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..as an 'aside'   if the angle is 6.2 degrees north of west, the answer comes out to a = 2.87 m/s^2

  Maybe there is a typo in the angle of the 8N force ??

well done thats correct

This was all that was given to me to solve the problem which is why i am struggling @ cphill

1/2 x 10000m^2 / 25 m^2/tree = 200 trees

Could you post the original problem???

1 1/2 −( 5/6 x+ 1/3 )= 8/9       note.....1 + 1/2  =  3/2  ....so we have....

3/2 - (5/6)x - 1/3  = 8/9         multiply through by the common denominator of 2,6,3,9 = 18

And we have

27 - 15x - 6 = 16

21 - 15x  =  16      add 15x to both sides, subtract 16 from both sides

5 = 15x        divide both sides by 15

5/15  = x  = 1/3

the forces acting laterally are

+9  N to the east

and  -8cos62 = -3.755 N

Net force is +5.244  east      f= ma     5.244 = 3 kg a       a = 1.75 m/s^2 east

Either the book has the wrong answer or the wrong info in the question!     

Good job !

Maybe you just need to put the times sign in?

\(\sin\beta -\cos^2 \beta=0\)

\(\sqrt{1-cos^2\beta}=cos^2\beta\)

\(cos^2\beta=z\)

\(\sqrt{1-z}=z\)

\(z^2+z-1=0\)

\(z=cos^2\beta=-\frac{1}{2}+\sqrt{(\frac{1}{2})^2+1}=\frac{-1+\sqrt{1+4}}{2}\)

\(cos^2\beta=\frac{\sqrt{5}-1}{2}\)   (equal with hectictar)

MaxWong, please tell us how to continue.

  ?

From the diagram, x= tower height

tan 30 = x / d

tan 60 = (x+5)/d      re arrange   d= (x+5)/tan60 = .57735x + 2.8868    Substtitute this into the first equation

tan30 = .57735 = x/(.57735x+2.8868)  solve for x

          .33333x +1.6666=  x

            x= 2.5 m 'tower' height    (short !)

no.

I found all these ones, that CPhill answered,  but the one you are asking now does not seem to be there.

Hi Sarah,

Do you have the address of your original question so that we can go to it?

(do you know what I mean? - just copy the address and paste it into a new post here)

It is good that you say when you do not understand but it is important that you say it on the original thread because the answerer want your feedback.  

People often give trivial anwers because they do not think anyone is really trying to learn from it.  We get so little feedback here.  Many askers only want to copy the answer and do not really want to learn.

Anyway, if you have that original question can you please copy the address and post it in here and then I will help you some more. :)

prime factor tree of 72

\(72=2^3*3^2\)

Thank you CPhill so much for answering my Questions! i can't thank you enough! I am so glad for this amazing website that you guys have created and thank you all so much! ;_)

Live a life you love

Love the life you live

Aced two tests and did a good Anime drawing which turned out really good! 

Hello :)

You question is incomplete and as it stands it is total nonsense.

Are you refering to a triangle, a square or an elephant perhaps?

Is it a special triangle that you are talking about, I suspect it is, what is special about the triangle?? 

Sixty per cent of students applying for admissions at NGASCE are female. 30 applications were received on a particular day. What is the probability that exactly 15 of the applications will be from females?

\(\binom{30}{15} * 0.6^{15}*0.4^{15}\approx 0.078 \)

nCr(30,15)*0.6^15*0.4^15 = 0.078312209686080141065170452

What is the probability that fewer than 10 of the applications will be from females?

nCr(30,0)*0.6^0*0.4^30+nCr(30,1)*0.6^1*0.4^29+nCr(30,2)*0.6^2*0.4^28+nCr(30,3)*0.6^3*0.4^27+nCr(30,4)*0.6^4*0.4^26+nCr(30,5)*0.6^5*0.4^25+nCr(30,6)*0.6^6*0.4^24+nCr(30,7)*0.6^7*0.4^23+nCr(30,8)*0.6^8*0.4^22+nCr(30,9)*0.6^9*0.4^21 = 0.0008563919557253

\(\text{P(less than 10 females)} \approx 0.000856\)

Also, calculate the expected number and variance of the number of applications from females?

The expected value is np = 30*0.6 = 18 females

The variance = np(1-p) = 30*0.6*0.4 = 7.2

Do you know the measure of an angle?

Actually you can throw out the option of \(\sin \beta = \frac{-\sqrt{5}-1}{2}\)because sin can never be bigger than 1 or less than -1.

Then you can use the pythagorean theorem to figure out that

\((\cos \beta)^2 + (\sin \beta)^2 = 1^2 \\ (\cos \beta)^2 = 1 - (\sin \beta)^2 \\ \cos \beta = \sqrt{1 - (\sin \beta)^2} \\~\\ \cos \beta = \sqrt{1 - (\frac{-1+\sqrt{5}}{2})^2} \\~\\ \cos \beta = \sqrt{\frac{\sqrt{5}-1}{2}}\)

Just go here http://www.howmanydays.net/how-many-days-i-lived/ and type you birth date

At first I was thinking like asinus that these equations were true for all values of beta, but then I realized that can't be because if beta were 90° that would make it 1 - 0 = 0, which is not true. So then I thought about this:

\(\sin \beta - \cos^2 \beta = 0 \\ - \cos^2 \beta = - \sin \beta \\ - \cos^2 \beta + 1 = - \sin \beta + 1 \\ 1 - \cos^2 \beta = - \sin \beta + 1 \\ \sin^2 \beta = - \sin \beta + 1 \\ 0 = - \sin^2 \beta - \sin \beta + 1 \\ 0 = \sin^2 \beta + \sin \beta - 1\)

Then use the quadratic formula or complete the square to find out that

\(\sin \beta = \frac{\sqrt{5}-1}{2}\)

and

\(\sin \beta = \frac{-\sqrt{5}-1}{2}\)

So

\( \beta =\arcsin( \frac{\sqrt{5}-1}{2} )\)

and

\(\beta = \arcsin (\frac{-\sqrt{5}-1}{2})\)

I don't really know if that helps figure out the rest of the problem or not.

I've used cos β = x = 1, but I'm not progressing. Please give a tip.