The 11th term in a geometric sequence is 48 and the common ratio is −0.8. The 12th term is and the 10th term is .

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'.

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

gimme brainliest > : 3

one way is to plot 2 points and connect them which is what the answer choices seem to be doing

it seems to be using the y intercept and the slope

y=mx+b, m=slope, b=y-intercept (where the line hits the y axis)

slope=rise/run

y-intercept is where the line croses the y axis which is the y value where x=0

you seem to have given the wrong slope (you didn't put a number in front of te x, it should be 2/3 or 3/2 looking at the answer options)

y=? x-1

m=? because of your fault

b=-1

so it has a slope of ? and y intercept of -1, meaning it pases through (0,-1)

so draw point (0,-1) (so answer is one of the first 2 option)

if slope is 2/3, then go up 2 and right 3 then draw line through those 2 points

if slope is 3/2, then go up 3 and right 2 then draw line through those 2 points