Set the whole expression = to 0 and solve for x.
3x^(5/3) - 4x^(7/3) = 0. Factor out x^(5/3): x^(5/3) [3 - 4x^(2/3)] = 0
Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.
In the latter case, 4x^(2/3) = 3.
To solve this: mult. both sides by x^(-2/3). Then we have
4x^(2/3)x^(-2/3) = 3x^(-2/3), or 4 = 3x^(-2/3). It'd be easier to work with this if we rewrote it as
4 3
--- = --------------------
1 x^(+2/3)
Then
4
--- = x^(-2/3). Then, x^(2/3) = (3/4), and x = (3/4)^(3/2). According to my 3 calculator, that comes out to x = 0.65 (approx.)
Check this result! subst. 0.65 for x in the given equation. Is the equation then true?
My method here was a bit roundabout, and longer than it should have been. Can you think of a more elegant (and shorter) solution?
Answer:
Gallons of water remaining = 16,962 gallons
Step-by-step explanation:
Total capacity of the pool = 17,897 gallons
Rate of leakage per day = 5 gallons
Number of days = 187 days
how many gallons of water will remain in the pool after 187 days?
Gallons of water remaining = Total capacity of the pool - (Rate of leakage per day × Number of days
= 17,897 - (5 × 187)
= 17,897 - 935
= 16,962 gallons
Gallons of water remaining = 16,962 gallons
Answer:
6 sqrt(3) = y
Step-by-step explanation:
We can use the leg rule to find y
hyp leg
----- = -------
leg part
9+3 y
----- = -------
y 9
Using cross products
12*9 = y^2
108 = y^2
Taking the square root of each side
sqrt(108) = sqrt(y^2)
sqrt(36 *3) = y
6 sqrt(3) = y
Answer:
400
Step-by-step explanation:
