• 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
  • 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 · 1 day 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
  • 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
  • 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
  • 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
  • VERY CHALLENGING MATH PROBLEM, PLEASE HELP! GUARANTEED 5 STARS!!!?

    Best answer: This problem involves a few steps. First you have to calculate the payment on the 30 year mortgage, the remaining balance at 10 years on the 30 year mortgage and the payment on the 15 year mortgage. The formula for a payment on a mortgage is: p = rL(1 + r)^t/((1 + r)^t - 1) Where: r = monthly rate L = Loan amount t = length of the... show more
    Best answer: This problem involves a few steps. First you have to calculate the payment on the 30 year mortgage, the remaining balance at 10 years on the 30 year mortgage and the payment on the 15 year mortgage. The formula for a payment on a mortgage is: p = rL(1 + r)^t/((1 + r)^t - 1) Where: r = monthly rate L = Loan amount t = length of the loan For the 30 year mortgage: r = 0.007 ---> (.084/12) L = $127,262.85 ( $149,721.00*.85 ---> Amount after 15% down) t = 360 (30 years * 12 months/year) The payment for the 30 year mortgage is $969.54. Now you need to calculate the balance of this loan at 10 years. The balance, B, is given by the following eqaution: B = L(1 + r)^t - p/r*((1 + r)^t - 1) Where: r = 0.007 ---> (.084/12) L = $127,262.85 ( $149,721.00*.85 ---> Amount after 15% down) t = 120 (10 years * 12 months/year) p = $969.54 (as calculated previously) After 10 years the balance of the loan is $112,539.90 and there are 240 payments left. So this person will have to make 240 payments of $969.54. This means this person will pay $232,688.72 (240 * $969.54) over the next 20 years, but only $112,539.90 is principal (the remaining amount owed). So that means that the difference of $232,688.72 and $112,539.90 is the amount of interest this person will pay over the next 20 years. The amount of interest they will pay is $120,148.81 ($232,688.72 - $112,539.90). Now calculate the payment on a 15 year mortgage using the payment formula above and the following amounts: L = $112,539.90 (the remaining balance on the mortgage after 10 years) r = 0.0045 (.054/12) t = 180 (15 years * 12 months/year) You should get a payment of $913.58 for the 15 year loan. Again, the person will make 180 payments of $913.58 for a total of $164,445.09 (180 * $913.58). But only $112,539.90 of that amount is principal, the rest is interest. The total amount of interest paid on the 15 year mortgage is $51,905.19 ($164,445.09 - $112,539.90). Now to calculate the interest savings, subtract the 2 interest amounts. $120,148.81 - $51,905.19 = $68,243.62 This means that the person would save $68,243.62 by switching to the 15 year mortgage.
    3 answers · Mathematics · 8 months ago
  • What does ||c'(t)|| mean?

    Best answer: c(t) is probably a vector valued function. c'(t) is the derivative of that function. The bars indicate the magnitude (the length of the vector) of that function.
    Best answer: c(t) is probably a vector valued function. c'(t) is the derivative of that function. The bars indicate the magnitude (the length of the vector) of that function.
    2 answers · Mathematics · 9 months ago
  • A closed box with square base is to be built to house an ant colony...?

    Best answer: The picture should help
    Best answer: The picture should help
    1 answer · Mathematics · 1 year ago
  • Cos3x=rad2/2?

    Best answer: Hope this helps
    Best answer: Hope this helps
    2 answers · Mathematics · 1 year ago
  • What is an even rational function and what is an odd rational function? How do you know which one it is based on the equation and/or graph?

    Best answer: The easiest way to tell if a function is even, odd, or neither is to look at it's equation. Even functions only have terms with even exponents. Odd functions have terms that only have odd exponents. Equations that contain terms with both odd and even exponents are neither odd nor even. For example: 3x² + 7 → even (remember that... show more
    Best answer: The easiest way to tell if a function is even, odd, or neither is to look at it's equation. Even functions only have terms with even exponents. Odd functions have terms that only have odd exponents. Equations that contain terms with both odd and even exponents are neither odd nor even. For example: 3x² + 7 → even (remember that the 7 is really 7x⁰) exponents are 0 and 2, both even numbers, function is even 5x³ - 9x → odd since exponents are 1 and 3 5x³ + 3x² - 9x - 7 → neither odd nor even Now for rational functions, there'e one more layer to consider. For rational functions: Even/Even → even function Odd/Odd → even function Even/Odd → odd function Odd/Even → odd function
    4 answers · Mathematics · 1 year ago
  • How can I show with calculus (not integrals or derivativ stuff) that the surface of sphere is: 4 pi r^2?

