Remember, order of operations
exponent before multiply
so 4x^5/6=4 times x^5/6
simplify the x^5/6 first
remember
![x^\frac{m}{n}=\sqrt[n]{s^m}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7Bm%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Bs%5Em%7D)
so
![x^\frac{5}{6}=\sqrt[6]{x^5}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7B5%7D%7B6%7D%3D%5Csqrt%5B6%5D%7Bx%5E5%7D)
so
![4x^\frac{5}{6}=4\sqrt[6]{x^5}](https://tex.z-dn.net/?f=4x%5E%5Cfrac%7B5%7D%7B6%7D%3D4%5Csqrt%5B6%5D%7Bx%5E5%7D)
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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
4(73)+3
273+3
276
Because you have to subsititue 73 for x. So you have to multiply 73 by 4 then what you get out of that you have to add it by 3.
Step-by-step explanation:
60 because .7 times 60 equals 42 therefore the answer is 60
