Consider j(x) = 2x - 5
Find j^-1(x) algebraically.

j^-1(x) = 1/2(x+5)

Step-by-step explanation:

1. Replace j(x) with y to make it easier

y = 2x - 5

2. Switch the places of x and y

x = 2y - 5

3. Solve for y

x + 5 = 2y

y = 1/2(x+5)

4. Replace y with j^-1(x)

j^-1(x) = 1/2(x+5)

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

the rocket cannot penetrate the ground, in a contextual sense, therefore the roots of this equation should be the domain of this function. i'll let you find the roots of this quadratic equation (hint: use the quadratic formula:  

1- c

2-a

step-by-step explanation:

mark brainliest!