Your slope would be -1/4x. If you look at the y-intercept on your graph, which is -2, you would start to count up one box and over 4 boxes to the left. We call this rise/run. So knowing what the slope is now, your entire equation would be y= -1/4x - 2. (Also, if the line is going up towards the left it is a negative slope, if it's to the right it is positive.) I hope this helps love! :)
Answer:
B: (2,[infinity]]
Step-by-step explanation:
because you will take all numbers bigger than 2 but 2 is not included
Answer:
y = 14
Step-by-step explanation:
move 5y over to the left and you should get -2y +6=-18. subtract 6from both sides and you will now have -2y=-24. divide bothe sides from -2 and you will get y=14
<u>Answer-</u>
The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

<u>Solution-</u>
Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)
The applying the distance formula,








Two equations are formed because one will be upper from the the given line and other will be below it.