Answer:
Solution given:
South distance :base[b]=80milea
East distance :perpendicular [p]=35miles
Now
<S=?
we have


=24°
24° bearing should be taken from south airport to East airport.
Answer:
13. B. 24 cm²
14. y = 5⅝ (none of the options is correct)
Step-by-step explanation:
13. The figure is a trapezoid
Area of the trapezoid = ½(a + b)h
Where,
a = 3 cm
b = 5 cm
h = 6 cm
Area = ½(3 + 5)6
= ½(8)6
= 4*6
= 24 cm²
14. Area of the triangle = ½*b*h
The area is given to be 45 cm²
base (b) = 16 cm
height (h) = y
Plug in the values and solve for y
45 = ½*16*y
45 = 8y
Divide both sides by 8
45/8 = 8y/8
45/8 = y
5⅝ = y
y = 5⅝ (none of the options is correct)
Answer:
In the triangle shown, AB = 11, BC = 61. Find AC. Right Triangle ABC v2. 3,600; 3,842; 60; 62. 4. Using the following measurements, find the length of the leg of the right triangle. leg = 5
Step-by-step explanation:
Not sure if that's helpful, but hope it is.
-13 to 0 = 13 degrees changed. 0 to 22 is 22 degrees changed. 13 + 22 = 35 degree change.
10-3 = 7.
So there was a change of 35 degrees across 7 hours
35/7 = 5 degrees per hour.
The velocity v and maximum height h of the water being pumped into the air are related by the equation
v= 
where g = 32
(a) To find the equation that will give the maximum height of the water , solve the equation for h
v= 
Take square root on both sides
= 2gh
Divide by 2g on both sides
= h
So maximum height of the water h = 
(b) Maximum height h= 80
velocity v= 75 ft/sec
Given g = 32
h = 
h = 
h= 87.89 ft
The pump withe the velocity of 75 ft/sec reaches the maximum height of 87.89 feet. 87.86 is greater than the maximum height 80 feet.
So the pump will meet the fire department needs.