These are the formulas that will help you determine which type of triangle they are:
a^2+b^2 < c^2 ----> Obtuse Triangle
a^2+b^2 > c^2 ----> Actue Triangle
a^2+b^2 = c^2 ----> Right Triangle
Okay so now that you know that information, lets get into it :)
a. 5 in, 6 in, 7 in
You're going to take the smallest numbers, 5 and 6, and add them, if it equals a larger number than 7 then its a triangle and you have to determine if its an obtuse, right or acute triangle. In this case it is a triangle because 5 + 6 = 11 aka larger than 7.
The way you'll set this up is:
5^2 + 6^2 = 7^2
solve 25+36=49 -----> 25+36=61
61 > 49 or a^2 + b^2 > c^2
61 > is greater than 49
If you look ate the formulas that are above, this is an acute triangle.
b. 18 in, 9 in, 12 in
In this question, 9 and 12 are the smallest numbers that equal 21 and 21 is larger than 18 so, this is a triangle.
9^2 + 12^2 = 18^2
Solve 81 + 144 = 324 ----> 81 + 144 = 225
225 < 324 or a^2+b^2 < c^2
225 < is less than 324
If you look ate the formulas that are above, this is an obtuse triangle.
Something to just remember:
Sometimes you'll get a question which is like, 4 in, 5 in, 10 in In this situation, if you add the smallest numbers which are, 4 and 5, you get 9, which is less than the larger number you have, 10. That means it is not a triangle. Just something to be aware about :)
First, you distrubute your numbers. On the left side, multiply the parentheses by '5', and on the right side multiply the perentheses by '-1'. It should look something like this:
x-5x+5 = x-2x+2
Now, you need to combine like terms. On the left side combine the 'x' and the '-5x'. On the right side combine the 'x' and the '-2x'. It should now look like this:
-4x+5 = -x+2
You need to isolate your variable. Working with positive variables is preferable and can decrease chances of confusion, so add '4x' to both sides, therefore canceling the '-4x' on the left and moving it to the right. It will now look like this:
5 = 3x+2
Subtract '-2' from both sides to get the '3x' by itself. It looks like this:
3 = 3x
For the final step, you want the x by itself. To do this, divide both sides by '3'. This will effectively cancel the '3' on the right side, moving it away from your variable. Your equation should end up being this:
Is none of the above an option? After solving, and graphing the equations none of these are perpendicular to the original, or first, line. After messing around a little the equation would have to be closer too y=-x+-6 and that's still not perfect. Sorry I couldn't be of more help.