<span>Which set of line segments could create a right triangle? a 15, 30, 35 b 15, 36, 39 c 15, 20, 29 d 5, 15, 30 = </span>b 15, 36, 39
Answer:
First one: Function
Second one: not a function (a function cannot have two outputs)
Third one: Function
Last one: Not a function (doesn't pass vertical line test)
Step-by-step explanation:
Hope it helps!
Answer:
(x, y) = (2, 5)
Step-by-step explanation:
I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...
3x' -y' = 13/10
x' +2y' = 9/10
Adding twice the first equation to the second, we get ...
2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)
7x' = 35/10 . . . . . . simplify
x' = 5/10 = 1/2 . . . . divide by 7
Using the first equation to find y', we have ...
y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5
So, the solution is ...
x = 1/x' = 1/(1/2) = 2
y = 1/y' = 1/(1/5) = 5
(x, y) = (2, 5)
_____
The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.
The product of 3p and q-3 would be put in to equation form like this: 3p(q-3)
To find your answer. You have to distribute the 3p to by individually multiplying it by q and -3
It should now look like this:

So your answer is:
Answer:
sorry this question is tough