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Help me solve a math equation please!?

Help me solve a math equation please!? Topic: Help me solve a math equation please!?
July 18, 2019 / By Bret
Question: Suppose that the width of a rectangle is 5 inches shorter than the length that the perimeter of the rectangle is 50 inches. The formula for the perimeter of the rectangle is P=2L+2W. a) Set up an equation involving only L, the length of the rectangle. b) What is the length? What is the width? (Please help me out! I'm swamped in work and this problem kinda stumped me >.< please and thank you)
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Best Answers: Help me solve a math equation please!?

Aherin Aherin | 10 days ago
a. W=L-5. P = 2L + 2W. Substitute W=L-5 and you get: P=2L + 2(L-5). Multiply 2 through the bracket and you get: P=2L + 2L - 10. Add the Ls and you get: P=4L-10. Add 10 to both sides and you get: P+10 = 4L. Divide both sides by 4 and L = (P+10)/4. b. P=50. Using the equation above for L: Put 50 in for P. L = (50+10)/4 = 60/4 = 15. From a, above, W=L-5 = 15-5 = 10. Answer. L=15. W=10 Proof of your work. P = 2L + 2W = 2x15 + 2x10 = 30 + 20 = 50
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Aherin Originally Answered: Can you solve this math equation?
What you have there is a cubic equation. The quadratic formula is no more use here than the solution of a linear equation would be for solving a quadratic equation. There is a formula like the quadratic formula for solving a general cubic equation, but it's much more complex (literally, because it involves complex numbers even when all the roots are real), and not really much use except in certain areas of theory. The best approach is usually to get it into lowest-integer-coefficient form, see whether the rational root theorem will make factoring possible, and if not (probably not in this case), then use Newton's Method to home in on a solution. x(-25): f(x) = 756900x³ + 25x² - 18350x + 338343 = 0 Personally, I wouldn't bother trying to factor the first and last coefficients in order to apply the rational root theorem. As a cubic with real coefficients, it has at least one real root. The other two are either both real, or complex conjugates of each other. To use Newton's Method, you need a good first guess, and the derivative of f: f'(x) = 2270700x² + 50x - 18350 For the first guess, notice that when x is relatively small in magnitude, the outer two terms dominate the two inner ones. f(0) = 338343 f(-1) = -756900 + 25 + 18350 + 338343 = -436932 This means there is definitely a root between 0 and -1. Because those two middle terms are a lot smaller than the outer ones, a good first guess is x³ = -338343/756900; x = -∛(338343/756900) = -0.7646... Then iterate: x[next] = x[last] - f(x[last])/f'(x[last]) until the value of x stabilizes. Using x[1st] as above, I get x[2] = -.77533796... x[3] = -.7751888724... x[4] = -.7751888439339311... x[5] = -.7751888439339311... It looks pretty certain that the other two roots will be complex, because it looks like the "squiggle" (inflection point) in the graph of the cubic occurs far above the x-axis, and the local min and max are likewise at large positive values of f(x), so that the graph doesn't cross the x-axis a 2nd or 3rd time.
Aherin Originally Answered: Can you solve this math equation?
you have to first factor out an X and leave it aside. then you do 734-13533.72. then you plug it into the quadratic formula. x(-30276x^2-x-12799.7)=0. s then you would get X=0 and the two other numbers you get from the quadratic formula

Aherin Originally Answered: Can you help me solve this math equation?
So a/(x+5) +b/(x-5)=x-273/(x^2-25) or a(x-5)+b(x+5)=x-273 so a+b=1 and 5b-5a=-273 Solving the first of these equations, a=1-b: substitute this into the second equation. 5b-5(1-b)=-273 or 10b-5=-273 or 10b=-268 so b=-26.8 and a = 27.8
Aherin Originally Answered: Can you help me solve this math equation?
Your graphics didn't seem to come through very clear. I'm assuming your equation is [(x-2)/(x^2-25)] = [7/(x+5)] + [3/(x-5)]. If this is the case, then you have the difference of two squares on the left side so we can get common denominators on the right by multiplying each term by the opposite denominator. Thus, [(x-2)/(x^2-25)] = [7(x-5)/(x+5)(x-5)] + [3(x+5)/(x-5)(x+5)]. Note, that we haven't changed the value of the terms because (x-5)/(x-5) and (x+5)/(x+5) both equal one. Simplifying the right side gives us [(x-2)/(x^2-25)] = [(7x-35)/(x^2-25)] + [(3x+15)/(x^2-25)] [(x-2)/(x^2-25)] = [(10x-20)/(x^2-25)] Multiplying both sides by x^2-25 to clear the fractions leaves us with x-2=10x-20 Add 20 to and subtract x from both sides. 18=9x Divide both sides by 9. 2=x.
Aherin Originally Answered: Can you help me solve this math equation?
Next time use / when you want to divide, not -------- Partial fractions (x-2) / (x^2 - 25) = (x-2) / (x+5)(x-5) = A/(x+5) + B/(x-5) = [ A*(x-5) + B*(x+5)] / (x-5)(x+5) Numerator = (A+B)x + 5*(B-A) = x-2 So A+B = 1 B-A = -2/5 B = 3/10 A = 7/10

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