Every year the state department of education gathers statistical information from all schools to award a letter grade showing th
1 answer:
You might be interested in
Refer to the diagram shown below.
Let x = m∠ADB.
Because m∠BDC is 32° greater than m∠ADB, therefore
m∠BDC = x + 32°
Each angle of a rectangle is 90°, therefore
x + (x+32) = 90
2x + 32 = 90
2x = 58
x = 29°
x+32 = 61°
Answer:
m∠BDC = 61°
m∠ADB = 29°
Let x = m∠ADB.
Because m∠BDC is 32° greater than m∠ADB, therefore
m∠BDC = x + 32°
Each angle of a rectangle is 90°, therefore
x + (x+32) = 90
2x + 32 = 90
2x = 58
x = 29°
x+32 = 61°
Answer:
m∠BDC = 61°
m∠ADB = 29°
Refer to the figure shown below.
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m