• Hard math problem, please help, show all work?

    Start with ac³. 135 = 3x3x3x5 = 3³x5 This means that a = 5 and c = 3. Now go to ba³ = 1375 We already know that a = 5 and a³ = 125. So to find b find 1375/125. You'll find that b = 11 and a + b + c = 5 + 11 + 3 = 19
    Start with ac³. 135 = 3x3x3x5 = 3³x5 This means that a = 5 and c = 3. Now go to ba³ = 1375 We already know that a = 5 and a³ = 125. So to find b find 1375/125. You'll find that b = 11 and a + b + c = 5 + 11 + 3 = 19
    3 answers · Mathematics · 7 hours ago
  • Hard math problem, please help, show all work?

    You have √((b/a)*√((b/a)*√((b/a)))) = (b/a)^x Square both sides: (b/a)^(2x) = (b/a)*√((b/a)*√((b/a))) Square both sides again: (b/a)^(4x) = (b/a)^2(b/a)√(b/a) Square both sides one last time: (b/a)^(8x) = (b/a)^4(b/a)^2(b/a) Simplify the right side to get: (b/a)^(8x) = (b/a)^7 Now this boils down to: 8x = 7 x = 7/8 As a general rule if... show more
    You have √((b/a)*√((b/a)*√((b/a)))) = (b/a)^x Square both sides: (b/a)^(2x) = (b/a)*√((b/a)*√((b/a))) Square both sides again: (b/a)^(4x) = (b/a)^2(b/a)√(b/a) Square both sides one last time: (b/a)^(8x) = (b/a)^4(b/a)^2(b/a) Simplify the right side to get: (b/a)^(8x) = (b/a)^7 Now this boils down to: 8x = 7 x = 7/8 As a general rule if you're taking the k-th root (in your case k = 2) n times (in your case n = 3), then: x = (kⁿ - 1)/(kⁿ⁺¹ - kⁿ) This formula works so long as you have the (b/a) part on each side and you have (b/a)^x on one side.
    2 answers · Mathematics · 7 hours ago
  • If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8 m from the dock?

    This is the picture that you should have to aid in solving this problem. From it, you get: x² + y² = L² Differentiate with respect to time: d/dt[x² + y²] - d/d[L²] 2x*dx/dt + 2y*dy/dt = 2L*dL/dt Now you want to find how fast the boat is approaching the dock which is dx/dt. Solve for dx/dt: x*dx/dt = L*dL/dt - y*dy/dt dx/dt = (L*dL/dt -... show more
    This is the picture that you should have to aid in solving this problem. From it, you get: x² + y² = L² Differentiate with respect to time: d/dt[x² + y²] - d/d[L²] 2x*dx/dt + 2y*dy/dt = 2L*dL/dt Now you want to find how fast the boat is approaching the dock which is dx/dt. Solve for dx/dt: x*dx/dt = L*dL/dt - y*dy/dt dx/dt = (L*dL/dt - y*dy/dt)/x x = 8 (Given) y = 1 (Given) dy/dt = 0 (the height of the pulley is constant) dL/dt = 1 (Given) L = √65 (Calculated from Pythagorean theorem) dx/dt = (√65*1 - 1*0)/8 dx/dt = √65/8 m/s ≈ 1.007782219 m/s
    1 answer · Mathematics · 19 hours ago
  • Prove (explain how) the sum of harmonics y = a cos(2πvt) + b sin(2πvt) can be expressed as a single harmonic oscillator, that is?

