Answer:
28. 120 degrees
29. 30 degrees
30. 56 degrees & 124 degrees
31. 72 degrees, 108 degrees, and 18 degrees
Step-by-step explanation:
We assign variable x for the answer we are looking for (28-29).
28.
Supplement means x + y = 180 degrees. We also know x = 2y. Substitution gives us 3y = 180 degrees, so y = 60 degrees and x = 120 degrees.
29.
Complement means x + y = 90 degrees. We are given 2x = y. Substitution brings us 3x = 90 degrees, x = 30 degrees.
30.
Supplement means x + y = 180 degrees. We are told that y = 2x + 12, so we substitute. This gives 3x + 12 = 180 degrees, x = 56 degrees. Substituting that back into the equation for y, we get 124 degrees.
31.
Supplement means x + y = 180 degrees. Complement means x + z = 90 degrees. Using our given info, we know y = 6z. We can substitute that in to get x + 6z = 180. Subtracting our second and third equations, we get 5z = 90, z = 18 degrees. Therefore, x = 72 degrees, y = 108 degrees.
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.
Answer:
x =68
Step-by-step explanation:
We know that x+22 = 90 because the sum of the two angles is a vertical angle to the right angle and vertical angles are equal.
x+22 =90
Subtract 22 from each side
x+22-22 = 90-22
x =68
Due to the distance formula the answer should be root 113 or 10.63
I'm calculated and I figure that it's graphing
