Answer:
Step-by-step explanation:
P = 480
P = 2 length + 2 width
but we also know that
2width = length
so plug that into the length
P = 2(2width) + 2 width
480 = 4 width + 2 width
480 = 6 width
80 = width
160= length
see?
Answer:
Step-by-step explanation:

The answer is 5 5/6 simplified to the max.
rewrite the equation with separated parts
-3 1/2 - 2 1/3
-3 - 2 = -5
-1/2 - 1/3? u need to find the LEAST common denominator
-3/6 - 2/6 = -5/6
how did I get 6?
Rewriting input as fractions if necessary:
1/2, 1/3
For the denominators (2, 3) the least common multiple (LCM) is 6.
Therefore, the least common denominator (LCD) is 6.
Now lastly combine them total and its -5 - 5/6 = -5 5/6.
<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.