    Best answer: First thing, linear combinations of sinusoids can only be combined if the frequency of each sinusoid is the same. For instance: 5 sin⁡(2x)+4cos⁡(2x-5) sin⁡(4x-5)-9 cos⁡(4x+2) can all be transformed into the form A sin⁡(Bx-C). However: 3 sin⁡(3x)+8cos⁡(x+3) 6sin⁡(3x-5)-9 cos⁡(8x+1) cannot be transformed into the form A sin⁡(Bx-C). ... show more
    Best answer: First thing, linear combinations of sinusoids can only be combined if the frequency of each sinusoid is the same. For instance: 5 sin⁡(2x)+4cos⁡(2x-5) sin⁡(4x-5)-9 cos⁡(4x+2) can all be transformed into the form A sin⁡(Bx-C). However: 3 sin⁡(3x)+8cos⁡(x+3) 6sin⁡(3x-5)-9 cos⁡(8x+1) cannot be transformed into the form A sin⁡(Bx-C). To combine linear combinations of sine and cosine functions you need to know the difference formula for sine: sin⁡(x-y)=sin⁡(x) cos⁡(y)-cos⁡(x) sin⁡(y) Begin with the linear combination: a sin⁡(Bx)+b cos⁡(Bx) You want to write this as a single sinusoid: a sin⁡(Bx)+b cos⁡(Bx)=A sin⁡(Bx-C) Begin by re-writing the right hand side using the difference formula: a sin⁡(Bx)+b cos⁡(Bx)=A(sin⁡(Bx) cos⁡(C)-cos⁡(Bx) sin⁡(C) ) a sin⁡(Bx)+b cos⁡(Bx)=A sin⁡(Bx) cos⁡(C)-A cos⁡(Bx) sin⁡(C) a sin⁡(Bx)+b cos⁡(Bx)=A 〖 cos⁡(C)sin〗⁡(Bx)-A sin⁡(C) cos⁡(Bx) Since both sides are equal, the coefficients of sin⁡(Bx) and cos⁡(Bx) are also equal: b=-A sin⁡(C) a=A cos⁡(C) Divide the top equation by the bottom equation to get: b/a=-(A sin⁡(C))/(A cos⁡(C) ) This reduces to: -b/a=tan⁡(C) Solve for C: C=tan^(-1)⁡(-b/a) Return to the equations equating the coefficients: b=-A sin⁡(C) a=A cos⁡(C) Square both sides of both equations: b^2=(-A sin⁡(C) )^2 a^2=(A cos⁡(C) )^2 This turns into: b^2=A^2 sin^2⁡(C) a^2=A^2 cos^2⁡(C) Add both equations to get: a^2+b^2=A^2 sin^2⁡(C)+A^2 cos^2⁡(C) a^2+b^2=A^2 (sin^2⁡(C)+cos^2⁡(C) ) Remember that sin^2⁡(C)+cos^2⁡(C)=1: A^2=a^2+b^2 A=√(a^2+b^2 ) To recap, given: a sin⁡(Bx)+b cos⁡(Bx) This can be re-written as: A sin⁡(Bx-C) A=√(a^2+b^2 ) C=tan^(-1)⁡(-b/a) Now for when the phase shift is different: Suppose you're given: A sin⁡(ωt+α)+B cos⁡(ωt+β) To write this as a single sinusoid you want: A sin⁡(ωt+α)+B cos⁡(ωt+β)=C sin⁡(ωt+θ) To solve this, you need to know the sum formulas for both sine and cosine: sin⁡(x+y)=sin⁡(x) cos⁡(y)+cos⁡(x) sin⁡(y) cos⁡〖(x+y)=cos⁡(x) cos⁡(y)-sin⁡(x) sin⁡(y) 〗 Also to help keep track of which item belongs to which trig function on the left hand side we'll redefine some of them: A=A_s B=A_c α=θ_s β=θ_c Re-write using the modifications: A_s sin⁡(ωt+θ_s )+A_c cos⁡(ωt+θ_c )=A sin⁡(ωt+θ) Re-write both sides using sum and difference formulas: A_s cos⁡(θ_s ) sin⁡(ωt)+A_s sin⁡(θ_s ) cos⁡(ωt)+A_c cos⁡(θ_c ) cos⁡(ωt)-A_c sin⁡(θ_c ) sin⁡(ωt)=A cos⁡(θ) sin⁡(ωt)+A sin⁡(θ) cos⁡(ωt) Re-write the left hand side by grouping sin⁡(ωt) and cos⁡(ωt): (A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) ) sin⁡(ωt)+(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) ) cos⁡(ωt)=A cos⁡(θ) sin⁡(ωt)+A sin⁡(θ) cos⁡(ωt) Equate the coefficients of sin⁡(ωt) and cos⁡(ωt): A sin⁡(θ)=A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) A cos⁡(θ)=A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) Divide the top equation by the bottom top to get: (A sin⁡(θ))/(A cos⁡(θ) )=(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) ) Simplify and solve for θ: tan⁡(θ)=(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) ) θ=tan^(-1)⁡((A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) )) Now return to the equations that equated the coefficients: A sin⁡(θ)=A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) A cos⁡(θ)=A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) Square both sides: (A sin⁡(θ) )^2=(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) )^2 (A cos⁡(θ) )^2=(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) )^2 A^2 sin^2⁡(θ)=〖A_s〗^2 sin^2⁡(θ_s )+2A_s A_c sin⁡〖(θ_s ) cos⁡(θ_c ) 〗+〖A_c〗^2 cos^2⁡(θ_c ) A^2 cos^2⁡(θ)=〖A_c〗^2 sin^2⁡(θ_c )-2A_s A_c sin⁡〖(θ_c ) cos⁡(θ_s ) 〗+〖A_s〗^2 cos^2⁡(θ_s ) Add both equations and reduce (remember that sin^2⁡(x)+cos^2⁡(x)=1): A^2 sin^2⁡(θ)+A^2 cos^2⁡(θ)=〖A_s〗^2 sin^2⁡(θ_s )+2A_s A_c sin⁡〖(θ_s ) cos⁡(θ_c ) 〗+〖A_c〗^2 cos^2⁡(θ_c )+〖A_c〗^2 sin^2⁡(θ_c )-2A_s A_c sin⁡〖(θ_c ) cos⁡(θ_s ) 〗+〖A_s〗^2 cos^2⁡(θ_s ) A^2=〖A_c〗^2+〖A_s〗^2+2A_s A_c (sin⁡〖(θ_s ) cos⁡(θ_c ) 〗-cos⁡(θ_s ) sin⁡(θ_c ) ) Remember the difference formula for sine: sin⁡(x-y)=sin⁡(x) cos⁡(y)-cos⁡(x) sin⁡(y) Reduce the right hand side and solve for A: A^2=〖A_c〗^2+〖A_s〗^2+2A_s A_c (sin⁡〖(θ_s ) cos⁡(θ_c ) 〗-cos⁡(θ_s ) sin⁡(θ_c ) ) A^2=〖A_c〗^2+〖A_s〗^2+2A_s A_c sin⁡(θ_s-θ_c ) A=√(〖A_c〗^2+〖A_s〗^2+2A_s A_c sin⁡(θ_s-θ_c ) ) To recap, given: A_s sin⁡(ωt+θ_s )+A_c cos⁡(ωt+θ_c ) This can be re-written as: A sin⁡(ωt+θ) A=√(〖A_c〗^2+〖A_s〗^2+2A_s A_c sin⁡(θ_s-θ_c ) ) θ=tan^(-1)⁡((A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) )) Hope all of this helps
    1 answer · Mathematics · 22 hours ago
  • Y= x/x, why doesn't it have a vertical asymptote at x = 0?

