m∠FDE = 52°
Solution:
Given data:
DE ≅ DF, CD || BE, BC || FD and m∠ABF = 116°
<em>Sum of the adjacent angles in a straight line = 180°</em>
m∠ABF + m∠CBF = 180°
116° + m∠CBF = 180°
m∠CBF = 64°
If CD || BE, then CD || BF.
Hence CD || BE and BE || FD.
Therefore BFCD is a parallelogam.
<em>In parallelogram, Adjacent angles form a linear pair.</em>
m∠CBF + m∠BFD = 180°
64° + m∠BFD = 180°
m∠BFD = 116°
<em>Sum of the adjacent angles in a straight line = 180°</em>
m∠BFD + m∠DFE = 180°
116° + m∠DFE = 180°
m∠DFE = 64°
we know that DE ≅ DF.
<em>In triangle, angles opposite to equal sides are equal.</em>
m∠DFE = m∠DEF
m∠DEF = 64°
<em>sum of all the angles of a triangle = 180°</em>
m∠DFE + m∠DEF + m∠FDE = 180°
64° + 64° + m∠FDE = 180°
m∠FDE = 52°
Ax - c = 2x + 5
ax - c + c = 2x + 5 + c
ax = 2x + 5 + c
ax - 2x = 2x - 2x + 5 + c
ax - 2x = c + 5
X(a - 2) = c + 5
X(a - 2)/(a - 2) = c+5/(a-2)
X = c+5/a-2
I believe this would be the solution.
12 action films are shown.
This is because if the ratio is 1:2, that means that exactly half of the total amount of movies are action. Thus, to evaluate this problem, you would find half of 24.
