Search results

  1. MarkFL

    what have I done wrong? "if [(sinx)^4]/2 + [(cosx)^4]/3 = 1/5 then prove that (tanx)^2=2/3"

    15\tan^4(x)+10=6\left(\tan^4(x)+2\tan^2(x)+1\right) 15\tan^4(x)+10=6\tan^4(x)+12\tan^2(x)+6 9\tan^4(x)-12\tan^2(x)+4=0 \left(3\tan^2(x)-2\right)^2=0 \tan^2(x)=\frac{2}{3}
  2. MarkFL

    what have I done wrong? "if [(sinx)^4]/2 + [(cosx)^4]/3 = 1/5 then prove that (tanx)^2=2/3"

    Dividing through by the 4th power of the cosine function, multiplying by 30 and using a Pythagorean identity, you may write: 15\tan^4(x)+10=6\left(\tan^2(x)+1\right)^2 Arrange this as a quadratic in the square of the tangent function...
  3. MarkFL

    L'Hopital's Rule Problem: Use l'Hopital's rule to find lim_{x->0} (e^(5/x) - 3x)^(x/2)

    Let's assume we are taking the limit from the right. Doing the division I suggested, the expression becomes: \frac{5+3x^2e^{-\frac{5}{x}}}{2\left(1-3xe^{-\frac{5}{x}}\right)}
  4. MarkFL

    L'Hopital's Rule Problem: Use l'Hopital's rule to find lim_{x->0} (e^(5/x) - 3x)^(x/2)

    I do agree that the problem should have stated from the right. I simplified what you have to: \ln(L)=\lim_{x\to0}\left(\frac{5e^{\frac{5}{x}}+3x^2}{2\left(e^{\frac{5}{x}}-3x\right)}\right) Now, divide numerator and denominator by: e^{\frac{5}{x}}...
  5. MarkFL

    L'Hopital's Rule Problem: Use l'Hopital's rule to find lim_{x->0} (e^(5/x) - 3x)^(x/2)

    Yes, I would write the expression as: \frac{\ln\left(e^{\frac{5}{x}}-3x\right)}{\frac{2}{x}} Now you have a form to which L'Hopital can be applied.
  6. MarkFL

    L'Hopital's Rule Problem: Use l'Hopital's rule to find lim_{x->0} (e^(5/x) - 3x)^(x/2)

    I do get what they worksheet key shows. You are letting a still indeterminate form go to zero. You did not apply L'Hopital.
  7. MarkFL

    Q_r = (1/R) sqrt[L/C]; show that dQ_r/dC = -(1/(2R)) sqrt[L/C^3]

    For example, write: Q_r=\frac{\sqrt{L}}{R}C^{-\frac{1}{2}}
  8. MarkFL

    Q_r = (1/R) sqrt[L/C]; show that dQ_r/dC = -(1/(2R)) sqrt[L/C^3]

    Try using rational exponents instead of radicals, then apply the power rule, then convert back to radical notation.
  9. MarkFL

    Probability Question on dice: What is the probability that the sum of the two dice is a prime number?

    How many total outcomes are there when two dice are rolled?
  10. MarkFL

    Is my math book wrong? sin(180° ‒ α) · sin(180° + α) ‒ cos(180° + α) · cos(180° ‒ α)

    Your book is correct. You made a sign error, taking 3 negatives to be positive when you stated simplify further.
  11. MarkFL

    Developing Algebraic Equations (number of squares in "staircase")

    The factorial is a product, not a sum. n!=\prod_{k=1}^{n}(k)
  12. MarkFL

    prove trig identity cos(2x) / [3 - sin(2x) - 4sin^2(x)] = [tan(x) + 1] / [tan(x) + 3]

    This is what I had in mind. Begin with the lhs: \frac{\cos(2x)}{3-\sin(2x)-4\sin^2(x)} Apply double-angle identities: \frac{\cos^2(x)-\sin^2(x)}{3-2\sin(x)\cos(x)-4\sin^2(x)} Divide numerator and denominator by \cos^2(x) \frac{1-\tan^2(x)}{3\sec^2(x)-2\tan(x)-4\tan^2(x)} In the...
  13. MarkFL

    prove trig identity cos(2x) / [3 - sin(2x) - 4sin^2(x)] = [tan(x) + 1] / [tan(x) + 3]

    Try using double angle identities, then divide numerator and denominator by the square of the cosine function , apply a Pythagorean identity in the denominator to get rid of the secant function, then factor and cancel.
  14. MarkFL

    Substitution in fraction: Given that x/y = 2/7, evaluate (7x + y) / (x - (y/7))

    The given equation implies: 7x=2y And having x exclusively with coefficients of 7 makes substitution simple.
  15. MarkFL

    How to solve w^2=-3-4i

    I would write: w^2=-3-4i=(1-4)-4i=1-4i+4i^2=(1-2i)^2 Thus: w=\pm(1-2i)
  16. MarkFL

    Substitution in fraction: Given that x/y = 2/7, evaluate (7x + y) / (x - (y/7))

    I would write the given expression as: \frac{7(7x+y)}{7x-y} Then with: 7x=2y We have; \frac{7(2y+y)}{2y-y}=\frac{21y}{y}=21
  17. MarkFL

    How to solve this limit

    Do you mean you do not know how to apply L'Hopital or you have been instructed not to use it?
  18. MarkFL

    How to solve this limit

    Oops, I obviously did not read your initial post well.
  19. MarkFL

    How to solve this limit

    Have you tried L'Hopital?
  20. MarkFL

    Developing Algebraic Equations (number of squares in "staircase")

    The nth triangular number is: T_n=\sum_{k=1}^{n}(k) Perhaps the simplest way to find this closed form is to write: T_n=1+2+3+\dots+(n-2)+(n-1)+n T_n=n+(n-1)+(n-2)\dots+3+2+1 Adding, we find we have n summands of n + 1: 2T_n=n(n+1) Thus: T_n=\frac{n(n+1)}{2}
Top