# Search results

1. ### 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. ### 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. ### 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. ### 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. ### Practice Completing the Square Math Help?

See this thread, for the exact same problem word for word: Completing the Square Math 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### How did this solution come about?

Both solutions work when substituted into the original equation.
17. ### 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. ### 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. ### Challenge Find the sum to infinity of a sequence

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