Find the range of the function f(x)=5-4x given the domain = {-1,0,1}
1 answer:
You might be interested in
Refer to the diagram shown below.
Given:
m∠A = 19°
c = 15
By definition,
sin A = a/c
Therefore
a = c*sin A = 15*sin(19°) = 4.8835
cos A = b/c
Therefore
b = c*cos A = 15*cos(19°) =14.1828
Answer:
The lengths are 4.88, 14.18, and 15.00 (nearest hundredth)
Given:
m∠A = 19°
c = 15
By definition,
sin A = a/c
Therefore
a = c*sin A = 15*sin(19°) = 4.8835
cos A = b/c
Therefore
b = c*cos A = 15*cos(19°) =14.1828
Answer:
The lengths are 4.88, 14.18, and 15.00 (nearest hundredth)
Answer:
a) see below
b) 32 minutes after turn-on
c) 52 minutes after turn-on
d) 20 minutes
e) 6856.6 ft²
Step-by-step explanation:
a) We have elected to put the origin at the point where the hose crosses the south edge of the sidewalk. Units are feet. Then the sprinkler starts at (0, -100). After 1 hour, 3600 seconds, the sprinkler is 1800 inches, or 150 ft north of where it started, so stops at (0, 50).
The lines forming the sidewalk boundaries are y=0 and y=10.
__
b) Water will first strike the sidewalk when the sprinkler is 20 feet south of it, or 80 feet north of where it started. The sprinkler travels that distance in ...
(80 ft)(12 in/ft)/(1/2 in/s)(1 min/(60 s)) = 32 min . . . time to start sprinkling sidewalk
__
c) The sprinkler has to travel to a point 130 ft north of its starting position for the water to fall north of the sidewalk. That distance is traveled in ...
(130 ft)(2/5 min/ft) = 52 min . . . time until end of sprinkling sidewalk
Note that we have combined the scale factors in the expression of part b into one scale factor of (2/5 min/ft).
__
d) The difference of times in parts b and c is the time water falls on the sidewalk: 20 minutes.
__
e) In one hour, the sprinkler travels a distance of ...
(60 min)(5/2 ft/min) = 150 ft
Of that distance, 10 feet is sidewalk. So, the sprinkler covers an area of grass that is a 140 ft by 40 ft rectangle and a circle of 20 ft radius. The total area of that is ...
A = LW + πr² = (140 ft)(40 ft) +π(20 ft)² = (14+π)(400) ft² ≈ 6856.6 ft²
The area of grass watered in 1 hour is about 6856.6 ft².
