For one line to be perpendicular to another, their slopes must multiply to -1. So, we really want the line with slope [tex]$-1/(1/2)$ = -2$[/tex] that passes through the point [tex](1, 7)[/tex].
Recall the point-slope form of a line. If you have a point [tex]$(x_{1}, y_{1})$[/tex] and a slope [tex]m[/tex], the line through that point with the given slope is [tex]$y-y_{1}=m(x-x_{1})$[/tex]. Plugging in our given point and slope gives our line: [tex]y-7=-2(x-1)[/tex](point-slope form).
We can convert this line to slope-intercept form, the most common form by expanding the right side and adding [tex]7[/tex] to get [tex]y[/tex] by itself:
chase is right because he already been to the swings and he know how far it will be to get there
step-by-step explanation:
[tex]y=-2x+5[/tex]
Step-by-step explanation:
For one line to be perpendicular to another, their slopes must multiply to -1. So, we really want the line with slope [tex]$-1/(1/2)$ = -2$[/tex] that passes through the point [tex](1, 7)[/tex].
Recall the point-slope form of a line. If you have a point [tex]$(x_{1}, y_{1})$[/tex] and a slope [tex]m[/tex], the line through that point with the given slope is [tex]$y-y_{1}=m(x-x_{1})$[/tex]. Plugging in our given point and slope gives our line: [tex]y-7=-2(x-1)[/tex](point-slope form).
We can convert this line to slope-intercept form, the most common form by expanding the right side and adding [tex]7[/tex] to get [tex]y[/tex] by itself:
[tex]y-7=-2x+2[/tex]
[tex]y=-2x+5[/tex]
x =±sqrt(1/3)
step-by-step explanation:
x+4/6x=1/x
multiply each side by 6x to get rid of the fractions
6x*(x+4/6x)=1/x*6x
distribute
6x^2 +4 = 6
subtract 4 from each side
6x^2 +4-4 = 6-4
6x^2 =2
divide by 6
6x^2/6 = 2/6
x^2 = 1/3
take the square root of each side
sqrt(x^2) = ±sqrt(1/3)
x =±sqrt(1/3)