    x/x has a hole at x = 0. If as x approaches 0, x/x got larger or smaller (more negative) then it would have an asymptote. But for any value of x that approaches 0, x/x = 1. Since this is always true except when x =0, then you get a hole not an asymptote.
    x/x has a hole at x = 0. If as x approaches 0, x/x got larger or smaller (more negative) then it would have an asymptote. But for any value of x that approaches 0, x/x = 1. Since this is always true except when x =0, then you get a hole not an asymptote.
    6 answers · Mathematics · 3 days ago
  • Math help?

    If you disregard the x = 10 portion you get 8.54. You might try that.
    If you disregard the x = 10 portion you get 8.54. You might try that.
    3 answers · Mathematics · 3 days ago
  • Calculus problem?

    I'm not sure of the syntax on these, but I can provide the basic logic and you'd have to translate it into the language of your choice input n S = 0 for i = 1 to n S = S + 3*cos(0+i*π/(2*n))*π/(2*n) end display S Now you could prompt for the interval [a,b] (in your problem it was [0,π/2]): input n input a input b S = 0 for i = 1 to n ... show more
    I'm not sure of the syntax on these, but I can provide the basic logic and you'd have to translate it into the language of your choice input n S = 0 for i = 1 to n S = S + 3*cos(0+i*π/(2*n))*π/(2*n) end display S Now you could prompt for the interval [a,b] (in your problem it was [0,π/2]): input n input a input b S = 0 for i = 1 to n S = S + 3*cos(a+i*b/n)*b/n end display S Hope this helps
    2 answers · Mathematics · 3 days ago
  • Can someone please help me asap with this math question!!???

