(5,4)
On a system of 2 perpendicular axis with O as origine, the pair (5,4) means:
5 is the distance from the origin O and situated on x-axis (on the right of O)
4 is the distance from the origin O and situated on y-axis (above O)
Then the pair (5,4) is situated in the 1st Quadrant
We would divide the shape into a rectangle and a trapezium
The rectangle is of length, 24.5 ft and width 10.75 ft
Area = 24.5 * 10.75 = 263.375 square feet.
Trapezium : Parallel sides are 18.5 ft and 24.5 ft
Height: 12 - 10.75 = 1.25ft
Area of Trapezium = (1/2)*(Sum of Parallel sides)*height
= 1/2 * (18.5+24.5)* 1.25
= 1/2 * 43* 1.25 = 26.875
Total Area = Area of Rectangle + Area of Trapezium
= 263.375 + 26.875 = 290.25
Area = 290.25 ft² (D)
We see that the area is 159 3/8 and the length is 18 3/4.
Because the area is the length*width, to find the width, we must divide the area by the length.
Think of it this way:
Divide l on both sides:
And w is what we're looking for.
So, divide 159 3/8 by 18 3/4.
You should get that the width is 8.5 feet :)
Answer: 50 minutes
Step-by-step explanation: a quarter to 5 is always 15 minutes from 5, 4:45. So 5:35-4:45 is 50.
Answer:
b) 690 - 7.5*t
c) 0 < t < 92s time (t) is independent quantity
d) 0 < s < 690ft distance from bus stop (s) is dependent quantity
e) f(0) = 690 ft away from bus stop , f(60.25) = 238.125 ft away from bus stop
Step-by-step explanation:
Part a - see diagram
part b
initial distance from bus stop s0 = 690 ft
distance covered = 7.5*t
s = s0 - distance covered
s = 690 - 7.5*t = f(t)
part c
s = 0 or s = 690
0 = 690 -7.5*t
t = 92 s
Hence domain : 0 < t < 92s time (t) is independent quantity
part d
s = 0 or s = 690
Hence range : 0 < s < 690ft distance from bus stop (s) is dependent quantity because it depends on time (t)
part e
f(0) is s @t = 0
f(0) = 690 ft away from bus stop
f(60.25) is s @t = 60.25
f(60.25) = 690 - 7.5*60.25 = 238.125 ft away from bus stop.