Answer:
a) ∠TVZ= 95°
Step-by-step explanation:
3a) ∠STY= ∠QTV (vert. opp. ∠s)
3x°= 120°
x°= 120° ÷3
x°= 40
∠TVZ= ∠RVW (vert. opp. ∠s)
∠TVZ= (2x +15)°
∠TVZ= [2(40) +15]°
∠TVZ= (80 +15)°
∠TVZ= 95°
b) Since ∠TVZ and ∠WVZ lies on a straight line,
∠TVZ +∠WVZ= 180° (adj. ∠s on a str. line) -----(1)
∠WVZ= (2x +5)°
∠WVZ= [2(40) +5]°
∠WVZ= (80 +5)°
∠WVZ= 85°
Substitute ∠WVZ= 85° into (1):
∠TVZ +85°= 180°
∠TVZ= 180° -85°
∠TVZ= 95°
Thus, ∠TVZ is indeed 95°.
Notes:
• What is vert. opp. ∠s?
It is an abbreviation used for a property of angles, vertically opposite angles. When two lines intersect each other, the angles facing each other (or the angles on the opposite side of each other) are equal.
• What is adj. ∠s on a str. line?
It is an abbreviation for 'adjacent angles on a straight line'. The sum of all the angles on a straight line is 180°.
90 degrees because you can see at the part that is in the middle it is 90 degrees
Answer:
W=7 and L=11
Step-by-step explanation:
We have two unknowns so we must create two equations.
First the problem states that length of a rectangle is 10 yd less than three times the width so: L= 3w-10
Next we are given the area so: L X W = 77
Then solve for the variable algebraically. It is just a system of equations.
3W^2 - 10W - 77 = 0
(3W + 11)(W - 7) = 0
W = -11/3 and/or W=7
Discard the negative solution as the width of the rectangle cannot be less then 0.
So W=7
Plug that into the first equation.
3(7)-10= 11 so L=11
Answer:
Step-by-step explanation:
