apakah rumus bagi LUAS segitiga ini

1 2 ( a × b ) \frac{1}{2}(a\times b) 2 1 ( a × b )

1 2 + ( a × b ) \frac{1}{2}+(a\times b) 2 1 + ( a × b )

a × b × c a\times b\times c a × b × c

1 2 ( a × b × c ) \frac{1}{2}(a\times b\times c) 2 1 ( a × b × c )

Jadikan pembolehubah dalam tanda kurungan sebagai perkara rumus m = 2 k + h [ k ] m=2k+h\ \ \ \ \ \ \ \ \ \ \ \ [k] m = 2 k + h [ k ]

k = ( m − h ) 2 k=\frac{(m-h)}{2} k = 2 ( m − h )

k = ( m + h ) × 2 k=(m+h)\times2 k = ( m + h ) × 2

k = ( m + h ) 2 k=\frac{(m+h)}{2} k = 2 ( m + h )

k = ( m − h ) × 2 k=(m-h)\times2 k = ( m − h ) × 2

Jadikan pembolehubah dalam tanda kurungan sebagai perkara rumus 2 m = 7 n − 6 p [ p ] 2m=7n-6p\ \ \ \ \ \ \ \ \ \ \ \left[p\right] 2 m = 7 n − 6 p [ p ]

p = ( 7 n − 2 m ) 6 p=\frac{(7n-2m)}{6} p = 6 ( 7 n − 2 m )

p = ( 2 m − 7 n ) 6 p=\frac{(2m-7n)}{6} p = 6 ( 2 m − 7 n )

p = ( 7 n + 2 m ) 6 p=\frac{(7n+2m)}{6} p = 6 ( 7 n + 2 m )

p = ( 9 m n ) 6 p=\frac{(9mn)}{6} p = 6 ( 9 m n )

Jadikan pembolehubah dalam tanda kurungan sebagai perkara rumus a = 2 r − b 3 [ r ] a=\frac{2r-b}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[r\right] a = 3 2 r − b [ r ]

r = ( 3 a + b ) 2 r=\frac{(3a+b)}{2} r = 2 ( 3 a + b )

r = ( 3 a − b ) 2 r=\frac{(3a-b)}{2} r = 2 ( 3 a − b )

r = ( a − b ) 6 r=\frac{(a-b)}{6} r = 6 ( a − b )

r = 3 ( a + b ) 2 r=\frac{3(a+b)}{2} r = 2 3 ( a + b )

Jadikan pembolehubah dalam tanda kurungan sebagai perkara rumus Y = − 1 − 2 x x + 1 [ x ] Y=\frac{-1-2x}{x+1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[x\right] Y = x + 1 − 1 − 2 x [ x ]

x = ( − 1 − Y ) ( Y + 2 ) x=\frac{(-1-Y)}{(Y+2)} x = ( Y + 2 ) ( − 1 − Y )

x = ( Y − 2 ) ( Y − 1 ) x=\frac{(Y-2)}{(Y-1)} x = ( Y − 1 ) ( Y − 2 )

x = − 2 y x=\frac{-2}{y} x = y − 2

x = − 2 Y Y + 2 x=\frac{-2Y}{Y+2} x = Y + 2 − 2 Y

x + 6 3 + 2 x − 5 6 = \frac{x+6}{3}+\frac{2x-5}{6}= 3 x + 6 + 6 2 x − 5 =

4 x + 7 6 \frac{4x+7}{6} 6 4 x + 7

3 x + 1 6 \frac{3x+1}{6} 6 3 x + 1

3 x + 11 6 \frac{3x+11}{6} 6 3 x + 1 1

4 x + 17 6 \frac{4x+17}{6} 6 4 x + 1 7

Diberi 8 m − 2 n 3 = m n + n \frac{8m-2n}{3}=mn+n 3 8 m − 2 n = m n + n , ungkapkan m dalam sebutan n.

m = 5 n 8 − 3 n m=\frac{5n}{8-3n} m = 8 − 3 n 5 n

m = 3 n 8 − 3 n m=\frac{3n}{8-3n} m = 8 − 3 n 3 n

m = n 8 − 3 n m=\frac{n}{8-3n} m = 8 − 3 n n

m = 5 n 8 m − 3 n m=\frac{5n}{8m-3n} m = 8 m − 3 n 5 n

Ungkapkan 2 + k 3 k − 2 h − 1 6 h \frac{2+k}{3k}-\frac{2h-1}{6h} 3 k 2 + k − 6 h 2 h − 1 sebagai satu pecahan tunggal dalam bentuk termudah.

4 h + k 6 h k \frac{4h+k}{6hk} 6 h k 4 h + k

4 h − k 6 h k \frac{4h-k}{6hk} 6 h k 4 h − k

4 + 3 k − 2 h k 6 h k \frac{4+3k-2hk}{6hk} 6 h k 4 + 3 k − 2 h k

2 h − h k + k 6 h k \frac{2h-hk+k}{6hk} 6 h k 2 h − h k + k

− 2 p q r − 5 p q − ( − p q r ) + 3 p q = -2pqr-5pq-(-pqr)+3pq= − 2 p q r − 5 p q − ( − p q r ) + 3 p q =

− p q r – 2 p q -pqr–2pq − p q r – 2 p q

− p q r + 2 p q -pqr+2pq − p q r + 2 p q

− 3 p q r – 2 p q -3pqr–2pq − 3 p q r – 2 p q

− 3 p q r – 8 p q -3pqr–8pq − 3 p q r – 8 p q

algebra

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10th Grade

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Practice Problem

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Hard

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CCSS

7.EE.A.1, 6.EE.A.1, HSA.APR.A.1

Standards-aligned

NORALIAH Moe

Used 35+ times

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25 questions

Show all answers

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

yang mana antara berikut adalah pemalar

jumlah hari dalam seminggu

bilangan pengunjung di sebuah majlis

tinggi paras air di sungai

suhu di malaysia

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

yang mana antara berikut adalah pemboleh ubah

takat didih air tulen

bilangan sisi sebuah oktagon

bilangan sisi poligon

jumlah warna dalam pelangi

MULTIPLE CHOICE QUESTION

1 min • 1 pt

y + y + y =

$y^3$

y + 3

3 + y

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

2x + 2x =

x + 4

2 + x

2 + 2

$-4y-2y=$

-6y

y + 2

y - 2

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$6c+d+2c+3d$ =

8c+4d

7c+5d

$-7xy+2xy-xy$ =

8xy

-4xy

-6xy

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