"y>-2x+2 and y>-2x+5" are illustrated graphically by 2 straight dashed lines with slope -2. The lines are parallel because their slopes (-2) are the same. One line has y intercept 2 and the other has y intercept 5. The latter is above the former. Since both inequality signs are " > " we must shade the area ABOVE each of the 2 lines. The solution set is the area of the graph that has been shaded twice, once for y>-2x+2 and again for y>-2x+5. It's y>-2x+5 that has been shaded twice; this area is immediately above the line y>-2x+5.
Below is a labeled rectangle:
length = w + 20
we know length is 36 so we plug that in:
36 = w + 20
We need to isolate w and to do that we must subtract 20 from both sides:
(36 - 20) = w + (20 - 20)
16 = w + 0
w = 16
width is 16 feet
Hope this helped!
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°