    Best answer: You can't. You can't say how do I use calculus to solve a problem but I can't use calculus. It would be like saying How do I drive my car across the country without using gas (assuming you have a car that has a gasoline engine)? You would say you can't. Same thing. You can't say solve the problem using calculus. ... show more
    Best answer: You can't. You can't say how do I use calculus to solve a problem but I can't use calculus. It would be like saying How do I drive my car across the country without using gas (assuming you have a car that has a gasoline engine)? You would say you can't. Same thing. You can't say solve the problem using calculus. Then say oh by the way you can't use calculus to solve this problem. I've included the links to 2 videos that show how Archimedes solved the surface area of a sphere. It is strictly a geometric derivation, which is what he had at his disposal. No calculus involved. https://www.youtube.com/watch?v=5Rrjbeuo... (watch first) https://www.youtube.com/watch?v=xCAWa757... (watch 2nd)
    3 answers · Mathematics · 1 year ago
  • Y=arc csc(2sqrt3/3) give the exact value of y in radians. Rmember that each inverse trig function has a certain interval.?

    Best answer: Given y = csc⁻¹(x), that is the same as y = sin⁻¹(1/x) You have y = csc⁻¹((2√3)/3). This becomes: y = sin⁻¹(3/(2√3)) 3/(2√3) = (3√3)/6 = √3/2 y = sin⁻¹(√3/2) The range of csc⁻¹(x) is [−π/2,0) ∪ (0,π/2] (sin⁻¹(x) is [−π/2,π/2], nearly identical). So you need to find when sin is √3/2 and is in the range [−π/2,π/2]. That occurs at π/3. show more
    Best answer: Given y = csc⁻¹(x), that is the same as y = sin⁻¹(1/x) You have y = csc⁻¹((2√3)/3). This becomes: y = sin⁻¹(3/(2√3)) 3/(2√3) = (3√3)/6 = √3/2 y = sin⁻¹(√3/2) The range of csc⁻¹(x) is [−π/2,0) ∪ (0,π/2] (sin⁻¹(x) is [−π/2,π/2], nearly identical). So you need to find when sin is √3/2 and is in the range [−π/2,π/2]. That occurs at π/3.
    4 answers · Mathematics · 1 year ago
  • Maths help please?

    Best answer: Given a + bi, to find θ, use tan⁻¹(b/a) to find the reference angle. Then based upon the angle you get, you have to put it into the correct quadrant. Think of a as x and b as y. So in the first example, θ = tan⁻¹(2/2), where both x and y are positive (QI). In the second example you have θ = tan⁻¹(4/-3) where x is negative and y is... show more
    Best answer: Given a + bi, to find θ, use tan⁻¹(b/a) to find the reference angle. Then based upon the angle you get, you have to put it into the correct quadrant. Think of a as x and b as y. So in the first example, θ = tan⁻¹(2/2), where both x and y are positive (QI). In the second example you have θ = tan⁻¹(4/-3) where x is negative and y is positive (QII). Then to find r use √(a² + b²). For a) θ = tan⁻¹(2/2) = π/4, and since it is in QI, that's the end of it. For r you get r = √(2² + 2²) = √8 = 2√2 θ = π/4 r = 2√2 For c) θ = tan⁻¹(4/-3). There is no exact value, you get θ = -0.9272952, which means the reference angle is 0.9272952. You want it in QII, so that means that θ = tan⁻¹(4/-3) + π. That is as exact as you can get, or you can write it as θ = 2.2142974. As for t, you get r = √((-3)² + 4²) = √(25) = 5 θ = tan⁻¹(4/-3) + π ≈ 2.2142974 r = 5
    2 answers · Mathematics · 1 year ago
  • How do I interpret this equation?

    Best answer: 1.0985 = initial rating .0704 = rating per age
    Best answer: 1.0985 = initial rating .0704 = rating per age
    2 answers · Mathematics · 1 year ago
  • Normal Distribution Curve basic question?

    Best answer: 1.76 is 2 standard deviations below the mean. 95% of the values fall between 2 standard deviations. That means that 2.5% falls outside of each side of 2 standard deviations. Since you want the percentage that are above 1.76, that means 97.5% is above 1.76.
    Best answer: 1.76 is 2 standard deviations below the mean. 95% of the values fall between 2 standard deviations. That means that 2.5% falls outside of each side of 2 standard deviations. Since you want the percentage that are above 1.76, that means 97.5% is above 1.76.
    2 answers · Mathematics · 1 year ago
  • Last question wasnt clear, so Solve r=1+3cos(pi/6) The solution's presented as 1 fraction: (2+3√3)/(2) How? Isn't it 1+(3√3)/(2)?

    Best answer: Both of those answers are correct, just different representations. 1 + (3√3)/2 is correct. But if you multiply 1 by 2/2 you get 2/2 + (3√3)/2 which turns into (2 + 3√3)/2
    Best answer: Both of those answers are correct, just different representations. 1 + (3√3)/2 is correct. But if you multiply 1 by 2/2 you get 2/2 + (3√3)/2 which turns into (2 + 3√3)/2
    2 answers · Mathematics · 1 year ago
  • Integral, mathematics I need help understanding this?

    Best answer: I'm not sure where the dp + ρVdV = 0 comes from, but moving on from there, to get rid of dp and dV you have to integrate. A few rules about integration: ∫kf(x)dx = k∫f(x)dx, where k is a constant ∫(f(x) + g(x))dx = ∫f(x)dx + ∫g(x)dx ∫dx = x + c, c is a constant and usually some initial value, a lot of times c works out to be... show more
    Best answer: I'm not sure where the dp + ρVdV = 0 comes from, but moving on from there, to get rid of dp and dV you have to integrate. A few rules about integration: ∫kf(x)dx = k∫f(x)dx, where k is a constant ∫(f(x) + g(x))dx = ∫f(x)dx + ∫g(x)dx ∫dx = x + c, c is a constant and usually some initial value, a lot of times c works out to be 0. ∫xdx = ¹/₂x² + c, same as c above Given F(x) = ∫f(x)dx b ∫f(x)dx = F(b) - F(a) a So when you integrate dp from p₁ to p₂, you get p₂ - p₁. When you integrate ρVdV from V₁ to V₂ you get ρ((V₂)²/2 - (V₁)²/2)
    1 answer · Mathematics · 1 year ago
  • Need help with calculus, answers please!?

    Best answer: I'll do the first one for you and give you a formula and you can do the second. For the left Riemann sum you have the following formula: n-1 ∑ f(a+kΔx)Δx k=0 on the interval [a,b] and Δx = (b - a)/n For your first problem, f(x) = 35 - 105x, a = -1, b = 1, Δx = 1/2 Let S be the sum then you get: S = f(-1 +0*.5)*.5 + f(-1 +... show more
    Best answer: I'll do the first one for you and give you a formula and you can do the second. For the left Riemann sum you have the following formula: n-1 ∑ f(a+kΔx)Δx k=0 on the interval [a,b] and Δx = (b - a)/n For your first problem, f(x) = 35 - 105x, a = -1, b = 1, Δx = 1/2 Let S be the sum then you get: S = f(-1 +0*.5)*.5 + f(-1 + 1*.5)*.5 + f(-1 + 2*.5) + f(-1 + 3*.5)*.5 Let's clear this up a bit: S = f(-1)*.5 + f(-.5)*.5 + f(0)*.5 + f(.5)*.5 f(-1) = 140 f(-.5) = 87.5 f(0) = 35 f(.5) = -17.5 S = 140*.5 + 87.5*.5 + 35*.5 + -17.5*.5 S = 70 + 43.75 + 17.5 + -8.75 S = 122.5000 Now for your next problem use f(x) = 13x², a = -2, b = 2. You should get 78.0000
    1 answer · Mathematics · 1 year ago