    Best answer: You have: cos(x)tan(x) + √3cos(x) = 0 Factor out cos(x) cos(x)(tan(x) + √3) = 0 This means that cos(x) = 0 or tan(x) + √3 = 0 cos(x) = 0 when x = 0 or x = π These occur with a period of 2π, but since it is at multiples of π, then cos(x) = 0, when x = 0 + πk tan(x) + √3 = 0 tan(x) = -√3 tan(x) = -√3 when x = 2π/3 or x =... show more
    Best answer: You have: cos(x)tan(x) + √3cos(x) = 0 Factor out cos(x) cos(x)(tan(x) + √3) = 0 This means that cos(x) = 0 or tan(x) + √3 = 0 cos(x) = 0 when x = 0 or x = π These occur with a period of 2π, but since it is at multiples of π, then cos(x) = 0, when x = 0 + πk tan(x) + √3 = 0 tan(x) = -√3 tan(x) = -√3 when x = 2π/3 or x = 5π/3 tan(x) has a period of π, so then tan(x) = --√3 (tan(x) + √3 = 0) when x = 2π/3 + πk or x = 5π/3 + πk You could actually write this as tan(x) = --√3 (tan(x) + √3 = 0) when x = 2π/3 + πk This would actually encompass both 2π/3 + πk and 5π/3 + πk It's probably looking for: x = 0 + 2πk x = 2π/3 + πk x = π + 2πk
    1 answer · Mathematics · 3 days ago
  • Rationalize the denominator?

    Best answer: In this instance, 4/(1-√3) you multiply by the conjugate of the denominator which in this case is 1+√3 to get: 4/(1-√3) x (1+√3)/(1+√3) = 4(1+√3)/(1 - 3) = 4(1+√3)/-2 = -2(1+√3)
    Best answer: In this instance, 4/(1-√3) you multiply by the conjugate of the denominator which in this case is 1+√3 to get: 4/(1-√3) x (1+√3)/(1+√3) = 4(1+√3)/(1 - 3) = 4(1+√3)/-2 = -2(1+√3)
    2 answers · Mathematics · 2 weeks ago
  • 9 – 5 ÷ (8 – 3) × 3 + 6?

    Best answer: PEMDAS is a bit of "marketing". You don't strictly do multiplication before any division. Think of it more as steps. On the first step is parenthesis. The next step is exponents. On the next step is multiplication and division. On the last step is addition and subtraction. Since multiplication and division are on the... show more
    Best answer: PEMDAS is a bit of "marketing". You don't strictly do multiplication before any division. Think of it more as steps. On the first step is parenthesis. The next step is exponents. On the next step is multiplication and division. On the last step is addition and subtraction. Since multiplication and division are on the same step you do them simultaneously. Same with addition and subtraction. You do them simultaneously from left to right as you read. So that is why you divide -5 by 5 before multiplying 5 by 3.
    6 answers · Mathematics · 2 weeks ago
  • If you are given odds of 2 to 3 in favor of winning a​ bet, what is the probability of winning the​ bet?

    Odds of 2 to 3 in favor means that there are 2 outcomes in your favor versus 3 outcomes not in your favor. That means that there are 5 outcomes. Since 2 are in you favor and there are 5 possible outcomes. The probability of winning is 2/5 or 40%.
    Odds of 2 to 3 in favor means that there are 2 outcomes in your favor versus 3 outcomes not in your favor. That means that there are 5 outcomes. Since 2 are in you favor and there are 5 possible outcomes. The probability of winning is 2/5 or 40%.
    10 answers · Mathematics · 2 weeks ago
  • Can anyone help me with this.....How many six-digit palindromic numbers are there which are greater than 250000?pls explain it ty?

