Answer:
The speed in still water is 5 miles per hour.
Step-by-step explanation:
distance with current = 90 miles
distance against current = 10 miles
speed in still water = s
speed of current = 4 mph
speed with current = s + 4
speed against current = s - 4
time = t
speed = distance/time
distance = speed * time
With current:
90 = (s + 4) * t
Against the current:
10 = (s - 4) * t
We have a system of equations:
90 = (s + 4) * t
10 = (s - 4) * t
90 = ts + 4t
10 = ts - 4t
Subtract the second equation from the first equation.
80 = 8t
10 = t
t = 10
10 = t(s - 4)
10 = 10(s - 4)
1 = s - 4
s = 5
Answer: The speed in still water is 5 miles per hour.
Answer:
B.
Step-by-step explanation:
Well we know that

so we can get the 2 outside of the radical

and we can get the x^2 outside too.

and we also can get y outside.
so we have:
![2x^{2}y\sqrt[5]{7xy^3}](https://tex.z-dn.net/?f=2x%5E%7B2%7Dy%5Csqrt%5B5%5D%7B7xy%5E3%7D)
Answer:
put a more clear pic
Step-by-step explanation:
75+ 15x ≥ 140
⇒ 15x ≥ 140 -75
⇒ 15x ≥ 65
⇒ x ≥ 65/15
⇒ x ≥ 13/3
⇒ x ≥ 4 1/3
The final answer is x ≥ 4 1/3~
<em>The</em><em> </em><em>clock</em><em> </em><em>says</em><em> </em><em>4</em><em>:</em><em>3</em><em>o</em><em> </em><em>am</em><em>/</em><em>pm</em><em> </em><em>as</em><em> </em><em>big</em><em> </em><em>one</em><em> </em><em>is</em><em> </em><em>at</em><em> </em><em> </em><em>6</em><em> </em><em>and</em><em> </em><em>small</em><em> </em><em>near</em><em> </em><em>4</em>