Yes most likely a variable square can be combined with a variable I believe
Answer:
Function:
c = f(w) = 0.49, 0 < w ≤ 1
= 0.70, 1 < w ≤ 2
= 0.91, 2 < w ≤ 3
Step-by-step explanation:
Yes, the relation described can be interpreted as a function.
Here, c is the cost of a mail letter. c depends upon w, which is the weights of the mail letter.
As described in the question, the relation can be expressed as a function.
c can be expressed as a function of w in the following manner:
c(cost of mail) = f(w), where w is the independent variable and c is the dependent variable
c = f(w) = 0.49, 0 < w ≤ 1
= 0.70, 1 < w ≤ 2
= 0.91, 2 < w ≤ 3
where, c is in dollars and w is in ounces.
The answer is 50.
i came to this conclusion when using this equation: 15x+5=2+16x
for x, i got 3.
using my x-value, i replace “x” with “3” when solving angle measure, “15(3)+5”
15(3)+5= 50
there is your answer!
That would be 15/80=3/16 so it would be 3/16 or unlikely because it is less than half of 16