    Best answer: Consider the number abcdef, where each of the digits may be different or the same. So in order for abcdef to be a palindrome you have def = cba. Basically the first 3 numbers would necessarily determine the last 3 numbers. But since you want the number to be greater than 250000, the first 3 numbers must be at least 250. So this now... show more
    Best answer: Consider the number abcdef, where each of the digits may be different or the same. So in order for abcdef to be a palindrome you have def = cba. Basically the first 3 numbers would necessarily determine the last 3 numbers. But since you want the number to be greater than 250000, the first 3 numbers must be at least 250. So this now turns into the problem of determining how many 3 digit numbers are greater than or equal to 250. So you have all 3 digit numbers from 250 to 999. If you take the difference of 999 and 250 you get 749 but that only includes one of the end points. So you need to add 1 to that to get both end points. So there are 750 6 digit palindromes greater than 250000.
    3 answers · Mathematics · 2 weeks ago
  • What can be the highest power of 5 in 100!?

    From 1 to 100, there are 20 numbers that are divisible by 5. 16 of these numbers are divisible by 5 and 4 are divisible by 25 (5^2) So there are 16 + 4*2 5's that compose the number 100!. That means that 100! is divisible by 5^24.
    From 1 to 100, there are 20 numbers that are divisible by 5. 16 of these numbers are divisible by 5 and 4 are divisible by 25 (5^2) So there are 16 + 4*2 5's that compose the number 100!. That means that 100! is divisible by 5^24.
    4 answers · Mathematics · 2 weeks ago
  • Can you solve this math problem?

    Best answer: This is the picture that you should have: A couple of things to note: 475 = x + y tan(18°) = x/L tan(2°) = y/L Solve for y: y = 475 - x Substitute: tan(18°) = x/L tan(2°) = (475 - x)/L Since you want to solve for L, solve for x in one of the equations: x = Ltan(18°) Substitute this into tan(2°) = (475 - x)/L: tan(2°) = (475 -... show more
    Best answer: This is the picture that you should have: A couple of things to note: 475 = x + y tan(18°) = x/L tan(2°) = y/L Solve for y: y = 475 - x Substitute: tan(18°) = x/L tan(2°) = (475 - x)/L Since you want to solve for L, solve for x in one of the equations: x = Ltan(18°) Substitute this into tan(2°) = (475 - x)/L: tan(2°) = (475 - Ltan(18°))/L Now solve for L: tan(2°) = (475 - Ltan(18°))/L Ltan(2°) = 475 - Ltan(18°) 475 = Ltan(2°) + Ltan(18°) 475 = L(tan(2°) + tan(18°)) L = 475/(tan(2°) + tan(18°)) L ≈ 1320.029416 feet
    3 answers · Mathematics · 3 weeks ago
  • Find all degree solutions of 2sin^2 10theta + sin10 theta - 1 = 0 I can't figure out how to do this problem. Please help!?

    Let x = sin(10θ) This gives: 2x² + x - 1 = 0 Using quadratic formula you get: x = 1/2 x = -1 This means: sin(10θ) = -1 or sin(10θ) = 1/2 First sin(10θ) = -1 sin(10θ) = -1 10θ = 270° + 360n°, where n is an integer Now divide by 10: θ = 27° + 36n°, where n is an integer Now for sin(10θ) = 1/2 10θ = 30° + 360n°, where n is an... show more
    Let x = sin(10θ) This gives: 2x² + x - 1 = 0 Using quadratic formula you get: x = 1/2 x = -1 This means: sin(10θ) = -1 or sin(10θ) = 1/2 First sin(10θ) = -1 sin(10θ) = -1 10θ = 270° + 360n°, where n is an integer Now divide by 10: θ = 27° + 36n°, where n is an integer Now for sin(10θ) = 1/2 10θ = 30° + 360n°, where n is an integer or 10θ = 150° + 360n°, where n is an integer Now divide by 10: θ = 3° + 36n°, where n is an integer or θ = 15° + 36n°, where n is an integer
    4 answers · Mathematics · 3 weeks ago
  • Having trouble with an inverse trig problem, any ideas?

