If my math is right, not even a full day since it’s 2 2/3 and they use 8
Answer:
It’s 60miles
Step-by-step explanation:
We assume the trip is "d" miles and that the "extra hour" refers to the additional time that a current of 2 mph would add. That is, we assume the reference time is for a current of 0 mph.
The time with no current is ...
time1 = distance/speed
time1 = d/12 . . . . hours
With a current of 2 mph in the opposite direction, the time is ...
time2 = d/(12 -2) = d/10
The second time is 1 hour longer than the first, so we have ...
time2 = 1 + time1
d/10 = 1 + d/12
6d = 60 + 5d . . . . multiply by 60
d = 60 . . . . . . . . . subtract 5d
The one-way distance is 60 miles.
Answer:
Final answer is
.
Step-by-step explanation:
Given problem is
.
Now we need to simplify this problem.
![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
Hence final answer is
.
Answer:
66 cm
Step-by-step explanation:
Diameter (d) = 21 cm

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