Answer:
∠RST = 120°
Step-by-step explanation:
We assume the positions of the lines and angles will match the attached figure. The angle addition theorem gives a relation that can be solved for x, then for the value of angle RST.
∠RSU +∠UST = ∠RST
__
78° + (3x -12)° = (6x +12)° . . . . . substitute given values into the above
54 = 3x . . . . . . . . . . . . . . . . divide by °, subtract 3x+12
108 = 6x . . . . . . . . . . . multiply by 2
120° = (6x +12)° = ∠RST . . . . add 12, show units
The measure of angle RST is 120 degrees.
_____
<em>Additional comment</em>
Note that we don't actually need to know the value of x (18) in this problem. We only need to know the value of 6x.
So you will need to solve for x and y before evaluating 2x+y....
2x-y=9, y=2x-9 now this will make 4x^2-y^2=171 become:
4x^2-(2x-9)^2=171
4x^2-(4x^2-36x+81)=171
36x-81=171
36x=252
x=7, now we can use 2x-y=9 to solve for y...
2(7)-y=9
14-y=9
-y=-5
y=5
now we know that x=7 and y=5, 2x+y becomes:
2(7)+5
14+5
19
a= 34 degrees
b= 28 degrees
c= 62 degrees
Step-by-step explanation:
First you know that b is 1/2 of 56 degrees or 28.
The triangle with the a in it is isoceles because the two sides are both radii.
In the triangle the top angle = 112 because it is a centeral angle to the 112 arc.
Angle a and opposite to a are equal and then have to be 34 degrees to equal 180.
We know two arc lengths are 112 and 56 and the one with angle a has to be 34x2 or 68.
a whole circle equals 360.
360-56-68-112 = 124
Angle c = 1/2 of 124, or 62 degrees
No solution
Ex:
Simply 2.40+1.8 to 4.2
Answer:
Step-by-step explanation:
Let 
Subbing in:
a = 9, b = -2, c = -7
The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:
Group together in groups of 2:
Now factor out what's common within each set of parenthesis:
We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:
Remember that 
so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.
The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring 
gives us that x = 1 and -1. The other set is a bit more tricky. If
then
and
You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:
±
which will simplify down to
±
Those are the 4 solutions to the quartic equation.
