MHB Solve 2000 = 4š¾^0.75 * L^0.25 - 5000(10/3)^.75

  • Thread starter Thread starter markosheehan
  • Start date Start date
AI Thread Summary
The equation 2000 = 4K^0.75 * L^0.25 is analyzed, with K defined as K = 3/10 L. By substituting K into the equation and simplifying, it is determined that 500 = (3/10)^(3/4) * L. Further manipulation leads to the solution for L, which is L = 500 * (10/3)^(3/4). The correct answer confirms the relationship between L and K in the context of the given equation.
markosheehan
Messages
133
Reaction score
0
can someone solve 2000 = 4š¾^0.75 * L^0.25 where k=3/10 L. i tryed to solve this by adding the powers 2000=4(3/10 L^1) however i get an answer of 5000/3 which is not the correct answer the correct answer is 5000(10/3)^.75
 
Mathematics news on Phys.org
Re: equation

We are given to solve for $L$:

$$2000=4K^{\frac{3}{4}}L^{\frac{1}{4}}$$

Divide through by 4:

$$500=K^{\frac{3}{4}}L^{\frac{1}{4}}$$

We are told:

$$K=\frac{3}{10}L$$:

$$500=\left(\frac{3}{10}L\right)^{\frac{3}{4}}L^{\frac{1}{4}}=\left(\frac{3}{10}\right)^{\frac{3}{4}}L$$

Hence:

$$L=500\left(\frac{10}{3}\right)^{\frac{3}{4}}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top