Answer:
C + 88 = 180
-88 -88
C and B = 92
Since C is vertical to B that means both values are the same. Therefore the values of c and b is 92 degrees.
Answer:
see explanation
Step-by-step explanation:
(1)
x + 20 and 45 are vertically opposite angles and are congruent, so
x + 20 = 45 ( subtract 20 from both sides )
x = 25
Similarly 5y and 135 are vertically opposite angles and are congruent
5y = 135 ( divide both sides by 5 )
y = 27
(2)
x + 30 and 70 are vertically opposite angles and are congruent, then
x + 30 = 70 ( subtract 30 from both sides )
x = 40
Similarly y + 50 and 110 are vertically opposite and congruent , so
x + 50 = 110 ( subtract 50 from both sides )
x = 60
Seria 25/5 por que si lo divides seria 5X5=25 ok a ver si te sirve
Answer:
x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>
The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true
x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>
Another way to solve this is using substitution for one of the radicals. We choose ...

Solutions to this equation are ...
u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
x = u² -2 = 2² -2
x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.