<em>Answer:</em>
<em>∠x = 70°</em>
<em>Step-by-step explanation:</em>
<em>Given that PQ||BR</em>
<em>Line PQ and BR with transversal AC,</em>
<em>∠x + ∠ACR = 180° {Co-interior angles}</em>
<em>⇒∠x + 110° = 180° {From what I can see it looks like 110.}</em>
<em>⇒∠x = 180° - 110°</em>
<em>⇒</em><em>∠x = 70°</em>
WX and ZY make up the legs, as illustrated in the diagram.
Since it is an isosceles trapezoid, the legs are congruent; that means
6<em>x</em> + 5 = 8<em>x</em> - 3
When solving this equation, we do not want a variable on both sides. We will cancel the 6<em>x</em> to avoid negatives:
6<em>x</em> + 5 - 6<em>x</em> = 8<em>x</em> - 3 - 6<em>x</em>
5 = 8<em>x</em> - 3 - 6<em>x</em>
Combine like terms:
5 = 2<em>x</em> - 3
Cancel the 3 by adding:
5 + 3 = 2<em /><em>x</em> - 3 + 3
8 = 2<em>x</em>
Cancel the 2 by dividing:
8/2 = 2<em>x</em>/2
4 = <em>x</em><em />
C(x)=rx+s we are told that x=3, c(x)=21, and s=3
21=3r+3 divide both sides by 3
7=r+1 subtract 1 from both sides
6=r so now c(x) becomes:
c(x)=6x+3
We want an equation which equals
0
at the given points
6
and
−
10
.
Our quadratic equation should be a product of expressions which are zero at the specified roots.
Consider
(
x
−
6
)
⋅
(
x
+
10
)
=
0
This equality holds if
x
=
6
since
(
6
−
6
)
⋅
(
6
+
10
)
=
0
⋅
16
=
0
And the equality holds if
x
=
−
10
since
(
−
10
−
6
)
⋅
(
−
10
+
10
)
=
−
16
⋅
0
=
0
Expanding this equation by the FOIL method, we get:
x
2
+
10
x
−
6
x
−
60
Combining like terms, we find our solution:
The sum of the angles of a triangle must equal 180.
A + B + C = 180
50 + B + C = 180
B + C = 180 - 50
B + C = 130
If B and C are equal to each other, we can just divide 130 by 2 to find the other two angles.
130 / 2 = 65
Angles B and C are both 65 degrees