-15 is in fact less than -7. Here is a number line:
-15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5...
Let me know if you need clarification :)
Answer:
<u>6</u>
10
=<u>3</u>
5
Step-by-step explanation:
3/5 you just need to divide
The answer is two is to three
first see the common multiple of 2&4which is 4 so in end u end up with 2/4 & 3/4 hence answer is 2:3
^ ^
Answer:

Step-by-step explanation:
Let x be the distance driven, d-distance and C our constant.
Our information can be presented as:

#Subtracting equation 2 from 1:

Hence the fixed cost per mile driven,
is $0.20
To find the constant,
we substitute
in any of the equations:

Now, substituting our values in the linear equation:
#y=cost of driving, x=distance driven
Hence the linear equation for the cost of driving is y+0.2x+284