Answer:
a) No
b) [tex]174^{\circ}-17x[/tex]
Step-by-step explanation:
a)
If [tex]x=5[/tex], we have that [tex]\angle c=9(5)-3=45-3=42[/tex], and [tex]\angle d=8(5)+9=40+9=49[/tex].
[tex]42+49=91\neq 90[/tex], so [tex]\angle c[/tex] and [tex]\angle d[/tex] are not complementary when [tex]x=5[/tex]
b)
If [tex]\angle c[/tex], [tex]\angle d[/tex], and [tex]\angle e[/tex] form half a circle, that means they sum to [tex]180^{\circ}[/tex].
We now need to solve [tex]\angle c + \angle d + \angle e = 180^{\circ}[/tex] for [tex]\angle e[/tex].
Plugging in for [tex]\angle c[/tex] and [tex]\angle d[/tex] gives [tex](9x-3)+(8x+9)+\angle e = 180[/tex].
Combining like terms on the left side gives [tex]17x+6+\angle e = 180[/tex].
Subtracting a [tex]17x[/tex] and a [tex]6[/tex] from both sides gives [tex]\angle e = 174-17x[/tex].
This means that [tex]\angle e = 174^{\circ}-17x[/tex] and we're done.
