Matematika

Matematika

12th Grade

40 Qs

quiz-placeholder

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Matematika

Matematika

Assessment

Quiz

Mathematics

12th Grade

Practice Problem

Hard

Created by

ujian Online

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40 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Bentuk sederhana dari (x2y3z6)2(x2y6z3)3\frac{\left(x^2y^3z^{-6}\right)^2}{\left(x^{-2}y^6z^3\right)^3}  adalah...

 x6y11z2\frac{x^6}{y^{11}z^2}  

 x10y12z21\frac{x^{10}}{y^{12}z^{21}}  

 x12y10z21\frac{x^{12}}{y^{10}z^{21}}  

 x10y12z20\frac{x^{10}y^{12}}{z^{20}}  

 x12z21y10\frac{x^{12}z^{21}}{y^{10}}  

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Hasil dari 12513180+4520\sqrt{125}-\frac{1}{3}\sqrt{180}+\sqrt{45}-\sqrt{20}  adalah...


 555\sqrt{5}  

 454\sqrt{5}  

 353\sqrt{5}  

 252\sqrt{5}  

 5\sqrt{5}  

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Persamaan fungsi kuadrat dari grafik diatas adalah ….

y=x26x+5y=-x^2-6x+5

y=x2+6x5y=-x^2+6x-5

y=x2+6x+5y=-x^2+6x+5

y=x26x+5y=x^2-6x+5

y=x2+6x+5y=x^2+6x+5

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Suatu fungsi kuadrat memotong sumbu X di titik (−1, 0) dan (4, 0) serta memotong sumbu Y di titik (0, 8). Fungsi kuadrat tersebut adalah …..

𝑓(𝑥) = −2𝑥2 − 6𝑥 − 8

𝑓(𝑥) = −2𝑥2 − 6𝑥 + 8

𝑓(𝑥) = −2𝑥2 + 6𝑥 +8

𝑓(𝑥) = 2𝑥2 + 6𝑥 +8

𝑓(𝑥) = 2𝑥2 + 6𝑥 - 8

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Koordinat titik balik dari grafik fungsi kuadrat

𝑦 = −𝑥2 − 𝑥 + 6 adalah ….

(5,2) \left(-5,-2\right)\

(2 12, 5)\left(2\ \frac{1}{2},\ -5\right)

(5,25)\left(-5,\frac{2}{5}\right)

(2,254)\left(2,\frac{25}{4}\right)

(12,6 14)\left(-\frac{1}{2},6\ \frac{1}{4}\right)

6.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Bentuk sederhana dari 232  5\frac{2\sqrt{3}}{\sqrt{2\ -\ \sqrt{5}}}  adalah

 23(6+15)-\frac{2}{3}\left(\sqrt{6}+\sqrt{15}\right)  

 23(615)-\frac{2}{3}\left(\sqrt{6}-\sqrt{15}\right)  

 32(15+6)\frac{3}{2}\left(\sqrt{15}+\sqrt{6}\right)  

 23(6+15)\frac{2}{3}\left(\sqrt{6}+\sqrt{15}\right)  

 32(615)\frac{3}{2}\left(\sqrt{6}-\sqrt{15}\right)  

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Jika 𝛼 dan 𝛽 merupakan akar-akar dari persamaan kuadrat 2𝑥2 − 3𝑥 − 4 = 0, persamaan kuadrat baru yang akar-akarnya (𝛼 − 3) dan (𝛽 − 3) adalah ….

2𝑥2 + 5𝑥 + 5 = 0

2𝑥2 + 6𝑥 − 5 = 0

2𝑥2 − 7𝑥 + 5 = 0

2𝑥2 − 8𝑥 − 5 = 0

2𝑥2 + 9𝑥 + 5 = 0

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