$\frac{ {x}^{2} - 7x + 12}{ {x}^{2} - x - 12 } \\ = \frac{ {x}^{2} - 4x - 3x + 12 }{ {x}^{2} - 4x + 3x - 12 } \\ = \frac{x(x - 4) - 3(x - 4)}{x(x - 4) + 3(x - 4)} \\ = \frac{(x - 3)(x - 4)}{(x + 3)(x - 4)} \\ = \frac{x - 3}{x + 3}$

x² - x - 12 ≠ 0

x² - 4x + 3x - 12 ≠ 0

x (x - 4) + 3 (x - 4) ≠ 0

(x + 3)(x - 4) ≠ 0

x ≠ -3 or x ≠ 4

(B)

3 0

2 years ago

Evaluate the following expression: p/-3 + (-8)= -8

aalyn [17]

P/-3+(-8)=-8
p/-3-8=-8
Add 8 to each side
p/-3-8+8=-8+8
p/-3=0
Multiply both side by 3
p/-3(3)=0(3)
p=0
Check:
p/-3+(-8)=-8
Substitute p with 0
0/-3+(-8)=-8
0-8=-8
-8=-8. As a result, p=0. Hope it help!

7 0

2 years ago

Simplify (Are all division) 2+4i 3i 3+2i 4+i 2i^11

Alla [95]

$We\ know:\ i=\sqrt{-1}\to i^2=-1$

$\dfrac{2+4i}{3i}=\dfrac{2+4i}{3i}\cdot\dfrac{3i}{3i}=\dfrac{6i+12i^2}{9i^2}=\dfrac{6i+12(-1)}{9(-1)}\\\\=\dfrac{6i-12}{-9}=\dfrac{2i-4}{-3}=\dfrac{4}{3}-\dfrac{2}{3}i$

$\dfrac{3+2i}{4+i}=\dfrac{3+2i}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{(3+2i)(4-i)}{4^2-i^2}\\\\=\dfrac{(3)(4)+(3)(-i)+(2i)(4)+(2i)(-i)}{16-(-1)}=\dfrac{12-3i+8i-2i^2}{16+1}\\\\=\dfrac{12+5i-2(-1)}{17}=\dfrac{12+5i+2}{17}=\dfrac{14+5i}{17}=\dfrac{14}{17}+\dfrac{5}{17}i$

$2i^{11}=2i^{10+1}=2i^{10}i^1=2i^{2\cdot5}i=2i(i^2)^5=2i(-1)^5=2i(-1)=-2i$

$Used:\ (a+b)(a-b)=a^2-b^2$

6 0

2 years ago