Recent content by youngstudent16

It was an online homework computer might have been off or the approach was wrong

The correct answer was 4.6 I got something different using those numbers

Homework Statement Calculate the approximate number of board feet (one board foot is defined as a volume of ##1 in \times 1 ft \times1 ft## of lumber that would be available in a log that has a diameter of ##34 cm## and a length of ##3 m##. Assume the log is a right cylinder and that the saw...

Ah perfect thank you I got 254 as the correct solution that was a lot of computation

Its multiplying not summing and yes I tried brute force now using wolframalpha for several numbers and its going up very slowly like 1.01 1..., 1..., 1... etc

I tried putting them with decimal and using wolframalpha I went to 10 numbers and still it was tiny

Ok trying to just work with the variables no numbers I'm getting this pattern ##\frac{\sqrt[3]{k-1)(k+1)^2}(k-1)}{k}## Now I simplify and set up the ineqaulity ##\frac{\sqrt[3]{(2k^2)(k+1)^2}}{2k}>4## Now what

Homework Statement For a positive integer ##n##, let ##a_n=\frac{1}{n} \sqrt[3]{n^{3}+n^{2}-n-1}## Find the smallest positive integer ##k \geq2## such that ##a_2a_3\cdots a_k>4## Homework Equations The restrictions are the only relevant thing I can think of The Attempt at a Solution I have...

I feel sure now looking over it in the moment I felt unsure. I have practiced a few other system of equations problems now and feel more adjusted to it. Thanks for all the feedback everyone. I'm sure this year I'll be asking more questions as the physics class picks up.

Hmm at that point though I have the issue with ##m_2g+ \mu m_1g \cos\theta=m_2a+m_1a## When I tried to get rid of either ##m_2## or ##m_1## I had problems Or would this be ok to do ##m_2g+ \mu m_1g \cos\theta=\left(m_2+m_1\right)(a)## ##\frac{m_2g+ \mu m_1g \cos\theta}{m_2+m_1}=a##? Thanks for...

I'm sorry that was a typo I will fix it

Homework Statement Below are four equations, with the known quantities listed. Solve these equations to obtain an expression for ##T## in terms of known quantities only. Do the same to obtain an expression for ##a## ##T-f=m_1a\hspace{5mm}N-m_1g\cos\theta=0## ##m_2g-T=m_2a \hspace{5mm} f=\mu N##...

So if know that the triangle will either have 2 vertices in the top row and 1 in the bottom row, or vice versa. If I have 2 vertices in the top row, how many ways can I choose a vertex in the bottom row so the triangle is isosceles?

I have to account for all the repeats how would I use the choose function so I don't count them all because I feel it will be quite a large number. Like what are my constraints is what I'm having trouble seeing. I can't count them by hand I see now. Like with those 30 points can I use them as...

So I plot the points do I have to worry about using a point more then once such as 123 345 I thought it read equilateral for some reason so yeah I know two points will be equal distance to another point. Where do I go from that thought? Plot them all would be a guess.