Answer:
a) distance covered by hare d1 = 8t
b) distance covered by tortoise d2 = 5t + 550
c) ∆d = 550 - 3t
Step-by-step explanation:
Given;
Speed of hare u = 8m/s
Speed of tortoise v = 5 m/s
Initial distance of tortoise d0 = 550 m
a) using the equation of motion;
distance covered = speed × time + initial distance
d = vt + d0
For hare;
d0 = 0
Substituting the values;
d1 = 8t + 0
d1 = 8t
b)using the equation of motion;
distance covered = speed × time + initial distance
d2 = vt + d0
For tortoise;
d0 = 550m
Substituting the values;
d2 = 5t + 550
d2 = 5t + 550 m
c) the number of meters the tortoise is ahead of the hare.
∆d = distance covered by tortoise - distance covered by hare
∆d = d2 - d1
Substituting the values;
∆d = (5t + 550) - 8t
∆d = 550 - 3t
Answer:
No
Step-by-step explanation:
Should start at -7 and move 9 units towards right (in the positive direction).
Should stop at 2
Congruent
SSS (because of Pythagoras)
Answer:426m^2
Step-by-step explanation:
Answer:
Joey has 18 nickles and 27 dimes in his piggy bank.
Step-by-step explanation:
1 nickle = 5 cents
1 dime = 10 cents
$1 = 100 cents
$3.15 = 135 cents
Let
n represent the number of nickles, n>=0
d represent the number of dimes, d>=0
Joey counted a total of 45 coins that added up to $3.15:
n + d = 45
5n + 10d = 315
n = 18 nickles
d = 27 dimes
