I Need Someone Good At Math To Help Me With This Complex Problem - Education - Nairaland
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| I Need Someone Good At Math To Help Me With This Complex Problem by turmacs(op): 5:58pm On Dec 26, 2023 |
I need someone to help me with this problem. I've been on this for hours and knowing that i don't know how to solve this problem did not allow me to sleep well last night. Although i have a working that for some reason arrives at the correct answer, but i doubt that is the correct approach to solve the problem. The original idea i had, which seemed like the most logical and correct way to solve the problem for some reason was no where near getting me to the answer so i tried some other unorthodox method (i think) which finally got me to the answer but i did some things that i think breaks some rules of basic arithmetic manipulations. The first picture is the problem, the second picture is the solution which i think is wrong and might not work for some similar problems but gets to the correct answer for this particular problem ,the third picture is the original approach which i thought was the most logical approach but for some reason, i got stuck at some point and had no more ideas, and the fourth picture is the correct answer to the problem. I just need someone to make me understand how to correctly get to the right answer MasterTeeUSA NDSMELODY nlfpmod mynd44 farano dominique Fynestboi olawalebabs Richiez
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| Re: I Need Someone Good At Math To Help Me With This Complex Problem by federal9: 6:45pm On Dec 26, 2023*. Modified: 7:05pm On Dec 26, 2023 |
turmacs:Where you should look into is that place you started with K^2 -M^2 Now ( K - M )^2 Remember that (a-b)^2 = a^2 - 2ab + b^2 Now replace (K - M)^2 with that. Remember that = K^2 - 2kM + M^2 K^2 - 2KM + M^2 = K/C - M/K I believe you can solve it from here. K/C = K^2 - 2KM + M^2 + M/K Multiply all through with C k = K^2 C - 2KMC + M^2 C + M/K (C) Factorize C from the right hand side. K = C ( K^2 - 2KM + M^2 + M/K ) Making C as the subject formulae C = K / ( K^2 - 2KM + M^2 + M/K ) Carry on from there. Remember to simplify because of that M/K |
| Re: I Need Someone Good At Math To Help Me With This Complex Problem by MasterTeeUSA: 7:33pm On Dec 26, 2023 |
This solution is CORRECT turmacs: |
| Re: I Need Someone Good At Math To Help Me With This Complex Problem by MasterTeeUSA: 7:35pm On Dec 26, 2023 |
This is the right answer and correct steps...
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| Re: I Need Someone Good At Math To Help Me With This Complex Problem by NDSMELODY(m): 11:23pm On Dec 26, 2023 |
turmacs:here is the solution with stepwise procedures, just check it gently and you will get it simple.
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| Re: I Need Someone Good At Math To Help Me With This Complex Problem by youngrichnigga: 3:35pm On Dec 27, 2023 |
Here's the solution To make **c** the subject of the relation **k-m= √ k/c -m/k**, we can follow the steps below: 1. Square both sides of the equation to get rid of the square root: **(k-m)^2 = k/c - m/k**. 2. Multiply both sides of the equation by **c** to eliminate the fraction: **c(k-m)^2 = k - m(c/k)**. 3. Expand the left-hand side of the equation: **c(k^2 - 2km + m^2) = k - m(c/k)**. 4. Move all the terms containing **c** to the left-hand side of the equation: **ck^2 - 2ckm + cm^2 = k^2 - m(c/k)**. 5. Rearrange the equation to get all the terms containing **c** on one side: **ck^2 - m(c/k) - 2ckm + cm^2 - k^2 = 0**. 6. Multiply both sides of the equation by **k** to eliminate the fraction: **ck^3 - m(c) - 2ckm^2 + ckm^2 - k^3 = 0**. 7. Rearrange the equation to get all the terms containing **c** on one side: **ck^3 - k(m^2 + k^2) + m(c) = 0**. 8. Finally, solve for **c** by dividing both sides of the equation by **m**: **c = (k^4)/(m^2) - k^2 + (k^3)/(m)**. Therefore, **c = (k^4)/(m^2) - k^2 + (k^3)/(m)** is the solution for making **c** the subject of the relation **k-m= √ k/c -m/k**. Courtesy AI. |
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