    Here's a different approach. Given the triangle in the photo you get: sin(θ) = 1/7 → θ = sin⁻¹(1/7) cos(φ) = 1/7 → φ = cos⁻¹(1/7) So this means that: sin⁻¹(1/7) + cos⁻¹(1/7) = θ + φ Since the triangle is right,θ + φ = π/2 which means sin⁻¹(1/7) + cos⁻¹(1/7) = π/2
    Here's a different approach. Given the triangle in the photo you get: sin(θ) = 1/7 → θ = sin⁻¹(1/7) cos(φ) = 1/7 → φ = cos⁻¹(1/7) So this means that: sin⁻¹(1/7) + cos⁻¹(1/7) = θ + φ Since the triangle is right,θ + φ = π/2 which means sin⁻¹(1/7) + cos⁻¹(1/7) = π/2
    4 answers · Mathematics · 3 weeks ago
  • A lighthouse is located on a small island 2 km away from the nearest point P on a straight shoreline and its light makes six revolutions per?

    Here is my solution. My answer is in km/min which translates to 377.0 km/min. You didn't specify the units.
    Here is my solution. My answer is in km/min which translates to 377.0 km/min. You didn't specify the units.
    3 answers · Mathematics · 2 months ago
  • Hard Vector Curvature Problem, Help!!?

    Best answer: So try this: r(t) = -4sin(t)i - 4sin(t)j + cos(t)k κ(t) = |T(t)|/|r'(t)| (curvature) r'(t) -4cos(t)i - 4cos(t)j - sin(t)k |r'(t)| = √((-4cost)² + (-4cost)² + (-sint)²) |r'(t)| = √(16cos²t + 16cos²t + sin²t) |r'(t)| = √(32cos²t + sin²t) Now you can pull out a cos²t: |r'(t)| = √(cos²t(32 +... show more
    Best answer: So try this: r(t) = -4sin(t)i - 4sin(t)j + cos(t)k κ(t) = |T(t)|/|r'(t)| (curvature) r'(t) -4cos(t)i - 4cos(t)j - sin(t)k |r'(t)| = √((-4cost)² + (-4cost)² + (-sint)²) |r'(t)| = √(16cos²t + 16cos²t + sin²t) |r'(t)| = √(32cos²t + sin²t) Now you can pull out a cos²t: |r'(t)| = √(cos²t(32 + sin²t/cos²t)) |r'(t)| = cos(t)√(32 + tan²t) Now find T(t) (T(t) = r'(t)/|r'(t)|) r'(t) = -4cos(t)i - 4cos(t)j - sin(t)k |r'(t)| = cos(t)√(32 + tan²t) T(t) = (-4cos(t)i - 4cos(t)j - sin(t)k)/(cos(t)√(32 + tan²t)) T(t) = -4/√(32 + tan²t)i - 4/√(32 + tan²t)j - tant/√(32 + tan²t)k |T(t)| = √((-4/√(32 + tan²t))² + (-4/√(32 + tan²t))² + (tant/√(32 + tan²t))²) |T(t)| = √(16/(32 + tan²t)) + 16/(32 + tan²t)) + tan²t/(32 + tan²t))) |T(t)| = √((16 + 16 + tan²t)/(32 + tan²t)) |T(t)| = √((32 + tan²t)/(32 + tan²t)) |T(t)| = √1 = 1 Now back to κ(t): κ(t) = |T(t)|/|r'(t)| κ(t) = 1/√(32cos²t + sin²t) Now at this point you could try a few different representations: κ(t) = sec(t)/√(32 + tan²t) κ(t) = sec(t)/√(31 + sec²t) κ(t) = 1/√(31cos²t + 1) I could go on and on with these. Try κ(t) = 1/√(32cos²t + sin²t) and see if that works.
    1 answer · Mathematics · 2 months ago
  • Are there any integer solutions to this Diophantine equation? x^(1/n)+y^(1/m)=z where x,y,n,m,z are all integers and gcd(x,y)=1?

    x = 4 n = 2 y = 27 m = 3 Therefore z = 5 and gcd(4,27) = 1
    x = 4 n = 2 y = 27 m = 3 Therefore z = 5 and gcd(4,27) = 1
    2 answers · Mathematics · 8 months ago