In quadrant 2 is where it lies.
Answer:
m=3/4
Step-by-step explanation:
first, let's put the line 4x+3y=9 from standard form (ax+by=c) into slope-intercept form (y=mx+b)
we have the equation 4x+3y=9
subtract 4x from both sides
3y=-4x+9
divide by 3
y=-4/3x+3
perpendicular lines have slopes that are negative and reciprocal. If the slopes are multiplied together, the result is -1
so to find the slope of the line perpendicular to the line y=-4/3x+3, we can take the slope of y=-4/3x+3 (-4/3) multiply it by a variable (this is our unknown value), and have that set to -1
(m is the slope value)
-4/3m=-1
multiply by -3/4
m=3/4
therefore the slope of the perpendicular line is 3/4
hope this helps!! :)
Answer:
s=8
Step-by-step explanation:
lmk if you want an in depth explanation
Step-by-step explanation:
x+y=5
3x-2y+5=0
3x-2y= -5
x+y=5...... equation 1
3x-2y= -5....... equation 2
there fore make x the subject of the formula in equation 1
x=5-y.. equation 3
substitute equation three in equation 2
3(5-y)-2y=-5
15y-3y-2y=-5
10y=-5
divide both sides by 10
y=-5/10
y=-½
substitute y into equation 1
x+(-½)=5
x=5+½
x=5½
x=5½,y= -½
It is not differentiable at x=1 since the slope of the tangent line as x -> 1 from the right is 1 while the slope of the tangent line as x->1 from the left is -1
