S = 4LW+2WH; S = 200, L = 8, W = 5. (Surface area of a special rectangular box) H=
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Answer is 0= -2
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To solve this problem, we need to know 2 relationships:
<h2>1. AC = AB + BC</h2>The distance of AC is the sum of AB and BC.
We know this since the distance of going from A to C (AC) is the same as going from A to B (AB), then B to C (BC).
<h2>2. AB = BC</h2>The distance of AB is the same as AC.
We know this since B is in the middle of AC, so the distance from B to A (BA) is the same as the distance from B to C (BC).
You can see the attached image (at the bottom) for a visualization of this.
<h2>Putting them together</h2>Since we know the values of AB and BC...
...we can put these values into our 2nd equation and solve for x:
Add 7 to both sides:
Subtract x from both sides:
Divide both sides by 2:
Knowing x, we can find the distance of AC using our first equation.
Let's put in the values of AB and BC:
Before we put in x = 8, we can simplify this:
We group x and 3x and add those together. Then we subtract 7 from 9.
With this equation, we can put in x = 8:
Since 4 * 8 = 32:
Finally, we have found both x and AC.
<h2>Answer</h2>x = 8
AC = 34
