C=8h+12
the 8h represents the $8 per hour for renting the boat, and h represents however many hours you rent the boat for. The +12 is the initial $12 fee. This is in slope intercept form, just using different variables than y=mx+b
In -2x
, the exponent will be dealt with first. With (-2x)
, the multiplication is going to be done first. This is determined because of Order of operations.
You forgot to say what the question actually is !
But I've seen this problem before, in the last few days, here on Brainly.
I think the problem has three parts: 1). Solve the equation for 'a';
2). Solve it for 'b'; and 3). Solve it for 'c'.
I'll slog through that, and I'll try to explain what I'm doing clearly enough
so that eventually, you can do it on your own ... which is really the whole
idea behind this website.
1). Solve the equation for 'a'. That means you have to wind up with something
that says a = everything else.
<u>D = (a + b + c) / c</u>
Split the right side into 3 fractions: D = a/c + b/c + c/c
But c/c =1 , so the equation says D = a/c + b/c + 1
Subtract (b/c +1) from each side: D - b/c - 1 = a/c
Multiply each side by 'c' : <em>Dc - b - c = a</em>
========================
2). Solve the equation for 'b'. That means you have to wind up with something
that says b = everything else.
<u>D = (a + b + c) / c</u>
Split the right side into 3 fractions: D = a/c + b/c + c/c
But c/c =1 , so the equation says D = a/c + b/c + 1
Subtract (a/c +1) from each side: D - a/c - 1 = b/c
Multiply each side by 'c' : <em>Dc - a - c = b</em>
==========================
3). Solve the equation for 'c'. That means you have to wind up with
something that says c = everything else.
<u>D = (a + b + c) / c</u>
Split the right side into 3 fractions: D = a/c + b/c + c/c
But c/c =1 , so the equation says D = a/c + b/c + 1
Subtract 1 from each side: D - 1 = a/c + b/c
The two fractions on the right can be added/combined: D - 1 = (a + b) / c
Multiply each side by 'c' : c(D - 1) = (a + b)
Divide each side by (D - 1) : <em>c = (a + b) / (D - 1)</em>
I can help you, as long as it's something I've already learned so as long as it anything less than stats. I'll pm you.
