Can somone please help me? I need help with 7 different problems!! and fasttttt

Select the pair of equations whose graphs are perpendicular.

viktelen [127]

$k:\ y=m_1x+b_1\ \ \ \ and\ \ \ \ l:\ y=m_2x+b_2\\ \\the\ line\k\ is\ perpendicular\ to\ the\ line\ l\ \ \ \Leftrightarrow\ \ \ m_1\cdot m_2=-1\\----------------------------\\\\A.\\2y=-3x+5\ \ \Rightarrow\ \ \ y=- \frac{3}{2} x+2.5\ \ \Rightarrow\ \ m_1=- \frac{3}{2} \\\\2x+3y=4\ \ \Rightarrow\ \ 3y=-2x+4\ \ \Rightarrow\ \ y=- \frac{2}{3}x+1 \frac{1}{3} \ \ \Rightarrow\ \ m_2=- \frac{2}{3} \\\\m_1\cdot m_2=- \frac{3}{2} \cdot (- \frac{2}{3})=1 \neq -1$

$B.\\5x-8y=9\ \ \Rightarrow\ \ -8y=-5x+9\ \ \Rightarrow\ \ \ y= \frac{5}{8} x-1 \frac{1}{8} \ \Rightarrow\ \ m_1= \frac{5}{8} \\\\12x-5y=7\ \Rightarrow\ \ -5y=-12x+7\ \ \Rightarrow\ \ y= \frac{12}{5}x-1 \frac{2}{5} \ \ \Rightarrow\ \ m_2= \frac{12}{5} \\\\m_1\cdot m_2= \frac{5}{8} \cdot \frac{12}{5}= \frac{3}{2} \neq -1$

$C.\\y=2x-7\ \ \Rightarrow\ \ m_1=2 \\\\x+2y=3\ \Rightarrow\ \ 2y=-x+3\ \ \Rightarrow\ \ y= -\frac{1}{2}x+1 \frac{1}{2}\ \ \Rightarrow\ \ m_2=- \frac{1}{2} \\\\m_1\cdot m_2= 2 \cdot (- \frac{1}{2})= -1\ \ \Rightarrow\ \ y=2x-7\ \ \ \ \perp\ \ \ \ x+2y=3$

$
D.\\x+6y=8\ \ \Rightarrow\ \ 6y=-x+8\ \ \Rightarrow\ \ \ y=- \frac{1}{6} x+1 \frac{1}{3}\ \ \Rightarrow\ \ m_1= -\frac{1}{6} \\\\y=2x-8\ \Rightarrow\ \ m_2= 2 \\\\m_1\cdot m_2= -\frac{1}{6} \cdot2=- \frac{1}{3} \neq -1$

7 0

3 years ago

Pls help I’m failing I’ll brainlest

LenKa [72]

Answer:

6x

Step-by-step explanation:

Since 2 sides are equal to 3x-3, multiply 3x-3 by 2.

=6x-6

Subtract 18x+6 - 6x-6

= 12x

Since 2 there are 2 sides missing, 12x divided by 2