczb dwucyfrowych niepodzielnych przez 5?

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

-1

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

2³⁰

2⁵⁷

2⁶³

2¹¹²

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Wyznacz wzór funkcji liniowej przechodzącej przez punkt A=(0;6), której wykres jest równoległy do wykresu funkcji y=3x+2

y=3x+6

y=-3x+6

y=2

y=6x-2

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Współrzędne punktu przecięcia wykresu funkcji y=-x+6 z osią OY wynoszą:

(0,-6)

(6,0)

(-6,0)

(0,6)

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Wierzchołkiem funkcji kwadratowej y=2(x-3)²-5 jest punkt:

(3;-5)

(3;5)

(-3;-5)

(-3;5)

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Zadania maturalne

Ciąg (1,x,16) jest ciągiem geometrycznym dla:

Ile jest liczb dwucyfrowych niepodzielnych przez 5?

Wyznacz wzór funkcji liniowej przechodzącej przez punkt A=(0;6), której wykres jest równoległy do wykresu funkcji y=3x+2

Współrzędne punktu przecięcia wykresu funkcji y=-x+6 z osią OY wynoszą:

Wierzchołkiem funkcji kwadratowej y=2(x-3)²-5 jest punkt:

Wykres funkcji f(x) = (x - 2)² + 1 otrzymamy, przesuwając parabolę y = x² o:

Które z poniższych zdań jest prawdziwe?

Suma pierwiastków równania
(2x-5)(4x+6)(x+1)(1-x³)=0
jest równa:

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Zadania maturalne

Ciąg (1,x,16) jest ciągiem geometrycznym dla:

Ile jest liczb dwucyfrowych niepodzielnych przez 5?

Wyznacz wzór funkcji liniowej przechodzącej przez punkt A=(0;6), której wykres jest równoległy do wykresu funkcji y=3x+2

Współrzędne punktu przecięcia wykresu funkcji y=-x+6 z osią OY wynoszą:

Wierzchołkiem funkcji kwadratowej y=2(x-3)²-5 jest punkt:

Wykres funkcji f(x) = (x - 2)2 + 1 otrzymamy, przesuwając parabolę y = x2 o:

Które z poniższych zdań jest prawdziwe?

Suma pierwiastków równania(2x-5)(4x+6)(x+1)(1-x3)=0 jest równa:

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Wykres funkcji f(x) = (x - 2)² + 1 otrzymamy, przesuwając parabolę y = x² o:

Suma pierwiastków równania
(2x-5)(4x+6)(x+1)(1-x³)=0
jest równa: