Answer:
Distance = rate x time or d = rt...that means time t = d/r...and we have
t = 21 km / .5 km/hr = 42 hr
Step-by-step explanation:
Step-by-step explanation:
Here we can see it is a quadrilateral so
x + 108° + 65° + 53° = 360° ( being sum of angles of quadrilateral)
226° + x =360°
x = 360° - 226°
Therefore x = 134°
Answer:
(1,1) (2,2) (1,2) (1,5) (4,4) (4,5)
there are a lot more but those are some.
Step-by-step explanation:
Answer:
20 meters
Step-by-step explanation:
The track is circular so it means that after Patrick raced the entire track he is back at the starting point. In other words, every 440 meters he is back to the beginning.
So we would have that, if he races round the track twice, he would run 440(2) = 880 meters and he would be back at the starting point.
The problem asks us how far is he from the starting point at the 900 meter mark. If at 880 meters he is at the starting point, then at 900 meters he would be 
meters from the starting point.
Answer: none
Step-by-step explanation:
(A)
(16÷32/10) ×2 + 0.2×(90)
Using bodmas principle ; solve bracket
(16×10/32)×2 + (2/10×90)
10+18 =28
(B)
{(16÷32/10) × (2+2/10)} ×90
Open brackets
{(16×10/32) × (22/10)} ×90
(5×11/5) ×90
11×90 = 990
(C)
16÷{(32/10×2) + (2/10×8)} +82
Open brackets, solve division first, dolled by addition
16÷(32/5 + 8/5) +82
16÷(40/5) +82
16÷8 +82
2+82= 84
(D)
[16÷(32/10 ×2) + 0.2× (90)]
16÷ (32/5) + 2/10 ×90
Solve division
16×5/32 + 18
5/2 + 18
L.c.m of denominator (2&1) =2
(5+36) / 2 = 41/2
=20.5
