Let L be the length
Let w be the width
Let p be the perimeter
L+w+L+w=p
L=w+20
3L+2w+3L+2w=240
Sub the first equation in for L in the second equation and solve for w
3(w+20)+2w+3(w+20)+2w=240
3w+60+2w+3w+60+2w=240
10w+120=240
10w=240-120
10w=120
W=120/10
W=12
Sub w into the first equation and solve for L
L=w+20
L=12+20
L=32
Hope this helps!
Answer:

Step-by-step explanation:
Since ED is parallel to CA, the two triangles in the figure share all 3 angles and therefore must be similar. By definition, the corresponding sides of similar polygons are in a constant proportion.
Therefore, we have:

Answer:
m∠PQR = 133
Step-by-step explanation:
∠PQR and ∠RQS are supplementary angles, meaning they have a sum of 180°.
3x - 5 + x + 1 = 180
4x - 4 = 180
4x = 184
x = 46
Now we simply substitute 46 in for x:
3(46) - 5 = 133
2x is the answer to that question it's simple math just