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  1. MarkFL

    Need Help with Trigonometry Problems

    For the third one, apply the double angle identity for sine to the RHS...what do you then have? In the fourth one, the tangent function has no argument...can you clarify?
  2. MarkFL

    Need Help with Trigonometry Problems

    For the second problem, I would use these two product-to-sum identities: \cos(\alpha)\cos(\beta)=\frac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}{2} \sin(\alpha)\sin(\beta)=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2} And so, the equation then becomes...
  3. MarkFL

    Probability of Independent Event - 2

    After having computed the probability of the ball being picked from the second bad is grey, I would use the complement rule to find the probability that it is orange. ;)
  4. MarkFL

    Probability of Independent Event

    Okay, so you get: P(X)=\frac{29}{51} If I were to reason my way though this one, I would first look at the case where the ball moved from the first bag is orange: P(A)=\frac{2}{3}\cdot\frac{6}{17}=\frac{4}{17} Then the case where the ball moved from the first bag is grey...
  5. MarkFL

    Practice Completing the Square Math Help?

    See this thread, for the exact same problem word for word: Completing the Square Math Help?
  6. MarkFL

    Geometry math problem please help?

    If we apply a bit of trigonometry here, and we let \(h\) be the altitude we seek, we may state: \sin\left(42^{\circ}\right)=\frac{h}{10}\implies h=10\sin\left(42^{\circ}\right)\approx6.7 Does this make sense?
  7. MarkFL

    Integration by Parts - What to choose?

    I generally follow the LIATE rule for choosing which will be \(u\), and which will be \(v\,du\). Integration by parts - Wikipedia
  8. MarkFL

    Completing the Square Math Help?

    Let's let \(x\) be the one side of the dog yard parallel to the barn and \(y\) be the two sides perpendicular. Let \(F\) be the length of fence the farmer has and \(A\) be the desired area of the yard. And so we have: x+2y=F xy=A Now, the first equation implies: x=F-2y And so substituting...
  9. MarkFL

    Challenge Determine the minimum value

    I would first examine the roots of the first derivative for turning points: f'(x)=4x^3-12x+8=4\left(x^3-3x+2\right)=0 We can see that \(x=1\) is a root of the cubic factor, and so let's use synthetic division to fully factor: \begin{array}{c|rr}& 1 & 0 & -3 & 2 \\ 1 & & 1 & 1 & -2 \\ \hline...
  10. MarkFL

    Challenge Evaluate the sum

    I would begin by writing: 1+\frac{4}{x}+\frac{4}{x^2}=1+4\frac{x+1}{x^2}=1+4\frac{ab}{(a-b)^2}=\left(\frac{a+b}{a-b}\right)^2
  11. MarkFL

    Challenge Quadratic equation that has two roots

    If we expand and equate the two expressions, we obtain: x^2+(k+1991)x+1991k+1=x^2+(p+q)x+pq Equating coefficients, we get: k+1991=p+q 1991k+1=pq Now, the second equation implies: q=\frac{1991k+1}{p} Substituting into the first equation, we obtain: k+1991=p+\frac{1991k+1}{p} Multiply by...
  12. MarkFL

    New Buckeye Enters The Fold

    Hello, and welcome! Although I now live in Florida, I was raised in Evansville, IN., and have been to the great state of Ohio many times. I am looking forward to your participation, and if you have any suggestions for site improvement, I'd be more than happy to hear them.
  13. MarkFL

    Challenge Simplify a sum

    If we use a co-function identity, then the product \(P\) may be written: P=\prod_{x=0}^{89}\left(\frac{\cos\left((89-x)^{\circ})\right)}{\cos(x^{\circ})}\right)=1
  14. MarkFL

    Challenge Simplify a sum

    I would write: \tan(x)\cos\left(1^{\circ}\right)+\sin\left(1^{\circ}\right)=\frac{\sin(x)\cos\left(1^{\circ}\right)+\cos(x)\sin\left(1^{\circ}\right)}{\cos(x)} Applying the angle-sum identity for sine, we may write...
  15. MarkFL

    Challenge Find the sum of squares

    Adding the two given equations, we have: x^3+y^3=19x+19y Factor: (x+y)\left(x^2-xy+y^2\right)=19(x+y) If \(x+y=0\) then the equations become: 1.) x^3=5x x(x^2-5)=0 (x,y)=(0,0),\,(\pm\sqrt{5},\mp\sqrt{5}) And so \(x^2+y^2\) has the possible values of 0 and 10. We get the same solutions...
  16. MarkFL

    How did this solution come about?

    Both solutions work when substituted into the original equation.
  17. MarkFL

    Challenge Solve for x

    In the original equation the RHS has 100 in the denominator. \frac{1}{100}=\frac{1}{10^2}=10^{-2}
  18. MarkFL

    Challenge Solve for x

    Let: x=10^r And so the equation becomes: 10^{\log\left(10^{r^2}\right)}=10^{3r-2} Or: 10^{r^2}=10^{3r-2} This implies: r^2=3r-2 r^2-3r+2=0 (r-1)(r-2)=0 Hence: x\in\{10,100\}
  19. MarkFL

    Challenge Find the sum to infinity of a sequence

    It looks like we used essentially the same method. (Yes)
  20. MarkFL

    Challenge Find the sum to infinity of a sequence

    Okay, we are being asked to find the sum \(S\), where: S=\sum_{k=1}^{\infty}\left(\frac{1}{(2k-1)^2}\right) Consider the following: S_1=\sum_{k=1}^{\infty}\left(\frac{1}{(2k)^2}\right)=\frac{1}{4}\sum_{k=1}^{\infty}\left(\frac{1}{k^2}\right)=\frac{1}{4}\frac{\pi^2}{6}=\frac{\pi^2}{24} Now...