HELP!!

In 5-8, write an equation of the trend line in slope-intercept form.

First, find out the y-intercept(s).

For 5, it is 6. For 6, it is 150. For 7, it is 20. For 8, it is 0.

Now we need to find the slopes.

For 5, it goes down one and right 2, so it would be -1/2 (rise/run).

For 6, it goes up 100 and right 2, so it would be 100/2 (rise/run) or 50.

For 7, it goes up 10 and right 3, so it would be 10/3 (rise/run).

For 8, it goes up 25 and right 1, so it would be 25/1 (rise/run) or 25.

Then you plug into the equation: y = mx + b

5: y = -1/2x + 6.

6: y = 50x + 150.

7: y = 10/3 + 20.

8: y = 25x + 0 which would be y = 25x (since add 0 is useless).

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.