# Search results

1. ### 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. ;)
2. ### 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...
3. ### 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?
5. ### 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
6. ### 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...
7. ### 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...
8. ### 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
9. ### 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...
10. ### 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.
11. ### Are hate crimes still a real problem?

I celebrate the birthday of Isaac Newton on December 25. :) But, I do think hate crimes are an issue, and need to be punished as such, when it can be reasonably demonstrated that they are indeed hate crimes.
12. ### 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
13. ### 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...
14. ### 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...
15. ### How did this solution come about?

Both solutions work when substituted into the original equation.
16. ### Tyranny of the Majority

I'm speaking in general...I don't know what situations you're talking about.
17. ### Tyranny of the Majority

I don't think people should be bashed for their beliefs. But, people shouldn't feel bashed simply because their beliefs don't stand up to rational scrutiny either. I have absolutely no belief that I can't be dissuaded from if it can be demonstrated to me that I have no good reason to hold that...
18. ### Challenge Solve for x

In the original equation the RHS has 100 in the denominator. \frac{1}{100}=\frac{1}{10^2}=10^{-2}
19. ### 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\}
20. ### Challenge Find the sum to infinity of a sequence

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