ADDITIONAL MATHEMATICS FORM 5

ADDITIONAL MATHEMATICS FORM 5

10th - 11th Grade

10 Qs

quiz-placeholder

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ADDITIONAL MATHEMATICS FORM 5

ADDITIONAL MATHEMATICS FORM 5

Assessment

Quiz

Mathematics

10th - 11th Grade

Practice Problem

Hard

CCSS
HSA-REI.B.4B, HSF.IF.A.2, HSF-BF.A.1C

Standards-aligned

Created by

FARAH Moe

Used 133+ times

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Suatu persamaan kuadratik mempunyai dua punca yang sama. Apakah rumus yang patut digunakan untuk menyelesaikannya?

b24ac=0b^2-4ac=0

b24ac>0b^2-4ac>0

b24ac<0b^2-4ac<0

b24ac0b^2-4ac\ge0

Tags

CCSS.HSA-REI.B.4B

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Diberi  g(x)=3x1g(x)=3x-1  , cari nilai  g(4)g\left(4\right)  

10

11

12

13

Tags

CCSS.HSF.IF.A.2

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Persamaan kuadratik     (k+3)x212x+2k=0(k+3)x^2-12x+2k=0  mempunyai dua punca yang berbeza. Cara untuk mencari julat k adalah:

 (12)24(k+3)(2k)=0\left(-12\right)^2-4\left(k+3\right)\left(2k\right)=0  

 (12)24(k+3)(2k)>0\left(12\right)^2-4\left(k+3\right)\left(2k\right)>0  

 (12)24(k+3)(2k)>0\left(-12\right)^2-4\left(k+3\right)\left(2k\right)>0  

 (12)24(k+3)(2k)<0\left(-12\right)^2-4\left(k+3\right)\left(2k\right)<0  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Rumus untuk membentuk persamaan kuadratik adalah

x2+(HTP)x+HDP=0x^2+(HTP)x+HDP=0

x2(HTP)xHDP=0x^2-(HTP)x-HDP=0

x2(HTP)x+HDP=0x^2-(HTP)x+HDP=0

x2+(HTP)xHDP=0x^2+(HTP)x-HDP=0

Tags

CCSS.HSA-REI.B.4B

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Diberi fungsi kuadratik     f(x)=a(x+p)2+9f(x)=a\left(x+p\right)^2+9  , apakah paksi simetri bagi fungsi kuadratik tersebut?

 pp  

 x=px=p  

 x=px=-p  

 p-p  

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Diberi    f:x4x1f:x→4x-1  dan   g:xx2+3g:x→x^2+3  , carikan  fg(3)fg\left(3\right)  

11

12

47

48

Tags

CCSS.HSF-BF.A.1C

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 Diberi    2x217x+30=02x^2-17x+30=0  , kaedah pemfaktoran adalah

 (2x+5)(x+6)=0\left(2x+5\right)\left(x+6\right)=0  

  (x+52)(x+6)=0\left(x+\frac{5}{2}\right)\left(x+6\right)=0  

 (x52)(x6)=0\left(x-\frac{5}{2}\right)\left(x-6\right)=0  

 (2x5)(x6)=0\left(2x-5\right)\left(x-6\right)=0  

Tags

CCSS.HSA-REI.B.4B

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